Unique solvability of compressible micropolar viscous fluids

Boundary Value Problems20122012:32

DOI: 10.1186/1687-2770-2012-32

Received: 4 December 2011

Accepted: 20 March 2012

Published: 20 March 2012

Abstract

In this article, we consider the compressible micropolar viscous flow in a bounded or unbounded domain Ω ⊆ ℝ3. We prove the existence of unique local strong solutions for large initial data satisfying some compatibility conditions. The key point here is that the initial density need not be positive and may vanish in an open set.

1 Introduction

Compressible micropolar viscous fluids study the viscous compressible fluids with randomly oriented (or spherical) particles suspended in the medium, when the deformation of fluid particles is ignored. The theory can help us consider some physical phenomena which cannot be treated by the compressible viscous baratropic flows, due to the effect of microparticles. The microstructure of the polar fluids is mechanically significant. The governing system of equations of compressible micropolar viscous fluids expresses the balance of mass, momentum, and moment of momentum see [1, 2], that is,
ρ t + div ( ρ u ) = 0 , ( ρ u ) t + div ( ρ u u ) + p ( ρ ) = ( μ + ζ ) Δ u + ( μ + λ - ζ ) div u + 2 ζ rot w , ( ρ w ) t + div ( ρ u w ) + 4 ζ w = μ Δ w + ( μ + λ ) div w + 2 ζ rot u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ1_HTML.gif
(1)

Here the density ρ = ρ(t, x), the velocity u = [u1(t, x), u2(t, x), u3(t, x)], the microrotational velocity w = [w1(t, x), w2(t, x), w3(t, x)], and the pressure p(ρ) = γ(a > 0, γ > 1) are functions of the time t ∈ (0, T) and the spatial coordinate xΩ, where Ω is either a bounded open subset in ℝ3 with smooth boundary or a usual unbounded domain such as the whole space ℝ3, the half space + 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq1_HTML.gif and an exterior domain with smooth boundary. μ, λ, μ', λ', and ζ are positive constants, which describe the viscosity of the fluids, satisfying the additional condition: μ + λ - ζ ≥ 0.

We prescribe the initial boundary value conditions:
( ρ , u , w ) ( 0 , x ) = ( ρ 0 , u 0 , w 0 ) in Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ2_HTML.gif
(2)
( u , w ) = ( 0 , 0 ) on ( 0 , T ) × Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ3_HTML.gif
(3)
( ρ , u , w ) ( t , x ) ( 0 , 0 , 0 ) , as x , ( t , x ) ( 0 , T ) × Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ4_HTML.gif
(4)

There are lots of literature on the well-posedness and ill-posedness of the incompressible micropolar viscous fluid, i.e., the system (1) with ρ ≡ Constant. Yamaguchi [3] considered the global strong solution in a bounded domain with small initial data. Lukaszewicz [4] showed the existence theorem for the incompressible micropolar viscous fluid with sufficiently regular initial data. And with the effect of Magnetic fields, the system of magneto-micropolar fluid, in [5], Yuan gave the local smooth solution without the smallness of the initial data; and also gave the blow-up criterion for the smooth solutions. Amirat and Hamdache [6] considered the global weak solutions with finite energy and establish the long-time behavior of the solution. And Amirat et al. [7] also proved the global in time weak solution with finite energy of an initial-boundary value problem and long-time behavior of such solution with the effect of magnetic field. We also refer the reader to [8] for local strong solution in bounded domain of ℝ3 and references therein.

For the compressible case, in the absence of vacuum, there also have been lots of studies on the full viscous compressible micropolar fluids (which include also the conservation law of energy) since Eringen [1]. The one-dimensional problem studied by Mujaković [912], and Dražić and Mujaković [13] and references therein. For the multidimensional case, we refer the readers to [2, 1417] and references therein. If the vacuum occur initially, Chen [18, 19] studied the global strong solutions of compressible micropolar viscous fluids in 1-D. Recently, Amirat and Hamdache [20] studied the micropolar fluids with the effect of magnetic field, they prove the global weak solution in a bounded domain in ℝ3 with initial vacuum.

Classical compressible viscous flows, i.e., w = 0 and ζ = 0 in (1), have been studied by many authors. In [21], Matsumura and Nishida considered the global smooth solutions under the condition that initial data close to a non-vacuum equilibrium. For the arbitrary initial data, the major breakthrough is due to Lions [22], where he established global existence of weak solutions for the whole space, periodic domain or bounded domains with Dirichlet boundary conditions if γ ≥ 9/5. The restriction on γ was improved to γ > 3/2 by Feireisl etal. in [23].

In [24], Choe and Kim established local in time strong solution of isentropic compressible fluids with initial density ρ0 may vanish in an open subset. After that, Cho and Kim [25] studied the local existence of strong solutions of viscous polytropic fluids with vacuum. Cho et al. [26] considered the unique solvability of the initial boundary value problems for compressible viscous fluids that the initial density need not be bounded away from zero.

Hoff and Serre [27] presented an example which showed the failure of continuous dependence of weak solution, so it should be noticed that contrary to the case of strong solutions, weak solutions with vacuum need not depend continuously on their initial data.

