# Local Strong Solution for a Class of Shear Thickening Fluids with Non-Newtonian Potential and Heat-Conducting

Authors: Yunliang Zhang; Zhidong Guo
Journal: Advances in Mathematical Physics
 (2015)

---

## Abstract

The aim of this paper is to discuss the model for a class of shear thickening fluids with non-Newtonian potential and heat-conducting. Existence and uniqueness of local strong solutions for the model are proved. In this paper, there exist two difficulties we have to overcome. One is the strong nonlinearity of the system. The other is that the state function is not fixed.


---

## Body

## 1. Introduction

With the development of the technology, we find that there are a lot of fluids which belong to non-Newtonian fluids in the process of product and nature. The classical non-Newtonian fluids are macromolecule melt and macromolecule liquor and all kinds of slurry and suspend liquor, paint, dope, palette, and biology fluids, for example, in the body of people and animals, the blood, the synovia of arthrosis cavity, lymph liquor, cell liquor, and brain liquor, which are provided with the property of non-Newtonian fluids. So the non-Newtonian fluids exist widely in nature (see [1–6]).In the absence of gravitational potential term, the mathematical model of non-Newtonian fluids reduces to the Navier-Stokes equations. There have been many results concerning the existence and nonexistence of solutions for such equations (such as [7–10]). In [11], Xin studied the blowup of smooth solutions to the compressible Navier-Stokes equations in arbitrary space dimensions with initial density of compact support. Choe and Kim in [12, 13] proved an existence result when Ω is either a bounded domain or the whole space with the compatibility condition. For other results one can refer to [14–17] and the references cited therein.In thermodynamics, a state function, function of state, state quantity, or state variable is a property of a system that depends only on the current state of the system, not on the way in which the system acquired that state. A state function describes the equilibrium state of a system. For example, internal energye, temperature θ, density ρ, pressure π, and so on are state quantities because they describe quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state. If two quantities are known, then we can get other quantities. For example, temperature θ and density ρ are known, and we can get internal energy e, pressure π, and so on.In mathematics, a state functionz
=
f
(
x
,
y
) has the following properties:(i) A state functionz
=
f
(
x
,
y
) has continuous first partial derivatives.(ii) There are the tiny variabled
z and the limited variable ∫
d
z. The integral value depends on the size of the integral upper limit and lower limit and has nothing to do with other factors. Thus, (1)
∫
1
2
d
z
=
f
x
2
,
y
2
-
f
x
1
,
y
1
=
z
2
-
z
1
.(iii) Consider(2)
∂
2
z
∂
x
∂
y
=
∂
2
z
∂
y
∂
x
.The following are some examples of state equations:(i) Polytropic gases:ϕ
=
A
ρ
γ (A is a constant and γ is gas thermal insulation coefficient and is a constant) (see [15–17]).(ii) State equation of ideal gas:ψ
=
R
ρ
θ (R is a constant) (see [18]).(iii) Isentropic process:φ
=
θ (see [18]).In this paper, we consider generalized state equation. Energy increases with the increase of temperature. In mathematics,π
=
φ
(
θ
,
ρ
) and φ
1
=
∂
φ
/
∂
θ
>
0 (we define φ
1
=
∂
φ
/
∂
θ, φ
2
=
∂
φ
/
∂
ρ, ψ
1
=
∂
ψ
/
∂
θ, and ψ
2
=
∂
ψ
/
∂
ρ). Ignore the molecular potential energy and all of the gas fluid, for example, methane, propane, and butane from the oil.In this paper, we consider a class of heat-conducting non-Newtonian fluids of Navier-Stokes system:(3)
ρ
t
+
ρ
u
x
=
0
,
x
,
t
∈
Ω
T
,
ρ
e
t
+
ρ
u
e
x
-
θ
x
2
+
μ
p
-
2
/
2
θ
x
x
+
π
u
x
=
u
x
2
,
x
,
t
∈
Ω
T
,
ρ
u
t
+
ρ
u
2
x
-
u
x
2
+
μ
2
q
-
2
/
2
u
x
x
+
π
x
=
0
,
x
,
t
∈
Ω
T
,
e
=
φ
θ
,
ρ
,
x
,
t
∈
Ω
T
,
π
=
ψ
θ
,
ρ
,
x
,
t
∈
Ω
T
,with the initial and boundary conditions(4)
ρ
,
u
,
θ
t
=
0
=
ρ
0
,
u
0
,
θ
0
,
x
∈
0,1
,
u
0
,
t
=
u
1
,
t
=
0
,
t
∈
0
,
T
,
θ
0
,
t
=
θ
1
,
t
=
0
,
t
∈
0
,
T
,where Ω
T
=
I
×
(
0
,
T
), I
=
(
0,1
), the initial density ρ
0
≥
0, p
,
q
>
2, and μ
1
,
μ
2
>
0 are given constants. The unknown variables ρ, u, and θ denote the density, velocity, temperature, and the non-Newtonian gravitational potential, respectively.Definition 1.
The triple(
ρ
,
u
,
θ
) is called a strong solution to the initial boundary value problem (3)-(4), if the following conditions are satisfied:
(i) Consider(5)
0
≤
ρ
∈
C
0
,
T
;
H
1
I
,
u
,
θ
∈
C
0
,
T
;
H
0
1
I
∩
L
∞
0
,
T
;
H
2
I
,
ρ
t
∈
C
0
,
T
;
L
2
I
,
u
t
,
θ
t
∈
L
2
0
,
T
;
H
0
1
I
,
ρ
u
t
,
ρ
θ
t
∈
L
∞
0
,
T
;
L
2
I
.
(ii) For allχ
∈
C
(
[
0
,
T
]
;
H
1
(
I
)
), χ
t
∈
L
∞
(
0
,
T
;
L
2
(
I
)
), for a.e. t
∈
(
0
,
T
), we have(6)
∫
0
1
ρ
χ
x
,
t
d
x
-
∫
0
t
∫
0
1
ρ
χ
t
+
ρ
u
χ
x
x
,
s
d
x
d
s
=
∫
0
1
ρ
0
χ
x
,
0
d
x
.
(iii) For allϕ
∈
C
(
[
0
,
T
]
;
H
0
1
(
I
)
)
∩
L
∞
(
0
,
T
;
H
2
(
I
)
), ϕ
t
∈
L
2
(
0
,
T
;
H
0
1
(
I
)
), for a.e. t
∈
(
0
,
T
), we have(7)
∫
0
1
ρ
e
ϕ
x
,
t
d
x
-
∫
0
t
∫
0
1
ρ
e
ϕ
t
+
ρ
u
e
-
θ
x
2
+
μ
1
p
-
2
/
2
θ
x
ϕ
x
x
,
s
d
x
d
s
+
∫
0
t
∫
0
1
-
u
x
2
+
π
u
x
ϕ
x
,
s
d
x
d
s
=
∫
0
1
ρ
0
e
0
ϕ
x
,
0
d
x
.(iv) For allϕ
∈
C
(
[
0
,
T
]
;
H
0
1
(
I
)
)
∩
L
∞
(
0
,
T
;
H
2
(
I
)
), ϕ
t
∈
L
2
(
0
,
T
;
H
0
1
(
I
)
), for a.e. t
∈
(
0
,
T
), we have(8)
∫
0
1
ρ
u
ϕ
x
,
t
d
x
-
∫
0
t
∫
0
1
ρ
u
ϕ
t
+
ρ
u
2
ϕ
x
-
u
x
2
+
μ
2
q
-
2
/
2
u
x
ϕ
x
+
π
x
ϕ
x
,
s
d
x
d
s
=
∫
0
1
ρ
0
u
0
ϕ
x
,
0
d
x
.


### 1.1. Main Results

Theorem 2.
Assume that(
ρ
0
,
u
0
,
θ
0
) satisfies the following conditions:(9)
0
<
δ
≤
ρ
0
∈
H
1
I
,
u
0
,
θ
0
∈
H
0
1
I
∩
H
2
I
,
0
<
δ
≤
φ
1and further that there exists a constant c
0
>
0, such that(10)
1
+
ρ
0
H
1
+
θ
0
H
1
+
u
0
H
1
≤
c
0
.Then there exist a time T
∗
∈
(
0
,
+
∞
) and a unique strong solution (
ρ
,
u
,
θ
) to (3)-(4) such that(11)
ρ
∈
C
0
,
T
∗
;
H
1
I
,
ρ
t
∈
C
0
,
T
∗
;
L
2
I
,
u
t
,
θ
t
∈
L
2
0
,
T
∗
;
H
0
1
I
,
u
,
θ
∈
C
0
,
T
∗
;
H
0
1
I
∩
Ł
∞
0
,
T
∗
;
H
2
I
,
ρ
u
t
,
ρ
θ
t
∈
L
∞
0
,
T
∗
;
L
2
I
.The rest of this paper is organized as follows. In Section2, we present some elementary lemmas. In Section 3, we devote ourselves to the study of the problem with positive density. In Section 4 we give the proof of the main theorem.


## 1.1. Main Results

Theorem 2.
Assume that(
ρ
0
,
u
0
,
θ
0
) satisfies the following conditions:(9)
0
<
δ
≤
ρ
0
∈
H
1
I
,
u
0
,
θ
0
∈
H
0
1
I
∩
H
2
I
,
0
<
δ
≤
φ
1and further that there exists a constant c
0
>
0, such that(10)
1
+
ρ
0
H
1
+
θ
0
H
1
+
u
0
H
1
≤
c
0
.Then there exist a time T
∗
∈
(
0
,
+
∞
) and a unique strong solution (
ρ
,
u
,
θ
) to (3)-(4) such that(11)
ρ
∈
C
0
,
T
∗
;
H
1
I
,
ρ
t
∈
C
0
,
T
∗
;
L
2
I
,
u
t
,
θ
t
∈
L
2
0
,
T
∗
;
H
0
1
I
,
u
,
θ
∈
C
0
,
T
∗
;
H
0
1
I
∩
Ł
∞
0
,
T
∗
;
H
2
I
,
ρ
u
t
,
ρ
θ
t
∈
L
∞
0
,
T
∗
;
L
2
I
.The rest of this paper is organized as follows. In Section2, we present some elementary lemmas. In Section 3, we devote ourselves to the study of the problem with positive density. In Section 4 we give the proof of the main theorem.


## 2. Preliminaries

We first give some known facts for later use.Lemma 3 (see [15]).
Assume thatf
=
0 on ∂
Ω, where Ω
⊂
R
1 is bounded and open, f
∈
C
2
+
α
(
Ω
¯
), 0
<
α
<
1. Then (12)
f
′
L
∞
Ω
≤
d
1
/
2
Ω
f
′
′
L
2
Ω
,where d
(
Ω
) denotes the length of Ω.Lemma 4 (see [15]).
LetH be a Hilbert space with a scalar product (
·
,
·
)
H and let X be a Banach space such that (13)
X
↪
H
≅
H
∗
↪
X
∗
,and X is dense in H, p
>
1. Then (14)
W
≡
u
∈
L
p
I
;
X
;
d
u
d
t
∈
L
p
′
I
;
X
∗
↪
C
I
;
H
.Lemma 5.
π
=
φ
(
θ
,
ρ
) and e
=
ϕ
(
θ
,
ρ
) is state function. According to the mathematical properties of state function, we can get that ∃
M
>
0, (15)
φ
+
φ
1
+
φ
2
+
φ
11
+
φ
22
≤
M
,
ψ
+
ψ
1
+
ψ
2
+
ψ
11
+
ψ
22
≤
M
.


