Open Access

An approximation of the axially symmetric flow through a pipe-like domain with a moving part of a boundary

Boundary Value Problems20132013:241

https://doi.org/10.1186/1687-2770-2013-241

Received: 6 August 2013

Accepted: 28 August 2013

Published: 18 November 2013

Abstract

The purpose of this work is to study the existence of solutions for approximation of an unsteady fluid-structure interaction problem. We consider a perturbed Navier-Stokes equation in the cylindrical coordinate system assuming axially symmetric flow. A priori unknown part of the boundary (that interacts with the fluid) is governed with a linear equation of fifth order. We prove the existence of at least one weak solution as long as the boundary does not touch the axis of symmetry. An explicit expression for a class of divergence-free functions is given.

MSC:35Q30, 35Q35, 35D05, 74F10, 76D03.

Keywords

perturbed Navier-Stokes equations artificial pressure fluid-structure interaction cylindrical coordinate system Schauder fixed point theorem

1 Introduction

This paper is devoted to the perturbed Navier-Stokes equations in a cylindrical coordinate system assuming rotationally symmetric flow
ρ v x t + ρ v x v x x + ρ v r v x r + ρ 2 v x ( v x x + v r r + 1 r v r ) = μ { 2 2 v x x 2 + 1 r r ( r ( v x r + v r x ) ) } p x , ρ v r t + ρ v x v r x + ρ v r v r r + ρ 2 v r ( v x x + v r r + 1 r v r ) = μ { 2 r r ( r v r r ) + x ( v r x + v x r ) 2 r 2 v r } p r }
(1.1)
with an artificial equation for the pressure
ε ( p t 2 p x 2 1 r r ( r p r ) ) + v x x + v r r + 1 r v r = 0
(1.2)
( 0 < ε 1 ) in an a priori unknown domain
Ω ( h ) { ( x , r , t ) : L < x < L + , 0 < r < h ( x , t ) , 0 < t < T } ,
(1.3)
where
h ( x , t ) = { χ ( x ) for  L x 0  and  L x L + , χ ( x ) + η ( x , t ) for  0 x L
(1.4)
with the given
χ ( x ) = { R , L x 0 , smooth and  χ 0  on [ L , L + ] , R + , L x L +

and ρ, μ, L, L + , T, R , R + are given positive constants, L 0 , R + R .

Unknown are the velocity field ( v x , v r ) , the pressure p and the interface η on [ 0 , L ] . We shall require from η to satisfy the following conditions:
η t ( x , t ) = v r ( x , h ( x , t ) , t ) ,
(1.5)
E [ 2 η t 2 + e 5 η t 3 x 2 + d 5 η t x 4 + f 4 η x 4 c 3 η t x 2 a 2 η x 2 + b η ] ( x , t ) = h ( x , t ) [ 2 μ v r r + μ ( v r x + v x r ) ( h x ) p + p w ] ( x , h ( x , t ) , t )
(1.6)
for 0 < x < L , 0 < t < T ,
η ( 0 , t ) = η x ( 0 , t ) = η ( L , t ) = η x ( L , t ) = 0
(1.7)
for any 0 < t < T and, for convenience, also
η ( x , 0 ) = 0 , η t ( x , 0 ) = 0 , 0 x L .
(1.8)
Here a, b, c, f are given nonnegative numbers and d, e, E are supposed to be positive. Moreover, p w = p w ( x , t ) is a given function. Of course, (1.1) is to be complemented with additional boundary and initial conditions. We shall consider
v x ( x , h ( x , t ) , t ) = 0
(1.9)
for any L < x < L + , 0 < t < T ,
v r ( x , h ( x , t ) , t ) = 0
(1.10)
for any L x 0 and L x L + , 0 < t < T ,
( 2 μ v x x p p out ρ 2 | v x | 2 ) ( L + , r , t ) = 0 , v r ( L + , r , t ) = 0
(1.11)
for any 0 < r < R + , 0 < t < T ,
( 2 μ v x x p + p in ρ 2 | v x | 2 ) ( L , r , t ) = 0 , v r ( L , r , t ) = 0
(1.12)
for any 0 < r < R , 0 < t < T ,
v x r ( x , 0 , t ) = 0 , v r ( x , 0 , t ) = 0
(1.13)
for any L < x < L + , 0 < t < T and, finally,
( v x ( x , r , 0 ) , v r ( x , r , 0 ) ) = ( 0 , 0 )
(1.14)

for any L < x < L + , 0 < r < h ( x , 0 ) . Here p in and p out are given functions of two variables ( r , t ) for 0 r R and 0 r R + , respectively, 0 t T .

The aim of this paper is to extend the existence result of [1] concerning the two-dimensional fluid-structure interaction problem to the radially symmetric case expressed in terms of the cylindrical coordinate system. Since we have not been able yet to handle this nonlinear problem modelling the flow of an incompressible viscous fluid in a cylinder with deformable wall, here nonlinear perturbed Navier-Stokes equations coupled with the artificial equation (1.2) and assuming e > 0 in the equation of the wall (1.6) are studied. As it turns out, we have not been able to let
ε 0

up to now due to the lack of regularity in the time variable.

We can give a justification of this limit, but only for a linearised perturbed Navier-Stokes equations and by including the additional term
g 6 η x 6 ( g > 0 )
(1.15)
in (1.6). Let us illustrate some arguments for the presence of this term in (1.6). To prove our existence result, we shall transform Ω ( h ) for h given by (1.4) on a fixed domain D = { ( x , y ) R 2 : L < x < L + , 0 < y < 1 } by setting
u ( x , y , t ) = ( v x ( x , h ( x , t ) y , t ) , v r ( x , h ( x , t ) y , t ) ) , q ( x , y , t ) = p ( x , h ( x , t ) y , t ) . }
(1.16)
The incompressibility condition takes then the form
div h ( u ) = u 1 x y h ( x , t ) h x ( x , t ) u 1 y + 1 h ( x , t ) u 2 y + 1 h ( x , t ) y u 2 = 0 .
In the final stage of the proof, we are to send ε 0 in an integral identity that includes also terms
0 T D { μ [ h ε u ε + ( h ε u ε ) T ] : h ε ψ ε q ε div h ε ψ ε } ε d ( x , y ) d t ,
(1.17)
where
h ε u i ε = ( u i ε x y h ε ( x , t ) h ε x ( x , t ) u i ε y , 1 h ε ( x , t ) u i ε y ) ,
( u ε , q ε , η ε ) solve an appropriate collection of approximating problems and ψ ε are test functions. The only way to overcome the lack of regularity for { q ε } is to consider divergence-free test functions, i.e.
div h ε ( ψ ε ) = 0 .
(1.18)
This motivates us to find an explicit expression for a class of functions satisfying (1.18). For ϕ C 3 ( D ¯ × [ 0 , T ] ) , let us define
ψ 1 ε ( x , y , t ) = 1 ( h ε ( x , t ) ) 2 0 x ϕ y ( σ , y , t ) h ε ( σ , t ) d σ , ψ 2 ε ( x , y , t ) = ϕ ( x , y , t ) 1 y 0 y ϕ ( x , z , t ) d z ψ 2 ε ( x , y , t ) = y ( h ε ( x , t ) ) 2 h ε x ( x , t ) 0 x ϕ y ( σ , y , t ) h ε ( σ , t ) d σ }
(1.19)
and the desired equality (1.18) is valid for ψ ε = ( ψ 1 ε , ψ 2 ε ) . But now another difficulty arises, for, by inserting (1.19) into (1.17), the term
μ 0 T D [ u 2 ε x y h ε h ε x u 2 ε y + 1 h ε u 1 ε y ] y ( h ε ) 2 2 h ε x 2 0 x ϕ y h ε d σ ε d ( x , y ) d t
appears. The problem is that we cannot deduce more than weak convergence of the velocity gradient in L 2 . What saves us are just the strong convergences of
2 h ε x 2 2 h x 2 in  L 2 ( ( 0 , L ) × ( 0 , T ) ) and h ε h in  W 1 , ( ( 0 , L ) × ( 0 , T ) ) .

