Compressible hydrodynamic flow of liquid crystals in 1-D

Transcription

1 Compressible hydrodynamic flow of liquid crystals in 1-D Shijin Ding Junyu Lin Changyou Wang Huanyao Wen Abstract We consider the equation modeling the compressible hydrodynamic flow of liquid crystals in one dimension. When the initial density function ρ has a positive lower bound, we obtain the existence and uniqueness of global classical, and strong solutions and the existence of weak solutions. For ρ, we obtain the existence of global strong solutions. Key Words: Liquid crystal, compressible hydrodynamic flow, global solutions. 1 ntroduction n this paper, we consider the one dimensional initial-boundary value problem for (ρ, u, n) : [, 1] [, + ) R + R S 2 : ρ t + (ρu) x =, (ρu) t + (ρu 2 ) x + (P (ρ)) x = µu xx λ( n x 2 ) x, (1.1) n t + un x = θ(n xx + n x 2 n), for (x, t) (, 1) (, + ), with the initial condition: (ρ, u, n) t= = (ρ, u, n ) in [, 1], (1.2) School of Mathematical Sciences, South China Normal University, Guangzhou, 51631, China, Department of Mathematics, South China University of Technology, Guangzhou, 51641, China Department of Mathematics, University of Kentucky, Lexington, KY 456 Corresponding author. huanyaowen@163.com 1

2 where n : [, 1] S 2 and the boundary condition: (u, n x ) = (, ), t >, (1.3) where ρ denotes the density function, u denotes the velocity field, n denotes the optical axis vector of the liquid crystal that is a unit vector (i.e., n = 1), µ >, λ >, θ > are viscosity of the fluid, competition between kinetic and potential energy, and microscopic elastic relaxation time respectively. P = Rρ γ, for some constants γ > 1 and R >, is the pressure function. The hydrodynamic flow of compressible (or incompressible) liquid crystals was first derived by Ericksen [2] and Leslie [3] in 196 s. However, its rigorous mathematical analysis was not taken place until 199 s, when Lin [4] and Lin-Liu [5, 6, 7] made some very important progress towards the existence of global weak solutions and partial regularity of the incompressible hydrodynamic flow equation of liquid crystals. When the Ossen-Frank energy configuration functional reduces to the Dirichlet energy functional, the hydrodynamic flow equation of liquid crystals in Ω R d can be written as follows (see Lin [4]): ρ t + div(ρu) =, (ρu) t + div(ρu u) + (P (ρ)) = µ u λdiv( n n n 2 2 d ), n t + u n = θ( n + n 2 n), ( ) where u u = (u i u j ) 1 i,j d, and n n=(n xi n xj ) 1 i,j d. Observe that for d = 1, the system ( ) reduces to (1.1). When the density function ρ is a positive constant, then ( ) becomes the hydrodynamic flow equation of incompressible liquid crystals (i.e., div u = ). n a series of papers, Lin [4] and Lin-Liu [5, 6, 7] addressed the existence and partial regularity theory of suitable weak solution to the incompressible hydrodynamic flow of liquid crystals of variable length. More precisely, they considered the approximate equation of incompressible hydrodynamic flow of liquid crystals: (i.e., ρ = 1, and n 2 in ( ) 3 is replaced by (1 n 2 )n ), and proved [5], among many other results, the local existence of ɛ 2 classical solutions and the global existence of weak solutions in dimension two and 2

3 three. For any fixed ɛ >, they also showed the existence and uniqueness of global classical solution either in dimension two or dimension three when the fluid viscosity µ is sufficiently large; in [7], Lin and Liu extended the classical theorem by Caffarelli- Kohn-Nirenberg [1] on the Navier-Stokes equation that asserts the one dimensional parabolic Hausdorff measure of the singular set of any suitable weak solution is zero. See also [9, 1, 18] for relevant results. For the incompressible case ρ = 1 and div u =, it remains to be an open problem that for ɛ whether a sequence of solutions (u ɛ, n ɛ ) to the approximate equation converges to a solution of the original equation ( ). t is also a very interesting question to ask whether there exists a global weak solution to the incompressible hydrodynamic flow equation ( ) similar to the Leray-Hopf type solutions in the context of Naiver-Stokes equation. We answer this question in [8] for d = 2. When dealing with the compressible hydrodynamic flow equation ( ), there seems very few results available. This motivates us to address the existence and uniqueness of global classical, strong solutions and the existence of weak solutions for < c 1 ρ c and the existence of strong solutions for ρ when the dimension d = 1. We remark that when the optical axis n is a constant unit vector, (1.1) becomes the Navier-Stokes equation for compressible isentropic flow with density-independent viscosity, which has been well studied recently. For example, the existence of global strong solutions to the compressible Navier-Stokes equation for ρ was obtained by Choe-Kim [17] in one dimension and by Choe-Kim [16] in higher dimensions. Notice that ρ = corresponds to the vacuum state, whose existence makes the analysis much more complicated. Okada [12] investigated the free boundary problem for one-dimensional Navier-Stokes equations with one boundary fixed and the other connected to vacuum and proved the existence of global weak solutions. Luo, Xin, and Yang [13] studied the free boundary value problem of the one-dimensional viscous gas which expands into the vacuum and established the regularity, behaviors of weak solutions near the interfaces (separating the gas and vacuum) and expanding rate of the interfaces. The reader can also refer to [14, 15] for related works. Notations: (1) = [, 1], = {, 1}, Q T = [, T ] for T >. 3

4 (2) For p 1, denote L p = L p () as the L p space with the norm L p. For k 1 and p 1, denote W k,p = W k,p () for the Sobolev space, whose norm is denoted as W k,p, H k = W k,2 (), and W k,p (, S 2 ) = { } u W k,p (, R 3 ) u(x) = 1 a.e. x. k+α k+α, (3) For an even integer k and < α < 1, let C 2 (Q T ) denote the Schauder space of functions on Q T, whose derivatives up to kth order in x-variable and up to k 2 th order in t-variable are Hölder continuous with exponents α and α 2 respectively, with the norm C k+α, k+α. 2 Our first main theorem is concerned with the existence of global classical solutions. Theorem 1.1 For < α < 1, assume that ρ C 1,α satisfies < c 1 ρ c for some c, u C 2,α and n C 2,α (, S 2 ). Then there exists a unique global classical solution (ρ, u, n) : R + R + R S 2 to the initial boundary value problem (1.1)-(1.3) satisfying that for any T > there exists c = c(c, T ) > such that (ρ x, ρ t ) C α, α 2 (QT ), c 1 2+α 2+α, ρ c, (u, n) C 2 (QT ). Our second main theorem is concerned with the existence of global strong solutions and weak solutions under the assumption that ρ H 1 () satisfies < c 1 ρ c. More precisely, Theorem 1.2 (i) f ρ H 1, < c 1 ρ c for some c, u L 2 (), and n H 1 (, S 2 ), then there exists a global weak solution (ρ, u, n) : R + R + R S 2 to (1.1)-(1.3) such that for any T > there exists c = c(c, T ) > such that ρ x L (, T ; L 2 ), ρ t L 2 (, T ; L 2 ), < c 1 ρ c, u L (, T ; L 2 ) L 2 (, T ; H 1 ), n L (, T ; H 1 ) L 2 (, T ; H 2 ). (ii) f ρ H 1, < c 1 ρ c for some c, u H 1, and n H 2 (, S 2 ), then there exists a unique global strong solution to the initial boundary value problem 4

