Asymptotic stability of viscous contact discontinuity to an inflow problem for compressible Navier–Stokes equations

Nonlinear Analysis 74 (2011) 6617–6639

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Asymptotic stability of viscous contact discontinuity to an inflow problem for compressible Navier–Stokes equations✩ Tingting Zheng a,b , Jianwen Zhang b,∗ , Junning Zhao b a

Department of Computer and Information Science, Fujian Agriculture and Forestry University, Fuzhou 350002, PR China

b

School of Mathematical Sciences, Xiamen University, Xiamen 361005, PR China

article

info

Article history: Received 4 February 2010 Accepted 27 June 2011 Communicated by Ravi Agarwal MSC: 35B40 35B45 76N10 76N17

abstract This paper is concerned with an initial-boundary value problem for one-dimensional full compressible Navier–Stokes equations with inflow boundary conditions in the half space R+ = (0, +∞). The asymptotic stability of viscous contact discontinuity is established under the conditions that the initial perturbations and the strength of contact discontinuity are suitably small. Compared with the free-boundary and the initial value problems, the inflow problem is more complicated due to the additional boundary effects and the different structure of viscous contact discontinuity. The proofs are given by the elementary energy method. © 2011 Elsevier Ltd. All rights reserved.

Keywords: Inflow problem Contact discontinuity Asymptotic stability Compressible Navier–Stokes equations The half space

1. Introduction This paper is concerned with an ‘‘inflow problem’’ for one-dimensional compressible viscous heat-conducting flows in the half space R+ = (0, ∞), which is governed by the following initial-boundary value problem in Eulerian coordinate (˜x, t ):

ρ˜ + (ρ˜ u˜ ) = 0, (˜x, t ) ∈ R × R , t + + x˜   (ρ˜ u˜ )t + (ρ˜ u˜ 2 + p˜ )x˜ = µ˜ux˜ x˜ ,          u˜ 2 u˜ 2 ρ˜ e˜ + + ρ˜ u˜ e˜ + + p˜ u˜ = κ θ˜x˜ x˜ + (µ˜uu˜ x˜ )x˜ , 2 2  t x˜    ˜ )|x˜ =0 = (ρ− , ub , θ− ) with ub > 0, ˜ ( ρ, ˜ u , θ   (ρ, ˜ u˜ , θ˜ )|t =0 = (ρ˜ 0 , u˜ 0 , θ˜0 )(˜x) → (ρ+ , ub , θ+ ) as x˜ → ∞,

(1.1)

where ρ˜ , u˜ and θ˜ are the density, velocity and absolute temperature, respectively; µ > 0 is the viscosity coefficient and κ > 0 is the heat-conductivity coefficient. Throughout the paper, it is assumed that ρ± , ub and θ± are prescribed positive

✩ This work is partially supported by the NSFC (Grant Nos 10801111 and 10971171), the Fundamental Research Funds for the Central Universities (Grant No. 2010121006), and the Natural Science Foundation of Fujian Province of China (Grant No. 2010J05011). ∗ Corresponding author. E-mail addresses: [email protected] (T. Zheng), [email protected] (J. Zhang), [email protected] (J. Zhao).

0362-546X/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2011.06.044

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constants. We shall only consider the case of viscous polytropic ideal gases, so that the pressure p˜ = p˜ (ρ, ˜ θ˜ ) and the internal energy e˜ = e˜ (ρ, ˜ θ˜ ) satisfy the second law of thermodynamics: p˜ (ρ, ˜ θ˜ ) = Rρ˜ θ˜ ,

e˜ (ρ, ˜ θ˜ ) =

R

γ −1

θ˜ + const.,

(1.2)

where γ > 1 is the adiabatic exponent and R > 0 is the gas constant. The boundary condition (1.1) 4 implies that, through the boundary x˜ = 0, the fluid with density ρ− flows into the region R+ at the speed ub > 0. So, the initial-boundary value problem (1.1) is the so-called inflow problem. On the other hand, in the case that ub = 0 (resp. ub < 0), the problem is called the impermeable wall (resp. outflow) problem in which the boundary condition of density cannot be imposed. In terms of various boundary values, Matsumura [1] classified all possible large-time behaviors of the solutions for the one-dimensional (isentropic) compressible Navier–Stokes equations. Our main purpose is to study the asymptotic stability of the contact discontinuity for the inflow problem (1.1). It is well known that there are three basic wave patterns for the 1D compressible Euler equations, including two nonlinear waves, say shock and rarefaction waves, and a linearly degenerate wave, say contact discontinuity. There have been a lot of works on the asymptotic behaviors of solutions to the initial/initial-boundary value problem for the Navier–Stokes equations toward these basic waves or their viscous versions, see, for example, [1–22] and the references therein. In what follows, we briefly recall some related references. Concerning the inflow problem, Matsumura and Nishihara [18] considered an inflow problem for the one-dimensional isentropic model system of compressible viscous gas (i.e. the 1D isentropic Navier–Stokes (1.1) 1 –(1.1) 2 with p˜ = Rρ˜ γ ) and established the stability theorems on both the boundarylayer solution and the superposition of a boundary-layer solution and a rarefaction wave. We also refer to the paper due to Huang et al. [3] in which the asymptotic stability on both the viscous shock wave and the superposition of a viscous shock wave and a boundary-layer solution are studied. On the other hand, the problem of stability of contact discontinuities is more subtle due to the fact that contact discontinuities are associated with linear degenerate fields and are less stable than the nonlinear waves for the inviscid system (Euler equations). It was observed in [23,24], where the metastability of contact waves was studied for viscous conservation laws with artificial viscosity, that the contact discontinuity cannot be the asymptotic state for the viscous system, and a diffusive wave, which approximated the contact discontinuity on any finite time interval, actually dominates the large-time behavior of solutions. The nonlinear stability of contact discontinuity for the (full) compressible Navier–Stokes equations was then investigated in [4,7] for the free-boundary value problem and [5,6] for the Cauchy problem. It is worth pointing out that the analysis in [5,6] is more complicated than that in [4,7], since a Poincaré-type inequality, which plays an essential role in the proof of [4,7], is not valid for the Cauchy problem due to the unboundedness of the domain considered. To overcome this difficulty, the authors introduced the anti-derivative of the perturbation (ϕ, ψ, ζ ) as dependent variables and worked with the integrated error system (cf. [5,6]). Very recently, Huang et al. [2] cleverly exploited an elementary inequality concerning the heat kernel to prove the stability of the combination wave of viscous contact wave with rarefaction waves, since the anti-derivative method used in [5,6] is not available any more due to the combination of rarefaction wave. Moreover, their results (see, Theorem 1 in [2]) improve those obtained in [5,6], and the method is somewhat simpler. However, to our best knowledge, there is no any mathematical literature known for the large-time behaviors of solutions to the inflow problem of the full compressible viscous heat-conducting Naiver–Stokes equations due to various difficulties. So, the aim of this paper is to show that the contact discontinuities are metastability wave patterns for the inflow problem (1.1) of the full Navier–Stokes system. To be more precise, as that in [1,18], we first transform problem (1.1) to the one in Lagrangian coordinate (ξ , t ) and then make use of the coordinate transformation x = ξ − st to reformulate the origin problem in the form:

 vt − svx − ux = 0, (x, t ) ∈ R+ × R+ ,     u    Rθ x   u − su + = µ ,  t x  v x v x    R R Rθ θx u2 (1.3) θ − s θ + u = κ +µ x,  t x x  γ − 1 γ − 1 v v v  x     (v, u, θ )|x=0 = (v− , ub , θ− ), t > 0, (v, u, θ )|t =0 = (v0 , u0 , θ0 ) → (v+ , ub , θ+ ) as x → ∞, where v = v(x, t ), u = u(x, t ) and θ = θ (x, t ) are the specific volume, velocity and temperature, respectively; v± , ub and θ± are given positive constants, while µ, κ , R and γ are the same physical constants as in (1.1). Here, it should be noted that s = −ub /v− < 0, the sign of which is important for the boundary analysis. The corresponding Euler system of (1.3) with Riemann initial data reads as follows:

 vt − svx − ux = 0,   ut − sux + p(v, θ )x = 0,    R R Rθ θt − sθx + ux = 0, γ −1 γ −1 v      (v, u, θ )(x, 0) = (v− , ub , θ− ) if x < 0, (v, u, θ )(x, 0) = (v+ , ub , θ+ ) if x > 0,

(1.4)

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where v±, ub and θ ± are the same ones as in (1.3). It is well known (cf. [25]) that the Riemann problem (1.4) admits a contact discontinuity defined by



V , U, Θ =





(v− , ub , θ− ), (v+ , ub , θ+ ),

x < −st , x > −st ,

(1.5)

provided there holds p− =

Rθ−

v−

Rθ+

= p+ =

v+

.

(1.6)

As was mentioned above, the contact discontinuity (V , U , Θ ) in (1.5) cannot be the asymptotic state for the compressible Navier–Stokes equations. Therefore, we aim to construct a viscous contact wave (V , U , Θ ), which dominates the large-time behavior of solutions of problem (1.3) and approximates (V , U , Θ ) on any finite time interval as the heat conductivity κ goes to zero. However, due to the presence of the shift x = −st, the mathematical structure of the viscous contact discontinuity constructed here is more complicated and is different from the previous works, see, for example, [2,3,5–7] in which the shift x = 0 is independent of t. Moreover, the viscous contact discontinuity in these references is a self-similarity solution with properties known from [26–28]. But, the viscous contact discontinuity presented here needs to be analyzed carefully, since it cannot be found (at least for the authors) in any literature. For the construction, by (1.6) we first conjecture that P (V , Θ ) =

RΘ V

= p+ .

(1.7)

˜ =Θ ˜ (ξ ) of the following ODE system: Analogous to that in [3,7], we now seek a self-similarity solution Θ  ′ ′   1 Θ  ′   − ξ Θ = a  , 2 Θ    st  Θ  √ = θ− , 1+t

κ p+ (γ − 1) > 0, γ R2

a=

(1.8)

 (+∞) = θ+ Θ

with x + st

ξ= √

d and ′ = . dξ

1+t

√

It should be noted that the boundary condition of this ODE system is imposed on ξ = st / 1 + t, and hence, the selfsimilarity solution depends strongly on the time variable and is more complicated than that for the free-boundary and the initial value problems (cf. [2,3,5–7]). To solve (1.8), we first rewrite the first equation (1.8) 1 as 1

 ′ )′ = (ln Θ  )′ − (ln Θ

, ξΘ

2a which, integrated over (0, ξ ∗ ), gives

 ′ (ξ ∗ ) − ln Θ  ′ (0) = ln Θ

∫

ξ∗



 )′ − (ln Θ

0

1 2a

  (ξ˜ )dξ˜ , ξ˜ Θ

from which it follows that

 ′ (ξ ∗ ) = Θ  ′ (0) exp Θ

ξ∗

∫



 )′ − (ln Θ

0

1 2a

  ˜ξ Θ  (ξ˜ )dξ˜ .

(1.9)

√

Integrating (1.9) over (st / 1 + t , ξ ) with respect to ξ ∗ , we get

 (ξ ) = Θ  ′ (0) Θ

ξ

∫

ξ∗

∫



exp √ st 1+t

 )′ − (ln Θ

0

1 2a





 (ξ˜ )dξ˜ dξ ∗ + θ− . ξ˜ Θ

(1.10)

√

On the other hand, if we integrate (1.9) over (st / 1 + t , +∞), then

θ+ − θ− = Θ (0) ′

∫

ξ∗

∫

∞

exp √ st 1+t



) − (ln Θ ′

0

1 2a









 (ξ˜ )dξ˜ dξ ∗ , ξ˜ Θ

which, inserted into (1.10), yields

 (ξ ) = Θ

θ+ − θ− I

∫

ξ

∫ exp

√ st 1+t

0

ξ∗



 )′ − (ln Θ

1 2a

 (ξ˜ )dξ˜ dξ ∗ + θ− ξ˜ Θ

(1.11)

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with I = I (t ) =

∫

∞

∫

ξ∗



exp √ st 1+t

) − (ln Θ ′

0

1 2a





 (ξ˜ )dξ˜ dξ ∗ > 0. ξ˜ Θ

(1.12)

With the help of (1.11) and (1.7), we put

 (ξ ), Θ (x, t ) = Θ

V (x, t ) =

R p+

Θ (x, t ).

