Global classical large solutions to 1D compressible Navier–Stokes equations with density-dependent viscosity and vacuum

J. Differential Equations 251 (2011) 1696–1725

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Global classical large solutions to 1D compressible Navier–Stokes equations with density-dependent viscosity and vacuum Shijin Ding a , Huanyao Wen a,b , Changjiang Zhu b,∗ a

School of Mathematical Sciences, South China Normal University, Guangzhou 510631, PR China The Hubei Key Laboratory of Mathematical Physics, Laboratory of Nonlinear Analysis, School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, PR China

b

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 24 November 2010 Revised 12 May 2011 Available online 25 May 2011

In this paper, we investigate an initial boundary value problem for 1D compressible isentropic Navier–Stokes equations with large initial data, density-dependent viscosity, external force, and vacuum. Making full use of the local estimates of the solutions in Cho and Kim (2006) [3] and the one-dimensional properties of the equations and the Sobolev inequalities, we get a unique global classical solution (ρ , u ) where ρ ∈ C 1 ([0, T ]; H 1 ([0, 1])) and u ∈ H 1 ([0, T ]; H 2 ([0, 1])) for any T > 0. As it is pointed out in Xin (1998) [31] that the smooth solution (ρ , u ) ∈ C 1 ([0, T ]; H 3 ( R 1 )) (T is large enough) of the Cauchy problem must blow up in finite time when the initial density is of nontrivial compact support. It seems that the regularities of the solutions we obtained can be improved, which motivates us to obtain some new estimates with the help of a new test function ρ 2 utt , such as Lemmas 3.2–3.6. This leads to further regularities of (ρ , u ) where ρ ∈ C 1 ([0, T ]; H 3 ([0, 1])), u ∈ H 1 ([0, T ]; H 3 ([0, 1])). It is still open whether the regularity of u could be improved to C 1 ([0, T ]; H 3 ([0, 1])) with the appearance of vacuum, since it is not obvious that the solutions in C 1 ([0, T ]; H 3 ([0, 1])) to the initial boundary value problem must blow up in finite time. Crown Copyright © 2011 Published by Elsevier Inc. All rights reserved.

MSC: 76D05 35B65 35B45 Keywords: Compressible Navier–Stokes equations Density-dependent viscosity Vacuum Global classical solution

*

Corresponding author. E-mail address: [email protected] (C. Zhu).

0022-0396/$ – see front matter Crown Copyright doi:10.1016/j.jde.2011.05.025

©

2011 Published by Elsevier Inc. All rights reserved.

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

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Contents 1. Introduction . . . . 2. Proof of Theorem 3. Proof of Theorem Acknowledgments . . . . References . . . . . . . . . .

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1697 1700 1717 1724 1725

1. Introduction In this paper, we consider the initial boundary value problem of compressible isentropic Navier– Stokes equations in one dimension:



ρt + (ρ u )x = 0, ρ  0,       (ρ u )t + ρ u 2 x + P (ρ ) x = μ(ρ )u x x + ρ f ,

(1.1)

for (x, t ) ∈ (0, 1) × (0, +∞), with the initial condition:

  (ρ , u )|t =0 = ρ0 (x), u 0 (x) for x ∈ [0, 1],

(1.2)

and the boundary condition:

u |x=0,1 = 0,

t  0,

(1.3)

where ρ and u, denoting density and velocity of the fluids respectively, are unknown functions; P (ρ ) = K ρ γ , for some constants γ > 1 and K > 0, is the pressure function; f is a given external force. Our main concern here is to show the existence, uniqueness and further regularity of global classical large solutions to (1.1)–(1.3). It worths mentioning that the initial density may vanish (i.e., the initial vacuum may exist), and the viscosity μ depends on the density ρ . The restrictions on μ in the following will imply that the case for constant viscosity is included. We begin with a rough review in this direction. When the viscosity μ is constant, the local classical solutions to the Navier–Stokes equations with heat-conducting fluid in Hölder spaces was obtained respectively by Itaya [11] for the Cauchy problem and by Tani [28] for IBVP with inf ρ0 > 0, where the spatial dimension N = 3. Using delicate energy methods in Sobolev spaces, Matsumura and Nishida showed in [21,22] that the global classical solutions exist provided that the initial data is small in some sense and away from vacuum and the spatial dimension N = 3. For large initial data, Kazhikhov, Shelukhi in [17] (for polytropic perfect gas with constant viscosity) and Kawohl in [15] (for real gas with μ = μ(ρ )) got global classical solutions in dimension N = 1 with inf ρ0 > 0, respectively. The viscosity μ in [15] satisfies

0 < μ0  μ(ρ )  μ1 ,

for ρ  0,

(1.4)

where μ0 and μ1 are constants. In fact, such the condition includes the case μ(ρ ) ≡ const. There are also some results about the existence of global strong (classical) solutions to the Navier–Stokes equations for isentropic fluid with inf ρ0 > 0, refer for instance to [1,16,29]. In the presence of vacuum, the existence of global weak solutions with large initial data in R N was first obtained by Lions in [18], where γ  N3N +2 for N = 2 or 3. Feireisl et al. in [9] extended the work in [18] to the case γ > 32 for N = 3. For solutions with spherical symmetry, Jiang and Zhang in [14] relaxed the restriction on γ in [18] to the case γ > 1, and got the global existence of the weak

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solutions for N = 2 or 3. On the existence and regularity of weak solutions with density connecting to vacuum continuously in 1D, please refer to [20]. During the pass two decades, Salvi, Choe, Kim and Jiang et al made great progress towards the local or global existence of strong (classical) solutions with vacuum, see [27,2,4,5,7,3]. Particularly, Choe and Kim in [5] showed that the radially symmetric strong solutions exist globally in time for γ  2 and N  2. This result had been generalized to the case γ  1 by Fan, Jiang and Ni in [7]. Note that [27,2,4,5,7] all considered the existence and uniqueness of strong solutions. On the classical solutions, Cho and Kim in [3] in 2006 got the local existence and uniqueness result for N = 3 by using successive approximations, based on some a priori estimates for the solutions to the corresponding linearized problems. The existence of global classical solutions with vacuum is still open. It is well known that the physically important case related to vacuum is the case when μ is not constant. It can be seen from the derivation of the Navier–Stokes equations from the Boltzmann equation through Chapman–Enskog expansion to the second order, cf. [10], where the viscosity coefficient depends on the temperature. For isentropic flow, this dependence is translated into the dependence on the density by the laws of Boyle and Gay–Lussac for ideal gas as discussed in [19]. For the case μ(ρ ) = αρ θ with α and θ being positive constants, when the initial density was assumed to be connected to vacuum discontinuously, the global existence of weak solutions for isentropic flow in one ˙ dimension was obtained by Okada, Matušu-Neˇ casová and Makino in [24] for 0 < θ < 13 and by Yang,

Yao and Zhu in [32] for 0 < θ < 12 and by Jiang, Xin and Zhang in [13] for 0 < θ < 1. Qin and Yao in [26] relaxed the restriction on θ in [13] to the case 0 < θ  1 with more regular initial data. On the global existence of classical solutions to the Navier–Stokes equations with heat-conducting fluid, inf ρ0 > 0, and viscosity μ(ρ ) = αρ θ , please refer to [12,25] where Jiang in [12] and Qin, Yao in [25] considered the case θ ∈ (0, 14 ) and θ ∈ (0, 12 ) respectively. For the Cauchy problem of (1.1)–(1.3), Mellet and Vasseur in [23] showed the existence of global strong solutions when μ satisfies

⎧ θ ⎪ for any ρ  1 and θ ∈ [0, 12 ), ⎨ μ(ρ )  αρ , μ(ρ )  α , for ρ  1, ⎪ ⎩ μ(ρ )  c + c ρ γ , for any ρ  0,

