An arbitrary lagrangian eulerian discontinuous galerkin approach to fluid-structure interaction and its application to cardiovascular problem

Transcription

1 An arbitrary lagrangian eulerian discontinuous galerkin approach to fluid-structure interaction and its application to cardiovascular problem Yifan Wang University of Houston, Department of Mathematics Joint work with: S. Canic and A. Quaini Acknowledgements: NSF DMS , DMS

2 Motivation Outline Motivation for Applications in Hemodynamics

3 Motivation Main motivation We are interested in study the blood flow inside the Abdominal Aortic Aneurysm with stent-graft treated. Provide new information and give insight in new treatment Serve as a potential predictor of AAA arterial wall rupture Why Discontinuous Galerkin? High order and less dissipative solution Ideal for problems contain discontinuous solution

4 Fluid model Thin structure model The coupling conditions Fluid model We treat blood as an incompressible, viscous Newtonian fluid, and its motion is described by the Navier-Stokes equations. tu + (u )u = 1 ρ ( σ) = 1 p + ν u ρ u = 0 in Ω(t), in Ω(t), where u is the fluid velocity, σ is the Cauchy stress tensor, ρ is the fluid density, and ν is the fluid kinematic viscosity. The Cauchy stress tensor is given by σ = pi + 2µD(u), where p is the pressure and D(u) = ( u + ( u) T )/2 is the strain rate tensor.

5 Fluid model Thin structure model The coupling conditions Thin structure model The elastodynamics of the thin structure will be described in general by ρ sh s ttη + Lη + γv tη = f s in Γ 0, Operator L is a linear operator acting on vector function η describing the elastic properties of the structure, and V is a linear operator acting on vector function tη describing the viscoelastic properties of the structure. Γ 0 denote the fixed reference structure domain, ρ s is the structure density, h s is the structure thickness, γ is a viscoelastic parameter, and f s is the force acting on the structure. We consider two different forms of operators L and V.

6 Fluid model Thin structure model The coupling conditions Thin structure model - I We consider the generalized string model for an elastic tube with the reference configuration as a cylinder of radius R 0 and length L (Quarteroni 2000). And assuming that the longitudinal displacement η x and the azimuthal displacement η z are negligible. ( 2 η y Lη = 0, kgh s x 2 + Ehs (1 νs 2 ) η y R 2 0 ) T, 0, V η ( ) T t = 0, 3 η y x 2 t, 0, where k is the Timoshenko shear correction factor, G is the shear modulus, E is the Young s modulus and ν s is the Poisson ratio of the structure. Structure model is endowed with absorbing boundary conditions of first order: η y t η y t + kg η y x ρ s kg η y ρ s x = 0 at x = 0, = 0 at x = L.

7 Fluid model Thin structure model The coupling conditions Thin structure model - II The second model we consider involves no spatial derivatives and no viscoelasticity. Operators L and V are given by: Lη = (C x η x, C y η y, 0) T, V = (0, 0, 0) T, where C x and C y are spring constants. Notice that the displacement components are independent. The structure model is supplemented with homogeneous Dirichlet boundary conditions: η = 0 on Γ 0.

8 Fluid model Thin structure model The coupling conditions The coupling conditions The fluid and the structure are coupled via two boundary conditions: - Kinematic coupling condition: it describes the continuity of velocity at the fluid-structure interface (no-slip condition) + u A t = tη on Γ 0 (0, T ); - Dynamic coupling condition: it describes the continuity of the stress at the fluid-structure interface + J σn Γ(t) = f s on Γ 0 (0, T ), where J denotes the Jacobian of the transformation from Eulerian to Lagrangian coordinates, and σn Γ(t) denotes the normal fluid stress at the deformed fluid-structure interface, evaluated with respect to the reference configuration. Vector n is the outward normal to the deformed fluid domain.

9 Fluid model Thin structure model The coupling conditions The coupling conditions Coupling conditions above can be written in the equivalent form: α f u A t J σn Γ(t) = α f tη f s on Γ 0 (0, T ), α su A t + J σn Γ(t) = α s tη + f s on Γ 0 (0, T ), where α f > 0 and α s > 0 (α f α s) are constants.

