71
Initial-boundary Value Problems to the One-
dimensional Compressible Navier-Stokes-Poisson
Equations with viscosity and heat conductivity
coefficients
Li WANG1,*
, Lei JIN2
1School of Applied Mathematics, Xiamen University of Technology, Xiamen 361024,
China
2School of Environment Science and Engineer, Xiamen University of Technology,
Xiamen 361024, China
* Corresponding author, E-mail: [email protected]
Abstract
In this paper, the global, non-vacuum solutions with large amplitude to the initial-
boundary value problem of the one-dimensional compressible Navier-Stokes-Poisson
system with viscosity and heat conductivity coefficients are considered. The proof is
based on the analysis on the positive lower and upper bounds on the specific volume and
the absolute temperature.
SCIREA Journal of Mathematics
http://www.scirea.org/journal/Mathematics
October 20, 2016
Volume 1, Issue1, October 2016
72
Key words: compressible Navier-Stokes-Poisson system; global, non-vacuum
solutions with large amplitude; viscosity and heat conductivity coefficients
1. Introduction
Magnetohydrodynamics, which combines the environmental fluid mechanics and
electrodynamics theories to study the interaction discipline between the conduction fluids
and electromagnetic, is the theory of the macroscopic, and it has spanned a very large
range of applications[1]-{3}
. The motion of compressible, viscous self-gravitating fluids
can be expressed by Navier-Stokes-Poisson equations. In this paper, we consider the one-
dimensional compressible Navier-Stokes-Poisson system with viscous coefficient and
heat conductivity:
2
,
0,
0,2
1 .
t x
xt x
x
t
x
x
v u
uv
ue u q
vv
(1)
Where , , , ,v u e q and denote the specific volume, velocity, stress, internal energy,
heat flux, and the electrostatic potential function, respectively. For Newtonian fluid and
Fourier’s law
( , )( , , ) ( , )x x
vv u p v u
v
,
( , )( , , )x x
vq v u
v
. (2)
Here, p and are the pressure and the absolute temperature respectively. ( , ) 0v
denotes the viscosity coefficient and ( , ) 0v is heat conductivity coefficient.
Physically, the Navier–Stokes–Poisson equations describe the motion of compressible
viscous isentropic gas flow under the self-gravitational force. Compressible Navier-
Stokes type equation with density and temperature dependent transport coefficients arise
in many applied sciences, such as certain class of solid-like materials [4], gases at very
73
high temperatures [5], etc. Recently, there are many studies on the non-vacuum solutions
to the one-dimensional compressible Navier-Stokes equations with density and
temperature dependent transportation coefficients in various forms, cf. [6, 7, 8, 9, 10, 11,
12]. However, there is a gap between the physical models and the existence theory.
This manuscript is concerned with the construction of global, non-vacuum, large, smooth
solutions to the one-dimensional compressible Navier-Stokes equation (1) in the domain
( , ) | [0,1], 0x t x I t with prescribed initial condition
0 0 0 0( ( ,0), ( ,0), ( ,0), ( ,0)) ( ( ), ( ), ( ), ( ))x xv x v x x x v x u x x x , [0,1]x (3)
and boundary condition
(0, ) (1, ) 0,
(0, ) (1, ) 0,
(0, ) (1, ) 0.x x
t t
q t q t
t t
(4)
Throughout this manuscript, we will focus on the ideal, polytrophic gases which contain
the case of gases for which kinetic theory provides constitutive relations, cf. [13, 14]
1
, ( , ) exp1
v
R Re C p v Av s
v R
(5)
where the specific gas constant R and the specific heat at constant volume vC are positive
constant and 1 is the adiabatic constant. And our main interest concerns the case
when the transport coefficients and may depend on the specific volume and/or the
absolute temperature which are degenerate in the sense that and /or are not
uniformly bounded from below or above some positive constants for all 0v and 0 .
2. Results
To simplify the presentation, the result is concerned with the case
( ) , ( , )a bv v v , (6)
74
Now we turn to state the main results obtained in this paper. The transport coefficients
and are assumed to satisfy the following condition:
● is a positive constant and ( , )v is a smooth function of v and satisfying
( , ) 0v
for 0, 0v and there exist positive constants 0 and ( , )v such that
0
0, 0( , ) 0, min ( , ) ( , ) 0
v vv v v
(7)
hold true for each given positive constants 0v and 0 ,which is stated as follows.
