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Article

Local Solvability for a Compressible Fluid Model of Korteweg Type on General Domains

1
Mathematics Department, Faculty of Science and Technology, Universitas Islam Negeri Syarif Hidayatullah Jakarta, Jl. Ir. H. Juanda No. 95, Ciputat 15412, Indonesia
2
Graduate School of Informatics and Engineering, The University of Electro-Communications, 5-1 Chofugaoka 1-chome, Chofu, Tokyo 182-8585, Japan
*
Author to whom correspondence should be addressed.
Mathematics 2023, 11(10), 2368; https://doi.org/10.3390/math11102368
Submission received: 1 March 2023 / Revised: 17 May 2023 / Accepted: 18 May 2023 / Published: 19 May 2023

Abstract

:
In this paper, we consider a compressible fluid model of the Korteweg type on general domains in the N-dimensional Euclidean space for N 2 . The Korteweg-type model is employed to describe fluid capillarity effects or liquid–vapor two-phase flows with phase transition as a diffuse interface model. In the Korteweg-type model, the stress tensor is given by the sum of the standard viscous stress tensor and the so-called Korteweg stress tensor, including higher order derivatives of the fluid density. The local existence of strong solutions is proved in an L p -in-time and L q -in-space setting, p ( 1 , ) and q ( N , ) , with additional regularity of the initial density on the basis of maximal regularity for the linearized system.

1. Introduction

This paper is concerned with a compressible fluid model of the Korteweg type presented in (1) below. In order to model fluid capillarity effects, Korteweg formulated a constitutive equation in 1901 for stress tensors that included gradients of the fluid density ρ . The Korteweg stress tensor K ( ρ ) , see (2) or (3) below, was introduced by Dunn and Serrin [1] (p. 107) on the basis of the thermodynamics of interstitial workings.
The Korteweg-type model was employed to analyze not only fluid capillarity effects but also a liquid–vapor phase transition; see, e.g., Liu, Landis, Gomez, and Hughes [2].
Let us introduce a short history of mathematical studies of the Korteweg-type model.
There are many studies of the Korteweg-type model such as the existence of weak solutions, the local and global well-posedness for strong solutions, large time decay of solutions, time periodic solutions, the vanishing capillarity limit, and maximal regularity; see, e.g., ref. [3] and references therein for more details. Concerning strong and weak solutions for other kind of fluids, we refer, e.g., to [4,5,6,7]. We focus on well-posedness results for strong solutions of the Korteweg-type model in what follows.
Let us start with problems in the whole space. Hattori and Li [8,9] proved local and global unique existence theorems on smooth solutions in L 2 -based Sobolev spaces. On the other hand, Dancian and Desjardins [10] used critical Besov spaces to relax the regularity of initial data and proved unique existence theorems on local and global strong solutions. Furthermore, Murata and Shibata [11] proved the global well-posedness in an L p -in-time and L q -in-space setting by means of maximal regularity and time decay estimates of an analytic C 0 -semigroup associated with a linearized system. Let P ( ρ ) be the pressure function on [ 0 , ) and let P ( ρ ) be the derivative of P ( ρ ) with respect to ρ . Recently, the asymptotic stability of the constant steady state ( ρ , u ) = ( ρ , 0 ) satisfying ρ > 0 and P ( ρ ) = 0 is actively studied by Kobayashi and his collaborators; see, e.g., [12,13,14].
Boundary value problems of the Korteweg-type model can be found in Bresch, Desjardins, and Lin [15]. Kotschote [16] considered (1) below for Γ S = in the case where Ω is a bounded domain or an exterior domain, and proved the local well-posedness for strong solutions in an L p setting with p > N + 2 for both space and time. This result was extended to a non-isothermal case in [17] and to a non-Newtonian case in [18]. Notice that [18] considered not only the Dirichlet boundary condition but also the slip boundary condition and that [17,18] treated inhomogeneous boundary data. Furthermore, Kotschote [19] proved the asymptotic stability of non-trivial steady states when Ω is a bounded domain with Γ S = .
The present paper aims to extend the local well-posedness result given by [16] to the case where Γ S and Ω is a general domain, which is also called a uniform C 3 domain, see Definition 1 below.
Furthermore, our result is in an L p -in-time and L q -in-space setting, p ( 1 , ) and q ( N , ) , with additional regularity of the initial density, which also gives us an extension of [16]; see Theorem 1 and Remark 2 below for more details. Theorem 1 is the main result of this paper, and is proved by the contraction mapping theorem with the help of the maximal regularity stated in Section 5.2, below.
This paper is organized as follows. The next section introduces our problem setting. Section 3 first introduces the notation used throughout this paper, and then the main result of this paper is stated. Section 4 treats resolvent problems in the whole space and in the half space, and then one introduces the existence of R -bounded solution operator families, also called R -solvers, for the resolvent problems. Based on these results, we next demonstrate that a resolvent problem in a general domain admits an R -solver. Section 5 demonstrates our linear theory, i.e., the generation of an analytic C 0 -semigroup and maximal regularity for some linearized system, which is obtained from the R -solver in a general domain given by Section 4. Section 6 proves the main result of this paper.

2. Problem Setting

Let Ω be a domain in the N-dimensional Euclidean space R N , N 2 , and let the boundary of Ω consist of two hypersurfaces Γ D and Γ S . Throughout this paper, we assume
dist ( Γ D , Γ S ) = inf { | x y | : x Γ D , y Γ S } d > 0 ,
provided that Γ D and Γ S . Notice that Γ D = or Γ S = is admissible in the present paper. Let n = n ( x ) = ( n 1 ( x ) , , n N ( x ) ) T be the unit outward normal vector on Γ D Γ S , where M T denotes the transpose of M .
We consider the motion of a compressible barotropic viscous fluid of the Korteweg type in Ω with the Dirichlet boundary condition on Γ D and the slip boundary condition on Γ S . Such a motion is governed by the following set of equations:
t ρ + div ( ρ u ) = 0 in Ω × ( 0 , T ) , ρ ( t u + u · u ) = Div ( S ( u ) + K ( ρ ) P ( ρ ) I ) + ρ b in Ω × ( 0 , T ) , n · ρ = 0 , u = 0 on Γ D × ( 0 , T ) , n · ρ = 0 , ( D ( u ) n ) τ = 0 , u · n = 0 on Γ S × ( 0 , T ) , ( ρ , u ) | t = 0 = ( ρ 0 + ρ , u 0 ) in Ω ,
where T is a positive constant. Throughout this paper, we assume that ρ > 0 denotes a constant reference density. The initial data
ρ 0 = ρ 0 ( x ) , u 0 = u 0 ( x ) = ( u 01 ( x ) , , u 0 N ( x ) ) T
are given functions of x Ω , and also the body force b = b ( x , t ) = ( b 1 ( x , t ) , , b N ( x , t ) ) T is a given function of ( x , t ) Ω × ( 0 , T ) .
Here, ρ = ρ ( x , t ) and u = u ( x , t ) = ( u 1 ( x , t ) , , u N ( x , t ) ) T are, respectively, the density of the fluid and the velocity of the fluid at position x = ( x 1 , , x N ) Ω and time t > 0 . Let t = / t and j = / x j for j = 1 , , N . The doubled deformation rate tensor is denoted by D ( u ) , i.e., D ( u ) = u + ( u ) T for
u = 1 u 1 N u 1 1 u N N u N ,
while I is the N × N identity matrix. The pressure P : ( 0 , ) R is a given smooth function. For a = ( a 1 , , a N ) T and b = ( b 1 , , b N ) T , we set a · b = j = 1 N a j b j and a b = ( a i b j ) 1 i , j N . In addition,
a τ = a n ( n · a ) .
Let v = ( v 1 ( x ) , , v N ( x ) ) T and w = ( w 1 ( x ) , , w N ( x ) ) T . Then
v · w = j = 1 N v j j w 1 , , j = 1 N v j j w N T
and
2 v = { i j v k : i , j , k = 1 , , N } .
For an N × N matrix-valued function M = ( M i j ( x ) ) 1 i , j N , we set
Div M = j = 1 N j M 1 j , , j = 1 N j M N j T .
Let us introduce two stress tensors S ( u ) and K ( ρ ) . One denotes the standard viscous stress tensor by S ( u ) , i.e.,
S ( u ) = μ D ( u ) + ( ν μ ) div u I
for the viscosity coefficients μ = μ ( x , t ) , ν = μ ( x , t ) and div u = j = 1 N j u j . On the other hand, K ( ρ ) is called the Korteweg stress tensor and given by
K ( ρ ) = κ 2 ( Δ ρ 2 | ρ | 2 ) I κ ρ ρ
for the capillary coefficient κ = κ ( x , t ) , where
ρ = ( 1 ρ , , N ρ ) T , | ρ | 2 = ρ · ρ = j = 1 N ( j ρ ) 2 .
Since
Δ ρ 2 = j = 1 N j 2 ρ 2 = 2 j = 1 N ( j ρ ) 2 + ρ j 2 ρ = 2 | ρ | 2 + 2 ρ Δ ρ ,
(2) is equivalent to
K ( ρ ) = κ ρ Δ ρ + | ρ | 2 2 I κ ρ ρ .

3. Notation and Main Result

This section first introduces the notation used throughout this paper, and then the main result of this paper is stated.

3.1. Notation

Let N be the set of all positive integers and N 0 = N { 0 } . Define R + = ( 0 , ) , C + , δ = { z C : z > δ } for δ R , and C + = C + , 0 .
Let p [ 1 , ] and G be a domain in R N . Then L p ( G ) and H p m ( G ) , m N , stands for the Lebesgue space on G and the Sobolev space on G, respectively. The norm of L p ( G ) is denoted by · L p ( G ) , while the norm of H p m ( G ) is denoted by · H p m ( G ) . Let H p 0 ( G ) = L p ( T ) . In addition, B q , p s ( G ) is the Besov space on G for q ( 1 , ) and s > 0 , and its norm is denoted by · B q , p s ( G ) .
Let X be a Banach space. Then X M denotes the M-product space of X for M N , while the norm of X M is usually denoted by · X instead of · X M for short. Let Y be another Banach space, and then L ( X , Y ) stands for the Banach space of all bounded linear operators from X to Y. In addition, L ( X ) is the abbreviation of L ( X , X ) . For a domain U in C , Hol ( U , L ( X , Y ) ) is the set of all L ( X , Y ) -valued holomorphic functions on U.
Let p [ 1 , ] and I be an interval of R . Then L p ( I , X ) and H p 1 ( I , X ) are the X-valued Lebesgue space on I and the X-valued Sobolev space on I, respectively. The norm of L p ( I , X ) is given by
f L p ( I , X ) = I f ( t ) X p d t 1 / p for p [ 1 , ) , ess sup t I f ( t ) X for p = ,
while the norm of H p 1 ( I , X ) is given by
f H p 1 ( I , X ) = f L p ( I , X ) p + t f L p ( I , X ) p 1 / p .
We denote the set of all continuous functions f : I X by C ( I , X ) . Furthermore, we set for T > 0 or T =
0 H p 1 ( ( 0 , T ) , X ) = { f H p 1 ( ( 0 , T ) , X ) : f | t = 0 = 0 in X }
with the norm · 0 H p 1 ( ( 0 , T ) , X ) : = · H p 1 ( ( 0 , T ) , X ) .
We now introduce the definition of uniform C 3 domains.
Definition 1.
Let D be a domain in R N with boundary D . Then D is called a uniform C 3 domain, if there exist positive constants α, β, and K such that the following assertions hold: for any x 0 = ( x 01 , , x 0 N ) D there exist a coordinate number j and a C 3 function h ( x )   ( x = ( x 1 , , x j 1 , x j + 1 , , x N ) ) on B α ( x 0 ) , with x 0 = ( x 01 , , x 0 j 1 , x 0 j + 1 , , x 0 N ) ,
B α ( x 0 ) = { x R N 1 : | x x 0 | < α } , h H 3 ( B α ( x 0 ) ) K ,
such that
D B β ( x 0 ) = { x R N : x j > h ( x ) , x B α ( x 0 ) } B β ( x 0 ) , D B β ( x 0 ) = { x R N : x j = h ( x ) , x B α ( x 0 ) } B β ( x 0 ) .
Here B β ( x 0 ) = { x R N : | x x 0 | < β } .
Example 1.
(1)
If Ω is a bounded domain or an exterior domain in R N , N 2 , whose boundary is of class C 3 , then Ω is a uniform C 3 domain.
(2)
Let h + ( x ) and h ( x ) , x = ( x 1 , , x N 1 ) be of class C 3 and have compact supports with h ± L ( R N 1 ) 1 / 2 . Then
Ω = { ( x , x N ) : x R N 1 , 1 + h ( x ) x N 1 + h + ( x ) }
becomes a uniform C 3 domain with boundary Γ S Γ D , where
Γ D = { ( x , x N ) : x R N 1 , x N = 1 + h ( x ) } , Γ S = { ( x , x N ) : x R N 1 , x N = 1 + h + ( x ) } .
Let Ω be a uniform C 3 domain and let C 0 , 1 ( Ω ¯ ) be the Banach space of all bounded and uniformly Lipschitz continuous functions on Ω ¯ = Ω Γ D Γ S with the norm:
f C 0 , 1 ( Ω ¯ ) = f L ( Ω ) + sup x , y Ω , x y | f ( x ) f ( y ) | | x y | .
Remark 1.
It follows from [20] (Theorem 3.14) that C 0 , 1 ( Ω ¯ ) = H 1 ( Ω ) . This fact is often used throughout this paper.
Let T > 0 or T = , and let p , q ( 1 , ) . Define
K p , q ; T 1 = H p 1 ( ( 0 , T ) , H q 1 ( Ω ) ) L p ( ( 0 , T ) , H q 3 ( Ω ) ) , ρ K p , q ; T 1 = ρ H p 1 ( ( 0 , T ) , H q 1 ( Ω ) ) + ρ L p ( ( 0 , T ) , H q 3 ( Ω ) ) ,
and also
K p , q ; T 2 = H p 1 ( ( 0 , T ) , L q ( Ω ) N ) L p ( ( 0 , T ) , H q 2 ( Ω ) N ) , u K p , q ; T 2 = u H p 1 ( ( 0 , T ) , L q ( Ω ) N ) + u L p ( ( 0 , T ) , H q 2 ( Ω ) N ) .
Furthermore,
0 K p , q ; T 1 = 0 H p 1 ( ( 0 , T ) , H q 1 ( Ω ) ) L p ( ( 0 , T ) , H q 3 ( Ω ) ) , 0 K p , q ; T 2 = 0 H p 1 ( ( 0 , T ) , L q ( Ω ) N ) L p ( ( 0 , T ) , H q 2 ( Ω ) N ) .
We now set
Z p , q ; T = Z p , q ; T 1 × Z p , q ; T 2 for Z { K , 0 K }
and
( ρ , u ) K p , q ; T = ρ K p , q ; T 1 + u K p , q ; T 2 .
Let L > 0 . Then, 0 K p , q ; T ( L ) is defined by
0 K p , q ; T ( L ) = { ( ρ , u ) 0 K p , q ; T : ( ρ , u ) K p , q ; T L } .

