Volume 2010, Article ID 796065,21pages doi:10.1155/2010/796065
Research Article
One-Dimensional Compressible Viscous
Micropolar Fluid Model: Stabilization of
the Solution for the Cauchy Problem
Nermina Mujakovi ´c
Department of Mathematics, University of Rijeka, Omladinska 14, 51000 Rijeka, Croatia
Correspondence should be addressed to Nermina Mujakovi´c,[email protected]
Received 8 November 2009; Revised 24 May 2010; Accepted 1 June 2010
Academic Editor: Salim Messaoudi
Copyrightq2010 Nermina Mujakovi´c. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We consider the Cauchy problem for nonstationary 1D flow of a compressible viscous and heat-conducting micropolar fluid, assuming that it is in the thermodynamical sense perfect and polytropic. This problem has a unique generalized solution onR×0, Tfor eachT >0. Supposing that the initial functions are small perturbations of the constants we derive a priori estimates for the solution independent ofT, which we use in proving of the stabilization of the solution.
1. Introduction
In this paper we consider the Cauchy problem for nonstationary 1D flow of a compressible viscous and heat-conducting micropolar fluid. It is assumed that the fluid is thermody-namically perfect and polytropic. The same model has been considered in 1, 2, where the global-in-time existence and uniqueness for the generalized solution of the problem on
R×0, T, T >0, are proved. Using the results from1,3we can also easily conclude that the mass density and temperature are strictly positive.
Stabilization of the solution of the Cauchy problem for the classical fluid where microrotation is equal to zerohas been considered in4,5. In4was analyzed the H ¨older continuous solution. In5is considered the special case of our problem. We use here some ideas of Kanel’4and the results from1,5as well.
The case of nonhomogeneous boundary conditions for velocity and microrotation which is called in gas dynamics “problem on piston” is considered in6.
2. Statement of the Problem and the Main Result
Letρ, v, ω, andθdenote, respectively, the mass density, velocity, microrotation velocity, and temperature of the fluid in the Lagrangean description. The problem which we consider has the formulation as follows1:
∂ρ ∂t ρ
2∂v
∂x 0, 2.1
∂v ∂t
∂ ∂x
ρ∂v ∂x
−K ∂
∂x
ρθ, 2.2
ρ∂ω ∂t A
ρ ∂ ∂x
ρ∂ω ∂x
−ω
, 2.3
ρ∂θ ∂t −Kρ
2θ∂v
∂xρ
2
∂v ∂x
2
ρ2
∂ω ∂x
2
ω2Dρ ∂
∂x
ρ∂θ ∂x
2.4
inR×R, whereK, A, andDare positive constants. Equations2.1–2.4are, respectively, local forms of the conservation laws for the mass, momentum, momentum moment, and energy. We take the following nonhomogeneous initial conditions:
ρx,0 ρ0x,
vx,0 v0x,
ωx,0 ω0x,
θx,0 θ0x
2.5
forx∈R, whereρ0, v0, ω0, andθ0are given functions. We assume that there existm, M∈R,
such that
m≤ρ0x≤M, m≤θ0x≤M, x∈R. 2.6
In the previous papers1,2we proved that for
ρ0−1, v0, ω0, θ0−1∈H1R 2.7
the problem2.1–2.5has, for eachT ∈R, a unique generalized solution:
with the following properties:
ρ−1∈L∞0, T;H1R ∩H1Π,
v, ω, θ−1∈L∞0, T;H1R ∩H1Π∩L20, T;H2R .
2.9
Using the results from1,3we can easily conclude that
θ, ρ >0 inΠ. 2.10
We denote byBkR, k∈N
0,the Banach space
BkR
u∈CkR: lim
|x| → ∞|D
nux|0,0≤n≤k
, 2.11
whereDnisnth derivative; the norm is defined by
uBkRsup n≤k
sup
x∈R
|Dnux|
. 2.12
From Sobolev’s embedding theorem7, Chapter IVand the theory of vector-valued distributions8, pages 467–480one can conclude that from2.9one has
ρ−1∈L∞0, T;B0R ∩C0, T;L2R , 2.13
v, ω, θ−1∈L20, T;B1R ∩C0, T;H1R ∩L∞0, T;B0R , 2.14
and hence
v, ω, θ−1∈C0, T;B0R , ρ∈L∞Π. 2.15
From2.7and2.6it is easy to see that there exist the constantsE1, E2, E3, M1∈R, M1>1,
such that
1 2
R
v20dx 1 2A
R
ω20dxK
R
1 ρ0 −
ln 1 ρ0 −
1
dx
R
θ0−lnθ0−1dxE1, 2.16
1 2
R
v02
1 Aω
2 0 θ02
dxE2, 2.17
1 2
R
1 ρ20ρ
2 0dx
R
v0ln
1
ρ0dx≤E3,
2.18
sup
As in5, we can find out the real numbersηandη, η <0< η, such that
0
η
eη−1−η dη
η
0
eη−1−η dηE
5, 2.20
where
E52
E1E4
K
1/2
, E42μE1
1M1E3
E1
, μmax
K 2D,1
. 2.21
Usingηandηwe construct the quantities
uexpη, uexpη. 2.22
The aim of this work is to prove the following theorem.