(1.5)

where α and c are some positive constants. Moreover, if μ(ρ )  α > 0 for all ρ  0, μ(ρ ) is uniformly Lipschitz and γ  2, then this global strong solution is unique in the large class of weak solutions satisfying the usual entropy inequality. When the density function connects to vacuum continuously, please refer to [8,30,33] for the isentropic case, where the global existence of weak solutions was obtained for μ(ρ ) = αρ θ . To sum up, the existence of global classical solutions of the Navier–Stokes equations with vacuum and large initial data is still open whether the viscosity μ is constant or depends on the density. This is our motivation in this paper. There are two main theorems in the paper. In Theorem 1.1, we get the existence and uniqueness of global classical solutions with relatively weak initial data. In Theorem 1.2, we get further regularities of the solutions with more regular initial data. The difficulties in Theorem 1.1 compared with [3] are that we need some a priori estimates globally in time, and that we have to handle the density-dependent viscosity. To handle these, we make full use of one-dimensional properties of the equations and the Sobolev inequalities. Since the initial density  may vanish, similarly to [3], we need the compatibility condition (1.6) to handle I ρ ut2 |t =0 and I u 2xt |t =0 in Lemma 2.6 and Lemma 2.10 respectively. The main difficulties in Theorem 1.2 lie in the higher order estimates of u. We get them by some new useful estimates with the help of a new test function ρ 2 utt , such as Lemmas 3.2–3.5. Among them, Lemma 3.3 is the most important one, where we take ρ 2 utt instead of utt as the test function in order to handle the vanishing effect of the density. We have to emphasize that the smooth solution (ρ , u ) ∈ C 1 ([0, T ]; H 3 ( R 1 ))(T is large enough) to the Cauchy problem of compressible isentropic Navier–Stokes equations in 1D must blow up in finite time when the initial density is of nontrivial compact support (please refer to [31]). In Theorem 1.2, we get ρ ∈ C 1 ([0, T ]; H 3 ([0, 1])), u ∈ H 1 ([0, T ]; H 3 ([0, 1])). It is open whether regularity of u could

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

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be improved to C 1 ([0, T ]; H 3 ([0, 1])) as vacuum appears, since it is not obvious that the solutions in C 1 ([0, T ]; H 3 ([0, 1])) to the initial boundary value problem must blow up in finite time. As in [3], we assume that the initial data satisfies the compatibility condition:













μ(ρ0 )u 0x x (x) − p (ρ0 ) x (x) = ρ0 (x) − f (x, 0) + g (x) , x ∈ [0, 1],

for a given function g ∈ H 01 , and the viscosity

(1.6)

μ satisfies 



μ ∈ C 2 [0, ∞), 0 < M 1  μ(ρ )  M 2 1 + ρ γ , for any ρ  0.

(1.7)

We give a remark on (1.6) and (1.7). Remark 1.1. (i) The function g in (1.6) plays a crucial role in handling

1

1 0

ρ ut2 (0),

1 0

u 2xt (0) and

3 2 utt (0)

ρ when we use the Gronwall inequality in Lemma 2.6, Lemma 2.10 and Lemma 3.3, re0 spectively. (ii) The restrictions on μ in (1.7) cover the condition (1.4) in [15]. The upper bound of μ in (1.7) is the same as (1.5)3 in [23]. Unfortunately, the case μ(0) = 0 is excluded by (1.7). The lower bound in (1.7) is imposed here because of the requirement of analysis. For the case μ(0) = 0, we will consider in the next future. Notations. (1) I = [0, 1], ∂ I = {0, 1}, Q T = I × [0, T ] for T > 0. (2) For p  1, L p = L p ( I ) denotes the L p space with the norm  ·  L p . For k  1 and p  1, W k, p = W k, p ( I ) denotes the Sobolev space, whose norm is denoted as  ·  W k, p , H k = W k,2 ( I ). (3) For an integer k  0 and 0 < α < 1, let C k+α ( I ) denote the Schauder space of functions on I , whose kth order derivative is Hölder continuous with exponents α , with the norm  · C k+α . Our main results are stated as follows. γ

Theorem 1.1. Assume that ρ0  0, ρ0 ∈ H 2 , ρ0 ∈ H 2 , u 0 ∈ H 3 ∩ H 01 , f ∈ C ([0, ∞); H 1 ), f t ∈ L 2loc ([0, ∞); L 2 ), and that the initial data and μ satisfy (1.6) and (1.7) respectively. Then for any T > 0 there exists a unique global classical solution (ρ , u ) to (1.1)–(1.3) satisfying











    ∈ C [0, T ]; H 1 , ρtt ∈ C [0, T ]; L 2 ,       ρ  0, ρ γ tt ∈ L ∞ [0, T ]; L 2 , (ρ u )t ∈ C [0, T ]; H 1 ,       u ∈ C [0, T ]; H 3 ∩ H 01 , ut ∈ L ∞ [0, T ]; H 01 ∩ L 2 [0, T ]; H 2 .

ρ , ρ γ ∈ C [0, T ]; H 2 ,



ρt , ρ γ

 t

Remark 1.2. (i) The classical solution here means that it satisfies the equations and the corresponding initial and boundary conditions everywhere. γ γ (ii) If γ  2, the assumption ρ0 ∈ H 2 is not necessary, since ρ0 ∈ H 2 implies ρ0 ∈ H 2 . For γ ∈ γ

(1, 2), such the conditions of ρ0 , ρ0 and u 0 in Theorem 1.1 are not empty. For example, ρ0 = xβ (β > 32 ), and u 0 is the solution to the elliptic equation (1.6) with zero boundary condition. By the standard elliptic theory, we get u 0 ∈ H 3 ∩ H 01 . (iii) By the Sobolev embedding theorems, we have 1

H k ( I ) → C k− 2 ( I ),

for k = 1, 2, 3.

This together with regularities of (ρ , u ) gives







1



ρ , ρ γ ∈ C [0, T ]; C 1+ 2 ( I ) ,







1





1



ρt , (ρ u )t ∈ C [0, T ]; C 2 ( I ) , u ∈ C [0, T ]; C 2+ 2 ( I ) ,

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which means (ρ , u ) is the classical solution to (1.1)–(1.3). Note that we get the continuity of (ρ u )t with respect to space–time variables instead of getting that of ut . Anyway, ρ and u satisfy (1.1)–(1.3) everywhere. The next result shows further regularities of the solutions. Theorem 1.2. Consider the same assumptions as in Theorem 1.1, and in addition assume that ρ0 ∈ H 4 , √ √ ρ0γ ∈ H 4 , ( ρ0 )x ∈ L ∞ , ( ρ0 g x )x ∈ L 2 , u 0 ∈ H 4 and ( f xt , f tt ) ∈ L 2loc ([0, ∞); L 2 ), f ∈ L 2loc ([0, ∞); H 3 ),

μ ∈ C 4 [0, ∞). Then the regularities of the solutions obtained in Theorem 1.1 can be improved as follows: 







√      √  ( ρ )x , ( ρ )t ∈ L ∞ ( Q T ), ρt , ρ γ t ∈ C [0, T ]; H 3 ,             ρtt , ρ γ tt ∈ C [0, T ]; H 1 ∩ L 2 [0, T ]; H 2 , ρttt , ρ γ ttt ∈ L 2 ( Q T ),         u ∈ C [0, T ]; H 4 ∩ L 2 [0, T ]; H 5 , ut ∈ L ∞ [0, T ]; H 01 ∩ L 2 [0, T ]; H 3 ,   √ ρ u xxt ∈ L ∞ [0, T ]; L 2 .

ρ , ρ γ ∈ C [0, T ]; H 4 ,

Remark 1.3. (i) The conditions of xβ (β

7 ), 2

ρ0 , ρ0γ and u 0 in Theorem 1.2 are not empty. For example, ρ0 =

> and u 0 is the solution to the elliptic equation (1.6) with zero boundary condition. By the standard elliptic theory, we get u 0 ∈ H 4 ∩ H 01 . (ii) From Theorem 1.2, we get

  (ρ , u ) ∈ C [0, T ]; H 4 ( I ) ,









ρ ∈ C 1 [0, T ]; H 3 ( I ) , u ∈ H 1 [0, T ]; H 3 ( I ) .

Whether could regularity of (ρ , u ) here be improved further? This will be considered in our forthcoming paper. The constant K in the pressure function P (ρ ) doesn’t play any role in the analysis, we assume henceforth that K = 1. The rest of the paper is organized as follows. In Section 2, we prove Theorem 1.1 by giving the initial density a lower bound δ > 0, getting a sequence of approximate solutions to (1.1)–(1.3) and taking δ → 0 after making some estimates uniformly for δ . The regularities guarantee the uniqueness of the solutions. In Section 3, based on some new useful estimates such as Lemmas 3.2–3.6, we prove Theorem 1.2 by the similar arguments as in Section 2. 2. Proof of Theorem 1.1 In this section, we get a unique classical solution of (1.1)–(1.3) with initial density bounded below away from zero by using some a priori estimates of the solutions based on the local existence. γ In this section, we denote “c ( T )” to be a generic constant depending on ρ0  H 2 , ρ0  H 2 , u 0  H 3 , T and some other known constants but independent of δ , for any δ ∈ (0, 1). Before proving Theorem 1.1, we need the following auxiliary theorem. Theorem 2.1. Consider the same assumptions as in Theorem 1.1, and in addition assume that ρ0  δ > 0. Then for any T > 0 there exists a unique global classical solution (ρ , u ) to (1.1)–(1.3) satisfying













ρ ∈ C [0, T ]; H 2 , ρt ∈ C [0, T ]; H 1 , ρtt ∈ C [0, T ]; L 2 , ρ  



utt ∈ L 2 [0, T ]; L 2 ,





u ∈ C [0, T ]; H 3 ∩ H 01 ,







δ c(T )

, 

ut ∈ C [0, T ]; H 01 ∩ L 2 [0, T ]; H 2 .

