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Article

Finite Element Iterative Methods for the Stationary Double-Diffusive Natural Convection Model

College of Mathematics and System Sciences, Xinjiang University, Urumqi 830017, China
*
Author to whom correspondence should be addressed.
Entropy 2022, 24(2), 236; https://doi.org/10.3390/e24020236
Submission received: 4 January 2022 / Revised: 27 January 2022 / Accepted: 31 January 2022 / Published: 3 February 2022

Abstract

:
In this paper, we consider the stationary double-diffusive natural convection model, which can model heat and mass transfer phenomena. Based on the fixed point theorem, the existence and uniqueness of the considered model are proved. Moreover, we design three finite element iterative methods for the considered problem. Under the uniqueness condition of a weak solution, iterative method I is stable. Compared with iterative method I, iterative method II is stable with a stronger condition. Moreover, iterative method III is stable with the strongest condition. From the perspective of viscosity, iterative method I displays well in the case of a low viscosity number, iterative method II runs well with slightly low viscosity, and iterative method III can deal with high viscosity. Finally, some numerical experiments are presented for testing the correctness of the theoretic analysis.

1. Introduction

The double-diffusive natural convection model, which does not only incorporate the velocity vector field as well as the pressure field, but also contains the temperature field and the concentration field, has been widely used in scientific, engineering and industrial applications such as nuclear design, cooling of electronic equipment, aircraft cabins, insulation with double pane windows, and so on. For greater understanding of the physical background, authors can refer to [1,2,3]. In recent years, the impact of nanofluid on free convection heat transfer was investigated by researchers in [4]. The free convective flow of a Nano-Encapsulated Phase Change Material (NEPCM) suspension in an eccentric annulus was investigated numerically in [5]. The authors obtained that the volume fraction of the NEPCM particles and Stefan number effect the thermal and hydrodynamic characteristics of the suspension. The effect of the arrangement of the tubes in a multi-tube heat exchanger during the solidification process was considered in [6], which focused on the natural convection effect in phase change material in this research.
Let Ω R 2 be a open bounded domain with a Lipschitz continuous boundary Ω and Γ is a subset of Ω , u = ( u 1 , u 2 ) denotes the velocity field, p is the fluid pressure, T is the temperature, C is the concentration, g = ( 0 , 1 ) is the gravitational acceleration vector, f i is the forcing function, i = 1 , 2 . Moreover, n represents the outer normal vector, ν > 0 is the viscosity, D a is the Darcy number, γ > 0 is the heat diffusivity, D c is the mass diffusivity, β T and β C are the thermal and solutal expansion coefficients.
The governing equations of this double-diffusive natural convection model are presented as follows [7].
ν 2 u 1 x 2 + 2 u 1 y 2 + u 1 u 1 x + u 2 u 1 y = D a 1 u 1 p x , in Ω , ν 2 u 2 x 2 + 2 u 2 y 2 + u 1 u 2 x + u 2 u 2 y = β T T + β C C D a 1 u 2 p y , in Ω , u 1 x + u 2 y = 0 , in Ω , γ 2 T x 2 + 2 T y 2 + u 1 T x + u 2 T y = f 2 in Ω , D c 2 C x 2 + 2 C y 2 + u 1 T x + u 2 T y = f 2 in Ω , u = 0 , on Ω , T = 0 , C = 0 , on Γ , T n = C n = 0 , on Ω \ Γ .
Many numerical studies were made concerning the double-diffusive natural convection model. A projection-based stabilized finite element method for steady-state natural convection problem was considered in [8]. A stabilized finite element error analysis for the Darcy–Brinkman model of double-diffusive convection in a porous medium was discussed in [9]. An efficient two-step algorithm for the steady-state natural convection problem was presented in [10]. The melting process of a nano-enhanced phase change material in a square cavity was investigated in [11]. In numerical test, the author used the Galerkin finite element method to solve the dimensionless partial differential equations. Based on the idea of curvature stabilization, Çıbık et al. [12] discussed a family of second order time stepping methods for the Darcy–Brinkman equations. A decoupled finite element method called the modified characteristics method was considered in [13]. Rajabi et al. performed the detailed uncertainty propagation analysis and variance-based global sensitivity analysis on the widely adopted double-diffuse convection benchmark problem of a square porous cavity with horizontal temperature and concentration gradients in [14]. In [15], the mixed convection heat transfer of AL2O3 nanofuid in a horizontal channel subjected with two heat sources was considered. In [16], the curvature-based stabilization method was considered for double-diffusive natural convection flows in the presence of a magnetic field and unconditionally stable and optimally accurate second order approximations were obtained. There are several works devoted to the efficient numerical methods for the treatment of nonlinear problems. For example, several iterative methods for the 2D steady penalty Navier–Stokes equations were presented and discussed in [17]. He et al. [18] discussed a combination of two-level methods and iterative methods for solving the 2D/3D steady Navier–Stokes equations. Some iterative finite element methods for steady Navier–Stokes equations with different viscosities were discussed in [19]. Furthermore, the authors refer to the Oseen method [20], the Newton method [21] and the Euler implicit-explicit methods [22]. Recently, Huang et al. [23] have considered and analyzed the Oseen, Newton and Stokes iterative methods for the 2D steady Navier–Stokes equations. He et al. [24] considered and analyzed three iterative methods for the 3D steady MHD equations.
The main work in this paper is to design, analyze, and compare three iteration methods to solve nonlinear equations based on the finite element discretization. Then, we will show the performance of these numerical methods in both theoretical analysis and numerical experiments. By setting σ = m α 2 ν 2 N ( γ 1 f 1 1 + D c 1 f 2 1 ) + m α 2 ν 1 N ¯ ( γ 2 f 1 1 + D c 2 f 2 1 ) , we obtain the conclusion that the three iterative methods are stable and convergent as σ ( 0 , 1 4 ) . Iterative method I and II are valid as σ [ 1 4 , 1 3 ) and only iterative method I runs well as σ [ 1 3 , 1 ) .
In this paper, by developing some techniques and using some ideas in [7], we prove the existence and uniqueness with a different method, then we obtain a different uniqueness condition. Furthermore, we propose and analyze iterative methods I and III. In addition to this, we use iterative method II to computer a smaller viscosity than them in numerical experiments. Compared with He et al. [24], although the iterative methods are the same, the considered problems are different.
The paper is organized as follows. In Section 2, we describe the considered problem and some mathematical preliminaries. In the next section, we prove the existence and uniqueness of the weak solution to the considered equations. Then, we analyze stability and iterative error estimates of three iterative methods in Section 4. In Section 5, we show some numerical experiments to verify the correctness of theoretical results. In the last section, conclusions are presented.