Theorem 2.1. Suppose that the initial functions satisfy2.6,2.7, and the following conditions:
E1
R
v02dx <
D 24
u u
2
, 2.23
E1A
R
ω02dx <
D 12
u u
2
, 2.24
2E1
48E24M1E1
u u
2
89M1 2KM1E4
E1K2M21
D
u u
3M1
2
E1M1u2
13AE1 D
u uu
2
E2
<min
⎧ ⎨ ⎩
D 12A
u u
2
,
D 24
u u
2
,
M1
1
s−1−lns ds
2
,
1
0
s−1−lns ds
2⎫⎬
⎭;
2.25
then, whent → ∞,
ρx, t−→1, vx, t−→0, ωx, t−→0, θx, t−→1, 2.26
uniformly with respect to allx∈R.
Remark 2.2. Conditions 2.23–2.25mean that the constants E1, E2, E3, and M1 are suffi
-ciently small. In other words the initial functionsρ0, v0, ω0, andθ0are small perturbations of
In the proof ofTheorem 2.1, we apply some ideas of4and obtain the similar results as in5where a stabilisation of the generalized solution was proved for the classical model
whereω0.
3. A Priori Estimates for
ρ, v, ω,
and
θ
Considering stabilization problem, one has to prove some a priori estimates for the solution independent of the time variable T, which is the main difficulty. When we derive these estimates we use some ideas from 4, 5. First we construct the energy equation for the solution of problem2.1–2.4under the conditions indicated above and we estimate the function 1/ρ.
Lemma 3.1. For eacht >0it holds that
1 2
R
v2dx 1 2A
R
ω2dx
R
θ−lnθ−1dxK
R
1 ρ−ln
1 ρ−1
dx
t
0
R
ρ θ
∂v ∂x
2
ρ
θ
∂ω ∂x
2
ω2
ρθ D ρ θ2
∂θ ∂x
2
dx dτ E1,
3.1
whereE1is defined by2.16.
Proof. Multiplying2.1,2.2,2.3, and2.4, respectively, byKρ−11−ρ−1, v, A−1ρ−1ω, and
ρ−11−θ−1, integrating by parts overRand over0, t, and taking into account2.13and
2.14, after addition of the obtained equations we easily get equality3.1independently of t.
If we multiply2.2 by∂/∂xln1/ρ, integrate it over Rand 0, t, and use some equalities and inequalities which hold by 2.1 and 2.13–2.15 together with Young’s inequality we get, as in5, the following formula:
1 4
R
ρ2
∂ ∂x
1 ρ
2
dxK 2
t
0
R
θρ3
∂ ∂x
1 ρ
2
dx dτ
≤
R
v2dxK 2
t
0
R
ρ θ
∂θ ∂x
2
dx dτ
t
0
R
ρ
∂v ∂x
2
dx dτ
R
v0ln
1 ρ0
dx
1
2
R
1 ρ20ρ
2 0dx
3.2
for eacht >0. Using3.1,2.16,2.18, and2.21we get easily
1 4
R
ρ2
∂ ∂x
1 ρ
2
dxK 2
t
0
R
θρ3
∂ ∂x
1 ρ
2
dx dτ ≤K1
where
θt sup
x,τ∈R×0,tθx, τ, K1
θt 2μE1
1θt E3 E1
. 3.4
As in5, we introduce the increasing function
ψη
η
0
eξ−1−ξ dξ 3.5
that satisfies the following inequality:
ψ
ln1 ρ
≤
R
1
ρ −1−ln 1 ρ
dx
1/2
R
ρ2
∂ ∂x
1 ρ
2
dx
1/2
≤K2
θt , 3.6
where
K2
θt 2E1
2μ K
1θt E3 E1
1/2
. 3.7
We can easily conclude that there exist the quantitiesη
1θt<0 andη1θt>0, such that
0
η 1θt
eη−1−η dη
η1θt
0
eη−1−η dηK
2
θt . 3.8
Comparing3.8and3.6we obtain, as in5, Lemma 3.2, the following result.
