EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 1, 2021, 82-111
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Existence of weak solution of Navier-Stokes-Fourier
System with a new successive approximation method

Rabe Bade 1,∗, Hedia Chaker2

1 Department of Mathematics and Computer Sciences Faculty of Science and Technology,
Abdou Moumouni University, BP: 10662 Niamey, Niger
2 LAMSIN, National School of Engineers of Tunis, University of Tunis El-Manar, BP:37,
1002-Tunis, Tunisia

Abstract. In this paper we prove the existence of a weak solution of the complete compressible
Navier-Stokes system. We follow an previous work where we added an artificial viscosity in the
continuity equation and then rewrite the system in hyperbolic and symmetric form. Our study
is based on the symmetric hyperbolic theory. We use for this aim a successive approximation in
time to show the existence of the hyperbolic system solution and by the fixed point theorem the
compacity property of some appropriate sobolev spaces and some established a priori estimates we
can pass to several limits to prove our result. As state law, we use the Stiffened gas law.

2020 Mathematics Subject Classifications: 76N06, 35D30, 47J25

Key Words and Phrases: Fluid dynamic, Weak solution, Iterative procedures.

1. Introduction

The compressible Navier-Stokes equations describe a viscous fluid flow in a bounded
tridimensional domain, they are given by:

∂ρ

∂t
+∇ · (ρu) = 0,

∂ρu

∂t
+∇ · (ρu⊗u)− µ4u− (

µ

3
+ λ)∇(∇ · u) +∇P = f ,

∂ρE

∂t
+∇ ·

((
ρE + P

)
u

)
= −∇ ·

(
q− [µ

(
∇u +∇uT

)
+
(
λ− 2

3
µ
)
Id∇ · u]u

)
+u · f .

(1)

Where ρ is the density, u = (u1, u2, u3) are the velocity components, P is the pressure, E
is the tolal energy, f is the external force and q is the heat flux.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i1.3903

Email addresses: baderabe@yahoo.fr (R. Bade), hedia.chaker@enit.rnu.tn (H. Chaker)

http://www.ejpam.com 82 c© 2021 EJPAM All rights reserved.



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 83

The mathematical analysis of these equations was made only recently with the works of
[16], in particular in the case of barotropic fluids P (ρ) = aργ . The author had established
a regularity result of weak solution under certain conditions related to γ. Studying a more
general case, [9] had looked the case of non-monotone pressure-law. [7, 8] was interested
in the system in dimension two where he refined the conditions on the initial data. The
author established a regularity result with periodic boundary conditions and especially in
prevision of possible appearance of the vacuum. In the case of the complete Navier-Stokes
equation, with heat conduction, [16] had sketched the existence of global weak solution
by compactness arguments and some a priori estimates. However, the author indicated
that these a priori estimates are very difficult to be establish and sometimes unavailable
except in a very restrictive cases. More recently, [19] made a major contribution in the
development of the general mathematical theory of this system without any limitation on
the data size. The authors proved the existence of a variational solution using a series
of approximations: artificial pressure, relaxation in the continuity equation and finally a
regularized thermal energy equation. Passing to the limit they obtain their variational
solution of the Navier-Stokes equations. Although the theory is very nice, the question of
the physical sense of these approximations could possibly arise. With another approach,
[9, 10] justify the existence of a variational solution using in addition some a priori esti-
mates.
In this work, we establish the existence of a weak solution for the compressible Navier-
Stokes system using the theory of the symmetric hyperbolic system [11, 13, 17]. For that
purpose, we add an artificial viscosity term −ε4ρ in the continuity equation. Classically
this artificial viscosity is added to regularize the solution, but here we add it in the aim of
rewriting and obtaining a hyperbolic system from compressible Navier-Stokes equations.
After adding −ε4ρ, we make a change of variable similar to that used by [5]. With this
change of variable, we come down to a system of order 1.
A semi-discretization in time and a linearization of the obtained system allow us to use
the results of [13]. By a fixed point theorem we show the existence of the solution of the
stationary, nonlinear system. Then by some a priori estimates we can pass to limit in time
(4t→ 0) and after when ε→ 0 we justify the existence of Navier-Stokes weak solution.

The outline of this paper is organized as follow. In Section 2 we will present the added
system and our main results. In section 3 we will prove that the density in the added
system remain strictly positive when its initial value is positive. In section 4 we show
how we obtain the hyperbolic system and we will present the successive approximations
which allows us to construct and prove the existence of the weak solution of the obtained
system. Finally in Section 5 we pass to limit in the artificial viscosity and then we prove
the existence of the weak solution for the compressible Navier-Stokes system.

2. Preliminaries and main results

Let us consider a bounded domain Ω of R3 with the boundary ∂Ω is supposed to be
enough regular, [0, T ] be a time interval with T > 0. We assume that heat flux q is



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 84

determined by the Fourier’s law (q = −k∇θ, with k is the thermal conductivity of the
fluid) and the pressure P , the internal energy e are described by the following state laws
[1, 3, 18, 20]:

P (ρ, e) = (γ − 1)ρe− γP∞ and e(ρ, θ) =
P∞
ρ

+ Cvθ. (2)

Where γ and P∞ are given constants depending on the fluid, Cv is the specific heat at
constant volume and θ is the temperature. The relationship between the total energy E

and the internal energy is: E = ρe+ ‖u‖2
2 .

We add an artificial viscosity term −ε4ρ in the continuity equation of the system (1), we
obtain: (see [4]):

∂ρε
∂t

+∇ · (ρεuε)− ε4ρε = 0,

ρε
∂uε
∂t

+ ρε(uε · ∇)uε − µ4uε − (
µ

3
+ λ)∇(∇ · uε) + Cv(γ − 1)θ∇ρε

+(γ − 1)ρε∇θε = f ,

ρεCv
∂θε
∂t

+ ρεCvuε · ∇θε − k4θε −
(
λ− 2

3
µ
)
|∇ · uε|2 −

µ

2
|∇uε +∇uTε |2

+ρεuε · ∇(
uε · uε

2
) + (γ − 1)ρεCvθε∇ · uε = 0,

ρε|t=0 = ρ0, uε|t=0 = u0, θε|t=0 = θ0,

uε|∂Ω = ub, θε|∂Ω = θb.

(3)

To close the system (3), we need an additional boundary condition on ρε given by:

ε
∂ρε
∂n

= 0. (4)

In all the sequel we assume that the following assumptions are hold:

(H1)

{
the viscosities verify : 0 < λ+

2

3
µ 6

4

3
µ,

s ∈ R such that s ≥ 5
2 ,

(H2)


f ∈ (L∞(0, T ;Hs(Ω)3))3,

ub ∈ (W 1,∞(0, T ;Hs+ 3
2 (∂Ω)3))3,

θb ∈W 1,∞(0, T ;Hs+ 3
2 (∂Ω)),

(H3)


ρ0 ∈ Hs(Ω) ∩W 1,∞(Ω),
u0 ∈ (Hs(Ω))3 ∩ (W 1,∞(Ω))3,
θ0 ∈ Hs(Ω) ∩W 1,∞(Ω),
ρ0 > ρ̄ with ρ̄ ∈ R+

∗ .

We denote L2(·), L∞(·), Hs(·) and W 1,∞(·) the usual Lebesgue and Sobolev spaces and
|| · ||L2 , || · ||L∞ , || · ||Hs and || · ||W 1,∞ their corresponding norms (see [2]). In all the following
we will adopte the following notation:
||·||

L2
(

0,T ;Hs(Ω)
) = ||·||2,s , ||·||

L∞
(

0,T ;Hs(Ω)
) = ||·||∞,s , ||·||

L∞
(

0,T ;L∞(Ω)
) = ||·||∞,∞

We are now able to announce our main results:



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 85

Theorem 1. Under assumptions (H1), (H2) and (H3), suppose that the state law is
given by (2). Then for any fixed time T > 0 the system (3) has a solution (ρε, uε, θε)
such that:

(ρε, θε) ∈
(
L∞(0, T ;Hs(Ω))

)2
and (

∂ρε
∂t

,
∂θε
∂t

) ∈
(
L2(0, T ;Hs(Ω))

)2
uε ∈

(
L∞
(
0, T ;Hs(Ω)

))3
and

∂uε
∂t
∈
(
L2
(
0, T ;Hs(Ω)

))3
.

In addition we have:

ρε > ρ̄ exp−KT and ‖uε‖L∞
(

0,T ;W 1,∞(Ω)
) ≤ K,

where K is a constant independent of ε.

