pdf file - Purdue Math
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 ρ0 has a positive lower bound, we obtain the existence and uniqueness of global classical, and strong solutions and the existence of weak solutions. For ρ0 ≥ 0, we obtain the existence of global strong solutions.
Key Words: Liquid crystal, compressible hydrodynamic flow, global solutions.
1
Introduction
In this paper, we consider the one dimensional initial-boundary value problem for (ρ, u, n) : [0, 1] × [0, +∞) → R+ × R × S 2 : ρt + (ρu)x = 0, (ρu)t + (ρu2 )x + (P (ρ))x = µuxx − λ(|nx |2 )x , nt + unx = θ(nxx + |nx |2 n),
(1.1)
for (x, t) ∈ (0, 1) × (0, +∞), with the initial condition: (ρ, u, n) t=0 = (ρ0 , u0 , n0 ) in [0, 1], ∗
(1.2)
School of Mathematical Sciences, South China Normal University, Guangzhou, 510631, China, Department of Mathematics, South China University of Technology, Guangzhou, 510641, China ‡ Department of Mathematics, University of Kentucky, Lexington, KY 40506 § Corresponding author. Email: [email protected] †
1
where n0 : [0, 1] → S 2 and the boundary condition: (u, nx ) ∂I = (0, 0), t > 0,
(1.3)
where ρ ≥ 0 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 > 0, λ > 0, θ > 0 are viscosity of the fluid, competition between kinetic and potential energy, and microscopic elastic relaxation time respectively. P = Rργ , for some constants γ > 1 and R > 0, is the pressure function. The hydrodynamic flow of compressible (or incompressible) liquid crystals was first derived by Ericksen [2] and Leslie [3] in 1960’s. However, its rigorous mathematical analysis was not taken place until 1990’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 ⊂ Rd can be written as follows (see Lin [4]): ρt + div(ρu) = 0, (ρu)t + div(ρu ⊗ u) + ∇(P (ρ)) = µ∆u − λdiv(∇n ∇n − nt + u · ∇n = θ(∆n + |∇n|2 n),
|∇n|2 2 Id ),
(?)
where u ⊗ u = (ui uj )1≤i,j≤d , and ∇n ∇n=(nxi · nxj )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 = 0). In 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 (1 − |n|2 )n by ), and proved [5], among many other results, the local existence of 2 classical solutions and the global existence of weak solutions in dimension two and 2
three. For any fixed > 0, 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 CaffarelliKohn-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, 10, 18] for relevant results. For the incompressible case ρ = 1 and div u = 0, it remains to be an open problem that for ↓ 0 whether a sequence of solutions (u , n ) to the approximate equation converges to a solution of the original equation (?). It 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 0 < c−1 0 ≤ ρ0 ≤ c0 and the existence of strong solutions for ρ0 ≥ 0 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 ρ0 ≥ 0 was obtained by Choe-Kim [17] in one dimension and by Choe-Kim [16] in higher dimensions. Notice that ρ = 0 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) I = [0, 1], ∂I = {0, 1}, QT = I × [0, T ] for T > 0. 3
(2) For p ≥ 1, denote Lp = Lp (I) as the Lp space with the norm k · kLp . For k ≥ 1 and p ≥ 1, denote W k,p = W k,p (I) for the Sobolev space, whose norm is denoted as k · kW k,p , H k = W k,2 (I), and n o W k,p (I, S 2 ) = u ∈ W k,p (I, R3 ) | |u(x)| = 1 a.e. x ∈ I . (3) For an even integer k ≥ 0 and 0 < α < 1, let C k+α,
k+α 2
(QT ) denote the Schauder
space of functions on QT , whose derivatives up to kth order in x-variable and up to k 2 th
order in t-variable are H¨ older continuous with exponents α and
with the norm k · k
C k+α,
k+α 2
α 2
respectively,
.
Our first main theorem is concerned with the existence of global classical solutions. Theorem 1.1 For 0 < α < 1, assume that ρ0 ∈ C 1,α satisfies 0 < c−1 0 ≤ ρ0 ≤ c0 for some c0 , u0 ∈ C 2,α and n0 ∈ C 2,α (I, S 2 ). Then there exists a unique global classical solution (ρ, u, n) : I × R+ → R+ × R × S 2 to the initial boundary value problem (1.1)-(1.3) satisfying that for any T > 0 there exists c = c(c0 , T ) > 0 such that α
(ρx , ρt ) ∈ C α, 2 (QT ), c−1 ≤ ρ ≤ c, (u, n) ∈ C 2+α,
2+α 2
(QT ).
Our second main theorem is concerned with the existence of global strong solutions and weak solutions under the assumption that ρ0 ∈ H 1 (I) satisfies 0 < c−1 0 ≤ ρ0 ≤ c0 . More precisely, 2 Theorem 1.2 (i) If ρ0 ∈ H 1 , 0 < c−1 0 ≤ ρ0 ≤ c0 for some c0 , u0 ∈ L (I), and n0 ∈
H 1 (I, S 2 ), then there exists a global weak solution (ρ, u, n) : I × R+ → R+ × R × S 2 to (1.1)-(1.3) such that for any T > 0 there exists c = c(c0 , T ) > 0 such that ρx ∈ L∞ (0, T ; L2 ), ρt ∈ L2 (0, T ; L2 ), 0 < c−1 ≤ ρ ≤ c, u ∈ L∞ (0, T ; L2 ) ∩ L2 (0, T ; H 1 ), n ∈ L∞ (0, T ; H 1 ) ∩ L2 (0, T ; H 2 ). 1 2 2 (ii) If ρ0 ∈ H 1 , 0 < c−1 0 ≤ ρ0 ≤ c0 for some c0 , u0 ∈ H0 , and n0 ∈ H (I, S ), then
there exists a unique global strong solution to the initial boundary value problem 4
(1.1)-(1.3) satisfying that for any T > 0 there exists c = c(c0 , T ) > 0 such that ρ ∈ L∞ (0, T ; H 1 ),
ρt ∈ L∞ (0, T ; L2 ),
0 < c−1 ≤ ρ ≤ c,
u ∈ L∞ (0, T ; H01 ) ∩ L2 (0, T ; H 2 ),
ut ∈ L2 (0, T ; L2 ),
n ∈ L∞ (0, T ; H 2 ) ∩ L2 (0, T ; H 3 ),
nt ∈ L2 (0, T ; H 1 ) ∩ L∞ (0, T ; L2 ).
When ρ0 is only nonnegative, we establish the existence of global strong solutions to (1.1)-(1.3). Theorem 1.3 Assume 0 ≤ ρ0 ∈ H 1 , u0 ∈ H01 and n0 ∈ H 2 (I, S 2 ). Then there exists a global strong solution to the initial boundary value problem (1.1)-(1.3) such that for any T > 0, ρ ∈ L∞ (0, T ; H 1 ), ρt ∈ L∞ (0, T ; L2 ), ρ ≥ 0, √ u ∈ L∞ (0, T ; H01 ) ∩ L2 (0, T ; H 2 ), ρut ∈ L2 (0, T ; L2 ), n ∈ L∞ (0, T ; H 2 ) ∩ L2 (0, T ; H 3 ), nt ∈ L2 (0, T ; H 1 ) ∩ L∞ (0, T ; L2 ). Remark 1.1 (i) It is unknown whether the strong solution obtained in Theorem 1.3 is unique. If, in additions, u0 ∈ H 2 satisfies the compatibility condition: 1
(µu0x )x − (P (ρ0 ))x − λ(|n0x |2 )x = ρ02 g, for some g ∈ L2 , 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∞ (0, T ; H 2 ),
√
ρut ∈ L∞ (0, T ; L2 ), ut ∈ L2 (0, 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 0 ≤ ρ0 ∈ Lγ , u0 ∈ L2 , and n0 ∈ H 1 (I, S 2 ). This will be discussed in a forthcoming paper. 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
The paper is organized as follows. In section 2, we prove the existence of the short time classical solutions of (1.1). In 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 ρ0 ≥ c0 > 0. In section 4, we prove the existence of strong solution for ρ0 ≥ 0.
2
Existence of local classical solutions
In 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 ρ0 has a positive lower bound. The main result of this section can be stated as follows. Theorem 2.1 For α ∈ (0, 1), assume that ρ0 ∈ C 1,α satisfies 0 < c−1 0 ≤ ρ0 ≤ c0 , u0 ∈ C 2,α , and n0 ∈ C 2,α (I, S 2 ). Then there exists T > 0 depending on ρ0 , u0 , n0 such that the initial boundary value problem (1.1)-(1.3) has a unique classical solution (ρ, u, n) : I × [0, T ) → R+ × R × S 2 satisfying α
(ρx , ρt ) ∈ C α, 2 (QT ), c−1 ≤ ρ ≤ c, (u, n) ∈ C 2+α,
2+α 2
(QT ).
We may assume throughout this section that 1
Z
ρ0 (ξ)dξ = 1.
(2.1)
0
To prove Theorem 2.1, we introduce for any T > 0 the Lagrangian coordinate (y, τ ) on I × [0, T ): Z y(x, t) =
x
ρ(ξ, t)dξ, τ (x, t) = t. 0
It is easy to check that (x, t) → (y, τ ) is a C 1 -bijective map from I × [0, T ) → R1 I × [0, T ), provide ρ(x, t) ∈ C 1 (I × [0, T )) is positive and 0 ρ(ξ, t) dξ = 1 for all t ∈ [0, T ). Direct calculations imply ∂ ∂ ∂ ∂ ∂ = −ρu + , =ρ , ∂t ∂y ∂τ ∂x ∂y
6
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: ρτ + ρ2 uy = 0, uτ + (P (ρ))y = (ρuy )y − (ρ2 |ny |2 )y , nτ = ρ(ρny )y + ρ2 |ny |2 n,
(2.2)
and (ρ, u, n)|τ =0 = (ρ0 , u0 , n0 ), in I
(2.3)
(u, ny )|∂I = 0, τ > 0.
(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 > 0 and (ρ0 , u0 , n0 ) be given by Theorem 2.1. For K > 0 and T > 0, to be determined later, define the space X = X(T, K) by o n 2+α (v, m) : QT → R × R3 | (v, m) ∈ C 2+α, 2 , (v, m) τ =0 = (u0 , n0 ), k(v, m)kX ≤ K where k(v, m)kX := kvk
C 2+α,
2+α 2 (QT )
+ kmk
C 2+α,
2+α 2 (QT )
.
It is evident that (X, k·kX ) is a Banach space. For any (v, m) ∈ X, we let (ρ, u, n) : I × [0, T ) → R+ × R × R3 solve the following system: ρτ + ρ2 vy = 0, uτ + (P (ρ))y = (ρuy )y − (ρ2 |ny |2 )y , nτ = ρ(ρny )y + ρ2 |my |2 n.
(2.2)0
along with the initial-boundary condition: (ρ, u, n) τ =0 = (ρ0 , u0 , n0 ); (ρ, u, nx ) ∂I = (ρ0 , 0, 0), τ > 0.
(2.5)
The first equation of (2.2)0 yields that ρ(y, τ ) =
ρ0 (y) Rτ . 1 + ρ0 (y) 0 vy (y, s)ds 7
(2.6)
Since (v, m) ∈ X, we have kvkC 1 (QT ) ≤ K and hence (2.6) implies ρ≤ ρ≥
c0 ρ Rτ 0 ≤ ≤ 2c0 , 1 − c0 KT 1 − |ρ0 0 vy (y, s)ds|
(2.7)
c0 c0 ρ Rτ 0 ≥ ≥ , 1 + c KT 2 v (y, s)ds| 0 0 y
(2.8)
1 + |ρ0
provided T ≤ T0 ≡ Since v ∈ C 2+α,
2+α 2
Since m ∈ C 2+α,
1 . 2c0 K α
(QT ) and ρ0 ∈ C 1+α (I), (2.6) implies that ρ, ρy ∈ C α, 2 (QT ).
2+α 2
α
(QT ), ρ, ρy ∈ C α, 2 (QT ), it follows from (2.7), (2.8), and the α
Schauder theory that there is a unique solution (ρ, u, n), with (ρ, u, n) ∈ C α, 2 (QT )× C 2+α,
2+α 2
(QT ) × C 2+α,
2+α 2
(QT ) to (2.2)0 and (2.5). Define the solution map:
H(v, m) = (u, n) : X → C 2+α,
2+α 2
(QT , R × R3 ).
Claim. There exist sufficiently large K > 0 and sufficiently small T > 0 such that H : X → X is a contraction map. Proof of Claim. We will first prove that H maps X into X. Set C1 = kρ0 kC 1,α + ku0 kC 2,α + kn0 kC 2,α . Direct differentiation of (2.6) implies that Rτ Rτ ρ0 ρ0y 0 vy (y, s)ds + ρ20 0 vyy (y, s)ds ρ0y Rτ Rτ ρy = − 1 + ρ0 0 vy (y, s)ds (1 + ρ0 0 vy (y, s)ds)2
(2.9)
It follows from (2.6) and (2.9) that n o max kρkC α, α2 (Q ) , kρy kC α, α2 (Q ) ≤ c(C1 ), T
(2.10)
T
2
where T ≤ T1 := min{ 2c10 K , ( K1 ) 2−α }. Apply the Schauder theory to (2.1)03 , we obtain that for any T ≤ T1 , knk
C 2+α,
2+α 2 (QT )
≤ C(kn0 kC 2+α (I) + kρ2 |my |2 nkC α, α2 (QT ) ).
(2.11)
Direction calculations imply knkC 0 (QT ) ≤ 1 + KT, [n]C α, α2 (Q
T)
α
≤ c(C1 )(1 + KT 2 ),
(2.12) α
kmy kC 0 (QT ) ≤ kmy − m0y kC 0 (QT ) + km0y kC 0 (QT ) ≤ c(C1 )(1 + KT 2 ), 8
(2.13)
[my ]C α, α2 (Q
T)
≤ [my − m0y ]C α, α2 (Q
T)
+ [m0y ]C α, α2 (Q
α
T)
≤ c(C1 )(1 + KT 2 ).
(2.14)
2
Hence if we choose T ≤ T2 = min{T1 , ( K1 ) α }, then kρ2 |my |2 nkC 0 (QT ) ≤ c(C1 ),
(2.15)
and [ρ2 |my |2 n]C α, α2 (Q
T)
≤ kρk2C 0 (QT ) kmy k2C 0 (QT ) [n]C α, α2 (Q +kmy k2 α, α2 C
(QT )
T)
+ kmy k2C 0 (QT ) knkC 0 (QT ) kρk2 α, α2 C
(QT )
kρk2C 0 (QT ) knkC 0 (QT )
≤ c(C1 ).
(2.16)
It follows from (2.11), (2.15) and (2.16) that for T ≤ T2 , knk
C 2+α,
Now we need to estimate kuk
2+α 2 (QT )
C 2+α,
2+α 2 (QT )
≤ c(C1 ).
(2.17)
as follows. It follows from (2.1)02 and
the Schauder theory that kuk
C 2+α,
2+α 2 (QT )
≤ C(ku0 kC 2+α (I) + k(ργ )y kC α, α2 (Q
T)
≤ c(C1 )[1 +
+ k(ρ2 |ny |2 )y kC α, α2 (Q
T)
kρkγ−1 kρy kC α, α2 (Q ) α T C α, 2 (QT )
+kρkC α, α2 (Q ) kρy kC α, α2 (Q ) kny k2 α, α2 T
T
C
(QT )
kny kC α, α2 (Q ) knyy kC α, α2 (Q ) ] +kρk2 α, α2 (QT ) C T T ≤ c(C1 ), provide T ≤ T2 . Thus we conclude that if K > 0 is sufficiently large and T ≤ T2 , then H maps X into X. Next we want to show H : X → X is a contraction map. For i = 1, 2, let (vi , mi ) ∈ X, and (ui , ni ) = H(vi , mi ). Set ρ = ρ1 − ρ2 , v = v1 − v2 , m = m1 − m2 , u = u1 − u2 and n = n1 − n2 . Then it is easy to see (
ρ )τ = −v y . ρ1 ρ2
Since ρ|τ =0 = 0, integrating (2.18) with respect to τ yields Z τ ρ = −ρ1 ρ2 v y ds. 0
9
(2.18)
(2.19)
Since both ρ1 and ρ2 satisfy (2.10), we obtain α
max{kρkC α, α2 (Q ) , kρy kC α, α2 (Q ) } ≤ c(C1 )T 1− 2 kvk T
T
C 2+α,
2+α 2 (QT )
.
(2.20)
For n ¯ , we have nτ
= ρ21 nyy + ρ(ρ1 + ρ2 )n2yy + ρρ1y n1y + ρ2 ρy n1y + ρ2 ρ2y ny +ρ(ρ1 + ρ2 )|m1y |2 n1 + ρ22 my (m1y + m2y )n1 + ρ22 |m2y |2 n.
