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Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics Nikolaj Takata Mücke Technical University of Munich 19/1 - 2018 Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 1 / 60

Overview Introduction 1 Infinite Domain Problems Can We Force Boundary Conditions? Acoustics 2 Linear vs Non-Linear Absorbing Boundary Conditions 3 The Approach Finite Element Approximation Perfectly Matched Layer 4 The Approach ABC vs. PML 5 Non-Linear Absorbing Boundary Condition 6 Pseudo-Differential Operators Linearisation Pseudo-Differential Representation Absorbing Boundary Conditions Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 2 / 60

Introduction Infinite Domain Problems PDE Formulation The general problem can be formulated as ∂ 2 u ∂ 2 u ∂ t 2 − 1 ∂ x 2 = f ( x , t ) , x ∈ Ω , t ∈ [0 , T ] , c 2 u ( x , 0) = g ( x ) Note: NO boundary conditions! Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 3 / 60

Introduction Infinite Domain Problems PDE Formulation - Problems! ∂ 2 u ∂ 2 u ∂ t 2 − 1 ∂ x 2 = f ( x , t ) , x ∈ Ω , t ∈ [0 , T ] , c 2 u ( x , 0) = g ( x ) Physical Reality vs. Mathematics Well-posed? Does solution make sense? Computational Simulations Computation time Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 4 / 60

Introduction Can We Force Boundary Conditions? Can We Force Boundary Conditions? To make the problem well-posed we need to truncate Ω and add boundary conditions! ∂ 2 u ∂ 2 u ∂ t 2 − 1 ∂ x 2 = f ( x , t ) , x ∈ Ω , t ∈ [0 , T ] , c 2 u ( x , 0) = g ( x ) u ( x , t ) =? , x ∈ ∂ Ω ∂ ∂ nu ( x , t ) =? , x ∈ ∂ Ω Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 5 / 60

Introduction Can We Force Boundary Conditions? Dirichlet or Neumann BC? Neumann Dirichlet ∂ ∂ nu ( x , t ) = 0 , x ∈ ∂ Ω u ( x , t ) = 0 , x ∈ ∂ Ω Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 6 / 60

Introduction Can We Force Boundary Conditions? But Then What?! Absorbing Boundary condition Truncate domain and add BC Perfectly Matched Layer Analytical continuation of the solution Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 7 / 60

Acoustics Acoustics What is sound? Compression and expansion of a fluid How do we express sound propagation mathematically? Linear: The Linear Wave Equation Non-Linear: Navier-Stokes Equation (and variations thereof) Figure 1: [ ? ] Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 8 / 60

Acoustics Linear vs Non-Linear Linear vs Non-Linear Linear Wave Equation ∂ 2 u ∂ 2 u ∂ t 2 − 1 ∂ x 2 = 0 , c 2 Westervelt Equation � � 2 c − 2 ∂ 2 u ∂ t 2 − ∂ 2 u ∂ x 2 − β ∂ 3 u ∂ 2 u ∂ t ∂ x 2 = γ ∂ t 2 Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 9 / 60

Acoustics Linear vs Non-Linear Linear vs Non-Linear Figure 2: [ ? ] Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 10 / 60

Absorbing Boundary Conditions Absorbing Boundary Conditions Absorbing boundary conditions (ABC) works by truncating the infinite domain to a suitable size and impose artificial boundary conditions that let waves pass through . Figure 3: [ ? ] Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 11 / 60

Absorbing Boundary Conditions The Wave Equation We consider the one-dimensionless case. ∂ 2 u ∂ 2 u ∂ t 2 − 1 ∂ x 2 = 0 , x > 0 , t > 0 (1a) c 2 u ( x , 0) = U ( x ) (1b) ∂ ∂ t u ( x , 0) = V ( x ) (1c) u (0 , t ) = 0 , (1d) This formulation leads to solutions with the following properties: Reflects at the left boundary, x = 0 Propagates to the right towards infinity Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 12 / 60

