Asymmetric equilibrium configurations of a body immersed in a 2d laminar flow

We study the equilibrium configurations of a possibly asymmetric fluid–structure interaction problem. The fluid is confined in a bounded planar channel and is governed by the stationary Navier–Stokes equations with laminar inflow and outflow. A body is immersed in the channel and is subject to both the lift force from the fluid and to some external elastic force. Asymmetry, which is motivated by natural models, and the possibly non-vanishing velocity of the fluid on the boundary of the channel require the introduction of suitable assumptions to prevent collisions of the body with the boundary. With these assumptions at hand, we prove that for sufficiently small inflow/outflow there exists a unique equilibrium configuration. Only if the inflow, the outflow and the body are all symmetric, the configuration is also symmetric. A model application is also discussed.


Introduction
Let L > H > 0 and consider the rectangle R = (−L, L) × (−H, H).Let B ⊂ R be a closed smooth domain having barycenter at the origin (x 1 , x 2 ) = (0, 0) and such that diam(B) ≪ L, H.We study the behavior of a stationary laminar (horizontal) fluid flow going through R and filling the domain Ω h = R \ B h , where B h = B + he 2 for some h (a vertical translation of B), see Figure 1.Note that B 0 = B. Here, µ > 0 is the kinematic viscosity, u is the velocity vector field, p is the scalar pressure.
The body B is subject to two vertical forces.The first force (the lift) is due to the fluid flow and tends to move B away from its original position B 0 , it is expressed through a boundary integral over ∂B, see (3.1) below.The second force is mechanical (elastic) and acts as a restoring force tending to maintain B in B 0 .When there is no inflow/ouflow, the body is only subject to the restoring force and remains in B 0 which is the unique equilibrium position.But, as soon as there is a fluid flow, these two forces start competing and one may wonder if the body remains in B 0 or, at least, if the equilibrium position remains unique.We show that, if the inflow/ouflow is sufficiently small, then the equilibrium position of B remains unique and coincides with B h for some h close to zero.We point out that, contrary to [3,8,10], we make no symmetry assumptions neither on B nor on the laminar inflow/outflow.Therefore, not only the overall configuration will be asymmetric but also some of the techniques developed in these papers do not work and B h may be different from B 0 .The motivation for studying asymmetric configurations comes from nature.Only very few bodies are perfectly symmetric and most fluid flows, although laminar in the horizontal direction, are asymmetric in the vertical direction: think of an horizontal wind depending on the altitude or the water flow in a river depending on the distance from the banks.Figure 2 shows two front waves in sandstorms that have no vertical symmetry although the wind is (almost) horizontally laminar.In Section 2 we give a detailed description of our model and we prove that, for small Reynolds numbers, the Navier-Stokes equations are uniquely solvable in any Ω h , see Theorem 2.2.The related a priori bounds depend on h, and this is one crucial difference compared to the (symmetric) Poiseuille inflow/outflow considered in [3].It is well-known [5] that to solve inhomogeneous Dirichlet problems for the Navier-Stokes equations, one needs to find a solenoidal extension of the boundary data and to transform the original problem in an homogeneous Dirichlet problem with an additional source term.For the existence issue, one can use the classical Hopf extension, but there are infinitely many other possible choices for the solenoidal extension.One of them, introduced in [12], was used in [3] to write the lift force as a volume integral by means of the solution of an auxiliary Stokes problem.For asymmetric flows, the same solenoidal extension does not allow to estimate all the boundary terms and, in order to obtain refined bounds for the solution to the Navier-Stokes equations in Ω h , we build a new explicit solenoidal extension that also plays a fundamental role in the analysis of the subsequent fluid-structure-interaction (FSI) problem.
The main physical interest in FSI problems is to determine the ω-limit of the associated evolution equations because this allows to forecast the long-time behavior of the structure.Since the evolution Navier-Stokes equations are dissipative, one is led to investigate if the global attractor exists, see [7,15]: the main difficulty is that the corresponding phase space is time-dependent and semigroup theory does not apply.