Theorem 2. Let (ρε, uε, θε) be a solution given by the theorem 1. Then, up to extracting
subsequence, we have the following convergence when ε goes to zero:

ρε ⇀ ρ weakly-∗ in L∞
(
0, T ;Hs(Ω)

)
,

∂ρε
∂t

⇀
∂ρ

∂t
weakly in L2

(
0, T ;L2(Ω)

)
,

θε ⇀ θ weakly-∗ in L∞
(
0, T ;Hs(Ω)

)
,

∂θε
∂t

⇀
∂θ

∂t
weakly in L2

(
0, T ;L2(Ω)

)
,

uε ⇀ u weakly-∗ in
(
L∞
(
0, T ;Hs(Ω)

))3
,

∂uε
∂t

⇀
∂u

∂t
weakly in

(
L2
(
0, T ;L2(Ω)

))3
,

where (ρ, u, θ) is a solution of the system (1).

The proof of these two theorems will be done in several steps. In section 4 we shortly
recall how to rewrite the system (3) in the hyperbolic form, his symmetrization and the
study of deduced operators that we be used in the demonstrations (see [4] for details).
The proof of Theorem 1 will be given in Section 4 and for the Theorem 2 in section 5.

3. Positivity of the density

Before we begin our demonstrations, we have to verify that the density ρε remains
strictly positive in time if its initial value is greater than a positive quantity. For this, we
take the first equation of the system (3):

∂ρε
∂t

+∇ · (ρεuε)− ε4ρε = 0. (5)

An implicit semi-discretization in time of the equation (5) gives:

1

4t
ρn+1
ε +∇ · (ρn+1

ε un+1
ε )− ε4ρn+1

ε =
1

4t
ρε
n. (6)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 86

To justify this strict positivity, we considered a more general convection-diffusion equation.
So, we have the following results:

Lemma 1. Let v ∈ (W1,∞(Ω))3, h ∈ L∞(Ω), K a constant as ‖v‖W1,∞ 6 K. Then the
solution ρ of the problem:

σρ+∇ · (ρv)− ε4ρ = σh in Ω,

ε
∂ρ

∂n
= 0, v = 0 on ∂Ω,

(7)

verify: for all ρ̄ > 0 such as h > ρ̄ we have ρ > ρ̄ exp−
K
σ .

Proof. Using the same idea like in [6], we set: ρ̃ = ρ − ρ̄∞, with ρ̄∞ = ρ̄ exp−
K
σ . ρ̃ is

then solution of:

σρ̃+∇ · (ρv)− ε4ρ̃ = σ
(
h− ρ̄∞

)
. (8)

Using the same decomposition like in [19]: ρ̃ = ρ̃+ − ρ̃+, where:

ρ̃+ =

{
ρ̃ if ρ̃ > 0,
0 if not .

ρ̃− =

{
−ρ̃ if ρ̃ < 0,
0 if not .

(9)

∂j ρ̃
+ =

{
∂j ρ̃ if ρ̃ > 0,
0 if not .

∂j ρ̃
− =

{
−∂j ρ̃ if ρ̃ < 0,
0 if not .

(10)

Multiplying (8) by the test function η̃− = −(ρ̃− + l)β with β ∈ [0, 1], one obtain:

−
∫

Ω
σρ̃(ρ̃− + l)β −

∫
Ω
ε∇ρ̃∇(ρ̃− + l)β +

∫
Ω
σ(h− ρ̄∞)(ρ̃− + l)β = −

∫
Ω
ρv∇(ρ̃− + l)β (11)

The different terms of equality (11) are estimated by:

ε
∫

Ω∇ρ̃∇(ρ̃− + l)β = −ε
∫

Ω∇ρ̃
−∇(ρ̃− + l)β

= −ε
∫

Ω∇(ρ̃− + l)∇(ρ̃− + l)β

= −ε
∫

Ω β(ρ̃− + l)β−1|∇(ρ̃− + l)|2.
(12)

Setting ρ = (ρ̃+ l) + (ρ̄∞ − l) one can found:∣∣ ∫
Ω ρv∇(ρ̃−+ l)β

∣∣ =
∣∣ ∫

Ω v(ρ̃− + l)∇(ρ̃− + l)β − (ρ̄∞ − l)
∫

Ω(∇ · v)(ρ̃− + l)β
∣∣

≤ Kβ
∫

Ω(ρ̃− + l)β∇(ρ̃− + l) +K(ρ̄∞ + l)
∫

Ω(ρ̃− + l)β

≤ K
∫

Ω β(ρ̃− + l)
β+1
2 (ρ̃− + l)

β−1
2 ∇(ρ̃− + l) +K(ρ̄∞ + l)

∫
Ω(ρ̃− + l)β

≤ Kβ‖ρ̃− + l‖
β+1
2

Lβ+1

√∫
Ω(ρ̃− + l)β−1|∇(ρ̃− + l)|2 +K(ρ̄∞ + l)

∫
Ω(ρ̃− + l)β

≤ βK2

2ε ‖ρ̃
− + l‖β+1

Lβ+1 + εβ
2

∫
Ω(ρ̃− + l)β−1|∇(ρ̃− + l)|2 +K(ρ̄∞ + l)

∫
Ω(ρ̃− + l)β.

(13)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 87

Since the left terms in (11) being positive, using (12) and (13) we get:

−
∫

Ω σρ̃(ρ̃− + l)β ≤ βK2

2ε ‖ρ̃
− + l‖β+1

Lβ+1 +K(l + ρ̄∞)
∫

Ω(ρ̃− + l)β − σ
∫

Ω(h− ρ̄∞)(ρ̃− + l)β

≤ βK2

2ε ‖ρ̃
− + l‖β+1

Lβ+1 − σ
∫

Ω

(
h− ρ̄∞(1 + K

σ )
)(
ρ̃− + l

)β
+Kl

∫
Ω(ρ̃− + l)β.

(14)

With the Taylor polynomial of order 1 one obtain:

ρ̄ > ρ̄∞ exp(
K

σ
) > ρ̄∞(1 +

K

σ
),

then we have:

0 < h− ρ̄ 6 h− ρ̄∞(1 +
K

σ
).

Then, the inequality (14) becomes:

−
∫

Ω
σρ̃(ρ̃− + l)β ≤ βK2

2ε
‖ρ̃− + l‖β+1

Lβ+1 +Kl

∫
Ω

(ρ̃− + l)β.

If l −→ 0, we get: (
σ − βK2

2ε

)
‖ρ̃−‖β+1

Lβ+1 ≤ 0.

Choosing β < inf(1,
2σε

K2
), one obtain ρ̃− = 0 which means that ρ > ρ̄ exp−

K
σ . �

Lemma 2. For s ≥ 5
2 , let v ∈ (Hs+2(Ω))3, h ∈ L∞(Ω), v∗ ∈ (Hs+ 3

2 (∂Ω))3. For all

σ >
K2

ε
+ K, where K is a constant such as ‖v‖Hs+2 6 K. Then, the solution ρ of the

problem: 
σρ+∇ · (ρv)− ε4ρ = σh in Ω,

ε
∂ρ

∂n
= 0, v = v∗ on ∂Ω,

(15)

verify the following properties:

i) if h > 0 then ρ > 0,

ii) let ρ̄ > 0, if h > ρ̄ then ρ > ρ̄ exp−
K
σ .

Proof. i): We have v∗ ∈ (Hs+ 3
2 (∂Ω))3, then there exist ṽ∗ ∈ (Hs+2(Ω))3 such that

ṽ∗|∂Ω = v∗; let set ṽ = v − ṽ∗. According to the Sobolev embendding theorem (see [2])
and the fact that s > 5

2 >
3
2 − 1; we have ṽ ∈ (W 1,∞(Ω))3, so replacing v = ṽ + ṽ∗ in

(15), ρ is then solution of the following system:
σρ+∇ · (ρṽ) + ε4ρ = σh−∇ · (ρṽ∗) in Ω,

ε
∂ρ

∂n
= 0, ṽ = 0 on ∂Ω.

(16)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 88

Considering the same test function η− = −(ρ−+ l)β and also use the same decomposition
of the density ρ = ρ+ + ρ−, we get:

−σ
∫

Ω
ρ(ρ− + l)β + ε

∫
Ω
∇ρ∇(ρ− + l)β + σ

∫
Ω
h(ρ− + l)β =

−
∫

Ω
ρṽ∇(ρ− + l)β +

∫
Ω
ρ(∇ · ṽ∗)(ρ− + l)β +

∫
Ω

ṽ∗(∇ρ)(ρ− + l)β.

