Stability and convergence analysis of the kinematically coupled scheme for the fluid-structure interaction

Transcription

1 Stability and convergence analysis of the kinematically coupled scheme for the fluid-structure interaction Boris Muha Department of Mathematics, Faculty of Science, University of Zagreb 2018 Modeling, Simulation and Optimization of the Cardiovascular System Magdeburg, October 2018 Joined work with M. Bukač (Notre Dame) and S. Čanić (UC Berkeley)

2 Talk summary 1. Fluid-structure interaction. 2. Kinematically coupled scheme. 3. Applications.

3 Motivation Fluid-structure interaction (FSI) problems describe the dynamics of a multiphysics system involving fluid and solid components. They are everyday phenomena in nature and arise in various applications ranging from biomedicine to engineering. Examples: blood flow in vessels, artificial heart valves, vocal cords, valveless pumping, airway closure in lungs, geophysics (underground flows, hydraulic fracturing), classical industrial applications (aeroelasticity, offshore structures), artificial micro-swimmers in body liquids, micro-(and nano-)electro-mechanical systems (MEMS), various sports equipment

4 Motivation II Main motivation for our work comes from biofluidic applications. Main example in this talk will be blood flow through compliant vessel. Densities of the structure and the fluid are comparable (unlike in e.g. aeroelasticity) - highly nonlinear coupling.

5 Formulation of an example of FSI problem 3D fluid is coupled with 3D nonlinear elasticity. Ω F - the fluid reference domain. Ω S - the structure reference domain. Ω F - the fluid-structure interface.

6 The structure equations Mathematically, FSI systems are described in terms of continuum mechanics, which gives rise to a system of partial differential equations (PDEs). More precisely, a non-linear system of partial differential equations of mixed parabolic-hyperbolic type with a moving boundary, i.e. part of the domain is also an unknown of the system. Unknowns: u - the fluid velocity and η - the structure displacement. The elastodynamics equations: ϱ s 2 t 2 η = T( η) in (0, T ) Ω S, Constitutive relation: First Piola-Kirchhoff stress tensor: T(F ) = F W (F ), where W : M 3 (R) R is a stored energy function.

7 The fluid equations The physical fluid domain is given by Ω f (t) = ϕ f (t, Ω f ), t (0, T ), where φ f the fluid domain displacement (i.e. an arbitrary extension of η to Ω F ). The fluid equations: ρ f ( t u + u u) = σ( u, p), u = 0, } in Ω f (t) = ϕ f (t, Ω f ), t (0, T ). Constitutive relation: Cauchy stress tensor is given by relation σ( u, p) = pi + 2µD(u)

8 Coupling and boundary condition Dynamic coupling condition (balance of forces): Tn = (σ ϕ) ϕ τ n on (0, T ) Γ Kinematic coupling condition (no-slip): t η(t,.) = u(t,.) Γ ϕ f, on Γ, t (0, T ). In some physical situations different kinematic boundary condition might be more appropriate (slip, Signorini type BC). Boundary data: dynamic pressure/stress free or periodic.

9 An FSI problem - summary find (u, η) such that ϱ s 2 t 2 η = T( η) in (0, T ) Ω s, ρ f ( t u + u u) = σ( u, p), u = 0, } in Ω f (t) = ϕ f (t, Ω f ), t (0, T ), t 0 u Γ ϕ = ϕ Γ, Tn = (σ ϕ) ϕ τ n on Γ, where u is the fluid velocity, η the structure deformation, ϕ f the fluid domain displacement (1)

10 Energy inequality We formally multiply the structure equations by t η and integrate over Ω S. Then formally multiply the fluid equations by u and integrate over Ω F.

11 Energy inequality We formally multiply the structure equations by t η and integrate over Ω S. Then formally multiply the fluid equations by u and integrate over Ω F. By adding resulting equalities, integrating by parts and using the coupling conditions we obtain formal energy inequality: d ( t η 2 L dt 2 (Ω + s) u 2 L 2 (Ω f (t) + W ( η) ) + µ D(u) 2 L 2 (Ω f (t)) Ω s C(data).

