Numerical Simulations of the Fluid-Structure Coupling in Physiological Vessels Mini-Workshop I

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1 Numerical Simulations of the Fluid-Structure Coupling in Physiological Vessels INRIA Rocquencourt, projet REO CRM Spring School, Montréal May 16, 2005

2 Motivations It is generally admitted that local blood flow behavior (shear stress) plays a key role in the understanding of several cardiovascular diseases Artery walls are compliant We aim at simulating the mechanical interaction between the blood flow and the arterial wall [press] [defor] [caro] [anev] Difficulties: Large displacements: geometrical non-linearities, moving computational domain Numerical stability: blood and wall densities are close Computational cost: strong research effort to improve the coupling algorithms FSI cost fluid cost + solid cost

3 Outline 1 Motivations 2 Part I: Model Equations The Structure Equations The Fluid Equations: ALE formulation The Coupled Problem Global Energy Balance 3 Part II: Explicit and Implicit Coupling Explicit Coupling Numerical Stability: Added Mass Effect Implicit Coupling Fixed-Point and Newton Methods 4 Part III: Semi-Implicit Coupling Remarks on Explicit and Implicit Coupling A New Coupling Scheme Numerical Stability

4 Part I Model Equations: Kinematics

5 Problem Setting Ω s (t) Γ in Γ w (t) Ω f (t) Γ out Ω f (t): fluid domain (filled with a viscous fluid, blood), Ω s (t): solid current configuration (arterial wall) Γ w (t): fluid-solid interface, where we enforce continuity of velocity equilibrium of stress

6 Problem Setting Ω s (t) Γ in Γ w (t) Ω f (t) Γ out Ω f (t): fluid domain (filled with a viscous fluid, blood), Ω s (t): solid current configuration (arterial wall) Γ w (t): fluid-solid interface, where we enforce continuity of velocity equilibrium of stress The problem Determine Ω f (t), velocity and stress within the fluid and the solid.

7 Solid Kinematics: Lagrange Formulation (I) Let Ω s 0 be reference material configuration, for instance Ωs 0 = Ωs (0). x s (, t) Ω s 0 Γ w 0 Γ w (t) Ω s (t) We introduce the solid deformation x s as an application that maps Ω s 0 into Ωs (t), i.e. x s (Ω s 0 ) = Ωs (t). x s : Ω s 0 R+ R 3. x def = x s (x 0, t) gives the position, at time t, of the material point x 0 Ω s 0. In other words, x s tracks the material points motion.

8 Solid Kinematics: Lagrange Formulation (II) We introduce the following notations: displacement: d s (x 0, t) def = x s (x 0, t) x 0 velocity: acceleration: d s def = ds t d s def = 2 d s t 2 deformation gradient: F (d s ) def = 0 x s = I + 0 d s jacobian: J(d s ) = det F (d s ) solid density in the reference configuration: ρ s (x 0 ) second Piola-Kirchoff stress tensor S(d s ), related to d s through a appropriate consitutive law. For instance, for an hyper-elastic material we have S(d s ) = W (E(ds )), E(d s ) def = 1 F (d s ) T F (d s ) I, E 2 with W a given elastic energy density and E(d s ) the Green-Lagrange stress tensor. unit normal vector on Ω s 0 : ns 0

9 Solid Equations: Lagrange Formulation From the Momentum conservation, we have that d s solves: ρ s ds div 0 (F (d s )S(d s )) = 0, in Ω s 0, Boundary conditions: d s = 0, on Γ D 0, F (d s )S(d s )n s 0 = 0, on ΓN 0, F (d s )S(d s )n s 0 = g fluid, on Γ w 0 (equilibrium of stress). Γ N 0 Γ D 0 Ω s 0 Γ w 0 Γ D 0 Γ N 0

10 Solid Equations: Variational Formulation Test functions space: V s def = H 1 Γ D (Ω s 0 ) def = 0 n v s H 1 (Ω s 0 ) : vs Γ D 0 o = 0 Solid variational formulation Find d s (t) V s such that ρ s d s v s + F (d s )S(d s ) : 0 d s = Ω s 0 Ω s 0 Γ w 0 g fluid v s, v s V s. Idea of the proof. Multiply by v s V s, integrate by parts and take into account the boundary conditions.

11 Fluid Equations: Some Notation We assume the fluid to be homogeneous, Newtonian and incompressible. We introduce the notations: density: ρ f dynamic viscosity: µ velocity: u(x, t), x Ω f (t) pressure: p(x, t) Cauchy stress tensor: σ(u, p), given by strain rate tensor: ε(u), defined as unit normal vector on Ω f (t): n unit normal vector on Ω f 0 : n0 σ(u, p) = pi + 2µε(u) ε(u) def = 1 h u + ( u) Ti 2

12 Fluid Equations: Eulerian Formulation (I) Γ in Ω f (t) Γ w (t) Γ out Incompressible Navier-Stokes equations in Eulerian formulation: «u ρ f t + u u 2µ div ε(u) + p = 0, in Ω f (t), Boundary conditions: σ(u, p)n = g in out, on Γ in out, div u = 0, in Ω f (t). u = d s, on Γ w (t) (continuity of velocities),

13 Fluid Equations: Eulerian Formulation (II) Γ w (t n+1 ) Γ w (t n ) j i i j Ω f (t n ) Remarks Ω f (t n+1 ) In the sequel we will deal with fluid meshes following the interface motion in order (accurate interface fluid load). In such case, the approximation of u (x, t), t becomes troublesome. This can be overcome by the introduction of a change of variables (the ALE map)

14 Fluid Domain Description (I) We set Ω f 0 def = Ω f (0) as a reference fluid domain. x f (, t) Ω f 0 Γ w 0 Ω f (t) Γ w (t) Assume we know the interface motion through the solid deformation x s Γ w. 0 We introduce the fluid domain deformation x f as an injective application x f : Ω f 0 R + R 3, such that x f Γ w 0 = x s Γ w 0, xf (Ω f 0 ) = Ωf (t). In other words x f = Ext(x s Γ w ), with Ext a given extension operator. 0 Remark Inside Ω f 0 the map xf is arbitrary

15 Fluid Domain Description (II) Thus, x f tracks the motion of the fluid domain. But, in general, it doesn t track the motion of the fluid particles. It is useful to introduce: the fluid domain displacement: d f (x 0, t) def = x f (x 0, t) x 0 Example of extension operator: the fluid domain velocity: w, defined by 8 d >< = 0, in Ω f 0, Harmonic lift: d f = d s, on Γ w 0 >: d f = 0, on Γ in out, w def = df t x 0 = xf t x 0 the deformation gradient: F (d f ) def = 0x f = I + 0d f and the jacobian: J(d f ) = det F (d f )

16 Fluid Domain Description (III) Remarks In general, w(x 0, t) u(x f (x 0, t), t) for x 0 Ω f 0. However, since by definition x f = x s on Γ w 0, we have and, therefore, w = u on Γ w 0 d f t x 0 = ds t x 0, on Γ w 0,

17 Time Derivatives: Eulerian vs. ALE (I) Let u = u(x, t) be a vector field defined on Ω f (t), t > 0. For x Ω f (t), the Arbitrary Lagrange-Euler (ALE) derivative of u is defined by with x 0 = ˆx f (, t) 1 (x), i.e. x = x f (x 0, t). u t x 0 (x, t) def = t u(xf (x 0, t), t), Proposition (ALE vs. Eulerian) Let x Ω f (t) with x = x f (x 0, t) and x 0 Ω f 0. Then with w def = d f. u u (x, t) = w(x 0, t) u(x, t) + (x, t), t x 0 t {z } {z } ALE derivative Eulerian derivative Idea of the proof. Use chain rule.

18 Why the terminology ALE? If w = 0, i.e. the fluid domain does not changes in time, we have u u (x, t) = (x, t), t x 0 t {z } {z } ALE derivative Eulerian derivative hence we recover the classical Eulerian derivative if w = u, i.e the map x f tracks the fluid particles motion, we have u (x, t) = u(x, t) u(x, t) + u (x, t), t x 0 t {z } {z } ALE derivative material or Lagrangian derivative hence we recover the classical Lagrangain or material derivative

19 Fluid Equations: ALE Formulation Γ in Ω f (t) Γ w (t) Γ out Incompressible Navier-Stokes equations in ALE formulation: «u ρ f + (w u) u 2µ div ε(u) + p = 0, t x 0 in Ω f (t), Boundary conditions: σ(u, p)n = g in, on Γ in σ(u, p)n = g out, on Γ in u = w, on Γ w (t) div u = 0, in Ω f (t).