This example of ψ ε implies that we in general need more regularity for h ε , and therefore we will assume (1.15). Section 7 provides a detailed analysis of a limit process as ε 0 in the linearised perturbed NS equations.

Our study was originally initiated by the papers of Quarteroni [24] and it is in some sense a continuation of the paper [5]. Due to the fact that our main contribution is in introducing the cylindrical coordinate system to the fluid-structure interaction-like problem, we do not provide a representative list of references for 2-D problems. For a full summary of the research on solvability of the fluid-structure interaction problem, we refer the reader to the introduction in [1] and references therein. See also [68] and [9]. The methods that we apply borrow some material from [5, 1012] and [13]; nevertheless, their application to our problem seems to be not straightforward.

The paper is organised as follows. In Section 2 we regularise once more our problem by considering it on a domain without the axis of symmetry (see, e.g. [14]) and then we transform it on the fixed domain. The main result of this section is the identity (2.21). Sections 3 and 5 provide necessary a priori estimates including the delicate identity (3.12). Sections 4 and 6 present main results of the paper. In Section 4, using the Schauder fixed point theorem, we prove the existence of the solution for the unknown interface, and in Section 6 we go to a limit with our regularised parameters. Finally, in Section 7 we let ε 0 in (1.2) but only for the linearised version of our problem, see (7.2).

Throughout the paper we will always assume that there are positive constants α, K, 0 < α 1 , such that
0 < 2 α h ( x , t ) α 1 , | h x ( x , t ) | , | h t ( x , t ) | K
(1.20)

for all ( x , t ) [ L , L + ] × [ 0 , T ] . Note that at the end we have to prove the validity of (1.20).

2 Regularisation

Our approach is to consider an approximate problem first. Given numbers
0 < ε 1 , 0 < δ α , κ 1
(2.1)
and given a function h C 1 ( [ L , L + ] × [ 0 , T ] ) satisfying (1.20), consider the system
ρ v x t + ρ v x v x x + ρ v r v x r + ρ 2 v x ( v x x + v r r + 1 r v r ) = μ { 2 2 v x x 2 + 1 r r ( r ( v x r + v r x ) ) } p x , ρ v r t + ρ v x v r x + ρ v r v r r + ρ 2 v r ( v x x + v r r + 1 r v r ) = μ { 2 r r ( r v r r ) + x ( v r x + v x r ) 2 r 2 v r } p r , ε ( p t 2 p x 2 1 r r ( r p r ) ) + v x x + v r r + 1 r v r = 0 }
(2.2)
in
Ω δ ( h ) = { ( x , r , t ) : L < x < L + , δ < r < h ( x , t ) , 0 < t < T } ,
(2.3)
[ 2 η t 2 + e 5 η t 3 x 2 + d 5 η t x 4 + f 4 η x 4 c 3 η t x 2 a 2 η x 2 + b η ] ( x , t ) = κ E ( η t ( x , t ) v r ( x , h ( x , t ) , t ) ) , 0 < x < L , 0 < t < T ,
(2.4)
equipped with the following boundary and initial conditions:
v x ( x , h ( x , t ) , t ) = 0
for any L < x < L + , 0 < t < T ,
v r ( x , h ( x , t ) , t ) = 0
for any L x 0 and L x L + , 0 < t < T ,
h ( x , t ) [ 2 μ v r r + μ ( v r x + v x r ) ( h x ) p + p w ρ 2 v r ( v r h t ) ] ( x , h ( x , t ) , t ) = κ ( η t ( x , t ) v r ( x , h ( x , t ) , t ) )
(2.5)
for any 0 < x < L , 0 < t < T ,
ε [ p x ( h x ) + p r ] ( x , h ( x , t ) , t ) = ε 2 h t ( x , t ) p ( x , h ( x , t ) , t )
for any L < x < L + , 0 < t < T ,
( 2 μ v x x p p out ρ 2 | v x | 2 ) ( L + , r , t ) = 0 , v r ( L + , r , t ) = 0 , ε p x ( L + , r , t ) = 0
for any δ < r < R + , 0 < t < T ,
( 2 μ v x x p + p in ρ 2 | v x | 2 ) ( L , r , t ) = 0 , v r ( L , r , t ) = 0 , ε p x ( L , r , t ) = 0
for any δ < r < R , 0 < t < T ,
v x r ( x , δ , t ) = 0 , v r ( x , δ , t ) = 0 , ε p r ( x , δ , t ) = 0
for any L < x < L + , 0 < t < T ,
( v x ( x , r , 0 ) , v r ( x , r , 0 ) ) = ( 0 , 0 ) , ε p ( x , r , 0 ) = 0
for any δ < r < h ( x , 0 ) , L < x < L + ,
η ( x , 0 ) = 0 , η t ( x , 0 ) = 0
for any 0 < x < L and, finally,
η ( 0 , t ) = η x ( 0 , t ) = η ( L , t ) = η x ( L , t ) = 0
(2.6)

for any 0 < t < T . Note that in this section h is given and ( v x , v r ) , p and η are to be found (h and η are therefore not related by (1.4)).

Assume for a moment that ( v x , v r , p , η ) is in fact a smooth solution of problem (2.2)-(2.6) above. Then
u ( x , y , t ) = def ( v x ( x , ( h ( x , t ) δ ) y + δ , t ) , v r ( x , ( h ( x , t ) δ ) y + δ , t ) ) , q ( x , y , t ) = def p ( x , ( h ( x , t ) δ ) y + δ , t ) }
(2.7)
for 0 < x < L , 0 < y < 1 , 0 < t < T and
z ( x , t ) = def η t ( x , t ) , Z ( x , t ) = 0 t z ( x , s ) d s
(2.8)
for 0 < x < L , 0 < t < T solve the following problem:
ρ ( u ) t ρ h t y ( y h δ u ) + ρ ( u h ) u + ρ 2 u div h u μ div h ( h u + ( h u ) T ) + μ 2 ( h δ ) 2 ( 0 u 2 ) = h q
(2.9)
in D × ( 0 , T ) , where D = ( L , L + ) × ( 0 , 1 ) ,
h u i = ( u i x + m u i y , 1 h δ u i y ) , m ( x , y , t ) = y h ( x , t ) δ h x ( x , t ) , div h ( u ) = u 1 x + m u 1 y + 1 h δ u 2 y + h δ u 2 , ( x , y , t ) = [ ( h ( x , t ) δ ) y + δ ] ( h ( x , t ) δ ) , }
(2.10)
ε ( ( h q ) t h t ( y q ) y h Δ h q ) + h div h u = 0 , Δ h q = div h ( h q ) ,
(2.11)
in D × ( 0 , T ) ,
z t ( x , t ) + e 4 z x 2 t 2 ( x , t ) + d 4 z x 4 ( x , t ) + f 4 Z x 4 ( x , t ) c 2 z x 2 ( x , t ) a 2 Z x 2 ( x , t ) + b Z ( x , t ) = κ E ( z ( x , t ) u 2 ( x , 1 , t ) )
(2.12)
for any 0 < x < L , 0 < t < T with the boundary and initial conditions listed below:
u 1 ( x , 1 , t ) = 0 for any  L < x < L + , 0 < t < T , u 2 ( x , 1 , t ) = 0 for any  L < x 0  and  L x < L + , 0 < t < T , h [ 2 μ h δ u 2 y μ h x ( u 2 x + m u 2 y + 1 h δ u 1 y ) q + q w ρ u 2 2 ( u 2 h t ) ] ( x , 1 , t ) = κ ( z ( x , t ) u 2 ( x , 1 , t ) ) , [ 1 h δ q y h x ( q x + m q y ) + 1 2 h t q ] ( x , 1 , t ) = 0
(2.13)
for any 0 < x < L , 0 < t < T , q w ( x , t ) = p w ( x , t ) ,
[ 2 μ ( u 1 x + m u 1 y ) q + q out 1 2 | u | 2 ] ( L + , y , t ) = 0 , u 2 ( L + , y , t ) = 0 , ( q x + m q y ) ( L + , y , t ) = 0
(2.14)
for any 0 < y < 1 , 0 < t < T , q out ( y , t ) = p out ( R + y , t ) ,
[ 2 μ ( u 1 x + m u 1 y ) q + q in 1 2 | u | 2 ] ( L , y , t ) = 0 , u 2 ( L , y , t ) = 0 , ( q x + m q y ) ( L , y , t ) = 0
(2.15)
for any 0 < y < 1 , 0 < t < T , q in ( y , t ) = p in ( R y , t ) ,
μ u 1 y ( x , 0 , t ) = 0 , u 2 ( x , 0 , t ) = 0 , ε q y ( x , 0 , t ) = 0
for any L < x < L + , 0 < t < T ,
u ( x , y , 0 ) = 0 , q ( x , y , 0 ) = 0
(2.16)
for any L < x < L + , 0 < y < 1 and, finally,
z ( x , 0 ) = z ( 0 , t ) = z x ( 0 , t ) = z ( L , t ) = z x ( L , t ) = 0
(2.17)