5 (1.1)-(1.3) satisfying that for any T > there exists c = c(c, T ) > such that ρ L (, T ; H 1 ), ρ t L (, T ; L 2 ), < c 1 ρ c, u L (, T ; H 1 ) L 2 (, T ; H 2 ), u t L 2 (, T ; L 2 ), n L (, T ; H 2 ) L 2 (, T ; H 3 ), n t L 2 (, T ; H 1 ) L (, T ; L 2 ). When ρ is only nonnegative, we establish the existence of global strong solutions to (1.1)-(1.3). Theorem 1.3 Assume ρ H 1, u H 1 and n H 2 (, S 2 ). Then there exists a global strong solution to the initial boundary value problem (1.1)-(1.3) such that for any T >, ρ L (, T ; H 1 ), ρ t L (, T ; L 2 ), ρ, u L (, T ; H 1 ) L 2 (, T ; H 2 ), ρu t L 2 (, T ; L 2 ), n L (, T ; H 2 ) L 2 (, T ; H 3 ), n t L 2 (, T ; H 1 ) L (, T ; L 2 ). Remark 1.1 (i) t is unknown whether the strong solution obtained in Theorem 1.3 is unique. f, in additions, u H 2 satisfies the compatibility condition: (µu x ) x (P (ρ )) x λ( n x 2 ) x = ρ 1 2 g, for some g L 2, then, by a method similar to [16] or [17], we can prove that the strong solution (ρ, u, n) obtained in Theorem 1.2 satisfies u L (, T ; H 2 ), ρu t L (, T ; L 2 ), u t L 2 (, T ; H 1 ) and hence the uniqueness of solutions can be shown by the argument similar to Theorem 1.2 and that of [16] and [17]. (ii) We believe that there exists a global weak solution (ρ, u, n) to (1.1)-(1.3) under the assumption that ρ L γ, u L 2, and n H 1 (, S 2 ). discussed in a forthcoming paper. This will be Since the constant R and µ, λ, θ in (1.1) don t play any role in the analysis, we assume henceforth that µ = λ = θ = R = 1. 5

6 The paper is organized as follows. n section 2, we prove the existence of the short time classical solutions of (1.1). n section 3, we derive some a priori estimates for classical solutions of (1.1), and prove the existence and uniqueness for both classical and strong solutions and the existence of weak solutions to (1.1)-(1.3) for ρ c >. n section 4, we prove the existence of strong solution for ρ. 2 Existence of local classical solutions n this section, we employ the contraction mapping theorem to prove that there exists a unique short time classical solution to (1.1)-(1.3) when ρ has a positive lower bound. The main result of this section can be stated as follows. Theorem 2.1 For α (, 1), assume that ρ C 1,α satisfies < c 1 ρ c, u C 2,α, and n C 2,α (, S 2 ). Then there exists T > depending on ρ, u, n such that the initial boundary value problem (1.1)-(1.3) has a unique classical solution (ρ, u, n) : [, T ) R + R S 2 satisfying (ρ x, ρ t ) C α, α 2 (QT ), c 1 2+α 2+α, ρ c, (u, n) C 2 (QT ). We may assume throughout this section that 1 ρ (ξ)dξ = 1. (2.1) To prove Theorem 2.1, we introduce for any T > the Lagrangian coordinate (y, τ) on [, T ): y(x, t) = x ρ(ξ, t)dξ, τ(x, t) = t. t is easy to check that (x, t) (y, τ) is a C 1 -bijective map from [, T ) [, T ), provide ρ(x, t) C 1 ( [, T )) is positive and 1 ρ(ξ, t) dξ = 1 for all t [, T ). Direct calculations imply t = ρu y + τ, x = ρ y, 6

7 and (ρ, u, n)(x, t) solves (1.1)-(1.3) is equivalent to (ρ, u, n)(y, τ) := (ρ, u, n)(x, t) solves the following system: ρ τ + ρ 2 u y =, u τ + (P (ρ)) y = (ρu y ) y (ρ 2 n y 2 ) y, (2.2) n τ = ρ(ρn y ) y + ρ 2 n y 2 n, and (ρ, u, n) τ= = (ρ, u, n ), in (2.3) (u, n y ) =, τ >. (2.4) Now we use the contraction mapping theorem to establish the existence and uniqueness of local, classical solutions to (2.2)-(2.4). Proof of Theorem 2.1: Let α > and (ρ, u, n ) be given by Theorem 2.1. For K > and T >, to be determined later, define the space X = X(T, K) by {(v, m) : Q T R R 3 2+α } 2+α, (v, m) C 2, (v, m) τ= = (u, n ), (v, m) X K where (v, m) X := v C 2+α, 2+α + m 2 2+α, 2+α. (Q T ) C 2 (Q T ) t is evident that (X, X ) is a Banach space. (ρ, u, n) : [, T ) R + R R 3 solve the following system: ρ τ + ρ 2 v y =, u τ + (P (ρ)) y = (ρu y ) y (ρ 2 n y 2 ) y, n τ = ρ(ρn y ) y + ρ 2 m y 2 n. along with the initial-boundary condition: For any (v, m) X, we let (2.2) (ρ, u, n) τ= = (ρ, u, n ); (ρ, u, n x ) = (ρ,, ), τ >. (2.5) The first equation of (2.2) yields that ρ(y, τ) = ρ (y) 1 + ρ (y) τ v y(y, s)ds. (2.6) 7

8 Since (v, m) X, we have v C 1 (Q T ) K and hence (2.6) implies provided ρ ρ ρ 1 ρ τ v y(y, s)ds c 1 c KT 2c, (2.7) ρ τ 1 + ρ v y(y, s)ds c 1 + c KT c 2, (2.8) T T 1 2c K. 2+α 2+α, Since v C 2 (Q T ) and ρ C 1+α (), (2.6) implies that ρ, ρ y C α, α 2 (Q T ). 2+α 2+α, Since m C 2 (Q T ), ρ, ρ y C α, α 2 (Q T ), it follows from (2.7), (2.8), and the Schauder theory that there is a unique solution (ρ, u, n), with (ρ, u, n) C α, α 2 (Q T ) 2+α 2+α 2+α, 2+α, C 2 (Q T ) C 2 (Q T ) to (2.2) and (2.5). Define the solution map: 2+α 2+α, H(v, m) = (u, n) : X C 2 (QT, R R 3 ). Claim. There exist sufficiently large K > and sufficiently small T > such that H : X X is a contraction map. Proof of Claim. We will first prove that H maps X into X. Set C 1 = ρ C 1,α + u C 2,α + n C 2,α. Direct differentiation of (2.6) implies that ρ y = ρ y 1 + ρ τ v y(y, s)ds ρ ρ y t follows from (2.6) and (2.9) that where T T 1 := min{ 1 2c K, ( 1 K ) 2 2 α }. τ v y(y, s)ds + ρ 2 τ v yy(y, s)ds (1 + ρ τ v y(y, s)ds) 2 (2.9) { } max ρ C α, α 2 (Q T ), ρ y C α, α c(c 2 (Q T ) 1 ), (2.1) Apply the Schauder theory to (2.1) 3, we obtain that for any T T 1, n C 2+α, 2+α 2 (Q T ) C( n C 2+α () + ρ 2 m y 2 n C α, α 2 (Q T ) ). (2.11) Direction calculations imply n C (Q T ) 1 + KT, [n] C α, α 2 (Q T ) c(c 1)(1 + KT α 2 ), (2.12) m y C (Q T ) m y m y C (Q T ) + m y C (Q T ) c(c 1 )(1 + KT α 2 ), (2.13) 8