(1.13)

√

Note that the function Θ (x, t ) defined in (1.13) depends not only on ξ , but also on st / 1 + t. Thus, keeping this in mind, we compute directly from (1.11) and (1.12) that

Θx =

 ′ (ξ ) Θ , (1 + t )1/2



Θx Θ





1

=

1+t

x

′ Θ  Θ

′

x + st

(ξ ),

ξ= √

1+t

,

and

Θt = −

θ+ − θ−



ξ

∞

∫

exp ξ

1

−

(1 + t )1/2 ∫ ∗ 

I2

×

s

st

2 (1 + t )3/2

 )′ − (ln Θ

0

√ st 1+t

∫

 exp



 )′ − (ln Θ

0

1 2a

  ˜ξ Θ  (ξ˜ )dξ˜



   1 ξ s  ′ (ξ ).  (ξ˜ )dξ˜ dξ ∗ + √ − Θ ξ˜ Θ 2a 21+t 1+t 1

Consequently, it follows from (1.8) and (1.11) that Θ = Θ (x, t ) satisfies

    Θ − s Θ = a Θx + K , t x Θ x  Θ (0, t ) = θ− , Θ (+∞, t ) = θ+ , κ p+ (γ −1) γ R2

> 0 and K = K (x, t ) is defined by ∫ st      √ θ+ − θ− s 1 st 1 1+t ′ ˜ ˜ ˜   K = − − exp (ln Θ ) − ξ Θ (ξ )dξ √ I2 2 (1 + t )3/2 2a 1+t 0 ∫ ∗    ∫ ∞ ξ 1 x + st ′ ˜ ˜ ˜   (ln Θ ) − ξ Θ (ξ )dξ dξ ∗ with ξ = √ . × exp 2a 1+t 0 ξ

where a =

(1.14)

(1.15)

As for U (x, t ), we put U ( x, t ) =

κ(γ − 1) Θx R − γR Θ p+

∞

∫

K (y, t )dy + ub ,

(1.16)

x

then it follows from (1.7), (1.13) and (1.14) that Vt − sVx − Ux = 0, Ut − sUx + p(V , Θ )x = µ

(1.17)



Ux V



+ F,

(1.18)

x

where p(V , Θ ) = RΘ /V = p+ and F = F (x, t ) is defined by F (x, t ) =

     ∫ ∞ κ(γ − 1) (ln Θ )xx R K R (ln Θ )xt − s(ln Θ )xx − µ −µ − (Kt − sKx )dx. γR V p+ V x p+ x x

(1.19)

To sum up, we have constructed a pair of functions (V , U , Θ )(x, t ) such that

V − sV = U , t x x     Ux    U − sUx + p(V , Θ )x = µ + F,   t V x   R R Θ Θx γR   Θ − s Θ + R U = κ + K, t x x   γ − 1 γ − 1 V V γ −1  x   Θ (0, t ) = θ− , Θ (+∞, t ) = θ+ ,

(1.20)

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where RΘ /V = p+ , and K and F are the functions defined in (1.15) and (1.19) respectively. We shall show in the next section that (U , V , Θ ) approximates (V , U , Θ ) in Lp norm with p ≥ 1 on any finite time interval as the heat conductivity κ goes to zero. So, we call (V , U , Θ ) the viscous contact wave for the Navier–Stokes system (1.3). Our main purpose is to justify that the solution (v, u, θ ) of the Navier–Stokes system (1.3) asymptotically tends to the viscous contact discontinuity (V , U , Θ ). Roughly speaking, the main result is: ‘‘if the initial perturbations and the strength of the contact discontinuity are sufficiently small, then the viscous contact discontinuity is asymptotic stable’’. For the proof, we shall utilize the analogous technique developed in [2], since the Poincaré-type inequality used for the freeboundary value problem in [4,7] cannot work for the inflow problem, see, Remark 2.1. However, comparing with the Cauchy problem considered in [2,5,6], we have to deal with some additional terms caused by the boundary effects and the complicated/different structure of viscous contact discontinuity. The rest of this paper is outlined as follows. In Section 2, we study the properties of the viscous continuity discontinuity (U , V , Θ ) in (1.20). In Section 3, we reformulate the problem and give the precise statement of our main theorem. Finally, we complete the proof of the main result by the global a priori estimates established in Section 4. 2. Preliminaries This section is devoted to the study of the viscous contact discontinuity (V , U , Θ ) in (1.20). First, it is easy to get that Lemma 2.1. Assume that δ := |θ+ − θ− | ∈ (0, 1). Then for i = 1, 2, 3, it holds that x + st

 i  ∂ Θ (x, t ) ≤ C δ(1 + t )−i/2 exp{−C0 ξ 2 },

ξ= √

x

1+t

.

(2.1)

Here and hereafter, C and Ci (i = 0, 1, 2, . . .) denote various positive constants which depend only on θ± and the generic physical constants µ, κ , etc., but not on t. Proof. Keeping in mind that ξ = √x+st , we deduce from (1.11) and (1.13) that 1 +t

 ′ (ξ )ξx = (1 + t )−1/2 Θ  ′ (ξ ) Θx (x, t ) = Θ

(2.2)

and

 ′ (ξ ) = Θ

θ+ − θ− I

ξ

∫



exp

 )′ − (ln Θ

0

1 2a

   (ξ˜ )dξ˜ , ξ˜ Θ

(2.3)

 (ξ ) are increasing (resp. decreasing) with respect to x and ξ , respectively, if it holds from which it follows that Θ (x, t ) and Θ that θ+ > θ− > 0 (resp. 0 < θ+ < θ− ). For simplicity and without loss of generality, we only consider the case of θ+ > θ− , since the other one can be dealt with in the same way. Due to the monotonicity, we have 0 < θ− ≤ Θ (x, t ),

 (ξ ) ≤ θ+ < ∞, Θ

(2.4)

so that

  ∫ ξ      θ− θ+ ξ 2 1 θ θ ξ2  )′ − ξ˜ Θ  (ξ˜ )dξ˜ ≤ + exp − − exp − ≤ exp (ln Θ , θ+ 4a 2a θ− 4a 0 which implies that C2 exp{−C1 ξ 2 } ≤ exp

ξ

∫



 )′ − (ln Θ

0

1 2a

   (ξ˜ )dξ˜ ≤ C4 exp{−C3 ξ 2 }. ξ˜ Θ

(2.5)

On the other hand, recalling the definition of I in (1.12), we infer from (2.5) that C −1 ≤ C2

∫

∞ √ st 1+t

exp{−C1 ξ 2 }dξ ≤ I ≤ C4

∫

∞ √ st 1+t

exp{−C3 ξ 2 }dξ ≤ C .

(2.6)

Putting (2.5) and (2.6) into (2.3), we see that

 ′  Θ  (ξ ) ≤ C δ exp{−C0 ξ 2 }, which combined with (2.2) leads to the desired estimate of Θx . By virtue of (2.3), (2.4) and (2.7), we have from a direct computation that (0 < δ < 1)

 ′    (ξ )  ′′   Θ   ′  1 Θ  (ξ ) =   (ξ ) Θ  ′ (ξ ) ≤ C (1 + |ξ |) Θ  (ξ ) , − ξ Θ  Θ   (ξ ) 2a

(2.7)

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which, together with (2.7), immediately leads to

 2   ′′   ′  ∂ Θ  = (1 + t )−1 Θ  (ξ ) ≤ C (1 + t )−1 (1 + |ξ |) Θ  (ξ ) ≤ C δ(1 + t )−1 exp{−C0 ξ 2 }, x since for any positive constants α1 and α2 , it holds that (1 +|ξ |α2 )e−α1 ξ ≤ γ e−βξ with two positive constants β, γ depend ′ (ξ ), Θ  ′′ (ξ ) and the fact that δ ∈ (0, 1), we obtain in a similar manner ing only on α1 and α2 . Finally, using the estimates of Θ that 2

2

 3   ′′′   ′  ∂ Θ  = (1 + t )−3/2 Θ  (ξ ) ≤ C (1 + t )−3/2 (1 + |ξ |2 ) Θ  (ξ ) ≤ C δ(1 + t )−3/2 exp{−C0 ξ 2 }. x The proof of Lemma 2.1 is therefore complete.



Remark 2.1. In the case that s = 0, it is easy to see from Lemma 2.1 that ∞

∫

x(|Θxx | + Θx2 )dx ≤ C

∫

∞

1+t

0

0

x2



x

exp −C0



1+t

dx ≤ C δ.

This estimate plays an important role in the analysis of [4,7] when a Poincaré-type inequality was used to derive some basic estimates. However, when s ̸= 0, the above estimate is not valid any more. So, the technique used in [4,7] does not work for the present problem. The next lemma is concerned with the relations between the viscous contact discontinuity and the contact discontinuity. We shall show that as the heat conductivity κ goes to zero, (U , V , Θ ) will approximate (V , U , Θ ) in Lp (p ≥ 1) norm on any finite time interval. Lemma 2.2. For any given T ∈ (0, ∞), there is a positive constant M, independent of κ , such that for any p ≥ 1 and t ∈ [0, T ],

  (V − V , U − U , Θ − Θ )(t ) p → 0 as κ → 0. L For simplicity, here and in what follows, we denote the usual Sobolev spaces by Lp = Lp (0, ∞) and H m = H m (0, ∞), and the norms by ‖ · ‖Lp and ‖ · ‖H m . Proof. Keeping in mind that s = −ub /v− < 0 and letting

Ω1 = (0, −st ) and Ω2 = (−st , ∞).

(2.8)

By the definition of Θ in (1.5), to estimate ‖Θ − Θ ‖Lp , it suffices to prove

‖Θ − θ+ ‖Lp (Ω2 ) ≤ M δ (κ(1 + t ))1/(2p) ,

‖Θ − θ− ‖Lp (Ω1 ) ,

p ≥ 1,

(2.9)

where and in what follows we shall denote by M, Mi (i = 0, 1, 2, . . .) the positive constants independent of κ . Note that since a = κ p+ (γ − 1)/(γ R2 ), it holds that κ = O(a) as κ → 0. To deal with ‖Θ − θ− ‖Lp (Ω1 ) , we first exploit (2.4) to get that M2 exp{−M1 ξ 2 /a} ≤ exp

ξ

∫

   1  )′ − ξ˜ Θ  (ξ˜ )dξ˜ ≤ M4 exp{−M3 ξ 2 /a}, (ln Θ 2a

0

(2.10)

from which and (1.12) we find M −1

√

∫

a ≤ M2

∞ √ st 1+t

exp{−M1 ξ 2 /a}dξ ≤ I ≤ M4

∞

∫

√ st 1+t

exp{−M3 ξ 2 /a}dξ ≤ M

√

a.

(2.11)

 and Θ given respectively in (1.11) and (1.13) that Thus, we deduce from the definitions of Θ δ

0 ≤ Θ (x, t ) − θ− ≤ M √ a

∫

ξ √ st 1+t

exp{−M0 ξ 2 /a}dξ .