3.2. Main Result

To state our main result for (1), we write (1) as an equivalent system in what follows. We replace ρ by ρ + ρ in (1) in order to obtain
t ρ + div ( ( ρ + ρ ) u ) = 0 in Ω × ( 0 , T ) , ( ρ + ρ ) ( t u + u · u ) = Div ( S ( u ) + K ( ρ + ρ ) P ( ρ + ρ ) I ) + ( ρ + ρ ) b in Ω × ( 0 , T ) , n · ρ = 0 , u = 0 on Γ D × ( 0 , T ) , n · ρ = 0 , ( D ( u ) n ) τ = 0 , u · n = 0 on Γ S × ( 0 , T ) , ( ρ , u ) | t = 0 = ( ρ 0 , u 0 ) in Ω .
Let us define
μ 0 ( x ) = μ ( x , t ) | t = 0 , ν 0 ( x ) = ν ( x , t ) | t = 0 , κ 0 ( x ) = κ ( x , t ) | t = 0 , r 0 ( x ) = ρ 0 ( x ) + ρ , S 0 ( u ) = μ 0 ( x ) D ( u ) + ( ν 0 ( x ) μ 0 ( x ) ) div u I .
The first equation of (4) is then written as
t ρ + r 0 div u = u · ρ ( ρ ρ 0 ) div u = : D ( ρ , u ) .
We next consider the second equation of (4). Recalling (3), we observe that
K ( ρ + 1 ) = ( κ κ 0 ) ( ρ + 1 ) Δ ρ I + κ 0 ( ρ ρ 0 ) Δ ρ I + κ 0 r 0 Δ ρ I + κ | ρ | 2 2 I κ ρ ρ .
The second equation of (4) is thus written as
r 0 t u Div ( S 0 ( u ) + κ 0 r 0 Δ ρ I ) = ( ρ ρ 0 ) t u ( ρ + ρ ) u · u + Div ( S ( u ) S 0 ( u ) ) P ( ρ + ρ ) ρ + Div ( ( κ κ 0 ) ( ρ + ρ ) Δ ρ I + κ 0 ( ρ ρ 0 ) Δ ρ I + κ | ρ | 2 2 I κ ρ ρ ) + ( ρ + ρ ) b = : F ˜ ( ρ , u ) ,
where P ( ρ ) = ( d P / d ρ ) ( ρ ) . Furthermore, since
Div ( S 0 ( u ) + κ 0 r 0 Δ ρ I ) = κ 0 Div ( κ 0 1 S 0 ( u ) + r 0 Δ ρ I ) + ( κ 0 1 S 0 ( u ) + r 0 Δ ρ I ) κ 0 ,
(6) is reduced to
t u r 0 1 κ 0 Div ( κ 0 1 S 0 ( u ) + r 0 Δ ρ I ) = r 0 1 F ˜ ( ρ , u ) + r 0 1 ( κ 0 1 S 0 ( u ) + r 0 Δ ρ I ) κ 0 = : F ( ρ , u ) .
Summing up the above calculations, we have achieved the following equivalent system of (1):
t ρ + r 0 div u = D ( ρ , u ) in Ω × ( 0 , T ) , t u r 0 1 κ 0 Div ( κ 0 1 S 0 ( u ) + r 0 Δ ρ I ) = F ( ρ , u ) in Ω × ( 0 , T ) , n · ρ = 0 , u = 0 on Γ D × ( 0 , T ) , n · ρ = 0 , ( D ( u ) n ) τ = 0 , u · n = 0 on Γ S × ( 0 , T ) , ( ρ , u ) | t = 0 = ( ρ 0 , u 0 ) in Ω .
We further reduce (7) to some system with ( ρ 0 , u 0 ) = ( 0 , 0 ) . Let ( ρ ^ , u ^ ) be a unique solution to the following linear system:
t ρ ^ + r 0 div u ^ = 0 in Ω × ( 0 , T ) , t u ^ r 0 1 κ 0 Div ( κ 0 1 S 0 ( u ^ ) + r 0 Δ ρ ^ I ) = 0 in Ω × ( 0 , T ) , n · ρ ^ = 0 , u ^ = 0 on Γ D × ( 0 , T ) , n · ρ ^ = 0 , ( D ( u ^ ) n ) τ = 0 , u ^ · n = 0 on Γ S × ( 0 , T ) , ( ρ ^ , u ^ ) | t = 0 = ( ρ 0 , u 0 ) in Ω ,
see Section 5.1 below for more details on ( ρ ^ , u ^ ) . Replace ( ρ , u ) by ( ρ + ρ ^ , u + u ^ ) in (7), and the resultant system becomes
t ρ + r 0 div u = D ( ρ + ρ ^ , u + u ^ ) in Ω × ( 0 , T ) , t u r 0 1 κ 0 Div ( κ 0 1 S 0 ( u ) + r 0 Δ ρ I ) = F ( ρ + ρ ^ , u + u ^ ) in Ω × ( 0 , T ) , n · ρ = 0 , u = 0 on Γ D × ( 0 , T ) , n · ρ = 0 , ( D ( u ) n ) τ = 0 , u · n = 0 on Γ S × ( 0 , T ) , ( ρ , u ) | t = 0 = ( 0 , 0 ) in Ω .
Our main result of this paper is then stated as follows.
Theorem 1.
Let N 2 and Ω be a uniform C 3 domain in R N . Let p ( 1 , ) and q ( N , ) . Suppose that R, R 1 , R 2 , T 0 , and ρ are positive constants with R 1 R 2 . Then, there exist constants L 1 and T ( 0 , T 0 ] such that (9) admits a unique solution ( ρ , u ) in 0 K p , q ; T ( L ) if ρ 0 , u 0 , b , P, μ, ν, and κ satisfy the following conditions:
(a)
( ρ 0 , u 0 ) D q , p ( Ω ) with ( ρ 0 , u 0 ) D q , p ( Ω ) R , where D q , p ( Ω ) is given by Section 5.1, below, as well as a subspace of B q , p 3 2 / p ( Ω ) × B q , p 2 2 / p ( Ω ) N ;
(b)
r 0 = ρ 0 + ρ C 0 , 1 ( Ω ¯ ) with r 0 C 0 , 1 ( Ω ¯ ) R and
ρ 2 r 0 ( x ) 2 ρ ( x Ω ¯ ) ;
(c)
b L p ( ( 0 , T 0 ) , L q ( Ω ) N ) with b L p ( ( 0 , T 0 ) , L q ( Ω ) N ) R ;
(d)
P is a C 1 function on [ ρ / 4 , 4 ρ ] with P C 0 , 1 ( [ ρ / 4 , 4 ρ ] ) and P C 0 , 1 ( [ ρ / 4 , 4 ρ ] ) R , where P ( s ) = ( d P / d s ) ( s ) ;
(e)
μ , ν , κ C ( [ 0 , T 0 ] , C 0 , 1 ( Ω ¯ ) ) with
sup t [ 0 , T 0 ] μ ( t ) C 0 , 1 ( Ω ¯ ) R , sup t [ 0 , T 0 ] ν ( t ) C 0 , 1 ( Ω ¯ ) R , sup t [ 0 , T 0 ] κ ( t ) C 0 , 1 ( Ω ¯ ) R ;
(f)
μ 0 ( x ) = μ ( x , t ) | t = 0 , ν 0 ( x ) = ν ( x , t ) | t = 0 , and κ 0 ( x ) = κ ( x , t ) | t = 0 satisfy
R 1 μ 0 ( x ) R 2 , R 1 μ 0 ( x ) + ν 0 ( x ) R 2 , R 1 κ 0 ( x ) R 2 ( x Ω ¯ ) .
Remark 2.
(1)
If p, q satisfy 2 / p + N / q < 2 additionally, then B q , p 3 2 / p ( Ω ) is continuously embedded into H 1 ( Ω ) , which is equivalent to C 0 , 1 ( Ω ¯ ) , as stated in Remark 1; see, e.g., Remark 1 ( b ) of Subsection 2.8.1 in [21]. In this case, the additional regularity of the initial density, i.e., r 0 C 0 , 1 ( Ω ¯ ) , may be removed.
(2)
Our linear theory requires that r 0 belongs to C 0 , 1 ( Ω ¯ ) ; see Section 5 below for more details.

4. R -Solvers for Resolvent Problems

In this section, we consider resolvent problems and prove the existence of R -bounded solution operator families, also called R -solvers, for the resolvent problems. The main result of this section, as shown in Theorem 2 below, gives us a generation of an analytic C 0 -semigroup and maximal regularity for the linearized system of (9) in the next section.

4.1. R -Solver in the Whole Space

Let us first introduce the definition of R -boundedness.
Definition 2.
Let X and Y be Banach spaces, and let r n ( t ) be the Rademacher functions on [0, 1], i.e.,
r n ( t ) = sign ( sin ( 2 n π t ) ) ( n N , 0 t 1 ) .
A family of operators T L ( X , Y ) is called R -bounded on L ( X , Y ) , if there exist constants p [ 1 , ) and C > 0 such that the following assertion holds: for each m N , { T j } j = 1 m T , and { f j } j = 1 m X ,
0 1 j = 1 m r j ( t ) T j f j Y p d t 1 / p C 0 1 j = 1 m r j ( t ) f j X p d t 1 / p .
The smallest such C is called R -bound of T on L ( X , Y ) and denoted by R L ( X , Y ) ( T ) .
Remark 3.
(1)
The constant C in Definition 2 may depend on p.
(2)
It is known that T is R -bounded for any p [ 1 , ) , provided that T is R -bounded for some p [ 1 , ) . This fact follows from Kahane’s inequality; see, e.g., [22] (Theorem 2.4).
(3)
The R -boundedness implies the uniform boundedness. In fact, taking m = 1 in the definition of the R -boundedness yields that T f Y C f X for any T T and f X .
This subsection considers the following resolvent problem in the whole space:
λ ρ + γ 1 div u = d in R N , λ u γ 4 1 Div ( γ 2 ( D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) = f in R N .
Let q ( 1 , ) . For the right member ( d , f ) of (10), we set
X q 1 ( R N ) = H q 1 ( R N ) × L q ( R N ) N , X q 1 ( R N ) = L q ( R N ) N + 1 + N
and set for F 1 = ( d , f ) X q 1 ( R N ) and λ C \ ( , 0 ]
F λ 1 F 1 = ( d , λ 1 / 2 d , f ) X q 1 ( R N ) .
On the other hand, for the solution ( ρ , u ) of (10), we set
A q 0 ( G ) = L q ( G ) N 3 + N 2 + N + 1 , S λ 0 ρ = ( 3 ρ , λ 1 / 2 2 ρ , λ ρ , λ 3 / 2 ρ ) , B q ( G ) = L q ( G ) N 3 + N 2 + N , T λ u = ( 2 u , λ 1 / 2 u , λ u ) ,
where G is a domain in R N . The following lemma then holds.
Lemma 1.
Let q ( 1 , ) and let γ 1 , γ 2 , γ 3 , and γ 4 be constants satisfying
γ i > 0 ( i = 1 , 2 , 4 ) , γ 2 + γ 3 > 0 .
Then, the following assertions hold.
(1)
For any λ C + there exist operators A 1 ( λ ) , B 1 ( λ ) , with
A 1 ( λ ) Hol ( C + , L ( X q 1 ( R N ) , H q 3 ( R N ) ) ) , B 1 ( λ ) Hol ( C + , L ( X q 1 ( R N ) , H q 2 ( R N ) N ) ) ,
such that for any F 1 = ( d , f ) X q 1 ( R N )
( ρ , u ) = ( A 1 ( λ ) F λ 1 F 1 , B 1 ( λ ) F λ 1 F 1 )
is a unique solution to (10).
(2)
There exists a positive constant C = C ( N , q , γ 1 , γ 2 , γ 3 , γ 4 ) , such that for n = 0 , 1
R L ( X q 1 ( R N ) , A q 0 ( R N ) ) λ d d λ n S λ 0 A 1 ( λ ) : λ C + C , R L ( X q 1 ( R N ) , B q ( R N ) ) λ d d λ n T λ B 1 ( λ ) : λ C + C ,
where A q 0 ( R N ) , B q ( R N ) , S λ 0 , and T λ are given by (11) with G = R N .
Proof. 
The proof is similar to [23] (Theorem 2.1), so that the detailed proof may be omitted. □

