Parameter-dependent Stokes problems in vector-valued spaces and applications

Boundary Value Problems20132013:172

DOI: 10.1186/1687-2770-2013-172

Received: 8 March 2013

Accepted: 28 June 2013

Published: 23 July 2013

Abstract

The stationary and instationary Stokes equations with operator coefficients in abstract function spaces are studied. The problems are considered in the whole space, and equations include small parameters. The uniform separability of these problems is established.

MSC:35Q30, 76D05, 34G10, 35J25.

Keywords

Stokes systems Navier-Stokes equations differential equations with small parameters semigroups of operators boundary value problems differential-operator equations maximal L p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq1_HTML.gif regularity

Dedication

Dedicated to International Conference on the Theory, Methods and Applications of Nonlinear Equations in Kingsville, TX-USA, Texas A&M University-Kingsville-2012

1 Introduction

We consider the initial value problem (IVP) for the following Stokes equation with small parameter:
u t Δ ε u + Au + φ = f ( x , t ) , x R n , t ( 0 , T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ1_HTML.gif
(1.1)
div u = 0 , u ( x , 0 ) = a ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ2_HTML.gif
(1.2)
where Δ ε u = k = 1 n ε k 2 u x k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq2_HTML.gif, A is a linear operator in a Banach space E and ε k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq3_HTML.gif are a small positive parameters. Here u = ( u 1 ( x , t ) , u 2 ( x , t ) , , u n ( x , t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq4_HTML.gif, φ = φ ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq5_HTML.gif are E-valued unknown solutions, f = ( f 1 ( x , t ) , f 2 ( x , t ) , , f n ( x , t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq6_HTML.gif is a given function and a = ( a 1 ( x ) , a 2 ( x ) , , a n ( x ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq7_HTML.gif is an initial date. This problem is characterized by the presence of an abstract operator A and small terms ε k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq3_HTML.gif which correspond to the inverse of Reynolds number Re very large. We prove that problem (1.1)-(1.2) has a unique strong maximal regular solution u on a time interval [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq8_HTML.gif independent of ε k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq3_HTML.gif. For ε k = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq9_HTML.gif, E = C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq10_HTML.gif, A = b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq11_HTML.gif, problem (1.1)-(1.2) is reduced to the Stokes problem
u t Δ u + b u + φ = f ( x , t ) , div u = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ3_HTML.gif
(1.3)
u ( x , 0 ) = a ( x ) , x R n , t ( 0 , T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ4_HTML.gif
(1.4)

where ℂ is the set of complex numbers and b is a positive constant.

Note that the existence of weak or strong solutions and regularity properties for the classical Stokes problems has been extensively studied, e.g., in [110]. There is an extensive literature on the solvability of the IVPs for the Stokes equation (see, e.g., [1, 3, 10] and further papers cited there). Solonnikov [8] proved that for every f L p ( Ω × ( 0 , T ) ; R 3 ) = B ( p ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq12_HTML.gif, p ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq13_HTML.gif, the time-dependent Stokes problem
u t Δ u + φ = f ( x , t ) , div u = 0 , u | Ω = 0 , u ( x , 0 ) = 0 , x Ω , t ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ5_HTML.gif
(1.5)
has a unique solution ( u , φ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq14_HTML.gif so that
u t B ( p ) + 2 u B ( p ) + φ B ( p , q ) C f B ( p , q ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equa_HTML.gif
Then Giga and Sohr [3] improved the result of Solonnikov for spaces with different exponents in space and time, i.e., they proved that for f L p ( 0 , T ; ( L q ( Ω ) ) n ) = B ( p , q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq15_HTML.gif there is a unique solution ( u , φ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq16_HTML.gif of problem (1.5) so that
u t B ( p , q ) + 2 u B ( p , q ) + φ B ( p , q ) C f B ( p , q ) , p , q ( 1 , ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ6_HTML.gif
(1.6)

Moreover, the estimate obtained was global in time, i.e., the constant C = C ( Ω , p , q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq17_HTML.gif is independent of T and f. To derive global L p L q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq18_HTML.gif estimates (1.6), Giga and Sohr used the abstract parabolic semigroup theory in UMD-type Banach spaces. Estimate (1.6) allows to study the existence of a solution and regularity properties of the corresponding Navier-Stokes problem (see, e.g., [5]).

In this paper, first we consider the following differential operator equation (DOE) with small parameters:
Δ ε u + ( A + λ ) u = f ( x ) , x R n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ7_HTML.gif
(1.7)

where A is a linear operator in a Banach space E, ε k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq19_HTML.gif are positive and λ is a complex parameter.

We show that for f W m , q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq20_HTML.gif, λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq21_HTML.gif, problem (1.7) has a unique solution u belonging to W 2 + m , q ( R n ; E ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq22_HTML.gif and the uniform coercive estimate holds
k = 1 n i = 0 m + 2 ε k i m + 2 | λ | 1 i m + 2 i u x k i L q ( R n ; E ) + Au L q ( R n ; E ) C f W m , q ( R n ; E ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equb_HTML.gif

where C ( q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq23_HTML.gif is independent of ε 1 , ε 2 , , ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq24_HTML.gif, λ and f.

We consider, then, the stationary abstract Stokes problem with small parameters
Δ ε u + Au + φ = f ( x ) , div u = 0 , x R n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ8_HTML.gif
(1.8)
where f = ( f 1 ( x ) , f 2 ( x ) , , f n ( x ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq25_HTML.gif is data and u = ( u 1 ( x ) , u 2 ( x ) , , u n ( x ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq26_HTML.gif is a solution. By applying the projection transformation P, equation (1.8) can be reduced to the following problem:
P Δ ε u + Au = f ( x ) , x R n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ9_HTML.gif
(1.9)
Let O ε , q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq27_HTML.gif denote the operator generated by problem (1.9), i.e., O ε , q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq27_HTML.gif is a Stokes operator in solenoidal space L σ q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq28_HTML.gif defined by
D ( O ε , q ) = ( W σ 2 , q ( R n ; E ( A ) , E ) ) n = { u ( W 2 , q ( R n ; E ( A ) , E ) ) n , div u = 0 } , O ε , q u = P Δ ε u + Au . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equc_HTML.gif

We prove that the operator O ε , q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq27_HTML.gif is uniformly positive and also is a generator of an analytic semigroup in L σ q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq29_HTML.gif. Finally, the instationary Stokes problem (1.1)-(1.3) is considered and the well-posedness of this problem is derived. In application we show the separability properties of the anisotropic stationary Stokes operator in mixed L p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq30_HTML.gif spaces and maximal regularity properties of infinity system of instationary Stokes equations in L p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq1_HTML.gif spaces.