for any 0 < x < L and for any 0 < t < T .

We continue this section by making precise the meaning of the solution of problem (2.9)-(2.17). To this end, we first define
V V × H 1 ( D ) × H 0 2 ( 0 , L ) ,
(2.18)
where
V { w H 1 ( D ) 2 : w 1 = 0  on  S w , w 2 = 0  on  S in S out S c Γ }
(2.19)
and
S w = { ( x , 1 ) : L < x < L + } , S in = { ( L , y ) : 0 < y < 1 } , S out = { ( L + , y 2 ) : 0 < y < 1 } , S c = { ( x , 0 ) : L < x < L + } , Γ = { ( x , 1 ) : x ( L , 0 ) ( L , L + ) } S w .
Throughout this and the next sections, we assume
q in , q out L 2 ( 0 , T ; L 2 ( 0 , 1 ) ) , q w L 2 ( 0 , T ; L 2 ( 0 , L ) ) .
(2.20)

The function spaces we use are rather familiar, and we adopt the notation of [15].

Definition 2.1 We call ( u , q , z ) L 2 ( 0 , T ; V ) a weak solution of the initial boundary value problem (2.9)-(2.17) if the following two properties are fulfilled:
  1. (1)

    u L ( 0 , T ; L 2 ( D ) 2 ) , q L ( 0 , T ; L 2 ( D ) ) , ( q ) t L 2 ( 0 , T ; H 1 ( D ) ) , ( u ) t ( L 2 ( 0 , T ; V ) L 4 ( 0 , T ; L 4 ( D ) 2 ) ) = L 2 ( 0 , T ; V ) + L 4 / 3 ( 0 , T ; L 4 / 3 ( D ) 2 ) and z L ( 0 , T ; H 0 2 ( 0 , L ) ) H 0 1 ( 0 , T ; H 0 1 ( 0 , L ) ) .

     
  2. (2)
    ( u , q , z ) satisfies the system of differential equations, that is,
    0 T { ρ ( u ) t , ψ + ε ( q ) t , ω } d t ρ 2 0 T D { ( u ψ ) t + [ u y ψ u ψ y ] [ ( h δ ) y + δ ] h t y } d ( x , y ) d t ε 2 0 T D { q ω t + [ q y ω q ω y ] [ ( h δ ) y + δ ] h t y } d ( x , y ) d t + 0 T D { ρ 2 [ ( ( u h ) u ) ψ ( ( u h ) ψ ) u ] + μ ( [ h u + ( h u ) T ] : h ψ + 2 ( h δ ) 2 2 u 2 ψ 2 ) + ε ( h q h ω ) + ( ω div h u q div h ψ ) } d ( x , y ) d t + E 0 T 0 L { z t ξ + d 2 z x 2 2 ξ x 2 + e 2 z x t 2 ξ x t + f 2 Z x 2 2 ξ x 2 + c z x ξ x + a Z x ξ x + b Z ξ } d x d t + 0 T 0 L { q w ψ 2 h ( h δ ) + κ ( u 2 z ) ( ψ 2 ξ ) } d x d t = 0 T 0 1 { q in ( y , t ) ( ψ 1 ) ( L , y , t ) + q out ( y , t ) ( ψ 1 ) ( L + , y , t ) } d y d t
    (2.21)
     

for every ( ψ , ω , ξ ) L 2 ( 0 , T ; V ) , ψ L 4 ( 0 , T ; L 4 ( D ) 2 ) , ξ H 1 ( 0 , T ; H 1 ( 0 , L ) ) L 2 ( 0 , T ; H 0 2 ( 0 , L ) ) .

Remark 2.1 Note that
0 T ( u ) t , ζ d t + 0 T D u ζ t d t = 0
(2.22)

for every test function ζ L 2 ( 0 , T ; V ) L 4 ( 0 , T ; L 4 ( D ) 2 ) H 1 , 1 ( 0 , T ; L 2 ( D ) 2 ) with ζ ( T ) = 0 .

At the end of this section, we discuss the existence and uniqueness of weak solutions ( u , q , z ) to problem (2.9)-(2.17) for given h, q w , q in and q out satisfying (1.20) and (2.20).

Theorem 2.1 Assume (2.1), (2.20), and let η ¯ C 1 ( [ 0 , L ] × [ 0 , T ] ) satisfying (1.7) and (1.8) be given. Hereafter h = h ( η ¯ ) denotes h defined by (1.4) with the given η ¯ , and we assume that (1.20) holds. Then there exists a unique solution
L ( η ¯ ) = ( u , q , z ) = ( u ε , κ , δ , q ε , κ , δ , z ε , κ , δ )
of problem (2.9)-(2.17) in the sense of Definition  2.1 such that
D | u | 2 ( t ) ( t ) d ( x , y ) + 0 t D | u | 2 t d ( x , y ) d s = 2 0 t ( u ) t , u d s ,
(2.23)
where the function in (2.23) is defined by (2.10) for h = h ( η ¯ ) , i.e.
( x , y , t ) = [ ( h ( x , t ) δ ) y + δ ] ( h ( x , t ) δ ) .

