1687-2770-2012-321687-2770 Research <p>Unique solvability of compressible micropolar viscous fluids</p> ChenMingtaomtchen@sdu.edu.cn

School of Mathematics and Statistics, Shandong University at Weihai, Weihai 264209, PR China

 Boundary Value Problems 1687-2770 2012 2012 1 32 http://www.boundaryvalueproblems.com/content/2012/1/32 10.1186/1687-2770-2012-32 412201120320122032012 2012Chen; licensee Springer.This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 .

Here the density ρ = ρ(t, x), the velocity u = [u 1(t, x), u 2(t, x), u 3(t, x)], the microrotational velocity w = [w 1(t, x), w 2(t, x), w 3(t, x)], and the pressure p(ρ) = aρ γ(a > 0, γ > 1) are functions of the time t ∈ (0, T) and the spatial coordinate x ∈ Ω, 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 ℝ + 3 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 ,

( u , w ) = ( 0 , 0 ) on ( 0 , T ) × ∂ Ω ,

( ρ , u , w ) ( t , x ) → ( 0 , 0 , 0 ) , as x → ∞ , ( t , x ) ∈ ( 0 , T ) × Ω .

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 9 10 11 12 , and Dražić and Mujaković 13 and references therein. For the multidimensional case, we refer the readers to 2 14 15 16 17 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 ρ 0 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 .

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 .

In spirit of 26 , our aim of this article is to establish the local strong solution of (1)-(4). Assume the initial data (ρ 0, w 0, u 0) satisfying the regularity

ρ 0 ∈ H 1 ∩ W 1 , q and ( w 0 , u 0 ) ∈ D 0 1 ∩ D 2 ,

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 ,

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 ,

for some (g 1, g 2) ∈ L 2.

Now, we state our main result in this article:

Theorem 1 Assume the data (ρ 0, u 0, w 0) 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 ) ,

where q 0 = 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 ,

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 ,

where v is a known vector fields. If the initial density ρ 0 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 (ρ 0, u 0, w 0) satisfies the regularity conditions:

0 ≤ ρ 0 ∈ H 1 ∩ W 1 , q , ( u 0 , w 0 ) ∈ D 0 1 ∩ D 2 ,

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 ,

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 ,

for some (g 1, g 2) ∈ L 2. 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 ) ,

where q 0 = min{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 ) .

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 in ℝ3 with smooth boundary, and let (ρ 0, u 0, w 0) 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 ) , v t ∈ L 2 ( 0 , T ; D 0 1 ) , ρ 0 ∈ H 2 and ρ 0 ≥ 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 ̄ .

Proof Due to the classical embedding result, that υ ∈ C([0, T];H 2). 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 ,

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 ̄ .

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 ,

for ( t , x ) ∈ [ 0 , T ] × Ω ̄ . 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 .

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 ) ,

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 ) , and ρ 0 D 2 .

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 ρr -1(r = 2 or q 0) and integrating (by parts) over Ω, we obtain:

d d t ∫ ρ r d x ≤ C ∫ ∇ v ρ r d x .

Then, using Sobolev inequality, we get

d d t ρ L r r ≤ C ∇ v W 1 , q 0 ρ L r r .

Differentiating (8)1 with respect to xi , 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 .

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 .

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 .

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 .

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 .

Multiplying this equation by wt 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 .

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 .

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 .

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 .

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 .

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

Multiplying the above equation by wt , 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 . )

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 ;

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 ;

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 ;

I 4 = 2 ζ ∫ ∇ v t w t d x ≤ ζ ∇ v t L 2 2 + ζ w t L 2 2 .

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 ) .

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 .

To estimate ρ w t τ L 2 2 , 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 ,

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 ) ) ,

where C 1 ( ρ 0 , u 0 , w 0 ) 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 .

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 ) ) .

To obtain further estimates, observe that since for each t ∈ [0, T], w = w ( t ) ∈ D 0 1 ∩ D 2 is a solution of the elliptic system:

μ ′ Δ w + ( μ ′ + λ ′ ) ∇ div w - 4 ζ w = F 1 ,

where F 1 = - ρ w t - ρ v ⋅ w + 2 ζ rot v ∈ L 2 ∩ L q 0 . 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 .

