Weak Solutions to a Fluid-Structure Interaction Model of a Viscous Fluid with an Elastic Plate under Coulomb Friction Coupling

We consider the behavior of a viscous fluid within a container that has an elastic upper, free boundary. The movement of the upper boundary is described by a combination of a plate equation and a boundary condition of friction type that quantifies the elasticity of the boundary. We show the local existence of weak solutions to this coupled system in three dimensions, by applying the Galerkin method to a regularized version of the problem and using a fixed-point argument afterwards.


Introduction
Let ω denote a bounded Lipschitz domain in R 2 . We consider a domain, Ω h 0 = {(x, y, z) ∈ R 3 | (x, y) ∈ ω, 0 < z < 1 + h 0 (x, y)}, filled with a viscous fluid, where h 0 is a given vertical displacement of the elastic, upper part of the boundary from the reference state ω × {1}. We denote the rigid part of the boundary of Ω h 0 by Γ 0 := (ω × {0}) ∪ (∂ω × [0, 1]), which consists of the lower part of the boundary as well as of the lateral part. Only Γ h 0 = ∂Ω h 0 \ Γ 0 , the upper part of ∂Ω h 0 , is subject to change in time which is described by a function h that is to be determined. So, while Ω h 0 denotes the domain occupied by the fluid at the initial time t = 0, Ω h(t) = {(x, y, z) ∈ R 3 | (x, y) ∈ ω, 0 < z < 1 + h(t, x, y)} is the domain occupied by the fluid at time t. The equations that describe the interplay between the fluid flow and the elastic boundary, are expressed by (u, p) and h, respectively. They are defined on the non-cylindrical open space-time domain and satisfy the coupled non-linear system u(t, x, y, 1 + h(t, x, y)) = (0, 0, ∂ t h(t, x, y)) on (0, T) × ω, (h, ∂ t h) |t=0 = (h 0 , h 1 ) in ω, (8) with Du := 1 2 (∇u + ∇u ) and n ω ∈ R 2 denoting the outer unit normal on ∂ω. On ω × (0, T) we have for given λ, κ > 0 that for α = α(t, x, y) ≥ 0 which is equivalent to Here, α = |∂ t h| κ in (9) 2 and α = 0 in case (9) 1 is an unknown function of the system; furthermore, µ > 0 denotes the viscosity of the fluid. Equations (2) and (3) are the incompressible Navier-Stokes equations with a given external force f , where (3) and (5) imply that the flow is volume preserving; see Remark 1 (ii). Moreover, (4), describes that the fluid does not move on the rigid part Γ 0 . Since the fluid adheres to the plate, we postulate the kinematic boundary condition (5). Moreover, T f = T f (t, x, y) is the surface force exerted by the fluid on the structure given by ω T f (t, x, y) · v(t, x, y, 1 + h(t, x, y)) dx dy = Γ h(t) (−2µD(u) · n t + pn t ) · v dσ (11) for all v ∈ C ∞ c (Ω h(t) ) 3 , where n t = (−∇ h, 1) 1 + |∇ h| 2 is the t-dependent outer normal at Γ h(t) .
Instead of the boundary condition (9), often an equation of the form is used in the literature. For instance, in Reference [1], Chambolle et al. have constructed weak solutions for the corresponding problem; the authors remark that the damping term ∆ 2 ∂ t h can be replaced by the term ∆∂ 2 t h. Similar results have been shown by Grandmont [2] in the case of a three-dimensional cavity, where one part of the boundary is assumed to be elastic and the other one rigid. However, the focus in Reference [2] is on the behavior of solutions in the limit λ → 0 and a uniform positive lower bound of the time of existence.