The global attractor contains stationary solutions of the evolution FSI problem that we call equilibrium configurations, which are investigated in the present work.
In Section 3 we introduce the lift force and the restoring force and we set up the steady-state FSI problem.Our main result, namely Theorem 3.1, states that, for small Reynolds numbers, the equilibrium position is unique and may differ from B 0 .By exploiting the strength of the restoring force, uniqueness for the FSI problem is obtained without assuming uniqueness for (1.1).To prove this result, we need some bounds on the lift force in proximity of collisions of B h with ∂R: these bounds are collected in Theorem 3.2 and proved in Section 4 by using the very same solenoidal extension introduced in Section 2. The remaining part of the proof of Theorem 3.1 is divided in two steps.In Subsection 5.1 we prove some properties of the global force exerted on the body B. These properties are then used in Subsection 5.2 to complete the proof by means of an implicit function argument, combined with some delicate bounds involving derivatives of moving boundary integrals.We emphasize that for our FSI problem we cannot use the explicit expression of the lift derivative as in [17] because the displacements B h within R do not follow the normal of ∂B h , in particular if ∂B h contains some vertical segments.Instead, based on the general approach introduced in [2] (see also the previous work [14]), we compute with high precision the lift variation with respect to the vertical displacement parameter h of B h by acting directly on the strong form of the FSI problem.Section 6 contains the symmetric version of Theorem 3.1, see Theorem 6.1 which states that, under symmetry assumptions on the inflow/outflow and on B, for small Reynolds numbers the equilibrium position is unique and coincides with B 0 .This extends former results in [3,8,10] to a wider class of symmetric frameworks.
As an application of our results, in Section 7 we consider a model where B h represents the cross-section of the deck of a suspension bridge [6], while Ω h is filled by the air and represents either a virtual box around the deck or a wind tunnel around a scaled model of the bridge.Since the deck may have a nonsmooth boundary, we also explain how to extend our results to the case where B is merely Lipschitz.

Fluid boundary-value problem
Let R and B be as in Section 1 (Figure 1) with On the one hand, ( Since we consider vertical displacements B h within R, we have h The bottom and top parts of ∂R are respectively while its lateral left and right parts are, respectively, For some λ ≥ 0, we consider the boundary-value problem Note that u | ∂R ∈ C 0 (∂R) and (2.3)-(2.4)are compatible with the Divergence Theorem.The role of λ ≥ 0 in the boundary conditions is to measure with a unique parameter the strength of both the inflow and the outflow.Hence, λ ≍ Re where Re is the Reynolds number.
We now state an apparently classical existence and uniqueness result which, however, has some novelties.First, since the domain Ω h is only Lipschitzian, the regularity of the solution is obtained through a geometric reflection.More important, the explicit upper bound for the blow-up of the H 1 -norm of the unique solution to (2.4) in proximity of collision: when B approaches Γ t the norm remains bounded while when B approaches Γ b we estimate its blow-up.This refined bound requires the construction of a suitable solenoidal extension of the boundary data.Note that, up to normalization, we can reduce to the cases where In order to state the result, we define the distances of the body B h to Γ b and Γ t respectively by Throughout the paper, any (positive) constant depending only on µ, B 0 , L, H will be denoted by C and, when it depends also on h, by C h .We may now state Theorem 2.2.Let h ∈ (−H + δ b , H − δ t ) and assume (2.3) with (2.5).Then (2.4) admits a strong solution (u, p) for any λ ≥ 0 and there exists Λ = Λ(h) > 0 such that the solution is unique if λ ∈ [0, Λ(h)); if U = 0, Λ(h) can be chosen independent of h, i.e.Λ(h) ≡ Λ > 0.Moreover, there exist C > 0 and C h > 0 such that the unique solution (when λ < Λ(h)) satisfies A priori bounds such as (2.7) and (2.8) are available for any λ ≥ 0 and any strong solution of (2.4), but with different powers of λ.