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

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Proof. The local solutions in Theorem 2.1 can be obtained by the successive approximations like in [3]. We omit it here for simplicity. The regularities guarantee the uniqueness (refer for instance to [4]). Based on the local existence of the solutions, Theorem 2.1 can be proved by some a priori estimates globally in time. For any given T ∈ (0, ∞), let (ρ , u ) be the classical solutions to (1.1)–(1.3) as in Theorem 2.1. Then we have the following basic energy estimate. Lemma 2.1. For any 0  t  T , it holds





t



ρ u 2 + ρ γ (t ) +

I

u 2x  c ( T ). 0

(2.1)

I

Proof. Multiplying (1.1)2 by u, integrating the resulting equation over I , and using integration by parts, (1.1)1 , (1.7), the Cauchy inequality and the Gronwall inequality, we get (2.1). This completes the proof of Lemma 2.1. 2 Lemma 2.2. For any ( y , s) ∈ Q T , it holds

δ c(T ) Proof. Step 1: we claim Denote

 ρ ( y , s)  c ( T ).

(2.2)

ρ ( y , s)  c ( T ), for ( y , s) ∈ Q T . t

w (x, t ) =



 μ(ρ )u x − ρ u − ρ γ +

x

2

0

t x

ρ0 u 0 + 0

ρf. 0

(2.3)

0

Differentiating (2.3) with respect to x, and using (1.1)2 , we have

w x = ρ u. This together with Lemma 2.1 and the Cauchy inequality gives

| w x |  c ( T ). I

It follows from (2.3), (1.7), and Lemma 2.1 that

| w |  c ( T ). I

Since W 1,1 → L ∞ , we get

 w L ∞ ( Q T )  c ( T ). For any ( y , s) ∈ Q T , let x(t , y ) satisfy

(2.4)

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S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

⎧ ⎨ dx(t , y ) ⎩

  = u x(t , y ), t ,

0  t < s,

dt x(s, y ) = y .

(2.5)

Denote

ρ (x,t ) F (x, t ) = exp

μ(ξ ) ξ

 dξ + w (x, t ) .

1

It is easy to verify

dF (x(t , y ), t ) dt

= ∂t F + u ∂x F   μ(ρ ) μ(ρ ) =F ρt + w t + ρx u + ρ u 2

ρ

ρ



x (t , y )

= −ρ γ +



ρf F 0

x (t , y )

F

ρf, 0

which implies

d dt

 t x  (τ , y ) F exp − ρ f (ξ, τ ) dξ dτ  0. 0

0

Integrating (2.6) over (0, s), we have

 s x  (τ , y )   F ( y , s) exp − ρ f (ξ, τ ) dξ dτ  F x(0, y ), 0 , 0

0

which implies

 s x (τ , y )







ρ f (ξ, τ ) dξ dτ F x(0, y ), 0  c ( T ),

F ( y , s)  exp 0

0

where we have used (1.7). This gives

 ρ ( y ,s) exp

μ(ξ ) ξ

 dξ

   c ( T ) exp − w ( y , s)  c ( T ),

1

where we have used (2.4). This together with (1.7) completes Step 1.

(2.6)

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

1703

ρ ( y , s)  c(δT ) , for ( y , s) ∈ Q T . x = μ(ρ )u x − ρ u 2 − ρ γ + 0 ρ f , ∂∂wx = ρ u, we have

Step 2: we claim Since ∂∂wt

dw (x(t , y ), t ) dt

∂w ∂w +u ∂t ∂x

=





  = μ(ρ )u x x(t , y ), t − ρ γ x(t , y ), t +

x (t , y )

ρf.

(2.7)

0

By (1.1)1 , we deduce







μ(ρ )u x x(t , y ), t = μ(ρ ) − =−

d



ρx u ρt − ρ ρ

 ln ρ ( x(t , y ),t )







μ exp{ξ } dξ .

dt 0

This together with (2.7) gives



d dt



ln ρ ( x(t , y ),t )



w x(t , y ), t +







μ exp{ξ } dξ 0

  = −ρ γ x(t , y ), t +

x (t , y )

ρf.

(2.8)

0

Integrating (2.8) over (0, s) with respect to t, we get ln ρ ( y ,s)





μ exp{ξ } dξ 0

  = w x(0, y ), 0 − w ( y , s) +

ln ρ0 (x(0, y ))





0

ln δ  −2 w L ∞ ( Q T ) +

s

s x (t , y )

ργ +

μ exp{ξ } dξ − 0

ρf 0

0





μ exp{ξ } dξ − c ( T ) 0

 M 1 ln δ − c ( T ),

(2.9)

where we have used Step 1, (1.7) and (2.4). If ρ ( y , s) < 1, we have ln ρ ( y ,s)





μ exp{ξ } dξ  M 1 ln ρ ( y , s). 0

(2.10)

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S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

(2.9) and (2.10) give

ρ ( y , s)  If

δ c(T )

(2.11)

.

ρ ( y , s)  1, (2.11) is obviously true. Step 2 is obtained.

2

By Steps 1 and 2, the proof of Lemma 2.2 is complete. Lemma 2.3. For any 0  t  T , it holds



u 2x

ρ ut2  c ( T ).

+

I

QT

Proof. Using (1.1)1 , we rewrite (1.1)2 as



ρ ut + ρ uu x + ρ γ



  = μ(ρ )u x x + ρ f .

x

(2.12)

Multiplying (2.12) by ut , integrating it over I , and using integration by parts and the Cauchy inequality, we have



1 d

ρ ut2 +



2 dt

I



d

μ(ρ )u 2x =

ργ ux +

dt

I

I



−γ

d



2



+γ

μ (ρ )ρt u 2x −

I

I

dt



ρ γ −1 ρt u x +

ργ ux − I

1

1

ρ uu x ut I



ρ f ut I





μ (ρ )(ρ

2



I

1

ρ γ −1 (ρ u )x u x +

+

1





ρ

8 I



ut2

I

ρ u 2 u 2x

+ c(T ) I



ρ ut2 + c ( T )

8

I

u )x u 2x

ρ f 2. I

This deduces

3



ρ ut2 +

4

1 d



μ(ρ )u 2x 

2 dt

I

I

d



ργ ux −

dt I



+γ

1



2

μ (ρ )(ρ u )x u 2x + c ( T )

I

ρ γ −1 (ρ u )x u x + c ( T )

I



ρ u 2 u 2x I



ρ f 2. I

By (1.7), Lemma 2.2, the Sobolev inequality and integration by parts, we have

3



ρ ut2 +

4 I



d



1 d



μ(ρ )u 2x

2 dt I

ργ ux −

dt I

1



2 I

μ (ρ )ρ u 3x −

1



2 I

2



μ (ρ )ρx uu 2x + c ( T )

μ(ρ )u 2x I

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725



ρ γ −1 ρx uu x + γ

+γ I



d



dt I

−



ρ γ u 2x + c ( T ) I

2    ρ γ u x + c(T ) μ(ρ )u 2x + c ( T )μ(ρ )u x L ∞ μ(ρ )u 2x I



1

1705

μ (ρ )ρx uu 2x + γ

2 I

I



ρ γ −1 ρx uu x + c ( T ).

I

We will estimate the fourth and the fifth terms of the right side above.