12 Energy inequality We formally multiply the structure equations by t η and integrate over Ω S. Then formally multiply the fluid equations by u and integrate over Ω F. By adding resulting equalities, integrating by parts and using the coupling conditions we obtain formal energy inequality: d ( t η 2 L dt 2 (Ω + s) u 2 L 2 (Ω f (t) + W ( η) ) + µ D(u) 2 L 2 (Ω f (t)) Ω s C(data). From the analysis point of view such a FSI problem is still out of reach (some results Coutand, Shkoller ( 06), Grandmont ( 02), Galdi, Kyed 09, Boulakia, Guerrero ( 16), Čanić, BM 16) - various simplified models in the literature.

13 Main challenges Nonlinear elastodynamics. Navier-Stokes equations. Nonlinear coupling - geometrical nonlinearity. Fluid domain deformation (injectivity, regularity). Hyperbolic-parabolic coupling.

14 Main challenges Nonlinear elastodynamics. Navier-Stokes equations. Nonlinear coupling - geometrical nonlinearity. Fluid domain deformation (injectivity, regularity). Hyperbolic-parabolic coupling. It is natural to consider various simplifications of the general FSI model. Which simplifications are physically relevant? Simplifications are usually obtained by neglecting some terms (physically - some small parameters)

15 Introduction Monolithic and partitioned approach. Kinematically coupled scheme is a partitioned scheme introduced by Guidoboni, Glowinski, Cavallini, Čanić (JCP 2009). This lecture will be mostly based on convergence analysis in Bukač, BM (SINUM 2016). The scheme is based on the Lie operator splitting, where the fluid and the structure subproblems are fully decoupled and communicate only via the interface conditions. Advantages are modularity, stability, and easy implementation. Several extensions and implementations by different groups.

16 Lie-Trotter formula u (t) = Au(t) u(x) = e At u 0, A M n Let us decompose A = A 1 + A 2? e A+B = e A e B AB = BA. However: e A+B = lim N (ea/n e B/N ) N. This can be generalized to certain unbounded operators. However it is not directly applicable to FSI problems.

17 Lie (Marchuk-Yanenko) operator splitting scheme We consider initial value problem d dt φ + A(φ) = 0, φ(0) = φ 0. We suppose that A = A 1 + A 2. Let k = T /N be time-dicretization step and t n = nk. Then we define: d dt φ n+ i + A i (φ 2 n+ i ) = 0 in (t n, t n+1 ), 2 i 1 n+ φ n+ i (t n ) = φ 2, n = 0,..., N 1, i = 1, 2, 2 where φ n+ i 2 = φ n+ i (t n+1 ). 2

18 Simplified linear model ρ f t u = σ(u, p), u = 0 in (0, T ) Ω, σ(u, p)n = p in/out (t)n on (0, T ) Σ, ρ s ɛ t 2 η + L s η = σ(u, p)n on (0, T ) Γ, t η = u on (0, T ) Γ, η(., 0) = η 0, t η(., 0) = v 0 on Γ, u(., 0) = u 0 in Ω.

19 Monolithic formulation Given t (0, T ) find (u, η, p) V f V s Q f with u = t η on Γ, such that for all (ϕ, ξ, q) V fsi Q f ρ f t u ϕdx + 2µ D(u) : D(ϕ)dx p ϕdx Ω Ω Ω + ( u)qdx + ρ s ɛ tt η ξdx + L s η ξds = p in/out (t)ϕ nds. Ω Γ Γ Σ V f = (H 1 (Ω)) d, Q f = L 2 (Ω), V s = (H 1 0 (Γ)) d, V fsi = {(ϕ, ξ) V f V s ϕ Γ = ξ},

20 Fluid and structure sub-problems Step 1: The structure sub-problem. Find ṽ n+1, and η n+1 such that ρ s ɛ ṽ n+1 v n + L S η n+1 = βσ(u n, p n )n t on Γ, (2) d t η n+1 = ṽ n+1 on Γ, (3) Step 2. The fluid sub-problem. Find u n+1, p n+1 and v n+1 such that ρ f d t u n+1 = σ(u n+1, p n+1 ) in Ω, (4) u n+1 = 0 in Ω, (5) ρ s ɛ v n+1 ṽ n+1 = σ(u n+1, p n+1 )n + βσ(u n, p n )n t on Γ, (6) u n+1 = v n+1 on Γ, (7)

21 Improving the accuracy Original kiniematically couple scheme β = 0 Splitting error: ṽ n+1 v n+1 - accuracy O(( t) 1/2 ).