(2.12)

In order to estimate the right-hand side of (2.12), we observe that for any β ≥ α ≥ 0, β

∫ 0 ≤

2

e−x dx

2

α π

∫

2

≤ 0

β

∫ =

2

e−x dx

α √

∫

dω √

2β

2α

β

∫

2

e−y dy =

α



2 2 e−(x +y ) dxdy

Π

re−r dr ≤ M e−2α − e−2β 2

∫∫

2

 2

≤ Me−2α , 2

(2.13)

where Π = {(x, y) : α ≤ x ≤ β, α ≤ y ≤ β} ⊂ D = {(x, y) : 2α 2 ≤ x2 + y2 ≤ 2β 2 , x ≥ 0, y ≥ 0}. Similarly, using the transformation argument, we also have

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

α

∫ 0≤

2

2

e−x dx

≤ Me−2α

2

for any β ≤ α ≤ 0,

β

6623

(2.14)

from which and (2.12) we find that

√ ξ = (x + st )/ 1 + t ,

0 ≤ Θ (x, t ) − θ− ≤ M2 δ exp{−M1 ξ 2 /a},

√

since x ∈ Ω1 implies ξ ∈ (st / 1 + t , 0). Consequently, −st

∫

‖Θ (·, t ) − θ− ‖pLp (Ω1 ) ≤ M δ p

exp{−M0 ξ 2 /a}dx 0 1/2

= M δ (a(1 + t )) p

∫

ξ

0 √ st 1+t

exp{−M0 ξ 2 /a}d √ a

≤ M δ p (a(1 + t ))1/2 .

(2.15)

On the other hand, using (2.4), (2.10), (2.11) and (2.13), we get from (1.11) and (1.13) that 0 ≤ θ+ − Θ (x, t ) =

δ ≤ M2 √

θ+ − θ−

∞

∫

I

ξ

  ∫ ∗ ξ  (ξ ∗ ) Θ 1  (ξ˜ )dξ˜ dξ ∗ ξ˜ Θ exp −  (0) 2a Θ 0

∞

∫

a ξ

exp{−M1 ξ 2 /a}dξ ≤ M2 δ exp{−M0 ξ 2 /a},

ξ ≥ 0,

from which it follows that

∫ ∞ ‖Θ (·, t ) − θ+ ‖pLp (Ω2 ) ≤ M δ p exp{−M0 ξ 2 /a}dx −st ∫ ∞ ξ p 1/2 = M δ (a(1 + t )) exp{−M0 ξ 2 /a}d √ ≤ M δ p (a(1 + t ))1/2 . a

0

This, together with (2.15), completes the proof of (2.9). As an immediate result, we also obtain the estimate of ‖V − V ‖Lp due to the facts that V = RΘ /p+ and V = RΘ /p+ . It remains to estimate ‖U − U ‖Lp . To do so, we first deduce from (1.16) and (1.5) that U −U =

κ(γ − 1) Θx R − γR Θ p+

∞

∫

K (y, t )dy,

(2.16)

x

where K = K (x, t ) is the function defined in (1.15). By virtue of (2.2), (2.3), (2.10) and (2.11), we find

 ′ (ξ ) ≤ M2 δ(a(1 + t ))−1/2 exp{−M1 ξ 2 /a}, 0 ≤ Θx (x, t ) = (1 + t )−1/2 Θ so that,

‖Θx ‖Lp ≤ M δ(a(1 + t ))−1/2

1/p

∞

∫

exp{−M0 ξ 2 /a}dx 0

= M δ(a(1 + t )) ≤ M δ(a(1 + t ))

∫

1−p 2p

∞ √ st 1+t

1−p 2p

ξ exp{−M0 ξ /a}d √

1/p

2

a

,

from which and (2.4) we immediately infer that for any p ≥ 1,

  1−p 1+p κ Θ −1 Θx Lp ≤ M δ(1 + t ) 2p κ 2p → 0 as κ → 0.

(2.17)

Recalling the definition of K in (1.15) and using (2.4), (2.10) and (2.11), we deduce from (2.13) and (2.14) that

   ξ2 s2 t 2 −1/2   M δ( a ( 1 + t )) exp − M − M , ξ ≥ 0, 0 0  1 a( 1 + t ) a |K | ≤    s2 t 2 st  M1 δ(a(1 + t ))−1/2 exp −M0 , √ ≤ ξ ≤ 0. a( 1 + t ) 1+t Putting K (x, t ) :=

∞

∫

|K (y, t )|dy, x

(2.18)

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T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

similar to the proof of (2.9), we split ‖K (t )‖Lp into the following two terms:

‖K (t )‖Lp = ‖K (t )‖Lp (Ω1 ) + ‖K (t )‖Lp (Ω2 ) , where Ω1 and Ω2 are √ the same as in (2.8). Since a = O(κ) as κ → 0, it follows from (2.18) and (2.13) that (keeping in mind that ξ = (x + st )/ 1 + t and s < 0)

‖K (t )‖

p Lp (Ω1 )

−st

∫

p

∞

∫

|K |dy

≤

dx ≤ −st (1 + t )p/2

∫

√ st 1+t

x

0

≤ −st (1 + t )

p/2

∫

p

0 √ st 1+t

|K |dξ

p

∞

− st (1 + t )p/2

|K |dξ ∞

∫

|K |dξ

p

0

  (−st )p+1 s2 t 2 s2 t 2 p exp − M − M δ ( st ) exp − M 0 1 0 (a(1 + t ))p/2 a(1 + t ) a( 1 + t )   2 2 s t ≤ −M2 δ p (st ) exp −M → 0, as κ → 0. a(1 + t ) 

≤ M1 δ p



On the other hand, using (2.13) and (2.18) 1 , we have

‖K (t )‖

p Lp (Ω2 )

∫

∞

p

∞

∫

≤

|K |dy −st

(p+1)/2

dx = (1 + t )

x

∞

∫

∞

∫

0

ξ

p  |K |dξ dξ

∫ ∞  s2 t 2 p 1/2 exp{−M ξ 2 /a}dξ ≤ M1 δ (1 + t ) exp −M a( 1 + t ) 0   s2 t 2 p 1/2 → 0, as κ → 0. ≤ M2 δ (a(1 + t )) exp −M a(1 + t ) So, for any fixed t ≥ 0 and p ≥ 1, we have proved that ‖K (t )‖Lp → 0 as κ → 0. This, combined with (2.16) and (2.17), implies that ‖(U − U )(t )‖Lp → 0 as κ → 0. The proof of Lemma 2.2 is therefore complete.  Finally, we present some auxiliary estimates on K , U and F which play an important role in the proof of global a priori estimates. Lemma 2.3. Let K , U and F be the functions defined in (1.15), (1.16) and (1.19), respectively, and assume that δ = |θ+ − θ− | ∈ (0, 1). Then there exist two positive constants C0 and C , independent of t, such that



|K | ≤ C δ(1 + t )−1/2 exp{−C0 t 2 /(1 + t )} exp{−C0 ξ 2 }, |Kx |, |Kt | ≤ C δ(1 + t )−1 exp{−C0 t 2 /(1 + t )} exp{−C0 ξ 2 },

|Ux | ≤ C δ(1 + t )−1 exp{−C0 ξ 2 },

|Uxx | ≤ C δ(1 + t )−3/2 exp{−C0 ξ 2 },

(2.19) (2.20)

and t

∫ 0

‖F (t )‖4L1/3 dt ≤ C δ,

t

∫ 0

‖F (t )‖2L2 dt ≤ C δ 2 .

(2.21)

√

Here, as in the above, ξ = (x + st )/ 1 + t. Proof. First, from (2.18) it is easy to see that for any ξ ≥ 0 and t ≥ 0,

|K | ≤ C δ(1 + t )−1/2 exp{−C0 t 2 /(1 + t )} exp{−C0 ξ 2 }, √ while for any ξ ∈ [st / 1 + t , 0],   s2 t 2 −1/2 |K | ≤ C δ(1 + t ) exp −C1 1+t     s2 t 2 C1 2 C1 2 −1/2 ≤ C δ(1 + t ) exp −C1 + ξ exp − ξ 1+t 2 2 ≤ C δ(1 + t )−1/2 exp{−C0 t 2 /(1 + t )} exp{−C0 ξ 2 }. √ Thus, for any x ∈ [0, ∞), i.e. ξ ∈ [st / 1 + t , ∞), it holds that |K (x, t )| ≤ C δ(1 + t )−1/2 exp{−C0 t 2 /(1 + t )} exp{−C0 ξ 2 }.

(2.22)

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

6625

By virtue of (2.4)–(2.6) we can easily deduce from (1.15) that

|Kx | = (1 + t )−1/2 |Kξ | ≤ C δ(1 + t )−1 exp{−C0 t 2 /(1 + t )} exp{−C0 ξ 2 }.

(2.23)

Similar to the derivation of (2.22), after some simple computations we have from (1.15) that

|Kt | ≤ C δ(1 + t )−1 exp{−C0 t 2 /(1 + t )} exp{−C0 ξ 2 }.

(2.24)

So, the estimates indicated in (2.19) readily follows from (2.22)–(2.24). By (2.4) and (2.19) 1 and Lemma 2.1, a direct computation from (1.16) gives (0 < δ < 1)

  |Ux | ≤ C (|Θxx | + Θx2 + |K |) ≤ C δ(1 + t )−1 exp −C0 ξ 2 ,

(2.25)

where the elementary inequality t α e−t ≤ C (α), α > 0 was used to get from (2.19) 1 that

|K | ≤ C δ(1 + t )−1 exp{−C0 ξ 2 }(1 + t )1/2 exp{−C0 t 2 /(1 + t )} ≤ C1 δ(1 + t )−1 exp{−C0 ξ 2 }. Similarly, we also have for 0 < δ < 1 that

  |Uxx | ≤ C |Θxxx | + |Θxx ||Θx | + |Θx |3 + |Kx | ≤ C δ(1 + t )−3/2 exp{−C0 ξ 2 },

(2.26)

which, together with (2.25), proves (2.20). Finally, to obtain (2.21), we observe that (1.14) implies

[

(ln Θ )xt − s(ln Θ )xx = a

1

Θ



Θx Θ

]

 +

x x

K



Θ

,

x

so that, using (2.4) and (2.19) and Lemma 2.1, we infer from (1.19) (i.e. the definition of F ) that

  |F | ≤ C |Θxxx | + |Θxx ‖Θx | + |Θx |3 + C (|K | + |Kx | + |K ‖Θx |) + C

∞

∫

|Kt |dx x

≤ C (1 + t )−3/2 exp{−C0 ξ 2 } + C

∫

∞

|Kt |dx,

(2.27)

x

where we have dealt with the second term in the first inequality in a similar manner as that used for the derivation of (2.25), using Lemma 2.1, (2.19) and the inequality t α e−t ≤ C (α). To complete the proof, we are now in a task of estimating the integral of |Kt |. Indeed, in view of (2.19), we can utilize √ (2.13) and (2.14) to get that (noting that ξ = (x + st )/ 1 + t) ∞

∫

|Kt |dx ≤ C δ(1 + t )

−1

exp{−C0 t /(1 + t )} 2

∞

∫

x

exp{−C0 ξ 2 }dx x

= C δ(1 + t )−1/2 exp{−C0 t 2 /(1 + t )}

∞

∫

exp{−C0 ξ 2 }dξ

ξ

= C2 δ(1 + t )−1/2 exp{−C0 t 2 /(1 + t )} exp{−C1 ξ 2 }

(2.28)

for x ∈ Ω2 (i.e., ξ ≥ 0), and ∞

∫

|Kt |dx ≤ C δ(1 + t )−1 exp{−C0 t 2 /(1 + t )}

∞

∫

x

exp{−C0 ξ 2 }dx x

≤ C δ(1 + t )−1/2 exp{−C0 t 2 /(1 + t )}

∫

∞ √ st 1+t

exp{−C0 ξ 2 }dξ

≤ C1 δ(1 + t )−1/2 exp{−C0 t 2 /(1 + t )} (2.29) √ for x ∈ Ω1 (i.e., st / 1 + t ≤ ξ ≤ 0). Here, Ω1 and Ω2 are the same as in (2.8). Hence, we can make use of (2.28) and (2.29) to get that

∫   

∞ x

 ∫    |Kt |dx ≤ C   p L

∞ x

  |Kt |dx p

L (Ω1 )

∫  +C 

≤ C δ exp{−C0 t /(1 + t )}, 2

∞ x

  |Kt |dx p

p ≥ 1.

