Pointwise Estimates for Solutions to a
System of Radiating Gas
Shikuan Mao, Yongqin Liu
School of Mathematics and Physics, North China Electric Power University, Beijing, China Email: [email protected], [email protected]
Received August 22, 2013; revised September 8, 2013; accepted September 13, 2013
Copyright © 2013 Shikuan Mao, Yongqin Liu. 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.
ABSTRACT
In this paper we focus on the initial value problem of a hyperbolic-elliptic coupled system in multi-dimensional space of a radiating gas. By using the method of Green function combined with Fourier analysis, we obtain the pointwise decay estimates of solutions to the problem.
Keywords: Radiating Gas; Initial Value Problem; Pointwise Estimates
1. Introduction
In this paper we consider the initial value problem
2
0
div 0, , 0,
div 0, , 0,
,0 , ,
n t
n
n
u a u q x t
q q u x t
u x u x x
(1.1)
here is a constant vector, and are unknown functions of
n
a
1, , nq q
, ,
x x
,u u x t
0. t
,q x t
n
x
1 n n 1 and Typically,
represent the velocity and radiating heat flux of the gas respectively.
, u q
The system (1.1) is a simplified version of the model for the motion of radiating gas in n-dimensional space. More precisely, in a certain physical situation, the system (1.1) gives a good approximation to the following system describing the motion of radiating gas, which is a quite general model for compressible gas dynamics where heat radiative transfer phenomena are taken into account,
2 2
4
1 2
div 0,
div 0,
div 0,
2 2
div 0,
t
t
t
u
u u u pI
u u
e u e pu
q a q a
q (1.2)
where ρ, u, p, e and θ are respectively the mass density, velocity, pressure, internal energy and absolute tempera-ture of the gas, while q is the radiative heat flux, and a1
and a2 are given positive constants depending on the gas itself. The first three equations are motivated by the usual Euler system, which describe the in-viscid flow of a compressible fluid and express conservation of mass, momentum and energy respectively. We refer to the book of Courant and Friedrichs [1] for a detailed derivation of several models in compressible gas-dynamics. The physical motivation of the fourth equation, which takes into account of heat radiation phenomena, is given in [2]. Moreover, the simplified model (1.1) was first recovered by Hamer (see [3]), and for the reduction of system (1.2) to system (1.1), see [2-4].
Concerning the investigation on the hyperbolic-elliptic coupled system in one-dimensional radiating gas, we refer to [5,6]. In the case of the muti-dimensional case, Francesco in [7] obtained the global well-posedness of the system (1.1) and analyzed the relaxation limits. Re-cently, in [8], Liu and Kawashima investigated the decay rate to diffusion wave for the initial value problem (1.1) in n(n≥ 1)-dimensional space by using a time-weighted energy method.
The rest of the paper is arranged as follows. Section 2 gives the full statement of our main theorem. In Section 3, we give estimates on the Green function by Fourier analysis which will be used in Section 5. Section 4 gives the global existence of solutions to the problem (2.3). In Section 5, we obtain the pointwise decay estimates of solutions.
2
· 1
ˆ : e
2 n n
ix d , f f
f x x
and we de- note its inverse transform as 1.
1p p n
L L p is the usual Lebesgue space with the norm ·Lp. Let be a nonnegative integer,
s
then HsHs
n denotes the Sobolev space offunctions, equipped with the norm
2
L
1 2
2 2
0
: .
s
s k x
H L
k
f f
For k0,
1 1
1
: n; 0 1 ,
n
n k
k k
x x x i
i
k i n k
k s n
.
i k
For non- negative integer , denotes the space of -times continuously differentiable functions on the interval
k C I H
;
k
I with values in the Sobolev space
n .s s H H
Finally, in this paper, we denote every positive con-stant by the same symbol or c without confusion. [·] is the Gauss’s symbol.
C
2. Main Theorems and Proof
For simplicity, without loss of generality, we choose 1, ,12 2
n
a
in (1.1). That is, we will con-
sider the following initial value problem:
1
0
div 0, , 0,
div 0, , 0,
,0 , .
i
n
n
t x
i
n
n
u uu q x t
q q u x t
u x u x x
(2.3)
Our Main results are the following:
Theorem 2.1. Let
3, 1,2 2, 2,
n s
n n
be an
inte-ger.