20 Fluid Equations: ALE Variational Formulation (I) Test functions space o V f (t) = nv f : Ω f (t) R 3 : v f = ˆv [x f (, t)] 1, ˆv HΓ 1 w0 (Ωf 0 ) In particular, for each v f V f we have v f t x 0 = 0. Fluid variational formulation Find u(t) H 1 (Ω f (t)) with u Γ w (t) = w Γ w (t) and p(t) L 2 (Ω f (t)) such that d d t + 2µ Ω f (t) Ω f (t) Ω f (t) ρ f u v f Ω f (t) ε(u) : ε(v f ) q div u = 0, ρ f (div w)u v f + Ω f (t) q L 2 (Ω f (t)), Ω f (t) p div v f = g v f, Γ in out ρ f (w u) u v f v f V f (t),

21 Fluid Equations: ALE Variational Formulation (II) Idea of the proof. By multiplying by v f V f, integrating by parts and taking into account the boundary conditions. Here we only consider the mass term: ρ f u v f = ρ f J(d f ) u v f t x 0 t x 0 Ω f (t) = Ω f 0 Ω f 0 = d d t = d d t ρ f J(d f ) v f J(d f ) ρ f u v f t x 0 Ω f t x 0 0 J(d f )ρ f u v f J(d f )(div w)ρ f u v f Ω f 0 Ω f (t) Ω f 0 ρ f u v f (div w)ρ f u v f. Ω f (t) where we used the fact that J(d f ) t x 0 = J(d f ) div w.

22 Fluid Load at the Interface: Variational Residual Proposition Let (u, p) solution of the previous fluid sub-problem, v s V s and L : H 1 2 (Γ w 0 ) H1 Ω f (Ω f 0 Γw 0 ), 0 a given lift operator. Let v f = L(v s ) [x f (, t)] 1 (obviously v f / V f (t)), then Γ w 0 g fluid v s = σ(u, p)n v s = Γ w (t) Idea of the proof. Integration by parts. Remarks: d ρ f u v f d t Ω f (t) + ρ f (w u) u v f + 2µ Ω f (t) Ω f (t) Ω f (t) ρ f (div w)u v f ε(u) : ε(v f ) We solve the fluid equations with test functions vanishing on the interface. Ω f (t) We compute the forcing term with test functions not vanishing on the interface. This explains the terminology variational residual This approach has a straightforward discrete counterpart. p div v f.

23 The Coupled Problem: Strong Formulation Fluid: Solid: «u ρ f + (u w) u 2µ div ε(u) + p = 0, in Ω f (t), t x 0 div u = 0, in Ω f (t), σ(u, p)n = g in out, on Γ in out, ρ s d s div 0 (F (d s )S(d s )) = 0, in Ω s 0, Coupling conditions: geometry, velocity and stress d s = 0, on Γ D 0, F (d s )S(d s )n s 0 = 0, on ΓN 0, d f = Ext(d s Γ w 0 ), in Ωf 0, Ωf (t) = (I + d f )(Ω f 0 ), u = w(d f ), on Γ w (t), F (d s )S(d s )n 0 = J(d f )σ(u, p)f (d f ) T n 0, on Γ w 0.

24 The Coupled Problem: Variational Formulation Fluid domain motion: d f = Ext(d s Γ w 0 ), w = d f, Ω f (t) = (I + d f )(Ω f 0 ). Fluid: Find u(t) H 1 (Ω f (t)) with u Γ w (t) = w Γ w (t) and p(t) L 2 (Ω f (t)) such that d d t + 2µ Ω f (t) Ω f (t) Ω f (t) ρ f u v f Ω f (t) ε(u) : ε(v f ) q div u = 0, ρ f (div w)u v f + Ω f (t) q L 2 (Ω f (t)), Ω f (t) p div v f = g v f, Γ in out ρ f (w u) u v f v f V f (t), Solid: Find d s (t) V s such that ρ s ds v s + F (d s )S(d s ) : 0 d s = Ω s 0 Ω s 0 Γ w 0 g fluid v s, v s V s.

25 The Coupled Problem: Variational Formulation Fluid domain motion: d f = Ext(d s Γ w 0 ), w = d f, Ω f (t) = (I + d f )(Ω f 0 ). Fluid: Find u(t) H 1 (Ω f (t)) with u Γ w (t) = w Γ w (t) and p(t) L 2 (Ω f (t)) such that d d t + 2µ Ω f (t) Ω f (t) Ω f (t) ρ f u v f Ω f (t) ε(u) : ε(v f ) q div u = 0, ρ f (div w)u v f + Ω f (t) q L 2 (Ω f (t)), Ω f (t) p div v f = g v f, Γ in out ρ f (w u) u v f v f V f (t), Solid: Find d s (t) V s such that ρ s ds v s + F (d s )S(d s ) : 0 d s = R f`u, p, d f ; L(v s ), v s V s. Ω s 0 Ω s {z } 0 variational fluid residual

26 Global Energy Equality Proposition Assume that u = 0 on Γ in out and that the solid is an hyper-elastic material. Then, we have the following energy equality d ρ f ρ s! d t Ω f (t) 2 u 2 + Ω s 2 d s 2 + W (E(d s )) + 2µ ε(u) 2 = 0 0 Ω s 0 Ω f (t)

27 Global Energy Equality Proposition Assume that u = 0 on Γ in out and that the solid is an hyper-elastic material. Then, we have the following energy equality d d t ρ f ρ s Ω f (t) 2 u 2 + Ω s 2 d s 2 0 {z } Kinetic energy + Ω s 0 W (E(d s )) {z } Elastic energy «+ Ω f (t) 2µ ε(u) 2 = 0 {z } Dissipated power

28 Global Energy Equality Proposition Assume that u = 0 on Γ in out and that the solid is an hyper-elastic material. Then, we have the following energy equality d d t ρ f ρ s Ω f (t) 2 u 2 + Ω s 2 d s 2 0 {z } Kinetic energy + Ω s 0 W (E(d s )) {z } Elastic energy «+ Ω f (t) 2µ ε(u) 2 = 0 {z } Dissipated power Idea of the proof. Multiply the fluid and solid equations by u(t) and d s, respectively, integrate by parts and take into account the boundary conditions.

29 Part II Time Discretization: Explicit and Implicit Coupling

30 Time Discretization Fluid domain motion: d f = Ext(d s Γ w 0 ), w = d f, Ω f (t) = (I + d f )(Ω f 0 ). Fluid : Find u(t) H 1 (Ω f (t)) with u Γ w (t) = w Γ w (t) and p(t) L 2 (Ω f (t)) such that d d t + 2µ Ω f (t) Ω f (t) Ω f (t) ρ f u v f Ω f (t) ε(u) : ε(v f ) q div u = 0, ρ f (div w)u v f + Ω f (t) q L 2 (Ω f (t)), Ω f (t) p div v f = g v f, Γ in out ρ f (w u) u v f v f V f (t), Solid : Find d s (t) V s such that ρ s ds v s + F (d s )S(d s ) : 0 d s = R f `u, p, d f ; L(v s ), v s V s. Ω s 0 Ω s 0

31 Time Discretization Fluid domain motion: d f,n+1 = Ext( d sn+1 Γ w ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 0 ). Fluid : Find u(t) H 1 (Ω f (t)) with u Γ w (t) = w Γ w (t) and p(t) L 2 (Ω f (t)) such that d d t + 2µ Ω f (t) Ω f (t) Ω f (t) ρ f u v f Ω f (t) ε(u) : ε(v f ) q div u = 0, ρ f (div w)u v f + Ω f (t) q L 2 (Ω f (t)), Ω f (t) p div v f = g v f, Γ in out ρ f (w u) u v f v f V f (t), Solid : Find d s (t) V s such that ρ s ds v s + F (d s )S(d s ) : 0 d s = R f `u, p, d f ; L(v s ), v s V s. Ω s 0 Ω s 0