Before giving the proof, let us explain qualitatively the main differences between the cases U = 0 and U = 1.For U = 0, the a priori bound (2.7) is independent of h, so that the graph of Λ(h) looks like Figure 3 (left).For U = 1, (2.7) depends on h and Λ(h) itself may depend on h, see Figure 3   Proof.Existence of weak solutions.For later use, we first define weak solution for the forced Navier-Stokes equations (2.9) which reduces to (2.4) when f = 0. We say that u ∈ H 1 (Ω h ) is a weak solution to (2.9) with f ∈ L 2 (Ω h ) if u is a solenoidal vector field satisfying the boundary conditions in the trace sense and (2.10) For any weak solution u, there exists a unique associated p ∈ L 2 0 (Ω h ) (i.e. with zero mean value), satisfying (2.11) [5]).In (2.24) below, we introduce an ad-hoc solenoidal extension matching our geometric framework which is not optimal for our current purpose.This is why we use here the well-known Hopf's extension s that reduces the effect of the nonlinearity and allows to prove existence for any λ ≥ 0. Hence, we recast (2.4) as (2.9) with homogeneous boundary conditions, namely (2.12) Then there exists v ∈ W (Ω h ) satisfying (2.10) for any λ ≥ 0 (Theorem IX.4.1, [5]).This is equivalent to say that the vector field u = v + s ∈ H 1 (Ω h ) and the associated pressure (2.13) In these bounds and the ones below we only emphasize the smallest and largest powers of λ, as for any polynomial.These bounds are not part of the statement but they will be used later in the present proof.
Regularity.We claim that any weak solution (u, p) to (2.4) satisfies (u, p) ∈ H 2 (Ω h ) × H 1 (Ω h ).This would be straightforward if Ω h ∈ W 2,∞ , see [14], but R is only Lipschitzian.Here, we take advantage of the particular shape of R and use a reflection argument as in [9].We construct a new domain Therefore, the couple which satisfies the corresponding of (2.15) in Ω b h .Thanks to these two vertical reflections, we obtain a solution in Ω s With the same principle, we then perform two horizontal reflections of Ω s h with respect to x 1 = ±L.At the end of this procedure, let Ω h = (−3L, 3L) × (−3H, 3H) \ {B h and its eight reflections} and ( u, p) : Ω h → R 2 × R be the extension of (u, p), so that (2.16) and ũ satisfies further boundary conditions that we do not need to make explicit.After introducing a suitable solenoidal extension, we can proceed as in the first part of the proof and obtain the existence of a solution ( u, p) (2.17) By applying [14] and [5, Theorems IV.4.1 and IV.5.1] to the Stokes problem (2.16), we infer that ( u, p) This also proves (2.8) whenever λ < Λ(h).