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 ) ) .

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 .

Multiplying this equation by ut 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 .

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 .

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 .

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 .

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 .

Multiplying the above equation by ut , 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 .

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 ;

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 ;

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 ;

I I 4 = ∫ p ( ρ ) t ∇ u t d x ≤ C p ( ρ ) t L 2 2 + ε ∇ u t L 2 2 ;

I I 5 = 2 ζ ∫ w t ∇ u t d x ≤ ζ w t L 2 2 + ζ ∇ u t L 2 2 .

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 ) .

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 .

To estimate ρ u t ( τ ) L 2 2 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

and thus

lim sup τ → 0 ρ u t ( τ ) L 2 2 ≤ C ( 1 + C 2 ( ρ 0 , u 0 , w 0 ) )

where C 2 ( ρ 0 , u 0 , w 0 ) 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 .

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 ) ) .

To obtain further estimates, observe that for each t ∈ [0, T], u = u ( t ) ∈ D 0 1 ∩ D 2 is a solution of the following elliptic system:

- μ + λ Δ u - ( μ + λ - ζ ) ∇ div u = F 2

where F 2 = - ρ u t - ρ v ⋅ ∇ u + 2 ζ rot w - ∇ p ( ρ ) ∈ L 2 ∩ L q 0 . 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 .

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 ) ) .

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 pδ = pδ (⋅) 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 ) .

Then, for each δ ∈ (0, 1), let ρ 0 δ = ρ 0 + δ and let ( u 0 δ , w 0 δ ) ∈ D 0 1 ∩ D 2 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 .

It follows from the elliptic regularity result that ( u 0 δ , w 0 δ ) → ( u 0 , w 0 ) in D 0 1 ∩ D 2 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 ) and (p, v) replaced by (pδ, vδ ), 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 , C 2 ( ρ 0 , u 0 , w 0 ) = g 2 L 2 2 ≤ C . 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 .

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 ) .

To show the strong continuity in H 1 ∩ W 1 , q 0 , 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

and thus lim sup t → 0 + ∂ i ρ ( t ) L r r ≤ ∂ i ρ ( 0 ) L r r . Hence it follows from a well-known criterion on the strong conver-gence for the space Lr (See 32 ) that

lim t → 0 + ∂ i ρ ( t ) - ∂ i ρ ( 0 ) L r r = 0 .

Therefore, the continuity of ∇ρ in Lr(r = 2, q 0) follows from the result and the observation that for each fixed t 0 ∈ [0, t], the function ρ ̃ = ρ ̃ ( t , x ) = ρ ̃ ( ± t + t 0 , x ) is a unique strong solution to the similar initial value problem:

ρ ̃ t + div ( ρ ̃ ṽ ) = 0 , and ρ ̃ ( 0 ) = ρ ( t 0 ) ,

where υ ̃ = υ ̃ ( t , x ) = ± υ ( ± t + t 0 , x ) .

To show the continuity of w, we first observe that

w , v ∈ C ( [ 0 , T ] ; D 0 1 ) ∩ C ( [ 0 , T ] ; D 2 − weak) .

We now prove the continuity of ρwt in L 2. For a.e. t ∈ (0, T) and φ ∈ H 0 1 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 ,

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 .

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 for some positive function A 1(t) ∈ L 2(0, T). Hence it follows, from the well-known result (see 31 ) that (ρwt ) t ∈ L 2(0, T; H -1) and d d t ( ρ w t , φ ) L 2 = ( ρ w t ) t , φ for all φ ∈ H 0 1 where 〈⋅, ⋅〉 denotes the dual pairing of H -1 and H 0 1 . Then since ρ w t ∈ L 2 ( 0 , T ; H 0 1 ) , it follows from a standard embedding result ρwt ∈ C([0, T]; L 2). Therefore, we conclude that for each t∈[0, T], w = w ( t , x ) ∈ D 0 1 ∩ D 2 is a solution of the elliptic system:

- μ ′ Δ w - ( μ ′ + λ ′ ) ∇ div w = G 1 - ρ v ⋅ ∇ w ,

where G 1 = -ρwt -4ζw ∈ C([0, T]; L 2). Now we turn to show that w ∈ C([0, T]; D 2). 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 .