Before stating the main result, we introduce the notions used throughout this article. We denote
f d x = Ω f d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equa_HTML.gif
For 1 ≤ r ≤ ∞, we denote the standard homogeneous and inhomogeneous Sobolev spaces as follows:
L r = L r ( Ω ) , D k , r = { u L loc 1 ( Ω ) : k u L r < } , W k , r = L r D k , r , H k = W k , 2 , D k = D k , 2 , D 0 1 = { u L 6 : u L 2 < , and ( 3 ) or ( 4 ) holds } , H 0 1 = L 2 D 0 1 , u D k , r = k u L r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equb_HTML.gif
In spirit of [26], our aim of this article is to establish the local strong solution of (1)-(4). Assume the initial data (ρ0, w0, u0) satisfying the regularity
ρ 0 H 1 W 1 , q and ( w 0 , u 0 ) D 0 1 D 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ5_HTML.gif
(5)
where 3 < q < ∞, and the compatibilities
- ( μ + ζ ) Δ u 0 - ( μ + λ - ζ ) div u 0 - 2 ζ rot w 0 + p ( ρ 0 ) = ρ 0 1 / 2 g 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ6_HTML.gif
(6)
and
- μ Δ w 0 - ( λ + μ ) div w 0 + 4 ζ w 0 - 2 ζ rot u 0 = ρ 0 1 / 2 g 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ7_HTML.gif
(7)

for some (g1, g2) ∈ L2.

Now, we state our main result in this article:

Theorem 1 Assume the data ( ρ0, u0, w0) satisfy the regularity conditions (5) and the compatibility conditions (6) and (7).

Then there exists a time T * ∈ (0, T) and a unique strong solutions (ρ, u, w) to the initial boundary value problem (1)-(4) such that
ρ C ( [ 0 , T * ] ; H 1 W 1 , q 0 ) , ρ t C ( [ 0 , T * ] ; L 2 L q 0 ) ; ( u , w ) C ( [ 0 , T * ] ; D 0 1 D 2 ) L 2 ( 0 , T * ; D 2 , q 0 ) ; ( u t , w t ) L 2 ( [ 0 , T * ] ; D 0 1 ) and ( ρ u t , ρ w t ) L ( 0 , T * ; L 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equc_HTML.gif

where q0 = min{6, q}.

The article is organized as follow. In Section 2, we prove the global existence and regularity of the unique strong solutions to a linearized problem of the nonlinear problem (1)-(4). The result is used in Section 3 to construct approximate solutions to the original nonlinear problem. In Section 3, we derive some uniform bounds of the higher derivatives independent of the lower bound of the density. Moreover, we prove the convergence and obtain the local existence of strong solutions. In Section 4, we finish the proof of Theorem 1.

Remark 1 In this article, we use the following fact frequently later:
Ω rot w u d x = Ω rot u w d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equd_HTML.gif

under the boundary condition of (3) or (4).

2 Global existence for the linearized equations

In this section, we consider the following linearized system:
ρ t + div ( ρ v ) = 0 , ( ρ w ) t + div ( ρ v w ) + 4 ζ w = μ Δ w + ( μ + λ ) div w + 2 ζ rot v , ( ρ u ) t + div ( ρ v u ) + p ( ρ ) = ( μ + ζ ) Δ u + ( μ + λ - ζ ) div u + 2 ζ rot w , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ8_HTML.gif
(8)

where v is a known vector fields. If the initial density ρ0 is bounded away from zero, then we can apply standard arguments to prove the global existence of a unique strong solution to the initial boundary value problem (2)-(4) and (8), since the system can be uncoupled into a linear transport equation and two linear parabolic equations.

In this section, we prove the following existence result for the general case of nonnegative initial densities.

Theorem 2 Assume that the data (ρ0, u0, w0) satisfies the regularity conditions:
0 ρ 0 H 1 W 1 , q , ( u 0 , w 0 ) D 0 1 D 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ9_HTML.gif
(9)
for some 3 < q < ∞ and the compatibility conditions:
- μ Δ w 0 - ( μ + λ ) div w 0 + 2 ζ rot v 0 + 4 ζ w 0 = ρ 0 1 / 2 g 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ10_HTML.gif
(10)
and
- ( μ + ζ ) Δ u 0 - ( μ + λ - ζ ) div u 0 - 2 ζ rot w 0 + p ( ρ 0 ) = ρ 0 1 / 2 g 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ11_HTML.gif
(11)
for some (g1, g2) ∈ L2. If in addition, v satisfies the regularity conditions:
v L ( 0 , T ; D 0 1 D 2 ) L 2 ( 0 , T ; D 2 , q 0 ) and v t L 2 ( 0 , T ; D 0 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Eque_HTML.gif
where q0 = m in{6, q}. Then there exists a unique strong solution (ρ, u, w) to the initial boundary value problems (2)-(4) and (8) such that
ρ C ( [ 0 , T ] ; H 1 W 1 , q 0 ) , ρ t L ( 0 , T ; L 2 L q 0 ) , ( u , w ) C ( [ 0 , T ] ; D 0 1 D 2 ) L 2 ( 0 , T ; D 2 , q 0 ) , ( u t , w t ) L 2 ( 0 , T ; D 0 1 ) , ( ρ u t , ρ w t ) L ( 0 , T ; L 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ12_HTML.gif
(12)

Here, we emphasize that we focus on the bounded open domain Ω with smooth boundary condition. As for the unbounded domain, we can deal the same problem with the standard domain expansion technique that derived in [22]. We also refer the reader to [26]. The key of this technique is that the a priori estimates do not depend on the size of the domain. So, we here emphasize that the a priori estimates deduced in this section are independent of the size of the domain.

