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### 2D Navier-Stokes Equations in a Time Dependent Domain with Neumann Type Boundary Conditions

J´an Filo and Anna Zauˇskov´a

Department of Mathematical Analysis and Numerical Mathematics Comenius University, Mlynsk´a dolina

SK–842 48 Bratislava, Slovakia

Abstract. In this paper the two-dimensional Navier-Stokes system for incompressible fluid coupled with a parabolic equation through the Neumann type boundary condition for the second component of the velocity is considered. Navier-Stokes equations are defined on a given time dependent domain. We prove the existence of a weak solution for this system. In addition, we prove the continuous dependence of solutions on the data for a regularized version of this system. For a special case of this regularized system also a problem with an unknown interface is solved.

The problem under consideration is an approximation of the fluid-structure interac- tion problem proposed by A. Quarteroni in [19]. We conjecture that our approach is useful also for the numerical treatment of the problem and at the end we shortly present our numerical experiments.

Mathematics Subject Classification. 35Q30, 65M20, 74F10, 76D03

Keywords. Navier-Stokes equations, Time dependent domain, Fluid-structure inter- action, Neumann boundary condition

## 1. Introduction

We consider the two-dimensional Navier–Stokes system for incompressible fluid ρ∂vi

∂t +ρ

2

X

j=1

vj

∂vi

∂xj

=µ∆vi− ∂p

∂xi

i= 1,2, divv = 0 (1.1) in a domain

Ω(h)≡ {(x1, x2, t) : 0< x1< L, 0< x2< h(x1, t), 0< t < T} for a givenC1 functionhsatisfying

0< α≤h(x1, t)≤α−1 , h(x1,0) =h(0, t) =h(L, t) =R >0. (1.2) On the upper part of the boundary of Ω(h), that we shall denote Γw= Γwall, we requirev1= 0, i.e.

v1(x1, h(x1, t), t) = 0 for 0< x1< L, 0< t < T (1.3)

Supported by the Grants of the Comenius University 66/2004 and 321/2005

Both authors were supported by the VEGA Grant Agency of the Slovak Republic No. 1/0260/03

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and we impose the following Neumann type boundary condition for the second component of the velocityv

µ∂v2

∂x1

−∂h

∂x1

+µ∂v2

∂x2 −p+pw−ρ 2v2

v2−∂h

∂t

(x1, h(x1, t), t)

=ρκ

λ∂η

∂t(x1, t) + (1−λ)∂h

∂t(x1, t)−v2(x1, h(x1, t), t)

(1.4) for a given function pw = pw(x1, t), 0 < λ ≤ 1 and some κ ≫ 1 . In this boundary condition an unknown function η =η(x1, t) appears. We require from η to satisfy the following differential equation

−E ∂2η

∂t2 −a∂2η

∂x21 +bη−c ∂3η

∂t∂x21

(x1, t) (1.5)

λ∂η

∂t(x1, t) + (1−λ)∂h

∂t(x1, t)−v2(x1, h(x1, t), t)

for any 0< x1< L, 0< t < T equipped with the boundary and initial condition forη,

η(0, t) =η(L, t) = 0 and η(x1,0) = ∂η

∂t(x1,0) = 0. (1.6) Hereρ, µ, L, T, α, R, E, a, b, care given positive constants.

According to our motivation, described in Section 2 below, we complete the Navier–Stokes system (1.1) with the following boundary and initial conditions.

On a part of the boundary which we shall denote Γin= Γinf low we put v2(0, x2, t) = 0,

µ∂v1

∂x1 −p+pin−ρ 2|v|2

(0, x2, t) = 0 (1.7) for any 0< x2< R, 0< t < T and for a given functionpin=pin(x2, t). On the opposite part of the boundary Γout = Γoutf low we put

v2(L, x2, t) = 0,

µ∂v1

∂x1 −p+pout−ρ 2|v|2

(L, x2, t) = 0 (1.8) for any 0< x2< R, 0< t < T and for a given functionpout=pout(x2, t). Finally, on the rest part of the boundary Γc we set

v2(x1,0, t) = 0, µ∂v1

∂x2

(x1,0, t) = 0 (1.9) for any 0< x1< L, 0< t < T and

v(x1, x2,0) =0 for any 0< x1< L, 0< x2< h(x1,0). (1.10) Problem (1.1)–(1.10) is an approximation of the fluid–structure interaction model proposed by A. Quarteroni in [19]. In the original proposalhis not known, instead, v,pand a free boundary

h=R+η (1.11)

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are to be found such that (1.4) and (1.5) are replaced by µ∂v2

∂x1

−∂η

∂x1

+µ∂v2

∂x2 −p+pw = −Eρ ∂2η

∂t2 −a∂2η

∂x21 +bη−c ∂3η

∂t∂x21

(1.12)

and ∂η

∂t(x1, t) =v2(x1, R+η(x1, t), t) (1.13) for any 0< x1< L, 0 < t < T. Our original aim was to solve just this problem with an unknown interfaceη, but our attempt has failed.

