Note on evolutionary free piston problem for Stokes equations with slip boundary conditions Boris Muha∗

Zvonimir Tutek† Abstract

In this paper we study a free boundary uid-rigid body interaction problem, the free piston problem. We consider an evolutionary incompressible viscous uid ow through a junction of two pipes. Inside the "vertical" pipe there is a heavy piston which can freely slide along the pipe. On the lateral boundary of the "vertical" pipe we prescribe Navier's slip boundary conditions. We prove the existence of a local in time weak solution. Furthermore, we show that the obtained solution is a strong solution.

Keywords: solution

uid-rigid body interaction, moving boundary, Navier-Stokes equations, weak

AMS Subject Classication: 1

74F10, 35Q30, 76D05, 76D03

Introduction

The main goal of this paper is to prove the existence of a solution of the free piston problem for viscous uid ow. Unlike the classical piston problem where ow is generated by the known movement of the piston, in the free piston problem movement of the piston is unknown. Therefore, this is an example of a free boundary uid-structure interaction problem. We consider uid ow through the junction of two pipes in the gravity eld. Inside the "vertical" pipe there is a piston which can slide along that pipe. The piston is modeled as a rigid body and the uid is modeled as an incompressible Newtonian uid. Fluid-rigid body (and solids in general) interaction problems have been intensively studied from late 90s (see for example [3], [8], [6], [7], [4] and references within). However, in these papers the rigid body is fully immersed in the uid, so the rigid body is not a part of the boundary of uid-solid domain. Indeed, there are results that state that no contact will occur in nite time if the rigid body is smooth and no-slip boundary condition for the uid are prescribed (see [9] and references within). However Neustupa and Penel ([16]) showed that if no-slip boundary conditions are replaced with Navier's slip boundary conditions, ∗ †

Department of Mathematics, University of Houston, Houston, Texas 77204-3476, [email protected] Department

of

Mathematics,

University

of

Zagreb,

[email protected]

1

Bijeni£ka

cesta

30,

10000

Zagreb,

Croatia,

collision may occur. In our case the rigid body (the piston) is part of the boundary, so there is permanent contact between the rigid body and the rigid boundary. In analogy, we also prescribe Navier's slip boundary condition on the lateral boundary of the "vertical" pipe. The main dierence between the free piston problem and the already considered uid-rigid body interaction problems is permanent contact between the rigid body and the rigid boundary and use of Navier's slip boundary conditions. Therefore, dierent techniques for mathematical analysis are required. Evolutionary free piston problem for gas dynamics in 1D case has been studied by Takeno ([18]) and D'Acunto and Rionero ([5]). A stationary free piston problem for an incompressible viscous uid ow and its numerical analysis have been considered by authors ([13], [14]). The authors have also studied the evolutionary free piston problem for irrotational ideal uid ow and have analyzed the qualitative properties of solutions ([15]). This paper has four sections. In Section two we describe geometry of the problem and give a precise mathematical formulation of the problem. The existence theorem is proven using a xed point argument. Therefore, in Section three we analyze a linearized problem, i.e. a Stokes problem in a domain with a prescribed moving boundary with mixed boundary conditions coupled with ODE. Finally, in Section four we state and prove the existence theorem using the xed point argument. At the end we show that the obtained solution is a strong solution. 2 2.1

Formulation of the problem Notation and geometry of the problem

First we will precisely describe the geometry of the problem. We consider a system of two pipes of constant cross-section in the gravity eld. The system consists of an innite horizontal pipe F1 and a semi-innite "vertical" pipe F2 . The angle between the horizontal and the vertical pipe is denoted by α; by denition, α is the angle between direction opposite to gravity and the vertical pipe. We consider only the control volume of pipe F1 with two articial boundaries, Σin and Σout . Inside the "vertical" pipe we have a heavy piston. The upper and lower face of the piston are horizontal, while its lateral surface aligns to the pipe. It can only translate along the pipe. The piston is modeled as a rigid body, so its motion is given by Newton's second law. The uid is modeled as a Newtonian uid and enters the "vertical" pipe only up to a height of the lower face of the piston. Let us now introduce some notations and precise assumptions on the geometry of the control volume (see Figure 1). The coordinate frame is chosen in such a way that the lower end of the "vertical" pipe Σ0 is a subset of x3 = 0 plane. The coordinate x1 is along the horizontal pipe and x3 is in the opposite direction of the acceleration of gravity. Let h be the height of the piston in the selected coordinate frame. By Ωαh ⊂ Rn , n = 2, 3 we denote the domain occupied by the uid. More precisely, let S1 be the cross-section of F1 and let Ω1 = {(x1 , x2 , x3 ) ; −L ≤ x1 ≤ L, (x2 , x3 ) ∈ S1 }, S1 ⊂ R2 . 2