2.1 Existence of theorem 2

We begin with an existence result for the case of positive initial densities.

Lemma 1 Let Ω be a bounded domain in3with smooth boundary, and let (ρ0, u0, w0) be a given data satisfying the regularity condition (9)-(11). Assume further that v L ( 0 , T ; D 0 1 D 2 ) L 2 ( 0 , T ; D 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq2_HTML.gif, v t L 2 ( 0 , T ; D 0 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq3_HTML.gif, ρ0H2and ρ0≥ δ in Ω for some constant δ > 0. Then there exists a unique strong solution (ρ, u, w) to the initial boundary value problems (2), (3) and (8) such that
ρ C ( [ 0 , T ] ; H 2 ) , ( u , w ) C ( [ 0 , T ] ; D 0 1 D 2 ) L 2 ( 0 , T ; D 3 ) , ρ t C ( [ 0 , T ] ; H 1 ) , ( u t , w t ) L 2 ( 0 , T ; D 0 1 ) C ( [ 0 , T ] ; L 2 ) , and ρ > 0 o n [ 0 , T ] × Ω ̄ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ13_HTML.gif
(13)
Proof Due to the classical embedding result, that υC([0, T];H2). Hence the existence and regularity of a unique solution of the linearized continuity Equation (8)1 have been well-known. Moreover, the unique solution ρ can be expressed by:
ρ ( t , x ) = ρ 0 ( U ( 0 , t , x ) ) exp - 0 t div v ( s , U ( s , t , x ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ14_HTML.gif
(14)
where U(t, s, x) is the solution to
t U ( t , s , x ) = v ( t , U ( t , s , x ) ) , 0 t T , U ( s , s , x ) = x , 0 s T , x Ω ̄ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equf_HTML.gif
We refer the readers to [28, 29] for a detailed proof. As a consequence of (14) and Sobolev inequality, we have
ρ ( t , x ) δ exp - 0 T v ( s ) L d s > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ15_HTML.gif
(15)
for ( t , x ) [ 0 , T ] × Ω ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq4_HTML.gif. Hence the linearized moment of momentum Equation (8)2 can be written as a linear parabolic system
w t + v w + 4 ρ - 1 ζ w - ρ - 1 μ Δ w - ρ - 1 ( μ + λ ) div w - 2 ρ - 1 ζ rot v = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equg_HTML.gif
The existence and regularity of the unique solution w can be proved by applying classical methods, for instance, the method of continuity (see [29]). Similarly, the linearized momentum Equation (8)3 also as a linear parabolic system
u t + v u - ρ - 1 ( μ + ζ ) Δ u - ρ - 1 ( μ + λ - ζ ) div u - 2 ρ - 1 ζ rot w = - ρ - 1 p ( ρ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equh_HTML.gif

and can be solved using the same method.

Now we prove the existence of strong solutions. Then thanks to Lemma 1, there exists a unique strong solution (ρ, u, w) satisfying the regularity (13). To remove the additional hypotheses in Lemma 1, we will derive some uniform estimates independent of δ, v L 2 ( 0 , T ; D 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq5_HTML.gif, and ρ 0 D 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq6_HTML.gif.

