# On the Stream Function-Vorticity Finite Element Formulation for Incompressible Flow in Porous Media

Authors: Abdellatif Agouzal; Karam Allali; Siham Binna
Journal: Advances in Numerical Analysis
 (2014)

---

## Abstract

Stream function-vorticity finite element formulation for incompressible flow in porous media is presented. The model consists of the heat equation, the equation for the concentration, and the equations of motion under the Darcy law. The existence of solution for the discrete problem is established. Optimal a priori error estimates are given.


---

## Body

## 1. Introduction

The stream function-vorticity formulation is widely used to perform the numerical simulation of an incompressible fluid flow in porous media [1, 2]. The main advantage of the formulation is a reduction of the numerical problem unknowns with the fact that the continuity equation remains satisfied. In this paper, we are interested in studying the finite element of stream function-vorticity formulation for propagating reaction front in porous media. The model considered is a system of reaction-diffusion equations coupled with the hydrodynamic under the Darcy-Boussinesq approximation in the open bounded convex domain Ω
⊂
R
2 [3–5]:
(1)
(
P
)
{
∂
t
T
-
λ
Δ
T
+
u
·
∇
T
=
α
C
exp
⁡
(
-
E
R
T
)
,
∂
t
C
-
η
Δ
C
+
u
·
∇
C
=
-
α
C
exp
⁡
(
-
E
R
T
)
,
∂
t
u
+
μ
K
u
+
∇
p
=
β
(
T
-
T
0
)
g
γ
,
div
⁡
u
=
0
,
where T is the temperature, C is the concentration, u is the velocity, p is the pressure, λ is the thermal diffusivity, η is the diffusion, μ is the viscosity, K is the permeability, E is the activation energy, R is the universal gas constant, α is the Arrhenius preexponential factor, T
0 is a mean value of temperature, g is the gravity, γ is the upward unit vector, and β is the coefficient of the thermal expansion of the fluid.The boundary conditions are of Dirichlet-Neumann type for the temperature and the concentration and of impermeability type for the normal component of velocity:(2)
T
|
Γ
1
=
C
|
Γ
1
=
0
,
∂
T
∂
n
|
Γ
2
=
∂
C
∂
n
|
Γ
2
=
0
,
u
·
n
|
∂
Ω
=
0
,
u
|
t
=
0
=
u
0
,
T
|
t
=
0
=
T
0
,
C
|
t
=
0
=
C
0
,
where Γ
1 and Γ
2 are disjoint open parts of ∂
Ω such that Γ
1
¯
∪
Γ
2
¯
=
∂
Ω.Due to the incompressibility of the fluid, we introduce the stream functionψ:
(3)
u
=
(
∂
ψ
∂
y
-
∂
ψ
∂
x
)
,
and by introducing also the vorticity ω
=
curl
⁡
(
u
), the problem (
P
) becomes
(4)
(
P
)
{
∂
t
T
-
λ
Δ
T
+
rot
⁡
(
ψ
)
·
∇
T
-
C
g
(
T
)
=
0
,
∂
t
C
-
η
Δ
C
+
rot
⁡
(
ψ
)
·
∇
C
+
C
g
(
T
)
=
0
,
∂
t
ω
+
μ
p
ω
=
f
(
T
)
,
-
Δ
ψ
=
ω
,
where g
(
T
)
=
α
exp
⁡
(
-
E
/
R
T
), f
(
T
)
=
β
g
(
∂
T
/
∂
x
), and μ
p
=
μ
/
K.Because there is no flow of fluid through the boundary (impermeability boundary), we will have the following zero-flux condition:(5)
ψ
|
∂
Ω
=
0
.The paper is organized as follows. We present the variational formulation of the problem in the next section. We establish the existence result in Section3. A priori error estimates are given in Section 4. We conclude in the last section.


## 2. The Variational Formulation

In order to state the problem in the variational form, we first describe the functional framework needed for our study. We set(6)
X
=
L
2
(
0
,
t
,
H
0
1
(
Ω
)
)
,
W
=
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
,
M
=
L
2
(
0
,
t
,
L
2
(
Ω
)
)
.
We introduce now the constant of Friedrichs-Poincaré which depends on the geometry of the domain Ω:
(7)
ρ
=
sup
⁡
u
∈
H
0
1
(
Ω
)
∥
u
∥
L
2
(
Ω
)
∥
∇
u
∥
L
2
(
Ω
)
.
The variational form of the continued problem (
P
) can be written as the following form:
(8)
(
P
v
)
{
Find
a
(
ψ
,
ω
,
T
,
C
)
∈
X
×
M
¯
×
W
¯
2
,
such
as
,
∀
(
ζ
,
v
,
ϕ
,
ξ
)
∈
X
×
M
×
W
2
,
we
have:
(
∂
t
T
,
ϕ
)
+
λ
(
∇
T
,
∇
ϕ
)
+
1
2
∫
Ω
‍
(
rot
⁡
(
ψ
)
∇
T
)
ϕ
-
1
2
∫
Ω
‍
(
rot
⁡
(
ψ
)
∇
ϕ
)
T
=
∫
Ω
‍
C
g
(
T
)
ϕ
,
(
∂
t
C
,
ξ
)
+
η
(
∇
C
,
∇
ξ
)
+
1
2
∫
Ω
‍
(
rot
⁡
(
ψ
)
∇
C
)
ξ
-
1
2
∫
Ω
‍
(
rot
⁡
(
ψ
)
∇
ξ
)
C
=
-
∫
Ω
‍
C
g
(
T
)
ξ
,
(
∂
t
ω
,
v
)
+
μ
p
(
ω
,
v
)
=
∫
Ω
‍
f
(
T
)
v
,
(
∇
ψ
,
∇
ζ
)
=
(
ω
,
ζ
)
,
where
(9)
M
¯
=
{
ω
∈
M
,
∂
ω
∂
t
∈
L
2
(
0
,
t
,
L
2
(
Ω
)
)
}
,
W
¯
=
{
T
∈
W
,
∂
T
∂
t
∈
L
2
(
0
,
t
,
(
H
0
,
Γ
1
1
(
Ω
)
)
*
)
}
.
Here, (
H
0
,
Γ
1
1
(
Ω
)
)
* is the dual of the space H
0
,
Γ
1
1
(
Ω
). We can see clearly that the parameters and the functions f and g of the problem verify the following conditions:
(10)
The
reals
μ
p
,
η
,
and
λ
are
strictly
positives
,
g
∈
W
1
,
∞
(
R
)
,
∥
g
′
∥
L
∞
(
Ω
)
=
E
α
R
T
i
2
,
g
≥
0
,
∥
g
∥
L
∞
(
Ω
)
=
α
,
f
∈
W
1
,
∞
(
R
)
,
f
(
T
0
)
=
0
,
∀
(
T
1
,
T
2
)
∈
(
H
0
,
Γ
1
1
(
Ω
)
)
2
,
∥
f
(
T
1
)
-
f
(
T
2
)
∥
L
2
(
Ω
)
≤
β
g
∥
∇
(
T
1
-
T
2
)
∥
L
2
(
Ω
)
,
where T
i is the temperature of the unburned mixture. For convenience, for all (
u
,
v
)
∈
X
2 and (
T
,
ϕ
,
C
,
ψ
)
∈
W
4, we introduce the forms defined by
(11)
a
(
u
,
v
)
=
(
∫
Ω
‍
∇
u
∇
v
d
x
)
,
a
1
(
u
,
T
,
ϕ
)
=
1
2
(
∫
Ω
‍
(
rot
⁡
(
u
)
∇
)
T
ϕ
d
x
-
∫
Ω
‍
(
rot
⁡
(
u
)
∇
)
ϕ
T
d
x
)
,
j
(
T
,
ψ
)
=
∫
Ω
‍
∇
T
∇
ψ
d
x
,
k
(
C
,
T
,
ψ
)
=
∫
Ω
‍
C
g
(
T
)
ψ
d
x
.
The variational formulation (
P
v
) is rewritten as follows:
(12)
(
P
v
)
{
Find
a
(
ψ
,
ω
,
T
,
C
)
∈
X
×
M
¯
×
W
¯
2
,
such
as
,
∀
(
v
,
u
,
ϕ
,
ξ
)
∈
M
×
X
×
W
2
,
we
have:
(
∂
t
T
,
ϕ
)
+
λ
j
(
T
,
ϕ
)
+
a
1
(
ψ
,
T
,
ϕ
)
-
k
(
C
,
T
,
ϕ
)
=
0
,
∀
ϕ
∈
W
,
(
∂
t
C
,
ξ
)
+
η
j
(
C
,
ξ
)
+
a
1
(
ψ
,
C
,
ξ
)
+
k
(
C
,
T
,
ξ
)
=
0
,
∀
ξ
∈
W
,
(
∂
t
ω
,
v
)
+
μ
p
(
ω
,
v
)
=
(
f
(
T
)
,
v
)
,
∀
v
∈
M
,
a
(
ψ
,
u
)
=
(
ω
,
u
)
,
∀
u
∈
X
.