Next, let s = sin αe1 + cos αe3 be the direction of the "vertical" pipe F2 and Σ ⊂ R2 the lower face of the piston. Then Ωh,α is a set which in non-orthogonal coordinate frame 2 (e1 , e2 , s) has the following form:

Ωh,α = {(z1 , z2 , z3 ) ; 0 ≤ z3 ≤ h/ cos α, (z1 , z2 ) ∈ Σ}, 2 Now

Σ ⊂ R2 .

Ωαh = Ω1 ∪ Ω0 ∪ Ωh,α 2 .

Note that only the "vertical" pipe Ωh,α depends on h and α. The lower face of the 2 piston Σh is a subset of the x3 = const plane. Ω0 is an extension of the vertical pipe up to the boundary of Ω1 ; its shape is complicated in general in 3D case, in 2D case it is an empty set. Inow and outow regions are denoted by Σin and Σout respectively; they are articial boundaries of the uid domain Ωαh . Γh,α = ∂Ωαh \ Σin ∪ Σout ∪ Σh is a rigid boundary. The rigid boundary consists of two parts: Γ1 = Γh,α ∩ {x3 < 0} is the lateral boundary of the horizontal pipe and is independent of h and α and Γh,α = ∂Ωh,α 2 2 \ Σh is the lateral boundary of the "vertical" pipe. We suppose that angles between pipes are smoothened.

F2

Γ1

Γ2h Γ1

-L

F1 Σin

Γ1

Σout

Figure 1: Ωαh

In the sequel, let α ∈ (−π/2, π/2) be xed. Since proofs do not depend on choice of α, we will omit superscript α for simplicity of notation. We will consider only the 3D case; however, formulation of the problem in the 2D case is straightforward and all proven results are also valid in the 2D case with analogous proofs. Now we can give dierential formulation of considered problem:

3

nd (u, p, h) such that

∂t u − ν△u + ∇p = −gϱe3

in Ωh(t) ,

div u = 0 in Ωh(t) , u = 0 on Γ1 , h(t)

(T n) · τ + µf u · τ = 0 on Γ2 , h(t)

u · n = 0 on Γ2 , h′ (t) u= s on Σh(t) , cos α u × n = 0, p = Pin/out (t) − gρx3 on Σin/out , ( ) ∫ mh′′ (t) = cos α − P0 − Σh(t) Tn · s − µs h′ (t) ,

(1)

u(0, .) = u0 , h(0) = h0 , h′ (0) = h′0 . Here ν is kinematic viscosity of the uid, τ unit tangent and n unit outer normal. Boundary h(t) conditions (1)4,5 are Navier's slip boundary condition on Γ2 , where µf ≥ 0 is friction coecient between the uid and the rigid boundary and µs ≥ 0 is friction coecient between the piston and the rigid boundary. Since the piston is a rigid body and the uid is viscous, the velocity of the uid at the contact with the piston is equal to the velocity of the piston. Therefore, we have boundary condition (1)6 . Boundary conditions (1)7 are articial boundary conditions involving pressure at the articial boundary (see e.g. [2], [10]), where Pin (t) and Pout (t) are given. Notice that ow is driven by given pressure dierence Pin (t)−Pout (t). For brevity of notation we write Pin/out (t, x3 ) = Pin/out (t)−gρx3 , gρx3 being hydrostatic pressure. Equation (1)8 is the second Newton's law that describes motion of the piston. Since the piston can move only in direction s, its motion is one dimensional and we have a scalar equation. Let h1 (t) = h(t)/ cos α be the height of the piston in a non-orthogonal coordinate frame (e1 , e2 , s). Then∫ the acceleration of the piston in direction s is given by h′′1 (t) = h′′ (t)/ cos α. The term − Σh(t) Tn · s is the total uid force in direction s on the piston and P0 is the total outer force in direction −s on the piston, where T is the Cauchy stress tensor of the uid, T = −pI + µ(∇u + (∇u)τ ). The last term on the right-hand side h(t) of (1)8 comes from the friction between the piston and the lateral wall Γ2 . Friction acts in the opposite direction of the piston movement and the friction between the piston and h(t) the lateral wall Γ2 is proportional to the velocity of the piston. The system is completed with initial conditions (1)9 . 2.2