(2.21)
Since ρi , mi , ni (i = 1, 2) satisfy (2.11)-(2.12), (2.13)-(2.14), we have by the Schauder theory that for T ≤ T2 , knk
C 2+α,
2+α 2 (QT )
≤ c(C1 )[kρkC α, α2 (Q +kny kC α, α2 (Q ≤ c(C1 )[T
1− α 2
T)
+ kρy kC α, α2 (Q
+ knkC α, α2 (Q
T)
+ kmy kC α, α2 (Q ) ]
T)
T
kvk
C 2+α,
+kny kC α, α2 (Q
T)
+ knkC α, α2 (Q
2+α 2 (QT )
T)
+ kmy kC α, α2 (Q ) ]
T)
T
Since m|τ =0 = n|τ =0 = 0, we see that α
max{knkC α, α2 (Q ) , kny kC α, α2 (Q ) } ≤ CT 2 knk T
T
C 2+α,
2+α 2 (QT )
,
and kmy kC α, α2 (Q
α
≤ CT 2 kmk
T)
C 2+α,
2+α 2 (QT )
.
Thus we obtain knk
α
C
2+α, 2+α 2
(QT )
≤ C(C1 )T 2 [kvk
C 2+α,
Finally, we need to estimate kuk
C 2+α,
2+α 2 (QT )
2+α 2 (QT )
+ kmk
C 2+α,
2+α 2 (QT )
].
(2.22)
as follows. For u, we have that
uτ + (ργ1 − ργ2 )y = (ρu1y )y + (ρ2 uy )y − [ρ(ρ1 + ρ2 )|n1y |2 ]y − [ρ22 (n1y + n2y ).ny ]y . (2.23) Hence, by the Schauder theory, we have kuk
C 2+α,
2+α 2 (QT )
≤ c(C1 )[kρkC α, α2 (Q +kny kC α, α2 (Q
T)
α 2
≤ c(C1 )T [kvk
T)
+ kρy kC α, α2 (Q
T)
+ k(ργ1 − ργ2 )y kC α, α2 (Q
T)
+ knyy kC α, α2 (Q ) ] T
C 2+α,
2+α 2 (QT )
10
+ kmk
C 2+α,
2+α 2 (QT )
],
(2.24)
where we have used both (2.11) and (2.22) in the last step. It follows from (2.22) 2
1 and (2.24) that if T ≤ T3 = min{T2 , ( 4c(C ) α }, then 1)
kuk
C 2+α,
2+α 2 (QT )
+ knk
C 2+α,
2+α 2 (QT )
1 ≤ (kvk 2+α, 2+α + kmk 2+α, 2+α ). 2 (QT ) 2 (QT ) C C 2
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). Furthermore, there is a unique ρ, with α
ρy , ρτ ∈ C α, 2 (QT3 ), solving ρτ + ρ2 uy = 0 Hence (ρ, u, n) is the unique classical solution to problem (2.2)-(2.4) on I × [0, T3 ]. Multiplying (2.2)3 by n and applying the Gronwall’s inequality, we can obtain |n| = 1 for (y, τ ) ∈ QT3 . The proof of Theorem 2.1 is completed.
3
2
Existence of global classical, strong, and weak solutions
In 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 0 < T < +∞, let (ρ, u, n) : I → [0, 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 0 ≤ t < T , it holds Z Z tZ 2 ρu2 ργ 2 ( + + |nx | )(t) + (u2x + 2 nxx + |nx |2 n ) 2 γ−1 I 0 I Z γ 2 ρ0 u0 ρ = ( + 0 + |(n0 )x |2 ). 2 γ −1 I
(3.1)
Proof. Multiplying (1.1)2 by u and integrating the resulting equation over I, we get Z Z Z Z d ρu2 γ 2 − ρ ux + ux = |nx |2 ux . (3.2) dt I 2 I I I 11
Now we claim that Z
d − ρ ux = dt I γ
Z I
ργ . γ−1
(3.3)
In fact, by (1.1)1 , we have Z Z ργ−1 (ρt + ρx u) − ργ ux = I I Z γ Z γ d ρ ρ = + ( )x u dt I γ γ Z γ ZI γ d ρ ρ = − ux . dt I γ I γ This clearly yields (3.3). Multiplying (1.1)3 by (nxx + |nx |2 n) and integrating it over I, we obtain Z Z Z 2 d 2 2 |nx | + |nx | ux + 2 nxx + |nx |2 n = 0. dt I I I
(3.4)
It follows from (3.2), (3.3), and (3.4) that Z Z Z 2 d ρu2 ργ 2 2 ( + + |nx | ) + ux + 2 nxx + |nx |2 n = 0. dt I 2 γ−1 I I This clearly implies (3.1). The proof is complete. Lemma 3.2 For any 0 ≤ t < T , it holds Z tZ |nxx |2 ≤ c(E0 )(1 + t), 0
I
where Z E0 :=
( I
ρ0 u20 ργ + 0 + |(n0 )x |2 ) 2 γ−1
denotes the total energy of the initial data. Proof. By the Gagliardo-Nirenberg inequality, we have Z Z Z 1 4 3 2 |nx | ≤ cknx kL2 (I) knxx kL2 (I) ≤ |nxx | + c( |nx |2 )3 . 2 I I I Since |n| = 1, this implies Z Z Z 2 2 2 |nxx | = |nxx + |nx | n| + |nx |4 I IZ I Z Z 1 2 2 2 |nxx | + |nxx + |nx | n| + c( |nx |2 )3 . ≤ 2 I I I 12
2
(3.5)
Hence Z
Z
2
|nxx | ≤ 2
Z |nxx + |nx | n| + c( |nx |2 )3 . 2
2
I
I
I
This, combined with (3.1), yields (3.5). The proof of the Lemma is complete.
2
Now we want to estimate knxx kL∞ ([0,T ],L2 (I)) in terms of both E0 and kn0 kH 2 (I) as follows. Lemma 3.3 For any 0 ≤ t < T , it holds Z tZ Z (|nxt |2 + |nxxx |2 ) ≤ c(E0 , kn0 kH 2 (I) ) exp{c(E0 )t}. |nxx |2 (t) + 0
I
(3.6)
I
Proof. Differentiating (1.1)3 with respect to x, multiplying the resulting equation by nxt , and integrating it over I, we have Z Z d 1 2 |nxt | + |nxx |2 dt 2 IZ I Z Z 2 = [|nx | nx · nxt + 2(nx · nxx )n · nxt ] − ux nx · nxt − unxx · nxt IZ I I Z 1 2 6 2 2 2 2 2 ≤ |nxt | + c (|nx | + |nx | |nxx | + ux |nx | + u |nxx |2 ). 2 I I Thus Z
Z d |nxt | dx + |nxx |2 dt I I Z ≤ c (|nx |6 + |nx |2 |nxx |2 + u2x |nx |2 + u2 |nxx |2 ) I Z Z Z ≤ c knx k4L∞ |nx |2 + knx k2L∞ u2x + (knx k2L∞ + kuk2L∞ ) |nxx |2 I I ZI Z Z ≤ c(1 + E0 )( |nxx |2 )2 + c u2x |nxx |2 , 2
I
I
I
where we have used the Sobolev embedding inequality: knx kL∞ (I) ≤ cknxx kL2 (I) , kukL∞ (I) ≤ ckux kL2 (I) . This, combined with Lemma 3.1, 3.2 and the Gronwall inequality, implies that any t ∈ [0, T ), Z
2
Z tZ
|nxx | (t) + I
0
|nxt |2 ≤ c(E0 , kn0 kH 2 (I) ) exp{c(E0 )t}.
I
Since nxxx = nxt + ux nx + unxx − 2(nx · nxx )n − |nx |2 nx , 13
we also obtain Z tZ 0
|nxxx |2 ≤ c(E0 , kn0 kH 2 (I) ) exp{c(E0 )t}.
I
2
The proof is now completed.
Now we want to improve the estimation of both lower and upper bounds of ρ in terms of E0 , kρ0 kH 1 , and the upper and lower bounds of ρ0 . This turns out to be the most difficult step. Lemma 3.4 There exist c = c(c0 , kρ0 kH 1 , E0 , γ) > 0 and C = C(c0 , kρ0 kH 1 , E0 ) > 0 depending only on c0 , kρ0 kH 1 , γ, and E0 such that for any 0 ≤ t < T , we have Z
1 ρ|( )x |2 (t) + ρ I
Z tZ 0
ργ−3 ρ2x ≤ C exp(Ct),
(3.7)
I
and 1 ≤ ρ(x, t) ≤ c exp(Ct), x ∈ I. c exp(Ct)
(3.8)
Proof. Using (1.1)1 , we have d dt
Z
1 ρ|( )x |2 = ρ I = = = =
=
Z
Z 1 2 1 1 ρt |( )x | + 2 ρ( )x ( )xt ρ ρ ρ IZ I Z 1 ρt 1 − (ρu)x |( )x |2 + 2 ρ( )x (− 2 )x ρ ρ ρ ZI ZI 1 2 1 (ρu)x − (ρu)x |( )x | + 2 ρ( )x ( 2 )x ρ ρ ρ ZI ZI 1 1 ux 1 2 − (ρu)x |( )x | + 2 ρ( )x [((− )x u)x + ( )x ] ρ ρ ρ ρ ZI ZI 1 2 1 2 1 1 2 − (ρu)x |( )x | + 2 [−ρ|( )x | ux + |( )x | ρx u ρ ρ 2 ρ I Z I Z 1 1 2 1 2 1 + |( )x | ρux ] + 2 ρ|( )x | ux + 2 ( )x uxx 2 ρ ρ I I ρ Z 1 2 ( )x uxx . ρ I
Thus we obtain 1d 2 dt
Z
1 ρ|( )x |2 = ρ I
Z
1 ( )x uxx . ρ I
(3.9)
Multiplying (1.1)2 by ( ρ1 )x , integrating the resulting equation over I, and utilizing
14
(3.9), we obtain Z Z Z 1 2 1 1d d γ−3 2 ρ|( )x | + γ ρ ρx + ρu(− )x 2 dt I ρ dt I ρ Z ZI Z 1 2 1 d 1 1d = ρ|( )x | − ( )x (ργ )x − ρu( )x 2 dt I ρ ρ dt I ρ Z ZI Z 1 1 1 = ( )x (|nx |2 )x − ρu( )xt + (ρu2 )x ( )x ρ ρ IZ ρ IZ IZ ρt ρx 1 = 2 ( )x nx · nxx − (ρu)x 2 − (ρu2 )x 2 ρ ρ ρ I Z I Z ZI 1 1 1 ≤ |nxx |2 + ρ|( )x |2 |nx |2 + [ 2 (|(ρu)x |2 − (ρu2 )x ρx )] ρ ρ ρ I I I Z Z Z 1 1 2 2 2 ≤ k kL∞ |nxx | + knx kL∞ ρ|( )x | + u2x ρ ρ I I Z I Z Z 1 1 2 (3.10) ≤ [k kL∞ + ρ|( )x | ] |nxx |2 + u2x . ρ ρ I I I R R Since (1.1)1 implies I ρ(ξ, t) = I ρ0 (ξ) = 1, there exists a(t) ∈ I such that R ρ(a(t), t) = I ρ(ξ, t)dξ = 1. Hence we have 1 ρ(x, t)
= = ≤ ≤
Z x 1 1 + ( )ξ ρ(a(t), t) ρ(ξ, t) a(t) Z x −ρξ 1+ 2 a(t) ρ Z 1 1 |ρx |2 1 1 + k kL2 ∞ ( )2 3 ρ I ρZ 1 1 1 1 1 + k kL∞ + ρ|( )x |2 . 2 ρ 2 I ρ
Taking the supremum over x ∈ I, this implies 1 k kL∞ ≤ 2 + ρ
Z
1 ρ|( )x |2 . ρ I
(3.11)
Plugging (3.11) into (3.10), we obtain Z Z Z 1 2 d 1 1d γ−3 2 ρ|( )x | + γ ρ ρx + ρu(− )x 2 dt I ρ dt I ρ IZ Z Z Z 1 ≤ c[ |nxx |2 + |nxx |2 ρ|( )x |2 ] + u2x . ρ I I I I
15
(3.12)
Integrating (3.12) over (0, t), we obtain Z Z tZ 1 2 1 ρ|( )x | + γ ργ−3 ρ2x 2 I ρ Z Z Z0 I 1 1 1 1 ≤ ρu( )x + ρ0 |( )x |2 − ρ0 u0 ( )x ρ 2 I ρ0 ρ0 I I Z Z tZ Z tZ 1 +c (u2x + |nxx |2 ) + c [ |nxx |2 ρ|( )x |2 ] ρ I 0 I 0 I Z Z Z tZ 1 1 1 ≤ ρ|( )x |2 + c [ |nxx |2 ρ|( )x |2 ] 4 I ρ ρ I 0 I Z tZ Z Z Z 1 1 1 2 2 2 2 (ux + |nxx | )] + ρ0 |( )x | − ρ0 u0 ( )x . +c[ ρu + 2 I ρ0 ρ0 0 I I I Since Z tZ Z Z Z 1 1 1 (u2x + |nxx |2 )] + c[ ρu2 + ρ0 |( )x |2 − ρ0 u0 ( )x 2 I ρ0 ρ0 0 I I I 3 1 1 1 ≤ c(E0 )(1 + t) + k k3L∞ kρ0 k2H 1 + k kL2 ∞ kρ0 u20 kL1 kρ0 kH 1 2 ρ0 ρ0 ≤ C(E0 , c0 , kρ0 kH 1 )(1 + t), we obtain Z
1 ρ|( )x |2 + γ ρ I
Z tZ 0
ργ−3 ρ2x
I
Z ≤ C(E0 , c0 , kρ0 kH 1 )(1 + t) + c Since Lemma 3.2 implies
0
I
Z
2
|nxx |
( 0
RtR
t
I
Z
1 ρ|( )x |2 ). ρ I
|nxx |2 ≤ c(E0 )(1 + t), we have, by the Gronwall’s
inequality, that Z tZ Z Z tZ 1 2 2 γ−3 2 ρ|( )x | + ρ ρx ≤ C(E0 , c0 , kρ0 kH 1 )(1 + t ) exp(c |nxx |2 ) ρ 0 I I 0 I ≤ C(E0 , c0 , kρ0 kH 1 ) exp{c(E0 )t}. This yields (3.7). It follows from (3.11) and (3.7) that ρ≥
1 , ∀(x, t) ∈ I × [0, T ). c exp{c(E0 )t}
Since γ > 1, we can write γ = 1 + 2δ for some δ > 0. Hence we have Z Z δ δ kρ kL∞ ≤ ρ + δ ρδ−1 |ρx | I Z ZI Z δ 1 1 1 γ γ ≤ ( ρ ) + δ( ργ ) 2 ( ρ|( )x |2 ) 2 ρ I I I ≤ c(E0 , c0 , kρ0 kH 1 , γ) exp{c(E0 )t}, 16
where we have used (3.7) in the last step. This clearly yields (3.8). The proof is 2
now complete.
Lemma 3.5 There exists C = C(γ, E0 , c0 , kρ0 kH 1 , ku0 kH 1 , kn0 kH 2 ) > 0 such that for any 0 ≤ t < T , Z
u2x (t)
Z tZ
(u2t + u2xx ) ≤ C exp{C exp(Ct)}.
+
I
0
(3.13)
I
Proof. It follows from (1.1)1 and (1.1)2 that ρut + ρuux + γργ−1 ρx = uxx − 2nx · nxx
(3.14)
Multiplying (3.14) by ut , integrating the resulting equation over I and employing integration by parts, we have Z Z 1d ρu2t + u2x 2 dt IZ ZI Z = − ρuux ut − γργ−1 ρx ut − 2 nx · nxx ut I I Z Z ZI = −2 nx · nxx ut − ρuux ut − γ ργ−1 ρx ut I Z I Z Z I Z 1 1 2 2 2 2 2 |nx | |nxx | + ρu ux + ρ2γ−3 ρ2x ] ≤ ρu + c[ 2 I t ρ I I Z Z ZI Z 1 1 γ 2 2 2 2 2 ≤ ρu + c[k kL∞ ( |nxx | ) + kρkL∞ ( ux ) ] + ckρkL∞ ργ−3 ρ2x . 2 I t ρ I I I This, combined with Lemma 3.3 and 3.4, implies Z Z Z d 2 2 ux ≤ C exp(Ct) + C exp(Ct)( u2x )2 ρut + dt I I I for some C = C(γ, E0 , c0 , kρ0 kH 1 , ku0 kH 1 , kn0 kH 2 ) > 0. Thus, by Lemma 3.1 and the Gronwall’s inequality, we have Z tZ Z ρu2t + u2x ≤ C exp{C exp(Ct)}. 0
Since ρ ≥
I
I
1 , we have C exp(Ct) Z Z tZ 2 ux (t) + u2t ≤ C exp{C exp(Ct)}. I
0
I
By Lemma 3.3, 3.4, (3.14) and (3.15), we also have Z tZ u2xx ≤ C exp{C exp(Ct)}. 0
I
17
(3.15)
2
This completes the proof.