9 [m y ] C α, α 2 (Q T ) [m y m y ] C α, α 2 (Q T ) + [m y] C α, α 2 (Q T ) c(c 1)(1 + KT α 2 ). (2.14) Hence if we choose T T 2 = min{t 1, ( 1 K ) 2 α }, then ρ 2 m y 2 n C (Q T ) c(c 1 ), (2.15) and [ρ 2 m y 2 n] C α, α 2 (Q T ) ρ 2 C (Q T ) m y 2 C (Q T ) [n] C α, α 2 (Q T ) + m y 2 C (Q T ) n C (Q T ) ρ 2 C α, α 2 (Q T ) + m y 2 C α, α 2 (Q T ) ρ 2 C (Q T ) n C (Q T ) c(c 1 ). (2.16) t follows from (2.11), (2.15) and (2.16) that for T T 2, n C 2+α, 2+α 2 (Q T ) c(c 1). (2.17) Now we need to estimate u C 2+α, 2+α 2 (Q T ) as follows. t follows from (2.1) 2 and the Schauder theory that u C 2+α, 2+α 2 (Q T ) C( u C 2+α () + (ρ γ ) y C α, α 2 (Q T ) + (ρ2 n y 2 ) y C α, α 2 (Q T ) c(c 1 )[1 + ρ γ 1 C α, α 2 (Q T ) ρ y C α, α 2 (Q T ) + ρ C α, α 2 (Q T ) ρ y C α, α 2 (Q T ) n y 2 C α, α 2 (Q T ) + ρ 2 C α, α n y 2 (Q T ) C α, α 2 (Q T ) n yy C α, α 2 (Q T ) ] c(c 1 ), provide T T 2. Thus we conclude that if K > is sufficiently large and T T 2, then H maps X into X. Next we want to show H : X X is a contraction map. For i = 1, 2, let (v i, m i ) X, and (u i, n i ) = H(v i, m i ). Set ρ = ρ 1 ρ 2, v = v 1 v 2, m = m 1 m 2, u = u 1 u 2 and n = n 1 n 2. Then it is easy to see ρ ( ) τ = v y. (2.18) ρ 1 ρ 2 Since ρ τ= =, integrating (2.18) with respect to τ yields τ ρ = ρ 1 ρ 2 v y ds. (2.19) 9

10 Since both ρ 1 and ρ 2 satisfy (2.1), we obtain max{ ρ C α, α 2 (Q T ), ρ y C α, α 2 (Q T ) } c(c 1)T 1 α 2 v C 2+α, 2+α 2 (Q T ). (2.2) For n, we have n τ = ρ 2 1n yy + ρ(ρ 1 + ρ 2 )n 2yy + ρρ 1y n 1y + ρ 2 ρ y n 1y + ρ 2 ρ 2y n y +ρ(ρ 1 + ρ 2 ) m 1y 2 n 1 + ρ 2 2m y (m 1y + m 2y )n 1 + ρ 2 2 m 2y 2 n. (2.21) Since ρ i, m i, n i (i = 1, 2) satisfy (2.11)-(2.12), (2.13)-(2.14), we have by the Schauder theory that for T T 2, n C 2+α, 2+α 2 (Q T ) c(c 1 )[ ρ C α, α 2 (Q T ) + ρ y C α, α 2 (Q T ) + n C α, α 2 (Q T ) + n y C α, α 2 (Q T ) + m y C α, α 2 (Q T ) ] c(c 1 )[T 1 α 2 v C 2+α, 2+α 2 (Q T ) + n C α, α 2 (Q T ) + n y C α, α 2 (Q T ) + m y C α, α 2 (Q T ) ] Since m τ= = n τ= =, we see that and Thus we obtain max{ n C α, α 2 (Q T ), n y C α, α 2 (Q T ) } CT α 2 n C 2+α, 2+α 2 (Q T ), m y C α, α 2 (Q T ) CT α 2 m C 2+α, 2+α 2 (Q T ). n C 2+α, 2+α C(C 1)T α 2 [ v C 2 2+α, (Q T ) 2+α + m 2 2+α, 2+α ]. (2.22) (Q T ) C 2 (Q T ) Finally, we need to estimate u C 2+α, 2+α as follows. For u, we have that 2 (Q T ) u τ +(ρ γ 1 ργ 2 ) y = (ρu 1y ) y +(ρ 2 u y ) y [ρ(ρ 1 +ρ 2 ) n 1y 2 ] y [ρ 2 2(n 1y +n 2y ).n y ] y. (2.23) Hence, by the Schauder theory, we have u C 2+α, 2+α 2 (Q T ) c(c 1 )[ ρ C α, α 2 (Q T ) + ρ y C α, α 2 (Q T ) + (ργ 1 ργ 2 ) y C α, α 2 (Q T ) + n y C α, α 2 (Q T ) + n yy C α, α 2 (Q T ) ] c(c 1 )T α 2 [ v C 2+α, 2+α + m 2 2+α, 2+α ], (2.24) (Q T ) C 2 (Q T ) 1

11 where we have used both (2.11) and (2.22) in the last step. t follows from (2.22) and (2.24) that if T T 3 = min{t 2, ( 1 4c(C 1 ) ) 2 α }, then u C 2+α, 2+α + n 2 2+α, 2+α 1 (Q T ) C 2 (Q T ) 2 ( v 2+α, 2+α + m C 2 2+α, 2+α ). (Q T ) C 2 (Q T ) Therefore H is a contraction map. By the fixed point theorem, there exists a unique (u, n) X such that H(u, n) = (u, n). ρ y, ρ τ C α, α 2 (Q T3 ), solving ρ τ + ρ 2 u y = Furthermore, there is a unique ρ, with Hence (ρ, u, n) is the unique classical solution to problem (2.2)-(2.4) on [, T 3 ]. Multiplying (2.2) 3 by n and applying the Gronwall s inequality, we can obtain n = 1 for (y, τ) Q T3. The proof of Theorem 2.1 is completed. 3 Existence of global classical, strong, and weak solutions n this section, we first prove that the short time classical solution, obtained in Theorem 2.1, can be extended to a global classical solution. This is based on several integral estimates for the short time classical solutions. The global weak, and strong solution are then achieved by both the approximation scheme and integral type a priori estimates for classical solutions. For < T < +, let (ρ, u, n) : [, T ) R + R S 2 be the classical solutions obtained by Theorem 2.1. The first estimate we have is the energy inequality. Lemma 3.1 For any t < T, it holds = ( ρu2 2 + ργ γ 1 + n x 2 )(t) + ( ρ u 2 2 t (u 2 x + 2 n xx + n x 2 n 2 ) + ργ γ 1 + (n ) x 2 ). (3.1) Proof. Multiplying (1.1) 2 by u and integrating the resulting equation over, we get d ρu 2 dt 2 ρ γ u x + u 2 x = n x 2 u x. (3.2) 11

12 Now we claim that n fact, by (1.1) 1, we have This clearly yields (3.3). ρ γ u x = d dt ρ γ u x = ρ γ 1 (ρ t + ρ x u) = d ρ γ dt γ + ( ργ γ ) xu = d ρ γ dt γ ρ γ γ u x. ρ γ γ 1. (3.3) Multiplying (1.1) 3 by (n xx + n x 2 n) and integrating it over, we obtain d n x 2 + n x 2 u x + 2 n xx + n x 2 n 2 =. (3.4) dt t follows from (3.2), (3.3), and (3.4) that d ( ρu2 dt 2 + ργ γ 1 + n x 2 ) + u 2 x + 2 n xx + n x 2 n 2 =. This clearly implies (3.1). The proof is complete. Lemma 3.2 For any t < T, it holds where t E := n xx 2 c(e )(1 + t), (3.5) ( ρ u 2 2 denotes the total energy of the initial data. + ργ γ 1 + (n ) x 2 ) Proof. By the Gagliardo-Nirenberg inequality, we have Since n = 1, this implies n x 4 c n x 3 L 2 () n xx L 2 () 1 2 n xx 2 = 1 2 n xx 2 + c( n x 2 ) 3. n xx + n x 2 n 2 + n x 4 n xx 2 + n xx + n x 2 n 2 + c( 12 n x 2 ) 3.

