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: lwang@xmut.edu.cn

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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