4.2. R -Solver in the Half Space

Let us first consider the following resolvent problem with the Dirichlet boundary condition for the fluid velocity:
λ ρ + γ 1 div u = d in R + N , λ u γ 4 1 Div ( γ 2 ( D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) = f in R + N , n · ρ = g , u = h on R 0 N ,
where
R + N = { ( x , x N ) : x = ( x 1 , , x N 1 ) R N 1 , x N > 0 } , R 0 N = { ( x , x N ) : x = ( x 1 , , x N 1 ) R N 1 , x N = 0 } .
Let q ( 1 , ) . For the right member ( d , f , g , h ) , we set
X q 2 ( R + N ) = H q 1 ( R + N ) × L q ( R + N ) N × H q 2 ( R + N ) × H q 2 ( R + N ) N , X q 2 ( R + N ) = L q ( R + N ) ( N + 1 ) + N + ( N 2 + N + 1 ) + ( N 3 + N 2 + N )
and set for F 2 = ( d , f , g , h ) X q 2 ( R + N ) and λ C \ ( , 0 ]
F λ 2 F 2 = ( d , λ 1 / 2 d , f , 2 g , λ 1 / 2 g , λ g , 2 h , λ 1 / 2 h , λ h ) X q 2 ( R + N ) .
The following lemma then holds.
Lemma 2.
Let q ( 1 , ) and γ i ( i = 1 , 2 , 3 , 4 ) be constants satisfying (12). Then, the following assertions hold.
(1)
For any λ C + , there exist operators A 2 ( λ ) , B 2 ( λ ) , with
A 2 ( λ ) Hol ( C + , L ( X q 2 ( R + N ) , H q 3 ( R + N ) ) ) , B 2 ( λ ) Hol ( C + , L ( X q 2 ( R + N ) , H q 2 ( R + N ) N ) ) ,
such that for any F 2 = ( d , f , g , h ) X q 2 ( R + N )
( ρ , u ) = ( A 2 ( λ ) F λ 2 F 2 , B 2 ( λ ) F λ 2 F 2 )
is a unique solution to (13).
(2)
There exists a positive constant C = C ( N , q , γ 1 , γ 2 , γ 3 , γ 4 ) such that for n = 0 , 1
R L ( X q 2 ( R + N ) , A q 0 ( R + N ) ) λ d d λ n S λ 0 A 2 ( λ ) : λ C + C , R L ( X q 2 ( R + N ) , B q ( R + N ) ) λ d d λ n T λ B 2 ( λ ) : λ C + C ,
where A q 0 ( R + N ) , B q ( R + N ) , S λ 0 , and T λ are given by (11) with G = R + N .
Proof. 
This lemma was proved by [24] (Theorem 1.4) when h = 0 and γ i ( i = 1 , 2 , 3 , 4 ) are positive constants. Define
X ˜ q 2 ( R + N ) = H q 1 ( R + N ) × L q ( R + N ) N × H q 2 ( R + N ) , X ˜ q 2 ( R + N ) = L q ( R + N ) ( N + 1 ) + N + ( N 2 + N + 1 ) ,
and set for F ˜ 2 = ( d , f , g ) X ˜ q 2 ( R + N ) and λ C \ ( , 0 ]
F ˜ λ 2 F ˜ 2 = ( d , λ 1 / 2 d , f , 2 g , λ 1 / 2 g , λ g ) X ˜ q 2 ( G ) .
Then, ref. [24] (Theorem 1.4) can be extended to the case where h = 0 and γ i ( i = 1 , 2 , 3 , 4 ) are constants satisfying (12) by slightly modifying its proof, i.e., for any λ C + there exist operators A ˜ 2 ( λ ) , B ˜ 2 ( λ ) , with
A ˜ 2 ( λ ) Hol ( C + , L ( X ˜ q 2 ( R + N ) , H q 3 ( R + N ) ) ) , B ˜ 2 ( λ ) Hol ( C + , L ( X ˜ q 2 ( R + N ) , H q 2 ( R + N ) N ) ) ,
such that for any F ˜ 2 = ( d , f , g ) X ˜ q 2 ( R + N )
( σ , v ) = ( A ˜ 2 ( λ ) F ˜ λ 2 F ˜ 2 , B ˜ 2 ( λ ) F ˜ λ 2 F ˜ 2 )
is a unique solution to
λ σ + γ 1 div v = d in R + N , λ v γ 4 1 Div ( γ 2 ( D ( v ) + ( γ 3 γ 2 ) div v I + γ 1 Δ σ I ) = f in R + N , n · σ = g , v = 0 on R 0 N ,
where γ i ( i = 1 , 2 , 3 , 4 ) are constants satisfying (12). In addition, for n = 0 , 1 ,
R L ( X ˜ q 2 ( R + N ) , A q 0 ( R + N ) ) λ d d λ n S λ 0 A ˜ 2 ( λ ) : λ C + C , R L ( X ˜ q 2 ( R + N ) , B q ( R + N ) ) λ d d λ n T λ B ˜ 2 ( λ ) : λ C + C ,
with a positive constant C = C ( N , q , γ 1 , γ 2 , γ 3 , γ 4 ) .
Let us now consider
λ u Δ u = 0 in R + N , u = h on R 0 N .
It is well-known that (15) admits an R -solver, i.e., there exists an operator
U ( λ ) Hol ( C + , L ( L q ( R + N ) N 2 + N + 1 , H q 2 ( R + N ) ) )
such that u = U ( λ ) T λ h , h H q 2 ( R + N ) is a solution to (15) and
R L ( L q ( R + N ) N 2 + N + 1 ) λ d d λ n T λ U ( λ ) : λ C + C
for n = 0 , 1 with a positive constant C = C ( N , q ) . This enables us to define
V ( λ ) Hol ( C + , L ( L q ( R + N ) N 3 + N 2 + N ) , H q 2 ( R + N ) N )
by V ( λ ) T λ h = ( U ( λ ) T λ h 1 , , U ( λ ) T λ h N ) for h = ( h 1 , , h N ) T , and then
R L ( L q ( R + N ) N 3 + N 2 + N ) λ d d λ n T λ V ( λ ) : λ C + C
for n = 0 , 1 with a positive constant C = C ( N , q ) .
Let u = w + V ( λ ) T λ h in (13). Then, (13) is reduced to
λ ρ + γ 1 div w = d ˜ in R + N , λ w γ 4 1 Div ( γ 2 ( D ( w ) + ( γ 3 γ 2 ) div w I + γ 1 Δ ρ I ) = f ˜ in R + N , n · ρ = g , w = 0 on R 0 N ,
together with
d ˜ = d γ 1 div V ( λ ) T λ h , f ˜ = f λ V ( λ ) T λ h + γ 4 1 ( γ 2 Δ V ( λ ) T λ h + γ 3 div V ( λ ) T λ h ) ,
where one has used the fact that
Div ( γ 2 D ( V ( λ ) T λ h ) + ( γ 3 γ 2 ) div V ( λ ) T λ h I ) = γ 2 Δ V ( λ ) T λ h + γ 3 div V ( λ ) T λ h .
From this viewpoint, we set for H = ( H 1 , , H 9 ) X q 2 ( R + N )
A 2 ( λ ) H = A ˜ 2 ( λ ) ( H 1 γ 1 div V ( λ ) ( H 7 , H 8 , H 9 ) , H 2 γ 1 λ 1 / 2 div V ( λ ) ( H 7 , H 8 , H 9 ) , H 3 λ V ( λ ) ( H 7 , H 8 , H 9 ) + γ 4 1 γ 2 Δ V ( λ ) ( H 7 , H 8 , H 9 ) + γ 4 1 γ 3 div V ( λ ) ( H 7 , H 8 , H 9 ) , H 4 , H 5 , H 6 ) , B 2 ( λ ) H = B ˜ 2 ( λ ) ( H 1 γ 1 div V ( λ ) ( H 7 , H 8 , H 9 ) , H 2 γ 1 λ 1 / 2 div V ( λ ) ( H 7 , H 8 , H 9 ) , H 3 λ V ( λ ) ( H 7 , H 8 , H 9 ) + γ 4 1 γ 2 Δ V ( λ ) ( H 7 , H 8 , H 9 ) + γ 4 1 γ 3 div V ( λ ) ( H 7 , H 8 , H 9 ) , H 4 , H 5 , H 6 ) + V ( λ ) ( H 7 , H 8 , H 9 ) ,
where H 1 , H 2 , H 3 , H 4 , H 5 , and H 6 , H 7 , H 8 , and H 9 are, respectively, corresponding to d , λ 1 / 2 d , f , 2 g , λ 1 / 2 g , λ g , 2 h , λ 1 / 2 h , and λ h . It is then clear that ( ρ , u ) = ( A 2 ( λ ) F λ 2 F 2 , B 2 ( λ ) F λ 2 F 2 ) is a solution to (13) for F 2 = ( d , f , g , h ) X q 2 ( R + N ) and that (14), (16), and the definition of the R -boundedness give us for n = 0 , 1
R L ( X q 2 ( R + N ) , A q 0 ( R + N ) ) λ d d λ n S λ 0 A 2 ( λ ) : λ C + C , R L ( X q 2 ( R + N ) , B q ( R + N ) ) λ d d λ n T λ B 2 ( λ ) : λ C + C ,
with a positive constant C = C ( N , q , γ 1 , γ 2 , γ 3 , γ 4 ) . This completes the proof of Lemma 2. □
We next consider the following resolvent problem with the slip boundary condition for the fluid velocity:
λ ρ + γ 1 div u = d in R + N , λ u γ 4 1 Div ( γ 2 ( D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) = f in R + N , n · ρ = g , ( D ( u ) n ) τ = k τ , u · n = l on R 0 N .
Let q ( 1 , ) . For the right member ( d , f , g , k , l ) , we set
X q 3 ( R + N ) = H q 1 ( R + N ) × L q ( R + N ) N × H q 2 ( R + N ) × H q 1 ( R + N ) N × H q 2 ( R + N ) , X q 3 ( R + N ) = L q ( R + N ) ( N + 1 ) + N + ( N 2 + N + 1 ) + ( N 2 + N ) + ( N 2 + N + 1 )
and set for F 3 = ( d , f , g , k , l ) X q 3 ( R + N ) and λ C ( , 0 ]
F λ 3 F 3 = ( d , λ 1 / 2 d , f , 2 g , λ 1 / 2 g , λ g , k , λ 1 / 2 k , 2 l , λ 1 / 2 l , λ l ) X q 3 ( R + N ) .
The following lemma then holds.
Lemma 3.
Let q ( 1 , ) and γ i ( i = 1 , 2 , 3 , 4 ) be constants satisfying (12). Then, the following assertions hold.
(1)
For any λ C + there exist operators A 3 ( λ ) , B 3 ( λ ) , with
A 3 ( λ ) Hol ( C + , L ( X q 3 ( R + N ) , H q 3 ( R + N ) ) ) , B 3 ( λ ) Hol ( C + , L ( X q 3 ( R + N ) , H q 2 ( R + N ) N ) ) ,
such that for any F 3 = ( d , f , g , k , l ) X q 3 ( R + N )
( ρ , u ) = ( A 3 ( λ ) F λ 3 F 3 , B 3 ( λ ) F λ 3 F 3 )
is a unique solution to (17).
(2)
There exists a positive constant C = C ( N , q , γ 1 , γ 2 , γ 3 , γ 4 ) such that for n = 0 , 1
R L ( X q 3 ( R + N ) , A q 0 ( R + N ) ) λ d d λ n S λ 0 A 3 ( λ ) : λ C + C , R L ( X q 3 ( R + N ) , B q ( R + N ) ) λ d d λ n T λ B 3 ( λ ) : λ C + C ,
where A q 0 ( R + N ) , B q ( R + N ) , S λ 0 , and T λ are given by (11) with G = R + N .
Proof. 
Let ρ ˜ = ρ / γ 1 , d ˜ = d / γ 1 , and g ˜ = g / γ 1 . Then, (17) is equivalent to
λ ρ ˜ + div u = d ˜ in R + N , λ u μ Δ u ν div u κ Δ ρ ˜ = f in R + N , n · ρ ˜ = g ˜ , ( D ( u ) n ) τ = k τ , u · n = l on R 0 N ,
where
μ = γ 2 γ 4 , ν = γ 3 γ 4 , κ = γ 1 2 γ 4 .
By [25] (Theorem 1.3), we observe that (18) admits R -solvers satisfying the desired properties under the condition that μ , ν , and κ are positive constants satisfying
μ + ν 2 κ 2 1 κ 0 and κ μ ν .
This result can be extended to the case where μ , ν , and κ are any constants satisfying μ > 0 , μ + ν > 0 , and κ > 0 by direct calculations in the same manner as in [24]. The uniqueness of solutions follows from the existence of solutions; see, e.g., ([3] Subsection 3.3). This completes the proof of Lemma 3. □

4.3. R -Solver in a General Domain

This subsection considers the resolvent problem in a uniform C 3 domain Ω :
λ ρ + γ 1 div v = d in Ω , λ u γ 4 1 Div ( γ 2 ( D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) = f in Ω , n · ρ = g D , u = h on Γ D , n · ρ = g S , ( D ( u ) n ) τ = k τ , u · n = l on Γ S .
We introduce an assumption about the coefficients γ 1 , γ 2 , γ 3 , and γ 4 .
Assumption 1.
The coefficients γ i = γ i ( x ) , i = 1 , 2 , 3 , 4 , are real valued uniformly Lipschitz continuous functions on Ω ¯ = Ω Γ D Γ S , i.e., there exists a positive constant γ L , such that | γ i ( x ) γ i ( y ) | γ L | x y | for any x , y Ω ¯ and for i = 1 , 2 , 3 , 4 . In addition, there exist positive constants γ , γ , such that γ γ i ( x ) γ ( i = 1 , 2 , 4 ) and γ γ 2 ( x ) + γ 3 ( x ) γ for any x Ω ¯ .
Let q ( 1 , ) . For the right member ( d , f , g D , h , g S , k , l ) of (19), we set
X q ( Ω ) = H q 1 ( Ω ) × L q ( Ω ) N × H q 2 ( Ω ) × H q 2 ( Ω ) N × H q 2 ( Ω ) × H q 1 ( Ω ) N × H q 2 ( Ω ) X q 0 ( Ω ) = L q ( Ω ) ( N + 1 ) + N + ( N 2 + N + 1 ) + ( N 3 + N 2 + N ) + ( N 2 + N + 1 ) + ( N 2 + N ) + ( N 2 + N + 1 )
and set for F = ( d , f , g D , h , g S , k , l ) X q ( Ω ) and λ C \ ( , 0 ]
F λ 0 F = ( d , λ 1 / 2 d , f , 2 g D , λ 1 / 2 g D , λ g D , 2 h , λ 1 / 2 h , λ h , 2 g S , λ 1 / 2 g S , λ g S , h , λ 1 / 2 h , 2 l , λ 1 / 2 l , λ l ) X q 0 ( Ω ) .
By Lemmas 1–3, we can prove the following proposition on the basis of the standard localization technique; see, e.g., [3].
Proposition 1.
Let Ω be a uniform C 3 domain in R N . Let q ( 1 , ) and suppose that Assumption 1 holds. Then, there exists a constant λ 1 1 , depending solely on N, q, γ L , γ , and γ , such that the following assertions hold.
(1)
For any λ C + , λ 1 , there exist operators A 0 ( λ ) , B 0 ( λ ) , with
A 0 ( λ ) Hol ( C + , λ 1 , L ( X q 0 ( Ω ) , H q 3 ( Ω ) ) ) , B 0 ( λ ) Hol ( C + , λ 1 , L ( X q 0 ( Ω ) , H q 2 ( Ω ) N ) ) ,
such that for any F = ( d , f , g D , h , g S , k , l ) X q ( Ω )
( ρ , u ) = ( A 0 ( λ ) F λ 0 F , B 0 ( λ ) F λ 0 F )
is a unique solution to (19).
(2)
There exists a positive constant C, depending solely on N, q, γ L , γ , and γ , such that for n = 0 , 1
R L ( X q 0 ( Ω ) , A q 0 ( Ω ) ) λ d d λ n S λ 0 A 0 ( λ ) : λ C + , λ 1 C , R L ( X q 0 ( Ω ) , B q ( Ω ) ) λ d d λ n T λ B 0 ( λ ) : λ C + , λ 1 C ,
where A q 0 ( Ω ) , B q ( Ω ) , S λ 0 , and T λ are given by (11) with G = Ω .
Proposition 1 is not enough to obtain our linear theory in the next section due to λ 3 / 2 ρ and λ 1 / 2 d . To eliminate these terms, we construct another R -solver for (19) based on A 0 ( λ ) , B 0 ( λ ) in what follows. We start with
λ R + θ 1 div U = D in R N , λ U θ 4 1 Div ( θ 2 D ( U ) + ( θ 3 θ 2 ) div U I + θ 1 Δ R I ) = F in R N .
Let us define for q ( 1 , ) and λ C ( , 0 ]
A q ( G ) = L g ( G ) N 3 + N 2 × H q 1 ( G ) , S λ ρ = ( 3 ρ , λ 1 / 2 2 ρ , λ ρ ) ,
where G is a domain in R N . The following proposition follows from [23] (Theorem 2.1) and the standard localization technique; see also [3] (Theorem 7.1).
Proposition 2.
Let q ( 1 , ) , and let θ i = θ i ( x ) ( i = 1 , 2 , 3 , 4 ) be real valued uniformly Lipschitz continuous functions on R N , i.e., there exists a positive constant θ L , such that | θ i ( x ) θ i ( y ) | θ L | x y | for any x , y R N and for i = 1 , 2 , 3 , 4 . Assume that there exist positive constants θ and θ , such that for any x R N
θ θ i ( x ) θ ( i = 1 , 2 , 4 ) , θ θ 2 ( x ) + θ 3 ( x ) θ .
Then, there exists λ 2 1 , depending on at most N, q, θ L , θ , and θ , such that the following assertions hold.
(1)
For any λ C + , λ 2 , there exist operators Φ ( λ ) and Ψ ( λ ) , with
Φ ( λ ) L ( C + , λ 2 , L ( H q 1 ( R N ) × L q ( R N ) N , H q 3 ( R N ) ) ) , Ψ ( λ ) L ( C + , λ 2 , L ( H q 1 ( R N ) × L q ( R N ) N , H q 2 ( R N ) N ) ) ,
such that for any ( D , F ) H q 1 ( R N ) × L q ( R N ) N
( R , U ) = ( Φ ( λ ) ( D , F ) , Ψ ( λ ) ( D , F ) )
is a unique solution to (20).
(2)
There exists a positive constant C, depending on at most N, q, θ L , θ , and θ , such that for n = 0 , 1
R L ( H q 1 ( R N ) × L q ( R N ) N , A q ( R N ) ) λ d d λ n S λ Φ ( λ ) : λ C + , λ 2 C , R L ( H q 1 ( R N ) × L q ( R N ) N , B q ( R N ) ) λ d d λ n T λ Ψ ( λ ) : λ C + , λ 2 C .
Here, A q ( R N ) and S λ are given by (21) for G = R N , while B q ( R N ) and T λ are given by (11) for G = R N .
To use Proposition 2, we extend the coefficients γ i ( i = 1 , 2 , 3 , 4 ) satisfying Assumption 1 to ones defined on R N by the following lemma.
Lemma 4.
Let Ω be a uniform C 3 domain in R N , and let f be a real valued uniformly Lipschitz continuous function on Ω ¯ , i.e., there exists a positive constant L, such that | f ( x ) f ( y ) | L | x y | for any x , y Ω ¯ . Assume that there exist positive constants c and c , such that c f ( x ) c for any x Ω ¯ . Then, there exists a real valued uniformly Lipschitz continuous function F on R N and a positive constant M, depending solely on c and L, such that the following assertions hold.
(1)
F ( x ) = f ( x ) for any x Ω ¯ .
(2)
| F ( x ) F ( y ) | ( M + ( c / 2 ) ) | x y | for any x , y R N .
(3)
c / 2 F ( x ) M + ( c / 2 ) for any x R N .
Proof. 
See [3] (Lemma 7.2 and Appendix A). □
Let us define
X q ( Ω ) = H q 1 ( Ω ) × L q ( Ω ) N × L q ( Ω ) ( N 2 + N + 1 ) + ( N 3 + N 2 + N ) + ( N 2 + N + 1 ) + ( N 2 + N ) + ( N 2 + N + 1 )
and set for F = ( d , f , g D , h , g S , k , l ) X q ( Ω ) and λ C \ ( , 0 ]
F λ F = ( d , f , 2 g D , λ 1 / 2 g D , λ g D , 2 h , λ 1 / 2 h , λ h , 2 g S , λ 1 / 2 g S , λ g S , h , λ 1 / 2 h , 2 l , λ 1 / 2 l , λ l ) X q ( Ω ) .
We are now in a position to construct a new R -solver for (19).
Theorem 2.
Let Ω be a uniform C 3 domain in R N . Let q ( 1 , ) and suppose that Assumption 1 holds. Then, there exists a constant λ 3 1 , depending solely on N, q, γ L , γ , and γ , such that the following assertions hold.
(1)
For any λ C + , λ 3 , there exist operators A ( λ ) and B ( λ ) , with
A ( λ ) Hol ( C + , λ 3 , L ( X q ( Ω ) , H q 3 ( Ω ) ) ) , B ( λ ) Hol ( C + , λ 3 , L ( X q ( Ω ) , H q 2 ( Ω ) N ) ) ,
such that for any F = ( d , f , g D , h , g S , k , l ) X q ( Ω )
( ρ , u ) = ( A ( λ ) F λ F , B ( λ ) F λ F )
is a unique solution to (19).
(2)
There exists a positive constant C, depending solely on N, q, γ L , γ , and γ , such that for n = 0 , 1
R L ( X q ( Ω ) , A q ( Ω ) ) λ d d λ n S λ A ( λ ) : λ C + , λ 3 C , R L ( X q ( Ω ) , B q ( Ω ) ) λ d d λ n T λ B ( λ ) : λ C + , λ 3 C .
Here, A q ( Ω ) and S λ are given by (21) for G = Ω , while B q ( Ω ) and T λ are given by (11) for G = Ω .
Proof. 
Define δ ( x ) = γ 2 ( x ) + γ 3 ( x ) . By Lemma 4, we extend γ i ( x ) ( i = 1 , 2 , 4 ) and δ ( x ) on Ω ¯ to γ ˜ i ( x ) ( i = 1 , 2 , 4 ) and δ ˜ ( x ) on R N , respectively. They are real valued uniformly Lipschitz continuous functions on R N and satisfy
γ 2 γ ˜ i ( x ) M + γ 2 ( i = 1 , 2 , 4 ) , γ 2 δ ˜ ( x ) M + γ 2
for any x R N with a positive constant M = M ( γ , γ L ) . Define γ ˜ 3 ( x ) = δ ˜ ( x ) γ ˜ 2 ( x ) . This shows that γ ˜ 3 ( x ) = γ 3 ( x ) for x Ω ¯ and that γ ˜ 3 ( x ) is a real valued uniformly Lipschitz continuous function on R N with
γ 2 γ ˜ 2 ( x ) + γ ˜ 3 ( x ) M + γ 2 for any x R N .
Furthermore, the Lipschitz constants of γ ˜ 1 , γ ˜ 2 , γ ˜ 3 , and γ ˜ 4 are bounded above by 2 ( M + γ / 2 ) .
We use Proposition 2 with θ i = γ ˜ i for i = 1 , 2 , 3 , 4 . Let ( d , f , g D , h , g S , k , l ) X q ( Ω ) in what follows. Let E be an extension operator from H q 1 ( Ω ) to H q 1 ( R N ) , while E 0 f is the zero extension of f , i.e., E 0 f = f in Ω and E 0 f = 0 in R N Ω . We define
( R , U ) = ( Φ ( λ ) ( E d , E 0 f ) , Ψ ( λ ) ( E d , E 0 f ) ) .
Then, ( R , U ) satisfies
λ R + γ ˜ 1 div U = E d in R N , λ U γ ˜ 4 1 Div ( γ ˜ 2 D ( U ) + ( γ ˜ 3 γ ˜ 2 ) div U I + γ ˜ 1 Δ R I ) = E 0 f in R N .
Setting ρ = R + σ in (19) yields
λ σ + γ 1 div u = d ˜ in Ω , λ u γ 4 1 Div ( γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ σ I ) = f ˜ in Ω , n · σ = g ˜ D , u = h on Γ D , n · σ = g ˜ S , ( D ( u ) ) τ = k τ , u · n = l on Γ S ,
where
d ˜ = γ 1 div U , f ˜ = λ U γ 4 1 Div ( γ 2 D ( U ) + ( γ 3 γ 2 ) div U I ) , g ˜ D = g D n ˜ · R , g ˜ S = g S n ˜ · R .
Notice that n ˜ is an extension of n with n ˜ H 2 ( R N ) , see [26] (Corollary A.3) for more details. From (22) and Proposition 1, we observe that the solution ( ρ , u ) of (19) can be written as
ρ = R + σ = Φ ( λ ) ( E d , E 0 f ) + A 0 ( λ ) F λ 0 ( d ˜ , f ˜ , g ˜ D , h , g ˜ S , k , l ) , u = B 0 ( λ ) F λ 0 ( d ˜ , f ˜ , g ˜ D , h , g ˜ S , k , l ) .
Let us recall
F λ 0 ( d ˜ , f ˜ , g ˜ D , h , g ˜ S , k , l ) = ( d ˜ , λ 1 / 2 d ˜ , f ˜ , 2 g ˜ D , λ 1 / 2 g ˜ D , λ g ˜ D , 2 h , λ 1 / 2 h , λ h , 2 g ˜ S , λ 1 / 2 g ˜ S , λ g ˜ S , k , λ 1 / 2 k , 2 l , λ 1 / 2 l , λ l ) .
In view of this formula, for H = ( H 1 , , H 16 ) X q ( Ω ) and ( Z , Z ) { ( A , A ) , ( B , B ) } , we set
Z ( λ ) H = Z 0 ( λ ) ( ( γ 1 div Ψ ( λ ) ( E H 1 , E 0 H 2 ) ) , λ 1 / 2 γ 1 div Ψ ( λ ) ( E H 1 , E 0 H 2 ) , λ Ψ ( λ ) ( E H 1 , E 0 H 2 ) γ 4 1 Div ( γ 2 D ( Ψ ( λ ) ( E H 1 , E 0 H 2 ) ) + ( γ 3 γ 2 ) div Ψ ( λ ) ( E H 1 , E 0 H 2 ) I ) , H 3 2 n ˜ · Φ ( λ ) ( E H 1 , E 0 H 2 ) , H 4 λ 1 / 2 n ˜ · Φ ( λ ) ( E H 1 , E 0 H 2 ) , H 5 λ n ˜ · Φ ( λ ) ( E H 1 , E 0 H 2 ) , H 6 , H 7 , H 8 , H 9 2 n ˜ · Φ ( λ ) ( E H 1 , E 0 H 2 ) , H 10 λ 1 / 2 n ˜ · Φ ( λ ) ( E H 1 , E 0 H 2 ) , H 11 λ n ˜ · Φ ( λ ) ( E H 1 , E 0 H 2 ) , H 12 , H 13 , H 14 , H 15 , H 16 ) .
We also set for H = ( H 1 , , H 16 ) X q ( Ω )
A ( λ ) H = Φ ( λ ) ( E H 1 , E 0 H 2 ) + A ( λ ) H , B ( λ ) H = B ( λ ) H .
It then follows from (23) that ( ρ , u ) = ( A ( λ ) F λ F , B ( λ ) F λ F ) . In addition, A ( λ ) and B ( λ ) satisfy the desired estimates from the definition of the R -boundedness and Propositions 1 and 2. The uniqueness of solutions is already discussed in Proposition 1. This completes the proof of Theorem 2. □

