Asymptotic stability of viscous contact wave for the 1D radiation hydrodynamics system

J. Differential Equations 251 (2011) 1030–1055

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Asymptotic stability of viscous contact wave for the 1D radiation hydrodynamics system Jing Wang a , Feng Xie b,∗ a b

Department of Mathematics, Shanghai Normal University, Shanghai 200234, PR China Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, PR China

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 31 August 2010 Revised 17 March 2011 Available online 30 March 2011 Keywords: Compressible radiation hydrodynamics Asymptotic stability Contact discontinuity wave Energy estimates

This paper is concerned with the large time behavior of solutions to a radiating gas model, which is represented mathematically as a Cauchy problem for a one-dimensional hyperbolic–elliptic coupled system, with suitably given far field states. Suppose the corresponding Riemann problem for the Euler system admits a contact discontinuity wave, then we can construct a “viscous contact wave” for such a hyperbolic–elliptic system. Based on the energy methods and the ellipticity of the radiation flux equation, we prove that the “viscous contact wave” is asymptotically stable provided that the strength of contact discontinuity and the perturbation of the initial data are suitably small. © 2011 Elsevier Inc. All rights reserved.

1. Introduction The fundamental system of equations in Lagrangian coordinates, describing the one-dimensional motion of a radiating gas, can be written in the following form (see [15,16,18,20,24]):

⎧ v t − u x = 0, ⎪ ⎪ ⎪ ⎪ ⎪ u t + P x = 0, ⎪ ⎪ ⎪  ⎨ u2 e+ + ( P u ) x + q x = 0, ⎪ 2 t ⎪ ⎪ ⎪   ⎪ ⎪  qx ⎪ ⎪ + avq + b θ 4 x = 0, − ⎩ v

*

x

Corresponding author. E-mail addresses: [email protected] (J. Wang), [email protected] (F. Xie).

0022-0396/$ – see front matter doi:10.1016/j.jde.2011.03.011

© 2011

Elsevier Inc. All rights reserved.

(1.1)

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

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where x ∈ R, t > 0, and a, b are two positive constants related to the absorption coefficient and the Stefan–Boltzmann constant. The unknown is ( v , u , θ, e , P , q) with v > 0 the specific volume, u the velocity, θ > 0 the absolute temperature, e > 0 the internal energy, P the pressure and q the radiative flux. Here we focus on the ideal polytropic gas, that is, the state equations take the form

P=R

θ v

= A v −γ e

γ −1 R

s

e=

,

R

γ −1

θ,

where s is the entropy, γ > 1 is the adiabatic exponent and A, R are two positive constants. We impose the following initial data and far field states



( v , u , θ)(x, 0) = ( v 0 , u 0 , θ0 ),

x ∈ R,

( v , u , θ, q)(±∞, t ) = ( v ± , u ± , θ± , 0), t > 0.

(1.2)

We are concerned with the global existence of solutions of the initial value problem (1.1)–(1.2) and their large time behavior in relation with the spatial asymptotic states ( v ± , u ± , θ± ). If the radiation effect is neglected in the fluids, the system (1.1) is reduced to the following compressible Euler system

⎧ v t − u x = 0, ⎪ ⎪ ⎪ ⎨ u + P = 0, t x  2 ⎪ u ⎪ ⎪ + ( P u ) x = 0, ⎩ e+ 2

(1.3)

t

with the asymptotic states ( v ± , u ± , θ± ) being the Riemann initial data

( v , u , θ)(x, 0) =

( v − , u − , θ− ), x < 0, ( v + , u + , θ+ ), x > 0.

(1.4)

It is known that the contact discontinuity solution of the Riemann problem (1.3)–(1.4) takes the form [21,22]

¯ x, t ) = ( V¯ , U¯ , Θ)(



( v − , u − , θ− ), x < 0, t > 0, ( v + , u + , θ+ ), x > 0, t > 0,

(1.5)

provided that

u− = u+ ,

P− 

R θ− v−

=

R θ+ v+

 P +.

(1.6)

In the setting of the hyperbolic–elliptic coupled system (1.1), the corresponding wave to the contact discontinuity becomes smooth and behaves like a diffusion wave due to the radiation effect. We call this wave “viscous contact wave”. Now, we construct the “viscous contact wave” ( V , U , Θ, Q ) in this paper as follows. Due to (1.6), the pressure of “viscous contact wave” ( V , U , Θ, Q ) is expected to be almost constant, that is

P≡

RΘ V

≈ P +,

(1.7)

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J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

and the radiation flux Q is expected to act as a diffusion term

Q ≈−

b  aV

Θ 4 x.

(1.8)

Then the leading order of the energy Eq. (1.1)3 is



R

γ −1

Θt + P + U x =

4b Θ 3 Θx a



V

(1.9)

. x

Then using (1.1)1 and (1.7) we have the following nonlinear diffusion equation



Θt =

γ − 1 4b P + 2 Θ Θx Rγ aR

 ,

Θ(±∞, t ) = θ± ,

(1.10)

x

which has a unique C ∞ self-similar solution Θ(x, t ) = Θ(ξ ), ξ = √ x

1+t

due to [1,4,23]. Furthermore,

Θ(ξ ) is a monotone function, increasing if θ+ > θ− and decreasing if θ+ < θ− . On the other hand, there exists some positive constant δ¯ , such that for δ = |θ+ − θ− |  δ¯ , Θ(ξ ) satisfies

  2 (1 + t )l/2 ∂xl Θ  + |Θ − θ± |  c 1 δ e −c2 x /(1+t ) ,

|x| → ∞, l  1,

(1.11)

where c 1 , c 2 are two positive constants depending only on θ+ and δ¯ . Once Θ is determined, then the “viscous contact wave” profile ( V , U , Θ, Q )(x, t ) is then defined as

V =

R P+

Θ,

U = u− +

(γ − 1) 4b Rγ

a

Θ 2 Θx ,

Θ = Θ,

Q =−

4b P + aR

Θ 2 Θx .

(1.12)

It is straightforward to check that the “viscous contact wave” ( V , U , Θ, Q )(x, t ) solves the system (1.1) time asymptotically, that is

⎧ V − U = 0, t x ⎪ ⎪   ⎪ ⎪ RΘ ⎪ ⎪ = R1, ⎪ Ut + ⎪ ⎪ V ⎪ x ⎨ 

 U2 ⎪ e ( V , Θ) + + P ( V , Θ)U x + Q x = R 2 , ⎪ ⎪ 2 t ⎪ ⎪ ⎪   ⎪ ⎪  Q ⎪ x ⎪ ⎩− + aV Q + b Θ 4 x = R 3 , V

(1.13)

x

where

 R1 = Ut =

(γ − 1) 4b Rγ

a

 2

Θ Θx t

−3/2 −c 2 x2 /(1+t )

= O (δ)(1 + t ) e , as |x| → ∞,    (γ − 1) 4b 2 (γ − 1) 4b 2 R 2 = U U t = u− + Θ Θx Θ Θx Rγ

−3/2 −c 2 x /(1+t )

= O (δ)(1 + t )

e

Rγ

a

2

,

as |x| → ∞,

a

t

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

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and

  2  4b P + Qx 2 R3 = − = 2(Θx )2 + ΘΘxx x = O (δ)(1 + t )−3/2 e −c2 x /(1+t ) , 2 V

as |x| → ∞.

aR

x

Where in the estimates of R i (i = 1, 2, 3) we used (1.10), (1.11) and the definition (1.12). Below, we will use the following function space, for any 0  T  ∞,