2. Preliminaries

In this section, we present some basic notations and properties for the stationary double-diffusive natural convection problem. First, for 1 q and m N + , we use standard notations for Sobolev space W m , q ( Ω ) and Lebegue space L q ( Ω ) . In particular, L 2 ( Ω ) norm and its inner product are denoted by · 0 and ( · , · ) . Moreover, for f, an element in the dual space of H 1 ( Ω ) , its norm is defined by
f 1 = sup ψ H 1 ( Ω ) | ( f , ψ ) | ψ 0 .
Next, we introduce the functional spaces associated with the velocity, the pressure, the temperature, and the concentration:
X = { u H 1 ( Ω ) 2 : u | Ω = 0 } , W = { ψ H 1 ( Ω ) : ψ | Γ = 0 } , Q = { s H 1 ( Ω ) : s | Γ = 0 } , M = q L 2 ( Ω ) : Ω q d x = 0 .
Then, we define the following particular subspace of the velocity space X
V = v X : Ω q div v d Ω = 0 , q M .
Moreover, define several continuous bilinear forms a 0 ( · , · ) , a 1 ( · , · ) , a 2 ( · , · ) and d ( · , · ) on X × X , W × W , Q × Q and X × M , respectively,
a 0 ( u , v ) = ν ( u , v ) , u , v X , a 1 ( T , ψ ) = γ ( T , ψ ) , T , ψ W , a 2 ( C , s ) = D c ( C , s ) , C , s Q , d ( q , v ) = ( q , div v ) , v X , q M .
Further, denote three skew-symmetric trilinear forms:
c 0 ( u , v , w ) = ( ( u · ) v , w ) + 1 2 ( ( div u ) v , w ) = 1 2 ( ( u · ) v , w ) 1 2 ( ( u · ) w , v ) , u , w , v X , c 1 ( u , T , ψ ) = ( ( u · ) T , ψ ) + 1 2 ( ( div u ) T , ψ ) = 1 2 ( ( u · ) T , ψ ) 1 2 ( ( u · ) ψ , T ) , u X , T , ψ W , c 2 ( u , C , s ) = ( ( u · ) C , s ) + 1 2 ( ( div u ) C , s ) = 1 2 ( ( u · ) C , s ) 1 2 ( ( u · ) s , C ) , u X , C , s Q .
Please note that the bilinear form d ( · , · ) is continuous on X × M and satisfies the inf-sup condition [25]: there exists a positive constant β 0 such that
sup v V | d ( q , v ) | v 0 β 0 q 0 , q M .
The trilinear forms [18] satisfy
c 0 ( u , v , w ) = c 0 ( u , w , v ) , | c 0 ( u , v , w ) | N 0 u 0 v 0 w 0 ,
and
c 1 ( u , T , ψ ) = c 1 ( u , ψ , T ) , c 2 ( u , C , s ) = c 2 ( u , s , C ) , | c 1 ( u , T , ψ ) | N 1 u 0 T 0 ψ 0 , | c 2 ( u , C , s ) | N 2 u 0 C 0 s 0 ,
where N i > 0 , i = 0 , 1 , 2 , are three constants defined as N 0 = sup u , v , w X | c 0 ( u , v , w ) | u 0 v 0 w 0 ,   N 1 = sup u X , T , ψ W | c 1 ( u , T , ψ ) | u 0 T 0 ψ 0 , and N 2 = sup u X , C , s Q | c 2 ( u , C , s ) | u 0 C 0 s 0 .
Furthermore, we recall the Poincaré inequality [25]
u 0 α u 0 , u H 1 ( Ω ) ,
where α is a positive constant depending on Ω .
The variational form of the model (1) is presented as follows: find ( u , p , T , C ) X × M × W × Q such that for all ( v , q , ψ , s ) X × M × W × Q
a 0 ( u , v ) + c 0 ( u , u , v ) + D a 1 ( u , v ) d ( p , v ) + d ( q , u ) = ( β T T g + β C C g , v ) , a 1 ( T , ψ ) + c 1 ( u , T , ψ ) = ( f 1 , ψ ) , a 2 ( C , s ) + c 2 ( u , C , s ) = ( f 2 , s ) .

3. Existence and Uniqueness

This section gives the existence and uniqueness of (5), which is crucial to consider the discrete scheme.
Theorem 1.
There exists at least a solution pair ( u , p , T , C ) X × M × W × Q which satisfies (5) and
T 0 γ 1 f 1 1 , C 0 D c 1 f 2 1 , u 0 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) .
Proof
First, for u X , it is easy to see that a 1 ( · , · ) + c 1 ( u , · , · ) and a 2 ( · , · ) + c 2 ( u , · , · ) are continuous, elliptic bilinear forms of W × W and Q × Q , respectively. Hence, according to the Lax–Milgram theorem, there exists a unique solution T W to the second equation of (5), and a unique solution C Q to the third equation of (5). The theorem will be proved if we can show that there is at least a solution u X in the first equation of (5).
Secondly, a 0 ( · , · ) is a continuous and elliptic bilinear form on X × X . Using (2) and (4) we obtain
| c 0 ( u , u , v ) + ( β T T g + β C C g , v ) | N 0 u 0 2 + m α 2 ( T 0 + C 0 ) v 0 ,
where m = | g | max { | β T | , | β C | } . Then, we define a mapping A : X X by A ( u ) = w 1 where
a 0 ( w 1 , v ) + D a 1 ( w 1 , v ) = c 0 ( u , u , v ) + ( β T T g + β C C g , v ) , v V .
Clearly, u is a solution of the first equation of (5) with v V , if it is a solution of A ( u ) = u . Using the Leray-Schauder Principle [26], A ( u ) = u has at least one solution u X , if (a) A is completely continuous; (b) there exists M 1 > 0 such that for every λ [ 0 , 1 ] and v X with λ A v = v , v satisfies the bound v 0 M 1 .
Assume u 1 , u 2 X and subtract the equations obtained from (7) with u = u 1 and u = u 2 . Then, set w = A ( u 2 ) A ( u 1 ) and choose v = w to yield
a 0 ( w , w ) + D a 1 ( w , w ) = c 0 ( u 2 u 1 , u 2 , w ) c 0 ( u 1 , u 2 u 1 , w ) + ( β T ( T 2 T 1 ) g + β C ( C 2 C 1 ) g , w ) .
Moreover, in order to estimate ( T 2 T 1 ) 0 and ( C 2 C 1 ) 0 , we substitute T 1 and T 2 in the second equation of (5) and subtract the ensuing equations to obtain
a 1 ( T 2 T 1 , ψ ) = c 1 ( u 2 u 1 , T 2 , ψ ) c 1 ( u 1 , T 2 T 1 , ψ ) .
Taking ψ = T 2 T 1 and using (3) we obtain
( T 2 T 1 ) 0 γ 1 N 1 ( u 2 u 1 ) 0 T 2 0 .
Analogously, we have
( C 2 C 1 ) 0 D c 1 N 2 ( u 2 u 1 ) 0 C 2 0 .
Further, combining (9) and (10), we obtain the bound of (8) as follows
w 0 ν 1 N 0 u 1 0 + N 0 u 2 0 + m α 2 ( γ 1 N 1 T 2 0 + D c 1 N 2 C 2 0 ) ( u 2 u 1 ) 0 .
Hence, A is completely continuous.
Now, we prove (b). If λ = 0 , then v = 0 and v 0 = 0 . Assume λ ( 0 , 1 ] and v X satisfies λ A v = v . Then, from (7), we have
λ 1 a 0 ( v , v ) + λ 1 D a 1 ( v , v ) = c 0 ( v , v , v ) + ( β T T g + β C C g , v ) .
Using (2) and (4), we arrive at
v 0 ν 1 λ m α 2 ( T 0 + C 0 ) .
Thirdly, setting ψ = T in the second equation of (5), we have
γ T 0 2 + c 1 ( u , T , T ) f 1 1 T 0 .
Thus, applying (3) leads to
T 0 γ 1 f 1 1 .
Similarly, taking s = C in the third equation of (5), we obtain
C 0 D c 1 f 2 1 .
Moreover, choosing v = u in the first equation of (5) and using (4), we arrive at
ν u 0 2 + c 0 ( u , u , u ) m α 2 ( T 0 + C 0 ) u 0 ,
which combines with the above two equations to give
u 0 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) .
The proof is completed. □
Theorem 2.
Assume that ( u , p , T , C ) X × M × W × Q is a solution pair of (5). If ν, D c , γ , C and T satisfy the following uniqueness condition
0 < σ : = m α 2 ν 2 N 0 ( γ 1 f 1 1 + D c 1 f 2 1 ) + m α 2 ν 1 ( γ 2 N 1 f 1 1 + D c 2 N 2 f 2 1 ) < 1 ,
then ( u , p , T , C ) is unique solution pair of (5).
Proof. 
Suppose ( u 1 , p 1 , T 1 , C 1 ) is also a solution pair of (5) and u 1 u , p 1 p , T 1 T , C 1 C , then
a 0 ( u 1 , v ) + c 0 ( u 1 , u 1 , v ) + D a 1 ( u 1 , v ) d ( p 1 , v ) + d ( q , u 1 ) = ( β T T 1 g + β C C 1 g , v ) , a 0 ( u , v ) + c 0 ( u , u , v ) + D a 1 ( u , v ) d ( p , v ) + d ( q , u ) = ( β T T g + β C C g , v ) ,
for all ( v , q ) X × M .
Now, choosing v = u u 1 and q = p p 1 , we obtain
a 0 ( u u 1 , u u 1 ) + D a 1 ( u u 1 , u u 1 ) = c 0 ( u u 1 , u , u u 1 ) + ( β T ( T T 1 ) g + β C ( C C 1 ) g , u u 1 ) .
Hence, applying (4), (9), (10), Theorem 1 and the uniqueness condition, we have
ν ( u u 1 ) 0 2 N 0 u 0 + m α 2 ( γ 1 N 1 T 0 + D c 1 N 2 C 0 ) ( u u 1 ) 0 2 < ν ( u u 1 ) 0 2 ,
a contradiction. Hence, u 1 = u , T 1 = T , C 1 = C .  □