Statement of the main result

Before we state the main result of this paper we need to introduce some notation and function spaces. Let T , hmin > 0 and h : [0, T ] → [hmin , ∞). We dene a non-cylindrical space-time domain by ∪ Ωh(t) × {t}. QhT = 0≤t≤T

4

Next we dene the following function spaces:

V h (t) = {v ∈ H 1 (Ωh(t) ; R3 ); div v = 0, v = 0 on Γ1 , v × n = 0 on Σin/out , h(t)

v · n = 0 on Γ2 , ∃C ∈ R such that v = Cs on Σh }, t ∈ (0, T ), VTh = {v ∈ C ∞ (QhT ; R3 ); div v = 0, v = 0 on Γ1 , v × n = 0 on Σin/out , v · n = 0 on Γ2 , ∃C ∈ C ∞ [0, T ] such that v(t, .) = C(t)s on Σh , t ∈ [0, T ]}. h(t)

For each t ∈ (0, T ), V h (t) is a Banach space with the H 1 norm. As usual, if X(t), t ∈ [0, T ], is a family of Banach spaces, then we dene the following norms:

∫ ∥.∥2L2 (0,T ;X(t))

∥.∥L∞ (0,T ;X(t)) = esssupt∈(0,T ) ∥.∥X(t) ,

T

∥.∥2X(t) , t ∈ (0, T ).

= 0

Furthermore, we dene function spaces

Y h (0, T ) = (L2 (0, T ; V h (t)) ∩ L∞ (0, T ; L2 (Ωh(t) )) × W 1,∞ (0, T ), X h (0, T ) = {(v, g) ∈ Y h (0, T ) : v(t, .) = g(t)s on Σh(t) , t ∈ (0, T )}. Y h (0, T ) and X h (0, T ) are Banach spaces with the usual norms. Now we can dene a weak solution of our problem: (u, h) ∈ Y h (0, T ) is weak solution of problem (1) if u|Σh(t) = h′ (t)/ cos α, (∂t u, h) ∈ L2 (QhT ) × H 2 (0, T ), and the following equalities hold

∫ Qh T

∫

∫

T

∫

m ∂t u · v + 2ν sym∇u : sym∇v + µf u·v+ h(t) cos α Qh 0 Γ2 T ∫ T∫ ∫ T∫ ∫ ∫ T = Pout v1 − Pin v1 − gϱ v3 − P0 f, 0

Σout

0

Qh T

Σin

∫

T

′′

∫

h f + µs 0

T

h′ f

0

(v, f ) ∈ X h (0, T ), (2)

0

u(0, .) = u0 , h(0) = h0 , h′ (0) = h′0 . The main result of this paper is:

Theorem 1. Let α ∈ (−π/2, π/2) and ν , g, ϱ, µs , µf , h0 > 0. Furthermore, let h′0 ∈ R, Pin/out , P0 ∈ L∞ (0, T ) and u0 ∈ V h . Then there exists T ′ ≤ T such that problem (1) has at least one weak solution dened on (0, T ′ ). 0

5

3

Linearized problem

In this section we linearize problem (1) by prescribing the movement of the piston, i.e. we consider a problem with prescribed moving boundary. Analysis of the linearized problem will be later used in the proof of Theorem 1. We assume that movement of the piston is described by a known function f ∈ W 1,∞ (0, T ) such that:

0 < fmin ≤ f (t) ≤ fmax , t ∈ [0, T ],

′ ∥f ′ ∥L∞ (0,T ) ≤ fmax .