First, we consider the solution ρ of the linearized continuity Equation (8)1. Since (8)1 is a linear transport equation, so we need to prove the estimates. Multiply (8)1 by ρ r -1(r = 2 or q0) and integrating (by parts) over Ω, we obtain:
d d t ρ r d x C v ρ r d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equi_HTML.gif
Then, using Sobolev inequality, we get
d d t ρ L r r C v W 1 , q 0 ρ L r r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ16_HTML.gif
(16)
Differentiating (8)1 with respect to x i , then multiplying the resultant equation by i ρ |∂ i ρ | r -2, i = 1, 2, 3 and then integrating over Ω, we have
d d t i ρ r d x C ( v ρ r + ρ ρ r - 1 2 v ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equj_HTML.gif
By virtue of Sobolev inequality, we get:
d d t ρ L r r C v H 1 D 1 , q 0 ρ H 1 W 1 , q 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ17_HTML.gif
(17)
Using (16) and (17) together with Gronwall's inequality, one yields:
sup 0 t T ρ ( t ) H 1 W 1 , q 0 ρ 0 H 1 W 1 , q 0 exp C 0 T v ( s ) H 1 D 1 , q 0 d s C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ18_HTML.gif
(18)
Since ρ t = -υ ·∇ρ - ρ div υ, p = p(ρ) and p(0) = 0, we can easily get
sup 0 t T ( ρ t L 2 L q 0 + p ( t ) H 1 W 1 , q 0 + p t ( t ) L 2 L q 0 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equk_HTML.gif
Then, we consider the solution w of the linearized Equation (8)2. Rewrite (8)2, with the help of (8)1, as:
ρ w t + ρ v w + 4 ζ w = μ Δ w + ( μ + λ ) div w + 2 ζ rot v . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ19_HTML.gif
(19)
Multiplying this equation by w t and integrating (by parts) over Ω, we have
ρ w t 2 d x + d d t μ 2 w 2 + μ + λ 2 div w 2 + 2 ζ w 2 d x = - ( ρ v w w t + 2 ζ rot v w t ) d x = - ( ρ v w w t + 2 ζ rot v t w ) d x + d d t 2 ζ rot v w d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ20_HTML.gif
(20)
Then, using Young's inequality, we obtain:
1 2 ρ w t 2 d x + d d t μ 2 w 2 + μ + λ 2 div w 2 + 2 ζ w 2 d x ( ρ v 2 w 2 + ζ v t 2 + ζ w 2 ) d x + d d t 2 ζ rot v w d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ21_HTML.gif
(21)
Integrating the above inequality over (0, t), we get:
0 t ρ w t L 2 2 d s + w L 2 2 + 2 ζ w L 2 2 C + C 0 t w L 2 2 d s + 0 t ρ v 2 w 2 d x d s + 2 ζ rot v w d x C + C 0 t ( w L 2 2 + w L 2 2 ) d s + ζ v L 2 2 + ζ w L 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ22_HTML.gif
(22)
Thus, we have
0 t ρ w t L 2 2 d s + w L 2 2 + w L 2 2 C 0 t ( w L 2 2 + w L 2 2 ) d s + C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equl_HTML.gif
Therefore, in view of Gronwall's inequality, we have:
0 T ρ w t L 2 2 d s + sup 0 t T ( w L 2 2 + w L 2 2 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ23_HTML.gif
(23)
To derive higher regularity estimates, we differentiate (19) with respect to t and obtain:
ρ w t t + ρ v w t + 4 ζ w t - μ Δ w t - ( μ + λ ) div w t = - ρ t w t - ρ t v w - ρ v t w + 2 ζ rot v t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equm_HTML.gif
Multiplying the above equation by w t , and integrating over Ω, we have:
1 2 d d t ρ w t L 2 2 + 4 ζ w t L 2 2 + μ w t L 2 2 ( ρ v w t w t + ρ t v w w t + ρ v t w w t + 2 ζ v t w t d x = i = 1 4 I i . ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ24_HTML.gif
(24)
Now, we estimate each term on the right-hand side of (24)
I 1 = ρ v w t w t d x ρ L 1 / 2 v L ρ w t L 2 w t L 2 C ρ L v L 2 ρ w t L 2 2 + ε w t L 2 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equn_HTML.gif
I 2 = ρ t v w w t d x v L ρ t L 3 w L 2 w t L 6 C v L ρ t L 2 α ρ t L q 0 1 - α w L 2 w t L 2 C v L 2 ρ t L 2 2 α ρ t L q 0 2 ( 1 - α ) w L 2 2 + ε w t L 2 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equo_HTML.gif
I 3 = ρ v t w w t d x ρ L 1 / 2 v t L 6 w L 2 ρ w t L 3 ρ L 1 / 2 v t L 2 w L 2 ρ w t L 2 1 / 2 ρ w t L 6 1 / 2 C ρ L 3 / 4 v t L 2 w L 2 ρ w t L 2 1 / 2 w t L 2 1 / 2 C ρ L 3 w L 2 4 ρ w t L 2 4 + C v t L 2 2 + ε w t L 2 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equp_HTML.gif
I 4 = 2 ζ v t w t d x ζ v t L 2 2 + ζ w t L 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equq_HTML.gif
Choosing ε= μ'/6, substitute the above estimates into (24), we can get
d d t ρ w t L 2 2 + μ 2 w t L 2 2 + 2 ζ w t L 2 2 C ( 1 + v t L 2 2 + ρ w t L 2 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ25_HTML.gif
(25)
Integrating over (τ, t) for some fixed τ ∈ (0, T), we deduce that
ρ w t L 2 2 + τ T ( w t L 2 2 + 2 ζ w t L 2 2 ) d s C + C ρ w t ( τ ) L 2 2 for τ t T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ26_HTML.gif
(26)
To estimate ρ w t τ L 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq7_HTML.gif, due to(19), we see
ρ w t 2 d x 4 ρ v 2 w 2 d x + ρ - 1 - μ Δ w - ( μ + λ ) div w - 2 ζ rot v + 4 ζ w 2 d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equr_HTML.gif
and thus we have
lim sup τ 0 ρ w t ( τ ) L 2 2 C ( 1 + C 1 ( ρ 0 , u 0 , w 0 ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equs_HTML.gif
where C 1 ( ρ 0 , u 0 , w 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq8_HTML.gif is defined by
C 1 ( ρ 0 , u 0 , w 0 ) = ρ 0 - 1 - μ Δ w 0 - ( μ + λ ) div w 0 - 2 ζ rot v 0 + 4 ζ w 0 2 d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equt_HTML.gif
Therefore, letting τ → 0 in (26), we conclude that
sup 0 t T ρ w t ( t ) L 2 2 + 0 T ( w t L 2 2 + w t L 2 2 ) d t C ( 1 + C 1 ( ρ 0 , u 0 , w 0 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ27_HTML.gif
(27)
To obtain further estimates, observe that since for each t ∈ [0, T], w = w ( t ) D 0 1 D 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq9_HTML.gif is a solution of the elliptic system:
μ Δ w + ( μ + λ ) div w - 4 ζ w = F 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equu_HTML.gif
where F 1 = - ρ w t - ρ v w + 2 ζ rot v L 2 L q 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq10_HTML.gif. And then
w D 2 , r C - ρ w t - ρ v w + 2 ζ rot v L r + C w D 1 , r , r = 2 , q 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ28_HTML.gif
(28)
Therefore, using the previous estimates, we can deduce from (28) that
sup 0 t T w ( t ) D 2 2 + 0 T w ( t ) D 2 , q 0 2 d t C ( 1 + C 1 ( ρ 0 , u 0 , w 0 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ29_HTML.gif
(29)
Now, we consider the solution u of the linearized Equation (8)3. Rewrite (8)3 as
ρ u t + ρ v u + p ( ρ ) = ( μ + ζ ) Δ u + ( μ + λ - ζ ) div u + 2 ζ rot w . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ30_HTML.gif
(30)
Multiplying this equation by u t and integrating over Ω, we have
ρ u t 2 d x + d d t μ + ζ 2 u 2 + μ + λ - ζ 2 div u 2 - p ( ρ ) - 2 ζ rot w u d x = ( ρ v u u t - p ( ρ ) t div u - 2 ζ rot w t u ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equv_HTML.gif