32 Time Discretization Fluid domain motion: d f,n+1 = Ext( d sn+1 Γ w ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 0 ). Fluid (Implicit Euler): Find u n+1 H 1 (Ω f,n+1 ) with u n+1 Γw,n+1 = wn+1 Γw,n+1 and p n+1 L 2 (Ω f,n+1 ) such that 1 ρ f u n+1 v f 1 ρ f u n v f + ρ f (div w n+1 )u n+1 v f Ω f,n+1 Ω f,n Ω f,n+1 + ρ f (w n+1 u n+1 ) u n+1 v f + 2µ ε(u n+1 ) : ε(v f ) Ω f,n+1 Ω f,n+1 p n+1 div v f = g n+1 v f, v f V f (t n+1 ), Ω f,n+1 Γ in out Solid : Find d s (t) V s such that ρ s ds v s + F (d s )S(d s ) : 0 d s = R f `u, p, d f ; L(v s ), v s V s. Ω s 0 Ω s 0

33 Time Discretization Fluid domain motion: d f,n+1 = Ext( d sn+1 Γ w ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 0 ). Fluid (Implicit Euler): Find u n+1 H 1 (Ω f,n+1 ) with u n+1 Γw,n+1 = wn+1 Γw,n+1 and p n+1 L 2 (Ω f,n+1 ) such that 1 ρ f u n+1 v f 1 ρ f u n v f + ρ f (div w n+1 )u n+1 v f Ω f,n+1 Ω f,n Ω f,n+1 + ρ f (w n+1 u n+1 ) u n+1 v f + 2µ ε(u n+1 ) : ε(v f ) Ω f,n+1 Ω f,n+1 p n+1 div v f = g n+1 v f, v f V f (t n+1 ), Ω f,n+1 Γ in out Solid (Leap-Frog): Ω s 0 Find d s,n+1 V s such that ρ s ds,n+1 2d s,n + d s,n 1 () 2 v s + F (d s,n+1 )S(d s,n+1 ) : 0 d s Ω s 0 = R f`u n+1, p n+1, d f,n+1 ; L(v s ), v s V s.

34 Coupling Strategies: Explicit/Implicit Key point: Give a prediction d sn+1 Γ w 0 of the interface displacement.

35 Coupling Strategies: Explicit/Implicit Key point: Give a prediction d sn+1 Γ w 0 of the interface displacement. Two possibilities:

36 Coupling Strategies: Explicit/Implicit Key point: Give a prediction d sn+1 Γ w 0 of the interface displacement. Two possibilities: Explicit coupling: d sn+1 Γ w 0 time-steps: d s,n, d s,n 1,... 0th-order interface prediction: 1st-order interface prediction:... given in terms of the interface displacement at the previous d sn+1 Γ w 0 = d s,n Γ w 0 d sn+1 Γ w 0 = d s,n Γ w. 0 d s,n Γ w 0 + d s,n 1 Γ w 0.

37 Coupling Strategies: Explicit/Implicit Key point: Give a prediction d sn+1 Γ w 0 of the interface displacement. Two possibilities: Explicit coupling: d sn+1 Γ w 0 time-steps: d s,n, d s,n 1,... 0th-order interface prediction: 1st-order interface prediction:... given in terms of the interface displacement at the previous d sn+1 Γ w 0 = d s,n Γ w 0 d sn+1 Γ w 0 Implicit coupling: No prediction, we simply set d sn+1 Γ w 0 = d s,n Γ w. 0 d s,n Γ w 0 + = d s,n+1 Γ w. 0 d s,n 1 Γ w 0.

38 Explicit Coupling: Detailed Algorithm 1 Compute fluid domain displacement and velocity: d f,n+1 = Ext(d s,n Γ w 0 ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 ). 2 Solve fluid in the new domain Ω f,n+1 : 1 ρ f u n+1 v f 1 ρ f u n v f + ρ f (div w n+1 )u n+1 v f Ω f,n+1 Ω f,n Ω f,n+1 + ρ f (w n+1 u n+1 ) u n+1 v f + 2µ ε(u n+1 ) : ε(v f ) Ω f,n+1 Ω f,n+1 p n+1 div v f = g n+1 v f, v f V f (t n+1 ), Ω f,n+1 Γ in out u n+1 Γw,n+1 = wn+1 Γ w,n+1. 3 Solve solid with the new fluid forcing term R f Ω s 0 4 Next time step ρ s ds,n+1 2d s,n + d s,n 1 () 2 v s + F (d s,n+1 )S(d s,n+1 ) : 0 d s Ω s 0 = R f`u n+1, p n+1, d f,n+1 ; L(v s ), v s V s.

39 Explicit Coupling (II) Some observations are in order:

40 Explicit Coupling (II) Some observations are in order: This algorithm is very attractive: FSI cost fluid cost + solid cost

41 Explicit Coupling (II) Some observations are in order: This algorithm is very attractive: FSI cost fluid cost + solid cost We have a fully uncoupled problem.

42 Explicit Coupling (II) Some observations are in order: This algorithm is very attractive: We have a fully uncoupled problem. FSI cost fluid cost + solid cost Velocity continuity is not exactly enforced (due to extrapolation) u n+1 Γ w 0 = d s,n Γ w 0 d s,n 1 Γ w 0 {z } extrapolated interface velocity d s,n+1 Γ w 0 d s,n Γ w 0 {z } interface velocity.

43 Explicit Coupling (II) Some observations are in order: This algorithm is very attractive: We have a fully uncoupled problem. FSI cost fluid cost + solid cost Velocity continuity is not exactly enforced (due to extrapolation) u n+1 Γ w 0 = d s,n Γ w 0 d s,n 1 Γ w 0 {z } extrapolated interface velocity d s,n+1 Γ w 0 d s,n Γ w 0 {z } interface velocity. Two major questions: Is energy well balanced at the discrete level?. Is this scheme stable?. [movie1] [movie2]

44 Explicit Coupling: A Priori Energy Estimate (I) Proposition Assume that u n+1 = 0 on Γ in out, the structure is an hyper-elastic material and» 1 q q = div w n+1 q, Ω f,n+1 Ω f,n Ω f,n+1 Then, the following energy inequality holds " 1 ρ f Ωf,n Ω f,n+1 2 un+1 2 ρ f 2 un 2 + " + 1 W (E(d s,n+1 )) Ω s 0 Ω s 0 Ω s 0 ρ s 2 d s,n+1 d s,n 2 # W (E(d s,n )) q with q t x 0 = 0. ρ s Ω s µ ε(u n+1 ) 2 Ω f,n+1 d s,n d s,n 1 2 # R f `u n+1, p n+1, d f,n+1 ; L(u n+1 ds,n+1 d s,n ) 0

45 Explicit Coupling: A Priori Energy Estimate (I) Proposition Assume that u n+1 = 0 on Γ in out, the structure is an hyper-elastic material and» 1 q q = div w n+1 q, Ω f,n+1 Ω f,n Ω f,n+1 Then, the following energy inequality holds " 1 ρ f Ωf,n Ω f,n+1 2 un+1 2 ρ f 2 un 2 + " + 1 W (E(d s,n+1 )) Ω s 0 Ω s 0 Ω s 0 ρ s 2 d s,n+1 d s,n 2 # W (E(d s,n )) q with q t x 0 = 0. ρ s Ω s µ ε(u n+1 ) 2 Ω f,n+1 d s,n d s,n 1 2 # R f `u n+1, p n+1, d f,n+1 ; L(u n+1 ds,n+1 d s,n ) 0 {z } Γw,n+1 σ(u n+1, p n+1 )n (u n+1 ds,n+1 d s,n )

46 Explicit Coupling: A Priori Energy Estimate (I) Proposition Assume that u n+1 = 0 on Γ in out, the structure is an hyper-elastic material and» 1 q q = div w n+1 q, Ω f,n+1 Ω f,n Ω f,n+1 Then, the following energy inequality holds q with q t x 0 = 0. " 1 ρ f Ωf,n Ω f,n+1 2 un+1 2 ρ f 2 un 2 ρ s + d s,n+1 d s,n 2 ρ s Ω s 2 d s,n d s,n 1 2 # 0 Ω s 2 0 " + 1 # W (E(d s,n+1 )) W (E(d s,n )) + 2µ ε(u n+1 ) 2 Ω s 0 Ω s 0 Ω f,n+1 j artificial R f `u n+1, p n+1, d f,n+1 ; L(u n+1 ds,n+1 d s,n ) 0 power {z } Γw,n+1 σ(u n+1, p n+1 )n (u n+1 ds,n+1 d s,n )

47 Explicit Coupling: A Priori Energy Estimate (II) Fluid domain displacement and velocity: d f,n+1 = Ext(d s,n Γ w 0 ), w n+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 ). Fluid subproblem: Find u n+1 H 1 (Ω f,n+1 ) with u n+1 Γw,n+1 = wn+1 Γ and w,n+1 pn+1 L 2 (Ω f,n+1 ) such that 1 ρ f u n+1 v f 1 ρ f u n v f + ρ f (div w n+1 )u n+1 v f Ω f,n+1 Ω f,n+1 Ω f,n+1 + ρ f (w n+1 u n+1 ) u n+1 v f + 2µ ε(u n+1 ) : ε(v f ) Ω f,n+1 Ω f,n+1 p n+1 div v f = g n+1 v f, v f V f (t n+1 ), Ω f,n+1 Γ in out Solid subproblem: Find d s,n+1 V s such that Ω s 0 ρ s ds,n+1 2d s,n + d s,n 1 () 2 v s + F (d s,n+1 )S(d s,n+1 ) : 0 d s Ω s 0 = R f`u n+1, p n+1, d f,n+1 ; L(v s ), v s V s.