(3)

Before formulating the linearized problem we prove a few auxiliary results. We will need to compare solutions of certain problems posed on dierent domains. One way to do it is to transform equations from one domain to another. So we dene an appropriate change of variables. Let h1 , h2 > 0 and let g : [0, h2 / cos α] → [0, h1 / cos α] be a smooth strictly increasing function such that

g(0) = 0, g(h2 / cos α) = h1 / cos α, g ′ (0) = 1. Then we dene a change of variables β(.; h1 , h2 ) : Ωh2 → Ωh1 with { (x1 , x2 , x3 ) x3 < 0, ) ( β(x1 , x2 , x3 ; h1 , h2 ) = x3 x3 ) , x1 + g( cos α ) sin α − x3 tan α, x2 , cos αg( cos α

(4)

x3 > 0.

(5)

Since we are dealing with divergence-free vector elds, we dene a transformation that preserves this property. Let vh1 ∈ H 1 (Ωh1 ; R3 ). Then we dene its transformation vh1 (h2 ) by the formula:   h1 h1 x3 ′ x3 )v ( x ˜ , x ˜ , x ˜ ) + tan αv ( x ˜ , x ˜ , x ˜ )(1 − g ( )) g ′ ( cos 1 2 3 1 2 3 3 α 1 cos α   h1 h1 ′ x3  . (6)  v (h2 )(x1 , x2 , x3 ) =  g ( cos α )v2 (x˜1 , x˜2 , x˜3 )  h1 v3 (x˜1 , x˜2 , x˜3 ) Here (x˜1 , x˜2 , x˜3 ) = β(x1 , x2 , x3 ; h1 , h2 ). Now if div vh1 = 0, straightforward calculation gives:

x3 )(∂x˜1 v1h1 + ∂x˜2 v2h1 ) cos α x x3 3 + ∂x˜1 v3h1 tan α(g ′ ( ) − 1) + g ′ ( )∂x˜ v h1 cos α cos α 3 3 x3 = g′( ) div vh1 = 0. cos α

div vh1 (h2 ) = ∂x1 v1h1 (h2 ) + ∂x2 v2h1 (h2 ) + ∂x3 v3h1 (h2 ) = g ′ ( + tan α(1 − g ′ (

x3 ))∂x˜1 v3h1 cos α

Lemma 1. There exists a divergence free function φ0 ∈ VTf Ωf ) × [0, T ].

max

such that φ0 = s on (Ωfmax \

min

Proof.

f (t)

On (Ωfmax \ Ωfmin ) × [0, T ] we dene φ0 = s. Notice that s · n = 0 on Γ2 , t ∈ (0, T ). and Next, let θ ∈ C ∞ (R) be such that θ(x) = 1, x ≥ fmin and θ(x) = 0, x < fmin 2 6

dene Z(x1 , x2 , x3 ) = θ(x3 )s. Now we dene φ0 as a divergence-free function on Ωfmin with boundary condition φ0 (t, .) = Z(t, .) on ∂Ωhc \ Σin/out ; for proof of existence of such function see [17], Chap. 1, Tm. 3.1. Now we dene a weak solution of the linearized problem and prove the corresponding existence theorem.

Theorem 2. Let assumptions of Theorem 1. hold and let f Then there exists unique (uf , hf ) ∈ Y f (0, T ) such that

∈ W 1,∞

be such that (3) holds.

= (hf )′ (t)/ cos α, t ∈ (0, T ), (∂t u, hf ) ∈ L2 (QfT ) × H 2 (0, T )

uf|Σ

f (t)

and the following equalities hold           

∫

∫ ∂t u · v + 2ν

∫

f

∫

sym∇u : sym∇v + µf 0 ∫ T ∫ T m + (hf )′′ q + µs (hf )′ q cos α 0 0 ∫ ∫ T∫ ∫ T Pout v1 − Pin v1 − gϱ v3 − P0 q,

QfT

∫ T∫      =    0 Σout  

T

f

0

QfT

QfT

Σin

uf · v (7)

(v, q) ∈ X f (0, T ),

0

u(0, .) = u0 , h (0) = h0 , (h ) (0) = h′0 . f

f ′

f (t) Γ2

Furthermore, we have the estimate: ′ ∥(uf , hf )∥Y f (0,T ) +∥∂t uf ∥L2 (Qf ) +∥hf ∥H 2 (0,T ) ≤ C(Pin/out (t), P0 (t), ν, ϱg, u0 , h′0 , fmax ). (8) T

Proof.