4. Several Iterative Methods Based on the Finite Element Discretization

In this section, we propose three iterative methods for the double-diffusive natural convection model. Then the stability and convergence of these iterative methods are considered. First, let 0 < h < 1 denote the mesh size which is a real positive parameter and K h = { K : K Ω K ¯ = Ω ¯ } be a uniform partition of Ω ¯ into non-overlapping triangles. Next, given a K h , we consider the finite element spaces X h , M h , W h and Q h
V h = { v h V C 0 ( Ω ¯ ) 2 : v h | K P 2 ( K ) 2 , K K h } , M h = { q h M C 0 ( Ω ¯ ) : q h | K P 1 ( K ) , K K h } , W h = { ψ h W C 0 ( Ω ¯ ) : ψ h | K P 2 ( K ) , K K h } , Q h = { s h Q C 0 ( Ω ¯ ) : s h | K P 2 ( K ) , K K h } ,
where P i ( K ) represents the space of the order polynomial on the set K h , i = 1 , 2 . Please note that the Taylor-Hood element X h × M h satisfies the following discret inf-sup condition
sup v X h | d ( q , v ) | v 0 β q 0 , q M h ,
where the constant β > 0 is independent of h.
With the above notations, the finite element scheme for the natural convection problem is defined as follows: find ( u h , p h , T h , C h ) X × M × W × Q such that
a 0 ( u h , v ) + c 0 ( u h , u h , v ) + D a 1 ( u h , v ) d ( p h , v ) + d ( q , u h ) = ( β T T h g + β C C h g , v ) , a 1 ( T h , ψ ) + c 1 ( u h , T h , ψ ) = ( f 1 , ψ ) , a 2 ( C h , s ) + c 2 ( u h , C h , s ) = ( f 2 , s ) ,
for all ( v , q , ψ , s ) X h × M h × W h × Q h . The following stability and convergence results of the numerical solutions to (12) are showed.
Theorem 3.
([7,8,26,27]) Let ( u , p , T , C ) ( H 3 ( Ω ) 2 X ) × ( H 2 ( Ω ) M ) × ( H 3 ( Ω ) W ) × ( H 3 ( Ω ) Q ) . Under the assumption of Theorem 2, the numerical solution pair ( u h , p h , T h , C h ) to (12) satisfies
T h 0 γ 1 f 1 1 , C h 0 D c 1 f 2 1 ,
and
u h 0 ν 1 m α ( γ 1 f 1 1 + D c 1 f 2 1 ) .
Moreover, the following error estimate holds
( u u h ) 0 + ( p p h ) 0 + ( T T h ) 0 + ( C C h ) 0 c h 2 ( u 3 + p 2 + T 3 + C 3 ) ,
where c is a positive constant depending on h.
In the following part of this section, we propose and analyse three iterative methods.
Iterative method I. Find ( u h n , p h n , T h n , C h n ) X h × M h × W h × Q h such that
a 0 ( u h n , v ) + c 0 ( u h n 1 , u h n , v ) + D a 1 ( u h n , v ) d ( p h n , v ) + d ( q , u h n ) = ( β T T h n g + β C C h n g , v ) , a 1 ( T h n , ψ ) + c 1 ( u h n 1 , T h n , ψ ) = ( f 1 , ψ ) , a 2 ( C h n , s ) + c 2 ( u h n 1 , C h n , s ) = ( f 2 , s ) ,
for all ( v h , q , ψ , s ) X h × M h × W h × Q h .
Iterative method II. Find ( u h n , p h n , T h n , C h n ) X h × M h × W h × Q h such that
a 0 ( u h n , v ) + c 0 ( u h n 1 , u h n , v ) + c 0 ( u h n , u h n 1 , v ) c 0 ( u h n 1 , u h n 1 , v ) + D a 1 ( u h n , v ) d ( v , p h n ) + d ( u h n , q ) = ( β T T h n g + β C C h n g , v ) , a 1 ( T h n , ψ ) + c 1 ( u h n 1 , T h n , ψ ) + c 1 ( u h n , T h n 1 , ψ ) c 1 ( u h n 1 , T h n 1 , ψ ) = ( f 1 , ψ ) , a 2 ( C h n , s ) + c 2 ( u h n 1 , C h n , s ) + c 2 ( u h n , C h n 1 , s ) c 2 ( u h n 1 , C h n 1 , s ) = ( f 2 , s ) ,
for all ( v , q , ψ , s ) X h × M h × W h × Q h .
Iterative method III. Find ( u h n , p h n , T h n , C h n ) X h × M h × W h × Q h such that
a 0 ( u h n , v ) + c 0 ( u h n 1 , u h n 1 , v ) + D a 1 ( u h n , v ) d ( p h n , v ) + d ( q , u h n ) = ( β T T h n g + β C C h n g , v ) , a 1 ( T h n , ψ ) + c 1 ( u h n 1 , T h n 1 , ψ ) = ( f 1 , ψ ) , a 2 ( C h n , s ) + c 2 ( u h n 1 , C h n 1 , s ) = ( f 2 , s ) ,
for all ( v , q , ψ , s ) X h × M h × W h × Q h .
For the above three iterative methods, the initial guess ( u h 0 , p h 0 , T h 0 , C h 0 ) X h × M h × W h × Q h is defined by solving the following equations
a 0 ( u h 0 , v ) + D a 1 ( u h 0 , v ) d ( p h 0 , v ) + d ( q , u h 0 ) = ( β T T h 0 g + β C C h 0 g , v ) , a 1 ( T h 0 , ψ ) = ( f 1 , ψ ) , a 2 ( C h 0 , s ) = ( f 2 , s ) ,
for all ( v , q , ψ , s ) X h × M h × W h × Q h .
Now, we will establish the stability and iterative error estimates of the presented iterative methods for the double-diffusive natural convection model. For the sake of simplicity, let ( e n , η n , ξ n , δ n ) = ( u h u h n , p h p h n , T h T h n , C h C h n ) .
Theorem 4.
Under the assumptions of Theorem 3, ( u h n , p h n , T h n , C h n ) defined by iterative method I satisfies
u h n 0 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , T h n 0 γ 1 f 1 1 , C h n 0 D c 1 f 2 1 ,
for all n 0 . Furthermore, the following iterative error bounds hold
e n 0 σ n ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , η n 0 4 β 1 σ n m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , ξ n 0 σ n γ 1 f 1 1 , δ n 0 σ n D c 1 f 2 1 ,
for all n 0 .
Proof. 
First, the induction method is used to consider the stability of iterative method I. Setting ( v , q , ψ , s ) = ( u h 0 , p h 0 , T h 0 , C h 0 ) in (16) leads to
T h 0 0 γ 1 f 1 1 , C h 0 0 D c 1 f 2 1 , u h 0 0 ν 1 m α 2 ( T h 0 0 + C h 0 0 ) ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) .
which shows that (17) holds for n = 0 .
Next, assuming that it holds for n = k , we prove that it is valid for n = k + 1 . Taking ( v , q , ψ , s ) = ( u h k + 1 , p h k + 1 , T h k + 1 , C h k + 1 ) in (13) with n = k + 1 and applying (2), (3) and (4) yield
T h k + 1 0 γ 1 f 1 1 , C h k + 1 0 D c 1 f 2 1 , u h k + 1 0 ν 1 m α 2 ( T h k + 1 0 + C h k + 1 0 ) .
Hence, we finish the induction method.
Moreover, we consider the iterative error estimates of iterative method I. Making use of (12) and (13) yields the error equations
a 0 ( e n , v ) + c 0 ( u h n 1 , e n , v ) + c 0 ( e n 1 , u h , v ) + D a 1 ( e n , v ) d ( η n , v ) + d ( q , e n ) = ( β T ξ n g + β C δ n g , v ) , a 1 ( ξ n , ψ ) + c 1 ( u h n 1 , ξ n , ψ ) + c 1 ( e n 1 , T h , ψ ) = 0 , a 2 ( δ n , s ) + c 2 ( u h n 1 , δ n , s ) + c 2 ( e n 1 , C h , s ) = 0 .
Setting ψ = ξ n , s = δ n in the second and the third equation of (20) and using (3), (17), and Theorem 3, we obtain
ξ n 0 N 1 γ 2 f 1 1 e n 1 0 , n 1 , δ n 0 N 2 D c 2 f 2 1 e n 1 0 , n 1 .
Then, taking ( v , q ) = ( e n , η n ) in the first equation of (20) and using (2), (4), (17), (21) and Theorem 3, we arrive at