In order to prove the existence of global classical solutions, we also need the following estimate. Lemma 3.6 There exists C = C(γ, c0 , E0 , kρ0 kH 1 , ku0 kH 2 , kn0 kH 2 ) > 0 such that for any 0 ≤ t < T , Z tZ Z 2 2 u2xt ≤ C exp{C exp[C exp(Ct)]}. (ut + uxx )(t) + 0
I
(3.16)
I
Proof. Differentiating (3.14) with respect to t, multiplying the resulting equation by ut , integrating it over I, and using integration by parts, we have Z Z 1d 2 ρu + u2xt 2 dt I t I Z Z Z Z Z 1 2 (|nx | )t uxt − = ρt u2t − (ρt uux + ρut ux )ut − ρuuxt ut + (ργ )t uxt 2 I I I IZ I Z Z Z Z 2 2 γ−1 = 2 nx · nxt uxt + (ρu)x ut + (ρu)x uux ut − ρut ux − γ ρ (ρu)x uxt I I I I Z I = (2nx · nxt − 2ρuut − ρu2 ux − γργ−1 ρx u − γργ ux )uxt IZ Z Z 2 2 − ρuut ux − ρu uxx ut − ρu2t ux I Z ZI Z I 1 2 2 2 2 2 2 2 4 2 2 (ρ2γ−2 ρ2x u2 + ρ2γ u2x ) u + c (|nx | |nxt | + ρ u ut + ρ u ux ) + cγ ≤ 2 I xt I I Z Z 2 4 4 2 +c (ρu ux + ρu uxx ) + c(1 + kux kL∞ ) ρu2t . I
I
Thus we have Z ρu2t + u2xt I Z I Z 2 ≤ cknx kL∞ |nxt |2 + c(1 + kux kL∞ + kρkL∞ kuk2L∞ ) ρu2t I ZI 2γ + c(γ 2 kρkL∞ + kρk2L∞ kuk4L∞ + kρkL∞ kuk2L∞ kux k2L∞ ) u2x I Z Z 2 + ckρkL∞ kuk4L∞ u2xx + cγ 2 kρkγ+1 ργ−3 ρ2x . L∞ kukL∞ d dt
Z
I
I
It follows from Lemma 3.3 and 3.5 that max{knxx kL∞ ([0,t],L2 (I) , kux kL∞ ([0,t],L2 (I) } ≤ C exp{C exp(Ct)}. Thus we have max{knx kL∞ , kukL∞ } ≤ C exp{C exp(Ct)}. 18
Also observe that kux k2L∞ (I)
Z ≤ I
|uxx |2 + kux k2L2 (I) .
Therefore, by the Gronwall’s inequality, we have Z
ρu2t (t) +
Z tZ 0
I
u2xt ≤ C(γ, c0 , E0 , kρ0 kH 1 , ku0 kH 2 , kn0 kH 2 ) exp{C exp[C exp(Ct)]}.
I
1 , this together with (3.14) and Lemma 3.3-3.5 yields (3.16). The c exp(Ct) proof is complete. 2
Since ρ ≥
Lemma 3.7 ([19]) Suppose that sup |u(x, t1 ) − u(x, t2 )| ≤ µ1 |t1 − t2 |α , ∀t1 , t2 ∈ [0, T ), x∈I
and sup |ux (x1 , t) − ux (x2 , t)| ≤ µ2 |x1 − x2 |β , ∀x1 , x2 ∈ I. t∈[0,T )
Then sup |ux (x, t1 ) − ux (x, t2 )| ≤ µ|t1 − t2 |δ , ∀t1 , t2 ∈ [0, T ), x∈I
where δ =
αβ 1+β ,
and µ depends only on α, β, µ1 , µ2 .
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 0 < T∗ < +∞ such that there exists a unique classical solution (ρ, u, n) : I × [0, T∗ ) → R+ × R × S 2 of (1.1)-(1.3), but at least one the following properties fails: α
(i) (ρx , ρt ) ∈ C α, 2 (QT∗ ), (ii) 0 < c−1 ≤ ρ ≤ c < +∞, ∀(x, t) ∈ QT∗ , (iii) (u, n) ∈ C 2+α,
2+α 2
(QT ∗ ).
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 n max knk
1 C 1, 2 (QT ∗ )
, kuk
o 1 C 1, 2 (QT ∗ )
≤ C(E0 , c0 , kρ0 kH 1 , ku0 kH 2 , kn0 kH 2 , T∗ ) < +∞.
19
Hence Lemma 3.3 and (1.1)3 imply that n and nx is uniformly H¨older continuous in t and x respectively with exponent 21 . Lemma 3.7 then implies nx is also uniformly H¨older continuous in t with exponent 61 . Thus max{knx k
1 1
C 3 , 6 (QT ∗ )
, kux k
1 1
C 3 , 6 (QT ∗ )
} ≤ C(E0 , c0 , kρ0 kH 1 , ku0 kH 2 , kn0 kH 2 , T∗ ) < +∞.
It follows from (1.1)3 and the Schauder theory that knk
1
1
C 2+ 3 ,1+ 6 (QT ∗ )
hence knx k
1
C 1, 2 (QT ∗ )
< +∞ and
< +∞. Hence, applying the Schauder theory to (1.1)3 again,
we have knkC 2+α,1+ α2 (Q
T∗)
< +∞.
Write F (x, t) = −(|nx |2 )x . Then kF kC α, α2 (Q
T∗)
< +∞. In term of the La-
grangian coordinate, (1.1)1 and (1.1)2 become ρτ + ρ2 uy = 0,
(3.17)
u + (ργ ) = (ρu ) + F, τ y y y Moreover, the estimates obtained by Lemma 3.4-3.6, in the Lagrangian coordinate, become 0 < c−1 ≤ ρ ≤ c < +∞, Z ρ2y ≤ c < +∞,
(3.18) (3.19)
I
Z
u2y + u2yy ≤ c < +∞.
(3.20)
I
Combining (3.17)1 with (3.18)-(3.20), we conclude kρk
1 1
C 2 , 4 (QT ∗ )
< +∞. On the
other hand, we have n max kF kC α, α2 (Q
T
, kuk ∗)
1 1 C 3 , 6 (QT ∗ )
, kuy k
o C
1, 1 2
(QT ∗ )
< +∞.
Now we claim that kρy k
1 1
C 2 , 4 (QT ∗ )
< +∞.
(3.21)
In fact, (3.17)1 and (3.17)2 imply (u + (ln ρ)y )τ = F − γργ [u + (ln ρ)y ] + γργ u. Hence we have −γ
u + (ln ρ)y = (u0 + (ln ρ0 )y )e
Rτ 0
ργ
Z + 0
20
τ
(F + γργ u)e−γ
Rτ s
ργ
ds.
This clearly implies (3.21). Thus, applying the Schauder theory to (3.17)2 , we conclude that kuk
1
1
C 2+ 2 ,1+ 4 (QT ∗ )
< +∞.
In particular, kukC α, α2 (Q
T∗)
+ kuy kC α, α2 (Q
T∗)
< +∞.
Applying these estimates to (3.17)1 , we obtain that kρkC α, α2 (Q the above argument once again, we see that max kuk 2+α, 2+α , kρy k 2+α, 2+α C
2
(QT ∗ )
C
2
(QT ∗ )
T∗)
< +∞. Repeating
, kρτ k
2+α C 2+α, 2 (QT ∗ )
< +∞.
This contradicts the choice of T ∗ . Hence T ∗ = ∞. The proof of Theorem 1.1 is now 2
complete. Now we recall the following well-known Lemma.
Lemma 3.8 [10]. Assume X ⊂ E ⊂ Y are Banach spaces and X ,→,→ E. Then the following embedding are compact: ∂ϕ q 1 (i) ϕ : ϕ ∈ L (0, T ; X), ∈ L (0, T ; Y ) ,→,→ Lq (0, T ; E), if 1 ≤ q ≤ ∞; ∂t ∂ϕ ∞ r (ii) ϕ : ϕ ∈ L (0, T ; X), ∈ L (0, T ; Y ) ,→,→ C([0, T ]; E), if 1 < r ≤ ∞. ∂t Proof of Theorem 1.2. Part (i): First, by the standard mollification, we may assume that for any α ∈ (0, 1), there exist a sequence of initial data (ρ0 , u0 , n0 ) ∈ C 1,α (I) × C 2,α (I) × C 2,α (I, S 2 ) such that (i) 0 < c−1 0 ≤ ρ0 ≤ c0 < +∞ on I,
(ii) lim [kρ0 − ρ0 kH 1 + ku0 − u0 kL2 + kn0 − n0 kH 1 ] = 0. ↓0
(To assure |n0 | = 1, we construct n0 =
η ∗n0 |η ∗n0 | )
Now let (ρ , u , n ) be the unique global classical solution of (1.1) along with
the initial condition (ρ0 , u0 , n0 ) and the boundary condition (u , ∂n ∂x )|∂I = (0, 0) for t > 0. It follows from Lemma 3.1, Lemma 3.2, and Lemma 3.4 that for any
21
0 < T < +∞, the following properties hold: 1 ≤ ρ ≤ c exp(cT ), in I × [0, T ], c exp(cT ) kρ kL∞ (0,T ;H 1 (I)) + kρt kL2 (0,T ;L2 (I)) ≤ C(T ), ku kL∞ (0,T ;L2 (I)) + ku kL2 (0,T ;H01 (I)) ≤ C(T ), kn kL∞ (0,T ;H 1 (I)) + kn kL2 (0,T ;H 2 (I)) + knt kL2 (0,T ;L2 (I)) ≤ C(T ). After taking possible subsequences, we may assume that as → 0, (ρ , ρx ) * (ρ, ρx ) weak∗ L∞ (0, T ; L2 (I)),
(3.22)
ρt * ρt weakly in L2 (0, T ; L2 (I)),
(3.23)
ρ → ρ strongly in C(QT ),
(3.24)
u * u weak∗ L∞ (0, T ; L2 (I)) and weakly in L2 (0, T ; H01 (I)),
(3.25)
(n , nx , nxx ) * (n, nx , nxx ) weakly in L2 (0, T ; L2 (I)),
(3.26)
(n , nx ) * (n, nx ) weak∗ L∞ (0, T ; L2 (I)),
(3.27)
nt * nt weakly in L2 (0, T ; L2 (I)),
(3.28)
n → n strongly in C(QT ) ∩ L2 (0, T ; C 1 (I)),
(3.29)
where we have used Lemma 3.8. It is easy to see that (3.24) and (3.25) imply ρt + (ρ u )x → ρt + (ρu)x in D0 (QT ) so that ρt + (ρu)x = 0. Since (ρ u )t = uxx − (|nx |2 )x − (ρ (u )2 )x − ((ρ )γ )x ∈ L2 (0, T ; H −1 (I)) and ρ u * ρu weak∗ L∞ (0, T ; L2 (I)), we have from Lemma 3.8 that ρ u → ρu strongly in C(0, T ; H −1 (I)).
22
This, combined with (3.25), implies that ρ (u )2 → ρu2 in D0 (QT ). Since uxx − (|nx |2 )x − ((ρ )γ )x → uxx − (|nx |2 )x − (ργ )x in D0 (QT ), we see that (ρu)t + (ρu2 )x + (ργ )x = uxx − (|nx |2 )x in D0 (QT ). Since u nx → unx , |nx |2 n → |nx |2 n in D0 (QT ), we also have that nt + unx = nxx + |nx |2 n in D0 (QT ). This completes the proof of (i). To prove part (ii). Observe that since u0 ∈ H01 and n0 ∈ H 2 , we have lim [ku0 − u0 kH 1 (I) + kn0 − n0 kH 2 (I) ] = 0.
→0
Thus, Lemma 3.3, and Lemma 3.5 imply Z Z 2 2 2 sup (|ux | + |nt | + |nxx | ) + 0≤t≤T
I
(|uxx |2 + |ut |2 + |nxt |2 + |nxxx |2 )
QT
≤ C(T ).
(3.30)
It follows from (3.30) that u ∈ L∞ (0, T ; H01 (I)) ∩ L2 (0, T ; H 2 ), ut ∈ L2 (0, T ; L2 (I)), n ∈ L∞ (0, T ; H 2 (I)) ∩ L2 (0, T ; H 3 (I)), and nt ∈ L2 (0, T ; H 1 (I)) ∩ L∞ (0, T ; L2 (I)). This proves the existence. To prove the uniqueness, let (ρi , ui , ni ) be two solutions to (1.1)-(1.3) obtained as above. Denote ρe = ρ1 − ρ2 , u e = u1 − u2 , n e = n1 − n2 , Then ρet + (e ρu1 )x + (ρ2 u e)x = 0, ρ1 u et − u exx = −e ρu2t − ρeu2 u2x − ρ1 u eu2x − ρ1 u1 u ex − (ργ1 − ργ2 )x −2n1x · n exx − 2e nx · n2xx , n et + u1 n ex + u en2x = n exx + |n1x |2 n e + [e nx · (n1x + n2x )]n2 , for (x, t) ∈ (0, 1) × (0, +∞), with the initial and boundary condition: (e ρ, u e, n e) t=0 = 0 in [0, 1], (e u, n ex ) ∂I = 0 t > 0. 23
(3.31)
Multiplying (3.31)1 by ρe, integrating the resulting equation over I, and using the integration by parts, we have 1d 2 dt
Z
Z
2
|e ρ|
Z
ρeu1 ρex − (ρ2x u e + ρ2 u ex )e ρ I Z Z 1 = − e + ρ2 u ex )e ρ |e ρ|2 u1x − (ρ2x u 2 I I Z 1 ≤ ρ|2 + ke ukL∞ kρ2x kL2 ke ρkL2 + kρ2 kL∞ ke ux kL2 ke ρkL2 . ku1x kL∞ |e 2 I Ry Since u e(0, t) = 0, we have u e(y, t) = 0 u ex (x, t)dx for (y, t) ∈ QT and hence =
I
I
ke ukL∞ ≤ ke ux kL2 , t ∈ [0, T ].
(3.32)
It follows from (3.32), the regularities of (ρi , ui ), H¨older inequality, and Sobolev inequality that d dt
Z
2
|e ρ|
Z ≤ cku1 kH 2
I
≤ c(ku1 kH 2
|e ρ|2 + cke ux kL2 ke ρkL2 I Z Z + 1) |e ρ|2 + |e ux |2 . I
(3.33)
I
Multiplying (3.31)2 by u e, integrating the resulting equation over I, and using the integration by parts, we have Z Z 1d ρ1 |e u|2 + |e ux |2 2 dt I Z I Z Z Z Z 1 2 ρ1t |e u| − ρ1 u1 u ex u e − ρeu eu2t − ρeu eu2 u2x − ρ1 |e u|2 u2x = 2 I I Z I IZ Z Z I γ γ ux + 2 (n1x · n ex )e ux + 2 (n1xx · n ex )e u − 2 (e nx · n2xx )e u. + (ρ1 − ρ2 )e I
I
I
I
Since ρ1t + (ρ1 u1 )x = 0, we have 1 2
Z
ρ1t |e u|2 −
I
Z I
ρ1 u1 u ex u e=
1 2
Z
ρ1t |e u|2 +
I
1 2
Z
|e u|2 (ρ1 u1 )x = 0.
I
Therefore, 1d 2 dt
Z
2
Z
ρ1 |e u| + I
|e ux |2
I
Z ρkL2 ku2t kL2 + ke ρkL2 ke ukL∞ ku2 kL∞ ku2x kL2 + ku2x kL∞ ≤ ke ukL∞ ke
ρ1 |e u|2
I
+ c[ke ρkL2 ke ux kL2 + ke ux kL2 ke nx kL2 kn1x kL∞ 24
n1xx + n2xx √ + k ρ1 u ekL2 ke nx kL2 k kL∞ ]. √ ρ1
It follows from (3.32), H¨ older inequality, Sobolev’s inequality, and the regularities of (ρi , ui ) that Z Z 1d 2 ux |2 ρ1 |e u| + |e 2 dt I I Z ≤ cke ux kL2 (ke ρkL2 ku2t kL2 + ke ρkL2 + ke nx kL2 ) + cku2 kH 2 ρ1 |e u|2 I √ +c(kn1 kH 3 + kn2 kH 3 )k ρ1 u ekL2 ke nx kL2 Z 1 2 2 2 2 2 u|2 ≤ ke ux kL2 + cke ρkL2 (1 + ku2t kL2 ) + c(ku2 kH 2 + kn1 kH 3 + kn2 kH 3 ) ρ1 |e 2 I +cke nx k2L2 . Thus we have Z Z Z Z d 2 2 2 2 ρ| + c |e nx |2 ρ1 |e u| + |e ux | ≤ c(1 + ku2t kL2 ) |e dt I I I I Z 2 u|2 . + c(ku2 kH 2 + kn1 kH 3 + kn2 k2H 3 ) ρ1 |e
(3.34)
I
Multiplying (3.31)3 by n e, integrating the resulting equation over I, and using the integration by parts, we have Z Z 1d 2 |e n| + |e nx |2 2 dt I IZ Z Z Z 2 2 = − u1 n ex · n e− u en2x · n e + |n1x | |e n| + [e nx · (n1x + n2x )]n2 · n e I
I
I
≤ ku1 kL∞ ke nx kL2 ke nkL2 +
I √ − 12 kn2x kL∞ kρ1 kL∞ k ρ1 u ekL2 ke nkL2
+
kn1x k2L∞
Z
|e n|2
I
nx kL2 . +kn2 kL∞ kn1x + n2x kL∞ ke nkL2 ke By H¨ older inequality and Sobolev’s inequality, we get Z Z 1d 2 |e n| + |e nx |2 2 dt I I Z √ n|2 nkL2 + ck ρ1 u nkL2 + c |e ≤ cke nx kL2 ke ekL2 ke I
≤
1 √ ke nx k2L2 + cke nk2L2 + ck ρ1 u ek2L2 . 2
Hence d dt
Z I
|e n|2 +
Z I
√ ek2L2 . |e nx |2 ≤ cke nk2L2 + ck ρ1 u
Adding together (3.33), (3.34), and (3.35), we obtain Z Z d 2 2 2 (|e ρ| + ρ1 |e u| + c|e n| ) ≤ A(t) (|e ρ|2 + ρ1 |e u|2 + c|e n|2 ), dt I I 25
(3.35)
(3.36)
where Z A(t) = c[2 + c + I
|u2t |2 (t) + ku1 (·, t)kH 2 + ku2 (·, t)kH 2 + kn1 (·, t)k2H 3 + kn2 (·, t)k2H 3 ].