(17)

The differents terms in the equality (17) can be estimated as follow:

|
∫

Ω
ρṽ∇(ρ− + l)β| ≤ Kβ

∫
Ω(ρ− + l)β∇(ρ− + l) + l|

∫
Ω ṽ∇(ρ− + l)β|

≤ βK2

2ε ‖ρ
− + l‖β+1

Lβ+1 + βε
2

∫
Ω(ρ− + l)β−1|∇(ρ− + l)|2+

lK|Ω|
1

β+1 ‖ρ− + l‖β
Lβ+1 ,

(18)

|
∫

Ω
ρ(∇ · ṽ∗)(ρ− + l)β| = |

∫
Ω

(ρ− + l)(∇ · ṽ∗)(ρ− + l)β − l
∫

Ω
(∇ · ṽ∗)(ρ− + l)β|

≤ K
∫

Ω(ρ− + l)β+1 + lK
∫

Ω(ρ− + l)β

≤ K‖ρ− + l‖β+1
Lβ+1 + lK|Ω|

1
β+1 ‖ρ− + l‖β

Lβ+1 ,

(19)

|
∫

Ω
ṽ∗(∇ρ)(ρ− + l)β| ≤ K

∫
Ω(∇(ρ− + l))(ρ− + l)

β−1
2 (ρ− + l)

β+1
2

≤ εβ
2

∫
Ω |∇(ρ− + l)|2(ρ− + l)β−1 + K2

2εβ‖ρ
− + l‖β+1

Lβ+1 .
(20)

Using inequalities (18) to (20) into (17) and h being positive we get:

−σ
∫

Ω
ρ(ρ− + l)β ≤

(
βK2

2ε + K2

2εβ +K

)
‖ρ− + l‖β+1

Lβ+1 + 2lK|Ω|
1

β+1 ‖ρ− + l‖β
Lβ+1 . (21)

So when l −→ 0, we obtain:(
σ −K2β

2 + 1

2βε
−K

)
‖ρ−‖β+1

Lβ+1 ≤ 0. (22)

let set:

σ −K2 β2+1
2βε −K = f(β)

2εβ , (23)

where, f(β) = −K2β2 + 2εβ(σ −K)−K2.

Since σ >
K2

ε
+K, we have f ′(β) = −2K2β + 2ε(σ −K) is strictly positive on [0, 1] and



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 89

f(0) = −K2 < 0, f(1) = −2K2 + 2ε(σ − K) > 0, then there exist an unique β1 ∈]0, 1[

such as f(β1) = 0. Choosing β ∈]β1, 1[, we have σ −K2 β2+1
2βε −K > 0. We conclude that

ρ− = 0, so ρ > 0.
ii): By setting ρ̃ = ρ− ρ̄∞; ρ̃ is then solution of:

σρ̃+∇ · (ρṽ)− ε4ρ̃ = σ(h− ρ̄∞)−∇ · (ρṽ∗),

ε
∂ρ̃

∂n
= 0, ṽ = 0 on ∂Ω.

(24)

We use the decomposition (9) and (10). So multiplying (24) by the test function
η̃− = −(ρ̃− + l)β, we obtain after integration by parts:

−σ
∫

Ω
ρ̃(ρ̃− + l)β − ε

∫
Ω
∇ρ̃∇(ρ̃− + l)β + σ

∫
Ω

(h− ρ̄∞)(ρ̃− + l)β = −
∫

Ω
ρṽ∇(ρ̃− + l)β

+

∫
Ω
ρ(∇ · ṽ∗)(ρ̃− + l)β +

∫
Ω

ṽ∗(∇ρ)(ρ̃− + l)β
(25)

By taking the modulus, we follow the same procedure as above to justify that ρ̃− = 0.
This allows us to conclude. �

4. Hyperbolization, demonstration of the theorem 1

the rewriting of the system (3) in hyperbolic form was presented in [4] and inspired by
[5]. Its consist to set:

Wε = (Uε, D1Uε, D2Uε, D3Uε), (26)

with

Uε = (ρε, uε, θε) and Di =
∂

∂xi
for i = 1, 2, 3. (27)

Let set

α = λ+
4

3
µ , β = λ+

µ

3
and δ = λ− 2

3
µ. (28)

One can found in [4] the proof of the following result:

Proposition 1. ([4]) Under assumption (H1) and using the change of variables (26)-(27),
the system (3) can be rewrite as:

A0(Wε)
∂Wε

∂t
+

3∑
i=1

Aiε
∂Wε

∂xi
+ K(Wε)Wε = F,

Wε|t=0 = (ρ0, u0, θ0, 0, . . . , 0),

Wε|∂Ω = (0, ub, θb, 0, . . . , 0),(
(Wε)6n1 + (Wε)11n2 + (Wε)16n3

)∣∣∣∣
∂Ω

= 0.

(29)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 90

Where n = (n1, n2, n3)T is the outward normal vector, A0(Wε), Aiε (i=1..3), Kε(Wε)
are matrix in M20(R) and Kε(Wε) contains all non-linear terms.
In addition there exist a positive definte matrix S which symmetrizes the system (29). The
corresponding boundary conditions are also rewrite like in [13]. we obtain:

(SA0(Wε))
∂Wε

∂t
+

3∑
i=1

(SAiε)
∂Wε

∂xi
+ (SKε)(Wε)Wε = SF,

(Bε −Mε)Wε = g,

(30)

with Bε =
3∑
i=1

ni(SAiε). the matrix Mε ∈ M20(R) and the vector g ∈ R20 are obtained

in accordance with the change of variables and:

gi = 0 for i = 1, 6, g11 = g16 = 0,

g7 = −2(αn1u
1
b + βn2u

2
b + βn3u

3
b), g8 = −2µn1u

2
b ,

g9 = −2µn1u
3
b , g10 = −2kn1θb,

g12 = −2µn2u
1
b , g13 = −2(βn1u

1
b + αn2u

2
b + βn3u

3
b),

g14 = −2µn2u
3
b , g15 = −2kn2θb,

g17 = −2µn3u
1
b , g18 = −2µn3u

2
b ,

g19 = −2(βn1u
1
b + βn2u

2
b + αn3u

3
b), g20 = −2kn3θb.

Notation: For simplicity, in all the following we will denote by:

1) (SA0(W)) = A0(W), (SAiε) = Aiε, (SKε) = Kε

2) A0 =

(
Id5 0
0 0

)
, Id5 is an identity matrix in M5(R).

Before we present our approach for the construction of the weak solution for the system
(30), note that by the assumption (H2) we have g ∈W1,∞(0, T ;Hs+ 3

2 (∂Ω))20, and by the

trace theorem (see [15]), for ub ∈W1,∞(0, T ;Hs+ 3
2 (∂Ω)3) and θb ∈W1,∞(0, T ;Hs+ 3

2 (∂Ω))
there exist (Wg)

2, (Wg)
3, (Wg)

4, (Wg)
5 in W1,∞(0, T ;Hs+2(Ω)) such as:

(Wg)
2|∂Ω = u1

b , (Wg)
3|∂Ω = u2

b , (Wg)
4|∂Ω = u3

b , (Wg)
5|∂Ω = θb.

Setting: Zg = (0, (Wg)
2, (Wg)

3, (Wg)
4, (Wg)

5), we obtain:

Wg =
(
Zg,D1Z

g,D2Z
g,D3Z

g
)
∈ L∞(0, T ; Hs+1(Ω)20) (31)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 91

4.1. Construction of successive approximations

For the existence study of a weak solution of the system (30), we use a process of
successive construction of a solution. One can found the used of this approach in [5, 21],
or in the continuous version in [22]. Thus in the first step, we perform a implicite semi-
discretization in time. So, let N be a given integer, we subdivide [0, T ] into N intervals
with the time step is 4t = T

N , we denote Wn(x) = W(tn,x).
The essential idea of our study is based on use of the following algorithm:

• Initialization: Wn = W0

• By using the semi-discretization in time of the system (30), we construct the sequence
Wn+1 solution of:

A0(Wn)

4t
Wn+1 +

3∑
i=1

Aiε
∂Wn+1

∂xi
+ Kε(W

n+1)Wn+1 = Fn+1 +
A0(Wn)

4t
Wn,

(Bε −Mε)W
n+1 = gn+1·

(32)

• By linearizing le system (32) we construct the sequence Wn+1
k+1 solution of:

A0(Wn)

4t
Wn+1

k+1 +
3∑
i=1

Aiε

∂Wn+1
k+1

∂xi
+ Kε(W

n+1
k )Wn+1

k+1 = Fn+1 +
A0(Wn)

4t
Wn,

(Bε −Mε)W
n+1
k+1 = gn+1·

(33)

• Convergences

− With the fixed point theorem, we prove:
Wn+1

k+1 →Wn+1, when k → +∞ and where Wn+1 is a solution of (32).

− By some a priori estimations, we prove:
Wn+1 →Wε, when n→ +∞ and where Wε is a solution of (30).

4.1.1. Existence of W1
1

To prove the existence of the sequence (W1
k), we begin with the existence of W1

1 by
initializing with W1

0 = W0. Thus W1
1 will be solution of:

A0(W0)

4t
W1

1 +
3∑
i=1

Aiε
∂W1

1

∂xi
+ Kε(W

0)W1
1 = F1 +

A0(W0)

4t
W0,

(Bε −Mε)W
1
1 = g1.