During the last two decades, however, a substantial beginning has been made on mathematical analysis of the equations governing the motion of compressible and incompressible fluids with free boundary, see e.g. [1], [4], [7], [8], [12], [14], [18], [19], [20], and references therein. Since this free boundary value problem seems to be very difficult to treat, here the Navier–Stokes system on a given time dependent domain coupled with a parabolic equation through the Neumann type boundary condition for the second component of the velocity is studied. We establish an existence result for Problem (1.1)-(1.10) assuming that his known. For fixed h we also prove an existence result for the limit case letting κ → ∞. We reflect the free boundary problem twice. In numerical experiments as well as in proof of local existence result for a special case of the regularized system (Section 3), which includes the problem of finding an unknown interfaceh=R+η. The methods, that we apply, borrow material from [21], [10], [2], [11] and the references therein, nevertheless, their application to our problem seems to be not straightforward.

The paper is organized as follows. Our study initially, in Section 2, recalls the original fluid-structure interaction model [19] and explains our main assumptions leading to Problem (1.1)-(1.10). In Section 3 we transform our problem to the problem on a fix domain and we conclude this section by making precise the meaning of a weak solution. To prove the existence of the solution we use the method of semi-discretization in time. The existence of a solution in one time step is determined in Section 4. Sections 5, 6 provide necessary a priori estimates and a convergence theorem to prove the existence of the regularized problem. The uniqueness and continuous dependence on the data for the regularized problem is demonstrated in Section 7. Finally, Section 8 presents the main result of this paper where the existence of a weak solution to Problem (1.1)-(1.10) is proved.

The case of fixed κ and also the case of κ → ∞ are considered. In Section 9 the way of our approximation is utilized to perform some numerical experiments.

At the end, Section 10 provides an explanation of our failed attempt to solve the original problem. Nevertheless, using the result of Section 7 and Banach’s fixed point theorem we are able to solve a free boundary value problem for the special case of the regularized system.

## 2. Motivation and derivation of the model

Should we restrict ourselves to the two-dimensional fluid–structure interaction model proposed by A. Quarteroni in [19],[12], it reads as follows. The problem is

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to find a functionη(x1, t) defined for 0< x1< L, 0< t < T and functions v(x1, x2, t) = (v1(x1, x2, t), v2(x1, x2, t)), p(x1, x2, t)

defined in the domain Ω(R+η) , with the following properties:

ρ∂v

∂t +ρ(v· ∇)v= div µ(∇v+∇vT)

− ∇p , divv = 0 (2.1) in Ω(R+η),

2η

∂t2 −a∂2η

∂x21 +bη−c ∂3η

∂t∂x21 =H (2.2)

for any 0< x1< L, 0< t < T, where H= 1

ρE p−pw−µ (∇v+∇vT)·n

·e2 , n=

−∂η

∂x1

,1

(2.3) and

v(x1, R+η(x1, t), t) = ∂η

∂t n

|n|(x1, t). (2.4) Problem (2.1)–(2.4) is in [19] equipped with the following conditions:

η(0, t) =α(t), η(L, t) =β(t) t∈(0, T), η(x1,0) =η0(x1), ∂η∂t(x1,0) =η1(x1) x1∈(0, L),

v=g on Γin,

µ

∂v1

∂x2 +∂v∂x21

= 0, 2µ∂v∂x11−p+pout = 0 on Γout, v2= 0, µ

∂v1

∂x2 +∂v∂x21

= 0 on Γc,

v(·,0) =v0(·)

















 (2.5)

for given functions α, β, η0, η1, g, pout and v0.

In what follows we shall simplify and regularize Problem (2.1)–(2.5) in several steps in order to arrive at our approximation given by (1.1)–(1.10).

1. We replace the operator div µ(∇v+∇vT)

in (2.1) by µ∆v and corre- spondingly modifyH in (2.2) putting

H = 1 ρE

p−pw −µ∂v2

∂x1

−∂η

∂x1

−µ∂v2

∂x2

. (2.6)

We take

v1(x1, R+η(x1, t), t) = 0 and v2(x1, R+η(x1, t), t) =∂η

∂t(x1, t) (2.7) instead of (2.4).

2. Now, following [11], we takeκ≫1 and we approximate (2.2) and the second condition of (2.7) by

E ∂2η

∂t2 −a∂2η

∂x21 +bη−c ∂3η

∂t∂x21

v2−∂η

∂t

(2.8)

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and (2.6) by µ∂v2

∂x1

− ∂η

∂x1

+µ∂v2

∂x2 −p+pw −ρ 2v2

v2−∂η

∂t

=ρκ ∂η

∂t −v2

. (2.9) Note that by

κ−→ ∞

we get, at least formally, that the second relation of (2.7) is satisfied. (1.12) holds then as well. This is the key point of our approximation.

One of the possible physical interpretations for introducing finite κcomes from mathematical modelling of semipervious boundary where a boundary condition of the third type occurs. This boundary condition is encountered, for example, when a clogged (e.g. by a thin layer of silt or clay) river bed serves as a boundary of the flow domain.

In our case, the boundary Γwall seems to be partly permeable for finiteκ, but lettingκ→ ∞it becomes impervious.