5. Linear Theory

This section considers the following time-dependent linear system:
t ρ + γ 1 div u = d in Ω × R + , t u γ 4 1 Div ( γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) = f in Ω × R + , n · ρ = 0 , u = 0 on Γ D × R + , n · ρ = 0 , ( D ( u ) n ) τ = 0 , u · n = 0 on Γ S × R + , ( ρ , u ) | t = 0 = ( ρ 0 , u 0 ) in Ω ,
where the coefficients γ i = γ i ( x ) , i = 1 , 2 , 3 , 4 , satisfy Assumption 1. In the following subsections, we first introduce an analytic C 0 -semigroup associated with (24), and then we state the maximal regularity for (24) with ( ρ 0 , u 0 ) = ( 0 , 0 ) .

5.1. An Analytic C 0 -Semigroup

Let us define for q ( 1 , )
X q = H q 1 ( Ω ) × L q ( Ω ) N , ( ρ , u ) X q = ρ H q 1 ( Ω ) + u L q ( Ω ) .
Furthermore, the operator A q is defined by
A q ( ρ , u ) = ( γ 1 div u , γ 4 1 Div ( γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) )
with the domain
D ( A q ) = { ( ρ , u ) H q 3 ( Ω ) × H q 2 ( Ω ) N : n · ρ = 0 , u = 0 on Γ D , n · ρ = 0 , ( D ( u ) n ) τ = 0 , u · n = 0 on Γ S } .
Noting Remark 3(3) and following [3] (Remark 2.10 (1)), we observe from Theorem 2 that A q generates an analytic C 0 -semigroup ( e A q t ) t 0 on X q , as follows.
Proposition 3.
Suppose that Ω is a uniform C 3 domain in R N , and Assumption 1 holds. Let q ( 1 , ) . Then, the following assertions hold.
(1)
A q is a densely defined closed operator on X q .
(2)
A q generates an analytic C 0 -semigroup ( e A q t ) t 0 on X q . In addition, there exist constants δ 1 = δ 1 ( N , q , γ L , γ , γ ) 1 and C = C ( N , q , γ L , γ , γ ) > 0 , such that for any t > 0
e A q t ( ρ 0 , u 0 ) X q C e ( δ 1 / 2 ) t ( ρ 0 , u 0 ) X q ( ( ρ 0 , u 0 ) X q ) , t e A q t ( ρ 0 , u 0 ) X q C e ( δ 1 / 2 ) t t 1 ( ρ 0 , u 0 ) X q ( ( ρ 0 , u 0 ) X q ) , t e A q t ( ρ 0 , u 0 ) X q C e ( δ 1 / 2 ) t ( ρ 0 , u 0 ) D ( A q ) ( ( ρ 0 , u 0 ) D ( A q ) ) ,
where · D ( A q ) denotes the graph norm of A q .
Let ( · , · ) θ , p be the real interpolation functor for θ ( 0 , 1 ) and p ( 1 , ) ; see, e.g., [27] (Definition 1.37). We set
D q , p ( Ω ) = ( X q , D ( A q ) ) 1 1 / p , p .
Then D q , p ( Ω ) B q , p 3 2 / p ( Ω ) × B q , p 2 2 / p ( Ω ) N . The next proposition immediately follows from Proposition 3 in the same manner as in [28] (Theorem 3.9).
Proposition 4.
Suppose that Ω is a uniform C 3 domain in R N and Assumption 1 holds. Let p , q ( 1 , ) . Then, for any ( ρ 0 , u 0 ) D q , p ( Ω ) , ( ρ , u ) = e A q t ( ρ 0 , u 0 ) is a unique solution to (24) with ( d , f ) = ( 0 , 0 ) and satisfies
e δ 1 t t ρ L p ( R + , H q 1 ( Ω ) ) + e δ 1 t ρ L p ( R + , H q 3 ( Ω ) ) + e δ 1 t t u L p ( R + , L q ( Ω ) N ) + e δ 1 t u L p ( R + , H q 2 ( Ω ) N ) C ( ρ 0 , u 0 ) D q , p ( Ω ) ,
where δ 1 is given by Proposition 3 and C = C ( N , p , q , γ L , γ , γ ) is a positive constant.
Let us now consider (8). Recall that S 0 ( u ) , μ 0 , ν 0 , κ 0 , and r 0 are given by (5). Define
γ 1 : = r 0 = ρ 0 + ρ , γ 2 : = μ 0 κ 0 , γ 3 : = ν 0 κ 0 , γ 4 : = r 0 κ 0 = ρ 0 + ρ κ 0 .
We assume that r 0 , μ 0 , ν 0 , and κ 0 satisfy the conditions (b), (e), and (f) of Theorem 1. Then, γ 1 , γ 2 , γ 3 , and γ 4 satisfy Assumption 1, and γ L , γ , and γ in Assumption 1 become constants depending only on R, R 1 , and R 2 . We therefore obtain the following corollary of Proposition 4.
Corollary 1.
Let Ω be a uniform C 3 domain in R N . Let p , q ( 1 , ) and let R, R 1 , R 2 , and ρ be positive constants with R 1 R 2 . Suppose that ( ρ 0 , u 0 ) D q , p ( Ω ) and that r 0 = ρ 0 + ρ , μ 0 , ν 0 , and κ 0 satisfy b, e, and f of Theorem 1. Then, (8) admits a unique solution ( ρ ^ , u ^ ) , which satisfies
e η 1 t t ρ ^ L p ( R + , H q 1 ( Ω ) ) + e η 1 t ρ ^ L p ( R + , H q 3 ( Ω ) ) + e η 1 t t u ^ L p ( R + , L q ( Ω ) N ) + e η 1 t u ^ L p ( R + , H q 2 ( Ω ) N ) C ( ρ 0 , u 0 ) D q , p ( Ω )
with positive constants η 1 = η 1 ( N , q , R , R 1 , R 2 , ρ ) and C = C ( N , p , q , R , R 1 , R 2 , ρ ) .