13 Hence n xx 2 2 n xx + n x 2 n 2 + c( n x 2 ) 3. This, combined with (3.1), yields (3.5). The proof of the Lemma is complete. Now we want to estimate n xx L ([,T ],L 2 ()) in terms of both E and n H 2 () as follows. Lemma 3.3 For any t < T, it holds t n xx 2 (t) + ( n xt 2 + n xxx 2 ) c(e, n H 2 ()) exp{c(e )t}. (3.6) Proof. Differentiating (1.1) 3 with respect to x, multiplying the resulting equation by n xt, and integrating it over, we have n xt 2 + d 1 dt 2 n xx 2 = [ n x 2 n x n xt + 2(n x n xx )n n xt ] u x n x n xt un xx n xt 1 n xt 2 + c ( n x 6 + n x 2 n xx 2 + u 2 2 x n x 2 + u 2 n xx 2 ). Thus n xt 2 dx + d n xx 2 dt c ( n x 6 + n x 2 n xx 2 + u 2 x n x 2 + u 2 n xx 2 ) c [ n x 4L n x 2 + n x 2 L u 2 x + ( n x 2 L + u 2 L ) c(1 + E )( n xx 2 ) 2 + c n xx 2, u 2 x where we have used the Sobolev embedding inequality: n x L () c n xx L 2 (), u L () c u x L 2 (). n xx 2 ] This, combined with Lemma 3.1, 3.2 and the Gronwall inequality, implies that any t [, T ), Since n xx 2 (t) + t n xt 2 c(e, n H 2 ()) exp{c(e )t}. n xxx = n xt + u x n x + un xx 2(n x n xx )n n x 2 n x, 13

14 we also obtain t n xxx 2 c(e, n H 2 ()) exp{c(e )t}. The proof is now completed. Now we want to improve the estimation of both lower and upper bounds of ρ in terms of E, ρ H 1, and the upper and lower bounds of ρ. This turns out to be the most difficult step. Lemma 3.4 There exist c = c(c, ρ H 1, E, γ) > and C = C(c, ρ H 1, E ) > depending only on c, ρ H 1, γ, and E such that for any t < T, we have ρ ( 1 ρ ) x 2 (t) + t ρ γ 3 ρ 2 x C exp(ct), (3.7) and 1 ρ(x, t) c exp(ct), x. (3.8) c exp(ct) Proof. Using (1.1) 1, we have d ρ ( 1 dt ρ ) x 2 = Thus we obtain = = = = = 2 ρ t ( 1 ρ ) x (ρu) x ( 1 ρ ) x (ρu) x ( 1 ρ ) x (ρu) x ( 1 ρ ) x (ρu) x ( 1 ρ ) x (1 ρ ) x 2 ρu x ] + 2 ( 1 ρ ) xu xx. 1 d 2 dt ρ( 1 ρ ) x( 1 ρ ) xt ρ( 1 ρ ) x( ρ t ρ 2 ) x ρ( 1 ρ ) x( (ρu) x ρ 2 ) x ρ( 1 ρ ) x[(( 1 ρ ) xu) x + ( u x ρ ) x] [ ρ ( 1 ρ ) x 2 u x (1 ρ ) x 2 ρ x u ρ ( 1 ρ ) x 2 u x + 2 ( 1 ρ ) xu xx ρ ( 1 ρ ) x 2 = ( 1 ρ ) xu xx. (3.9) Multiplying (1.1) 2 by ( 1 ρ ) x, integrating the resulting equation over, and utilizing 14

15 (3.9), we obtain Since (1.1) 1 1 d ρ ( 1 2 dt ρ ) x 2 + γ ρ γ 3 ρ 2 x + d ρu( 1 dt ρ ) x = 1 d ρ ( 1 2 dt ρ ) x 2 ( 1 ρ ) x(ρ γ ) x d ρu( 1 dt ρ ) x = ( 1 ρ ) x( n x 2 ) x ρu( 1 ρ ) xt + (ρu 2 ) x ( 1 ρ ) x = 2 ( 1 ρ ) ρ t xn x n xx (ρu) x ρ 2 (ρu 2 ρ x ) x ρ 2 1 ρ n xx 2 + ρ ( 1 ρ ) x 2 n x 2 + [ 1 ρ 2 ( (ρu) x 2 (ρu 2 ) x ρ x )] 1 ρ L n xx 2 + n x 2 L ρ ( 1 ρ ) x 2 + u 2 x [ 1 ρ L + ρ ( 1 ρ ) x 2 ] n xx 2 + u 2 x. (3.1) implies ρ(ξ, t) = ρ (ξ) = 1, there exists a(t) such that ρ(a(t), t) = ρ(ξ, t)dξ = 1. Hence we have 1 ρ(x, t) = = ρ(a(t), t) + x a(t) ρ ξ ρ 2 x a(t) 1 ( ρ(ξ, t) ) ξ ρ 1 ρ 2 x 2 L ( ρ 3 ) ρ L + 1 ρ ( 1 2 ρ ) x 2. Taking the supremum over x, this implies 1 ρ L 2 + Plugging (3.11) into (3.1), we obtain ρ ( 1 ρ ) x 2. (3.11) 1 d ρ ( 1 2 dt ρ ) x 2 + γ ρ γ 3 ρ 2 x + d ρu( 1 dt ρ ) x c[ n xx 2 + n xx 2 ρ ( 1 ρ ) x 2 ] + u 2 x. (3.12) 15

16 ntegrating (3.12) over (, t), we obtain 1 ρ ( 1 t 2 ρ ) x 2 + γ ρ γ 3 ρ 2 x ρu( 1 ρ ) x + 1 ρ ( 1 ) x 2 ρ u ( 1 ) x 2 ρ ρ t t +c (u 2 x + n xx 2 ) + c [ n xx 2 ρ ( 1 ρ ) x 2 ] 1 ρ ( 1 t 4 ρ ) x 2 + c [ n xx 2 ρ ( 1 ρ ) x 2 ] t +c[ ρu 2 + (u 2 x + n xx 2 )] + 1 ρ ( 1 ) x 2 ρ u ( 1 ) x. 2 ρ ρ Since we obtain c[ ρu 2 + t (u 2 x + n xx 2 )] ρ ( 1 ) x 2 ρ u ( 1 ) x ρ ρ c(e )(1 + t) ρ 3 L ρ 2 H ρ 3 2 L ρ u 2 L 1 ρ H 1 C(E, c, ρ H 1)(1 + t), ρ ( 1 ρ ) x 2 + γ t ρ γ 3 ρ 2 x C(E, c, ρ H 1)(1 + t) + c t ( n xx 2 ρ ( 1 ρ ) x 2 ). Since Lemma 3.2 implies t n xx 2 c(e )(1 + t), we have, by the Gronwall s inequality, that ρ ( 1 ρ ) x 2 + t ρ γ 3 ρ 2 x C(E, c, ρ H 1)(1 + t 2 ) exp(c C(E, c, ρ H 1) exp{c(e )t}. This yields (3.7). t follows from (3.11) and (3.7) that ρ 1, (x, t) [, T ). c exp{c(e )t} Since γ > 1, we can write γ = 1 + 2δ for some δ >. Hence we have ρ δ L ρ δ + δ ρ δ 1 ρ x ( ρ γ ) δ γ + δ( ρ γ ) 1 2 ( ρ ( 1 ρ ) x 2 ) 1 2 c(e, c, ρ H 1, γ) exp{c(e )t}, 16 t n xx 2 )