3. Now we split the problem by decoupling the fluid - domain problem assuming that a wall deformationη(k)(k)(x1, t) and thus, the domain

(k)≡Ω(R+η(k)) is given. We look then for a solution

(v, p, η) = (v(k+1), p(k+1), η(k+1)) of the following problem:

ρ∂v

∂t +ρ(v· ∇)v =µ∆v− ∇p , divv = 0 in Ω(k), (2.10)

µ∂v2

∂x1

−∂η(k)

∂x1

+µ∂v2

∂x2 −p+pw (2.11)

−ρ 2v2

v2−∂η(k)

∂t

(x1, R+η(k)(x1, t), t)

=ρκ

λ∂η

∂t(x1, t) + (1−λ)∂η(k)

∂t (x1, t)−v2(x1, R+η(k)(x1, t), t)

and

−E ∂2η

∂t2 −a∂2η

∂x21 +bη−c ∂3η

∂t∂x21

(x1, t) (2.12)

λ∂η

∂t(x1, t) + (1−λ)∂η(k)

∂t (x1, t)−v2(x1, R+η(k)(x1, t), t)

for any 0< x1< L, 0< t < T and givenλ∈(0,1] . Note that in the right hand side of (2.9) we approximate ∂η

∂t with λ∂η(k+1)

∂t + (1−λ)∂η(k)

∂t .

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4. Boundary and initial conditions for η we simplify by putting α(t) =β(t)≡0 and η0(x1) =η1(x1)≡0, see (2.5). The inflow condition (2.5) given by the non homogeneous Dirichlet boundary condition we replace by Neumann type boundary conditions involving pressure as it seems to be more natural to have given pressure impulsespin(·, t) on Γinflow than the prescribed velocityg.

As concerns boundary conditions on Γin and Γout, in both cases we prescribe the second component of velocity v2 = 0 and we consider the Neumann type boundary conditions for the first component of the velocity involving the dynamic pressure

p+ρ

2|v|2 instead of static pressure p .

This type of boundary conditions are considered in [13] and discussed in [19]. Note that in [6] the problem of 3D-flow in a network of pipes is considered and on Γin

they prescribed boundary conditions involving the pressure v×n=0 and p+ρ

2|v|2=p0 .

Boundary conditions on Γc correspond to the assumptions of symmetry.

5. Finally, the incompressibility condition divv = 0 in Ω(h) we approximate in the next section by

ε ∂p

∂t −∆p

+ divv= 0 in Ω(h) (2.13)

for small 0< ε <1 and we letε→0 in Section 8. We overcome the difficulties con- nected with this constraint following [21], where the approximationε∂tp+divv= 0 was used. Since the divergence free condition does not hold by this approximation, we follow [21] and add the term ρ2vidivv to the nonlinear convective term, see (3.1) below. Later, after passingε→0 we show that divv = 0 holds and this term dissapears.

6. Our idea for constructively solving the original problem (1.1)–(1.10) with (1.4) and (1.5) replaced by (1.12) and (1.13) is to select some h(0) = R+η(0) and then iteratively solve the system on a given time dependent domain Ω(h(k−1)) for an unknown (v(k), p(k), η(k)) , k ∈ N. In addition, in the items above we regularize the problem by introducing positive parameters ε, κ and by adding some additional terms that should disappear if we let ε→0 and κ→ ∞. We hope that physically and mathematically correct solutions should arise as the limit of solutions to our regularized system. Unfortunately, we are not able to prove that {(v(k), p(k), η(k))}k=1 converges to a solution of the original problem requiring also that ε(k)→0 and κ(k)→ ∞ if k→ ∞. Therefore we denote

h(x1, t) =R+η(k)(x1, t) (2.14) and we arrive at our problem (1.1)–(1.10). Numerical experiments indicate, how- ever, that the process converges, see Section 9 below. If 0< ε≪1 and κ≫ 1 are fixed, we show in Section 10 that {(v(k), p(k), η(k))}k=1 converges for small data.

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## 3. Auxiliary problem

Given numbers κ ≫ 1, 0 < ε ≪ 1 and a smooth function h, consider now the system

ρ∂vi

∂t +ρ

2

X

i=1

vj∂vi

∂xj

+ ρ

2 vidivv =µ∆vi− ∂p

∂xi

, i= 1,2 (3.1) together with (2.13) in Ω(h),

2η

∂t2 −a∂2η

∂x21 +bη−c ∂3η

∂t∂x21

(x1, t) (3.2)

=−κ E

λ∂η

∂t(x1, t) + (1−λ)∂h

∂t(x1, t)−v2(x1, h(x1, t), t)

for any 0< x1< L, 0< t < T, equipped with the following boundary and initial conditions:

v1(x1, h(x1, t), t) = 0

µ∂v2

∂x1

−∂h

∂x1

+µ∂v2

∂x2 −p+pw−ρ 2v2

v2−∂h

∂t

(x1, h(x1, t), t)

=ρκ

λ∂η

∂t(x1, t) + (1−λ)∂h

∂t(x1, t)−v2(x1, h(x1, t), t)

∂p

∂x1

−∂h

∂x1

+ ∂p

∂x2

(x1, h(x1, t), t) =−ρ 2

∂h

∂t(x1, t)p(x1, h(x1, t), t) for any 0< x1< L, 0< t < T ,

µ∂v1

∂x1 −p+pout−ρ 2|v|2

(L, x2, t) = 0 v2(L, x2, t) = 0, ∂p

∂x1

(L, x2, t) = 0 for any 0< x2< R, 0< t < T,

µ∂v1

∂x1 −p+pin−ρ 2|v|2

(0, x2, t) = 0 v2(0, x2, t) = 0, ∂p

∂x1(0, x2, t) = 0 for any 0< x2< R, 0< t < T,

µ∂v1

∂x2

(x1,0, t) = 0, v2(x1,0, t) = 0, ∂p

∂x2

(x1,0, t) = 0

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for any 0< x1< L, 0< t < T ,

v(x1, x2,0) =0, p(x1, x2,0) = 0 for any 0< x2< R, 0< x1< L,

η(x1,0) = 0, ∂η

∂t(x1,0) = 0 for any 0< x1< Land, finally,

η(0, t) =η(L, t) = 0

for any 0< t < T . Note that h(0, t) =h(L, t) =h(x1,0) =R.