L (Ω2 )

(2.30)

From (2.27) and (2.30), it is easy to obtain the estimates stated in (2.21). Therefore, the proof of Lemma 2.3 is complete. 

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T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

3. Reformulation and main result Let (v, u, θ )(x, t ) be the solution of problem (1.3), and let (V , U , Θ )(x, t ) be the viscous contact discontinuity constructed in (1.20). In order to reformulate the problem, we put

(ϕ, ψ, ζ )(x, t ) = (v − V , u − U , θ − Θ )(x, t ).

(3.1)

Then, it follows from (1.3) and (1.20) that (ϕ, ψ, ζ ) satisfies

ϕt − sϕx = ψx ,         ζ Θϕ ψx Ux ϕ ψt − sψx + R −R =µ −µ − F, v x vV x v x vV x      2  θ ψx Ux ζ Ux Θ ϕ ζx Θx ϕ ux Ux2 cv (ζt − sζx ) + R +R −R =κ −κ +µ − − G, v v vV v x vV x v V

(3.2) (3.3)

(3.4)

with the following initial-boundary conditions

(ϕ, ζ )(0, t ) = 0, t ≥ 0, (ϕ, ψ, ζ )(x, 0) = (ϕ0 , ψ0 , ζ0 )(x),

(3.5) x ≥ 0,

(3.6)

where cv = R/(γ − 1), u = U + ψ , v = V + ϕ , θ = Θ + ζ and G := G(x, t ) = −µ

Ux2 V

+

γR K. γ −1

(3.7)

Here, K , U and F are the functions defined in (1.15), (1.16) and (1.19). From (1.6), (1.7), (1.11), (1.13) and (1.16), it is easy to verify that the initial-boundary data in (3.5) and (3.6) satisfies the compatible condition, and

(ϕ0 , ψ0 , ζ0 )(x) = (v − V , u − U , θ − Θ )(x, 0) → (0, 0, 0) as x → ∞. To state our main result, we assume throughout of this section that

(ϕ0 , ζ0 )(x) ∈ H01 (0, ∞),

ψ0 (x) ∈ H 1 (0, ∞).

(3.8)

Moreover, for an interval I ⊂ [0, ∞), we define the function space X (I ) = (ϕ, ψ, ζ ) ∈ C I ; H 1 |ϕx ∈ L2 I ; L2 , ψx , ζx ∈ L2 I ; H 1















.

(3.9)

The main result of this paper now reads as follows. Theorem 3.1. There exist two positive constants δ0 and ε0 such that if δ = |θ+ − θ− | ≤ δ0 and ‖(ϕ0 , ψ0 , ζ0 )‖H 1 ≤ ε0 , then problem (3.2)–(3.6) has a unique global solution (ϕ, ψ, ζ ) satisfying (ϕ, ψ, ζ ) ∈ X ([0, ∞)) and sup |(ϕ, ψ, ζ )(x, t )| → 0

as t → ∞.

(3.10)

x≥0

We shall prove Theorem 3.1 by combining the local existence and the global-in-time a priori estimates. Since the local existence of the solution is well known (see, for example, [4]), we omit it here for brevity. To prove the global existence part of Theorem 3.1, it is sufficient to establish the following a priori estimates. Proposition 3.1. Let (ϕ, ψ, ζ ) ∈ X ([0, T ]) be a solution of problem (3.2)–(3.6) for some T > 0. Then there exist positive constants δ0 , ε0 and C , independent of T , such that if N (T ) = sup ‖(ϕ, ψ, ζ )(t )‖H 1 ≤ ε0 ,

δ = |θ+ − θ− | ≤ δ0 ,

(3.11)

0 ≤t ≤T

then it holds that sup ‖(ϕ, ψ, ζ )(t )‖2H 1 +

0≤t ≤T

T

∫ 0

    ‖ϕx (t )‖2L2 + ‖(ψx , ζx )(t )‖2H 1 dt ≤ C δ + ‖(ϕ0 , ψ0 , ζ0 )‖2H 1 .

(3.12)

of Proposition 3.1 is based on the elementary L2 -energy method. The main step is to estimate the quantity  The proof  (1 + τ )−1 exp −C0 (x + sτ )2 /(1 + τ ) (ϕ 2 + ζ 2 )dxdτ , which comes from the viscous contact discontinuity. This will be done by modifying the argument in [2] where the Cauchy problem was considered. However, comparing with [2], we have to deal with some additional terms due to the boundary effects and the (more) complicated structure of viscous contact discontinuity. We mention here that the asymptotic behavior of (ϕ, ψ, ζ ) stated in (3.10) can be proved by using (3.2)–(3.4) and (3.12) and Sobolev inequality, see, the end of Section 4.

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

6627

4. A priori estimates This section is devoted to the proof of Proposition 3.1, which will be done by a series of lemmas. Lemma 4.1. If N (T ) and δ defined in (3.11) are suitably small, then

‖(ϕ, ψ, ζ )(t )‖2L2 +

t

∫ 0

  ∫ t δ2 ‖(ψx , ζx )(τ )‖2L2 dτ ≤ C δ + ψx2 (0, τ )dτ + ‖(ϕ0 , ψ0 , ζ0 )‖2L2 + η η 0   ∫ t∫ ∞ δ (x + st )2 +C exp −C0 (ϕ 2 + ζ 2 )dxdτ , 1+t 1+t 0 0

where η ∈ (0, 1) is a small constant to be chosen later. Proof. Similar to the proof of Lemma 3.2 in [4], using (1.20) 1 , (1.20) direct but tiresome computations that



ψ2 2

+ RΘΦ

+µ

v V

+ cv ΘΦ



θ Θ



 −s

t

Θψ Θζ +κ + Hx + Q = µ vθ vθ 2 x

ψ2

2 x 2



+ RΘΦ

2

ψψx v



v V

− Fψ − x

2

and (3.1), we deduce from (3.2)–(3.4) after some

+ cv ΘΦ



θ Θ

 x

ζG , θ

(4.1)

where Φ (z ) = z − ln z − 1 satisfying Φ (1) = Φ ′ (1) = 0 is a strictly convex function around z = 1, H =R

Θ ϕψ Ux ϕψ ζ ζx Θx ϕζ ζψ −R +µ −κ +κ , v vV vV vθ vθ V

and

ζ Ux Ux − (p+ − p)Ux − µ ϕψx γ −1 θ vV Θx ΘΘx Ux Θ2 U2 − κ 2 ζ ζx − κ 2 ϕζx − 2µ ζ ψx + κ 2x ϕζ + µ x ϕζ . vθ vθ V vθ vθ V vθ V Note that p = Rθ /v , p+ = RΘ /V , and |p+ − p| ≤ C (|ϕ| + |ζ |) if N (T ) is small enough. Using the smallness condition of N (T ) and Cauchy–Schwarz inequality, we easily get that Q = p+ Φ

  V

v

p+

Ux +

Φ



Θ θ



|Q | ≤ ε(ψx2 + ζx2 ) + C ε −1 (|Ux | + Ux2 + Θx2 )(ϕ 2 + ζ 2 ),

ε ∈ (0, 1),

from which and Lemmas 2.1 and 2.3 it follows that

  C ‖Q ‖L1 ≤ ε ‖ϕx ‖2L2 + ‖ζx ‖2L2 + ε

∞

∫ 0

   δ (x + st )2  2 exp −C0 ϕ + ζ 2 dx. 1+t 1+t

(4.2)

On the other hand, using Sobolev and Young inequalities, we have for small ε ∈ (0, 1) that

  ‖F ψ‖L1 + ‖Gζ ‖L1 ≤ C ‖ψ‖L∞ ‖F ‖L1 + ‖ζ ‖L∞ ‖G‖L1   1/2 1/2 1/2 1/2 ≤ C ‖ψ‖L2 ‖ψx ‖L2 ‖F ‖L1 + ‖ζ ‖L2 ‖ζx ‖L2 ‖G‖L1     4/3 4/3 ≤ ε ‖ψx ‖2L2 + ‖ζx ‖2L2 + C ε −1 ‖F ‖L1 + ‖G‖L1 .

(4.3)

Integrating (4.1) over R+ × (0, t ) and choosing ε suitably small, by virtue of (4.2) and (4.3) we obtain

‖(ϕ, ψ, ζ )(t )‖2L2 +

t

∫

‖(ψx , ζx )(τ )‖2L2 dτ   ∫ t∫ ∞  δ (x + st )2  2 2 ≤ C ‖(ϕ0 , ψ0 , ζ0 )‖L2 + C exp −C0 ϕ + ζ 2 dxdτ 1 + t 1 + t 0 0 ∫ t ∫ t   2  4/3 4/3 +C ‖F (τ )‖L1 + ‖G(τ )‖L1 dτ + C ψ + |ψψx | (0, τ )dτ , 0

0

(4.4)

0

since the integration of Hx vanishes due to the boundary conditions (3.5). We still need to estimate the last two terms on the right-hand side of (4.4). First, recalling the definition of G in (3.7) and using (2.19)–(2.21), we have

∫ t  4/3 4/3 ‖F (τ )‖L1 + ‖G(τ )‖L1 dτ ≤ C δ. 0

(4.5)

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T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

In order to deal with the boundary terms, we first observe from (1.16) that

κ(γ − 1) ψ(0, t ) = u(0, t ) − U (0, t ) = − γR



Θx Θ



(0, t ) +

∞

∫

R p+

K (y, t )dy 0

√

so that, using (2.1), (2.4) and (2.19), we find (noting that ξ = (x + st )/ 1 + t)

|ψ(0, t )| ≤ √

Cδ 1+t

 exp −

C0 t 2



∫

∞

exp −C ξ 2 dx

1+t





1+



0

  C0 t 2 ≤ C δ exp − . 1+t

Consequently, t

∫

ψ 2 (0, τ )dτ ≤ C δ 2

∫

t

 exp −

0

0

C1 τ 2 1+τ



dτ ≤ C δ 2 ,

from which and Cauchy–Schwarz inequality it follows that t

∫



 ψ 2 + |ψψx | (0, τ )dτ ≤ η

t

∫

0

ψx2 (0, τ )dτ + 0

C

η

δ2 ,

η ∈ (0, 1).

This, together with (4.4) and (4.5), completes the proof Lemma 4.1.



Lemma 4.2. If δ > 0 and N (T ) are sufficiently small, then for any t ∈ [0, T ] we have

∫ t ∫ t   ‖(ϕ, ψ, ζ )(t )‖2L2 + ‖ϕx (t )‖2L2 + ‖(ϕx , ψx , ζx )(τ )‖2L2 dτ ≤ C δ + ‖ϕ0 ‖2H 1 + ‖(ψ0 , ζ0 )‖2L2 + C ψx2 (0, τ )dτ 0 0   ∫ t∫ ∞ δ (x + st )2 +C exp −C0 (ϕ 2 + ζ 2 )dxdτ . 1+t 1+t 0 0 Proof. Similar to that in [16], we introduce a new variable v¯ = v/V . Then we can rewrite (3.3) as follows, using (1.3) 1 , (1.20) 2 and the facts that p = Rθ /v and p+ = RΘ /V .