5.2. Maximal Regularity

From Proposition 5 to Corollary 2 below, we discuss the maximal regularity for (24) with ( ρ 0 , u 0 ) = ( 0 , 0 ) . Concerning the theory of maximal regularity in L p -in-time and L q -in-space settings, we refer to [29] (Chapter 3), written by Shibata.
Combining Theorem 2 with the operator-valued Fourier multiplier theorem introduced by Weis [30] yields the following proposition.
Proposition 5.
Suppose that Ω is a uniform C 3 domain in R N and Assumption 1 holds. Let p , q ( 1 , ) . Then, there exists a constant δ 2 = δ 2 ( N , q , γ L , γ , γ ) 1 , such that the following assertions hold.
(1)
For any e δ 2 t d L p ( R + , H q 1 ( Ω ) ) and e δ 2 t f L p ( R + , L q ( Ω ) N ) , (24) with ( ρ 0 , u 0 ) = ( 0 , 0 ) admits a unique solution ( ρ , u ) with
ρ H p , loc 1 ( R + , H q 1 ( Ω ) ) L p , loc ( R + , H q 3 ( Ω ) ) , u H p , loc 1 ( R + , L q ( Ω ) N ) L p , loc ( R + , H q 2 ( Ω ) N ) .
(2)
The solution ( ρ , u ) satisfies
e δ 2 t t ρ L p ( R + , H q 1 ( Ω ) ) + e δ 2 t ρ L p ( R + , H q 3 ( Ω ) ) + e δ 2 t t u L p ( R + , L q ( Ω ) N ) + e δ 2 t u L p ( R + , H q 2 ( Ω ) N ) C e δ 2 t d L p ( R + , H q 1 ( Ω ) ) + e δ 2 t f L p ( R + , L q ( Ω ) N )
for some positive constant C = C ( N , p , q , γ L , γ , γ ) .
Proof. 
See [29] (Subsection 3.4.6) for the proof of the existence of solutions satisfying the desired estimate.
Let us prove the uniqueness of solutions in what follows. Let ( ρ , u ) satisfy the regularity stated in (1) and the following homogeneous system:
t ρ + γ 1 div u = 0 in Ω × R + , t u γ 4 1 Div ( γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) = 0 in Ω × R + , n · ρ = 0 , u = 0 on Γ D × R + , n · ρ = 0 , ( D ( u ) n ) τ = 0 , u · n = 0 on Γ S × R + , ( ρ , u ) | t = 0 = ( 0 , 0 ) in Ω .
Let φ C 0 ( Ω × R + ) N , where C 0 ( Ω × R + ) is the set of all C functions whose supports are compact and contained in Ω × R + . Let T be a positive constant such that supp φ Ω × ( 0 , T ) and define φ T ( x , t ) = φ ( x , T t ) . Then, supp φ T Ω × ( 0 , T ) and φ T L p ( R + , L q ( Ω ) N ) for p = p / ( p 1 ) and q = q / ( q 1 ) . Thus, there exists ( σ , v ) , with
σ H p , loc 1 ( R + , H q 1 ( Ω ) ) L p , loc ( R + , H q 3 ( Ω ) ) , v H p , loc 1 ( R + , L q ( Ω ) N ) L p , loc ( R + , H q 2 ( Ω ) N ) ,
such that
t σ + γ 1 div v = 0 in Ω × R + , t v γ 4 1 Div ( γ 2 D ( v ) + ( γ 3 γ 2 ) div v I + γ 1 Δ σ I ) = γ 4 1 φ T in Ω × R + , n · σ = 0 , v = 0 on Γ D × R + , n · σ = 0 , ( D ( v ) n ) τ = 0 , v · n = 0 on Γ S × R + , ( σ , v ) | t = 0 = ( 0 , 0 ) in Ω .
Let a = ( a 1 ( x ) , , a N ( x ) ) T , b = ( b 1 ( x ) , , b N ( x ) ) T , A = ( A i j ( x ) ) 1 i , j N , and B = ( B i j ( x ) ) 1 i , j N . Define
( a , b ) Ω = j = 1 N Ω a j ( x ) b j ( x ) d x , ( A , B ) Ω = i , j = 1 N Ω A i j ( x ) B i j ( x ) d x ,
and set for Γ = Γ D or Γ = Γ S
( a , b ) Γ = i = 1 N Γ a j ( x ) b j ( x ) d S ,
where d S is the surface element of Γ . In addition, for f = f ( x , t ) , g = g ( x , t ) , f = ( f 1 ( x , t ) , , f N ( x , t ) ) T , g = ( g 1 ( x , t ) , , g N ( x , t ) ) T , F = ( F i j ( x , t ) ) 1 i , j N , and G = ( G i j ( x , t ) ) 1 i , j N ,
( f , g ) Ω × ( 0 , T ) = 0 T Ω f ( x , t ) g ( x , t ) d x d t , ( f , g ) Ω × ( 0 , T ) = j = 1 N 0 T Ω f j ( x , t ) g j ( x , t ) d x d t , ( F , G ) Ω × ( 0 , T ) = i , j = 1 N 0 T Ω F i j ( x , t ) G i j ( x , t ) d x d t .
Let M = ( M i j ( x ) ) 1 i , j N with M T = M . Integration by parts then shows
( Div M , a ) Ω = 1 2 ( M , D ( a ) ) Ω + ( M n , a ) Γ D Γ S ,
which, combined with
M n = ( M n ) τ + n ( n · M n ) ,
furnishes
( Div M , a ) Ω = 1 2 ( M , D ( a ) ) Ω + ( M n , a ) Γ D + ( ( M n ) τ , a ) Γ S + ( n · M n , a · n ) Γ S .
Let us define for ( x , t ) Ω × ( 0 , T )
τ ( x , t ) = σ ( x , T t ) , w ( x , t ) = v ( x , T t ) .
It then follows from the second equation of (27) that for ( x , t ) Ω × ( 0 , T )
γ 4 t w + Div ( γ 2 D ( w ) + ( γ 3 γ 2 ) div w I + γ 1 Δ τ I ) = { γ 4 t v Div ( γ 2 D ( v ) + ( γ 3 γ 2 ) div v I + γ 1 Δ σ I ) } ( x , T t ) = φ ( x , t ) .
Since u | t = 0 = 0 and w | t = T = 0 , one observes by integration by parts that
( γ 4 t u , w ) Ω × ( 0 , T ) = ( u , γ 4 t w ) Ω × ( 0 , T ) .
Together with the boundary condition of (26) and (27), we use (28) with a = w and M = γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I in order to obtain
( Div ( γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) , w ) Ω × ( 0 , T ) = 1 2 ( γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I , D ( w ) ) Ω × ( 0 , T ) .
It holds that
( ( γ 3 γ 2 ) div u I , D ( w ) ) Ω × ( 0 , T ) = ( D ( u ) , ( γ 3 γ 2 ) div w I ) Ω × ( 0 , T ) .
In addition,
( γ 1 Δ ρ I , D ( w ) ) Ω × ( 0 , T ) = ( Δ ρ , γ 1 div w ) Ω × ( 0 , T ) = : ( RHS ) 1 ,
which, combined with t τ = γ 1 div w , furnishes
( RHS ) 1 = ( Δ ρ , t τ ) Ω × ( 0 , T ) = : ( RHS ) 2 .
Together with ρ | t = 0 = 0 , τ | t = T = 0 , and the boundary condition of (26) and (27), we observe by integration by parts that
( RHS ) 2 = ( t ρ , Δ τ ) Ω × ( 0 , T ) = ( γ 1 div u , Δ τ ) Ω × ( 0 , T ) = ( D ( u ) , γ 1 Δ τ I ) Ω × ( 0 , T ) .
Thus
( γ 1 Δ ρ I , D ( w ) ) Ω × ( 0 , T ) = ( D ( u ) , γ 1 Δ τ I ) Ω × ( 0 , T ) .
Summing up this Equations (31) and (32), we have
( Div ( γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) , w ) Ω × ( 0 , T ) = 1 2 ( D ( u ) , γ 2 D ( w ) + ( γ 3 γ 2 ) div w I + γ 1 Δ τ I ) Ω × ( 0 , T ) .
Let us use (28) with a = u and M = γ 2 D ( w ) + ( γ 3 γ 2 ) div w I + γ 1 Δ τ I together with the boundary condition of (26) and (27), and then
( Div ( γ 2 D ( w ) + ( γ 3 γ 2 ) div w I + γ 1 Δ τ I ) , u ) Ω × ( 0 , T ) = 1 2 ( γ 2 D ( w ) + ( γ 3 γ 2 ) div w I + γ 1 Δ τ I , D ( u ) ) Ω × ( 0 , T ) .
The last two equations give us
( Div ( γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) , w ) Ω × ( 0 , T ) = ( u , Div ( γ 2 D ( w ) + ( γ 3 γ 2 ) div w I + γ 1 Δ τ I ) ) Ω × ( 0 , T ) ,
which, combined with (29) and (30), and the second equation of (26), shows that
0 = ( γ 4 t u Div ( γ 2 D ( u ) + ( γ 3 γ 2 ) div u I + γ 1 Δ ρ I ) , w ) Ω × ( 0 , T ) = ( u , γ 4 t w + Div ( γ 2 D ( w ) + ( γ 3 γ 2 ) div w I + γ 1 Δ τ I ) ) Ω × ( 0 , T ) = ( u , φ ) Ω × ( 0 , T ) = ( u , φ ) Ω × R + .
Thus, u = 0 . It then follows from the first equation of (26) that t ρ = 0 , which, combined with ρ | t = 0 = 0 , furnishes for ( x , t ) Ω × R +
0 = 0 t s ρ ( x , s ) d s = ρ ( x , t ) ρ ( x , 0 ) = ρ ( x , t ) .
Thus ρ = 0 . This shows the uniqueness of solutions to (24) and completes the proof of Proposition 5. □
Let T be a positive constant. We next consider the following time-dependent linear system on ( 0 , T ) :
t ρ + r 0 div u = d in Ω × ( 0 , T ) , t u r 0 1 κ 0 Div ( κ 0 1 S 0 ( u ) + r 0 Δ ρ I ) = f in Ω × ( 0 , T ) , n · ρ = 0 , u = 0 on Γ D × ( 0 , T ) , n · ρ = 0 , ( D ( u ) n ) τ = 0 , u · n = 0 on Γ S × ( 0 , T ) , ( ρ , u ) | t = 0 = ( 0 , 0 ) in Ω ,
where S 0 ( u ) , μ 0 , ν 0 , κ 0 , and r 0 are given by (5). As a corollary of Proposition 5, we obtain
Corollary 2.
Let Ω be a uniform C 3 domain in R N . Let p , q ( 1 , ) and let R, R 1 , R 2 , and ρ be positive constants with R 1 R 2 . Let T 0 ( 0 , ) and T ( 0 , T 0 ] . Suppose that r 0 = ρ 0 + ρ , μ 0 , ν 0 , and κ 0 satisfy (b), (e), and (f) of Theorem 1. Then, the following assertions hold.
(1)
For any d L p ( ( 0 , T ) , H q 1 ( Ω ) ) and f L p ( ( 0 , T ) , L q ( Ω ) N ) , (33) admits a unique solution ( ρ , u ) with
ρ H p 1 ( ( 0 , T ) , H q 1 ( Ω ) ) L p ( ( 0 , T ) , H q 3 ( Ω ) ) , u H p 1 ( ( 0 , T ) , L q ( Ω ) N ) L p ( ( 0 , T ) , H q 2 ( Ω ) N ) .
(2)
The solution ( ρ , u ) satisfies
t ρ L p ( ( 0 , T ) , H q 1 ( Ω ) ) + ρ L p ( ( 0 , T ) , H q 3 ( Ω ) ) + t u L p ( ( 0 , T ) , L q ( Ω ) N ) + u L p ( ( 0 , T ) , H q 2 ( Ω ) N ) M 1 d L p ( ( 0 , T ) , H q 1 ( Ω ) ) + f L p ( ( 0 , T ) , L q ( Ω ) N )
for some positive constant M 1 = M 1 ( N , p , q , R , R 1 , R 2 , T 0 , ρ ) . In particular, M 1 is independent of T.
Proof. 
We apply Proposition 5 with (25)–(33). Notice that Assumption 1 is satisfied by our assumption about r 0 , μ 0 , ν 0 , and κ 0 .
Let d ˜ and f ˜ be the zero extensions of d and f , respectively, i.e.,
d ˜ = d for t ( 0 , T ) , 0 for t ( T , ) , f ˜ = f for t ( 0 , T ) , for t ( T , ) .
Then
e δ 2 t d ˜ L p ( R + , H q 1 ( Ω ) ) , e δ 2 t f ˜ L p ( R + , L q ( Ω ) N ) ,
where δ 2 is given by Proposition 5, which yields the solution ( ρ ˜ , u ˜ ) to
t ρ ˜ + r 0 div u ˜ = d ˜ in Ω × R + , t u ˜ r 0 1 κ 0 Div ( κ 0 1 S 0 ( u ˜ ) + r 0 Δ ρ ˜ I ) = f ˜ in Ω × R + , n · ρ ˜ = 0 , u ˜ = 0 on Γ D × R + , n · ρ ˜ = 0 , ( D ( u ˜ ) n ) τ = 0 , u ˜ · n = 0 on Γ S × R + , ( ρ ˜ , u ˜ ) | t = 0 = ( 0 , 0 ) in Ω .
In addition, ( ρ ˜ , u ˜ ) satisfies
e δ 2 t t ρ ˜ L p ( R + , H q 1 ( Ω ) ) + e δ 2 t ρ ˜ L p ( R + , H q 3 ( Ω ) ) + e δ 2 t t u ˜ L p ( R + , L q ( Ω ) N ) + e δ 2 t u ˜ L p ( R + , H q 2 ( Ω ) N ) C e δ 2 t d ˜ L p ( R + , H q 1 ( Ω ) ) + e δ 2 t f ˜ L p ( R + , H q 1 ( Ω ) ) .
Combining this inequality with
e δ 2 t d ˜ L p ( R + , H q 1 ( Ω ) ) d L p ( ( 0 , T ) , H q 1 ( Ω ) ) , e δ 2 t f ˜ L p ( R + , L q ( Ω ) N ) f L p ( ( 0 , T ) , L q ( Ω ) N ) ,
shows that
e δ 2 t t ρ ˜ L p ( R + , H q 1 ( Ω ) ) + e δ 2 t ρ ˜ L p ( R + , H q 3 ( Ω ) ) + e δ 2 t t u ˜ L p ( R + , L q ( Ω ) N ) + e δ 2 t u ˜ L p ( R + , H q 2 ( Ω ) N ) C d L p ( ( 0 , T ) , H q 1 ( Ω ) ) + f L p ( ( 0 , T ) , L q ( Ω ) N ) ,
where C is a positive constant independent of T.
Let ( ρ , u ) be the restriction of ( ρ ˜ , u ˜ ) to ( 0 , T ) . Then, ( ρ , u ) becomes a solution to (33). In addition, since
t ρ L p ( ( 0 , T ) , H q 1 ( Ω ) ) + ρ L p ( ( 0 , T ) , H q 3 ( Ω ) ) e δ 2 T e δ 2 t t ρ L p ( ( 0 , T ) , H q 1 ( Ω ) ) + e δ 2 t ρ L p ( ( 0 , T ) , H q 3 ( Ω ) ) e δ 2 T 0 e δ 2 t t ρ ˜ L p ( R + , H q 1 ( Ω ) ) + e δ 2 t ρ ˜ L p ( R + , H q 3 ( Ω ) ) ,
(34) gives us
t ρ L p ( ( 0 , T ) , H q 1 ( Ω ) ) + ρ L p ( ( 0 , T ) , H q 3 ( Ω ) ) C e δ 2 T 0 d L p ( ( 0 , T ) , H q 1 ( Ω ) ) + f L p ( ( 0 , T ) , L q ( Ω ) N ) .
Analogously,
t u L p ( ( 0 , T ) , L q ( Ω ) N ) + u L p ( ( 0 , T ) , H q 2 ( Ω ) N ) C e δ 2 T 0 d L p ( ( 0 , T ) , H q 1 ( Ω ) ) + f L p ( ( 0 , T ) , L q ( Ω ) N ) .
The last two estimates demonstrate that ( ρ , u ) satisfies the desired estimate. The uniqueness of solutions can be proved in the same manner as in Proposition 5. This completes the proof of Corollary 2. □

6. Local Solvability of the Nonlinear Problem

This section proves our main result of this paper, i.e., Theorem 1. To this end, we first introduce several embedding properties. We next estimate nonlinear terms. Finally, we prove Theorem 1. Throughout this section, we assume that Ω is a uniform C 3 domain in R N for N 2 .