−

1







μ (ρ )ρ

2 I

=−

ρ γ −1 ρx uu x

+γ I



1

2 x uu x

μ (ρ )



μ2 (ρ )

2



2

ρx u μ(ρ )u x + γ

I

=−

I



1

ρ γ −1 ρ x u μ(ρ )u x μ(ρ )

μ (ρ )



μ2 (ρ )

2

2 1 ρx u μ(ρ )u x − ρ γ +

ρ γ −1 ρ

+γ

=



1



1

μ(ρ )

2

=



1

I

−

1

 ρ

2 I

ρ 2 γ −1 ρ x u μ(ρ )



I

  1 μ (ρ ) γ ρ ρx u μ(ρ )u x − ρ γ − 2 μ2 (ρ )

I

 ds

u





μ(s) I

2 u μ(ρ )u x − ρ γ − x

 ρ I

μ (s) 2γ s ds μ2 (s)

s 2 γ −1

μ(ρ )u x − ρ γ + γ





0

u+γ x

μ (ρ ) 2γ ρ ρx u μ2 (ρ )



0

μ(s)

 ds



u

u x

0

μ (s) γ s ds μ2 (s)

 ρ γ −1 s I

I

 ρ

x

0

μ(ρ )

2



I

μ(ρ )u x − ρ γ + γ

2 u μ(ρ )u x − ρ γ −

μ(s) I





 ρ γ −1 s



1

u



x

I

+γ

x

μ(ρ ) I

  μ (ρ ) ρx u −2μ(ρ )u x ρ γ + ρ 2γ μ2 (ρ )

2

I







μ(ρ )u x − ρ γ

x



 u μ(ρ )u x − ρ γ + γ

ds x

0

  ρ

s 2 γ −1

μ(s) I

 ds

0

Using integration by parts, we have

−

1



2

μ (ρ )ρx uu 2x + γ

I

=−

1

ρ γ −1 ρx uu x

I



1

μ(ρ )

2



ux



μ(ρ )u x − ρ γ

I

−

1

μ(ρ )

u





μ(ρ )u x − ρ γ μ(ρ )u x − ρ γ

 x

I

 ρ + I

2

0



  μ (s) γ s ds u x μ(ρ )u x − ρ γ + μ2 (s)

 ρ I

0



  μ (s) γ s ds u μ(ρ )u x − ρ γ x μ2 (s)

u. x

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S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

+

1

 ρ

2 I

−γ



 ρ γ −1 s

μ (s) 2γ s ds u x − γ μ2 (s)

μ(s) I

0

 ρ γ −1 s



I

ds u x



μ(ρ )u x − ρ γ



0

 ρ

 ds u μ(ρ )u x − ρ γ x − γ

μ(s)





s 2 γ −1

μ(s) I

0

 ds u x .

0

This together with Lemma 2.2, (2.12), (1.7), the Cauchy inequality, and the Sobolev inequality gives

−

1



2

μ (ρ )ρx uu 2x + γ

I

   c ( T )μ(ρ )u x 





ρ γ −1 ρx uu x

I



μ(ρ )u 2x + c ( T )

L∞ I

μ(ρ )u 2x + c ( T ) I

 ρ

ρ γ −1 μ (s)sγ s (μ(ρ )u x − ρ γ )u + + ds u − γ ds u (ρ ut + ρ uu x − ρ f ) − μ(ρ ) μ(s) μ2 (s) I

0

0



   c ( T )μ(ρ )u x  L ∞

μ(ρ )u 2x +

1



8

I

 2 ρ ut2 + c ( T ) μ(ρ )u 2x + c ( T ).

I

I

From above, we get

5



ρ ut2 +

8

1 d



μ(ρ )u 2x 

2 dt

I

I

d





ρ γ u x + c(T )

dt I

2

μ(ρ )u 2x I

  + c ( T )μ(ρ )u x  L ∞



μ(ρ )u 2x + c ( T ).

(2.13)

I

By Lemma 2.2, (2.12), W 1,1 → L ∞ , and the Cauchy inequality, we get

  μ(ρ )u x 

L∞

   μ(ρ )u x − ρ γ L ∞ + c ( T )      μ(ρ )u x − ρ γ  +  μ(ρ )u x − ρ γ  + c ( T )  c(T ) x I

 c(T )



μ(ρ )|u x | + c ( T ) I

|ρ ut + ρ uu x − ρ f | + c ( T ) I



√

μ(ρ )u 2x + c ( T ) ρ ut L 2 ( I ) + c ( T ).

 c(T )

(2.14)

I

By (2.13)–(2.14) and the Cauchy inequality, we get

5



ρ ut2 +

8 I

1 d



μ(ρ )u 2x 

2 dt I

d



dt I

2



ρ γ u x + c(T )

μ(ρ )u 2x I

+

1



ρ ut2 + c ( T ).

8 I

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

1707

Thus,

1



2



1 d

ρ ut2 +

μ(ρ )u 2x 

2 dt

I





d

2

ρ γ u x + c(T )

dt

I

μ(ρ )u 2x

I

+ c ( T ).

(2.15)

I

Integrating (2.15) over (0, t ) and using the Cauchy inequality, we have

1

t

ρ

2 0

ut2

+

1

μ(ρ

2

I

t 



)u 2x

ρ γ u x + c(T )



I

I



+ c(T )

I

0

1

2

μ(ρ )u 2x t 



μ(ρ )u 2x + c ( T )

4 I

2

μ(ρ )u 2x

+ c ( T ),

I

0

which implies

1

t

ρ

2

ut2

+

μ(ρ

4

I

0

t 



1

)u 2x

I

2

μ(ρ )u 2x

 c(T ) 0

+ c ( T ).

I

Using the Gronwall inequality and Lemma 2.1, we complete the proof of Lemma 2.3.

2

Lemma 2.4. For any 0  t  T , it holds







ρx2 + ρt2  c ( T ).

I

Proof. Differentiating (1.1)1 with respect to x, multiplying the resulting equation by over I , and using integration by parts, we have

1 d



ρx2 = −

2 dt I

3





ρx2 u x −

2 I

ρρx [μ(ρ )u x ]x + μ(ρ )

I



ρx , integrating

ρρx2 μ (ρ )u x . μ(ρ )

I

By (2.12), (2.14), Lemmas 2.2–2.3 and the Cauchy inequality, we have

1 d



  ρ  c ( T )μ(ρ )u x L ∞

2 dt I





ρ ut2

 c(T ) I





2 x

2 x

ρ − I



ρx2 + c ( T ) I

   ρρx  ρ ut + ρ uu x + ρ γ x − ρ f μ(ρ )

I



ρx2 + c ( T ) I

By the Gronwall inequality, and Lemma 2.3, we get



ρx2  c ( T ). I

ρ ut2 + c ( T ). I

1708

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

This together with (1.1)1 and Lemma 2.3 gives



ρt2  c ( T ). I

2

The proof of Lemma 2.4 is complete. Lemma 2.5. For any 0  t  T , we have

u 2xx  c ( T ). QT

Proof. It follows from (2.12), (2.14), (1.7), Lemmas 2.2–2.4 that

T 



u 2xx

 c(T )

QT

ρ

ut2

QT



ρx2 + c ( T ) I

0

ρx2

QT

u x 2L ∞

+ c(T )

I

0

T + c(T )

2 u 2x

+ c(T )

f2 I

 c ( T ). This proves Lemma 2.5.

2

Lemma 2.6. For any 0  t  T , it holds





ρ ut2 + I

u 2xt  c ( T ). QT

Proof. Differentiating (2.12) with respect to t, we have



ρ utt + ρt ut + ρt uu x + ρ ut u x + ρ uu xt + ρ γ

 xt

  = μ(ρ )u x xt + ρt f + ρ f t .

Multiplying (2.16) by ut , integrating over I , and using (1.1)1 and integration by parts, we have

1 d





ρ ut2 +

2 dt I



= −2 I

I

I



ρ uut u xt −

−

μ(ρ )u 2xt

I

μ (ρ )ρt u x u xt +





ρ ut2 u x +

ρt uu x ut − I

I



ρt f ut + I

γρ γ −1 ρt utx

ρ f t ut I

√ √  2 ρ ut L 2  ρ u L ∞ u xt L 2 + ut L ∞ u L ∞ ρt L 2 u x L 2

(2.16)

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725



1709

ρ ut2 + γ ρ γL ∞−1 ρt L 2 u xt L 2 + c ( T )ρt L 2 u x L ∞ u xt L 2

+ u x  L ∞ I

+ ut L ∞ ρt L 2  f L 2 + ut L ∞ ρ L 2  f t L 2 . This, combining Lemmas 2.2–2.4 and the Cauchy inequality, gives

1 d





ρ ut2 +

2 dt

μ(ρ )u 2xt

I

I

  √     c ( T )  ρ ut L 2 + u xx L 2 +  f t L 2 + 1  μ(ρ )u xt L 2 + c ( T ) u xx L 2 + 1 

1



2

  μ(ρ )u 2xt + c ( T ) u xx L 2 + 1

I





ρ ut2 + c ( T ) I

I

ρ ut2 I

u 2xx + c ( T )



f t2 + c ( T ), I

which implies

d





ρ

dt

ut2

+

I

μ(ρ

   c ( T ) u xx L 2 + 1

)u 2xt

I



ρ ut2 I





u 2xx + c ( T )

+ c(T ) I

f t2 + c ( T ).