Theorem 1 suppose
(i) 1
0 0 0( ), ( ), ( ) ( );v x u x x H I
(ii)0 0inf ( ) 0,inf ( ) 0
x I x Iv x x
, the initial data 0 0 0( ), ( ), ( )v x u x x are compatible with
the boundary condition (4);
(iii) and are assumed to satisfies one of the following two conditions
● is a positive constant and are assumed to satisfies ( , ) 0v for
0, 0v and
10 ( , ) ( )(1 ), 0cv C V V v V (8)
hold for some positive constant ( ) 0C V and 0 sufficiently large. Here
0 1c
is constant and 0V is any given positive constant;
● and are given by (6) with a and b satisfying
1
0 , 2.5
a b (9)
Then the initial-boundary value problem (1), (3), (4) exists a unique global solution
( ), ( ), ( )v x u x x which satisfies
75
0 1( , ), ( , ), ( , ) (0, ; ( )),v x t u x t x t C T H I
2 1( , ), ( , ) (0, ; ( ))x xu x t x t L T H I , (10)
, , ( , ) [0, ]V v V x t I T .
Here 0T is any given positive constant and , , ,V V are some positive constants
which may depend on T and the initial data 0 0 0 0( ), ( ), ( ), ( )v x u x x x .
Remark The initial-boundary value problem (1), (3), (4) has also been studied in []. To
deduced the desired lower and upper bound on the specific volume v , the viscosity
coefficient ( )v is assumed to satisfy
0 10 ( )v
and the entropy ( , )s v and the internal energy ( , )e v are assumed to satisfy
0
( )( , ) 1 ( , )
rv z
s v dz e vz
(11)
in []. Here 0 1, , and 2r are positive constant. For the ideal polytrophic gas, if the
transport coefficients and are assumed to satisfy (6), (11) holds only if 0a .
Notations: Throughout this manuscript, 1C stands for a generic positive constant
which may depend on 0 0inf ( ),inf ( ), ,
x I x Iv x x T
and 10 0 0 ( )
, ,H I
v u . Note that all these
constant may vary in different places. ( )sH I represents the usual Sobolev spaces on I
with the standard norm ( )sH I
and for 1 , (I)pp L denotes the usual pL spaces
equipped with the usual norm ( )pL I
. For simplicity, we use to denote the norm in
( [0, ])L I T .
76
3. Methods
Proof the Theorem 1
This section is devoted to the proof of Theorem 1 based on the continuation argument.
Since the local solvability of the initial-boundary value problem is well-established [15],
if we suppose that the local solution ( , ), ( , ), ( , )v x t u x t x t to the initial-boundary value
problem (1), (3), (4) has been extended to the time step t T >0 for some 0T , then to
extend such a solution ( , ), ( , ), ( , )v x t u x t x t step by step to a global one.
Lemma 1 (Basic energy estimates) Let the conditions in Lemma 1 hold and suppose
that the local solution ( , ), ( , ), ( , ), ( , )v x t u x t x t x t constructed in Lemma 1 satisfies the
a priori assumption (H), then we have for 0 t T that
2 2 21 1
20 0 0
( ) ( , )1( ( , , ) ) ( )
2
tx x xv u v
v u dx dxdsv v v
21
00 0, 0
00
1( ( , ) )
2
xv u dxv
. (12)
Lemma 2 (Estimate on the total energy) Let the conditions stated in Theorem 1 hold
and suppose that ( ( , ), ( , ), ( , ))v x t u x t x t is a solution to the initial-boundary value
problem (1), (2), (4) defined on [0, ]I T for some 0T . If we assume further that
( ( , ), ( , ), ( , ))v x t u x t x t satisfies the a priori assumption (H), then we have for 0 t T
that
22
1 10
00 02 2
v v
uuC dx C dx
(13)
First we consider the case when the transport coefficients and satisfy (6) and (7).
Lemma 3 Under the condition listed in Lemma 1, we have
1 1
, ( , ) [0, ]( , ) ( )
C C x t I Tx t v v
. (14)
Proof: First of all, (1)3 implies
77
2
2 2
( ) ( , )1 1x x xv
t x
v u Ru vC
v v v
. (15)
From (15), we can get for each 1p that
2 2
2 2
2 1 2
2 1
2 (2 1) ( , )1
2 ( , )1 ( )2 .