## 3. The Existence Result

### 3.1. The Semidiscrete Problem

In order to give the semidiscrete problem, we will need the following spaces:(13)
X
h
⊂
H
0
1
(
Ω
)
,
M
h
⊂
L
2
(
Ω
)
,
W
h
⊂
H
0
,
Γ
1
1
(
Ω
)
,
where h is a strictly positive constant. We assume that the spaces X
h, M
h, and W
h satisfy the following assumptions.(1)
For all0
<
σ
≤
1, there exists a linear continuous operator R
h from W
1
+
σ
,
4
(
Ω
)
∩
W
∘
1,4
(
Ω
) onto X
h such that, for all ψ
∈
W
1
+
σ
,
4
(
Ω
)
∩
W
∘
1,4
(
Ω
),
(14)
∥
ψ
-
R
h
ψ
∥
1,4
≲
h
σ
|
ψ
|
1
+
σ
,
4
,
∫
Ω
‍
∇
(
ψ
-
R
h
ψ
)
∇
v
h
=
0
∀
v
h
∈
X
h
.
(2)
For all0
<
σ
≤
1, there exists a linear continuous operator i
h from H
1
+
σ
(
Ω
)
∩
H
0
1
(
Ω
) onto W
h such that, for all T
∈
H
1
+
σ
(
Ω
)
∩
H
0
1
(
Ω
),
(15)
∥
T
-
i
h
T
∥
1
,
Ω
≲
h
σ
|
T
|
1
+
σ
,
Ω
,
∫
Ω
‍
∇
(
T
-
i
h
T
)
∇
v
h
=
0
∀
v
h
∈
X
h
.Examples of such spaces verifying these conditions are given in [6, 7].The discretized form of the problem(
P
v
) is given as follows:
(16)
(
P
h
)
{
Find
ω
h
∈
C
1
(
0
,
t
,
M
h
)
,
ψ
h
∈
C
0
(
0
,
t
,
X
h
)
,
and
(
C
h
,
T
h
)
∈
(
C
1
(
0
,
t
,
W
h
)
)
2
such
that:
(
∂
t
T
h
,
ϕ
h
)
+
λ
j
(
T
h
,
ϕ
h
)
+
a
1
(
ψ
h
,
T
h
,
ϕ
h
)
-
k
(
C
h
,
T
h
,
ϕ
h
)
=
0
,
∀
ϕ
h
∈
W
h
,
(
∂
t
C
h
,
ξ
h
)
+
η
j
(
C
h
,
ξ
h
)
+
a
1
(
ψ
h
,
C
h
,
ξ
h
)
+
k
(
C
h
,
T
h
,
ξ
h
)
=
0
,
∀
ξ
h
∈
W
h
,
(
∂
t
ω
h
,
v
h
)
+
μ
p
(
ω
h
,
v
h
)
=
(
f
(
T
h
)
,
v
h
)
,
∀
v
h
∈
M
h
,
a
(
ψ
h
,
u
h
)
=
(
ω
h
,
u
h
)
,
∀
u
h
∈
X
h
.
With the initial conditions,
(17)
ω
h
(
t
=
0
)
=
ω
h
0
∈
M
h
,
T
h
(
t
=
0
)
=
T
h
0
∈
W
h
,
C
h
(
t
=
0
)
=
C
h
0
∈
W
h
.
In the sequel, we assume for simplicity that
(18)
ω
0
=
ω
h
0
,
T
0
=
T
h
0
,
C
0
=
C
h
0
.The main result of the paper is written as follows.Theorem 1.
The problem(
P
h
) admits a unique solution. Moreover, for ψ, ω, T, and C solutions of the problem (
P
v
) and for ψ
h, ω
h, T
h, and C
h solutions of the problem (
P
h
), one has
(19)
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
(
W
1,4
(
Ω
)
)
)
+
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
+
∥
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
+
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
≲
h
σ
(
∥
ψ
∥
L
2
(
0
,
t
,
W
1
+
σ
,
4
(
Ω
)
)
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
L
2
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ψ
∥
L
2
(
0
,
t
,
W
1
+
σ
,
4
(
Ω
)
)
)
,
under the following conditions:
(
H
1
)
δ
1
=
1
-
5
N
C
α
E
ρ
2
R
T
b
2
λ
>
0
,
(
H
2
)
δ
2
=
1
-
2
5
ρ
4
α
2
E
N
C
δ
1
λ
η
R
T
b
2
>
0
,
(
H
3
)
δ
2
δ
1
>
2
3
β
g
N
λ
μ
p
(
2
10
α
ρ
2
N
C
η
+
5
N
T
)
,
where
(20)
N
T
=
sup
⁡
(
∥
T
h
∥
L
∞
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
,
∥
T
∥
L
∞
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
)
,
N
C
=
sup
⁡
(
∥
C
h
∥
L
∞
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
,
∥
C
∥
L
∞
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
)
,
M
1
=
sup
⁡
(
∥
ψ
h
∥
L
∞
(
0
,
t
,
(
W
1,4
(
Ω
)
)
)
,
∥
ψ
∥
L
∞
(
0
,
t
,
(
W
1,4
(
Ω
)
)
)
)
,
N
=
sup
⁡
ψ
∈
W
°
(
Ω
)
,
v
∈
(
H
0
1
)
(
Ω
)
,
w
∈
(
H
0
1
)
(
Ω
)
a
1
(
ψ
,
v
,
w
)
∥
v
∥
H
0
1
(
Ω
)
∥
ψ
∥
W
°
1,4
(
Ω
)
∥
w
∥
H
0
1
(
Ω
)
,
0
<
σ
≤
1
.Throughout the paper, we often use the following notation.For eachζ
,
η
>
0
:
ζ
≲
η
⇔
∃
C
⋆
>
0
:
ζ
≤
C
⋆
η; without further specification, the constant C
⋆ is independent of the mesh size and the solutions.