Uniqueness.Let u 1 and u 2 be two weak solutions to (2.4), let w = u 1 − u 2 , then Then take φ = w so that the latter yields where we used Hölder, Ladyzhenskaya and Poincaré inequalities and (2.13).Hence, there exists Λ = Λ(h) > 0 (uniformly upper-bounded with respect to h) such that Refined bounds.For λ ∈ [0, Λ(h)), in all the above bounds we can drop the largest power of λ and they all become linear upper bounds.We treat separately the cases U = 1 and U = 0 and we make explicit the dependence of the constant C h in (2.13) on h.When U = 1, we claim that the unique strong solution u to (2.4) satisfies with C > 0 independent of h.To this end, we introduce a different (and explicit) solenoidal extension.Consider the cut-off functions 1, defined piece-wise in the rectangles of Figure 4 by 2 , H], where ζ 0 l is a function only of x 1 , and where ζ 0 r is a function only of x 1 .Then, letting ∇ ⊥ = (−∂ 2 , ∂ 1 ), consider the vector field s : R → R 2 defined by which is solenoidal and satisfies the boundary conditions in (2.4).Rewriting s as its partial derivatives read We need to quantify the dependence of C h > 0 on ε b (h) and ε t (h).On the one hand, we notice that, by construction, both ζ l and ζ r depend on x 2 only in In this domain the x 1 -derivatives of ζ l and ζ r are uniformly bounded with respect to h while the x 2 -derivatives blow-up as ε t (h) goes to zero, for instance we have On the other hand, the cut-off functions depend only on x 1 in Ω h \ Ω εt(h) and their x 1 and x 2 -derivatives are uniformly bounded with respect to h.Therefore, in Gathering all together, we refine the bounds in (2.25) as with all the constants C > 0 independent of h.Then, testing (2.12) with v = u − s we obtain (2.28) We want to estimate, when possible, only s and not ∇s since the bounds for s are less singular in terms of ε t (h).Hence, since ∇ • v = ∇ • s = 0 and using integration by parts, we rewrite (2.28) as We split the first integral in the right-hand side over Ω εt(h) and Ω h \ Ω εt(h) .On the one hand, since v and Hölder inequality yield where we used that ∥s∥ On the other hand, since v | Γ l ,Γr = 0, Poincaré and Hölder inequalities yield where we used that ∥s∥ L ∞ (Ω h \Ω ε t (h) ) ≤ Cλ.Therefore, from (2.27) and (2.29) we infer the cut-off functions depend only on x 1 , we infer that s, ∂ 1 s and ∂ 2 s are uniformly bounded with respect to h in all Ω h and (2.32) Repeating the same computations as in the case U = 1 and using (2.32), we obtain (2.31) for h ≤ 0. If h > 0, we make a vertical reflection x 2 → −x 2 and we consider the new cut-off functions defined piece-wise in the rectangles of Figure 5, where ε b (0) = H − δ b .Then, we consider the vector field s : R → R 2 defined by which is solenoidal and satisfies the boundary conditions in (2.4).By the same argument used when h ≤ 0, s, ∂ 1 s and ∂ 2 s are uniformly bounded with respect to h in Ω h .Therefore, using again (2.32), we obtain (2.31) for h < 0. □ Remark 2.3.We stated (2.7) and (2.8)only in case of uniqueness because, in what follows, λ will be taken small and higher powers of λ can be upper estimated with the first power.The reflection method used to obtain the regularity result has its own interest.The rectangular shape of the domain is crucial and the technique fails for other polygons.However, in the case of convex polygons, in particular also for a rectangle, one can obtain the more C ∞ -regularity result by using Theorem 2 in [13], see also [11,Section 7.3.3]and [4].

Equilibrium configurations of a FSI problem
By Theorem 2.2, for any (λ, h) ∈ [0, +∞) × (−H + δ b , H − δ t ) there exists at least a strong solution (u, p) = (u(λ, h), p(λ, h)) to (2.4).The fluid described by (u, p) in Ω h exerts on B h a force perpendicular to the direction of the inflow, called lift (see [16]).Since the inflow in (2.4) is horizontal, the lift is vertical and given by where T is the fluid stress tensor, namely and n is the unit outward normal vector to ∂Ω h , which, on ∂B h , points towards the interior of B h .In fact, L(λ, h) is a multi-valued function when uniqueness for (2.4) fails.However, we keep this simple notation instead of writing L(λ, h, u(λ, h), p(λ, h)), in which also the dependence on the particular solution (u, p) is emphasized.The regularity of the solution (see Theorem 2.2) and the smoothness of ), hence the integral in (3.1) is finite.In fact, the lift can also be defined for merely weak solutions, see (7.5) in Section 7. Note that (3.1) holds for any λ ≥ 0 and any solution to (2.4) but our main result on the FSI problem focuses on small inflows, see Theorem 3.1.Aiming to model, in particular, a wind flow hitting a suspension bridge, the body B may also be subject to a (possibly nonsmooth) vertical restoring force f tending to maintain B in the equilibrium position B 0 (for h = 0); see Section 7. We assume that f depends only on the position h, that f ∈ C 0 (−H + δ b , H − δ t ) with f (0) = 0 and Moreover, we assume that there exists K > 0 such that The assumption (3.3) is somehow technical and prevents collisions of B with the horizontal boundary Γ b ∪ Γ t , at least for small inflow/outflow.It can probably be relaxed but, so far, only few (numerical) investigations on the effect of proximity to collisions of hydrodynamic forces (such as the lift), acting on non-spherical bodies, have been tackled, see [20] Our main result concerns the existence and uniqueness of the solution to (3.4) for small values of λ, that we expect to be stable.