2 ( ) 2 ( )
p
xv p
t
p
x x
p
x
p p vC
v
u p vv R Rp
v v v v
(16)
Integrating (16) with respect to x over I, we have
22
2 2 1 2 121
0
1 1 1 12 2
4 ( ) ( ) pp
p p p
v
LLt
RC p dx pC
v v v v
, (17)
which implies
2 2
1
00
1 1(inf ( ))
( )p p
t
x IL L
C x C dsv v
. (18)
Letting p in (18), we can deduce (14) immediately. This completes the proof of
lemma 3.
Lemma 4 Under the conditions listed in lemma 2 and assume that the transport
coefficients and satisfy (6) and (7), there exist positive constants3 3,V V , and 3
depending only on T and the initial data 0 0 0 0( ( , ), ( , ), ( , ), ( , ))v x t u x t x t x t such that
3 3( , )V v x t V , ( , ) [0, ]x t I T , (19)
and
3( , )x t , ( , ) [0, ]x t I T . (20)
Proof: We first define
1
( )( ) :
v
g v d
. (21)
Then we get
78
( ) ( ) v
[ ( )]x t xxt t x
x x
v u vg v u p
v v v
. (22)
Integrating (22) over [ , ] [0, t]y x yields
0
00 0
( ( , )) ( , )
( ( ) ( , ) ( ( , )) ( ( ,0)) ( (y,0)) (y, ) .
t
x t t xz
y y
g v x t p x s ds
u z u z t dz g v y t g v x g v p s ds dzdsv
(23)
Set 0y in (23), then involving the boundary condition (4), we have
0 0 0 00 0 0 0
log ( , ) ( , ) ( ) ( , ) log ( )t x t x
zv x t p x s ds u z u z t dz v x dzdsv
. (24)
(24) together with the fact that ( , ) 0p x t and the estimate (13), we can easily get the
lower bound of ( , )v x t with (15). That is,
3( , )v x t V , 3( , )x t , ( , ) [0, ]x t I T . (25)
The assumption (6) together with the estimates (25) imply that
( , )x t K (26)
hold for some positive constant K depending on 3V and 3 for all v and under our
consideration.
To deduce an upper bound on ( , )v x t by exploiting the argument used in Lemma 3, we
only need to recover the dissipative estimates2 2
1
20 0
tx xu
dxdsv v
. For this purpose,
multiplying (1)3 by 1 and integrating the resulting identity with respect to x and t
over [0, ]I t , one has
2 21 1
0
20 0 0 0
1 1 1 1
0 00 0 0 0
1
0
( , )
log log log log
log ,
t tx x
v v
u vdxds dxds
v v
C dx C dx R vdx R v dx
C R vdx
(27)
79
where (13) and (25) are used.
As for the last term on the right hand side of (27), we have by integrating (24) with
respect to x over [0, 1] that
1 1
00 0 0
log ( , )t
vdx C p x s dxds C , (28)
which together with (27) implies that
2
1 10
20 0 0 0
( , )t txu vdxds dxds C
v v
. (29)
Having obtained (29), we can deduce the upper bound on ( , )v x t with the aid of the
Gronwall inequality and (12). This completes the proof of Lemma 4.
Now we turn to deduce the upper bound on ( , )x t for the case when the transport
coefficients and satisfy (6) and (7).
Corollary 1 Under the conditions listed in Lemma 4, we have for 0 t T that
0 ( )
( )t
L Is ds C
(30)
and
1
2
0 0( , )
t
x s dxds C . (31)
By (30), we can obtain
Lemma 5 Under the conditions listed in Lemma 4, we have for 0 t T that
1 1
2 2
0 0 0
t
xu dx u dxds C . (32)
Proof: Multiplying (1)2 by u and integrating the resulting equation with respect to x
and t over [0, ]I t , one has
1 1
0 0 0 0
( )( )
t tx x
t x
x
v uuu up dxdt u u dxdt
v v
(33)
and
80
2
22 21 1 12
00 0 0 0 02
t tx
L
uudx dxds C u C dxds
v v
, (34)
where 2 2
1 1 1
0 0 0 0 0 0
1 1
2 2
t t tx xu u
dxds dxds dxdsv v v
,
21 1 1
2
0 0 0 0 0 0
21 1
2
0 0 0
1 1
2 2
1 1 .
2 2
t t tx x
tx
udxds u dxds dxds
v v
T u dx dxdsv
Thus applying (22) and (31), we get (32). This proves Lemma 5.