(34)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 92

Definition 1. W1
1 is called a weak solution of (34) if W1

1 ∈ L2(Ω)20 and for all test
function V ∈ H1

0 (Ω)20 we have

< W1
1,Ψ

∗V >=< G, V >, (35)

where G = F1 +
A0(W0)

4t
W0 and Ψ∗ is the adjoint operator of

Ψ(·) =
A0(W0)

4t
(·) +

3∑
i=1

Aiε
∂(·)
∂xi

+ Kε(W
0)(·).

The existence of W1
1 is given by following result:

Proposition 2. Let that assumptions (H2) and (H3) hold. Then the system (34) admit
a weak solution W1

1 ∈ L2(Ω)20. In addition if W1
1 ∈ H1(Ω)20 we have uniqueness.

Moreover if F1 +
A0(W0)

4t
W0 ∈ Hs(Ω)20, g1 ∈ Hs+ 3

2 (∂Ω)20, then the system (34)

admit a weak solution W1
1 ∈ Hs(Ω)20.

In addition we have (W1
1)1 > ρ̄ exp(−4tK), with K given by (49).

Proof. By the assumptions (H2) and (H3), we have F1 +
A0(W0)

4t
W0 ∈ L2(Ω)20. In

the aim to homogeneous the boundary conditions of the system (34), let us set:

W1∗
1 = W1

1 −W1
g, (36)

W1∗
1 is then solution of:

(
A0(W0)

4t
+ Kε(W

0)

)
W1∗

1 +

3∑
i=1

Aiε
∂W1∗

1

∂xi
= F1 +

A0(W0)

4t
W0

−
(

A0(W0)

4t
+ Kε(W

0)

)
W1

g −
3∑
i=1

Aiε

∂W1
g

∂xi
,

(B−M)W1∗
1 = 0.

(37)

Let define the operator:

<(W0,W0)(·) =

(
A0(W0)

4t
+ Kε(W

0)

)
(·) +

d∑
i=1

Aiε
∂(·)
∂xi

. (38)

(< + <∗) is positive (see for instance [4], Proposition 2). Do to [13], we can deduce the
existence of a weak solution in L2(Ω)20. Moreover if the solution is in H1(Ω)20 we have



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 93

uniqueness and the following estimate:

‖W1∗
1 ‖L2 ≤

2

Υ(ε,4t)

(
‖F1‖L2 +

1

4t
‖A0(W0)W0‖L2 +

1

4t
‖A0(W0)W1

g‖L2

+‖Kε(W
0)W1

g +
3∑
i=1

Aiε

∂W1
g

∂xi
‖L2

)
≤ 2

Υ(ε,4t)

(
‖F1‖L2 +

C03(W0)

4t
‖A0W

0‖L2 +
(
1 +

1

4t
)
C1(W0)‖W1

g‖L2 + C̃‖W1
g‖H1

)
·

Υ(ε,4t) is the constant positivity of (<+ <∗). Using (36), we obtain:

‖W1
1‖L2 ≤

2

Υ(ε,4t)

(
‖F1‖L2 +

C03(W0)

4t
‖A0W

0‖L2 +
(
1 +

1

4t
)
C1(W0)‖W1

g‖L2

+C̃‖W1
g‖H1

)
+ ‖W1

g‖L2

where
C03(W0) = sup(1, ‖W0‖L∞ , Cv‖W0‖L∞)

C04(W0) = max
16j620

5∑
i=1

|(Kε)ij(W
0)| , C1(W0) = sup(C03(W0), C04(W0))

C̃ = sup(µ, β, α, k) a constant independente to ε. By construction of Wg (equality (31)),
have: (Wg)

1, (Wg)
6, (Wg)

11, and (Wg)
16 are all equal to zero.

For the second part of the proposition, let us choose a test function test V ∈ V with

V =
{
V ∈ D(Ω)20/Vi+5 = D1Vi, Vi+10 = D2Vi, Vi+15 = D3Vi, for i = 1, 5

}
.

One can remark that: < AiεW
1∗
1 ,V >= 0 for i = 1 · · · 3. Thus, for V ∈ V we obtain

<
A0(W0)

4t
W1∗

1 , V > + < Kε(W
0)W1∗

1 , V >=< F1 +
A0(W0)

4t
W0∗, V >

+ < Hε(W
0)W1

g, V > − < A0(W0)

4t
(W1

g −W0
g), V >

(39)

where

Hε(W
0)W1

g = Kε(W
0)W1

g +

d∑
i=1

Aiε

∂W1
g

∂xi
· (40)

Using the fact that in the expression of Hε(W
0)W1

g only elements at the five first lines
are not equal to zero. We decompose the matrix Kε(W

0) as follow:

Kε(W
0) = D−A0 + R1(W0) + R2(W0), (41)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 94

where: D =


0 0 0 0
0 K22 K23 K24

0 K32 K33 K34

0 K42 K43 K44

 ,

R1 =


K11 0 0 0

0 0 0 0
0 0 0 0
0 0 0 0

 , R2 =


0 K12 K13 K14

0 0 0 0
0 0 0 0
0 0 0 0

 ,

with

K11 =


w7 + w13 + w19 w6 w11 w16 0
(γ − 1)Cvw10 w1w7 w1w12 w1w17 (γ − 1)cvw6

(γ − 1)Cvw15 w1w8 w1w13 w1w18 (γ − 1)Cvw11

(γ − 1)Cvw20 w1w9 w1w14 w1w19 (γ − 1)Cvw16

T1 T2 T3 T4 T5


and

T1 = Cv(w2w10 + w3w15 + w4w20),
T2 = w1(w2w7 + w3w8 + w4w9),
T3 = w1(w2w12 + w3w13 + w4w14),
T4 = w1(w2w17 + w3w18 + w4w19),
T5 = Cvw1(w7 + w13 + w19).

(K12)ij =


−δ(w7 + w13 + w19)− µw7 if (i, j) = (5, 2) ,

−µ
2 (w8 + w12) if (i, j) = (5, 3) ,

−µ
2 (w9 + w17) if (i, j) = (5, 4) ,

0 if not .

(K13)ij =


−µ

2 (w8 + w12) if (i, j) = (5, 2) ,

−δ(w7 + w13 + w19)− µw13 if (i, j) = (5, 3) ,

−µ
2 (w14 + w18) if (i, j) = (5, 4) ,

0 if not .

(K14)ij =


−µ

2 (w9 + w17) if (i, j) = (5, 2) ,

−µ
2 (w8 + w14) if (i, j) = (5, 3) ,

−δ(w7 + w13 + w19)− µw19 if (i, j) = (5, 4) ,

0 if not .

K22 =


ε 0 0 0 0
0 α 0 0 0
0 0 µ 0 0
0 0 0 µ 0
0 0 0 0 k

 , K33 =


ε 0 0 0 0
0 µ 0 0 0
0 0 α 0 0
0 0 0 µ 0
0 0 0 0 k

 , K44 =


ε 0 0 0 0
0 µ 0 0 0
0 0 µ 0 0
0 0 0 α 0
0 0 0 0 k

 ,



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 95

(K23)ij =

{
β if (i, j) = (2, 3) ,

0 if not .
(K32)ij =

{
β if (i, j) = (3, 2) ,

0 if not .

(K42)ij =

{
β if (i, j) = (4, 2) ,

0 if not .
(K24)ij =

{
β if (i, j) = (2, 4) ,

0 if not .

(K34)ij =

{
β if (i, j) = (3, 4) ,

0 if not .
(K43)ij =

{
β if (i, j) = (4, 3) ,

0 if not .

Multiplying the equality (39) by 4t and using the decomposition (41) we have:

< A0(W0)W1∗
1 ,A0V > +4t < DW1∗

1 , V > +4t < R2(W0)W1∗
1 , V >= 4t < F1,A0V >

+ < A0(W0)W0∗,A0V > +4t < (A0 −R1(W0))W1∗
1 ,A0V >

−4t < Hε(W
0)W1

g,A0V > − < A0(W0)(W1
g −W0

g),A0V >

One can note that all elements of R2(·) are equal to zero except somes of the fifth line,
then we get

sup
V ∈V

| < A0(W0)W1∗
1 ,A0V > |

‖A0V ‖H−s
6 sup

V ∈V

∣∣∣∣< A0(W0)W1∗
1 ,A0V >

‖A0V ‖H−s
+4t< DW1∗

1 , V >

‖V ‖H−s+1

+4t< R2(W0)W1∗
1 ,A0V >

‖A0V ‖H−s

∣∣∣∣
6 4t‖F1‖Hs + C03(W0)‖A0W0∗‖Hs+ 4t(C + C‖W0‖3L∞)‖A0W1∗

1 ‖Hs

+4t(C + C‖W0‖3L∞)‖W1
g‖Hs+1+ C03(W0)‖A0(W1

g −W0
g)‖Hs

where
C = sup

(
3, 3Cv, (γ − 1)Cv, µ, β, α, k

)
By definition of the Hs norm given in [12], we finally deduce:

‖A0W1∗
1 ‖Hs ≤ 4t

C01(ρ̄)
‖F1‖Hs +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs +4t

C + C‖W0‖3L∞
C01(ρ̄)

‖A0W1∗
1 ‖Hs

+4t
C + C‖W0‖3L∞

C01(ρ̄)
‖W1

g‖Hs+1 +
C03(W0)

C01(ρ̄)
‖A0(W1

g −W0
g)‖Hs ,

(42)

with C01(ρ̄) is a strict positive constant independent to ε given by:

C01(ρ̄∞) = inf(1, ρ̄∞, Cvρ̄∞)· (43)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 96

The discrete Grönwall’s inequality allows us to deduct:ä

‖A0W1∗
1 ‖Hs ≤M exp

4t
C + C‖W0‖3L∞

C01(ρ̄) , (44)

where

M =
4t

C01(ρ̄)
‖F1‖Hs +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs +4t

C + C‖W0‖3L∞
C01(ρ̄)

‖W1
g‖Hs+1

+
C03(W0)

C01(ρ̄)
‖A0(W1

g −W0
g)‖Hs .