ν e n 0 N 0 e n 1 0 u h 0 + m α 2 ( ξ n 0 + δ n 0 ) N 0 e n 1 0 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + m α 2 ( N 1 γ 2 f 1 1 + N 2 D c 2 f 2 1 ) e n 1 0 .
Hence, using uniqueness condition, we have
e n 0 σ e n 1 0 σ n e 0 0 , n 1 .
Furthermore, subtracting (16) from (12), we obtain
a 0 ( e 0 , v ) + c 0 ( u h , u h , v ) + D a 1 ( e 0 , v ) d ( η 0 , v ) + d ( q , e 0 ) = ( β T ξ 0 g + β C δ 0 g , v ) , a 1 ( ξ 0 , ψ ) + c 1 ( u h , T h , ψ ) = 0 , a 2 ( δ 0 , s ) + c 2 ( u h , C h , s ) = 0 .
Applying (4), the Theorem 2 and the Theorem 3, we obtain
ξ 0 0 N 1 γ 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) f 1 1 γ 1 f 1 1 , δ 0 0 N 2 D c 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) f 2 1 D c 1 f 2 1 , e 0 0 N 0 ν 3 m 2 α 4 ( γ 1 f 1 1 + D c 1 f 2 1 ) 2 + N 1 γ 2 ν 1 m 2 α 4 ( γ 1 f 1 1 + D c 1 f 2 1 ) f 1 1 + N 2 D c 2 ν 1 m 2 α 4 ( γ 1 f 1 1 + D c 1 f 2 1 ) f 2 1 , ν 1 m α ( γ 1 f 1 1 + D c 1 f 2 1 ) ,
which combines with (21) and (22), we arrive at
e n 0 σ n ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , ξ n 0 N 1 γ 2 f 1 1 σ n 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) σ n γ 1 f 1 1 , δ n 0 N 2 D c 2 f 2 1 σ n 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , σ n D c 1 f 2 1 ,
for all n 0 .
Finally, applying the discrete inf-sup condition, from the first equation of (20) with q = 0 , the error estimate of the pressure can be stated as follows.
η n 0 β 1 ν e n 0 + N 0 u h n 1 0 e n 0 + N 0 e n 1 0 u h 0 + β 1 m α 2 ( ξ n 0 + δ n 0 ) β 1 ( σ n m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + N 0 ν 2 m 2 α 4 ( σ n + σ n 1 ) ( γ 1 f 1 1 + D c 1 f 2 1 ) 2 + m α 2 σ n ( γ 1 f 1 1 + D c 1 f 2 1 ) ) 4 β 1 σ n m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) ,
for all n 0 .  □
Theorem 5.
Under the assumptions of Theorem 3, suppose that the following condition (the strong uniqueness condition)
0 < σ < 1 3 ,
holds. Then ( u h n , p h n , T h n , C h n ) generated by iterative method II satisfies
T h n 0 4 3 γ 1 f 1 1 , C h n 0 4 3 D c 1 f 2 1 , u h n 0 4 3 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) ,
for all n 0 . Furthermore, the following iterative error bounds hold
e n 0 9 5 σ 2 n 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , η n 0 119 45 β 1 9 5 σ 2 n 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , ξ n 0 9 5 σ 2 n 1 γ 1 f 1 1 , δ n 0 9 5 σ 2 n 1 D c 1 f 2 1 ,
for all n 0 .
Proof. 
Combining with (19) and (23), it is found that (25) and (26) hold for n = 0 . Supposing that (25) and (26) hold for n = k , we shall prove that they are valid for n = k + 1 .
Subtracting (14) from (12), we obtain the error equations
a 0 ( e n , v ) + c 0 ( u h n 1 , e n , v ) + c 0 ( e n , u h n 1 , v ) + c 0 ( e n 1 , e n 1 , v ) + D a 1 ( e n , v ) d ( v , η n ) + d ( e n , q ) = ( β T ξ n g + β C δ n g , v ) , a 1 ( ξ n , ψ ) + c 1 ( u h n 1 , ξ n , ψ ) + c 1 ( e n , T h n 1 , ψ ) + c 1 ( e n 1 , ξ n 1 , ψ ) = 0 , a 2 ( σ n , s ) + c 2 ( u h n 1 , δ n , s ) + c 2 ( e n , C h n 1 , s ) + c 2 ( e n 1 , δ n 1 , s ) = 0 .
Setting ( v , q , ψ , s ) = ( e h n , η h n , ξ h n , δ h n ) in (27) with n = k + 1 and applying (2), (3), (4) and the assumptions on n = k , we have
ξ k + 1 0 N 1 γ 1 e k + 1 0 T h k 0 + N 1 γ 1 e k 0 ξ k 0 4 3 N 1 γ 2 f 1 1 e k + 1 0 + N 1 γ 1 e k 0 ξ k 0 , δ k + 1 0 N 2 D c 1 e k + 1 0 C h k 0 + N 2 D c 1 e k 0 δ k 0 4 3 N 2 D c 2 f 2 1 e k + 1 0 + N 2 D c 1 e k 0 δ k 0 ,
and
ν e k + 1 0 N 0 u h k 0 e k + 1 0 + N 0 e k 0 2 + m α 2 ( ξ k + 1 0 + δ k + 1 0 ) 4 3 N 0 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) e k + 1 0 + N e k 0 2 + m α 2 ( 4 3 N 1 γ 2 f 1 1 e k + 1 0 + N 1 γ 1 e k 0 ξ k 0 ) + m α 2 ( 4 3 N 2 D c 2 f 2 1 e k + 1 0 + N 2 D c 1 e k 0 δ k 0 ) .
Moreover, imply the strong uniqueness condition (24) on (29), we obtain
e k + 1 0 9 5 N 0 ν 1 9 5 σ 2 k + 1 2 ν 2 m 2 α 4 ( γ 1 f 1 1 + D c 1 f 2 1 ) 2 + 9 5 ν 1 m α 2 N 1 9 5 σ 2 k + 1 2 ν 1 m α 2 γ 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) f 1 1 + 9 5 ν 1 m α 2 N 2 9 5 σ 2 k + 1 2 ν 1 m α 2 D c 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) f 2 1 9 5 σ 2 k + 1 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) .
Hence, making use of (30), we rewrite (28) as
ξ k + 1 0 4 3 N 1 γ 2 f 1 1 9 5 σ 2 k + 1 1 ν 1 m α ( γ 1 f 1 1 + D c 1 f 2 1 ) + N 1 γ 1 9 5 σ 2 k + 1 2 ν 1 m α ( γ 1 f 1 1 + D c 1 f 2 1 ) γ 1 f 1 1 9 5 σ 2 k + 1 1 γ 1 f 1 1 . δ k + 1 0 4 3 N 2 D c 2 f 2 1 9 5 σ 2 k + 1 1 ν 1 m α ( γ 1 f 1 1 + D c 1 f 2 1 ) + N 2 D c 1 9 5 σ 2 k + 1 2 ν 1 m α ( γ 1 f 1 1 + D c 1 f 2 1 ) D c 1 f 2 1 9 5 σ 2 k + 1 1 D c 1 f 2 1 .
Combining the first equation of (27) with n = k + 1 and q = 0 and the discrete inf-sup condition, we have
η k + 1 0 β 1 ( ν e k + 1 0 + N 0 e k + 1 0 u h k 0 + N 0 e k 0 2 ) + β 1 m α 2 ( ξ k + 1 0 + δ k + 1 0 ) β 1 ( ν 9 5 σ 2 k + 1 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + N 0 9 5 σ 2 k + 1 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) × 4 3 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + N 0 9 5 σ 2 k + 1 2 ν 2 m 2 α 4 ( γ 1 f 1 1 + D c 1 f 2 1 ) 2 + m α 2 9 5 σ 2 k + 1 1 γ 1 f 1 1 + 9 5 σ 2 k + 1 1 D c 1 f 2 1 ) 119 45 β 1 9 5 σ 2 k + 1 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) .
Furthermore, subtracting (16) from (14) with n = 1 that
a 0 ( u h 1 u h 0 , v ) + c 0 ( u h 0 , u h 1 u h 0 , v ) + c 0 ( u h 1 , u h 0 , v ) + D a 1 ( u h 1 u h 0 , v ) d ( v , p h 1 p h 0 ) + d ( u h 1 u h 0 , q ) = ( β T ( T h 1 T h 0 ) g + β C ( C h 1 C h 0 ) g , v ) , a 1 ( T h 1 T h 0 , ψ ) + c 1 ( u h 0 , T h 1 T h 0 , ψ ) + c 1 ( u h 1 , T h 0 , ψ ) = 0 , a 2 ( C h 1 C h 0 , s ) + c 2 ( u h 0 , C h 1 C h 0 , s ) + c 2 ( u h 1 , C h 0 , s ) = 0 .
Then, taking ψ = T h 1 T h 0 in the second equation of (33), we observe that
( T h 1 T h 0 ) 0 N 1 γ 1 u h 1 0 T h 0 0 ,
and
( C h 1 C h 0 ) 0 N 2 D c 1 u h 1 0 C h 0 0 .
Moreover, setting v = u h 1 u h 0 in the first equation of (33), we obtain
( u h 1 u h 0 ) 0 ν 1 N 0 u h 1 0 u h 0 0 + ν 1 m α 2 ( ( T h 1 T h 0 ) 0 + ( C h 1 C h 0 ) 0 ) N 0 ν 2 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) u h 1 0 + ν 1 m α 2 ( N 1 γ 2 f 1 1 + N 2 D c 2 f 2 1 ) u h 1 0 σ u h 1 0 .
Combining (14) with n = 1 and using (34), we obtain