2 Notations and background

Let E be a Banach space and let L p ( Ω ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq31_HTML.gif denote the space of strongly measurable E-valued functions that are defined on the measurable subset Ω R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq32_HTML.gif with the norm
f L p = f L p ( Ω ; E ) = ( Ω f ( x ) E p d x ) 1 p , 1 p < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equd_HTML.gif
The Banach space E is called a UMD-space if the Hilbert operator
( H f ) ( x ) = lim ε 0 | x y | > ε f ( y ) x y d y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Eque_HTML.gif

is bounded in L p ( R , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq33_HTML.gif, p ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq13_HTML.gif (see, e.g., [11]). UMD spaces include, e.g., L p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq34_HTML.gif, l p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq35_HTML.gif spaces and Lorentz spaces L p q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq36_HTML.gif, p , q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq37_HTML.gif.

Let E 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq38_HTML.gif and E 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq39_HTML.gif be two Banach spaces. B ( E 1 , E 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq40_HTML.gif denotes the space of bounded linear operators from E 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq38_HTML.gif into E 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq39_HTML.gif endowed with the usual uniform operator topology. For E 1 = E 2 = E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq41_HTML.gif, it is denoted by B ( E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq42_HTML.gif. Now ( E 1 , E 2 ) θ , p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq43_HTML.gif, 0 < θ < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq44_HTML.gif, 1 p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq45_HTML.gif, denotes interpolation spaces defined by the K method [[12], §1.3.1].

Let
S ψ = { λ C , | arg λ | φ { 0 } , 0 ψ < π } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equf_HTML.gif
A linear operator A is said to be ψ-positive in a Banach space E with bound M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq46_HTML.gif if the domain D ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq47_HTML.gif is dense on E and ( A + λ I ) 1 B ( E ) M ( 1 + | λ | ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq48_HTML.gif for any λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif, 0 ψ < π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq50_HTML.gif, where I is the identity operator in E. It is known [[12], §1.15.1] that there exist the fractional powers A θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq51_HTML.gif of the positive operator A. Let E ( A θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq52_HTML.gif denote the space D ( A θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq53_HTML.gif with the norm
u E ( A θ ) = ( u p + A θ u p ) 1 p , 1 p < , 0 < θ < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equg_HTML.gif
ℕ denotes the set of natural numbers. A set G B ( E 1 , E 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq54_HTML.gif is called R-bounded (see, e.g., [11]) if there is a positive constant C such that for all T 1 , T 2 , , T m G http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq55_HTML.gif and u 1 , u 2 , , u m E 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq56_HTML.gif, m N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq57_HTML.gif,
Ω j = 1 m r j ( y ) T j u j E 2 d y C Ω j = 1 m r j ( y ) u j E 1 d y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equh_HTML.gif

where { r j } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq58_HTML.gif is a sequence of independent symmetric { 1 , 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq59_HTML.gif-valued random variables on Ω. The smallest C, for which the estimate above holds, is called an R-bound of the collection G and denoted by R ( G ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq60_HTML.gif.

A set G h B ( E 1 , E 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq61_HTML.gif depending of parameter h Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq62_HTML.gif is called uniform R-bounded with respect to h if there is a constant C, independent of h Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq62_HTML.gif such that for all T 1 ( h ) , T 2 ( h ) , , T m ( h ) G h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq63_HTML.gif and u 1 , u 2 , , u m E 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq56_HTML.gif, m N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq57_HTML.gif,
Ω j = 1 m r j ( y ) T j ( h ) u j E 2 d y C Ω j = 1 m r j ( y ) u j E 1 d y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equi_HTML.gif

It implies that sup h Q R ( G h ) C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq64_HTML.gif.

The ψ-positive operator A is said to be R-positive in a Banach space E if the set L A = { ξ ( A + ξ ) 1 : ξ S ψ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq65_HTML.gif, 0 ψ < π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq50_HTML.gif, is R-bounded.

The operator A ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq66_HTML.gif is said to be ψ-positive in E uniformly with respect to t with bound M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq46_HTML.gif if D ( A ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq67_HTML.gif is independent of t, D ( A ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq68_HTML.gif is dense in E and ( A ( t ) + λ ) 1 M ( 1 + | λ | ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq69_HTML.gif for all λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif, 0 ψ < π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq50_HTML.gif, where M does not depend of t and λ.

Let E 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq70_HTML.gif and E be two Banach spaces and let E 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq70_HTML.gif be continuously and densely embedded into E. Let Ω be a measurable set in R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq71_HTML.gif and m be a positive integer. W p , m ( Ω ; E 0 , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq72_HTML.gif denotes the class of all functions u L p ( Ω ; E 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq73_HTML.gif that have the generalized derivatives m u x k m L p ( Ω ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq74_HTML.gif with the norm
u W p , m ( Ω ; E 0 , E ) = u L p ( Ω ; E 0 ) + k = 1 n m u x k m L p ( Ω ; E ) < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equj_HTML.gif

For n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq75_HTML.gif, Ω = ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq76_HTML.gif, a , b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq77_HTML.gif, the space W p , m ( Ω ; E 0 , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq72_HTML.gif is denoted by W p , m ( a , b ; E 0 , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq78_HTML.gif. For E 0 = E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq79_HTML.gif the space W p , m ( Ω ; E 0 , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq72_HTML.gif is denoted by W p , m ( Ω ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq80_HTML.gif.