22 Improving the accuracy Original kiniematically couple scheme β = 0 Splitting error: ṽ n+1 v n+1 - accuracy O(( t) 1/2 ). We improve the accuracy by taking into account fluid stress in the structure sub-problem (β = 1): ṽ n+1 v n+1 = t ( ) σ n+1 n βσ n n ϱ s ɛ = t [ β ( σ n+1 n σ n ) ] n + (1 β)σ n+1 n on Γ. ϱ s ɛ

23 Improving the accuracy Original kiniematically couple scheme β = 0 Splitting error: ṽ n+1 v n+1 - accuracy O(( t) 1/2 ). We improve the accuracy by taking into account fluid stress in the structure sub-problem (β = 1): ṽ n+1 v n+1 = t ( ) σ n+1 n βσ n n ϱ s ɛ = t [ β ( σ n+1 n σ n ) ] n + (1 β)σ n+1 n on Γ. ϱ s ɛ Optimal first order accuracy for β = 1.

24 Improving the accuracy Original kiniematically couple scheme β = 0 Splitting error: ṽ n+1 v n+1 - accuracy O(( t) 1/2 ). We improve the accuracy by taking into account fluid stress in the structure sub-problem (β = 1): ṽ n+1 v n+1 = t ( ) σ n+1 n βσ n n ϱ s ɛ = t [ β ( σ n+1 n σ n ) ] n + (1 β)σ n+1 n on Γ. ϱ s ɛ Optimal first order accuracy for β = 1. Extension to second order accuracy - Oyekole, Trenchea, Bukač (SINUM 2018)

25 Stability estimate We obtain stability estimate in the standard way by taking ṽ n+1 and (u n+1, v n+1 ) and using identity (a b)a = 1 2 a2 1 2 b (a b)2.

26 Stability estimate We obtain stability estimate in the standard way by taking ṽ n+1 and (u n+1, v n+1 ) and using identity (a b)a = 1 2 a2 1 2 b (a b)2. Only splitting term is non-standard: I = t σ(u n h, (v pn n+1 h )n h ṽ n+1 ) h ds Γ

27 Stability estimate We obtain stability estimate in the standard way by taking ṽ n+1 and (u n+1, v n+1 ) and using identity (a b)a = 1 2 a2 1 2 b (a b)2. Only splitting term is non-standard: I = t σ(u n h, (v pn n+1 h )n h ṽ n+1 ) h ds Γ t 2 σ(u n h ρ s ɛ, pn h )n (σ(u n h, pn h )n σ(un+1 h, p n+1 h )n) ds Γ ) = t2 2ρ s ɛ ( σ(u n h, pn h )n 2 L 2 (Γ) σ(un+1 h, p n+1 h )n 2 L 2 (Γ) + ρ sɛ n+1 v 2 h ṽ n+1 h 2 L 2 (Γ).

28 Stability analysis We define discrete energies: E f (u n h ) = ρ f 2 un h 2 L 2 (Ω), E v (v n h ) = ρ sɛ 2 v n h 2 L 2 (Γ), E s(η n h ) = 1 2 ηn h 2 S. (8) Theorem (Bukač, BM, SINUM 16) Let {(u n h, pn h, ṽ n h, v n h, ηn h } 0 n N obtained by the numerical scheme. E f (u N h ) + E v (v N h ) + E s(η N h ) + t2 2ρ s ɛ σ(un h, pn h )n 2 L 2 (Γ) + ρ f t 2 N 1 d t u n+1 2 h 2 L + t2 2 N 1 n=0 d t η n+1 h N 1 2 S + µ t n=0 u n+1 h 2 F + ρ N 1 sɛ 2 n=0 ṽ n+1 h n=0 v n h 2 L 2 (Γ) E f (u 0 h ) + E v (v 0 h ) + E s(η 0 h ) + t2 N 1 2ρ s ɛ σ(u0 h, p0 h )n 2 L 2 (Γ) + t p in/out (t n+1 ) n=0

29 Error analysis Theorem Consider the numerical solution (u h, p h, ṽ h, v h, η h ) with discrete initial data (u 0 h, p0 h, ṽ 0 h, v 0 h, η0 h ) = (S hu 0, Π h p 0, I h ṽ 0, I h v 0, R h η 0 ). Assume that β = 1 and the exact solution satisfies additional regularity assumptions. Furthermore, we assume that γ t < 1, γ 1 < ρ sɛ 8 t, γ 2 < 1 4, where γ > 0, γ 1 > 0, γ 2 > 0. Let γ = max{γ, γ 2, γ 3 }. u N u N h L 2 (Ω) + u N u N h L 2 (0,T ;F ) + v N v N h L 2 (Γ) + η N η N h S + σ(u N, p N )n σ(u N h, pn h )n L 2 (Γ) e ( ta γt 1 + t 2( t 1/ ) + γ 1 t A 2 γ 2 γ 1 +h k B 1 + h k+1 B 2 + h s+1 B 3