It follows from the regularities of ui and ni that
RT 0
A(t)dt < +∞. (3.36) and the
Gronwall’s inequality imply Z
(|e ρ|2 + ρ1 |e u|2 + c|e n|2 )(x, t)dx Z Z 2 2 2 ρ| + ρ1 |e u| + c|e n| )(x, 0)dx exp{ ≤ (|e I
T
A(s)ds} = 0.
0
I
Since ρ1 > 0 and c > 0, we have (e ρ, u e, n e) = 0. This proves the uniqueness. Hence 2
the proof of Theorem 1.2 (ii) is complete.
Global strong solutions for ρ0 ≥ 0.
4
In this section, we establish the existence of global strong solutions for ρ0 ≥ 0. 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. η ∗n0 |η ∗n0 | .
For a small > 0, let ρ0 = η ? ρ0 + , u0 = η ? u0 , n0 =
Then ρ0 ≥ and
lim [kρ0 − ρ0 kH 1 + ku0 − u0 kH 1 + kn0 − n0 kH 2 ] = 0. ↓0
Let (ρ , u , n ) be the unique global classical solution of (1.1) along with the initial
condition (ρ0 , u0 , n0 ) and the boundary condition (u , ∂n ∂x )|∂I = (0, 0). Now we outline several integral estimates for (ρ , u , n ). For simplicity, we write (ρ, u, n) = (ρ , u , n ). 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 > 0, it holds Z sup 0≤t≤T
I
(ρu2 + ργ + |nx |2 + |nxx |2 ) +
Z 0
T
Z
(u2x + |nxt |2 + |nxxx |2 ) ≤ C(E0 , T ). (4.1)
I
Proof. It is exactly as same as that of Lemma 3.1, Lemma 3.2, and Lemma 3.3. 2 The next Lemma is concerned with the upper bound estimate of ρ.
26
Lemma 4.2 For any T > 0, there exists C > 0 independent of such that kρkL∞ (I×(0,T )) ≤ C.
(4.2)
Proof. Set Z
t 2
2
γ
Z
x
(ux − |nx | − ρu − ρ ) +
w(x, t) = 0
(ρ0 u0 )(ξ). 0
Then we have wt = ux − |nx |2 − ρu2 − ργ , wx = ρu. It follows from Lemma 4.1 that Z (|w + |wx |) ≤ C I
and hence Z kwkL∞ (I) ≤ C
(|w| + |wx |) ≤ C. I
Since ρ ≥ 0, it suffices to prove ρ(y, s) ≤ C for any (y, s) ∈ I × (0, T ). Let x(y, t) solve
dx(y, t) = u(x(y, t), t), 0 ≤ t < s; dt x(y, s) = y, 0 ≤ y ≤ 1.
.
Denote f = exp w. Then we have d ((ρf )(x(y, t), t)) = (ρt + ρx u)f + ρf (wt + uwx ) dt = [ρt + ρx u + ρux − ρ|nx |2 − ρ2 u2 − ργ+1 + (ρu)2 ]f = (−ρ|nx |2 − ργ+1 )f ≤ 0. Thus ρ(y, s)f (y, s) = ρ(x(y, s), s)f (x(y, s), s) ≤ ρ0 (x(y, 0))f (x(y, 0), 0) ≤ C. Therefore ρ(y, s) ≤ C exp(−w(y, s)) ≤ C exp(kwkL∞ (I×(0,T )) ) ≤ C. 2
The proof is completed. Next we want to estimate kux kL∞ (0,T ;L2 (I)) and kρu2t kL1 (I×[0,T ]) . 27
Lemma 4.3 For any T > 0, there exists C > 0 independent of such that Z sup 0≤t≤T
T
Z
u2x
Z
+
I
0
ρu2t ≤ C
(4.3)
I
Proof. It is similar to Lemma 3.5. Multiplying (3.14) by ut , integrating the resulting equation over I, and employing integration by parts, we have Z
ρu2t
I
Z
1d + 2 dt
u2x
Z
2
IZ
1 2
I
ρu2t
d + dt
Z I
ρ uxt −
ρuux ut Z Z Z 2 2 2 2 ρut dx + kρkL∞ kukL∞ ux + |nx | uxt + ργ uxt .
= = = ≤ − ≤ + ≤ ≤
I
I
u2x
I
I
I
|ux |2 , we obtain
R I
Z Z Z 2 2 γ ≤ 2[( ux ) + ρ uxt + |nx |2 uxt ] I
= I + II + III.
II =
Z
I
Thus, using Lemma 4.2 and kuk2L∞ ≤ Z
γ
|nx | uxt +
=
I
≤
Z
I
I
(4.4)
Z Z d γ ρ ux − γργ−1 ρt ux dt I Z ZI Z d γ γ ρ ux + (ρ )x uux + γ ργ u2x dt I I I Z Z Z d γ γ 2 ρ ux + (γ − 1) ρ ux − ργ uuxx dt I Z ZI ZI d γ γ 2 ρ ux + (γ − 1) ρ ux − ργ u(ρut + ρuux + (ργ )x + (|nx |2 )x ) dt I I Z ZI Z Z Z d 1 γ γ 2 2 2 ρ ux + (γ − 1) ρ ux + ρu + C(1 + |ux | ) u2x dt I 4 I t I I I Z Z ργ (ργ )x u − 2 ργ unx · nxx I I Z Z Z Z Z d 1 1 γ 2 2 2 ρ ux + C(1 + ux ) ux + ρut + ρ2γ ux dt I 4 2 I I I Z Z I 2 2 2 C u |nx | + C |nxx | I ZI Z Z Z Z Z Z d 1 γ 2 2 2 2 2 ρ ux + C(1 + ux ) ux + ρut + C( |nx | ux + |nxx |2 ) dt I 4 I I I I Z ZI Z I d 1 ργ ux + C[1 + ( u2x )2 ] + ρu2 . dt I 4 I t I
28
Z Z d 2 III = |nx | ux − 2 nx · nxt ux dt I Z Z ZI d 2 2 2 ≤ |nx | ux + knx kL∞ ux + |nxt |2 dt I ZI ZI Z d 2 2 2 ≤ |nx | ux + knxx kL2 ux + |nxt |2 dt I I ZI Z Z d 2 2 2 ≤ |nx | ux + ux + |nxt | . dt I I I Combining (4.4) with the estimates of II and III, we obtain Z
Z d + u2x dt I I Z Z Z Z d d 2 2 ≤ C[1 + ( ux ) ] + ργ ux + |nx |2 ux + C |nxt |2 . dt I dt I I I ρu2t
(4.5)
Integrating (4.5) over [0, T ] yields Z
T
Z
ρu2t
Z
u2x 0 I I Z Z Z T Z Z γ 2 2 2 γ ≤ u0x + C(T + ( ux ) ) + (ρ ux − ρ0 u0x ) + (|nx |2 ux − |n0x |2 u0x ) I I 0 I I Z T Z 1 Z Z Z Z 1 1 2 2 2 2γ 2 ux ) + ( ≤ C(1 + T ) + C ux + ρ + ux + C |nx |4 8 8 0 0 I I I I Z T Z Z Z Z 1 ≤ C(1 + T ) + C ( u2x )2 + u2x + C |nx |2 |nxx |2 . 4 0 I I I I +
Thus we have Z 0
T
Z
ρu2t
Z
u2x
+
I
Z
T
≤ C(1 + T ) + C
I
Z ( u2x )2 .
0
This, combined with the inequality
RT R 0
I
I
u2x ≤ C and the Gronwall inequality, im2
plies (4.3). This completes the proof. We also need to estimate kρkL∞ (0,T ;L2 (I)) as follows. Lemma 4.4 For any T > 0, there exists C > 0 independent of such that Z sup 0≤t≤T
I
ρ2x +
Z
T
0
29
kux k2L∞ ≤ C.
(4.6)
Proof. By (3.14), we have kux k2L∞
≤ 2kux − ργ − |nx |2 k2L∞ + 2kργ + |nx |2 k2L∞ ≤ C[kux − ργ − |nx |2 k2L2 + kuxx − (ργ )x − (|nx |2 )x k2L2 ] Z γ 2 + C[kρ kL∞ + ( |nxx |2 )2 ] ZI ≤ C[1 + knx k2L∞ |nx |2 ] + Ckρut + ρuux k2L2 I Z Z 1 Z Z 2 2 2 2 |nxx | ] + C ρu2t . ≤ C[1 + ( ux ) + |nx | I
I
I
0
Thus we obtain T
Z
kux k2L∞ ≤ C.
0
To estimate
R I
ρ2x , take the derivative of (1.1)1 with respect to x, multiply the
resulting equation by ρx , and integrate it over I and employ integration by parts. Then we have Z Z 1 d 2 ρx = 2 (ρu)x ρxx − (ρu)x ρx 0 dt I Z I 1 = (ρx uρxx + ρux ρxx ) − ρρx ux 0 I Z Z Z 1 1 1 2 2 ρx ux + ρux ρx 0 − ρx ux − ρρx uxx − ρρx ux 0 = − 2 I I I Z Z 3 ρ2 ux − ρρx uxx = − 2 I x I Z Z 2 ≤ Ckux kL∞ ρx − ρρx [ρut + ρuux + (ργ )x + (|nx |2 )x ]dx I Z I Z Z Z ≤ C(1 + kux kL∞ ) ρ2x + C ρu2t + C( u2x )2 + C( |nxx |2 )2 I I I I Z Z 2 2 ≤ C + C(1 + kux kL∞ ) ρx + C ρut . I
I
(4.6) follows from the Gronwall inequality. The proof is now completed. Finally we need to estimate kuxx kL2 . Lemma 4.5 For any T > 0, there exists C > 0 independent of such that Z TZ u2xx ≤ C. 0
I
Proof. It follows from (3.14) that uxx = ρut + ρuux + (ργ )x + (|nx |2 )x . 30
2
Hence Z
u2xx ≤ C
QT
Z
ρu2t + C
QT T
Z
T
kuk2L∞
0
Z
Z
u2x + C
I
Z
ρ2x
QT
Z
knx k2L∞ |nxx |2 0 I Z T Z Z T Z 2 2 ( |nxx |2 )2 ] ≤ C. ( ux ) + ≤ C[1 + + C
0
0
I
I
2
This completes the proof.
Proof of Theorem 1.3. It is similar to that of Theorem 1.2, we sketch it here. It follows from Lemma 4.1-4.5 that sup (kρ kH 1 + kρt kL2 + ku kH 1 + kn kH 2 + knt kL2 )
0≤t≤T Z T
+ 0
(k(ρ u )t k2L2 + kuxx k2L2 + knxxx k2L2 + knt k2H 1 )dt ≤ C.
After taking possible subsequences, we may assume (ρ , ρx , ρt ) * (ρ, ρx , ρt ) weak∗ L∞ 0, T ; L2 (I) , (u , ux ) * (u, ux ) weak∗ L∞ 0, T ; L2 (I) , uxx * uxx weakly in L2 0, T ; L2 (I) , (ρ u )t * v weakly in L2 0, T ; L2 (I) , (n , nx , nxx , nt ) * (n, nx , nxx , nt ) weak∗ L∞ 0, T ; L2 (I) , (nxt , nxxx ) * (nxt , nxxx ) weakly in L2 0, T ; L2 (I) . Since ρ is bounded in L∞ 0, T ; H 1 (I) , and ρt bounded in L∞ 0, T ; L2 (I) , Lemma 3.8 implies that as → 0 ρ → ρ strongly in C(QT ). Hence, as → 0, we have ρ u * ρu weak∗ L∞ (0, T ; L2 (I)). Thus v = (ρu)t . In fact, Lemma 3.8 yields ρ u → ρu strongly in C(QT ), 31
as ρu and (ρu)t are bounded in L∞ (0, T ; H 1 (I)) and L2 (0, T ; L2 (I)) respectively. Combining these convergence together, we have ρ (u )2 * ρu2 , weak∗ L∞ (0, T ; L2 (I)). Since (ρ (u )2 )x is bounded in L∞ (0, T ; L2 (I)), it follows (ρ (u )2 )x * (ρu2 )x , weak∗ L∞ (0, T ; L2 (I)). Lemma 3.8 also implies (n )x → nx strongly in C(QT ). Since (|nx |2 )x is bounded in L∞ (0, T ; L2 (I)), we have (|nx |2 )x * (|nx |2 )x , weak∗ L∞ (0, T ; L2 (I)). Similarly, we can get (ρ u )x * (ρu)x , weak∗ L∞ (0, T ; L2 (I)), P (ρ )x * P (ρ)x , weak∗ L∞ (0, T ; L2 (I)), n → n strongly in C(QT ), |nx |2 n → |nx |2 n strongly in C(QT ), u nx * unx weak∗ L∞ (0, T ; L2 (I)). 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 2
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.10471050), by the National 973 Project of China (Grant No.2006CB805902), by Guangdong Provincial Natural Science Foundation (Grant No.7005795) and by University Special Research Foundation for Ph.D Program (Grant. No. 20060574002). The third author is partially supported by NSF 0601162. 32
References [1] L. Caffarelli, R. Kohn, L. Nirenberg, Partial regularity of suitable weak solutions of the Navier-Stokes equations. Comm. Pure Appl. Math. 35 (1982), 771-831. [2] J. Ericksen, Hydrostatic Theory of Liquid Crystal, Arch. Rational Mech. Anal. 9 (1962), 371-378. [3] F. Leslie, Some Constitute Equations for Anisotropic Fluids, Q. J. Mech. Appl. Math. 19 (1966), 357-370. [4] F. H. Lin, Nonlinear theory of defects in nematic liquid crystal: phase transition and flow phenomena, Comm. Pure Appl. Math. 42, 1989, 789-814. [5] F. H. Lin, C. Liu, Nonparabolic dissipative systems modeling the flow of liquid crystals, Comm. Pure Appl. Math. Vol. XLV III, 1995, 501-537. [6] F. H. Lin, C. Liu, Existence of solutions for the Ericksen-Leslie system, Arch. Rational Mech. Anal. 154(2000) 135-156. [7] F. H. Lin, C. Liu, Partial regularities of the nonlinear dissipative systems modeling the flow of liquid crystals, DCDS, 2(1996) 1-23. [8] F. H. Lin, J. Y. Lin, C. Y. Wang, Liquid crystal flows in dimensions two. Preprint (2009). [9] C. Liu, N. J. Walkington, Mixed methods for the approximation of liquid crystal flows, Math. Modeling and Numer. Anal., 36(2002), 205-222. [10] J. Simon, Nonhomogeneous viscous incompressible fluids: existence of vecocity, density and pressure, SIAM J. Math. Anal., 21(5)(1990), 1093-1117. [11] S. Jiang, On initial boundary value problems for a viscous, heat-conducting, one-dimensional real gas, J. Diff. Equa. 110(1994), 157-181. [12] M. Okada, Free boundaryvalue problems for the equation of one-dimensional motion of viscous gas, Japan J. Appl. Math. 6 (1989) 161-177.
33
[13] T. Luo, Z. Xin, T. Yang, Interface behavior of compressible Navier-Stokes equations with vacuum, SIAM J. Math. Anal. 31 (2000) 1175-1191. [14] H. Beir˜ ao da Veiga, Long time behavior for one-dimensional motion of a general barotropic viscous fluid, Arch. Ration. Mech. Anal. 108 (1989) 141-60. [15] A. V. Kazhikhov, Stabilization of solutions of an initial-boundary-value problem for the equations of motion of a barotropic viscous fluid, Differ. Equ. 15 (1979) 463-467. [16] H. J. Choe, H. Kim, Strong solutions of the Navier-Stokes equations for isentropic compressible fluids, J. Diff. Equa. 190 (2003) 504-523. [17] H. J. Choe, H. Kim. Global existence of the radially symmetric solutions of the Navier-Stokes equations for the isentropic compressible fuids, Math. Meth. Appl. Sci. 28(2005) 1-28. [18] Blanca Climent-Ezquerra, Francisco Guill´ en-Gonz´ alez and Marko Rojas-Medar, Reproductivity for a nematic liquid crystal model, Z. angew. Math. Phys.(2006) 984-998. [19] O. A. Ladyzenskaja, V. A. Solonnikov, and N. N. Ural’ceva, Linear and quasilinear equations of parabolic type, Amer. Math. Soc., Providence RI, 1968.