### 3.2. Existence of Semidiscrete Solutions

In order to prove the existence result of the problem(
P
h
), we need the following lemmas; first for the concentration, we have the following.Lemma 2.
For any local solutionC
h of the problem (
P
h
), one has the a priori estimate
(21)
∥
C
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
2
η
∥
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
2
∥
C
h
0
∥
L
2
(
Ω
)
2
.Proof.
By choosingξ
h
=
C
h, as test function in the third equation of the problem (
P
h
), we have
(22)
1
2
d
d
t
∥
C
h
∥
L
2
(
Ω
)
2
+
η
∥
∇
C
h
∥
L
2
(
Ω
)
2
+
a
1
(
C
h
,
T
h
,
C
h
)
=
0
.
Integrating the last equality and noticing that a
1
(
C
h
,
T
h
,
C
h
) is positive, we obtain
(23)
2
η
∥
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
∥
C
h
0
∥
L
2
(
Ω
)
2
,
(24)
∥
C
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
≤
∥
C
h
0
∥
L
2
(
Ω
)
2
.
By using the two last inequalities, we obtain the result of the lemma.Also, for the temperature, we have the following result.Lemma 3.
For any local solutionT
h of the problem (
P
h
), one has the a priori estimate
(25)
∥
T
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
λ
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
(
α
ρ
)
2
λ
η
∥
C
h
0
∥
L
2
(
Ω
)
2
+
2
∥
T
h
0
∥
L
2
(
Ω
)
2
.Proof.
By choosingϕ
h
=
T
h, as test function in the second equation of the problem (
P
h
), we have
(26)
1
2
d
d
t
∥
T
h
∥
L
2
(
Ω
)
2
+
λ
∥
∇
T
h
∥
L
2
(
Ω
)
2
=
k
(
C
h
,
T
h
,
T
h
)
≤
ρ
α
∥
C
h
∥
L
2
(
Ω
)
∥
∇
T
h
∥
L
2
(
Ω
)
≤
(
α
ρ
)
2
2
λ
∥
C
h
∥
L
2
(
Ω
)
2
+
λ
2
∥
∇
T
h
∥
L
2
(
Ω
)
2
.
Via integration of the last inequality, it follows that
(27)
∥
T
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
+
λ
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
(
α
ρ
)
2
λ
∥
C
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
+
∥
T
h
0
∥
L
2
(
Ω
)
2
.
Thus, using inequality (23), we get
(28)
∥
T
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
+
λ
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
(
α
ρ
)
2
2
λ
η
∥
C
h
0
∥
L
2
(
Ω
)
2
+
∥
T
h
0
∥
L
2
(
Ω
)
2
.
It leads to
(29)
∥
T
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
(
α
ρ
)
2
2
λ
η
∥
C
h
0
∥
L
2
(
Ω
)
2
+
∥
T
h
0
∥
L
2
(
Ω
)
2
.
From (28) and (29), it follows that
(30)
∥
T
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
λ
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
(
α
ρ
)
2
λ
η
∥
C
h
0
∥
L
2
(
Ω
)
2
+
2
∥
T
h
0
∥
L
2
(
Ω
)
2
.For the vorticity, we have the following.Lemma 4.
For any local solutionω
h of the problem (
P
h
), one has the estimate
(31)
∥
ω
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
μ
p
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
2
α
ρ
2
λ
2
η
μ
p
∥
C
h
0
∥
L
2
(
Ω
)
2
+
4
λ
μ
p
∥
T
h
0
∥
L
2
(
Ω
)
2
+
2
∥
ω
h
0
∥
L
2
(
Ω
)
2
.Proof.
By choosingv
h
=
ω
h, as test function in the first equation of the problem (
P
h
), we have
(32)
1
2
d
d
t
∥
ω
h
∥
L
2
(
Ω
)
2
+
μ
p
∥
ω
h
∥
L
2
(
Ω
)
2
≤
∥
∇
T
h
∥
L
2
(
Ω
)
∥
ω
h
∥
L
2
(
Ω
)
.
Then, by using triangular inequality, we get
(33)
1
2
d
d
t
∥
ω
h
∥
L
2
(
Ω
)
2
+
μ
p
2
∥
ω
h
∥
L
2
(
Ω
)
2
≤
1
2
μ
p
∥
∇
T
h
∥
L
2
(
Ω
)
2
.
While integrating the last inequality, we obtain
(34)
∥
ω
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
+
μ
p
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
1
μ
p
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
+
∥
ω
h
0
∥
L
2
(
Ω
)
2
.
However, using Lemma3, we have
(35)
∥
ω
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
+
μ
p
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
α
ρ
2
λ
2
η
μ
p
∥
C
h
0
∥
L
2
(
Ω
)
2
+
2
λ
μ
p
∥
T
h
0
∥
L
2
(
Ω
)
2
+
∥
ω
h
0
∥
L
2
(
Ω
)
2
.
It leads to
(36)
∥
ω
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
α
ρ
2
λ
2
η
μ
p
∥
C
h
0
∥
L
2
(
Ω
)
2
+
2
λ
μ
p
∥
T
h
0
∥
L
2
(
Ω
)
2
+
∥
ω
h
0
∥
L
2
(
Ω
)
2
.
From (35) and (36), we obtain
(37)
∥
ω
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
μ
p
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
2
α
ρ
2
λ
2
η
μ
p
∥
C
h
0
∥
L
2
(
Ω
)
2
+
4
λ
μ
p
∥
T
h
0
∥
L
2
(
Ω
)
2
+
2
∥
ω
h
0
∥
L
2
(
Ω
)
2
.Finally, for the stream function, we have the following.Lemma 5.
For any local solutionψ
h of the problem (
P
h
), one has the estimate
(38)
∥
ψ
h
∥
L
2
(
0
,
t
,
W
1,4
(
Ω
)
)
2
≲
2
α
ρ
2
λ
2
η
μ
p
2
∥
C
h
0
∥
L
2
(
Ω
)
2
+
4
λ
μ
p
2
∥
T
h
0
∥
L
2
(
Ω
)
2
+
2
μ
p
∥
ω
h
0
∥
L
2
(
Ω
)
2
.Proof.
Letψ
h be the solution of the problem (
P
h
). From [8, 9], we have
(39)
|
ψ
h
|
W
1,4
(
Ω
)
≲
sup
⁡
v
∈
W
°
1
,
4
/
3
(
Ω
)
〈
∇
(
ψ
h
)
,
∇
v
〉
|
v
|
W
1
,
4
/
3
(
Ω
)
.
However,
(40)
〈
∇
(
ψ
h
)
,
∇
v
〉
=
〈
∇
(
ψ
h
)
,
∇
(
Π
h
v
)
〉
,
where Π
h is the projection operator defined from W
1
,
4
/
3
(
Ω
) onto X
h, such as
(41)
∀
v
∈
W
1
,
4
/
3
(
Ω
)
,
〈
∇
w
h
,
∇
(
Π
h
v
)
-
∇
v
〉
=
0
,
∀
w
h
∈
X
h
.
By choosing v
h
=
ψ
h, as test function in the last equation of the problem (
P
h
), we get
(42)
〈
∇
ψ
h
,
∇
Π
h
v
〉
=
〈
ω
h
,
Π
h
v
〉
.
Therefore,
(43)
〈
∇
ψ
h
,
∇
v
〉
≤
∥
ω
h
∥
L
2
(
Ω
)
∥
Π
h
v
∥
L
2
(
Ω
)
.
Due to the embedding of W
1
,
4
/
3
(
Ω
) onto L
2
(
Ω
), we have
(44)
∥
Π
h
v
∥
L
2
(
Ω
)
≲
|
Π
h
v
|
W
1
,
4
/
3
(
Ω
)
,
and also, due to the propriety of the operator Π
h, we have
(45)
∥
Π
h
v
∥
W
1
,
4
/
3
(
Ω
)
≲
|
v
|
W
1
,
4
/
3
(
Ω
)
,
so
(46)
sup
⁡
v
∈
W
°
1
,
4
/
3
(
Ω
)
〈
∇
ψ
h
,
∇
v
〉
|
v
|
W
1
,
4
/
3
(
Ω
)
≲
∥
ω
h
∥
L
2
(
Ω
)
.
We conclude that
(47)
|
ψ
h
|
W
1,4
(
Ω
)
≲
∥
ω
h
∥
L
2
(
Ω
)
.
Then, we get
(48)
∥
ψ
h
∥
L
2
(
0
,
t
,
(
W
1,4
(
Ω
)
)
)
≲
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
.
Using Lemma4, we conclude that
(49)
∥
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
2
≲
(
2
α
ρ
2
λ
2
η
μ
p
2
∥
C
h
0
∥
L
2
(
Ω
)
2
+
4
λ
μ
p
2
∥
T
h
0
∥
L
2
(
Ω
)
2
+
2
μ
p
∥
ω
h
0
∥
L
2
(
Ω
)
2
)
.Now, we are able to prove the main theorem of this section more precisely.Theorem 6.
The problem(
P
h
) admits at least a solution
(50)
(
ω
h
,
ψ
h
,
C
h
,
T
h
)
∈
H
1
(
0
,
t
,
M
h
)
×
L
2
(
0
,
t
,
X
h
)
×
(
C
1
(
0
,
t
,
W
h
)
)
2
.Proof.
Indeed, it is obvious that the problem(
P
h
) admits a local solution in the interval (
0
,
t
h
). For t
h which is rather small, Lemmas 2, 3, 4, and 5 show that this solution can be defined on the interval (
0
,
t
) for t
>
0.