By performing the tedious but straightforward formal manipulations one can see that if (v, p, η) solves our problem, then

u(y1, y2, t)def= v(y1, h(y1, t)y2, t), (3.3) q(y1, y2, t)def= ρ−1p(y1, h(y1, t)y2, t) (3.4) and

u(y1, t)def= ∂η

∂t(y1, t) (3.5)

for 0< y1< L, 0< y2<1, 0< t < T solve the following problem:

∂(hu1)

∂t −∂h

∂t

∂(y2u1)

∂y2 +hu1

∂u1

∂y1 −y2

h

∂h

∂y1

∂u1

∂y2

+u2∂u1

∂y2 (3.6)

+h

2u1divhu− ∂

∂y1

νh

∂u1

∂y1 −y2

h

∂h

∂y1

∂u1

∂y2

−hq

− ∂

∂y2

ν h

∂u1

∂y2 −νy2

∂h

∂y1

∂u1

∂y1 −y2

h

∂h

∂y1

∂u1

∂y2

+y2

∂h

∂y1

q

= 0 in D×(0, T),

∂(hu2)

∂t −∂h

∂t

∂(y2u2)

∂y2

+hu1

∂u2

∂y1 −y2

h

∂h

∂y1

∂u2

∂y2

+u2

∂u2

∂y2

(3.7) +h

2u2divhu− ∂

∂y1

νh

∂u2

∂y1 −y2

h

∂h

∂y1

∂u2

∂y2

− ∂

∂y2

ν h

∂u2

∂y2 −νy2

∂h

∂y1

∂u2

∂y1 −y2

h

∂h

∂y1

∂u2

∂y2

−q

= 0 in D×(0, T),

ε

∂(hq)

∂t −∂h

∂t

∂(y2q)

∂y2

−ε ∂

∂y1

h

∂q

∂y1−y2

h

∂h

∂y1

∂q

∂y2

(3.8)

−ε ∂

∂y2

1 h

∂q

∂y2 −y2

∂h

∂y1

∂q

∂y1 −y2

h

∂h

∂y1

∂q

∂y2

+ hdivhu= 0

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in D×(0, T), where D= (0, L)×(0,1) and ν =ρ−1µ, divhu def= ∂u1

∂y1 −y2

h

∂h

∂y1

∂u1

∂y2

+ 1 h

∂u2

∂y2

, (3.9)

∂u

∂t(y1, t)−c∂2u

∂y12(y1, t)−a ∂2

∂y12 Z t

0

u(y1, s)ds+b Z t

0

u(y1, s)ds

=−κ E

λu(y1, t) + (1−λ)∂h

∂t(y1, t)−u2(y1,1, t)

(3.10) for any 0< y1 < L, 0 < t < T with the boundary and initial conditions listed below:

u1(y1,1, t) = 0, (3.11)

ν h

∂u2

∂y2 −ν ∂h

∂y1

∂u2

∂y1 −y2

h

∂h

∂y1

∂u2

∂y2

−q

(y1,1, t)

=

−qw+1 2u2

u2−∂h

∂t

−κ

u2−λu−(1−λ)∂h

∂t

(y1,1, t), 1

h

∂q

∂y2 − ∂h

∂y1

∂q

∂y1− 1 h

∂h

∂y1

∂q

∂y2

(y1,1, t) = −1 2

∂h

∂t(y1, t)q(y1,1, t) for any 0< y1< L, 0< t < T, qw(y1, t) =ρ−1pw(y1, t),

ν ∂u1

∂y1 −y2

h

∂h

∂y1

∂u1

∂y2 −q

(L, y2, t) =

−qout+1 2|u|2

(L, y2, t), u2(L, y2, t) = 0,

∂q

∂y1−y2

h

∂h

∂y1

∂q

∂y2

(L, y2, t) = 0 (3.12) for any 0< y2<1, 0< t < T, qout(y2, t) =ρ−1pout(Ry2, t),

ν ∂u1

∂y1 −y2

h

∂h

∂y1

∂u1

∂y2 −q

(0, y2, t) =

−qin+1 2|u|2

(0, y2, t) u2(0, y2, t) = 0,

∂q

∂y1−y2

h

∂h

∂y1

∂q

∂y2

(0, y2, t) = 0 (3.13) for any 0< y2<1, 0< t < T, qin(y2, t) =ρ−1pin(Ry2, t),

ν∂u1

∂y2

(y1,0, t) = 0, u2(y1,0, t) = 0, ε∂q

∂y2

(y1,0, t) = 0 (3.14) for any 0< y1< L, 0< t < T ,

u(y1, y2,0) =0, q(y1, y2,0) = 0 (3.15) for any 0< y1< L, 0< y2<1 and, finally,

u(y1,0) =u(0, t) =u(L, t) = 0 (3.16)

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for any 0< y1< Land for any 0< t < T .