Let H q , s ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq81_HTML.gif, < s < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq82_HTML.gif denote an E-valued Liouville space of order s, i.e.,
H q , s ( R n ; E ) = { u L q ( R n ; E ) , u H q , s ( R n ; E ) = F 1 ( 1 + | ξ | 2 ) s 2 F u L q ( R n ; E ) < } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equk_HTML.gif

where F and F 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq83_HTML.gif denote the Fourier and inverse Fourier transforms, respectively.

Let H q , s ( R n ; E 0 , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq84_HTML.gif be a Liouville-Lions type space, i.e.,
H q , s ( R n ; E 0 , E ) = { u H q , s ( R n ; E ) L q ( R n ; E 0 ) , u H q , s ( R n ; E 0 , E ) = u L q ( R n ; E 0 ) + u H q , s ( R n ; E ) < } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equl_HTML.gif
For ε = ( ε 1 , ε 2 , , ε n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq85_HTML.gif we define the parameter-dependent norm in H q , s ( R n ; E 0 , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq86_HTML.gif such that
u H ε q , s ( R n ; E 0 , E ) = u L q ( R n ; E 0 ) + F 1 ( 1 + k = 1 n ε k ξ k 2 ) s 2 F u L q ( R n ; E ) < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equm_HTML.gif

Sometimes we use one and the same symbol C without distinction to denote positive constants which may differ from each other even in a single context. When we want to specify the dependence of such a constant on a parameter, say α, we write C α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq87_HTML.gif.

3 Boundary value problems for abstract elliptic equations

In this section, we derive the maximal regularity properties of problem (1.7).

BVPs for DOEs were studied, e.g., in [9, 11, 1319]. For references, see, e.g., [19]. From [[18], Theorem 4.1] we have the following result.

Theorem 3.1 Let E be a UMD space and let A be an R-positive operator in E for 0 ψ < π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq50_HTML.gif. Then problem (1.7) has a unique solution u W q , 2 ( R n ; E ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq88_HTML.gif for f L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq89_HTML.gif and λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif. Moreover, the following uniform coercive estimate holds:
k = 1 n i = 0 2 ε k i 2 | λ | 1 i 2 i u x k i L q ( R n ; E ) + Au L q ( G ; E ) C f L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ10_HTML.gif
(3.1)

with C ( q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq23_HTML.gif independent of ε 1 , ε 2 , , ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq90_HTML.gif, λ and f.

Consider the differential operator Q ε = Q ε q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq91_HTML.gif in L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq92_HTML.gif generated by problem (3.1), i.e.,
D ( Q ε ) = W q , 2 ( R n ) , Q ε u = Δ ε u + Au . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equn_HTML.gif

Let B q = B ( L q ( R n ; E ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq93_HTML.gif. From Theorem 3.1 we obtain the following.

Result 3.1 For λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif, there is a resolvent ( Q ε + λ ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq94_HTML.gif satisfying the uniform estimate
k = 1 n i = 0 2 | λ | 1 i 2 ε k i 2 i x k i ( Q ε + λ ) 1 B q + A ( Q ε + λ ) 1 B q C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equo_HTML.gif

Next we show the smoothness of problem (3.1). The main result is the following.

Theorem 3.2 Assume that E is a UMD space, A is an R-positive operator in E, q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif and m is a positive integer.

Then, for all f W q , m ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq96_HTML.gif, λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif, problem (3.1) has a unique solution u that belongs to W q , 2 + m ( R n ; E ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq97_HTML.gif and the following uniform coercive estimate holds:
k = 1 n i = 0 m + 2 ε k i m + 2 | λ | 1 i m + 2 i u x k i W q , m ( R n ; E ) + Au L q ( R n ; E ) C f W q , m ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ11_HTML.gif
(3.2)

with C = C ( q , A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq98_HTML.gif independent of ε 1 , ε 2 , , ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq99_HTML.gif, λ and f.