17 where we have used (3.7) in the last step. This clearly yields (3.8). The proof is now complete. Lemma 3.5 There exists C = C(γ, E, c, ρ H 1, u H 1, n H 2) > such that for any t < T, u 2 x(t) + t Proof. t follows from (1.1) 1 and (1.1) 2 that (u 2 t + u 2 xx) C exp{c exp(ct)}. (3.13) ρu t + ρuu x + γρ γ 1 ρ x = u xx 2n x n xx (3.14) Multiplying (3.14) by u t, integrating the resulting equation over and employing integration by parts, we have ρu 2 t + 1 d u 2 x 2 dt = ρuu x u t γρ γ 1 ρ x u t 2 n x n xx u t = 2 n x n xx u t ρuu x u t γ ρ γ 1 ρ x u t 1 ρu 2 1 t + c[ 2 ρ n x 2 n xx 2 + ρu 2 u 2 x + ρ 2γ 3 ρ 2 x] 1 ρu 2 t + c[ 1 2 ρ L ( n xx 2 ) 2 + ρ L ( u 2 x) 2 ] + c ρ γ L This, combined with Lemma 3.3 and 3.4, implies ρu 2 t + d u 2 x C exp(ct) + C exp(ct)( dt u 2 x) 2 ρ γ 3 ρ 2 x. for some C = C(γ, E, c, ρ H 1, u H 1, n H 2) >. Thus, by Lemma 3.1 and the Gronwall s inequality, we have t ρu 2 t + Since ρ 1, we have C exp(ct) u 2 x(t) + t u 2 x C exp{c exp(ct)}. By Lemma 3.3, 3.4, (3.14) and (3.15), we also have t u 2 xx C exp{c exp(ct)}. u 2 t C exp{c exp(ct)}. (3.15) 17

18 This completes the proof. n order to prove the existence of global classical solutions, we also need the following estimate. Lemma 3.6 There exists C = C(γ, c, E, ρ H 1, u H 2, n H 2) > such that for any t < T, (u 2 t + u 2 xx)(t) + t u 2 xt C exp{c exp[c exp(ct)]}. (3.16) Proof. Differentiating (3.14) with respect to t, multiplying the resulting equation by u t, integrating it over, and using integration by parts, we have 1 d ρu 2 t + u 2 xt 2 dt = x ( n 2 ) t u xt 1 ρ t u 2 t (ρ t uu x + ρu t u x )u t ρuu xt u t + (ρ γ ) t u xt 2 = 2 n x n xt u xt + (ρu) x u 2 t + (ρu) x uu x u t ρu 2 t u x γ ρ γ 1 (ρu) x u xt = (2n x n xt 2ρuu t ρu 2 u x γρ γ 1 ρ x u γρ γ u x )u xt ρuu t u 2 x ρu 2 u xx u t ρu 2 t u x 1 u 2 xt + c ( n x 2 n xt 2 + ρ 2 u 2 u 2 t + ρ 2 u 4 u 2 2 x) + cγ 2 (ρ 2γ 2 ρ 2 xu 2 + ρ 2γ u 2 x) +c (ρu 2 u 4 x + ρu 4 u 2 xx) + c(1 + u x L ) ρu 2 t. Thus we have d ρu 2 t + u 2 xt dt c n x 2 L n xt 2 + c(1 + u x L + ρ L u 2 L ) + c(γ 2 ρ 2γ L + ρ 2 L u 4 L + ρ L u 2 L u x 2 L ) + c ρ L u 4 L ρ γ 3 ρ 2 x. t follows from Lemma 3.3 and 3.5 that Thus we have u 2 xx + cγ 2 ρ γ+1 L u 2 L ρu 2 t u 2 x max{ n xx L ([,t],l 2 (), u x L ([,t],l 2 ()} C exp{c exp(ct)}. max{ n x L, u L } C exp{c exp(ct)}. 18

19 Also observe that u x 2 L () Therefore, by the Gronwall s inequality, we have ρu 2 t (t) + t u xx 2 + u x 2 L 2 (). u 2 xt C(γ, c, E, ρ H 1, u H 2, n H 2) exp{c exp[c exp(ct)]}. 1 Since ρ, this together with (3.14) and Lemma yields (3.16). The c exp(ct) proof is complete. Lemma 3.7 ([19]) Suppose that and Then sup u(x, t 1 ) u(x, t 2 ) µ 1 t 1 t 2 α, t 1, t 2 [, T ), x sup u x (x 1, t) u x (x 2, t) µ 2 x 1 x 2 β, x 1, x 2. t [,T ) sup u x (x, t 1 ) u x (x, t 2 ) µ t 1 t 2 δ, t 1, t 2 [, T ), x where δ = αβ 1+β, and µ depends only on α, β, µ 1, µ 2. Now we are ready to prove Theorem 1.1. Proof of Theorem 1.1. Suppose it were false. Then there exists a maximal time interval < T < + such that there exists a unique classical solution (ρ, u, n) : [, T ) R + R S 2 of (1.1)-(1.3), but at least one the following properties fails: (i) (ρ x, ρ t ) C α, α 2 (Q T ), (ii) < c 1 ρ c < +, (x, t) Q T, 2+α 2+α, (iii) (u, n) C 2 (Q T ). Notice that (3.8) of Lemma 3.4 implies (ii) holds. Hence either (i) or (iii) fails. On the other hand, it follows from Lemma 3.3, Lemma 3.5, Lemma 3.6, and the Sobolev embedding Theorem that { } max n C 1, 1, u 2 (Q T ) C 1, 2 1 C(E, c, ρ (Q T ) H 1, u H 2, n H 2, T ) < +. 19

20 Hence Lemma 3.3 and (1.1) 3 imply that n and n x is uniformly Hölder continuous in t and x respectively with exponent 1 2. Lemma 3.7 then implies n x is also uniformly Hölder continuous in t with exponent 1 6. Thus max{ n x C 1 3, 1 6 (Q T ), u x C 1 3, 1 6 (Q T ) } C(E, c, ρ H 1, u H 2, n H 2, T ) < +. t follows from (1.1) 3 and the Schauder theory that n C ,1+ 1 < + and 6 (Q T ) hence n x C 1, 2 1 < +. Hence, applying the Schauder theory to (1.1) 3 again, (Q T ) we have n C 2+α,1+ α 2 (Q T ) < +. Write F (x, t) = ( n x 2 ) x. Then F C α, α 2 (Q T ) < +. n term of the Lagrangian coordinate, (1.1) 1 and (1.1) 2 become ρ τ + ρ 2 u y =, u τ + (ρ γ ) y = (ρu y ) y + F, (3.17) Moreover, the estimates obtained by Lemma , in the Lagrangian coordinate, become < c 1 ρ c < +, (3.18) ρ 2 y c < +, (3.19) u 2 y + u 2 yy c < +. (3.2) Combining (3.17) 1 with (3.18)-(3.2), we conclude ρ C 1 2, 1 4 (Q T ) < +. other hand, we have Now we claim that { } max F C α, α 2 (Q T ), u C 1 3, 1, u y 6 (Q T ) C 1, 1 < +. 2 (Q T ) n fact, (3.17) 1 and (3.17) 2 imply On the ρ y 1 C 2, 1 < +. (3.21) 4 (Q T ) (u + (ln ρ) y ) τ = F γρ γ [u + (ln ρ) y ] + γρ γ u. Hence we have u + (ln ρ) y = (u + (ln ρ ) y )e γ τ ργ + 2 τ (F + γρ γ u)e γ τ s ργ ds.