Lemma 6 Under the conditions listed in Lemma 4, we have for 0 t T that
1 2 2
( ) ( ) ( )0( ( ) ( ) )xL I L I L I
C C u s s ds . (35)
Proof: From (1)3, we can easily deduce for each 1p that
2 2
2 2 2 2 1 2 1 2 1( ) 2 (2 1) 2 2 2p P p p px x x xv t
x
u R uC p p p p p
v v v v
. (36)
Integrating (36) with respect to x over I, one has
2
21 12 2 1 2 1
0 0( ( ) ) 2 2
p p px xv tL
u R uC t p dx p dx
v v
. (37)
By exploiting the Holder inequality and letting p , we get from (37) that
2
0( ) ( ) 0( )( )
( )t
x x
L I L IL IL I
u ut C C ds
v v
. (38)
Then with the help of Cauchy’s inequality, we can deduce (35) and the proof of Lemma 6
is complete.
Lemma 7 Under the conditions listed in Lemma 4, we have for 0 t T that
2
1
10 0
( , )t rx
r
vdxds C C
, (0,1)r . (39)
81
Proof: Multiplying (1)3 by r and integrating the resulting equation with respect to x
over I yield
2
1 11
10 0 0
( , )tr x
v r
r vC dx dxds
v
2 11 1 1
1 00
0 0 0 0 0
1 11 2 2
00 0 0 0
,
r rt t
r x xv
t tr r
x
r
u uC dx dxds R dxds
v v
C C dxds u dxds
C C
(40)
where (30) and (31) are used. This is (39) and completes the proof of Lemma 7.
A direct consequence of (37) is
Lemma 8 Under the conditions listed in Lemma 4, we have for 0 t T that
1
22
( )0( ) .
t
L Is ds C C
(41)
Proof: Observe that (31) imply
2 2
( )
11
2 21 11 22
1( ) 0 0
1
2 2112
1( ) 0
( , ) ( ( ), ) 2 ( , ) ( , )
( ) ( , ) ( , )
( ) ( , ) .
x
yb t
r
x
rL I
r
x
rL I
x t b t t y t y t dy
C C t x t dx x t dx
C C t x t dx
(42)
From the above inequality together with the estimates (39) and (40), we can get that
82
1
2 21122
1( ) ( )0 0 0
11
2 212 2
1( )0 0 0
121
22
1( )0
( ) ( ) ( , )
( ) ( , )
( )
rt t
x
rL I L I
t tr x
rL I
r tx
rL I
s C C s x s dx ds
C C s x s dx
C C s
1
21
0 0
1
2
( , )
.
t
x s dx
C C
(43)
This is exactly (41) and the proof of Lemma 8 is complete.
Lemma 9 Under the conditions listed in Lemma 4, we have for 0 t T that
1 1
2 2
0 0 0
t r
x xv dx v dxds C C
, (0,1)r . (44)
Proof: As in (38), we can rewrite (1)2 as
x xt
x t
vRu
v v v
. (45)
Multiplying the identity (45) by xv
v
, we get that
2 2 2
2 22
x x x x x x x
x xtt
v u u p v vu up
v v v v v
. (46)
Integrating (46) with respect to x and t over [0, ]I t , with the help of (12), Cauchy’s
inequality, and the fact (0, ) (1, ) 0t t , we have
21 1 1
2 2 2 2 2
0 0 0 0 0
21
10 0 ,
t tx
x x x
tx
r
v dx v dxds C u u dxds
C C dxds
(47)
where (32)-(33) are used. Then by (39), we can easily get (43).
83
By employing the arguments used in [6, 9, 16], we can control 1
4
0 0
t
xu dxds as in the
following lemma
Lemma 10 Under the conditions listed in Lemma 4, we have for 0 t T that
1 24
0 01
t
xu dxds C
. (48)
Proof: Set 0
( , ) ( , )x
U x t u y t dy . (49)
Under the boundary condition
(0, ) (1, ) 0t t , (50)
We can get by integrating (1)2 over (0, x) and by using (50) that
0
00
1
00
( , ) ,
( ,0) ( ) ,
(0, ) 0,
(1, ) ( ) .
t y
t xx
x
U U p x t dyv v
U x u y dy
U t
U t u x dx
(51)
Hence the standard pL -estimates for solutions to the linear problem (51), cf. [16], yields
2
1 1 14 4 4
0 ( )0 0 0 0 0 0
t t t
xx L IU dxds C C u C p dxds C C dxds . (52)
Thus by (31), we get (48) and the proof of Lemma 10 is complete.