Assume that
1
and put0 s n n ,
u H L
1
0: 0 Hs 0 L . Then there is a small positive con-stant such that if 0 0
E u u
0 1
E ,
u xthen the problem (2.3) has a unique global solution ,t
with
0, ; s n
, 2
0,
; s1
n
,u C H u L H
0, ; s1 n
2
0,
; s1
n .q C H L H
Moreover, if s
n 2 6, and for any multi-indexes with s
n 2 1, there exists some constant2 n
r such that 0
0
1 2
rx
D u x CE x , then for
any s
n 2 4, the solution to Equation (2.3)has the following decay estimate,
, 0
1
2 1 2 . We also have1
r n
x
x
D u x t CE t
t
the following corollary by using Theorem 2.1.
Corollary 2.2. Under the same assumptions in
Theo-rem 2.1, the solution satisfies the following decay esti-mates:
4 22 0 1 ,
n k
k
xu t L CE t
with satisfying k
0 4
2
n k s ;
2
1 4 2
0 1 ,
n k k
xq t L CE t
with k satisfying 0 5 2
n k s .
Remark. In Theorem 2.1, we do not need to assume
that 1
0 ,
n
u L if The results in Corollary 2.2 is similar to those in [7].
2.
n
3. The Global Existence of Solution
This section is devoted to prove the global existence re-sult stated in Theorem 2.1. In [7], the global existence of solutions to the problem (2.3) is obtained, but for the completeness of this paper, here we give the sketch of the proof.
Since a local existence result can be obtained by the standard method based on the successive approximation sequence, we omit its details and only derive the desired a priori estimates of solutions.
From
2.31
2.31div 2.3 ,
2 we get that1 1
0.
i i
n n
t t x x
i i
u u u uu uu
(3.4)Now we make energy estimates by using (3.4) under the following a priori estimate:
1,xu t L
(3.5) here 1 1 is a given constant.
Multiplying (3.4) by and integrating with respect to
u
x, by integration by parts we have that
1
2
2 2
d .
d x 2
2
H L L
u t u t C u t u t
t L (3.6)
Multiplying l
3.4x
by and integrating with respect to
1
l xu l
x, by integration by parts we have that
1
1
2 2
2 d
d
, 1
l l
x H x L
l
x L x H
u t u t
t
C u t u t l
2
.
1
(3.7)
1 2
1 2
2 2
2
d d
.
s s
s
x H x H
x L x H
u t u t
t
C u t u t
(3.8)
Combining (3.6) with (3.8) we have that
1
1
2 2
2 d
d
.
s
s
x
H H
x L x H
u t u t
t
C u t u t
s
(3.9)
In view of (3.5), (3.9) yields that
12 2 2
0
0 d s.
s s
t
x H
H H
u t
u C u (3.10)From 2 we have that thus (3.10) yields that
2.3 q
1
1u,
1
12 2 2
0
0 d s.
s s
t
x H
H H
q t
q C u (3.11)By the continuity argument, we have the following result.
Theorem 3.1. Let 2
2
n s
be an integer. Assume
that u0Hs
n , and put E0: u0 Hs. Then there isa small positive constant 0
1 such that if E00, then the problem (2.3) has a unique global solutionwith
u x,t
0, ; s n
, 2
0,
; s1
n
, u C H u L H
0, ; s1 n
2
0,
; s1
n
. q C H L H 4. Estimates on Green Function
In order to study the problem (2.3), we start with the Green function (or the fundamental solution) to the linear problem corresponding to the Equation (3.4), which sat-isfies
0, , 0,
, 0 , .
n
t t
n
G G G x t
G x x x
(4.12)
By Fourier transform we get that,
ˆ ˆ ˆ 0, , 0,
ˆ ,0 1, .
n t t
n
G G G t
G
(4.13)
By direct calculation we have that
22 1
ˆ , e
t
G t
.
L e t 1
3
1, , 1, ,
0, 2 , 0, 1,
R R
b e
the smooth cut-off functions, where and are any fixed positive numbers satisfying
R
0 1 R. Set
2 1 1 3 ,
and
ˆ , ˆ ,
Gi t i G t , i1, 2,3. We are going to study G x ti
, , which is the inverse Fourier transform of
ˆ , .
i
G t
First we give a lemma which is important for us to make estimates on the low frequency part.