Proof of Theorem 2.1 By analogy with the approach taken in [[5], the proof of Theorem 6.3], we can construct our weak solution by the implicit time discretisation method. The paper [5] presents all the technicalities of the proof, therefore we omit the proof here. □

We conclude this section by some interpolation inequalities which are needed especially for estimating the nonlinear item. For this purpose, we complete the notations of function spaces (2.18) and (2.19) by appropriate weighted spaces. Let
L y p ( D ) { ϕ : D R : D | ϕ ( x , y ) | p y d ( x , y ) < } , H y 1 , p ( D ) { ω L y p ( D ) : ω L y p ( D ) } , H y 1 ( D ) H y 1 , 2 ( D )
(2.24)
and
V y { w L y 2 ( D ) 2 : | w | L y 2 ( D ) , y 1 / 2 w 2 L y 2 ( D ) , w 1 = 0  on  S w , w 2 = 0  on  S in S out S c } , V y V y × H y 1 ( D ) × H 0 2 ( 0 , L ) .
(2.25)
Proposition 2.1 (i) Let φ be any function in H 1 ( D ) such that φ = 0 on S w or φ = 0 on S c . Then, for any p 2 and for any number θ with
p 2 p θ 1 ,
there exists a constant C = C ( p , θ ) such that
φ L p ( D ) C φ L 2 ( D ) θ φ L 2 ( D ) 1 θ .
(2.26)
Moreover, if φ L 2 ( 0 , T ; H 1 ( D ) ) L ( 0 , T ; L 2 ( D ) ) such that φ = 0 on S w or φ = 0 on S c for almost all t [ 0 , T ] , then the following holds for any p 2 :
φ L 2 p p 2 ( 0 , T ; L p ( D ) ) C ( φ L ( 0 , T ; L 2 ( D ) ) ) 2 / p ( φ L 2 ( 0 , T ; H 1 ( D ) ) ) ( p 2 ) / p .
(2.27)
  1. (ii)
    Let now p 2 and θ 1 fulfil the condition
    3 2 p 2 p θ 1 .
     
Then there exists a constant C = C ( p , θ ) such that
w L y p ( D ) 2 C w L y 2 ( D ) 2 θ w L y 2 ( D ) 2 1 θ for any w V y .
(2.28)
If, moreover, w L 2 ( 0 , T ; H y 1 ( D ) 2 ) L ( 0 , T ; L y 2 ( D ) 2 ) such that w 1 = 0 on S c and w 2 = 0 on S w for almost all t [ 0 , T ] , then the following holds for any p [ 2 , 6 ] :
w L 4 p 3 ( p 2 ) ( 0 , T ; L y p ( D ) 2 ) C ( w L ( 0 , T ; L y 2 ( D ) 2 ) ) ( 6 p ) / 2 p ( w L 2 ( 0 , T ; H y 1 ( D ) 2 ) ) ( 3 p 6 ) / 2 p .
(2.29)

Proof (i) The form of Nirenberg-Gagliardo inequality (2.26) can be found, e.g. in [[16], Theorem 2.2]. Then (2.27) follows from (2.26) for θ = ( p 2 ) / p by integration over ( 0 , T ) .

(ii) To prove (2.28), we define ψ ˜ ( x , y 2 , y 3 ) : = ψ ( x , y 2 2 + y 3 2 ) and transform the integrals on the domain D into integrals on the cylinder D ˜ = { ( x , y 2 , y 3 ) R 3 : L < x < L + , y 2 2 + y 3 2 < 1 } . Then the weighted interpolation inequality (2.28) is equivalent to an unweighted interpolation inequality in R 3 for ψ ˜ given on D ˜ , which yields the assertion (cf. also [[17], Theorem 19.9] for θ = 1 ). Inequality (2.29) again follows by integration of (2.28) over t ( 0 , T ) with θ = 3 ( p 2 ) / 2 p . □

3 The first a priori estimate

Our ultimate goal is to show that a subsequence of solutions ( u ε , κ , δ , q ε , κ , δ , z ε , κ , δ ) of the approximate problems (2.9)-(2.17) converges to a weak solution of (1.1)-(1.14). For this we will need some uniform estimates. Let us start with the following theorem.

Theorem 3.1 Let the hypotheses of Theorem  2.1 be satisfied, and let L ( η ¯ ) = ( u , q , z ) be the corresponding solution of problem (2.9)-(2.17).

Then there exists a constant C such that
D ( ρ | u ( t ) | 2 + ε | q ( t ) | 2 ) ( t ) d ( x , y ) + α 2 2 ( K 2 + 2 ) 0 t D ( μ | u | 2 + ε | q | 2 ) d ( x , y ) d s + 2 μ 0 t D ( h δ ) 2 | u 2 | 2 d ( x , y ) d s + E 0 L ( | Z t | 2 + f | Z x x | 2 + a | Z x | 2 + b | Z | 2 ) ( x , t ) d x + 2 E 0 t 0 L ( d | Z t x x | 2 + e | Z t x t | 2 + c | Z t x | 2 ) d x d s + 2 κ 0 t 0 L | u 2 ( x , 1 , s ) z ( x , s ) | 2 d x d s α 2 M q b 2 ( t ) [ 1 + e F t F t ] }
(3.1)
for a.e. time 0 t T , where
M q b 2 ( t ) = 0 t 0 L | q w | 2 d x d s + 0 t 0 1 ( | q in | 2 + | q out | 2 ) y d y d s , F = C ( K 2 + 2 ) α 2 μ ρ and Z t = Z t = z , Z x x = 2 Z x 2 , .

The constant C does not depend on α, K, ε, κ and δ.

Before proving Theorem 3.1, we pause to introduce some of its consequences.

Theorem 3.2 Let u = ( u 1 , u 2 ) = u ε , κ , δ and = δ . According to the energy estimates (3.1), we see that the set
{ σ u } is bounded in L ( 0 , T ; L 2 ( D ) 2 ) L 2 ( 0 , T ; V )
(3.2)
for any σ 3 / 2 , the set
{ u 2 } is bounded in L ( 0 , T ; L 2 ( D ) ) L 2 ( 0 , T ; H 1 ( D ) )
(3.3)
and the set
{ u 1 } is bounded in L ( 0 , T ; L 2 ( D ) )
(3.4)
for all ε, κ, δ. Moreover, we obtain that
{ ( λ / 2 ) u } and { y ( λ / 2 ) u } are bounded in L 2 ( 0 , T ; L 2 ( D ) 2 )
(3.5)

for any fixed λ > 1 and all ε, κ, δ satisfying (2.1).

Proof of Theorem 3.2 It remains to prove (3.5) and it suffices to prove the assertion for the first component u 1 of u since for u 2 it follows from estimate (3.1). For a moment, let us fix x and t and set u ( y ) = u 1 ( x , y , t ) , ( y ) = ( x , y , t ) . As u ( 1 ) = 0 , we have
| u ( y ) | 2 = | y 1 u ( s ) s s d s | 2 ln y 0 1 | u ( s ) | 2 s d s
(3.6)
and
α 2 y ( y ) 1 + α 2 .
(3.7)
Now we shall distinguish two cases. In order to derive (3.5) in the case 1 < λ < 0 , let us multiply (3.6) by α 2 λ y λ and integrate it over ( 0 , 1 ) . Due to (3.7), this gives the relation
0 1 | u ( y ) | 2 λ ( y ) d y α 2 λ 0 1 | u ( y ) | 2 y λ d y α 2 λ ( λ + 1 ) 2 0 1 | u ( s ) | 2 s d s α 2 ( λ 1 ) ( λ + 1 ) 2 0 1 | u ( s ) | 2 ( s ) d s
(3.8)
and (3.5) follows by integration over ( L , L + ) × ( 0 , T ) . If λ 0 , then
α 2 λ 0 1 | u ( y ) | 2 y λ d y 0 1 | u ( y ) | 2 λ ( y ) d y ( 1 + α 2 ) λ ( 0 1 ln y d y ) ( 0 1 | u ( s ) | 2 s d s )
(3.9)

and (3.5) follows easily. □

Note that (3.8) is a special case of the Hardy inequality [[17], Theorem 6.2]. Moreover, we remark that this argument also applies to q, which in analogue to (3.5) yields that
{ ( λ / 2 ) q ε , κ , δ } and { y ( λ / 2 ) q ε , κ , δ } are bounded in  L 2 ( 0 , T ; L 2 ( D ) )
(3.10)

for any fixed λ > 1 .