From inequality (44) we obtain:

‖A0W1
1‖Hs ≤M exp

4t
C + C‖W0‖3L∞

C01(ρ̄) + ‖A0Wg‖Hs . (45)

So with the compactly embedded of Hs−1 into L∞, we obtain:

‖W1∗
1 ‖L∞ ≤ ‖A0W1∗

1 ‖Hs ≤MT exp
T
C + C‖W0‖3L∞

C01(ρ̄) , (46)

where MT is a constant independent of ε and of 4t, given by:

MT =
T

C01(ρ̄)
‖F‖∞,s +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs + T

C + C‖W0‖3L∞
C01(ρ̄)

‖Wg‖∞,s+1

+
2C03(W0)

C01(ρ̄)
‖A0Wg‖∞,s

(47)

Finally we have the estimate:

‖W1
1‖L∞ ≤ K (48)

with

K = MT exp
T
C + C‖W0‖3L∞

C01(ρ̄) + ‖Wg‖∞,s. � (49)

4.1.2. Existence of W1
k+1

In this part, we will establish the existence of W1
k+1, weak solution of the following system:

A0(W0)

4t
W1

k+1 +
3∑
i=1

Aiε

∂W1
k+1

∂xi
+ Kε(W

1
k)W1

k+1 = F1 +
A0(W0)

4t
W0,

(B−M)W1
k+1 = g1.

(50)

In fact we have the following result:



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 97

Proposition 3. Let the assumptions (H2) and (H3) hold, then the system (50) admit a
weak solution W1

k+1 ∈ L2(Ω)20. In addition if W1
k+1 ∈ H1(Ω)20 this solution is unique.

If F1 +
A0(W0)

4t
W0 ∈ Hs(Ω)20, g1 ∈ Hs+ 3

2 (∂Ω)20, then the weak solution of the system

(34): W1
k+1 ∈ Hs(Ω)20.

We have (W1
k+1)1 > ρ̄ exp(−K4t).

Proof. We use recurrence and follow the same approach as in the Proposition 2. So in
the second step we will construct W1

2 starting from W1
1 given by the Proposition 2. Then

we will get the following estimates:

‖A0W1∗
2 ‖Hs ≤

(
4t

C01(ρ̄)
‖F1‖Hs +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs +4t

C + C‖W0‖3L∞
C01(ρ̄)

‖W1
g‖Hs+1

+
C03(W0)

C01(ρ̄)
‖A0(W1

g −W0
g)‖Hs

)
exp
4t

C + C‖W1∗
1 ‖3L∞

C01(ρ̄) ,

(51)

‖W1∗
2 ‖L∞ ≤

(
T

C01(ρ̄)
‖F‖∞,s +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs + T

C + C‖W0‖3L∞
C01(ρ̄)

‖Wg‖∞,s+1

+
C03(W0)

C01(ρ̄)
‖A0(W1

g −W0
g)‖∞,s

)
exp

T
C + C‖W1∗

1 ‖3L∞
C01(ρ̄) .

(52)

Note here that we cannot continue the recurrence only if these estimates are uniform in the

sense that T
C + C‖W1∗

1 ‖3L∞
C01(ρ̄)

remains lower than a quantity independent of the iteration.

For this aim let us consider the term M defined by:

M = T̃
C + C‖W0‖3L∞

C01(ρ̄)
, (53)

where T̃ is an enough time to have T
C + C‖W1∗

1 ‖3L∞
C01(ρ̄)

6M. We will note in the sequel

C01(ρ̄) as:

C01(ρ̄) = inf
(
1, ρ̄ exp(−T̃ K̃), Cvρ̄ exp(−T̃ K̃)

)
, (54)

with

K̃ = MT̃ expT̃M+ ‖Wg‖L∞(0,T̃ ;L∞(Ω)20)· (55)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 98

Using the estimation (46) and the equality (47) we get:

T

C01(ρ̄)

(
C + C‖W1∗

1 ‖3L∞
)
6

T

C01(ρ̄)

(
C + CK3

)
6

T

C01(ρ̄)

[
C + C

(
T

C01(ρ̄)
‖F‖L∞(0,T̃ ;Hs(Ω)20 +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs

+M‖Wg‖L∞(0,T̃ ;Hs+1(Ω)20) +
2C03(W0)

C01(ρ̄)
‖A0Wg‖L∞(0,T̃ ;Hs(Ω)20)

)3

exp3M
]
,

(56)

where we have set:

K = MT expM . (57)

We wish that the right term in the estimate (56) be majoreted by M. Thus, we rewrite
the difference between the right term of (56) and M as a polynomial function in T :

gM(T ) =
a3

1

C01(ρ̄)
exp3M T 4 + 3

a2
1a2

C01(ρ̄)
exp3M T 3 + 3

a1a
2
2

C01(ρ̄)
exp3M T 2 +

C + a3
2 exp3M

C01(ρ̄)
T −M,

where:

a1 = C
‖F‖L∞(0,T̃ ;Hs(Ω))

C01(ρ̄)
,

a2 = C

(
C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs +M‖Wg‖L∞(0,T̃ ;Hs+1(Ω)20)

+
2C03(W0)

C01(ρ̄)
‖A0Wg‖L∞(0,T̃ ;Hs(Ω)20)

)
.

We expect to have:
gM(T ) 6 0·

Let E be a subset of R+ containing the zeros of the function gM:

E =

{
T ∈ R+ such as gM(T ) = 0

}
.

First one can note that the E is not empty, indeed gM(T ) is a polynomial function of
degree four which is negative at T = 0, and is tends towards +∞ when T → +∞. Thus
there exists at least one T ∈ R such as gM(T ) = 0. Thus we takes T0 as follows:

T0 = inf
T∈[0,+∞[

E (58)

Consequently, for any T ∈ [0, T0[ we have gM(T ) 6 0. We then obtain the following
uniform estimate:

‖A0W1∗
2 ‖Hs ≤

(
4t

C01(ρ̄)
‖F1‖Hs +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs +M‖W1

g‖Hs+1

+
C03(W0)

C01(ρ̄)
‖A0(W1

g −W0
g)‖Hs

)
expM

(59)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 99

and

‖W1
2‖L∞ ≤

(
T

C01(ρ̄)
T‖F‖∞,s +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs +M‖Wg‖∞,s+1

+
C03(W0)

C01(ρ̄)
‖A0(W1

g −W0
g)‖Hs

)
expM+‖Wg‖∞,∞.

(60)

Finally we have by recurrence:

‖A0W1
k‖Hs ≤

(
4t

C01(ρ̄)
‖F1‖Hs +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs +M‖W1

g‖Hs+1

+
C03(W0)

C01(ρ̄)
‖A0(W1

g −W0
g)‖Hs

)
expM+‖A0W1

g‖Hs

(61)

‖W1
k‖L∞ ≤

(
T

C01(ρ̄)
‖F‖∞,s +

C03(W0)

C01(ρ̄)
‖A0W0∗‖Hs +M‖Wg‖∞,s+1

+
2C03(W0)

C01(ρ̄)
‖A0Wg‖∞,s

)
expM+‖Wg‖∞,∞. �

(62)

4.1.3. Passing to the limit k →∞

We have shown in the paragraphs 4.1.1 and 4.1.2 the existence of the sequence (W1
k+1),

here we will show that this sequence converge and its limit is solution of the following
system:

A0(W0)

4t
W1 +

3∑
i=1

Aiε
∂W1

∂xi
+ Kε(W

1)W1 = F1 +
A0(W0)

4t
W0,

(B−M)W1 = g1.