ψt − sψx + px = µ



ux

v

−

Ux



V

−F =µ x



v¯ x v¯



− sµ



t

v¯ x v¯



− F, x

which, multiplied by v¯ x /¯v , yields

    2     2   v¯ x 2 v¯ x v¯ x Rθ v¯ x v¯ t ∂t − s ∂x − ψ + + ψ 2 v¯ v¯ v¯ t v v¯ v¯ x   ψ2 Ux ϕψx Rζx R Θx ζ v¯ x = x − + − +F := RHS , v vV v vΘ v¯

µ

(4.6)

since from (1.3) 1 , (1.7) and (1.20) 1 and (3.1) it follows that

 2 v¯ x Rζx v¯ x RΘx ζ v¯ x Rθ v¯ x = − − , and V v x v¯ v v¯ v Θ v¯ v v¯     v¯ t Vt Ux + sVx ψ2 Ux ϕψx v¯ x vt ux + s v x ψx = ψx − = ψx − = x − + sψx . v¯ v V v V v vV v¯ Vx

=

Θx , Θ



Rθ



Using the smallness condition of N (T ) and the Cauchy–Schwarz inequality, we see that

 2 v¯ x v¯ x C ≤ε + ψ 2 , ε ∈ (0, 1), v¯ v¯ ε  2    v¯ x C  2 |RHS | ≤ ε + ζx + Θx2 ζ 2 + F 2 + C ψx2 + Ux2 ϕ 2 , v¯ ε ψ

while, the boundary conditions v|x=0 = V |x=0 = v− , together with (1.3) 1 and (1.20) 1 , imply



 v¯ t  ψ = 0, v¯ x=0



v¯ x v¯

2    

x =0

 =

vx Vx − v V

2    

x =0

 =

ux − U x sv−

2    

x =0

=

ψx2 (0, t ) . 2 s2 v−

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

6629

Now, choosing ε > 0 appropriately small and integrating (4.6) over R+ × (0, t ), we have from the smallness assumption of N (T ) that

‖ϕx (t )‖2L2 +

t

∫ 0

∫ t∫

  ‖ϕx (τ )‖2L2 dτ ≤ C ‖ϕ0 ‖2H 1 + ‖(ϕ, ψ)(t )‖2L2 + C

∞



+C

 ζx2 + ψx2 dxdτ + C

ψx2 (0, τ )dτ 0

∞

∫ t∫ 0

0

0

t

∫

 2 2  Θx ζ + Ux2 ϕ 2 + Vx2 ϕ 2 + F 2 dxdτ ,

(4.7)

0

since it follows from the definition of v¯ and the smallness condition of N (T ) that C

ϕx −

−1 2

CVx2

ϕ ≤ 2



v¯ x v¯

2

 =

ϕx Vx ϕ − v vV

2

≤ C (ϕx2 + Vx2 ϕ 2 ).

With the help of Lemmas 2.3 and 4.1, we immediately obtain Lemma 4.2 from (4.7).



The next lemma is concerned with the estimates of the derivatives of ψx and ζx . Lemma 4.3. If δ > 0 and N (T ) are small enough, then it holds that t

t   (0, τ )dτ + ‖ϕx (τ )‖2L2 + ‖(ψx , ζx )(τ )‖2H 1 dτ 0 0   ∫ t∫ ∞   (x + st )2 δ 2 exp −C0 (ϕ 2 + ζ 2 )dxdτ . ≤ C δ + ‖(ϕ0 , ψ0 , ζ0 )‖H 1 + C 1+t 1+t 0 0

‖(ϕ, ψ, ζ )(t )‖

2 H1

∫



+

ψ +ζ 2 x

2 x

∫



Proof. Multiplying (3.3) and (3.4) by ψxx and ζxx , respectively, and summing up the resulting equations, we find



 ζ2 ψ2 ζx2 + µ xx + κ xx 2 2 2 2 v v t  x  ψx ψxx ζx ζxx Ux ϕ Θϕ ζ = (ϕx + Vx ) µ 2 + κ 2 + µ −R +R ψxx v v vV vV v x     2   Θx ϕ u U2 θ ψx Ux ζ Ux Θ ϕ + κ −µ x − x +R +R −R ζxx vV x v V v v vV + (F ψxx + Gζxx ) + (ψt ψx + cv ζt ζx )x = I1 + I2 + I3 + I4 + I5 .

1

ψx2 +

cv

ζx2





+s

1

ψx2 +

cv

which, integrated over R+ × (0, t ), gives

‖(ψx , ζx )(t )‖ + 2 L2

t

∫

 2  ψx + ζx2 (0, τ )dτ +

t

∫

0

0 5 ∫ t ∫ −   

≤ C ‖(ψ0x , ζ0x )‖2L2 + C

i =1

0

∞ 0

‖(ψxx , ζxx )(τ )‖2L2 dτ

 

Ii dxdτ  .

(4.8)

where we have used the fact that s = −ub /v− < 0.  We are now in a task of estimating each integral Ii dxdτ (i = 1, . . . , 5). First, using the smallness condition of N (T ) and Cauchy–Schwarz inequality, we have ∞

∫ t∫ ∞ C ‖(ψxx , ζxx )‖2L2 dτ + (Vx2 + ϕx2 )(ψx2 + ζx2 )dxdτ ε 0 0 0 0 0 ∫ t ∫ t   C 2 ‖Vx ‖2L∞ ‖(ψx , ζx )‖2L2 + ‖(ψx , ζx )‖2L∞ ‖ϕx ‖2L2 dτ , ≤ε ‖(ψxx , ζxx )‖L2 dτ + ε 0 0

∫ t∫

|I1 |dxdτ ≤ ε

∫

t

which, combined with (2.1), the Sobolev inequality ‖(ψx , ζx )‖2L∞ ≤ C ‖(ψx , ζx )‖L2 ‖(ψxx , ζxx )‖L2 , and the Cauchy–Schwarz inequality, leads to

∫ t∫ 0

∞

|I1 |dxdτ ≤ 2ε

0

t

∫ 0

‖(ψxx , ζxx )(τ )‖2L2 dτ +

C 

ε2

 δ 2 + N 4 (T )

t

∫ 0

‖(ψx , ζx )(τ )‖2L2 dτ .

Observing that

|I2 | ≤ C (|Uxx ||ϕ| + |Ux ||ϕx | + |Ux ||ϕ||ϕx | + |Ux ||Vx ||ϕ| + |Θx ||ϕ| + |ϕx | + |ϕ||ϕx | + |Vx ||ϕ| + |ζx | + |ζ ||ϕx | + |Vx ||ζ |) |ψxx | ≤ C (|ϕx | + |ζx | + |Ux ||ϕx | + |Ux ||ϕ||ϕx | + |ϕ||ϕx | + |ζ ||ϕx |) |ψxx | + C (|Uxx | + |Ux ||Vx | + |Θx | + |Vx |) (|ϕ| + |ζ |)|ψxx |,

(4.9)

6630

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

so that, using the smallness condition of N (T ) and Lemma 2.3, we infer from Cauchy–Schwarz and Sobolev inequalities that ∞

∫ t∫

|I2 |dxdτ ≤ ε

‖ψxx (τ )‖2L2 dτ +

0

0

0

t

∫

∞

∫ t∫

C

+

ε

0

0

C

∫

ε

t

0

‖(ϕx , ζx )(τ )‖2L2 dτ

  δ (x + st )2 exp −C0 (ϕ 2 + ζ 2 )dxdτ . 1+t 1+t

(4.10)

In a similar manner, we also have ∞

∫ t∫

|I3 |dxdτ ≤ ε

‖(ψxx , ζxx )(τ )‖2L2 dτ +

0

0

0

t

∫

+

∞

∫ t∫

C

ε

0

0

C

t

∫

ε

‖(ϕx , ψx )(τ )‖2L2 dτ

0

  δ (x + st )2 exp −C0 (ϕ 2 + ζ 2 )dxdτ , 1+t 1+t

(4.11)

since the Sobolev and Young inequalities imply that 3/2

1/2

‖ψx2 ζxx ‖L1 ≤ ‖ψx ‖L∞ ‖ψx ‖L2 ‖ζxx ‖L2 ≤ C ‖ψx ‖L2 ‖ψxx ‖L2 ‖ζxx ‖L2 ≤ ε‖(ψxx , ζxx )‖2L2 +

C

ε

N 4 (T )‖ψx ‖2L2 .

Recalling the definition of G in (3.7), we have from Lemma 2.3 that t

∫

 0

 ‖F ‖2L2 + ‖G‖2L2 dτ ≤ C

t

∫

  ‖F ‖2L2 + ‖Ux ‖4L4 + ‖K ‖2L2 dτ ≤ C δ 2 ,

0

consequently, we easily get from Cauchy–Schwarz inequality that ∞

∫ t∫ 0

|I4 |dxdτ ≤ ε

t

∫

0

‖(ψxx , ζxx )(τ )‖2L2 dτ +

0

C

ε

δ2 .

(4.12)

Finally, due to the boundary condition ζ |x=0 = 0, we have (ζt ζx )|x=0 = 0. Moreover, recalling the definition of U in (1.16), we find

ψt (0, t ) = −Ut (0, t ) = −

κ(γ − 1) Θxt (0, t ) R + γR θ− p+

∫

∞

Kt (y, t )dy.

(4.13)

0

With the help of (2.28) and (2.29), we have ∞

∫ t ∫ 0

2 ∫ t   |Kt |dy dτ ≤ C δ 2 (1 + τ )−1 exp −C τ 2 /(1 + τ ) dt ≤ C δ 2 ,

0

0

while, by virtue of Lemmas 2.1 and 2.3, we deduce t

∫

‖Θxt ‖2L∞ dτ ≤ C δ 2

t

∫

0

1

(1 + τ )2

0

dτ ≤ C δ 2 ,

since it follows from (1.20) 3 and the smallness condition of N (T ) that

  |Θxt | ≤ C |Θxxx | + |Θxx ||Θx | + |Θx |3 + |Θx ||Ux | + |Uxx | + |Θxx | + |Kx | . Putting the above estimates into (4.13), we get t

∫

ψt2 (0, τ )dτ ≤ C δ 2 , 0

which, together with Cauchy–Schwarz inequality, gives

∫ t ∫    0

0

∞

 

I5 dxdτ  ≤ ε

t

∫

ψx2 (0, τ )dτ + 0

C

ε

δ2 .

Therefore, inserting (4.9)–(4.12) and (4.14) into (4.8) and choosing ε > 0 suitably small, we conclude that

∫ t  2  ‖(ψxx , ζxx )(τ )‖2L2 dτ ψx + ζx2 (0, τ )dτ + 0 0 ∫ t  2  2 ≤ C δ + ‖(ψ0x , ζ0x )‖L2 + C ‖(ϕx , ψx , ζx )(τ )‖2L2 dτ

‖(ψx , ζx )(t )‖2L2 +

∫

t

0

(4.14)

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

+C

 exp −C0

1+t

0

0

δ

∞

∫ t∫

(x + st ) 1+t

 2

6631

(ϕ 2 + ζ 2 )dxdτ ,

which, combined with Lemma 4.2, completes the proof of Lemma 4.3.