## 3.1. The Semidiscrete Problem

In order to give the semidiscrete problem, we will need the following spaces:(13)
X
h
⊂
H
0
1
(
Ω
)
,
M
h
⊂
L
2
(
Ω
)
,
W
h
⊂
H
0
,
Γ
1
1
(
Ω
)
,
where h is a strictly positive constant. We assume that the spaces X
h, M
h, and W
h satisfy the following assumptions.(1)
For all0
<
σ
≤
1, there exists a linear continuous operator R
h from W
1
+
σ
,
4
(
Ω
)
∩
W
∘
1,4
(
Ω
) onto X
h such that, for all ψ
∈
W
1
+
σ
,
4
(
Ω
)
∩
W
∘
1,4
(
Ω
),
(14)
∥
ψ
-
R
h
ψ
∥
1,4
≲
h
σ
|
ψ
|
1
+
σ
,
4
,
∫
Ω
‍
∇
(
ψ
-
R
h
ψ
)
∇
v
h
=
0
∀
v
h
∈
X
h
.
(2)
For all0
<
σ
≤
1, there exists a linear continuous operator i
h from H
1
+
σ
(
Ω
)
∩
H
0
1
(
Ω
) onto W
h such that, for all T
∈
H
1
+
σ
(
Ω
)
∩
H
0
1
(
Ω
),
(15)
∥
T
-
i
h
T
∥
1
,
Ω
≲
h
σ
|
T
|
1
+
σ
,
Ω
,
∫
Ω
‍
∇
(
T
-
i
h
T
)
∇
v
h
=
0
∀
v
h
∈
X
h
.Examples of such spaces verifying these conditions are given in [6, 7].The discretized form of the problem(
P
v
) is given as follows:
(16)
(
P
h
)
{
Find
ω
h
∈
C
1
(
0
,
t
,
M
h
)
,
ψ
h
∈
C
0
(
0
,
t
,
X
h
)
,
and
(
C
h
,
T
h
)
∈
(
C
1
(
0
,
t
,
W
h
)
)
2
such
that:
(
∂
t
T
h
,
ϕ
h
)
+
λ
j
(
T
h
,
ϕ
h
)
+
a
1
(
ψ
h
,
T
h
,
ϕ
h
)
-
k
(
C
h
,
T
h
,
ϕ
h
)
=
0
,
∀
ϕ
h
∈
W
h
,
(
∂
t
C
h
,
ξ
h
)
+
η
j
(
C
h
,
ξ
h
)
+
a
1
(
ψ
h
,
C
h
,
ξ
h
)
+
k
(
C
h
,
T
h
,
ξ
h
)
=
0
,
∀
ξ
h
∈
W
h
,
(
∂
t
ω
h
,
v
h
)
+
μ
p
(
ω
h
,
v
h
)
=
(
f
(
T
h
)
,
v
h
)
,
∀
v
h
∈
M
h
,
a
(
ψ
h
,
u
h
)
=
(
ω
h
,
u
h
)
,
∀
u
h
∈
X
h
.
With the initial conditions,
(17)
ω
h
(
t
=
0
)
=
ω
h
0
∈
M
h
,
T
h
(
t
=
0
)
=
T
h
0
∈
W
h
,
C
h
(
t
=
0
)
=
C
h
0
∈
W
h
.
In the sequel, we assume for simplicity that
(18)
ω
0
=
ω
h
0
,
T
0
=
T
h
0
,
C
0
=
C
h
0
.The main result of the paper is written as follows.Theorem 1.
The problem(
P
h
) admits a unique solution. Moreover, for ψ, ω, T, and C solutions of the problem (
P
v
) and for ψ
h, ω
h, T
h, and C
h solutions of the problem (
P
h
), one has
(19)
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
(
W
1,4
(
Ω
)
)
)
+
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
+
∥
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
+
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
≲
h
σ
(
∥
ψ
∥
L
2
(
0
,
t
,
W
1
+
σ
,
4
(
Ω
)
)
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
L
2
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ψ
∥
L
2
(
0
,
t
,
W
1
+
σ
,
4
(
Ω
)
)
)
,
under the following conditions:
(
H
1
)
δ
1
=
1
-
5
N
C
α
E
ρ
2
R
T
b
2
λ
>
0
,
(
H
2
)
δ
2
=
1
-
2
5
ρ
4
α
2
E
N
C
δ
1
λ
η
R
T
b
2
>
0
,
(
H
3
)
δ
2
δ
1
>
2
3
β
g
N
λ
μ
p
(
2
10
α
ρ
2
N
C
η
+
5
N
T
)
,
where
(20)
N
T
=
sup
⁡
(
∥
T
h
∥
L
∞
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
,
∥
T
∥
L
∞
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
)
,
N
C
=
sup
⁡
(
∥
C
h
∥
L
∞
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
,
∥
C
∥
L
∞
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
)
,
M
1
=
sup
⁡
(
∥
ψ
h
∥
L
∞
(
0
,
t
,
(
W
1,4
(
Ω
)
)
)
,
∥
ψ
∥
L
∞
(
0
,
t
,
(
W
1,4
(
Ω
)
)
)
)
,
N
=
sup
⁡
ψ
∈
W
°
(
Ω
)
,
v
∈
(
H
0
1
)
(
Ω
)
,
w
∈
(
H
0
1
)
(
Ω
)
a
1
(
ψ
,
v
,
w
)
∥
v
∥
H
0
1
(
Ω
)
∥
ψ
∥
W
°
1,4
(
Ω
)
∥
w
∥
H
0
1
(
Ω
)
,
0
<
σ
≤
1
.Throughout the paper, we often use the following notation.For eachζ
,
η
>
0
:
ζ
≲
η
⇔
∃
C
⋆
>
0
:
ζ
≤
C
⋆
η; without further specification, the constant C
⋆ is independent of the mesh size and the solutions.