6.1. Embedding Properties

Recall that ( · , · ) θ , p is the real interpolation functor for θ ( 0 , 1 ) and p ( 1 , ) . We then have the following lemma; see, e.g., [20] (Section 1.4).
Lemma 5.
Let p ( 1 , ) . Let X , Y be Banach spaces so that Y is a dense subspace of X and Y is continuously embedded into X. Then,
H p 1 ( R + , X ) L p ( R + , Y ) C ( [ 0 , ) , ( X , Y ) 1 1 / p , p )
and
sup t [ 0 , ) f ( t ) ( X , Y ) 1 1 / p , p f H p 1 ( R + , X ) p + f L p ( R + , Y ) p 1 / p
for any f H p 1 ( R + , X ) L p ( R + , Y ) .
Let us recall
( H q 1 ( Ω ) , H q 3 ( Ω ) ) 1 1 / p , p = B q , p 3 2 / p ( Ω ) , ( L q ( Ω ) , H q 2 ( Ω ) ) 1 1 / p , p = B q , p 2 2 / p ( Ω ) .
Lemma 5 gives us
Lemma 6.
Let p , q ( 1 , ) . Then, the following assertions hold.
(1)
There holds
H p 1 ( R + , H q 1 ( Ω ) ) L p ( R + , H q 3 ( Ω ) ) C ( [ 0 , ) , B q , p 3 2 / p ( Ω ) )
and
sup t [ 0 , ) f ( t ) B q , p 3 2 / p ( Ω ) C f H p 1 ( R + , H q 1 ( Ω ) ) + f L p ( R + , H q 3 ( Ω ) )
for any f H p 1 ( R + , H q 1 ( Ω ) ) L p ( R + , H q 3 ( Ω ) ) with a positive constant C.
(2)
There holds
H p 1 ( R + , L q ( Ω ) ) L p ( R + , H q 2 ( Ω ) ) C ( [ 0 , ) , B q , p 2 2 / p ( Ω ) )
and
sup t [ 0 , ) f ( t ) B q , p 2 2 / p ( Ω ) C f H p 1 ( R + , L q ( Ω ) ) + f L p ( R + , H q 2 ( Ω ) )
for any f H p 1 ( R + , L q ( Ω ) ) L p ( R + , H q 2 ( Ω ) ) with a positive constant C.
We next prove.
Lemma 7.
Let p ( 1 , ) and q ( N , ) . Suppose that T > 0 or T = . Then L p ( ( 0 , T ) , H q 1 ( Ω ) ) L p ( ( 0 , T ) , L ( Ω ) ) and
f L p ( ( 0 , T ) , L ( Ω ) ) C f L p ( ( 0 , T ) , L q ( Ω ) ) N q f L p ( ( 0 , T ) , L q ( Ω ) ) 1 N q + f L p ( ( 0 , T ) , L q ( Ω ) )
for any f L p ( ( 0 , T ) , H q 1 ( Ω ) ) , where C is a positive constant independent of T.
Proof. 
Since q > N , it holds that H q 1 ( Ω ) L ( Ω ) and
f L ( Ω ) C f L q ( Ω ) N q f L q ( Ω ) 1 N q + f L q ( Ω )
for any f H q 1 ( Ω ) . This inequality shows that for θ = N / q ( 0 , 1 )
f L p ( ( 0 , T ) , L ( Ω ) ) p = 0 T f ( t ) L ( Ω ) p d t C 0 T f ( t ) L q ( Ω ) p θ f ( t ) L q ( Ω ) p ( 1 θ ) d t + 0 T f ( t ) L q ( Ω ) p d t .
On the other hand, H o ¨ lder’s inequality gives us
0 T f ( t ) L q ( Ω ) p θ f ( t ) L q ( Ω ) p ( 1 θ ) d t 0 T f ( t ) L q ( Ω ) p θ 0 T f ( t ) L q ( Ω ) p d t 1 θ .
Summing up the last two inequalities, we have
f L p ( ( 0 , T ) , L ( Ω ) ) p C f L p ( ( 0 , T ) , L q ( Ω ) ) p θ f L p ( ( 0 , T ) , L q ( Ω ) ) p ( 1 θ ) + f L p ( ( 0 , T ) , L q ( Ω ) ) p .
This yields the desired inequality and completes the proof of Lemma 7. □
Let us next prove:
Lemma 8.
Suppose that T > 0 or T = . Then, the following assertions hold.
(1)
Let p , q ( 1 , ) . Then, for any f L p ( ( 0 , T ) , H q 2 ( Ω ) )
f L p ( ( 0 , T ) , H q 1 ( Ω ) ) C f L p ( ( 0 , T ) , H q 2 ( Ω ) ) 1 2 f L p ( ( 0 , T ) , L q ( Ω ) ) 1 2 ,
where C is a positive constant independent of T.
(2)
Let p ( 1 , ) and q ( N , ) . Then, for any f L p ( ( 0 , T ) , H q 2 ( Ω ) )
f L p ( ( 0 , T ) , H 1 ( Ω ) ) C f L p ( ( 0 , T ) , H q 2 ( Ω ) ) 1 2 × f L p ( ( 0 , T ) , H q 2 ( Ω ) ) N 2 q f L p ( ( 0 , T ) , L q ( Ω ) ) 1 2 1 N q + f L p ( ( 0 , T ) , L q ( Ω ) ) 1 2 ,
where C is a positive constant independent of T.
Proof. 
(1) Let [ · , · ] θ be the complex interpolation functor for θ ( 0 , 1 ) ; see, e.g., [27] (Definition 1.38). It follows from Remark 2 (d) of Subsection 2.4.2 in [21] that [ L q ( Ω ) , H q 2 ( Ω ) ] 1 / 2 = H q 1 ( Ω ) , which, combined with Theorem 1.9.3 (f) in [21], demonstrates that for any g H q 2 ( Ω )
g H q 1 ( Ω ) C g H q 2 ( Ω ) 1 2 g L q ( Ω ) 1 2 .
Using this inequality, we observe that
f L p ( ( 0 , T ) , H q 1 ( Ω ) ) p = 0 T f ( t ) H q 1 ( Ω ) p d t C 0 T f ( t ) H q 2 ( Ω ) p 2 f ( t ) L q ( Ω ) p 2 d t C 0 T f ( t ) H q 2 ( Ω ) p d t 1 2 0 T f ( t ) L q ( Ω ) p d t 1 2 .
This yields the desired inequality.
(2) Let us first consider j f for j = 1 , , N . Lemma 7 gives us
j f L p ( ( 0 , T ) , L ( Ω ) ) C j f L p ( ( 0 , T ) , L q ( Ω ) ) N q j f L p ( ( 0 , T ) , L q ( Ω ) ) 1 N q + j f L p ( ( 0 , T ) , L q ( Ω ) ) C f L p ( ( 0 , T ) , H q 2 ( Ω ) ) N q f L p ( ( 0 , T ) , H q 1 ( Ω ) ) 1 N q + f L p ( ( 0 , T ) , H q 1 ( Ω ) ) ,
which, combined with (1), demonstrates
j f L p ( ( 0 , T ) , L ( Ω ) ) C f L p ( ( 0 , T ) , H q 2 ( Ω ) ) 1 2 f L p ( ( 0 , T ) , H q 2 ( Ω ) ) N 2 q f L p ( ( 0 , T ) , L q ( Ω ) ) 1 2 1 N q + f L p ( ( 0 , T ) , L q ( Ω ) ) 1 2 .
This estimate holds with j f L p ( ( 0 , T ) , L ( Ω ) ) replaced by f L p ( ( 0 , T ) , L ( Ω ) ) . The desired estimate thus holds. This completes the proof of Lemma 8 □
From Lemma 8, we obtain
Lemma 9.
Let p ( 1 , ) and q ( N , ) , and let T ( 0 , ) . Then there exists a positive constant C, independent of T, such that the following assertions hold.
(1)
For any f 0 H p 1 ( ( 0 , T ) , L q ( Ω ) ) L p ( ( 0 , T ) , H q 2 ( Ω ) )
f L p ( ( 0 , T ) , H q 1 ( Ω ) ) C T 1 2 f H p 1 ( ( 0 , T ) , L q ( Ω ) ) + f L p ( ( 0 , T ) , H q 2 ( Ω ) ) .
(2)
For any f 0 H p 1 ( ( 0 , T ) , L q ( Ω ) ) L p ( ( 0 , T ) , H q 2 ( Ω ) )
f L p ( ( 0 , T ) , H 1 ( Ω ) ) C T 1 2 1 N q + T 1 2 × f H p 1 ( ( 0 , T ) , L q ( Ω ) ) + f L p ( ( 0 , T ) , H q 2 ( Ω ) ) .
(3)
For any g 0 H p 1 ( ( 0 , T ) , H q 1 ( Ω ) ) L p ( ( 0 , T ) , H q 3 ( Ω ) )
g L p ( ( 0 , T ) , H q 2 ( Ω ) ) C T 1 2 g H p 1 ( ( 0 , T ) , H q 1 ( Ω ) ) + g L p ( ( 0 , T ) , H q 3 ( Ω ) ) .
(4)
For any g 0 H p 1 ( ( 0 , T ) , H q 1 ( Ω ) ) L p ( ( 0 , T ) , H q 3 ( Ω ) )
g L p ( ( 0 , T ) , H 2 ( Ω ) ) C T 1 2 1 N q + T 1 2 × g H p 1 ( ( 0 , T ) , H q 1 ( Ω ) ) + g L p ( ( 0 , T ) , H q 3 ( Ω ) ) .
Proof. 
(1) Let f 0 H p 1 ( ( 0 , T ) , L q ( Ω ) ) L p ( ( 0 , T ) , H q 2 ( Ω ) ) . Since
f L p ( ( 0 , T ) , L q ( Ω ) ) T 1 / p f L ( ( 0 , T ) , L q ( Ω ) ) ,
Lemma 8(1) shows that
f L p ( ( 0 , T ) , H q 1 ( Ω ) ) T 1 2 p f L p ( ( 0 , T ) , H q 2 ( Ω ) ) 1 2 f L ( ( 0 , T ) , L q ( Ω ) ) 1 2 .
On the other hand, f | t = 0 = 0 gives us
f ( x , t ) = 0 t s f ( x , s ) d s ,
which implies
f L ( ( 0 , T ) , L q ( Ω ) ) 0 T s f ( s ) L q ( Ω ) d s T 1 / p f H p 1 ( ( 0 , T ) , L q ( Ω ) )
for p = p / ( p 1 ) . Combining this with the last estimate of f L p ( ( 0 , T ) , H q 1 ( Ω ) ) yields the desired inequality.
(2) The desired inequality follows from Lemma 8b in the same manner as in the proof of (1), so that the detailed proof may be omitted.
(3), (4) Let j = 1 , , N . Since j g , g 0 H p 1 ( ( 0 , T ) , L q ( Ω ) ) L p ( ( 0 , T ) , H q 2 ( Ω ) ) , the desired inequalities of (3) and (4) immediately follow from (1) and (2), respectively. This completes the proof of Lemma 9. □
Recall K p , q ; T = K p , q ; T 1 × K p , q ; T 2 and 0 K p , q ; T = 0 K p , q ; T 1 × 0 K p , q ; T 2 given by SubSection 3.1. We then have
Proposition 6.
Let p ( 1 , ) , q ( N , ) , and T ( 0 , ) . Then, there exists a positive constant C, independent of T, such that the following assertions hold.
(1)
ρ L p ( ( 0 , T ) , H q 2 ( Ω ) ) C T 1 / 2 ρ K p , q ; T 1 for any ρ 0 K p , q ; T 1 .
(2)
ρ L p ( ( 0 , T ) , H 2 ( Ω ) ) C ( T ( 1 N / q ) / 2 + T 1 / 2 ) ρ K p , q ; T 1 for any ρ 0 K p , q ; T 1 .
(3)
ρ L ( ( 0 , T ) , H q 1 ( Ω ) ) C T 1 1 / p ρ K p , q ; T 1 for any ρ 0 K p , q ; T 1 .
(4)
ρ L ( ( 0 , T ) , L ( Ω ) ) C T 1 1 / p ρ K p , q ; T 1 for any ρ 0 K p , q ; T 1 .
(5)
u L p ( ( 0 , T ) , H q 1 ( Ω ) N ) C T 1 / 2 u K p , q ; T 2 for any u 0 K p , q ; T 2 .
(6)
u L p ( ( 0 , T ) , H 1 ( Ω ) N ) C ( T ( 1 N / q ) / 2 + T 1 / 2 ) u K p , q ; T 2 for any u 0 K p , q ; T 2 .
(7)
u L ( ( 0 , T ) , L q ( Ω ) N ) C T 1 1 / p u K p , q ; T 2 for any u 0 K p , q ; T 2 .
Proof. 
The desired inequalities of (1), (2), (5), and (6) follow from Lemma 9 immediately. The proofs of (3) and (7) are similar to (35) and (36), so that the detailed proof may be omitted. Since H q 1 ( Ω ) is continuously embedded into L ( Ω ) by the assumption q > N , the desired inequality of (4) follows from (3). This completes the proof of Proposition 6. □
Let T 0 ( 0 , ) and T ( 0 , T 0 ] . Since
T 1 / 2 T 0 1 / 2 , T ( 1 N / q ) / 2 T 0 ( 1 N / q ) / 2 , T 1 1 / p T 0 1 1 / p
for p ( 1 , ) and q ( N , ) , the next proposition follows from Proposition 6 immediately.
Proposition 7.
Let p ( 1 , ) and q ( N , ) . Let T 0 ( 0 , ) and T ( 0 , T 0 ] . Then, there exists a positive constant C T 0 depending on T 0 , but independent of T, such that the following assertions hold.
(1)
ρ L p ( ( 0 , T ) , H 2 ( Ω ) ) C T 0 ρ K p , q ; T 1 for any ρ 0 K p , q ; T 1 .
(2)
ρ L ( ( 0 , T ) , H q 1 ( Ω ) ) C T 0 ρ K p , q ; T 1 for any ρ 0 K p , q ; T 1 .
(3)
ρ L ( ( 0 , T ) , L ( Ω ) ) C T 0 ρ K p , q ; T 1 for any ρ 0 K p , q ; T 1 .
(4)
u L p ( ( 0 , T ) , H 1 ( Ω ) N ) C T 0 u K p , q ; T 2 for any u 0 K p , q ; T 2 .
(5)
u L ( ( 0 , T ) , L q ( Ω ) N ) C T 0 u K p , q ; T 2 for any u 0 K p , q ; T 2 .
We finally introduce embedding properties for the solution ( ρ ^ , u ^ ) of (8).
Proposition 8.
Let p ( 1 , ) and q ( N , ) , and let R, R 1 , R 2 , and ρ be positive constants with R 1 R 2 . Let T 0 ( 0 , ) and T ( 0 , T 0 ] . Suppose that ( ρ 0 , u 0 ) D q , p ( Ω ) and that r 0 = ρ 0 + ρ , μ 0 , ν 0 , and κ 0 satisfy b, e, and f of Theorem 1. Then, there exists a positive constant C T 0 depending on N, p, q, R, R 1 , R 2 , T 0 , and ρ , but independent of T, such that the following assertions hold.
(1)
( ρ ^ , u ^ ) K p , q ; T C T 0 ( ρ 0 , u 0 ) D q , p ( Ω ) .
(2)
( ρ ^ , u ^ ) L ( ( 0 , T ) , B q , p 3 2 / p ( Ω ) × B q , p 2 2 / p ( Ω ) N ) C T 0 ( ρ 0 , u 0 ) D q , p ( Ω ) .
(3)
( ρ ^ , u ^ ) L ( ( 0 , T ) , H q 1 ( Ω ) × L q ( Ω ) N ) C T 0 ( ρ 0 , u 0 ) D q , p ( Ω ) .
(4)
( ρ ^ , u ^ ) L p ( ( 0 , T ) , H 2 ( Ω ) × H 1 ( Ω ) N ) C T 0 ( ρ 0 , u 0 ) D q , p ( Ω ) .
(5)
ρ ^ L ( ( 0 , T ) , L ( Ω ) ) C T 0 ( ρ 0 , u 0 ) D q , p ( Ω ) .
(6)
ρ ^ ρ 0 L ( ( 0 , T ) , H q 1 ( Ω ) ) C T 0 T 1 1 / p ( ρ 0 , u 0 ) D q , p ( Ω ) .
(7)
ρ ^ ρ 0 L ( ( 0 , T ) , H q 1 ( Ω ) ) C T 0 ( ρ 0 , u 0 ) D q , p ( Ω ) .
(8)
ρ ^ ρ 0 L ( ( 0 , T ) , L ( Ω ) ) C T 0 T 1 1 / p ( ρ 0 , u 0 ) D q , p ( Ω ) .
(9)
ρ ^ ρ 0 L ( ( 0 , T ) , L ( Ω ) ) C T 0 ( ρ 0 , u 0 ) D q , p ( Ω ) .
Proof. 
(1), (2) The desired inequalities follow from Corollary 1 and Lemma 6 in the same manner as in the proof of Corollary 2.
(3) Since B q , p 3 2 / p ( Ω ) and B q , p 2 2 / p ( Ω ) are continuously embedded into H q 1 ( Ω ) and L q ( Ω ) , respectively, the desired inequality follows from (2).
(4), (5) By the assumption q > N , we observe for m N that H q m ( Ω ) is continuously embedded into H m 1 ( Ω ) . Thus, the desired inequality of (4) follows from (1), while one of (5) follows from (3).
(6) Since
ρ ^ ( x , t ) ρ 0 ( x ) = 0 t s ρ ^ ( x , s ) d s ,
we observe by H o ¨ lder’s inequality that
ρ ^ ρ 0 L ( ( 0 , T ) , H q 1 ( Ω ) ) 0 T s ρ ^ ( s ) H q 1 ( Ω ) d s T 1 1 / p t ρ ^ L p ( ( 0 , T ) , H q 1 ( Ω ) ) .
Combining this inequality with (1) demonstrates the desired inequality.
(7) Since T 1 1 / p T 0 1 1 / p , the desired inequality follows from (6) immediately.
(8) and (9) Since H q 1 ( Ω ) is continuously embedded into L ( Ω ) as mentioned above, the desired inequalities of (8) and (9) follow from (6) and (7), respectively. This completes the proof of Proposition 8. □