Using Lemma 4.3 and taking η > 0 in Lemma 4.1 suitably small, we can eliminate the boundary effect as follows. Indeed, an application of Sobolev inequality and Cauchy–Schwarz inequality yields

‖ψx ‖2L∞ ≤ C ‖ψx ‖L2 ‖ψxx ‖L2 ≤ η1 ‖ψxx ‖2L2 +

C

‖ψx ‖2L2

η1

for some η1 > 0 suitably small. From this and Lemma 4.3, we infer that

∫ t    ψx2 + ζx2 (0, τ )dτ + ‖ϕx (τ )‖2L2 + ‖(ψx , ζx )(τ )‖2H 1 dτ 0 0 ∫ t   2 ‖ψx (τ )‖2L2 dτ ≤ C1 δ + ‖(ϕ0 , ψ0 , ζ0 )‖H 1 + C1 0   ∫ t∫ ∞ δ (x + st )2 + C1 (ϕ 2 + ζ 2 )dxdτ exp −C0 1+t 1+t 0 0 for some positive constant C1 independent of η. This, combined with Lemma 4.1, gives ∫ t ∫ t  2    ‖(ϕ, ψ, ζ )(t )‖2H 1 + ψx + ζx2 (0, τ )dτ + ‖ϕx (τ )‖2L2 + ‖(ψx , ζx )(τ )‖2H 1 dτ 0 0   ∫ t δ2 ≤ CC1 δ + + ‖(ϕ0 , ψ0 , ζ0 )‖2H 1 + C1 η ψx2 (0, τ )dτ η 0   ∫ t∫ ∞ δ (x + st )2 + CC1 exp −C0 (ϕ 2 + ζ 2 )dxdτ . 1+t 1+t 0 0 Thus, if η is taken to be such that 0 < η < 1/(2C1 ), then the boundary integral on the right-hand side of the above ‖(ϕ, ψ, ζ )(t )‖2H 1 +

t

∫



inequality can be well controlled. As a result, we have Lemma 4.4. If N (T ) and δ are sufficiently small, then

∫ t    ψx2 + ζx2 (0, τ )dτ + ‖ϕx (τ )‖2L2 + ‖(ψx , ζx )(τ )‖2H 1 dτ 0 0   ∫ t∫ ∞   (x + st )2 δ 2 ≤ C δ + ‖(ϕ0 , ψ0 , ζ0 )‖H 1 + C exp −C0 (ϕ 2 + ζ 2 )dxdτ . 1+t 1+t 0 0

‖(ϕ, ψ, ζ )(t )‖2H 1 +

t

∫



In order to get the desired estimates from Lemma 4.4, we still need to deal with the second term on the right-hand side. This will be achieved by using the technique developed in [2]. For this purpose, let C0 be the fixed positive constant in Lemmas 2.1, 2.3 and 4.4. We also put

(x + st )2 f (x, t ) = (1 + t ) w (y, t )dy with w(x, t ) = exp −α 1+t 0 for some positive constant α ∈ (0, C0 /2). It is easy to see that −1

‖f (t )‖L∞ ≤

∫

x

C

(1 + t )1/2



2

,

‖fx (t )‖L2 ≤

C

(1 + t )3/4

,

‖fx (t )‖L∞ ≤

 (4.15)

C 1+t

(4.16)

and C

‖ft − sfx ‖L∞ ≤

(1 + t )3/2

+

C 1+t



exp −2α

s2 t 2



1+t

≤ C (1 + t )−3/2 ,

(4.17)

since it holds that ft − sfx =

∫ x

1

(1 + t )2

0

2α

   (y + st )2 s s2 t 2 − 1 w 2 (y, t )dy − exp −2α . 1+t 1+t 1+t

Lemma 4.5. If N (T ) and δ are sufficiently small, then t

∫ 0

1 1+τ

∞

∫

 0

 (Rζ − p+ ϕ)2 + ψ 2 w 2 dxdτ ≤ C + C

t

∫ 0

‖(ϕx , ψx , ζx )(τ )‖2L2 dτ .

(4.18)

6632

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

Proof. Since p+ = RΘ /V , multiplying (3.3) by (Rζ − p+ ϕ)v f gives [ψ(Rζ − p+ ϕ)v f ]t − s [ψ(Rζ − p+ ϕ)v f ]x − ψv f [(Rζ − p+ ϕ)t − s(Rζ − p+ ϕ)x ]  1 − ψ f (Rζ − p+ ϕ) (vt − svx ) − ψv(Rζ − p+ ϕ) (ft − sfx ) + (Rζ − p+ ϕ)2 f x 2 1 1 2 2 − (Rζ − p+ ϕ) fx − (Rζ − p+ ϕ) vx f 2 v 1 = µ [ψx (Rζ − p+ ϕ)f ]x − µψx (Rζ − p+ ϕ)x f − µ ψx (Rζ − p+ ϕ)vx f

v

− µψx (Rζ − p+ ϕ)fx − µ



Ux ϕ



vV

(Rζ − p+ ϕ)v f − F (Rζ − p+ ϕ)v f .

(4.19)

x

Due to fx = w 2 /(1 + t ), the seventh term on the left-hand side of (4.19) admits the form 1

1

2

2

− (Rζ − p+ ϕ)2 fx = − (1 + t )−1 (Rζ − p+ ϕ)2 w 2 , so that, integrating (4.19) over R+ , we obtain

∫ ∞ ( 1 + t ) −1 (Rζ − p+ ϕ)2 w 2 dx 2 0 ∫ ∞ ∫ ∞ d = ψv f (Rζ − p+ ϕ)dx − ψv f [(Rζ − p+ ϕ)t − s(Rζ − p+ ϕ)x ] dx dt 0 0 ∫ ∞ ∫ ∞ − ψ f (Rζ − p+ ϕ)(vt − svx )dx − ψv(Rζ − p+ ϕ)(ft − sfx )dx ∫0 ∞ ∫ ∞ 0 vx − f (Rζ − p+ ϕ)2 dx + µ ψx f (Rζ − p+ ϕ)x dx v 0 ∫ ∞ ∫0 ∞ vx +µ ψx f (Rζ − p+ ϕ) + µ ψx fx (Rζ − p+ ϕ)dx v 0 0  ∫ ∞ ∫ ∞ 10 − Ux ϕ Ii , +µ v f (Rζ − p+ ϕ)dx + F v f (Rζ − p+ ϕ)dx := vV x 0 0 i =1 1

(4.20)

where we have used the fact that (Rζ − p+ ϕ)(0, t ) = f (0, t ) = 0 because of (3.5) and (4.15). We are now in a task of estimating each term on the right-hand side of (4.20). First, since it follows from (3.2) and (3.4) that 1

γ −1 =κ

[(Rζ − p+ ϕ)t − s(Rζ − p+ ϕ)x ]



ζx Θx ϕ − v vV



 − p+ x

p+

γ −1



ψx − R

Ux ζ

v

+ p+

Ux ϕ

v

+µ



u2x

v

−

Ux2



V

− G,

and thus, integrating by parts and using the smallness condition of N (T ), we infer that

 ∫ ∞ ζx Θx ϕ − (ψv f )x + ((γ − 1)p + p+ ) v f ψψx dx v vV 0 0  2  ] ∫ ∞[ 3 − Ux ϕ u U2 Ux ζ + (γ − 1) − p+ − µ x − x + G ψv fdx = I2i . R v v v V 0 i=1

I2 = κ(γ − 1)

∫

∞



Thanks to (2.1), (4.16) and the smallness condition of N (T ), we have from Cauchy–Schwarz inequality that

|I21 | ≤ C

∫

∞

∫0 ∞

  |ζx | + (1 + t )−1/2 |ϕ| (|ψx v f | + |ψvx f | + |ψv fx |) dx

  |ζx | + (1 + t )−1/2 |ϕ| [(|ψx f | + |ψϕx f |) + |ψ| (|Vx f | + |fx |)] dx 0   ≤ C ‖ζx ‖L2 + (1 + t )−1/2 ‖ϕ‖L2 (1 + t )−1/2 ‖(ψx , ϕx )‖L2 + (1 + t )−1 ‖ψ‖L2   ≤ C ‖(ϕx , ψx , ζx )‖2L2 + (1 + t )−3/2 ‖(ϕ, ψ)‖2L2 ≤ C ‖(ϕx , ψx , ζx )‖2L2 + C (1 + t )−3/2 . ≤C

(4.21)

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

6633

Keeping in mind that p = Rθ /v and p+ = RΘ /V , we have

((γ − 1)p + p+ ) v = R(γ − 1)(Θ + ζ ) + p+ (V + ϕ) = Rγ Θ + R(γ − 1)ζ + p+ ϕ, so that, by virtue of (2.5), (2.1) and (4.16), we find I22

∞

∫

(R(γ − 1)ζ + p+ ϕ) f ψψx dx + Rγ

= ∫0 ∞

Rγ

∞

∫

Θ f ψψx dx

∫0

∞

0

2

Θx f ψ 2 dx −

Rγ

∞

∫

Θ fx ψ 2 dx ∫ ∞ ∫ ∞ 2 −1 w 2 ψ 2 dx |Θx |ψ dx − C1 (1 + t ) ≤ C ‖f ‖L∞ ‖(ϕ, ζ )‖L∞ ‖ψ‖L2 ‖ψx ‖L2 + C ‖f ‖L∞ 0 0   ∫ ∞ (x + st )2 1/2 −1/2 −1 ≤ C (1 + t ) ‖(ϕx , ζx )‖L2 ‖ψx ‖L2 + C δ(1 + t ) exp −C0 ψ 2 dx 1+t 0   ∫ ∞ (x + st )2 exp −2α ψ 2 dx − C1 (1 + t )−1 1 + t 0   ∫ ∞ (x + st )2 C1 2 ≤ C ‖(ϕx , ψx , ζx )‖L2 + C (1 + t )−2 − (1 + t )−1 exp −2α ψ 2 dx, 2 1 + t 0

(R(γ − 1)ζ + p+ ϕ) f ψψx dx −

=

0

2

0

since 0 < α < C0 /2 and δ is sufficiently small. Note that, here, we have also used the Sobolev and Cauchy–Schwarz inequalities. Using (4.16), Lemma 2.3 and the smallness condition of N (T ), we have from Sobolev and Cauchy–Schwarz inequalities that

∫ ∞  ψx2 + Ux2 + |G| |ψ|fdx + C |Ux | (|ζ | + |ϕ|) |ψ|fdx 0 0   ≤ C ‖ψx ‖2L2 + C (1 + t )−1/2 ‖Ux ‖2L2 + ‖K ‖L1 + C (1 + t )−3/2 ‖(ϕ, ψ, ζ )‖2L2

I23 ≤ C

∫

∞



≤ C ‖ψx ‖2L2 + C (1 + t )−3/2 , since it follows from (3.7) and Lemma 2.3 that

  ‖Ux ‖2L2 + ‖G‖L1 ≤ C ‖Ux ‖2L2 + ‖K ‖L1 ≤ C (1 + t )−1 . Putting the estimates of I2i (i = 1, 2, 3) into (4.21) gives I2 ≤ C ‖(ϕx , ψx , ζx )‖2L2 + C (1 + t )−3/2 − C1 (1 + t )−1

∞

∫

w 2 ψ 2 dx.

(4.22)

0

Since (1.3) 1 and (3.1) imply that vt − svx = ux = Ux + ψx , we can thus estimate I3 as follows, using (2.20), (4.16), Sobolev inequality, Cauchy–Schwarz inequality as well as the smallness condition of N (T ). ∞

∫

|ψ||f | (|ϕ| + |ζ |) |Ux |dx + C

|I3 | ≤ C 0

∞

∫

|ψ||f | (|ϕ| + |ζ |) |ψx |dx 0

≤ C (1 + t )−3/2 ‖(ϕ, ψ, ζ )‖2L2 + C ‖f ‖L∞ ‖(ϕ, ζ )‖L∞ ‖ψ‖L2 ‖ψx ‖L2 1/2

≤ C (1 + t )−3/2 + C (1 + t )−1/2 ‖(ϕx , ζx )‖L2 ‖ψx ‖L2 ≤ C ‖(ϕx , ψx , ζx )‖2L2 + C (1 + t )−3/2 .

(4.23)

By the smallness condition of N (t ), it is easy to deduce from (4.17) that

|I4 | ≤ C ‖ft − sfx ‖L∞ ‖(ϕ, ψ, ζ )‖2L2 ≤ C ‖ft − sfx ‖L∞ ≤ C (1 + t )−3/2 .