Proof A solution of equation (1.7) is given by
u ( x ) = F 1 L 1 ( λ , ε , ξ ) F f = 1 ( 2 π ) n / 2 R n e i ξ x L 1 ( λ , ε , ξ ) f ˆ ( ξ ) d ξ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equp_HTML.gif
where L ( λ , ε , ξ ) = A + k = 1 n ε k ξ k 2 + λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq100_HTML.gif and f ˆ ( ξ ) = F f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq101_HTML.gif. It follows from the expression above that
k = 1 n i = 0 m + 2 ε k i m + 2 | λ | 1 i m + 2 i u x k i L q ( R n ; E ) + Au L q ( R n ; E ) = k = 1 n i = 0 m + 2 ε k i m + 2 | λ | 1 i m + 2 F 1 ξ k i L 1 ( λ , ε , ξ ) f ˆ L q ( R n ; E ) + F 1 AL 1 ( λ , ε , ξ ) f ˆ L q ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ12_HTML.gif
(3.3)
It is sufficient to show that the operator-functions
Ψ ξ λ ( ξ ) = AL 1 ( λ , ε , ξ ) ( 1 + k = 1 n | ξ k | m ) 1 , σ ε λ ( ξ ) = k = 1 n i = 0 m + 2 ε k i m + 2 | λ | 1 i m + 2 ξ k i ( 1 + k = 1 n | ξ k | m ) 1 L 1 ( λ , ε , ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equq_HTML.gif
are uniform Fourier multipliers in L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq92_HTML.gif. Actually, due to positivity of A, we have
L 1 ( λ , ε , ξ ) M ( 1 + k = 1 n ε k ξ k 2 + | λ | ) 1 , Ψ ε , λ ( ξ ) = AL 1 ( λ , ε , ξ ) ( 1 + k = 1 n | ξ k | m ) 1 C ( 1 + k = 1 n | ξ k | m ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ13_HTML.gif
(3.4)
It is clear to observe that
ξ k ξ k Ψ ε λ ( ξ ) = 2 ε k ξ k 2 AL 2 ( λ , ε , ξ ) = [ 2 ε k ξ k 2 L 1 ( λ , ε , ξ ) ] AL 1 ( λ , ε , ξ ) , k = 1 , 2 , , n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equr_HTML.gif
Due to R-positivity of the operator A, the sets
{ [ 2 ε k ξ k 2 L 1 ( λ , ε , ξ ) ] : ξ R n { 0 } } , { AL 1 ( λ , ε , ξ ) : ξ R n { 0 } } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equs_HTML.gif
are R-bounded. So, in view of Kahane’s contraction principle and from the product properties of the collection of R-bounded operators (see, e.g., [11], Lemma 3.5, Proposition 3.4), we obtain
R { ξ k ξ k Ψ ε λ ( ξ ) ξ R n { 0 } } C k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equt_HTML.gif
Namely, the R-bounds of sets { ξ k ξ k Ψ ε λ ( ξ ) ξ R n { 0 } } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq102_HTML.gif are independent of ε and λ. Next, let us consider σ ε λ ( ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq103_HTML.gif. It is clear to see that
σ ε λ ( ξ ) B ( E ) C | λ | k = 1 n i = 0 m + 2 [ ε k 1 m + 2 | ξ k | | λ | 1 m + 2 ] i ( 1 + k = 1 n | ξ k | m ) 1 L 1 ( λ , ε , ξ ) B ( E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equu_HTML.gif
By using the well-known inequality
k = 1 n y k α k C ( 1 + k = 1 n y k l ) , α k , y k 0 , | α | l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equv_HTML.gif
for y k = ( ε k 1 2 | λ | 1 2 | ξ k | ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq104_HTML.gif and l = m + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq105_HTML.gif, we get the uniform estimate
| k = 1 n i = 0 m + 2 [ ε k i m + 2 | ξ k | | λ | i m + 2 ] | C | λ | ( 1 + k = 1 n | ξ k | m ) ( 1 + | λ | 1 k = 1 n ε k | ξ k | m + 2 ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ14_HTML.gif
(3.5)
From (3.4) and (3.5) we have the uniform estimate
σ ε λ ( ξ ) B ( E ) C | λ | ( 1 + | λ | 1 k = 1 n ε k | ξ k | m + 2 ) ( 1 + k = 1 n | ξ k | m ) 1 L 1 ( λ , ε , ξ ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equw_HTML.gif
Due to R-positivity of the operator A, the set
{ ( | λ | + k = 1 n ε k ξ k 2 ) L 1 ( λ , ε , ξ ) : ξ R n { 0 } } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equx_HTML.gif
is R-bounded. By using this fact, in view of (3.4) and Kahane’s contraction principle, we obtain the R-boundedness of the set { σ ε λ ( ξ ) : ξ R { 0 } } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq106_HTML.gif. In a similar way, we obtain the uniform estimates
ξ k Ψ ε , λ ( ξ ) B ( E ) C 1 , ξ k σ ε λ ( ξ ) B ( E ) C 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equy_HTML.gif
Let
σ k ε λ ( ξ ) = { ξ k ξ k σ ε λ ( ξ ) : ξ R n { 0 } } , Ψ k ε λ ( ξ ) = { ξ k ξ k Ψ ε λ ( ξ ) : ξ R n { 0 } } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equz_HTML.gif
By the aid of the estimates above, due to R-positivity of the operator A, in view of estimate (3.4), by virtue of Kahane’s contraction principle, from the additional and product properties of the collection of R-bounded operators, for ξ 1 , ξ 2 , , ξ μ R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq107_HTML.gif, u 1 , u 2 , , u μ E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq108_HTML.gif and independent symmetric { 1 , 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq109_HTML.gif-valued random variables r j ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq110_HTML.gif, j = 1 , 2 , , μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq111_HTML.gif, μ N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq112_HTML.gif, we obtain the uniform estimate
Ω j = 1 μ r j ( y ) σ k ε λ ( ξ ( j ) ) u j E d y C k = 1 n Ω j = 1 μ σ k ε λ ( ξ ( j ) ) r j ( y ) u j E d y C sup ε , λ R ( { ξ k ξ σ k ε λ ( ξ ) : ξ R { 0 } } ) Ω j = 1 μ r j ( y ) u j E d y C k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equaa_HTML.gif

The same estimates are obtained for Ψ k ε λ ( ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq113_HTML.gif in a similar way. Hence, by virtue of [[11], Theorem 3.4] it follows that Ψ ε , λ ( ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq114_HTML.gif and σ ε λ ( ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq115_HTML.gif are the uniform collection of multipliers in L p ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq116_HTML.gif. Then, by using equality (3.3), we obtain the assertion. □

4 The stationary Stokes system with small parameters

In this section we derive the maximal regularity properties of the stationary abstract Stokes problem (1.8).

Let H q , s ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq81_HTML.gif, < s < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq82_HTML.gif denote the E-valued Liouville space of order s such that H q , 0 ( R n ; E ) = L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq117_HTML.gif. It is known that if E is a UMD space, then H q , m ( R n ; E ) = W q , m ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq118_HTML.gif for a positive integer m (see, e.g., [[20], §15]). For q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif let X q = ( L q ( R n ; E ) ) n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq119_HTML.gif denote the space of an E-valued system of functions f = ( f 1 ( x ) , f 2 ( x ) , , f n ( x ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq120_HTML.gif with the norm
f X q = ( i = 1 n f i L q ( R n ; E ) q ) 1 q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equab_HTML.gif
X q σ = L σ q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq121_HTML.gif denotes the E-valued solenoidal space, i.e., closure of ( C 0 σ ( R n ; E ) ) n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq122_HTML.gif in ( L q ( R n ; E ) ) n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq123_HTML.gif, where
C 0 σ ( R n ; E ) = { u C 0 ( R n ; E ) , div u = 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equac_HTML.gif
Let
X q , s = ( H q , s ( R n ; E ) ) n , X q , s ( A ) = ( H q , s ( R n ; E ( A ) , E ) ) n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equad_HTML.gif
Let E be a Banach space. Consider the space
Y q = { u X q , div u L q ( R n ; E ) } , u Y q = ( u X q q + div u L q ( R n ; E ) q ) 1 q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equae_HTML.gif

Y q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq124_HTML.gif becomes a Banach space with this norm.