30 2 2 Fluid-structure interaction Kinematically coupled scheme. Applications. Error analysis II + th k( t γ 1 t 2) C 1 + th s+1( t γ 1 t 2) ) C 2, γ 2 γ 1 γ 2 γ 1 A 1 = tt u L 2 t (L 2 (Ω)) + 1 γ ttv L 2 t (L 2 (Γ)) + 1 γ ttη L 2 t (H 1 (Γ)) + 1 γ 1 t σn L 2 t (L 2 x ) A 2 = t σn L 2 (0,T ;L 2 (Γ)), B 1 = 1 γ v L 2 t (Hk+1 (Γ)) + tu L 2 t (H k+1 (Ω)) + u L 2 t (Hk+1 (Ω)) + 1 u γ L 2 t (H k+1 (Γ)) + u L t (H k+1 (Ω)) + u L t (H k+1 (Γ)) + η L t (H k+1 (Γ)), 1 ) B 2 = (1 + 1γ1 t v L 2 t (H k+1 (Γ)) + v L t (H k+1 (Γ)), B 3 = p 2 L 2 t (Hs+1 (Ω)) + 1 γ 1 p 2 L 2 t (Hs+1 (Γ)) + p L t (H s+1 (Γ)),

31 Thick structure - simplified problem Ω F = (0, 1) 2 ( 1, 0), Ω S = (0, 1) 2 (0, 1), Γ = (0, 1) 2 {0}. S(η) = 2µ s D(η) + λ s ( η)i.

32 Thick structure - simplified problem Ω F = (0, 1) 2 ( 1, 0), Ω S = (0, 1) 2 (0, 1), Γ = (0, 1) 2 {0}. S(η) = 2µ s D(η) + λ s ( η)i. ρ f t u = σ(u, p), u = 0 in (0, T ) Ω, ϱ s t 2 η = S(η) in (0, T ) Ω S, S(η)n = σ(u, p)n on (0, T ) Γ, t η = u on (0, T ) Γ,

33 Thick structure - simplified problem Ω F = (0, 1) 2 ( 1, 0), Ω S = (0, 1) 2 (0, 1), Γ = (0, 1) 2 {0}. S(η) = 2µ s D(η) + λ s ( η)i. ρ f t u = σ(u, p), u = 0 in (0, T ) Ω, ϱ s t 2 η = S(η) in (0, T ) Ω S, S(η)n = σ(u, p)n on (0, T ) Γ, t η = u on (0, T ) Γ, If we consider coupling via elastic interface, coupling condition reads: ρ s ɛ t 2 η + L s η = S(η)n σ(u, p)n

34 Function spaces: Splitting for the thick structure V f = H 1 (Ω F ) 3, Q = L 2 (Ω F ), V s = H 1 (Ω S ) 3, V fsi = {(ϕ, ξ) V f V s Step 1: Find (ũ n+1 h, ṽ n+1 h ) Vh fsi, ηn+1 h Vh s such that for every (ϕ h, ξ h ) Vh fsi the following equality holds: n+1 ṽ h v n h ρ s ξ Ω S t h + a ts (η n+1 h, ξ h ) n+1 ũ h u n h +ρ f ϕ Ω F t h = σ n h n ξ, Γ ṽ n+1 h = ηn+1 h η n h, (ṽ n+1 t h ) Γ = (ũ n+1 h ) Γ, where σ n h = σ(un h, pn h ). (9)

35 Step 2 Find (u n+1 h, v n+1 h, p n+1 ) Vh fsi the following equality holds: ρ f Ω F u n+1 h ũ n+1 h ϕ t h + ρ s Ω S Q h, such that for (ϕ h, ξ h, q h ) V fsi h Q h, v n+1 h ṽ n+1 h ξ t h +a f (u n+1 h, ϕ h ) b(p n+1 h, ϕ h ) + b(q h, u n+1 h ) = σ n h n ϕ h, Γ (v n+1 h ) Γ = (u n+1 h ) Γ. (10)