Lemma 3.2. For eacht > 0there exist the strictly positive quantitiesu1 expη
1θtandu1
expη1θtsuch that
u1≤ρ−1x, τ≤u1, x, τ∈R×0, t. 3.9
Lemma 3.3. For eacht >0it holds that
1 2
R
∂v ∂x
2
1
A
∂ω ∂x
2
∂θ ∂x
2
dx
6u11/2 Du1
2t
0
R
∂v ∂x
2
dx
2
K3
θτ −
R
∂v ∂x
2
dx
dτ
2
3u1
Du1
2t
0
R
∂ω ∂x
2
dx
2
K4
θτ −
R
∂ω ∂x
2
dx
dτ
D
12
t
0
R
ρ
∂2θ
∂x2
2
dx dτ ≤K5
θt ,
3.10
where
K3
θt
Du1 24u1E11/2
2
, 3.11
K4
θt
Du1 12u1A1/2E11/2
2
, 3.12
K5
θt 48K12
θt θtE1
u1
u1
2
89θt 2KθtK1
θt
E1K2θ 2
t D
u1
u1 3θt
2
E1θtu21
1 3AE1 D
u1
u1 u
2 1
E2.
3.13
Proof. Multiplying2.2,2.3, and2.4, respectively, by−∂2v/∂x2,−A−1ρ−1∂2ω/∂x2,and
−ρ−1∂2θ/∂x2, integrating overR×0, t, and using the following equality:
− t
0
R
∂v ∂t
∂2v
∂x2dx dτ
1 2
R
∂v ∂x
2
dx|t0 3.14
that is satisfied for the functionsθandωas well, after addition of the obtained equalities we find that
1 2
R
∂v ∂x
2
1
A
∂ω ∂x
2
∂θ ∂x
2
dx|t0
t
0
R
ρ
∂2v
∂x2
2
dx dτ
t
0
R
ρ
∂2ω
∂x2
2
dx dτD
t
0
R
ρ
∂2θ
∂x2
2
− t 0 R ∂ρ ∂x ∂v ∂x
∂2v
∂x2dx dτK
t 0 R ρ∂θ ∂x
∂2v
∂x2dx dτK
t 0 R θ∂ρ ∂x
∂2v
∂x2dx dτ
− t 0 R ∂ρ ∂x ∂ω ∂x
∂2ω
∂x2dx dτ
t 0 R ω ρ
∂2ω
∂x2dx dτK
t 0 R ρθ∂v ∂x
∂2θ
∂x2dx dτ
− t 0 R ρ ∂v ∂x
2∂2θ
∂x2dx dτ−
t 0 R ρ ∂ω ∂x
2∂2θ
∂x2dx dτ−
t 0 R ω2 ρ ∂2θ
∂x2dx dτ
−D t 0 R ∂ρ ∂x ∂θ ∂x
∂2θ
∂x2dx dτ.