Proof of Theorem 3.1 Fix now a positive t < T and substitute ( ψ , ω , ξ ) = ( u , q , z ) into the identity (2.21). This is legitimate since u L 4 ( 0 , T ; L 4 ( D ) 2 ) . Then, with the assistance of (2.23), we write the resulting expression as
ρ D | u ( t ) | 2 ( t ) d ( x , y ) + ε D | q ( t ) | 2 ( t ) d ( x , y ) + 2 μ 0 t D { [ h u + ( h u ) T ] : h u + 2 ( h δ u 2 ) 2 } d ( x , y ) d s + 2 ε 0 t D | h q | 2 d ( x , y ) d s + E 0 L { | z | 2 + f | Z x x | 2 + a | Z x | 2 + b | Z | 2 } ( t ) d x + 2 E 0 t 0 L { d | z x x | 2 + e | z x t | 2 + c | z x | 2 } d x d s + 2 κ 0 t 0 L | u 2 ( x , 1 , s ) z ( x , s ) | 2 d x d s = 2 0 t 0 L q w ( x , s ) u 2 ( x , 1 , s ) ( x , 1 , s ) d x d s + 2 0 t 0 1 [ ( q in u 1 ) ( L , y , s ) + ( q out u 1 ) ( L + , y , s ) ] d y d s .
(3.11)

Our plan now is to check the ellipticity condition that follows from the following two propositions.

Proposition 3.1 Suppose
u V = { w H 1 ( D ) 2 : w 1 = 0 on S w , w 2 = 0 on S in S out S c Γ } ,
see (2.19). Then
2 μ D { [ h u + ( h u ) T ] : h u + 2 ( h δ u 2 ) 2 } d ( x , y ) d t = μ D { 3 ( u 1 x + m u 1 y ) 2 + ( 1 h δ u 1 y ) 2 + ( u 2 x + m u 2 y ) 2 + 2 ( 1 h δ u 2 y ) 2 + 2 ( h δ u 2 ) 2 + ( div h u ) 2 + ( h δ u 2 1 h δ u 2 y ) 2 + ( u 2 x + m u 2 y + 1 h δ u 1 y ) 2 } d ( x , y ) .
(3.12)
Proof of Proposition 3.1 Let us write the left-hand side of (3.12) as a sum of two integrals
μ D { h u : h u + 2 ( h δ u 2 ) 2 + ( h u ) T : h u } d ( x , y ) d t + μ D { [ h u + ( h u ) T ] : h u + 2 ( h δ u 2 ) 2 } d ( x , y ) d t
and we deduce that it equals
μ D { h u : h u + 2 ( h δ u 2 ) 2 + ( u 1 x + m u 1 y ) 2 + 2 h δ u 1 y ( u 2 x + m u 2 y ) + ( 1 h δ u 2 y ) 2 } d ( x , y ) + μ D { 2 ( u 1 x + m u 1 y ) 2 + 1 h δ u 1 y ( u 2 x + m u 2 y + 1 h δ u 1 y ) + ( u 2 x + m u 2 y ) ( u 2 x + m u 2 y + 1 h δ u 1 y ) + 2 ( 1 h δ u 2 y ) 2 + 2 ( h δ u 2 ) 2 } d ( x , y ) .
(3.13)

Now, let us focus on the first term in the second line of (3.13).

Lemma 3.3 Assume u V , then
D h δ u 1 y u 2 x d ( x , y ) = D h δ u 1 x u 2 y d ( x , y ) + D ( u 1 x + m u 1 y ) ( h δ ) u 2 d ( x , y ) .
(3.14)
Proof of Lemma 3.3 Assume for a moment that u 1 H 2 ( D ) . After integrating by parts, we obtain
D h δ u 1 y u 2 x d ( x , y ) = D y h x u 1 y u 2 d ( x , y ) D h δ 2 u 1 y x u 2 d ( x , y ) = D ( y h x u 1 y u 2 + y ( h δ u 2 ) u 1 x ) d ( x , y )

and (3.14) follows easily. The validity of (3.14) for u V is a consequence of the approximation argument.  □

We apply now Lemma 3.3 to the first term on the second line of (3.13) and the validity of (3.12) can be verified by direct computations. Let us note, however, that it is rather tedious. □

The proof of the next proposition can be found in [[5], Lemma 4.3].

Proposition 3.2 Let the hypotheses of Theorem  2.1 be satisfied, and let u V , then
i = 1 2 μ D [ ( u i x + m u i y ) 2 + ( 1 h δ u i y ) 2 ] d ( x , y ) α 2 μ K 2 + 2 D | u | 2 d ( x , y ) .
(3.15)
We would now like to finish the proof of Theorem 3.1. Returning to the relation (3.11), we estimate its second line with the assistance of (3.12), (3.15) and its right-hand side by Hölder’s inequality to find
ξ ( t ) + α 2 μ K 2 + 2 0 t D | u | 2 d ( x , y ) d s α 2 M q b 2 ( t ) + 0 t 0 L ( | u 2 | 2 ) ( x , 1 , s ) d x d s + 0 t 0 1 [ ( | u 1 | 2 ) ( L , y , s ) + ( | u 1 | 2 ) ( L + , y , s ) ] d y d s
(3.16)
for ξ ( t ) = ρ D ( | u | 2 ) ( x , y , t ) d ( x , y ) . Next, observe that for u V , the last two terms on the right-hand side of (3.16) are bounded by
ι 0 t D | u | 2 d ( x , y ) d s + C ι 0 t ξ ( s ) d s
for any 0 < ι 1 and therefore also for ι = α 2 μ / ( 2 ( K 2 + 2 ) ) . Consequently,
ξ ( t ) α 2 M q b 2 ( t ) + F 0 t ξ ( s ) d s
(3.17)
for a.e. 0 t T . Thus Gronwall’s inequality yields the estimate
0 t ξ ( s ) d s e F t 0 t α 2 M q b 2 ( s ) d s ,

that due to (3.16) and (3.17) implies (3.1). □

4 A free boundary value problem

So far we have studied problem (2.9)-(2.17) for h = h ( η ¯ ) with the given function η ¯ , and Theorem 2.1 gives the solution L ( η ¯ ) = ( u , q , z ) . For this section, keeping still ε, κ, δ fixed, we turn our attention to the problem of finding a function η such that
η ( x , t ) = 0 t z ( x , s ) d s for  z  given by  L ( η ) = ( u , q , z ) .

Our plan is hereafter to formulate it as a fixed point problem.

Thus, let L ( η ¯ ) = ( u , q , z ) be the corresponding solution of problem (2.9)-(2.17). Then the first a priori estimate (3.1) suggests that bounds (1.20) should be valid for h = h ( Z ) if positive T or the norms of outer pressure are chosen sufficiently small provided h = h ( η ¯ ) satisfies (1.20). Indeed, let us denote
M ( T ) = H 0 1 ( 0 , T ; H 0 2 ( 0 , L ) ) H 0 2 ( 0 , T ; H 0 1 ( 0 , L ) ) ,
(4.1)
and for ξ M ( T ) ,
ξ T 2 = 2 E 0 T 0 L ( d | ξ x t x | 2 + e | ξ t x t | 2 ) d x d t .
(4.2)
In view of Theorem 3.1 and the embeddings
M ( T ) C 1 , 1 / 2 ( S ¯ T ) C 1 ( S ¯ T ) , ξ C 1 , 1 / 2 ( S ¯ T ) C e ( T ) ξ T
(4.3)

(note that C e ( T ) is increasing in T), we have the following theorem.