36 Thick structure - stability theorem Theorem Let {(u n h, ṽ n h, v n h, ηn h } 0 n N be the numerical solution. Then, the following estimate holds: E f (u N h ) + E v (v N h ) + E s(η N h ) + t2 ρ s h σ(un h, pn h )n 2 L 2 (Γ) + ρ f t 2 2 N 1 n=0 +µ t d t u n+1 h 2 L 2 (Ω F ) + t2 N 1 2 N 1 n=0 u n+1 h 2 F + ρ N 1 s 2 n=0 n=0 a s (d t η n+1 h, d t η n+1 h ) ṽ n+1 h v n h 2 L 2 (Ω S ) E f (u 0 h ) + E v (v 0 h ) + E s(η 0 h ) + t2 N 1 ρ s h σ(u0 h, p0 h )n 2 L 2 (Γ) + t p in/out (t n+1 ) n=0

37 Numerical convergence - thin structure

38 Numerical convergence - thick structure

39 Nonlinear moving boundary problem η denote the vertical displacement of the deformable boundary.

40 Nonlinear moving boundary problem η denote the vertical displacement of the deformable boundary. Since domain is symmetric we consider only upper half of the domain.

41 Nonlinear moving boundary problem η denote the vertical displacement of the deformable boundary. Since domain is symmetric we consider only upper half of the domain. Fluid domain at time t: Ω η (t) = {(z, r) : 0 < z < L, 0 < r < R + η(t, z)}

42 Nonlinear moving boundary problem η denote the vertical displacement of the deformable boundary. Since domain is symmetric we consider only upper half of the domain. Fluid domain at time t: Ω η (t) = {(z, r) : 0 < z < L, 0 < r < R + η(t, z)} Γ(t) = {(z, R + η(t, z)) : 0 < z < 1} is deformable boundary.

43 Nonlinear moving boundary problem η denote the vertical displacement of the deformable boundary. Since domain is symmetric we consider only upper half of the domain. Fluid domain at time t: Ω η (t) = {(z, r) : 0 < z < L, 0 < r < R + η(t, z)} Γ(t) = {(z, R + η(t, z)) : 0 < z < 1} is deformable boundary. Longitudinal displacement is neglected.

44 Full FSI problem Find u = (u z (t, z, r), u r (t, z, r)), p(t, z, r), and η(t, z) such that ( ρ f t u + (u )u ) = σ u = 0 } in Ω η (t), t (0, T ), } u = t ηe r ρ s h t 2 η + C 0 η C 1 z 2 η + C 2 z 4 on (0, T ) (0, L), η = Jσn e r } u r = 0, on (0, T ) Γ r u z = 0 b, p + ρ f 2 u 2 = P in/out (t), u r = 0, u(0,.) = u 0, η(0,.) = η 0, t η(0,.) = v 0. } on (0, T ) Γ in/out, at t = 0.

45 Energy inequality By formally taking solution (u, t η) as test function in the weak formulation we get following energy estimate: where d dt E + D C(P in/out),

46 Energy inequality By formally taking solution (u, t η) as test function in the weak formulation we get following energy estimate: where d dt E + D C(P in/out), E = ϱ f 2 u 2 L 2 (Ω) + ϱ sh 2 η t 2 L 2 (Γ) + 1 ( C0 η 2 L + C z η 2 L + C 2 2 z 2 η 2 ) L, 2 D = µ D(u) 2 L 2 (Ω).

47 ALE formulation on reference domain We want to rewrite problem in the reference configuration Ω = (0, L) (0, 1).

48 ALE formulation on reference domain We want to rewrite problem in the reference configuration Ω = (0, L) (0, 1). Since we consider control domain, we can not use Lagrangian coordinates.

49 ALE formulation on reference domain We want to rewrite problem in the reference configuration Ω = (0, L) (0, 1). Since we consider control domain, we can not use Lagrangian coordinates. We use ALE mapping A η (t) : Ω Ω η (t), ( A η (t)( x, z) = x (R + η(t, x)) z ), ( x, z) Ω.

50 ALE formulation on reference domain We want to rewrite problem in the reference configuration Ω = (0, L) (0, 1). Since we consider control domain, we can not use Lagrangian coordinates. We use ALE mapping A η (t) : Ω Ω η (t), ( A η (t)( x, z) = x (R + η(t, x)) z ), ( x, z) Ω. u η (t,.) = u(t,.) A η (t), p η (t,.) = p(t,.) A η (t).