3.15
Using3.3,3.9, the inequality
∂v ∂x 2 ≤2 R ∂v ∂x
21/2⎛
⎝
R
∂2v
∂x2
2⎞
⎠
1/2
, 3.16
that holds for the functions∂θ/∂x, ∂ω/∂x, v, andωas well, and applying Young’s inequality with a sufficiently small parameter on the right-hand side of3.15, we find, similarly as in
5, the following estimates:
t 0 R ∂ρ ∂x ∂v ∂x
∂2v
∂x2dx dτ
≤ t 0 R 1 ρ ∂ρ ∂x 2 ∂v ∂x 2
dx dτ1 4 t 0 R ρ
∂2v
∂x2
2
dx dτ
≤ 2u
1/2 1 u1 t 0 R 1 ρ2 ∂ρ ∂x 2 dx R ∂v ∂x 2 dx
1/2⎛
⎝
R
ρ
∂2v
∂x2
2
dx
⎞ ⎠
1/2
dτ 1 4 t 0 R ρ
∂2v ∂x2 2 dx dτ ≤ 5 16 t 0 R ρ
∂2v
∂x2
2
dx dτ16K1
θt 2u1 u21
t 0 R ∂v ∂x 2 dx dτ ≤ 5 16 t 0 R ρ
∂2v
∂x2 2 dx dτ 16K1
θt u1 u1
2
θt
K t 0 R ρ∂θ ∂x
∂2v
∂x2dx dτ
≤K2
t 0 R ρ ∂θ ∂x 2
dx dτ 1 4 t 0 R ρ
∂2v
∂x2
2
dx dτ
≤ K2u1θ 2 t u1 t 0 R ρ θ2 ∂θ ∂x 2
dx dτ1 4 t 0 R ρ
∂2v
∂x2 2 dx dτ, 3.18 K t 0 R θ∂ρ ∂x
∂2v
∂x2dx dτ
≤K2
t 0 R θ2 ρ ∂ρ ∂x 2
dx dτ1 4 t 0 R ρ
∂2v
∂x2
2
dx dτ
≤K2θt
t 0 R θρ3 ∂ ∂x 1 ρ 2
dx dτ1 4 t 0 R ρ
∂2v
∂x2 2 dx dτ, 3.19 t 0 R ∂ρ ∂x ∂ω ∂x
∂2ω
∂x2dx dτ
≤ t 0 R 1 ρ ∂ρ ∂x 2 ∂ω ∂x 2
dx dτ 1 4 t 0 R ρ
∂2ω
∂x2
2
dx dτ
≤ 2u
1/2 1 u1 t 0 R 1 ρ2 ∂ρ ∂x 2 dx R ∂ω ∂x 2 dx
1/2⎛
⎝
R
ρ
∂2ω
∂x2
2
dx
⎞ ⎠
1/2
dτ 1 4 t 0 R ρ
∂2ω
∂x2 2 dx dτ ≤ 3 8 t 0 R ρ
∂2ω
∂x2
2
dx dτ2
⎛ ⎜
⎝8K1
θt u11/2 u1 ⎞ ⎟ ⎠ 2 t 0 R ∂ω ∂x 2 dx dτ ≤ 3 8 t 0 R ρ
∂2ω
∂x2
2
dx dτ2
8K1
θt u1 u1
2
θt
t 0 R ρ θ ∂ω ∂x 2 dx dτ, 3.20 t 0 R ω ρ
∂2ω
∂x2dx dτ
≤ t 0 R ω2
ρ3dx dτ
1 4 t 0 R ρ
∂2ω
∂x2
2
dx dτ
≤u21θt
t
0
R
ω2
ρθdx dτ 1 4 t 0 R ρ
∂2ω
∂x2
2
dx dτ,
K t 0 R ρθ∂v ∂x
∂2θ
∂x2dx dτ
≤ 3K2
2D
t
0
R
θ2ρ
∂v ∂x
2
dx dτD 6 t 0 R ρ
∂2θ
∂x2
2
dx dτ
≤ 3K2θ
3 t 2D t 0 R ρ θ ∂v ∂x 2
dx dτ D 6 t 0 R ρ
∂2θ
∂x2 2 dx dτ, 3.22 t 0 R ρ ∂v ∂x
2∂2θ
∂x2dx dτ
≤ 3 2D t 0 R ρ ∂v ∂x 4
dx dτ D 6 t 0 R ρ
∂2θ
∂x2
2
dx dτ
≤ 3u
1/2 1 Du1 t 0 R ∂v ∂x 2 dx
3/2⎛
⎝
R
ρ
∂2v
∂x2
2
dx
⎞ ⎠
1/2
dτD 6 t 0 R ρ
∂2θ
∂x2
2
dx dτ
≤4
3u11/2 Du1
2t