Theorem 4.1 Suppose that the data satisfy (2.20), and assume 0 < α 1 such small that
4 α < R + + R < α 1 , and let K = R + 2 α + χ .
Fix any point η ¯ C 1 ( S ¯ T ) such that h = h ( η ¯ ) satisfies (1.20) with α and K given above. Let us recall that L ( η ¯ ) = ( u , q , z ) is the solution of problem (2.9)-(2.17) with h = h ( η ¯ ) in the sense of Theorem  2.1. If we recall the notation of (2.8), then
Z C 1 , 1 / 2 ( S ¯ T ) C e ( T ) Z T R + 2 α
(4.4)

provided M q b ( T ) is sufficiently small. Moreover, h = h ( Z ) satisfies (1.20) with the given α and K.

Note that smallness of M q b ( T ) (cf. Theorem 3.1) can be achieved by small outer pressure or a small time interval, but from now on, let T be fixed.

Proof It remains to prove that h = h ( Z ) satisfies (1.20). Because of 0 < R + R and | Z | R + 2 α , the following holds:
2 α R + | Z ( x , t ) | h ( x , t ) R + | Z ( x , t ) | R + R + α 1 .

The second estimate in (1.20) obviously follows from (4.4) by the definition of h and the choice of K. □

Denote now X = C 1 ( S ¯ T ) and
K = { φ M ( T ) : φ C 1 , 1 / 2 ( S ¯ T ) R + 2 α } .
Then K X is compact and convex. Now we define a mapping
A : K K
in the following way: We start with η ¯ K and define h = h ( η ¯ ) corresponding to formula (1.4), i.e. by
h ( η ¯ ) ( x , t ) = { R , L x 0 , χ ( x ) + η ¯ ( x , t ) , 0 x L , R + , L x L + .

Now, let L ( η ¯ ) = ( u , q , z ) be the solution of problem (2.9)-(2.17) according to Definition 2.1 with h = h ( η ¯ ) , and let us define a mapping P ( u , q , z ) : = Z .

Then our fixed point operator A is defined by A = P L , i.e.
A [ η ¯ ] = Z .

Corollary 4.2 A is well defined on K and maps K into itself.

Proof Note that h = h ( η ¯ ) fulfils (1.20) if η ¯ K in view of Theorem 4.1. Then Theorem 2.1 implies that the mapping L : C 1 ( S ¯ T ) K L 2 ( 0 , T ; V ) is well defined and unique. Moreover, due to Theorem 3.1, Z = P ( u , q , z ) belongs to M ( T ) with the finite norm Z T defined by (4.2). Finally, estimate (4.4) in Theorem 4.1 yields Z = ( P L ) [ η ¯ ] K . □

An application of the Schauder fixed point theorem requires the continuity of our mapping A . This means that we need continuous dependence of solutions on the data.

Theorem 4.3 Assume (2.1). Let ( u 1 , q 1 , z 1 ) and ( u 2 , q 2 , z 2 ) be weak solutions of the initial boundary value problem (2.9)-(2.17) in the sense of Definition  2.1 with given functions h 1 = h ( η ¯ 1 ) , q in 1 , q w 1 , q out 1 and h 2 = h ( η ¯ 2 ) , q in 2 , q w 2 , q out 2 , respectively, and suppose that both h 1 , h 2 satisfy (1.20). Let, moreover, τ , t [ 0 , T ] , 0 τ < t , with ( u 1 , q 1 , z 1 ) ( , s ) = ( u 2 , q 2 , z 2 ) ( , s ) and η ¯ 1 ( , s ) = η ¯ 2 ( , s ) for all s [ 0 , τ ] . Then there exist positive constants ν u and ν q such that
ρ D | 1 u 1 2 u 2 | 2 ( t ) d ( x , y ) + ν u τ t D | ( 1 u 1 2 u 2 ) | 2 d ( x , y ) d s + ε ( D | 1 q 1 2 q 2 | 2 ( t ) d ( x , y ) + ν q τ t D | ( 1 q 1 2 q 2 ) | 2 d ( x , y ) d s ) + E 0 L { | z 1 z 2 | 2 + f | ( Z 1 Z 2 ) x x | 2 + a | ( Z 1 Z 2 ) x | 2 + b | Z 1 Z 2 | 2 } ( t ) d x + E τ t 0 L { d | ( z 1 z 2 ) x x | 2 + e | ( z 1 z 2 ) x t | 2 + c | ( z 1 z 2 ) x | 2 } d x d s η ¯ 1 η ¯ 2 W 1 , ( S τ , t ) 2 ω ( τ , t ) + C τ t ( q out 1 q out 2 L 2 ( S out ) 2 + q in 1 q in 2 L 2 ( S in ) 2 + q w 1 q w 2 L 2 ( S w ) 2 ) d s
(4.5)
for almost all t [ 0 , T ] , where S τ , t = ( 0 , L ) × ( τ , t ) , C > 0 , and ω is a positive function ω ( τ , t ) 0 as t τ . Moreover, for any 0 < q < 1 , there is ι > 0 such that
ω ( τ , t ) q < 1 if t τ ι .
(4.6)

Unfortunately, ι depends on the regularising parameter δ in such a way that ι = ι ( δ ) 0 as δ 0 .

Proof of Theorem 4.3 Due to tedious calculations, we postpone the proof to the Appendix of this paper. □

Consider next the special case τ = 0 , t = T , q out 1 = q out 2 , q in 1 = q in 2 , q w 1 = q w 2 , and remember z = Z t , then for η ¯ 1 , η ¯ 2 K , estimate (4.5) gives
Z 1 Z 2 T 2 2 η ¯ 1 η ¯ 2 C 1 ( S ¯ T ) 2 ω ( 0 , T ) .

Together with the embedding inequality (4.3) this yields continuity of the operator A : X X on K . Hence, we arrive at the following result.

Theorem 4.4 Suppose that the data satisfy (2.20) and assume 0 < α 1 such small that
4 α < R + + R < α 1 , and let K = R + 2 α + χ .
In addition, let our regularised parameters be given and satisfy (2.1). Moreover, let
C e ( T ) α M q b ( T ) 1 + e F T F T R + 2 α .
(4.7)
Then A has a fixed point A [ η ] = η K , i.e. there is a solution ( u , q , z ) to problem (2.9)-(2.17) with
h = h ( η ) and z = η t

on the time interval [ 0 , T ] . This solution is unique.

Proof Because of Theorem 4.1, Corollary 4.2 and Theorem 4.3, the operator A : X X is continuous on K and maps the convex and compact subset K into itself. Then the existence of η = A [ η ] K follows from the Schauder fixed point theorem. This fixed point is unique due to Theorem 4.3. Indeed, let η 1 η 2 K be two functions with η 1 = A [ η 1 ] and η 2 = A [ η 2 ] , and let
τ : = sup { s [ 0 , T ] : η 1 ( , s ) = η 2 ( , s ) } .
Choose now t ( τ , T ] . Since the solution ( u , q , z ) = L ( η ) is unique on [ 0 , τ ] [ 0 , T ] , we have ( u 1 , q 1 , z 1 ) ( , s ) = ( u 2 , q 2 , z 2 ) ( , s ) for s [ 0 , τ ] . Hence, estimate (4.5), the fact that z 1 = η t 1 , z 2 = η t 2 together with (4.3) imply
η 1 η 2 C 1 ( S ¯ τ , t ) 2 η 1 η 2 C 1 , 1 / 2 ( S ¯ τ , t ) 2 C e ( T τ ) 2 η 1 η 2 τ , t 2 2 C e ( T τ ) 2 η 1 η 2 C 1 ( S ¯ τ , t ) 2 ω ( τ , t ) ,

where ξ τ , t denotes the norm (4.2) with integration over [ τ , t ] instead of [ 0 , T ] . For 0 < t τ small enough such that 2 C e ( T τ ) 2 ω ( τ , t ) < 1 , finally, we obtain η 1 η 2 C 1 ( S ¯ τ , t ) = 0 , which is a contradiction to the choice of τ. □

Note that our solution ( u , q , η t ) with h = h ( η ) depends on δ, ε, κ; however, the maximal time interval [ 0 , T ] given by (4.7) does not depend on these regularising parameters.