(4.24)

Observing that vx = Vx + ϕx and V = RΘ /p+ , by virtue of Sobolev and Young inequalities we have from (2.1) and (4.16) that ∞

∫

|Vx | (Rζ − p+ ϕ)2 dx + C ‖f ‖L∞

|I5 | ≤ C ‖f ‖L∞ 0

≤ C δ(1 + t )−1

∞

∫

|ϕx | (Rζ − p+ ϕ)2 dx 0

∞

∫

 exp −C0

0

 (x + st )2 (Rζ − p+ ϕ)2 dx + C (1 + t )−1/2 ‖(ϕ, ζ )‖L∞ ‖(ϕ, ζ )‖L2 ‖ϕx ‖L2 1+t

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T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

≤ C δ(1 + t )−1

∞

∫ 0

≤ C δ(1 + t )−1

1/2

(Rζ − p+ ϕ)2 w 2 dx + C (1 + t )−1/2 ‖(ϕx , ζx )‖L2 ‖ϕx ‖L2

∞

∫

(Rζ − p+ ϕ)2 w 2 dx + C ‖(ϕx , ψx , ζx )‖2L2 + C (1 + t )−2 ,

0

(4.25)

where we have used the fact that 0 < α ≤ C0 /2. Note that, the first term on the right-hand side of (4.25) can be absorbed into the left-hand side of (4.20) due to the smallness of δ . Using Cauchy–Schwarz inequality again, we infer from (4.16) and (2.1) that

|I6 | + |I7 | + |I8 | ≤ C ‖f ‖L∞ ‖(ϕx , ψx , ζx )‖2L2 + C ‖f ‖L∞ ‖(ϕ, ζ )‖L∞ ‖(ϕx , ψx )‖2L2 + C ‖f ‖L∞ ‖Vx ‖L∞ ‖ψx ‖L2 ‖(ϕ, ζ )‖L2 + C ‖fx ‖L∞ ‖ψx ‖L2 ‖(ϕ, ζ )‖L2 ≤ C ‖(ϕx , ψx , ζx )‖2L2 + C (1 + t )−2 .

(4.26)

By the smallness condition of N (T ) and the fact that ‖(ϕ, ζ )‖L2 ∩L∞ ≤ CN (T ) ≤ C , using (2.1), (2.20), (4.16) and Cauchy– Schwarz inequality, we find ∞

∫

(|Ux ||ϕx | + |Uxx ||ϕ| + |Ux ||ϕ||Vx | + |Ux ||ϕ||ϕx |) (|ϕ| + |ζ |)fdx

|I9 | ≤ C 0

≤ C (1 + t )−3/2

∞

∫

 0

 ϕx2 + ϕ 2 + ζ 2 dx ≤ C ‖ϕx ‖2L2 + C (1 + t )−3/2 .

(4.27)

Finally, it readily follows from (4.16) and the smallness condition of N (T ) that 4/3

|I10 | ≤ C (1 + t )−1/2 ‖F ‖L1 ≤ C ‖F ‖L1 + C (1 + t )−2 .

(4.28)

Since the smallness condition of N (T ) and (4.16) also imply that t

∫

I1 dτ =

∞

∫

0

0

t    ψv f (Rζ − p+ ϕ)dx ≤ C ‖(ϕ0 , ψ0 , ζ0 )‖2L2 + ‖(ϕ, ψ, ζ )(t )‖2L2 ≤ C , 0

putting (4.22)–(4.28) into (4.20) and integrating it over (0, t ), by the smallness condition of δ and (2.21), we immediately arrive at the desired estimate of Lemma 4.5.  In view of Lemma 4.5, to get the desired estimate from Lemma 4.4, it suffices to estimate the following quantity t

∫

1+τ

0

∞

∫

1

(cv ζ + p+ ϕ)2 w 2 dxdτ . 0

To this end, let w = w(x, t ) be the function in (4.15) and set g = g (x, t ) = (1 + t )−1/2

h = h(x, t ) = cv ζ + p+ ϕ,

∫

x

w(y, t )dy.

(4.29)

0

By the definition of w in (4.15), we easily see that g (0, t ) = 0,

‖gx ‖L∞ ≤ C (1 + t )−1/2 ,

‖g ‖L∞ ≤ C ,

‖gx ‖L2 ≤ C (1 + t )−1/4 .

(4.30)

Lemma 4.6. Let w(x, t ), g (x, t ) and h(x, t ) be the functions defined in (4.15) and (4.29), and assume that N (T ) and δ are sufficiently small. Then we have t

∫ 0

∞

∫

1 1+τ

≤C+

(cv ζ + p+ ϕ)2 w 2 dxdτ 0

C

ε

t

∫ 0

‖(ϕx , ψx , ζx )(τ )‖2L2 dτ + C (δ + ε)

t

∫ 0

1 1+τ

∞

∫

(ϕ 2 + ζ 2 )w 2 dxdτ , 0

where ε ∈ (0, 1) is a small positive constant. Proof. It is obvious that the function h = cv ζ + p+ ϕ satisfies

‖h(τ )‖2L2 ≤ C ‖(ϕ, ζ )(τ )‖2L2 ,

‖hx (τ )‖2L2 ≤ C ‖(ϕx , ζx )(τ )‖2L2 .

(4.31)

Moreover, from (3.2) and (3.4) it follows that ht − shx = (cv ζ + p+ ϕ)t − s(cv ζ + p+ ϕ)x

=κ



V ζ x − Θx ϕ

vV



+µ x



u2x

v

−

Ux2 V



− G − (p − p+ ) (Ux + ψx ).

(4.32)

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

6635

and hence,

⟨ht − shx , hg 2 ⟩ = ⟨(cv ζ + p+ ϕ)t − s(cv ζ + p+ ϕ)x , hg 2 ⟩   ∫ ∞ 2 ∫ ∞ ∫ ∞ ux V ζ x − Θx ϕ U2 Ghg 2 dx hg 2 dx + µ − x hg 2 dx − =κ vV v V 0 0 0 x ∫ ∞ ∫ ∞ 5 − Ji , ψx (p − p+ ) hg 2 dx := Ux (p − p+ ) hg 2 dx − − 0

0

(4.33)

i=1

where ⟨·, ·⟩ denotes the usual L2 -inner product. Next we shall estimate the right-hand side of (4.33) term by term. First, in view of (4.15) and Lemma 2.1, we infer from the fact α ∈ (0, C0 /2) that

|Θx | ≤ C δ(1 + t )−1/2 e−C0 ξ ≤ C δ(1 + t )−1/2 w(x, t ), 2

(4.34)

so that, after integrating by part and using the smallness condition of N (T ) we have from (4.30), (4.31) and (4.34) that ∞

∫

(|ζx | + |Θx ||ϕ|)(|hx |g 2 + g |h||gx |)dx ∫ ∞ ∫ ∞ C 2 2 2 2 2 ≤ ‖(ϕx , ζx )‖L2 + ε (Θx + gx )(ϕ + ζ )dx + C |Θx ||gx |(ϕ 2 + ζ 2 )dx ε 0 0 ∫ ∞ C ≤ ‖(ϕx , ζx )‖2L2 + C (δ + ε)(1 + t )−1 (ϕ 2 + ζ 2 )w 2 dx ε 0

|J1 | ≤ C

0

(4.35)

for some small constant ε ∈ (0, 1) to be determined later. Here, we have also used the Cauchy–Schwarz inequality and the fact that gx = (1 + t )−1/2 w . Thanks to (2.20), we have ‖Ux ‖2L2 ≤ C (1 + t )−1 . So, the second term J2 can be estimated from above as follows, using the Sobolev and Young inequalities. ∞

∫



|J2 | ≤ C 0

 ψx2 + Ux2 hg 2 dx ≤ C ‖ψx ‖2L2 + C ‖h‖L∞ ‖Ux ‖2L2 1/2

1/2

≤ C ‖ψx ‖2L2 + C (1 + t )−1 ‖h‖L2 ‖hx ‖L2 ≤ C ‖(ϕx , ψx , ζx )‖2L2 + C (1 + t )−4/3 . Since it follows from (3.7), (2.19) and (2.20) that ‖G‖L1 ≤ C (‖ J2 , we have 1/2

(4.36)

‖ + ‖K ‖L1 ) ≤ C (1 + t ) , similar to the treatment of

Ux 2L2

−1

1/2

|J3 | ≤ C ‖h‖L∞ ‖G‖L1 ≤ C (1 + t )−1 ‖h‖L2 ‖hx ‖L2 ≤ C ‖(ϕx , ζx )‖2L2 + C (1 + t )−4/3 .

(4.37)

Observing that p = Rθ /v , p+ = RΘ /V , and |p − p+ | ≤ C (|ϕ| + |ζ |) due to the smallness condition of N (T ), we thus infer from Cauchy–Schwarz inequality, (4.30) and (2.20) that ∞

∫

|Ux |(ϕ 2 + ζ 2 )g 2 dx ≤ C δ(1 + t )−1

|J4 | ≤ C 0

∞

∫

(ϕ 2 + ζ 2 )w 2 dx,

(4.38)

0

since (2.20), together with (4.15), implies

  (x + st )2 |Ux | ≤ C δ(1 + t )−1 exp −C0 ≤ C δ(1 + t )−1 w2 (x, t ), α ∈ (0, C0 /2). 1+t The last term J5 needs more works. Indeed, since p+ = RΘ /V and cv = R/(γ − 1), it follows that Rθ − p+ v = Rζ − p+ ϕ = (γ − 1)h − γ p+ ϕ . Thus, by (3.2) we see that ∫ ∞ J5 = v −1 (γ p+ ϕ − (γ − 1)h)(ϕt − sϕx )hg 2 dx 0 ∫ ∫ ∞  2 γ p + ∞ −1  2 2 = v (ϕ )t − s(ϕ )x hg dx − (γ − 1) v −1 (ϕt − sϕx )h2 g 2 dx 2

=

J51

To deal with J51 =

0

0

+ J52 . J51 ,

(4.39)

we first rewrite it as

∫  γ p + ∞ −2 2 2 (v −1 hg 2 ϕ 2 )t − s(v −1 hg 2 ϕ 2 )x dx + v hg ϕ (vt − svx )dx 2 2 0 0 ∫ ∞ ∫ ∞ 4 − γ p+ 1 ,i − v −1 g 2 ϕ 2 (ht − shx )dx − γ p+ v −1 ghϕ 2 (gt − sgx )dx = J5 . γ p+

∫

2

∞



0

0

i =1

(4.40)

6636

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

Using (4.30) and Sobolev inequality, we have from the smallness condition of N (T ) that t

∫

1 ,1

J5 dτ ≤ C ‖h‖L∞ ‖ϕ‖2L2 + ‖ϕ0 ‖2L2 ≤ CN 3 (T ) ≤ C ,

0





where we have used the boundary condition h(0, t ) = (cv ζ + p+ ϕ)(0, t ) = 0. Secondly, from (1.3) 1 and (3.1) it follows that vt − svx = ψx + Ux , and thus, we have ∞

∫

|J51,2 | ≤ C

|h||ψx |ϕ 2 dx + C

∞

∫

|h||Ux |ϕ 2 dx 0

0

1/2

1/2

≤ C ‖(ϕ, ζ )‖L2 ‖ψx ‖L2 ‖ϕ‖L2 ‖ϕx ‖L2 + C (1 + t )−1 ‖(ϕ, ζ )‖L2 ‖(ϕx , ζx )‖L2 ‖ϕ‖2L2 ≤ C ‖(ϕx , ψx , ζx )‖2L2 + C (1 + t )−4/3 , where we have also used (2.20), (4.31), the Sobolev and Young inequalities. 1,3 Thirdly, in view of (4.32), the term J5 can be rewritten as 1,3