It is known that (see, e.g., Fujiwara and Morimoto [4]) the vector field u ( L q ( R n ) ) n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq125_HTML.gif has a Helmholtz decomposition. In the following theorem we generalize this result for an E-valued function space X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq126_HTML.gif.

Theorem 4.1 Let E be a UMD space and q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif. Then u X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq127_HTML.gif has a Helmholtz decomposition, i.e., there exists a bounded linear projection operator P q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq128_HTML.gif from X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq126_HTML.gif onto X q σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq129_HTML.gif with the null space N ( P q ) = { φ X q : φ L loc q ( R n ; E ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq130_HTML.gif. In particular, all u X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq127_HTML.gif has the unique decomposition u = u 0 + φ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq131_HTML.gif with u 0 X q σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq132_HTML.gif, u 0 = P q u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq133_HTML.gif so that
φ L q ( B ; E ) + u 0 X q C u X q , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ15_HTML.gif
(4.1)

for any open ball B R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq134_HTML.gif. Moreover, ( X q σ ) = X q σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq135_HTML.gif, 1 q + 1 q = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq136_HTML.gif.

To prove Theorem 4.1, we need some lemmas. Consider the problem
Δ ε u + ( A + λ ) u = f ( x ) , x R n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ16_HTML.gif
(4.2)
Lemma 4.1 Let E be a UMD space, let A be an R-positive operator in E, q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif and 2 < s < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq137_HTML.gif. Then, for f H q , s ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq138_HTML.gif, λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif, problem (4.2) has a unique solution u H q , 2 + s ( R n ; E ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq139_HTML.gif and the following uniform coercive estimate holds:
u H ε q , s + 2 ( R n ; E ( A ) , E ) + Au L q ( R n ; E ) + | λ | u L q ( R n ; E ) C f H q , s ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ17_HTML.gif
(4.3)
Proof By using the Fourier transform, we see that estimate (4.3) is equivalent to the following estimate:
F 1 ( 1 + k = 1 n ε k ξ k 2 ) s + 2 2 L 1 ( λ , ε , ξ ) f ˆ L q ( R n ; E ) + F 1 AL 1 ( λ , ε , ξ ) f ˆ L q ( R n ; E ) + | λ | F 1 L 1 ( λ , ε , ξ ) f ˆ L q ( R n ; E ) C F 1 ( 1 + | ξ | 2 ) s 2 f ˆ L q ( R n ; E ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ18_HTML.gif
(4.4)
To prove (4.4) it is sufficient to show that the operator functions
( 1 + k = 1 n ε k ξ k 2 ) L 1 ( λ , ε , ξ ) , A ( 1 + | ξ | 2 ) s 2 L 1 ( λ , ε , ξ ) , | λ | ( 1 + | ξ | 2 ) s 2 L 1 ( λ , ε , ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equaf_HTML.gif

are multipliers in L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq92_HTML.gif uniformly in λ and ε. This fact is derived as the step in the proof of Theorem 3.2. □

Now consider the system of equations
Δ ε u + Au + λ u = f ( x ) , x R n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ19_HTML.gif
(4.5)

where f = ( f 1 ( x ) , f 2 ( x ) , , f n ( x ) ) X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq140_HTML.gif and u = ( u 1 ( x ) , u 2 ( x ) , , u n ( x ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq26_HTML.gif is a solution of (4.5).

We define in X q , s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq141_HTML.gif the following parameter-dependent norm:
u X ε , q , s = F 1 ( 1 + k = 1 n ε k ξ k 2 ) s 2 F u X q < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equag_HTML.gif
Lemma 4.2 Let E be a UMD space, let A be an R-positive operator in E, 2 < s < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq137_HTML.gif and q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif. Then, for f X q , s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq142_HTML.gif, λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif, problem (4.5) has a unique solution u X q , s + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq143_HTML.gif and the following coercive uniform estimate holds:
u X ε , q , s + 2 + Au X q + | λ | u X q C f X q , s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ20_HTML.gif
(4.6)
Proof Problem (4.5) can be expressed as the following system:
Δ ε u j + Au j + λ u j = f j , x R n , j = 1 , 2 , , n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ21_HTML.gif
(4.7)
By Lemma 4.1 we obtain that for f j H q , s ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq144_HTML.gif, λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif, equation (4.7) has a unique solution u = ( u 1 , u 2 , , u n ) ( X q , s + 2 ( A ) ) n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq145_HTML.gif and the following uniform coercive estimate holds:
u j X ε , q , s + 2 + Au j X q + | λ | u j X q C f j X q , s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equah_HTML.gif

Hence, we get that u = ( u 1 , u 2 , , u n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq146_HTML.gif is a unique solution of problem (4.5) and (4.3) implies (4.6). □

By reasoning as in [[6], Lemma 2], we get the following lemma.

Lemma 4.3 C ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq147_HTML.gif is dense in Y p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq148_HTML.gif.

Consider the problem
Δ φ + A φ + λ φ = div f ( x ) , x R n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ22_HTML.gif
(4.8)

From Lemma 4.2 we obtain the following results.

Result 4.1 Let E be a UMD space, let A be an R-positive operator in E and q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif. Then, for f L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq149_HTML.gif, λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif, problem (4.8) has a unique solution φ H q , 1 ( R n ; E ( A ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq150_HTML.gif and the following coercive uniform estimate holds:
u X ε , q , 1 + Au X q + | λ | u X q C div f X q , 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equai_HTML.gif
Consider the operator P = P q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq151_HTML.gif defined by
D ( P ) = L q ( R n ; E ) , P f = f grad φ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equaj_HTML.gif

where φ is a solution of problem (4.8).

Result 4.2 Let E be a UMD space, let A be an R-positive operator in E and q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif. Then P q X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq152_HTML.gif is a closed subspace of X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq126_HTML.gif.