We use the Galerkin method; our approach is similar to that in [1] and [12]. First we need to construct an appropriate Galerkin basis. Let V 1 denote the function space V h (t) for a constant function h(t) = 1; it is independent of t. Next, let V01 be its subspace dened by

V01 = {v ∈ H 1 (Ω1 ; R3 ); div v = 0, v = 0 on Γ1 ∪ Σ1 , v × n = 0 on Σin/out , v · n = 0 on Γ12 }. Notice that V01 is a subspace of V 1 of codimension 1. Now let (ψi )i∈N be a smooth Galerkin basis of V01 such that ∂3 ψi (x1 , x2 , 1) = 0 and let φ0 be the function constructed in Lemma 1. for function f . Then (φ0|Ωf , (ψi )i∈N ) is a basis for V 1 . Now we will use transformations min

(6) to dene a Galerkin basis of V f (t) for t ∈ (0, T ). We use the same notation as in (6) and dene

ϕi (t, .) = (ψi )(f (t)), i ∈ N and ϕ0 (t, .) = (φ0 )|Ωf (t) , t ∈ (0, T ). It is immediate that (ϕi (t, .))i∈N∪{0} is a basis of V f (t), t ∈ (0, T ). We proceed in a standard way, see for example [19], Chap III. Therefore, we will emphasize only the main steps. 7

For every m ∈ N we seek an approximate solution (um , hm ) in the form:

um (t, .) =

m ∑

αim (t)ϕi (t, .) +

i=1

(hm )′ (t) ϕ0 (t, .). cos α

Then (u , hm ) are dened as solutions of the following approximate problem: ∫ ∫ ∫ T∫ m m ∂t u · v + 2ν sym∇u : sym∇v + µf um · v f (t) f f QT QT 0 Γ2 ∫ T ∫ T m h′m q h′′m q + µs + cos α∫ 0 ∫ 0 ∫ T∫ ∫ ∫ T T = Pout v1 − Pin v1 − gϱ v3 − P0 q, m

0

where v(t, .) =

m ∑

Σout

0

Σin

QfT ′

(9)

0

u(0, .) = (u0 )m , hm (0) = h0 , (hm ) (0) = h′0 ,

βi (t)ϕi (t, .) + q(t)ϕ0 (t, .), β ∈ Cc∞ (0, T ) and (u0 )m is a projection of

i=0

initial uid velocity u0 on the space spanned with (ϕi (0, .))0≤i≤m . Note that (9) is a linear rst order system of ODE's for (um , h′m ) with the following mass matrix: ∫ ∫ ( ) m/cos2 α + Ωf (t) ϕ0 (t, .)ϕ0 (t, .) Ωf (t) ϕi (t, .)ϕ0 (t, .) ∫ , M (t) ∈ Mm (R), ϕ (t, .)ϕ0 (t, .) M (t) Ωf (t) i ∫ (M (t))ij = Ωf (t) ϕi (t, .)ϕj (t, .). M (t) is smooth, symmetric and positive denite for every t ∈ (0, T ) and therefore mass matrix is also smooth, symmetric and positive denite for every t. Now local in time existence of (um , hm ) on some time-interval (0, Tm ) follows from the existence theorem for linear system of ODE's. However, we will obtain estimates independent of m and thus we can take Tm = T , m ∈ N. Now we take um as a test function to obtain energy estimates. We use the following formula: ∫ ∫ 1d 1 2 ∥um ∥L2 (Ωf (t) ) = |um |2 f ′ (t) ∂t um · um − 2 dt 2 Σf (t) Ωf (t) to obtain the inequality ( ) 1 d m 2 ′ 2 ∥u ∥ + (h (t)) + ν∥sym∇um ∥2L2 (Ωf (t) ) + µf ∥um ∥2L2 (Γ2 ) + m L2 (Ωf (t) ) m 2 dt cos2 α ′ )(h′m (t))2 . ≤ C(Pin/out (t), P0 (t), ν, ϱg, u0 , h′0 ) + C(fmax

µs (h′m (t))2 cos α

(10) Now from Gronwall's and Korn's inequalities follows that sequence (um , hm )m∈N is bounded in Y f (0, T ). To get estimates for higher order derivatives we take ∂t um as a test function. Notice that this is an admissible test function because of our specic construction of the Galerkin basis. Furthermore, we use the following inequality: ∫ |Pin/out (t) v1 | ≤ C|Pin/out (t)|(∥v∥L2 (Ωh(t) ) + ∥ div v∥L2 (Ωh(t) ) ). Σin/out