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 .

Substituting this into (40), we conclude that

w ( t ) - w ( s ) D 2 → 0 as t → s .

The continuity of ρut in L 2 is similar as the proof of ρwt . From (30), for a.e. t ∈ (0, T), and all ψ ∈ H 0 1 , 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

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 .

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 for some positive function A 2(t) ∈ L 2(0, T). Hence by the similar argument of ρwt , we see that (ρut ) t ∈ L 2(0, T; H -1), and then ρut ∈ C([0, T]; L 2). Therefore, we conclude that for each t ∈ [ 0 , T ] , u = u ( t ) ∈ D 0 1 ∩ D 2 is a solution of the elliptic system:

- ( μ + λ ) Δ u - ( μ + λ - ζ ) ∇ div u = G 2 - ρ v ⋅ ∇ u ,

where G 2 = -ρu t - ∇p(ρ) + 2ζ rot w ∈ C([0, T]; L 2).

Now, we will show that u ∈ C([0, T]; D 2). 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 .

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 .

Substituting this into (43), we conclude the continuity of u in D 2. This completes the proof of the continuity.

Finally, we prove the uniqueness of solutions satisfying the regularity (38). Let (ρ 1, u 1, w 1) and (ρ 2, u 2, w 2) 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 .

Then, it follows from (8)1 that

d d t ∫ ρ ̄ 2 d x ≤ ∫ div v ρ ̄ 2 d x .

Since ∇ v ∈ L 2 ( 0 , T ; W 1 , q 0 ) and ρ ̄ ( 0 ) = 0 , we conclude from Gronwall's inequality that ρ ̄ L 2 = 0 , i.e., ρ 1 = ρ 2 in (0, T) × Ω. Next, we choose a cut-off function φ ∈ C c ∞ ( ℝ 3 ) such that

φ ( x ) = 1 , if x ≤ 1 , 0 , if x ≥ 2 .

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 ̄ .

Then multiplying the above equation by φ R 2 w ̄ , 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 .

Hence, we deduce that ρ 1 w ̄ L 2 = 0 , w ̄ L 2 = 0 and ∇ w ̄ L 2 = 0 in (0, T), due to w ̄ ( 0 ) = 0 . Therefore, we conclude that w ̄ = 0 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 .

Then multiplying it by φ R 2 ū 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 .

Due to ū(0) = 0, we get ρ 1 ū L 2 = 0 , ∇ ū L 2 = 0 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 = aρ γ (a > 0, γ > 1), and the data (ρ 0, u 0, w 0) satisfies the regularity conditions:

ρ 0 ∈ H 1 ∩ W 1 , q , ( u 0 , w 0 ) ∈ D 0 1 ∩ D 2 ,

for some q with 3 < q < ∞ and compatibilities (6)-(7). Assume further that ρ 0 ≥ 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 ) ,

where q0 = 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 u t , ρ w t ) L 2 + ∫ 0 T * ( u ( t ) , w ( t ) ) D 2 , q 0 2 + ( u t , w t ) D 0 1 2 d t ≤ C .

The constant C and the local time T * in (44) are independent of δ.

To prove the proposition, we first construct approximate solutions, inductively, as follows:

- first define u 0 = 0, and

- assume that uk -1 was defined for k ≥ 1, let (ρ k , u k , w k ) be the unique solution to the following initial boundary value problem:

ρ t k + u k - 1 ⋅ ∇ ρ k + ρ k div u k - 1 = 0 ,

ρ k w t k + ρ k u k - 1 ⋅ ∇ w k + 4 ζ w k = μ ′ Δ w k + ( μ ′ + λ ′ ) ∇ div w k + 2 ζ rot u k - 1 ,

ρ k u t k + ρ k u k - 1 ⋅ ∇ u k + ∇ p k = ( μ + ζ ) Δ u k + ( μ + λ - ζ ) ∇ div u k + 2 ζ rot w k ,

ρ | t = 0 = ρ 0 ,   u k | t = 0 = u 0 ,   w k | t = 0 = w 0 ,   u k | ∂ Ω = w k | ∂ Ω = 0 ,

ρ k ( t , x ) → 0 , u k ( t , x ) → 0 , w k ( t , x ) → 0 as x → ∞ , ( t , x ) ∈ ( 0 , T ) × Ω .