Then, using Young's inequality, we obtain:
1 2 ρ u t 2 d x + d d t μ + ζ 2 u 2 + μ + λ - ζ 2 div u 2 - p ( ρ ) - 2 ζ rot w u d x 1 2 ρ v 2 u 2 + p ( ρ ) t u + ζ w t 2 + ζ u 2 d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equw_HTML.gif
Integrating the above inequality over (0, t) using Young's inequality, we get
0 t ρ u t L 2 2 d s + u L 2 2 C 1 + 0 t u L 2 2 d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equx_HTML.gif
Therefore, in view of Gronwall's inequality, we have
0 T ρ u t L 2 2 d t + sup 0 t T u ( t ) D 0 1 2 C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ31_HTML.gif
(31)
To derive higher regularity estimates, differentiate (30) with respect to t and obtain:
ρ u t t + ρ v u t + p ( ρ ) t - ( μ + ζ ) Δ u t - ( μ + λ - ζ ) div u t = - ρ t u t - ρ t v u - ρ v t u + 2 ζ rot w t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equy_HTML.gif
Multiplying the above equation by u t , integrating (by parts) over Ω, and using (8)1, we obtain:
1 2 d d t ρ u t 2 d x + ( u + ζ ) u t 2 + ( μ + λ - ζ ) div u t 2 d x ρ v u t u t d x + ρ t v u u t d x + ρ v t u u t d x + p ( ρ ) t u t d x + 2 ζ w t u t d x = i = 1 5 I I i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ32_HTML.gif
(32)
Using Sobolev inequality, interpolation inequality and Young's inequality, we obtain:
I I 1 = ρ v u t u t d x ρ L 1 / 2 v L ρ u t L 2 u t L 2 C ρ L v L 2 ρ u t L 2 2 + ε u t L 2 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equz_HTML.gif
I I 2 = ρ t v u u t d x v L ρ t L 3 u L 2 u t L 6 v L ρ t L 2 α ρ t L q 0 1 - α u L 2 u t L 2 C v L 2 ρ t L 2 2 α ρ t L q 0 2 ( 1 - α ) u L 2 2 + ε u t L 2 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equaa_HTML.gif
I I 3 = ρ v t u u t d x ρ L 1 / 2 u t L 6 u L 2 ρ u t L 3 C ρ L 1 / 2 v t L 2 u L 2 ρ u t L 2 1 / 2 ρ u t L 6 1 / 2 C ρ L 3 / 4 v t L 2 u L 2 ρ u t L 2 1 / 2 u t L 2 1 / 2 C ρ L 3 u L 2 4 ρ u t L 2 2 + C v t L 2 2 + ε u t L 2 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equab_HTML.gif
I I 4 = p ( ρ ) t u t d x C p ( ρ ) t L 2 2 + ε u t L 2 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equac_HTML.gif
I I 5 = 2 ζ w t u t d x ζ w t L 2 2 + ζ u t L 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equad_HTML.gif
Substituting the above estimates into (32), we can easily show that:
d d t ρ u t L 2 2 + u t L 2 2 C ( 1 + v t L 2 2 + ρ u t L 2 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ33_HTML.gif
(33)
Now, for fixed τ ∈ (0, T). Since the right-hand side of (33) is integrable in (0, T), we deduce that
ρ u t ( t ) L 2 2 + τ t u t L 2 2 d s C + C ρ u t ( τ ) L 2 2 for τ t T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ34_HTML.gif
(34)
To estimate ρ u t ( τ ) L 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq11_HTML.gif from (8)3, we see that
ρ u t ( τ ) L 2 2 4 ρ v 2 u 2 d x ρ - 1 - ( μ + λ ) Δ u - ( μ + λ - ζ ) Δ div u - 2 ζ rot w + p ( ρ ) 2 d x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equae_HTML.gif
and thus
lim sup τ 0 ρ u t ( τ ) L 2 2 C ( 1 + C 2 ( ρ 0 , u 0 , w 0 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equaf_HTML.gif
where C 2 ( ρ 0 , u 0 , w 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq12_HTML.gif is defined by
C 2 ( ρ 0 , u 0 , w 0 ) = ρ 0 - 1 - ( μ + λ ) Δ u 0 - ( μ + λ - ζ ) div u 0 - 2 ζ rot w 0 + p ( ρ 0 ) 2 d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equag_HTML.gif
Therefore, letting τ → 0 in (34), we conclude that
sup 0 t T ρ u t L 2 2 + 0 T u t L 2 2 d t C ( 1 + C 2 ( ρ 0 , u 0 , w 0 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ35_HTML.gif
(35)
To obtain further estimates, observe that for each t ∈ [0, T], u = u ( t ) D 0 1 D 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq13_HTML.gif is a solution of the following elliptic system:
- μ + λ Δ u - ( μ + λ - ζ ) div u = F 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equah_HTML.gif
where F 2 = - ρ u t - ρ v u + 2 ζ rot w - p ( ρ ) L 2 L q 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq14_HTML.gif. It follows from the elliptic regularity theory, we get
u D 2 , r C - ρ u t - ρ v u + 2 ζ rot w - p ( ρ ) L r + C u D 1 , r , r = 2 , q 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ36_HTML.gif
(36)
Therefore, using the previous estimates, we get from (36) that
sup 0 t T u ( t ) D 2 2 + 0 T u ( t ) D 2 , q 0 2 d t C ( 1 + C 2 ( ρ 0 , u 0 , w 0 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ37_HTML.gif
(37)
Since the estimates we have derived, we prove the existence result. First, we consider for the case of bounded domain. Using standard regularization techniques, we choose p δ = p δ (⋅) and υ δ , 0 < δ ≪ 1, so that
p δ ( ) C 2 [ 0 , ) , p δ p in C 1 [ 0 , ) v δ L ( 0 , T ; D 0 1 D 2 ) L 2 ( 0 , T ; D 3 ) , v t δ L 2 ( 0 , T ; D 0 1 ) , ( v δ , v t δ ) ( v , v t ) in L ( 0 , T ; D 0 1 D 2 ) L 2 ( 0 , T ; D 2 , q 0 ) × L 2 ( 0 , T ; D 0 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equai_HTML.gif
Then, for each δ ∈ (0, 1), let ρ 0 δ = ρ 0 + δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq15_HTML.gif and let ( u 0 δ , w 0 δ ) D 0 1 D 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq16_HTML.gif is the solution to the boundary value problem:
- ( μ + ζ ) Δ u 0 δ - ( μ + λ - ζ ) div u 0 δ - 2 ζ rot w 0 δ = - p δ ( ρ 0 δ ) + ( ρ 0 δ ) 1 / 2 g 1 , - μ Δ w 0 δ - ( μ + λ ) div w 0 δ + 4 ζ w 0 δ = ( ρ 0 δ ) 1 / 2 g 2 , ( u 0 δ , w 0 δ ) = 0 , on Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equaj_HTML.gif
It follows from the elliptic regularity result that ( u 0 δ , w 0 δ ) ( u 0 , w 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq17_HTML.gif in D 0 1 D 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq18_HTML.gif as δ → 0. Hence, if we denote by (ρ δ , u δ , w δ ) the solution of (8) with the initial data ( ρ 0 δ , u 0 δ , w 0 δ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq19_HTML.gif and (p, v) replaced by (p δ , v δ ), it satisfies the estimates (35), (37), (31), (23), (18), (27), (29), where C 1 ( ρ 0 , u 0 , w 0 ) = g 1 L 2 2 C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq20_HTML.gif, C 2 ( ρ 0 , u 0 , w 0 ) = g 2 L 2 2 C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq21_HTML.gif. Therefore, we conclude that a subsequence of solutions (ρ δ , u δ , w δ ) converges to a limit (ρ, u, w) in a weak sense. Then it can easy to show that (ρ, u, w) is a weak solution to the original problem (8). Moreover, due to the lower semi-continuity of various norms, we have the following regularity estimates for (ρ, u, w):
ess  sup 0 t T ρ H 1 W 1 , q 0 + ρ t L 2 L q 0 + ( u , w ) D 0 1 D 2 + ( ρ u t , ρ w t ) L 2 + 0 T ( u , w ) D 2 , q 0 2 + ( u t , w t ) D 0 1 d t C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ38_HTML.gif
(38)