(63)

Let us set:

Rs = MT expM+‖Wg‖∞,∞ (64)

Lemma 3. Let B be a ball of Hs(Ω)20 with radius Rs given by (64). Then the function
G defined by:

G :

(
B −→ B
W1

k −→ G(W1
k) = W1

k+1

)
admits an unique fixed point W1 which is solution of the system (63). In addition we have
W1 ∈ L∞(Ω)20 with (W1)1 > ρ̄ exp(−4tRs) ·



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 100

Proof. Let W1
k and W1

k+1 be two successive solutions of (50) then we have:

A0(W0)

4t
W1

k+1 +
3∑
i=1

Aiε

∂W1
k+1

∂xi
+ Kε(W

1
k)W1

k+1 = F1 +
A0(W0)

4t
W0, (65)

A0(W0)

4t
W1

k +
3∑
i=1

Aiε
∂W1

k

∂xi
+ Kε(W

1
k−1)W1

k = F1 +
A0(W0)

4t
W0. (66)

The difference between (65) and (66) gives:

[A0(W0)

4t
+ Kε(W

1
k)
](

W1
k+1 −W1

k

)
+

3∑
i=1

Aiε
∂

∂xi

(
W1

k+1 −W1
k

)
=
[
Kε(W

1
k−1)−Kε(W

1
k)
]
(W1

k).

A scalar product with W1
k+1−W1

k and using the fact that: the matrix Aiε are symmetric,
W1

k+1 −W1
k = W1∗

k+1 −W1∗
k and the operator <+ <∗ is positive we obtain:

< (<+ <∗)(W0,W1
k)(W

1
k+1 −W1

k),W1
k+1 −W1

k >> 2 Υ(ε,4t)‖W1
k+1 −W1

k‖2L2 (67)

and

< (<+ <∗)(W0,W1
k)(W

1
k+1 −W1

k),W1
k+1 −W1

k >>
C01(ρ̄)

4t
‖A0

(
W1

k+1 −W1
k

)
‖2L2 . (68)

By the obtained result in [4] (lemma 2) we have:

<
[
Kε(W

1
k−1)−Kε(W

1
k)
]
W1

k,W
1
k+1 −W1

k >6 C
W
1,∞‖W1

k −W1
k−1‖L2‖A0

(
W1

k+1 −W1
k

)
‖L2 , (69)

where the constant CW
1,∞ is defined as follow:

CW
1,∞ = max

16j620

20∑
i=1

sup
‖W1

k−1
‖L∞6Rs

‖W1
k
‖L∞6Rs

∣∣∣∣(N [W1
k−1,W

1
k,W

1
k]

)
ij

∣∣∣∣. (70)

Using the inequality (69) into (67) we get:

2Υ(ε,4t)‖W1
k+1 −W1

k‖2L2 6 C(Rs)‖W1
k −W1

k−1‖L2‖A0

(
W1

k+1 −W1
k

)
‖L2 . (71)

The inequality (68) allows us to deduce:

‖A0

(
W1

k+1 −W1
k

)
‖L2 6 C(Rs)

4t
C01(ρ̄)

‖W1
k −W1

k−1‖L2 . (72)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 101

The time step 4t being intended to goes to zero before ε, taking (72) into (71) it follow:

‖W1
k+1 −W1

k‖2L2 ≤
4tC2(Rs)

4εC01(ρ̄)
‖W1

k −W1
k−1‖2L2 . (73)

Choosing the time step such that4t < 4εC01(ρ̄)
C2(Rs)

, we deduce that G is strictly L2-contraction.

Hence there exist an unique W1 ∈ L2(Ω)20 such as G(W1) = W1.
It remains to prove that W1 ∈ Hs(Ω). By Sobolev spaces interpolation:

for all reels s, s′ with 0 < s′ < s,∃ C(s,Ω), such that

‖W1
k −W1

l ‖Hs′ 6 C(s,Ω)‖W1
k −W1

l ‖
1− s

′
s

L2 ‖W1
k −W1

l ‖
s′
s
Hs

6 2R
s′
s
s C(s,Ω)‖W1

k −W1
l ‖

1− s
′
s

L2 .

(74)

According to (73), (W1
k) is a Cauchy sequence in Hs′(Ω). So choosing s′ ≥ 5

2 we have

W1
k −→W1 in Hs′(Ω). By duality product of Sobolev spaces we have:

< Φ,W1
k >−→< Φ,W1 > ∀ Φ ∈ H−s′(Ω),

and by density of H−s
′
(Ω) in H−s(Ω), we have:

< Φ,W1
k >−→< Φ,W1 > ∀ Φ ∈ H−s(Ω). �

Now we are able to pass to the limit in the system (50). By term by term convergence
we obtain that W1 is a solution of the system (63). By construction of the matrix A0,
Aiε(i = . . . 3) and Kiε, the last fifteen lines of the system (63) gives that W1 have exactly
the following form:

W1 = (U1, D1U
1, D2U

1, D3U
1)

with U1 =

(
(W1)1,W

1)2, (W1)3, (W1)4, W1)5

)
and Di =

∂

∂xi
for i = 1, 2, 3

(75)

The estimate (48) being true for the sequence (W1
k+1), the limit also verify the same

estimate. Then we have v =
(
(W1)2, (W1)3, (W1)4

)
∈ W1,∞(Ω)3 and according to

lemma 2 we deduce that (W1)1 > ρ̄ exp(−4tRs).

4.1.4. Existence of Wn+1
k+1 and passing to limit k →∞

In the paragraph 4.1.2, we have proved the existence of (W1
k+1) solution of (50), whose

limit W1 is solution of the system (63) (paragraph 4.1.3).
Now we construct the sequence (W2

k+1) whose first term W2
0 = W1. According to the

lemma 3 and the positivity of the operator <+<∗, one obtain the existence of (W2
1), and

moreover we have (W2
1)1 > ρ̄ exp(−24tRs). Then, by recurrence we have the existence of

(W2
k+1). We repeat the process in the same way, so successively we construct the sequence

(Wn+1
k+1) solution of the sytem (33) and whose existence is given by the following result:



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 102

Proposition 4. Suppose that Fn+1 +
A0(Wn)

4t
Wn ∈ L2(Ω)20 and gn+1 ∈ Hs+ 3

2 (∂Ω)20,

then the system (33) admit a weak solution Wn+1
k+1 ∈ L2(Ω)20. In addition if Wn+1

k+1 ∈
H1(Ω)20, then this solution is unique.

If Fn+1 +
A0(Wn)

4t
Wn ∈ Hs(Ω)20, gn+1 ∈ Hs+ 3

2 (∂Ω)20 then the system (33) admit a

weak solution Wn+1
k+1 ∈ H

s(Ω)20. In addition, we have (Wn+1
k+1)1 > ρ̄ exp(−(n+ 1)4tRs)

Proof. Use the same demarche as in the existence of W1
1. �

At the limit with the same argument of the fixed point theorem one can get Wn+1

solution of the system (32).

4.2. A priori estimations

Lemma 4. Under assumptions (H2) and (H3), the following estimations hold:

i) ‖A0Wn‖Hs ≤ C(T,F,Wg),

ii) 4t
n∑
i=0

‖A0(Wi)
Wi+1 −Wi

4t
‖2L2 ≤ C(T,F,Wg),

iii) 4t
n∑
i=0

‖ 1

4t
A0

(
Wi+1 −Wi

)
‖2L2 ≤ C(T,F,Wg),

(76)

with C(T,F,Wg) is a constante independente of ε and 4t.

Proof. i): Let Wn+1 be the solution of the following system:

A0(Wn)

4t
(Wn+1 −Wn) +

3∑
i=1

Aiε
∂W(n+1)∗

∂xi
+ Kε(W

n+1)W(n+1)∗ = Fn+1 −Hε(W
n+1)Wn+1

g ,

with

Hε(W
n+1)Wn+1

g = Kε(W
n+1)Wn+1

g +
3∑
i=1

Aiε

∂Wn+1
g

∂xi
.

Using the same decomposition of Kε as previously, we get:

<
A0(Wn)

4t
(Wn+1 −Wn),A0V > + < DWn+1, V > + < R2(Wn+1)Wn+1,A0V >

=< Fn+1,A0V > + <
[
A0 −R1(Wn+1)

]
W(n+1)∗,A0V > + < Hε(W

n+1)Wn+1
g ,A0V > .

(77)

It follows that:

sup
V ∈V

| < A0(Wn)

4t
(Wn+1 −Wn),A0V > |

‖A0V ‖H−s
6 ‖Fn+1‖Hs + (C + CR3

s)‖A0W
n+1‖Hs

+(C + CR3
s)‖Wn+1

g ‖Hs + C̃‖Wn+1
g ‖Hs+1 ,



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 103

then,

‖A0(Wn)(Wn+1 −Wn)‖Hs 6 4t‖Fn+1‖Hs + (C + CR3
s)

(
4t‖A0W

n+1‖Hs

+4t‖Wn+1
g ‖Hs

)
+ C̃4t‖Wn+1

g ‖Hs+1 .