On the other hand, noticing that
1 1
2 2
0 0( , ) ( , ) 2 ( , ) ( , )x x x xxu y t u x t dx u x t u x t dx , (53)
We have from (31) and Holder’s inequality that
Lemma 11 Under the conditions listed in Lemma 4, we have for 0 t T that
1
12 22
( )0 0 0( ) ( , )
t t
x xxL Iu s ds C C u x s dxds . (54)
Lemma 12 Under the conditions listed in Lemma 4, we have for 0 t T that
84
1 max{2 ,1, 1}2
0 0
t r c
xxu dxds C C
. (55)
Proof: By differentiating (1)2 with respect to x and multiplying the resulting equation by
0
x
Ru
, we have
2
0 0 0 0 02
x x x xx x x
xt t x x
u R u vR v Ru u
v
. (56)
Integrating (56) with respect to x and t over [0, ]I t , one has
2
1 1 1 1
0 0 0 0 0 0 00 0 0 0
1 .2
t t tx x
x t x x
x
u R u R v Rdx C u dxds dxds v u dxds
(57)
Since by (8), (32), (39), (41), (44), (54) and (55), we have
1
0 00
1 123
0 0 0 00
1 12 2 23
0 0 0 00
1 1 12 2 2 2 20
2 0 0 0 0 0 03
1 220 3
2 ( )0 03
2
( )4
( ) (4
t
x
x
t t
x x x
t t
x x x
t t t
xx x x x
t
xx x L I
vdxds
Vdxds C v dxds
Vdxds C v dxds
u dxds C u v dxds C dxdsV
Vu dxds C u s
V
2
212 2 1
1( ) ( )0 0 0
11 max{2 , ,1}20 3 22 0 0
3
) ( )
,8
t tr xx rL I L I
t r r
xx
s v s C dxds
Vu dxds C
V
(58)
and
85
1
0 00
2
0
00
1 12 2 2 3 20 3
2 0 0 0 03
1 12
1 1 1 12 22 4 2 20 3
2 0 0 0 0 0 03
( , )16
( , )
16
t
x t
tx x x
x
xv
t t
xx x x x
t t t c r xxx x x
Ru dxds
u R uRu dxds
C v v v
Vu dxds C v u u dxds
V
V vu dxds C u dxds u dxds C
V
1
10 0
1 max{1, 1}20 3
2 0 03
.16
t
r
t c
xx
dxds
Vu dxds C
V
(59)
The above two estimates together with (13), (57) and Cauchy’s inequality, we get (55).
This completes the proof of the lemma 12.
Having obtained (35), (41), (54), and (55), we can obtain the upper bound on ( , )x t if the
parameter c is chosen such that c<1. Here we have used the fact that r>0 can be chosen
as small as wanted.
Now we consider the case when the transport coefficients and satisfy (5) with
10
5a and 2b . For such a case, (24) should be replaced by
1
0 00 0 0 0
( ( , )) ( , ) ( ( ) (z, )) ( ( ))t x t
zg v x t p x t ds u z u t dz g v x dzdsv
(60)
with
1, 0,
( )
ln , 0.
ava
g v a
v a
With (58) in hand, we can deduce by repeating the argument used in the proof of Lemma
4, especially the way to deduce (23)-(25), that there exist some positive constants 3 0V
and 3 0 such that
3 3( , ) , ( , )v x t V x t
86
hold for all ( , ) [0, ]x t I T . But since the boundary condition (4) does not yield any pL -
estimates on v , we can deduce from the fact ln v v
for any 0 that
2 2
1
0 0
bt
x dxds C C vv
. (61)
To deduce as upper bound on ( , )v x t , we try to recover the 1L -estimates on ( , )v x t , which
plays an important role in deriving the upper bound on ( , )v x t for the case when the
transport coefficients and satisfy (7). To do so, integrating (1)1 with respect to x
and t over [0, ]I t , we get
1 1 1
00 0 0 0
11
2 21 1 22
10 0 0 0
21 1
10 0 0 0
.
