anunbounded2Dchannel fluid-structure ...
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Existence and uniqueness result for a fluid-structure-interaction evolution problem in an unbounded 2D channel Clara Patriarca, Dipartimento di Matematica, Politecnico di Milano [email protected] Abstract In an unbounded 2D channel, we consider the vertical displacement of a rectangular obstacle, modelling the interaction between the cross-section of the deck of a suspension bridge and the wind. We prove well-posedness for a fluid-structure-interaction evolution problem set in this channel, where at infinity the velocity field of the fluid has a Poiseuille flow profile. We introduce a suitable definition of weak solutions and we make use of a penalty method. In order to prevent collisions of the obstacle with the boundary of the channel, we introduce a strong force f in the differential equation governing the motion of the rigid body and we find a unique global-in-time solution. Motivation Dynamic response: 1. one degree and two degrees of freedom instability 2. buffeting 3. vortex shedding Idea of the proof: existence Step 1 Solenoidal extension for the non- homogeneous condition at infinity. Step 2 Reformulation into an equivalent problem. Step 3 The penalty method. Step 1: Solenoidal extension Consider the problem div z (x)=ζ 0 (x 1 )(L 2 - x 2 2 ) in A ∩{2 < |x 1 | < 3}, z =0 on ∂A ∩{2 < |x 1 | < 3}. with ζ cut-off function. The function s(x)= λ ζ (x 1 )(L 2 - x 2 2 )ˆ e 1 - z (x) satisfies the following properties ∇· s =0 in Ω h , s = λ(L 2 - x 2 2 )ˆ e 1 in Ω h,i . Moreover s ∈ W 1,∞ (Ω h ) ∩ H 2 loc (Ω h ). Idea of the proof: uniqueness Any two solutions (v 1 ,h 1 ) and (v 2 ,h 2 ) are not defined on the same domain: the domain of the solution ˜ Ω(t) depends on the solution itself. Idea: Build ψ t : ˜ Ω 2 (t) → ˜ Ω 1 (t), ϕ t = ψ -1 t : ˜ Ω 1 (t) → ˜ Ω 2 (t) and define the pullback of v 2 by such map, w 2 . For any y =(y 1 ,y 2 ) ∈ ˜ Ω 1 (t): w 2 = ∇ψ t (y ) · v 2 (t, ϕ t (y )). Then, one can define ¯ v := v 1 - w 2 , ¯ h := h 1 - h 2 , and prove that ¯ v = ¯ h =0. References [1] C. Patriarca. Existence and uniqueness result for a fluid-structure-interaction evolution problem in an unbounded 2D channel, preprint. 2021. [2] D. Bonheure, G.P. Galdi and F.Gazzola. Equilibrium configuration of a rectangular obstacle immersed in a channel flow, Comptes Rendus Acad. Sci. Paris. 2020. [3] C. Conca, H.J. San Martin, and M. Tucsnak. Ex- istence of solutions for the equations modelling the motion of a rigid body in a viscous fluid. Communi- cations in Partial Differential Equations, 25, 2000. [4] O. Glass and F. Sueur. Uniqueness results for weak solutions of two-dimensional fluid-solid sys- tems. Arch. Rational Mech. Anal.,, 218, 2015. Model and main result We study the following two-dimensional fluid-structure-interaction evolution problem u t = μ Δu - (u ·∇)u -∇p, div u =0 in Ω h × (0,T ) u =0 on Γ= R × {-L, L}, u = h 0 ˆ e 2 on ∂B lim |x 1 |→∞ u(x 1 ,x 2 )= λ(L 2 - x 2 2 )ˆ e 1 (F ) h 00 + f (h) = - ˆ e 2 · Z S T (u, p) · ˆ n in (0,T ) (S ) B =[-d, d] × [-δ, δ ], B h = B + h ˆ e 2 ∀|h| <L - δ Ω h = R × (-L, L) \ B h = A \ B h . Main Theorem Assume that |h 0 | <L - δ and that u 0 satisfying u 0 (x)=¯ u 0 (x)+ ζ (x 1 ) λ(L 2 - x 2 2 )ˆ e 1 , with ¯ u 0 (x) ∈ L 2 (Ω h ), is such that u 0 · ˆ n| B h 0 = h 1 ˆ e 2 · ˆ n. Then, problem (F)-(S) admits a unique weak solution (u, h), defined in a suitable sense, for any T< ∞. Moreover the energy of (u, h) is bounded. Step 2: An equivalent problem We set y = x - h(t)ˆ e 2 , and we denote v (y,t)= u(y + h(t)ˆ e 2 ,t), q (y,t)= p(y + h(t)ˆ e 2 ,t), a(y )= a h (y )= s(y + h(t)ˆ e 2 ), ˜ Ω(t)=Ω h - h(t)ˆ e 2 , A h(t) = A - h(t)ˆ e 2 , ˜ Γ(t)=Γ - h(t)ˆ e 2 , B = B h - h(t)ˆ e 2 . (F)-(S) becomes (F’)-(S’) v t = μ Δv - (v ·∇) v -∇q +(h 0 ˆ e 2 ·∇) v div v =0 in ˜ Ω(t) × (0,T ) v =0 on ˜ Γ(t), v = h 0 ˆ e 2 on ∂B = S lim |y 1 |→∞ v (y ) := λ(L 2 - (y 2 + h(t)) 2 )ˆ e 1 (F 0 ) h 00 + f (h)= - ˆ e 2 · Z S T (v,q ) · ˆ n in (0,T ). (S 0 ) We look for solutions to the problem (F’)-(S’) of the form v =ˆ v + a. Step 3: The penalty method The crucial idea of the method implies introducing an auxiliary fixed, infinite domain ˜ A given by: ˜ A = A - A = {x - y | x ∈ A, y ∈ A}, such that ˜ Ω(t) ⊂ A h(t) ⊂ ˜ A and to build a penalized problem PP as follows. We set (F’)-(S’) on ˜ A and we add to the NS equations the term nχ E h v, where E h = ˜ A \ A h and n ≥ 1 is fixed. How do we exploit the penalized problem PP? 1. Prove existence of weak solutions to PP 2. Prove an energy estimate ⇒ nkv k L 2 (E h ) ≤ M< +∞ 3. Let n →∞
Transcript of anunbounded2Dchannel fluid-structure ...
Existenceanduniqueness result forafluid-structure-interactionevolutionproblemin
anunbounded2DchannelClara Patriarca, Dipartimento di Matematica, Politecnico di Milano
AbstractIn an unbounded 2D channel, we consider the vertical displacement of a rectangular obstacle, modelling the interaction between the cross-section of thedeck of a suspension bridge and the wind. We prove well-posedness for a fluid-structure-interaction evolution problem set in this channel, where at infinitythe velocity field of the fluid has a Poiseuille flow profile. We introduce a suitable definition of weak solutions and we make use of a penalty method.In order to prevent collisions of the obstacle with the boundary of the channel, we introduce a strong force f in the differential equation governing themotion of the rigid body and we find a unique global-in-time solution.
Motivation
Dynamic response:1. one degree and two degrees of freedominstability2. buffeting3. vortex shedding
Idea of the proof: existenceStep 1 Solenoidal extension for the non-homogeneous condition at infinity.Step 2 Reformulation into an equivalentproblem.Step 3 The penalty method.
Step 1: Solenoidal extensionConsider the problem
div z(x) =ζ′(x1)(L2 − x22) inA ∩ 2 < |x1| < 3,z = 0 on ∂A ∩ 2 < |x1| < 3.
with ζ cut-off function. The function
s(x) = λ
[ζ(x1)(L2 − x22)e1 − z(x)
]satisfies the following properties
∇ · s = 0 in Ωh, s = λ(L2 − x22) e1 in Ωh,i.
Moreover s ∈W 1,∞(Ωh) ∩H2loc(Ωh).