48 Explicit Coupling: A Priori Energy Estimate (III) Idea of the proof. We take appropriate test functions and then we add the fluid and the solid variational formulations. In particular, for the solid discrete variational formulation: v s = ds,n+1 d s,n, Remark v s Γ w 0 = 0 and hence v s V s. for the fluid discrete variational formulation: v f = u n+1 L(v s Γ w 0 ) L(un+1 Γ w 0 v s Γ w 0 ) V f (t n+1 ). Remark 1 v f Γ w 0 = 0 and hence v f V f. 2 Let us recall that u n+1 Γ w 0 v Γ s w. 0

49 Explicit Coupling: Let s Summarize

50 Explicit Coupling: Let s Summarize Observations: Explicit coupling is cheap FSI cost fluid cost + solid cost

51 Explicit Coupling: Let s Summarize Observations: Explicit coupling is cheap FSI cost fluid cost + solid cost Numerical experiments (haemodynamics) show hat explicit coupling can be unstable Energy estimate with an artificial power term Γw,n+1 σ(u n+1, p n+1 )n (u n+1 ds,n+1 d s,n ).

52 Explicit Coupling: Let s Summarize Observations: Explicit coupling is cheap FSI cost fluid cost + solid cost Numerical experiments (haemodynamics) show hat explicit coupling can be unstable Energy estimate with an artificial power term Γw,n+1 σ(u n+1, p n+1 )n (u n+1 ds,n+1 d s,n ). Explicit coupling is stable, and widely used, in aeroelasticity (see Farhat, Piperno)

53 Explicit Coupling: Let s Summarize Observations: Explicit coupling is cheap FSI cost fluid cost + solid cost Numerical experiments (haemodynamics) show hat explicit coupling can be unstable Energy estimate with an artificial power term Γw,n+1 σ(u n+1, p n+1 )n (u n+1 ds,n+1 d s,n ). Explicit coupling is stable, and widely used, in aeroelasticity (see Farhat, Piperno) Major questions: What is the source of instabilities? Is there a hidden stability condition?

54 A 2D Simplified Model Γ w η Γ in Ω f symmetry axis Γ out

55 A 2D Simplified Model Γ w η Γ in Ω f symmetry axis Γ out Solid: string model (small displacements) ρ s h s η + Lη = p Γ w, in Γ w,

56 A 2D Simplified Model Γ w η Γ in Ω f symmetry axis Γ out Solid: string model (small displacements) ρ s h s η + Lη = p Γ w, in Γ w, with η: vertical displacement h s : thickness L: linear operator (for instance Lη = aη b 2 η x )

57 A 2D Simplified Model Γ w η Γ in Ω f symmetry axis Γ out Solid: string model (small displacements) ρ s h s η + Lη = p Γ w, in Γ w, Fluid: we keep a fixed fluid domain and we neglect viscous and convective effects 8 ρ f u >< t + p = 0, in Ωf, div u = 0, in Ω f, u n = η, on Γ >: w, p = p in out, on Γ in out.

58 A 2D Simplified Model Γ w η Γ in Ω f symmetry axis Γ out Solid: string model (small displacements) ρ s h s η + Lη = p Γ w, in Γ w, Fluid: we keep a fixed fluid domain and we neglect viscous and convective effects 8 ρ f u >< t + p = 0, in 8 Ωf, p = 0, in Ω f, >< div u = 0, in Ω f, p u = = ρf u n = η, on Γ >: w, n t n = ρf η, on Γ w, >: p = p in out, on Γ in out. p = p in out, on Γ in out.

59 A 2D Simplified Model Γ w η Γ in Ω f symmetry axis Γ out Solid: string model (small displacements) ρ s h s η + Lη = p Γ w, in Γ w, Fluid: we keep a fixed fluid domain and we neglect viscous and convective effects 8 ρ f u >< t + p = 0, in 8 Ωf, p = 0, in Ω f, >< div u = 0, in Ω f, p u = = ρf u n = η, on Γ >: w, n t n = ρf η, on Γ w, >: p = p in out, on Γ in out. p = p in out, on Γ in out. Why a new model? Physics: reproduces propagation phenomena [movie] Numerics: explicit coupling unstable [movie]

60 The Added Mass Effect The simplified coupled problem: Fluid: 8 p = 0, in Ω f, >< p n = ρf η, on Γ w, >: p = 0, on Γ in out. (1) Solid: ρ s h s η + Lη = p Γ w, in Γ w,

61 The Added Mass Effect The simplified coupled problem: Fluid: 8 p = 0, in Ω f, >< p n = ρf η, on Γ w, >: p = 0, on Γ in out. (1) Solid: ρ s h s η + Lη = p Γ w, in Γ w, Definition (Inverse Steklov-Poincaré operator) The operator M A : H 1 2 (Γ w ) H 1 2 (Γ w ) is defined as follows: for each g H 1 2 (Γ w ) we define M A (g) = q Γ w, with q H 1 (Ω f ) the solution of 8 q = 0, in Ω f, >< q n = g, on Γw, >: q = 0, on Γ in out.

62 The Added Mass Effect The simplified coupled problem: Fluid: 8 p = 0, in Ω f, >< p n = ρf η, on Γ w, >: p = 0, on Γ in out. (1) Solid: ρ s h s η + Lη = p Γ w, in Γ w, Definition (Inverse Steklov-Poincaré operator) The operator M A : H 1 2 (Γ w ) H 1 2 (Γ w ) is defined as follows: for each g H 1 2 (Γ w ) we define M A (g) = q Γ w, with q H 1 (Ω f ) the solution of Using this definition we have: 8 q = 0, in Ω f, >< q n = g, on Γw, >: q = 0, on Γ in out. p Γ w = M A ( ρ f η).

63 The Added Mass Effect The simplified coupled problem: Fluid: 8 p = 0, in Ω f, >< p n = ρf η, on Γ w, >: p = 0, on Γ in out. (1) Solid: ρ s h s η + Lη = p Γ w, in Γ w, Condensed solid problem: we have p Γ w = M A ( ρ f η), thus ρ s h s η + Lη = M A ( ρ f η).

64 The Added Mass Effect The simplified coupled problem: Fluid: 8 p = 0, in Ω f, >< p n = ρf η, on Γ w, >: p = 0, on Γ in out. (1) Solid: ρ s h s η + Lη = p Γ w, in Γ w, Condensed solid problem: we have p Γ w = M A ( ρ f η), thus `ρs h s + ρ f M A η + Lη = 0, in Γ w. (2)

65 The Added Mass Effect The simplified coupled problem: Fluid: 8 p = 0, in Ω f, >< p n = ρf η, on Γ w, >: p = 0, on Γ in out. (1) Solid: ρ s h s η + Lη = p Γ w, in Γ w, Condensed solid problem: we have p Γ w = M A ( ρ f η), thus `ρs h s + ρ f M A {z } added mass η + Lη = 0, in Γ w. (2)

66 The Added Mass Effect The simplified coupled problem: Fluid: 8 p = 0, in Ω f, >< p n = ρf η, on Γ w, >: p = 0, on Γ in out. (1) Solid: ρ s h s η + Lη = p Γ w, in Γ w, Condensed solid problem: we have p Γ w = M A ( ρ f η), thus `ρs h s + ρ f M A {z } added mass η + Lη = 0, in Γ w. (2) Remarks: This new equation looks like a structure equation, except for the extra term ρ f M A The fluid-structure coupling can be condensed into an extra mass acting on the structure (hence the terminology added mass effect )

67 The Added Mass Effect The simplified coupled problem: Fluid: 8 p = 0, in Ω f, >< p n = ρf η, on Γ w, >: p = 0, on Γ in out. (1) Solid: ρ s h s η + Lη = p Γ w, in Γ w, Condensed solid problem: we have p Γ w = M A ( ρ f η), thus `ρs h s + ρ f M A {z } added mass η + Lη = 0, in Γ w. (2) Remarks: This new equation looks like a structure equation, except for the extra term ρ f M A The fluid-structure coupling can be condensed into an extra mass acting on the structure (hence the terminology added mass effect ) Major question: What kind of time integration scheme of (2) arises from the a explicit coupling of (1)?