The existence of a global strong solution (ρ k , u k , w k ) with the regularity (12) to the linearized problem (45)-(49) was proved in the previous section.

From now on, we derive uniform bounds on the approximate solutions and then prove the convergence of the approximate solutions to a strong solutions of the original nonlinear problem.

3.1 Uniform bounds

Let K ≥ 1 be a fixed large integer, and let us introduce an auxiliary function Φ K (t), defined by:

Φ K ( t ) = max 1 ≤ k ≤ K sup 0 ≤ s ≤ t 1 + ρ k ( s ) H 1 ∩ W 1 , q 0 2 + ( ρ k u t k , ρ k w t k ) L 2 2 + w k ( s ) L 2 2 + ( u k ( s ) , w k ( s ) ) D 0 1 2 .

Then we estimate each term of ΦK in terms of some integrals of ΦK , apply arguments of Gronwall-type and thus prove that ΦK is locally bounded. First, notice that

ρ k ( t ) L q + p k ( t ) L q ≤ C Φ K ( t ) , for all 1 ≤ q ≤ ∞ .

Then, we estimate ( u k , w k ) H 2 by ΦK .

• Estimate w k H 2

Observe that for any t ∈ [0, T], w k = w k ( t ) ∈ D 0 1 ∩ D 2 is a solution of the elliptic system:

- μ ′ Δ w k - ( μ ′ + λ ) ∇ div w k = - ρ k w t k - ρ k μ k - 1 ⋅ ∇ w k - 4 ζ w k + 2 ζ rot u k - 1 .

Hence we deduce from the elliptic regularity result that

∇ 2 w 2 L 2 ≤ C ρ k w t k L 2 + ρ k u k - 1 ⋅ ∇ w k L 2 + w k L 2 + ∇ u k - 1 L 2 + ∇ w k L 2 ≤ C ρ k L ∞ ρ k w t k L 2 + ρ k L ∞ ∇ u k - 1 L 2 ∇ w k L 2 1 / 2 ∇ w k H 1 1 / 2 + ∇ u k - 1 L 2 + ∇ w k L 2 ≤ C Φ K 5 / 2 + 1 2 ∇ w k H 1 .

Thus we conclude

∇ w k H 1 ≤ C Φ K 5 / 2 .

• Estimate u k H 2

Similar as the estimate of w k H 2 . We see that for any t ∈ [0, T], u k = u k ( t ) ∈ D 0 1 ∩ D 2 is a solution of the elliptic system:

( μ + ζ ) Δ u k + ( μ + λ - ζ ) ∇ div u k = - ρ u t k - ρ u k - 1 ⋅ ∇ u k - ∇ p k + 2 ζ rot w k .

Hence, we have

∇ 2 u k L 2 ≤ C ρ k u t k L 2 + ρ k u k - 1 ⋅ ∇ u k L 2 + ∇ p k L 2 + ∇ w k L 2 + ∇ u k L 2 ≤ C ρ k L ∞ 1 / 2 ρ k u t k L 2 + ρ k L ∞ u k - 1 L 6 ∇ u k L 3 + ρ k L ∞ γ - 1 ∇ ρ k L 2 + ∇ w k L 2 + ∇ u k L 2 ≤ C ρ k L ∞ 1 / 2 ρ k u t k L 2 + ρ k L ∞ ∇ u k - 1 L 2 ∇ u k L 2 1 / 2 ∇ u k H 1 1 / 2 + ρ k L ∞ γ - 1 ∇ ρ k L 2 + ∇ w 2 L 2 + ∇ u k L 2 ≤ C Φ K γ + 5 / 2 + 1 2 ∇ u k H 1 .

Thus we conclude

∇ u k H 1 ≤ C Φ K γ + 5 / 2 .

• Estimate w k D 0 1

Multiplying (46) by w t k , and integrating over Ω, using Young's inequality we obtain:

  1 2 ∫ ​ ρ k | w t k | 2 d x + d d t ∫ ​ ( μ ' 2 | ∇ w k | 2 + μ ' + λ ' 2 | d i v w k | + 2 ζ | w k | 2 ) d x ≤ d d t ∫ ​ 2 ζ r o t u k − 1 w k d x + ∫ ​ ( ρ | u k − 1 | 2 | ∇ w k | 2 + 2 ζ | ∇ u t k − 1 | | w k | d x .