(2.17)

I

By (1.6) and (2.12), we have

ut (x, 0) =

1





ρ0 (x)

μ(ρ0 )u 0x − ρ0γ x (x) − (u 0 u 0x )(x) + f (x, 0)

= g (x) − (u 0 u 0x )(x).

(2.18)

Integrating (2.17) over (0, t ), and using Lemma 2.5 and (2.18), we have



t

ρ I

ut2

+

t

μ(ρ 0

)u 2xt

 c(T )

I



u xx L 2 + 1

0





ρ ut2 + c ( T ). I

Using the Gronwall inequality together with Lemma 2.5 and (1.7), we complete the proof of Lemma 2.6. 2 Lemma 2.7. For any 0  t  T , it holds

u 2xx  c ( T ). I

Proof. First, we claim

u x L ∞ ( Q T )  c ( T ). In fact, this can be obtained directly by (2.14), Lemmas 2.3 and 2.6.

(2.19)

1710

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

It follows from (2.12), (2.19), (1.7), Lemmas 2.2–2.4, and Lemma 2.6 that

u 2xx  c ( T ). I

This completes the proof of Lemma 2.7.

2

Lemma 2.8. For any 0  t  T , it holds







2 2 ρxx + ρxt +  ργ

 2  γ  2   + ρ  + xx



xt

I





ρtt2 +  ρ γ

 2    c ( T ). tt

QT

Proof. From (1.1)1 , we have

ρxxt = −ρxxx u − 3ρxx u x − 3ρx u xx − ρ u xxx . Multiplying (2.20) by

ρxx , integrating over I , and using integration by parts, we have



1 d

2 ρxx =−

2 dt

(2.20)

5



2

I



2 ρxx ux − 3

I

ρx ρxx u xx − I



ρρxx u xxx I

2 ρxx + 3ρx L ∞ ρxx L 2 u xx L 2 + ρ L ∞ ρxx L 2 u xxx L 2 .

 c ( T )u x L ∞ I

By the Sobolev inequality, the Cauchy inequality, Lemmas 2.2–2.4, Lemma 2.7 and (2.19), we have

1 d



2 ρxx  c(T )

2 dt I

I



2 ρxx + c(T )

 c(T ) I

Multiplying (1.1)1 by





2 ρxx + c ( T ) 1 + ρxx L 2 ρxx L 2 + c ( T )ρxx L 2 u xxx L 2

u 2xxx + c ( T ).

(2.21)

I

γρ γ −1 , we have  γ  γ ρ t + ρ x u + γρ γ u x = 0,

(2.22)

which implies

 γ ρ

xxt

      + ρ γ xxx u + (γ + 2) ρ γ xx u x + (2γ + 1) ρ γ x u xx + γρ γ u xxx = 0.

Similarly to (2.21), we get

1 d



2 dt I

 γ  2  ρ   c(T ) xx

I

 γ  2  ρ  + c(T ) xx

u 2xxx + c ( T ). I

(2.23)

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

1711

(2.21) and (2.23) imply

d



dt



  2  ρ +  ρ γ xx   c ( T )





2 xx

I

  2  ρ +  ρ γ xx  + c ( T )



2 xx

I

u 2xxx + c ( T ).

(2.24)

I



The next step is to estimate the term I u 2xxx . Differentiating (2.12) with respect to x, we have



μ(ρ )u xxx = −2μ (ρ )ρx u xx + ρx ut + ρ u xt + ρx uu x + ρ u 2x + ρ uu xx + ρ γ



− μ (ρ )ρ

2 x ux

 xx



− μ (ρ )ρxx u x − ρx f − ρ f x .

(2.25)

(2.25), combining (1.7), Lemmas 2.2–2.4, Lemma 2.7 and the Sobolev inequality, gives







u 2xxx  c ( T ) I



2 ρxx + c(T ) + c(T )

u 2xt + c ( T )

I

I

 γ  2  ρ  + c ( T ) f 2 1 . xx H

I

Since f ∈ C ([0, T ]; H 1 ), we have



u 2xxx

  2  ρ +  ρ γ xx  + c ( T )



 c(T )

I



2 xx

I

u 2xt + c ( T ). I

By (2.24) and (2.26), we get

d



dt



  2  ρ +  ρ γ xx   c ( T )



2 xx

I



  2  ρ +  ρ γ xx  + c ( T )



2 xx

I

u 2xt + c ( T ). I

By Lemma 2.6 and the Gronwall inequality, we get







2 ρxx +  ργ

 2    c ( T ). xx

I

This, combining (1.1)1 , (2.22), Lemmas 2.2–2.4, Lemmas 2.6–2.7, immediately gives







ρxt2 +  ρ γ

 2   + xt

I

The proof of Lemma 2.8 is complete.







ρtt2 +  ρ γ

 2    c ( T ). tt

QT

2

From (2.26), Lemmas 2.6 and 2.8, we immediately get the next result. Lemma 2.9. For any 0  t  T , it holds

u 2xxx  c ( T ). QT

(2.26)

1712

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

Lemma 2.10. For any 0  t  T , it holds



u 2xt +

I

ρ utt2  c ( T ). QT

Proof. Multiplying (2.16) by utt , integrating over I , and using integration by parts, (1.7), Lemma 2.2 and the Cauchy inequality, we have



1 d

ρ utt2 + I

=

μ(ρ )u 2xt

2 dt I



1



μ (ρ )ρt u 2xt −

2 I

μ (ρ )ρt u x u xtt +

(ρt f + ρ f t )utt

I



I







ρt ut utt + ρt uu x utt + ρ ut u x utt + ρ uu xt utt − ρ γ t u xtt

− I





 c ( T )ρt  H 1

μ(ρ I

)u 2xt

−





μ (ρ )ρt u x u xtt − I



+ c ( T )u x 2L ∞



I



ρ ut2 + c ( T )u 2L ∞ I

(ρt ut utt + ρt uu x utt ) +

μ(ρ )u 2xt + I

 γ ρ t u xtt +

I

1



ρ utt2

4 I



f t2 .

ρt f utt + c ( T ) I

I

By Lemmas 2.3–2.4, Lemmas 2.6–2.8, (1.7), the Cauchy inequality and the Sobolev inequality, we have

3



ρ utt2 +

4 I

1 d

I



μ (ρ )ρt u x u xtt −

μ(ρ )u 2xt −

 c(T ) I

+

I

 c(T )

μ(ρ I

1 d



dt

−



ρt ut2 +

2 dt d

)u 2xt

I



ρtt ut2 −



d

d

dt I

dt

I

I





I

ρtt uu x ut + ρt ut2 u x + ρt uu xt ut



ρt f ut − I

f t2 + c ( T )

I

I

  μ (ρ )ρt u x u xt + ρt ut2 + ρt uu x ut − ρ γ t u xt − ρt f ut 2

I

I



(ρtt f + ρt f t )ut + c ( T )

1

ρtt2 + c ( T )ut 4L ∞ + c ( T )

+ c(T )

 γ ρ t u xtt

μ

(ρ )ρt2 u x u xt + μ (ρ )ρtt u x u xt + μ (ρ )ρt u 2xt

ρt uu x ut +

 γ d ρ tt u xt + 



I



dt

I

μ(ρ )u 2xt −





I

I

 γ ρ t u xt −



ρt uu x utt + I

μ (ρ )ρt u x u xt +

2

I

I



dt

1

 c(T )

d

I



ρt ut utt −

f t2 + c ( T ) I







ρt f utt + c ( T ) I

+

μ(ρ )u 2xt

2 dt



−



 γ  2  ρ  + c(T ) tt

f t2 + c ( T ). I





S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725



Since ut 4L ∞  (



u 2xt )2  c ( T )(

I



3

1 d

ρ utt2 +

4 I

−

d dt

μ(ρ )u 2xt )2 , we have



μ(ρ )u 2xt

2 dt



I

1713

I



1



μ (ρ )ρt u x u xt + ρt ut2 + ρt uu x ut − ρ γ t u xt − ρt f ut



2

I

2   2  γ  2  2   μ(ρ )u xt + c ( T ) ρtt + ρ tt + c ( T ) f t2 + c ( T ). + c(T ) I

I

(2.27)

I

Integrating (2.27) over (0, t ), and using Lemma 2.8, we have

3

t

ρ utt2 +

4 0



I

1

1



μ(ρ )u 2xt

2 I





2 0 )u xt (x, 0) −

I



+

2 t ut

μ (ρ )ρt u x u xt + ρ

μ(ρ

2

1



2

  + ρt uu x ut − ρ γ t u xt − ρt f ut

I



1







μ (ρ )ρt u x u xt + ρt ut2 + ρt uu x ut − ρ γ t u xt − ρt f ut (x, 0) 2

I

t 

2

μ(ρ )u 2xt

+ c(T )

+ c ( T ).