8

Now we get ) 1 µs m d( ν∥sym∇um ∥2L2 (Ωf (t) ) + (h′m (t))2 +µf ∥um ∥2L2 (Γ2 ) + ∥∂t um ∥2L2 (Ωf (t) ) + 2 (h′′m (t))2 dt 2 cos α 2 cos α ′ ≤ C(Pin/out (t), P0 (t), ν, ϱg, u0 , h′0 ) + C(fmax )(h′m (t))2 .

To pass to the limit we rst transform equation (7) to the cylindrical domain Ω1 ×(0, T ). We can do that by using a change of variables (5). We omit the details for now because we will do it explicitly in the last section. Since f ∈ W 1,∞ satises (3), transformed solutions are bounded in the same norms as the original ones. Now we can pass to the limit in a standard way since system (9) is linear. At the end, we get solutions by transforming limits back to the original domain. The uniqueness follows directly from the energy estimate (8) and linearity of the considered problem. 4

Proof of Theorem 1.

We prove Theorem 1. by using the Schauder xed point theorem. For f ∈ W 1,∞ (0, T ) satisfying (3) we dene N (f ) = hf , where hf is a unique solution of the linearized problem given by Theorem 2. Suppose that h is a xed point of N . Then it is immediate that (uh , h) is a weak solution of (1). Therefore, it remains to prove that N has a xed point. Let

AR (0, T ) = {f ∈ W 1,∞ (0, T ) : f (0) = h0 , f (t) ≥ hm > 0, t ∈ [0, T ], ∥f ∥W 1,∞ (0,T ) ≤ R}, where h0 > hm are given. Set AR (0, T ) is a closed convex set in W 1,∞ (0, T ). From Theorem 2. we know that N (f ) is bounded in H 2 (0, T ) by some constant C(R) that depends only on R and the data. Because of compactness of the imbedding H 2 (0, T ) ,→ W 1,∞ (0, T ) we conclude that N (AR (0, T )) is relatively compact in W 1,∞ (0, T ). Now we prove continuity of N in W 1,∞ norm. Let fn be a convergent sequence in AR (0, T ), i.e. fn → f in W 1,∞ and let (ufn , hfn ), (uf , hf ) be corresponding solutions of the linearized problem. We can not use Theorem 2. for comparison of these solutions directly because the solutions are dened on dierent domains. Therefore, we again use transformations (6) to transform all equations to the reference domain QfT . Since the transformed solutions will be dened on the same domain QfT we can use Theorem 2. to compare them. Now, let us dene function

gn (x3 , t) = x3 + cos α

fn (t) − f (t) 2 x3 , f 2 (t)

x3 ∈ [0, f (t)/ cos α], t ∈ [0, T ], n ∈ N.

It is easy to verify that, provided that fn (t) is uniformly close to f (t), for every t ∈ [0, T ] function gn (., t) : [0, f (t)/ cos α] → [0, fn (t)/ cos α] satises conditions (4) and therefore we can dene divergence free vector eld u(fn )(t, .) = ufn (f )(t, .) as in (6). Furthermore, straightforward computation shows that

gn (x3 , t) → x3 , ∂t gn (x3 , t) → 0, ∂x3 gn (x3 , t) → 1, ∂x23 gn (x3 , t) → 0, ∂t ∂x3 gn (x3 , t) → 0 as n → ∞ 9

uniformly in x3 and t. Now we dene un (t, x1 , x2 , x3 ) = ufn (t, x˜1 , x˜2 , x˜3 ); note that ∇·un ̸= 0 in general. Next, denote     ∂x1 ∂x˜1  ∂x2 ∇n =  ∂x˜2  =  1 1 ∂ + tan α( ∂x g(x3 / cos α,t) − 1)∂x1 ∂x˜3 ∂x3 g(x3 / cos α,t) x3 3

∂t˜ = ∂t − (∂t g(x3 / cos α, t) sin α + tan α(

1 ∂t g(x3 / cos α, t) − 1))∂x1 − cos α ∂x . ∂x3 g(x3 / cos α, t) ∂x3 g(x3 / cos α, t) 3