Lemma 4.4 Let E be a UMD space, let A be an R-positive operator in E and q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif. Then the operator P q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq153_HTML.gif is a bounded linear operator in X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq126_HTML.gif and P f = f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq154_HTML.gif if div f ( x ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq155_HTML.gif.

Proof The linearity of the operator P is clear by construction. Moreover, by Result 4.1 we have
P f X q f X q + grad φ X q C f X q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ23_HTML.gif
(4.9)

If div f ( x ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq156_HTML.gif, then by Lemma 4.2 we get that φ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq157_HTML.gif, i.e., P f = f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq154_HTML.gif. □

Let E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq158_HTML.gif denote the dual space of E.

Lemma 4.5 Assume that E is a UMD space and q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq159_HTML.gif. Then the conjugate of P q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq128_HTML.gif is defined as P q = P q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq160_HTML.gif, 1 q + 1 q = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq161_HTML.gif and is bounded linear in ( L q ( R n ; E ) ) n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq162_HTML.gif.

Proof It is known (see, e.g., [13, 20]) that the dual space of L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq92_HTML.gif is L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq163_HTML.gif. Since C 0 ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq164_HTML.gif is dense in L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq165_HTML.gif, we only have to show P q φ = P q φ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq166_HTML.gif for any φ ( C 0 ( R n ; E ) ) n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq167_HTML.gif. But this is derived by reasoning as in [[4], Lemma 5]. Moreover, by Lemma 4.4, the dual operator P q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq168_HTML.gif is bounded linear in L q ( R n ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq169_HTML.gif.

Let
G q = { φ : φ W q , 1 ( R n ; E ( A ) ; E ) } , ( P q X q ) = { f ( L q ( R n ; E ) ) n , f , υ = 0  for any  υ P q X q } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equak_HTML.gif

 □

From Lemmas 4.4, 4.5 we obtain the following result.

Result 4.3 Assume that E is a UMD space, A is an R-positive operator in E and q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif. Then any element f X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq170_HTML.gif uniquely can be expressed as the sum of elements of P q X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq152_HTML.gif and G q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq171_HTML.gif.

In a similar way to Lemmas 6, 7 of [4] we obtain, respectively, the following lemmas.

Lemma 4.6 Assume E is a UMD space and q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq159_HTML.gif. Then
( P q X q ) = G q , 1 q + 1 q = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equal_HTML.gif
Lemma 4.7 Assume E is a UMD space and q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq159_HTML.gif. Then
X q σ = G q , 1 q + 1 q = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equam_HTML.gif

Now we are ready to prove Theorem 4.1.

Proof of Theorem 4.1 From Lemmas 4.6, 4.7 we get that X q σ = ( P q X q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq172_HTML.gif. Then, by construction of P q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq128_HTML.gif, we have X q = X q σ G q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq173_HTML.gif. By Lemmas 4.2, 4.4, we obtain estimate (4.1). Moreover, by Result 4.2, G q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq171_HTML.gif is a close subspace of X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq126_HTML.gif. It is known that the dual space of the quotient space X q / G q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq174_HTML.gif is G q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq175_HTML.gif. By the first assertion we have X q / G q = X q σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq176_HTML.gif, and by Lemma 4.7 we obtain the second assertion. □

Theorem 4.2 Let E be a UMD space, let A be an R-positive operator in E, q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq95_HTML.gif. Then, for f X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq170_HTML.gif, φ X q , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq177_HTML.gif, λ S ψ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq49_HTML.gif, problem (1.8) has a unique solution u X q , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq178_HTML.gif and the uniform coercive estimate holds
k = 1 n i = 0 2 ε k i 2 | λ | 1 i 2 i u x k i X q + Au X q + φ X q C f X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ24_HTML.gif
(4.10)

with C = C ( q , A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq98_HTML.gif independent of ε 1 , ε 2 , , ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq99_HTML.gif, λ and f.

Proof By applying the operator P q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq128_HTML.gif to equation (1.8), we get problem (1.9). It is clear to see that
D ( Q ε q ) = D ( B ε ) X q σ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equan_HTML.gif
where Q ε q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq179_HTML.gif is the Stokes operator and B ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq180_HTML.gif is an operator in X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq126_HTML.gif generated by problem (4.5) for λ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq181_HTML.gif, i.e.,
D ( B ε ) = X q , 2 , B ε u = Δ u + Au . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equao_HTML.gif

 □

Then by Lemma 4.2 we obtain the assertion.

Result 4.4 From Theorem 4.2 we get that Q ε = Q ε q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq91_HTML.gif is a positive operator in X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq126_HTML.gif and it also generates a bounded holomorphic semigroup S ε ( t ) = exp ( Q ε t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq182_HTML.gif for t > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq183_HTML.gif.

In a similar way to that in [21] we show the following.

Proposition 4.1 The following estimate holds
Q ε α S ε ( t ) C t α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equap_HTML.gif

uniformly in ε = ( ε 1 , ε 2 , , ε n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq184_HTML.gif for α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq185_HTML.gif and t > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq183_HTML.gif.

Proof From Theorem 4.2 we obtain that the operator Q ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq186_HTML.gif is uniformly positive in X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq126_HTML.gif, i.e., the following estimate holds
( Q ε + λ ) 1 M | λ | 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equaq_HTML.gif

for λ S ψ , ϰ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq187_HTML.gif, 0 < ψ < π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq188_HTML.gif, where the constant M is independent of λ and ε. Hence, by using Danford integral and operator calculus (see, e.g., in [11]) we obtain the assertion. □

5 Well-posedness of the instationary parameter-dependent Stokes problem

In this section, we show the uniform well-posedness of problem (1.1)-(1.2).

Theorem 5.1 Assume that E is a UMD space, A is an R-positive operator in E and 0 < ε k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq189_HTML.gif. Then, for f L p ( 0 , T ; X q ) = B ( p , q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq190_HTML.gif and a ( X q , 2 ( A ) , X q ) 1 p , p = G ( p , q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq191_HTML.gif, p , q ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq192_HTML.gif, there is a unique solution ( u , φ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq193_HTML.gif of problem (1.1)-(1.2) and the following uniform estimate holds:
u t B ( p , q ) + k = 1 n ε k 2 u x k 2 B ( p , q ) + Au B ( p , q ) + φ B ( p , q ) C ( f B ( p , q ) + a G ( p , q ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ25_HTML.gif
(5.1)

with C = C ( T , p , q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq194_HTML.gif independent of f and ε.