6.2. Estimates of Nonlinear Terms

This subsection estimates the nonlinear terms D ( ρ , u ) and F ( ρ , u ) given by Section 3.2. Throughout this subsection, we assume
Assumption 2.
Let p ( 1 , ) , q ( N , ) , and T 0 ( 0 , ) . Let R, R 1 , R 2 , and ρ be positive constants with R 1 R 2 . Suppose that ( ρ 0 , u 0 ) D q , p ( Ω ) and that (b), (e), and (f) of Theorem 1 hold.
Let us define for T ( 0 , T 0 ]
φ ( T ) = T 1 2 1 N q + T 1 2 + T 1 1 p , ψ ( T ) = sup t [ 0 , T ] ( μ ( t ) μ 0 H 1 ( Ω ) + ν ( t ) ν 0 H 1 ( Ω ) + κ ( t ) κ 0 H 1 ( Ω ) ) .
We then observe that
lim T 0 φ ( T ) = 0 , lim T 0 ψ ( T ) = 0 ,
where we note Remark 1.
Let us decompose the nonlinear terms into the lower order terms and the the highest order terms. Define
D 1 ( ρ , u ) = u · ρ , D 2 ( ρ , u ) = ( ρ ρ 0 ) div u .
One then sees that
D ( ρ , u ) = D 1 ( ρ , u ) D 2 ( ρ , u ) ,
where D 1 ( ρ , u ) and D 2 ( ρ , u ) are corresponding to the lower order and the highest order, respectively.
We next consider F ( ρ , u ) . Let us observe that
Div ( S ( u ) S 0 ( u ) ) = ( μ μ 0 ) Δ u + ( ν ν 0 ) div u + D ( u ) ( μ μ 0 ) + ( div u ) ( ( ν ν 0 ) ( μ μ 0 ) )
and that
Div ( κ κ 0 ) ( ρ + ρ ) Δ ρ I + κ 0 ( ρ ρ 0 ) Δ ρ I + κ | ρ | 2 2 I κ ρ ρ = ( ( κ κ 0 ) ) ( ρ + ρ ) Δ ρ + ( κ κ 0 ) ( ρ ) Δ ρ + ( κ κ 0 ) ( ρ + ρ ) Δ ρ + ( κ 0 ) ( ρ ρ 0 ) Δ ρ + κ 0 ( ( ρ ρ 0 ) ) Δ ρ + κ 0 ( ρ ρ 0 ) Δ ρ + ( κ ) | ρ | 2 2 ( ρ ρ ) κ κ ( ρ ) Δ ρ .
Define
F 1 ( ρ , u ) = ( ρ + ρ ) u · u + D ( u ) ( μ μ 0 ) + ( div u ) ( ( ν ν 0 ) ( μ μ 0 ) ) + ( ( κ κ 0 ) ) ( ρ + ρ ) Δ ρ + ( κ κ 0 ) ( ρ ) Δ ρ + ( κ 0 ) ( ρ ρ 0 ) Δ ρ + κ 0 ( ( ρ ρ 0 ) ) Δ ρ + ( κ ) | ρ | 2 2 ( ρ ρ ) κ κ ( ρ ) Δ ρ + ( ρ + ρ ) b + ( κ 0 1 ( ν 0 μ 0 ) div u I + r 0 Δ ρ I ) κ 0
and
F 2 ( ρ , u ) = ( ρ ρ 0 ) t u + ( μ μ 0 ) Δ u + ( ν ν 0 ) div u + ( κ κ 0 ) ( ρ + ρ ) Δ ρ + κ 0 ( ρ ρ 0 ) Δ ρ .
These give us
F ( ρ , u ) = r 0 1 F 1 ( ρ , u ) + r 0 1 F 2 ( ρ , u ) r 0 1 P ( ρ + ρ ) ρ ,
where F 1 ( ρ , u ) and F 2 ( ρ , u ) are corresponding to the lower order and the highest order, respectively.
Let us now estimate D 1 ( ρ , u ) .
Lemma 10.
Suppose that Assumption 2 holds. Then, there exists a positive constant C = C ( N , p , q , R , R 1 , R 2 , T 0 , ρ ) , such that for any T ( 0 , T 0 ] and ( ρ i , u i ) 0 K p , q ; T , i = 1 , 2 ,
D 1 ( ρ 1 + ρ ^ , u 1 + u ^ ) D 1 ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) C φ ( T ) ( ρ 2 ρ 1 , u 2 u 1 ) K p , q ; T × ( ρ 1 , u 1 ) K p , q ; T + ( ρ 2 , u 2 ) K p , q ; T + ( ρ 0 , u 0 ) D q , p ( Ω ) .
Proof. 
Let us write
D 1 ( ρ 1 + ρ ^ , u 1 + u ^ ) D 1 ( ρ 2 + ρ ^ , u 2 + u ^ ) = ( u 1 u 2 ) · ( ρ 1 + ρ ^ ) + ( u 2 + u ^ ) · ( ρ 1 ρ 2 ) = : I 1 + I 2 .
For k = 1 , , N
k I 1 = k ( u 1 u 2 ) · ( ρ 1 + ρ ^ ) + ( u 1 u 2 ) · k ( ρ 1 + ρ ^ ) ,
which gives us
k I 1 L p ( ( 0 , T ) , L q ( Ω ) ) k ( u 1 u 2 ) L p ( ( 0 , T ) , L ( Ω ) N ) ( ρ 1 + ρ ^ ) L ( ( 0 , T ) , L q ( Ω ) N ) + u 1 u 2 L ( ( 0 , T ) , L q ( Ω ) N ) k ( ρ 1 + ρ ^ ) L p ( ( 0 , T ) , L ( Ω ) N ) u 1 u 2 L p ( ( 0 , T ) , H 1 ( Ω ) N ) ρ 1 + ρ ^ L ( ( 0 , T ) , H q 1 ( Ω ) ) + u 1 u 2 L ( ( 0 , T ) , L q ( Ω ) N ) ρ 1 + ρ ^ L p ( ( 0 , T ) , H 2 ( Ω ) ) .
By Propositions 6–8, we observe that
k I 1 L p ( ( 0 , T ) , L q ( Ω ) ) C φ ( T ) u 1 u 2 K p , q ; T 2 ρ 1 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) .
Analogously, we obtain from Propositions 6–8
I 1 L p ( ( 0 , T ) , L q ( Ω ) ) C φ ( T ) u 1 u 2 K p , q ; T 2 ρ 1 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) , I 2 L p ( ( 0 , T ) , L q ( Ω ) ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 u 2 K p , q ; T 2 + ( ρ 0 , u 0 ) D q , p ( Ω ) , k I 2 L p ( ( 0 , T ) , L q ( Ω ) ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 u 2 K p , q ; T 2 + ( ρ 0 , u 0 ) D q , p ( Ω ) .
Summing up the above estimates of I 1 , I 2 , k I 1 , and k I 2 , we complete the proof of Lemma 10. □
In the same manner as in Lemma 10, we have
Lemma 11.
Suppose that Assumption 2 holds. Then, for any T ( 0 , T 0 ]
D 1 ( ρ ^ , u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) C ( ρ 0 , u 0 ) D q , p ( Ω ) 2 ,
where C = C ( N , p , q , R , R 1 , R 2 , T 0 , ρ ) is a positive constant independent of T.
We next consider F 1 ( ρ , u ) .
Lemma 12.
Suppose that Assumption 2 holds and b L p ( ( 0 , T 0 ) , L q ( Ω ) N ) . Then, there exists a positive constant C = C ( N , p , q , R , R 1 , R 2 , T 0 , ρ ) , such that for any T ( 0 , T 0 ] and ( ρ i , u i ) 0 K p , q ; T , i = 1 , 2 ,
F 1 ( ρ 1 + ρ ^ , u 1 + u ^ ) F 1 ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ( ρ 2 ρ 1 , u 2 u 1 ) K p , q ; T { b L p ( ( 0 , T 0 ) , L q ( Ω ) N ) + j = 0 2 ( ρ 1 , u 1 ) K p , q ; T + ( ρ 2 , u 2 ) K p , q ; T + ( ρ 0 , u 0 ) D q , p ( Ω ) j } .
Proof. 
Notice that the quadratic terms of F 1 ( ρ , u ) :
u · u , ( ( κ κ 0 ) ) ρ Δ ρ , ( κ κ 0 ) ( ρ ) Δ ρ , ( κ ) | ρ | 2 2 , ( ρ ρ ) κ , κ ( ρ ) Δ ρ
can be treated in the same manner as in the proof of Lemma 10. In this proof, we focus on estimating the following terms:
F 1 ( u ) = D ( u ) ( μ μ 0 ) , F 2 ( u ) = ( div u ) ( ( ν ν 0 ) ( μ μ 0 ) ) , F 3 ( ρ ) = ( ( κ κ 0 ) ) Δ ρ , F 4 ( ρ ) = ( κ 0 ) ( ρ ρ 0 ) Δ ρ , F 5 ( ρ ) = κ 0 ( ( ρ ρ 0 ) ) Δ ρ , F 6 ( ρ ) = ( ρ + ρ ) b ,
and also
F 7 ( ρ , u ) = ( κ 0 1 ( ν 0 μ 0 ) div u I + r 0 Δ ρ I ) κ 0 , F 8 ( ρ , u ) = ρ u · u .
Let us first consider F 1 ( u ) . It can be written as
F 1 ( u 1 + u ^ ) F 1 ( u 2 + u ^ ) = D ( u 1 u 2 ) ( μ μ 0 ) .
It thus follows from Proposition 6 that
F 1 ( u 1 + u ^ ) F 1 ( u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) μ L ( ( 0 , T 0 ) , H 1 ( Ω ) ) + μ 0 H 1 ( Ω ) u 1 u 2 L p ( ( 0 , T ) , H q 1 ( Ω ) ) C φ ( T ) u 1 u 2 K p , q ; T 2 .
Analogously,
F 2 ( u 1 + u ^ ) F 2 ( u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) u 1 u 2 K p , q ; T 2 , F 3 ( ρ 1 + ρ ^ ) F 4 ( ρ 2 + ρ ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 ,
and
F 7 ( ρ 1 + ρ ^ , u 1 + u ^ ) F 7 ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ( ρ 1 ρ 2 , u 1 u 2 ) K p , q ; T .
We next consider F 4 ( ρ ) . It can be written as
F 4 ( ρ 1 + ρ ^ ) F 4 ( ρ 2 + ρ ^ ) = ( κ 0 ) ( ρ 1 ρ 2 ) Δ ( ρ 1 + ρ ^ ) + ( κ 0 ) ( ρ 2 + ρ ^ ρ 0 ) Δ ( ρ 1 ρ 2 ) = : I 1 + I 2 .
Propositions 6–8 show that
I 1 L p ( ( 0 , T ) , L q ( Ω ) N ) κ 0 L ( Ω ) ρ 1 ρ 2 L ( ( 0 , T ) , L q ( Ω ) ) Δ ( ρ 1 + ρ ^ ) L p ( ( 0 , T ) , L ( Ω ) ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 ρ 1 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) , I 2 L p ( ( 0 , T ) , L q ( Ω ) N ) κ 0 L ( Ω ) ρ 2 L ( ( 0 , T ) , L q ( Ω ) ) + ρ ^ ρ 0 L ( ( 0 , T ) , L q ( Ω ) ) × Δ ( ρ 1 ρ 2 ) L p ( ( 0 , T ) , L ( Ω ) ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 ρ 2 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) .
Thus
F 4 ( ρ 1 + ρ ^ ) F 4 ( ρ 2 + ρ ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 ρ 1 K p , q ; T 1 + ρ 2 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) .
Analogously,
F 5 ( ρ 1 + ρ ^ ) F 5 ( ρ 2 + ρ ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 ρ 1 K p , q ; T 1 + ρ 2 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) .
We next consider F 6 ( ρ ) . Since
F 6 ( ρ 1 + ρ ^ ) F 6 ( ρ 2 + ρ ^ ) = ( ρ 1 ρ 2 ) b ,
it follows from Proposition 6 that
F 6 ( ρ 1 + ρ ^ ) F 6 ( ρ 2 + ρ ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) ρ 1 ρ 2 L ( ( 0 , T ) , L ( Ω ) ) b L p ( ( 0 , T 0 ) , L q ( Ω ) N ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 b L p ( ( 0 , T 0 ) , L q ( Ω ) N ) .
We finally consider F 8 ( ρ , u ) . It can be written as
F 8 ( ρ 1 + ρ ^ , u 1 + u ^ ) F 8 ( ρ 2 + ρ ^ , u 2 + u ^ ) = ( ρ 1 ρ 2 ) ( u 1 + u ^ ) · ( u 1 + u ^ ) + ( ρ 2 + ρ ^ ) ( u 1 u 2 ) · ( u 1 + u ^ ) + ( ρ 2 + ρ ^ ) ( u 2 + u ^ ) · ( u 1 u 2 ) = : I 3 + I 4 + I 5 .
By Propositions 6–8, we observe that
I 3 L p ( ( 0 , T ) , L q ( Ω ) N ) ρ 1 ρ 2 L ( ( 0 , T ) , L ( Ω ) ) u 1 + u ^ L ( ( 0 , T ) , L q ( Ω ) N ) × ( u 1 + u ^ ) L p ( ( 0 , T ) , L ( Ω ) N × N ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 u 1 K p , q ; T 2 + ( ρ 0 , u 0 ) D q , p ( Ω ) 2 .
Analogously,
I 4 L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) u 1 u 2 K p , q ; T 2 ρ 2 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) × u 1 K p , q ; T 2 + ( ρ 0 , u 0 ) D q , p ( Ω ) , I 5 L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) u 1 u 2 K p , q ; T 2 ρ 2 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) × u 2 K p , q ; T 2 + ( ρ 0 , u 0 ) D q , p ( Ω ) .
It thus holds that
F 8 ( ρ 1 + ρ ^ , u 1 + u ^ ) F 8 ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ( ρ 1 ρ 2 , u 1 u 2 ) K p , q ; T × ( ρ 1 , u 1 ) K p , q ; T + ( ρ 2 , u 2 ) K p , q ; T + ( ρ 0 , u 0 ) D q , p ( Ω ) 2 .
Summing up the above estimates of the quadratic terms and F 1 , , F 8 , we have obtained the desired inequality. This completes the proof of Lemma 12. □
In the same manner as in Lemma 12, we have
Lemma 13.
Suppose that Assumption 2 holds and b L p ( ( 0 , T 0 ) , L q ( Ω ) N ) . Then, for any T ( 0 , T 0 ]
F 1 ( ρ ^ , u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C ( ρ 0 , u 0 ) D q , p ( Ω ) + 1 b L p ( ( 0 , T 0 ) , L q ( Ω ) N ) + j = 1 3 ( ρ 0 , u 0 ) D q , p ( Ω ) j ,
where C = C ( N , p , q , R , R 1 , R 2 , T 0 , ρ ) is a positive constant independent of T.
We next consider the highest order terms D 2 ( ρ , u ) and F 2 ( ρ , u ) .
Lemma 14.
Suppose that Assumption 2 holds. Then, there exists a positive constant C = C ( N , p , q , R , R 1 , R 2 , T 0 , ρ ) , such that for any T ( 0 , T 0 ] and ( ρ i , u i ) 0 K p , q ; T , i = 1 , 2 ,
D 2 ( ρ 1 + ρ ^ , u 1 + u ^ ) D 2 ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) C φ ( T ) ( ρ 2 ρ 1 , u 2 u 1 ) K p , q ; T × ( ρ 1 , u 1 ) K p , q ; T + ( ρ 2 , u 2 ) K p , q ; T + ( ρ 0 , u 0 ) D q , p ( Ω ) , F 2 ( ρ 1 + ρ ^ , u 1 + u ^ ) F 2 ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C ( ψ ( T ) + φ ( T ) ) ( ρ 2 ρ 1 , u 2 u 1 ) K p , q ; T × j = 0 1 ( ρ 1 , u 1 ) K p , q ; T + ( ρ 2 , u 2 ) K p , q ; T + ( ρ 0 , u 0 ) D q , p ( Ω ) j .
Proof. 
Let us first consider D 2 ( ρ , u ) . It can be written as
D 2 ( ρ 1 + ρ ^ , u 1 + u ^ ) D 2 ( ρ 2 + ρ ^ , u 2 + u ^ ) = ( ρ 1 ρ 2 ) div ( u 1 + u ^ ) + ( ρ 2 + ρ ^ ρ 0 ) div ( u 1 u 2 ) .
Since H q 1 ( Ω ) is a Banach algebra by the assumption q > N , we observe that
D 2 ( ρ 1 + ρ ^ , u 1 + u ^ ) D 2 ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) C { ρ 1 ρ 2 L ( ( 0 , T ) , H q 1 ( Ω ) ) div ( u 1 + u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) + ρ 2 L ( ( 0 , T ) , H q 1 ( Ω ) ) + ρ ^ ρ 0 L ( ( 0 , T ) , H q 1 ( Ω ) ) × div ( u 1 u 2 ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) } .
Combining this inequality with Propositions 6 and 8 demonstrates the desired inequality of D 2 ( ρ , u ) .
We next consider F 2 ( ρ , u ) . To this end, we set
G 1 ( ρ , u ) = ( ρ ρ 0 ) t u , G 2 ( u ) = ( μ μ 0 ) Δ u , G 3 ( u ) = ( ν ν 0 ) div u , G 4 ( ρ ) = ( κ κ 0 ) ρ Δ ρ , G 5 ( ρ ) = ( κ κ 0 ) Δ ρ , G 6 ( ρ ) = κ 0 ( ρ ρ 0 ) Δ ρ .
We observe that
G 1 ( ρ 1 + ρ ^ , u 1 + u ^ ) G 1 ( ρ 2 + ρ ^ , u 2 + u ^ ) = ( ρ 1 ρ 2 ) t ( u 1 + u ^ ) + ( ρ 2 + ρ ^ ρ 0 ) t ( u 1 u 2 ) ,
which yields
G 1 ( ρ 1 + ρ ^ , u 1 + u ^ ) G 1 ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) ρ 1 ρ 2 L ( ( 0 , T ) , L ( Ω ) ) t ( u 1 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) + ρ 2 L ( ( 0 , T ) , L ( Ω ) ) + ρ ^ ρ 0 L ( ( 0 , T ) , L ( Ω ) ) × t ( u 1 u 2 ) L p ( ( 0 , T ) , L q ( Ω ) N ) .
Combining this inequality with Propositions 6 and 8 shows
G 1 ( ρ 1 + ρ ^ , u 1 + u ^ ) G 1 ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ( ρ 1 ρ 2 , u 1 u 2 ) K p , q ; T × u 1 K p , q ; T 2 + ρ 2 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) .
Analogously, we have for j = 4 , 6
G j ( ρ 1 + ρ ^ ) G j ( ρ 2 + ρ ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 ρ 1 K p , q ; T 1 + ρ 2 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) .
On the other hand, it is clear that for j = 2 , 3
G j ( u 1 + u ^ ) G j ( u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C ψ ( T ) u 1 u 2 K p , q ; T 2 , G 5 ( ρ 1 + ρ ^ ) G 5 ( ρ 2 + ρ ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C ψ ( T ) ρ 1 ρ 2 K p , q ; T 2 .
Summing up the above inequalities of G 1 , , G 6 , we have obtained the desired inequality of F 2 ( ρ , u ) . This completes the proof of Lemma 14. □
In the same manner as in Lemma 14, we have
Lemma 15.
Suppose that Assumption 2 holds. Then for any T ( 0 , T 0 ]
D 2 ( ρ ^ , u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) C ( ρ 0 , u 0 ) D q , p ( Ω ) 2 , F 2 ( ρ ^ , u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C j = 1 2 ( ρ 0 , u 0 ) D q , p ( Ω ) j ,
where C = C ( N , p , q , R , R 1 , R 2 , T 0 , ρ ) is a positive constant independent of T.
The pressure term is next estimated.
Lemma 16.
Suppose that Assumption 2 and (d) of Theorem 1 hold. Let ( ρ 0 , u 0 ) D q , p ( Ω ) R . Then, there exists a positive constant T 1 ( 0 , T 0 ] , depending on N, p, q, R, R 1 , R 2 , T 0 , and ρ , such that for any T ( 0 , T 1 ]
P ( ρ ^ + ρ ) ρ ^ L p ( ( 0 , T ) , L q ( Ω ) N ) C ( ρ 0 , u 0 ) D q , p ( Ω ) ,
where C = C ( N , p , q , R , R 1 , R 2 , T 0 , ρ ) is a positive constant independent of T.
Proof. 
Since ρ ^ + ρ = ρ ^ ρ 0 + ρ 0 + ρ , it holds that
ρ 0 + ρ L ( Ω ) ρ ^ ρ 0 L ( ( 0 , T ) , L ( Ω ) ) ρ ^ + ρ L ( ( 0 , T ) , L ( Ω ) ) ρ 0 + ρ L ( Ω ) + ρ ^ ρ 0 L ( ( 0 , T ) , L ( Ω ) ) ,
which, combined with (b) of Theorem 1 and Propositions 8, furnishes
ρ 2 C 1 T 1 1 / p R ρ ^ + ρ L ( ( 0 , T ) , L ( Ω ) ) 2 ρ + C 2 T 1 1 / p R
for any T ( 0 , T 0 ] with positive constants C 1 and C 2 depending on N, p, q, R, R 1 , R 2 , T 0 , and ρ , but independent of T. Choosing T 1 ( 0 , T 0 ] so small that
C 1 T 1 1 1 / p R ρ 8 , C 2 T 1 1 1 / p R ρ ,
we obtain for any T ( 0 , T 1 ]
3 8 ρ ρ ^ ( x , t ) + ρ 3 ρ ( x , t ) Ω ¯ × [ 0 , T ] .
It thus holds for any T ( 0 , T 1 ] that
P ( ρ ^ + ρ ) ρ ^ L p ( ( 0 , T ) , L q ( Ω ) N ) P L ( [ ρ / 4 , 4 ρ ] ) ρ ^ L p ( ( 0 , T ) , L q ( Ω ) N ) ,
which, combined with Propositions 8, furnishes the desired inequality. This completes the proof of Lemma 16. □
Summing up Lemmas 11, 13, 15 and 16, we obtain
Proposition 9.
Suppose that all the assumptions of Theorem 1 hold. Let T 1 be the positive constant given by Lemma 16. Then there exists a positive constant L, depending on N, p, q, R, R 1 , R 2 , T 0 , and ρ , such that for any T ( 0 , T 1 ]
D ( ρ ^ , u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) L 4 M 1 , F ( ρ ^ , u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) L 4 M 1 ,
where M 1 is the positive constant given by Corollary 2.
We continue to estimate the pressure term.
Lemma 17.
Suppose that Assumption 2 and (d) of Theorem 1 hold. Let ( ρ 0 , u 0 ) D q , p ( Ω ) R . Let T 1 and L be the positive constants given by Lemma 16 and Proposition 9, respectively. Then, the following assertions hold.
(1)
There exists a constant T 2 = T 2 ( N , p , q , R , R 1 , R 2 , T 0 , ρ , L ) ( 0 , T 1 ] such that for any T ( 0 , T 2 ] and ρ 0 K p , q ; T 1 with ρ K p , q ; T 1 L
ρ 4 ρ ( x , t ) + ρ ^ ( x , t ) + ρ 4 ρ for ( x , t ) Ω ¯ × [ 0 , T ] .
(2)
For any T ( 0 , T 2 ] and ρ i 0 K p , q ; T 1 , i = 1 , 2 , with ρ i K p , q ; T 1 L
P ( ρ 1 + ρ ^ + ρ ) ( ρ 1 + ρ ^ ) P ( ρ 2 + ρ ^ + ρ ) ( ρ 2 + ρ ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 j = 0 1 ρ 1 K p , q ; T 1 + ρ 2 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) j ,
where C = C ( N , p , q , R , R 1 , R 2 , T 0 , ρ ) is a positive constant independent of T.
Proof. 
(1) The proof is similar to one of Lemma 16; thus, the detailed proof may be omitted.
(2) Let us write
P ( ρ 1 + ρ ^ + ρ ) ( ρ 1 + ρ ^ ) P ( ρ 2 + ρ ^ + ρ ) ( ρ 2 + ρ ^ ) = ( P ( ρ 1 + ρ ^ + ρ ) P ( ρ 2 + ρ ^ + ρ ) ) ( ρ 1 + ρ ^ ) + P ( ρ 2 + ρ ^ + ρ ) ( ρ 1 ρ 2 ) = : I 1 + I 2 .
It holds by (1) that for any T ( 0 , T 2 ]
P ( ρ 1 + ρ ^ + ρ ) P ( ρ 2 + ρ ^ + ρ ) L ( ( 0 , T ) , L ( Ω ) ) P C 0 , 1 ( [ ρ / 4 , 4 ρ ] ) ρ 1 ρ 2 L ( ( 0 , T ) , L ( Ω ) ) ,
which, combined with Propositions 6 and 8, demonstrates
I 1 L p ( ( 0 , T ) , L q ( Ω ) N ) P C 0 , 1 ( [ ρ / 4 , 4 ρ ] ) ρ 1 ρ 2 L ( ( 0 , T ) , L ( Ω ) ) ( ρ 1 + ρ ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 ρ 1 K p , q ; T 1 + ( ρ 0 , u 0 ) D q , p ( Ω ) .
On the other hand, (1) and Proposition 6 shows that for any T ( 0 , T 2 ]
I 2 L p ( ( 0 , T ) , L q ( Ω ) N ) P L ( [ ρ / 4 , 4 ρ ] ) ( ρ 1 ρ 2 ) L p ( ( 0 , T ) , L q ( Ω ) N ) C φ ( T ) ρ 1 ρ 2 K p , q ; T 1 .
The desired inequality thus holds. This completes the proof of Lemma 17. □
From Lemmas 10, 12, 14 and 17, we obtain the following proposition.
Proposition 10.
Suppose that all the assumptions of Theorem 1 hold. Let L and T 2 be the positive constants given by Proposition 9 and Lemma 17, respectively. Then, there exist a positive constant M 2 , depending on N, p, q, R, R 1 , R 2 , T 0 , ρ , and L, such that for any T ( 0 , T 2 ] the following assertions hold.
(1)
For any ( ρ i , u i ) 0 K p , q ; T ( L ) , i = 1 , 2 ,
D ( ρ 1 + ρ ^ , u 1 + u ^ ) D ( ρ 2 + ρ ^ , u 2 + ρ ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) M 2 ( φ ( T ) + ψ ( T ) ) ( ρ 1 ρ 2 , u 1 u 2 ) K p , q ; T , F ( ρ 1 + ρ ^ , u 1 + u ^ ) F ( ρ 2 + ρ ^ , u 2 + ρ ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) M 2 ( φ ( T ) + ψ ( T ) ) ( ρ 1 ρ 2 , u 1 u 2 ) K p , q ; T .
(2)
For any ( ρ , u ) 0 K p , q ; T ( L )
D ( ρ + ρ ^ , u + u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) M 2 ( φ ( T ) + ψ ( T ) ) ( ρ , u ) K p , q ; T + L 4 M 1 , F ( ρ + ρ ^ , u + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) M 2 ( φ ( T ) + ψ ( T ) ) ( ρ , u ) K p , q ; T + L 4 M 1 ,
where M 1 is the positive constant given by Corollary 2.
Proof. 
(1) The desired inequalities follow from Lemmas 10, 12, 14 and 17, immediately.
(2) The desired inequalities follow from (1) and Proposition 9 immediately. This completes the proof of Proposition 10. □