ˆ , C2N
n
f tLemma 4.1. If has compact
sup-port in the variables , is a positive integer, and there exists a constant such that
N 0,
b fˆ
,t satis-fies
2| | 2
2 2
,
1 1
1 e ,
k
k k
b t
t
C t t t
2
ˆ
m
D f
t
for any multi-indexes , with 2N, then
, 2
,n k
x N N
D f x t C t B x t
, where k and are any fixed integers, m
a x 0,
a , and
2
, 1
1 .
N
N
x
B x t
t
ma
Proof. If k , by direct calculation we have that
2
1
| | | | 2 | |
2 2
2 2
2
ˆ
, e , d
1 1
1 e d 1 .
n n
x
k
k k
n k m
b t
x D f x t C D f t
C t t
t C t t
t
If k , we also have
2
1
2
2 2
2 2
2
ˆ
, e ,
1 1 1
1 e d 1
n n
x
k k
n k m
b t
x D f x t C D f t
C t t t
t C t
d
. t
Let 0 when x2 1 t, and 2N when 2
1 ,
x t we obtain from the above estimates that
22 1
, min 1,
N n k
t D f x t Ct
x
.
Since
2 2
2 2
1, 1 ,
1 2
1 , 1
1 ,
x t
x
x
t x t
t
large such that m 1 12
22
1 2
min 1, 2 , .
1 1
N
N
N N N
t B x t
x x
t
.
Since
2
suppG ; R , we have that
Thus we complete the proof.
By using Lemma 4.1 we can get the following propo-sition about the estimates on G x t1
, .Proposition 4.2. For sufficiently small , there exists
constant C0 such that
2
1 , ,
n
x N N
D G x t C t B x t
.
Proof. For being sufficiently small, by noticing that Gˆ ,t is a smooth function of near 0 and using Taylor expansion we have that
2
2 4
2 1 ,
ˆ t e t e t O t
G
. It yields that
22
2 2
2 2
ˆ ,
1 1
1 e .
t
D G t
C t t
t
t
Since
1 2
1 2 1
1 1 2
,
! ˆ
, ,
! !
D G t
D D G t
and
1
, we have that
21
| | 2
2 2
2 2
,
1 1
1 e .
t
D G t
C t t
t
t
From Lemma 4.1 we obtain that
2
1 , ,
n
x N N
D G x t C t B x t
.
Thus we complete the proof of Proposition 4.2.
As for G x t2
, we have the following estimates.Proposition 4.3. For fixed and , there exist
positive numbers and such that
R
m C
2
2 , e ,
t m
x N
D G x t C B x t
2
ˆ , e mt
G t C . (4.14)
It yields that 2
, e t m xD G x t C . (4.15) Now we shall give an estimate on x G x t 2
, by induction on . Assume that, if l 1, then
2
ˆ , 1 e mt
D G t C t
, (4.16) which is true as 0 by (4.14)
By using (4.13), we have the following problem for
2 2
2 2
0
1,
ˆ , ˆ , , ˆ
1 1
ˆ , 0.
t
D G t D G t D G t
D G t
, ,
(4.17) By multiplying (4.17), whose variables are now changed to
,s by Gˆ
,t s
and integrating over the region s
0,t , we have that
0
22
ˆ , ˆ , , ˆ ,
1
t
D G t G t s D G s s
d .In view of (4.16) for l 1, it yields that
1
0
ˆ , te t sm 1 e dms 1 t
R
D G t C t s C t
e ,m
which shows that (4.16) is valid as l. This implies that, for 1 l,
2
· 2
{ }
,
ˆ , e d
e 1 1
1 e .
n
h x t
ix
t m
R
t m
x D G x t
C D G t
C t
C t
d
(4.18)
By using (4.15) when x2 1 t and (4.18) with
2N
when x2 1 t, as well as the fact that 2
2
2 2
2, 1 ,
1 2
1 , 1
1 ,
x t
x
x
t x t
t
we get that
.
2
2 , e ,
t m
x N
D G x t C B x t .
Thus we complete the proof of Proposition 4.3.
Next we will come to consider First we give a lemmas which is useful in dealing with the high fre-quency part.
2 ,
G x t
Lemma 4.4. If supp fˆ
OR:
; R
, and
ˆf satisfies
1ˆ , ˆ ,
f C D f C
1,
,
then there exist distributions f x f x1
, 2 and con-stant C0 such that f x
f1
x f x2
C0
x , where
x is the Dirac function. Furthermore, for positive integer 2N n ,
2
1 1 ,
N x
D f x C x
1
2 L ,supp2 ; 2 ,
f C f x x x 0
with 0 being sufficiently small.