E [0, T ] = (φ, ψ, ζ, ω)  (φ, ψ, ζ, ω, ωx ) ∈ L ∞ 0, T ; H 2 (R) , (φx , ψx , ζx ) ∈ L 2 0, T ; H 1 (R) ,





ω ∈ L 2 0, T ; H 3 (R) . Thus, we are ready to state our main results. Theorem 1.1. For any given ( v − , u − , θ− ), suppose that ( v + , u + , θ+ ) satisfies (1.6). Let ( V , U , Θ, Q ) be the “viscous contact wave” defined in (1.12) with strength δ = |θ+ − θ− |. Then there exist suitably small positive constants δ0 and σ0 , such that if 0 < δ < δ0 and the initial data ( v 0 , u 0 , θ0 ) satisfies

   v 0 (·) − V (·, 0), u 0 (·) − U (·, 0), θ0 (·) − Θ(·, 0) 

H2

 σ0 ,

then the Cauchy problem (1.1)–(1.2) admits a unique global solution ( v , u , θ, q) satisfying ( v − V , u − U , θ − Θ, q − Q ) ∈ E [0,∞] . Moreover,





lim sup( v − V , u − u − , θ − Θ, q − Q )(x, t ) = 0.

t →∞ x∈R

(1.14)

We now recall some related topics in this aspect and make some remarks on the analysis in this paper. It is known that there are many interesting results on the asymptotic behavior of solutions to the Navier–Stokes system, which are characterized by those of the solutions of the corresponding Riemann problems for the Euler system. Here, we only mention the contact discontinuity wave case for the Euler system (see [3,5,6,27] and references therein). In [3,5], the authors studied the asymptotic behavior of solutions to the Navier–Stokes system by the viscous contact wave constructed there. Furthermore, the decay estimates are given by the anti-derivative methods in [5]. In our case, there are no viscosity and heat-conductivity terms but an additional term q x in the energy Eq. (1.1)3 , so the system is of less dissipation. Thus, how to construct an ansatz and how to derive the suitable a priori estimates will be worth investigating. Since the higher order space derivatives of q are expected to decay faster in time, thus, the approximation (1.8) is proposed here. This is different from the construction of the “viscous contact wave” in [3,5]. Then, in order to show the stability of our “viscous contact wave” defined in (1.12), we need to prove the global existence of solutions to (1.1)–(1.2) near the “viscous contact wave”. The key step is to close the a priori estimates (2.4)–(2.5). In this way elaborate energy estimates are needed. For example, how to control the terms t  t 2 (1 + s)−1 R (ζ 2 + φ 2 )e −c2 x /(1+s) dx ds and 0 (φx , ψx , ζx )(·, s) 2 ds (see (2.1)) uniformly in t? For 0 the first term we will use some properties of heat kernel and the structure of the system (1.1) (Lemma 3.2). And for the second term, we can not use the compensating matrix method used in [8,10,11], which is an effective tool for the small perturbation problem near constant states. However, for the linearized system around the viscous contact wave profile ( V , U , Θ, Q ), one can not assure the existence and differentiability of compensating matrix by the theories developed by Kawashima [8]. Consequently,  t we need to perform direct  t energy estimates. By finding some suitable multipliers, we can control 0 φx (·, s), ψx (·, s) 2 ds by 0 ζx (·, s), ω(·, s), ωx (·, s) 2 ds, then by use of the internal relation between the difference of temperature ζ and the difference of radiation flux ω , we can bounded the L t2 ( L 2x )-norms of (φx , ψx , ζx ) by L t2 ( L 2x )-norms of ω, ωx . which is carried out in details in Section 3 of our paper. By such a (tricky) energy estimate process, we can close the a priori estimates (2.4) and (2.5).

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J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

We should remark that for the study of the non-equilibrium radiation hydrodynamics, most papers are concerned with the “baby model” which is the third order approximation of (1.1) (see [2,9,12,14, 25] and references therein). Recently, many mathematicians have started to study the hyperbolic– elliptic coupled system (1.1) [10,11,13,15–17,19]. In [10,11] Kawashima and Nishibata established the global existence of classical solutions for the initial value problem, which is the small perturbation problem near constant states. And in [15] the authors proved the global existence of shock profile for (1.1), and then Lattanzio, Mascia and Serre [13] obtained the global existence of shock profile for a general hyperbolic–elliptic coupled system. In [16], Lin, Coulombel and Goudon studied the stability of shock profile obtained in [15] under the zero mass assumption. Then, Nguyen, Zumbrun and collaborators also studied the stability of shock profile without the zero mass assumption by using Green’s function in [19]. recently, Lin considered the asymptotic stability of viscous rarefaction wave in [17]. But the results of the stability of “viscous contact discontinuity” for the hyperbolic– elliptic coupled system (1.1) are much more limited, which is also one motivation of this paper. In addition, we should remark that the methods in this paper can also be used to deal with singular limit problem considered in [26], when the strength of the contact discontinuity wave is small. This paper is organized as follows. We reformulate the problem into a small perturbation problem near the “viscous contact wave” and give some preliminaries in Section 2. Section 3 is devoted to obtain a priori estimates, which are the crucial parts to establish the global existence of solutions and large time behavior. By the energy estimates in Section 3, we prove our main theorem in Section 4. Notation. Throughout this paper, several positive generic constants are denoted by C , c without confusion, which are independent of the time T . And H l (R) denotes the lth order Sobolev space with its norm

f l =

l   j  ∂x f ,

where ·  · L 2 (R) .

j =0

2. Reformulation of the problem and preliminaries To establish the global existence of solutions to the system (1.1) with the asymptotic behavior (1.14), it suffices to show that there exists an exact solution to (1.1) in a neighborhood of the approximate solution ( V , U , Θ, Q ). Suppose ( v , u , θ, q) is the exact solution to (1.1), we set

φ = v − V,

ψ = u − U,

ζ = θ − Θ,

ω=q− Q.

(2.1)

Thus, using (1.13), we obtain

⎧ φt − ψx = 0, ⎪ ⎪   ⎪ ⎪ ⎪ Rζ − P +φ ⎪ ⎪ ψt + = F, ⎪ ⎪ v ⎪ x ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ R ζt + P u x − P + U x = −ωx , γ −1 ⎪       ⎪ ⎪ ζ Θ ζ Θx qx aΘ vq VQ ⎪ ⎪ − − , =− + − ⎪ ⎪ ⎪ θ x θΘ b 4bθ 5 v x 4θ 5 4Θ 5 ⎪ ⎪ ⎪ ⎪ ⎪ (φ, ψ, ζ, ω)(±∞, t ) = 0, ⎪ ⎪ ⎩ (φ, ψ, ζ )(x, 0) = (φ0 , ψ0 , ζ0 ),

(2.2)

F = −U t .

(2.3)

where

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

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The local existence of the solution to (2.2) is similar as in [7,16], which relies on the linear iteration scheme and fixed point theorem. To prove the main theorem, it suffices for us to close the following a priori estimates. Precisely speaking, for any time T , we define the following two quantities and we want to show that there exists a small positive parameter ε0 such that





N ( T ) = sup (φ, ψ, ζ, ω, ωx )(·, t )2  ε0 ,

(2.4)

0t  T

T M 2 (T ) =

    (φx , ψx , ζx )(·, s)2 + ω(·, s)2 ds  ε 2 , 1

(2.5)

0

3

0

where ε0 < 1 is independent of T , which is to be determined later. First, we assume that ε0 is suitably small, then we can show the solutions satisfy the estimates in the Proposition 4.1. In this way, we can find that the a priori estimates (2.4) and (2.5) are indeed valid by choosing the initial perturbation and the strength δ small enough. In order to obtain the essential a priori estimates (2.4)–(2.5), the following lemma on the heat kernel will be used, which is first used in [3]. For α > 0, we define

−1/2

β(x, t ) = (1 + t )

exp −

α x2



1+t

x g (x, t ) =

,

β( y , t ) dy .