## 3. Existence of Solutions with Positive Density

### 3.1. A Priori Estimates for Smooth Solutions

We first construct approximate solutions of problem (3)-(4). By using the iterative scheme, inductively, consider the following:(i) Firstly defineu
0
=
0, θ
0
=
0.(ii) Assuming thatu
k
-
1, θ
k
-
1 were defined for k
≥
1, we can obtain (
ρ
k
,
u
k
,
θ
k
,
e
k
,
π
k
) as the unique smooth solution to the following system:(16)
ρ
t
k
+
u
k
-
1
ρ
x
k
+
u
x
k
-
1
ρ
k
=
0
,
x
,
t
∈
Ω
T
,
(17)
π
k
=
ϕ
θ
k
,
ρ
k
,
x
,
t
∈
Ω
T
,
(18)
e
k
=
φ
θ
k
,
ρ
k
,
x
,
t
∈
Ω
T
,
(19)
ρ
k
φ
1
k
θ
t
k
+
ρ
k
φ
1
k
u
k
-
1
θ
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
+
H
p
μ
1
θ
k
+
ψ
k
u
x
k
-
1
-
u
x
k
-
1
2
=
0
,
x
,
t
∈
Ω
T
,
(20)
ρ
k
u
t
k
+
ρ
k
u
k
-
1
u
x
k
-
H
q
μ
2
θ
k
+
ψ
x
=
0
,
x
,
t
∈
Ω
T
,
(21)
ρ
k
x
,
0
=
ρ
0
,
u
k
x
,
0
=
u
0
,
θ
k
x
,
0
=
θ
0
,
u
k
0
,
t
=
u
k
1
,
t
=
0
,
θ
k
0
,
t
=
θ
k
1
,
t
=
0
,where m
0
=
∫
0
1
ρ
0
(
x
)
d
x
>
0 is the initial mass, for a given function ω and numbers s
>
2, μ
>
0,(22)
H
s
μ
ω
=
-
ω
x
2
+
μ
s
-
2
/
2
ω
x
x
.With this process, the nonlinear coupled system has been deduced into a sequence of decoupled problems and each problem admits a smooth solution. Moreover, we findρ
k from (16), (21) with the smooth function u
k
-
1; that is, (23)
ρ
t
k
+
u
k
-
1
ρ
x
k
-
1
+
u
x
k
-
1
ρ
k
-
1
=
0
,
ρ
k
t
=
0
=
ρ
0
.Obviously there is a unique solutionρ
k to the above initial value problem and also by a standard argument, we could get (24)
ρ
k
x
,
t
≥
δ
exp
⁡
-
∫
0
T
∗
u
x
k
-
1
·
,
s
L
∞
d
s
>
0
,
∀
t
∈
0
,
T
∗
.Then by (19) and (21) we get θ
k. Lastly, with ρ
k, θ
k, we obtain u
k from (20) and (21).Moreover, assume that for a givenT
1, 0
<
T
1
≤
T
(25)
sup
0
≤
t
≤
T
1
⁡
u
x
x
k
-
1
L
2
+
∫
0
T
1
u
x
t
k
-
1
L
2
2
d
s
≤
c
1
,
sup
0
≤
t
≤
T
1
⁡
u
x
x
k
-
1
L
2
≤
c
2
.For some, c
2
>
c
1
>
c
0
>
1 will be fixed later. We will prove that(26)
sup
0
≤
t
≤
T
1
⁡
u
x
k
L
2
+
∫
0
T
1
u
x
t
k
L
2
2
d
s
≤
c
1
,
sup
0
≤
t
≤
T
1
⁡
u
x
x
k
L
2
≤
c
2
,and some more useful uniform estimate about (
ρ
k
,
u
k
,
θ
k
).(i) Estimation forρ
k.Then multiplying (16) by ρ
k and integrating it over (
0,1
) with respect to x, we have (27)
1
2
d
d
t
∫
0
1
ρ
k
2
d
s
+
∫
0
1
ρ
k
-
1
u
k
-
1
x
ρ
k
d
x
=
0
.Integrating by parts, using the Sobolev inequality and Lemma3, we deduce that(28)
d
d
t
ρ
k
t
L
2
2
≤
u
x
x
k
-
1
L
2
ρ
k
L
2
2
.Then, considering differential equation (16) with respect to x and multiplying it by ρ
x
k, integrating it over (
0,1
) on x, and using the Sobolev inequality, we have(29)
d
d
t
∫
0
1
ρ
x
k
2
d
x
=
-
∫
0
1
3
2
u
x
k
-
1
ρ
x
k
2
+
ρ
k
ρ
x
k
u
x
x
k
-
1
t
d
x
≤
3
2
u
x
k
-
1
L
∞
ρ
x
k
L
2
2
+
ρ
k
L
∞
u
x
x
k
-
1
L
2
≤
3
ρ
x
k
L
2
2
u
x
x
k
-
1
L
2
.From (28) and (29), by Gronwall’s inequality, it follows that(30)
sup
0
≤
t
≤
T
⁡
ρ
k
t
H
1
2
≤
ρ
0
k
H
1
2
exp
⁡
4
∫
0
t
u
x
x
k
-
1
L
2
d
s
,and, taking T
2
=
c
2
-
1, we get(31)
sup
0
≤
t
≤
T
⁡
ρ
k
t
H
1
≤
ρ
0
k
H
1
exp
⁡
4
∫
0
t
u
x
x
k
-
1
L
2
d
s
1
/
2
≤
C
c
0
.And using (16) we can also obtain(32)
ρ
t
k
t
L
2
≤
ρ
x
k
-
1
L
2
u
x
k
-
1
L
2
+
ρ
k
-
1
L
2
u
x
x
k
-
1
L
2
≤
C
c
1
2
,where C is a positive constant.(ii) Estimation forθ
k
(
t
).Differentiating (19) with respect to t, multiplying it by θ
t
k, and integrating it over (
0,1
) on x, we derive(33)
1
2
d
d
t
∫
0
1
ρ
k
φ
1
k
θ
t
k
2
d
x
+
∫
0
1
θ
x
2
+
μ
1
p
-
2
/
2
θ
x
t
θ
x
t
k
d
x
=
1
2
∫
0
1
ρ
k
φ
1
k
t
θ
t
k
2
d
x
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
x
k
t
θ
t
k
d
x
-
∫
0
1
ρ
k
2
φ
2
k
u
x
k
-
1
t
θ
t
k
d
x
-
∫
0
1
ψ
k
u
x
k
-
1
t
θ
t
k
d
x
+
∫
0
1
u
x
k
-
1
t
2
θ
t
k
d
x
=
∑
j
=
1
5
I
j
.Using the Sobolev inequality, Young’s inequality, Lemma3, and (25), (30) and (31), we obtain (34)
I
1
=
1
2
∫
0
1
ρ
k
φ
1
k
t
θ
t
k
2
d
x
=
1
2
∫
0
1
ρ
t
k
φ
1
k
θ
t
k
2
d
x
+
1
2
∫
0
1
ρ
k
φ
11
k
θ
t
k
θ
t
k
2
d
x
+
1
2
∫
0
1
ρ
k
φ
12
k
ρ
t
k
θ
t
k
2
d
x
=
∑
j
=
1
3
I
1
j
.Here, (35)
I
11
=
1
2
∫
0
1
ρ
t
k
φ
1
k
θ
t
k
2
d
x
≤
C
∫
0
1
ρ
t
k
θ
t
k
2
d
x
≤
C
∫
0
1
ρ
k
u
k
-
1
θ
t
k
θ
x
t
k
d
x
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
u
k
-
1
L
∞
θ
x
t
k
L
2
≤
C
c
0
1
/
2
c
1
ρ
k
θ
t
k
L
2
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
3
ρ
k
θ
t
k
L
2
2
,
I
12
=
1
2
∫
0
1
ρ
k
φ
11
k
θ
t
k
θ
t
k
2
d
x
≤
C
∫
0
1
ρ
k
θ
t
k
θ
t
k
2
d
x
≤
C
θ
t
k
L
∞
ρ
k
θ
t
k
L
2
2
≤
η
θ
x
t
k
L
2
2
+
C
η
ρ
k
θ
t
k
L
2
4
,
I
13
=
1
2
∫
0
1
ρ
k
φ
12
k
-
1
ρ
t
k
θ
t
k
2
d
x
≤
C
ρ
k
L
∞
∫
0
1
ρ
t
k
θ
t
k
2
d
x
≤
C
c
0
3
/
2
c
1
ρ
k
θ
t
k
L
2
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
5
ρ
k
θ
t
k
L
2
2
.In the same way, we have(36)
I
2
=
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
x
k
t
θ
t
k
d
x
=
-
∫
0
1
ρ
t
k
φ
1
k
u
k
-
1
θ
x
k
θ
t
k
d
x
-
∫
0
1
ρ
k
φ
11
k
θ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
-
∫
0
1
ρ
k
φ
12
k
ρ
t
k
u
k
θ
x
k
θ
t
k
d
x
-
∫
0
1
ρ
k
φ
1
k
u
t
k
-
1
θ
x
k
θ
t
k
d
x
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
x
t
k
θ
t
k
d
x
=
∑
j
=
1
5
I
2
j
.Here, (37)
I
21
=
-
∫
0
1
ρ
t
k
φ
1
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
∫
0
1
ρ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
ρ
t
k
L
2
u
k
-
1
L
∞
θ
x
k
L
2
θ
t
k
L
∞
≤
C
ρ
t
k
L
2
u
x
k
-
1
L
2
θ
x
k
L
2
θ
x
t
k
L
2
≤
C
c
1
3
θ
x
k
L
2
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
6
θ
x
k
L
2
2
,
I
22
=
-
∫
0
1
ρ
k
φ
11
k
θ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
θ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
ρ
k
L
∞
1
/
2
u
k
-
1
L
∞
θ
t
k
L
∞
θ
x
k
L
2
ρ
k
θ
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
3
θ
x
k
L
2
2
ρ
k
θ
t
k
L
2
2
,
I
23
=
-
∫
0
1
ρ
k
φ
12
k
ρ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
ρ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
ρ
k
L
∞
ρ
t
k
L
2
u
k
-
1
L
∞
θ
x
k
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
8
θ
x
k
L
2
2
,
I
24
=
-
∫
0
1
ρ
k
φ
1
k
u
t
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
u
t
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
u
t
k
-
1
L
∞
θ
x
k
L
2
≤
C
c
0
θ
x
k
L
2
2
ρ
k
θ
t
k
L
2
2
+
1
2
u
x
t
k
-
1
L
2
2
,
I
25
=
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
x
t
k
θ
t
k
d
x
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
u
k
-
1
L
∞
θ
x
t
k
L
2
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
u
k
-
1
L
∞
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
6
ρ
k
θ
t
k
L
2
2
.Following the same method, we obtain(38)
I
3
=
-
∫
0
1
ρ
k
2
φ
2
k
u
x
k
-
1
t
θ
t
k
d
x
=
-
∫
0
1
2
ρ
k
ρ
t
k
φ
2
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ρ
k
2
φ
12
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ρ
k
2
φ
22
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ρ
k
2
φ
2
k
u
x
t
k
-
1
θ
t
k
d
x
=
∑
j
=
1
4
I
3
j
.Here, (39)
I
31
=
-
∫
0
1
2
ρ
k
ρ
t
k
φ
2
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
ρ
k
L
∞
ρ
t
k
L
2
u
x
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
8
,
I
32
=
-
∫
0
1
ρ
k
2
φ
12
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
2
θ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
ρ
k
L
∞
u
x
k
-
1
L
∞
ρ
k
θ
t
k
L
2
2
≤
C
η
c
1
2
ρ
k
θ
t
k
L
2
2
,
I
33
=
-
∫
0
1
ρ
k
2
φ
22
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
2
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
ρ
k
L
∞
2
ρ
t
k
L
2
u
x
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
10
,
I
34
=
-
∫
0
1
ρ
k
2
φ
2
k
u
x
t
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
2
u
x
t
k
-
1
θ
t
k
d
x
≤
C
ρ
k
L
∞
2
u
x
t
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
4
u
x
t
k
-
1
L
2
2
.In a similar way,(40)
I
4
=
-
∫
0
1
ψ
k
u
x
k
-
1
t
θ
t
k
d
x
=
-
∫
0
1
ψ
1
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ψ
2
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ψ
k
-
1
u
x
t
k
-
1
θ
t
k
d
x
=
∑
j
=
1
3
I
4
j
.Here, (41)
I
41
=
-
∫
0
1
ψ
1
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
ρ
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
u
x
k
-
1
L
∞
ρ
k
θ
t
k
L
2
2
≤
C
c
1
ρ
k
θ
t
k
L
2
2
,
I
42
=
-
∫
0
1
ψ
2
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
ρ
t
k
L
2
u
x
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
10
,
I
43
=
-
∫
0
1
ψ
k
u
x
t
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
u
x
t
k
-
1
θ
t
k
d
x
≤
C
u
x
t
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
2
.Similarly,(42)
I
5
=
∫
0
1
u
x
k
-
1
t
2
θ
t
k
d
x
≤
2
u
x
k
-
1
L
∞
u
x
t
k
-
1
L
2
θ
k
L
∞
≤
2
u
x
x
k
-
1
L
2
u
x
t
k
-
1
L
2
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
2
u
x
t
k
-
1
L
2
2
.SubstitutingI
j
(
j
=
1,2
,
…
,
5
) into (43), after choosing η
=
μ
1
(
p
-
2
)
/
2
/
26, we get(43)
1
2
d
d
t
ρ
k
φ
1
k
θ
t
k
t
L
2
2
+
1
2
μ
1
p
-
2
/
2
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
10
+
C
c
1
8
ρ
k
θ
t
k
L
2
2
+
C
ρ
k
θ
t
k
L
2
4
+
C
c
1
6
θ
x
k
L
2
2
+
C
c
1
3
θ
x
k
L
2
2
ρ
k
θ
t
k
L
2
2
+
C
c
1
4
u
x
t
k
-
1
L
2
2
,and, integrating it over (
τ
,
t
)
⊂
(
0
,
T
), we have(44)
1
2
ρ
k
φ
1
k
θ
t
k
t
L
2
2
+
1
2
μ
1
p
-
2
/
2
∫
τ
t
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
10
∫
τ
t
1
+
θ
x
k
L
2
2
1
+
ρ
k
θ
t
k
L
2
2
2
s
d
s
+
C
c
1
4
∫
τ
t
u
x
t
k
-
1
L
2
2
s
d
s
+
ρ
k
φ
1
k
θ
t
k
τ
L
2
2
,where C is a positive constant.We estimateρ
k
φ
1
k
θ
t
k
τ
L
2
2 as follows: multiplying (17) by θ
t
k and integrating it over (
0,1
), we obtain (45)
∫
0
1
ρ
k
φ
1
k
θ
t
k
2
d
x
≤
∫
0
1
φ
1
k
-
1
ρ
k
3
φ
2
k
2
u
x
k
-
1
2
+
ρ
k
-
1
ψ
k
2
u
x
k
-
1
2
+
ρ
k
-
1
H
p
μ
1
θ
k
-
u
x
k
2
2
d
x
+
∫
0
1
ρ
k
φ
1
k
u
k
-
1
2
θ
x
k
2
d
x
.Next, we get (46)
∫
0
1
ρ
k
φ
1
k
θ
t
k
2
d
x
≤
C
∫
0
1
ρ
k
u
k
-
1
2
θ
x
k
2
+
ρ
k
3
u
x
k
-
1
2
+
ρ
k
-
1
u
x
k
-
1
2
+
ρ
k
-
1
H
p
μ
1
θ
k
-
u
x
k
2
2
d
x
,and, according to the smoothness of (
ρ
k
,
u
k
,
θ
k
), we have(47)
lim
τ
→
0
⁡
∫
0
1
ρ
k
u
k
-
1
2
θ
x
k
2
+
ρ
k
3
u
x
k
-
1
2
+
ρ
k
-
1
u
x
k
-
1
2
+
ρ
k
-
1
H
p
μ
1
θ
k
-
u
x
k
2
2
d
x
≤
C
c
0
5
,where C is a positive constant.We have(48)
d
d
t
θ
x
k
L
2
2
≤
2
∫
0
1
θ
x
k
θ
x
t
k
d
x
≤
2
θ
x
k
L
2
θ
x
t
k
L
2
≤
θ
x
k
L
2
2
+
θ
x
t
k
L
2
2
;by Gronwall’s inequality, it follows that(49)
θ
x
k
L
2
2
≤
exp
⁡
t
θ
x
k
0
L
2
2
+
∫
0
t
θ
x
t
k
L
2
2
d
sfor 0
<
t
<
min
⁡
{
T
1
,
T
2
},(50)
θ
x
k
L
2
2
≤
C
θ
x
k
0
L
2
2
+
C
∫
0
t
θ
x
t
k
L
2
2
d
s
.Due to (44), (47), and (50), we get(51)
θ
x
k
L
2
2
+
ρ
k
θ
t
k
t
L
2
2
+
∫
τ
t
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
10
∫
τ
t
1
+
θ
x
k
L
2
2
1
+
ρ
k
θ
t
k
L
2
2
2
d
s
+
C
c
1
5
.DenoteΓ
(
t
)
=
1
+
θ
x
k
L
2
2
+
ρ
k
θ
t
k
L
2
2; we get(52)
Γ
t
≤
C
c
1
10
∫
0
t
Γ
t
3
d
s
+
C
c
1
5
;by Gronwall’s inequality, ∃
T
3
≤
T
2 (T
3 has nothing to do with k), it follows that(53)
Γ
t
≤
C
c
1
5
.Note that, for a given functionω, by a simple computation, we have(54)
θ
x
k
2
+
μ
1
p
-
2
/
2
θ
x
k
t
θ
x
t
k
=
θ
x
k
2
+
μ
1
p
-
4
/
2
r
-
1
θ
x
k
2
+
μ
1
θ
x
t
k
2
≥
θ
x
k
2
+
μ
1
p
-
4
/
2
θ
x
k
2
+
μ
1
θ
x
t
k
2
=
θ
x
k
2
+
μ
1
p
-
2
/
2
θ
x
t
k
2
≥
μ
1
p
-
2
/
2
θ
x
t
k
2
.Firstly, we need to do the following estimates: by (19)(55)
θ
x
2
+
μ
1
p
-
2
/
2
θ
x
x
=
ρ
k
φ
1
k
θ
t
k
+
ρ
k
φ
1
k
u
k
-
1
θ
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
+
ψ
k
u
x
k
-
1
-
u
x
k
-
1
2
,and we have (56)
θ
x
k
2
+
μ
1
p
-
2
/
2
θ
x
k
x
=
θ
x
k
2
+
μ
1
p
-
4
/
2
p
-
1
θ
x
k
2
+
μ
1
θ
x
x
k
≥
θ
x
k
2
+
μ
1
p
-
4
/
2
θ
x
k
2
+
μ
1
θ
x
x
k
=
θ
x
k
2
+
μ
1
p
-
2
/
2
θ
x
x
k
≥
μ
1
p
-
2
/
2
θ
x
x
k
.taking it by L
2-norm, we get(57)
θ
x
x
k
L
2
≤
μ
1
2
-
p
/
2
ρ
k
φ
1
k
θ
t
k
L
2
+
ρ
k
φ
1
k
u
k
-
1
θ
x
k
L
2
+
ρ
k
2
φ
2
k
u
x
k
-
1
L
2
+
ψ
k
u
x
k
-
1
L
2
+
u
x
k
-
1
2
L
2
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
+
ρ
k
L
∞
u
k
-
1
L
∞
θ
x
k
L
2
+
ρ
k
L
∞
2
u
x
k
-
1
L
2
+
u
x
k
-
1
L
2
+
u
x
k
-
1
L
∞
u
x
k
-
1
L
2
≤
C
ρ
x
k
L
2
1
/
2
ρ
k
θ
t
k
L
2
+
ρ
x
k
L
2
u
x
k
-
1
L
2
θ
x
k
L
2
+
ρ
x
k
L
2
2
u
x
k
-
1
L
2
+
u
x
k
-
1
L
2
+
u
x
x
k
-
1
L
2
u
x
k
-
1
L
2
≤
C
c
1
4
c
2
.(iii) Estimation foru
k
(
t
).Differentiating (20) with respect to t, multiplying it by u
t
k, and integrating it over (0,1) on x, we derive(58)
1
2
d
d
t
∫
0
1
ρ
k
u
t
k
2
d
x
+
∫
0
1
u
x
2
+
μ
2
q
-
2
/
2
u
x
t
u
x
t
k
d
x
=
1
2
∫
0
1
ρ
t
k
u
t
k
2
d
x
-
∫
0
1
ρ
t
k
u
k
-
1
u
x
k
+
ρ
k
u
t
k
-
1
u
x
k
+
ρ
k
u
k
-
1
u
x
t
k
u
t
k
d
x
+
∫
0
1
ψ
t
k
u
x
t
k
d
x
=
∑
j
=
1
5
I
j
.Then, using the Sobolev inequality, Young’s inequality, Lemma3, and (25), (30), and (31), we obtain (59)
I
1
=
1
2
∫
0
1
ρ
t
k
u
t
k
2
d
x
≤
∫
0
1
ρ
k
u
k
-
1
u
t
k
u
x
t
k
d
x
≤
ρ
k
L
∞
1
/
2
u
k
-
1
L
∞
ρ
k
u
t
k
L
2
u
x
t
k
L
2
≤
η
1
u
x
t
k
L
2
2
+
C
η
1
c
1
3
ρ
k
u
t
k
L
2
2
,
I
2
=
-
∫
0
1
ρ
t
k
u
k
-
1
u
x
k
u
t
k
d
x
≤
∫
0
1
ρ
t
k
u
k
-
1
u
x
k
u
t
k
d
x
≤
ρ
t
k
L
2
u
k
-
1
L
∞
2
u
x
k
L
2
u
t
k
L
∞
≤
η
1
u
x
t
k
L
2
2
+
C
η
1
c
1
6
u
x
k
L
2
2
,
I
3
=
-
∫
0
1
ρ
k
u
t
k
-
1
u
x
k
u
t
k
d
x
≤
∫
0
1
ρ
k
u
t
k
-
1
u
x
k
u
t
k
d
x
≤
ρ
k
L
∞
1
/
2
u
t
k
-
1
L
∞
u
x
k
L
2
ρ
k
u
t
k
L
2
≤
c
0
1
/
2
u
x
t
k
-
1
L
2
u
x
k
L
2
ρ
k
u
t
k
L
2
≤
c
1
-
1
u
x
t
k
-
1
L
2
2
ρ
k
u
t
k
L
2
2
+
1
4
c
0
c
1
-
1
u
x
k
L
2
2
,
I
4
=
-
∫
0
1
ρ
k
u
k
-
1
u
x
t
k
u
t
k
d
x
≤
∫
0
1
ρ
k
u
k
-
1
u
t
k
u
x
t
k
d
x
≤
ρ
k
L
∞
1
/
2
u
k
-
1
L
∞
ρ
k
u
t
k
L
2
u
x
t
k
L
2
≤
η
1
u
x
t
k
L
2
2
+
C
η
1
c
1
3
ρ
k
u
t
k
L
2
2
,
I
5
=
∫
0
1
ψ
t
k
u
x
t
k
d
x
=
∫
0
1
ψ
1
k
θ
t
k
u
x
t
k
d
x
+
∫
0
1
ψ
2
k
ρ
t
k
u
x
t
k
d
x
≤
C
ρ
k
θ
t
k
L
2
u
x
t
k
L
2
+
C
ρ
t
k
L
2
u
x
t
k
L
2
≤
2
η
1
u
x
t
k
L
2
2
+
C
η
1
c
1
4
+
C
η
1
c
1
5
.SubstitutingI
j
(
j
=
1,2
,
…
,
5
) into (33), after choosing η
=
μ
2
(
q
-
2
)
/
2
/
10
(60)
1
2
d
d
t
ρ
k
u
t
k
t
L
2
2
+
1
2
μ
2
q
-
2
/
2
u
x
t
k
L
2
2
≤
C
c
1
3
ρ
k
u
t
k
L
2
2
+
C
c
1
6
u
x
k
L
2
2
+
c
1
-
1
u
x
t
k
-
1
L
2
ρ
k
u
t
k
L
2
+
1
4
c
0
c
1
-
1
u
x
k
L
2
+
C
c
1
5
.We estimateρ
k
u
t
k
τ
L
2
2 as follows: multiplying (20) by u
t
k and integrating it over (
0,1
), we obtain (61)
∫
0
1
ρ
k
u
t
k
2
d
x
≤
2
∫
0
1
ρ
k
u
k
-
1
2
u
x
k
2
+
ρ
k
-
1
H
q
μ
2
u
k
-
ψ
x
2
d
x
.According to the smoothness of(
ρ
k
,
u
k
,
θ
k
), we have(62)
lim
τ
→
0
⁡
∫
0
1
ρ
k
u
k
-
1
2
u
x
k
2
+
ρ
k
-
1
L
p
u
k
+
ψ
x
2
d
x
≤
ρ
0
L
∞
u
0
L
∞
2
u
0
x
L
2
2
+
g
L
2
2
≤
C
c
0
5
,where C is a positive constant.In the same way,(63)
u
x
k
L
2
2
≤
C
u
x
k
0
L
2
2
+
C
∫
0
t
u
x
t
k
L
2
2
d
s
.Integrating (60) over (
0
,
t
)
⊂
(
0
,
T
), combining with (62) and (63) we have(64)
u
x
k
L
2
2
+
ρ
k
u
t
k
t
L
2
2
+
∫
0
t
u
x
t
k
L
2
2
≤
C
∫
0
t
c
1
6
+
c
1
-
1
u
x
t
k
-
1
L
2
ρ
k
u
t
k
t
L
2
2
+
u
x
k
L
2
2
d
s
+
C
c
1
5
t
+
C
c
0
5
,and for 0
<
t
<
min
⁡
{
T
1
,
T
4
}, T
4
≤
c
0
5
/
c
1
5, we get(65)
u
x
k
L
2
2
+
ρ
k
u
t
k
t
L
2
2
+
∫
0
t
u
x
t
k
L
2
2
≤
C
∫
0
t
c
1
6
+
c
1
-
1
u
x
t
k
-
1
L
2
ρ
k
u
t
k
t
L
2
2
+
u
x
k
L
2
2
d
s
+
C
c
0
5
,and, by Gronwall’s inequality, ∃
T
5
≤
T
4 (T
5 has nothing to do with k), it follows that(66)
u
x
k
L
2
2
+
ρ
k
u
t
k
t
L
2
2
+
∫
0
t
u
x
t
k
L
2
2
s
d
s
≤
C
c
0
5
=
c
1
.Secondly, we need to do the following estimates. By (20)(67)
u
x
2
+
μ
2
q
-
2
/
2
u
x
x
=
ρ
k
u
t
k
+
ρ
k
u
k
-
1
u
x
k
-
1
+
ϕ
1
k
-
1
θ
x
k
-
1
+
ϕ
2
k
-
1
ρ
x
k
,we have (68)
u
x
k
2
+
μ
2
q
-
2
/
2
u
x
k
x
=
u
x
k
2
+
μ
2
q
-
4
/
2
q
-
1
u
x
k
2
+
μ
2
u
x
x
k
≥
u
x
k
2
+
μ
2
q
-
4
/
2
u
x
k
2
+
μ
2
u
x
x
k
=
u
x
k
2
+
μ
2
q
-
2
/
2
u
x
x
k
≥
μ
2
q
-
2
/
2
u
x
x
k
,and, taking it by L
2-norm, using Lemma 3, we get(69)
u
x
x
k
L
2
≤
μ
2
2
-
q
/
2
ρ
k
u
t
k
L
2
+
ρ
k
u
k
-
1
u
x
k
L
2
+
ψ
1
k
θ
x
k
L
2
+
ψ
2
k
ρ
x
k
L
2
≤
C
ρ
k
L
∞
1
/
2
ρ
k
u
t
k
L
2
+
ρ
k
L
∞
u
k
-
1
L
∞
u
x
k
L
q
+
θ
x
k
L
2
+
ρ
x
k
L
2
≤
C
ρ
x
k
L
2
1
/
2
ρ
k
u
t
k
L
2
+
ρ
x
k
L
2
u
x
k
-
1
L
q
u
x
k
L
q
+
θ
x
k
L
2
+
ρ
x
k
L
2
≤
C
c
1
3
=
c
2
=
C
c
0
15
,where C is a positive constant. For T
6
=
min
⁡
[
T
1
,
T
2
,
T
3
,
T
4
,
T
5
], we get (70)
sup
0
≤
s
≤
T
6
⁡
u
x
k
L
2
s
+
∫
0
T
6
u
x
t
k
L
2
2
s
d
s
≤
c
1
,
sup
0
≤
s
≤
T
6
⁡
u
x
x
k
L
2
s
≤
c
2
,
sup
0
≤
s
≤
T
6
⁡
θ
x
k
L
2
2
s
+
ρ
k
θ
t
k
t
L
2
2
s
+
∫
0
T
6
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
5
,
sup
0
≤
s
≤
T
6
⁡
u
x
k
L
2
2
s
+
ρ
k
u
t
k
t
L
2
2
s
+
∫
0
T
6
u
x
t
k
L
2
2
s
d
s
≤
c
1
,
sup
0
≤
s
≤
T
6
⁡
θ
x
x
k
L
2
s
<
C
c
1
4
c
2
,
sup
0
≤
s
≤
T
6
⁡
ρ
k
t
H
1
s
≤
C
c
0
,
sup
0
≤
s
≤
T
6
⁡
ρ
t
k
t
L
2
s
≤
C
c
1
2
.