0 R ∂v ∂x 2 dx 3
dτ 1 16 t 0 R ρ
∂2v
∂x2 2 dx dτ D 6 t 0 R ρ
∂2θ
∂x2 2 dx dτ, 3.23 t 0 R ρ ∂ω ∂x
2∂2θ
∂x2dx dτ
≤ 3 2D t 0 R ρ ∂ω ∂x 4
dx dτD 6 t 0 R ρ
∂2θ
∂x2
2
dx dτ
≤ 3u
1/2 1 Du1 t 0 R ∂ω ∂x 2 dx
3/2⎛
⎝
R
ρ
∂2ω
∂x2
2
dx
⎞ ⎠
1/2
dτD 6 t 0 R ρ
∂2θ
∂x2
2
dx dτ
≤2
3u11/2 Du1
2t
0 R ∂ω ∂x 2 dx 3
dτ1 8 t 0 R ρ
∂2ω ∂x2 2 dx dτ D 6 t 0 R ρ
∂2θ
∂x2
2
dx dτ,
t 0 R ω2 ρ ∂2θ
∂x2dx dτ
≤ 3 2D t 0 R ω4
ρ3dx dτ
D 6 t 0 R ρ
∂2θ
∂x2
2
dx dτ
≤ 6AE1u31
D
t
0
R
ω2dx
1/2
R ρ ∂ω ∂x 2 dx
1/2
dτD 6 t 0 R ρ
∂2θ ∂x2
2
dx dτ
≤ 6AE1u31
D
θt 2u1 t 0 R ω2
ρθx dτ u1θt
2 t 0 R ρ θ ∂ω ∂x 2 dx dτ D 6 t 0 R ρ
∂2θ
∂x2 2 dx dτ, 3.25 D t 0 R ∂ρ ∂x ∂θ ∂x
∂2θ
∂x2dx dτ
≤ 3D 2 t 0 R 1 ρ ∂ρ ∂x 2 ∂θ ∂x 2
dx dτD 6 t 0 R ρ
∂2θ
∂x2
2
dx dτ
≤ 3Du
1/2 1 u1 t 0 ⎛ ⎝ R ρ
∂2θ
∂x2
2
dx
⎞ ⎠
1/2
R ∂θ ∂x 2 dx
1/2
R 1 ρ2 ∂ρ ∂x 2 dx dτ D 6 t 0 R ρ
∂2θ
∂x2 2 dx dτ ≤ ⎛ ⎜
⎝12DK1
θt θtu1
u1 ⎞ ⎟ ⎠ 2 3 D t 0 R ρ θ2 ∂θ ∂x 2
dx dτ D 4 t 0 R ρ
∂2θ
∂x2
2
dx dτ.
3.26
Inserting3.17–3.26into3.15and using estimates3.1,3.3, and2.17we obtain
1 2 R ∂v ∂x 2 1 A ∂ω ∂x 2 ∂θ ∂x 2
dx1 8 t 0 R ρ
∂2v
∂x2 2 dx dτ 1 4 t 0 R ρ
∂2ω
∂x2
2
dx dτ D 12 t 0 R ρ
∂2θ
∂x2
2
dx dτ
≤K5
θt
6u11/2 Du1
2t
0 R ∂v ∂x 2 dx 3
dτ2
3u11/2 Du1
2t
whereK5θtis defined by3.10. We also use the following inequality:
R
∂v ∂x
2
dx−
R
v∂
2v
∂x2dx≤u 1/2 1
R
v2dx
1/2⎛
⎝
R
ρ
∂2v
∂x2
2
dx
⎞ ⎠
1/2
≤2E1u11/2
⎛
⎝
R
ρ
∂2v
∂x2
2
dx
⎞ ⎠
1/2 3.28
that is satisfied for the function∂ω/∂xas well. Therefore we have
R
ρ
∂2v ∂x2
2
dx≥2E1u1−1
R
∂v ∂x
2
dx
2
,
R
ρ
∂2ω
∂x2
2
dx≥2AE1u1−1
R
∂ω ∂x
2
dx
2
.
3.29
Inserting3.29into3.27we obtain
1 2
R
∂v ∂x
2
1
A
∂ω ∂x
2
∂θ ∂x
2
dx D 12
t
0
R
ρ
∂2θ
∂x2
2
dxdτ
2
3u11/2 Du1
2t
0
R
∂ω ∂x
2
dx
2⎛
⎝
Du1 12u1A1/2E11/2
2
−
R
∂ω ∂x
2
dx
⎞
⎠dτ
6u11/2 Du1
2t
0
R
∂v ∂x
2
dx
2⎛
⎝
Du1 24u1E11/2
2
−
R
∂v ∂x
2
dx
⎞
⎠dτ ≤K5θt ,
3.30
and3.10is satisfied.