51 ALE formulation on reference domain We want to rewrite problem in the reference configuration Ω = (0, L) (0, 1). Since we consider control domain, we can not use Lagrangian coordinates. We use ALE mapping A η (t) : Ω Ω η (t), ( A η (t)( x, z) = u η (t,.) = u(t,.) A η (t), x (R + η(t, x)) z ), ( x, z) Ω. p η (t,.) = p(t,.) A η (t). We have problem on a fixed domain, but with coefficients that depend on the solution.

52 ALE formulation on reference domain We want to rewrite problem in the reference configuration Ω = (0, L) (0, 1). Since we consider control domain, we can not use Lagrangian coordinates. We use ALE mapping A η (t) : Ω Ω η (t), ( A η (t)( x, z) = u η (t,.) = u(t,.) A η (t), x (R + η(t, x)) z ), ( x, z) Ω. p η (t,.) = p(t,.) A η (t). We have problem on a fixed domain, but with coefficients that depend on the solution. Test functions still depend on the solution (because of divergence-free condition).

53 Few remarks η is gradient in ALE coordinates: η = z η z r R + η r 1 R + η r.

54 Few remarks η is gradient in ALE coordinates: η = z η z r R + η r 1 R + η r. w η = t η re r is domain (ALE) velocity.

55 Few remarks η is gradient in ALE coordinates: η = z η z r R + η r 1 R + η r. w η = t η re r is domain (ALE) velocity. In numerical simulations, one can use the ALE transformation A η n to transform the problem back to domain Ω η n and solve it there, thereby avoiding the un-necessary calculation of the transformed gradient. The ALE velocity is the only extra term that needs to be included with that approach.

56 Weak solution of the reference domain We say that (u, η) is weak solution of 1. u L (L 2 ) L 2 (H 1 ), η L (H 2 ) W 1, (L 2 ), 2. η u = 0, u(t, z, 1) = t η(t, z)e r, 3. for every (q, ψ) such that q(t, z, 1) = ψ(t, z)e r, η v = 0 following equality holds: ( T ρ f 0 Ω(R + η)u t q T 0 Ω (R + η) ( ((u w η ) η )u q ((u w η ) η )q u )) T + 2µ (R + η)d η (u) : D η (q) 0 Ω ρ f 2 T 0 Ω +C 2 2 z η 2 z ψ ) = ±R T ( t η)u q ρ s h 0 T 0 L 0 T t η t ψ + 0 P in/out Γ in/out q z + ρ f L 0 ( C0 ηψ + C 1 z η z ψ Ω η0 u 0 q(0) + ρ s h L 0 v 0 ψ(0)

57 Weak formulation of Step 1 u n+ 1 2 = u n. Then we define (v n+ 1 2, η n+ 1 2 ) H0 2(0, L) H2 0 (0, L) as a solution of the following problem, written in weak form: L 0 η n+ 1 2 η n φ = t L 0 v n+ 1 2 φ, φ L 2 (0, L), T + 0 L 0 L ρ s h 0 v n+ 1 2 v n ψ t ( C0 η n+ 1 2 ψ + C1 z η n+ 1 2 z ψ + C 2 2 z η n z ψ ) = 0, ψ H 2 0 (0, L).

58 Ω Weak formulation of Step 2 for every (q, ψ), such that q(t, z, 1) = ψ(t, z), ηn q = 0 ( ρ f (R + η n u n+1 u n+ 1 2 ) q + 1 [ (u n v n+ 1 2 rer ) ηn] u n+1 q t [(u n v n+ 1 2 rer ) ηn] q u n+1 ) + ρ f 2 +ρ s h s L 0 +2µ (R + η n )D ηn (u) : D ηn (q) Ω v n+1 v n+ 1 2 t ψ = R ( P n in Ω Γ in q z P n out v n+ 1 2 u n+1 q Γ out q z ), with ηn u n+1 = 0, u n+1 Γ = v n+1 e r, η n+1 = η n+ 1 2.