T h 1 0 γ 1 N 1 ( u h 1 u h 0 ) 0 T h 0 0 + γ 1 f 1 1 γ 2 N 1 σ f 1 1 u h 1 0 + γ 1 f 1 1 , C h 1 0 D c 1 N 2 ( u h 1 u h 0 ) 0 C h 0 0 + D c 1 f 2 1 D c 2 N 2 σ f 2 1 u h 1 0 + D c 1 f 2 1 , u h 1 0 ν 1 N 0 ( u h 1 u h 0 ) 0 u h 0 0 + ν 1 m α 2 ( T h 1 0 + C h 1 0 ) ν 2 N 0 σ u h 1 0 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + ν 1 m α 2 ( γ 2 N 1 σ f 1 1 u h 1 0 + γ 1 f 1 1 + D c 2 N 2 σ f 2 1 u h 1 0 + D c 1 f 2 1 ) σ 2 u h 1 0 + ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) .
In view of the strong uniqueness condition (24), we arrive at
u h 1 0 9 8 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , T h 1 0 9 8 γ 1 f 1 1 , C h 1 0 9 8 D c 1 f 2 1 .
Next, taking ( v , q , ψ , s ) = ( u h n , p h n , T h n , C h n ) in (14) with n 2 , and using (2), (3) and (26), we obtain
T h n 0 γ 1 c 1 ( u h n u h n 1 , T h n 1 T h n , ψ ) + γ 1 f 1 1 γ 1 N 1 ( e n 1 e n ) 0 ( ξ n 1 ξ n ) 0 + γ 1 f 1 1 γ 1 N 1 9 5 σ 3 + 9 5 σ 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) γ 1 f 1 1 + γ 1 f 1 1 γ 1 N 1 3 5 3 + 3 5 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) γ 1 f 1 1 + γ 1 f 1 1 4 3 γ 1 f 1 1 .
Similarly, we obtain
C h n 0 4 3 D c 1 f 2 1 .
Finally, it has
u h n 0 ν 1 N 0 ( e n 1 e n ) 0 2 + ν 1 m α 2 ( T h n 0 + C h n 0 ) ν 1 N 0 ( e n 1 e n ) 0 2 + ν 1 m α 2 ( γ 1 N 1 ( e n 1 e n ) 0 ( ξ n 1 ξ n ) 0 + γ 1 f 1 1 ) + ν 1 m α 2 ( D c 1 N 2 ( e n 1 e n ) 0 ( δ n 1 δ n ) 0 + D c 1 f 2 1 ) ν 1 N 0 3 5 3 + 3 5 2 ν 2 m 2 α 4 ( γ 1 f 1 1 + D c 1 f 2 ) 1 ) 2 + ν 1 m α 2 ( γ 1 N 1 3 5 3 + 3 5 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) γ 1 f 1 1 + γ 1 f 1 1 ) + ν 1 m α 2 ( D c 1 N 2 3 5 3 + 3 5 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) D c 1 f 2 1 + D c 1 f 2 1 ) 4 3 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) .
The proof is completed. □
Theorem 6.
Under the assumptions of Theorem 3, suppose that the following condition (the stronger uniqueness condition),
0 < σ < 1 4 ,
holds. Then ( u h n , p h n , T h n , C h n ) defined by the iterative method III satisfies
u h n 0 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , T h n 0 2 γ 1 f 1 1 , C h n 0 2 D c 1 f 2 1 ,
for all n 0 . Furthermore, the following iterative error bounds hold
e n 0 ( 3 σ ) n ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , η n 0 5 β 1 ( 3 σ ) n m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , ξ n 0 ( 3 σ ) n γ 1 f 1 1 , δ n 0 ( 3 σ ) n D c 1 f 2 1 ,
for all n 0 .
Proof. 
From (19) and (23), it is obvious that (36) and (37) hold for n = 0 . Supposing that (36) and (37) hold for n = k , we shall prove that they are valid for n = k + 1 .
Setting ( v , q , ψ , s ) = ( u h n , p h n , T h n , C h n ) in (15) with n = k + 1 and using (2), (3), (4) and (36), we obtain that
T h k + 1 0 γ 1 N 1 u h k 0 T h k 0 + γ 1 f 1 1 γ 1 N 1 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) 2 γ 1 f 1 1 + γ 1 f 1 1 2 γ 1 f 1 1 , C h k + 1 0 D c 1 N 2 u h k 0 C h k 0 + D c 1 f 1 1 D c 1 N 2 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) 2 D c 1 f 2 1 + D c 1 f 2 1 2 D c 1 f 2 1 , u h k + 1 0 ν 1 N 0 u h k 0 2 + ν 1 m α 2 ( T h k + 1 0 + C h k + 1 0 ) ν 1 N 0 u h k 0 2 + ν 1 m α 2 ( γ 1 N 1 u h m 0 T h k 0 + γ 1 f 1 1 ) + ν 1 m α 2 ( D c 1 N 2 u h k 0 C h k 0 + D c 1 f 1 1 ) ν 1 N 0 4 ν 2 m 2 α 4 ( γ 1 f 1 1 + D c 1 f 2 1 ) 2 + 4 ν 2 γ 2 m 2 α 4 N 1 ( γ 1 f 1 1 + D c 1 f 2 1 ) f 1 1 + ν 1 m α 2 γ 1 f 1 1 + ν 1 m α 2 D c 1 f 2 1 + 4 ν 2 D c 2 m 2 α 4 N 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) f 2 1 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) .
Hence, (36) is valid for n = k + 1 . Consequently, subtracting (15) from (12) yields
a 0 ( e n , v ) + c 0 ( u h n 1 , e n 1 , v ) + c 0 ( e n 1 , u h , v ) + D a 1 ( e n , v ) d ( η n , v ) + d ( q , e n ) = ( β T ξ n g + β C δ n g , v ) , a 1 ( ξ n , ψ ) + c 1 ( u h n 1 , ξ n 1 , ψ ) + c 1 ( e n 1 , T h , ψ ) = 0 , a 2 ( δ n , s ) + c 2 ( u h n 1 , δ n 1 , s ) + c 2 ( e n 1 , C h , s ) = 0 .
Now, choosing ψ = ξ n , in the second equation of (38) and using (3), (36), (37) and Theorem 3, we can deduce that
ξ n 0 2 N 1 γ 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) ξ n 1 0 + N 1 γ 2 f 1 1 e n 1 0 2 N 1 γ 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) ( 3 σ ) n 1 γ 1 f 1 1 + N 1 γ 2 f 1 1 ( 3 σ ) n 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) ( 3 σ ) n γ 1 f 1 1 , n 1 .
Similarly, one has
δ n 0 2 N 2 D c 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) δ n 1 0 + N 2 D c 2 f 2 1 e n 1 0 2 N 2 D c 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) ( 3 σ ) n 1 D c 1 f 2 1 + N 2 D c 2 f 2 1 ( 3 σ ) n 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) ( 3 σ ) n D c 1 f 2 1 , n 1 .
Moreover, taking ( v , q ) = ( e n , η n ) in the first equation of (38) and using (2), (4), (36), (37) and the Theorem 3, we find that
e n 0 ν 1 N 0 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) e n 1 0 + 2 ν 1 N 0 e n 1 0 ( ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + m α 2 ν 1 ( 2 γ 1 ν 1 m α 2 ( γ 1 N 1 f 1 1 + D c 1 N 2 f 2 1 ) ξ n 1 0 + N 1 γ 2 f 1 1 e n 1 0 ) + m α 2 ν 1 ( 2 D c 1 ν 1 m α 2 ( γ 1 N 1 f 1 1 + D c 1 N 2 f 2 1 ) δ n 1 0 + N 2 D c 2 f 2 1 e n 1 0 ) 3 ν 1 N 0 ( 3 σ ) n 1 ν 2 m 2 α 4 ( γ 1 f 1 1 + D c 1 f 2 1 ) 2 + m α 2 ν 1 ( N 1 γ 2 f 1 1 ( 3 σ ) n 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + 2 γ 1 ν 1 m α 2 ( γ 1 N 1 f 1 1 + D c 1 N 2 f 2 1 ) ( 3 σ ) n 1 γ 1 f 1 1 ) + m α 2 ν 1 ( N 2 D c 2 f 2 1 ( 3 σ ) n 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + 2 D c 1 ν 1 m α 2 ( γ 1 N 1 f 1 1 + D c 1 N 2 f 2 1 ) ( 3 σ ) n 1 D c 1 f 2 1 ) ( 3 σ ) n ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) , n 1 .
Finally, combining the first equation of (38) with q = 0 and the discrete inf-sup condition, the error estimate for the pressure can be stated as follows
η n 0 β 1 ( ν e n 0 + N 0 u h n 1 0 e n 1 0 + N 0 e n 1 0 u h 0 + m α 2 ( ξ n 0 + δ n 0 ) ) β 1 ( ν ( 3 σ ) n ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + N 0 ( 3 σ ) n 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) 2 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + N 0 ( 3 σ ) n 1 ν 1 m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) + m α 2 ( 3 σ ) n ( γ 1 f 1 1 + D c 1 f 2 1 ) ) 5 β 1 ( 3 σ ) n m α 2 ( γ 1 f 1 1 + D c 1 f 2 1 ) .