Absorbing Boundary Conditions But... We need to restrict the domain to the right by adding a boundary. Hence, we need another boundary condition: B ( u ) = b , x = L , t > 0 The goal is now to find an operator B and a function b such that: Waves don’t reflect The solution in the desired domain is unaffected The problem stays well-posed We can implement it on a computer (it has to be computationally efficient) Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 13 / 60

Absorbing Boundary Conditions The Approach The Approach We define two new functions v = ∂ u ∂ t + 1 ∂ u w = ∂ u ∂ t − 1 ∂ u (2) ∂ x , c c ∂ x These two functions satisfy � ∂ u � � ∂ u � ∂ t − 1 ∂ t + 1 − 1 ∂ t + 1 ∂ v ∂ v ∂ x = ∂ ∂ u ∂ ∂ u ∂ t ∂ x ∂ x ∂ x c c c c = ∂ 2 u ∂ 2 u ∂ 2 u ∂ 2 u ∂ t 2 + 1 ∂ t ∂ x − 1 ∂ x ∂ t − 1 c 2 ∂ x 2 c c = ∂ 2 u ∂ 2 u ∂ t 2 − 1 ← − Original PDE c 2 ∂ x 2 = 0 And similarly ∂ w ∂ t + 1 ∂ w ∂ x = 0 c Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 14 / 60

Absorbing Boundary Conditions The Approach ∂ v ∂ t − 1 ∂ v ∂ w ∂ t + 1 ∂ w ∂ x = 0 , ∂ x = 0 c c Thus, we can rewrite (1a) as a system � � � � � � ∂ + 1 − 1 0 ∂ v v = 0 . (3) w 0 1 w ∂ t c ∂ x (3) has the general solution � � � � x + 1 x − 1 v ( x , t ) = φ c t , w ( x , t ) = ψ c t , (4) for some functions φ and ψ depending on the initial condition. Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 15 / 60

Absorbing Boundary Conditions The Approach v ( x , t ) = φ ( x + ct ) , w ( x , t ) = ψ ( x − ct ) . This solutions shows us that the solution of the wave equation consists of some waves propagating left and some waves propagating left. In this case: v ( x , t ) represents the waves propagating left . w ( x , t ) represents the waves propagating right . Main point: On the boundary, x = L , we only want waves propagating to the right! Hence, we want to make sure that v vanishes on the boundary. Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 16 / 60

Absorbing Boundary Conditions The Approach Main Point Main point: On the boundary, x = L , we only want waves propagating to the right! Hence, we want to make sure that v vanishes on the boundary: v ( x , t ) = 0 , x = L , t > 0 (5) which means B ( u ) = ∂ u ∂ t + 1 ∂ u ∂ x = 0 , x = L , t > 0 (6) c is the boundary conditions we were seeking! Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 17 / 60

Absorbing Boundary Conditions The Approach The Full Problem ∂ 2 u ∂ t 2 − ∂ 2 u ∂ x 2 = 0 , x > 0 , t > 0 (7a) u ( x , 0) = U ( x ) (7b) ∂ ∂ t u ( x , 0) = V ( x ) (7c) u (0 , t ) = 0 , (7d) ∂ t + 1 ∂ u ∂ u ∂ x = 0 , x = L , t > 0 (7e) c Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 18 / 60

Absorbing Boundary Conditions Finite Element Approximation Finite Element Approximation We want to be able to impose this boundary condition in our numerical scheme Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 19 / 60

Absorbing Boundary Conditions Finite Element Approximation Weak Formulation v is a test function. Multiplying and integrating (1a) gives: �� � � � � � u tt v dx − 1 u tt v dx + 1 ∂ u ∆ uv dx = ∇ u ∇ v dx − ∂ nv dx c 2 c 2 Ω Ω Ω Ω ∂ Ω We use the solution anzats ∞ ∞ � � u = N i u i ( t ) , v ( x , t ) = N j v j ( t ) , (8) i =0 j =0 with N i as our element basis. Consequently ∞ � u tt = N i ¨ u i ( t ) i =0 Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 20 / 60