68 Time Discretization: Explicit Coupling (I) Fluid: 8>< >: + p n+1 = 0, 8 >< div u n+1 = 0, = >: u n+1 n = ηn η n 1. ρ f un+1 u n p n+1 = 0, p n = ρf ηn 2η n 1 + η n 2 () 2.

69 Time Discretization: Explicit Coupling (I) Fluid: 8>< Solid: >: + p n+1 = 0, 8 >< div u n+1 = 0, = >: u n+1 n = ηn η n 1. ρ f un+1 u n p n+1 = 0, p n = ρf ηn 2η n 1 + η n 2 () 2. ρ s h s ηn+1 2η n + η n 1 () 2 + Lη n+1 = p n+1 Γ w.

70 Time Discretization: Explicit Coupling (I) Fluid: 8>< Solid: >: + p n+1 = 0, 8 >< div u n+1 = 0, = >: u n+1 n = ηn η n 1. ρ f un+1 u n Condensed solid problem: ρ s h s ηn+1 2η n + η n 1 () 2 p n+1 = 0, p n = ρf ηn 2η n 1 + η n 2 () 2. ρ s h s ηn+1 2η n + η n 1 () 2 + Lη n+1 = p n+1 Γ w. + ρ f M A η n 2η n 1 + η n 2 () 2 + Lη n+1 = 0.

71 Time Discretization: Explicit Coupling (I) Fluid: 8>< Solid: >: + p n+1 = 0, 8 >< div u n+1 = 0, = >: u n+1 n = ηn η n 1. ρ f un+1 u n Condensed solid problem: p n+1 = 0, p n = ρf ηn 2η n 1 + η n 2 () 2. ρ s h s ηn+1 2η n + η n 1 () 2 + Lη n+1 = p n+1 Γ w. ρ s h s ηn+1 2η n + η n 1 () 2 +ρ f η n 2η n 1 + η n 2 M A () 2 + Lη n+1 = 0. {z } implicit

72 Time Discretization: Explicit Coupling (I) Fluid: 8>< Solid: >: + p n+1 = 0, 8 >< div u n+1 = 0, = >: u n+1 n = ηn η n 1. ρ f un+1 u n Condensed solid problem: p n+1 = 0, p n = ρf ηn 2η n 1 + η n 2 () 2. ρ s h s ηn+1 2η n + η n 1 () 2 + Lη n+1 = p n+1 Γ w. ρ s h s ηn+1 2η n + η n 1 () 2 +ρ f η n 2η n 1 + η n 2 M A () 2 +Lη n+1 = 0. {z } {z } implicit explicit

73 Time Discretization: Explicit Coupling (I) Fluid: 8>< Solid: >: + p n+1 = 0, 8 >< div u n+1 = 0, = >: u n+1 n = ηn η n 1. ρ f un+1 u n Condensed solid problem: p n+1 = 0, p n = ρf ηn 2η n 1 + η n 2 () 2. ρ s h s ηn+1 2η n + η n 1 () 2 + Lη n+1 = p n+1 Γ w. ρ s h s ηn+1 2η n + η n 1 () 2 +ρ f η n 2η n 1 + η n 2 M A () 2 +Lη n+1 = 0. {z } {z } implicit explicit Observation: Explicit coupling leads to a explicit discretization of the added mass term.

74 Times discretization: Explicit Coupling (II) Theorem (Causin-Gerbeau-Nobile 04) Let λ max be the for largest eigenvalue of M A and assume that Lη = aη. Then, the previous explicit coupling scheme is unconditionally unstable whenever ρ f λ max ρ s h s 1. (3)

75 Times discretization: Explicit Coupling (II) Theorem (Causin-Gerbeau-Nobile 04) Let λ max be the for largest eigenvalue of M A and assume that Lη = aη. Then, the previous explicit coupling scheme is unconditionally unstable whenever ρ f λ max ρ s h s 1. (3) Remarks: The stability of the coupling scheme does not depend on the time step. The instability condition (3) confirms empirical observations: instabilities may occur when the structure is light, as in haemodynamic applications ρ f ρ s in aeroelasticity ρ f ρ s, hence explicit coupling is stable

76 Times discretization: Explicit Coupling (II) Theorem (Causin-Gerbeau-Nobile 04) Let λ max be the for largest eigenvalue of M A and assume that Lη = aη. Then, the previous explicit coupling scheme is unconditionally unstable whenever ρ f λ max ρ s h s 1. (3) Remarks: The stability of the coupling scheme does not depend on the time step. The instability condition (3) confirms empirical observations: Idea of the proof. instabilities may occur when the structure is light, as in haemodynamic applications ρ f ρ s in aeroelasticity ρ f ρ s, hence explicit coupling is stable Note that M A is a compact, postive and self-adjoint operator on L 2 (Γ w ). Expand η n+1, η n, η n 1 and η n 2 on a orthonormal basis made of eigen-vectors of M A. Conclude by proving that characteristic polynomial of the resulting difference equation, has a root with magnitude larger than 1 under condition (3).

77 Time Discretization: Implicit Coupling Fluid: Solid: Ω s 0 d f,n+1 = Ext(d s,n+1 Γ w ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 0 ), 1 ρ f u n+1 v f 1 ρ f u n v f + ρ f (div w n+1 )u n+1 v f Ω f,n+1 Ω f,n Ω f,n+1 + ρ f (w n+1 u n+1 ) u n+1 v f + 2µ ε(u n+1 ) : ε(v f ) Ω f,n+1 Ω f,n+1 p n+1 div v f = g n+1 v f, v f V f (t n+1 ), Ω f,n+1 Γ in out u n+1 Γw,n+1 = wn+1 Γ w,n+1 ρ s ds,n+1 2d s,n + d s,n 1 () 2 v s + F (d s,n+1 )S(d s,n+1 ) : 0 d s Ω s 0 = R f`u n+1, p n+1, d f,n+1 ; L(v s ), v s V s.

78 Time Discretization: Implicit Coupling Fluid: d f,n+1 = Ext(d s,n+1 Γ w ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 0 ), 1 ρ f u n+1 v f 1 ρ f u n v f + ρ f (div w n+1 )u n+1 v f Ω f,n+1 Ω f,n Ω f,n+1 + ρ f (w n+1 u n+1 ) u n+1 v f + 2µ ε(u n+1 ) : ε(v f ) Ω f,n+1 Ω f,n+1 p n+1 div v f = g n+1 v f, v f V f (t n+1 ), Ω f,n+1 Γ in out u n+1 Γw,n+1 = wn+1 Γ w,n+1 We formally note this problem as (d f,n+1, u n+1, p n+1 ) = F(d s,n+1 ). Solid: Ω s 0 ρ s ds,n+1 2d s,n + d s,n 1 () 2 v s + F (d s,n+1 )S(d s,n+1 ) : 0 d s Ω s 0 = R f`u n+1, p n+1, d f,n+1 ; L(v s ), v s V s.

79 Time Discretization: Implicit Coupling Fluid: d f,n+1 = Ext(d s,n+1 Γ w ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 0 ), 1 ρ f u n+1 v f 1 ρ f u n v f + ρ f (div w n+1 )u n+1 v f Ω f,n+1 Ω f,n Ω f,n+1 + ρ f (w n+1 u n+1 ) u n+1 v f + 2µ ε(u n+1 ) : ε(v f ) Ω f,n+1 Ω f,n+1 p n+1 div v f = g n+1 v f, v f V f (t n+1 ), Ω f,n+1 Γ in out u n+1 Γw,n+1 = wn+1 Γ w,n+1 We formally note this problem as (d f,n+1, u n+1, p n+1 ) = F(d s,n+1 ). Solid: Ω s 0 ρ s ds,n+1 2d s,n + d s,n 1 () 2 v s + F (d s,n+1 )S(d s,n+1 ) : 0 d s Ω s 0 We formally note this problem as d s,n+1 = S(d f,n+1, u n+1, p n+1 ). = R f`u n+1, p n+1, d f,n+1 ; L(v s ), v s V s.