(2.6)

−∞

It is easy to check that

4α g t = β x ,

   g (·, t )

L∞

=

√

π α −1/2 .

(2.7)

Then we have Lemma 2.1. For any 0 < T  +∞, suppose that f (x, t ) satisfies





f ∈ L ∞ 0, T ; L 2 (R) ,









f t ∈ L 2 0, T ; H −1 (R) .

f x ∈ L 2 0, T ; L 2 (R) ,

Then the following estimate holds:

T 

 2 f β dx dt  4π  f (·, 0) + 4πα −1

T

2 2

0 R

   f x (·, t )2 dt + 8α

0

T





f t , f g 2 dt ,

(2.8)

0

here ·,· denotes the inner product on H −1 × H 1 . Proof. For the details of proof, one refers to Lemma 1 in [3].

2

3. A priori estimates In this section, we will establish the a priori estimates (2.4)–(2.5) under the assumption that ε0 is suitably small, which are direct consequences of the following lemmas. From now on, we always assume that ε0 + δ0 < 1 where δ0 is the same parameter as in the Theorem 1.1.

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J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

Lemma 3.1. For any T > 0 there exists suitably small δ0 independent of T , such that if (φ, ψ, ζ, ω) ∈ E [0, T ] satisfy (2.2), (2.4) and (2.5) with δ = |θ+ − θ− | < δ0 then we have

 2 sup (φ, ψ, ζ )(·, t ) +

T

  ω(·, s)2 ds 1

0t  T

0

 2  C (φ0 , ψ0 , ζ0 ) + C δ 4/3 T 2 0

T 2

+ Cε

ψx , ζx ds + C δ 0



(1 + s)−1



2 ζ 2 + φ 2 e −c2 x /(1+s) dx ds.

(3.1)

R

0

Proof. Multiplying Eq. (2.2)1 by − R Θ( v −1 − V −1 ), (2.2)2 by ψ , and (2.2)3 by ζ θ −1 , we have



1 2

 ψ 2 + R ΘΦ

v

 +

V



R

γ −1

ΘΦ

θ Θ

 + Lx + Z = F ψ − t

ζ ωx , θ

(3.2)

where

Φ(s) = s − 1 − ln s, L=R

ζψ v

−

P +φ v

ψ,

and

 Z = P +Φ

V v

 Ux +



P+

γ −1

Φ

 ζ Θ U x − ( P + − P )U x θ θ

which satisfies

  | Z |  C ζ 2 + φ 2 |Θxx | + |Θx |2 , where we have used (1.12). We multiply (2.2)4 by obtain after integration by parts that

 −

ω R

(3.3)

ω and integrate the resulting equation over R to

     ζ ζ Θx Θ Θ 2 dx = − ω dx + ω dx + ωx Q x dx θ x θΘ 4bv θ 5 x 4bv θ 5 R

R



+   R

Notice that f L ∞ (R) 

ω

(ω + Q )x

R

Θ 4bv θ 5

√



v



ωx2 dx + R

Θ 4bθ 5



aΘ v 4bθ 5

 dx + x

R

R

aΘ v

aω Θ Q

4bθ

b

ω2 dx + 5



v 4θ 5

ω2 dx + H .

2 f 1/2 f x 1/2 , then it follows from (1.11) and (2.4) that

−

V 4Θ 5

 dx

(3.4)

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055



ω R

( Q x + ωx )



v

Θ 4bθ 5



 dx  C x

1037

  |ω| | Q x | + |ωx | |Θx | + |ζx | dx

R

  (η + C δ) ω 21 + C η Q x Θx 2 + Q x 2 + C η ε02 ζx 2 .

Together with (1.12) and (2.4), we have

    aΘ v 2 Θ 2 2 | H |  C (η + δ) ω dx + | ω | dx C ε |ζx |2 dx + x η 0 4bθ 5 4bv θ 5 R





+ Cη

R

| Q x |2 + | Q x Θx |2 dx + C η

R

From (3.2)–(3.5), choosing



R



 ζ 2 + φ 2 |Θx |2 + | Q |2 dx.

(3.5)

R

η, δ suitably small, we get

  (φ, ψ, ζ )(·, t )2 +

T

  ω(·, s)2 ds 1

0

 2  C (φ0 , ψ0 , ζ0 ) + C

T

4/3 F L 1 ds

T  2 0 0 R

0

T  +C



|ψx |2 dx ds

+ Cε

 ζ 2 + φ 2 |Θxx | + |Θx |2 + | Q |2 dx ds + C δ 2 + C ε02

0 R

T  |ζx |2 dx ds 0 R

 2  C (φ0 , ψ0 , ζ0 ) + C δ 4/3 + C ε02

T





|ψx |2 + |ζx |2 dx ds

0 R

T + Cδ 0

(1 + s)−1





2 ζ 2 + φ 2 e −c2 x /(1+s) dx ds,

(3.6)

R

where we have used

T 

T ψ 1/2 ψx 1/2 F L 1 (R)

ψ F dx ds  C 0 R

0

T  ε02

T ψx 2 ds + C

0

4/3

0

due to the fact that

F L 1  U t L 1  C δ(1 + t )−1 . Hence we finish the proof of Lemma 3.1.

2

T

F L 1 ds  ε02

ψx 2 ds + C δ 4/3 0

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J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

T  2 Then it requires us to estimate the term 0 (1 + s)−1 R (ζ 2 + φ 2 )e −c2 x /(1+s) dx ds, which relies strongly on the structure of the system and the decay properties of the heat kernel in Lemma 2.1. Lemma 3.2. For α ∈ (0, c 2 /4] and β defined in (2.6), there exists a positive constant C depending on α such that the following estimate holds

t 



2

2

2

t



2

φ + ψ + ζ β dx ds  C + C

0 R



φx 2 + ψx 2 + ζx 2 + ω 2 ds.

(3.7)

0

Proof. Similar as in [3], the proof of (3.7) is divided into the following two parts: For any η > 0

t 



( R ζ − P + φ)2 + ψ 2 β 2 dx ds

0 R

t  C + Cη



φx 2 + ψx 2 + ζx 2 + ω 2 ds

0

t  + Cδ



 φ 2 + ζ 2 β 2 dx ds + C δ 2 + η

0 R

t  ψ 2 β 2 dx ds,

(3.8)

0 R

and

t 



R ζ + (γ − 1) P + φ

2

t β 2 dx ds  C + C η

0 R



φx 2 + ψx 2 + ζx 2 + ω 2 ds

0

t  + C (δ + η)



φ 2 + ζ 2 β 2 dx ds.

(3.9)

0 R

If (3.8) and (3.9) hold true, then adding (3.8) to (3.9) and choosing Let’s prove (3.8) first. Setting

η and δ suitably small imply (3.7).

x h=

β 2 dy .

−∞

Then

  h(·, t )

L∞

 2α −1/2 (1 + t )−1/2 ,

  ht (·, t )

L∞

 4α −1/2 (1 + t )−3/2 .