5. Numerical Experiments

In this section, several numerical experiments are presented to compare these iterative methods for the considered equations. We use the public finite element software FreeFem++ [28].

5.1. An Analytical Solution Problem

For numerical implementations, the iterative tolerance is 1.0 × 10 5 . The first issue to be considered here is to compare these iterative methods for the stationary double-diffusive natural convection in the case of Ω = [ 0 , 1 ] × [ 0 , 1 ] R 2 , to reveal the relationship between the iterative methods and the viscosity. We consider the following exact solutions.
p ( x , y ) = cos ( π x ) cos ( π y ) , u 1 ( x , y ) = 2 π sin 2 ( π x ) sin ( π y ) cos ( π y ) , u 2 ( x , y ) = 2 π sin ( π x ) sin 2 ( π y ) cos ( π x ) , T ( x , y ) = u 1 ( x , y ) + u 2 ( x , y ) , C ( x , y ) = u 1 ( x , y ) u 2 ( x , y ) .
Set the Darcy number D a = 1 , the thermal expansion coefficient β T = 1 , the solutal expansion coefficient β C = 1 , the heat diffusivity γ = 1 , the mass diffusivity D c = 1 and u i = 0 , T = 0 , C = 0 on Ω , i = 1 , 2 . The forcing function f i can be calculated using (42), i = 1,2. We use a fixed value of mesh size h = 1 64 , and perform tests for the values of the viscosity coefficients going from ν = 1 to ν = 1.0 × 10 4 .
We compare the numbers of iteration and the computational time in Table 1. This table shows that all iterative methods run well in the case of ν = 1 . When the viscosity number increases to ν = 1.0 × 10 2 , iterative method III is divergent. Finally, iterative methods II and III can not export the data with ν = 1.0 × 10 4 , iterative method I is still convergent. From a computational point of view, the calculation time of iterative method I and iterative method II is similar. However, iterative method II saves about 30 % of calculation time compared iterative method III when ν = 1 . Iterative method II saves about 35 % of calculation time compared with iterative method I when ν = 1.0 × 10 2 . We can conclude that iterative method III is the simplest method for a high viscosity number. The iterative method II is a fast and high accuracy method for a slightly lower viscosity number. Iterative method I is stable under uniqueness condition of weak solutions in the case of the lowest viscosity number. For three iterative methods, the relative error estimates are presented in Table 2, Table 3 and Table 4.