t
x
a t tx
a
t ta x
a
vdx v dx u dxds
uC C v dxds vdxds
v
uC C v dxds vdxds
v
(62)
Then by the Gronwall inequality, we can easily deduce that
2
1 1
10 0 0
ta x
a
uvdx C C v dxds
v . (64)
Since 2b , we have
2
(I) ( )0 0
1 12
0 0
11
2 2 21 12
0 0 0 0
12 2
210
( ) ( )
,
bt t
L L I
bt
x
bt t
x
a tx
a
s ds C C s ds
C C dxds
C C vdxds dxdsv
uC C v dxds
v
(64)
which implies that
87
12 21 1
22
10 0 0 0
at tx
a
udxds C C v dxds
v
. (65)
Thus with the help of (33), we have
12 2 21 1
21 10 0 0 0
at tx x
a a
u udxds C C v dxds
v v
, (66)
Then by Cauchy’s inequality, we can easily obtain the following results
Lemma 13 Under the conditions listed in Lemma 3.4, we have for 0 t T that
2
1
10 0
t ax
a
udxds C C v
v
, (67)
1 2
0
avdx C C v
, (68)
( )0
( )t a
L Is ds C C v
, (69)
and
1
2
0 0
t adxds C C v
. (70)
To estimate 2 (I)( )x L
v t , we have by integrating (46) with respect to x and t over [0, ]I t
and with the help of (13) and Cauchy’s inequality that
2 21 1
2(1 ) 30 0 0
22 22 21 1
1 1 1 10 0 0 0
21
0 0
.
tx x
a a
t tx x x
a a a a
ta x
v vdx dxds
v v
u uC C dxds C dxds
v v v v v
C C v dxdsv
(71)
To control2
1
0 0
tx dxds
v
, we have by multiplying (1)3 byb , and integrating the resulting
identity over [0, ]I t that
88
2 2
1 1 12
10 0 0 0 0 0
at t txx x
a b
uudxds dxds C C dxds C C
v v v
, (72)
and the above estimate together with (71) imply
2
1
2(1 )0 0
t ax
a
vdxds C C v
v
. (73)
Since
1 1
0 0
11
21 21 122
2(1 )0 0
1 5
2 2
( , ) ( , )
,
x
a a x
a
a
v y t v x t dx v dx
vC C v C v vdx
v
C C v
(74)
for which and the assumption 1
05
a , we can deduce that
3( , ) , ( , ) [0, ]v x t V x t I T (75)
holds for some positive constant 3V which depends only on T and the initial data
0 0 0 0( ( , ), ( , ), ( , ), ( , ))v x t u x t x t x t . As a by-produce of the estimate (75), we can deduce
that the terms on the right-hand side of the inequalities in Lemma 13 and (73) can all be
bounded by some constant C depending only on T and the initial data
0 0 0 0( ( , ), ( , ), ( , ), ( , ))v x t u x t x t x t .
Now we turn to derive the upper bound on ( , )x t . For this purpose, we have by
multiplying (1)3 by
for some (0,1) and integrating the resulting identity over
[0, ]I t that
2 1 2
1 1
10 0 0 0
bt t
x x
a
udxds dxds C
v v
. (76)
Then by (52), we have
89
1 14 4
0 0 0 0
3
( )0
21
0 0
1max{2 ,0} 1 2
0 0
max{2 ,0}
( )
.
t t
x
t
L I
t
x
tb b
x
b
u dxds C C dxds
C C s ds
C C dx ds
C C dxds
C C
(77)
Now we set
1
2
0 0:
tb
tX dxds , 1
2 2
0: max b
xt
Y dx , 1
2
0: max xx
tZ u dx . (78)
Observe that
12 2 2 1
0
1 11
1 12 22 22
( ) 0 0
11
22
( )
,
b b
x
bb
xL I
b
L I
C C dx
C C dx dx
C C Y
(79)
which implies
1
2 3
( )
b
L IC CY
. (80)
Combining (53), the inequality
1 1
1 1 1 12 22 2 2 2
0 0 0 0x xxu dx C u dx C u dx u dx , (81)
and by (13), we have
11
2 2
0max x
tu dx C CZ , (82)
and
3
8
( )x L Iu C CZ . (83)
90
Our next result is to show that X and Y can be controlled by Z.