34
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 ρ0 has a positive lower bound, we obtain the existence and uniqueness of global classical, and strong solutions and the existence of weak solutions. For ρ0 ≥ 0, we obtain the existence of global strong solutions.
Key Words: Liquid crystal, compressible hydrodynamic flow, global solutions.
1
Introduction
In this paper, we consider the one dimensional initial-boundary value problem for (ρ, u, n) : [0, 1] × [0, +∞) → R+ × R × S 2 : ρt + (ρu)x = 0, (ρu)t + (ρu2 )x + (P (ρ))x = µuxx − λ(|nx |2 )x , nt + unx = θ(nxx + |nx |2 n),
(1.1)
for (x, t) ∈ (0, 1) × (0, +∞), with the initial condition: (ρ, u, n) t=0 = (ρ0 , u0 , n0 ) in [0, 1], ∗
(1.2)
School of Mathematical Sciences, South China Normal University, Guangzhou, 510631, China, Department of Mathematics, South China University of Technology, Guangzhou, 510641, China ‡ Department of Mathematics, University of Kentucky, Lexington, KY 40506 § Corresponding author. Email: [email protected] †
1
where n0 : [0, 1] → S 2 and the boundary condition: (u, nx ) ∂I = (0, 0), t > 0,
(1.3)
where ρ ≥ 0 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 > 0, λ > 0, θ > 0 are viscosity of the fluid, competition between kinetic and potential energy, and microscopic elastic relaxation time respectively. P = Rργ , for some constants γ > 1 and R > 0, is the pressure function. The hydrodynamic flow of compressible (or incompressible) liquid crystals was first derived by Ericksen [2] and Leslie [3] in 1960’s. However, its rigorous mathematical analysis was not taken place until 1990’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 ⊂ Rd can be written as follows (see Lin [4]): ρt + div(ρu) = 0, (ρu)t + div(ρu ⊗ u) + ∇(P (ρ)) = µ∆u − λdiv(∇n ∇n − nt + u · ∇n = θ(∆n + |∇n|2 n),
|∇n|2 2 Id ),
(?)
where u ⊗ u = (ui uj )1≤i,j≤d , and ∇n ∇n=(nxi · nxj )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 = 0). In 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 (1 − |n|2 )n by ), and proved [5], among many other results, the local existence of 2 classical solutions and the global existence of weak solutions in dimension two and 2
three. For any fixed > 0, 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 CaffarelliKohn-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, 10, 18] for relevant results. For the incompressible case ρ = 1 and div u = 0, it remains to be an open problem that for ↓ 0 whether a sequence of solutions (u , n ) to the approximate equation converges to a solution of the original equation (?). It 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 0 < c−1 0 ≤ ρ0 ≤ c0 and the existence of strong solutions for ρ0 ≥ 0 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 ρ0 ≥ 0 was obtained by Choe-Kim [17] in one dimension and by Choe-Kim [16] in higher dimensions. Notice that ρ = 0 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) I = [0, 1], ∂I = {0, 1}, QT = I × [0, T ] for T > 0. 3
(2) For p ≥ 1, denote Lp = Lp (I) as the Lp space with the norm k · kLp . For k ≥ 1 and p ≥ 1, denote W k,p = W k,p (I) for the Sobolev space, whose norm is denoted as k · kW k,p , H k = W k,2 (I), and n o W k,p (I, S 2 ) = u ∈ W k,p (I, R3 ) | |u(x)| = 1 a.e. x ∈ I . (3) For an even integer k ≥ 0 and 0 < α < 1, let C k+α,
k+α 2
(QT ) denote the Schauder
space of functions on QT , whose derivatives up to kth order in x-variable and up to k 2 th
order in t-variable are H¨ older continuous with exponents α and
with the norm k · k
C k+α,
k+α 2
α 2
respectively,
.
Our first main theorem is concerned with the existence of global classical solutions. Theorem 1.1 For 0 < α < 1, assume that ρ0 ∈ C 1,α satisfies 0 < c−1 0 ≤ ρ0 ≤ c0 for some c0 , u0 ∈ C 2,α and n0 ∈ C 2,α (I, S 2 ). Then there exists a unique global classical solution (ρ, u, n) : I × R+ → R+ × R × S 2 to the initial boundary value problem (1.1)-(1.3) satisfying that for any T > 0 there exists c = c(c0 , T ) > 0 such that α
(ρx , ρt ) ∈ C α, 2 (QT ), c−1 ≤ ρ ≤ c, (u, n) ∈ C 2+α,
2+α 2
(QT ).
Our second main theorem is concerned with the existence of global strong solutions and weak solutions under the assumption that ρ0 ∈ H 1 (I) satisfies 0 < c−1 0 ≤ ρ0 ≤ c0 . More precisely, 2 Theorem 1.2 (i) If ρ0 ∈ H 1 , 0 < c−1 0 ≤ ρ0 ≤ c0 for some c0 , u0 ∈ L (I), and n0 ∈
H 1 (I, S 2 ), then there exists a global weak solution (ρ, u, n) : I × R+ → R+ × R × S 2 to (1.1)-(1.3) such that for any T > 0 there exists c = c(c0 , T ) > 0 such that ρx ∈ L∞ (0, T ; L2 ), ρt ∈ L2 (0, T ; L2 ), 0 < c−1 ≤ ρ ≤ c, u ∈ L∞ (0, T ; L2 ) ∩ L2 (0, T ; H 1 ), n ∈ L∞ (0, T ; H 1 ) ∩ L2 (0, T ; H 2 ). 1 2 2 (ii) If ρ0 ∈ H 1 , 0 < c−1 0 ≤ ρ0 ≤ c0 for some c0 , u0 ∈ H0 , and n0 ∈ H (I, S ), then
there exists a unique global strong solution to the initial boundary value problem 4
(1.1)-(1.3) satisfying that for any T > 0 there exists c = c(c0 , T ) > 0 such that ρ ∈ L∞ (0, T ; H 1 ),
ρt ∈ L∞ (0, T ; L2 ),
0 < c−1 ≤ ρ ≤ c,
u ∈ L∞ (0, T ; H01 ) ∩ L2 (0, T ; H 2 ),
ut ∈ L2 (0, T ; L2 ),
n ∈ L∞ (0, T ; H 2 ) ∩ L2 (0, T ; H 3 ),
nt ∈ L2 (0, T ; H 1 ) ∩ L∞ (0, T ; L2 ).
When ρ0 is only nonnegative, we establish the existence of global strong solutions to (1.1)-(1.3). Theorem 1.3 Assume 0 ≤ ρ0 ∈ H 1 , u0 ∈ H01 and n0 ∈ H 2 (I, S 2 ). Then there exists a global strong solution to the initial boundary value problem (1.1)-(1.3) such that for any T > 0, ρ ∈ L∞ (0, T ; H 1 ), ρt ∈ L∞ (0, T ; L2 ), ρ ≥ 0, √ u ∈ L∞ (0, T ; H01 ) ∩ L2 (0, T ; H 2 ), ρut ∈ L2 (0, T ; L2 ), n ∈ L∞ (0, T ; H 2 ) ∩ L2 (0, T ; H 3 ), nt ∈ L2 (0, T ; H 1 ) ∩ L∞ (0, T ; L2 ). Remark 1.1 (i) It is unknown whether the strong solution obtained in Theorem 1.3 is unique. If, in additions, u0 ∈ H 2 satisfies the compatibility condition: 1
(µu0x )x − (P (ρ0 ))x − λ(|n0x |2 )x = ρ02 g, for some g ∈ L2 , 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∞ (0, T ; H 2 ),
√
ρut ∈ L∞ (0, T ; L2 ), ut ∈ L2 (0, 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 0 ≤ ρ0 ∈ Lγ , u0 ∈ L2 , and n0 ∈ H 1 (I, S 2 ). This will be discussed in a forthcoming paper. 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
The paper is organized as follows. In section 2, we prove the existence of the short time classical solutions of (1.1). In 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 ρ0 ≥ c0 > 0. In section 4, we prove the existence of strong solution for ρ0 ≥ 0.
2
Existence of local classical solutions
In 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 ρ0 has a positive lower bound. The main result of this section can be stated as follows. Theorem 2.1 For α ∈ (0, 1), assume that ρ0 ∈ C 1,α satisfies 0 < c−1 0 ≤ ρ0 ≤ c0 , u0 ∈ C 2,α , and n0 ∈ C 2,α (I, S 2 ). Then there exists T > 0 depending on ρ0 , u0 , n0 such that the initial boundary value problem (1.1)-(1.3) has a unique classical solution (ρ, u, n) : I × [0, T ) → R+ × R × S 2 satisfying α
(ρx , ρt ) ∈ C α, 2 (QT ), c−1 ≤ ρ ≤ c, (u, n) ∈ C 2+α,
2+α 2
(QT ).
We may assume throughout this section that 1
Z
ρ0 (ξ)dξ = 1.
(2.1)
0
To prove Theorem 2.1, we introduce for any T > 0 the Lagrangian coordinate (y, τ ) on I × [0, T ): Z y(x, t) =
x
ρ(ξ, t)dξ, τ (x, t) = t. 0
It is easy to check that (x, t) → (y, τ ) is a C 1 -bijective map from I × [0, T ) → R1 I × [0, T ), provide ρ(x, t) ∈ C 1 (I × [0, T )) is positive and 0 ρ(ξ, t) dξ = 1 for all t ∈ [0, T ). Direct calculations imply ∂ ∂ ∂ ∂ ∂ = −ρu + , =ρ , ∂t ∂y ∂τ ∂x ∂y
6
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: ρτ + ρ2 uy = 0, uτ + (P (ρ))y = (ρuy )y − (ρ2 |ny |2 )y , nτ = ρ(ρny )y + ρ2 |ny |2 n,
(2.2)
and (ρ, u, n)|τ =0 = (ρ0 , u0 , n0 ), in I
(2.3)
(u, ny )|∂I = 0, τ > 0.
(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 > 0 and (ρ0 , u0 , n0 ) be given by Theorem 2.1. For K > 0 and T > 0, to be determined later, define the space X = X(T, K) by o n 2+α (v, m) : QT → R × R3 | (v, m) ∈ C 2+α, 2 , (v, m) τ =0 = (u0 , n0 ), k(v, m)kX ≤ K where k(v, m)kX := kvk
C 2+α,
2+α 2 (QT )
+ kmk
C 2+α,
2+α 2 (QT )
.
It is evident that (X, k·kX ) is a Banach space. For any (v, m) ∈ X, we let (ρ, u, n) : I × [0, T ) → R+ × R × R3 solve the following system: ρτ + ρ2 vy = 0, uτ + (P (ρ))y = (ρuy )y − (ρ2 |ny |2 )y , nτ = ρ(ρny )y + ρ2 |my |2 n.
(2.2)0
along with the initial-boundary condition: (ρ, u, n) τ =0 = (ρ0 , u0 , n0 ); (ρ, u, nx ) ∂I = (ρ0 , 0, 0), τ > 0.
(2.5)
The first equation of (2.2)0 yields that ρ(y, τ ) =
ρ0 (y) Rτ . 1 + ρ0 (y) 0 vy (y, s)ds 7
(2.6)
Since (v, m) ∈ X, we have kvkC 1 (QT ) ≤ K and hence (2.6) implies ρ≤ ρ≥
c0 ρ Rτ 0 ≤ ≤ 2c0 , 1 − c0 KT 1 − |ρ0 0 vy (y, s)ds|
(2.7)
c0 c0 ρ Rτ 0 ≥ ≥ , 1 + c KT 2 v (y, s)ds| 0 0 y
(2.8)
1 + |ρ0
provided T ≤ T0 ≡ Since v ∈ C 2+α,
2+α 2
Since m ∈ C 2+α,
1 . 2c0 K α
(QT ) and ρ0 ∈ C 1+α (I), (2.6) implies that ρ, ρy ∈ C α, 2 (QT ).
2+α 2
α
(QT ), ρ, ρy ∈ C α, 2 (QT ), it follows from (2.7), (2.8), and the α
Schauder theory that there is a unique solution (ρ, u, n), with (ρ, u, n) ∈ C α, 2 (QT )× C 2+α,
2+α 2
(QT ) × C 2+α,
2+α 2
(QT ) to (2.2)0 and (2.5). Define the solution map:
H(v, m) = (u, n) : X → C 2+α,
2+α 2
(QT , R × R3 ).
Claim. There exist sufficiently large K > 0 and sufficiently small T > 0 such that H : X → X is a contraction map. Proof of Claim. We will first prove that H maps X into X. Set C1 = kρ0 kC 1,α + ku0 kC 2,α + kn0 kC 2,α . Direct differentiation of (2.6) implies that Rτ Rτ ρ0 ρ0y 0 vy (y, s)ds + ρ20 0 vyy (y, s)ds ρ0y Rτ Rτ ρy = − 1 + ρ0 0 vy (y, s)ds (1 + ρ0 0 vy (y, s)ds)2
(2.9)
It follows from (2.6) and (2.9) that n o max kρkC α, α2 (Q ) , kρy kC α, α2 (Q ) ≤ c(C1 ), T
(2.10)
T
2
where T ≤ T1 := min{ 2c10 K , ( K1 ) 2−α }. Apply the Schauder theory to (2.1)03 , we obtain that for any T ≤ T1 , knk
C 2+α,
2+α 2 (QT )
≤ C(kn0 kC 2+α (I) + kρ2 |my |2 nkC α, α2 (QT ) ).
(2.11)
Direction calculations imply knkC 0 (QT ) ≤ 1 + KT, [n]C α, α2 (Q
T)
α
≤ c(C1 )(1 + KT 2 ),
(2.12) α
kmy kC 0 (QT ) ≤ kmy − m0y kC 0 (QT ) + km0y kC 0 (QT ) ≤ c(C1 )(1 + KT 2 ), 8
(2.13)
[my ]C α, α2 (Q
T)
≤ [my − m0y ]C α, α2 (Q
T)
+ [m0y ]C α, α2 (Q
α
T)
≤ c(C1 )(1 + KT 2 ).
(2.14)
2
Hence if we choose T ≤ T2 = min{T1 , ( K1 ) α }, then kρ2 |my |2 nkC 0 (QT ) ≤ c(C1 ),
(2.15)
and [ρ2 |my |2 n]C α, α2 (Q
T)
≤ kρk2C 0 (QT ) kmy k2C 0 (QT ) [n]C α, α2 (Q +kmy k2 α, α2 C
(QT )
T)
+ kmy k2C 0 (QT ) knkC 0 (QT ) kρk2 α, α2 C
(QT )
kρk2C 0 (QT ) knkC 0 (QT )
≤ c(C1 ).
(2.16)
It follows from (2.11), (2.15) and (2.16) that for T ≤ T2 , knk
C 2+α,
Now we need to estimate kuk
2+α 2 (QT )
C 2+α,
2+α 2 (QT )
≤ c(C1 ).
(2.17)
as follows. It follows from (2.1)02 and
the Schauder theory that kuk
C 2+α,
2+α 2 (QT )
≤ C(ku0 kC 2+α (I) + k(ργ )y kC α, α2 (Q
T)
≤ c(C1 )[1 +
+ k(ρ2 |ny |2 )y kC α, α2 (Q
T)
kρkγ−1 kρy kC α, α2 (Q ) α T C α, 2 (QT )
+kρkC α, α2 (Q ) kρy kC α, α2 (Q ) kny k2 α, α2 T
T
C
(QT )
kny kC α, α2 (Q ) knyy kC α, α2 (Q ) ] +kρk2 α, α2 (QT ) C T T ≤ c(C1 ), provide T ≤ T2 . Thus we conclude that if K > 0 is sufficiently large and T ≤ T2 , then H maps X into X. Next we want to show H : X → X is a contraction map. For i = 1, 2, let (vi , mi ) ∈ X, and (ui , ni ) = H(vi , mi ). Set ρ = ρ1 − ρ2 , v = v1 − v2 , m = m1 − m2 , u = u1 − u2 and n = n1 − n2 . Then it is easy to see (
ρ )τ = −v y . ρ1 ρ2
Since ρ|τ =0 = 0, integrating (2.18) with respect to τ yields Z τ ρ = −ρ1 ρ2 v y ds. 0
9
(2.18)
(2.19)
Since both ρ1 and ρ2 satisfy (2.10), we obtain α
max{kρkC α, α2 (Q ) , kρy kC α, α2 (Q ) } ≤ c(C1 )T 1− 2 kvk T
T
C 2+α,
2+α 2 (QT )
.
(2.20)
For n ¯ , we have nτ
= ρ21 nyy + ρ(ρ1 + ρ2 )n2yy + ρρ1y n1y + ρ2 ρy n1y + ρ2 ρ2y ny +ρ(ρ1 + ρ2 )|m1y |2 n1 + ρ22 my (m1y + m2y )n1 + ρ22 |m2y |2 n.