59 Discrete energy inequality Lemma (The Uniform Energy Bounds) Let t > 0 and N = T / t > 0. There exists a constant C > 0 independent of t (and N), such that the following estimates hold: 1. E n+ 1 2 N C, E n+1 N C, for all n = 0,..., N 1, 2. N j=1 Dj N C, N 1 n=0 N 1 n=0 ( (R + η n ) u n+1 u n 2 + v n+1 v n L 2 (0,L) Ω ) + v n+ 1 2 v n 2 L 2 (0,L) ( (C 0 η n+1 η n 2 L 2 (0,L) + C 1 z (η n+1 η n ) 2 L 2 (0,L) C, ) +C 2 z 2 (η n+1 η n ) 2 L 2 (0,L) C.

60 Discrete energy inequality II E n+ 1 2 N, E n+1 N, and Dj N are discrete kinetic energy and dissipation: E n+ i 2 N = 1 (ρ f (R + η n ) u n+ i 2 2 N 2 + ρ s h s v n+ i 2 N 2 L 2 (0,L) Ω ) +C 0 η n+ i 2 N 2 L 2 (0,L) + C 1 z η n+ i 2 N 2 L 2 (0,L) + C 2 z 2 η n+ i 2 N 2 L 2 (0,L), D n+1 N = tµ (R + η n ) D ηn (u n+1 N ) 2, n = 0,..., N, i = 0, 1. Ω C depends only on the parameters in the problem, on the kinetic energy of the initial data E 0, and on the energy norm of the inlet and outlet data P in/out 2 L 2 (0,T )

61 Approximate solutions We define approximate solutions as piece-wise constant in time: u N (t,.) = u n N, η N(t,.) = η n N, v N(t,.) = v n N, v N (t,.) = v n 1 2 N, t ((n 1) t, n t], n = 1... N.

62 Approximate solutions We define approximate solutions as piece-wise constant in time: u N (t,.) = u n N, η N(t,.) = η n N, v N(t,.) = v n N, v N (t,.) = v n 1 2 N, t ((n 1) t, n t], n = 1... N. We show that approximate solutions are well defined for small T, i.e. R + η N > 0.

63 Approximate solutions We define approximate solutions as piece-wise constant in time: u N (t,.) = u n N, η N(t,.) = η n N, v N(t,.) = v n N, v N (t,.) = v n 1 2 N, t ((n 1) t, n t], n = 1... N. We show that approximate solutions are well defined for small T, i.e. R + η N > 0. Discrete energy inequality implies: 1. Sequence (η N ) n N is uniformly bounded in L (0, T ; H0 2 (0, L)). 2. Sequence (v N ) n N is uniformly bounded in L (0, T ; L 2 (0, L)). 3. Sequence (vn ) n N is uniformly bounded in L (0, T ; L 2 (0, L)). 4. Sequence (u N ) n N is uniformly bounded in L (0, T ; L 2 (Ω)) L 2 (0, T ; H 1 (Ω)).

64 Convergence of approximate solution Let us now summarize obtained convergence results: η N η weakly in L (0, T ; H0 2 (0, L)) v n v weakly in L (0, T ; L 2 (0, L)) u N u weakly in L (0, T ; L 2 (Ω)) u N u weakly in L 2 (0, T ; H 1 (Ω)) u N u in L 2 (0, T ; L 2 (Ω)), v N v in L 2 (0, T ; L 2 (0, L)), η N η in L (0, T ; H s (0, L)), 0 s < 2.

65 Convergence(existence) theorem Theorem Let ϱ f, ϱ s, µ, h s, C i > 0, D i 0, i = 1, 2, 3. Suppose that the initial data v 0 L 2 (0, L), u 0 L 2 (Ω η0 ), and η 0 H0 2(0, L) is such that (R + η 0(z)) > 0, z [0, L]. Furthermore, let P in, P out L 2 loc (0, ). Then there exist T > 0 and a weak solution of (u, η) of Considered FSI problem has at least one weak solution on (0, T ), which satisfies the following energy estimate: t E(t) + D(τ)dτ E 0 + C( P in 2 L 2 (0,T ) + P out 2 L 2 (0,T )), t [0, T ], 0 where C depends only on the coefficients, and E(t) and D(t) are given by

66 Convergence(existence) II E(t) = ρ f 2 u 2 L 2 (Ω η(t)) + ρ sh 2 tη 2 L 2 (0,L) + 1 2( C0 η 2 L 2 (0,L) + C 1 z η 2 L 2 (0,L) + C 2 2 z η 2 L 2 (0,L)), D(t) = µ D(u) 2 L 2 (Ω η(t))) Furthermore, one of the following is true: 1. T =, 2. lim min (R + η(z)) = 0. t T z [0,L]

67 Regularization by a thin structure The interface is linearly elastic Koiter shell: ρ s1 h tt η + L s η = η J(t, z)(σn) (t,z,r+η(t,z)) e r + S(t, z, R)e r e r. S - the structure (linearized 3d elasticity) Piola-Kirchhoff stress tensor.