10 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Partitioned methods for the FSI problem The FSI problem will be solved using two different partitioned strategies based on Domain Decomposition methods: the Dirichlet-Neumann (DN) and the Robin-Neumann (RN) algorithms. (A. Quarteroni and A. Valli 1999) The Dirichlet-Neumann: Dirichlet boundary conditions are imposed at the interface Γ(t) for the fluid problem, whereas Neumann boundary conditions are supplemented for the structure problem. The DN algorithm iterates over these two sub-problems until meeting the convergence criteria for the interface displacement. When the density of structure has the same magnitude as fluid density, relaxation step on the structure displacement is required (Nobile 2001). The Robin-Neumann: When the FSI problem contains a closure structure, the Dirichlet-Neumann algorithms fails due to the fact that the coupling conditions are not implemented synchronously. Robin-Neumann algorithm always converges without any relaxation and its convergence is almost insensitive to the added-mass effect.

11 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise We consider a smooth mapping: A t : Ω 0 Ω(t), A t : x 0 x, where x and x 0 are the coordinates in the physical domain Ω(t) and the reference domain Ω 0, respectively. The domain velocity w is given by w(t, ) = dat dt (t, At(t, ) 1 ). The ALE time derivative of the fluid velocity is defined as: tu x0 = D tu(t, A t(x 0 )) = tu(t, x) + w(t, x) u(t, x), for x = A t(x 0 ), x 0 Ω 0.

12 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise The incompressible Navier-Stokes equations in the ALE formulation: tu x0 + (u w) u = 1 p + ν u ρ u = 0 in Ω(t), in Ω(t),

13 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise We introduce a notation of F stands for the nonlinear flux vector. Given the velocity components u = (u, v) T for d = 2 and u = (u, v, w) T for d = 3, F is defined as: F = [ ] u 2 uv vu v 2 if d = 2, or u 2 uv uw vu v 2 vw if d = 3. wu wv w 2 Let J At = det( x x 0 ) be the Jacobian of the deformation gradient: tj At x0 = J At w. We can derive the fully conservative form of the NS equation in ALE: t(j At u) + J At (F wu) = J A t x0 ρ p + J A t ν u.

14 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Harmonic extension Describing the ALE map in terms of the fluid domain displacement d(x 0, t) = A t(x 0 ) x 0. Let η be the displacement of the deformable boundary Γ(t). Inside Ω 0 the displacement d is arbitrary: it can be any reasonable extension of η over Ω 0. A classical choice is to consider a harmonic extension in the reference domain, that is: with boundary conditions: d = 0, in Ω(t), d = η d = 0 on Γ(t), on Ω(t)/Γ(t). The fluid domain velocity w is obtained by differentiation: w = td.

15 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Temporal discretization of the fluid problem The second order semi-implicit splitting scheme is used to discretize NS equation (Karniadakis 1991). The NS equation in the reference domain Ω 0 is splitted to three sub-equations. Step 1: Using the Backward Difference Formula (BDF2) to obtain the first intermediate velocity field J n u k+ 1 by the explicit approximation of the convective 3 component (F wu): (3J n u k+ 1 4J n u n + J n 1 u n 1 ) 3 = 2J n (F wu) n + J n 1 (F wu) n 1. 2 t Step 2: Known the J n u k+ 1 3, we solve for the second intermediate velocity field J n u k+ 2 3 by including the pressure gradient term: J n u k+ 2 J n u 3 3 k t = 1 ρ Jn p k+1. We impose the divergence free requirement to J n u k+ 2 3, (J n International Workshop on Fluid-Structure Interaction Problemsu k+ 2 IMS ) = National 0. University of Singapore 30 May - 3 June 2016

16 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Temporal discretization of the fluid problem Taking the divergence of both sides of the equation: 1 ρ Jn p k+1 = 3 2 t (Jn u k+ 1 ), 3 and solve for the pressure J n p k+1. After obtaining the pressure field J n p k+1, the second intermediate velocity field J n u k+ 2 is updated from: 3 J n u k+ 2 3 = J n u k t 3ρ Jn p k+1. Step 3: Known the second intermediate velocity field J n u k+ 2 3, we implicitly integrate the viscous component ( u) to solve a Helmholtz problem for J n u k+1 : J n u k+1 J n u k t = νj n u k+1.