Integrating the above inequality we get

∫ 0 t ρ k w t k L 2 2 d s + ∇ w k L 2 2 + w k L 2 2 ≤ C ∇ u k - 1 L 2 2 + ∫ 0 t ρ k L ∞ u k - 1 L 6 2 ∇ w k L 3 2 + η ∇ u t k - 1 L 2 2 + C w k L 2 2 d s .

To estimate the right-hand side of (52), we first observe that ρ k L ∞ ≤ C Φ K 1/2 , u k - 1 L 6 2 ≤ C ∇ u k - 1 L 2 2 ≤ C Φ K , together with the estimate u k H 2 we conclude that:

∫ 0 t ρ k w t 2 L 2 2 d s + ∇ w k L 2 2 + w k L 2 2 ≤ C ∫ 0 t Φ K 4 d s + η ∇ u t k - 1 L 2 2 .

• Estimate ρ k w t k L 2

Differentiating (46) with respect to t, we obtain

ρ k w t t k + ρ k u k - 1 ⋅ ∇ w t k - 4 ζ w t k - μ ′ Δ w t k - ( μ ′ + λ ′ ) ∇ div w k = 2 ζ rot u t k - 1 - ρ t k w t k - ρ t k u k - 1 ⋅ ∇ w k - ρ k u k - 1 ⋅ ∇ w k .

Multiplying this by w t k and integrating over Ω, we obtain:

  d d t ∫ ρ k | w t k | 2 d x + ∫ ( 4 ζ | w t k | 2 + μ ' | ∇ w t k | 2 + ( μ ' + λ ' ) | div w t k | 2 ) d x ≤ 2 ζ ∫ ( | ∇ u t k − 1 | | w t k | d x + ∫ 2 ρ k | u k − 1 | | w t k | | ∇ w t k | + p k | u t k − 1 | | ∇ w k | | w t k |   + ρ k | u k − 1 | | ∇ u k − 1 | | ∇ w k | | w t k | + ρ k | u k − 1 | 2 | ∇ 2 w k | | w t k |   + ρ k | u k − 1 | 2 | ∇ w k | | ∇ w t k | d x ∑ i = 1 6 I I I i .

Using Sobolev inequality and Young's inequality (with ε) repeatedly, we have:

I I I 1 = 2 ζ ∫ ∇ u t k - 1 w t k d x ≤ ζ 2 ∇ u t k - 1 L 2 2 + 2 ζ w t k L 2 2 ;

I I I 2 = 2 ∫ ρ k u k - 1 w t k ∇ w t k d x ≤ 2 ρ k L ∞ 1 / 2 u k - 1 L 6 ρ k w t k L 3 ∇ w t k L 2 ≤ 2 ρ k L ∞ 1 / 2 ∇ u k - 1 L 2 ρ k w t k L 2 1 / 2 ρ k w t k L 6 1 / 2 ∇ w t k L 2 ≤ 2 ρ k L ∞ 3 / 4 ∇ u k - 1 L 2 ρ k w t k L 2 1 / 2 ∇ w t k L 2 3 / 2 ≤ C Φ K 9 / 2 + ε ∇ w t k L 2 2 ;

I I I 3 = ∫ ρ k u t k - 1 ∇ w k w t k d x ≤ ρ k L ∞ 1 / 2 u t k - 1 L 6 ρ k w t k L 3 ∇ w k L 2 ≤ ρ k L ∞ 1 / 2 ∇ u t k - 1 L 2 ρ k w t k L 2 1 / 2 ρ k w t k L 6 1 / 2 ∇ w k L 2 ≤ ρ k L ∞ 3 / 4 ∇ u t k - 1 L 2 ρ k w t k L 2 1 / 2 ∇ w t k L 2 1 / 2 ∇ w k L 2 ≤ C ρ k L ∞ 3 ρ k w t k L 2 2 ∇ w k L 2 4 + ε ∇ w t k L 2 2 + η ∇ u t k - 1 L 2 2 ≤ C Φ K 9 / 2 + ε ∇ w t k L 2 2 + η ∇ u t k - 1 L 2 2 ;