21 This clearly implies (3.21). conclude that Thus, applying the Schauder theory to (3.17) 2, we u C , (Q T ) < +. n particular, u C α, α 2 (Q T ) + u y C α, α 2 (Q T ) < +. Applying these estimates to (3.17) 1, we obtain that ρ C α, α < +. Repeating 2 (Q T ) the above argument once again, we see that { } max u C 2+α, 2+α, ρ y 2 (Q T ) C 2+α, 2+α, ρ τ 2 (Q T ) C 2+α, 2+α < +. 2 (Q T ) This contradicts the choice of T. Hence T =. The proof of Theorem 1.1 is now complete. Now we recall the following well-known Lemma. Lemma 3.8 [1]. Assume X E Y are Banach spaces and X E. Then the following embedding are compact: (i) (ii) { ϕ : ϕ L q (, T ; X), ϕ } t L1 (, T ; Y ) L q (, T ; E), if 1 q ; { ϕ : ϕ L (, T ; X), ϕ } t Lr (, T ; Y ) C([, T ]; E), if 1 < r. Proof of Theorem 1.2. Part (i): First, by the standard mollification, we may assume that for any α (, 1), there exist a sequence of initial data (ρ ɛ, uɛ, nɛ ) C1,α () C 2,α () C 2,α (, S 2 ) such that (i) < c 1 ρ ɛ c < + on, (ii) lim [ ρ ɛ ρ H 1 + u ɛ u L 2 + n ɛ n H 1] =. ɛ (To assure n ɛ = 1, we construct nɛ = ηɛ n η ɛ n ) Now let (ρ ɛ, u ɛ, n ɛ ) be the unique global classical solution of (1.1) along with the initial condition (ρ ɛ, uɛ, nɛ ) and the boundary condition (uɛ, nɛ x ) = (, ) for t >. t follows from Lemma 3.1, Lemma 3.2, and Lemma 3.4 that for any 21

22 < T < +, the following properties hold: 1 c exp(ct ) ρɛ c exp(ct ), in [, T ], ρ ɛ L (,T ;H 1 ()) + ρ ɛ t L 2 (,T ;L 2 ()) C(T ), u ɛ L (,T ;L 2 ()) + u ɛ L 2 (,T ;H 1 ()) C(T ), n ɛ L (,T ;H 1 ()) + n ɛ L 2 (,T ;H 2 ()) + n ɛ t L 2 (,T ;L 2 ()) C(T ). After taking possible subsequences, we may assume that as ɛ, (ρ ɛ, ρ ɛ x) (ρ, ρ x ) weak L (, T ; L 2 ()), (3.22) ρ ɛ t ρ t weakly in L 2 (, T ; L 2 ()), (3.23) ρ ɛ ρ strongly in C(Q T ), (3.24) u ɛ u weak L (, T ; L 2 ()) and weakly in L 2 (, T ; H 1 ()), (3.25) (n ɛ, n ɛ x, n ɛ xx) (n, n x, n xx ) weakly in L 2 (, T ; L 2 ()), (3.26) (n ɛ, n ɛ x) (n, n x ) weak L (, T ; L 2 ()), (3.27) n ɛ t n t weakly in L 2 (, T ; L 2 ()), (3.28) n ɛ n strongly in C(Q T ) L 2 (, T ; C 1 ()), (3.29) where we have used Lemma 3.8. t is easy to see that (3.24) and (3.25) imply ρ ɛ t + (ρ ɛ u ɛ ) x ρ t + (ρu) x in D (Q T ) so that ρ t + (ρu) x =. Since (ρ ɛ u ɛ ) t = u ɛ xx ( n ɛ x 2 ) x (ρ ɛ (u ɛ ) 2 ) x ((ρ ɛ ) γ ) x L 2 (, T ; H 1 ()) and ρ ɛ u ɛ ρu weak L (, T ; L 2 ()), we have from Lemma 3.8 that ρ ɛ u ɛ ρu strongly in C(, T ; H 1 ()). 22

23 This, combined with (3.25), implies that ρ ɛ (u ɛ ) 2 ρu 2 in D (Q T ). Since u ɛ xx ( n ɛ x 2 ) x ((ρ ɛ ) γ ) x u xx ( n x 2 ) x (ρ γ ) x in D (Q T ), we see that (ρu) t + (ρu 2 ) x + (ρ γ ) x = u xx ( n x 2 ) x in D (Q T ). Since u ɛ n ɛ x un x, n ɛ x 2 n ɛ n x 2 n in D (Q T ), we also have that n t + un x = n xx + n x 2 n in D (Q T ). This completes the proof of (i). To prove part (ii). Observe that since u H 1 and n H 2, we have Thus, Lemma 3.3, and Lemma 3.5 imply lim ɛ [ uɛ u H 1 () + n ɛ n H 2 ()] =. sup ( u ɛ x 2 + n ɛ t 2 + n ɛ xx 2 ) + t T ( u ɛ xx 2 + u ɛ t 2 + n ɛ xt 2 + n ɛ xxx 2 ) Q T C(T ). (3.3) t follows from (3.3) that u L (, T ; H 1 ()) L2 (, T ; H 2 ), u t L 2 (, T ; L 2 ()), n L (, T ; H 2 ()) L 2 (, T ; H 3 ()), and n t L 2 (, T ; H 1 ()) L (, T ; L 2 ()). This proves the existence. To prove the uniqueness, let (ρ i, u i, n i ) be two solutions to (1.1)-(1.3) obtained as above. Denote ρ = ρ 1 ρ 2, ũ = u 1 u 2, ñ = n 1 n 2, Then ρ t + ( ρu 1 ) x + (ρ 2 ũ) x =, ρ 1 ũ t ũ xx = ρu 2t ρu 2 u 2x ρ 1 ũu 2x ρ 1 u 1 ũ x (ρ γ 1 ργ 2 ) x 2n 1x ñ xx 2ñ x n 2xx, (3.31) ñ t + u 1 ñ x + ũn 2x = ñ xx + n 1x 2 ñ + [ñ x (n 1x + n 2x )]n 2, for (x, t) (, 1) (, + ), with the initial and boundary condition: ( ρ, ũ, ñ) t= = in [, 1], (ũ, ñ x ) = t >. 23

24 Multiplying (3.31) 1 by ρ, integrating the resulting equation over, and using the integration by parts, we have 1 d 2 dt ρ 2 = ρu 1 ρ x (ρ 2x ũ + ρ 2 ũ x ) ρ = 1 ρ 2 u 1x (ρ 2x ũ + ρ 2 ũ x ) ρ u 1x L ρ 2 + ũ L ρ 2x L 2 ρ L 2 + ρ 2 L ũ x L 2 ρ L 2. Since ũ(, t) =, we have ũ(y, t) = y ũx(x, t)dx for (y, t) Q T and hence ũ L ũ x L 2, t [, T ]. (3.32) t follows from (3.32), the regularities of (ρ i, u i ), Hölder inequality, and Sobolev inequality that d ρ 2 c u 1 dt H 2 ρ 2 + c ũ x L 2 ρ L 2 c( u 1 H 2 + 1) ρ 2 + ũ x 2. (3.33) Multiplying (3.31) 2 by ũ, integrating the resulting equation over, and using the integration by parts, we have 1 d ρ 1 ũ 2 + ũ x 2 2 dt = 1 ρ 1t ũ 2 ρ 1 u 1 ũ x ũ ρũu 2t ρũu 2 u 2x ρ 1 ũ 2 u 2x 2 + (ρ γ 1 ργ 2 )ũ x + 2 (n 1x ñ x )ũ x + 2 (n 1xx ñ x )ũ 2 (ñ x n 2xx )ũ. Since ρ 1t + (ρ 1 u 1 ) x =, we have Therefore, 1 d 2 dt 1 ρ 1t ũ 2 ρ 1 u 1 ũ x ũ = 1 ρ 1t ũ ũ 2 (ρ 1 u 1 ) x = ρ 1 ũ 2 + ũ x 2 ũ L ρ L 2 u 2t L 2 + ρ L 2 ũ L u 2 L u 2x L 2 + u 2x L ρ 1 ũ 2 + c[ ρ L 2 ũ x L 2 + ũ x L 2 ñ x L 2 n 1x L + ρ 1 ũ L 2 ñ x L 2 n 1xx + n 2xx L ]. ρ1 24