80 Explicit Coupling (II) Some observations are in order:

81 Explicit Coupling (II) Some observations are in order: Implicit coupling leads to stable schemes. Velocity continuity is exactly enforced u n+1 Γ w 0 = d s,n+1 Γ w 0 d s,n Γ w 0 {z } interface velocity We then recover an energy estimate without artificial power at the interface.

82 Explicit Coupling (II) Some observations are in order: Implicit coupling leads to stable schemes. Velocity continuity is exactly enforced u n+1 Γ w 0 = d s,n+1 Γ w 0 d s,n Γ w 0 {z } interface velocity We then recover an energy estimate without artificial power at the interface. Highly non-linear coupled problem (geometrical non-linearities): (d f,n+1, u n+1, p n+1 ) = F(d s,n+1 ) d s,n+1 = S(d f,n+1, u n+1, p n+1 )

83 Explicit Coupling (II) Some observations are in order: Implicit coupling leads to stable schemes. Velocity continuity is exactly enforced u n+1 Γ w 0 = d s,n+1 Γ w 0 d s,n Γ w 0 {z } interface velocity We then recover an energy estimate without artificial power at the interface. Highly non-linear coupled problem (geometrical non-linearities): Thus, by composition: (d f,n+1, u n+1, p n+1 ) = F(d s,n+1 ) d s,n+1 = S(d f,n+1, u n+1, p n+1 ) Fixed-point problem: d s,n+1 = S F(d s,n+1 )

84 Explicit Coupling (II) Some observations are in order: Implicit coupling leads to stable schemes. Velocity continuity is exactly enforced u n+1 Γ w 0 = d s,n+1 Γ w 0 d s,n Γ w 0 {z } interface velocity We then recover an energy estimate without artificial power at the interface. Highly non-linear coupled problem (geometrical non-linearities): Thus, by composition: (d f,n+1, u n+1, p n+1 ) = F(d s,n+1 ) d s,n+1 = S(d f,n+1, u n+1, p n+1 ) Fixed-point problem: d s,n+1 = S F(d s,n+1 ) Root-finding problem: R(d s,n+1 ) def = d s,n+1 S F(d s,n+1 ) = 0

85 Solution Methods: Fixed-Point Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Relaxation step: d s k+1 ω kd s k+1 + (1 ω k)d s k

86 Solution Methods: Fixed-Point Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Relaxation step: d s k+1 ω kd s k+1 + (1 ω k)d s k Acceleration via dynamic relaxation: Aitken s method (Mok-Wall-Ramm 01, Gerbeau-Vidrascu 03) ω k = (ds k ds k 1 ) (ds k S F(ds k ) ds k 1 + S F(ds k 1 )) d s k S F(ds k ) ds k 1 + S F(ds k 1 ) 2

87 Solution Methods: Fixed-Point Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Relaxation step: d s k+1 ω kd s k+1 + (1 ω k)d s k Acceleration via dynamic relaxation: Aitken s method (Mok-Wall-Ramm 01, Gerbeau-Vidrascu 03) ω k = (ds k ds k 1 ) (ds k S F(ds k ) ds k 1 + S F(ds k 1 )) d s k S F(ds k ) ds k 1 + S F(ds k 1 ) 2 Reducing the fluid computational cost via transpiration conditions (Deparis-MF-Formaggia 03)

88 Solution Methods: Fixed-Point Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Relaxation step: d s k+1 ω kd s k+1 + (1 ω k)d s k Acceleration via dynamic relaxation: Aitken s method (Mok-Wall-Ramm 01, Gerbeau-Vidrascu 03) ω k = (ds k ds k 1 ) (ds k S F(ds k ) ds k 1 + S F(ds k 1 )) d s k S F(ds k ) ds k 1 + S F(ds k 1 ) 2 Reducing the fluid computational cost via transpiration conditions (Deparis-MF-Formaggia 03) Observation Too expensive and may fail to converge!

89 Solution Methods: Newton Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Evaluate residual: R(d s k ) def = d s k S F(ds k ) = ds k ds k+1 (c) Solve tangent problem ˆ Dd s R(d s k ) δd s = R(d s k ) (solved via GMRES) (d) Update rule: d s k+1 ds k + δds

90 Solution Methods: Newton Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Evaluate residual: R(d s k ) def = d s k S F(ds k ) = ds k ds k+1 (c) Solve tangent problem ˆ Dd s R(d s k ) δd s = R(d s k ) (solved via GMRES) (d) Update rule: d s k+1 ds k + δds The bottleneck of Newton s method For a given solid variation z, how to compute ˆ D d s R(d s k ) z?

91 Solution Methods: Newton Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Evaluate residual: R(d s k ) def = d s k S F(ds k ) = ds k ds k+1 (c) Solve tangent problem ˆ Dd s R(d s k ) δd s = R(d s k ) (solved via GMRES) (d) Update rule: d s k+1 ds k + δds The bottleneck of Newton s method For a given solid variation z, how to compute ˆ D d s R(d s k ) z? Finite difference Jacobian approximations (Matthies-Steindorf 01, Heil 03), ˆ Dd s R(d s k ) z R(ds k + ɛz) R(ds k ), 0 < ɛ 1 ɛ

92 Solution Methods: Newton Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Evaluate residual: R(d s k ) def = d s k S F(ds k ) = ds k ds k+1 (c) Solve tangent problem ˆ Dd s R(d s k ) δd s = R(d s k ) (solved via GMRES) (d) Update rule: d s k+1 ds k + δds The bottleneck of Newton s method For a given solid variation z, how to compute ˆ D d s R(d s k ) z? Finite difference Jacobian approximations (Matthies-Steindorf 01, Heil 03), too expensive and may fail to converge!

93 Solution Methods: Newton Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Evaluate residual: R(d s k ) def = d s k S F(ds k ) = ds k ds k+1 (c) Solve tangent problem ˆ Dd s R(d s k ) δd s = R(d s k ) (solved via GMRES) (d) Update rule: d s k+1 ds k + δds The bottleneck of Newton s method For a given solid variation z, how to compute ˆ D d s R(d s k ) z? Finite difference Jacobian approximations (Matthies-Steindorf 01, Heil 03), too expensive and may fail to converge! Physically based Jacobian approximations (Tezduyar 01, Gerbeau-Vidrascu 03, Deparis-Gerbeau-Vaseur 04),

94 Solution Methods: Newton Based Methods 1 Initial guess d s 0, ds def 0 = d s,n 2 For k = 0, 1,... until convergence (a) Solve fluid and solid: d s k+1 = S F(ds k ) (b) Evaluate residual: R(d s k ) def = d s k S F(ds k ) = ds k ds k+1 (c) Solve tangent problem ˆ Dd s R(d s k ) δd s = R(d s k ) (solved via GMRES) (d) Update rule: d s k+1 ds k + δds The bottleneck of Newton s method For a given solid variation z, how to compute ˆ D d s R(d s k ) z? Finite difference Jacobian approximations (Matthies-Steindorf 01, Heil 03), too expensive and may fail to converge! Physically based Jacobian approximations (Tezduyar 01, Gerbeau-Vidrascu 03, Deparis-Gerbeau-Vaseur 04), cheaper but still may fail to converge!

95 Exact Jacobian Evaluation Given a solid displacement z we need to evaluate ˆ D d s R( d s ) z with R(d s ) def = d s S F(d s ).

96 Exact Jacobian Evaluation Given a solid displacement z we need to evaluate ˆ D d s R( d s ) z with R(d s ) def = d s S F(d s ). One application of the chain rule yields ˆ Dd s R( d s ) z = z ˆ D (d s,u,p) S(F( d s )) ˆ D d s F( d s ) z.

97 Exact Jacobian Evaluation Given a solid displacement z we need to evaluate ˆ D d s R( d s ) z with R(d s ) def = d s S F(d s ). One application of the chain rule yields ˆ Dd s R( d s ) z = z ˆ D (d f,u,p) S(F( d s )) ˆ D d s F( d s ) z {z } δu f = (δu, δp, δd f ).

98 Exact Jacobian Evaluation Given a solid displacement z we need to evaluate ˆ D d s R( d s ) z with R(d s ) def = d s S F(d s ). One application of the chain rule yields ˆ Dd s R( d s ) z = z ˆ D (d f,u,p) S(F( d s )) ˆ D d s F( d s ) z {z } δu f = (δu, δp, δd f ) {z } δz.