6.3. Proof of Theorem 1

Throughout this subsection, we assume that all the assumptions of Theorem 1 hold. Let M 1 be the positive constant given by Corollary 2. In addition, L, T 2 , and M 2 are the same positive constants as in the previous subsection. Recall that φ ( T ) and ψ ( T ) satisfy (37) and (38).
Let us choose T so small that
M 1 M 2 ( φ ( T ) + ψ ( T ) ) 1 4 .
Let ( ρ , u ) 0 K p , q ; T ( L ) . We consider
t σ + r 0 div v = D ( ρ + ρ ^ , u + u ^ ) in Ω × ( 0 , T ) , t v r 0 1 κ 0 Div ( κ 0 1 S 0 ( v ) + r 0 Δ σ I ) = F ( ρ + ρ ^ , u + u ^ ) in Ω × ( 0 , T ) , n · σ = 0 , v = 0 on Γ D × ( 0 , T ) , n · σ = 0 , ( D ( v ) n ) τ = 0 , v · n = 0 on Γ S × ( 0 , T ) , ( σ , v ) | t = 0 = ( 0 , 0 ) in Ω .
By Proposition 10(2), we observe that
D ( ρ + ρ ^ , u + u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) , F ( ρ + ρ ^ , u + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N )
and that
D ( ρ + ρ ^ , u + u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) L 2 M 1 , F ( ρ + ρ ^ , u + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) L 2 M 1 .
This enables us to apply Corollary 2 to (39), and then there exists a solution ( σ , v ) 0 K p , q ; T of (39) such that
( σ , v ) K p , q ; T M 1 L 2 M 1 + L 2 M 1 = L .
Thus the mapping Φ : 0 K p , q ; T ( L ) 0 K p , q ; T ( L ) can be defined by Φ ( ρ , u ) : = ( σ , v ) .
From now on, we prove that Φ is a contraction mapping on 0 K p , q ; T ( L ) . Let ( ρ i , u i ) 0 K p , q ; T ( L ) for i = 1 , 2 and set ( σ i , v i ) = Φ ( ρ i , u i ) . Define
τ = σ 1 σ 2 , w = v 1 v 2 .
Then ( τ , w ) satisfies
t τ + r 0 div w = D ( ρ 1 + ρ ^ , u 1 + u ^ ) D ( ρ 2 + ρ ^ , u 2 + u ^ ) in Ω × ( 0 , T ) , t w r 0 1 κ 0 Div ( κ 0 1 S 0 ( w ) + r 0 Δ τ I ) = F ( ρ 1 + ρ ^ , u 1 + u ^ ) F ( ρ 2 + ρ ^ , u 2 + u ^ ) in Ω × ( 0 , T ) , n · τ = 0 , w = 0 on Γ D × ( 0 , T ) , n · τ = 0 , ( D ( w ) n ) τ = 0 , w · n = 0 on Γ S × ( 0 , T ) , ( τ , w ) | t = 0 = ( 0 , 0 ) in Ω .
By Proposition 10(1), we observe that
D ( ρ 1 + ρ ^ , u 1 + u ^ ) D ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , H q 1 ( Ω ) ) 1 4 M 1 ( ρ 1 ρ 2 , u 1 u 2 ) K p , q ; T , F ( ρ 1 + ρ ^ , u 1 + u ^ ) F ( ρ 2 + ρ ^ , u 2 + u ^ ) L p ( ( 0 , T ) , L q ( Ω ) N ) 1 4 M 1 ( ρ 1 ρ 2 , u 1 u 2 ) K p , q ; T .
Applying Corollary 2 to (40) shows that
( τ , w ) K p , q ; T M 1 1 4 M 1 + 1 4 M 1 ( ρ 1 ρ 2 , u 1 u 2 ) K p , q ; T 1 2 ( ρ 1 ρ 2 , u 1 u 2 ) K p , q ; T .
This guarantees that Φ is a contraction mapping on 0 K p , q ; T ( L ) . The contraction mapping theorem thus yields a unique fixed point ( ρ , u ) of Φ in 0 K p , q ; T ( L ) , i.e., Φ ( ρ , u ) = ( ρ , u ) 0 K p , q ; T ( L ) . Then, ( ρ , u ) becomes a unique solution of (9) in 0 K p , q ; T ( L ) . This completes the proof of Theorem 1.

7. Conclusions and Future Works

In this paper, we have proved in Theorem 1 the local existence of strong solutions for the Navier–Stokes–Korteweg system in a general domain with the Dirichlet boundary condition or the slip boundary condition in an L p -in-time and L q -in-space setting, where p ( 1 , ) and q ( N , ) , based on the theory of maximal regularity. Our result demonstrates an extension of Kotschote [16] in view of domains and boundary conditions, and also extends the exponents p and q.
We will consider time periodic solutions of the Navier–Stokes–Korteweg system in a forthcoming paper by means of results of the present paper.

Author Contributions

Formal analysis, S.I.; original draft preparation, H.S. All authors have read and agreed to the published version of the manuscript.

Funding

This research was supported by the Research Center of Universitas Islam Negeri Syarif Hidayatullah Jakarta with Grant Number 22112100032 and JSPS KAKENHI Grant Number JP21K13817.

Data Availability Statement

Not applicable.

Acknowledgments

We thank the anonymous referees for informing us of the references [4,5,6,7] and for their helpful comments on our manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

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Inna, S.; Saito, H. Local Solvability for a Compressible Fluid Model of Korteweg Type on General Domains. Mathematics 2023, 11, 2368. https://doi.org/10.3390/math11102368

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Inna S, Saito H. Local Solvability for a Compressible Fluid Model of Korteweg Type on General Domains. Mathematics. 2023; 11(10):2368. https://doi.org/10.3390/math11102368

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Inna, Suma, and Hirokazu Saito. 2023. "Local Solvability for a Compressible Fluid Model of Korteweg Type on General Domains" Mathematics 11, no. 10: 2368. https://doi.org/10.3390/math11102368

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Inna, S., & Saito, H. (2023). Local Solvability for a Compressible Fluid Model of Korteweg Type on General Domains. Mathematics, 11(10), 2368. https://doi.org/10.3390/math11102368

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