(2.28)

ρt |t =0 = −(ρ0 u 0 )x ∈ H 1 .

(2.29)

I

0

By (1.1)1 , we get

(2.18), (2.28) and (2.29) show

3

t

ρ

4 0

I

2 utt



−

+

1



μ(ρ )u 2xt

2 I



1



μ (ρ )ρt u x u xt + ρt ut2 + ρt uu x ut − ρ γ t u xt − ρt f ut



2

I

t  + c(T )

2

μ(ρ 0

)u 2xt

+ c(T )

I

 c ( T )ρt L 2 u x L ∞ u xt L 2 +

1

(ρ u )x ut2 + ρt L 2 u L ∞ u x L 2 ut L ∞

2 I

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S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

   +  ρ γ t  L 2 u xt L 2 + ut L ∞  f L 2 ρt L 2 + c ( T )

t  0

   c ( T ) μ(ρ )u xt  L 2 −

ρ uut u xt + c ( T )ut L ∞ + c ( T )



μ(ρ

4

)u 2xt

I

2

μ(ρ 0

t 



+ c(T )

I

t 

I

1

2

μ(ρ )u 2xt

)u 2xt

+ c(T )

I

2

+ c(T )

μ(ρ

)u 2xt

+ c ( T ),

I

0

which implies



t 

t

μ(ρ )u 2xt + I

ρ utt2  c ( T ) I

0

2

μ(ρ )u 2xt 0

+ c ( T ).

(2.30)

I

Here we have used Lemmas 2.2–2.4, Lemmas 2.6–2.8, (1.7), the Hölder inequality, the Sobolev inequality, and the Cauchy inequality. By Lemma 2.6 and the Gronwall inequality, we can complete the proof of Lemma 2.10. 2 Lemma 2.11. For any 0  t  T , it holds

u 2xxx  c ( T ),

(2.31)

u 2xxt  c ( T ).

(2.32)

I



QT





ρtt2 +  ρ γ

 2    c ( T ). tt

(2.33)

I

Proof. By (2.26), Lemmas 2.8 and 2.10, we get (2.31). (2.32) can be obtained by (2.16), Lemmas 2.2– 2.4, Lemmas 2.7–2.8, Lemma 2.10. From (1.1)1 , Lemmas 2.2–2.4, Lemmas 2.8 and 2.10, we have (2.33). 2 From above, we get the following estimates

δ  ρ (x, t )  c ( T ), c(T )

(2.34)

and

           2 + u  3 + u t  1  ρ , ρ γ  2 +  ρt , ρ γ  1 +  ρtt , ρ γ H H H t H tt L  2  + ρ utt + u 2xxt  c ( T ),

(2.35)

QT

where c ( T ) is a positive constant, independent of δ and (·,·) X =  ·  X +  ·  X , for some Banach space X . The proof of Theorem 2.1 is complete. 2

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

1715

Proof of Theorem 1.1. To prove Theorem 1.1, we construct a sequence of approximate solutions to (1.1)–(1.3) by giving the initial density a positive lower bound δ and using Theorem 2.1. Then, we get some estimates uniformly on δ , and take limit δ → 0 (take a subsequence if necessary). After that, we get existence of classical solutions to (1.1)–(1.3). More precisely, we denote ρ0δ = ρ0 + δ . u δ0 is the unique solution of the following elliptic problems for each δ > 0:

 





 

μ ρ0δ u δ0x x (x) − p ρ0δ

 x

  (x) = ρ0δ (x) − f (x, 0) + g (x) ,

x ∈ (0, 1),

(2.36)

and



u δ0 x=0,1 = 0. Since

(2.37)

ρ0 , ρ0γ ∈ H 2 , we have as δ → 0

ρ0δ → ρ0 , in H 2 , 

 δ γ

ρ0

γ

in H 2 .

→ ρ0 ,

(2.38) (2.39)

From (2.36)–(2.39) and the elliptic theory (please refer to [6])

u δ0 → u 0 ,

in H 3 ,

(2.40)

as δ → 0. Consider (1.1), (2.36) with initial boundary data









ρ δ , u δ t =0 = ρ0δ , u δ0 , in I ,

and



u δ ∂ I = 0,

for t  0.

We can get a unique solution (ρ δ , u δ ) for each δ > 0 by Theorem 2.1, with the following estimates:

δ c(T )

 ρ δ (x, t )  c ( T ),

(2.41)

and

 δ  δ   ρ ,P ρ    + u δ  t

H1

H2



+

          +  ρtδ , P ρ δ t  H 1 +  ρttδ , P ρ δ tt  L 2 + u δ  H 3 



2



2 

ρ δ uttδ  + u δxxt   c ( T ),

QT

where c ( T ) is a positive constant, independent of δ .

(2.42)

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S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

Based on the estimates (2.41) and (2.42), we get a solution (ρ , u ) to (1.1)–(1.3) after taking limit

δ → 0 (take the subsequence if necessary):













ρ ∈ L ∞ [0, T ]; H 2 , ρt ∈ L ∞ [0, T ]; H 1 , ρtt ∈ L ∞ [0, T ]; L 2 , ρ  0,

 γ           ρ xx ∈ L ∞ [0, T ]; L 2 , ρ γ xt ∈ L ∞ [0, T ]; L 2 , ρ γ tt ∈ L ∞ [0, T ]; L 2 ,       u ∈ L ∞ [0, T ]; H 3 ∩ H 01 , ut ∈ L ∞ [0, T ]; H 01 ∩ L 2 [0, T ]; H 2 . Since u ∈ L ∞ ([0, T ]; H 3 ) and ut ∈ L ∞ ([0, T ]; H 01 ), we have u ∈ C ([0, T ]; H 2 ) (please refer to [6]). This together with (1.1)1 , (2.22) and [3] implies





ρ ∈ C [0, T ]; H 2 ,

(2.43)

and





ρ γ ∈ C [0, T ]; H 2 .

(2.44)

Denote G = [μ(ρ )u x − ρ γ ]x + ρ f . By (2.12), (1.7) and the regularities of (ρ , u ), we have



Gt =





G = ρ ut + ρ uu x ∈ L 2 [0, T ]; H 2 ,

μ(ρ )u x − ρ γ



xt

  + (ρ f )t ∈ L 2 [0, T ]; L 2 .

From the embedding theorem ([6]), we have G ∈ C ([0, T ]; H 1 ). Since



μ(ρ )u x − ρ γ

 x

ρ f ∈ C ([0, T ]; H 1 ), we get

  ∈ C [0, T ]; H 1 .

This means





μ(ρ )u x − ρ γ ∈ C [0, T ]; H 2 .

(2.45)

By (2.44) and (2.45), we get





μ(ρ )u x ∈ C [0, T ]; H 2 . This together with (1.7) and (2.43) implies









u ∈ C [0, T ]; H 3 .

(2.46)

By (1.1)1 , (2.43) and (2.46), we obtain

ρt ∈ C [0, T ]; H 1 .

(2.47)

It follows from (1.1)2 , (2.43)–(2.44), (2.46) and (1.7) that

  (ρ u )t ∈ C [0, T ]; H 1 . Differentiating (1.1)1 with respect to t, we have

ρtt = −(ρ u )tx .

(2.48)

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1717

This combining (2.48) gives





ρtt ∈ C [0, T ]; L 2 .

(2.49)

By (2.22), (2.44) and (2.46), we get

 γ   ρ t ∈ C [0, T ]; H 1 . The existence of global classical solutions as in Theorem 1.1 is obtained. The uniqueness of the solutions can be proved by the standard method like in [4], we omit it for simplicity. The proof of Theorem 1.1 is complete. 2 3. Proof of Theorem 1.2 Before proving Theorem 1.2, we need the following auxiliary theorem. Theorem 3.1. Consider the same assumptions as in Theorem 1.2, and in addition assume that ρ0  δ > 0. Then for any T > 0 there exists a unique global classical solution (ρ , u ) to (1.1)–(1.3) satisfying

















ρ ∈ C [0, T ]; H 4 , ρt ∈ C [0, T ]; H 3 , ρtt ∈ C [0, T ]; H 1 ∩ L 2 [0, T ]; H 2 , ρttt ∈ L 2 ( Q T ), ρ  



δ c(T )









u ∈ C [0, T ]; H 4 ∩ H 01 ∩ L 2 [0, T ]; H 5 ,

,





ut ∈ C [0, T ]; H 2 ∩ L 2 [0, T ]; H 3 ,









utt ∈ C [0, T ]; L 2 ∩ L 2 [0, T ]; H 01 .