Proof Problem (1.1)-(1.2) can be expressed as the following parabolic problem:
d u d t + Q ε u = f ( t ) , u ( 0 ) = a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ26_HTML.gif
(5.2)
If we put E = X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq195_HTML.gif, then by Proposition 4.1 operator Q ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq186_HTML.gif is uniformly positive and generates a bounded holomorphic semigroup in X q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq126_HTML.gif uniformly in ε. Moreover, by using [[15], Theorem 3.1] we get that the operator Q ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq196_HTML.gif is R-positive in E uniformly in ε. Since E is a UMD space, in a similar way to that in [[22], Theorem 4.2] we obtain that for f L p ( 0 , T ; E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq197_HTML.gif and a ( D ( Q ε ) , E ) 1 p , p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq198_HTML.gif, there is a unique solution u W 1 , p ( 0 , T , D ( Q ε ) , E ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq199_HTML.gif of problem (5.2) such that the following uniform estimate holds:
d u d t L p ( 0 , T ; E ) + Q ε u L p ( 0 , T ; E ) C ( f L p ( 0 , T ; E ) + a ( D ( A ε ) , E ) 1 p , p ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ27_HTML.gif
(5.3)

 □

From estimates (4.10) and (5.3) we obtain the assertion.

Result 5.1 It should be noted that if ε 1 = ε 2 = = ε n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq200_HTML.gif, then we obtain maximal regularity properties of an abstract Stokes problem without any parameters in principal part.

Remark 5.2 There are a lot of positive operators in concrete Banach spaces. Therefore, putting in (1.8) and (1.1) concrete Banach spaces instead of E and concrete positive differential, pseudo differential operators, or finite, infinite matrices, etc. instead of A, by virtue of Theorem 4.2 and Theorem 5.1, we can obtain the maximal regularity properties of a different class of stationary and instationary Stokes problems which occur in numerous physics and engineering problems.

6 Separability properties of anisotropic Stokes equations

Let Ω R m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq201_HTML.gif be an open connected set with compact C 2 l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq202_HTML.gif-boundary Ω. Consider the BVP for the following anisotropic Stokes equations with parameters:
( L + λ ) u = Δ ε u + φ + | α | 2 l a α ( y ) D y α u + λ u = f , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ28_HTML.gif
(6.1)
div x u = 0 , x R n , y Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ29_HTML.gif
(6.2)
B j u = | β | l j b j β ( y ) D y β u ( x , y ) = 0 , y Ω , j = 1 , 2 , , l , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ30_HTML.gif
(6.3)
where a α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq203_HTML.gif and b j β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq204_HTML.gif are complex-valued functions,
u = u ( x , y ) = ( u 1 ( x , y ) , u 2 ( x , y ) , , u n ( x , y ) ) , φ = φ ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equar_HTML.gif
are unknown solutions and
f = f ( x , y ) = ( f 1 ( x , y ) , f 2 ( x , y ) , , f n ( x , y ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equas_HTML.gif
is a given function;
Δ ε u = k = 1 n ε k 2 u x k 2 , D j = i y j , y = ( y 1 , , y m ) , x = ( x 1 , , x n ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equat_HTML.gif

ε = ( ε 1 , ε 2 , , ε n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq205_HTML.gif, ε k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq3_HTML.gif are positive and λ is a complex parameter.

If G = R n × Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq206_HTML.gif, p = ( p 1 , p ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq207_HTML.gif, L p ( G ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq208_HTML.gif will denote the space of all p-summable scalar-valued functions with mixed norm (see, e.g., [[12], §1], i.e., the space of all measurable functions f defined on G, for which
f L p ( G ) = ( R n ( Ω | f ( x , y ) | p 1 d y ) p p 1 d x ) 1 p < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equau_HTML.gif

Analogously, W p , 2 , 2 l ( G ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq209_HTML.gif denotes the anisotropic Sobolev space with a corresponding mixed norm [[12], §10]. Let X p = ( L p ( G ) ) n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq210_HTML.gif. From Theorem 4.2 we obtain the following result.

Theorem 6.1 Let the following conditions be satisfied:
  1. (1)

    a α C ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq211_HTML.gif for each | α | = 2 l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq212_HTML.gif and a α [ L + L γ k ] ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq213_HTML.gif for each | α | = k < 2 l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq214_HTML.gif with r k q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq215_HTML.gif and 2 l k > m r k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq216_HTML.gif;

     
  2. (2)

    b j β C 2 l l j ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq217_HTML.gif for each j, β and l j < 2 l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq218_HTML.gif, j = 1 l b j β ( y ) σ j 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq219_HTML.gif, for | β | = l j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq220_HTML.gif, y G http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq221_HTML.gif, where σ = ( σ 1 , σ 2 , , σ m ) R m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq222_HTML.gif is normal to Ω;

     
  3. (3)

    for y Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq223_HTML.gif, ξ R m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq224_HTML.gif, ν S ( φ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq225_HTML.gif, φ ( 0 , π ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq226_HTML.gif, | ξ | + | ν | 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq227_HTML.gif let ν + | α | = 2 l a α ( y ) ξ α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq228_HTML.gif;