Lemma 14 Under the conditions listed in Lemma 3.4, we have for 0 t T that
3
4X Y C CZ . (84)
Proof: Multiplying (1)3 by b
t , and integrating the resulting identity over [0, ]I t , one
has
1
1 2 2 2
0 0
tb b b
x t x t x xX Y C C u u u dxds . (85)
Since by Cauchy’s inequality and (75), (76), (77), we can get from (80) and (83) that
2
1 121 2 2 3
0 0 0 04 4
bt tbb b
x t x
X Xu dxds C u dxds CY
, (86)
max{ ,2 }
1 12 4 2 3
0 0 0 04 4
bt tbb b
x t x
X Xu dxds C u dxds CY
, (87)
and
1 31 112 2 1 2 2 3 8
0 0 0 01
bt tbb b b
x x x xu dxds u dxds CY Z
. (88)
Based on the above three estimates and (85) and by employing the Cauchy inequality, we
can get (84) immediately if we choose1
(0, )2
. This completes the proof of Lemma 14.
Our last result in this section is to show that Z can be bounded by X and Y.
Lemma 15 Under the conditions listed in Lemma 4, we have for 0 t T that
2 3
2 3 4bZ C CY CX CZ
. (89)
Proof: Using (1) 2, we can easily get the following identity
1
2
(1 )a ax xxx t x xa
a v uu v u p v
v
. (90)
Integrating (90) with respect to x and t over [0, ]I t yields
91
1 12 2 2 2 2 2 2
0 0 0 0
1 21 1
2
0 0 0 0
12 2 2
0 0
2 31
2 2 3 4
0 0
.
t t
xx t x x x x
bt t
xt
t
x x
tb
t
u C u v v u dxds
C u dxds C dxdsv
C u v dxds
C u dxds CY CZ
(91)
Next we need to estimate 1
2
0 0
t
tu dxds to complete the proof of this lemma. To this end,
we have by differentiating (1)2 with respect to t and multiplying the resulting identity by
tu that
2 2 2
1 2 2 2
(1 ).
2
t xt x xt t xt x xt xt t x t tt ta ax
t
u u a u u R u R u u u v uu
v v v v v v
(92)
Integrating (92) with respect to x and t over [0, ]I t and with the help of Cauchy’s
inequality, one has
21 1 1
2 2 4 2 2 2
0 0 0 0 0
2
2 3 .
t tx
t xt x t x
b
u dx u dxds C C u u dxdsv
C CY CX
(93)
(93)together with (91) implies (89) and the proof of Lemma 15 .
Combining (84) and (89), we can obtain Y C , then we derive the upper bounds on
( , )x t from (80).
In summary, we have obtained the desired lower and upper bounds on v and provide
that the transport coefficients and satisfy the conditions listed in Theorem 1 and
then Theorem 1 can be proved by employing the continuation argument.
4. Discussion
Compressible Navier-Stokes type equation with density and temperature dependent
transport coefficients arise in many applied sciences, such as certain class of solid-like
92
materials, gases at very high temperatures, etc. Such a dependence of and on v and
will obviously influence the solutions of the field equations as well as the mathematical
analysis and to establish the corresponding well-posedness theory has been the subject of
many recent researches. These studies indicate that temperature dependence of the
viscosity in is especially challenging but one can incorporate various forms of density
dependence in and also temperature dependence in . We note, however, that in all
the previous studies although the viscosity coefficient may depend on v and the heat
conductivity may depend on both v and , they ask that at least one of and is
non-degenerate. What we are interested in this paper focuses on the case when is a
function of v and depend on v and/or and both and are degenerate. We hope
the study here can shed some light on the construction of global classical solutions to the
fluid model derived from the Vlasov-Possion-Boltzmann system.
5. Conclusions
In this paper, we concern with the construction of global solutions with large amplitude
to the initial-boundary value problem of the one-dimensional compressible NSP system
with degenerate transport coefficients. The results can not only provide the global
existence of solutions, but also reduce the gap between the physical models and the
satisfactory existence theory. It implies that the viscous fluid under the influence of the
self-induced electric force will doesn’t affect construction of global, non-vacuum
solutions.
For the case when ( )v is a smooth function of v satisfying ( ) 0v for 0v and
( ) b , if the specific volume v is bounded both from below and from above and the
absolute temperature is bounded from below, i.e., there exist some positive constants
3 30, 0,V V and 3 0 such that
3 3( , ) ,V v x t V 3( , ) 0x t
93
hold for ( , ) [0, ]x t I T , then the argument used above can be employed to derive the
upper hound on ( , )x t provide that 0b .
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