In Reference [3], Denk and Saal exploit the damping term −∆∂ t h in a half space like domain with base ω = R n−1 and maximal regularity to get strong solutions; similar techniques are used by Celik and Kyed [4] in a spatially periodic setting (n = 3) to prove the existence of strong t-periodic solutions. Beiraõ da Veiga [5] proved the local existence of strong solutions in a one-dimensional periodic base ω = (0, L). Similar results are obtained by Badra and Takahashi [6] working with semigroups of Gevrey class. Fluid structure interaction problems with a classical non-linear von Kármán shallow shell of finite thickness that allow for both transversal and lateral displacements are considered by Chueshov and Ryzhkova [7]. Lengeler and Růžička [8] discussed the case of a Koiter shell instead of the flat plate and, hence, replaced the operator ∆ 2 by an operator better suited for non-flat boundaries.
These equations model the movement of the plate via a (damped) plate equation, where the damping term used here is λ∆ 2 ∂ t h and the forces causing the movement are given by (T f ) 3 and a prescribed external force g. Our boundary condition is a combination of this equation with a boundary condition of friction type, namely (for h = 0) Here, the subscripts n and τ denote the normal and tangential component of a vector, respectively. Condition (12) describes that the fluid adheres to the boundary, but if the normal component of the stress tensor becomes large enough, the fluid may start to "leak" and presses the upper elastic boundary in the opposite direction of (T f ) n . This boundary condition allowing for leaking and, in particular, its counterpart on slip where tangential components are controlled as in (12) (Coulomb's boundary or dry friction condition) have been studied extensively. For instance, the existence of strong solutions with respect to time for the Navier-Stokes equations coupled with this boundary condition, i.e., u ∈ W 1 2 (0, T; L 2 (Ω)), has been shown by Kashiwabara [9] for finite time intervals. On the other hand, a thorough analysis of the stationary Stokes case in bounded smooth domains of R d , d ≥ 3, including H 2 -results and the Stokes resolvent problem, is due to Saito [10]. Besides results on weak solutions, Bȃlilescu, San Martín, and Takahashi [11] present several numerical examples for the slip Coulomb boundary condition, whereas Jing et al. [12] propose a discontinuous Galerkin method for the non-linear leak boundary condition. In [13], Bȃlilescu et al. consider the translational and rotational motion of a rigid body immersed into a viscous fluid together with Coulomb's boundary condition. Consiglieri [14] discusses the stationary case of a non-Newtonian fluid with Coulomb's condition, whereas Baranovskii [15] puts the focus on a viscoelastic fluids of Oldroyd-B type.
In combination, the boundary condition (9) describes that, if the force g : 3 is smaller than the threshold κ, then h, and, therefore, the domain does not change in time. However, if the force becomes large enough, the upper boundary may start to change in time, and it does so in the opposite direction of the force. So, κ can be viewed as a constant measuring the elasticity of the plate, which is the upper boundary. If the material that the plate consists of is very elastic, κ is very small. If κ is very large, the plate is harder to move. For simplicity, we assume in the following that λ = 1.
In this work, we show the existence of weak solutions. First, we will formally derive an energy estimate to identify suitable solution spaces in Section 2. In Section 3, we rigorously define the solution spaces and gather results about trace operators associated with the domain Ω h(t) . In Section 4, we define weak solutions before we show their existence in Sections 5 and 6. Indeed, in Section 5, we describe several smoothing arguments and perform the Galerkin approximation method on a given domain Ω δ(t) , yielding a weak solution (u, h) which ignores the kinematic boundary condition (5) on the upper part of the boundary. Finally, in Section 6, a fixed point argument yields a solution u satisfying h = δ, i.e., the kinematic boundary condition and also (9).
We want to emphasize that we follow the work of Chambolle, Desjardins, Esteban, and Grandmont; see Reference [1]; however, we have an additional non-linear boundary term. In a first weak formulation, as in (19) or (20) in Section 4, we get a variational inequality in which both the solution and the test function are variables in the non-linear function | · |. By a mollification, the modulus is replaced by a smooth function G ε , as in (28) below, which is non-linear in the solution, but linear in the test function. These terms require from time to time significant changes from the mentioned paper. In order to keep this work self-contained and comprehensible, we still give a detailed proof of the existence of weak solution that is in some points even more detailed than that in [1].