Idea of the proof: uniquenessAny two solutions (v1, h1) and (v2, h2) are notdefined on the same domain: the domain of thesolution Ω(t) depends on the solution itself.Idea: Build
ψt : Ω2(t)→ Ω1(t), ϕt = ψ−1t : Ω1(t)→ Ω2(t)
and define the pullback of v2 by such map, w2.For any y = (y1, y2) ∈ Ω1(t):
w2 = ∇ψt(y) · v2(t, ϕt(y)).
Then, one can definev := v1 − w2, h := h1 − h2,
and prove that v = h = 0.
References[1] C. Patriarca. Existence and uniqueness result for a
fluid-structure-interaction evolution problem in anunbounded 2D channel, preprint. 2021.
[2] D. Bonheure, G.P. Galdi and F.Gazzola. Equilibriumconfiguration of a rectangular obstacle immersed ina channel flow, Comptes Rendus Acad. Sci. Paris.2020.
[3] C. Conca, H.J. San Martin, and M. Tucsnak. Ex-istence of solutions for the equations modelling themotion of a rigid body in a viscous fluid. Communi-cations in Partial Differential Equations, 25, 2000.
[4] O. Glass and F. Sueur. Uniqueness results forweak solutions of two-dimensional fluid-solid sys-tems. Arch. Rational Mech. Anal.,, 218, 2015.
Model and main resultWe study the following two-dimensional fluid-structure-interaction evolution problem
ut = µ∆u− (u · ∇)u−∇p, divu = 0 in Ωh × (0, T )
u = 0 on Γ = R× −L,L, u = h′ e2 on ∂Blim|x1|→∞ u(x1, x2) = λ(L2 − x22) e1
(F )
h′′ + f(h) = −e2 ·∫ST (u, p) · n in (0, T ) (S)
B = [−d, d]× [−δ, δ], Bh = B + he2 ∀ |h| < L− δΩh = R× (−L,L) \Bh = A \Bh.
Main Theorem Assume that |h0| < L− δ and that u0 satisfying
u0(x) = u0(x) + ζ(x1)λ(L2 − x22)e1, with u0(x) ∈ L2(Ωh),
is such that u0 · n|Bh0= h1e2 · n. Then, problem (F)-(S) admits a unique weak solution (u, h), defined
in a suitable sense, for any T <∞. Moreover the energy of (u, h) is bounded.
Step 2: An equivalent problem
We set y = x− h(t) e2, and we denote
v(y, t) = u(y + h(t) e2, t), q(y, t) = p(y + h(t) e2, t), a(y) = ah(y) = s(y + h(t) e2),
Ω(t) = Ωh − h(t)e2, Ah(t) = A− h(t)e2, Γ(t) = Γ− h(t)e2, B = Bh − h(t)e2.
(F)-(S) becomes (F’)-(S’)vt = µ∆v − (v · ∇) v −∇q + (h′e2 · ∇) v div v = 0 in Ω(t)× (0, T )
v = 0 on Γ(t), v = h′e2 on ∂B = S
lim|y1|→∞ v(y) := λ(L2 − (y2 + h(t))2) e1
(F ′)
h′′ + f(h) = −e2 ·∫ST (v, q) · n in (0, T ). (S′)
We look for solutions to the problem (F’)-(S’) of the form
v = v + a.
Step 3: The penalty method
The crucial idea of the method implies introducing an auxiliary fixed, infinite domain A given by:A = A−A = x− y |x ∈ A, y ∈ A,
such that Ω(t) ⊂ Ah(t) ⊂ A and to build a penalized problem PP as follows. We set (F’)-(S’) on Aand we add to the NS equations the term
nχEhv,
where Eh = A \Ah and n ≥ 1 is fixed. How do we exploit the penalized problem PP?
1. Prove existence of weak solutions to PP
2. Prove an energy estimate ⇒ n‖v‖L2(Eh) ≤M < +∞
3. Let n→∞