J5

=− +

κγ p+ 2

γ p+

∞

∫

V ζ x − Θx ϕ



vV

0

v −1 ϕ 2 g 2 dx − x

∞

∫

2



Gv −1 ϕ 2 g 2 dx + 0

γ p+

2

∞

∫



u2x

v

0

−

Ux2 V



v −1 ϕ 2 g 2 dx

∞

∫

2

µγ p+

(Ux + ψx ) (p − p+ ) v −1 ϕ 2 g 2 dx = 0

4 −

Ji .

i =1

Similar to the estimates of J1 –J4 , the terms on the right-hand side can be estimates as follows:

∫

∞

  (|ζx | + |Θx ||ϕ|) |Vx |ϕ 2 + |gx |ϕ 2 + |ϕ||ϕx | + ϕ 2 |ϕx | dx 0 ∫ ∞  2  C 2 −1 ≤ ‖(ϕx , ζx )‖L2 + C (δ + ε)(1 + t ) ϕ + ζ 2 w 2 dx, ε ∈ (0, 1), ε 0

J1 ≤ C

and

  J2 + J3 + J4 ≤ C ‖ψ ‖ + C ‖ϕ‖ ‖Ux ‖2L2 + ‖G‖L1 + C ‖ϕ‖2L∞ ‖ψx ‖L2 ‖(ϕ, ζ )‖L2 + C ∫ ∞ ≤ C ‖(ϕx , ψx )‖2L2 + C (1 + t )−1 ‖ϕ‖L2 ‖ϕx ‖L2 + C δ(1 + t )−1 ϕ 2 w2 dx 0 ∫ ∞ 2 −2 −1 ≤ C ‖(ϕx , ψx )‖L2 + C (1 + t ) + C δ(1 + t ) ϕ 2 w2 dx, 2 x L2

2 L∞

∞

∫

|Ux |ϕ 2 dx 0

0

from which we immediately obtain (0 < ε < 1) 1,3

J5

≤

C

ε

‖(ϕx , ψx , ζx )‖2L2 + C (1 + t )−2 + C (δ + ε)(1 + t )−1

∞

∫

 2  ϕ + ζ 2 w2 dx. 0

By the definitions of g and w (see, (4.15) and (4.29)), a straightforward calculation gives gt =

1

(1 + t )1/2 α = (1 + t )1/2

∫

x

1

1

∫

x

wt (y, t )dy − w(y, t )dy 2 (1 + t )3/2 0 0  ∫ x ∫ x (y + st )2 (y + st ) 1 1 − 2s w( y , t ) dy − w(y, t )dy, (1 + t )2 1+t 2 (1 + t )3/2 0 0

in which the first term on the right-hand side admits the following form

α (1 + t )1/2

   ∫ x (y + st )2 (y + st ) 1 (y + st )2 −3/2 − 2s w( y , t ) dy = − ( 1 + t ) ( y + st ) d exp −α (1 + t )2 1+t 2 1+t 0 0     ∫ x 2 2 (y + st ) (y + st ) + s(1 + t )−1/2 exp −α d −α 1+t 1+t 0     1 x + st (x + st )2 1 s(t + 2) s2 t 2 =− exp −α − exp −α 2 (1 + t )3/2 1+t 2 (1 + t )3/2 1+t     ∫ x 2 1 1 (y + st ) s (x + st )2 + exp −α dy + exp −α . 2 (1 + t )3/2 0 1+t (1 + t )1/2 1+t ∫ x

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

6637

Consequently, we have gt − sgx = −

=



x + st

exp −α 2(1 + t )3/2 1

4α(1 + t )

(x + st )2 1+t

s(t + 2)

s2 t 2 exp −α 2(1 + t )3/2 1+t

 −

s(t + 2)

s2 t 2 exp −α 3 / 2 2(1 + t ) 1+t

wx − 1/2







 (4.41)

and

  |gt − sgx | = 

  s(t + 2) s2 t 2  ≤ C . w − exp −α x 4α(1 + t )1/2 2(1 + t )3/2 1+t  1+t 1

(4.42)

So, we can utilize (4.31), (4.42), the Sobolev and Young inequalities as well as the smallness condition of N (T ) to estimate 1,4 J5 as follows. ∞

∫

|J51,4 | ≤ C ‖gt − sgx ‖L∞

|h|ϕ 2 dx ≤ C (1 + t )−1 ‖h‖L∞ ‖ϕ‖2L2

0 1/2

1/2

≤ C (1 + t )−1 ‖h‖L2 ‖hx ‖L2 ≤ C ‖(ϕx , ζx )‖2L2 + C (1 + t )−4/3 . 1,i

Putting the estimates of J5 (i = 1, . . . , 4) into (4.40), we conclude that (0 < ε < 1) t

∫

C

J51 dτ ≤ C +

ε

0

t

∫

‖(ϕx , ψx , ζx )(τ )‖2L2 dτ + C (δ + ε)

0

t

∫

1 1+τ

0

∞

∫

(ϕ 2 + ζ 2 )w 2 dxdτ .

(4.43)

0

Following the argument in the proof of (4.43) step by step, we obtain the same estimate for J52 . Thus, it follows from (4.39) that t

∫

C

J5 dτ ≤ C +

ε

0

t

∫

‖(ϕx , ψx , ζx )(τ )‖2L2 dτ + C (δ + ε)

0

t

∫

1 1+τ

0

∞

∫

(ϕ 2 + ζ 2 )w 2 dxdτ .

(4.44)

0

Due to h(0, t ) = 0, the left-hand side of (4.33) can be written as follows, using (4.41).

⟨ht − shx , hg 2 ⟩ = =

2 dt

h2 g 2 dx − 0

2 dt

s

∞

∫

2

h2 g 2 dx − 0

s(t + 2)

(h2 g 2 )x dx − ∫ ∞

0

∞

∫

1 d

+

∞

∫

1 d

1



h2 g (gt − sgx ) dx 0

h2 g wx dx

4α(1 + t )1/2

s2 t 2 exp −α 3 / 2 2(1 + t ) 1+t

∞

∫

0

∞

∫

h2 gdx = R1 + R2 + R3 ,

(4.45)

0

in which the third term on the right-hand side can be bounded by

|R3 | ≤



C 1+t

exp −α

s2 t 2



1+t

‖h‖2L2 ‖g ‖L∞ ≤



C

exp −α

1+t

s2 t 2 1+t



.

By integration by part and the fact that gx = (1 + t )−1/2 w , we see that the second term on the right-hand side of (4.45) can be treated in the following manner, using (4.31) and Cauchy–Schwarz inequality. R2 =

= ≥

∞

∫

1 1/2

4α(1 + t ) 1 4α(1 + t )

h2 gx w dx + 0

2α(1 + t )

∞

∫

h2 w 2 dx +

1

∫0 ∞

8α(1 + t )

0

∞

∫

1 1/2

∞

∫

1 2α(1 + t )1/2

hhx g w dx 0

hhx g w dx 0

h2 w 2 dx − C ‖(ϕx , ζx )‖2L2 .

Putting the above two estimates into (4.45) and integrating it over (0, t ), by (4.30) we find t

∫

⟨ht − shx , hg ⟩dτ ≥ 2

0

1 8α

t

∫ 0

1 1+τ

∞

∫

h w dxdτ − C 2

0

2



1 + ‖(ϕ0 , ζ0 )‖

2 L2

∫ + 0

t

‖(ϕx , ζx )(τ )‖



τ .

2 d L2

(4.46)

Therefore, taking the estimates of (4.35)–(4.38), (4.44) and (4.46) into account, after integrating (4.33) over (0, t ) we arrive at the desired estimate stated in Lemma 4.6. 

6638

T. Zheng et al. / Nonlinear Analysis 74 (2011) 6617–6639

Proofs of Proposition 3.1 and Theorem 3.1. With the help of Lemmas 4.4–4.6, we can now complete the proof of Proposition 3.1. Indeed, combining Lemmas 4.5 and 4.6, and choosing ε > 0 suitably small, we have t

∫

(1 + τ )−1

∞

∫

(ϕ 2 + ζ 2 )w2 dxdτ ≤ C + C

0

0

0

t

∫

‖(ϕx , ψx , ζx )(τ )‖2L2 dτ ,

(4.47)

since δ > 0 is sufficiently small. Moreover, since α ∈ (0, C0 /2), by virtue of (4.47) and the smallness of δ we immediately deduce from Lemma 4.4 that

‖(ϕ, ψ, ζ )(t )‖2H 1 + ∫ + 0

t

∫ 0

  ‖ϕx (τ )‖2L2 + ‖(ψx , ζx )(τ )‖2H 1 dτ

t

 2    ψx + ζx2 (0, τ )dτ ≤ C δ + ‖(ϕ0 , ψ0 , ζ0 )‖2H 1 .

(4.48)

This finishes the proof of Proposition 3.1. In particular, it also means that the smallness assumption of N (T ) is reasonable, and thus, the global-in-time estimates of the solution (ϕ, ψ, ζ ) in (4.48) hold, if ‖(ϕ0 , ψ0 , ζ0 )‖H 1 and δ are chosen to be sufficiently small. So, the global existence of a solution (ϕ, ψ, ζ ) ∈ X (0, ∞) follows immediately from the local existence theorem and the standard continuity argument. We still need to establish (3.10). To do so, we first differentiate (3.2) with respect to x and multiply the resulting equation by 2ϕx in L2 to get

− sϕx2 (0, t ) = −2

∞

∫

ϕx ψxx dx + 0

d dt

‖ϕx ‖2L2 ,

s < 0.

(4.49)

Thanks to Cauchy–Schwarz inequality, we have

 ∫  2 

∞ 0

    ϕx ψxx dx ≤ ‖ϕx ‖2L2 + ‖ψxx ‖2L2 ,

so that, it follows from (4.48) and (4.49) that ∞

∫

ϕx2 (0, t )dt ≤ C .

(4.50)

0

As an immediate result of (4.48) and (4.50), we infer from (4.49) that ∞

∫ 0

  ∫ d   ‖ϕx (t )‖22  dt ≤ C  dt L 

∞

ϕx2 (0, t )dt + C

∞

∫

0

 0

 ‖ϕx ‖2L2 + ‖ψxx ‖2L2 dt ≤ C .

(4.51)

Similarly, using Lemmas 2.1, 2.3 and (4.48) as well as Sobolev inequality, one easily gets from (3.3) and (3.4) that ∞

∫ 0

    d     ‖ψx (t )‖22  +  d ‖ζx (t )‖22  dτ ≤ C .  dt  dt L  L 

(4.52)

The estimates (4.48), (4.51) and (4.52) as well as the Sobolev inequality imply

‖(ϕ, ψ, ζ )(t )‖2L∞ ≤ 2‖(ϕ, ψ, ζ )(t )‖L2 ‖(ϕx , ψx , ζx )(t )‖L2 → 0 as t → ∞, which proves the asymptotic behavior of the solution (ϕ, ψ, ζ ) stated in (3.10). The proof of Theorem 3.1 is therefore complete.  Acknowledgment The authors would like to thank Prof. Jing Li for his helpful discussions. References [1] A. Matsumura, Inflow and outflow problems in the half space for a one-dimensional isentropic model system of compressible viscous gas, Methods and Appl. Anal. 8 (4) (2001) 645–666. [2] 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. [3] F.M. Huang, A. Matsumura, X. Shi, Viscous shock wave and boundary-layer solution to an inflow problem for compressible viscous gas, Comm. Math. Phys. 239 (2003) 261–285. [4] F.M. Huang, A. Matsumura, X. Shi, On the stability of contact discontinuity for compressible Navier–Stokes equations with free boundary, Osaka J. Math. 41 (1) (2004) 193–210. [5] F.M. Huang, A. Matsumura, Z.P. Xin, Stability of contact discontinuity for the 1-D compressible Navier–Stokes Equations, Arch. Raton. Mech. Anal. 179 (2005) 55–77.

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