We continue this section by making precise the meaning of the solution of the problem (3.6)–(3.16). To this end we first define

V ≡V ×H1(D)×H01(0, L) (3.17) where

V ≡

w∈H1(D)2: w1= 0 onSw, w2= 0 onSin∪Sout∪Sc (3.18) and

Sw ={(y1,1) : 0< y1< L}, Sin ={(0, y2) : 0< y2<1}, Sout ={(L, y2) : 0< y2<1}, Sc={(y1,0) : 0< y1< L}. Throughout this and the next sections we assume

h∈C1([0, L]×[0, T])∩W2,2((0, L)×(0, T)), (3.19) qin, qout ∈L2(0, T;L2(0,1)), qw ∈L2(0, T;L2(0, L)).

The function spaces we use are rather familiar and we omit the definition, see [22].

Definition 3.1 We call (u, q, u) ∈ L2(0, T;V) a weak solution of the initial boundary value problem (3.6)–(3.16) if the following two properties are fulfilled:

1) u∈L(0, T;L2(D)2), ∂(hu)

∂t ∈ L2(0, T;V)∩L4(0, T;L4(D)2)

=L2(0, T;V) +L4/3(0, T;L4/3(D)2), that is, Z T

0

∂(hu)

∂t ,ζ

dt+ Z T

0

Z

D

hu·∂ζ

∂t dt= 0 (3.20)

for every test function ζ ∈L2(0, T;V)∩L4(0, T;L4(D)2)∩H1,1(0, T;L2(D)2) with ζ(T) = 0 , q ∈ L(0, T;L2(D)) , ∂(hq)

∂t ∈L2(0, T;H−1(D)) and u ∈ L(0, T;H01(0, L))∩H1(0, T;L2(0, L)).

2) (u, q, u)satisfies the system of differential equations, that is,

− Z T

0

∂(hu)

∂t ,ψ

dt (3.21)

= Z T

0

Z

D

−∂h

∂t

∂(y2u)

∂y2 ·ψ+

hu1

∂u

∂y1 −y2

h

∂h

∂y1

∂u

∂y2

+u2∂u

∂y2

·ψ +h

2 u·ψdivhu−h qdivhψ

dy + ν((u,ψ))h

+R Z 1

0

qout−1 2|u1|2

ψ1(L, y2, t)−

qin−1 2|u1|2

ψ1(0, y2, t)

dy2

+ Z L

0

qw−1

2u2

u2−∂h

∂t

u2−λu−(1−λ)∂h

∂t

ψ2(y1,1, t)dy1

(11)

∂(hq)

∂t , φ

− Z

D

ε∂h

∂t

∂(y2q)

∂y2 φ − hdivh

dy

+ε((q, φ))h + ε 2

Z L 0

∂h

∂t(y1, t)qφ(y1,1, t)dy1

+ Z L

0

∂u

∂tξ+c ∂u

∂y1

∂ξ

∂y1

+a ∂

∂y1

Z t 0

u(y1, s)ds∂ξ

∂y1

+b Z t

0

u(y1, s)ds ξ+κ E

λu+ (1−λ)∂h

∂t −u2

ξ

(y1, t)dy1

dt

for every(ψ, φ, ξ)∈L2(0, T;V),ψ∈L4(0, T;L4(D)2), where ((u,ψ))h

def= ((u1, ψ1))h+ ((u2, ψ2))h and (3.22) ((q, φ))h

def= Z

D

h

∂q

∂y1 −y2

h

∂h

∂y1

∂q

∂y2

∂φ

∂y1

+ 1

h

∂q

∂y2 −y2

∂h

∂y1

∂q

∂y1 −y2

h

∂h

∂y1

∂q

∂y2

∂φ

∂y2

dy .

Note, that (3.20) implies Z τ

0

∂(hu)

∂t ,ζ

dt+ Z τ

0

Z

D

hu·∂ζ

∂t dy dt= Z

D

hu·ζ(τ)dy (3.23) and that (3.21) holds ifT is replaced byτ for almost allτ∈(0, T).

We will use the following form of the interpolation theorems, which play an important role by proving the existence of the weak solution.

Proposition 3.1 Let ϕ be any function in H1(D) such that ϕ = 0 on Sw or ϕ= 0 onSc. Then for any p≥2and for any numberθ in the interval

p−2

p ≤θ≤1 there exists a constant C=C(p, θ)such that

kϕkLp(D)≤Ck∇ϕkθL2(D)kϕk1−θL2(D). (3.24) Moreover, if ϕ be any function in L2(0, T;H1(D))∩L(0, T;L2(D)) such that ϕ= 0 onSw orϕ= 0on Sc for almost allt, then for anyp≥2

kϕk

L

2p

p−2(0,T;Lp(D)) ≤Ch

kϕkL(0,T;L2(D))

i2ph

kϕkL2(0,T;H1(D))

ip−2p

. (3.25) Proof. The form of Nierenberg–Gagliardo inequality (3.24) can be found e.g. in [15, Theorem 2.2]. Then (3.25) follows from (3.24) forθ= (p−2)/pby integration over (0, T).