2.2 Continuity and uniqueness

Now, turn our attention to the continuity of the solution (ρ, u, w). The continuity of ρ can be proved by a standard argument from the theory of hyperbolic equations. Since ρ satisfies the regularity (38), it follows from a result of DiPerna and Lions [30] and classical embedding results that (See [31]):
ρ C ( [ 0 , T ] ; L 2 L q 0 ) C ( [ 0 , T ] ; H 1 W 1 , q 0 - weak ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equak_HTML.gif
To show the strong continuity in H 1 W 1 , q 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq22_HTML.gif, observe from (17), (18) and i = 1, 2, 3,
i ρ ( t ) L r r i ρ ( 0 ) L r r + C 0 t v ( s ) H 1 D 1 , q 0 d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equal_HTML.gif
and thus lim sup t 0 + i ρ ( t ) L r r i ρ ( 0 ) L r r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq23_HTML.gif. Hence it follows from a well-known criterion on the strong conver-gence for the space L r (See [32]) that
lim t 0 + i ρ ( t ) - i ρ ( 0 ) L r r = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equam_HTML.gif
Therefore, the continuity of ∇ρ in L r (r = 2, q0) follows from the result and the observation that for each fixed t0 ∈ [0, t], the function ρ ̃ = ρ ̃ ( t , x ) = ρ ̃ ( ± t + t 0 , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq24_HTML.gif is a unique strong solution to the similar initial value problem:
ρ ̃ t + div ( ρ ̃ ) = 0 , and ρ ̃ ( 0 ) = ρ ( t 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equan_HTML.gif

where υ ̃ = υ ̃ ( t , x ) = ± υ ( ± t + t 0 , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq25_HTML.gif.