### 3.2. Convergence

In this subsection, the approximate solutions(
ρ
k
,
u
k
,
θ
k
) strongly converging to the solution of (3)-(4) will be proved. To this end, we define (71)
ρ
-
k
+
1
=
ρ
k
+
1
-
ρ
k
,
u
-
k
+
1
=
u
k
+
1
-
u
k
,
θ
-
k
+
1
=
θ
k
+
1
-
θ
k
,and then we can easily verify that ρ
-
k
+
1, u
-
k
+
1, θ
-
k
+
1 satisfy the system of equations(72)
ρ
-
t
k
+
1
+
ρ
-
k
+
1
u
k
x
+
ρ
k
u
-
k
x
=
0
,
(73)
ρ
k
+
1
φ
1
k
+
1
θ
t
k
+
1
+
ρ
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
-
ρ
k
+
1
2
φ
2
k
+
1
u
x
k
+
H
p
μ
1
θ
k
+
1
+
ψ
k
+
1
u
x
k
-
u
x
k
2
-
ρ
k
φ
1
k
θ
t
k
+
ρ
k
φ
1
k
u
k
-
1
θ
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
+
H
p
μ
1
θ
k
+
ψ
k
u
x
k
-
1
-
u
x
k
-
1
2
=
0
,
x
,
t
∈
Ω
T
,
(74)
ρ
k
+
1
u
-
t
k
+
1
+
ρ
k
+
1
u
k
u
-
x
k
+
1
+
H
q
μ
2
u
k
+
1
-
H
q
μ
2
u
k
+
ψ
k
+
1
x
-
ψ
k
x
=
ρ
-
k
+
1
-
u
t
k
-
u
k
-
1
u
x
k
-
ρ
k
+
1
u
-
k
u
x
k
,with the initial and boundary conditions (75)
u
-
k
+
1
=
0
,
θ
-
k
+
1
=
0
,
on
∂
Ω
×
0
,
T
,
ρ
-
k
+
1
x
,
0
=
0
,
u
-
k
+
1
x
,
0
=
0
,
θ
-
k
+
1
x
,
0
=
0
,
in
Ω
.Multiplying (72) by ρ
-
k
+
1 and integrating it over (
0,1
), we obtain(76)
d
d
t
ρ
-
k
+
1
L
2
2
≤
C
ρ
-
k
+
1
L
2
2
u
x
k
L
∞
+
ρ
k
H
1
u
-
x
k
L
2
ρ
-
k
+
1
L
2
≤
B
η
k
t
ρ
-
k
+
1
L
2
2
+
η
u
-
x
k
L
2
2
,where B
η
k
(
t
)
=
C
u
x
x
k
L
2
+
C
η
ρ
k
H
1
2, for all t
<
T
1 and k
≥
1.Multiplying (73) by θ
-
k
+
1 and integrating it over (
0,1
), we obtain(77)
1
2
d
d
t
∫
0
1
ρ
k
+
1
φ
1
k
+
1
θ
-
k
+
1
2
d
x
+
∫
0
1
H
p
μ
1
θ
k
+
1
-
H
p
μ
1
θ
k
θ
-
k
+
1
d
x
=
1
2
∫
0
1
ρ
k
+
1
φ
1
k
+
1
t
θ
-
k
+
1
2
d
x
-
∫
0
1
ρ
k
+
1
φ
1
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
∫
0
1
ρ
k
φ
1
k
θ
t
k
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
-
ρ
k
φ
1
k
u
k
-
1
θ
x
k
θ
-
k
+
1
d
x
-
∫
0
1
ψ
k
+
1
u
x
k
-
ψ
k
u
x
k
-
1
θ
-
k
+
1
d
x
+
∫
0
1
u
x
k
2
-
u
x
k
-
1
2
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
+
1
2
φ
2
k
+
1
u
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
θ
-
k
+
1
d
x
=
∑
j
=
1
6
I
j
.Using the Sobolev inequality, Young’s inequality, and Lemma3, we obtain (78)
I
1
=
1
2
∫
0
1
ρ
k
+
1
φ
1
k
+
1
t
θ
-
k
+
1
2
d
x
=
1
2
∫
0
1
ρ
t
k
+
1
φ
1
k
+
1
θ
-
k
+
1
2
d
x
+
1
2
∫
0
1
ρ
k
+
1
φ
11
k
+
1
θ
t
k
+
1
θ
-
k
+
1
2
d
x
+
1
2
∫
0
1
ρ
k
+
1
φ
12
k
+
1
ρ
t
k
+
1
θ
-
k
+
1
2
d
x
=
∑
j
=
1
3
I
1
j
.Here, (79)
I
11
=
1
2
∫
0
1
ρ
t
k
+
1
φ
1
k
+
1
θ
-
k
+
1
2
d
x
≤
C
∫
0
1
ρ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
C
∫
0
1
ρ
k
+
1
u
k
θ
-
k
+
1
θ
-
x
k
+
1
d
x
≤
C
ρ
k
+
1
L
∞
1
/
2
ρ
k
+
1
θ
-
k
+
1
L
2
u
k
L
∞
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
+
1
L
2
u
x
k
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
12
=
1
2
∫
0
1
ρ
k
+
1
φ
11
k
+
1
θ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
C
∫
0
1
ρ
k
+
1
θ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
C
θ
t
k
+
1
L
∞
ρ
k
+
1
θ
-
k
+
1
L
2
2
≤
C
θ
x
t
k
+
1
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
13
=
1
2
∫
0
1
ρ
k
+
1
φ
12
k
+
1
ρ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
C
ρ
k
+
1
L
∞
∫
0
1
ρ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
+
1
L
2
3
u
x
k
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
.In the same way, we have(80)
I
2
=
-
∫
0
1
ρ
k
+
1
φ
1
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
∫
0
1
ρ
k
φ
1
k
θ
t
k
θ
-
k
+
1
d
x
=
-
∫
0
1
ρ
-
k
+
1
φ
1
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
∫
0
1
ρ
k
φ
-
1
k
+
1
θ
t
k
θ
-
k
+
1
d
x
≤
C
∫
0
1
ρ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
C
∫
0
1
ρ
k
ρ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
C
∫
0
1
ρ
k
θ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
=
∑
j
=
1
3
I
2
j
.Here, (81)
I
21
=
C
∫
0
1
ρ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
≤
C
ρ
-
k
+
1
L
2
θ
x
t
k
L
2
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
θ
x
t
k
L
2
2
ρ
-
k
+
1
L
2
2
,
I
22
=
C
∫
0
1
ρ
k
ρ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
≤
C
ρ
x
k
L
2
ρ
-
k
+
1
L
2
θ
x
t
k
L
2
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
L
2
2
θ
x
t
k
L
2
2
ρ
-
k
+
1
L
2
2
,
I
23
=
C
∫
0
1
ρ
k
θ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
≤
C
ρ
x
k
L
2
θ
x
t
k
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
.Following the same method, we obtain(82)
I
3
=
-
∫
0
1
ρ
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
-
ρ
k
φ
1
k
u
k
-
1
θ
x
k
θ
-
k
+
1
d
x
=
-
∫
0
1
ρ
-
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
φ
-
1
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
φ
1
k
u
-
k
θ
x
k
+
1
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
-
x
k
+
1
θ
-
k
+
1
d
x
=
∑
j
=
1
4
I
3
j
.Here, (83)
I
31
=
-
∫
0
1
ρ
-
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
≤
C
ρ
-
k
+
1
L
2
u
x
k
L
2
θ
x
t
k
+
1
L
2
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
ρ
-
k
+
1
L
2
2
,
I
32
=
-
∫
0
1
ρ
k
φ
-
1
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
≤
C
∫
0
1
ρ
k
ρ
-
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
+
C
∫
0
1
ρ
k
θ
-
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
ρ
-
k
+
1
L
2
2
+
C
u
x
k
L
2
ρ
x
k
L
2
θ
x
t
k
+
1
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
33
=
∫
0
1
ρ
k
φ
1
k
u
-
k
θ
x
k
+
1
θ
-
k
+
1
d
x
≤
C
u
-
x
k
L
2
ρ
x
k
L
2
θ
x
t
k
+
1
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
≤
C
u
-
x
k
L
2
2
+
C
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
34
=
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
-
x
k
+
1
θ
-
k
+
1
d
x
≤
C
ρ
k
L
∞
ρ
k
+
1
θ
-
k
+
1
L
2
u
k
-
1
L
∞
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
L
2
2
+
C
η
ρ
x
k
L
2
2
u
x
k
-
1
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
.In a similar way, we have(84)
I
4
=
-
∫
0
1
ψ
k
+
1
u
x
k
-
ψ
k
u
x
k
-
1
θ
-
k
+
1
d
x
=
-
∫
0
1
ψ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
-
∫
0
1
ψ
k
u
-
x
k
θ
-
k
+
1
d
x
=
∑
j
=
1
2
I
4
j
.Here, (85)
I
41
=
-
∫
0
1
ψ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
≤
C
∫
0
1
ρ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
+
C
∫
0
1
θ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
ρ
-
k
+
1
L
2
2
+
C
u
x
k
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
42
=
-
∫
0
1
ψ
k
u
-
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
-
x
k
L
2
2
.Similarly,(86)
I
5
=
∫
0
1
u
x
k
2
-
u
x
k
-
1
2
θ
-
k
+
1
d
x
≤
u
-
x
k
L
2
θ
-
x
k
+
1
L
2
u
x
k
L
2
+
u
x
k
-
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
+
u
x
k
-
1
L
2
2
u
-
x
k
L
2
2
.Using the same method, we have(87)
I
6
=
-
∫
0
1
ρ
k
+
1
2
φ
2
k
+
1
u
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
θ
-
k
+
1
d
x
=
-
∫
0
1
ρ
-
k
+
1
ρ
k
+
1
+
ρ
k
φ
2
k
+
1
u
x
k
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
2
φ
-
2
k
+
1
u
x
k
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
2
φ
2
k
u
-
x
k
θ
-
k
+
1
d
x
=
∑
j
=
1
3
I
6
j
.Here, (88)
I
61
=
∫
0
1
ρ
-
k
+
1
ρ
k
+
1
+
ρ
k
φ
2
k
+
1
u
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
+
1
L
2
+
ρ
x
k
L
2
u
x
k
L
2
ρ
-
k
+
1
L
2
2
,
I
62
=
∫
0
1
ρ
k
2
φ
-
2
k
+
1
u
x
k
θ
-
k
+
1
d
x
=
C
∫
0
1
ρ
k
2
ρ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
+
C
∫
0
1
ρ
k
2
θ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
ρ
x
k
L
2
4
ρ
-
k
+
1
L
2
2
+
C
u
x
k
L
2
ρ
x
k
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
63
=
∫
0
1
ρ
k
2
φ
2
k
u
-
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
L
2
4
u
-
x
k
L
2
2
.SubstitutingI
j
(
j
=
1,2
,
…
,
6
) into (105), after choosing η
=
μ
1
(
p
-
2
)
/
2
/
26, we get(89)
1
2
d
d
t
ρ
k
+
1
φ
1
k
+
1
θ
-
k
+
1
L
2
2
+
1
2
μ
1
p
-
2
/
2
θ
-
x
k
+
1
L
2
2
≤
u
-
x
k
L
2
2
·
C
1
+
u
x
k
L
2
2
+
u
x
k
-
1
L
2
2
+
ρ
x
k
L
2
4
+
ρ
-
k
+
1
L
2
2
·
C
θ
x
t
k
L
2
2
+
ρ
x
k
L
2
2
θ
x
t
k
L
2
2
+
u
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
+
u
x
k
L
2
2
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
+
u
x
k
L
2
2
+
ρ
x
k
+
1
L
2
+
ρ
x
k
L
2
u
x
k
L
2
+
u
x
k
L
2
2
ρ
x
k
L
2
4
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
·
C
ρ
x
k
+
1
L
2
u
x
k
L
2
2
+
θ
x
t
k
+
1
L
2
+
ρ
x
k
+
1
L
2
3
u
x
k
L
2
2
+
ρ
x
k
L
2
θ
x
t
k
L
2
+
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
+
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
+
ρ
x
k
L
2
2
u
x
k
-
1
L
2
2
+
u
x
k
L
2
+
u
x
k
L
2
ρ
x
k
L
2
2
.Thus, we get(90)
δ
2
d
d
t
ρ
k
+
1
θ
-
k
+
1
L
2
2
+
1
2
μ
1
p
-
2
/
2
θ
-
x
k
+
1
L
2
2
≤
C
~
u
-
x
k
L
2
2
+
D
k
t
ρ
-
k
+
1
L
2
2
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
·There D
k
(
t
)
⊂
L
1
(
0
,
T
1
), ∫
0
1
D
k
(
t
)
≤
C
~
+
C
~
t.Multiplying (74) by u
-
k
+
1 and integrating it over (
0,1
), we obtain(91)
1
2
d
d
t
∫
0
1
ρ
k
+
1
u
-
k
+
1
2
d
x
+
∫
0
1
H
q
μ
2
u
k
+
1
-
H
q
μ
2
u
k
u
-
k
+
1
d
x
=
-
∫
0
1
ψ
x
k
+
1
-
ψ
x
k
u
-
k
+
1
d
x
+
∫
0
1
ρ
-
k
+
1
-
u
t
k
-
u
k
-
1
u
x
k
u
-
k
+
1
d
x
-
∫
0
1
ρ
k
+
1
u
-
k
u
x
k
u
-
k
+
1
d
x
.We use the Sobolev inequality, Young’s inequality, and Lemma3, and we obtain (92)
I
1
=
-
∫
0
1
ψ
x
k
+
1
-
ψ
x
k
u
-
k
+
1
d
x
d
x
=
-
∫
0
1
ψ
-
k
+
1
u
-
k
+
1
d
x
≤
C
∫
0
1
ρ
-
k
+
1
u
-
k
+
1
d
x
+
C
∫
0
1
θ
-
k
+
1
u
-
k
+
1
d
x
≤
η
u
-
x
k
+
1
L
2
2
+
C
η
ρ
-
k
+
1
L
2
2
+
η
u
-
x
k
+
1
L
2
2
+
C
η
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
2
=
-
∫
0
1
ρ
-
k
+
1
u
t
k
u
-
k
+
1
d
x
≤
ρ
-
k
+
1
L
2
u
t
k
L
∞
u
-
x
k
+
1
L
2
≤
η
u
-
x
k
+
1
L
2
2
+
C
η
u
x
t
k
L
2
2
ρ
-
k
+
1
L
2
2
,
I
3
=
-
∫
0
1
ρ
-
k
+
1
u
k
-
1
u
x
k
u
-
k
+
1
d
x
≤
ρ
-
k
+
1
L
2
u
k
-
1
L
∞
u
x
k
L
2
u
-
k
+
1
L
∞
≤
ρ
-
k
+
1
L
2
u
x
k
-
1
L
2
u
x
k
L
2
u
-
x
k
+
1
L
2
≤
η
u
-
x
k
+
1
L
2
2
+
C
η
u
x
k
-
1
L
2
2
u
x
k
L
2
2
ρ
-
k
+
1
L
2
2
,
I
4
=
-
∫
0
1
ρ
k
+
1
u
-
k
u
x
k
u
-
k
+
1
d
x
≤
C
ρ
k
+
1
L
∞
1
/
2
ρ
k
+
1
u
-
k
+
1
L
2
u
x
k
L
2
u
-
x
k
L
2
≤
ɛ
u
-
x
k
L
2
2
+
C
ɛ
ρ
x
k
+
1
L
2
u
x
k
L
2
2
ρ
k
+
1
u
-
k
+
1
L
2
2
.SubstitutingI
j
(
j
=
1,2
,
…
,
4
) into (91), after choosing η
=
μ
2
(
q
-
2
)
/
2
/
8,(93)
1
2
d
d
t
∫
0
1
ρ
k
+
1
u
-
k
+
1
2
d
x
+
1
2
μ
2
q
-
2
/
2
u
-
x
k
+
1
L
2
2
≤
ρ
-
k
+
1
L
2
2
·
C
1
+
u
x
t
k
L
2
2
+
u
x
k
-
1
L
2
2
u
x
k
L
2
2
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
·
C
ɛ
u
-
x
k
L
2
2
+
C
ɛ
ρ
x
k
+
1
L
2
u
x
k
L
2
2
ρ
k
+
1
u
-
k
+
1
L
2
2
.Thus, we get(94)
d
d
t
ρ
-
k
+
1
L
2
2
≤
C
ρ
-
k
+
1
L
2
2
u
x
k
L
∞
+
ρ
k
H
1
u
-
x
k
L
2
ρ
-
k
+
1
L
2
≤
B
η
k
t
ρ
-
k
+
1
L
2
2
+
η
u
-
x
k
L
2
2
,
∫
0
1
B
η
k
t
≤
C
~
t
,
1
2
d
d
t
ρ
k
+
1
θ
-
k
+
1
L
2
2
+
1
2
μ
1
p
-
2
/
2
∫
0
1
θ
-
x
k
+
1
L
2
2
d
x
≤
C
~
u
-
x
k
L
2
2
+
D
k
t
ρ
-
k
+
1
L
2
2
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
∫
0
1
D
k
t
≤
C
~
+
C
~
t
,
1
2
d
d
t
∫
0
1
ρ
k
+
1
u
-
k
+
1
2
d
x
+
1
2
μ
2
q
-
2
/
2
u
-
x
k
+
1
L
2
2
≤
E
ɛ
k
t
ρ
-
k
+
1
L
2
2
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
+
ρ
k
+
1
u
-
k
+
1
L
2
2
+
ɛ
u
-
x
k
L
2
2
.There E
ɛ
k
(
t
)
⊂
L
1
(
0
,
T
1
), ∫
0
1
E
ɛ
k
(
t
)
≤
C
~
+
C
~
ɛ
t. Using (94), denote(95)
Ψ
k
+
1
=
ρ
-
k
+
1
L
2
2
+
ɛ
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
+
1
2
ρ
k
+
1
u
-
k
+
1
L
2
2
.There F
ɛ
k
(
t
)
⊂
L
1
(
0
,
T
1
), ∫
0
1
F
ɛ
k
(
t
)
≤
C
~
+
C
~
ɛ
t.Thus, we get(96)
d
d
t
Ψ
k
+
1
t
+
ɛ
2
δ
μ
1
p
-
2
/
2
θ
-
x
k
+
1
L
2
2
+
1
2
μ
2
q
-
2
/
2
u
-
x
k
+
1
L
2
2
≤
F
ɛ
k
t
·
Ψ
k
+
1
t
+
C
~
·
ɛ
·
u
-
x
k
L
2
2
.By Gronwall’s inequality, noting that Ψ
k
+
1
(
t
)
=
0, we get(97)
Ψ
k
+
1
t
+
ɛ
2
δ
μ
1
p
-
2
/
2
∫
0
t
θ
-
x
k
+
1
L
2
2
s
d
s
+
1
2
μ
2
q
-
2
/
2
∫
0
t
u
-
x
k
+
1
L
2
2
s
d
s
≤
exp
⁡
∫
0
t
F
ɛ
k
t
d
t
·
∫
0
t
C
~
·
ɛ
·
u
-
x
k
L
2
2
s
d
s
≤
exp
⁡
C
~
+
C
~
ɛ
t
·
∫
0
t
C
~
·
ɛ
·
u
-
x
k
L
2
2
s
d
s
.We can obtain(98)
C
~
·
ɛ
·
exp
⁡
C
~
=
1
4
,
exp
⁡
C
~
ɛ
·
T
7
=
μ
2
q
-
2
/
2
,and thus we can get(99)
∑
k
=
1
K
Ψ
k
+
1
t
+
∑
k
=
1
K
∫
0
T
∗
θ
-
x
k
+
1
s
L
2
2
+
u
-
x
k
+
1
s
L
2
2
d
s
≤
C
~
<
+
∞
.There, T
∗
=
min
⁡
{
T
6
,
T
7
}.Therefore, we conclude that the full sequence(
ρ
k
,
u
k
,
θ
k
) converges to a limit (
ρ
,
u
,
θ
) in the following strong sense:(100)
ρ
k
⟶
ρ
in
L
∞
0
,
T
∗
;
L
2
,
u
k
⟶
u
in
L
∞
0
,
T
∗
;
L
2
∩
L
2
0
,
T
∗
;
H
0
1
,
θ
k
⟶
θ
in
L
∞
0
,
T
∗
;
L
2
∩
L
2
0
,
T
∗
;
H
0
1
,for T
∗
=
min
⁡
T
1
,
T
2
,
T
3
,
T
4
,
T
5
,
T
6
,
T
7, and we get (101)
sup
0
≤
s
≤
T
∗
⁡
u
x
L
2
s
+
∫
0
T
∗
u
x
t
k
L
2
2
s
d
s
≤
c
1
,
sup
0
≤
s
≤
T
∗
⁡
u
x
x
L
2
s
≤
c
2
,
sup
0
≤
s
≤
T
∗
⁡
θ
x
L
2
2
s
+
ρ
k
θ
t
k
t
L
2
2
s
+
∫
0
T
∗
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
5
,
sup
0
≤
s
≤
T
∗
⁡
u
x
L
2
2
s
+
ρ
k
u
t
k
t
L
2
2
s
+
∫
0
T
∗
u
x
t
L
2
2
s
d
s
≤
c
1
,
sup
0
≤
s
≤
T
∗
⁡
θ
x
x
L
2
s
<
C
c
1
4
c
2
,
sup
0
≤
s
≤
T
∗
⁡
ρ
t
H
1
s
≤
C
c
0
,
sup
0
≤
s
≤
T
∗
⁡
ρ
t
t
L
2
s
≤
C
c
1
2
.