(12)

Proposition 3.2 DenoteS ≡Sin∪Sc∪Sout ∪Sw and letϕbe any function in H1(D) such thatϕ= 0on Sw or ϕ= 0onSc. Then for any r >1there exists a constant C=C(r)such that

kϕkLr(S)≤Ck∇ϕk1−L2(D)1r kϕkL1r2(D). (3.26) Moreover, if ϕ be any function in L2(0, T;H1(D))∩L(0, T;L2(D)) such that ϕ= 0 onSw or ϕ= 0on Sc for almost allt then for any q >1

kϕkLq−12q (0,T;Lq

(S)) ≤Ch

kϕkL(0,T;L2(D))

i1qh

kϕkL2(0,T;H1(D))

iq−1q

. (3.27) Proof. The inequality (3.26) can be found e.g. in [15, Inequality (2.21), page 69].

(3.27) follows, similarly as above, from (3.26) by integration over (0, T).

## 4. Implicit time discretization

To prove the existence of the solution to (3.6)–(3.16) we shall approximate the problem by the perturbed stationary Navier–Stokes systems coupled with the el- liptic problems for the pressure and deformation of the wall. This approach is important also for the numerical treatment of the problem. For this we replace

∂(huk)

∂t , ∂(hq)

∂t and ∂u

∂t by the backward difference quotients

hiuik−hi−1ui−1k

∆t , hiqi−hi−1qi−1

∆t and ui−ui−1

∆t for ∆t≡T /n,n∈N,n≫1 andRt

0u(s)dsbyPi

k=1uk∆tfori∆t≤t <(i+ 1)∆t.

Hence, for each i ∈ {1, . . . , n} and given (uj, qj, uj), 0 ≤ j ≤ i−1 we get the perturbed stationary Navier–Stokes system for ui coupled with the elliptic problems for qi and ui, i= 1, . . . , n, which in the variational formulation reads as follows:

Z

D

hiui−hi−1ui−1

∆t −d∆thi∂(y2ui)

∂y2

·ω (4.1)

+

hiui1 ∂ui

∂y1 −y2

hi

∂hi

∂y1

∂ui

∂y2

+ui2∂ui

∂y2

·ω+hi

2 ui·ωdiviui

−hiqidiviω

dy + ν((ui,ω))hi + ε((qi, v))hi

+ Z

D

ε

hiqi−hi−1qi−1

∆t −d∆thi ∂(y2qi)

∂y2

v+hivdiviui

dy

+R Z 1

0

qouti −1 2

ui1

2

ω1(L, y2) −

qini −1 2

ui1

2

ω1 (0, y2)

dy2

(13)

+ Z L

0

qwi −1

2ui2 ui2−d∆thi ω2

+κ ui2−λui−(1−λ)d∆thi

ω2− ϑ E

2d∆thiqiv

(y1,1)dy1

+ Z L

0

ui−ui−1

∆t ϑ+c∂ui

∂y1

∂ϑ

∂y1

+a

i

X

k=1

∂uk

∂y1

∆t∂ϑ

∂y1

+b

i

X

k=1

uk∆t

! ϑ

! dy1

= 0

for any̟= (ω, v, ϑ)∈V, where the notation from (3.22) withhiinstead ofhis adopted. Here hi(y1) =h(y1, i∆t) ,

d∆thi= hi−hi−1

∆t and qi...(y1) = 1

∆t Z i∆t

(i−1)∆t

q...(y1, s)ds . In what follows we shall consider the following variational problem:

Findwi= (ui, qi, ui)∈V such that

ai(wi,̟) +bi(wi,wi,̟) =Li(̟) ∀̟∈V , (4.2) where ̟= (ω, v, ϑ),V is defined by (3.17) andai(·,·), bi(·,·,·),Li(·) are deter- mined by (4.1), i.e.

1. ai(·,·) : V ×V →Rbe the following bilinear continuous form onV: ai(wi,̟) = ν((ui,ω))hi + ε((qi, v))hi

+ 1

∆t Z

D

hiui·ωdy+ ε

∆t Z

D

hiqiv dy

+ Z L

0

(c+a∆t)∂ui

∂y1

∂ϑ

∂y1

+ 1

∆t+b∆t

uiϑ

dy1

− Z

D

d∆thi∂(y2ui)

∂y2 ·ωdy+ Z L

0

1

2ui2d∆thiω2(y1,1)dy1

−ε Z

D

d∆thi∂(y2qi)

∂y2 v dy+ε 2

Z L 0

d∆thiqiv(y1,1)dy1

+κ Z L

0

(λui−ui2) ϑ

E −ω2

(y1)dy1

+ Z

D

hivdiviui−hiqidiviω dy .

2. The trilinear formbi(·,·,·) is defined by

bi(·,·,·) : V ×V ×V −→R, bi(wi,mi,̟) = Bi(ui,zi,ω) +

Z

D

hi

2 zi·ωdiviuidy (4.3)

(14)

−R 2

Z 1 0

ui1zi1ω1(L, y2)dy2+R 2

Z 1 0

ui1z1iω1(0, y2)dy2

−1 2

Z L 0

ui2zi2ω2(y1,1)dy1, Bi(ui,zi,ω) =

Z

D

hiui1

∂zi

∂y1 −y2

hi

∂hi

∂y1

∂zi

∂y2

+ui2∂zi

∂y2

·ωdy

formi= (zi,·,·),wi = (ui,·,·),̟= (ω,·,·) andzi,ui,ω∈V. 3. Finally,Li(·) is the linear functional onV,

Li(̟) = 1

∆t Z

D

hi−1 ui−1·ω+ε qi−1v

dy+ 1

∆t Z L

0

ui−1ϑ dy1

+R Z 1

0

qini ω1(0, y2)−qouti ω1(L, y2) dy2

+ Z L

0 −qiwω2(y1,1)−

i−1

X

k=1

a∂uk

∂y1

∂ϑ

∂y1

+bukϑ

(y1)∆t

! dy1

+κ(1−λ) Z L

0

d∆thi2− ϑ

E)(y1)dy1.