I I I 4 = ∫ ρ k u k - 1 ∇ u k - 1 ∇ w k w t k d x ≤ ρ k L ∞ u k - 1 L 6 ∇ u k - 1 L 3 ∇ w k L 3 w t k L 6 ≤ C ρ k L ∞ ∇ u k - 1 L 2 3 / 2 ∇ u k - 1 H 1 1 / 2 ∇ w k L 2 1 / 2 ∇ w k H 1 1 / 2 ∇ w t k L 2 ≤ C ρ k L ∞ 2 ∇ u k - 1 L 2 3 ∇ u k - 1 H 1 ∇ w k L 2 ∇ w k H 1 + ε ∇ w t k L 2 2 ≤ C Φ K γ + 8 + ε ∇ w t k L 2 2 ;

I I I 5 = ∫ ρ k u k - 1 2 ∇ 2 w k w t k d x ≤ ρ k L ∞ u k - 1 L 6 2 ∇ 2 w k L 2 w t k L 6 ≤ C ρ k L ∞ 2 ∇ u k - 1 L 2 4 ∇ 2 w k L 2 2 + ε ∇ w t k L 2 2 ≤ C Φ K 8 + ε ∇ w t k L 2 2 ;

I I I 6 = ∫ ρ k u k - 1 2 ∇ w k ∇ w t k d x ≤ ρ k L ∞ ∇ u k - 1 L 2 2 ∇ w k H 1 ∇ w t k L 2 2 ≤ C Φ K 8 + ε ∇ w t k L 2 2 .

Substituting all these estimates into (54) and choosing ε, η > 0 small enough, we obtain that:

d d t ∫ ρ k w t k 2 d x + ζ w t k L 2 2 + ∇ w t k L 2 2 ≤ C Φ K γ + 8 + ζ 2 + η ∇ u t k - 1 L 2 2 .

Integrating over (τ, t) for fixed τ > 0, we have

∫ ρ k w t k 2 d x + ∫ τ t 2 ζ w t k L 2 2 + μ ′ 2 ∇ w t k L 2 2 d s ≤ C ∫ τ t Φ γ + 8 ζ 2 + η ∇ u t k - 1 L 2 2 d s + ∫ ρ k w t k 2 ( τ ) d x .

To estimate ∫ ρ k w t k 2 ( τ ) d x as τ → 0, we multiplying (46) by w t k , integrate over Ω, we have:

∫ ρ k w t k 2 d x = - ∫ ρ k u k - 1 ⋅ ∇ w k w t d x + ∫ ( μ ′ Δ w k + ( μ ′ + λ ′ ) ∇ div w k + 2 ζ rot u k - 1 - 4 ζ w k ) w t k d x .

Using Young's inequality, together with the compatibility condition (7), we have

lim τ → 0 1 2 ∫ ​ ρ k | w t k | 2 d x ≤ C ( ∫ ​ ρ k | u k − 2 | 2 | ∇ w k | 2 d x + ∫ ​ ρ − 1 ( μ ' Δ w k + ( μ ' + λ ' ) ∇ div + 2 ζ rot u k − 1 − 4 ζ w k ) 2 d x ) ≤ C .

Substituting the above estimate into (56), we conclude that

∫ ρ k w t k 2 d x = - ∫ ρ k u k - 1 ⋅ ∇ w k w t d x ≤ C 1 + ∫ 0 t Φ γ + 8 d s + ζ 2 + η ∫ 0 t ∇ u t k - 1 L 2 2 d s .

• Estimate u k D 0 1

Multiplying (47) by u t k and integrating over Ω, we obtain:

1 2 ∫ ρ k u t k 2 d x + d d t ∫ μ + ζ 2 ∇ u k 2 + μ + λ - ζ 2 div u k 2 d x ≤ d d t ∫ ( p k div u k + 2 ζ rot u k w k ) d x + ∫ ρ k u k - 1 2 ∇ u k 2 d x + ∫ p t k ∇ u k d x + 2 ζ ∫ w t k ∇ u k d x .

Using (45), we have

p t k + u k - 1 ⋅ ∇ p k + γ p k div u k - 1 = 0 .