Multiplying (2.2)2 by ( R ζ − P + φ) vh and integrating the resulting equation over R lead to

1 2





( R ζ − P + φ)2 β 2 dx = R



ψt ( R ζ − P + φ) vh dx − R

R



−

F ( R ζ − P + φ) vh dx R

v −1 ( R ζ − P + φ)2 v x h dx

(3.10)

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

1039

  ψ( R ζ − P + φ) vh dx − ψ( R ζ − P + φ)t vh dx

 =

t

R

R





−

ψ( R ζ − P + φ) v t h dx − R

ψ( R ζ − P + φ) vht dx R



v −1 ( R ζ − P + φ)2 v x h dx −

− R

 F ( R ζ − P + φ) vh dx R



 =

ψ( R ζ − P + φ) vh dx

+ t

R

5 

Ii,

(3.11)

i =1

where I i , i = 1, . . . , 5 can be treated as follows. Using Eq. (2.2)1 yields that

 I1 = −

ψ( R ζ − P + φ)t vh dx R





= −(γ − 1)

ψ vh

γ −1

R

 = (γ − 1)



R

=

ψ vhψx dx R

ψ h( R ζ − P + φ)(U x + ψx ) dx 

γ P+



vh ψ 2

2

3 

dx + γ P + t

R

+



ζ + P +φ



 dx + (γ − 1) x

R

ωx ψ vh dx R

I 1i ,

(3.12)

i =1

where in the second equality we have used





R

γ −1

ζ + P +φ

=− t

Rζ − P +φ (ψx + U x ) − ωx . v

(3.13)

Due to v t = u x (2.4) and (3.10), we have

          1  I  + | I 2 | = (γ − 1) ψ h( R ζ − P + φ)(U x + ψx ) dx +  ψ( R ζ − P + φ) v t h dx 1     R −1/2

 C (1 + t )

1/2

ψ

1/2

ψx



R

 ζ + φ U x + ψx

 C (1 + t )−5/4 ψx 1/2 + C (1 + t )−1/2 ψx 3/2  C ψx 2 + C (1 + t )−5/3 . Integrating by parts and choosing δ suitably small give that

(3.14)

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J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

I 12 = −

− −

γ P+ 2

v β 2 ψ 2 dx −

2

2

hψ 2 Θx dx −

γ P+

R





v β 2 ψ 2 dx + C δ R

γ P+



γR

R

γ P+

4



2

 hψ 2 φx dx R

β 2 ψ 2 dx + C (1 + t )−1/2 ψ 3/2 ψx 1/2 φx

R







v β 2 ψ 2 dx + C ψx 2 + φx 2 + C (1 + t )−2 .

(3.15)

R

And

         3     I  = −(γ − 1) ω(ψ vh)x dx = −(γ − 1) ω ψx vh + ψ( V x + φx )h + ψ vh x dx 1     R



 C ω 2 + ψx 2 + C δ 2

R

 β 2 ψ 2 dx R

+ C ψ 1/2 ψx 1/2 ω φx (1 + t )−1/2 + C η ω 2 + η    C η ω 2 + ψx 2 + φx 2 + C δ 2 + η



 β 2 ψ 2 dx R

β 2 ψ 2 dx + C (1 + t )−2 .

(3.16)

R

In view of (3.10), we get

 |I3| = | −

 ψ( R ζ − P + φ) vht dx|  C ψ φ + ζ ht L ∞  C (1 + t )−3/2 .

(3.17)

R

It follows from (1.11) and (3.10) that

     −1 2  | I 4 | = − v ( R ζ − P + φ) v x h dx R

     R  −1 2 −1 2  = v ( R ζ − P + φ) Θx h dx + v ( R ζ − P + φ) φx h dx P +

  Cδ

R

R



 φ 2 + ζ 2 β 2 dx + C ζx 2 + φx 2 + C (1 + t )−2 .

(3.18)

R

By (1.11) and (2.3) we obtain

      | I 5 | = − F ( R ζ − P + φ) vh dx R

  C F L 1 (1 + t )−1/2 ζ L ∞ + φ L ∞   C δ(1 + t )−2 + C δ φx 2 + ζx 2 .

(3.19)

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

1041

Integrating (3.11) over [0, t ], combining all estimates (3.14)–(3.19) and Lemma 3.1, we get (3.8). Now we are ready to prove (3.9). Set

f = R ζ + (γ − 1) P + φ in Lemma 2.1 and use (3.13) to derive



ft , f g2



 H −1 × H 1

= −(γ − 1) R

R ζ − P +φ f ψx g 2 dx v



3 

f U x g 2 dx − (γ − 1)

v

R

= (γ − 1)



Rζ − P +φ

− (γ − 1)

ωx f g 2 dx R

Ji.

(3.20)

i =1

Notice that |U x |  C δβ 2 , we get from (2.7) that

 | J 2|  C δ



ζ 2 + φ 2 β 2 dx.

(3.21)

R

Due to R ζ − P + φ = f − γ P + φ and ψx = φt , we have from (2.7), (1.1)1 and (3.13) that









v −1 f 2 − γ P + f φ φt g 2 dx

−2 J 1 = 2 R



=



2v −1 f 2 g 2 φt − γ P + v −1 f g 2 φ 2

R



= +

= +

v −1 f g φ(2 f − γ P + φ) gt dx

R

v −2 v t f g 2 φ(2 f − γ P + φ) dx −

v −1 f g 2 φ(2 f − γ P + φ) dx



v −1 g 2 φ(4 f − γ P + φ) f t dx

R

 − t

R



 −2

t

R



t



v −1 f g 2 φ(2 f − γ P + φ) dx

R





1 2α



v −1 f g φ(2 f − γ P + φ)βx dx

R





v −2 u x g 2 φ f (2 f − γ P + φ) + (γ − 1)(4 f − γ P + φ)( R ζ − P + φ) dx

R



+ (γ − 1)

v −1 g 2 φ(4 f − γ P + φ)ωx dx

R

 = R



v −1 f g 2 φ(2 f − γ P + φ) dx

+ t

3  i =1

J 1i ,

(3.22)

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J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

where J 1i , i = 1, 2, 3 can be estimated as follows. Since |βx |  C (1 + t )−1 , then

 1  J   C (1 + t )−1



1

 3 |ζ | + |φ|3 dx

R

  C (1 + t ) φ 5/2 φx 1/2 + ζ 5/2 ζx 1/2   C (1 + t )−4/3 + C ζx 2 + φx 2 . −1

(3.23)

Similarly, notice that u x = U x + ψx , we derive that

 2 J C





1

 |ζ |3 + |φ|3 |U x | + |ψx | dx

R



  C ζ 2 ζx + φ 2 φx U x + ψx   C ζx 2 + φx 2 + U x 2 + ψx 2   C ζx 2 + φx 2 + ψx 2 + C (1 + t )−3/2 .

(3.24)

And by V x  C δβ , we get

              −1 2  J 3 + J 3   (γ − 1) ω f g 2 dx + (γ − 1) v g φ(4 f − γ P + φ) x ω dx 1    x R

R

           (γ − 1) ω f x g 2 dx + 2(γ − 1) ω f g β dx R

R

     V x + φx  2  g φ( 4 f − γ P φ) ω dx + (γ − 1) +  2 R

v

     + 2(γ − 1) v −1 g βφ(4 f − γ P + φ)ω dx R

      −1 2  + (γ − 1) v g φ(4 f − γ P + φ) x ω dx  η

R



 ζ 2 + φ 2 β 2 dx + C η ζx 2 + φx 2 + ψx 2 + ω 2 .