25 t follows from (3.32), Hölder inequality, Sobolev s inequality, and the regularities of (ρ i, u i ) that 1 d 2 dt ρ 1 ũ 2 + ũ x 2 c ũ x L 2( ρ L 2 u 2t L 2 + ρ L 2 + ñ x L 2) + c u 2 H 2 ρ 1 ũ 2 +c( n 1 H 3 + n 2 H 3) ρ 1 ũ L 2 ñ x L ũ x 2 L + c ρ 2 2 L (1 + u 2 2t 2 L ) + c( u 2 2 H 2 + n 1 2 H + n H ) 3 +c ñ x 2 L 2. ρ 1 ũ 2 Thus we have d ρ 1 ũ 2 + ũ x 2 c(1 + u 2t 2 L ) ρ 2 + c ñ dt 2 x 2 + c( u 2 H 2 + n 1 2 H + n H ) ρ 3 1 ũ 2. (3.34) Multiplying (3.31) 3 by ñ, integrating the resulting equation over, and using the integration by parts, we have 1 d ñ 2 + ñ x 2 2 dt = u 1 ñ x ñ ũn 2x ñ + n 1x 2 ñ 2 + [ñ x (n 1x + n 2x )]n 2 ñ u 1 L ñ x L 2 ñ L 2 + n 2x L ρ L ρ 1 ũ L 2 ñ L 2 + n 1x 2 L + n 2 L n 1x + n 2x L ñ L 2 ñ x L 2. By Hölder inequality and Sobolev s inequality, we get 1 d ñ 2 + ñ x 2 2 dt c ñ x L 2 ñ L 2 + c ρ 1 ũ L 2 ñ L 2 + c Hence 1 2 ñ x 2 L 2 + c ñ 2 L 2 + c ρ 1 ũ 2 L 2. ñ 2 ñ 2 d ñ 2 + ñ x 2 c ñ 2 L + c ρ dt 2 1 ũ 2 L. (3.35) 2 Adding together (3.33), (3.34), and (3.35), we obtain d ( ρ 2 + ρ 1 ũ 2 + c ñ 2 ) A(t) ( ρ 2 + ρ 1 ũ 2 + c ñ 2 ), (3.36) dt 25

26 where A(t) = c[2 + c + u 2t 2 (t) + u 1 (, t) H 2 + u 2 (, t) H 2 + n 1 (, t) 2 H + n 3 2 (, t) 2 H ]. 3 t follows from the regularities of u i and n i that T Gronwall s inequality imply ( ρ 2 + ρ 1 ũ 2 + c ñ 2 )(x, t)dx A(t)dt < +. (3.36) and the T ( ρ 2 + ρ 1 ũ 2 + c ñ 2 )(x, )dx exp{ A(s)ds} =. Since ρ 1 > and c >, we have ( ρ, ũ, ñ) =. This proves the uniqueness. Hence the proof of Theorem 1.2 (ii) is complete. 4 Global strong solutions for ρ. n this section, we establish the existence of global strong solutions for ρ. The proof of Theorem 1.3 is based on several estimates of the approximate solutions without the hypothesis that the initial density function has a positive lower bound. For a small ɛ >, let ρ ɛ = η ɛ ρ + ɛ, u ɛ = η ɛ u, n ɛ = ηɛ n η ɛ n. Then ρɛ ɛ and lim [ ρ ɛ ρ H 1 + u ɛ u H 1 + n ɛ n H 2] =. ɛ Let (ρ ɛ, u ɛ, n ɛ ) be the unique global classical solution of (1.1) along with the initial condition (ρ ɛ, uɛ, nɛ ) and the boundary condition (uɛ, nɛ x ) = (, ). Now we outline several integral estimates for (ρ ɛ, u ɛ, n ɛ ). For simplicity, we write (ρ, u, n) = (ρ ɛ, u ɛ, n ɛ ). The first Lemma follows from the global energy inequality, Nirenberg inequality, and the second order estimate of (1.1) 3. Lemma 4.1 For any T >, it holds T sup (ρu 2 + ρ γ + n x 2 + n xx 2 ) + (u 2 x + n xt 2 + n xxx 2 ) C(E, T ). (4.1) t T Proof. t is exactly as same as that of Lemma 3.1, Lemma 3.2, and Lemma 3.3. The next Lemma is concerned with the upper bound estimate of ρ. 26

27 Lemma 4.2 For any T >, there exists C > independent of ɛ such that ρ L ( (,T )) C. (4.2) Proof. Set Then we have w(x, t) = t (u x n x 2 ρu 2 ρ γ ) + x (ρ u )(ξ). w t = u x n x 2 ρu 2 ρ γ, w x = ρu. t follows from Lemma 4.1 that ( w + w x ) C and hence w L () C ( w + w x ) C. Since ρ, it suffices to prove ρ(y, s) C for any (y, s) (, T ). Let x(y, t) solve dx(y, t) = u(x(y, t), t), t < s; dt x(y, s) = y, y 1.. Denote f = exp w. Then we have d dt ((ρf)(x(y, t), t)) = (ρ t + ρ x u)f + ρf(w t + uw x ) = [ρ t + ρ x u + ρu x ρ n x 2 ρ 2 u 2 ρ γ+1 + (ρu) 2 ]f = ( ρ n x 2 ρ γ+1 )f. Thus ρ(y, s)f(y, s) = ρ(x(y, s), s)f(x(y, s), s) ρ (x(y, ))f(x(y, ), ) C. Therefore ρ(y, s) C exp( w(y, s)) C exp( w L ( (,T ))) C. The proof is completed. Next we want to estimate u x L (,T ;L 2 ()) and ρu 2 t L 1 ( [,T ]). 27

28 Lemma 4.3 For any T >, there exists C > independent of ɛ such that T sup u 2 x + ρu 2 t C (4.3) t T Proof. t is similar to Lemma 3.5. Multiplying (3.14) by u t, integrating the resulting equation over, and employing integration by parts, we have ρu 2 t + 1 d 2 dt u 2 x = 1 2 n x 2 u xt + ρ γ u xt ρuu x u t ρu 2 t dx + ρ L u 2 L u 2 x + Thus, using Lemma 4.2 and u 2 L u x 2, we obtain ρu 2 t + d dt n x 2 u xt + ρ γ u xt. u 2 x 2[( u 2 x) 2 + ρ γ u xt + n x 2 u xt ] = + +. (4.4) = d ρ γ u x γρ γ 1 ρ t u x dt = d ρ γ u x + (ρ γ ) x uu x + γ ρ γ u 2 x dt = d ρ γ u x + (γ 1) ρ γ u 2 x ρ γ uu xx dt = d ρ γ u x + (γ 1) ρ γ u 2 x ρ γ u(ρu t + ρuu x + (ρ γ ) x + ( n x 2 ) x ) dt d ρ γ u x + (γ 1) ρ γ u 2 x + 1 ρu 2 t + C(1 + u x 2 ) u 2 x dt 4 ρ γ (ρ γ ) x u 2 ρ γ un x n xx d ρ γ u x + C(1 + u 2 dt x) u 2 x + 1 ρu 2 t + 1 ρ 2γ u x C u 2 n x 2 + C n xx 2 d ρ γ u x + C(1 + u 2 dt x) u 2 x + 1 ρu 2 t + C( n x 2 u 2 x + n xx 2 ) 4 d ρ γ u x + C[1 + ( u 2 dt x) 2 ] + 1 ρu 2 t. 4 28