In the continuation we use the above results and the conditions of Theorem 2.1. Similarly as in4,5, we derive the estimates for the solutionρ, v, ω, θof problem2.1–
2.7, defined by2.8–2.10in the domainΠ R×0, T, for arbitraryT >0.
Taking into account assumption2.19and the fact thatθ∈CΠ see2.15we have the following alternatives: either
sup
x,t∈Πθx, t θT≤M1, 3.31
or there existst1,0< t1< T, such that
Now we assume that3.32 is satisfied and we will show later that because of the choice E1, E2, E3, andM1the conditions ofTheorem 2.1, the property3.32is impossible.
Because K2θt, defined by 3.7, increases with increasing θt, we can easily
conclude that
K2
θt < K2M1 for 0≤t < t1, 3.33
andK2M1 E5. Therefore, comparing2.20and3.8, we obtain
u < u1θt , u > u1
θt , 3.34
where u, u, andu1θt, u1θt are defined by 2.22 and Lemma 3.2. It is important to
point out that the quantitiesK3θtand K4θt, defined by3.11-3.12, decrease with
increasingθtand forθt1 M1they become
K3M1
Du 24uE11/2
2
, K4M1
Du 12uA1/2E1/2
1
2
. 3.35
Now, using these facts we will obtain the estimates for R∂ω/∂x2dx and
R∂v/∂x
2dxon0, t
1. Taking into account the assumptions2.23and2.24ofTheorem 2.1
and the following inclusionsee2.14:
∂v ∂x,
∂ω ∂x ∈C
0, T;L2R , 3.36
we have again the following two alternatives: either
R
∂v ∂x
2
x, tdx≤K3M1,
R
∂ω ∂x
2
x, tdx≤K4M1 fort∈0, t1, 3.37
or there existst2,0< t2< t1, such that∂ω/∂xand∂v/∂xor converselyhave the following
properties:
R
∂ω ∂x
2
x, tdx < K4M1 for 0≤t < t2, 3.38
R
∂ω ∂x
2
x, t2dxK4M1 fort2< t1, 3.39
R
∂v ∂x
2
We assume that3.38–3.40are satisfied. Then we have
θt< M1, K4M1< K4
θt , K3M1< K3
θt 3.41
fort∈0, t2. Using3.41from3.10, fortt2, we obtain
R
∂ω ∂x
2
x, tdx≤2AK5
θt2 , 0≤t≤t2. 3.42
SinceK5θt, defined by3.13, increases with the increase ofθt, it holds that
K5
θt ≤K5M1, t∈0, t2. 3.43
Using condition2.25we get
2AK5
θt < K4M1, t∈0, t2 3.44
and conclude that
R
∂ω ∂x
2
x, t2dx < K4M1. 3.45
This inequality contradicts3.39. Consequently, the only case possible is when
t2t1, 3.46
and then for∂ω/∂x3.37is satisfied.
If in 3.38–3.40 the functions ∂ω/∂x and ∂v/∂x exchange positions, using assumption 2.25, in the same way as above we obtain that the function ∂v/∂x satisfies the inequality
R
∂v ∂x
2
x, tdx≤K3M1, t∈0, t1. 3.47
With the help of3.37from3.10we can easily conclude that
R
∂θ ∂x
2
x, tdx <2K5M1 for 0< t≤t1. 3.48
Now, as in5, we introduce the functionΨby
Ψθx, t
θx,t
1
Taking into account that from2.14follows thatθx, t → 1 as|x| → ∞, we have
Ψθx, t−→0 as |x| −→ ∞. 3.50
Consequently,
ψθx, t≤ψθx, t
θx,t
1
d
dsψsds
x
−∞
θx, t−1−lnθx, t∂θx, t ∂x dx
≤
R
θx, t−1−lnθx, tdx
1/2
R
∂θ ∂x
2
x, tdx
1/2
.
3.51
Using3.32,3.48, and3.1from3.51we get
max
0≤θx,t≤M1
ψθx, t ψθt1 ψM1≤2K5M1E11/2, 3.52
or
M1
1
s−1−lns ds−2K5M1E11/2≤0. 3.53
Since this inequality contradicts2.25, it remains to assume thatt1 T. Hence we have the
following lemma.