The proof of Lemma 4.4 can be seen in [9]. Choose sufficiently large R such that
2
2
1, 2 1
if R 1. By Taylor expansion, we
have that
2 41 1
1 ( )
3 3
ˆ , e O t
G t
.
It is obvious that
2 3ˆ , e t
G t C . Since
2 41 2
1 2
1 1
1
3 3
ˆ , e O t
D G t C D D
,by direct calculation, we have that for 1,
2 13
ˆ , e t
D G t C
.
By using Lemma 4.4 we have the following result.
Proposition 4.5. For being sufficiently large,
there exist distributions and con-stant such that
R
31
G
x, ,t G32
x t, , 0C
2
3 , e 31 , 32 , 0
t
G x t G x t G x t C x ,
where
x is the Dirac function. Furthermore, for positive integer 2N n , the following estimates hold:
2
31 , 1 ,
N
x
D G x t C x and
1
32 ., L ,supp 32 ., ; 2 0 ,
G t C G t x x
here 0 is sufficiently small.
Combining Proposition 4.2, Proposition 4.3, and Proposition 4.5, we have the following theorem on the Green function.
Theorem 4.6. For any multi-indexes , there exists a
distribution
2
32 0
, e ,
t
,
K x t G x t C x such
that the following estimate holds:
,
1
2
,n
x N
D G K x t C t B x t
,
here,
2
n
N is an arbitrary positive integer.
5. Pointwise Estimates
In this section, we focus on the pointwise estimates of solutions to the problem (2.3).
By Duhamel principle, the solution to the Equation (3.4) with initial datum can be ex-pressed as following,
0
,0
u x u x
0 0
1
, ., 1
: , , .
i
n t
x n
u x t G t u x G t uu
u x t u x t
d
Now we give a lemma which will be used in the fol-lowing analysis.
Lemma 5.1. When n n1, 2n2, and
3 min 1, ,
n n n2 we have that
1 3
2
2 2
2
1 1 d 1
1 1
n
n n
n
x y x
y y C
t t
.
The proof of Lemma 5.1 can be seen in [9]. Since u x t
, G K x t
, K x t
, :I1I2, by using Lemma 5.1 and Theorem 4.6 with , we have thatNr
2
1 0 1 , .
n
x r
D I CE t B x t
Noticing that if x y 2 ,0 then
2
1
11 y C 1 x2
, we have that
22 32 0 0
2 0
e ,
e , .
n
t
x x
t
r
D I G x y t C x y D u y y
CE B x t
dThus we obtain that
2 0
, 1
, , 1.
2
n x
r
D u x t CE t n
B x t s
(5.19)
Next we come to make estimates on To this end, we will use the following lemma.
, .
u x t
Lemma 5.2. Assume , then the following
ine-qualities hold,
1
n
2 1 2
1 2 1
1 1 1
n n n
n
A A
t t
Denote
1 2
, 0, , 4 2
, : 1 , ,
: sup , ,
n
n r
x
x T
n s
x t t B x t
M T D u x x
.
;
2t, then 1 2 1 2 1
n
n A
t
.
A
2) If
Now we come to make estimates to u x t
, by using Theorem 4.6 with Nr and Lemma 5.2. We decompose D u x tx
, as following,
2 2
0 ; 2 1
2 2
0 ; 2 1
2 ; 2
1 2
2 ; 2
1 2
0
,
1 , ,
1 , , d
1 1 ,
2 1
1 ,
2
i i
i i n
x
t n
x x y x y
i
t n
x x y x y
i n t
t y x y x y
i n t
t y x y x y
i
t
R
D u x t
D G K x y t u y y
D G K x y t u y y
G K x y t D u y y
G K x y t D u y y
K
d d
d
, d d
, d d
1
31 32 41 42 5
, 1 , d d
: .
i
n
y x
i
x y t D uu y y
I I I I I
Next we estimate I ii
1, 2,3, 4,5
respectively by using Theorem 4.6. By using Lemma 5.2 (1), we have that
1 2 2
2
31 0 ; 2
1 2
2 2
0 ; 2
| |
2 2
1 , 1
1 , 1 , d d
, 1
, d d
n
n
N r
y x y
t n
n
N y x y r
n r
t
CM T t B x y t B y y
CM T t B x t B y y
CM T B x t t
I
Now we estimate I32. in two cases.