(2.21)
Since ρi , mi , ni (i = 1, 2) satisfy (2.11)-(2.12), (2.13)-(2.14), we have by the Schauder theory that for T ≤ T2 , knk
C 2+α,
2+α 2 (QT )
≤ c(C1 )[kρkC α, α2 (Q +kny kC α, α2 (Q ≤ c(C1 )[T
1− α 2
T)
+ kρy kC α, α2 (Q
+ knkC α, α2 (Q
T)
+ kmy kC α, α2 (Q ) ]
T)
T
kvk
C 2+α,
+kny kC α, α2 (Q
T)
+ knkC α, α2 (Q
2+α 2 (QT )
T)
+ kmy kC α, α2 (Q ) ]
T)
T
Since m|τ =0 = n|τ =0 = 0, we see that α
max{knkC α, α2 (Q ) , kny kC α, α2 (Q ) } ≤ CT 2 knk T
T
C 2+α,
2+α 2 (QT )
,
and kmy kC α, α2 (Q
α
≤ CT 2 kmk
T)
C 2+α,
2+α 2 (QT )
.
Thus we obtain knk
α
C
2+α, 2+α 2
(QT )
≤ C(C1 )T 2 [kvk
C 2+α,
Finally, we need to estimate kuk
C 2+α,
2+α 2 (QT )
2+α 2 (QT )
+ kmk
C 2+α,
2+α 2 (QT )
].
(2.22)
as follows. For u, we have that
uτ + (ργ1 − ργ2 )y = (ρu1y )y + (ρ2 uy )y − [ρ(ρ1 + ρ2 )|n1y |2 ]y − [ρ22 (n1y + n2y ).ny ]y . (2.23) Hence, by the Schauder theory, we have kuk
C 2+α,
2+α 2 (QT )
≤ c(C1 )[kρkC α, α2 (Q +kny kC α, α2 (Q
T)
α 2
≤ c(C1 )T [kvk
T)
+ kρy kC α, α2 (Q
T)
+ k(ργ1 − ργ2 )y kC α, α2 (Q
T)
+ knyy kC α, α2 (Q ) ] T
C 2+α,
2+α 2 (QT )
10
+ kmk
C 2+α,
2+α 2 (QT )
],
(2.24)
where we have used both (2.11) and (2.22) in the last step. It follows from (2.22) 2
1 and (2.24) that if T ≤ T3 = min{T2 , ( 4c(C ) α }, then 1)
kuk
C 2+α,
2+α 2 (QT )
+ knk
C 2+α,
2+α 2 (QT )
1 ≤ (kvk 2+α, 2+α + kmk 2+α, 2+α ). 2 (QT ) 2 (QT ) C C 2
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). Furthermore, there is a unique ρ, with α
ρy , ρτ ∈ C α, 2 (QT3 ), solving ρτ + ρ2 uy = 0 Hence (ρ, u, n) is the unique classical solution to problem (2.2)-(2.4) on I × [0, T3 ]. Multiplying (2.2)3 by n and applying the Gronwall’s inequality, we can obtain |n| = 1 for (y, τ ) ∈ QT3 . The proof of Theorem 2.1 is completed.
3
2
Existence of global classical, strong, and weak solutions
In 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 0 < T < +∞, let (ρ, u, n) : I → [0, 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 0 ≤ t < T , it holds Z Z tZ 2 ρu2 ργ 2 ( + + |nx | )(t) + (u2x + 2 nxx + |nx |2 n ) 2 γ−1 I 0 I Z γ 2 ρ0 u0 ρ = ( + 0 + |(n0 )x |2 ). 2 γ −1 I
(3.1)
Proof. Multiplying (1.1)2 by u and integrating the resulting equation over I, we get Z Z Z Z d ρu2 γ 2 − ρ ux + ux = |nx |2 ux . (3.2) dt I 2 I I I 11
Now we claim that Z
d − ρ ux = dt I γ
Z I
ργ . γ−1
(3.3)
In fact, by (1.1)1 , we have Z Z ργ−1 (ρt + ρx u) − ργ ux = I I Z γ Z γ d ρ ρ = + ( )x u dt I γ γ Z γ ZI γ d ρ ρ = − ux . dt I γ I γ This clearly yields (3.3). Multiplying (1.1)3 by (nxx + |nx |2 n) and integrating it over I, we obtain Z Z Z 2 d 2 2 |nx | + |nx | ux + 2 nxx + |nx |2 n = 0. dt I I I
(3.4)
It follows from (3.2), (3.3), and (3.4) that Z Z Z 2 d ρu2 ργ 2 2 ( + + |nx | ) + ux + 2 nxx + |nx |2 n = 0. dt I 2 γ−1 I I This clearly implies (3.1). The proof is complete. Lemma 3.2 For any 0 ≤ t < T , it holds Z tZ |nxx |2 ≤ c(E0 )(1 + t), 0
I
where Z E0 :=
( I
ρ0 u20 ργ + 0 + |(n0 )x |2 ) 2 γ−1
denotes the total energy of the initial data. Proof. By the Gagliardo-Nirenberg inequality, we have Z Z Z 1 4 3 2 |nx | ≤ cknx kL2 (I) knxx kL2 (I) ≤ |nxx | + c( |nx |2 )3 . 2 I I I Since |n| = 1, this implies Z Z Z 2 2 2 |nxx | = |nxx + |nx | n| + |nx |4 I IZ I Z Z 1 2 2 2 |nxx | + |nxx + |nx | n| + c( |nx |2 )3 . ≤ 2 I I I 12
2
(3.5)
Hence Z
Z
2
|nxx | ≤ 2
Z |nxx + |nx | n| + c( |nx |2 )3 . 2
2
I
I
I
This, combined with (3.1), yields (3.5). The proof of the Lemma is complete.
2
Now we want to estimate knxx kL∞ ([0,T ],L2 (I)) in terms of both E0 and kn0 kH 2 (I) as follows. Lemma 3.3 For any 0 ≤ t < T , it holds Z tZ Z (|nxt |2 + |nxxx |2 ) ≤ c(E0 , kn0 kH 2 (I) ) exp{c(E0 )t}. |nxx |2 (t) + 0
I
(3.6)
I
Proof. Differentiating (1.1)3 with respect to x, multiplying the resulting equation by nxt , and integrating it over I, we have Z Z d 1 2 |nxt | + |nxx |2 dt 2 IZ I Z Z 2 = [|nx | nx · nxt + 2(nx · nxx )n · nxt ] − ux nx · nxt − unxx · nxt IZ I I Z 1 2 6 2 2 2 2 2 ≤ |nxt | + c (|nx | + |nx | |nxx | + ux |nx | + u |nxx |2 ). 2 I I Thus Z
Z d |nxt | dx + |nxx |2 dt I I Z ≤ c (|nx |6 + |nx |2 |nxx |2 + u2x |nx |2 + u2 |nxx |2 ) I Z Z Z ≤ c knx k4L∞ |nx |2 + knx k2L∞ u2x + (knx k2L∞ + kuk2L∞ ) |nxx |2 I I ZI Z Z ≤ c(1 + E0 )( |nxx |2 )2 + c u2x |nxx |2 , 2
I
I
I
where we have used the Sobolev embedding inequality: knx kL∞ (I) ≤ cknxx kL2 (I) , kukL∞ (I) ≤ ckux kL2 (I) . This, combined with Lemma 3.1, 3.2 and the Gronwall inequality, implies that any t ∈ [0, T ), Z
2
Z tZ
|nxx | (t) + I
0
|nxt |2 ≤ c(E0 , kn0 kH 2 (I) ) exp{c(E0 )t}.
I
Since nxxx = nxt + ux nx + unxx − 2(nx · nxx )n − |nx |2 nx , 13
we also obtain Z tZ 0
|nxxx |2 ≤ c(E0 , kn0 kH 2 (I) ) exp{c(E0 )t}.
I
2
The proof is now completed.
Now we want to improve the estimation of both lower and upper bounds of ρ in terms of E0 , kρ0 kH 1 , and the upper and lower bounds of ρ0 . This turns out to be the most difficult step. Lemma 3.4 There exist c = c(c0 , kρ0 kH 1 , E0 , γ) > 0 and C = C(c0 , kρ0 kH 1 , E0 ) > 0 depending only on c0 , kρ0 kH 1 , γ, and E0 such that for any 0 ≤ t < T , we have Z
1 ρ|( )x |2 (t) + ρ I
Z tZ 0
ργ−3 ρ2x ≤ C exp(Ct),
(3.7)
I
and 1 ≤ ρ(x, t) ≤ c exp(Ct), x ∈ I. c exp(Ct)
(3.8)
Proof. Using (1.1)1 , we have d dt
Z
1 ρ|( )x |2 = ρ I = = = =
=
Z
Z 1 2 1 1 ρt |( )x | + 2 ρ( )x ( )xt ρ ρ ρ IZ I Z 1 ρt 1 − (ρu)x |( )x |2 + 2 ρ( )x (− 2 )x ρ ρ ρ ZI ZI 1 2 1 (ρu)x − (ρu)x |( )x | + 2 ρ( )x ( 2 )x ρ ρ ρ ZI ZI 1 1 ux 1 2 − (ρu)x |( )x | + 2 ρ( )x [((− )x u)x + ( )x ] ρ ρ ρ ρ ZI ZI 1 2 1 2 1 1 2 − (ρu)x |( )x | + 2 [−ρ|( )x | ux + |( )x | ρx u ρ ρ 2 ρ I Z I Z 1 1 2 1 2 1 + |( )x | ρux ] + 2 ρ|( )x | ux + 2 ( )x uxx 2 ρ ρ I I ρ Z 1 2 ( )x uxx . ρ I
Thus we obtain 1d 2 dt
Z
1 ρ|( )x |2 = ρ I
Z
1 ( )x uxx . ρ I
(3.9)
Multiplying (1.1)2 by ( ρ1 )x , integrating the resulting equation over I, and utilizing
14
(3.9), we obtain Z Z Z 1 2 1 1d d γ−3 2 ρ|( )x | + γ ρ ρx + ρu(− )x 2 dt I ρ dt I ρ Z ZI Z 1 2 1 d 1 1d = ρ|( )x | − ( )x (ργ )x − ρu( )x 2 dt I ρ ρ dt I ρ Z ZI Z 1 1 1 = ( )x (|nx |2 )x − ρu( )xt + (ρu2 )x ( )x ρ ρ IZ ρ IZ IZ ρt ρx 1 = 2 ( )x nx · nxx − (ρu)x 2 − (ρu2 )x 2 ρ ρ ρ I Z I Z ZI 1 1 1 ≤ |nxx |2 + ρ|( )x |2 |nx |2 + [ 2 (|(ρu)x |2 − (ρu2 )x ρx )] ρ ρ ρ I I I Z Z Z 1 1 2 2 2 ≤ k kL∞ |nxx | + knx kL∞ ρ|( )x | + u2x ρ ρ I I Z I Z Z 1 1 2 (3.10) ≤ [k kL∞ + ρ|( )x | ] |nxx |2 + u2x . ρ ρ I I I R R Since (1.1)1 implies I ρ(ξ, t) = I ρ0 (ξ) = 1, there exists a(t) ∈ I such that R ρ(a(t), t) = I ρ(ξ, t)dξ = 1. Hence we have 1 ρ(x, t)
= = ≤ ≤
Z x 1 1 + ( )ξ ρ(a(t), t) ρ(ξ, t) a(t) Z x −ρξ 1+ 2 a(t) ρ Z 1 1 |ρx |2 1 1 + k kL2 ∞ ( )2 3 ρ I ρZ 1 1 1 1 1 + k kL∞ + ρ|( )x |2 . 2 ρ 2 I ρ
Taking the supremum over x ∈ I, this implies 1 k kL∞ ≤ 2 + ρ
Z
1 ρ|( )x |2 . ρ I
(3.11)
Plugging (3.11) into (3.10), we obtain Z Z Z 1 2 d 1 1d γ−3 2 ρ|( )x | + γ ρ ρx + ρu(− )x 2 dt I ρ dt I ρ IZ Z Z Z 1 ≤ c[ |nxx |2 + |nxx |2 ρ|( )x |2 ] + u2x . ρ I I I I
15
(3.12)
Integrating (3.12) over (0, t), we obtain Z Z tZ 1 2 1 ρ|( )x | + γ ργ−3 ρ2x 2 I ρ Z Z Z0 I 1 1 1 1 ≤ ρu( )x + ρ0 |( )x |2 − ρ0 u0 ( )x ρ 2 I ρ0 ρ0 I I Z Z tZ Z tZ 1 +c (u2x + |nxx |2 ) + c [ |nxx |2 ρ|( )x |2 ] ρ I 0 I 0 I Z Z Z tZ 1 1 1 ≤ ρ|( )x |2 + c [ |nxx |2 ρ|( )x |2 ] 4 I ρ ρ I 0 I Z tZ Z Z Z 1 1 1 2 2 2 2 (ux + |nxx | )] + ρ0 |( )x | − ρ0 u0 ( )x . +c[ ρu + 2 I ρ0 ρ0 0 I I I Since Z tZ Z Z Z 1 1 1 (u2x + |nxx |2 )] + c[ ρu2 + ρ0 |( )x |2 − ρ0 u0 ( )x 2 I ρ0 ρ0 0 I I I 3 1 1 1 ≤ c(E0 )(1 + t) + k k3L∞ kρ0 k2H 1 + k kL2 ∞ kρ0 u20 kL1 kρ0 kH 1 2 ρ0 ρ0 ≤ C(E0 , c0 , kρ0 kH 1 )(1 + t), we obtain Z
1 ρ|( )x |2 + γ ρ I
Z tZ 0
ργ−3 ρ2x
I
Z ≤ C(E0 , c0 , kρ0 kH 1 )(1 + t) + c Since Lemma 3.2 implies
0
I
Z
2
|nxx |
( 0
RtR
t
I
Z
1 ρ|( )x |2 ). ρ I
|nxx |2 ≤ c(E0 )(1 + t), we have, by the Gronwall’s
inequality, that Z tZ Z Z tZ 1 2 2 γ−3 2 ρ|( )x | + ρ ρx ≤ C(E0 , c0 , kρ0 kH 1 )(1 + t ) exp(c |nxx |2 ) ρ 0 I I 0 I ≤ C(E0 , c0 , kρ0 kH 1 ) exp{c(E0 )t}. This yields (3.7). It follows from (3.11) and (3.7) that ρ≥
1 , ∀(x, t) ∈ I × [0, T ). c exp{c(E0 )t}
Since γ > 1, we can write γ = 1 + 2δ for some δ > 0. Hence we have Z Z δ δ kρ kL∞ ≤ ρ + δ ρδ−1 |ρx | I Z ZI Z δ 1 1 1 γ γ ≤ ( ρ ) + δ( ργ ) 2 ( ρ|( )x |2 ) 2 ρ I I I ≤ c(E0 , c0 , kρ0 kH 1 , γ) exp{c(E0 )t}, 16
where we have used (3.7) in the last step. This clearly yields (3.8). The proof is 2
now complete.
Lemma 3.5 There exists C = C(γ, E0 , c0 , kρ0 kH 1 , ku0 kH 1 , kn0 kH 2 ) > 0 such that for any 0 ≤ t < T , Z
u2x (t)
Z tZ
(u2t + u2xx ) ≤ C exp{C exp(Ct)}.
+
I
0
(3.13)
I
Proof. It follows from (1.1)1 and (1.1)2 that ρut + ρuux + γργ−1 ρx = uxx − 2nx · nxx
(3.14)
Multiplying (3.14) by ut , integrating the resulting equation over I and employing integration by parts, we have Z Z 1d ρu2t + u2x 2 dt IZ ZI Z = − ρuux ut − γργ−1 ρx ut − 2 nx · nxx ut I I Z Z ZI = −2 nx · nxx ut − ρuux ut − γ ργ−1 ρx ut I Z I Z Z I Z 1 1 2 2 2 2 2 |nx | |nxx | + ρu ux + ρ2γ−3 ρ2x ] ≤ ρu + c[ 2 I t ρ I I Z Z ZI Z 1 1 γ 2 2 2 2 2 ≤ ρu + c[k kL∞ ( |nxx | ) + kρkL∞ ( ux ) ] + ckρkL∞ ργ−3 ρ2x . 2 I t ρ I I I This, combined with Lemma 3.3 and 3.4, implies Z Z Z d 2 2 ux ≤ C exp(Ct) + C exp(Ct)( u2x )2 ρut + dt I I I for some C = C(γ, E0 , c0 , kρ0 kH 1 , ku0 kH 1 , kn0 kH 2 ) > 0. Thus, by Lemma 3.1 and the Gronwall’s inequality, we have Z tZ Z ρu2t + u2x ≤ C exp{C exp(Ct)}. 0
Since ρ ≥
I
I
1 , we have C exp(Ct) Z Z tZ 2 ux (t) + u2t ≤ C exp{C exp(Ct)}. I
0
I
By Lemma 3.3, 3.4, (3.14) and (3.15), we also have Z tZ u2xx ≤ C exp{C exp(Ct)}. 0
I
17
(3.15)
2
This completes the proof.