Energy Estimate
Formally multiplying (2) with u and integrating over Ω h leads to Note that we will omit the infinitesimal increments, such as dx dy in R 2 , dσ for 2d-surfaces, and dτ for time to keep lengthy formulae short. Then, the Reynolds transport theorem, as in [16] (Theorem 3.5), implies that with v denoting the velocity of the area element which equals u. Hence, using div u = 0 and the divergence theorem, we deduce that Plugging this into (13), we get that Now, due to the definition of T f given by (11), we calculate for the boundary term We rewrite the last integral as Applying (10) to the second last integral, we get We also obtain the reverse inequality if we consider y = −∂ t h in (10). Hence, we conclude Plugging these identities into (14), we get the equality Integrating over time yields for all t > 0 Using Young's and Hölder's inequality, we deduce for all t > 0 Then, Grönwall's lemma yields for all t > 0. Hence, it is natural to assume for the given data that f ∈ L 2 (0, T; Due to the fact that we do not know how large Ω h(t) is, we defined f on R 3 , which is not restrictive given that the extension by zero preserves L 2 -integrability. Via (17) these properties, in turn, imply that For a rigorous definition of these spaces, see the next section. Due to u(t, ·) = 0 on Γ 0 , it is tempting to use Korn's inequality in order to get u ∈ L 2 (0, T; H 1 (Ω h(t) )). However, here, we encounter the main problem in solving Equations (2)-(9) in a weak sense. While h ∈ H 1 (0, T; H 2 0 (ω)) ⊂ C([0, T] × ω) is continuous, h does not need to be Lipschitz continuous. Hence, Ω h(t) is not necessarily a domain with Lipschitz boundary. Nevertheless, due to the continuity of h, Ω h(t) is open and uniformly bounded. Therefore, we can find an M ∈ N such that B M := ω × (0, M) ⊃ Ω h(t) for all t ∈ [0, T]. Now, we define Due to the interface condition (5) and the previous properties, we see that D(u) ∈ L 2 ((0, T) × B M ). Ergo, we can apply Korn's inequality in B M to get u ∈ L 2 (0, T; H 1 (B M )), and, by restriction, we still conclude that u ∈ L 2 (0, T; H 1 (Ω h(t) )).

Solution Spaces and Trace Operators
Let T > 0 and 0 < m < M. Moreover, let δ ∈ C([0, T] × ω) be a function such that Then, the set Even though Ω δ(t) might not necessarily be Lipschitz, one can still show that as in [17] (Theorem 3.22, p. 68), by using that Ω δ(t) is locally the graph of a function.
The following lemmata treat the definition and properties of the trace on Γ δ(t) . For these lemmata, always assume that δ satisfies (18).
Moreover, we have the existence of a lifting operator: The next lemma gives a meaning to the normal trace on the boundary: For every t ∈ (0, T), there exists a linear continuous operator The following lemma justifies Korn's inequality.
We also have Poincaré's inequality due to the boundedness of B M .

Weak Formulation
After the definition of solution and trace spaces and their basic properties, we are now in the position to define weak solutions. Let u 0 ∈ L 2 (Ω h 0 ), h 1 ∈ L 2 (ω) and h 0 ∈ H 2 0 (ω) ⊂ C 0 (ω).
holds for almost all t ∈ (0, T). (19) is explained by a formal application of Reynolds transport theorem: To explain the other terms, we formally calculate Applying (11) to the last term, we estimate Using the identity we can also write Later, it will become clear why we use (20) rather than (19).

Existence of Weak Solutions: The Galerkin Approximation
This section is dedicated to the proof of the main theorem.

Remark 1. (i)
The first assumption on h 0 , i.e., (21) 1 , is natural since we assume that there is no intersection of the free boundary with the rigid boundary at the start. In fact, the reason why the existence of a global solution for all times will not be shown is due to such a possible self-intersection of the boundary. The other conditions are compatibility conditions. The last one is due to (21) 2 and (21) 4 .
(ii) The conditions (21) 4,5 and the solenoidality of u formally imply that the flow is volume preserving. Actually, Proof.