Theorem 4.1 Let i∈ {1,2, . . . , n} and wj ∈V for j ≤i−1 be given. Assume there are nonnegative constants α, K, independent oni, such that

0< α≤hi(y1)≤α−1 (4.4)

and

∂hi

∂y1

(y1)

+

d∆thi(y1)

≤K (4.5)

for all0≤y1≤Landi= 1,2, . . . n. Moreover, assume that qini , qouti ∈L2(0,1), qiw ∈L2(0, L) and ∆t≤α/K . Then Problem (4.2) has at least one solution.

Proof. Following [6, Proof of Theorem 2.1], we use Galerkin’s method. V is a closed subspace of H1(D)3×H01(0, L) and it is thus possible to choose a basis {ζk}k=1⊂V. For everyℓ≥1 we define an approximate problem by:

Find ckℓ ∈R, 1≤k≤ℓsuch thatw=

X

k=1

ckℓζk is a solution of

ai(wk) +bi(w,wk) =Lik) ∀k= 1, . . . , ℓ . (4.6) To prove the existence of a solution to (4.6) we will use the following lemma (see [16, Lemma 1.4.3, p.53] or [21, Lemma 2.1.4, p. 164]).

(15)

Lemma 4.2 Let Y be a finite–dimensional Hilbert space with scalar product(·,·) and norm k · k. Let P be a continuous mapping from Y into itself such that, for a sufficiently large ρ >0,

(P(ζ), ζ)≥0 ∀ζ∈Y such that kζk=ρ . (4.7) Then there existsζ∈Y,kζk ≤ρsuch that P(ζ) = 0 .

In our case Y = Y = span{ζ1, . . . ,ζ} equipped with the scalar product of H1(D)3×H01(0, L), and for any ζ ∈ Y define P(ζ) =P(ζ) ∈ Y using Riesz’s Representation Theorem by

(P(ζ),z) =ai(ζ,z) +bi(ζ,ζ,z)−Li(z) ∀z∈Y .

It is easy to see thatPis continuous. In order to prove (4.7) let us first introduce the following lemma.

Lemma 4.3 Let (4.4)–(4.5) be satisfied. Then

((v, v))hi ≥ α 2 +K2

Z

D|∇v|2dy (4.8)

for anyv∈H1(D), where((·,·))hi is given by (3.22) withhiinstead ofh.

Proof. Note first that ((v, v))hi=

Z

D

( hi

∂v

∂y1 −y2

hi

∂hi

∂y1

∂v

∂y2

2 + 1

hi ∂v

∂y2

2)

dy . (4.9) For a moment, let us denote

A=√ hi ∂v

∂y1

, B= 1

√hi

∂v

∂y2

, z=y2∂h

∂y1

and rewrite

((v, v))hi= Z

D

nA2−2zAB+ (1 +z2)B2o

dy . (4.10)

Taking 0< δ <1/(K2+ 1) one easily obtains

A2−2zAB+ (1 +z2)B2≥δA2+1−δ(z2+ 1) 1−δ B2. As|z|< K,puttingδ= 1/(K2+ 2),(4.8) follows easily.

Next, note that

(P(ζ),ζ) =ai(ζ,ζ)−Li(ζ), as

bi(ζ,ζ,ζ) = 0 (4.11)

(16)

forζ= (v, p, λEv). To get (4.11) we have used the fact that bi(wi,mi,̟) =1

2Bi(ui,zi,ω)−1

2Bi(ui,ω,zi) (4.12) formi= (zi,·,·),wi= (ui,·,·),̟= (ω,·,·), whereBi is defined by (4.3). Some calculations have to be performed to obtain (4.12) and we omit it here. Next, it is easy to see that there exists a positive constantC such that

Li(ζ)

≤CkζkV ∀ζ∈V . (4.13)

Finally, one can easily verify that ai(ζ,ζ) = ν((v,v))hi+ε((p, p))hi

+ Z

D

hi

∆t +1 2d∆thi

|v|2+ε|p|2

dy (4.14)

+ Z L

0

λE(c+a∆t)

∂v

∂y1

2

+λE 1

∆t+b∆t

|v|2+κ(v2−λv)2dy1

and therefore, if ∆t is sufficiently small, say 0 < ∆t < α/K, there exists a constantδ >0 such that

ai(ζ,ζ)≥δkζk2V ∀ζ∈V . (4.15) Hence

(P(ζ),ζ)≥δkζkV

kζkV −C δ

∀ζ∈Y

which implies that P satisfies (4.7) with ρ = C/δ. Thus, for any ℓ ∈ N there exists a solutionw of (4.6) which verifies

kwkV ≤ρ .