## 3.1. A Priori Estimates for Smooth Solutions

We first construct approximate solutions of problem (3)-(4). By using the iterative scheme, inductively, consider the following:(i) Firstly defineu
0
=
0, θ
0
=
0.(ii) Assuming thatu
k
-
1, θ
k
-
1 were defined for k
≥
1, we can obtain (
ρ
k
,
u
k
,
θ
k
,
e
k
,
π
k
) as the unique smooth solution to the following system:(16)
ρ
t
k
+
u
k
-
1
ρ
x
k
+
u
x
k
-
1
ρ
k
=
0
,
x
,
t
∈
Ω
T
,
(17)
π
k
=
ϕ
θ
k
,
ρ
k
,
x
,
t
∈
Ω
T
,
(18)
e
k
=
φ
θ
k
,
ρ
k
,
x
,
t
∈
Ω
T
,
(19)
ρ
k
φ
1
k
θ
t
k
+
ρ
k
φ
1
k
u
k
-
1
θ
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
+
H
p
μ
1
θ
k
+
ψ
k
u
x
k
-
1
-
u
x
k
-
1
2
=
0
,
x
,
t
∈
Ω
T
,
(20)
ρ
k
u
t
k
+
ρ
k
u
k
-
1
u
x
k
-
H
q
μ
2
θ
k
+
ψ
x
=
0
,
x
,
t
∈
Ω
T
,
(21)
ρ
k
x
,
0
=
ρ
0
,
u
k
x
,
0
=
u
0
,
θ
k
x
,
0
=
θ
0
,
u
k
0
,
t
=
u
k
1
,
t
=
0
,
θ
k
0
,
t
=
θ
k
1
,
t
=
0
,where m
0
=
∫
0
1
ρ
0
(
x
)
d
x
>
0 is the initial mass, for a given function ω and numbers s
>
2, μ
>
0,(22)
H
s
μ
ω
=
-
ω
x
2
+
μ
s
-
2
/
2
ω
x
x
.With this process, the nonlinear coupled system has been deduced into a sequence of decoupled problems and each problem admits a smooth solution. Moreover, we findρ
k from (16), (21) with the smooth function u
k
-
1; that is, (23)
ρ
t
k
+
u
k
-
1
ρ
x
k
-
1
+
u
x
k
-
1
ρ
k
-
1
=
0
,
ρ
k
t
=
0
=
ρ
0
.Obviously there is a unique solutionρ
k to the above initial value problem and also by a standard argument, we could get (24)
ρ
k
x
,
t
≥
δ
exp
⁡
-
∫
0
T
∗
u
x
k
-
1
·
,
s
L
∞
d
s
>
0
,
∀
t
∈
0
,
T
∗
.Then by (19) and (21) we get θ
k. Lastly, with ρ
k, θ
k, we obtain u
k from (20) and (21).Moreover, assume that for a givenT
1, 0
<
T
1
≤
T
(25)
sup
0
≤
t
≤
T
1
⁡
u
x
x
k
-
1
L
2
+
∫
0
T
1
u
x
t
k
-
1
L
2
2
d
s
≤
c
1
,
sup
0
≤
t
≤
T
1
⁡
u
x
x
k
-
1
L
2
≤
c
2
.For some, c
2
>
c
1
>
c
0
>
1 will be fixed later. We will prove that(26)
sup
0
≤
t
≤
T
1
⁡
u
x
k
L
2
+
∫
0
T
1
u
x
t
k
L
2
2
d
s
≤
c
1
,
sup
0
≤
t
≤
T
1
⁡
u
x
x
k
L
2
≤
c
2
,and some more useful uniform estimate about (
ρ
k
,
u
k
,
θ
k
).(i) Estimation forρ
k.Then multiplying (16) by ρ
k and integrating it over (
0,1
) with respect to x, we have (27)
1
2
d
d
t
∫
0
1
ρ
k
2
d
s
+
∫
0
1
ρ
k
-
1
u
k
-
1
x
ρ
k
d
x
=
0
.Integrating by parts, using the Sobolev inequality and Lemma3, we deduce that(28)
d
d
t
ρ
k
t
L
2
2
≤
u
x
x
k
-
1
L
2
ρ
k
L
2
2
.Then, considering differential equation (16) with respect to x and multiplying it by ρ
x
k, integrating it over (
0,1
) on x, and using the Sobolev inequality, we have(29)
d
d
t
∫
0
1
ρ
x
k
2
d
x
=
-
∫
0
1
3
2
u
x
k
-
1
ρ
x
k
2
+
ρ
k
ρ
x
k
u
x
x
k
-
1
t
d
x
≤
3
2
u
x
k
-
1
L
∞
ρ
x
k
L
2
2
+
ρ
k
L
∞
u
x
x
k
-
1
L
2
≤
3
ρ
x
k
L
2
2
u
x
x
k
-
1
L
2
.From (28) and (29), by Gronwall’s inequality, it follows that(30)
sup
0
≤
t
≤
T
⁡
ρ
k
t
H
1
2
≤
ρ
0
k
H
1
2
exp
⁡
4
∫
0
t
u
x
x
k
-
1
L
2
d
s
,and, taking T
2
=
c
2
-
1, we get(31)
sup
0
≤
t
≤
T
⁡
ρ
k
t
H
1
≤
ρ
0
k
H
1
exp
⁡
4
∫
0
t
u
x
x
k
-
1
L
2
d
s
1
/
2
≤
C
c
0
.And using (16) we can also obtain(32)
ρ
t
k
t
L
2
≤
ρ
x
k
-
1
L
2
u
x
k
-
1
L
2
+
ρ
k
-
1
L
2
u
x
x
k
-
1
L
2
≤
C
c
1
2
,where C is a positive constant.(ii) Estimation forθ
k
(
t
).Differentiating (19) with respect to t, multiplying it by θ
t
k, and integrating it over (
0,1
) on x, we derive(33)
1
2
d
d
t
∫
0
1
ρ
k
φ
1
k
θ
t
k
2
d
x
+
∫
0
1
θ
x
2
+
μ
1
p
-
2
/
2
θ
x
t
θ
x
t
k
d
x
=
1
2
∫
0
1
ρ
k
φ
1
k
t
θ
t
k
2
d
x
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
x
k
t
θ
t
k
d
x
-
∫
0
1
ρ
k
2
φ
2
k
u
x
k
-
1
t
θ
t
k
d
x
-
∫
0
1
ψ
k
u
x
k
-
1
t
θ
t
k
d
x
+
∫
0
1
u
x
k
-
1
t
2
θ
t
k
d
x
=
∑
j
=
1
5
I
j
.Using the Sobolev inequality, Young’s inequality, Lemma3, and (25), (30) and (31), we obtain (34)
I
1
=
1
2
∫
0
1
ρ
k
φ
1
k
t
θ
t
k
2
d
x
=
1
2
∫
0
1
ρ
t
k
φ
1
k
θ
t
k
2
d
x
+
1
2
∫
0
1
ρ
k
φ
11
k
θ
t
k
θ
t
k
2
d
x
+
1
2
∫
0
1
ρ
k
φ
12
k
ρ
t
k
θ
t
k
2
d
x
=
∑
j
=
1
3
I
1
j
.Here, (35)
I
11
=
1
2
∫
0
1
ρ
t
k
φ
1
k
θ
t
k
2
d
x
≤
C
∫
0
1
ρ
t
k
θ
t
k
2
d
x
≤
C
∫
0
1
ρ
k
u
k
-
1
θ
t
k
θ
x
t
k
d
x
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
u
k
-
1
L
∞
θ
x
t
k
L
2
≤
C
c
0
1
/
2
c
1
ρ
k
θ
t
k
L
2
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
3
ρ
k
θ
t
k
L
2
2
,
I
12
=
1
2
∫
0
1
ρ
k
φ
11
k
θ
t
k
θ
t
k
2
d
x
≤
C
∫
0
1
ρ
k
θ
t
k
θ
t
k
2
d
x
≤
C
θ
t
k
L
∞
ρ
k
θ
t
k
L
2
2
≤
η
θ
x
t
k
L
2
2
+
C
η
ρ
k
θ
t
k
L
2
4
,
I
13
=
1
2
∫
0
1
ρ
k
φ
12
k
-
1
ρ
t
k
θ
t
k
2
d
x
≤
C
ρ
k
L
∞
∫
0
1
ρ
t
k
θ
t
k
2
d
x
≤
C
c
0
3
/
2
c
1
ρ
k
θ
t
k
L
2
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
5
ρ
k
θ
t
k
L
2
2
.In the same way, we have(36)
I
2
=
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
x
k
t
θ
t
k
d
x
=
-
∫
0
1
ρ
t
k
φ
1
k
u
k
-
1
θ
x
k
θ
t
k
d
x
-
∫
0
1
ρ
k
φ
11
k
θ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
-
∫
0
1
ρ
k
φ
12
k
ρ
t
k
u
k
θ
x
k
θ
t
k
d
x
-
∫
0
1
ρ
k
φ
1
k
u
t
k
-
1
θ
x
k
θ
t
k
d
x
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
x
t
k
θ
t
k
d
x
=
∑
j
=
1
5
I
2
j
.Here, (37)
I
21
=
-
∫
0
1
ρ
t
k
φ
1
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
∫
0
1
ρ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
ρ
t
k
L
2
u
k
-
1
L
∞
θ
x
k
L
2
θ
t
k
L
∞
≤
C
ρ
t
k
L
2
u
x
k
-
1
L
2
θ
x
k
L
2
θ
x
t
k
L
2
≤
C
c
1
3
θ
x
k
L
2
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
6
θ
x
k
L
2
2
,
I
22
=
-
∫
0
1
ρ
k
φ
11
k
θ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
θ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
ρ
k
L
∞
1
/
2
u
k
-
1
L
∞
θ
t
k
L
∞
θ
x
k
L
2
ρ
k
θ
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
3
θ
x
k
L
2
2
ρ
k
θ
t
k
L
2
2
,
I
23
=
-
∫
0
1
ρ
k
φ
12
k
ρ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
ρ
t
k
u
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
ρ
k
L
∞
ρ
t
k
L
2
u
k
-
1
L
∞
θ
x
k
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
8
θ
x
k
L
2
2
,
I
24
=
-
∫
0
1
ρ
k
φ
1
k
u
t
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
u
t
k
-
1
θ
x
k
θ
t
k
d
x
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
u
t
k
-
1
L
∞
θ
x
k
L
2
≤
C
c
0
θ
x
k
L
2
2
ρ
k
θ
t
k
L
2
2
+
1
2
u
x
t
k
-
1
L
2
2
,
I
25
=
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
x
t
k
θ
t
k
d
x
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
u
k
-
1
L
∞
θ
x
t
k
L
2
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
u
k
-
1
L
∞
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
6
ρ
k
θ
t
k
L
2
2
.Following the same method, we obtain(38)
I
3
=
-
∫
0
1
ρ
k
2
φ
2
k
u
x
k
-
1
t
θ
t
k
d
x
=
-
∫
0
1
2
ρ
k
ρ
t
k
φ
2
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ρ
k
2
φ
12
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ρ
k
2
φ
22
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ρ
k
2
φ
2
k
u
x
t
k
-
1
θ
t
k
d
x
=
∑
j
=
1
4
I
3
j
.Here, (39)
I
31
=
-
∫
0
1
2
ρ
k
ρ
t
k
φ
2
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
ρ
k
L
∞
ρ
t
k
L
2
u
x
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
8
,
I
32
=
-
∫
0
1
ρ
k
2
φ
12
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
2
θ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
ρ
k
L
∞
u
x
k
-
1
L
∞
ρ
k
θ
t
k
L
2
2
≤
C
η
c
1
2
ρ
k
θ
t
k
L
2
2
,
I
33
=
-
∫
0
1
ρ
k
2
φ
22
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
2
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
ρ
k
L
∞
2
ρ
t
k
L
2
u
x
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
10
,
I
34
=
-
∫
0
1
ρ
k
2
φ
2
k
u
x
t
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
2
u
x
t
k
-
1
θ
t
k
d
x
≤
C
ρ
k
L
∞
2
u
x
t
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
4
u
x
t
k
-
1
L
2
2
.In a similar way,(40)
I
4
=
-
∫
0
1
ψ
k
u
x
k
-
1
t
θ
t
k
d
x
=
-
∫
0
1
ψ
1
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ψ
2
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
-
∫
0
1
ψ
k
-
1
u
x
t
k
-
1
θ
t
k
d
x
=
∑
j
=
1
3
I
4
j
.Here, (41)
I
41
=
-
∫
0
1
ψ
1
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
k
ρ
k
θ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
u
x
k
-
1
L
∞
ρ
k
θ
t
k
L
2
2
≤
C
c
1
ρ
k
θ
t
k
L
2
2
,
I
42
=
-
∫
0
1
ψ
2
k
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