Then by the chain rule we have

∂t ufn = ∂t˜un , ∇ufn = ∇n un . By using change of variables (x, y, z) 7→ (˜ x, y˜, z˜) in (7) we see that un satises the following variational equation: ∫ T ∫ ∫ ∫ T∫ m n n n n n ∂x3 g∂t˜u · v + 2ν (hfn )′′ q ∂x3 gsym∇ u · sym∇ v + µf ∂x3 gu · v + f f cos α 0 QT QT 0 Γ1 ∫ T ∫ T∫ ∫ T∫ ∫ ∫ T +µs (hfn )′ q = Pout v1 − Pin v1 − gϱ v3 − P0 q, (v, q) ∈ X f (0, T ), 0

0

Σout

0

QfT

Σin

0

where ∂x3 g is a Jacobian of the transformation (x, y, z) 7→ (˜ x, y˜, z˜). After a long but straightforward calculation we get that (u(fn ), hn ) satises linearized problem (7) on QfT with an additional forcing term R(un ): ∫ ( n ⟨R(u ), v⟩ = (∂x3 g − 1)∂t un3 v3 + ∂t ∂x3 g(un1 v1 + un2 v2 ) QfT

+ tan α(∂t un3 (1 − ∂x3 g) + un3 ∂t ∂x g)v1 + ∂x3 ((∂t g sin α + tan α(

1 ))∂x1 un · v ∂x 3 g − 1

( ∂t g n ∂x u · v) + 2ν (1 − ∂x3 g)sym∇un · sym∇n v − sym(∇n − ∇)un · sym∇n v + cos α ∂x 3 g 3 −sym∇un · sym(∇n − ∇)v − diag(1 − ∂x3 g, 1 − ∂x3 g, 0)∇un · sym∇v +∂x23 g(un1 (∂x3 v1 + ∂x1 v3 ) + un2 (∂x3 v2 + ∂x2 v3 )) + (∂x3 g − 1)sym∇un · sym∇v )) ∫ T ∫ 1 n (1 − ∂x3 g)un · v. + ∇(tan αu3 (1 − ∂x3 g)) · (∇v1 + ∂x1 v) + f (t) 2 0 Γ2 Using properties of gn and the fact that ∥∂t u(hn )∥2L2 (Qf ) +∥u(hn )∥2L∞ (0,T ;L2 (Ωf (t) )) +∥u(hn )∥2L2 (0,T ;V f (t)) T

is bounded, we get that ∥R(un )∥L2 (0,T ;(V f )′ ) → 0. Since problem (7) is linear, (8) and (10) imply that hfn → hf in W 1,∞ (0, T ). Thus continuity of N is proved. Now it only remains to prove N (AR (0, T )) ⊂ AR (0, T ). We have √ √ ∥N (f )′ ∥L∞ (0,T ) ≤ T ∥N (f )∥H 2 (0,T ) ≤ C(R) T .

Therefore, we complete the proof by taking T small enough. 10

Remark 1. Extension of the Theorem 1. to the Navier-Stokes equations would require dierent techniques. First, in the Navier-Stokes case one can not prove estimate (8) in a way analogous to the proof of Theorem 2.. Namely, when taking ∂t um as a test function, one can not get suitable estimates for the nonlinear term. Note that the estimate (8) is crucial for the compactness in the proof of Theorem 1.. Therefore, one would need to linearize the Navier-Stokes equations. However, it seems that additional regularization of the linearized problem is needed to be able to obtain suitable estimates for the xed point argument (similar as in [1]). We plan to address this issues in a future work. Corollary 1. Every solution of problem (1), solution and satises

(uh , ph , h),

given by theorem 1. is a strong

(uh , ph , h) ∈ L2 (0, T ′ ; W 2,8/7 (Ωh(t) )) × L2 (0, T ′ ; W 1,8/7 (Ωh(t) )) × H 2 (0, T ′ ).

Proof.

We have shown that ∂t uh ∈ L2 (QhT ) ∩ L2 (0, T : V h (t)). Therefore, for almost every t is h(t) uh h(t) ∈ H 1/2 (Γ2 ) and function uh satises the stationary Stokes equation in Ωh(t) with |Γ2

right-hand side in L2 (Ωh(t) ). Furthermore, mixed boundary conditions are given on the boundaries (type (i)-(iii) in notation from [11]). Since s is perpendicular to the normal of h(t) Γ2 , compatibility conditions are satised. Now assertion of proposition follows directly from [11] (section 5.5).