To show the continuity of w, we first observe that
w , v C ( [ 0 , T ] ; D 0 1 ) C ( [ 0 , T ] ; D 2 weak) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equao_HTML.gif
We now prove the continuity of ρw t in L2. For a.e. t ∈ (0, T) and φ H 0 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq26_HTML.gif the from (19), we have
( ρ w t , φ ) L 2 = ( - ρ v w - 4 ζ w + μ w + ( μ + λ ) div w + 2 ζ rot v , φ ) L 2 = ( - ρ v w - 4 ζ w + 2 ζ rot v , φ ) L 2 - μ ( w , φ ) L 2 - ( μ + λ ) ( div w , div φ ) L 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equap_HTML.gif
and thus
d d t ( ρ w t , φ ) L 2 = ( ( - ρ v w - 4 ζ w + 2 ζ rot v ) t , φ ) L 2 - μ ( w t , φ ) L 2 - ( μ + λ ) ( div w t , div φ ) L 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ39_HTML.gif
(39)
Using the regularity (38) of (ρ, w), we show that the right-hand side of (39) is bounded above by A 1 ( t ) φ H 0 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq27_HTML.gif for some positive function A1(t) ∈ L2(0, T). Hence it follows, from the well-known result (see [31]) that (ρw t ) t L2(0, T; H-1) and d d t ( ρ w t , φ ) L 2 = ( ρ w t ) t , φ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq28_HTML.gif for all φ H 0 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq26_HTML.gif where 〈⋅, ⋅〉 denotes the dual pairing of H-1 and H 0 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq29_HTML.gif. Then since ρ w t L 2 ( 0 , T ; H 0 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq30_HTML.gif, it follows from a standard embedding result ρw t C([0, T]; L2). Therefore, we conclude that for each t∈[0, T], w = w ( t , x ) D 0 1 D 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq31_HTML.gif is a solution of the elliptic system:
- μ Δ w - ( μ + λ ) div w = G 1 - ρ v w , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equaq_HTML.gif
where G1 = -ρw t -4ζwC([0, T]; L2). Now we turn to show that wC([0, T]; D2). In view of the elliptic regularity estimate (36), we obtain
w ( t ) - w ( s ) D 2 C w ( t ) - w ( s ) D 0 1 + C G 1 ( t ) - G 1 ( s ) L 2 + C ρ v w ( t ) - ρ v w ( s ) L 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ40_HTML.gif
(40)
Using the estimate (36), we obtain:
C ρ v w ( t ) - ρ v w ( s ) L 2 C ( ρ ( t ) - ρ ( s ) ) v ( t ) w ( t ) L 2 + C ρ ( s ) ( v ( t ) - v ( s ) ) w ( t ) L 2 + C ρ ( s ) v ( s ) ( w ( t ) - w ( s ) ) L 2 C ρ ( t ) - ρ ( s ) L v ( t ) L 6 w ( t ) L 3 + C ρ ( s ) L v ( t ) - v ( s ) L 6 w ( t ) L 3 + C ρ ( s ) L v ( s ) L 6 w ( t ) - w ( s ) L 3 C ρ ( t ) - ρ ( s ) L v ( t ) L 2 w ( t ) L 2 1 / 2 2 w ( t ) L 2 1 / 2 + C ρ ( s ) L v ( t ) - v ( s ) L 2 w ( t ) L 2 1 / 2 2 w ( t ) L 2 1 / 2 + C ρ ( s ) L v ( s ) L 2 w ( t ) - w ( s ) L 2 1 / 2 2 w ( t ) - 2 w ( s ) L 2 1 / 2 C ( ρ ( t ) - ρ ( s ) L + v ( t ) - v ( s ) L 2 + w ( t ) - w ( s ) L 2 ) + 1 2 w ( t ) - w ( s ) D 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ41_HTML.gif
(41)
Substituting this into (40), we conclude that
w ( t ) - w ( s ) D 2 0 as t s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equar_HTML.gif
The continuity of ρu t in L2 is similar as the proof of ρw t . From (30), for a.e. t ∈ (0, T), and all ψ H 0 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq32_HTML.gif, we have
( ρ u t , ψ ) L 2 = ( - ρ u u + ( μ + λ ) Δ u + ( μ + λ - ζ ) div u + 2 ζ rot w - p ( ρ ) , ψ ) L 2 = ( - ρ u u + 2 ζ rot w , ψ ) L 2 - ( ( μ + ζ ) u , ψ ) L 2 - ( ( μ + λ - ζ ) div u - p ( ρ ) , div ψ ) L 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equas_HTML.gif
and thus
d d t ( ρ u t , ψ ) L 2 = ( ( ρ u u + 2 ζ r o t w ) t , ψ ) L 2 ( ( μ + ζ ) u t , ψ ) L 2 ( ( μ + λ ζ ) d i v u t p ( ρ ) t , d i v ψ ) L 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ42_HTML.gif
(42)
Using the regularity (38) of (ρ, u, w), we show the right-hand side of (42) is bounded above by A 2 ( t ) ψ H 0 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq33_HTML.gif for some positive function A2(t) ∈ L2(0, T). Hence by the similar argument of ρw t , we see that (ρu t ) t L2(0, T; H-1), and then ρu t C([0, T]; L2). Therefore, we conclude that for each t [ 0 , T ] , u = u ( t ) D 0 1 D 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq34_HTML.gif is a solution of the elliptic system:
- ( μ + λ ) Δ u - ( μ + λ - ζ ) div u = G 2 - ρ v u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equat_HTML.gif

where G2 = -ρu t - ∇p(ρ) + 2ζ rot wC([0, T]; L2).