99 Fluid Tangent Problem Using shape-derivative calculus (Sokolowski-olesio 91, Delfour-olesio 01), (δu, δp) solves the following linear problem: 8 ρ h δu + ρ div δu (u n w g( d i f )) σ(δu, δp) = ρ (div δdf )ū nh div ρū (u n w g( d i f )) σ(ū, p) hi div δd f ( δd f ) Tio >< + ρ n h div(ū δdf ) div µ ū δd f + ( δd f ) T ( ū) Tio, in Ω f ( d), div δu = div n ū h I div δd f ( δd f ) Tio, in Ω f ( d f ), δu = δdf, on Γw ( d f ), h >: σ(δu, δp)n = µ ū δd f + ( δd f ) T ( ū) Ti n, on Γ in out, with δd f = Ext ( d s )z (MF-Moubachir 03,04).

100 Fluid Tangent Problem Approximations An approximate fluid tangent problem can be derived by neglecting the shape derivative terms, yielding 8 ρ h δu + ρ div δu (u n w g( d i f )) σ(δu, δp) = 0, >< div δu = 0, in Ω f ( d f ), in Ω f ( d f ), δu = δdf, on Γw ( d f ), >: σ(δu, δp)n = 0, on Γ in out.

101 Fluid Tangent Problem Approximations An approximate fluid tangent problem can be derived by neglecting the shape derivative terms, yielding 8 ρ h δu + ρ div δu (u n w g( d i f )) σ(δu, δp) = 0, in Ω f ( d f ), >< div δu = 0, in Ω f ( d f ), δu = δdf, on Γw ( d f ), >: σ(δu, δp)n = 0, on Γ in out. On the other hand, neglecting here the convective and diffusive terms (Gerbeau-Vidrascu 2003), we get the following problem 8 ρ δu + δp = 0, in Ωf ( d f ), >< div δu = 0, in Ω f ( d f ), δu = δdf, on Γw ( d f ), >: δpn = 0, on Γ in out ( d f ).

102 Iterations History ( = 10 4 s) Newton: exact Newton: not exact FP-Aitken number of iterations time

103 Computational Cost ( = 10 4 s) Algorithm CPU time (dimensionless) Picard-Aitken 1.00 Quasi-Newton 0.55 Newton 0.60

104 Computational Cost ( = 10 4 s) Algorithm CPU time (dimensionless) Picard-Aitken 1.00 Quasi-Newton 0.55 Newton 0.60 Question What about increasing the time step?

105 Residuals History ( = 10 3 s) 0.01 Newton: exact Newton: not exact residual 1e-05 1e-06 1e-07 1e number of iterations

106 Some Remarks on Implicit Schemes Newton based methods perform better than fixed-point based methods Approximated Jacobians may lead to divergent quasi-newton methods Exact Jacobians may improve the convergence of the Newton s loop (moderate time steps) but are too expensive Realistic simulations with this kind of methods are very expensive

107 Part III Semi-Implicit Coupling

108 Let s Summarize Explicit coupling is cheap but unstable

109 Let s Summarize Explicit coupling is cheap but unstable Implicit coupling is stable but too expensive

110 Let s Summarize Explicit coupling is cheap but unstable Implicit coupling is stable but too expensive Added-mass effect seems to be the source of numerical instabilities in explicit coupling

111 Let s Summarize Explicit coupling is cheap but unstable Implicit coupling is stable but too expensive Added-mass effect seems to be the source of numerical instabilities in explicit coupling Geometrical non-linearities (moving domains), convective and viscous effects do not seem to affect the stability of a coupling algorithm. However, they are implicitly treated in fully implicit schemes (very expensive!)

112 Let s Summarize Explicit coupling is cheap but unstable Implicit coupling is stable but too expensive Added-mass effect seems to be the source of numerical instabilities in explicit coupling Geometrical non-linearities (moving domains), convective and viscous effects do not seem to affect the stability of a coupling algorithm. However, they are implicitly treated in fully implicit schemes (very expensive!) Major question: How to balance stability and computational cost?

113 Let s Summarize Explicit coupling is cheap but unstable Implicit coupling is stable but too expensive Added-mass effect seems to be the source of numerical instabilities in explicit coupling Geometrical non-linearities (moving domains), convective and viscous effects do not seem to affect the stability of a coupling algorithm. However, they are implicitly treated in fully implicit schemes (very expensive!) Major question: How to balance stability and computational cost? First idea: semi-implicit coupling Treat implicitly the added-mass effect Treat explicitly the geometrical non-linearities and the convective and viscous effects

114 Let s Summarize Explicit coupling is cheap but unstable Implicit coupling is stable but too expensive Added-mass effect seems to be the source of numerical instabilities in explicit coupling Geometrical non-linearities (moving domains), convective and viscous effects do not seem to affect the stability of a coupling algorithm. However, they are implicitly treated in fully implicit schemes (very expensive!) Major question: How to balance stability and computational cost? First idea: semi-implicit coupling Treat implicitly the added-mass effect Treat explicitly the geometrical non-linearities and the convective and viscous effects Second idea: semi-implicit coupling Perform this using a projection scheme (Chorin-Teman) within the fluid

115 The Chorin-Teman Projection Scheme Main feature Incompressibility and viscous/convective effects are decoupled Set u 0 = u 0, for n 0 compute (ũ n+1, u n+1, p n+1 ) Elliptic step: Projection step: 8 >< ρ f ũn+1 u n >: + ũ n+1 ũ n+1 «2µ div ε(ũ n+1 ) = f n+1, 8 >< ρ f un+1 ũ n+1 >: + p n+1 = 0, u n+1 = 0, u n+1 n Γ = 0. (see Guermond-Minev-Shen 04, Rannacher 91, Shen 92) ũ n+1 Γ = 0.

116 The Chorin-Teman Projection Scheme Main feature Incompressibility and viscous/convective effects are decoupled Set u 0 = u 0, for n 0 compute (ũ n+1, u n+1, p n+1 ) Elliptic step: 8 >< ρ f ũn+1 u n >: + ũ n+1 ũ n+1 «2µ div ε(ũ n+1 ) = f n+1, ũ n+1 Γ = 0. Projection step: 8 >< ρ f un+1 ũ n+1 >: + p n+1 = 0, u n+1 = 0, u n+1 n Γ = 0. 8 >< p n+1 = ρf ũn+1, = p >: n+1 = 0. n Γ (see Guermond-Minev-Shen 04, Rannacher 91, Shen 92)

117 Semi-Implicit Coupling Algorithm: Explicit Part Update fluid domain (mesh) displacement and velocity: d f,n+1 = Ext(d s,n Γ w 0 ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 ),

118 Semi-Implicit Coupling Algorithm: Explicit Part Update fluid domain (mesh) displacement and velocity: d f,n+1 = Ext(d s,n Γ w 0 ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 ), Fluid: Elliptic step in the new domain Ω f,n+1 : ρ f ũn+1 u n + (ũ n+1 w n+1 ) ũ n+1 «2µ div ε(ũ n+1 ) = 0, ũ n+1 Γ w,n+1 = wn+1 Γ w,n+1.

119 Semi-Implicit Coupling Algorithm: Explicit Part Update fluid domain (mesh) displacement and velocity: d f,n+1 = Ext(d s,n Γ w 0 ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 ), Fluid: Elliptic step in the new domain Ω f,n+1 : 1 ρ f ũ n+1 v f Ω f,n ρ f u n v f Ω f,n 1 + ρ f (div w n+1 )ũ n+1 v f Ω f,n ρ f (w n+1 ũ n+1 ) ũ n+1 v f + 2µ ε(ũ n+1 ) : ε(v f Ω f,n+1 Ω f,n+1 1 ) = 0, vf 1 V f (t n+1 ), ũ n+1 Γ w,n+1 = wn+1 Γ w,n+1.