Therefore, there existswi∈V and a subsequence{w} of{w}such that w →wi weakly inV as ℓ → ∞. (4.16) Due to the compact embedding ofV into corresponding Lebesgue spaces (Lp(·))3× L2(0, L) we get

w →wi strongly in (Lp(D))3×Lp(0, L) as ℓ→ ∞ (4.17) and

w →wi strongly in (Lp(S))3×Lp(0, L) as ℓ → ∞ (4.18) for any p≥ 1. Concerning test functions, let ̟ ∈V and a sequence {̟} be such that̟∈Y and

̟ →̟ strongly in V (4.19)

(17)

asℓ→ ∞. Note that it converges also in the spaces as in (4.17) and (4.18).

Finally, according to (4.2) for everyℓ we have

ai(w) +bi(w,w) =Li). (4.20) Using (4.16)–(4.19) we can pass to the limit in (4.20) and we obtain that

ai(wi,̟) +bi(wi,wi,̟) =Li(̟) ∀̟∈V , i.e. (4.2) holds, which completes the proof.

## 5. A priori estimates

We now ascertain the a priori estimates for ui, qi, ui, i= 1,2, . . . , n. In the first step, we test (4.1) with wi = (ui, qi, Eui) and sum overi = 1,2, . . . , r for some r ≤n. Before writing a final formula, let us focus on the separate terms, where some calculations were performed:

2

r

X

i=1

Z

D

hiui−hi−1ui−1

·uidy = Z

D

hr|ur|2dy

+

r

X

i=1

Z

D

1 hi

hiui−hi−1ui−1

2+hi−1

hi hi−hi−1 ui−1

2 dy ,

−2 Z

D

d∆thi ∂(y2ui)

∂y2 ·uidy

= Z L

0

d∆thi ui2

2(y1,1)dy1− Z

D

d∆thi ui

2dy ,

2

r

X

i=1

Z L 0

ui−ui−1

uidy1= Z L

0 |ur|2dy1+

r

X

i=1

Z L 0

ui−ui−1

2dy1,

U0≡0, Ui

i

X

k=1

uk∆t , Ui−Ui−1

∆t =ui, (5.1)

a

r

X

i=1

Z L 0

∂Ui

∂y1

∂ui

∂y1

dy1∆t=a 2

Z L 0

(

∂Ur

∂y1

2

dy1+

r

X

i=1

∂(Ui−Ui−1)

∂y1

2) dy1,

b

r

X

i=1

Z L 0

Uiuidy1∆t= b 2

Z L 0

(

|Ur|2dy1+

r

X

i=1

Ui−Ui−1

2

) dy1

and, finally, let us recall (4.11). (4.1) and (5.1) then easily yield Z

D

hr

|ur|2+ε|qr|2

dy+E Z L

0 |ur|2dy1

+

r

X

i=1

Z

D

1 hi

hiui−hi−1ui−1

2 + ε

hiqi−hi−1qi−1

2 dy

(18)

+

r

X

i=1

E

Z L 0

ui−ui−1

2dy1+ 2ν((ui,ui))hi + 2ε((qi, qi))hi

∆t

+ 2

r

X

i=1

Z L 0

n

(1−λ)κ ui2−d∆thi

ui2−ui +λκ

ui2−ui

2o dy1∆t

+ 2cE

r

X

i=1

Z L 0

∂ui

∂y1

2

dy1∆t+aE Z L

0

(

∂Ur

∂y1

2

+

r

X

i=1

∂(Ui−Ui−1)

∂y1

2) dy1

+bE Z L

0

(

|Ur|2dy1+

r

X

i=1

Z L 0

Ui−Ui−1

2

) dy1

=

r

X

i=1

2R

Z 1 0

qini (y2)ui1(0, y2)−qouti (y2)ui1(L, y2) dy2

+ 2 Z L

0

qwi(y1)ui2(y1,1)dy1− Z

D

hi−1

hi d∆thi ui−1

2+ε qi−1

2 dy

∆t . Now, with the assistance of Lemma 4.3 and Young’s inequality we obtain

Z

D

hr

|ur|2+ε|qr|2 dy+E

Z L

0 |ur|2dy1

+

r

X

i=1

2α 2 +K2

Z

D

ν ∇ui

2+ε ∇qi

2

dy+ 2cE Z L

0

∂ui

∂y1

2

dy1

∆t

+aE Z L

0

∂Ur

∂y1

2

dy1+bE Z L

0 |Ur|2dy1

r

X

i=1

0≤ymax1≤L

hi−1 (hi)2

−d∆thi

+

Z

D

hi |ui|2+ε|qi|2 dy

+C1

Z

D

∇ui

2dy 1/2

qiin

L2(0,1)+ qouti

L2(0,1)+ qwi

L2(0,L)

−1(1−λ)2κ Z L

0

ui2−d∆thi

2dy1

∆t , whereC1 depends only onD. Furthermore,

Z

D

hr

|ur|2+ε|qr|2 dy+E

Z L

0 |ur|2dy1 (5.2)

+

r

X

i=1

αν 2 +K2

Z

D

∇ui

2+ε ν ∇qi

2

dy+ 2cE Z L

0

∂ui

∂y1

2

dy1

∆t

r

X

i=1

Hn

Z

D

hi |ui|2+ε|qi|2

dy +C22(2 +K2) 2αν

qSi

2

+2(1−λ)2κ λ

d∆thi

2+C2(2 +K2)(1−λ)2κ α2λ

Z

D

hi ui

2dy

∆t

Abbildung

+4

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