5 The second a priori estimate

The aim of this section is to show compactness of our set ( u ε , κ , δ , q ε , κ , δ , ( η ε , κ , δ ) t ) with respect to δ and κ. Throughout this section let the assumptions of Theorem 4.4 be fulfilled. We start with some obvious estimates concerning the weight = ( ( h δ ) y + δ ) ( h δ ) based on condition (1.20):
α 2 ( y + δ ) α 2 ( y + δ ) ,
(5.1)
| x | 2 K α 1 , | t | 2 K α 1 , α 2 y α 2 ,
(5.2)
| m | K α 1 ( y + δ ) K α 3 .
(5.3)

For compactness in time, the following second a priori estimate is essential.

Theorem 5.1 Let ( u , q , z ) be the solution of free boundary value problem (2.9)-(2.17) given by Theorem  4.4. Then there is a constant C ( ε ) independent of κ and δ such that for all τ [ 0 , T ] the inequality
0 T τ D | u ( t + τ ) u ( t ) | 2 ( y + δ ) 7 / 2 d ( x , y ) d t + ε 0 T τ D | q ( t + τ ) q ( t ) | 2 ( y + δ ) d ( x , y ) d t C ( ε ) τ
(5.4)

holds.

Before starting the proof of the theorem, let us prepare some estimations of the nonlinear items.

Lemma 5.2 Let
0 T D ( | u | 2 + | u | 2 ) ( y + δ ) d ( x , y ) d t + sup ess t [ 0 , T ] D | u ( x , y , t ) | 2 ( y + δ ) d ( x , y ) C ,
(5.5)
then for every fixed λ 5 / 2 , we have
0 T D | u | 2 | u | ( y + δ ) λ + 1 d ( x , y ) d t C
(5.6)
with generic constants C independent of u and δ. If, moreover,
0 T D | u 2 | 2 ( y + δ ) 1 d ( x , y ) d t C ,
(5.7)
then
0 T D | u 2 | | u | 2 ( y + δ ) λ d ( x , y ) d t C
(5.8)

under the same condition on λ.

Proof of Lemma 5.2 Denote w : = u ( y + δ ) β , then w x = u x ( y + δ ) β and w y = u y ( y + δ ) β + β u ( y + δ ) β 1 , hence
| w | 2 c ( | u | 2 ( y + δ ) 2 β + | u | 2 ( y + δ ) 2 β 2 ) .
Then the interpolation inequality (2.27) with p = 4 implies
w L 4 ( 0 , T ; L 4 ( D ) 2 ) C w L ( 0 , T ; L 2 ( D ) 2 ) 1 / 2 w L 2 ( 0 , T ; L 2 ( D ) 2 ) 1 / 2 C ( sup ess t [ 0 , T ] D | u ( x , y , t ) | 2 ( y + δ ) 2 β d ( x , y ) ) 1 / 4 × ( 0 T D ( | u | 2 ( y + δ ) 2 β + | u | 2 ( y + δ ) 2 β 2 ) d ( x , y ) d t ) 1 / 4 C
(5.9)
because of (5.5) if 2 β > 2 β 2 1 , i.e. β 3 / 2 . This yields (5.6) since
0 T D | u | 2 | u | ( y + δ ) λ + 1 d ( x , y ) d t = 0 T D | u | 2 | ( y + δ ) λ + 1 / 2 u | ( y + δ ) 1 / 2 d ( x , y ) d t ( 0 T D | u | 4 | ( y + δ ) 2 λ + 1 d ( x , y ) d t ) 1 / 2 ( 0 T D | u | 2 ( y + δ ) d ( x , y ) d t ) 1 / 2 C w L 4 ( 0 , T ; L 4 ( D ) ) 2 C
due to assumption (5.5) if β = ( 2 λ + 1 ) / 4 , i.e. λ 5 / 2 . Inequality (5.8) follows from Hölder’s inequality
0 T D | u | 2 | u 2 | ( y + δ ) λ d ( x , y ) d t ( 0 T D | u | 4 ( y + δ ) 2 λ + 1 d ( x , y ) d t ) 1 / 2 × ( 0 T D | u 2 | 2 ( y + δ ) 1 d ( x , y ) d t ) 1 / 2 ,

where the first integral of the product on the right-hand side is bounded owing to (5.9) and the second integral is bounded by our assumption. □

Proof of Theorem 5.1 Let λ 0 be a fixed exponent that will be specified later, let t , t + τ [ 0 , T ] be fixed and χ [ t , t + τ ] be the characteristic function of the subinterval [ t , t + τ ] [ 0 , T ] . Replace for a moment the integration variable with respect to time in (2.21) by s instead of t and test it with χ [ t , t + τ ] ( s ) ( w , p , ζ ) , where ( w , p , ζ ) V is independent on s. Then the first line of (2.21) passes into
ρ D ( ( u ) ( x , y , t + τ ) ( u ) ( x , y , t ) ) w d ( x , y ) + ε D ( ( q ) ( x , y , t + τ ) ( q ) ( x , y , t ) ) p d ( x , y )
(this follows by approximation from (2.22)). The idea how to derive the estimate of the theorem is now to choose
w = t τ u ( y + δ ) λ : = ( u ( x , y , t + τ ) u ( x , y , t ) ) ( y + δ ) λ , p = t τ q : = q ( x , y , t + τ ) q ( x , y , t ) , ζ = t τ z ( 1 + δ ) λ
and integrate the resulting relation over t [ 0 , T τ ] . This leads to
ρ 0 T τ D ( ( u ) ( x , y , t + τ ) ( u ) ( x , y , t ) ) × ( u ( x , y , t + τ ) u ( x , y , t ) ) ( y + δ ) λ d ( x , y ) d t + ε 0 T τ D ( ( q ) ( x , y , t + τ ) ( q ) ( x , y , t ) ) ( q ( x , y , t + τ ) q ( x , y , t ) ) d ( x , y ) d t + 0 T τ D ( further items with  u q ) d ( x , y ) d t + 0 T τ 0 L ( further items with  Z z ) d x d t + 0 T τ 0 L τ ( [ q w h ( h δ ) ] τ ( x , t ) t τ u 2 + κ [ u 2 z ] τ ( x , t ) t τ ( u 2 z ) ) d x d t = 0 T 0 1 τ ( [ q in ] τ t τ u 1 ( L , y , t ) + [ q out ] τ t τ u 1 ( L + , y , t ) ) d y d t ,
(5.10)
where [ v ] τ = 1 τ t t + τ v ( s ) d s denotes the Steklov average of v : [ 0 , T ] X on the interval [ t , t + τ ] . Note that for any normed space X and 1 p , the estimate
[ v ] τ L p ( 0 , T τ ; X ) v L p ( 0 , T ; X )
(5.11)
holds. Owing to estimates (5.1) and (5.2), for λ = 5 / 2 , the difference of the left-hand side of (5.4) and the first two lines of (5.10) may be estimated by ; hence it remains to estimate the items in the other lines of (5.10) also by . Since these are tedious calculations, we restrict ourselves to some exemplary items.
  1. (i)
    In order to demonstrate the technique, we start with the first item occurring in the third line of (5.10) which is
    I 1 = | ρ 2 0 T τ D t t + τ u ( , s ) ( u ( , t + τ ) u ( , t ) ) t ( , s ) ( y + δ ) λ d s d ( x , y ) d t | .
    In view of (5.2), (5.1) and ( y + δ ) λ c , we may estimate
    I 1 c τ 0 T τ D 1 τ t t + τ | u ( s ) | d s ( y + δ ) 1 / 2 | u ( t + τ ) u ( t ) | ( y + δ ) 1 / 2 d ( x , y ) d t c τ [ | u | ] τ L 2 ( 0 , T τ ; L 2 , y + δ ( D ) ) u ( + τ ) u L 2 ( 0 , T τ ; L 2 , y + δ ( D ) 2 ) c τ u L 2 ( 0 , T ; L 2 , y + δ ( D ) 2 ) c τ

    for any λ 0 . In the last steps here, we have used inequality (5.11) and boundedness of the norm due to Theorem 3.1.