29 = d n x 2 u x 2 n x n xt u x dt d n x 2 u x + n x 2 L dt u 2 x + n xt 2 d n x 2 u x + n xx 2 L u 2 dt 2 x + n xt 2 d n x 2 u x + u 2 x + n xt 2. dt Combining (4.4) with the estimates of and, we obtain ρu 2 t + d u 2 x dt C[1 + ( u 2 x) 2 ] + d dt ntegrating (4.5) over [, T ] yields T Thus we have ρu 2 t + u 2 x u 2 x + C(T + C(1 + T ) + C C(1 + T ) + C T ( T 1 T u 2 x) 2 ) + ρ γ u x + d dt n x 2 u x + C (ρ γ u x ρ γ u x) + ( u 2 x) u 2 x + ρ 2γ + 1 u 2 x + C 8 8 ( u 2 x) u 2 x + C n x 2 n xx 2. 4 T ρu 2 t + u 2 x C(1 + T ) + C This, combined with the inequality T u2 x C and the Gronwall inequality, implies (4.3). This completes the proof. We also need to estimate ρ L (,T ;L 2 ()) as follows. T n xt 2. (4.5) ( n x 2 u x n x 2 u x ) ( u 2 x) 2. n x 4 Lemma 4.4 For any T >, there exists C > independent of ɛ such that T sup ρ 2 x + u x 2 L C. (4.6) t T 29

30 Proof. By (3.14), we have Thus we obtain u x 2 L 2 u x ρ γ n x 2 2 L + 2 ργ + n x 2 2 L C[ u x ρ γ n x 2 2 L + u 2 xx (ρ γ ) x ( n x 2 ) x 2 L ] 2 + C[ ρ γ 2 L + ( n xx 2 ) 2 ] C[1 + n x 2 L n x 2 ] + C ρu t + ρuu x 2 L 2 C[1 + ( u 2 x) 2 + T n x 2 u x 2 L C. n xx 2 ] + C 1 ρu 2 t. To estimate ρ2 x, take the derivative of (1.1) 1 with respect to x, multiply the resulting equation by ρ x, and integrate it over and employ integration by parts. Then we have d ρ 2 x = 2 (ρu) x ρ xx (ρu) x ρ x 1 dt = (ρ x uρ xx + ρu x ρ xx ) ρρ x u x 1 = 1 ρ 2 2 xu x + ρu x ρ x 1 ρ 2 xu x ρρ x u xx ρρ x u x 1 = 3 ρ 2 2 xu x ρρ x u xx C u x L ρ 2 x ρρ x [ρu t + ρuu x + (ρ γ ) x + ( n x 2 ) x ]dx C(1 + u x L ) ρ 2 x + C ρu 2 t + C( u 2 x) 2 + C( n xx 2 ) 2 C + C(1 + u x L ) ρ 2 x + C ρu 2 t. (4.6) follows from the Gronwall inequality. The proof is now completed. Finally we need to estimate u xx L 2. Lemma 4.5 For any T >, there exists C > independent of ɛ such that T u 2 xx C. Proof. t follows from (3.14) that u xx = ρu t + ρuu x + (ρ γ ) x + ( n x 2 ) x. 3

31 Hence This completes the proof. u 2 xx Q T C ρu 2 t + C Q T + C T C[1 + n x 2 L T T u 2 L n xx 2 ( u 2 x) 2 + T u 2 x + C Q T ρ 2 x ( n xx 2 ) 2 ] C. Proof of Theorem 1.3. t is similar to that of Theorem 1.2, we sketch it here. t follows from Lemma that sup t T T + ( ρ ɛ H 1 + ρ ɛ t L 2 + u ɛ H 1 + n ɛ H 2 + n ɛ t L 2) ( (ρ ɛ u ɛ ) t 2 L 2 + u ɛ xx 2 L 2 + n ɛ xxx 2 L 2 + n ɛ t 2 H 1 )dt C. After taking possible subsequences, we may assume (ρ ɛ, ρ ɛ x, ρ ɛ t) (ρ, ρ x, ρ t ) weak L (, T ; L 2 () ), (u ɛ, u ɛ x) (u, u x ) weak L (, T ; L 2 () ), u ɛ xx u xx weakly in L 2 (, T ; L 2 () ), (ρ ɛ u ɛ ) t v weakly in L 2 (, T ; L 2 () ), (n ɛ, n ɛ x, n ɛ xx, n ɛ t) (n, n x, n xx, n t ) weak L (, T ; L 2 () ), (n ɛ xt, n ɛ xxx) (n xt, n xxx ) weakly in L 2 (, T ; L 2 () ). Since ρ ɛ is bounded in L (, T ; H 1 () ), and ρ ɛ t bounded in L (, T ; L 2 () ), Lemma 3.8 implies that as ɛ Hence, as ɛ, we have ρ ɛ ρ strongly in C(Q T ). ρ ɛ u ɛ ρu weak L (, T ; L 2 ()). Thus v = (ρu) t. n fact, Lemma 3.8 yields ρ ɛ u ɛ ρu strongly in C(Q T ), 31

32 as ρu and (ρu) t are bounded in L (, T ; H 1 ()) and L 2 (, T ; L 2 ()) respectively. Combining these convergence together, we have ρ ɛ (u ɛ ) 2 ρu 2, weak L (, T ; L 2 ()). Since (ρ ɛ (u ɛ ) 2 ) x is bounded in L (, T ; L 2 ()), it follows (ρ ɛ (u ɛ ) 2 ) x (ρu 2 ) x, weak L (, T ; L 2 ()). Lemma 3.8 also implies (n ɛ ) x n x strongly in C(Q T ). Since ( n ɛ x 2 ) x is bounded in L (, T ; L 2 ()), we have ( n ɛ x 2 ) x ( n x 2 ) x, weak L (, T ; L 2 ()). Similarly, we can get (ρ ɛ u ɛ ) x (ρu) x, weak L (, T ; L 2 ()), P (ρ ɛ ) x P (ρ) x, weak L (, T ; L 2 ()), n ɛ n strongly in C(Q T ), n ɛ x 2 n ɛ n x 2 n strongly in C(Q T ), u ɛ n ɛ x un x weak L (, T ; L 2 ()). Based on these convergence, we can conclude that (ρ, u, n) is a strong solution of the system (1.1) along with the initial and boundary conditions. The proof of Theorem 1.3 is completed. Acknowledgment. The authors would like to thank Professor Hai-Liang Li for his suggestions. The first two and the fourth authors are supported by the National Natural Science Foundation of China (No.14715), by the National 973 Project of China (Grant No.26CB8592), by Guangdong Provincial Natural Science Foundation (Grant No.75795) and by University Special Research Foundation for Ph.D Program (Grant. No ). The third author is partially supported by NSF

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