ρ
t
k
u
x
k
-
1
θ
t
k
d
x
≤
C
ρ
t
k
L
2
u
x
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
10
,
I
43
=
-
∫
0
1
ψ
k
u
x
t
k
-
1
θ
t
k
d
x
≤
C
∫
0
1
u
x
t
k
-
1
θ
t
k
d
x
≤
C
u
x
t
k
-
1
L
2
θ
t
k
L
∞
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
2
.Similarly,(42)
I
5
=
∫
0
1
u
x
k
-
1
t
2
θ
t
k
d
x
≤
2
u
x
k
-
1
L
∞
u
x
t
k
-
1
L
2
θ
k
L
∞
≤
2
u
x
x
k
-
1
L
2
u
x
t
k
-
1
L
2
θ
x
t
k
L
2
≤
η
θ
x
t
k
L
2
2
+
C
η
c
1
2
u
x
t
k
-
1
L
2
2
.SubstitutingI
j
(
j
=
1,2
,
…
,
5
) into (43), after choosing η
=
μ
1
(
p
-
2
)
/
2
/
26, we get(43)
1
2
d
d
t
ρ
k
φ
1
k
θ
t
k
t
L
2
2
+
1
2
μ
1
p
-
2
/
2
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
10
+
C
c
1
8
ρ
k
θ
t
k
L
2
2
+
C
ρ
k
θ
t
k
L
2
4
+
C
c
1
6
θ
x
k
L
2
2
+
C
c
1
3
θ
x
k
L
2
2
ρ
k
θ
t
k
L
2
2
+
C
c
1
4
u
x
t
k
-
1
L
2
2
,and, integrating it over (
τ
,
t
)
⊂
(
0
,
T
), we have(44)
1
2
ρ
k
φ
1
k
θ
t
k
t
L
2
2
+
1
2
μ
1
p
-
2
/
2
∫
τ
t
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
10
∫
τ
t
1
+
θ
x
k
L
2
2
1
+
ρ
k
θ
t
k
L
2
2
2
s
d
s
+
C
c
1
4
∫
τ
t
u
x
t
k
-
1
L
2
2
s
d
s
+
ρ
k
φ
1
k
θ
t
k
τ
L
2
2
,where C is a positive constant.We estimateρ
k
φ
1
k
θ
t
k
τ
L
2
2 as follows: multiplying (17) by θ
t
k and integrating it over (
0,1
), we obtain (45)
∫
0
1
ρ
k
φ
1
k
θ
t
k
2
d
x
≤
∫
0
1
φ
1
k
-
1
ρ
k
3
φ
2
k
2
u
x
k
-
1
2
+
ρ
k
-
1
ψ
k
2
u
x
k
-
1
2
+
ρ
k
-
1
H
p
μ
1
θ
k
-
u
x
k
2
2
d
x
+
∫
0
1
ρ
k
φ
1
k
u
k
-
1
2
θ
x
k
2
d
x
.Next, we get (46)
∫
0
1
ρ
k
φ
1
k
θ
t
k
2
d
x
≤
C
∫
0
1
ρ
k
u
k
-
1
2
θ
x
k
2
+
ρ
k
3
u
x
k
-
1
2
+
ρ
k
-
1
u
x
k
-
1
2
+
ρ
k
-
1
H
p
μ
1
θ
k
-
u
x
k
2
2
d
x
,and, according to the smoothness of (
ρ
k
,
u
k
,
θ
k
), we have(47)
lim
τ
→
0
⁡
∫
0
1
ρ
k
u
k
-
1
2
θ
x
k
2
+
ρ
k
3
u
x
k
-
1
2
+
ρ
k
-
1
u
x
k
-
1
2
+
ρ
k
-
1
H
p
μ
1
θ
k
-
u
x
k
2
2
d
x
≤
C
c
0
5
,where C is a positive constant.We have(48)
d
d
t
θ
x
k
L
2
2
≤
2
∫
0
1
θ
x
k
θ
x
t
k
d
x
≤
2
θ
x
k
L
2
θ
x
t
k
L
2
≤
θ
x
k
L
2
2
+
θ
x
t
k
L
2
2
;by Gronwall’s inequality, it follows that(49)
θ
x
k
L
2
2
≤
exp
⁡
t
θ
x
k
0
L
2
2
+
∫
0
t
θ
x
t
k
L
2
2
d
sfor 0
<
t
<
min
⁡
{
T
1
,
T
2
},(50)
θ
x
k
L
2
2
≤
C
θ
x
k
0
L
2
2
+
C
∫
0
t
θ
x
t
k
L
2
2
d
s
.Due to (44), (47), and (50), we get(51)
θ
x
k
L
2
2
+
ρ
k
θ
t
k
t
L
2
2
+
∫
τ
t
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
10
∫
τ
t
1
+
θ
x
k
L
2
2
1
+
ρ
k
θ
t
k
L
2
2
2
d
s
+
C
c
1
5
.DenoteΓ
(
t
)
=
1
+
θ
x
k
L
2
2
+
ρ
k
θ
t
k
L
2
2; we get(52)
Γ
t
≤
C
c
1
10
∫
0
t
Γ
t
3
d
s
+
C
c
1
5
;by Gronwall’s inequality, ∃
T
3
≤
T
2 (T
3 has nothing to do with k), it follows that(53)
Γ
t
≤
C
c
1
5
.Note that, for a given functionω, by a simple computation, we have(54)
θ
x
k
2
+
μ
1
p
-
2
/
2
θ
x
k
t
θ
x
t
k
=
θ
x
k
2
+
μ
1
p
-
4
/
2
r
-
1
θ
x
k
2
+
μ
1
θ
x
t
k
2
≥
θ
x
k
2
+
μ
1
p
-
4
/
2
θ
x
k
2
+
μ
1
θ
x
t
k
2
=
θ
x
k
2
+
μ
1
p
-
2
/
2
θ
x
t
k
2
≥
μ
1
p
-
2
/
2
θ
x
t
k
2
.Firstly, we need to do the following estimates: by (19)(55)
θ
x
2
+
μ
1
p
-
2
/
2
θ
x
x
=
ρ
k
φ
1
k
θ
t
k
+
ρ
k
φ
1
k
u
k
-
1
θ
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
+
ψ
k
u
x
k
-
1
-
u
x
k
-
1
2
,and we have (56)
θ
x
k
2
+
μ
1
p
-
2
/
2
θ
x
k
x
=
θ
x
k
2
+
μ
1
p
-
4
/
2
p
-
1
θ
x
k
2
+
μ
1
θ
x
x
k
≥
θ
x
k
2
+
μ
1
p
-
4
/
2
θ
x
k
2
+
μ
1
θ
x
x
k
=
θ
x
k
2
+
μ
1
p
-
2
/
2
θ
x
x
k
≥
μ
1
p
-
2
/
2
θ
x
x
k
.taking it by L
2-norm, we get(57)
θ
x
x
k
L
2
≤
μ
1
2
-
p
/
2
ρ
k
φ
1
k
θ
t
k
L
2
+
ρ
k
φ
1
k
u
k
-
1
θ
x
k
L
2
+
ρ
k
2
φ
2
k
u
x
k
-
1
L
2
+
ψ
k
u
x
k
-
1
L
2
+
u
x
k
-
1
2
L
2
≤
C
ρ
k
L
∞
1
/
2
ρ
k
θ
t
k
L
2
+
ρ
k
L
∞
u
k
-
1
L
∞
θ
x
k
L
2
+
ρ
k
L
∞
2
u
x
k
-
1
L
2
+
u
x
k
-
1
L
2
+
u
x
k
-
1
L
∞
u
x
k
-
1
L
2
≤
C
ρ
x
k
L
2
1
/
2
ρ
k
θ
t
k
L
2
+
ρ
x
k
L
2
u
x
k
-
1
L
2
θ
x
k
L
2
+
ρ
x
k
L
2
2
u
x
k
-
1
L
2
+
u
x
k
-
1
L
2
+
u
x
x
k
-
1
L
2
u
x
k
-
1
L
2
≤
C
c
1
4
c
2
.(iii) Estimation foru
k
(
t
).Differentiating (20) with respect to t, multiplying it by u
t
k, and integrating it over (0,1) on x, we derive(58)
1
2
d
d
t
∫
0
1
ρ
k
u
t
k
2
d
x
+
∫
0
1
u
x
2
+
μ
2
q
-
2
/
2
u
x
t
u
x
t
k
d
x
=
1
2
∫
0
1
ρ
t
k
u
t
k
2
d
x
-
∫
0
1
ρ
t
k
u
k
-
1
u
x
k
+
ρ
k
u
t
k
-
1
u
x
k
+
ρ
k
u
k
-
1
u
x
t
k
u
t
k
d
x
+
∫
0
1
ψ
t
k
u
x
t
k
d
x
=
∑
j
=
1
5
I
j
.Then, using the Sobolev inequality, Young’s inequality, Lemma3, and (25), (30), and (31), we obtain (59)
I
1
=
1
2
∫
0
1
ρ
t
k
u
t
k
2
d
x
≤
∫
0
1
ρ
k
u
k
-
1
u
t
k
u
x
t
k
d
x
≤
ρ
k
L
∞
1
/
2
u
k
-
1
L
∞
ρ
k
u
t
k
L
2
u
x
t
k
L
2
≤
η
1
u
x
t
k
L
2
2
+
C
η
1
c
1
3
ρ
k
u
t
k
L
2
2
,
I
2
=
-
∫
0
1
ρ
t
k
u
k
-
1
u
x
k
u
t
k
d
x
≤
∫
0
1
ρ
t
k
u
k
-
1
u
x
k
u
t
k
d
x
≤
ρ
t
k
L
2
u
k
-
1
L
∞
2
u
x
k
L
2
u
t
k
L
∞
≤
η
1
u
x
t
k
L
2
2
+
C
η
1
c
1
6
u
x
k
L
2
2
,
I
3
=
-
∫
0
1
ρ
k
u
t
k
-
1
u
x
k
u
t
k
d
x
≤
∫
0
1
ρ
k
u
t
k
-
1
u
x
k
u
t
k
d
x
≤
ρ
k
L
∞
1
/
2
u
t
k
-
1
L
∞
u
x
k
L
2
ρ
k
u
t
k
L
2
≤
c
0
1
/
2
u
x
t
k
-
1
L
2
u
x
k
L
2
ρ
k
u
t
k
L
2
≤
c
1
-
1
u
x
t
k
-
1
L
2
2
ρ
k
u
t
k
L
2
2
+
1
4
c
0
c
1
-
1
u
x
k
L
2
2
,
I
4
=
-
∫
0
1
ρ
k
u
k
-
1
u
x
t
k
u
t
k
d
x
≤
∫
0
1
ρ
k
u
k
-
1
u
t
k
u
x
t
k
d
x
≤
ρ
k
L
∞
1
/
2
u
k
-
1
L
∞
ρ
k
u
t
k
L
2
u
x
t
k
L
2
≤
η
1
u
x
t
k
L
2
2
+
C
η
1
c
1
3
ρ
k
u
t
k
L
2
2
,
I
5
=
∫
0
1
ψ
t
k
u
x
t
k
d
x
=
∫
0
1
ψ
1
k
θ
t
k
u
x
t
k
d
x
+
∫
0
1
ψ
2
k
ρ
t
k
u
x
t
k
d
x
≤
C
ρ
k
θ
t
k
L
2
u
x
t
k
L
2
+
C
ρ
t
k
L
2
u
x
t
k
L
2
≤
2
η
1
u
x
t
k
L
2
2
+
C
η
1
c
1
4
+
C
η
1
c
1
5
.SubstitutingI
j
(
j
=
1,2
,
…
,
5
) into (33), after choosing η
=
μ
2
(
q
-
2
)
/
2
/
10
(60)
1
2
d
d
t
ρ
k
u
t
k
t
L
2
2
+
1
2
μ
2
q
-
2
/
2
u
x
t
k
L
2
2
≤
C
c
1
3
ρ
k
u
t
k
L
2
2
+
C
c
1
6
u
x
k
L
2
2
+
c
1
-
1
u
x
t
k
-
1
L
2
ρ
k
u
t
k
L
2
+
1
4
c
0
c
1
-
1
u
x
k
L
2
+
C
c
1
5
.We estimateρ
k
u
t
k
τ
L
2
2 as follows: multiplying (20) by u
t
k and integrating it over (
0,1
), we obtain (61)
∫
0
1
ρ
k
u
t
k
2
d
x
≤
2
∫
0
1
ρ
k
u
k
-
1
2
u
x
k
2
+
ρ
k
-
1
H
q
μ
2
u
k
-
ψ
x
2
d
x
.According to the smoothness of(
ρ
k
,
u
k
,
θ
k
), we have(62)
lim
τ
→
0
⁡
∫
0
1
ρ
k
u
k
-
1
2
u
x
k
2
+
ρ
k
-
1
L
p
u
k
+
ψ
x
2
d
x
≤
ρ
0
L
∞
u
0
L
∞
2
u
0
x
L
2
2
+
g
L
2
2
≤
C
c
0
5
,where C is a positive constant.In the same way,(63)
u
x
k
L
2
2
≤
C
u
x
k
0
L
2
2
+
C
∫
0
t
u
x
t
k
L
2
2
d
s
.Integrating (60) over (
0
,
t
)
⊂
(
0
,
T
), combining with (62) and (63) we have(64)
u
x
k
L
2
2
+
ρ
k
u
t
k
t
L
2
2
+
∫
0
t
u
x
t
k
L
2
2
≤
C
∫
0
t
c
1
6
+
c
1
-
1
u
x
t
k
-
1
L
2
ρ
k
u
t
k
t
L
2
2
+
u
x
k
L
2
2
d
s
+
C
c
1
5
t
+
C
c
0
5
,and for 0
<
t
<
min
⁡
{
T
1
,
T
4
}, T
4
≤
c
0
5
/
c
1
5, we get(65)
u
x
k
L
2
2
+
ρ
k
u
t
k
t
L
2
2
+
∫
0
t
u
x
t
k
L
2
2
≤
C
∫
0
t
c
1
6
+
c
1
-
1
u
x
t
k
-
1
L
2
ρ
k
u
t
k
t
L
2
2
+
u
x
k
L
2
2
d
s
+
C
c
0
5
,and, by Gronwall’s inequality, ∃
T
5
≤
T
4 (T
5 has nothing to do with k), it follows that(66)
u
x
k
L
2
2
+
ρ
k
u
t
k
t
L
2
2
+
∫
0
t
u
x
t
k
L
2
2
s
d
s
≤
C
c
0
5
=
c
1
.Secondly, we need to do the following estimates. By (20)(67)
u
x
2
+
μ
2
q
-
2
/
2
u
x
x
=
ρ
k
u
t
k
+
ρ
k
u
k
-
1
u
x
k
-
1
+
ϕ
1
k
-
1
θ
x
k
-
1
+
ϕ
2
k
-
1
ρ
x
k
,we have (68)
u
x
k
2
+
μ
2
q
-
2
/
2
u
x
k
x
=
u
x
k
2
+
μ
2
q
-
4
/
2
q
-
1
u
x
k
2
+
μ
2
u
x
x
k
≥
u
x
k
2
+
μ
2
q
-
4
/
2
u
x
k
2
+
μ
2
u
x
x
k
=
u
x
k
2
+
μ
2
q
-
2
/
2
u
x
x
k
≥
μ
2
q
-
2
/
2
u
x
x
k
,and, taking it by L
2-norm, using Lemma 3, we get(69)
u
x
x
k
L
2
≤
μ
2
2
-
q
/
2
ρ
k
u
t
k
L
2
+
ρ
k
u
k
-
1
u
x
k
L
2
+
ψ
1
k
θ
x
k
L
2
+
ψ
2
k
ρ
x
k
L
2
≤
C
ρ
k
L
∞
1
/
2
ρ
k
u
t
k
L
2
+
ρ
k
L
∞
u
k
-
1
L
∞
u
x
k
L
q
+
θ
x
k
L
2
+
ρ
x
k
L
2
≤
C
ρ
x
k
L
2
1
/
2
ρ
k
u
t
k
L
2
+
ρ
x
k
L
2
u
x
k
-
1
L
q
u
x
k
L
q
+
θ
x
k
L
2
+
ρ
x
k
L
2
≤
C
c
1
3
=
c
2
=
C
c
0
15
,where C is a positive constant. For T
6
=
min
⁡
[
T
1
,
T
2
,
T
3
,
T
4
,
T
5
], we get (70)
sup
0
≤
s
≤
T
6
⁡
u
x
k
L
2
s
+
∫
0
T
6
u
x
t
k
L
2
2
s
d
s
≤
c
1
,
sup
0
≤
s
≤
T
6
⁡
u
x
x
k
L
2
s
≤
c
2
,
sup
0
≤
s
≤
T
6
⁡
θ
x
k
L
2
2
s
+
ρ
k
θ
t
k
t
L
2
2
s
+
∫
0
T
6
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
5
,
sup
0
≤
s
≤
T
6
⁡
u
x
k
L
2
2
s
+
ρ
k
u
t
k
t
L
2
2
s
+
∫
0
T
6
u
x
t
k
L
2
2
s
d
s
≤
c
1
,
sup
0
≤
s
≤
T
6
⁡
θ
x
x
k
L
2
s
<
C
c
1
4
c
2
,
sup
0
≤
s
≤
T
6
⁡
ρ
k
t
H
1
s
≤
C
c
0
,
sup
0
≤
s
≤
T
6
⁡
ρ
t
k
t
L
2
s
≤
C
c
1
2
.