     
  2. (ii)
    We proceed with the estimation of the integral arising from the nonlinear items, cf. the forth line of (2.21). Recall the definition (2.10) of h , after inserting ψ = χ [ t , t + τ ] ( s ) ( u ( x , y , t + τ ) u ( x , y , t ) ) ( y + δ ) λ into (2.21) and integration over t [ 0 , T τ ] , the items to be integrated are
    ( u h ) u ψ = ( u 1 u 1 x + m u 1 u 1 y + 1 h δ u 2 u 1 y ) ψ 1 + ( u 1 u 2 x + m u 1 u 2 y + 1 h δ u 2 u 2 y ) ψ 2
    (5.12)
    and
    ( u h ) ψ u = ( u 1 ψ 1 x + m u 1 ψ 1 y + 1 h δ u 2 ψ 1 y ) u 1 + ( u 1 ψ 2 x + m u 1 ψ 2 y + 1 h δ u 2 ψ 2 y ) u 2
    (5.13)
    with
    u = u ( x , y , s ) and ψ = ( u ( x , y , t + τ ) u ( x , y , t ) ) ( y + δ ) λ if  s [ t , t + τ ] .
    Observing (5.1), (5.3) and, moreover,
    | ψ | ( | u ( t + τ ) | + | u ( t ) | ) ( y + δ ) λ , | ψ y | ( | u y ( t + τ ) | + | u y ( t ) | ) ( y + δ ) λ + λ ( | u ( t + τ ) | + | u ( t ) | ) ( y + δ ) λ 1 ,
    we have to show boundedness of integrals of the form
    0 T τ D [ | u | ] τ 2 | u | ( y + δ ) λ ˜ + 1 d ( x , y ) d t , 0 T τ D | u | [ | u | ] τ [ | u | ] τ ( y + δ ) λ ˜ + 1 d ( x , y ) d t , 0 T τ D [ | u | ] τ 2 | u | ( y + δ ) λ + 1 d ( x , y ) d t , 0 T τ D [ | u | ] τ [ | u 2 | ] τ | u | ( y + δ ) λ d ( x , y ) d t , }
    (5.14)

    where λ ˜ { λ , λ + 1 } and [ ] τ is the Steklov average defined after (5.10). Because of ( y + δ ) λ + 1 ( 1 + α ) ( y + δ ) λ , we can assume λ ˜ = λ . Moreover, due to inequality (5.11), we may omit the Steklov averages if the factors under the integrals are separated by Hölder’s inequality. Just this is the case in the proof of Lemma 5.2. Hence, we can apply Lemma 5.2 to the integrals (5.14). Since the assumptions of Lemma 5.2 are fulfilled due to Theorem 3.1, the integrals over the items (5.12) and (5.13) are estimated for λ 5 / 2 by with a constant c independent of κ, δ, ε.

     
  3. (iii)
    The integrals over the items μ ( [ h u + ( h u ) T ] : h ψ + 2 ( h δ ) 2 2 u 2 ψ 2 ) and ε ( h q h ω ) are estimated in a similar way as in (ii). We consider now
    ( ω div h u ) = ω ( u 1 x + m u 1 y + 1 h δ u 2 y + h δ u 2 )
    (5.15)
    and
    ( q div h ψ ) = q ( ψ 1 x + m ψ 1 y + 1 h δ ψ 2 y + h δ ψ 2 )
    (5.16)
    with
    u = u ( x , y , s ) , q = q ( x , y , s ) and ψ = ( u ( x , y , t + τ ) u ( x , y , t ) ) ( y + δ ) λ , ω = q ( x , y , t + τ ) q ( x , y , t )
    if s [ t , t + τ ] . Hence, in this case, we need boundedness of the integrals
    0 T τ D | q | [ | u | ] τ ( y + δ ) d ( x , y ) d t , 0 T τ D | q | [ | u 2 | ] τ d ( x , y ) d t , 0 T τ D [ | q | ] τ ( | u | + | u | ) ( y + δ ) λ + 1 d ( x , y ) d t , 0 T τ D [ | q | ] τ | u 2 | ( y + δ ) λ d ( x , y ) d t . }
    (5.17)
    This may be obtained for λ 0 again by means of Hölder’s inequality, (5.11) and the estimates of Theorem 3.1. However, since
    ( 0 T D | q | 2 ( y + δ ) d ( x , y ) d t ) 1 / 2 c ε ,

    the integrals over items (5.15) and (5.16) can only be estimated by c ε 1 / 2 τ .

     
  4. (iv)
    Since the integrals over the derivatives of z are obviously bounded due to the estimates of Theorem 3.1, finally we have a look at the last item on the left-hand side of (2.21),
    κ ( u 2 z ) ( ψ 2 ξ ) ,
    (5.18)
     
where u 2 = u 2 ( x , 1 , s ) , z = z ( x , s ) , ψ 2 = ( u 2 ( x , 1 , t + τ ) u 2 ( x , 1 , t ) ) ( 1 + δ ) λ , ξ = ( z ( x , t + τ ) z ( x , t ) ) ( 1 + δ ) λ if s [ t , t + τ ] . Hence, we obtain
τ 0 T τ 0 L κ [ | u 2 z | ] τ | ( u 2 ( t + τ ) z ( t + τ ) ) ( u 2 ( t ) z ( t ) ) | ( 1 + δ ) λ d x d t 2 τ κ 0 T 0 L | u 2 z | 2 d x d t ( 1 + α ) λ c τ

due to Theorem 3.1 with c independent of κ. Since the estimation of the right-hand side of (2.21) is obvious, this concludes the proof. □

Note that (5.4) depends on ε also if we test with χ [ t , t + τ ] ( s ) ( w , 0 , ζ ) and the corresponding integral in (5.4) disappears. The reason is that it remains to prove boundedness of the integral over item (5.16) which we are not able to estimate independent on ε. Define now
w ( x , y , t ) : = u ( x , y , t ) ( y + δ ) 7 / 4 .
(5.19)

Then from Theorems 3.1 and 5.1 we can derive the following estimates, which yield compactness of w in L 2 ( ( 0 , T ) × D ) .

Lemma 5.3 There are constants c independent of δ, ε, κ and w such that
0 T τ D | w ( t + τ ) w ( t ) | 2 d ( x , y ) d t c ε 1 / 2 τ
(5.20)
and
0 T D w : w d ( x , y ) d t c .
(5.21)
Proof The first estimate (5.20) is equivalent to (5.4) due to the definition of w. Hence, it remains to prove (5.21). By means of (5.1), we see that
w x = u x ( y + δ ) 7 / 4 | w x | 2 ( 1 + α ) 5 / 2 | u x | 2 ( y + δ ) α 2 ( 1 + α ) 5 / 2 | u x | 2 and w y = u y ( y + δ ) 7 / 4 + 7 4 u ( y + δ ) 3 / 4 | w y | 2 2 | u y | 2 ( y + δ ) ( 1 + α ) 5 / 2 + 49 8 | u | 2 ( y + δ ) ( 1 + α ) 1 / 2 c ( | u y | 2 + | u | 2 ) .

Theorem 3.1 then implies (5.21). □

To simplify the notation for the limit process δ