(3.25)

R

Using Lemma 2.1, together with estimates (3.21)–(3.25), gives (3.9). Then we finish the proof of Lemma 3.2. 2 As a consequence of Lemma 3.1 and Lemma 3.2, we obtain Lemma 3.3. For any T > 0 there exist suitably small ε0 , δ0 independent of T , such that if (φ, ψ, ζ, ω) ∈ E [0, T ] satisfy (2.2), (2.4) and (2.5) with δ = |θ+ − θ− | < δ0 then we have

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

 2 sup (φ, ψ, ζ )(·, t ) +

0t  T

T

1043

  ω(·, s)2 ds 1

0

 2   C (φ0 , ψ0 , ζ0 ) + C δ + C δ + ε02

T φx , ψx , ζx 2 ds.

(3.26)

0

Below, we will give the estimates of derivatives. Lemma 3.4. For any T > 0 there exist suitably small ε0 , δ0 independent of T , such that if (φ, ψ, ζ, ω) ∈ E [0, T ] satisfy (2.2), (2.4) and (2.5) with δ = |θ+ − θ− | < δ0 then we have

 2 sup (φx , ψx , ζx )(·, t ) +

0t  T

T ωx 21 ds 0

 2  C (φ0 , ψ0 , ζ0 )1 + C (ε0 + δ)

T φx , ψx , ζx 2 ds + C δ.

(3.27)

0

Proof. Set Λ = ( v , u , θ)t and rewrite (1.1)1,2,3 under the following symmetric form

A 0 (Λ)Λt + A (Λ)Λx = g (q x ),

(3.28)

where g (q x ) = (0, 0, −q x )t and

⎛ A (Λ) = ⎝ 0

−θ P v

0

0 0

θ 0

⎞



0 0 ⎠,

A (Λ) =

R

γ −1

0 θ Pv 0

θ Pv 0 P

0 P 0

 .

Similarly, setting Λ¯ = ( V , U , Θ)t , the system (1.13)1,2,3 is transformed into

¯ Λ¯ t + A (Λ) ¯ Λ¯ x = g ( Q x ) + F¯ , A 0 (Λ)

(3.29)

where F¯ = (0, −Θ U t , 0)t . From here to the end of this paper, we set W = Λ − Λ¯ = ( v , u , θ)t − ( V , U , Θ)t = (φ, ψ, ζ ), from (3.28) and (3.29), we have

A 0 (Λ) W t + A (Λ) W x = F˜ (Λ, Λx ) + g (q x ) − g ( Q x ) = F˜ (Λ, Λx ) + g (ωx ),

(3.30)

where









¯ − A 0 (Λ) Λ¯ t + A (Λ) ¯ − A (Λ) Λ¯ x − F¯ . F˜ (Λ, Λx ) = A 0 (Λ) In view of the structure of A 0 (Λ) and function g (ωx ), we multiply (3.30) by A 0 (Λ)−1 and differentiate with respect to x, then multiply the resulting system by (∂x W )t A 0 (Λ) and integrate over R to obtain

 R







A 0 (Λ)∂x W t , ∂x W dx − R







A (Λ)∂xx W , ∂x W dx = −



ωxx ζx dx + R

R

H˜ , ∂x W  dx, (3.31)

1044

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

with





˜ = − A 0 (Λ)∂x A 0 (Λ)−1 A (Λ)∂x W + A 0 (Λ)∂x A 0 (Λ)−1 F˜ , H and ·,· denotes the inner product in L 2 , then

 



1 2





0



1 

=

A (Λ)∂x W , ∂x W dx t

R

2

R







−



A 0 (Λ)t − A (Λ)x ∂x W , ∂x W dx

H˜ , ∂x W  dx.

ωxx ζx dx + R

(3.32)

R

Differentiating (2.2)4 with respect to x yields



av

−ζxx = Multiplying (3.33) by

4bθ

av

−

4bθ 3



aV

+

4bΘ 3



1

qx

4bθ 3

  (3.33)

.

v

x

x

ωx and integrating over R give  

 −



ω+Q 3

ωx ζxx dx = R

av 4bθ

R



ω+Q 3

av 4bθ 3

−



aV

−

4bΘ 3



1

qx

4bθ 3

v

 

ωx dx. x

(3.34)

x

Combining (3.32) and (3.34) leads to

  1 2







0

+

A (Λ)∂x W , ∂x W dx t

R



= R

1  2

 



av 4bθ 3

R



ω+Q

av 4bθ 3

−

4bΘ 3





A 0 (Λ)t − A (Λ)x ∂x W , ∂x W dx −



aV

 (ωx ζx )x dy +

R

−

1



4bθ 3

qx v



ωx dx x

x

H˜ , ∂x W  dx.

R

Then we have

  1 2





0



 +

A (Λ)∂x W , ∂x W dx t

R



1 

=

2

R

R

av 4bθ 3



ωx2 dx +





A 0 (Λ)t − A (Λ)x ∂x W , ∂x W dx +

1 4bv θ 3

2 ωxx dx

H˜ , ∂x W  dx + K˜ ,

(3.35)

R

with

K˜ = −

 

1 4bθ 3



Qx v

 



4bθ 3

R

+

av

ωωx + Q x



av 4bθ 3

−

aV 4bΘ 3



ωx + x

1 4bθ 3

  1 v

ωx ωxx x



ωxx dx. x

Next, we estimate the terms in the right-hand side of (3.35) separately. First, using (1.11), (1.12), (2.2)1,2,3 and (2.4), we have

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

1045

    1 0      |Λ¯ t | + | W t | | W x |2 dx  C (δ + ε0 ) ∂x W 2 , A (Λ) ∂ W , ∂ W dx C  t x x  2  R

(3.36)

R

    1       |Λ¯ x | + | W x | | W x |2 dx  C (δ + ε0 ) ∂x W 2 , A (Λ) ∂ W , ∂ W dx C x x x  2  R

(3.37)

R

and



H˜ , ∂x W  dx =

R











A 0 (Λ)∂x A 0 (Λ)−1 A (Λ)∂x W + A 0 (Λ)∂x A 0 (Λ)−1 F˜ + ∂x F˜ ∂x W dx

R

 I1 + I2 + I3.

(3.38)

where

| I 1 |  C (δ + ε0 ) ∂x W 2 .

(3.39)

Using (1.11) and (1.12) leads to

  | F˜ |  C |φ| + |ζ | | V t | + |U t | + |Θt | + | V x | + |U x | + |Θx | + |Θ U t |  C ε0 |Θx | + |Θ U t |,

(3.40)

then we have

 |I2| = R

 

C







A 0 (Λ)∂x A 0 (Λ)−1 F˜ ∂x W dx







ε0 |Θx | + |Θ U t | |Θx | + |ζx | + | V x | + |φx | |∂x W | dx

R

 2  C (δ + ε0 ) ∂x W 2 + C (Θx )2  + C U t 2 .

(3.41)

Similarly,

 |I3|  C



  

|φx | + | V x | + |ζx | + |Θx | |Λ¯ t | + |Λ¯ x | + |φ| + |ζ | |Λ¯ tx | + |Λ¯ xx | + |∂x F¯ | |∂x W | dx

R

 C δ ∂x W 2 + C δ −1  +C



 |Θx |2 Λ¯ t |2 + |Λ¯ x |2 + |Λ¯ xx |2 + |Λ¯ tx |2 dx

R



|∂x Θ U t | + |Θ∂x U t | |∂x W | dx

R

 C δ ∂x W 2 + C δ −1

 R

and





|Θx |4 + |Θxx |2 + |Θx U t |2 + |Θ U tx |2 dx,

(3.42)

1046

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

| K˜ |  C





 | V x | + |Θx | + |φx | + |ζx | |ωωx | + | Q x | |φ| + |ζ | |ωx |

R

  + | Q | | V x | + |Θx | + |φx | + |ζx | |ωx | + | V x | + |φx | |ωx ωxx |  + | Q xx ||ωxx | + | Q x | | V x | + |φx | |ωxx | dx   C η ω 21 + η ωxx 2 + C η Θx Q 2 + Q xx 2 + Q x 2 + Q x V x 2 + C η (δ + ε0 ) W x 2 .