For 0 < δ < 1, we define the scaling It is easy to check that u [1+δ] 0 is also divergence free. Due to h 0 ∈ C(ω), the set is open and a neighborhood of Γ h 0 . In particular, we have that u The latter can be assumed due to u Since the unknown domain Ω h(t) depends on the solution h, we first replace it with a regularised domain Ω δ ε (t) and use a fixed point argument in Step 6 below to find δ ε = h ε .
For any auxiliary Note that we use the same parameter ε introduced in the beginning of Step 1 for h ε 0 , h ε 1 , for δ ε , the sequence h ε , as well as for N ε ; moreover, δ depends on ε via h ε 0 . The operator N ε can be constructed as follows: Consider for each ε > 0 a regularization by considering ε > 0 small enough. For further technical details, we refer to Reference [18]. The properties (24) of N ε will be crucial in Step 6.3 below in the analysis of the fixed point operator F ε , as in (52). Furthermore, in order to regularize the non-linearity, we also introduce the space-time being an approximate identity with compact support in (0, T) × B 2M , and E 0 denotes the extension by zero on R 4 .
Step 2: The approximate, almost linearized problem Step 1, we consider the following, almost linearized, approximate problem: Here, we linearized the term However, G ε (∂ t h ε ) is still non-linear in the unknown h ε . In addition, we wrote Ω δ ε (s) ∇u ε : ∇φ ε as 1 2 Ω δ ε (s) Du ε : Dφ ε which is possible due to Lemma 6. Furthermore, comparing (28) with (20), we reversed the integration by parts with respect to time. For h ε , we used which explains the new term (28). For u ε , we used the Reynolds transport theorem to get This in combination with the replacement of the term − 1 (28). Note that we are allowed to take the solution (u ε , ∂ t h ε ) as test function (φ, b) in (28). This yields Here, we used that which is true since the upper boundary of Ω δ ε (t) moves at the velocity (0, 0, ∂ t δ ε ) and u ε (t, x, y, 1 + δ ε (t, x, y)) = (0, 0, ∂ t h ε (t, x, y)) on ω, see iii) above. Similar to the considerations in Section 2, we get with C > 0 depending only on the data and T but not on ε > 0, M or m. Due to ω being bounded, we also deduce by Poincaré's inequality, as in Lemma 7, that u ε is bounded in L 2 (0, T; H 1 (Ω δ ε (t) )) by a constant independent of ε > 0. Now, we transform the domain Ω δ ε (t) to the reference configuration, the cylinder Note that χ ε (t) is a smooth diffeomorphism with det ∇χ ε (t) = 1 + δ ε (t) for all t ∈ (0, T). For the sake of abbreviation, we set ρ ε (t) := ρ • χ ε (t) for a function ρ in the following.

Now, we consider the 2n equations
for j = 1, . . . , n, combined with the initial conditions here, u n ε,0 and h n ε,1 denote the orthogonal projections of u 0 ε and h ε 1 onto the finite dimensional spaces span(φ 0,ε i , φ 1,ε j ) i,j∈{1,...,n} and span(ξ j ) j∈{1,...,n} , respectively. Due to the smoothness of the involved functions with respect to time, we get the existence of a solution (u n ε , h n ε ) of the form on [0, T]. In order to prove the unique existence of the solution, we first plug (37) into (35) and (36) to have a system that we want to solve for the unknowns α := (α 1 , ..., α n ) and β := (β 1 , ..., β n ). This, in turn, can be done by reducing it to a system of first order with respect to time. This first order system for (β, α,β) (without initial conditions) reads Here, F ε,t denotes all lower-order terms that do not involveα orβ. Since the equation is linear, so is F ε,t ∈ R 2n×3n . Furthermore, F ε,t is smooth with respect to