## 3.2. Convergence

In this subsection, the approximate solutions(
ρ
k
,
u
k
,
θ
k
) strongly converging to the solution of (3)-(4) will be proved. To this end, we define (71)
ρ
-
k
+
1
=
ρ
k
+
1
-
ρ
k
,
u
-
k
+
1
=
u
k
+
1
-
u
k
,
θ
-
k
+
1
=
θ
k
+
1
-
θ
k
,and then we can easily verify that ρ
-
k
+
1, u
-
k
+
1, θ
-
k
+
1 satisfy the system of equations(72)
ρ
-
t
k
+
1
+
ρ
-
k
+
1
u
k
x
+
ρ
k
u
-
k
x
=
0
,
(73)
ρ
k
+
1
φ
1
k
+
1
θ
t
k
+
1
+
ρ
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
-
ρ
k
+
1
2
φ
2
k
+
1
u
x
k
+
H
p
μ
1
θ
k
+
1
+
ψ
k
+
1
u
x
k
-
u
x
k
2
-
ρ
k
φ
1
k
θ
t
k
+
ρ
k
φ
1
k
u
k
-
1
θ
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
+
H
p
μ
1
θ
k
+
ψ
k
u
x
k
-
1
-
u
x
k
-
1
2
=
0
,
x
,
t
∈
Ω
T
,
(74)
ρ
k
+
1
u
-
t
k
+
1
+
ρ
k
+
1
u
k
u
-
x
k
+
1
+
H
q
μ
2
u
k
+
1
-
H
q
μ
2
u
k
+
ψ
k
+
1
x
-
ψ
k
x
=
ρ
-
k
+
1
-
u
t
k
-
u
k
-
1
u
x
k
-
ρ
k
+
1
u
-
k
u
x
k
,with the initial and boundary conditions (75)
u
-
k
+
1
=
0
,
θ
-
k
+
1
=
0
,
on
∂
Ω
×
0
,
T
,
ρ
-
k
+
1
x
,
0
=
0
,
u
-
k
+
1
x
,
0
=
0
,
θ
-
k
+
1
x
,
0
=
0
,
in
Ω
.Multiplying (72) by ρ
-
k
+
1 and integrating it over (
0,1
), we obtain(76)
d
d
t
ρ
-
k
+
1
L
2
2
≤
C
ρ
-
k
+
1
L
2
2
u
x
k
L
∞
+
ρ
k
H
1
u
-
x
k
L
2
ρ
-
k
+
1
L
2
≤
B
η
k
t
ρ
-
k
+
1
L
2
2
+
η
u
-
x
k
L
2
2
,where B
η
k
(
t
)
=
C
u
x
x
k
L
2
+
C
η
ρ
k
H
1
2, for all t
<
T
1 and k
≥
1.Multiplying (73) by θ
-
k
+
1 and integrating it over (
0,1
), we obtain(77)
1
2
d
d
t
∫
0
1
ρ
k
+
1
φ
1
k
+
1
θ
-
k
+
1
2
d
x
+
∫
0
1
H
p
μ
1
θ
k
+
1
-
H
p
μ
1
θ
k
θ
-
k
+
1
d
x
=
1
2
∫
0
1
ρ
k
+
1
φ
1
k
+
1
t
θ
-
k
+
1
2
d
x
-
∫
0
1
ρ
k
+
1
φ
1
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
∫
0
1
ρ
k
φ
1
k
θ
t
k
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
-
ρ
k
φ
1
k
u
k
-
1
θ
x
k
θ
-
k
+
1
d
x
-
∫
0
1
ψ
k
+
1
u
x
k
-
ψ
k
u
x
k
-
1
θ
-
k
+
1
d
x
+
∫
0
1
u
x
k
2
-
u
x
k
-
1
2
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
+
1
2
φ
2
k
+
1
u
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
θ
-
k
+
1
d
x
=
∑
j
=
1
6
I
j
.Using the Sobolev inequality, Young’s inequality, and Lemma3, we obtain (78)
I
1
=
1
2
∫
0
1
ρ
k
+
1
φ
1
k
+
1
t
θ
-
k
+
1
2
d
x
=
1
2
∫
0
1
ρ
t
k
+
1
φ
1
k
+
1
θ
-
k
+
1
2
d
x
+
1
2
∫
0
1
ρ
k
+
1
φ
11
k
+
1
θ
t
k
+
1
θ
-
k
+
1
2
d
x
+
1
2
∫
0
1
ρ
k
+
1
φ
12
k
+
1
ρ
t
k
+
1
θ
-
k
+
1
2
d
x
=
∑
j
=
1
3
I
1
j
.Here, (79)
I
11
=
1
2
∫
0
1
ρ
t
k
+
1
φ
1
k
+
1
θ
-
k
+
1
2
d
x
≤
C
∫
0
1
ρ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
C
∫
0
1
ρ
k
+
1
u
k
θ
-
k
+
1
θ
-
x
k
+
1
d
x
≤
C
ρ
k
+
1
L
∞
1
/
2
ρ
k
+
1
θ
-
k
+
1
L
2
u
k
L
∞
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
+
1
L
2
u
x
k
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
12
=
1
2
∫
0
1
ρ
k
+
1
φ
11
k
+
1
θ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
C
∫
0
1
ρ
k
+
1
θ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
C
θ
t
k
+
1
L
∞
ρ
k
+
1
θ
-
k
+
1
L
2
2
≤
C
θ
x
t
k
+
1
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
13
=
1
2
∫
0
1
ρ
k
+
1
φ
12
k
+
1
ρ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
C
ρ
k
+
1
L
∞
∫
0
1
ρ
t
k
+
1
θ
-
k
+
1
2
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
+
1
L
2
3
u
x
k
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
.In the same way, we have(80)
I
2
=
-
∫
0
1
ρ
k
+
1
φ
1
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
∫
0
1
ρ
k
φ
1
k
θ
t
k
θ
-
k
+
1
d
x
=
-
∫
0
1
ρ
-
k
+
1
φ
1
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
∫
0
1
ρ
k
φ
-
1
k
+
1
θ
t
k
θ
-
k
+
1
d
x
≤
C
∫
0
1
ρ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
C
∫
0
1
ρ
k
ρ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
+
C
∫
0
1
ρ
k
θ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
=
∑
j
=
1
3
I
2
j
.Here, (81)
I
21
=
C
∫
0
1
ρ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
≤
C
ρ
-
k
+
1
L
2
θ
x
t
k
L
2
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
θ
x
t
k
L
2
2
ρ
-
k
+
1
L
2
2
,
I
22
=
C
∫
0
1
ρ
k
ρ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
≤
C
ρ
x
k
L
2
ρ
-
k
+
1
L
2
θ
x
t
k
L
2
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
L
2
2
θ
x
t
k
L
2
2
ρ
-
k
+
1
L
2
2
,
I
23
=
C
∫
0
1
ρ
k
θ
-
k
+
1
θ
t
k
θ
-
k
+
1
d
x
≤
C
ρ
x
k
L
2
θ
x
t
k
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
.Following the same method, we obtain(82)
I
3
=
-
∫
0
1
ρ
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
-
ρ
k
φ
1
k
u
k
-
1
θ
x
k
θ
-
k
+
1
d
x
=
-
∫
0
1
ρ
-
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
φ
-
1
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
φ
1
k
u
-
k
θ
x
k
+
1
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
-
x
k
+
1
θ
-
k
+
1
d
x
=
∑
j
=
1
4
I
3
j
.Here, (83)
I
31
=
-
∫
0
1
ρ
-
k
+
1
φ
1
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
≤
C
ρ
-
k
+
1
L
2
u
x
k
L
2
θ
x
t
k
+
1
L
2
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
ρ
-
k
+
1
L
2
2
,
I
32
=
-
∫
0
1
ρ
k
φ
-
1
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
≤
C
∫
0
1
ρ
k
ρ
-
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
+
C
∫
0
1
ρ
k
θ
-
k
+
1
u
k
θ
x
k
+
1
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
ρ
-
k
+
1
L
2
2
+
C
u
x
k
L
2
ρ
x
k
L
2
θ
x
t
k
+
1
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
33
=
∫
0
1
ρ
k
φ
1
k
u
-
k
θ
x
k
+
1
θ
-
k
+
1
d
x
≤
C
u
-
x
k
L
2
ρ
x
k
L
2
θ
x
t
k
+
1
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
≤
C
u
-
x
k
L
2
2
+
C
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
34
=
∫
0
1
ρ
k
φ
1
k
u
k
-
1
θ
-
x
k
+
1
θ
-
k
+
1
d
x
≤
C
ρ
k
L
∞
ρ
k
+
1
θ
-
k
+
1
L
2
u
k
-
1
L
∞
θ
-
x
k
+
1
L
2
≤
η
θ
-
x
k
L
2
2
+
C
η
ρ
x
k
L
2
2
u
x
k
-
1
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
.In a similar way, we have(84)
I
4
=
-
∫
0
1
ψ
k
+
1
u
x
k
-
ψ
k
u
x
k
-
1
θ
-
k
+
1
d
x
=
-
∫
0
1
ψ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
-
∫
0
1
ψ
k
u
-
x
k
θ
-
k
+
1
d
x
=
∑
j
=
1
2
I
4
j
.Here, (85)
I
41
=
-
∫
0
1
ψ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
≤
C
∫
0
1
ρ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
+
C
∫
0
1
θ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
ρ
-
k
+
1
L
2
2
+
C
u
x
k
L
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
42
=
-
∫
0
1
ψ
k
u
-
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
-
x
k
L
2
2
.Similarly,(86)
I
5
=
∫
0
1
u
x
k
2
-
u
x
k
-
1
2
θ
-
k
+
1
d
x
≤
u
-
x
k
L
2
θ
-
x
k
+
1
L
2
u
x
k
L
2
+
u
x
k
-
1
L
2
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
+
u
x
k
-
1
L
2
2
u
-
x
k
L
2
2
.Using the same method, we have(87)
I
6
=
-
∫
0
1
ρ
k
+
1
2
φ
2
k
+
1
u
x
k
-
ρ
k
2
φ
2
k
u
x
k
-
1
θ
-
k
+
1
d
x
=
-
∫
0
1
ρ
-
k
+
1
ρ
k
+
1
+
ρ
k
φ
2
k
+
1
u
x
k
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
2
φ
-
2
k
+
1
u
x
k
θ
-
k
+
1
d
x
-
∫
0
1
ρ
k
2
φ
2
k
u
-
x
k
θ
-
k
+
1
d
x
=
∑
j
=
1
3
I
6
j
.Here, (88)
I
61
=
∫
0
1
ρ
-
k
+
1
ρ
k
+
1
+
ρ
k
φ
2
k
+
1
u
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
+
1
L
2
+
ρ
x
k
L
2
u
x
k
L
2
ρ
-
k
+
1
L
2
2
,
I
62
=
∫
0
1
ρ
k
2
φ
-
2
k
+
1
u
x
k
θ
-
k
+
1
d
x
=
C
∫
0
1
ρ
k
2
ρ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
+
C
∫
0
1
ρ
k
2
θ
-
k
+
1
u
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
u
x
k
L
2
2
ρ
x
k
L
2
4
ρ
-
k
+
1
L
2
2
+
C
u
x
k
L
2
ρ
x
k
L
2
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
63
=
∫
0
1
ρ
k
2
φ
2
k
u
-
x
k
θ
-
k
+
1
d
x
≤
η
θ
-
x
k
+
1
L
2
2
+
C
η
ρ
x
k
L
2
4
u
-
x
k
L
2
2
.SubstitutingI
j
(
j
=
1,2
,
…
,
6
) into (105), after choosing η
=
μ
1
(
p
-
2
)
/
2
/
26, we get(89)
1
2
d
d
t
ρ
k
+
1
φ
1
k
+
1
θ
-
k
+
1
L
2
2
+
1
2
μ
1
p
-
2
/
2
θ
-
x
k
+
1
L
2
2
≤
u
-
x
k
L
2
2
·
C
1
+
u
x
k
L
2
2
+
u
x
k
-
1
L
2
2
+
ρ
x
k
L
2
4
+
ρ
-
k
+
1
L
2
2
·
C
θ
x
t
k
L
2
2
+
ρ
x
k
L
2
2
θ
x
t
k
L
2
2
+
u
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
+
u
x
k
L
2
2
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
+
u
x
k
L
2
2
+
ρ
x
k
+
1
L
2
+
ρ
x
k
L
2
u
x
k
L
2
+
u
x
k
L
2
2
ρ
x
k
L
2
4
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
·
C
ρ
x
k
+
1
L
2
u
x
k
L
2
2
+
θ
x
t
k
+
1
L
2
+
ρ
x
k
+
1
L
2
3
u
x
k
L
2
2
+
ρ
x
k
L
2
θ
x
t
k
L
2
+
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
+
ρ
x
k
L
2
2
θ
x
t
k
+
1
L
2
2
+
ρ
x
k
L
2
2
u
x
k
-
1
L
2
2
+
u
x
k
L
2
+
u
x
k
L
2
ρ
x
k
L
2
2
.Thus, we get(90)
δ
2
d
d
t
ρ
k
+
1
θ
-
k
+
1
L
2
2
+
1
2
μ
1
p
-
2
/
2
θ
-
x
k
+
1
L
2
2
≤
C
~
u
-
x
k
L
2
2
+
D
k
t
ρ
-
k
+
1
L
2
2
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
·There D
k
(
t
)
⊂
L
1
(
0
,
T
1
), ∫
0
1
D
k
(
t
)
≤
C
~
+
C
~
t.Multiplying (74) by u
-
k
+
1 and integrating it over (
0,1
), we obtain(91)
1
2
d
d
t
∫
0
1
ρ
k
+
1
u
-
k
+
1
2
d
x
+
∫
0
1
H
q
μ
2
u
k
+
1
-
H
q
μ
2
u
k
u
-
k
+
1
d
x
=
-
∫
0
1
ψ
x
k
+
1
-
ψ
x
k
u
-
k
+
1
d
x
+
∫
0
1
ρ
-
k
+
1
-
u
t
k
-
u
k
-
1
u
x
k
u
-
k
+
1
d
x
-
∫
0
1
ρ
k
+
1
u
-
k
u
x
k
u
-
k
+
1
d
x
.We use the Sobolev inequality, Young’s inequality, and Lemma3, and we obtain (92)
I
1
=
-
∫
0
1
ψ
x
k
+
1
-
ψ
x
k
u
-
k
+
1
d
x
d
x
=
-
∫
0
1
ψ
-
k
+
1
u
-
k
+
1
d
x
≤
C
∫
0
1
ρ
-
k
+
1
u
-
k
+
1
d
x
+
C
∫
0
1
θ
-
k
+
1
u
-
k
+
1
d
x
≤
η
u
-
x
k
+
1
L
2
2
+
C
η
ρ
-
k
+
1
L
2
2
+
η
u
-
x
k
+
1
L
2
2
+
C
η
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
I
2
=
-
∫
0
1
ρ
-
k
+
1
u
t
k
u
-
k
+
1
d
x
≤
ρ
-
k
+
1
L
2
u
t
k
L
∞
u
-
x
k
+
1
L
2
≤
η
u
-
x
k
+
1
L
2
2
+
C
η
u
x
t
k
L
2
2
ρ
-
k
+
1
L
2
2
,
I
3
=
-
∫
0
1
ρ
-
k
+
1
u
k
-
1
u
x
k
u
-
k
+
1
d
x
≤
ρ
-
k
+
1
L
2
u
k
-
1
L
∞
u
x
k
L
2
u
-
k
+
1
L
∞
≤
ρ
-
k
+
1
L
2
u
x
k
-
1
L
2
u
x
k
L
2
u
-
x
k
+
1
L
2
≤
η
u
-
x
k
+
1
L
2
2
+
C
η
u
x
k
-
1
L
2
2
u
x
k
L
2
2
ρ
-
k
+
1
L
2
2
,
I
4
=
-
∫
0
1
ρ
k
+
1
u
-
k
u
x
k
u
-
k
+
1
d
x
≤
C
ρ
k
+
1
L
∞
1
/
2
ρ
k
+
1
u
-
k
+
1
L
2
u
x
k
L
2
u
-
x
k
L
2
≤
ɛ
u
-
x
k
L
2
2
+
C
ɛ
ρ
x
k
+
1
L
2
u
x
k
L
2
2
ρ
k
+
1
u
-
k
+
1
L
2
2
.SubstitutingI
j
(
j
=
1,2
,
…
,
4
) into (91), after choosing η
=
μ
2
(
q
-
2
)
/
2
/
8,(93)
1
2
d
d
t
∫
0
1
ρ
k
+
1
u
-
k
+
1
2
d
x
+
1
2
μ
2
q
-
2
/
2
u
-
x
k
+
1
L
2
2
≤
ρ
-
k
+
1
L
2
2
·
C
1
+
u
x
t
k
L
2
2
+
u
x
k
-
1
L
2
2
u
x
k
L
2
2
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
·
C
ɛ
u
-
x
k
L
2
2
+
C
ɛ
ρ
x
k
+
1
L
2
u
x
k
L
2
2
ρ
k
+
1
u
-
k
+
1
L
2
2
.Thus, we get(94)
d
d
t
ρ
-
k
+
1
L
2
2
≤
C
ρ
-
k
+
1
L
2
2
u
x
k
L
∞
+
ρ
k
H
1
u
-
x
k
L
2
ρ
-
k
+
1
L
2
≤
B
η
k
t
ρ
-
k
+
1
L
2
2
+
η
u
-
x
k
L
2
2
,
∫
0
1
B
η
k
t
≤
C
~
t
,
1
2
d
d
t
ρ
k
+
1
θ
-
k
+
1
L
2
2
+
1
2
μ
1
p
-
2
/
2
∫
0
1
θ
-
x
k
+
1
L
2
2
d
x
≤
C
~
u
-
x
k
L
2
2
+
D
k
t
ρ
-
k
+
1
L
2
2
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
,
∫
0
1
D
k
t
≤
C
~
+
C
~
t
,
1
2
d
d
t
∫
0
1
ρ
k
+
1
u
-
k
+
1
2
d
x
+
1
2
μ
2
q
-
2
/
2
u
-
x
k
+
1
L
2
2
≤
E
ɛ
k
t
ρ
-
k
+
1
L
2
2
+
ρ
k
+
1
θ
-
k
+
1
L
2
2
+
ρ
k
+
1
u
-
k
+
1
L
2
2
+
ɛ
u
-
x
k
L
2
2
.There E
ɛ
k
(
t
)
⊂
L
1
(
0
,
T
1
), ∫
0
1
E
ɛ
k
(
t
)
≤
C
~
+
C
~
ɛ
t. Using (94), denote(95)
Ψ
k
+
1
=
ρ
-
k
+
1
L
2
2
+
ɛ
2
ρ
k
+
1
θ
-
k
+
1
L
2
2
+
1
2
ρ
k
+
1
u
-
k
+
1
L
2
2
.There F
ɛ
k
(
t
)
⊂
L
1
(
0
,
T
1
), ∫
0
1
F
ɛ
k
(
t
)
≤
C
~
+
C
~
ɛ
t.Thus, we get(96)
d
d
t
Ψ
k
+
1
t
+
ɛ
2
δ
μ
1
p
-
2
/
2
θ
-
x
k
+
1
L
2
2
+
1
2
μ
2
q
-
2
/
2
u
-
x
k
+
1
L
2
2
≤
F
ɛ
k
t
·
Ψ
k
+
1
t
+
C
~
·
ɛ
·
u
-
x
k
L
2
2
.By Gronwall’s inequality, noting that Ψ
k
+
1
(
t
)
=
0, we get(97)
Ψ
k
+
1
t
+
ɛ
2
δ
μ
1
p
-
2
/
2
∫
0
t
θ
-
x
k
+
1
L
2
2
s
d
s
+
1
2
μ
2
q
-
2
/
2
∫
0
t
u
-
x
k
+
1
L
2
2
s
d
s
≤
exp
⁡
∫
0
t
F
ɛ
k
t
d
t
·
∫
0
t
C
~
·
ɛ
·
u
-
x
k
L
2
2
s
d
s
≤
exp
⁡
C
~
+
C
~
ɛ
t
·
∫
0
t
C
~
·
ɛ
·
u
-
x
k
L
2
2
s
d
s
.We can obtain(98)
C
~
·
ɛ
·
exp
⁡
C
~
=
1
4
,
exp
⁡
C
~
ɛ
·
T
7
=
μ
2
q
-
2
/
2
,and thus we can get(99)
∑
k
=
1
K
Ψ
k
+
1
t
+
∑
k
=
1
K
∫
0
T
∗
θ
-
x
k
+
1
s
L
2
2
+
u
-
x
k
+
1
s
L
2
2
d
s
≤
C
~
<
+
∞
.There, T
∗
=
min
⁡
{
T
6
,
T
7
}.Therefore, we conclude that the full sequence(
ρ
k
,
u
k
,
θ
k
) converges to a limit (
ρ
,
u
,
θ
) in the following strong sense:(100)
ρ
k
⟶
ρ
in
L
∞
0
,
T
∗
;
L
2
,
u
k
⟶
u
in
L
∞
0
,
T
∗
;
L
2
∩
L
2
0
,
T
∗
;
H
0
1
,
θ
k
⟶
θ
in
L
∞
0
,
T
∗
;
L
2
∩
L
2
0
,
T
∗
;
H
0
1
,for T
∗
=
min
⁡
T
1
,
T
2
,
T
3
,
T
4
,
T
5
,
T
6
,
T
7, and we get (101)
sup
0
≤
s
≤
T
∗
⁡
u
x
L
2
s
+
∫
0
T
∗
u
x
t
k
L
2
2
s
d
s
≤
c
1
,
sup
0
≤
s
≤
T
∗
⁡
u
x
x
L
2
s
≤
c
2
,
sup
0
≤
s
≤
T
∗
⁡
θ
x
L
2
2
s
+
ρ
k
θ
t
k
t
L
2
2
s
+
∫
0
T
∗
θ
x
t
k
L
2
2
s
d
s
≤
C
c
1
5
,
sup
0
≤
s
≤
T
∗
⁡
u
x
L
2
2
s
+
ρ
k
u
t
k
t
L
2
2
s
+
∫
0
T
∗
u
x
t
L
2
2
s
d
s
≤
c
1
,
sup
0
≤
s
≤
T
∗
⁡
θ
x
x
L
2
s
<
C
c
1
4
c
2
,
sup
0
≤
s
≤
T
∗
⁡
ρ
t
H
1
s
≤
C
c
0
,
sup
0
≤
s
≤
T
∗
⁡
ρ
t
t
L
2
s
≤
C
c
1
2
.