## 3.2. Existence of Semidiscrete Solutions

In order to prove the existence result of the problem(
P
h
), we need the following lemmas; first for the concentration, we have the following.Lemma 2.
For any local solutionC
h of the problem (
P
h
), one has the a priori estimate
(21)
∥
C
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
2
η
∥
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
2
∥
C
h
0
∥
L
2
(
Ω
)
2
.Proof.
By choosingξ
h
=
C
h, as test function in the third equation of the problem (
P
h
), we have
(22)
1
2
d
d
t
∥
C
h
∥
L
2
(
Ω
)
2
+
η
∥
∇
C
h
∥
L
2
(
Ω
)
2
+
a
1
(
C
h
,
T
h
,
C
h
)
=
0
.
Integrating the last equality and noticing that a
1
(
C
h
,
T
h
,
C
h
) is positive, we obtain
(23)
2
η
∥
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
∥
C
h
0
∥
L
2
(
Ω
)
2
,
(24)
∥
C
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
≤
∥
C
h
0
∥
L
2
(
Ω
)
2
.
By using the two last inequalities, we obtain the result of the lemma.Also, for the temperature, we have the following result.Lemma 3.
For any local solutionT
h of the problem (
P
h
), one has the a priori estimate
(25)
∥
T
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
λ
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
(
α
ρ
)
2
λ
η
∥
C
h
0
∥
L
2
(
Ω
)
2
+
2
∥
T
h
0
∥
L
2
(
Ω
)
2
.Proof.
By choosingϕ
h
=
T
h, as test function in the second equation of the problem (
P
h
), we have
(26)
1
2
d
d
t
∥
T
h
∥
L
2
(
Ω
)
2
+
λ
∥
∇
T
h
∥
L
2
(
Ω
)
2
=
k
(
C
h
,
T
h
,
T
h
)
≤
ρ
α
∥
C
h
∥
L
2
(
Ω
)
∥
∇
T
h
∥
L
2
(
Ω
)
≤
(
α
ρ
)
2
2
λ
∥
C
h
∥
L
2
(
Ω
)
2
+
λ
2
∥
∇
T
h
∥
L
2
(
Ω
)
2
.
Via integration of the last inequality, it follows that
(27)
∥
T
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
+
λ
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
(
α
ρ
)
2
λ
∥
C
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
+
∥
T
h
0
∥
L
2
(
Ω
)
2
.
Thus, using inequality (23), we get
(28)
∥
T
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
+
λ
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
(
α
ρ
)
2
2
λ
η
∥
C
h
0
∥
L
2
(
Ω
)
2
+
∥
T
h
0
∥
L
2
(
Ω
)
2
.
It leads to
(29)
∥
T
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
(
α
ρ
)
2
2
λ
η
∥
C
h
0
∥
L
2
(
Ω
)
2
+
∥
T
h
0
∥
L
2
(
Ω
)
2
.
From (28) and (29), it follows that
(30)
∥
T
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
λ
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≤
(
α
ρ
)
2
λ
η
∥
C
h
0
∥
L
2
(
Ω
)
2
+
2
∥
T
h
0
∥
L
2
(
Ω
)
2
.For the vorticity, we have the following.Lemma 4.
For any local solutionω
h of the problem (
P
h
), one has the estimate
(31)
∥
ω
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
μ
p
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
2
α
ρ
2
λ
2
η
μ
p
∥
C
h
0
∥
L
2
(
Ω
)
2
+
4
λ
μ
p
∥
T
h
0
∥
L
2
(
Ω
)
2
+
2
∥
ω
h
0
∥
L
2
(
Ω
)
2
.Proof.
By choosingv
h
=
ω
h, as test function in the first equation of the problem (
P
h
), we have
(32)
1
2
d
d
t
∥
ω
h
∥
L
2
(
Ω
)
2
+
μ
p
∥
ω
h
∥
L
2
(
Ω
)
2
≤
∥
∇
T
h
∥
L
2
(
Ω
)
∥
ω
h
∥
L
2
(
Ω
)
.
Then, by using triangular inequality, we get
(33)
1
2
d
d
t
∥
ω
h
∥
L
2
(
Ω
)
2
+
μ
p
2
∥
ω
h
∥
L
2
(
Ω
)
2
≤
1
2
μ
p
∥
∇
T
h
∥
L
2
(
Ω
)
2
.
While integrating the last inequality, we obtain
(34)
∥
ω
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
+
μ
p
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
1
μ
p
∥
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
+
∥
ω
h
0
∥
L
2
(
Ω
)
2
.
However, using Lemma3, we have
(35)
∥
ω
h
(
t
=
s
)
∥
L
2
(
Ω
)
2
+
μ
p
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
α
ρ
2
λ
2
η
μ
p
∥
C
h
0
∥
L
2
(
Ω
)
2
+
2
λ
μ
p
∥
T
h
0
∥
L
2
(
Ω
)
2
+
∥
ω
h
0
∥
L
2
(
Ω
)
2
.
It leads to
(36)
∥
ω
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
α
ρ
2
λ
2
η
μ
p
∥
C
h
0
∥
L
2
(
Ω
)
2
+
2
λ
μ
p
∥
T
h
0
∥
L
2
(
Ω
)
2
+
∥
ω
h
0
∥
L
2
(
Ω
)
2
.
From (35) and (36), we obtain
(37)
∥
ω
h
∥
L
∞
(
0
,
t
,
L
2
(
Ω
)
)
2
+
μ
p
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
2
≤
2
α
ρ
2
λ
2
η
μ
p
∥
C
h
0
∥
L
2
(
Ω
)
2
+
4
λ
μ
p
∥
T
h
0
∥
L
2
(
Ω
)
2
+
2
∥
ω
h
0
∥
L
2
(
Ω
)
2
.Finally, for the stream function, we have the following.Lemma 5.
For any local solutionψ
h of the problem (
P
h
), one has the estimate
(38)
∥
ψ
h
∥
L
2
(
0
,
t
,
W
1,4
(
Ω
)
)
2
≲
2
α
ρ
2
λ
2
η
μ
p
2
∥
C
h
0
∥
L
2
(
Ω
)
2
+
4
λ
μ
p
2
∥
T
h
0
∥
L
2
(
Ω
)
2
+
2
μ
p
∥
ω
h
0
∥
L
2
(
Ω
)
2
.Proof.
Letψ
h be the solution of the problem (
P
h
). From [8, 9], we have
(39)
|
ψ
h
|
W
1,4
(
Ω
)
≲
sup
⁡
v
∈
W
°
1
,
4
/
3
(
Ω
)
〈
∇
(
ψ
h
)
,
∇
v
〉
|
v
|
W
1
,
4
/
3
(
Ω
)
.
However,
(40)
〈
∇
(
ψ
h
)
,
∇
v
〉
=
〈
∇
(
ψ
h
)
,
∇
(
Π
h
v
)
〉
,
where Π
h is the projection operator defined from W
1
,
4
/
3
(
Ω
) onto X
h, such as
(41)
∀
v
∈
W
1
,
4
/
3
(
Ω
)
,
〈
∇
w
h
,
∇
(
Π
h
v
)
-
∇
v
〉
=
0
,
∀
w
h
∈
X
h
.
By choosing v
h
=
ψ
h, as test function in the last equation of the problem (
P
h
), we get
(42)
〈
∇
ψ
h
,
∇
Π
h
v
〉
=
〈
ω
h
,
Π
h
v
〉
.
Therefore,
(43)
〈
∇
ψ
h
,
∇
v
〉
≤
∥
ω
h
∥
L
2
(
Ω
)
∥
Π
h
v
∥
L
2
(
Ω
)
.
Due to the embedding of W
1
,
4
/
3
(
Ω
) onto L
2
(
Ω
), we have
(44)
∥
Π
h
v
∥
L
2
(
Ω
)
≲
|
Π
h
v
|
W
1
,
4
/
3
(
Ω
)
,
and also, due to the propriety of the operator Π
h, we have
(45)
∥
Π
h
v
∥
W
1
,
4
/
3
(
Ω
)
≲
|
v
|
W
1
,
4
/
3
(
Ω
)
,
so
(46)
sup
⁡
v
∈
W
°
1
,
4
/
3
(
Ω
)
〈
∇
ψ
h
,
∇
v
〉
|
v
|
W
1
,
4
/
3
(
Ω
)
≲
∥
ω
h
∥
L
2
(
Ω
)
.
We conclude that
(47)
|
ψ
h
|
W
1,4
(
Ω
)
≲
∥
ω
h
∥
L
2
(
Ω
)
.
Then, we get
(48)
∥
ψ
h
∥
L
2
(
0
,
t
,
(
W
1,4
(
Ω
)
)
)
≲
∥
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
.
Using Lemma4, we conclude that
(49)
∥
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
2
≲
(
2
α
ρ
2
λ
2
η
μ
p
2
∥
C
h
0
∥
L
2
(
Ω
)
2
+
4
λ
μ
p
2
∥
T
h
0
∥
L
2
(
Ω
)
2
+
2
μ
p
∥
ω
h
0
∥
L
2
(
Ω
)
2
)
.Now, we are able to prove the main theorem of this section more precisely.Theorem 6.
The problem(
P
h
) admits at least a solution
(50)
(
ω
h
,
ψ
h
,
C
h
,
T
h
)
∈
H
1
(
0
,
t
,
M
h
)
×
L
2
(
0
,
t
,
X
h
)
×
(
C
1
(
0
,
t
,
W
h
)
)
2
.Proof.
Indeed, it is obvious that the problem(
P
h
) admits a local solution in the interval (
0
,
t
h
). For t
h which is rather small, Lemmas 2, 3, 4, and 5 show that this solution can be defined on the interval (
0
,
t
) for t
>
0.