time. Obviously, M ε (t) + N ∈ R 2n×2n is given by where i, j ∈ {1, ..., n}. Obviously, M ε (t) is smooth with respect to time, too, and it is easy to check that M ε (t) + N is positive definite and, hence, invertible. Therefore, we deduce the existence of a unique solution of the form as in (37) that satisfies the mentioned initial conditions. Now, we multiply equation (35) with α i and sum over i ∈ {1, ..., n}. By analogy, we multiply (36) withβ j and sum over j ∈ {1, ..., n}. Adding those two terms yields Since ∂ t (1 + δ ε ) = div(0, 0, z ∂ t δ ε ) , we get, with the exterior normal vector n Z of ω × (0, 1), that Therefore, we deduce Plugging this into (38), we get Similarly as done in Section 2, along with G ε (∂ t h n ε )∂ t h n ε ≥ 0 and using that A ε is elliptic with a constant independent of n, we can integrate in time and apply Grönwall's inequality to conclude that u n ε L ∞ (0,T;L 2 (Z)) + ∇u n ε L 2 (0,T;L 2 (Z)) + ∂ t h n ε L ∞ (0,T;L 2 (ω)) + ∆h n ε H 1 (0,T;L 2 (ω)) Here, the constant C > 0 depends on ε, m and M. However, if we return to the problem in the deformed configuration, we obtain-similarly as done in Section 2-that u n ε L ∞ (0,T;L 2 (Ω δ ε (t) ) + ∇u n ε L 2 (0,T;L 2 (Ω δ ε (t) ) + ∂ t h n ε L ∞ (0,T;L 2 (ω)) with C > 0 independent of ε, m, and M.
Step 4: Uniform estimates of time derivatives in n In order to obtain a solution for n → ∞, we still need additional estimates of the largest time derivatives. To be more specific, we need the estimate ∂ t u n ε L 2 (0,T;L 2 (Z)) + ∂ tt h n ε L 2 (0,T;L 2 (ω)) ≤C(ε), withC(ε) > 0 independent of n but possibly dependent on the given data, m, M, and ε.
In order to achieve this estimate, we multiply (35) withα i and sum over i ∈ {1, ..., n}. In a similar fashion, we multiply (36) withβ j and sum over j ∈ {1, ..., n}. For simplicity, we omit the indices ε and n in the following and obtain, after adding the two sums, Hence, we get that Then, an integration in time implies that Now, we have to estimate every term on the right-hand side of (41) by a constant independent of n, but possibly depending on ε, m, M, and on δ and v. The first three pose no problem since they depend on the initial data that are smooth. For the fourth term, Hölder's and Young's inequality imply that Let us estimate the second summand of the right-hand side of (42). Remembering that Concerning the second term, we recall that φ 1 j solves the Stokes-like problem (33), and, hence, ∑ n j=1β in the weak sense. Due to the duality estimate Now, using the equation solved by (φ 1 j , p j ) and their continuous dependence on the given data, as in (34), we can estimate further: Combining the last two estimates, we have The term (∂ t B − B in (43) is uniformly bounded on (0, T) × Z due to the smoothness and boundedness of δ . The bound may depend on ε though. Hence, we get that where we use the estimate (44) for the term involvingβ and (39) for the term involving u. Therefore, we get from (42) that is bounded independently of n since u is bounded uniformly in L 2 (0, t; H 1 (Z)), as in (39), and ∂ t A is bounded in L ∞ ((0, t) × ω × (0, 1)) due to δ being independent of n. Concerning the term we use again (43) and deduce that is bounded due to (46) and (45). For the next term in (41) involving and the boundedness of u in L 2 (0, t; H 1 (Z)), to get that .
For the second integral involving v , integration by parts implies that The first integral on the right-hand side can be estimated as , while, for the second one, we can write For the next term in (41), we easily have .