5.2. The Cavity Problem

In this numerical experiment, we assume that the boundary conditions satisfy [7,9]
T = 1 , C = 1 , u = 0 a t x 1 = 0 , T = 1 , C = 1 , u = 0 a t x 1 = 1 , T n = 0 , C n = 0 , u = 0 a t x 2 = 0 , T n = 0 , C n = 0 , u = 0 a t x 2 = 1 ,
and set D a = 1 , β T = 1 , β C = 1 , γ = 0.1 , D c = 0.01 , f i = 0 and the mesh size h = 1 64 , i = 1 , 2 . Moreover, the convergence tolerance is set to equal 1.0 × 10 6 . The domain with its boundary conditions is illustrated in Figure 1. We present the velocity streamlines, the pressure isobars, the isotherms and the isoconcentration lines for different values of the viscosity coefficients ν = 1.0 , ν = 1.0 × 10 3 , ν = 1.0 × 10 4 .
Then, we show numerical velocity streamlines, isobars of pressure, isotherms, and isoconcentration lines obtained by three iterative methods with different viscosity numbers. We plot these results in Figure 2, Figure 3, Figure 4 and Figure 5. From these graphs, we obtain that the values of viscosity not only heavily impact on the velocity streamlines and the isobars, but also affect the isotherms and the isoconcentration lines. In fact, three iterations run well with ν = 1.0 . However, iterative method III cannot run with ν = 1.0 × 10 3 while iterative method II cannot export the data with ν = 1.0 × 10 4 .
To consider the independency of mesh in a square cavity, we use iterative method I to calculate the model (1) under different mesh sizes. The results are presented in Figure 6. We can see that there is no difference in the calculation results under different mesh sizes, so we can verify the independence of the mesh size.
Figure 1. Velocity streamlines of iteration method I (the first line), iteration method II (the second line) and iteration method III (the third line) with different viscosity coefficients 1.0 (the first column), 1.0 × 10 3 (the second column) and 1.0 × 10 4 (the third column). D a = 1 , β T = 1 , β C = 1 , γ = 0.1 , D c = 0.01 .
Figure 1. Velocity streamlines of iteration method I (the first line), iteration method II (the second line) and iteration method III (the third line) with different viscosity coefficients 1.0 (the first column), 1.0 × 10 3 (the second column) and 1.0 × 10 4 (the third column). D a = 1 , β T = 1 , β C = 1 , γ = 0.1 , D c = 0.01 .
Entropy 24 00236 g001aEntropy 24 00236 g001b
Figure 2. Pressure isobars of iteration method I (the first line), iteration method II (the second line) and iteration method III (the third line) with different viscosity coefficients 1.0 (the first column), 1.0 × 10 3 (the second column) and 1.0 × 10 4 (the third column). D a = 1 , β T = 1 , β C = 1 , γ = 0.1 , D c = 0.01 .
Figure 2. Pressure isobars of iteration method I (the first line), iteration method II (the second line) and iteration method III (the third line) with different viscosity coefficients 1.0 (the first column), 1.0 × 10 3 (the second column) and 1.0 × 10 4 (the third column). D a = 1 , β T = 1 , β C = 1 , γ = 0.1 , D c = 0.01 .
Entropy 24 00236 g002aEntropy 24 00236 g002b
Figure 3. Isotherms of iteration method I (the first line), iteration method II (the second line) and iteration method III (the third line) with different viscosity coefficients 1.0 (the first column), 1.0 × 10 3 (the second column) and 1.0 × 10 4 (the third column). D a = 1 , β T = 1 , β C = 1 , γ = 0.1 , D c = 0.01 .
Figure 3. Isotherms of iteration method I (the first line), iteration method II (the second line) and iteration method III (the third line) with different viscosity coefficients 1.0 (the first column), 1.0 × 10 3 (the second column) and 1.0 × 10 4 (the third column). D a = 1 , β T = 1 , β C = 1 , γ = 0.1 , D c = 0.01 .
Entropy 24 00236 g003
Figure 4. The computational domain with its boundary conditions.
Figure 4. The computational domain with its boundary conditions.
Entropy 24 00236 g004
Figure 5. Isotherms of iteration method I (the first line), iteration method II (the second line), iteration method III (the third line) with different viscosity coefficients 1.0 (the first column), 1.0 × 10 3 (the second column) and 1.0 × 10 4 (the third column). D a = 1 , β T = 1 , β C = 1 , γ = 0.1 , D c = 0.01 .
Figure 5. Isotherms of iteration method I (the first line), iteration method II (the second line), iteration method III (the third line) with different viscosity coefficients 1.0 (the first column), 1.0 × 10 3 (the second column) and 1.0 × 10 4 (the third column). D a = 1 , β T = 1 , β C = 1 , γ = 0.1 , D c = 0.01 .
Entropy 24 00236 g005aEntropy 24 00236 g005b
Figure 6. Velocity streamlines (the first line) and pressure isobars (the second line) of iteration method I with different mesh size h = 1 16 (the first column), h = 1 32 (the second column) and h = 1 64 (the third column). ν = 1 , D a = 0.01 , β T = 100 , β C = 100 , γ = 0.1 , D c = 0.1 .
Figure 6. Velocity streamlines (the first line) and pressure isobars (the second line) of iteration method I with different mesh size h = 1 16 (the first column), h = 1 32 (the second column) and h = 1 64 (the third column). ν = 1 , D a = 0.01 , β T = 100 , β C = 100 , γ = 0.1 , D c = 0.1 .
Entropy 24 00236 g006

6. Conclusions

In conclusion, for solving stationary double-diffusive natural convection equations, three iterative methods have their own advantages under different viscosity numbers. In the case of 0 < σ < 1 4 , all methods can export data. Moreover, in the case of 1 4 σ < 1 3 , iterative method I and II can run well. Finally, in the case of 1 3 σ < 1 , only iterative method I can export data.
From the perspective of physical applications, these finite element iterative methods can be used to simulate different double-diffusive natural convection models, such as the aluminum oxide nanofluid natural convection heat transfer, the natural convection flow of a suspension containing nano-encapsulated. Furthermore, some different boundary conditions of these models with some different calculation areas should be considered, such as the T-geometry enclosure porous cavity, L-geometry cavity, and porous cavity.

Author Contributions

Investigation, Y.W. and P.H.; Supervision, P.H.; Writing—original draft, Y.W.; Writing—review and editing, P.H. All authors have read and agreed to the published version of the manuscript.

Funding

This research was funded by the Natural Science Foundation of Xinjiang Province grant number 2021D01E11.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Data sharing not applicable.

Acknowledgments

The authors sincerely thank the editor and anonymous reviewers for their valuable comments and suggestions which led to a large improvement of the paper.

Conflicts of Interest

The authors declare no conflict of interest.

Abbreviations

abilinear formwmapping difference
Aa mappingWtemperature space
ctrilinear formxdimensionless coordinate
CconcentrationXvelocity space
D a Darcy numberydimensionless coordinate
D c mass diffusivity Greeksymbols
eiterative error of velocity β T thermal expansion coefficient
fforcing function β C solutal expansion coefficient
ggravitational acceleration vector β positive constant
hmesh size γ heat diffusivity
Hdual space ψ test function for temperature
kiterative step σ uniqueness condition
Ktriangular region λ constant[0,1]
LLebegue space ν viscosity
m m = | g | max { | β T | , | β C | } α Poincaré constant
Mpressure space η iterative error of pressure
niterative step δ iterative error of concentration
Nconstant ξ iterative error of temperature
pfluid pressure Subscript
Ppolynomiali1,2
qtest function for pressurehfinite element discretization
Qconcentration space
stest function for concentration
Ttemperature
uvelocity field
vtest function for velocity
Vsubspace of the velocity space