Integrating (58) over (0, t), using (59) and Young's inequality, we have

1 2 ∫ 0 t ∫ ρ k u t k 2 d x d s + μ + ζ 2 ∇ u k L 2 2 ≤ C + C ∫ w k 2 d x + ζ 2 ∫ ∇ u k 2 d x + C ∫ ( p k ) 2 d x + ∫ 0 t ∫ ρ k u k - 1 2 ∇ u k 2 d x d s + ∫ 0 t ∫ u k ∇ p k ∇ u k d x d s + ∫ 0 t ∫ γ p k ∇ u k - 1 ∇ u k d x d s + 2 ζ ∫ 0 t ∫ w t k ∇ u k d x d s = C + C Φ K + ∑ i = 1 5 I V i .

To estimate the right-hand side of (60), with the help of Sobolev inequality and Young's inequality (with ε), we have:

I V 1 = ∫ ( p k ) 2 d x = ∫ p k ( 0 ) 2 d x + ∫ 0 t ∂ ∂ s ∫ p k ( s ) 2 d x d s ≤ C + 2 ∫ 0 t ∫ ( a 2 γ ( ρ k ) 2 γ - 1 ∇ ρ k u k - 1 + a 2 γ ( ρ k ) 2 ∇ u k - 1 ) d x d s ≤ C + C ∫ 0 t ρ k L ∞ 2 ( γ - 1 ) ρ k L 3 ∇ ρ k L 2 u k - 1 L 6 + ρ k L ∞ 2 γ - 1 ρ k L 2 ∇ u k - 1 L 2 d s ≤ C + C ∫ 0 t ρ k L ∞ 2 ( γ - 1 ) ρ k L 2 α ρ k L q 0 1 - α ∇ ρ k L 2 ∇ u k - 1 L 2 + ρ k L ∞ 2 γ - 1 ρ k L 2 ∇ u k - 1 L 2 d s ≤ C + C ∫ 0 t Φ K 2 γ + 1 / 2 d s .

I V 2 = ∫ 0 t ∫ ρ k u k - 1 2 ∇ u k 2 d x d s ≤ ∫ 0 t ρ k L ∞ u k - 1 L 6 2 ∇ u k L 3 2 ≤ C ∫ 0 t ρ k L ∞ ∇ u k - 1 L 2 2 ∇ u k L 2 ∇ u k H 1 d s ≤ C ∫ 0 t Φ K 9 / 2 d s .

I V 3 = ∫ 0 t ∫ u k ∇ p k ∇ u k d x d s ≤ ∫ 0 t u k L 6 ∇ p k L 3 ∇ u k L 2 d s ≤ C ∫ 0 t ∇ u k L 2 2 ρ k L ∞ γ - 1 ρ k L 2 α ρ k L q 0 1 - α d s ≤ C ∫ 0 t Φ K γ + 1 / 2 d s ;

I V 4 = ∫ 0 t ∫ γ ρ k ∇ u k - 1 ∇ u k d x d s ≤ C ∫ 0 t ρ k L ∞ γ ∇ u k - 1 L 2 ∇ u k L 2 d s ≤ C ∫ 0 t Φ K γ  + 1 d s ;

I V 5 = 2 ζ ∫ 0 t ∫ w t k ∇ u k d x d s ≤ ε ∫ 0 t w t k L 2 2 d s + C ∫ 0 t ∇ u k L 2 2 d s ≤ ε ∫ 0 t w t k L 2 2 d s + C ∫ 0 t Φ K d s .

Substitute all the above estimates into (60), we obtain:

∫ 0 t ρ k u t k L 2 2 d s + ∫ 0 t ∇ u k L 2 2 d s ≤ C + C ∫ 0 t Φ k max 2 γ + 1 / 2 , 9 / 2 d s + ε ∫ 0 t w t k L 2 2 d s .

• Estimate ρ K u t k L 2

Differentiating (47) with respect to t, we obtain:

ρ k u t t k + ρ k u k - 1 ⋅ ∇ u t k + ∇ p t k - μ + ζ Δ u t k - μ + λ - ζ ∇ div u t k = - ρ t k u t k - ρ t k u k - 1 ⋅ ∇ u k - ρ k u t k - 1 ⋅ ∇ u </