In order to prove the existence of global classical solutions, we also need the following estimate. Lemma 3.6 There exists C = C(γ, c0 , E0 , kρ0 kH 1 , ku0 kH 2 , kn0 kH 2 ) > 0 such that for any 0 ≤ t < T , Z tZ Z 2 2 u2xt ≤ C exp{C exp[C exp(Ct)]}. (ut + uxx )(t) + 0
I
(3.16)
I
Proof. Differentiating (3.14) with respect to t, multiplying the resulting equation by ut , integrating it over I, and using integration by parts, we have Z Z 1d 2 ρu + u2xt 2 dt I t I Z Z Z Z Z 1 2 (|nx | )t uxt − = ρt u2t − (ρt uux + ρut ux )ut − ρuuxt ut + (ργ )t uxt 2 I I I IZ I Z Z Z Z 2 2 γ−1 = 2 nx · nxt uxt + (ρu)x ut + (ρu)x uux ut − ρut ux − γ ρ (ρu)x uxt I I I I Z I = (2nx · nxt − 2ρuut − ρu2 ux − γργ−1 ρx u − γργ ux )uxt IZ Z Z 2 2 − ρuut ux − ρu uxx ut − ρu2t ux I Z ZI Z I 1 2 2 2 2 2 2 2 4 2 2 (ρ2γ−2 ρ2x u2 + ρ2γ u2x ) u + c (|nx | |nxt | + ρ u ut + ρ u ux ) + cγ ≤ 2 I xt I I Z Z 2 4 4 2 +c (ρu ux + ρu uxx ) + c(1 + kux kL∞ ) ρu2t . I
I
Thus we have Z ρu2t + u2xt I Z I Z 2 ≤ cknx kL∞ |nxt |2 + c(1 + kux kL∞ + kρkL∞ kuk2L∞ ) ρu2t I ZI 2γ + c(γ 2 kρkL∞ + kρk2L∞ kuk4L∞ + kρkL∞ kuk2L∞ kux k2L∞ ) u2x I Z Z 2 + ckρkL∞ kuk4L∞ u2xx + cγ 2 kρkγ+1 ργ−3 ρ2x . L∞ kukL∞ d dt
Z
I
I
It follows from Lemma 3.3 and 3.5 that max{knxx kL∞ ([0,t],L2 (I) , kux kL∞ ([0,t],L2 (I) } ≤ C exp{C exp(Ct)}. Thus we have max{knx kL∞ , kukL∞ } ≤ C exp{C exp(Ct)}. 18
Also observe that kux k2L∞ (I)
Z ≤ I
|uxx |2 + kux k2L2 (I) .
Therefore, by the Gronwall’s inequality, we have Z
ρu2t (t) +
Z tZ 0
I
u2xt ≤ C(γ, c0 , E0 , kρ0 kH 1 , ku0 kH 2 , kn0 kH 2 ) exp{C exp[C exp(Ct)]}.
I
1 , this together with (3.14) and Lemma 3.3-3.5 yields (3.16). The c exp(Ct) proof is complete. 2
Since ρ ≥
Lemma 3.7 ([19]) Suppose that sup |u(x, t1 ) − u(x, t2 )| ≤ µ1 |t1 − t2 |α , ∀t1 , t2 ∈ [0, T ), x∈I
and sup |ux (x1 , t) − ux (x2 , t)| ≤ µ2 |x1 − x2 |β , ∀x1 , x2 ∈ I. t∈[0,T )
Then sup |ux (x, t1 ) − ux (x, t2 )| ≤ µ|t1 − t2 |δ , ∀t1 , t2 ∈ [0, T ), x∈I
where δ =
αβ 1+β ,
and µ depends only on α, β, µ1 , µ2 .
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 0 < T∗ < +∞ such that there exists a unique classical solution (ρ, u, n) : I × [0, T∗ ) → R+ × R × S 2 of (1.1)-(1.3), but at least one the following properties fails: α
(i) (ρx , ρt ) ∈ C α, 2 (QT∗ ), (ii) 0 < c−1 ≤ ρ ≤ c < +∞, ∀(x, t) ∈ QT∗ , (iii) (u, n) ∈ C 2+α,
2+α 2
(QT ∗ ).
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 n max knk
1 C 1, 2 (QT ∗ )
, kuk
o 1 C 1, 2 (QT ∗ )
≤ C(E0 , c0 , kρ0 kH 1 , ku0 kH 2 , kn0 kH 2 , T∗ ) < +∞.
19
Hence Lemma 3.3 and (1.1)3 imply that n and nx is uniformly H¨older continuous in t and x respectively with exponent 21 . Lemma 3.7 then implies nx is also uniformly H¨older continuous in t with exponent 61 . Thus max{knx k
1 1
C 3 , 6 (QT ∗ )
, kux k
1 1
C 3 , 6 (QT ∗ )
} ≤ C(E0 , c0 , kρ0 kH 1 , ku0 kH 2 , kn0 kH 2 , T∗ ) < +∞.
It follows from (1.1)3 and the Schauder theory that knk
1
1
C 2+ 3 ,1+ 6 (QT ∗ )
hence knx k
1
C 1, 2 (QT ∗ )
< +∞ and
< +∞. Hence, applying the Schauder theory to (1.1)3 again,
we have knkC 2+α,1+ α2 (Q
T∗)
< +∞.
Write F (x, t) = −(|nx |2 )x . Then kF kC α, α2 (Q
T∗)
< +∞. In term of the La-
grangian coordinate, (1.1)1 and (1.1)2 become ρτ + ρ2 uy = 0,
(3.17)
u + (ργ ) = (ρu ) + F, τ y y y Moreover, the estimates obtained by Lemma 3.4-3.6, in the Lagrangian coordinate, become 0 < c−1 ≤ ρ ≤ c < +∞, Z ρ2y ≤ c < +∞,
(3.18) (3.19)
I
Z
u2y + u2yy ≤ c < +∞.
(3.20)
I
Combining (3.17)1 with (3.18)-(3.20), we conclude kρk
1 1
C 2 , 4 (QT ∗ )
< +∞. On the
other hand, we have n max kF kC α, α2 (Q
T
, kuk ∗)
1 1 C 3 , 6 (QT ∗ )
, kuy k
o C
1, 1 2
(QT ∗ )
< +∞.
Now we claim that kρy k
1 1
C 2 , 4 (QT ∗ )
< +∞.
(3.21)
In fact, (3.17)1 and (3.17)2 imply (u + (ln ρ)y )τ = F − γργ [u + (ln ρ)y ] + γργ u. Hence we have −γ
u + (ln ρ)y = (u0 + (ln ρ0 )y )e
Rτ 0
ργ
Z + 0
20
τ
(F + γργ u)e−γ
Rτ s
ργ
ds.
This clearly implies (3.21). Thus, applying the Schauder theory to (3.17)2 , we conclude that kuk
1
1
C 2+ 2 ,1+ 4 (QT ∗ )
< +∞.
In particular, kukC α, α2 (Q
T∗)
+ kuy kC α, α2 (Q
T∗)
< +∞.
Applying these estimates to (3.17)1 , we obtain that kρkC α, α2 (Q the above argument once again, we see that max kuk 2+α, 2+α , kρy k 2+α, 2+α C
2
(QT ∗ )
C
2
(QT ∗ )
T∗)
< +∞. Repeating
, kρτ k
2+α C 2+α, 2 (QT ∗ )
< +∞.
This contradicts the choice of T ∗ . Hence T ∗ = ∞. The proof of Theorem 1.1 is now 2
complete. Now we recall the following well-known Lemma.
Lemma 3.8 [10]. Assume X ⊂ E ⊂ Y are Banach spaces and X ,→,→ E. Then the following embedding are compact: ∂ϕ q 1 (i) ϕ : ϕ ∈ L (0, T ; X), ∈ L (0, T ; Y ) ,→,→ Lq (0, T ; E), if 1 ≤ q ≤ ∞; ∂t ∂ϕ ∞ r (ii) ϕ : ϕ ∈ L (0, T ; X), ∈ L (0, T ; Y ) ,→,→ C([0, T ]; E), if 1 < r ≤ ∞. ∂t Proof of Theorem 1.2. Part (i): First, by the standard mollification, we may assume that for any α ∈ (0, 1), there exist a sequence of initial data (ρ0 , u0 , n0 ) ∈ C 1,α (I) × C 2,α (I) × C 2,α (I, S 2 ) such that (i) 0 < c−1 0 ≤ ρ0 ≤ c0 < +∞ on I,
(ii) lim [kρ0 − ρ0 kH 1 + ku0 − u0 kL2 + kn0 − n0 kH 1 ] = 0. ↓0
(To assure |n0 | = 1, we construct n0 =
η ∗n0 |η ∗n0 | )
Now let (ρ , u , n ) be the unique global classical solution of (1.1) along with
the initial condition (ρ0 , u0 , n0 ) and the boundary condition (u , ∂n ∂x )|∂I = (0, 0) for t > 0. It follows from Lemma 3.1, Lemma 3.2, and Lemma 3.4 that for any
21
0 < T < +∞, the following properties hold: 1 ≤ ρ ≤ c exp(cT ), in I × [0, T ], c exp(cT ) kρ kL∞ (0,T ;H 1 (I)) + kρt kL2 (0,T ;L2 (I)) ≤ C(T ), ku kL∞ (0,T ;L2 (I)) + ku kL2 (0,T ;H01 (I)) ≤ C(T ), kn kL∞ (0,T ;H 1 (I)) + kn kL2 (0,T ;H 2 (I)) + knt kL2 (0,T ;L2 (I)) ≤ C(T ). After taking possible subsequences, we may assume that as → 0, (ρ , ρx ) * (ρ, ρx ) weak∗ L∞ (0, T ; L2 (I)),
(3.22)
ρt * ρt weakly in L2 (0, T ; L2 (I)),
(3.23)
ρ → ρ strongly in C(QT ),
(3.24)
u * u weak∗ L∞ (0, T ; L2 (I)) and weakly in L2 (0, T ; H01 (I)),
(3.25)
(n , nx , nxx ) * (n, nx , nxx ) weakly in L2 (0, T ; L2 (I)),
(3.26)
(n , nx ) * (n, nx ) weak∗ L∞ (0, T ; L2 (I)),
(3.27)
nt * nt weakly in L2 (0, T ; L2 (I)),
(3.28)
n → n strongly in C(QT ) ∩ L2 (0, T ; C 1 (I)),
(3.29)
where we have used Lemma 3.8. It is easy to see that (3.24) and (3.25) imply ρt + (ρ u )x → ρt + (ρu)x in D0 (QT ) so that ρt + (ρu)x = 0. Since (ρ u )t = uxx − (|nx |2 )x − (ρ (u )2 )x − ((ρ )γ )x ∈ L2 (0, T ; H −1 (I)) and ρ u * ρu weak∗ L∞ (0, T ; L2 (I)), we have from Lemma 3.8 that ρ u → ρu strongly in C(0, T ; H −1 (I)).
22
This, combined with (3.25), implies that ρ (u )2 → ρu2 in D0 (QT ). Since uxx − (|nx |2 )x − ((ρ )γ )x → uxx − (|nx |2 )x − (ργ )x in D0 (QT ), we see that (ρu)t + (ρu2 )x + (ργ )x = uxx − (|nx |2 )x in D0 (QT ). Since u nx → unx , |nx |2 n → |nx |2 n in D0 (QT ), we also have that nt + unx = nxx + |nx |2 n in D0 (QT ). This completes the proof of (i). To prove part (ii). Observe that since u0 ∈ H01 and n0 ∈ H 2 , we have lim [ku0 − u0 kH 1 (I) + kn0 − n0 kH 2 (I) ] = 0.
→0
Thus, Lemma 3.3, and Lemma 3.5 imply Z Z 2 2 2 sup (|ux | + |nt | + |nxx | ) + 0≤t≤T
I
(|uxx |2 + |ut |2 + |nxt |2 + |nxxx |2 )
QT
≤ C(T ).
(3.30)
It follows from (3.30) that u ∈ L∞ (0, T ; H01 (I)) ∩ L2 (0, T ; H 2 ), ut ∈ L2 (0, T ; L2 (I)), n ∈ L∞ (0, T ; H 2 (I)) ∩ L2 (0, T ; H 3 (I)), and nt ∈ L2 (0, T ; H 1 (I)) ∩ L∞ (0, T ; L2 (I)). This proves the existence. To prove the uniqueness, let (ρi , ui , ni ) be two solutions to (1.1)-(1.3) obtained as above. Denote ρe = ρ1 − ρ2 , u e = u1 − u2 , n e = n1 − n2 , Then ρet + (e ρu1 )x + (ρ2 u e)x = 0, ρ1 u et − u exx = −e ρu2t − ρeu2 u2x − ρ1 u eu2x − ρ1 u1 u ex − (ργ1 − ργ2 )x −2n1x · n exx − 2e nx · n2xx , n et + u1 n ex + u en2x = n exx + |n1x |2 n e + [e nx · (n1x + n2x )]n2 , for (x, t) ∈ (0, 1) × (0, +∞), with the initial and boundary condition: (e ρ, u e, n e) t=0 = 0 in [0, 1], (e u, n ex ) ∂I = 0 t > 0. 23
(3.31)
Multiplying (3.31)1 by ρe, integrating the resulting equation over I, and using the integration by parts, we have 1d 2 dt
Z
Z
2
|e ρ|
Z
ρeu1 ρex − (ρ2x u e + ρ2 u ex )e ρ I Z Z 1 = − e + ρ2 u ex )e ρ |e ρ|2 u1x − (ρ2x u 2 I I Z 1 ≤ ρ|2 + ke ukL∞ kρ2x kL2 ke ρkL2 + kρ2 kL∞ ke ux kL2 ke ρkL2 . ku1x kL∞ |e 2 I Ry Since u e(0, t) = 0, we have u e(y, t) = 0 u ex (x, t)dx for (y, t) ∈ QT and hence =
I
I
ke ukL∞ ≤ ke ux kL2 , t ∈ [0, T ].
(3.32)
It follows from (3.32), the regularities of (ρi , ui ), H¨older inequality, and Sobolev inequality that d dt
Z
2
|e ρ|
Z ≤ cku1 kH 2
I
≤ c(ku1 kH 2
|e ρ|2 + cke ux kL2 ke ρkL2 I Z Z + 1) |e ρ|2 + |e ux |2 . I
(3.33)
I
Multiplying (3.31)2 by u e, integrating the resulting equation over I, and using the integration by parts, we have Z Z 1d ρ1 |e u|2 + |e ux |2 2 dt I Z I Z Z Z Z 1 2 ρ1t |e u| − ρ1 u1 u ex u e − ρeu eu2t − ρeu eu2 u2x − ρ1 |e u|2 u2x = 2 I I Z I IZ Z Z I γ γ ux + 2 (n1x · n ex )e ux + 2 (n1xx · n ex )e u − 2 (e nx · n2xx )e u. + (ρ1 − ρ2 )e I
I
I
I
Since ρ1t + (ρ1 u1 )x = 0, we have 1 2
Z
ρ1t |e u|2 −
I
Z I
ρ1 u1 u ex u e=
1 2
Z
ρ1t |e u|2 +
I
1 2
Z
|e u|2 (ρ1 u1 )x = 0.
I
Therefore, 1d 2 dt
Z
2
Z
ρ1 |e u| + I
|e ux |2
I
Z ρkL2 ku2t kL2 + ke ρkL2 ke ukL∞ ku2 kL∞ ku2x kL2 + ku2x kL∞ ≤ ke ukL∞ ke
ρ1 |e u|2
I
+ c[ke ρkL2 ke ux kL2 + ke ux kL2 ke nx kL2 kn1x kL∞ 24
n1xx + n2xx √ + k ρ1 u ekL2 ke nx kL2 k kL∞ ]. √ ρ1
It follows from (3.32), H¨ older inequality, Sobolev’s inequality, and the regularities of (ρi , ui ) that Z Z 1d 2 ux |2 ρ1 |e u| + |e 2 dt I I Z ≤ cke ux kL2 (ke ρkL2 ku2t kL2 + ke ρkL2 + ke nx kL2 ) + cku2 kH 2 ρ1 |e u|2 I √ +c(kn1 kH 3 + kn2 kH 3 )k ρ1 u ekL2 ke nx kL2 Z 1 2 2 2 2 2 u|2 ≤ ke ux kL2 + cke ρkL2 (1 + ku2t kL2 ) + c(ku2 kH 2 + kn1 kH 3 + kn2 kH 3 ) ρ1 |e 2 I +cke nx k2L2 . Thus we have Z Z Z Z d 2 2 2 2 ρ| + c |e nx |2 ρ1 |e u| + |e ux | ≤ c(1 + ku2t kL2 ) |e dt I I I I Z 2 u|2 . + c(ku2 kH 2 + kn1 kH 3 + kn2 k2H 3 ) ρ1 |e
(3.34)
I
Multiplying (3.31)3 by n e, integrating the resulting equation over I, and using the integration by parts, we have Z Z 1d 2 |e n| + |e nx |2 2 dt I IZ Z Z Z 2 2 = − u1 n ex · n e− u en2x · n e + |n1x | |e n| + [e nx · (n1x + n2x )]n2 · n e I
I
I
≤ ku1 kL∞ ke nx kL2 ke nkL2 +
I √ − 12 kn2x kL∞ kρ1 kL∞ k ρ1 u ekL2 ke nkL2
+
kn1x k2L∞
Z
|e n|2
I
nx kL2 . +kn2 kL∞ kn1x + n2x kL∞ ke nkL2 ke By H¨ older inequality and Sobolev’s inequality, we get Z Z 1d 2 |e n| + |e nx |2 2 dt I I Z √ n|2 nkL2 + ck ρ1 u nkL2 + c |e ≤ cke nx kL2 ke ekL2 ke I
≤
1 √ ke nx k2L2 + cke nk2L2 + ck ρ1 u ek2L2 . 2
Hence d dt
Z I
|e n|2 +
Z I
√ ek2L2 . |e nx |2 ≤ cke nk2L2 + ck ρ1 u
Adding together (3.33), (3.34), and (3.35), we obtain Z Z d 2 2 2 (|e ρ| + ρ1 |e u| + c|e n| ) ≤ A(t) (|e ρ|2 + ρ1 |e u|2 + c|e n|2 ), dt I I 25
(3.35)
(3.36)
where Z A(t) = c[2 + c + I
|u2t |2 (t) + ku1 (·, t)kH 2 + ku2 (·, t)kH 2 + kn1 (·, t)k2H 3 + kn2 (·, t)k2H 3 ].