Lemma 3.4. For eachT >0it holds that
θx, t≤M1, x, t∈Π, 3.54
R
∂ω ∂x
2
x, tdx≤K4M1, 0≤t≤T, 3.55
R
∂v ∂x
2
x, tdx≤K3M1, 0≤t≤T, 3.56
R
∂θ ∂x
2
x, tdx≤2K5M1, 0≤t≤T. 3.57
Lemma 3.5. It holds that
0< u≤ 1
ρx, t ≤u, x, t∈Π, 3.58
sup
x,t∈Π|ωx, t| ≤8AE1K4M1 1/4
, 3.59
sup
x,t∈Π|vx, t| ≤8E1K3M1
1/4, 3.60
θx, t≥h >0, x, t∈Π, 3.61
where uanduare defined by2.20–2.22and a constanthdepends only on the data of problem
2.1–2.5.
Proof. Because the quantity u1θt in Lemma 3.2 decreases with increasing θt while u1θtincreases, it follows, in the same way as in 5, from3.9and 3.54that3.58is
satisfied. Using the inequalities
v22
x
−∞v
∂v ∂xdx≤2
R
v2dx
1/2
R
∂v ∂x
2
dx
1/2
,
ω22
x
−∞ω
∂ω ∂xdx≤2
R
ω2dx
1/2
R
∂ω ∂x
2
dx
1/2
3.62
and estimations3.1,3.55, and3.56we get immediately3.59and3.60. From3.50,
3.53,3.56, and3.1we have, as in5, forθx, t≤1 that
1
θx,t
s−1−lns ds≤2K5M1E11/2<
1
0
s−1−lns ds 3.63
Remark 3.6. Using the properties of the functions u1 expη
1θtand u1 expη1θt
defined inLemma 3.2, from3.55-3.56and3.59-3.60we get the following estimates:
R
∂ω ∂x
2
dx≤
D 12A1/2E1/2
1
2
exp{−2λt},
R
∂v ∂x
2
dx≤
D 24E11/2
2
exp{−2λt},
sup
x,t∈Π
|ωx, t| ≤
D2
18
1/4
exp
− 1
2λt
,
sup
x,t∈Π
|vx, t| ≤
D2
72
1/4
exp
− 1
2λt
,
3.64
whereλt η1θt−η
1θt>0.
Lemma 3.7. For eachT >0it holds that
T
0
R
∂v ∂x
2
dx dτ ≤K6, 3.65
T
0
R
∂θ ∂x
2
dx dτ ≤K7, 3.66
T
0
R
∂ω ∂x
2
dx dτ ≤K8, 3.67
T
0
R
ω2
ρθdx dτ ≤K9, 3.68
T
0
R
∂ρ ∂x
2
dx dτ ≤K10, 3.69
R
∂ρ ∂x
2
dx≤K11, t∈0, T, 3.70
T
0
R
∂2θ
∂x2
2
dx dτ ≤K12, 3.71
T
0
R
∂2v
∂x2
2
dx dτ ≤K13, 3.72
T
0
R
∂2ω
∂x2
2
dx dτ ≤K14, 3.73
Proof. Taking into account3.58,3.61, and3.54from3.1,3.3,3.10, and3.27we get all above estimates.
4. Proof of
Theorem 2.1
In the following we use the results ofSection 3. The conclusions ofTheorem 2.1are immediate consequences of the following lemmas.
Lemma 4.1. It holds that
R
∂v ∂x
2
x, tdx−→0,
R
∂ω ∂x
2
x, tdx−→0,
R
∂θ ∂x
2
x, tdx−→0,
4.1
whent → ∞.