Case 1. x2t. By using Lemma 5.2 (1), we have that
1 2 2
2
32 0 ; 2
1 2
2 2
2
0 ; 2
2
2
1 , 1
1
1 , 1 ,
1
, 1 .
t n
n
N r
y x y
n t
n
n
N r
y x y n r
I CM T t B x y t B y y
CM T t B x y t B x t y
t
CM T B x t t
, d d
d d
Case 2. x2t. By using Lemma 5.2 (2), we have that
1 2 2
2
32 0 ; 2
1 2
2 2
0
2
2 1
, 1 , d d
1 1
, d , 1 .
t n
y x y n
N r
t n
n
n
r r
I CM T t
B x y t B y y
CM T t
B y CM T B x t t
2
2
32 , 1
n r
I CM T B x t t .
As for I41, we also need to divide it into two cases. Case 1. x2t. By using Lemma 5.2 (1), we have that
1 2 241 ; 2
2 2 2 1 2 2 2 2 ; 2 2 2 2 2 1 ,
1 , d d
1 (1 )
1
, , d d
1
, 1 .
n t
t y x y N
n r
n n t
t y x y n
N r
n r
I CM T t B x y t
B y y
CM T t t
t
B x t B y t y
t CM T B x t t
Case 2. x2t. By using Lemma 5.2 (2), we have that
1 2 241 ; 2
2 2 2 1 2 2 2 2 ; 2 2 2 2 1
1 , d d
1 (1 )
, d d
, 1 .
n t
t y x y N
n r
n n t
t y x y
N
n r
I CM T t B x y t
B y y
CM T t t
B x y t y CM T B x t t
,
Combining the two cases, we have that
2
2
41 , 1
n
r
I CM T B x t t .
As for I42, by direct calculation, we have that
1 2 242 ; 2
2 2 2 2 2 2 1 2 ; 2 2 2 2 1 ,
1 , d d
1 ,
(1 ) ,
1 .
d d
,
n t
t y x y N
n r n r n t
t y x y N
n
r
I CM T t B x y t
B y y
CM T B x t
t B x y t y
CM T B x t t
To estimate I5, we will use the following result, which is obtained in [7].
Lemma 5.3. 1) If
3, 1, 2, 2, 2 n s n n and
10 ,
s n n
u H L put E0: u0 Hs u0 L1, then
the following estimate holds:
4 20 1 ,
s k
n k k
xu t H CE t
with 0 k s 1.
2) If 2, 2 n s
and 0
, 2 puts n
u H n , 0: 0 Hs,
E u then the following estimate holds:
20 1
s k
k k
xu t H CE t
, with 0 k s.
We estimate I5 as following,
2
5 0 32
1
2 0
1
e 1
e 1 , d .
i i t n t x x i t n t x x i
I C G t D uu
C D uu x
d
Notice that
2 consists of terms of 1 i n x x i D
u1 2
1 2 1 2
, , 0, 3.
k k
xu xu k k k k
Without loss of
generality, we assume that k1k2, then
1 max , 2 .
k Since 6
2 n s
and 4,
2 n s
we have that
1
1 , 1 2
n k k
xu y M T B y, .
By using Gagliardo-Nirenberg inequality and Lemma 5.3, we have that
22 2
0 1 , 2
2
k k
x L
n
u CE k s
1.
Thus we have that
1 21 2 2
0
, , 1 n k k .
k k
xu y x u y CE M T
Combined with Proposition 4.5 and the fact that
1 1
2 2
1 1 ,
1 1 y x C t t
if x y 2 ,1 it yields that
1
2 2
5 0 0
2 0
e 1 d
, 1 .
t n
t
n r
I CE M T
CE M T B x t t
Combining I31,I32,I41,I42 and I5, we have that
2
10 ,
, ,
x h
D u x t C E M T M T x t
.(5.20)Proof of Theorem 2.1. In view of (5.19) and (5.20),
we get that
2
10 0 ,
, ,
x h
D u x t C E E M T M T x t .
20 0 .
M T C E E M T M T
Thus if E0 is suitably small, we obtain M T
CE0 by the continuous dependence on the initial data. In view of Theorem 3.1, the proof of Theorem 2.1 is completed.
6. Acknowledgements
The first author is partially supported by the National Natural Science Foundation of China (Grant No. 11201142) and by the Fundamental Research Funds for the Central Universities (Grant No. 11QL40). The second author is partially supported by the National Natural Science Foundation of China (Grant No. 11201144).
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