## 4. Proof of the Main Theorem

### 4.1. Existence (k
→
∞)

In this part, we need to prove that(
ρ
,
u
,
θ
) is a solution of problem(102)
ρ
t
+
ρ
u
x
=
0
,
x
,
t
∈
Ω
T
,
ρ
e
t
+
ρ
u
e
x
-
θ
x
2
+
μ
p
-
2
/
2
θ
x
x
+
π
u
x
=
u
x
2
,
x
,
t
∈
Ω
T
,
ρ
u
t
+
ρ
u
2
x
-
u
x
2
+
μ
2
q
-
2
/
2
u
x
x
+
π
x
=
0
,
x
,
t
∈
Ω
T
,
π
=
φ
θ
,
ρ
,
x
,
t
∈
Ω
T
,
e
=
ψ
θ
,
ρ
,
x
,
t
∈
Ω
T
,with the initial and boundary conditions (103)
ρ
t
=
0
=
ρ
0
≥
0
,
u
t
=
0
=
u
0
,
θ
t
=
0
=
θ
0
,
u
0
,
t
=
u
1
,
t
=
0
,
θ
0
,
t
=
θ
1
,
t
=
0and also satisfy the integral identity after the k
→
∞ process. Finally, we need to show that the solution satisfies the regular properties in Definition 1. Since the abovementioned can be obtained by the same method as in [15], we omit it for brevity.


### 4.2. Uniqueness

Let(
ρ
1
,
u
1
,
θ
1
), (
ρ
2
,
u
2
,
θ
2
) be two solutions of problem (3)-(4). After substituting into (7), respectively, we define (104)
ρ
-
=
ρ
1
-
ρ
2
,
u
-
=
u
2
-
u
1
,
θ
-
=
θ
1
-
θ
2
,and then we can obtain(105)
1
2
d
d
t
∫
0
1
ρ
1
φ
1
1
θ
-
2
d
x
+
∫
0
1
H
p
μ
1
θ
1
-
H
p
μ
1
θ
2
θ
-
d
x
=
1
2
∫
0
1
ρ
1
φ
1
1
t
θ
-
2
d
x
-
∫
0
1
ρ
1
φ
1
1
θ
t
2
θ
-
d
x
+
∫
0
1
ρ
2
φ
1
2
θ
t
2
θ
-
d
x
-
∫
0
1
ρ
1
φ
1
1
u
1
θ
x
1
-
ρ
2
φ
1
2
u
2
θ
x
2
θ
-
d
x
-
∫
0
1
ψ
1
u
x
1
-
ψ
2
u
x
2
θ
-
d
x
+
∫
0
1
u
x
1
2
-
u
x
2
2
θ
-
d
x
+
∫
0
1
ρ
1
2
φ
2
1
u
x
1
-
ρ
2
2
φ
2
2
u
x
2
θ
-
d
x
=
∑
j
=
1
6
I
j
,since (106)
∫
0
t
∫
0
1
θ
x
1
p
-
2
θ
x
1
-
θ
x
2
p
-
2
θ
x
2
θ
-
x
d
x
≥
μ
1
p
-
2
/
2
∫
0
t
θ
-
x
L
2
2
d
s
.Using the Sobolev inequality, Young’s inequality, and Lemma 3, we obtain (107)
I
1
=
1
2
∫
0
1
ρ
1
φ
1
1
t
θ
-
2
d
x
=
1
2
∫
0
1
ρ
t
1
φ
1
1
θ
-
2
d
x
+
1
2
∫
0
1
ρ
1
φ
11
1
θ
t
1
θ
-
2
d
x
+
1
2
∫
0
1
ρ
1
φ
12
1
ρ
t
1
θ
-
2
d
x
=
∑
j
=
1
3
I
1
j
.Here, (108)
I
11
=
1
2
∫
0
1
ρ
t
1
φ
1
1
θ
-
1
2
d
x
≤
C
∫
0
1
ρ
t
1
θ
-
2
d
x
=
C
∫
0
1
ρ
1
u
1
θ
-
θ
-
x
d
x
≤
C
ρ
1
L
∞
1
/
2
ρ
1
θ
-
1
L
2
u
1
L
∞
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
1
L
2
u
x
1
L
2
2
ρ
1
θ
-
L
2
2
,
I
12
=
1
2
∫
0
1
ρ
1
φ
11
1
θ
t
1
θ
-
2
d
x
≤
C
∫
0
1
ρ
1
θ
t
1
θ
-
1
2
d
x
≤
C
θ
t
1
L
∞
ρ
1
θ
-
L
2
2
≤
C
θ
x
t
1
L
2
ρ
1
θ
-
L
2
2
,
I
13
=
1
2
∫
0
1
ρ
1
φ
12
1
ρ
t
1
θ
-
2
d
x
≤
C
ρ
1
L
∞
∫
0
1
ρ
t
1
θ
-
1
2
d
x
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
1
L
2
3
u
x
1
L
2
2
ρ
1
θ
-
L
2
2
.In the same way,(109)
I
2
=
-
∫
0
1
ρ
1
φ
1
1
θ
t
2
θ
-
d
x
+
∫
0
1
ρ
2
φ
1
2
θ
t
2
θ
-
d
x
=
-
∫
0
1
ρ
-
φ
1
1
θ
t
2
θ
-
d
x
+
∫
0
1
ρ
2
φ
-
1
1
θ
t
2
θ
-
d
x
≤
C
∫
0
1
ρ
-
1
θ
t
2
θ
-
d
x
+
C
∫
0
1
ρ
2
ρ
-
θ
t
2
θ
-
d
x
+
C
∫
0
1
ρ
2
θ
-
θ
t
2
θ
-
d
x
=
∑
j
=
1
3
I
2
j
.Here, (110)
I
21
=
C
∫
0
1
ρ
-
1
θ
t
2
θ
-
1
d
x
≤
C
ρ
-
L
2
θ
x
t
2
L
2
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
θ
x
t
2
L
2
2
ρ
-
L
2
2
,
I
22
=
C
∫
0
1
ρ
2
ρ
-
1
θ
t
2
θ
-
1
d
x
≤
C
ρ
x
2
L
2
ρ
-
L
2
θ
x
t
2
L
2
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
2
L
2
2
θ
x
t
2
L
2
2
ρ
-
L
2
2
,
I
23
=
C
∫
0
1
ρ
k
θ
-
θ
t
2
θ
-
d
x
≤
C
ρ
x
2
L
2
θ
x
t
2
L
2
ρ
1
θ
-
L
2
2
.By the same token,(111)
I
3
=
-
∫
0
1
ρ
1
φ
1
1
u
1
θ
x
1
-
ρ
2
φ
1
2
u
2
θ
x
2
θ
-
d
x
=
-
∫
0
1
ρ
-
φ
1
1
u
1
θ
x
1
θ
-
d
x
-
∫
0
1
ρ
2
φ
-
1
u
1
θ
x
1
θ
-
d
x
-
∫
0
1
ρ
2
φ
1
2
u
-
θ
x
1
θ
-
d
x
-
∫
0
1
ρ
2
φ
1
2
u
2
θ
-
x
θ
-
d
x
=
∑
j
=
1
4
I
3
j
.Here, (112)
I
31
=
-
∫
0
1
ρ
-
φ
1
1
u
1
θ
x
1
θ
-
d
x
≤
C
ρ
-
L
2
u
x
1
L
2
θ
x
t
1
L
2
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
θ
x
t
1
L
2
2
ρ
-
L
2
2
,
I
32
=
-
∫
0
1
ρ
2
φ
-
1
u
1
θ
x
1
θ
-
d
x
≤
C
∫
0
1
ρ
2
ρ
-
u
1
θ
x
1
θ
-
d
x
+
C
∫
0
1
ρ
2
θ
-
u
1
θ
x
1
θ
-
d
x
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
ρ
x
2
L
2
2
θ
x
t
1
L
2
2
ρ
-
L
2
2
+
C
u
x
1
L
2
ρ
x
2
L
2
θ
x
t
1
L
2
ρ
1
θ
-
L
2
2
,
I
33
=
∫
0
1
ρ
2
φ
1
2
u
-
θ
x
1
θ
-
d
x
≤
C
u
-
x
L
2
ρ
x
2
L
2
θ
x
t
1
L
2
ρ
1
θ
-
L
2
≤
C
u
-
x
L
2
2
+
C
ρ
x
2
L
2
2
θ
x
t
1
L
2
2
ρ
1
θ
-
L
2
2
,
I
34
=
∫
0
1
ρ
2
φ
1
2
u
2
θ
-
x
θ
-
d
x
ρ
1
ρ
1
≤
C
ρ
2
L
∞
ρ
1
θ
-
L
2
u
2
L
∞
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
2
L
2
2
u
x
2
L
2
2
ρ
1
θ
-
L
2
2
.In the similar way, we have(113)
I
4
=
-
∫
0
1
ψ
1
u
x
1
-
ψ
2
u
x
2
θ
-
d
x
=
-
∫
0
1
ψ
-
u
x
1
θ
-
d
x
-
∫
0
1
ψ
2
u
-
θ
-
d
x
=
∑
j
=
1
2
I
4
j
.Here, (114)
I
41
=
-
∫
0
1
ψ
-
1
u
x
1
θ
-
d
x
≤
C
∫
0
1
ρ
-
1
u
x
1
θ
-
1
d
x
+
C
∫
0
1
θ
-
u
x
1
θ
-
d
x
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
ρ
-
L
2
2
+
C
u
x
1
L
2
ρ
1
θ
-
L
2
2
,
I
42
=
-
∫
0
1
ψ
2
u
-
θ
-
1
d
x
≤
η
θ
-
x
L
2
2
+
C
η
u
-
x
L
2
2
.Similarly,(115)
I
5
=
∫
0
1
u
x
1
2
-
u
x
2
2
θ
-
d
x
≤
u
-
x
L
2
θ
-
x
L
2
u
x
1
L
2
+
u
x
2
L
2
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
+
u
x
2
L
2
2
u
-
x
L
2
2
.Using the same method, we obtain(116)
I
6
=
∫
0
1
ρ
1
2
φ
2
1
u
x
1
-
ρ
2
2
φ
2
2
u
x
2
θ
-
d
x
=
-
∫
0
1
ρ
-
ρ
1
+
ρ
2
φ
2
1
u
x
1
θ
-
d
x
-
∫
0
1
ρ
2
2
φ
-
2
u
x
1
θ
-
d
x
-
∫
0
1
ρ
2
2
φ
2
2
u
-
θ
-
d
x
=
∑
j
=
1
3
I
6
j
.Here, (117)
I
61
=
∫
0
1
ρ
-
ρ
2
+
ρ
1
φ
2
1
u
x
k
θ
-
d
x
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
1
L
2
+
ρ
x
2
L
2
u
x
1
L
2
ρ
-
L
2
2
,
I
62
=
∫
0
1
ρ
2
2
φ
-
2
u
x
1
θ
-
d
x
=
C
∫
0
1
ρ
2
2
ρ
-
u
x
1
θ
-
d
x
+
C
∫
0
1
ρ
2
2
θ
-
u
x
1
θ
-
d
x
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
ρ
x
2
L
2
4
ρ
-
L
2
2
+
C
u
x
1
L
2
ρ
x
2
L
2
2
ρ
1
θ
-
L
2
2
,
I
63
=
∫
0
1
ρ
2
2
φ
2
2
u
-
x
θ
-
k
+
1
d
x
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
2
L
2
4
u
-
x
L
2
2
.Thus, we have(118)
δ
2
ρ
1
θ
-
L
2
2
+
μ
1
p
-
2
/
2
2
∫
0
t
θ
-
x
L
2
2
d
s
≤
∫
0
t
E
t
ρ
-
L
2
2
+
ρ
1
θ
-
L
2
2
+
ρ
1
u
-
L
2
2
d
s
+
C
~
∫
0
t
u
-
x
k
L
2
2
d
s
.There E
(
t
)
∈
L
1
(
0
,
T
∗
), ∫
0
1
E
(
t
)
≤
C
~
+
C
~
t.In the same way,(119)
1
2
ρ
1
u
-
L
2
2
+
μ
2
p
-
2
/
2
2
∫
0
t
u
-
x
L
2
2
d
s
≤
∫
0
t
F
t
ρ
-
L
2
2
+
ρ
1
θ
-
L
2
2
+
ρ
1
u
-
L
2
2
d
s
.There F
(
t
)
∈
L
1
(
0
,
T
∗
), ∫
0
1
F
(
t
)
≤
C
~
+
C
~
t. Consider (120)
ρ
-
1
L
2
2
≤
∫
0
t
B
t
ρ
-
L
2
2
d
s
+
η
∫
0
t
u
-
x
k
L
2
2
d
s
.There B
(
t
)
∈
L
1
(
0
,
T
∗
), ∫
0
1
B
(
t
)
≤
C
~
t.With (119) multiplied by η, combining with (118) and (120),(121)
C
~
+
1
η
=
μ
2
p
-
2
/
2
4
.We can get(122)
ρ
-
L
2
2
+
ρ
1
θ
-
L
2
2
+
ρ
1
u
-
L
2
2
+
∫
0
t
θ
-
x
L
2
2
d
s
+
∫
0
t
u
-
x
L
2
2
d
s
≤
∫
0
t
G
t
ρ
-
L
2
2
+
ρ
1
θ
-
L
2
2
+
ρ
1
u
-
L
2
2
d
s
.There G
(
t
)
∈
L
1
(
0
,
T
∗
), ∫
0
1
G
(
t
)
≤
C
~
t
+
C
~.By Gronwall’s inequality, it yields(123)
u
1
=
u
2
,
ρ
1
=
ρ
2
,
θ
1
=
θ
2
.