Acknowledgments

The research of B. Muha has been supported by the Texas Higher Education Board, Advanced Research Project, post-doctoral support under grant number 003652-0023-2009 and MZOS grant number 0037-0693014-2765. The research of Z. Tutek has been supported by MZOS grant number 0037-0693014-2765. References

[1] A. Chambolle, B. Desjardins, M. J. Esteban, and C. Grandmont. Existence of weak solutions for the unsteady interaction of a viscous uid with an elastic plate. J. Math. Fluid Mech., 7(3):368404, 2005. [2] C. Conca, F. Murat, and O. Pironneau. The Stokes and Navier-Stokes equations with boundary conditions involving the pressure. Japan. J. Math. (N.S.), 20(2):279318, 1994. [3] C. Conca, J. San Martín H., and M. Tucsnak. Motion of a rigid body in a viscous uid. C. R. Acad. Sci. Paris Sér. I Math., 328(6):473478, 1999. [4] P. Cumsille and T. Takahashi. Wellposedness for the system modelling the motion of a rigid body of arbitrary form in an incompressible viscous uid. Czechoslovak Math. J., 58(133)(4):961992, 2008. 11

[5] B. D'Acunto and S. Rionero. A note on the existence and uniqueness of solutions to a free piston problem. Rend. Accad. Sci. Fis. Mat. Napoli (4), 66:7584, 1999. [6] B. Desjardins and M. J. Esteban. On weak solutions for uid-rigid structure interaction: compressible and incompressible models. Comm. Partial Dierential Equations, 25(7-8):13991413, 2000. [7] G. P. Galdi. Mathematical problems in classical and non-Newtonian uid mechanics. In Hemodynamical ows, volume 37 of Oberwolfach Semin., pages 121273. Birkhäuser, Basel, 2008. [8] M. Hillairet and D. Serre. Chute stationnaire d'un solide dans un uide visqueux incompressible le long d'un plan incliné. Ann. Inst. H. Poincaré Anal. Non Linéaire, 20(5):779803, 2003. [9] M. Hillairet and T. Takahashi. Collisions in three-dimensional uid structure interaction problems. SIAM J. Math. Anal., 40(6):24512477, 2009. [10] E. Maru²i¢-Paloka. Rigorous justication of the Kirchho law for junction of thin pipes lled with viscous uid. Asymptot. Anal., 33(1):5166, 2003. [11] V. Maz'ya and J. Rossmann. Lp estimates of solutions to mixed boundary value problems for the Stokes system in polyhedral domains. Math. Nachr., 280(7):751 793, 2007. [12] T. Miyakawa and Y. Teramoto. Existence and periodicity of weak solutions of the Navier-Stokes equations in a time dependent domain. Hiroshima Math. J., 12(3):513 528, 1982. [13] B. Muha and Z. Tutek. On a free piston problem for Stokes and Navier-Stokes equations. To appear in Glasnik Matemati£ki. [14] B. Muha and Z. Tutek. Numerical analysis of a free piston problem. 15(2):573585, 2010.

Math. Commun.,

[15] B. Muha and Z. Tutek. On a stationary and evolutionary free piston problem for potential ideal uid ow. To appear in Math. Meth. Appl. Sci., DOI: 10.1002/mma.2555 [16] J. Neustupa and P. Penel. A weak solvability of the Navier-Stokes equation with Navier's boundary condition around a ball striking the wall. In Advances in mathematical uid mechanics, pages 385407. Springer, Berlin, 2010. [17] V. G. Osmolovski. Linear and nonlinear perturbations of the operator div, volume 160 of Translations of Mathematical Monographs. American Mathematical Society, Providence, RI, 1997. Translated from the 1995 Russian original by Tamara Rozhkovskaya. [18] S. Takeno. Free piston problem for isentropic gas dynamics. Math., 12(2):163194, 1995. 12

Japan J. Indust. Appl.

[19] R. Temam. Navier-Stokes equations. Theory and numerical analysis. North-Holland Publishing Co., Amsterdam, 1977. Studies in Mathematics and its Applications, Vol. 2.

13