References

  1. Ghorayeb, K.; Mojtabi, A. Double-diffusive convection in a vertical rectangular cavity. Phys. Fluids 1997, 9, 2339–2348. [Google Scholar] [CrossRef] [Green Version]
  2. Mirouh, G.M.; Garaud, P.; Stellmach, S.; Traxler, A.L.; Wood, T.S. A new model for mixing by double-diffusive convection (semi-convection). I. The conditions for layer formation. Astrophys. J. 2012, 750, 61. [Google Scholar] [CrossRef]
  3. Tone, F.; Wang, X.M.; Wirosoetisno, D. Long-time dynamics of 2d double-diffusive convection: Analysis and/of numerics. Numer. Math. 2015, 130, 541–566. [Google Scholar] [CrossRef] [Green Version]
  4. Mehryan, S.A.M.; Izadpanahi, E.; Ghalambaz, M.; Chamkha, A.J. Mixed convection flow caused by an oscillating cylinder in a square cavity filled with Cu–Al 2O3/water hybrid nanofluid. J. Therm. Anal. Calorim. 2019, 137, 965–982. [Google Scholar] [CrossRef]
  5. Mehryan, S.A.M.; Ghalambaz, M.; Gargari, L.S.; Hajjar, A.; Sheremet, M. Natural convection flow of a suspension containing nano-encapsulated phase change particles in an eccentric annulus. J. Energy Storage 2020, 28, 101–236. [Google Scholar] [CrossRef]
  6. Tiji, M.E.; Mahdi, J.M.; Mohammed, H.I.; Majdi, H.S.; Ebrahimi, A.; Mahani, R.B.; Yaïci, W. Natural convection effect on solidification enhancement in a multi-tube latent heat storage system: Effect of tubes’ arrangement. Energies 2021, 14, 74–89. [Google Scholar] [CrossRef]
  7. Zhu, W.X.; Huang, P.Z.; Wang, K. Newton iterative method based on finite element discretization for the stationary Darcy-Brinkman equations. Comput. Math. Appl. 2020, 80, 3098–3122. [Google Scholar] [CrossRef]
  8. Çıbık, A.; Kaya, S. A projection-based stabilized finite element method for steady-state natural convection problem. J. Math. Anal. Appl. 2011, 381, 469–484. [Google Scholar] [CrossRef] [Green Version]
  9. Çıbık, A.; Kaya, S. Finite element analysis of a projection-based stabilization method for the Darcy-Brinkman equations in double-diffusive convection. Appl. Numer. Math. 2013, 64, 35–49. [Google Scholar] [CrossRef]
  10. Wu, J.; Huang, P.Z.; Feng, X.L.; Liu, D. An efficient two-step algorithm for steady-state natural convection problem. Int. J. Heat Mass Transf. 2016, 101, 387–398. [Google Scholar] [CrossRef]
  11. Chamkha, A.J.; Doostanidezfuli, A.; Izadpanahi, E.; Ghalambaz, M.J.A.P.T. Phase-change heat transfer of single/hybrid nanoparticles-enhanced phase-change materials over a heated horizontal cylinder confined in a square cavity. Adv. Powder Technol. 2017, 282, 385–397. [Google Scholar] [CrossRef]
  12. Çıbık, A.; Demir, M.; Kaya, S. A family of second order time stepping methods for the Darcy-Brinkman equations. J. Math. Anal. Appl. 2019, 472, 148–175. [Google Scholar] [CrossRef]
  13. Liao, C.; Huang, P.Z. The modified characteristics finite element method for time-dependent Darcy-Brinkman problem. Eng. Comput. 2019, 36, 356–376. [Google Scholar]
  14. Rajabi, M.M.; Fahs, M.; Panjehfouladgaran, A.; Ataie-Ashtiani, B.; Simmons, C.T.; Belfort, B. Uncertainty quantification and global sensitivity analysis of double-diffusive natural convection in a porous enclosure. Int. J. Heat Mass Transf. 2020, 162, 120–291. [Google Scholar] [CrossRef]
  15. Ghaneifar, M.; Raisi, A.; Ali, H.M.; Talebizadehsardari, P. Mixed convection heat transfer of AL2O3 nanofluid in a horizontal channel subjected with two heat sources. J. Therm. Anal. Calorim. 2021, 143, 2761–2774. [Google Scholar] [CrossRef]
  16. Cibik, A.; Eroglu, F.G.; Kaya, S. On the performance of curvature stabilization time stepping methods for double-diffusive natural convection flows in the presence of magnetic field. Numer. Algorithms 2021, 88, 475–498. [Google Scholar] [CrossRef]
  17. Codina, R. An iterative penalty method for the finite element solution of the stationary Navier–Stokes equations. Comput. Methods Appl. Mech. Eng. 1993, 110, 237–262. [Google Scholar] [CrossRef]
  18. He, Y.N.; Zhang, Y.; Shang, Y.Q.; Xu, H. Two-level Newton iterative method for the 2D/3D steady Navier–Stokes equations. Numer. Methods Partial. Differ. Equ. 2012, 28, 1620–1642. [Google Scholar] [CrossRef]
  19. Xu, H.; He, Y.N. Some iterative finite element methods for steady Navier–Stokes equations with different viscosities. J. Comput. Phys. 2013, 232, 136–152. [Google Scholar] [CrossRef]
  20. Shang, Y.Q. A three-step Oseen correction method for the steady Navier–Stokes equations. J. Eng. Math. 2018, 111, 145–163. [Google Scholar] [CrossRef]
  21. Oliver, H.J.; Reiman, A.H.; Monticello, D.A. Solving the 3D MHD equilibrium equations in toroidal geometry by Newton’s method. J. Comput. Phys. 2006, 211, 99–128. [Google Scholar] [CrossRef]
  22. He, Y.N. Euler implicit/explicit iterative scheme for the stationary Navier–Stokes equations. Numer. Math. 2013, 123, 67–96. [Google Scholar] [CrossRef]
  23. Huang, P.Z. Iterative methods in penalty finite element discretizations for the steady Navier–Stokes equations. Numer. Methods Partial. Differ. Equ. 2014, 30, 74–94. [Google Scholar] [CrossRef]
  24. He, Y.N.; Dong, X.J.; Feng, X.L. Uniform Stability and Convergence with Respect to (ν, μ, s, 1 − σ) of the Three Iterative Finite Element Solutions for the 3D Steady MHD Equations. J. Sci. Comput. 2022, 90, 1–28. [Google Scholar] [CrossRef]
  25. Girault, V.; Raviart, P.A. Finite Element Method for Navier—Stokes Equations: Theory and Algorithms; Springer: Berlin/Heidelberg, Germany, 1986. [Google Scholar]
  26. Boland, J.; Layton, W. Error analysis for finite element methods for steady natural convection problems. Numer. Funct. Anal. Optim. 1990, 11, 449–483. [Google Scholar] [CrossRef]
  27. Si, Z.Y.; Shang, Y.Q.; Zhang, T. New one- and two-level Newton iterative mixed finite element methods for stationary conduction-convection problems. Finite Elem. Anal. Des. 2011, 47, 175–183. [Google Scholar] [CrossRef]
  28. Hecht, F. FreeFem++ Version 4.2.1. 2019. Available online: http://www.freefem.org/ff++ (accessed on 3 January 2022).
Table 1. CPU-time in second (iterative step) needed to reach the convergence tolerance.
Table 1. CPU-time in second (iterative step) needed to reach the convergence tolerance.
Scheme ν = 1 ν = 1.0 × 10 2 ν = 1.0 × 10 4
I50.696 (4)174.857 (14)424.661  (41)
II49.432 (4)112.317 (6)
III78.703 (7)
Table 2. Comparison of three iterative methods using P 2 P 1 P 2 P 2 ( h = 1 64 and ν = 1 ).
Table 2. Comparison of three iterative methods using P 2 P 1 P 2 P 2 ( h = 1 64 and ν = 1 ).
Scheme ( u u h n ) 0 u 0 p p h n 0 p 0 ( T T h n ) 0 T 0 ( C C h n ) 0 C 0
I0.0007179120.0002063010.0003591320.00094964
II0.0007179120.0002063030.0003591320.00094964
III0.0007179120.0002062510.0003591450.000949645
Table 3. Comparison of three iterative methods using P 2 P 1 P 2 P 2 ( h = 1 64 and ν = 1.0 × 10 2 ).
Table 3. Comparison of three iterative methods using P 2 P 1 P 2 P 2 ( h = 1 64 and ν = 1.0 × 10 2 ).
Scheme ( u u h n ) 0 u 0 p p h n 0 p 0 ( T T h n ) 0 T 0 ( C C h n ) 0 C 0
I0.0007381370.0002009650.0003591320.00094964
II0.0007381360.000200960.0003591320.00094964
III
Table 4. Comparison of three iterative methods using P 2 P 1 P 2 P 2 ( h = 1 64 and ν = 1.0 × 10 4 ).
Table 4. Comparison of three iterative methods using P 2 P 1 P 2 P 2 ( h = 1 64 and ν = 1.0 × 10 4 ).
Scheme ( u u h n ) 0 u 0 p p h n 0 p 0 ( T T h n ) 0 T 0 ( C C h n ) 0 C 0
I0.007594370.0002032860.0003591330.000949641
II
III
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Wei, Y.; Huang, P. Finite Element Iterative Methods for the Stationary Double-Diffusive Natural Convection Model. Entropy 2022, 24, 236. https://doi.org/10.3390/e24020236

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Wei Y, Huang P. Finite Element Iterative Methods for the Stationary Double-Diffusive Natural Convection Model. Entropy. 2022; 24(2):236. https://doi.org/10.3390/e24020236

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Wei, Yaxin, and Pengzhan Huang. 2022. "Finite Element Iterative Methods for the Stationary Double-Diffusive Natural Convection Model" Entropy 24, no. 2: 236. https://doi.org/10.3390/e24020236

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