## 4.1. Existence (k
→
∞)

In this part, we need to prove that(
ρ
,
u
,
θ
) is a solution of problem(102)
ρ
t
+
ρ
u
x
=
0
,
x
,
t
∈
Ω
T
,
ρ
e
t
+
ρ
u
e
x
-
θ
x
2
+
μ
p
-
2
/
2
θ
x
x
+
π
u
x
=
u
x
2
,
x
,
t
∈
Ω
T
,
ρ
u
t
+
ρ
u
2
x
-
u
x
2
+
μ
2
q
-
2
/
2
u
x
x
+
π
x
=
0
,
x
,
t
∈
Ω
T
,
π
=
φ
θ
,
ρ
,
x
,
t
∈
Ω
T
,
e
=
ψ
θ
,
ρ
,
x
,
t
∈
Ω
T
,with the initial and boundary conditions (103)
ρ
t
=
0
=
ρ
0
≥
0
,
u
t
=
0
=
u
0
,
θ
t
=
0
=
θ
0
,
u
0
,
t
=
u
1
,
t
=
0
,
θ
0
,
t
=
θ
1
,
t
=
0and also satisfy the integral identity after the k
→
∞ process. Finally, we need to show that the solution satisfies the regular properties in Definition 1. Since the abovementioned can be obtained by the same method as in [15], we omit it for brevity.


## 4.2. Uniqueness

Let(
ρ
1
,
u
1
,
θ
1
), (
ρ
2
,
u
2
,
θ
2
) be two solutions of problem (3)-(4). After substituting into (7), respectively, we define (104)
ρ
-
=
ρ
1
-
ρ
2
,
u
-
=
u
2
-
u
1
,
θ
-
=
θ
1
-
θ
2
,and then we can obtain(105)
1
2
d
d
t
∫
0
1
ρ
1
φ
1
1
θ
-
2
d
x
+
∫
0
1
H
p
μ
1
θ
1
-
H
p
μ
1
θ
2
θ
-
d
x
=
1
2
∫
0
1
ρ
1
φ
1
1
t
θ
-
2
d
x
-
∫
0
1
ρ
1
φ
1
1
θ
t
2
θ
-
d
x
+
∫
0
1
ρ
2
φ
1
2
θ
t
2
θ
-
d
x
-
∫
0
1
ρ
1
φ
1
1
u
1
θ
x
1
-
ρ
2
φ
1
2
u
2
θ
x
2
θ
-
d
x
-
∫
0
1
ψ
1
u
x
1
-
ψ
2
u
x
2
θ
-
d
x
+
∫
0
1
u
x
1
2
-
u
x
2
2
θ
-
d
x
+
∫
0
1
ρ
1
2
φ
2
1
u
x
1
-
ρ
2
2
φ
2
2
u
x
2
θ
-
d
x
=
∑
j
=
1
6
I
j
,since (106)
∫
0
t
∫
0
1
θ
x
1
p
-
2
θ
x
1
-
θ
x
2
p
-
2
θ
x
2
θ
-
x
d
x
≥
μ
1
p
-
2
/
2
∫
0
t
θ
-
x
L
2
2
d
s
.Using the Sobolev inequality, Young’s inequality, and Lemma 3, we obtain (107)
I
1
=
1
2
∫
0
1
ρ
1
φ
1
1
t
θ
-
2
d
x
=
1
2
∫
0
1
ρ
t
1
φ
1
1
θ
-
2
d
x
+
1
2
∫
0
1
ρ
1
φ
11
1
θ
t
1
θ
-
2
d
x
+
1
2
∫
0
1
ρ
1
φ
12
1
ρ
t
1
θ
-
2
d
x
=
∑
j
=
1
3
I
1
j
.Here, (108)
I
11
=
1
2
∫
0
1
ρ
t
1
φ
1
1
θ
-
1
2
d
x
≤
C
∫
0
1
ρ
t
1
θ
-
2
d
x
=
C
∫
0
1
ρ
1
u
1
θ
-
θ
-
x
d
x
≤
C
ρ
1
L
∞
1
/
2
ρ
1
θ
-
1
L
2
u
1
L
∞
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
1
L
2
u
x
1
L
2
2
ρ
1
θ
-
L
2
2
,
I
12
=
1
2
∫
0
1
ρ
1
φ
11
1
θ
t
1
θ
-
2
d
x
≤
C
∫
0
1
ρ
1
θ
t
1
θ
-
1
2
d
x
≤
C
θ
t
1
L
∞
ρ
1
θ
-
L
2
2
≤
C
θ
x
t
1
L
2
ρ
1
θ
-
L
2
2
,
I
13
=
1
2
∫
0
1
ρ
1
φ
12
1
ρ
t
1
θ
-
2
d
x
≤
C
ρ
1
L
∞
∫
0
1
ρ
t
1
θ
-
1
2
d
x
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
1
L
2
3
u
x
1
L
2
2
ρ
1
θ
-
L
2
2
.In the same way,(109)
I
2
=
-
∫
0
1
ρ
1
φ
1
1
θ
t
2
θ
-
d
x
+
∫
0
1
ρ
2
φ
1
2
θ
t
2
θ
-
d
x
=
-
∫
0
1
ρ
-
φ
1
1
θ
t
2
θ
-
d
x
+
∫
0
1
ρ
2
φ
-
1
1
θ
t
2
θ
-
d
x
≤
C
∫
0
1
ρ
-
1
θ
t
2
θ
-
d
x
+
C
∫
0
1
ρ
2
ρ
-
θ
t
2
θ
-
d
x
+
C
∫
0
1
ρ
2
θ
-
θ
t
2
θ
-
d
x
=
∑
j
=
1
3
I
2
j
.Here, (110)
I
21
=
C
∫
0
1
ρ
-
1
θ
t
2
θ
-
1
d
x
≤
C
ρ
-
L
2
θ
x
t
2
L
2
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
θ
x
t
2
L
2
2
ρ
-
L
2
2
,
I
22
=
C
∫
0
1
ρ
2
ρ
-
1
θ
t
2
θ
-
1
d
x
≤
C
ρ
x
2
L
2
ρ
-
L
2
θ
x
t
2
L
2
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
2
L
2
2
θ
x
t
2
L
2
2
ρ
-
L
2
2
,
I
23
=
C
∫
0
1
ρ
k
θ
-
θ
t
2
θ
-
d
x
≤
C
ρ
x
2
L
2
θ
x
t
2
L
2
ρ
1
θ
-
L
2
2
.By the same token,(111)
I
3
=
-
∫
0
1
ρ
1
φ
1
1
u
1
θ
x
1
-
ρ
2
φ
1
2
u
2
θ
x
2
θ
-
d
x
=
-
∫
0
1
ρ
-
φ
1
1
u
1
θ
x
1
θ
-
d
x
-
∫
0
1
ρ
2
φ
-
1
u
1
θ
x
1
θ
-
d
x
-
∫
0
1
ρ
2
φ
1
2
u
-
θ
x
1
θ
-
d
x
-
∫
0
1
ρ
2
φ
1
2
u
2
θ
-
x
θ
-
d
x
=
∑
j
=
1
4
I
3
j
.Here, (112)
I
31
=
-
∫
0
1
ρ
-
φ
1
1
u
1
θ
x
1
θ
-
d
x
≤
C
ρ
-
L
2
u
x
1
L
2
θ
x
t
1
L
2
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
θ
x
t
1
L
2
2
ρ
-
L
2
2
,
I
32
=
-
∫
0
1
ρ
2
φ
-
1
u
1
θ
x
1
θ
-
d
x
≤
C
∫
0
1
ρ
2
ρ
-
u
1
θ
x
1
θ
-
d
x
+
C
∫
0
1
ρ
2
θ
-
u
1
θ
x
1
θ
-
d
x
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
ρ
x
2
L
2
2
θ
x
t
1
L
2
2
ρ
-
L
2
2
+
C
u
x
1
L
2
ρ
x
2
L
2
θ
x
t
1
L
2
ρ
1
θ
-
L
2
2
,
I
33
=
∫
0
1
ρ
2
φ
1
2
u
-
θ
x
1
θ
-
d
x
≤
C
u
-
x
L
2
ρ
x
2
L
2
θ
x
t
1
L
2
ρ
1
θ
-
L
2
≤
C
u
-
x
L
2
2
+
C
ρ
x
2
L
2
2
θ
x
t
1
L
2
2
ρ
1
θ
-
L
2
2
,
I
34
=
∫
0
1
ρ
2
φ
1
2
u
2
θ
-
x
θ
-
d
x
ρ
1
ρ
1
≤
C
ρ
2
L
∞
ρ
1
θ
-
L
2
u
2
L
∞
θ
-
x
L
2
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
2
L
2
2
u
x
2
L
2
2
ρ
1
θ
-
L
2
2
.In the similar way, we have(113)
I
4
=
-
∫
0
1
ψ
1
u
x
1
-
ψ
2
u
x
2
θ
-
d
x
=
-
∫
0
1
ψ
-
u
x
1
θ
-
d
x
-
∫
0
1
ψ
2
u
-
θ
-
d
x
=
∑
j
=
1
2
I
4
j
.Here, (114)
I
41
=
-
∫
0
1
ψ
-
1
u
x
1
θ
-
d
x
≤
C
∫
0
1
ρ
-
1
u
x
1
θ
-
1
d
x
+
C
∫
0
1
θ
-
u
x
1
θ
-
d
x
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
ρ
-
L
2
2
+
C
u
x
1
L
2
ρ
1
θ
-
L
2
2
,
I
42
=
-
∫
0
1
ψ
2
u
-
θ
-
1
d
x
≤
η
θ
-
x
L
2
2
+
C
η
u
-
x
L
2
2
.Similarly,(115)
I
5
=
∫
0
1
u
x
1
2
-
u
x
2
2
θ
-
d
x
≤
u
-
x
L
2
θ
-
x
L
2
u
x
1
L
2
+
u
x
2
L
2
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
+
u
x
2
L
2
2
u
-
x
L
2
2
.Using the same method, we obtain(116)
I
6
=
∫
0
1
ρ
1
2
φ
2
1
u
x
1
-
ρ
2
2
φ
2
2
u
x
2
θ
-
d
x
=
-
∫
0
1
ρ
-
ρ
1
+
ρ
2
φ
2
1
u
x
1
θ
-
d
x
-
∫
0
1
ρ
2
2
φ
-
2
u
x
1
θ
-
d
x
-
∫
0
1
ρ
2
2
φ
2
2
u
-
θ
-
d
x
=
∑
j
=
1
3
I
6
j
.Here, (117)
I
61
=
∫
0
1
ρ
-
ρ
2
+
ρ
1
φ
2
1
u
x
k
θ
-
d
x
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
1
L
2
+
ρ
x
2
L
2
u
x
1
L
2
ρ
-
L
2
2
,
I
62
=
∫
0
1
ρ
2
2
φ
-
2
u
x
1
θ
-
d
x
=
C
∫
0
1
ρ
2
2
ρ
-
u
x
1
θ
-
d
x
+
C
∫
0
1
ρ
2
2
θ
-
u
x
1
θ
-
d
x
≤
η
θ
-
x
L
2
2
+
C
η
u
x
1
L
2
2
ρ
x
2
L
2
4
ρ
-
L
2
2
+
C
u
x
1
L
2
ρ
x
2
L
2
2
ρ
1
θ
-
L
2
2
,
I
63
=
∫
0
1
ρ
2
2
φ
2
2
u
-
x
θ
-
k
+
1
d
x
≤
η
θ
-
x
L
2
2
+
C
η
ρ
x
2
L
2
4
u
-
x
L
2
2
.Thus, we have(118)
δ
2
ρ
1
θ
-
L
2
2
+
μ
1
p
-
2
/
2
2
∫
0
t
θ
-
x
L
2
2
d
s
≤
∫
0
t
E
t
ρ
-
L
2
2
+
ρ
1
θ
-
L
2
2
+
ρ
1
u
-
L
2
2
d
s
+
C
~
∫
0
t
u
-
x
k
L
2
2
d
s
.There E
(
t
)
∈
L
1
(
0
,
T
∗
), ∫
0
1
E
(
t
)
≤
C
~
+
C
~
t.In the same way,(119)
1
2
ρ
1
u
-
L
2
2
+
μ
2
p
-
2
/
2
2
∫
0
t
u
-
x
L
2
2
d
s
≤
∫
0
t
F
t
ρ
-
L
2
2
+
ρ
1
θ
-
L
2
2
+
ρ
1
u
-
L
2
2
d
s
.There F
(
t
)
∈
L
1
(
0
,
T
∗
), ∫
0
1
F
(
t
)
≤
C
~
+
C
~
t. Consider (120)
ρ
-
1
L
2
2
≤
∫
0
t
B
t
ρ
-
L
2
2
d
s
+
η
∫
0
t
u
-
x
k
L
2
2
d
s
.There B
(
t
)
∈
L
1
(
0
,
T
∗
), ∫
0
1
B
(
t
)
≤
C
~
t.With (119) multiplied by η, combining with (118) and (120),(121)
C
~
+
1
η
=
μ
2
p
-
2
/
2
4
.We can get(122)
ρ
-
L
2
2
+
ρ
1
θ
-
L
2
2
+
ρ
1
u
-
L
2
2
+
∫
0
t
θ
-
x
L
2
2
d
s
+
∫
0
t
u
-
x
L
2
2
d
s
≤
∫
0
t
G
t
ρ
-
L
2
2
+
ρ
1
θ
-
L
2
2
+
ρ
1
u
-
L
2
2
d
s
.There G
(
t
)
∈
L
1
(
0
,
T
∗
), ∫
0
1
G
(
t
)
≤
C
~
t
+
C
~.By Gronwall’s inequality, it yields(123)
u
1
=
u
2
,
ρ
1
=
ρ
2
,
θ
1
=
θ
2
.



---

*Source: 541027-2015-12-16.xml*