It follows from the regularities of ui and ni that
RT 0
A(t)dt < +∞. (3.36) and the
Gronwall’s inequality imply Z
(|e ρ|2 + ρ1 |e u|2 + c|e n|2 )(x, t)dx Z Z 2 2 2 ρ| + ρ1 |e u| + c|e n| )(x, 0)dx exp{ ≤ (|e I
T
A(s)ds} = 0.
0
I
Since ρ1 > 0 and c > 0, we have (e ρ, u e, n e) = 0. This proves the uniqueness. Hence 2
the proof of Theorem 1.2 (ii) is complete.
Global strong solutions for ρ0 ≥ 0.
4
In this section, we establish the existence of global strong solutions for ρ0 ≥ 0. 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. η ∗n0 |η ∗n0 | .
For a small > 0, let ρ0 = η ? ρ0 + , u0 = η ? u0 , n0 =
Then ρ0 ≥ and
lim [kρ0 − ρ0 kH 1 + ku0 − u0 kH 1 + kn0 − n0 kH 2 ] = 0. ↓0
Let (ρ , u , n ) be the unique global classical solution of (1.1) along with the initial
condition (ρ0 , u0 , n0 ) and the boundary condition (u , ∂n ∂x )|∂I = (0, 0). Now we outline several integral estimates for (ρ , u , n ). For simplicity, we write (ρ, u, n) = (ρ , u , n ). 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 > 0, it holds Z sup 0≤t≤T
I
(ρu2 + ργ + |nx |2 + |nxx |2 ) +
Z 0
T
Z
(u2x + |nxt |2 + |nxxx |2 ) ≤ C(E0 , T ). (4.1)
I
Proof. It is exactly as same as that of Lemma 3.1, Lemma 3.2, and Lemma 3.3. 2 The next Lemma is concerned with the upper bound estimate of ρ.
26
Lemma 4.2 For any T > 0, there exists C > 0 independent of such that kρkL∞ (I×(0,T )) ≤ C.
(4.2)
Proof. Set Z
t 2
2
γ
Z
x
(ux − |nx | − ρu − ρ ) +
w(x, t) = 0
(ρ0 u0 )(ξ). 0
Then we have wt = ux − |nx |2 − ρu2 − ργ , wx = ρu. It follows from Lemma 4.1 that Z (|w + |wx |) ≤ C I
and hence Z kwkL∞ (I) ≤ C
(|w| + |wx |) ≤ C. I
Since ρ ≥ 0, it suffices to prove ρ(y, s) ≤ C for any (y, s) ∈ I × (0, T ). Let x(y, t) solve
dx(y, t) = u(x(y, t), t), 0 ≤ t < s; dt x(y, s) = y, 0 ≤ y ≤ 1.
.
Denote f = exp w. Then we have d ((ρf )(x(y, t), t)) = (ρt + ρx u)f + ρf (wt + uwx ) dt = [ρt + ρx u + ρux − ρ|nx |2 − ρ2 u2 − ργ+1 + (ρu)2 ]f = (−ρ|nx |2 − ργ+1 )f ≤ 0. Thus ρ(y, s)f (y, s) = ρ(x(y, s), s)f (x(y, s), s) ≤ ρ0 (x(y, 0))f (x(y, 0), 0) ≤ C. Therefore ρ(y, s) ≤ C exp(−w(y, s)) ≤ C exp(kwkL∞ (I×(0,T )) ) ≤ C. 2
The proof is completed. Next we want to estimate kux kL∞ (0,T ;L2 (I)) and kρu2t kL1 (I×[0,T ]) . 27
Lemma 4.3 For any T > 0, there exists C > 0 independent of such that Z sup 0≤t≤T
T
Z
u2x
Z
+
I
0
ρu2t ≤ C
(4.3)
I
Proof. It is similar to Lemma 3.5. Multiplying (3.14) by ut , integrating the resulting equation over I, and employing integration by parts, we have Z
ρu2t
I
Z
1d + 2 dt
u2x
Z
2
IZ
1 2
I
ρu2t
d + dt
Z I
ρ uxt −
ρuux ut Z Z Z 2 2 2 2 ρut dx + kρkL∞ kukL∞ ux + |nx | uxt + ργ uxt .
= = = ≤ − ≤ + ≤ ≤
I
I
u2x
I
I
I
|ux |2 , we obtain
R I
Z Z Z 2 2 γ ≤ 2[( ux ) + ρ uxt + |nx |2 uxt ] I
= I + II + III.
II =
Z
I
Thus, using Lemma 4.2 and kuk2L∞ ≤ Z
γ
|nx | uxt +
=
I
≤
Z
I
I
(4.4)
Z Z d γ ρ ux − γργ−1 ρt ux dt I Z ZI Z d γ γ ρ ux + (ρ )x uux + γ ργ u2x dt I I I Z Z Z d γ γ 2 ρ ux + (γ − 1) ρ ux − ργ uuxx dt I Z ZI ZI d γ γ 2 ρ ux + (γ − 1) ρ ux − ργ u(ρut + ρuux + (ργ )x + (|nx |2 )x ) dt I I Z ZI Z Z Z d 1 γ γ 2 2 2 ρ ux + (γ − 1) ρ ux + ρu + C(1 + |ux | ) u2x dt I 4 I t I I I Z Z ργ (ργ )x u − 2 ργ unx · nxx I I Z Z Z Z Z d 1 1 γ 2 2 2 ρ ux + C(1 + ux ) ux + ρut + ρ2γ ux dt I 4 2 I I I Z Z I 2 2 2 C u |nx | + C |nxx | I ZI Z Z Z Z Z Z d 1 γ 2 2 2 2 2 ρ ux + C(1 + ux ) ux + ρut + C( |nx | ux + |nxx |2 ) dt I 4 I I I I Z ZI Z I d 1 ργ ux + C[1 + ( u2x )2 ] + ρu2 . dt I 4 I t I
28
Z Z d 2 III = |nx | ux − 2 nx · nxt ux dt I Z Z ZI d 2 2 2 ≤ |nx | ux + knx kL∞ ux + |nxt |2 dt I ZI ZI Z d 2 2 2 ≤ |nx | ux + knxx kL2 ux + |nxt |2 dt I I ZI Z Z d 2 2 2 ≤ |nx | ux + ux + |nxt | . dt I I I Combining (4.4) with the estimates of II and III, we obtain Z
Z d + u2x dt I I Z Z Z Z d d 2 2 ≤ C[1 + ( ux ) ] + ργ ux + |nx |2 ux + C |nxt |2 . dt I dt I I I ρu2t
(4.5)
Integrating (4.5) over [0, T ] yields Z
T
Z
ρu2t
Z
u2x 0 I I Z Z Z T Z Z γ 2 2 2 γ ≤ u0x + C(T + ( ux ) ) + (ρ ux − ρ0 u0x ) + (|nx |2 ux − |n0x |2 u0x ) I I 0 I I Z T Z 1 Z Z Z Z 1 1 2 2 2 2γ 2 ux ) + ( ≤ C(1 + T ) + C ux + ρ + ux + C |nx |4 8 8 0 0 I I I I Z T Z Z Z Z 1 ≤ C(1 + T ) + C ( u2x )2 + u2x + C |nx |2 |nxx |2 . 4 0 I I I I +
Thus we have Z 0
T
Z
ρu2t
Z
u2x
+
I
Z
T
≤ C(1 + T ) + C
I
Z ( u2x )2 .
0
This, combined with the inequality
RT R 0
I
I
u2x ≤ C and the Gronwall inequality, im2
plies (4.3). This completes the proof. We also need to estimate kρkL∞ (0,T ;L2 (I)) as follows. Lemma 4.4 For any T > 0, there exists C > 0 independent of such that Z sup 0≤t≤T
I
ρ2x +
Z
T
0
29
kux k2L∞ ≤ C.
(4.6)
Proof. By (3.14), we have kux k2L∞
≤ 2kux − ργ − |nx |2 k2L∞ + 2kργ + |nx |2 k2L∞ ≤ C[kux − ργ − |nx |2 k2L2 + kuxx − (ργ )x − (|nx |2 )x k2L2 ] Z γ 2 + C[kρ kL∞ + ( |nxx |2 )2 ] ZI ≤ C[1 + knx k2L∞ |nx |2 ] + Ckρut + ρuux k2L2 I Z Z 1 Z Z 2 2 2 2 |nxx | ] + C ρu2t . ≤ C[1 + ( ux ) + |nx | I
I
I
0
Thus we obtain T
Z
kux k2L∞ ≤ C.
0
To estimate
R I
ρ2x , take the derivative of (1.1)1 with respect to x, multiply the
resulting equation by ρx , and integrate it over I and employ integration by parts. Then we have Z Z 1 d 2 ρx = 2 (ρu)x ρxx − (ρu)x ρx 0 dt I Z I 1 = (ρx uρxx + ρux ρxx ) − ρρx ux 0 I Z Z Z 1 1 1 2 2 ρx ux + ρux ρx 0 − ρx ux − ρρx uxx − ρρx ux 0 = − 2 I I I Z Z 3 ρ2 ux − ρρx uxx = − 2 I x I Z Z 2 ≤ Ckux kL∞ ρx − ρρx [ρut + ρuux + (ργ )x + (|nx |2 )x ]dx I Z I Z Z Z ≤ C(1 + kux kL∞ ) ρ2x + C ρu2t + C( u2x )2 + C( |nxx |2 )2 I I I I Z Z 2 2 ≤ C + C(1 + kux kL∞ ) ρx + C ρut . I
I
(4.6) follows from the Gronwall inequality. The proof is now completed. Finally we need to estimate kuxx kL2 . Lemma 4.5 For any T > 0, there exists C > 0 independent of such that Z TZ u2xx ≤ C. 0
I
Proof. It follows from (3.14) that uxx = ρut + ρuux + (ργ )x + (|nx |2 )x . 30
2
Hence Z
u2xx ≤ C
QT
Z
ρu2t + C
QT T
Z
T
kuk2L∞
0
Z
Z
u2x + C
I
Z
ρ2x
QT
Z
knx k2L∞ |nxx |2 0 I Z T Z Z T Z 2 2 ( |nxx |2 )2 ] ≤ C. ( ux ) + ≤ C[1 + + C
0
0
I
I
2
This completes the proof.
Proof of Theorem 1.3. It is similar to that of Theorem 1.2, we sketch it here. It follows from Lemma 4.1-4.5 that sup (kρ kH 1 + kρt kL2 + ku kH 1 + kn kH 2 + knt kL2 )
0≤t≤T Z T
+ 0
(k(ρ u )t k2L2 + kuxx k2L2 + knxxx k2L2 + knt k2H 1 )dt ≤ C.
After taking possible subsequences, we may assume (ρ , ρx , ρt ) * (ρ, ρx , ρt ) weak∗ L∞ 0, T ; L2 (I) , (u , ux ) * (u, ux ) weak∗ L∞ 0, T ; L2 (I) , uxx * uxx weakly in L2 0, T ; L2 (I) , (ρ u )t * v weakly in L2 0, T ; L2 (I) , (n , nx , nxx , nt ) * (n, nx , nxx , nt ) weak∗ L∞ 0, T ; L2 (I) , (nxt , nxxx ) * (nxt , nxxx ) weakly in L2 0, T ; L2 (I) . Since ρ is bounded in L∞ 0, T ; H 1 (I) , and ρt bounded in L∞ 0, T ; L2 (I) , Lemma 3.8 implies that as → 0 ρ → ρ strongly in C(QT ). Hence, as → 0, we have ρ u * ρu weak∗ L∞ (0, T ; L2 (I)). Thus v = (ρu)t . In fact, Lemma 3.8 yields ρ u → ρu strongly in C(QT ), 31
as ρu and (ρu)t are bounded in L∞ (0, T ; H 1 (I)) and L2 (0, T ; L2 (I)) respectively. Combining these convergence together, we have ρ (u )2 * ρu2 , weak∗ L∞ (0, T ; L2 (I)). Since (ρ (u )2 )x is bounded in L∞ (0, T ; L2 (I)), it follows (ρ (u )2 )x * (ρu2 )x , weak∗ L∞ (0, T ; L2 (I)). Lemma 3.8 also implies (n )x → nx strongly in C(QT ). Since (|nx |2 )x is bounded in L∞ (0, T ; L2 (I)), we have (|nx |2 )x * (|nx |2 )x , weak∗ L∞ (0, T ; L2 (I)). Similarly, we can get (ρ u )x * (ρu)x , weak∗ L∞ (0, T ; L2 (I)), P (ρ )x * P (ρ)x , weak∗ L∞ (0, T ; L2 (I)), n → n strongly in C(QT ), |nx |2 n → |nx |2 n strongly in C(QT ), u nx * unx weak∗ L∞ (0, T ; L2 (I)). 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 2
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.10471050), by the National 973 Project of China (Grant No.2006CB805902), by Guangdong Provincial Natural Science Foundation (Grant No.7005795) and by University Special Research Foundation for Ph.D Program (Grant. No. 20060574002). The third author is partially supported by NSF 0601162. 32
References [1] L. Caffarelli, R. Kohn, L. Nirenberg, Partial regularity of suitable weak solutions of the Navier-Stokes equations. Comm. Pure Appl. Math. 35 (1982), 771-831. [2] J. Ericksen, Hydrostatic Theory of Liquid Crystal, Arch. Rational Mech. Anal. 9 (1962), 371-378. [3] F. Leslie, Some Constitute Equations for Anisotropic Fluids, Q. J. Mech. Appl. Math. 19 (1966), 357-370. [4] F. H. Lin, Nonlinear theory of defects in nematic liquid crystal: phase transition and flow phenomena, Comm. Pure Appl. Math. 42, 1989, 789-814. [5] F. H. Lin, C. Liu, Nonparabolic dissipative systems modeling the flow of liquid crystals, Comm. Pure Appl. Math. Vol. XLV III, 1995, 501-537. [6] F. H. Lin, C. Liu, Existence of solutions for the Ericksen-Leslie system, Arch. Rational Mech. Anal. 154(2000) 135-156. [7] F. H. Lin, C. Liu, Partial regularities of the nonlinear dissipative systems modeling the flow of liquid crystals, DCDS, 2(1996) 1-23. [8] F. H. Lin, J. Y. Lin, C. Y. Wang, Liquid crystal flows in dimensions two. Preprint (2009). [9] C. Liu, N. J. Walkington, Mixed methods for the approximation of liquid crystal flows, Math. Modeling and Numer. Anal., 36(2002), 205-222. [10] J. Simon, Nonhomogeneous viscous incompressible fluids: existence of vecocity, density and pressure, SIAM J. Math. Anal., 21(5)(1990), 1093-1117. [11] S. Jiang, On initial boundary value problems for a viscous, heat-conducting, one-dimensional real gas, J. Diff. Equa. 110(1994), 157-181. [12] M. Okada, Free boundaryvalue problems for the equation of one-dimensional motion of viscous gas, Japan J. Appl. Math. 6 (1989) 161-177.
33
[13] T. Luo, Z. Xin, T. Yang, Interface behavior of compressible Navier-Stokes equations with vacuum, SIAM J. Math. Anal. 31 (2000) 1175-1191. [14] H. Beir˜ ao da Veiga, Long time behavior for one-dimensional motion of a general barotropic viscous fluid, Arch. Ration. Mech. Anal. 108 (1989) 141-60. [15] A. V. Kazhikhov, Stabilization of solutions of an initial-boundary-value problem for the equations of motion of a barotropic viscous fluid, Differ. Equ. 15 (1979) 463-467. [16] H. J. Choe, H. Kim, Strong solutions of the Navier-Stokes equations for isentropic compressible fluids, J. Diff. Equa. 190 (2003) 504-523. [17] H. J. Choe, H. Kim. Global existence of the radially symmetric solutions of the Navier-Stokes equations for the isentropic compressible fuids, Math. Meth. Appl. Sci. 28(2005) 1-28. [18] Blanca Climent-Ezquerra, Francisco Guill´ en-Gonz´ alez and Marko Rojas-Medar, Reproductivity for a nematic liquid crystal model, Z. angew. Math. Phys.(2006) 984-998. [19] O. A. Ladyzenskaja, V. A. Solonnikov, and N. N. Ural’ceva, Linear and quasilinear equations of parabolic type, Amer. Math. Soc., Providence RI, 1968.
34