17 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Discontinuous Galerkin method Let us consider the local polynomial approximation of fluid velocity component of u(r, s): N N u(r, s) = û i Φ i (r, s) = u j l j (r, s), i=1 j=1 Here, we are assuming that the number of nodes and number of modes are the same, and are equal to N. û i is the expansion coefficient associating with the orthonormal modal base Φ i (r, s), while u j is the value of u(r, s) at grid point of (r, s) j and l j (r, s) is the corresponding nodal base. In the 2D case, we have N = (n + 1)(n + 2)/2 which represents the least number of terms required to have the completeness of expansion using the n th order polynomial.

18 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Discontinuous Galerkin method Modal bases: A set of orthonormal polynomial modes defined in standard isosceles right triangle Ω st (Gottlieb 1977, Dubiner 1991,Hesthaven 2008): Φ i (r, s) = 2P 0,0 k (2 1 + r 1 s 1)P(2k+1,0) l (s)(1 s) k, in which r and s are the coordinates of two short sides of an isosceles right triangle. Subscript i stands for the i th mode of the basis functions, and i = (N + 1)k + l + 1 k (k 1), (k, l) 0; k + l N. P(α,β) 2 k (x) is Jacobi polynomials. Nodal bases: Lagrangian polynomials built on Fekete points (GKarniadakis 2005)

19 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Discontinuous Galerkin method Nodal bases are expressed in terms of modal bases to simplify the evaluation of the numerical integration. And the generalized Vandermonde matrix V works as a role to connect the nodal bases with modal bases, and it is defined as: V ij = Φ j (r, s) i. Between modal Φ j (r, s) and nodal l j (r, s) bases, one can be transformed from the other via Vandermonde matrix: l(r, s) = (V T ) 1 Φ(r, s). In other words, nodal basis l i (r, s), it can be expressed as: N l i (r, s) = (V T ) 1 ik Φ k(r, s). k=1

20 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Discontinuous Galerkin method In an standard isosceles triangle, the local mass matrix is: M ij = (VV T ) 1 ij. And the local matrices for convective term are: where D r and D s can be computed from: S rij = (MD r ) ij, S sij = (MD s) ij. D r = V r V 1, D s = V sv 1. And V r and V s denote two Vandermonde matrices as below: V rij = dφ j (r, s) dr (r,s)i, V sij = dφ j (r, s) ds (r,s)i.

21 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Spatial discretization of the fluid problem Step 1: For the convective equation, we have: 3J n u k+ 1 4J n u n + J n 1 u n 1 M 3 e 2 t = 2J n S : (F wu) n + J n 1 S : (F wu) n 1 2J n n ((F wu) (F wu) ) n l(x, y)dxdy E + J n 1 n ((F wu) (F wu) ) n 1 l(x, y)dxdy. E Local Lax-Friedrichs flux is used and denoted by (F wu) : (F wu) = (F wu)+ + (F wu) + τ(u + u ), 2 in which the coefficient τ is the local maximum velocity magnitude.

22 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Spatial discretization of the fluid problem Step 2: For the pressure possion equation, we convert it into a system of first order equations (Cockburn and Shu 1998): J n ρ p k+1 = q q = 3 2 t (Jn u k+ 1 3 ) In each triangle E, we discretize (J n p k+1, q) as ((J n p k+1 ) e, q x, q y ) and approximate it by the n th order basis function l(x, y) constructed on that triangle. Internal penalty fluxes as below are used: q = Jn ( (pk+1 ) + e + (p k+1) e ρ 2 p k+1 = p+ k+1 + p k+1 2 where the τ is penalty parameter. τ [ n (p k+1 ) e. + n+ (p k+1 ) + ] ) e,

23 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Spatial discretization of the fluid problem Then we have: J n ( (S x ) ep k+1 ρ E J n ( (S y ) ep k+1 ρ E (S x ) eq x n (q x q )l(x, y)dxdy E + (S y ) eq y E ( n x pk+1 p ) ) k+1 l(x, y)dxdy = M eq x, ( n y pk+1 p ) ) k+1 l(x, y)dxdy = M eq y. n (q y q )l(x, y)dxdy = 3 2 t S (Jn u k+ 1 3 ).