Concerning integrals involving h, we start with the estimate The term t 0 ∆∂ t h 2 L 2 (ω) is bounded in view of (39). For the next term, we have, due to H 1 (0, T; L 2 (ω)) → L ∞ (0, T; L 2 (ω)) and (39), that Moreover, due to the boundedness of ω, Finally, the estimate holds. So, in total, we conclude with the second term bounded in L 2 (0, T; L 2 (Z)) because of (43), (45) and (46), and √ 1 + δ ≥ m 2 , we have shown-in the original notation, that is to say, with indices n and ε-that Step 5: Convergence of u n ε , h n to a weak solution The bounds from (39) and (48) imply that there is a subsequence of (u n ε , h n ε ) n∈N , which we denote by (u n ε , h n ε ) n∈N again such that as n → ∞ for some The last convergence in (49) is implied by the second to last (up to a subsequence), which, in turn, is implied by (49) 3,4 and the Aubin-Lions lemma. Using the above convergences, we let n tend to infinity in (35) and (36) after an integration in time. For the convergence for all j ∈ N, we use that |G ε (∂ t h n ε )| ≤ 1, the convergence of (∂ t h n ε ) n∈N to ∂ t h ε pointwise almost everywhere on (0, T) × ω and the dominated convergence theorem.
For the initial values, we have by design for n → ∞ Now, in order to combine the equations (35) and (36), and to solve the weak formulation (32) on Z as a whole, we take a relevant test function (φ ε , b) ∈ L 2 (0, T; H 1 0,Γ 0 (Z)) × L 2 (0, T; H 2 0 (ω)), as in (32). Let P N (b) = ∑ N j=1b j (t)ξ j denote the projection of b onto span(ξ j ) j∈{1,...,N} . As before, φ 1,ε j denotes the solution of the stationary Stokes-like system (33), and φ 1,ε b denotes the weak solution to the similar system but with boundary value (0, 0, b) instead of (0, 0, ξ j ) . Due to the linearity of this system and the continuous dependence of solutions on the data, we get This enables us to multiply (36) withb j (t), integrate in time, take the limit n to infinity for the solution, as justified above, take the sum over j ∈ {1, ..., N}, and then consider the limit N → ∞ for the test function. This allows us to basically replace ξ j with b and φ 1,ε j with φ 1,ε b in (36). Then, we consider the Dirichlet part φ 0 := φ ε − φ 1,ε b which satisfies φ 0 = 0 on ∂Z and div(B ε (φ 0 )) = 0. For this reason, we can consider the projection of ..,M} . Analogously to before, we multiply (35) withα i (t), integrate over time, take the limit n → ∞ for the solution, take the sum over i ∈ {1, ..., M}, and then take the limit M → ∞ for the test function.
Next using that φ ε = φ 0 + φ 1,ε b , we can add the two resulting equations to get that (u ε , h ε ) solves (28). Since the diffeomorphism χ ε is smooth, we obtain a weak solution for the problem (28), formulated on Ω δ ε , by setting (u ε , h ε ) = (u ε • χ −1 ε , h ε ). Furthermore, weak solutions of (28) are unique. For the proof, take two solutions (u 1 ε , h 1 ε ), (u 2 ε , h 2 ε ) of (28) to the same given data and initial values. In the equation that is solved by the difference (u 1 ε − u 2 ε , h 1 ε − h 2 ε ), we test with the difference itself. After integration by parts, we obtain Since G ε is the derivative of a convex function, it is monotonically increasing. Therefore, every term on the left-hand side is non-negative, and we conclude that

Existence of Weak Solutions: The Fixed Point Argument
Step 6: The fixed point problem of F ε Given initial values u ε 0 , h ε 0 , and h ε 1 , for any pair (δ, v) with δ ∈ H 1 (0, T; C 0 (ω) ∩ H 1 0 (ω)), M ≥ 1 + δ ≥ m > 0 on [0, T] × ω and v ∈ L 2 (0, T; L 2 (B 2M )), we have shown in Section 5 the existence of a weak unique solution (u ε , h ε ) to the "linearized" problem (28) with v ε = R (v) and δ ε = N (δ) where δ |t=0 = h ε 0 . This solution satisfies the estimates, cf. the a priori estimates (31) and (40) with a constant C > 0 independent of ε > 0 and ∂ t u ε L 2 (0,T;L 2 (Z)) + ∂ 2 t h ε L 2 (0,T;L 2 (ω)) ≤ C(ε), as in (48), the corresponding weak convergences in (49) and the weak lower semi-continuity of the norm. For T * ∈ (0, T], let Then, we consider the solution operator where and (u ε , h ε ) denotes the solution constructed in the previous steps for the data (δ, v) and with It is easy to check that B m,ε M is non-empty, closed and convex. If we can show that F ε (B m,ε M ) ⊂ B m,ε M , that F ε (B m,ε M ) is relatively compact and that F ε is continuous, we get the existence of at least one fixed point by Schauder's fixed-point theorem, as in Reference [20] (Corollary 10.2). Note that m > 0 was already chosen such that min w (1 + h 0 ) ≥ 2m > 0, but M and T * still have to be chosen large and small enough, respectively, such that F ε satisfies the mentioned properties, if C M is chosen large enough, as well.