## 4. A Priori Error Estimations

In this section, we prove some error estimates on the stream function, on the vorticity, on the temperature, and on the concentration. Subsequently, we assume that there exists0
<
σ
≤
1, such as
(51)
ω
∈
L
2
(
0
,
t
,
H
σ
(
Ω
)
)
∩
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
,
T
,
C
∈
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
∩
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
,
ψ
∈
W
°
1,4
(
Ω
)
∩
W
1
+
σ
,
4
(
Ω
)
.Lemma 7.
For anyω and ω
h solutions of the problems (
P
) and (
P
h
), respectively, one has
(52)
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
≲
h
σ
(
∥
ω
∥
L
2
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
+
3
β
g
μ
p
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
.Proof.
Forω and ω
h, respectively, solutions of the continuous and the semidiscrete problem, we have
(53)
(
∂
t
ω
-
∂
t
ω
h
,
v
h
)
κ
+
μ
p
(
ω
-
ω
h
,
v
h
)
κ
=
(
f
(
T
)
-
f
(
T
h
)
,
v
h
)
κ
,
∀
v
h
∈
X
h
,
where κ is an element of the uniform meshes family τ
h. Let ω
¯
h be the approximation of ω in X
h defined by
(54)
ω
¯
h
|
κ
=
1
mes
(
κ
)
∫
κ
‍
ω
d
x
.
Hence, for all 0
<
σ
<
1 and ω
∈
H
σ
(
Ω
), we have the following result:
(55)
∥
ω
-
ω
¯
h
∥
L
2
(
κ
)
≲
h
κ
σ
∥
ω
∥
H
σ
(
κ
)
,
where h
κ is the diameter of the element κ.
By using equality (53) and setting v
h
=
ω
¯
h
-
ω
h, we obtain
(56)
(
∂
t
ω
¯
h
-
∂
t
ω
h
,
ω
¯
h
-
ω
h
)
κ
+
μ
p
(
ω
¯
h
-
ω
h
,
ω
¯
h
-
ω
h
)
κ
=
-
(
∂
t
ω
-
∂
t
ω
¯
h
,
ω
¯
h
-
ω
h
)
κ
+
μ
p
(
ω
¯
h
-
ω
,
ω
¯
h
-
ω
h
)
κ
+
(
f
(
T
)
-
f
(
T
h
)
,
ω
¯
h
-
ω
h
)
κ
.
Therefore, we have
(57)
1
2
d
d
t
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
2
+
μ
p
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
2
≤
(
∥
∂
t
ω
-
∂
t
ω
¯
h
∥
L
2
(
κ
)
+
β
g
∥
∇
(
T
-
T
h
)
∥
L
2
(
κ
)
)
×
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
+
μ
p
∥
ω
-
ω
¯
h
∥
L
2
(
κ
)
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
.
Then,
(58)
1
2
d
d
t
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
2
+
μ
p
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
2
≤
(
∥
∂
t
ω
-
∂
t
ω
¯
h
∥
L
2
(
κ
)
+
β
g
∥
∇
(
T
-
T
h
)
∥
L
2
(
κ
)
)
×
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
+
μ
p
∥
ω
-
ω
¯
h
∥
L
2
(
κ
)
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
.
However, by using the triangular inequality, we can write
(59)
1
2
d
d
t
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
2
+
μ
p
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
2
≤
3
2
μ
p
∥
∂
t
ω
-
∂
t
ω
¯
h
∥
L
2
(
κ
)
2
+
3
(
β
g
)
2
2
μ
p
∥
∇
(
T
-
T
h
)
∥
L
2
(
κ
)
2
+
3
μ
p
2
∥
ω
¯
h
-
ω
∥
L
2
(
κ
)
2
+
μ
p
2
∥
ω
¯
h
-
ω
∥
L
2
(
κ
)
2
.
It follows that
(60)
1
2
d
d
t
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
2
+
μ
p
2
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
2
≤
3
2
μ
p
∥
∂
t
ω
-
∂
t
ω
¯
h
∥
L
2
(
κ
)
2
+
3
β
2
g
2
2
μ
p
∥
T
-
T
h
∥
H
0
,
Γ
1
1
(
κ
)
2
+
3
μ
p
2
∥
ω
¯
h
-
ω
∥
L
2
(
κ
)
2
.
However, for all 0
<
σ
<
1 and ω
∈
H
1
(
0
,
t
,
H
σ
(
κ
)
), we have
(61)
∥
∂
t
ω
-
∂
t
ω
¯
h
∥
L
2
(
κ
)
2
≲
h
κ
2
σ
∥
∂
t
ω
∥
H
σ
(
κ
)
2
.
While multiplying (60) by 2
/
μ and using (55) and (61), we obtain
(62)
∥
ω
¯
h
-
ω
h
∥
L
2
(
κ
)
2
≲
h
κ
2
σ
(
∥
ω
∥
H
σ
(
κ
)
+
∥
∂
t
ω
∥
H
σ
(
κ
)
)
2
+
3
(
β
g
)
2
μ
p
2
∥
T
-
T
h
∥
H
0
,
Γ
1
1
(
κ
)
.
Finally, by integrating and using Jensen’s inequality, we get
(63)
∥
ω
¯
h
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
≲
h
σ
(
∥
ω
∥
L
2
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
)
+
3
β
g
μ
p
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
.
However,
(64)
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
≤
∥
ω
¯
h
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
+
∥
ω
-
ω
¯
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
.
We conclude that
(65)
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
≲
h
σ
(
∥
ω
∥
L
2
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
)
+
3
β
g
μ
p
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
.Lemma 8.
For anyψ and ψ
h solution of the problems (
P
) and (
P
h
), respectively, one has
(66)
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
≲
h
σ
∥
ψ
∥
L
2
(
0
,
t
,
(
W
1
+
σ
,
4
(
Ω
)
)
)
+
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
.Proof.
Letψ and ψ
h be the solution of the problems (
P
v
) and (
P
h
), respectively. From [8, 9], we have
(67)
|
ψ
-
ψ
h
|
W
1,4
(
Ω
)
≲
sup
⁡
v
∈
W
°
1
,
4
/
3
(
Ω
)
〈
∇
(
ψ
-
ψ
h
)
,
∇
v
〉
|
v
|
W
1
,
4
/
3
(
Ω
)
.
However,
(68)
〈
∇
(
ψ
-
ψ
h
)
,
∇
v
〉
=
〈
∇
(
ψ
-
ψ
h
)
,
∇
(
v
-
Π
h
v
)
〉
+
〈
∇
(
ψ
-
ψ
h
)
,
∇
(
Π
h
v
)
〉
,
where Π
h is the projection operator verifying (41). Moreover, we have
(69)
〈
∇
(
R
h
ψ
-
ψ
h
)
,
∇
(
Π
h
v
-
v
)
〉
=
0
,
〈
∇
(
R
h
ψ
-
ψ
)
,
∇
(
Π
h
v
)
〉
=
0
.
Then, we get(70)
〈
∇
(
ψ
-
ψ
h
)
,
∇
v
〉
=
〈
∇
(
ψ
-
R
h
ψ
)
,
∇
(
v
-
Π
h
v
)
〉
+
〈
∇
(
ψ
-
ψ
h
)
,
∇
(
Π
h
v
)
〉
.
By choosing v
h
=
Π
h
v as test function in the last equation of the problem (
P
h
), we obtain
(71)
〈
∇
(
ψ
-
ψ
h
)
,
∇
v
〉
=
〈
∇
(
ψ
-
R
h
ψ
)
,
∇
v
〉
+
〈
ω
-
ω
h
,
Π
h
v
〉
.
It leads to
(72)
〈
∇
(
ψ
-
R
h
ψ
)
,
∇
v
〉
≤
|
ψ
-
R
h
ψ
|
W
1,4
(
Ω
)
|
v
|
W
1
,
4
/
3
(
Ω
)
,
〈
ω
-
ω
h
,
Π
h
v
〉
≤
∥
ω
-
ω
h
∥
L
2
(
Ω
)
∥
Π
h
v
∥
L
2
(
Ω
)
.
On the other hand, due to the embedding of W
1
,
4
/
3
(
Ω
) onto L
2
(
Ω
), we have
(73)
∥
Π
h
v
∥
L
2
(
Ω
)
≲
|
Π
h
v
|
W
1
,
4
/
3
(
Ω
)
.
In addition,
(74)
|
Π
h
v
|
W
1
,
4
/
3
(
Ω
)
≲
|
v
|
W
1
,
4
/
3
(
Ω
)
,
so
(75)
sup
⁡
v
∈
W
1
,
4
/
3
(
Ω
)
〈
ω
-
ω
h
,
Π
h
v
〉
|
v
|
W
1
,
4
/
3
(
Ω
)
≲
∥
ω
-
ω
h
∥
L
2
(
Ω
)
.
From inequalities (67) and (72), we have
(76)
|
ψ
-
ψ
h
|
W
1,4
(
Ω
)
≲
|
ψ
-
R
h
ψ
|
W
1,4
(
Ω
)
+
∥
ω
-
ω
h
∥
L
2
(
Ω
)
.
We conclude that(77)
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
≲
h