Now, we will show that uC([0, T]; D2). In view of the elliptic regularity estimate (36), we have
u ( t ) - u ( s ) D 2 C u ( t ) - u ( s ) L 2 + C G 2 ( t ) - G 2 ( s ) L 2 + C ρ v u ( t ) - ρ v u ( s ) L 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equ43_HTML.gif
(43)
To estimate the third term in the right-hand side of (43), using the estimate (38), similarly as (41), we get:
C ρ v u ( t ) - ρ v u ( s ) L 2 C ρ ( t ) - ρ ( s ) L + v ( t ) - v ( s ) L 2 + u ( t ) - u ( s ) L 2 + 1 2 u ( t ) - u ( s ) D 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equau_HTML.gif

Substituting this into (43), we conclude the continuity of u in D2. This completes the proof of the continuity.

Finally, we prove the uniqueness of solutions satisfying the regularity (38). Let (ρ1, u1, w1) and (ρ2, u2, w2) be two strong solutions to the problem (8) and (2)-(4). Denote
ρ ̄ = ρ 1 - ρ 2 , ū = u 1 - u 2 , w ̄ = w 1 - w 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equav_HTML.gif
Then, it follows from (8)1 that
d d t ρ ̄ 2 d x div v ρ ̄ 2 d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equaw_HTML.gif
Since v L 2 ( 0 , T ; W 1 , q 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq35_HTML.gif and ρ ̄ ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq36_HTML.gif, we conclude from Gronwall's inequality that ρ ̄ L 2 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq37_HTML.gif, i.e., ρ1 = ρ2 in (0, T) × Ω. Next, we choose a cut-off function φ C c ( 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq38_HTML.gif such that
φ ( x ) = 1 , if x 1 , 0 , if x 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equax_HTML.gif
Define φ R (x) = φ(x/R) for x ∈ ℝ3. From (8)2 and using the uniqueness of ρ, we deduce that
ρ 1 w ̄ t + ρ 1 u w ̄ + 4 ζ w ̄ = μ Δ w ̄ + ( μ + λ ) div w ̄ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equay_HTML.gif
Then multiplying the above equation by φ R 2 w ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq39_HTML.gif, integrating over (0, T) × Ω, and letting R → ∞, we easily get
1 2 ρ 1 w ̄ 2 ( t ) d x + 0 t ( μ w ̄ 2 + ( μ + λ ) div w ̄ 2 + 4 ζ w ̄ 2 ) d x d s = 1 2 ρ 1 w ̄ 2 ( 0 ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equaz_HTML.gif

Hence, we deduce that ρ 1 w ̄ L 2 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq40_HTML.gif, w ̄ L 2 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq41_HTML.gif and w ̄ L 2 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq42_HTML.gif in (0, T), due to w ̄ ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq43_HTML.gif. Therefore, we conclude that w ̄ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq44_HTML.gif in (0, T) × Ω.

Similarly, from the uniqueness of ρ, and (8)3, we get
ρ 1 ū t + ρ 1 v ū = ( μ + ζ ) Δ ū + ( μ + λ - ζ ) div ū . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equba_HTML.gif
Then multiplying it by φ R 2 ū http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq45_HTML.gif integrating over (0, T) × Ω, and letting R → ∞, we obtain
1 2 ρ 1 ū 2 ( t ) d x + 0 t ( μ + ζ ) ū 2 + ( μ + λ - ζ ) div ū 2 d x d s = 1 2 ρ 1 ū 2 ( 0 ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equbb_HTML.gif

Due to ū(0) = 0, we get ρ 1 ū L 2 = 0 , ū L 2 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_IEq46_HTML.gif in (0, T). Then ū = 0 in (0, T) × Ω.

This completes the proof of the Theorem 2.

3 A local existence result for positive densities

In this section, we assume also that Ω is a bounded domain in ℝ3 with smooth boundary and prove a local (in time) existence result on strong solutions with positive densities to the original nonlinear problem (1)-(4).

Proposition 1 Assume that p = γ (a > 0, γ > 1), and the data (ρ0, u0, w0) satisfies the regularity conditions:
ρ 0 H 1 W 1 , q , ( u 0 , w 0 ) D 0 1 D 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equbc_HTML.gif
for some q with 3 < q < ∞ and compatibilities (6)-(7). Assume further that ρ0≥ δ in Ω for some constant δ > 0. Then there exist a time T * ∈ (0, T) and a unique strong solution (ρ, u, w) to the nonlinear problem (1)-(3) such that
ρ C ( [ 0 , T ] ; H 1 W 1 , q 0 ) , ( u , w ) C ( [ 0 , T * ] ; D 0 1 D 2 ) L 2 ( 0 , T * ; D 2 , q 0 ) , ρ t C ( [ 0 , T * ] ; L 2 L q 0 ) , ( u t , w t ) L 2 ( 0 , T * ; D 0 1 ) , ( ρ u t , ρ w t ) L ( 0 , T * ; L 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-32/MediaObjects/13661_2011_Article_154_Equbd_HTML.gif
where q 0 = min(6, q). Furthermore, we have the following estimates:
ess  sup 0 < t < T * ρ ( t ) H 1 W 1 , q 0 + ρ t ( t ) L 2 L q 0 + ( u ( t ) , w ( t ) ) D 0 1 D 2 + ( ρ