3) with (2.5).There exists Λ 1 > 0 and a unique h ∈ C 0 [0, Λ 1 ) such that for λ ∈ [0, Λ 1 ) the FSI problem (3.4) admits a unique solution We emphasize that Theorem 3.1 ensures uniqueness of the equilibrium configuration for the FSI problem (3.4) in the uniform interval [0, Λ 1 ) even in absence of uniqueness for (2.4) that, instead, is only ensured in the possibly non-uniform interval [0, Λ(h)).The proof of Theorem 3.1 is given in Section 5.It is fairly delicate because if U = 0 (as for symmetric inflow/outflow), then from (2.21) we infer that the H 1 -norm is uniformly bounded with respect to h.However, if U = 1, the same norm obviously blows up when B h approaches Γ t , which affects the bounds for the lift in (3.1).As already mentioned, very little is known when a body approaches a collision, see again [20] and references therein.Therefore, the next statement has its own independent interest, it provides some upper bounds and shows that, probably, the lift behaves differently for homogeneous and inhomogeneous boundary data.
The proof of Theorem 3.2 is given in the next section.

Proof of Theorem 3.2
We rewrite the lift (3.1), which is a boundary integral, as a volume integral.This can be done by considering w ∈ H 1 (Ω h ) that satisfies (4.1) The Divergence Theorem ensures that (4.1) admits infinitely many solutions.Testing (2.4) with one such solution w (recall that ∇ and, using the boundary conditions on w, Among the infinitely many solutions of (4.1), we select one obtained by using a solenoidal extension similar to the ones introduced in Section 2. We consider a cut-off C ∞ -completion elsewhere.
We put w = ∇ ⊥ (x 1 χ).Clearly w ∈ H 1 (Ω h ) satisfies (4.1) and supp . Moreover, from the definition of χ it follows that w and its x 1 and x 2 -derivatives are uniformly bounded with respect to h in Ω w,c , while in Ω w,b and in Ω w,t (4.4) B h close to Γ b .We consider the case when h is close to −H + δ b , hence ε b (h) is close to zero.This implies that ε t (h) ≥ 1 and the bounds in (4.4) become uniform.
Choosing in (4.2) the previously constructed w, we observe that the integrals in the right-hand side are defined only on Ω w .Let us split these integrals over the regions Ω w,b , which is shrinking as ε b (h) goes to zero, and Ω w \ Ω w,b .On the one hand, Hölder inequality and (2.7) yield for ε b (h) close to zero, due to (4.3).Putting together (4.5)-(4.7),then there exists η b > 0 sufficiently small such that, for any (λ, h) We remark that the same blow-up rate in (4.8) could be obtained without taking advantage of Poincaré inequality in (4.6) but using directly u ∈ H 1 ⊂ L 4 .This idea, however, will be crucial to obtain a better blow-up rate for the lift in the case when the body is close to Γ t , that we now analyze.B h close to Γ t .We consider the case when h is close to H − δ t , hence ε t (h) is close to zero.Analogously to what done in the previous case, we split the integrals over the regions Ω w,t , which is shrinking as ε t (h) goes to zero, and Ω w \ Ω w,t .On the one hand, Hölder inequality yields ˆΩw\Ωw,t using that w and its derivatives are uniformly bounded with respect to h in Ω w \ Ω w,t .On the other hand, since Putting together (4.9)-(4.12),then there exists η t > 0 sufficiently small such that, for (λ, h) , ε b (h) and ε t (h) are uniformly bounded from below with respect to h.Therefore, by combining (4.8) and (4.13), there exists C > 0 independent of h such that, for any (λ, h) We first focus on the λ-dependence by maintaining h fixed and we prove the Lipschitzcontinuity of the map λ → ϕ(λ, h).