(3.43)

We integrate (3.35) over [0, T ], and choose

   W x (·, t )2 +

T

η suitably small to obtain from (3.36)–(3.43) that

    ωx (·, s)2 ds  C  W x (·, 0)2 + C (ε0 + δ)

T 0

+ Cηδ

−1

T

ω 21 ds

∂x W ds + C

1

0

T 2

0



Θx2 2 + Q xx 2 + Q x 2 + Θxx 2 + U t 2 ds



Q x U t 2 + Θ U tx 2 + Θx Q 2 + Q x V x 2 ds

0

+ Cηδ

−1

T 0

T  C (ε0 + δ)

 2 φx , ψx , ζx 2 ds + C δ + C (φ0 , ψ0 , ζ0 )1 ,

(3.44)

0

where in the second inequality, we used Lemma 3.3, (1.11) and (1.12). Thus we have finished the proof of Lemma 3.4. 2 Now, it requires us to estimate the term equations in (2.2) as

T 0

∂x W (·, s) 2 ds. First, we recast the second and third

⎧ Rθ R φΘx ζ Vx φ 2 + 2V φ ⎪ ⎪ +R 2 −R Θ V x − Ut , ⎨ ψt − 2 φx + ζx = Γ1 := R 2 2 v

v

⎪ ⎪ ⎩

R

γ −1

ζt +

Rθ v

Vv

v

V v

  ψx + ωx = Γ2 := − P ( v , θ) − P ( V , Θ) U x .

(3.45)

Then we linearize (3.45) around the viscous contact discontinuity wave,

⎧ RΘ R φζx ζ φx φ 2 + 2V φ ⎪ ⎪ +R 2 −R Θφx , ⎨ ψt − 2 φx + ζx = Γ¯1 := Γ1 + R V Vv V v V 2 v2   ⎪ R RΘ Θφ ζ ⎪ ⎩ ψx . ζt + ψx + ωx = Γ¯2 := Γ2 + R −R γ −1 V Vv v Multiplying (3.46)1 by − V φx , (3.46)2 by

2V 2 ψ RΘ x

and adding the resulting equations we have

(3.46)

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

RΘ V

 φx2 + V ψx2 − V ψt φ +

2V 2

R

γ − 1 RΘ

 + V ψx φ +

 ζ ψt

R

2V 2

1047

 ζ ψx

γ − 1 RΘ t   2V 2 R 2V 2 ζ = R φx ζx − ωx ψx − V x φ + ψt RΘ γ − 1 RΘ x  2 R 2V 2V 2 ψx Γ¯2 + V t ψx φ + ζ ψx − V φx Γ¯1 + γ −1 RΘ t RΘ   2     2V 2 R 2V R Θ ζ = R φx ζx − ωx ψx − V x φ + φx − ζx RΘ γ − 1 RΘ x V V   2    2 R 2V R 2V 2V 2 ¯ − V xφ + Γ1 + V t ψx φ + ζ ψx Γ¯2 ζ ψx − V φx Γ¯1 + γ − 1 RΘ x γ −1 RΘ t RΘ   2     2V 2 R 2V R Θ ζ = R φx ζx − ωx ψx − V x φ + φx − ζx RΘ γ − 1 RΘ x V V   2  2  R 2V R 2V 2V 2 ¯ Γ1 + V t ψx φ + − ( V φ)x + ζ ψx Γ¯2 ζ ψx + γ − 1 RΘ x γ −1 RΘ t RΘ 

=

7 



x

Ji.

(3.47)

i =1

Integrating (3.47) over [0, T ] × R and using Cauchy–Schwarz inequality give

T

 T    7    φx , ψx ds  C φ0 , ζ0 , ψ0x + C φ, ζ, ψx + J i  dx ds.    2

2

2

0 R

0

i =1

Below we estimate J i (i = 1, . . . , 7) respectively.

T  0 R

 T     2V 2   | J 1 + J 2 | dx ds =  R φx ζx − R Θ ωx ψx  dx ds 0 R



1

T

4



ψx 2 + φx 2 ds + C

T

0

 ζx 2 + ωx 2 ds.

0

Noticing that | V t , Θt |  C δβ , where β is defined as in Lemma 3.2, then we have

T  | J 5 + J 6 | dx ds = 0 R

 2  T      V t ψx φ + R ζ 2V ψx  dx ds  γ −1 RΘ t 0 R



1

T ψx ds + C δ

4 0



1

T  2

2

 β 2 |φ|2 + |ζ |2 dx ds

0 R

T

T ψx 2 ds + C δ 2 + C δ 2

4 0

0



φx , ψx , ζx , ω, ωx 2 ds.

(3.48)

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J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

In the last inequality, we have used Lemma 3.2.

T  0 R

 2     T     R 2V R Θ  | J 3 | dx ds = φx − ζx  dx ds  V xφ + γ − 1 R Θ ζ V V x 0 R

T 



C

 | V x φ| + | V x ζ | + |Θx ζ | + |ζx | |φx | + |ζx | dx ds

0 R

  Cδ +

1

 T

T 2

φx ds + C

4 0

T  2

ζx ds + C δ

 β 2 |φ|2 + |ζ |2 dx ds

0 R

0

  T T 1  Cδ + φx , ψx , ω, ωx 2 ds + C ζx 2 ds + C δ, 4

0

0

where in the second inequality | V x , Θx | < C δβ is used and in the last inequality Lemma 3.2 is applied.

T  0 R

 2   T     2V  ( V φ)x + R Γ¯1  dx ds ζ | J 4 | dx ds =  γ − 1 RΘ x 0 R

T  C



 | V x φ| + |φx | + | V x | + |Θx | |ζ | + |ζx |

0 R

 × |φζx | + |ζ φx | + |φφx | + |φΘx | + |ζ V x | + |φ V x | + |U t | dx ds T  2

 Cδ



2

β φ +ζ

2



T dx ds + C (δ + ε0 )

0 R

+ C δ −1



φx 2 + ζx 2 ds

0

T  |U t |2 dx ds 0 R

T φx , ψx , ζx , ω, ωx 2 ds + C δ.

 C (ε0 + δ) 0

In the second inequality we used the fact that | V x , Θx |  C δβ and the a priori assumption (2.4). In the third inequality, Lemma 3.2 and the definition of U are used. Similarly, by use of |U x | < C δβ and Lemma 3.2, we have

T  0 R

 T   2  2V    ¯ | J 7 | dx ds =  R Θ ψx Γ2  dx ds 0 R

T  C 0 R

  |ψx | |φ| + |ζ | |ψx | + |φ| + |ζ | |U x | dx ds

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

T

T 

 β 2 |φ|2 + |ζ |2 dx ds

2

 C (δ + ε0 )

1049

ψx ds + C δ 0 R

0

T φx , ψx , ζx , ω, ωx 2 ds + C δ.

 C (ε0 + δ) 0

Consequently, combining (3.48) and all above estimates of J i (i = 1, . . . , 7) and choosing δ, ε0 suitably small yields

T

T 2

2

2

ζx , ω, ωx 2 ds + C δ

φx , ψx ds  C φ0 , ζ0 , ψ0x + C φ, ζ, ψx + C 0

0

T  C φ0 , ζ0 , ψ0x 2 + C

ζx 2 ds + C δ.