σ
∥
ψ
∥
L
2
(
0
,
t
,
(
W
1
+
σ
,
4
(
Ω
)
)
)
+
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
.Also, we have the following error estimate for temperature.Lemma 9.
If we assume that the hypothesis(
H
1
) is verified, then one has
(78)
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
≲
h
σ
(
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
+
5
α
ρ
2
δ
1
λ
∥
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
+
5
N
N
T
δ
1
λ
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
.Proof.
ForT solution of the problem (
P
) and T
h solution of the problem (
P
h
), we have
(79)
(
∂
t
T
-
∂
t
T
h
,
ϕ
h
)
+
λ
a
(
T
-
T
h
,
ϕ
h
)
+
a
1
(
ψ
-
ψ
h
,
T
,
ϕ
h
)
+
a
1
(
ψ
h
,
T
-
T
h
,
ϕ
h
)
=
k
(
C
,
T
,
ϕ
h
)
-
k
(
C
h
,
T
h
,
ϕ
h
)
.
Therefore, using the propriety of the operator i
h, we have
(80)
a
(
i
h
T
-
T
,
ϕ
h
)
=
0
.
So,
(81)
(
i
h
∂
t
T
-
∂
t
T
h
,
ϕ
h
)
+
λ
a
(
i
h
T
-
T
h
,
ϕ
h
)
=
(
i
h
∂
t
T
-
∂
t
T
,
ϕ
h
)
-
a
1
(
ψ
-
ψ
h
,
T
,
ϕ
h
)
-
a
1
(
ψ
h
,
T
-
T
h
,
ϕ
h
)
+
k
(
C
,
T
,
ϕ
h
)
-
k
(
C
h
,
T
h
,
ϕ
h
)
.
By setting ϕ
h
=
i
h
T
-
T
h, we obtain
(82)
1
2
d
d
t
∥
i
h
T
-
T
h
∥
L
2
(
Ω
)
2
+
λ
∥
∇
(
i
h
T
-
T
h
)
∥
L
2
(
Ω
)
2
=
-
(
∂
t
T
-
i
h
∂
t
T
,
ϕ
h
)
-
a
1
(
ψ
-
ψ
h
,
T
,
ϕ
h
)
-
a
1
(
ψ
h
,
T
-
i
h
T
,
i
h
T
-
T
h
)
+
k
(
C
,
T
,
ϕ
h
)
-
k
(
C
h
,
T
h
,
ϕ
h
)
.
Then, we get
(83)
1
2
d
d
t
∥
i
h
T
-
T
h
∥
L
2
(
Ω
)
2
+
λ
∥
∇
(
i
h
T
-
T
h
)
∥
L
2
(
Ω
)
2
≤
(
ρ
∥
∂
t
T
-
i
h
∂
t
T
∥
L
2
(
Ω
)
+
N
N
T
∥
ψ
-
ψ
h
∥
W
°
1,4
(
Ω
)
+
N
M
1
∥
∇
(
T
-
i
h
T
)
∥
L
2
(
Ω
)
ρ
∥
∂
t
T
-
i
h
∂
t
T
∥
L
2
(
Ω
)
+
N
N
T
∥
ψ
-
ψ
h
∥
W
°
1,4
(
Ω
)
)
∥
∇
(
i
h
T
-
T
h
)
∥
L
2
(
Ω
)
+
|
k
(
C
,
T
,
ϕ
h
)
+
k
(
C
h
,
T
h
,
ϕ
h
)
|
.
We have also
(84)
|
k
(
C
,
T
,
ϕ
h
)
-
k
(
C
h
,
T
h
,
ϕ
h
)
|
≤
α
ρ
2
∥
C
-
C
h
∥
H
0
,
Γ
1
1
(
Ω
)
∥
ϕ
h
∥
H
0
,
Γ
1
1
(
Ω
)
+
N
C
α
E
R
T
i
2
ρ
2
∥
T
-
T
h
∥
H
0
,
Γ
1
1
(
Ω
)
∥
ϕ
h
∥
H
0
,
Γ
1
1
(
Ω
)
.
From (83) and (84), we get
(85)
1
2
d
d
t
∥
i
h
T
-
T
h
∥
L
2
(
Ω
)
2
+
λ
∥
∇
(
i
h
T
-
T
h
)
∥
L
2
(
Ω
)
2
≤
(
ρ
∥
∂
t
T
-
i
h
∂
t
T
∥
L
2
(
Ω
)
+
N
N
T
∥
ψ
-
ψ
h
∥
W
°
1,4
(
Ω
)
+
N
M
1
∥
∇
(
T
-
i
h
T
)
∥
L
2
(
Ω
)
)
∥
∇
(
i
h
T
-
T
h
)
∥
L
2
(
Ω
)
+
(
α
ρ
2
∥
C
-
C
h
∥
H
0
,
Γ
1
1
(
Ω
)
+
N
C
α
E
R
T
i
2
ρ
2
∥
T
-
T
h
∥
H
0
,
Γ
1
1
(
Ω
)
)
×
∥
∇
(
i
h
T
-
T
h
)
∥
L
2
(
Ω
)
.
Then, we deduce
(86)
1
2
d
d
t
∥
i
h
T
-
T
h
∥
L
2
(
Ω
)
2
+
λ
2
∥
∇
(
i
h
T
-
T
h
)
∥
L
2
(
Ω
)
2
≤
5
ρ
2
2
λ
∥
∂
t
T
-
i
h
∂
t
T
∥
L
2
(
Ω
)
2
+
5
N
2
M
1
2
2
λ
∥
∇
(
T
-
i
h
T
)
∥
L
2
(
Ω
)
2
+
5
α
2
ρ
4
2
λ
∥
∇
(
C
-
C
h
)
∥
L
2
(
Ω
)
2
+
5
N
2
N
T
2
2
λ
∥
ψ
-
ψ
h
∥
W
°
1,4
(
Ω
)
2
+
5
N
C
2
α
2
E
2
ρ
4
2
λ
R
2
T
i
4
∥
∇
(
T
-
T
h
)
∥
L
2
(
Ω
)
2
.
By multiplying the last inequality by 2
/
λ, we obtain
(87)
∥
i
h
T
-
T
h
∥
H
0
,
Γ
1
1
(
Ω
)
2
≤
5
ρ
2
λ
2
∥
∂
t
T
-
i
h
∂
t
T
∥
L
2
(
Ω
)
2
+
5
N
2
M
1
2
λ
2
∥
∇
(
T
-
i
h
T
)
∥
L
2
(
Ω
)
2
+
5
α
2
ρ
4
λ
2
∥
∇
(
C
-
C
h
)
∥
L
2
(
Ω
)
2
+
5
N
2
N
T
2
λ
2
∥
ψ
-
ψ
h
∥
W
°
1,4
(
Ω
)
2
+
5
N
C
2
(
α
E
)
2
ρ
4
R
2
T
i
4
λ
2
∥
∇
(
T
-
T
h
)
∥
L
2
(
Ω
)
2
.
Via integration of the last inequality, we obtain
(88)
∥
i
h
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≲
(
h
σ
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
h
σ
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
2
+
5
α
2
ρ
4
λ
2
∥
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
+
5
N
2
N
T
2
λ
2
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
2
+
5
N
C
2
(
α
E
)
2
ρ
4
R
2
T
i
4
λ
2
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
.
Therefore,
(89)
∥
i
h
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
≲
(
h
σ
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
h
σ
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
+
5
α
ρ
2
λ
∥
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
+
5
N
N
T
λ
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
+
5
N
C
α
E
ρ
2
R
T
i
2
λ
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
.
However,
(90)
∥
T
-
T
h
∥
H
0
,
Γ
1
1
(
Ω
)
≤
∥
T
-
i
h
T
∥
H
0
,
Γ
1
1
(
Ω
)
+
∥
i
h
T
-
T
h
∥
H
0
,
Γ
1
1
(
Ω
)
.
Then, we have
(91)
∥
T
-
T
h
∥
H
0
,
Γ
1
1
(
Ω
)
≲
h
σ
∥
T
∥
H
1
+
σ
(
Ω
)
+
∥
i
h
T
-
T
h
∥
H
0
,
Γ
1
1
(
Ω
)
.
Inequalities (89) and (91) allow us to obtain the following estimate:
(92)
(
1
-
5
N
C
α
E
ρ
2
R
T
i
2
λ
)
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
≲
h
σ
(
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
+
5
α
ρ
2
λ
∥
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
+
5
N
N
T
λ
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
.
Finally, if we set
(93)
δ
1
=
1
-
5
N
C
α
E
ρ
2
R
T
i
2
λ
>
0
,
we obtain the following estimate:
(94)
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
≲
h
σ
(
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
+
5
α
ρ
2
δ
1
λ
(
∥
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
)
+
5
N
N
T
δ
1
λ
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
.In addition, we have the following error estimate on concentration.Lemma 10.
ForC and C
h, respectively, solution of the problems (
P
) and (
P
h
), one has
(95)
∥
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
≲
h
σ
(
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
+
2
2
N
N
C
η
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
+
2
N
C
α
E
ρ
2
η
R
T
i
2
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
.Proof.
By choosingχ
h
=
i
h
C
-
C
h as test function, we have
(96)
1
2
d
d
t
∥
i
h
C
-
C
h
∥
0
,
t
,
L
2
(
Ω
)
2
+
η
∥
∇
(
i
h
C
-
C
h
)
∥
0
,
t
,
L
2
(
Ω
)
2
=
-
(
∂
t
C
-
i
h
∂
t
C
,
χ
h
)
-
a
1
(
ψ
-
ψ
h
,
C
,
χ
h
)
-
a
1
(
ψ
h
,
C
-
i
h
C
,
i
h
C
-
C
h
)
-
k
(
C
,
T
,
χ
h
)
+
k
(
C
h
,
T
h
,
χ
h
)
.
However,
(97)
k
(
C
,
T
,
χ
h
)
-
k
(
C
h
,
T
h
,
χ
h
)
=
k
(
C
,
T
,
χ
h
)
-
k
(
i
h
C
,
T
h
,
χ
h
)
+
k
(
χ
h
,
T
h
,
χ
h
)
,
k
(
C
,
T
,
χ
h
)
-
k
(
i
h
C
,
T