24 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Temporal discretization of the structure problem Consider the general string model for the structure, we introduce a new variable of U: U = ρ sh s η y t γ 2 η y x 2, which combines the temporal derivative terms. Such that the problem can be rewritten as: ( ρ sh s η y t t γ 2 η y x 2 ) = U t = 2 η y kghs x 2 Ehs (1 ν 2 ) η y R J σn Γ(t). We also introduce new variables W denote the second order spatial derivative term 2 η y x 2.

25 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Temporal discretization of the structure problem As a result, the original problem is transformed into a system of ODEs: η y ρ sh s = U + γw, t U t = kghsw Ehs η y (1 ν 2 ) R0 2 + J σn Γ(t). The initial Conditions of η and U can be reasonably obtained as following: η y t=0 = 0, η y t t=0 = 0, U t=0 = ρ sh s η y t t=0 γw t=0 = 0.

26 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Temporal discretization of the structure problem The above system can be discretized by using forward difference scheme. Given η n y, U n, W n and σ k+1, find the value of (η y ) k+1, U k+1 and W k+1, so that: (η y ) k+1 ηy n ρ sh s = U k+1 + γw n, t U k+1 U n = kgh sw n Ehs ηy n t 1 ν 2 R0 2 + J n σ k+1 n Γ(t n ), W k+1 = 2 (η y ) k+1 x 2.

27 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Spatial discretization of the structure problem We introduce an intermediate variable q to split W into a system of two first order differential equations which can later be easily solved by Local Discontinuous Galerkin (LDG) (Cockburn and Shu 1998): q = ηy x W = q x By choosing the fluxes as q = q and η y = η + y, we can write out the matrix form of eq. (1) as below: M eq = (S x ) eη y n x (η y η y )l(x, y)dxdy, E M ew = (S x ) eq n x (q q )l(x, y)dxdy; E.

28 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Element-wise The internal penalty flux also is used in solving the displacement equation in the mapping, our approach does not force two neighbor triangles to have the same displacement at the common edge but limit the differences between them and disallow large gaps in the displacement. We reconstruct harmonic extension equation as a system of two first order equations for each component of displacement vector: { d = q q = 0 where d corresponds to each component of the displacement vector. We then replace q with the following internal penalty fluxes: q = d+ e + d e 2 τ[n d + n + d + ], d = d+ e + d e 2 Here, d correspond to the vector of each component of displacement. Parameter τ is the penalty parameter..

29 Partitioned methods for the FSI problem Harmonic extension Discretization of the fluid problem Discretization of the structure problem Element-wise Element-wise Then the martix form are: M eq x = (S x ) ed n x (d d )l(x, y)dxdy, E M eq y = (S y ) ed n y (d d )l(x, y)dxdy; E (S x ) eq x + (S y ) eq y n (q x q )l(x, y)dxdy E n (q y q )l(x, y)dxdy = 0. E

30 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation Benchmark test 1: Compliant vessel Inlet Γ Γ Outlet (a) Computational domain (b) Inlet data over time (c) Mesh Figure: Test 1: (a) computational domain at the initial time, (b) the inlet pressure pulse, and (c) unstructured triangle mesh (fine mesh).

31 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation Benchmark test 1: Compliant vessel (a) t = 2 ms (b) t = 4 ms (c) t = 6 ms (d) t = 8 ms (e) t = 10 ms (f) t = 12 ms Figure: Test 1: pressure at different times: (a) t = 2 ms, (b) t = 4 ms, (c) t = 6 ms, (d) t = 8 ms, (e) t = 10 ms, (f) t = 12 ms. The unit for the pressure is dynes/cm 2.