(3.49)

0

The second inequality is due to Lemmas 3.3, 3.4. T  It remains to estimate 0 R |ζx |2 dx ds. First, by the difference equation of

−ζx = =



av 4bθ 3 av 4bθ 3

−

ω+Q 

ω+Q

1

Q xx

4bθ 3

v

av 4bθ 3 av 4bθ 3

+

− −

aV 4bΘ 3 aV

 −  −

4bΘ 3

Q x V x + φx 4bθ 3

v



1

ωx

4bθ 3

−

v

ωxx

1



4bθ 3 v

x

+

1 4bθ 3

ω (2.2)4 , we have 

Qx



v

x

ωx ( V x + φx ) 4bθ 3

v2 (3.50)

.

Then

    |ζx |  C |ω| + | Q | |φ| + |ζ | + |ωxx | + |ωx | | V x | + |φx | + | Q xx | + | Q x | | V x | + |φx | . Consequently, in view of | Q |2  C δ 2 β 2 , we get

T

T 2

ζx ds  C 0

 ω 2 + ωxx 2 ds + C

T 

T  2

| Q xx | dx ds + C 0 R

0

T  +C

0 R

 2 |ωx |2 | V x | + |φx | dx ds + C

0 R

C

2

φx ds + C δ + C δ 0

T

T  2

+ C (δ + ε0 )

0

2

 β 2 |φ|2 + |ζ |2 dx ds

0 R

T ω 22 ds + C (δ + ε0 )

0

 2 | Q x |2 | V x | + |φx | dx ds

T ω

C

T  0 R

T 22 ds

 2 | Q |2 |φ| + |ζ | dx ds

φx , ψx , ζx , ω, ωx 2 ds + C δ 2 , 0

(3.51)

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J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

Combining (3.49) and (3.51) and choosing

T

ε0 , δ suitably small lead to

 2 W x ds  C (φ0 , ψ0 , ζ0 )1 + C (ε0 + δ)

T

2

0

T 2

0

By using Lemma 3.3, Lemma 3.4 and choosing

ω 22 ds.

W x ds + C δ + C

(3.52)

0

ε0 , δ suitably small, we have the following lemma.

Lemma 3.5. For any T > 0 there exist suitably small ε0 , δ0 independent of T , such that if (φ, ψ, ζ, ω) ∈ E [0, T ] satisfy (2.2), (2.4) and (2.5) with δ = |θ+ − θ− | < δ0 then we have

 2 sup (φ, ψ, ζ )(·, t )1 +

0t  T

T



ω 22 + φx 2 + ψx 2 + ζx 2 (·, s) ds

0

 2  C (φ0 , ψ0 , ζ0 ) + C δ.

(3.53)

1

Next, we will give the estimate of sup0t  T ω 22 . Rewrite the difference equation of

ω as

     ωx Qx − + av ω = − aφ Q − 4bθ 3 ζx − 4bθ 3 − 4bΘ 3 Θx . v

Multiplying (3.54) by

 R

ωx2 v

v

x

ω and integrating by parts yield



 av ω2 dx = −

dx +

(3.54)

x

R

due to (1.12). Choosing

R



Qx v



ωx dx −



R

 4bθ 3 ζx ω dx +

aφ Q ω dx + R





4b θ 3 − Θ 3 Θx ω dx R

  η ω 2 + ωx 2 + C η Q x 2 + C η δ 2 φ 2 + ζ 2 + C η ζx 2

(3.55)

η suitably small and using (3.53) yield  2 ω 21  C (φ0 , ψ0 , ζ0 )1 + C δ.

(3.56)

Then by (3.54), we have

 2 ωxx 2  C (φ0 , ψ0 , ζ0 )1 + C δ. Therefore

 2 ω 22  C (φ0 , ψ0 , ζ0 )1 + C δ.

(3.57)

Then we have Lemma 3.6. For any T > 0 there exist suitably small ε0 , δ0 independent of T , such that if (φ, ψ, ζ, ω) ∈ E [0, T ] satisfy (2.2), (2.4) and (2.5) with δ = |θ+ − θ− | < δ0 then we have

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

 2 sup (φ, ψ, ζ, ω, ωx )(·, t )1 +

T

0t  T



1051

ω 22 + φx 2 + ψx 2 + ζx 2 (·, s) ds

0

 2  C (φ0 , ψ0 , ζ0 ) + C δ.

(3.58)

1

For the estimates of second order derivatives of (φ, ψ, ζ ) and

ωx , we have the following lemma.

Lemma 3.7. For any T > 0 there exist suitably small ε0 , δ0 independent of T , such that if (φ, ψ, ζ, ω) ∈ E [0, T ] satisfy (2.2), (2.4) and (2.5) with δ = |θ+ − θ− | < δ0 then we have

T

 2 sup (φxx , ψxx , ζxx , ωxxx )(·, t ) +

0t  T



ωxx 21 + φxx 2 + ψxx 2 + ζxx 2 (·, s) ds

0

 2  C (φ0 , ψ0 , ζ0 ) + C δ.

(3.59)

2

Proof. The proof is similar to the proof of Lemma 3.6. We should remark that the system (3.30) is a positive symmetric system. Multiplying (3.30) by A 0 (Λ)−1 , applying the second order differential operator ∂xx on the resulting equation. And then we multiply the resulting system by (∂xx W )t A 0 (Λ) and integrate over R to obtain











A 0 (Λ)∂xx W , ∂xx W t dx +

R

 =−



ωxxx ζxx dx + R



A 0 (Λ)∂xx W , A 0 (Λ)−1 A (Λ) W x

 xx

dx

R





A 0 (Λ)∂xx W , A 0 (Λ)−1 F˜

 xx

dx.

(3.60)

R

It is obvious that the first term on the LHS of (3.60) can be dealt with by the same argument as in Lemma 3.4 due to the fact that matrix A 0 (Λ) is symmetric. We only deal with the second term on the LHS of (3.60), other terms can be estimated similarly







A 0 (Λ)∂xx W , A 0 (Λ)−1 A (Λ) W x

R

 =2





A 0 (Λ)∂xx W , A 0 (Λ)−1



 x

R



+

xx

   ∂xx W , A (Λ) W x xx dx +

R

dx

A (Λ) W x





 x

dx



A 0 (Λ)∂xx W , A 0 (Λ)−1





xx

A (Λ) W x dx.

(3.61)

R

Three terms on RHS of (3.61) can be estimated respectively as follows,

 2 R



0



0

−1

A (Λ)∂xx W , A (Λ)

 x

A (Λ) W x

 x

 dx = 2



R



A 0 (Λ)∂xx W , A 0 (Λ)−1



+2 R









x

A (Λ) W xx dx

A 0 (Λ)∂xx W , A 0 (Λ)−1

 x





A (Λ) x W x dx,

1052

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

     0    0 −1   A (Λ)∂ W , A (Λ) A (Λ) W dx C (δ + ε ) | W xx |2 dx,  xx xx 0   x R

R

where we used the facts that

|Λ¯ xx | = | V xx , U xx , Θxx |  C δ,

|Λ| = | V + φ, U + ψ, Θ + ζ |  C ,

|Λx | = | V x + φx , U x + ψx , Θx + ζx |  C (δ + ε0 ).