h
,
χ
h
)
=
k
(
C
-
i
h
C
,
T
,
χ
h
)
-
k
(
i
h
C
,
T
-
T
h
,
χ
h
)
≤
α
ρ
2
∥
∇
(
C
-
i
h
C
)
∥
L
2
(
Ω
)
∥
∇
χ
h
∥
L
2
(
Ω
)
+
∫
‍
i
h
C
(
g
(
T
)
-
g
(
T
h
)
)
χ
h
.
Then, we have
(98)
k
(
C
,
T
,
χ
h
)
-
k
(
i
h
C
,
T
h
,
χ
h
)
≤
α
ρ
2
∥
C
-
i
h
C
∥
H
0
,
Γ
1
1
(
Ω
)
∥
χ
h
∥
H
0
,
Γ
1
1
(
Ω
)
+
N
C
α
E
R
T
i
2
ρ
2
∥
T
-
T
h
∥
H
0
,
Γ
1
1
(
Ω
)
∥
χ
h
∥
H
0
,
Γ
1
1
(
Ω
)
.
Using (96) and (98), we obtain
(99)
1
2
d
d
t
∥
i
h
C
-
C
h
∥
L
2
(
Ω
)
2
+
η
∥
∇
(
i
h
C
-
C
h
)
∥
L
2
(
Ω
)
2
≤
(
ρ
∥
∂
t
C
-
i
h
∂
t
C
∥
L
2
(
Ω
)
+
N
N
C
∥
ψ
-
ψ
h
∥
W
°
1,4
(
Ω
)
+
N
M
1
∥
∇
(
C
-
i
h
C
)
∥
L
2
(
Ω
)
ρ
∥
∂
t
C
-
i
h
∂
t
C
∥
L
2
(
Ω
)
+
N
N
C
∥
ψ
-
ψ
h
∥
W
°
1,4
(
Ω
)
)
∥
∇
(
i
h
C
-
C
h
)
∥
L
2
(
Ω
)
+
(
α
ρ
2
∥
C
-
i
h
C
∥
H
0
,
Γ
1
1
(
Ω
)
+
N
C
α
E
R
T
i
2
ρ
2
∥
∇
(
T
-
T
h
)
∥
L
2
(
Ω
)
)
×
∥
∇
χ
h
∥
L
2
(
Ω
)
.
By using Young’s inequality, we obtain
(100)
1
2
d
d
t
∥
i
h
C
-
C
h
∥
L
2
(
Ω
)
2
+
η
2
∥
∇
(
i
h
T
-
T
h
)
∥
L
2
(
Ω
)
2
≤
2
ρ
2
η
∥
∂
t
C
-
i
h
∂
t
C
∥
L
2
(
Ω
)
2
+
2
N
2
M
1
2
η
∥
∇
(
C
-
i
h
C
)
∥
L
2
(
Ω
)
2
+
2
α
2
ρ
4
η
∥
∇
(
C
-
i
h
C
)
∥
L
2
(
Ω
)
2
+
2
N
2
N
C
2
η
∥
ψ
-
ψ
h
∥
W
°
1,4
(
Ω
)
2
+
N
C
α
E
R
T
i
2
ρ
2
∥
∇
(
T
-
T
h
)
∥
L
2
(
Ω
)
∥
∇
χ
h
∥
L
2
(
Ω
)
.
While multiplying the last inequality by 2
/
η and via integration, we obtain
(101)
∥
i
h
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≲
(
h
σ
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
h
σ
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
2
+
4
N
2
N
C
2
η
2
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
2
+
2
N
C
α
E
ρ
2
η
R
T
i
2
∥
∇
(
T
-
T
h
)
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
∥
∇
χ
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
.
So, we get(102)
∥
i
h
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
≲
(
h
σ
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
h
σ
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
2
+
4
N
2
N
C
2
η
2
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
2
+
2
N
C
2
α
2
E
2
ρ
4
R
2
T
i
4
η
2
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
+
1
2
∥
i
h
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
2
.
Finally, by the triangular inequality, we obtain(103)
∥
C
-
C
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
≲
h
σ
(
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
+
2
2
N
N
C
η
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
+
2
N
C
α
E
ρ
2
η
R
T
i
2
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
.Lemma 11.
ForT and T
h solution of the problems (
P
) and (
P
h
), respectively, one has
(104)
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
≲
h
σ
(
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
)
+
(
2
10
α
ρ
2
N
N
C
δ
2
δ
1
λ
η
+
5
N
N
T
δ
2
δ
1
λ
)
×
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
.Proof.
By using Lemmas9 and 10, we have
(105)
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
≲
h
σ
(
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
)
+
(
2
10
α
ρ
2
N
N
C
δ
1
λ
η
+
5
N
N
T
δ
1
λ
)
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
+
2
5
ρ
4
α
2
E
N
C
δ
1
λ
η
R
T
i
2
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
.
If we set
(106)
δ
2
=
1
-
2
5
ρ
4
α
2
E
N
C
δ
1
λ
η
R
T
i
2
>
0
,
then we get
(107)
δ
2
∥
T
-
T
h
∥
L
2
(
0
,
t
,
H
0
,
Γ
1
1
(
Ω
)
)
≲
h
σ
(
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
)
+
(
2
10
α
ρ
2
N
N
C
δ
1
λ
η
+
5
N
N
T
δ
1
λ
)
×
∥
ψ
-
ψ
h
∥
L
2
(
0
,
t
,
W
°
1,4
(
Ω
)
)
.Lemma 12.
One assumes that the hypotheses(
H
1
), (
H
2
), and (
H
3
) are verified, so
(108)
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
≲
h
σ
(
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ψ
∥
L
2
(
0
,
t
,
W
1
+
σ
,
4
(
Ω
)
)
+
∥
ω
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
L
2
(
0
,
t
,
H
σ
(
Ω
)
)
)
.Proof.
First of all, by using Lemmas7, 8, and 11, we obtain
(109)
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
≲
h
σ
(
∥
ψ
∥
L
2
(
0
,
t
,
W
1
+
σ
,
4
(
Ω
)
)
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
L
2
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ψ
∥
L
2
(
0
,
t
,
W
1
+
σ
,
4
(
Ω
)
)
)
+
3
β
g
μ
p
(
2
10
α
ρ
2
N
N
C
δ
2
δ
1
λ
η
+
5
N
N
T
δ
2
δ
1
λ
)
×
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
.
If we set(110)
A
=
(
2
10
α
ρ
2
N
C
η
+
5
N
T
)
,
assuming that
(111)
3
β
g
N
δ
2
δ
1
λ
μ
p
A
≤
1
-
δ
7
with
0
<
δ
7
<
1
,
for example, if δ
7
=
1
/
2, we obtain
(112)
δ
2
δ
1
>
2
3
β
g
N
λ
μ
p
A
,
and then we obtain
(113)
∥
ω
-
ω
h
∥
L
2
(
0
,
t
,
L
2
(
Ω
)
)
≲
h
σ
(
∥
ψ
∥
L
2
(
0
,
t
,
W
1
+
σ
,
4
(
Ω
)
)
∥
T
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
T
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
C
∥
L
2
(
0
,
t
,
H
1
+
σ
(
Ω
)
)
+
∥
C
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
L
2
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ω
∥
H
1
(
0
,
t
,
H
σ
(
Ω
)
)
+
∥
ψ
∥
L
2
(
0
,
t
,
W
1
+
σ
,
4
(
Ω
)
)
)
.Finally, using Lemmas8, 10, 11, and 12, we obtain the a priori error estimate at the same time on the temperature, the concentration, the stream function, and the vorticity, so the main Theorem 1 is proved.


## 5. Conclusion

The propagation of reaction front in porous media is modelled by a system of equations, coupling hydrodynamic equations and the reaction-diffusion equations. We have taken into account Darcy-Boussinesq approximation. We have adopted the stream function-vorticity formulation of the Darcy equation and we have chosen the appropriate functional framework for our variational problem. We have proved the existence result for the semidiscrete solution. Furthermore, we have established an optimal a priori estimate on the temperature, the concentration, the stream function, and the vorticity.



---

*Source: 459761-2014-09-18.xml*