Proof.We prove the result in two steps, namely by analyzing the behavior of ϕ in two different subregions of [0, λ 0 ) × (−H + δ b , H − δ t ).We start by considering the case when |h| is close to 0. Let again h = H − max{δ b , δ t }.

Symmetric configuration
We consider here a symmetric framework for (3.4), that is, when and the boundary data are symmetric with respect to the line x 2 = 0. Therefore, the FSI problem (3.4) is modified on Γ b and reads (6.1) In this symmetric framework, δ b = δ t = δ and h ∈ (−H + δ, H − δ).Then, we prove that the unique curve h(λ) found in Theorem 3.1 reduces to h(λ) ≡ 0, namely that the unique equilibrium position is symmetric.Again, we expect this position to be stable, at least for small λ.
given by u 0 (λ, 0), p 0 (λ, 0), 0 , where (u 0 (λ, 0), p 0 (λ, 0)) is the unique solution to the first two lines in (6.1) for h = 0 and has the following symmetries: ). Proof.The first step is to obtain the counterpart of Theorem 2.2.The case U = 0 is already included in the original statement.When U = 1, we construct the cut-off functions ζ l and ζ r in a slightly different way with Figure 4 replaced by Figure 6.We define the solenoidal extension as in (2.24), which satisfies the boundary conditions in (6.1).With this construction the refined bound (2.7) is replaced by Hence, in both cases U ∈ {0, 1}, by arguing as in the proof of Theorem 2.2, we infer that there exists Λ = Λ(h) > 0 such that for λ ∈ [0, Λ(h)) the solution (u, p) to (6.3) is unique for any h ∈ (−H + δ, H − δ).This proves the counterpart of Theorem 2.2.
Then the sustaining cables are installed between the two couples of towers and the hangers are hooked to the cables.Once all these components are in position, they furnish a stable working base from which the deck can be raised from floating barges.We refer to [18,Section 15.23] for full details.The deck segments are put in position one aside the other (see Figure 7, left) and have the shape of rectangles while their cross-section resembles to smoothened irregular hexagons (see Figure 7, right) that satisfy (2.1).This cross-section B plays the role of the obstacle in (2.4) while Ω h is the region filled by the air.This region can be either be a virtual box around the deck of the bridge or a wind tunnel around a scaled model of the bridge.In both cases, we may refer to inflow and outflow also as windward and leeward respectively: λV in e 1 represents the laminar horizontal windward while λV out e 1 is the leeward.Typically, the higher is the altitude the stronger is the wind.Therefore, in this application we consider specific laminar shear flows, which are the Couette flows.Thus, the inflow and outflow now read (7.1) and satisfy (2.3).The windward creates both vertical and torsional displacements of the deck.However, the cross-section of the suspension bridge is also subject to some elastic restoring forces tending to maintain the deck in its original position B 0 .These forces are of three different kinds.There is an upwards restoring force due to the elastic action of both the hangers and the sustaining cables of the bridge.
The hangers behave as nonlinear springs which may slacken [1, 9-VI] so that they have no downwards action and they be nonsmooth.There is the weight of the deck which acts constantly downwards: this is why there is no odd requirement on the restoring force considered in the model.There is also a nonlinear resistance to both elastic bending and stretching of the whole deck for which B merely represents a cross-section.Moreover, since the boundary of the channel R is virtual and our physical model breaks down in case of collision of B with ∂R, we require that there exists an "unbounded force" preventing collisions.