(3.62)

Due to (1.11) and (2.4). By Cauchy–Schwarz inequality







A 0 (Λ)∂xx W , A 0 (Λ)−1

 x







R

 |Λx |2 | W xx |2 dx + C

A (Λ) x W x dx  C R



2

|Λx |2 | W x |2 dx R



 | W xx |2 dx + C δ 2 + ε02

 C δ 2 + ε0

R

 | W x |2 dx R

in the second inequality, (3.62) is used.

             3  ∂xx W , A (Λ) W x   ∂xx W , A (Λ) x W xx + A (Λ) xx W x dx dx =   xx 2 R

R



        | W xx | dx +  ∂xx W , A (Λ) xx W x dx. 2

 C (δ + ε0 ) R

R

Here in the first equality, we used the fact that matrix A (Λ) is   symmetric and integration by parts. In this way, the term R ∂xx W , A (Λ)∂xxx W  dx to the term − 12 R ∂xx W , ( A (Λ))x ∂xx W  dx. This is a key point for energy estimates of solutions to positive symmetric hyperbolic system. On the other hand, since

            A (Λ)    A (Λ)(Λx )2 +  A (Λ)|Λxx |   A (Λ) | W x | + |Λ¯ x | 2 +  A (Λ) | W xx | + |Λ¯ xx | xx where denotes the partial derivatives with respect to Λ = ( v , u , θ). Consequently, we have

          2 2  ∂xx W , A (Λ) W x dx  C (δ + ε0 ) | W | dx + | W | dx xx x   xx R

R

R

where (1.11) and | W x | L ∞  ε0 due to (2.4) and Cauchy–Schwarz inequality are used. Then we can obtain

          2 2  ∂xx W , A (Λ) W x  dx  C (δ0 + ε0 ) | W xx | dx + | W x | dx .  xx R

R

R

Similarly,

 R





A 0 (Λ)∂xx W , A 0 (Λ)−1







 | W xx |2 dx +

A (Λ) W x dx  C (δ0 + ε0 ) xx R

R

 | W x |2 dx .

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

1053

Consequently, since W = (φ, ψ, ζ )t , Integrating (3.60) over [0, T ] with respect to t and carry out a similar argument as in Lemma 3.4 yield





2 sup (φxx , ψxx , ζxx )(·, t ) +

0t  T

T ωxx 21 ds 0

 2  C (φ0 , ψ0 , ζ0 ) + C (ε0 + δ)

T φxx , ψxx , ζxx 2 ds + C δ.

2

(3.63)

0

Here, we should remark that we use integration by parts and the facts that matrices A 0 (Λ), 0 A (Λ) are symmetric to deal with the highest order derivative terms R A (Λ)∂xx W , ∂xx W t  dx and

∂ W , A (Λ)∂ W  dx as above. And the other low order space derivative terms can be estimated xxx R xx by using (1.11) and | W x , ωx , ωxx | L ∞  ε0 due to (2.4). T  Below, we will estimate 0 R | W xx |2 dx ds, we apply the differential operator ∂x on the system 2

(3.46) and multiply the first resulting equation by − V φxx and the second equation by 2V ψ . Then R Θ xx calculating the sum of the two resulting equations and integrate over R × [0, T ], we can obtain

T

T 2

φxx , ψxx ds 

C φ0 , ζ0 , ψ0x 21

+

C φ, ζ, ψx 21

ζx , ω, ωx 21 ds + C δ

+C

0

0

T  C φ0 , ζ0 , ψ0x 21 + C

ζx 21 ds + C δ,

(3.64)

0

where in the second inequality, we used (3.63) and Lemma 3.5. In this step, all estimates can be achieved by using Cauchy–Schwarz inequality and | W x , ωx , ωxx | L ∞  ε0 due to (2.4), which is similar as in (3.49). Then, we differentiate (3.50) with respect to x, A similar calculation as that in (3.51) yields

T

T 2

ζxx ds  C 0

T ω

23 ds

φx , ψx , ζx , ω, ωx 21 ds + C δ 2 .

+ C (δ + ε0 )

0

(3.65)

0

In this way, we have

 2 sup (φxx , ψxx , ζxx )(·, t ) +

0t  T

 2  C (φ0 , ψ0 , ζ0 ) + C δ.

T



ωxx 21 + φxx 2 + ψxx 2 + ζxx 2 (·, s) ds

0

2

And the elliptic estimate about

(3.66)

ωx is similar as (3.57)  2 ωx 22  C (φ0 , ψ0 , ζ0 )2 + C δ.

We finish the proof of Lemma 3.7.

2

(3.67)

1054

J. Wang, F. Xie / J. Differential Equations 251 (2011) 1030–1055

4. Proof of the main Theorem 1.1 Combining the estimates of Lemmas 3.1–3.7, we have the following proposition. Proposition 4.1. For any T > 0 there exist suitably small ε0 , δ0 independent of T , such that if (φ, ψ, ζ, ω) ∈ E [0, T ] satisfy (2.2), (2.4) and (2.5) with δ = |θ+ − θ− | < δ0 then we have





2 sup (φ, ψ, ζ, ω, ωx )(·, t )2 +

0t  T

 2  C (φ0 , ψ0 , ζ0 ) + C δ.

T



ω 23 + φx 21 + ψx 21 + ζx 21 ds

0

(4.1)

2

Then, it is the position for us to prove the main theorem. Proof of Theorem 1.1. Since Proposition 4.1 is proved, we can choose the initial perturbation (φ0 , ψ0 , ζ0 ) 2 and the strength of the contact discontinuity δ suitably small. Then we can close the uniform a priori estimates (2.4) and (2.5) by choosing (φ0 , ψ0 , ζ0 ) 22 and δ suitably small. In this way, we can extend the local solution to T = ∞. Moreover, the estimate (4.1) and (2.2) imply that

  ∞       (φx , ψx , ζx )(t )2 +  d (φx , ψx , ζx )(t )2  dt < ∞,  dt 

(4.2)

0

which as well as (4.1) and the Sobolev inequality lead to the following asymptotic behavior of solutions





lim sup( v − V , u − U , θ − Θ)(x, t ) = 0.

t →∞ x∈R

(4.3)

From (4.3), (1.11), (1.12) (3.54) and (4.2), we have (1.14) easily. Thus we finish the proof of the main theorem. 2 Acknowledgments The authors would like to thank Prof. Feimin Huang for his helpful discussion. And the authors also thank the referee(s) for their valuable suggestions for improving the presentation of this work. Jing Wang is supported by Science and Technology Commission of Shanghai Municipality (09PJ1408200), NSFC (No. 10926163) and Shanghai Municipal Education Commission Excellent Young Teacher Program (No. ssd10029). Feng Xie is supported by NSFC (No. 10801016), Shanghai Jiao Tong University ChenXing Project SMC-B and Alexander von Humboldt Foundation Germany. References [1] D.G. Aronson, The porous media equations, in: A. Fasano, M. Primicerio (Eds.), Nonlinear Diffusion Problem, in: Lecture Notes in Math., vol. 1224, Springer-Verlag, Berlin, 1986. [2] W. Gao, C.J. Zhu, Asymptotic decay toward the planar rarefaction waves for a model system of the radiating gas in two dimensions, Math. Models Methods Appl. Sci. 18 (4) (2008) 511–541. [3] F.M. Huang, J. Li, A. Matsumura, Asymptotic stability of combination of viscous contact wave with rarefaction waves for one-dimensional compressible Navier–Stokes system, Arch. Ration. Mech. Anal. 197 (2010) 89–116. [4] F.M. Huang, R.H. Pan, Convergence rate for compressible Euler equations with damping and vacuum, Arch. Ration. Mech. Anal. 166 (4) (2003) 359–376.

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