Step 7: The limit for ε → 0 In order to pass to the limit as ε → 0, we state the following: Claim 1. For all τ > 0 small enough, there holds the estimates and Actually, since the proof of this claim is almost word for word the same as the one in [1], Lemma 9, we only treat here the new, additional term The key idea in the proof is to test the equation solved by (u ε , h ε ) with the special test function (φ ε , b) depending on the solution (h ε , u ε ) as follows: for some δ > 0; see (23) for the definition of u [1+δ] . After inserting this test function, the resulting terms are estimated by bounds of the form C √ τ with C independent of ε. Here, the only new term is t−τ ∂ t h ε (s, x, y) ds , which we can estimate as follows where the uniform boundedness of ∂ t h ε L 2 (0,T * ;L 2 (ω)) is used; see (50).
From these considerations, as well as from the estimates in (50) where the constant does not depend on ε, we find functions h on [0, T * ] × ω and u on [0, T * ] × B 2M such that the following convergences (of a suitable subsequence which we identify with the sequence itself) follow as ε goes to 0: The last three convergences follow from the way we regularized h ε and u ε , as in (24) and (27), and from the corresponding convergences without regularization. Furthermore, (χ Ω h ε (t) ∇u ε ) ε>0 converges (up to a subsequence) weakly to a matrix-valued function q ∈ (L 2 (0, T; L 2 (B 2M ))) 3×3 due to (50) as ε → 0.
This shows that the interface condition is satisfied. Now, we take the limit in the equation solved by (u ε , h ε ). Recall that (u ε , h ε ) solves This equation is derived from (28) by replacing (v, δ) with the fixed point (u ε , h ε ) found in Section 6. In particular, δ ε = N ε (δ) becomes N ε (h ε ) = h ε and v ε = R ε (v) becomes R ε (u ε ) = u ε , but note that u ε and h ε will still differ from u ε and h ε , respectively. Since G ε is the derivative of the convex function j ε : R → R, x → √ x 2 + ε 2 , we get which is equivalent to If we plug this into (60) and use integration by parts with respect to time for the first term and for the term involving ∂ 2 t h ε , as in (30) and (29) (with δ replaced by h), we end up with the inequality for a.e. t ∈ (0, T * ) and for all (φ ε , b) ∈ V h ε × C 1 ([0, T * ]; H 2 0 (ω)) with φ ε (t, x, y, 1 + h ε ) = (0, 0, b) on [0, T] × ω.

Discussion
We presented the existence of weak solutions to a fluid-structure interaction problem for Navier-Stokes equations in a domain with a damped Kirchhoff plate as an upper lid. The interaction is given only in the normal direction, together with a threshold for the normal stress, so that, for small stresses, the plate stays immovable. The proof follows the line of Reference [1], where the damped plate is linearly coupled with the fluid. There appear the natural questions whether similar results can be obtained for unbounded domains of half space type and for bounded domains where the plate is fastened to the arbitrarily shaped upper brim of the lateral boundary of a cylindrical domain or even cannot be written as a graph.
Author Contributions: This article is part of the PhD thesis of A.S. [18]. Writing, review and editing were carried out by both authors in equal parts. All authors have read and agreed to the published version of the manuscript.

Conflicts of Interest:
The authors declare no conflict of interest.