68 Regularization by a thin structure The interface is linearly elastic Koiter shell: ρ s1 h tt η + L s η = η J(t, z)(σn) (t,z,r+η(t,z)) e r + S(t, z, R)e r e r. S - the structure (linearized 3d elasticity) Piola-Kirchhoff stress tensor. Numerical scheme is unconditionally stable and first order convergent.

69 Regularization by a thin structure The interface is linearly elastic Koiter shell: ρ s1 h tt η + L s η = η J(t, z)(σn) (t,z,r+η(t,z)) e r + S(t, z, R)e r e r. S - the structure (linearized 3d elasticity) Piola-Kirchhoff stress tensor. Numerical scheme is unconditionally stable and first order convergent. Convergence is proven also in the nonlinear case (but not order of convergence).

70 Regularization by a thin structure The interface is linearly elastic Koiter shell: ρ s1 h tt η + L s η = η J(t, z)(σn) (t,z,r+η(t,z)) e r + S(t, z, R)e r e r. S - the structure (linearized 3d elasticity) Piola-Kirchhoff stress tensor. Numerical scheme is unconditionally stable and first order convergent. Convergence is proven also in the nonlinear case (but not order of convergence). Fluid dissipation is acts stronger on the structure when the interface is shell - regularization mechanism.

71 Regularization by a thin structure The interface is linearly elastic Koiter shell: ρ s1 h tt η + L s η = η J(t, z)(σn) (t,z,r+η(t,z)) e r + S(t, z, R)e r e r. S - the structure (linearized 3d elasticity) Piola-Kirchhoff stress tensor. Numerical scheme is unconditionally stable and first order convergent. Convergence is proven also in the nonlinear case (but not order of convergence). Fluid dissipation is acts stronger on the structure when the interface is shell - regularization mechanism. As thickness h 0, the solution converges to the solution of FSI with thick structure (numerical proof).

72 FSI with stent Joined work with M. Bukač, S. Čanić, M. Galić, J. Tambača, Y. Wang. Stents are metallic mesh-like biomedical devices used to prop the elastic coronary arteries open. The idea is to model stent as a network of 1D inextensible rods (Tambača, Kosor, Čanić, Paniagua 10). The resulting model gives good approximation of 3D displacement and significantly reduces the computational cost. The 1D stent model is coupled with a 2D shell model and 3D fluid.

73 FSI with stent

74 FSI with stent Splitting strategy can be also applied to the FSI-stent problem. The approximate solutions constructed with the splitting scheme converge to a weak solution in both linear and nonlinear moving boundary case. The scheme is unconditionally stable. Solution is not regular near the stent.

75 Numerical results

76 Numerical results

77 Numerical results

78 Numerical results

79 References Martina Bukač, Sunčica Čanić, and Boris Muha. A nonlinear fluid-structure interaction problem in compliant arteries treated with vascular stents. Appl. Math. Optim., 73(3): , M. Bukac, S. Canic, and B. Muha. A partitioned scheme for fluid-composite structure interaction problems. Journal of Computational Physics, 281(0): , Martina Bukač and Boris Muha. Stability and Convergence Analysis of the Extensions of the Kinematically Coupled Scheme for the Fluid-Structure Interaction. SIAM J. Numer. Anal., 54(5): , Giovanna Guidoboni, Roland Glowinski, Nicola Cavallini, and Sunčica Čanić. Stable loosely-coupled-type algorithm for fluid-structure interaction in blood flow. J. Comput. Phys., 228(18): , 2009.

80 References II S. Čanić, M. Galić, B. Muha, J. Tambača Analysis Of A Linear 3d Fluid-stent-shell Interaction Problem submitted Boris Muha and Sunčica Čanić. Existence of a weak solution to a nonlinear fluid-structure interaction problem modeling the flow of an incompressible, viscous fluid in a cylinder with deformable walls. Arch. Ration. Mech. Anal., 207(3): , Boris Muha and Sunčica Čanić. Existence of a solution to a fluid-multi-layered-structure interaction problem. J. Differential Equations, 256(2): , 2014.

81 Thank you for your attention!