Automatic differentiation of a fluid-structure interaction problem - PowerPoint PPT Presentation

Automatic differentiation of a fluid-structure interaction problem Gabriel Balaban, Anders Logg, Marie E. Rognes Simula Research Laboratory University of Oslo FEniCS Worksop 2013, Cambridge 2013–03–18 March 20, 2013

Examples of Fluid–structure interaction Modelling Challenges ◮ model must integrate solid and fluid mechanics ◮ fluid geometry depends on structure deformation

Solving the FSI problem Issues: ◮ Continuum mechanics formulation ◮ Partitioned vs monolithic ◮ Fixed-pointed vs Newton ◮ Approximation of the Jacobian In this work we: ◮ derive a Newton’s method with exact Jacobians for FSI problems using the arbitrary Lagrangian Eulerian formulation ◮ implement a monolithic solver in Python (using FEniCS) ◮ investigate various optimizations and simplifications

Setup of the FSI problem Ω S D S t = 0 Ω F U F ω S ( t ) d S u F ω F ( t )

Mismatch in standard fluid and solid models ρ S ¨ D S − Div Σ S ( D S ) = B S ρ F (˙ u F + grad u F · u F ) − div σ F = b F div u F = 0

Mesh smoothing problem Mesh equation ˙ D M − Div Σ M ( D M ) = 0

Arbitrary Lagrangian-Eulerian framework u ( x ( X , t ) , t ) ALE time derivative: d ρ u + ρ ( grad u · ( u − ˙ dt ( ρ u ) = ˙ D M )) ALE fluid equation: u F + grad u F · ( u F − ˙ ρ F (˙ D M )) − div σ F ( u F , p F ) = b F in ω F ( t ) div u F = 0 in ω F ( t )

Interface conditions Γ FS γ FS ( t ) ◮ Stress continuity: σ S · n = σ F · n on γ FS ( t ) ◮ Kinematic continuity: u F = u S on γ FS ( t ) ◮ Domain continuity: d M = d S on γ FS ( t )

Linearization of the FSI problem Two challenges: ◮ Derivative of fluid equation with respect to geometry? d dt ( ρ u ) − div σ F ( u F , p F ) = b F in ω F ( t ) div u F = 0 in ω F ( t ) ◮ Linearization of essential BCs u F = u S on γ FS ( t ) d M = d S on γ FS ( t )

The reference domain approach ◮ Map the fluid problem to the reference domain ◮ Use standard techniques to differentiate ◮ Straightforward but tedious ◮ Can be automated!

Navier–Stokes pulled back to reference domain ◮ Equation: ρ F J M ( ˙ · ( U F − ˙ U F + Grad U F · F − 1 D M )) − Div Σ F = B F M Div ( J M F − 1 · U F ) = 0 M ◮ Pulled-back fluid stress: � � µ F ( Grad U F · F − 1 + F −⊤ · Grad U ⊤ · F −⊤ Σ F = J M F ) − P F I M M M

Interface conditions: How to linearize? Γ FS γ FS ( t ) Stress continuity: Σ S · N = Σ F · N on Γ FS Kinematic continuity: U F = U S on Γ FS Domain continuity: D M = D S on Γ FS

Linearization of essential boundary conditions Introduce Lagrange multipliers ( τ F , τ M ) and corresponding trial functions ( χ F , χ M ) Kinematic continuity: U F − U S τ F Γ FS + χ F v F Γ FS Domain continuity: D M − D S τ M Γ FS + χ M v M Γ FS

The linearized FSI operator (the Jacobian)

FEniCS implementation J = derivative(R, U)

An analytic test problem pdf/pdf/analyticproblem.pdf t ex Primary variables: U F = y (1 − y ) sin t P F = 2 C sin t (1 − x − Cxy (1 − y )(1 − cos t )) D S = Cy (1 − y )(1 − cos t ) U S = Cy (1 − y ) sin t D M = Cxy (1 − y )(1 − cos t )

Convergence for analytic test problem -2 10 P F D S -3 10 D F U F -4 10 0.99 -5 10 L2 error -6 10 -7 10 -8 10 2.67 2.00 -9 10 -1 0 -2 10 10 10 mesh size hmin

A two-dimensional blood vessel

Break-down of run-time Problem Routine Calls Time (s) Analytic problem Jacobian assembly 28 83.9s 90% mesh size = 231 Linear solve 28 1.86s 2% time steps = 10 Residual assembly 38 0.915s 1% Blood vessel Jacobian assembly 343 2980s 81% mesh size = 1271 Linear solve 343 254s 18% time steps = 140 Residual assembly 483 64.1s 1%

Effect of Jacobian reuse Problem Routine Calls Analytic problem Jacobian assembly 1 (-27) -95 % mesh size = 231 Linear solve 51 (+23) -96% time steps = 10 Residual assembly 61 (+23) +54% Total Runtime: -93 % Blood vessel Jacobian assembly 25 (-308) -93 % mesh size = 1271 Linear solve 1287 (+944) -92 % time steps = 140 Residual assembly 1427 (+944) +192% Total -91 %

Effect of Jacobian reuse Newton Iterations per time step 35 30 25 Number of Iterations 20 15 10 5 0 0 10 20 30 40 50 60 70 time

Optimization summary Optimization Runtime Memory Robustness Jacobian Reuse Jacobian Simplification Jacobian Buffering Reduced Quadrature Order

Summary: How to use the automatic derivative to solve FSI problems ◮ Map the fluid equation to the reference domain ◮ Impose essential BC’s using Lagrange multipliers ◮ Let the automatic derivative compute the Jacobian Challenges / work in progress: ◮ Long FFC compilation times ◮ Preconditioning