     
  4. (4)
    for each y 0 Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq229_HTML.gif, the local BVP in local coordinates corresponding to y 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq230_HTML.gif
    ν + | α | = 2 l a α ( y 0 ) D α ϑ ( y ) = 0 , B j 0 ϑ = | β | = l j b j β ( y 0 ) D β u ( y ) = h j , j = 1 , 2 , , l , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equav_HTML.gif
     
has a unique solution ϑ C 0 ( R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq231_HTML.gif for all h = ( h 1 , h 2 , , h m ) R m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq232_HTML.gif, and for ξ R m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq233_HTML.gif with | ξ | + | ν | 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq234_HTML.gif. Then for f X p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq235_HTML.gif, λ S ( φ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq236_HTML.gif with sufficiently large | λ | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq237_HTML.gif problem (6.1)-(6.3) has a unique solution u belonging to W p , 2 , 2 l ( G ; R n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq238_HTML.gif and the uniform coercive estimate holds
k = 1 n i = 0 2 ε k i 2 | λ | 1 i 2 i u x k i X p + | β | = 2 m D y β u L p ( G ) + φ X q C f X p . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equaw_HTML.gif
Proof Let E = L p 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq239_HTML.gif. By virtue of [[11], Theorem 3.6], E is a UMD space. Consider the operator A in L p 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq240_HTML.gif defined by
D ( A ) = W p 1 , 2 l ( Ω ; B j u = 0 ) , Au = | α | 2 l a α ( y ) D α u ( y ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equax_HTML.gif
Problem (6.1)-(6.3) can be rewritten in the form (1.8), where u ( x , y ) = u ( x , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq241_HTML.gif, f ( x , y ) = f ( x , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq242_HTML.gif are vector-functions with values in E = L p 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq239_HTML.gif. By virtue of [[11], Theorem 8.2] the problem
ν u ( y ) + | α | 2 l a α ( y ) D y α u ( y ) = f ( y ) , B j u = | β | l j b j β ( y ) D y β u ( y ) = 0 , j = 1 , 2 , , l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equay_HTML.gif

has a unique solution for f L p 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq243_HTML.gif and for ν S ( φ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq225_HTML.gif, | ν | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq244_HTML.gif. Moreover, the operator A is R-positive in L p 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq245_HTML.gif, i.e., all the conditions of Theorem 4.2 hold. So, we obtain the assertion. □

7 Infinite system of Stokes equations with small parameters

Consider the IVB for the following infinite system of instationary Stokes equations with small parameters:
u m t k = 1 n ε k 2 u m x k 2 + j = 1 g j u j + φ m = f m ( x , t ) , div u = 0 , u m ( x , 0 ) = 0 , x R n , t ( 0 , T ) , m = 1 , 2 , , , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equ31_HTML.gif
(7.1)
where ε k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq3_HTML.gif are positive parameters. Here u m = ( u m 1 ( x , t ) , u m 2 ( x , t ) , , u m n ( x , t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq246_HTML.gif, φ m = φ m ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq247_HTML.gif are unknown solutions, and f = ( f m 1 ( x , t ) , f m 2 ( x , t ) , , f m n ( x , t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq248_HTML.gif is a given function. Let
G = { g m } , g m > 0 , u = { u m } , G u = { g m u m } , m = 1 , 2 , , l q ( G ) = { u u l σ , u l σ ( G ) = G u l σ = ( m = 1 | g m u m | σ ) 1 σ < } , 1 < σ < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equaz_HTML.gif
X p , q , σ = L p ( 0 , T ; L q ( R n ; l σ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq249_HTML.gif is a class of functions
f = ( f 1 ( x , t ) , f 2 ( x , t ) , , f n ( x , t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equba_HTML.gif
with the norm
f X p , q , σ = ( i = 1 f i L p ( 0 , T ; L q ( R n ) ) σ ) 1 σ < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equbb_HTML.gif

Let X p , q , σ 2 = W p , 2 ( 0 , T ; L q ( R n ; l σ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq250_HTML.gif. From Theorem 5.1 we obtain the following.

Theorem 7.1 Let 0 < ε k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq189_HTML.gif and g j > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq251_HTML.gif. Then for f X p , q , σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq252_HTML.gif, p , q , σ ( 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq253_HTML.gif there is a unique solution ( u m , φ m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq254_HTML.gif of problem (7.1) belonging to X p , q , σ 2 × X p , q , σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq255_HTML.gif and the following uniform coercive estimate holds:
u t X p , q , σ + k = 1 n ε k 2 u x k 2 X p , q , σ + G u X p , q , σ + φ X p , q , σ C f X p , q , σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equbc_HTML.gif

with C = C ( T , p , q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq194_HTML.gif independent of f and ε.

Proof Really, let E = l σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq256_HTML.gif, A and be an infinite matrix, defined by
A = [ g m δ j m ] , m , j = 1 , 2 , , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equbd_HTML.gif
It is easy to see that
B ( λ ) = λ ( A + λ ) 1 = [ λ ( g m + λ ) 1 δ j m ] , m , j = 1 , 2 , , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Eqube_HTML.gif
For all u 1 , u 2 , , u μ l q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq257_HTML.gif, λ 1 , λ 2 , , λ μ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq258_HTML.gif, λ i g m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq259_HTML.gif, m = 1 , 2 , , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq260_HTML.gif and independent symmetric { 1 , 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq59_HTML.gif-valued random variables r i ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq261_HTML.gif, j = 1 , 2 , , μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq111_HTML.gif, μ N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq262_HTML.gif, we have
Ω i = 1 μ r i ( y ) B ( λ i ) u i l σ σ d y C Ω m = 1 | i = 1 μ λ i ( g m + λ i ) 1 r i ( y ) u i | σ d y sup m , i | λ i ( g m + λ i ) 1 | σ Ω m = 1 | i = 1 μ r i ( y ) u i | σ d y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equbf_HTML.gif
Since sup m , i | λ i ( g m + λ i ) 1 | σ C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq263_HTML.gif for λ i g m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq264_HTML.gif, from the above we get
Ω i = 1 μ r i ( y ) B ( λ i ) u i l σ σ d y C Ω i = 1 μ r i ( y ) u i l σ σ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_Equbg_HTML.gif

i.e., the operator A is R-positive in l σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-172/MediaObjects/13661_2013_Article_438_IEq265_HTML.gif. Therefore, all the conditions of Theorem 5.1 hold and we obtain the assertion. □

Declarations

Authors’ Affiliations

(1)
Department of Mechanical Engineering, Okan University
(2)
Institute of Mathematics and Mechanics, Azerbaijan National Academy of Sciences

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