(78)

Now we set:

CRs01 (ρ̄) = inf
(
1, ρ̄ exp(−T Rs), Cvρ̄ exp(−T Rs)

)
(79)

we get finally:

‖A0W
n+1‖Hs 6

4t
CRs01 (ρ̄)

‖Fn+1‖Hs +4tC + CR3
s

CRs01 (ρ̄)
‖A0W

n+1‖Hs +4tC + CR3
s

CRs01 (ρ̄)
‖Wn+1

g ‖Hs

+4t C̃

CRs01 (ρ̄)
‖Wn+1

g ‖Hs+1 + ‖A0W
n‖Hs .

(80)

By summation we have:

‖A0W
n+1‖Hs 6

4t
CRs01 (ρ̄)

n∑
i=1

‖Fi+1‖Hs +4tC + CR3
s

CRs01 (ρ̄)

n∑
i=1

‖A0W
i+1‖Hs

+4tC + CR3
s

CRs01 (ρ̄)

n∑
i=1

‖Wi+1
g ‖Hs +4t C̃

CRs01 (ρ̄)

n∑
i=1

‖Wi+1
g ‖Hs+1 + ‖A0W

0‖Hs .

(81)

By the Grönwall’s inequality we deduce:

‖A0W
n‖Hs 6

(
T

CRs01 (ρ̄)
‖F‖∞,s + CT‖Wg‖∞,s+1 + ‖A0W

0‖Hs

)
expM, (82)

where M is given by (53). By assumptions (H2) and (H3), we deduce i).
For the proof of ii), let us consider two successive solutions Wn+1 and Wn of the system

(32), using the same change of variable like (36) and setting: (∂W)n+1
n = W(n+1)∗−Wn∗,

we get:

A0(Wn)

4t
(∂W)n+1

n +
3∑
i=1

Aiε
∂

∂xi
((∂W)n+1

n ) + Kε(W
n+1)(∂W)n+1

n

+
(
Kε(W

n+1)−Kε(W
n)
)
Wn∗

= Fn+1 − Fn +
A0(Wn−1)

4t
(∂W)nn−1 −

(
Hε(W

n+1)−Hε(W
n)
)
Wn+1

g

−Hε(W
n)(Wn+1

g −Wn
g )·

(83)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 104

By the lemma 2, proved in [4], we rewrite (83) as follow:[
A0(Wn)

4t
+ Kε(W

n+1) +N [Wn+1,Wn,Wn∗]

]
(∂W)n+1

n +

3∑
i=1

Aiε
∂

∂xi
((∂W)n+1

n )

= Fn+1 − Fn +
A0(Wn−1)

4t
(∂W)nn−1 −Kε(W

n+1)(Wn+1
g

−Wn
g )−

(
Kε(W

n+1)−Kε(W
n)
)
Wn

g −
3∑
i=1

Aiε
∂

∂xi
(Wn+1

g −Wn
g )

(84)

Applying the Proposition 3 of [4] to the system (84) we have:

‖A0(Wn)

4t
(Wn+1 −Wn

)
‖L2 6 2‖Fn+1 − Fn‖L2 + 2‖A0(Wn−1)

4t
(Wn −Wn−1)‖L2

+2‖Kε(W
n)‖L∞‖Wn+1

g −Wn
g ‖L2 + 2‖N [Wn+1,Wn,Wn∗]‖L∞‖Wn+1

g −Wn
g ‖L2

+2‖A0(Wn−1)

4t
(Wn

g −Wn−1
g )‖L2 + ‖A0(Wn)

4t
(Wn+1

g −Wn
g

)
‖L2

or∥∥∥∥A0(Wn)

4t
(Wn+1 −Wn

)∥∥∥∥
L2

6 2‖Fn+1 − Fn‖L2 + 2

∥∥∥∥A0(Wn−1)
4t (Wn −Wn−1)

∥∥∥∥
L2

+2

∥∥∥∥A0(Wn−1)

4t
(Wn

g −Wn−1
g )

∥∥∥∥
L2

+

∥∥∥∥A0(Wn)

4t
(Wn+1

g −Wn
g

)∥∥∥∥
L2

+C(Rs)‖Wn+1
g −Wn

g ‖L2

Squared, one obtain:∥∥∥∥A0(Wn)

4t
(Wn+1 −Wn

)∥∥∥∥2

L2

6 C‖Fn+1 − Fn‖2L2 +
1

2

∥∥∥∥A0(Wn−1)

4t
(Wn −Wn−1)

∥∥∥∥2

L2

+C

∥∥∥∥A0(Wn−1)

4t
(Wn

g −Wn−1
g )

∥∥∥∥2

L2

+ C

∥∥∥∥A0(Wn)

4t
(Wn+1

g −Wn
g

)∥∥∥∥2

L2

+C(Rs)
2‖Wn+1

g −Wn
g ‖2L2

(85)

with C a positive constant. The inequality (85) remains true for i = 1, 2, · · ·n, by sum-
mation, we get:

n∑
i=1

∥∥∥∥A0(Wi−1)

4t
(Wi+1 −Wi

)∥∥∥∥2
L2

6 C
n∑
i

∥∥Fi+1 − Fi
∥∥2
L2 + C

n∑
i

∥∥∥∥A0(Wi−1)

4t
(Wi

g −Wi−1
g )

∥∥∥∥2
L2

+C

n∑
i

∥∥∥∥A0(Wi)

4t
(Wi+1

g −Wi
g

)∥∥∥∥2
L2

+ C(Rs)
2

n∑
i

∥∥Wi+1
g −Wi

g

∥∥2
L2 .

(86)



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 105

Multiplying by 4t we have:

4t
n∑
i=1

∥∥∥∥A0(Wi−1)

4t
(Wi+1 −Wi

)∥∥∥∥2

L2

6 4tC
n∑
i

∥∥Fi+1 − Fi
∥∥2

L2

+4t
n∑
i

∥∥∥∥A0(Wi−1)

4t
(Wi+1

g −Wi
g

)∥∥∥∥2

L2

+4t
n∑
i

∥∥∥∥A0(Wi)

4t
(Wi+1

g −Wi
g

)∥∥∥∥2

L2

+24tC(Rs)
2

n∑
i

∥∥Wi+1
g ‖2L2 + ‖Wi

g

∥∥2

L2 .

Finally we obtain ii), with:

4t
n∑
i=1

∥∥∥∥A0(Wi−1)

4t
(Wi+1 −Wi

)∥∥∥∥2

L2

6 TC

(
‖F‖2∞,2 + ‖A0Wg‖2W1,∞(0,T ;L2(Ω)20) + ‖Wg‖2∞,2

)
For iii), we multiply (86) by 4t, which implies

4t
n∑
i=1

∥∥∥∥A0

4t
(Wi+1 −Wi

)∥∥∥∥2

L2

6 4t C
C01

n∑
i

∥∥Fi+1 − Fi
∥∥2

L2 + 24tC03

C01

n∑
i

∥∥∥∥A0

4t
(Wi+1

g −Wi
g

)∥∥∥∥2

L2

+24tC(Rs)
2

C01

n∑
i

∥∥Wi+1
g ‖2L2 + ‖Wi

g

∥∥2

L2 ,

and finally we obtain:

4t
n∑
i=1

∥∥∥∥A0

4t
(Wi+1 −Wi

)∥∥∥∥2

L2

6 TC

(
‖F‖2∞,2 + ‖A0Wg‖2W1,∞(0,T ;L2(Ω)20) + ‖Wg‖2∞,2

)
. �

4.2.1. Passing to limit m→∞

The a priori estimates given by the lemma 4 allow us to pass to the limit in non-linear
terms. But to be able to do it, we introduce the approximate functions on [0, T ] and which
are defined as follows:

V∗m(t) = Wn +
1

k
(Wn+1 −Wn)(t− nk) ∀t ∈ [nk, (n+ 1)k[,

Vm(t) = Wn ∀t ∈ [nk, (n+ 1)k]
(87)

with k = T
m . So we have the following result:

Lemma 5. Let suppose that the assumptions (H2) and (H3) hold, then up to extract sub
sequences of V∗m and Vm, if necessary (which we will note in the same maner) we have
the following convergences:



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 106

a) A0V
∗
m ⇀ A0V weakly in L2(0, T ;Hs(Ω)20),

b) A0V
m ⇀ A0V weakly in L2(0, T ;Hs(Ω)20),

c) V∗m and Vm ⇀ V weakly-∗ in L∞(0, T ;Hs−1(Ω)20),

d)
d

dt
(A0V

∗
m) ⇀

d

dt
(A0V) weakly in L2(0, T ;L2(Ω)20).