Proof. Letε >0 be arbitrary. With the help of3.65–3.69we conclude that there existst0>0
such that
t
t0
R
∂v ∂x
2
dx dτ < ε,
t
t0
R
∂θ ∂x
2
dx dτ < ε,
t
t0
R
∂ω ∂x
2
dx dτ < ε,
t
t0
R
ω2
ρθdx dτ < ε,
t
t0
R
∂ρ ∂x
2
dx dτ < ε,
4.2
fort > t0, and
R
∂v ∂x
2
x, t0dx < ε,
R
∂θ ∂x
2
x, t0dx < ε,
R
∂ω ∂x
2
x, t0dx < ε. 4.3
Similarly to3.10, we have
1 2
R
∂v ∂x
2
1
A
∂ω ∂x
2
∂θ
∂x
2
dx D 12
t
t0
R
ρ
∂2θ ∂x2
2
dx dτ
6u1/2 Du
2t
t0
R
∂v ∂x
2
dx
2
K3
θτ −
R
∂v ∂x
2
dx
dτ
2
3u Du
2t
t0
R
∂ω ∂x
2
dx
2
K4
θτ −
R
∂ω ∂x
2
dx
≤ 1
2
R
∂v ∂x
2
1
A
∂ω ∂x
2
∂θ ∂x
2
x, t0dxK2
t
t0
R
ρ
∂θ ∂x
2
dx dτ
16K1
θt 2 u u2
t
t0
R
∂v ∂x
2
dx dτK2
t
t0
R
θ2
ρ
∂ρ ∂x
2
dx dτ
2
⎛ ⎜
⎝8K
2 1
θt u1/2 u
⎞ ⎟ ⎠
2
t
t0
R
∂ω ∂x
2
dx dτu2θt
t
t0
R
ω2
ρθdx dτ
3K2
2D
t
t0
R
θ2ρ
∂v ∂x
2
dx dτ3AE1θtu
3
Du
t
t0
R
ω2
ρθdx dτ
3AE1u3
D
t
t0
R
∂ω ∂x
2
dx dτ
⎛ ⎜
⎝12DK1
θt u1/2 u
⎞ ⎟ ⎠
2
t
t0
R
∂θ ∂x
2
dx dτ.
4.4
Taking into account3.54,3.58,3.61, and4.2–4.3from4.4we obtain
R
∂v ∂x
2
1
A
∂v ∂x
2
∂θ ∂x
2
dx≤K15ε fort > t0, 4.5
where K15 depends only on the data of our problem and does not depend on t0. Hence
relations4.1hold.
Lemma 4.2. It holds that
vx, t−→0, ωx, t−→0, θx, t−→1 4.6
whent → ∞, uniformly with respect to allx, x∈R.
Proof. We havesee3.51and3.62
v2x, t≤2
R
v2x, tdx
1/2
R
∂v ∂x
2
x, tdx
1/2
,
ω2x, t≤2
R
ω2x, tdx
1/2
R
∂ω ∂x
2
x, tdx
1/2
,
ψθx, t≤
R
θx, t−1−lnθx, tdx
1/2
R
∂θ ∂x
2
dx
1/2
.
Taking into account3.1from4.7we get
v2x, t≤22E11/2
R
∂v ∂x
2
x, tdx
1/2
,
ω2x, t≤22AE11/2
R
∂ω ∂x
2
x, tdx
1/2
,
ψθx, t≤E1/2 1
R
∂θ ∂x
2
dx
1/2
.
4.8
Using4.1and property3.50of the functionψwe can easily obtain that4.6holds.
Now, as in5, we analyze the behavior of the functionρast → ∞. From3.69and
3.72we conclude that forε >0 there existst0>0 such that
t
t0
R
∂ρ ∂x
2
dx dτ < ε,
t
t0
R
∂2v
∂x2
2
dx dτ < ε,
R
∂ρ ∂x
2
x, t0dx < ε 4.9
fort > t0. Deriving2.1with respect tox, multiplying by∂/∂x1/ρ, and integrating over
Randt0, t, after using3.58and Young’s inequality, we obtain
u4
2
R
∂ρ ∂x
2
dx≤u2
t
t0
R
∂ρ ∂x
2
dx dτu2
t
t0
R
∂2v
∂x2
2
dx dτ
u4
2
R
∂ρ ∂x
2
x, t0dx.
4.10
With the help of4.9we get easily the following result.
Lemma 4.3. It holds that
R
∂ρ ∂x
2
x, tdx−→0 4.11
whent → ∞.
Similarly as for the functionψθ, we have
ψ
1 ρ
1/ρ
1
s−1−lns ds
≤
R
1
ρ−1−ln 1 ρ
dx
1/2
R
1 ρ2
∂ρ ∂x
2
dx
1/2
.
Taking into account3.1from4.12one has
ψ
1 ρ
≤K−1E1 1/2
u
R
∂ρ ∂x
2
dx
1/2
. 4.13
Using4.11we obtain the following conclusion.
Lemma 4.4. It holds that
ρx, t−→1 4.14
whent → ∞, uniformly with respect tox∈R.
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3 S. N. Antontsev, A. V. Kazhikhov, and V. N. Monakhov, Boundary Value Problems in Mechanics of Nonhomogeneous Fluids, vol. 22 of Studies in Mathematics and Its Applications, North-Holland, Amsterdam, The Netherlands, 1990.
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