120 Semi-Implicit Coupling Algorithm: Explicit Part Update fluid domain (mesh) displacement and velocity: d f,n+1 = Ext(d s,n Γ w 0 ), wn+1 = df,n+1 d f,n, Ω f,n+1 = (I + d f,n+1 )(Ω f 0 ), Fluid: Elliptic step in the new domain Ω f,n+1 : 1 ρ f ũ n+1 v f Ω f,n ρ f u n v f Ω f,n 1 + ρ f (div w n+1 )ũ n+1 v f Ω f,n ρ f (w n+1 ũ n+1 ) ũ n+1 v f + 2µ ε(ũ n+1 ) : ε(v f Ω f,n+1 Ω f,n+1 1 ) = 0, vf 1 V f (t n+1 ), ũ n+1 Γ w,n+1 = wn+1 Γ w,n+1. Observations Geometrical non-linearities, viscous and convective effects are out of inner iterations

121 Semi-Implicit Coupling Algorithm: Implicit Part Fluid: Projection step in Ω f,n+1 (known domain) 8 >< ρ f un+1 ũ n+1 + p n+1 = 0, in Ω f,n+1 >: u n+1 = 0, in Ω f,n+1, Solid: u n+1 n = ds,n+1 d s,n n, on Γ w,n+1. 8 >< ρ s ds,n+1 2d s,n + d s,n 1 () 2 div 0 `F (d s,n+1 )S(d s,n+1 ) = 0, in Ω s 0, >: F (d s,n+1 )S(d s,n+1 )n 0 = J(d f,n+1 )σ(ũ n+1, p n+1 )F (d f,n+1 ) T n 0, on Γ w 0.

122 Semi-Implicit Coupling Algorithm: Implicit Part Fluid: Projection step in Ω f,n+1 (known domain) 1 ρ f u n+1 v f Ω f,n ρ f ũ n+1 v f Ω f,n+1 2 p n+1 div v f Ω f,n q div u n+1 = 0, Ω f,n+1 v f 2 H Γ w,n+1(div; Ωf,n+1 ), q L 2 (Ω f,n+1 ), Solid: u n+1 n = ds,n+1 d s,n n, on Γ w,n+1. 8 >< ρ s ds,n+1 2d s,n + d s,n 1 () 2 div 0 `F (d s,n+1 )S(d s,n+1 ) = 0, in Ω s 0, >: F (d s,n+1 )S(d s,n+1 )n 0 = J(d f,n+1 )σ(ũ n+1, p n+1 )F (d f,n+1 ) T n 0, on Γ w 0.

123 Semi-Implicit Coupling Algorithm: Implicit Part Fluid: Projection step in Ω f,n+1 (known domain) 1 ρ f u n+1 v f Ω f,n ρ f ũ n+1 v f Ω f,n+1 2 p n+1 div v f Ω f,n q div u n+1 = 0, Ω f,n+1 Solid: Ω s 0 v f 2 H Γ w,n+1(div; Ωf,n+1 ), q L 2 (Ω f,n+1 ), u n+1 n = ds,n+1 d s,n n, on Γ w,n+1. ρ s ds,n+1 2d s,n + d s,n 1 () 2 v s + F (d s,n+1 )S(d s,n+1 ) : 0 d s Ω s 0 = p n+1 v s n 2µ ε(ũ n+1 )n v s, v s V s. Γ w,n+1 Γ w,n+1

124 Semi-Implicit Coupling Algorithm: Implicit Part Fluid: Projection step in Ω f,n+1 (known domain) 1 ρ f u n+1 v f Ω f,n ρ f ũ n+1 v f Ω f,n+1 2 p n+1 div v f Ω f,n q div u n+1 = 0, Ω f,n+1 Solid: Ω s 0 v f 2 H Γ w,n+1(div; Ωf,n+1 ), q L 2 (Ω f,n+1 ), u n+1 n = ds,n+1 d s,n n, on Γ w,n+1. ρ s ds,n+1 2d s,n + d s,n 1 () 2 v s + F (d s,n+1 )S(d s,n+1 ) : 0 d s Ω s 0 = R f p`pn+1 ; L 2 (v s n ) Γ w R f u`ũn+1, d f,n+1 ; L 0 1 (v s ) Γ {z } w, v s V s. 0 {z } Projection residual Elliptic residual

125 Semi-Implicit Coupling Algorithm: Implicit Part Fluid: Projection step in Ω f,n+1 (known domain) 1 ρ f u n+1 v f Ω f,n ρ f ũ n+1 v f Ω f,n+1 2 p n+1 div v f Ω f,n q div u n+1 = 0, Ω f,n+1 Solid: Ω s 0 v f 2 H Γ w,n+1(div; Ωf,n+1 ), q L 2 (Ω f,n+1 ), u n+1 n = ds,n+1 d s,n n, on Γ w,n+1. ρ s ds,n+1 2d s,n + d s,n 1 () 2 v s + F (d s,n+1 )S(d s,n+1 ) : 0 d s Ω s 0 = R f p`pn+1 ; L 2 (v s n ) Γ w R f u`ũn+1, d f,n+1 ; L 0 1 (v s ) Γ {z } w, v s V s. 0 {z } Projection residual Elliptic residual Observations Projection sub-step in a fixed fluid domain (fixed matrix) Linear coupled problem if linear solid This coupled problem requires cheap inner iterations

126 Semi-Implicit Coupling: A Priori Energy Estimate (I) Proposition Assume that ũ n+1 = 0 on Γ in out, the structure is an hyper-elastic material and» 1 q q = div w n+1 q, Ω f,n+1 Ω f,n Ω f,n+1 Then, the following energy inequality holds " 1 ρ f Ωf,n Ω f,n+1 2 un+1 2 ρ f 2 un 2 + " + 1 W (E(d s,n+1 )) Ω s 0 Ω s 0 Ω s 0 ρ s 2 d s,n+1 d s,n 2 # W (E(d s,n )) q with q t x 0 = 0. ρ s Ω s µ ε(ũ n+1 ) 2 Ω f,n+1 d s,n d s,n 1 2 # R f `ũn+1 u, d f,n+1 ; L 2 (ũ n+1 ds,n+1 d s,n ) 0

127 Semi-Implicit Coupling: A Priori Energy Estimate (I) Proposition Assume that ũ n+1 = 0 on Γ in out, the structure is an hyper-elastic material and» 1 q q = div w n+1 q, Ω f,n+1 Ω f,n Ω f,n+1 Then, the following energy inequality holds " 1 ρ f Ωf,n Ω f,n+1 2 un+1 2 ρ f 2 un 2 + " + 1 W (E(d s,n+1 )) Ω s 0 Ω s 0 Ω s 0 ρ s 2 d s,n+1 d s,n 2 # W (E(d s,n )) q with q t x 0 = 0. ρ s Ω s µ ε(ũ n+1 ) 2 Ω f,n+1 d s,n d s,n 1 2 # R f `ũn+1 u, d f,n+1 ; L 1 (ũ n+1 ds,n+1 d s,n ) 0 {z } Γw,n+1 2µ ε(ũ n+1 )n (ũ n+1 ds,n+1 d s,n )

128 Semi-Implicit Coupling: A Priori Energy Estimate (I) Proposition Assume that ũ n+1 = 0 on Γ in out, the structure is an hyper-elastic material and» 1 q q = div w n+1 q, Ω f,n+1 Ω f,n Ω f,n+1 Then, the following energy inequality holds q with q t x 0 = 0. " 1 ρ f Ωf,n Ω f,n+1 2 un+1 2 ρ f 2 un 2 ρ s + d s,n+1 d s,n 2 ρ s Ω s 2 d s,n d s,n 1 2 # 0 Ω s 2 0 " + 1 # W (E(d s,n+1 )) W (E(d s,n )) + 2µ ε(ũ n+1 ) 2 Ω s 0 Ω s 0 Ω f,n+1 j artificial R f `ũn+1 viscous power u, d f,n+1 ; L 1 (ũ n+1 ds,n+1 d s,n ) 0 {z } Γw,n+1 2µ ε(ũ n+1 )n (ũ n+1 ds,n+1 d s,n )

129 Semi-Implicit Coupling: A Priori Energy Estimate (II) Idea of the proof. We take appropriate test functions and then we add the fluid and the solid variational formulations. In particular, for the solid: for the fluid (explicit part): for the fluid (implicit part): v f 1 = ũn+1 L 1 (v s Γ w 0 v s = ds,n+1 d s,n ) L 1 (ũ n+1 Γ w 0 V s. v Γ s w ) V f (t n+1 ). 0 v f 2 = un+1 L 2 (v s n Γ w 0 ) H Γ w,n+1(div; Ω f,n+1 ). Remarks: 1 v f 1 Γ w 0 = 0 and hence v f V f 2 v f 2 n Γ w 0 = 0 and hence v f 2 H Γ w,n+1(div; Ωf,n+1 ) 3 Let us recall that ũ n+1 Γ w 0 v s Γ w 0