32 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation Benchmark test 1: Compliant vessel (a) t = 2 ms (b) t = 4 ms (c) t = 6 ms (d) t = 8 ms (e) t = 10 ms (f) t = 12 ms Figure: Test 1: pressure along the centerline at different times: (a) t = 2 ms, (b) t = 4 ms, (c) t = 6 ms, (d) t = 8 ms, (e) t = 10 ms, (f) t = 12 ms. In the legend, A. Quaini denotes the results in (Quaini 2009), DG-IP HI and DG-IP LO denote the results obtained with our DG solver with IP fluxes with the fine and coarse mesh, respectively.

33 Introduction Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation Benchmark test 2: Oscillating immersed linearly elastic 1D membrane (a) coarse mesh (b) fine mesh Figure: Test 3: (a) coarse mesh and (b) fine mesh. International Workshop on Fluid-Structure Interaction Problems IMS National University of Singapore 30 May - 3 June 2016

34 Introduction Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation Benchmark test 2: Oscillating immersed linearly elastic 1D membrane (a) t = 0.1 s (b) t = 0.5 s (c) t = 1 s (e) t = 2 s (f) t = 8 s (d) t = 1.5 s Figure: Test 3: pressure contour and velocity vector field at time (a) t = 0.1 s, (b) t = 0.5 s, (c) t = 1 s, (d) t = 1.5 s, (e) t = 2 s, (f) t = 8 s. International Workshop on Fluid-Structure Interaction Problems IMS National University of Singapore 30 May - 3 June 2016

35 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation Benchmark test 2: Oscillating immersed linearly elastic 1D membrane (a) maximum x-coordinate over time (b) difference Figure: Test 3: (a) maximum x-coordinate over time, (b) difference in absolute value between the curves in (a).

36 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation Section of an Abdominal Aortic Aneurysm We consider a section of a patient-specific descending aorta affected by AAA with compliant wall. Inlet Triple point 4.03 cm Γ stent 11.2 cm Γ Outlet Outlet (a) Geometry (b) Mesh Figure: (a) Geometry of the section of aorta affected by AAA under consideration and (b) mesh used for the computations.

37 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation (a) Inflow Velocity (b) Outlet Pressure Figure: Rescaled (a) experimental average velocity (cm 2 /s) at the inlet section and (b) experimental pressure (Pa) at the outlet section.

38 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation (a) Pressure (b) Velocity magnitude (c) Displacement magnitude

39 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation (d) Pressure (e) Velocity magnitude (f) Displacement magnitude Figure: (a) and (d) Pressure, (b) and (e) velocity magnitude and vector field, (c) and (f) displacement magnitude on a section of aorta affected by AAA without (top) and with (bottom) stent-graft at time t = 2.3 s.

40 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation (a) Pressure (b) Velocity (c) Displacement

41 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation (d) Pressure (e) Velocity (f) Displacement Figure: (a) and (d) Pressure, (b) and (e) velocity magnitude and vector field, (c) and (f) displacement magnitude on a section of aorta affected by AAA without (top) and with (bottom) stent-graft at time t = 2.85 s.

42 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation Figure: Maximum displacement magnitude over time (3 cycles): blue line corresponds the case without stent-graft, red line corresponds the case with the stent-graft.

43 Benchmark test 1 Benchmark test 2 Section of an Abdominal Aortic Aneurysm Preliminary 3D simulation Preliminary 3D simulation (a) Inlet data over time Figure: The inlet pressure pulse. Pressure Propagation Vortex

44 We proposed an DG ALE approach for FSI problem. Benefit from allowing an element-wise mapping, the implementation of DG in ALE frame becomes straight forward. For certain FSI which contains discontinuity solution our DG-ALE approach will be a feasible approach. Our solver is a serial version only at the moment, parallelization is desired.

45 THANK YOU FOR YOUR ATTENTION!