Proof. Similarly to the proof of Theorem 2.1, Theorem 3.1 can be proved by some a priori estimates globally in time. Since (2.34) and (2.35) are also valid here, we need other a priori estimates about higher order derivatives of (ρ , u ). The generic positive constant c ( T ) may depend on the initial data presented in Theorem 1.2 and some other known constants but independent of δ . Lemma 3.1. For any 0  t  T , it holds







2 2 ρxxx + ρxxt +  ργ

 xxx

2  γ   + ρ

I

xxt

2   +







2 ρxtt +  ργ

 2  + u2



xxxx

xtt

 c ( T ).

QT

Proof. Differentiating (2.20) with respect to x, we have

ρxxxt = −ρxxxx u − 4ρxxx u x − 6ρxx u xx − 4ρx u xxx − ρ u xxxx . Multiplying (3.1) by we have

1 d

ρxxx , integrating the resulting equation over I , and using integration by parts,

2 ρxxx =−

2 dt

(3.1)

I

7



2 ρxxx ux − 6

2 I

7



I





ρxx ρxxx u xx − 4

ρx ρxxx u xxx − I

ρρxxx u xxxx I

2 ρxxx + 6u xx L ∞ ρxx L 2 ρxxx L 2

 u x  L ∞ 2

I

+ 4ρx L ∞ ρxxx L 2 u xxx L 2 + ρ L ∞ ρxxx L 2 u xxxx L 2 .

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S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

By the Sobolev inequality, (2.34)–(2.35) and the Cauchy inequality, we get





1 d



2 ρxxx  c(T )

2 dt

2 ρxxx + c(T )

I

u 2xxxx + c ( T ).

I

(3.2)

I

Similarly to (3.2), we get from (2.22)



1 d 2 dt

 γ   ρ

xxx

2   c(T )



I

 γ   ρ

xxx

2  + c(T )

I

u 2xxxx + c ( T ).

(3.3)

I

By (3.2) and (3.3), we have

d



dt





2 ρxxx +  ργ

 xxx

2    c(T )



I





2 ρxxx +  ργ

 xxx

2   + c(T )

u 2xxxx + c ( T ).

I

(3.4)

I

Differentiating (2.25) with respect to x, we have

μ(ρ )u xxxx = −3μ (ρ )ρx u xxx − 3μ (ρ )ρxx u xx − 3μ

(ρ )ρx2 u xx + ρxx ut + 2ρx u xt

    + ρ u xxt + (ρx uu x )x + ρ u 2x x + (ρ uu xx )x + ρ γ xxx − μ

(ρ )ρx3 u x − 3μ

(ρ )ρx ρxx u x − μ (ρ )ρxxx u x − ρxx f − 2ρx f x − ρ f xx .

(3.5)

By (3.5) and (2.34)–(2.35), we have



u 2xxxx  c ( T ) I





2 ρxxx +  ργ

 xxx

2   + c(T )

I



u 2xxt + c ( T )

I

2 f xx + c ( T ).

(3.6)

I

Combining (3.4) and (3.6), we get

d



dt





2 ρxxx +  ργ

 xxx

2    c(T )

I







2 ρxxx +  ργ

 xxx

2   + c(T )

I



u 2xxt + c ( T )

I

2 f xx + c ( T ). I

By the Gronwall inequality and (2.35), we obtain







2 ρxxx +  ργ

 xxx

2    c ( T ).

(3.7)

I

It follows from (2.20), (2.22), (1.1)1 , (2.34), (2.35), (3.6) and (3.7) that







2 ρxxt + + ρ γ

 xxt

2   +

I

The proof of Lemma 3.1 is complete.

QT

2





2 ρxtt +  ργ

 2  + u2 xtt

xxxx



 c ( T ).

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

1719

Lemma 3.2. For any T > 0, we have

√  ( ρ ) x 

L∞ ( Q T )

√  + ( ρ )t  L ∞ ( Q

T)

 c ( T ).

1 Proof. Multiplying (1.1)1 by 2√ ρ , we have

1√ √ √ ( ρ )t + ( ρ )x u + ρ u x = 0.

(3.8)

2

Differentiating (3.8) with respect to x, we get

3 √ 1√ √ √ ( ρ )xt + ( ρ )xx u + ( ρ )x u x + ρ u xx = 0. 2

√

Denote h = (

2

ρ )x , we have ht + h x u +

3

1√

hu x +

2

ρ u xx = 0,

2

which implies

d dt

 t 3

h exp







u x x(τ , y ), τ dτ

2

=−

 t

1√

ρ u xx exp

2

3





u x x(τ , y ), τ dτ ,

2

0



(3.9)

0

where x(t , y ) is the solution to (2.5). Integrating (3.9) over (0, s), we get

 h( y , s) = exp −

3

s









u x x(τ , y ), τ dτ h x(0, y ), 0

2



0

−

1 2

 exp −

3

s





 s

u x x(τ , y ), τ dτ

2 0

√

ρ u xx exp

0

 t 3





u x x(τ , y ), τ dτ

2

 dt .

0

This together with (2.35) implies

√  ( ρ ) x 

L∞ ( Q T )

 c ( T ).

From (2.34)–(2.35), (3.8) and (3.10), we get

√  ( ρ )t  ∞ L (Q The proof of Lemma 3.2 is complete.

T)

 c ( T ).

2

The next lemma plays the most important role in this section.

(3.10)

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S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

Lemma 3.3. For any 0  t  T , it holds









ρ 3 utt2 + ρ u 2xxt +

I





ρ 2 u 2xtt + u 2xxxt  c ( T ).

QT

Proof. Differentiating (2.16) with respect to t, multiplying the resulting equation by over I , and using integration by parts and the Cauchy inequality, we have

1 d





ρ 3 utt2 +

2 dt I

=−

1

I





ρ 2 ρt utt2 −

2 I

ρ

−2

3 2 utt u x



ρtt ut + ρtt uu x + 2ρt ut u x + 2ρt uu xt + 2ρ ut u xt + ρ γ

I



− I

μ(ρ )ρ 2 u 2xtt







I

I





μ

(ρ )ρt2 u x + μ (ρ )ρtt u x + 2μ (ρ )ρt u xt + μ(ρ )u xtt (ρρx utt )





ρ utt2 + c ( T ) ρ γ

 2  +1 xtt L 2

−4



μ(ρ )ρ 2 u 2xtt

3

I

I



μ(ρ )ρ u xtt

  √ ρ utt ( ρ )x + c ( T )  f t 2L 2 +  f tt 2L 2 + c ( T )

√

I



μ(ρ )ρ

3



I



 c(T )



ρ 2 utt

μ

(ρ )ρt2 u x + μ (ρ )ρtt u x + 2μ (ρ )ρt u xt ρ 2 u xtt

ρ uutt u xtt −



xtt

  (ρtt f + 2ρt f t + ρ f tt ) ρ 2 utt

+





3

−

 

I



2

ρ 2 utt , integrating

I



+ c(T )

2 2 u xtt

+ c(T )



ρ utt2 + c ( T ) ρ γ

I

√

2

ρ utt2 ( ρ )x  + c ( T )

I

 2 2



xtt L

f t2 + c ( T )

I

f tt2 + c ( T ). I

This together with Lemma 3.2 implies

1 d



ρ

2 dt I

3 2 utt

+

1



μ(ρ )ρ 2 u 2xtt

3 I





ρ utt2 + c ( T )

 c(T ) I

 γ  2  ρ  + c(T ) xtt

I







f t2 + f tt2 + c ( T ).

(3.11)

I

Integrating (3.11) over (0, t ), and using (2.35) and Lemma 3.1, we get

1



ρ 3 utt2 +

2 I

1

t

μ(ρ )ρ 2 u 2xtt 

3 0

I

1



ρ 3 utt2 (x, 0) + c ( T ).

2 I

(3.12)

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

1721

√

ρ0 g x )x ∈ L 2 , we have ρ 3 utt2 + ρ 2 u 2xtt  c ( T ).