Overall, the position of B depends on both the displacement parameter h and the angle of rotation θ with respect to the horizontal axis.With the addition of this second degree of freedom, we have B = B h,θ and Ω = Ω h,θ .A "plastic" regime leading to the collapse of the bridge is reached when θ = ± π 4 (see [1]) since the sustaining cables of the bridge attain their maximum elastic tension.The strong point of the analysis carried out in this paper is that it applies independently of the part of ∂B closest to ∂R.Therefore, for any θ ∈ (− π 4 , π 4 ), we can apply our general theory considering the family of bodies B h,θ simply by adapting it to the rotating scenario.The only difference now is that, when the body is free to rotate, the collision with Γ b and Γ t occurs at h = −H + δ b (θ) and h = H − δ t (θ), where δ b (θ) and δ t (θ) are positive functions of θ.For θ = 0, δ b (0) and δ t (0) are as in (2.2) while, for θ ̸ = 0, δ b (θ) := − min (x 1 ,x 2 )∈∂B 0,θ x 2 > 0, δ t (θ) := max (x 1 ,x 2 )∈∂B 0,θ x 2 > 0, both being independent of h.Due to the possible complicated shape of B, these functions are not easy to be determined explicitly.For this reason, we define the set of non-contact values of (h, θ) by In fact, the second line in (7.4) is not mathematically needed but, from a physical point of view, it states that the restoring force does not act at equilibrium and tends to maintain B in an horizontal position.A straightforward consequence of Theorem 3.1, in the case of the interaction between the wind and the deck of a suspension bridge, is the following:

Figure 1 .
Figure 1.The rectangle R and the body B with its vertical displacements B h .

Figure 2 .
Figure 2. Front wave of two wind storms.
and references therein.The presence of U in (3.3) highlights the different behavior of f when B is close to Γ t for U = 0 or U = 1.In the first case, f has the same strength close to Γ b and Γ t .Conversely, for U = 1, the asymmetry of the boundary conditions requires a different strength of f , which is stronger when B is close to Γ t than when B is close to Γ b .Overall, (3.2)-(3.3)model the fact that B is not allowed to go too far away from the equilibrium position B 0 .Since we are interested in the equilibrium configurations of the FSI problem, we consider the boundary-value problem (2.4) coupled with a compatibility condition stating that the restoring force balances the lift force, namely(3.4) for λ ∈ [0, Λ 0 ], using that w and its derivatives are uniformly bounded with respect to h in Ω w \Ω w,b .On the other hand, sincew ≡ 0 in Ω 0 w,b := [−2τ, 2τ ]×[−H, h−δ b − ε b (h) 2 ] and u | Γ b = 0, Poincaré inequality for u in Ω w,b ∪ Ω 0 w,b, the Hölder inequality and (2.7) yield(4.6)

Figure 6 .
Figure 6.The cut-off functions ζ l (left) and ζr (right) on R when U = 1 for the symmetric configuration.

Figure 7 .
Figure 7. Left: erection of a suspension bridge.Right: sketch of a cross-section.
.22)-(2.23) (see Figure5below) replacing ε t (h) with the distance of B 0 to Γ t , namely ε t (0) = H − δ t .The solenoidal extension s is then defined as in(2.24).By construction both ζ l and ζ r depend on x 2 only in Ω εt(0) , defined as in (2.26) with ε t (h) replaced by ε t (0).In this domain both x 1 and x 2 -derivatives of ζ l and ζ r are uniformly bounded with respect to h, for instance we have (2) ≤ Cλ with C > 0 independent of h, which will imply that Λ(h) ≡ Λ can be also taken independent of h.In this case, we shall define the cut-off functions and the solenoidal extension differently depending if h ≤ 0 or h > 0. If h ≤ 0, we define ζ l , ζ r as in(2