Proof. By construction of the functions V∗m, Vm and the estimation i) of the lemma
4, we have A0V∗m and A0Vm are bounded in L∞(0, T ;Hs(Ω)20) thus we have the weak-∗
convergence of the two sequences in L∞(0, T ;Hs(Ω))20. By compact embedded we get a)
and b).
Notice that A0V∗m and A0Vm represent the first five variables of V∗m and Vm thus,
V∗m et Vm are bounded in L∞(0, T ;Hs−1(Ω)20). Then we have c).
Finally d) come from iii) of the lemma 4. It remains to prove that V∗m and Vm have the
same limit when m→∞. For that, one can remark that:

A0V∗m(t)−A0Vm(t) =
t− nk
k

(A0Wn+1 −A0Wn) ∀t ∈ [nk, (n+ 1)k]

∫ (n+1)k

nk
|A0V∗m(t)−A0Vm(t)|2 dt = |A0Wn+1 −A0Wn|2

∫ (n+1)k

nk

(τ − nk
k

)2
dτ

=
k2

3
|A0Wn+1 −A0Wn|2,

so, by summation, we obtain:

‖A0V∗m −A0Vm‖2,2 ≤
Ck

3
.

Passing to the limit m → +∞ which means k → 0, we have: A0V∗ = A0V and conse-
quently V∗ = V. �

Now we can announce the result which allows us to pass to the limit in the nonlinear
term.

Lemma 6. Let us suppose that the assumptions (H1), (H2) and (H3) hold, then up to
subsequence, we have:

1) Vm
1 ⇀ W1 weakly-∗ in L∞(0, T ;L2(Ω)),

2) Vm
i →Wi strongly in L2(0, T ;L4(Ω)),

3) Vm
i →Wi strongly in L2(0, T ;L6(Ω)),

4) Vm
i Vm

j ⇀ WiWj weakly L2(0, T ;L2(Ω)),



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 107

5) Vm
i Vm

j Vm
r ⇀ WiWjWr weakly in L2(0, T ;L2(Ω)),

6) Vm
1 Vm

i Vm
j Vm

r ⇀ W1WiWjWr weakly in L2(0, T ;L2(Ω)),

with 1 ≤ i, j, r ≤ 20.

Proof. The point 1) is a direct deduction of the point c) of the lemma 5.
Form the apriori estimates, we have (Vm

i ) is bounded in L2(0, T ;Hs−1(Ω)), thus (Vm
i )

converge weakly in L2(0, T ;Hs−1(Ω)). For s ≥ 3
2 + 1, we have the compact embendding

of Hs−1(Ω) in L4(Ω) (see [2]), so by the compacity theorem do to [14], we have that (Vm
i )

converge strongly in L2(0, T ;L4(Ω)). So we have the point 2). By the same argument one
can obtain the point 3).
For the point 4), we have (Vm

i ) and (Vm
j ) are bounded in L2(0, T ;L4(Ω)), thus (Vm

i Vm
j )

is bounded in L2(0, T ;L2(Ω)) hence it converges weakly towards a limit that we note Ψij .
To justify the equality Ψij = WiWj , we takes Q = [0, T [×Ω, thus we have:

< Vm
i Vm

j , φ >=< Vm
i ,V

m
j φ >, ∀ φ ∈ D(Q)

By the point 2), we have Vm
i is bounded in L2(0, T ;L2(Ω)), then converge weakly, on the

other hand we have the strong convergence of Vm
j φ in L2(0, T ;L2(Ω)). It result that:∫

Q
Vm
i Vm

j φ dxdt −→
∫
Q

WiWjφ dxdt ∀ φ ∈ D(Q).

For the point 5), given that (3
2 , 3) are conjugate, we have the following estimate:

‖(Vm
i Vm

j Vm
r )(t)‖2L2 ≤ ‖(Vm

i Vm
j )(t)‖2L3‖Vr(t)‖2L6

≤
(
‖Vm

i (t)‖L6‖Vm
j (t)‖L6‖Vm

r (t)‖L6

)2
.

The sequence (Vm
i Vm

j Vm
r ) is then bounded in L2(0, T ;L2(Ω)) so it converges weakly

towards Φijr. For the equality Φijr = WiWjWr, we juste follow the same way as
previously.
The point 6) is a deduction of 1) and 5). �

Lemma 7. Under the same assumptions like in the lemma 5, the following convergence
hold:

A0V
m d

dt
(A0V

∗
m) ⇀ A0V

d

dt
(A0V) weakly in L2(0, T ;L2(Ω))20.

Proof. From the estimation ii) of the lemma 4 we deduce that A0V
m d

dt
(A0V

∗
m) is

bounded in L2(0, T ;L2(Ω)20), hence it follow the weak converge to Φ in the same space.

To justify the equality Φ = A0V
d

dt
(A0V), we have: firstly the estimate i) of lemma 4



R. Bade, H. Chaker / Eur. J. Pure Appl. Math, 14 (1) (2021), 82-111 108

guarantees the weak convergence of A0V
m in L∞(0, T ;Hs(Ω)) and due to the compact em-

bendding Hs(Ω) ↪→ L2(Ω), we have the strong convergence of A0V
m in L∞(0, T ;L2(Ω)),

and as T < ∞, the strong convergence also holds in L2(0, T ;L2(Ω)). And secondly, the

point d) of lemma 7 gives us the weak convergence of
d

dt
(A0V

∗
m) in L2(0, T ;L2(Ω))5. Thus,

equality is obtained in the same way as previously. �

Lemma 8. Kε(V
m)Vm ⇀ Kε(W)W weakly in L2(0, T ;L2(Ω))20 when m→∞

Proof. Performing the vector-matrix product Kε(V
m)Vm, the result is a consequence

of the lemma 6. �

We now be able to pass to the limit in the weak sense in the following system:

< A0(Vm)
d

dt

(
A0(V∗m)

)
, φ > + <

3∑
i=1

Aiε
∂V∗m
∂xi

, φ > + < Kε(V
m)V∗m, φ >=< F, φ >, ∀φ ∈ D(Q)20

with Q = [0, T [×Ω.
Applying the lemmas 7 and 8 we get: for all test function φ ∈ D(Q)20:

< A0(Wε)
∂

∂t

(
A0(Wε)

)
, φ > + <

3∑
i=1

Aiε
∂Wε

∂xi
, φ > + < Kε(W

ε)Wε, φ >=< F, φ >,

hence A0W
ε is a weak solution of the system (3). �

5. Passing to limit ε −→ 0, démonstration of the theorem 2

In this section, we will pass to the limit ε→ 0 in Wε solution of (3). By definition of
the matrix A0, we have: A0W

ε = (ρε,uε, θε, 0, . . . , 0)T . So by the theorem 1 we have:

‖A0W
ε‖∞,s 6 C and ‖D0W

ε‖2,s−1 6 C

where C is a positive constant independente to ε and D0 ∈M(R20) a matrix given by:

(D0)ij =

{
1 if i = j with i /∈ (6, 11, 16)
0 if not

(88)

Then, extracting subsequence if necessary, we have the convergence of Wε to W in
L∞(0, T,Hs(Ω))20. It remains now to verify that the limit A0W is solution of the system
(1) which means to show that:

∂Wi

∂xj
= Wi+5j for i = 1, . . . , 5 and j = 1, . . . , 3 (89)

We have (Wε)1 ∈ L2(0, T ;Hs(Ω))20, which gives:∥∥∥∥∂Wε
1

∂x1

∥∥∥∥
Hs−1

6 ‖Wε
1‖Hs 6 C∥∥∥∥∂Wε

1

∂x1

∥∥∥∥
L2

6 ‖Wε
1‖Hs 6 C



REFERENCES 109

(Wε)1 being bounded in L2(0, T ;Hs(Ω)), one can deduce that,
∂Wε

1

∂x1
= (Wε)6 is bounded

in L2(0, T ;Hs−1(Ω)). Consequently their exist subsequence noted again
∂Wε

1

∂x1
which con-

verges weakly to W in L2(0, T ;Hs−1(Ω)). On the other hand we have Wε
1 converge

strongly to −→W1 in L2(0, T ;Hs(Ω)), which gives W =
∂W1

∂x1
. This allows us to deduce

the following convergences:

∂(Wε)1

∂x1
−→ ∂W1

∂x1
in L2(0, T ;Hs−1(Ω))

(Wε)6 −→ W6 in L2(0, T ;Hs−1(Ω))

By equality of sequences and their limits, we have:
∂W1

∂x1
= W6.

We repeat the same calculation for
∂W1

∂x2
= W11, which finally gives:

A0W ∈ L∞(0, T ;Hs(Ω))

∂

∂t
(A0W) ∈ L2(0, T ;L2(Ω))

The verification of A0W solution of the compressible Navier-Stokes system (1) is done in
the same way as before.

6. Conclusion

In this work we have proved the existence of a weak solution of the Navier-Stokes-
Fourier system if the second member remains bounded in time and with a certain reg-
ularity in spaces. The justification is based on a reduction of the system order due to
the add of a diffusion in the continuity equation. However, we prove that this addition
preserves the positivity of the density as long as there is not a vacuum at the initial time.
The treatment of the nonlinearity of the Navier-Stokes-Fourier system is done by a suc-
cessive approximation and the passage to the limit was possible thanks to some a priori
estimations.

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