By (2.16), (2.18), (2.29), (3.12), (1.7) and (

I

(3.13)

QT

Differentiating (2.16) with respect to x, we get

μ(ρ )u xxxt = −2μ (ρ )ρx utxx − 2μ

(ρ )ρx ρt u xx − 2μ (ρ )ρxt u xx − 2μ

(ρ )ρxt ρx u x − μ (ρ )ρt u xxx − μ

(ρ )ρx2 ρt u x − μ

(ρ )ρt ρxx u x − μ (ρ )ρxxt u x √ √ − μ

(ρ )ρx2 u xt − μ (ρ )ρxx u xt + 2( ρ )x ρ utt + ρ u xtt + ρxt ut + ρt u xt + ρxt uu x + ρt u 2x + ρt uu xx + ρx ut u x + 2ρ u xt u x + ρ ut u xx + ρx uu xt   + ρ uu xxt + ρ γ xxt − ρxt f − ρt f x − ρx f t − ρ f xt . This together with (1.7), (2.35), (3.13) and Lemmas 3.1–3.2 implies

u 2xxxt  c ( T ). QT

By (2.16), (2.34)–(2.35), and (3.13), we have



ρ u 2xxt  c ( T ). I

2

The proof of Lemma 3.3 is complete. Lemma 3.4. For any 0  t  T , it holds





ρ

2 xxxx

  +  ργ

xxxx

2   +

I

u 2xxxxx  c ( T ). QT

Proof. Differentiating (3.1) with respect to x, multiplying the resulting equation by over I , and using integration by parts, we get

1 d

2 ρxxxx =−

2 dt

9



2

I

2 ρxxxx u x − 10

I

−5



ρxxx ρxxxx u xx − 10 I

I



 c(T )

ρ

2 xxxx

I

ρρxxxx u xxxxx I



u 2xxxx + c ( T )

+ c(T )

ρxx ρxxxx u xxx I



ρx ρxxxx u xxxx −

ρxxxx , integrating

I

u 2xxxxx + c ( T ).

(3.14)

I

Similarly, we have

1 d



2 dt I

 γ   ρ

xxxx

2   c(T )

I

 γ   ρ

xxxx

2  + c(T )



u 2xxxx + c ( T )

I

u 2xxxxx + c ( T ). I

(3.15)

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S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

By (3.14) and (3.15), we obtain

d



dt





2 ρxxxx +  ργ

 xxxx

2    c(T )

I





I



2 ρxxxx +  ργ

 xxxx

2   + c(T )

u 2xxxx I

u 2xxxxx

+ c(T )

+ c ( T ).

(3.16)

I

Now we estimate the third term of the right-hand side of (3.16). Differentiating (3.5) with respect to x, we have

μ(ρ )u xxxxx = −4μ (ρ )ρx u xxxx − 6μ

(ρ )ρx2 u xxx − 6μ (ρ )ρxx u xxx − 12μ

(ρ )ρx ρxx u xx − 4μ (ρ )ρxxx u xx − 4μ

(ρ )ρx3 u xx + ρxxx ut   + 3ρxx u xt + 3ρx u xxt + ρ u xxxt + (ρx uu x )xx + ρ u 2x xx   + (ρ uu xx )xx + ρ γ xxxx − μ

(ρ )ρx4 u x − 6μ

(ρ )ρx2 ρxx u x 2 − 3μ

(ρ )ρxx u x − 4μ

(ρ )ρx ρxxx u x − μ (ρ )ρxxxx u x

− ρxxx f − 3ρxx f x − 3ρx f xx − ρ f xxx . Therefore,



u 2xxxxx

 c(T )

I

u 2xxxx

I





+ c(T )

u 2xxt

+ c(T ) I

ρ

2 xxxx

  +  ργ

u 2xxxt

+ c(T ) I

xxxx

2   + c(T )



I





2 2 f xx + f xxx + c ( T ),

(3.17)

I

where we have used (1.7), (2.34)–(2.35), Lemma 3.1. By (3.16) and (3.17), we obtain

d



dt





2 ρxxxx +  ργ

 xxxx

2    c(T )

I





I



2 ρxxxx +  ργ





u 2xxxt

+ c(T )

I

u 2xxxx I

u 2xxt

+ c(T )

xxxx

2   + c(T )



+ c(T )

I





2 2 + c ( T ). f xx + f xxx

I

Using the Gronwall inequality, (2.35), Lemmas 3.1 and 3.3, we get







2 ρxxxx +  ργ

 xxxx

2    c ( T ).

(3.18)

I

It follows from (3.17), (3.18), (2.35), Lemmas 3.1 and 3.3 that

u 2xxxxx  c ( T ). QT

This proves Lemma 3.4.

2

(3.19)

S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

1723

Lemma 3.5. For any 0  t  T , it holds

u 2xxxx  c ( T ). I

Proof. From (2.35), Lemmas 3.1, 3.3, and 3.4, we get

u L 2 (0,T ; H 5 )  c ( T ),

(3.20)

ut L 2 (0,T ; H 3 )  c ( T ).

(3.21)

and

By the embedding theorem (cf. [6]), (3.20), (3.21) show

u C ([0,T ]; H 4 )  c ( T ). 2

This proves Lemma 3.5.

From (1.1)1 , (2.22), (2.34)–(2.35), Lemma 3.1, Lemmas 3.3–3.5, we immediately get the following lemma. Lemma 3.6. For any 0  t  T , it holds







2 ρxtt +  ργ

I

+



 2  + ρ2

xxxt

xtt



2 ρttt +  ργ

  2  +  ρ γ xxxt 

 2  + ρ2 ttt

xxtt

  2  +  ρ γ xxtt   c ( T ).

QT

Here we have used the following inequality

√

√ 2

ρx2 utt2 = 2 ( ρ )x ρ utt2  c ( T )ρ utt2 . From above estimates, we get

δ  ρ (x, t )  c ( T ), c(T )

(3.22)

and

√  ( ρ ) x 

√        + ( ρ )t  L ∞ +  ρ , ρ γ  H 4 +  ρt , ρ γ t  H 3   3     √ +  ρtt , ρ γ tt  H 1 + u  H 4 + ut  H 1 + ρ 2 utt  L 2 +  ρ u xxt L 2   2   2   2 2 2 2 + ρ u xtt + ρ utt2 + u 2xxt + u 2xxxt + u 2xxxxx + ρttt +  ρ γ ttt  + ρxxtt +  ρ γ xxtt  L∞

QT

 c ( T ).

(3.23)

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S. Ding et al. / J. Differential Equations 251 (2011) 1696–1725

From (3.22) and (3.23), we get

ρ  H 4 + ρt  H 3 + ρtt  H 1 + u  H 4 + ut  H 2 + utt L 2  2  2 2  c ( T , δ), + u xtt + u 2xxxt + u 2xxxxx + ρttt + ρxxtt

(3.24)

QT

where c ( T , δ) is a positive constant, and may depend on δ . The proof of Theorem 3.1 is complete. 2 Proof of Theorem 1.2. To prove Theorem 1.2, we follow the similar arguments as in Section 2. After taking δ → 0 (take subsequence if necessary), we get a solution (ρ , u ) to (1.1)–(1.3) satisfying









√       √ ( ρ )x , ( ρ )t ∈ L ∞ ( Q T ), ρt , ρ γ t ∈ L ∞ [0, T ]; H 3 ,             ρtt , ρ γ tt ∈ L ∞ [0, T ]; H 1 ∩ L 2 [0, T ]; H 2 , ρttt , ρ γ ttt ∈ L 2 ( Q T ),         u ∈ L ∞ [0, T ]; H 4 ∩ L 2 [0, T ]; H 5 , ut ∈ L ∞ [0, T ]; H 01 ∩ L 2 [0, T ]; H 3 ,   √ ρ u xxt ∈ L ∞ [0, T ]; L 2 .

ρ , ρ γ ∈ L ∞ [0, T ]; H 4 ,

Since







u ∈ L 2 [0, T ]; H 5 ,





ρtt , ρ γ

  tt



ut ∈ L 2 [0, T ]; H 3 ,

  ∈ L 2 [0, T ]; H 2 ,





ρttt , ρ γ

  ttt

  ∈ L 2 [0, T ]; L 2 ,

we apply the embedding theorem in [6] to get









ρtt , ρ γ

u ∈ C [0, T ]; H 4 ,

  tt

  ∈ C [0, T ]; H 1 .

(3.25)

By (3.25), u ∈ L 2 ([0, T ]; H 5 ), (1.1)1 , (2.22) and [3], we have





ρ ∈ C [0, T ]; H 4 ,





ρ γ ∈ C [0, T ]; H 4 ,

The proof of Theorem 1.2 is complete.





ρt , ρ γ

 t

  ∈ C [0, T ]; H 3 .

2

Acknowledgments The authors would like to thank the referee for his/her valuable comments and suggestions. The first and second authors were supported by the National Basic Research Program of China (973 Program) No. 2011CB808002, and by the National Natural Science Foundation of China No. 11071086. The third author was supported by the National Natural Science Foundation of China #10625105, #11071093, the PhD specialized grant of the Ministry of Education of China #20100144110001, and the Special Fund for Basic Scientific Research of Central Colleges #CCNU10C01001.

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