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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

1

Introduction Infinite Domain Problems Can We Force Boundary Conditions?

2

Acoustics Linear vs Non-Linear

3

Absorbing Boundary Conditions The Approach Finite Element Approximation

4

Perfectly Matched Layer The Approach

5

ABC vs. PML

6

Non-Linear Absorbing Boundary Condition 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 ∂2u ∂t2 − 1 c2 ∂2u ∂x2 = f (x, t), x ∈ Ω, t ∈ [0, T], 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!

∂2u ∂t2 − 1 c2 ∂2u ∂x2 = f (x, t), x ∈ Ω, t ∈ [0, T], 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! ∂2u ∂t2 − 1 c2 ∂2u ∂x2 = f (x, t), x ∈ Ω, t ∈ [0, T], 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?

Dirichlet u(x, t) = 0, x ∈ ∂Ω Neumann ∂ ∂nu(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 ∂2u ∂t2 − 1 c2 ∂2u ∂x2 = 0, Westervelt Equation c−2 ∂2u ∂t2 − ∂2u ∂x2 − β ∂3u ∂t∂x2 = γ

• ∂2u

∂t2

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: [?]

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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. ∂2u ∂t2 − 1 c2 ∂2u ∂x2 = 0, x > 0, t > 0 (1a) 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 c ∂u ∂x , w = ∂u ∂t − 1 c ∂u ∂x (2) These two functions satisfy ∂v ∂t − 1 c ∂v ∂x = ∂ ∂t

∂u

∂t + 1 c ∂u ∂x

• − 1

c ∂ ∂x

∂u

∂t + 1 c ∂u ∂x

• = ∂2u

∂t2 + 1 c ∂2u ∂t∂x − 1 c ∂2u ∂x∂t − 1 c2 ∂2u ∂x2 = ∂2u ∂t2 − 1 c2 ∂2u ∂x2 ← − Original PDE = 0 And similarly ∂w ∂t + 1 c ∂w ∂x = 0

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 c ∂v ∂x = 0, ∂w ∂t + 1 c ∂w ∂x = 0 Thus, we can rewrite (1a) as a system ∂ ∂t

• v

w

• + 1

c

• −1

1

• ∂

∂x

• v

w

• = 0.

(3) (3) has the general solution v(x, t) = φ

• x + 1

c t

• ,

w(x, t) = ψ

• x − 1

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 c ∂u ∂x = 0, x = L, t > 0 (6) 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

∂2u ∂t2 − ∂2u ∂x2 = 0, x > 0, t > 0 (7a) u(x, 0) = U(x) (7b) ∂ ∂t u(x, 0) = V (x) (7c) u(0, t) = 0, (7d) ∂u ∂t + 1 c ∂u ∂x = 0, x = L, t > 0 (7e)

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:

• Ω

uttv dx − 1 c2

• Ω

∆uv dx =

• Ω

uttv dx + 1 c2

• Ω

∇u∇v dx −

• ∂Ω

∂u ∂nv dx

• We use the solution anzats

u =

∞

• i=0

Niui(t), v(x, t) =

∞

• j=0

Njvj(t), (8) with Ni as our element basis. Consequently utt =

∞

• i=0

Ni¨ ui(t)

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 20 / 60

Absorbing Boundary Conditions Finite Element Approximation

Weak Formulation

Inserting our solution ansatz gives the infinite dimensional system to solve

• i,j
• Ω

NiNj dx

• =Mi,j

¨ uivj + 1 c2

• i,j

    

• Ω

∇Ni∇Nj dx

• Ki,j

−

• ∂Ω

∂Ni ∂n Nj dx

• =Bi,j

     uivj

Which gives us the folowing problem M¨ u + Ku − Bu = 0 (9)

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 21 / 60

Absorbing Boundary Conditions Finite Element Approximation

By approximating our solution u and the test function v by u(x, t) ≈ U(x, t) =

n

• i=0

Ni(x)Ui(t), (10) v(x, t) ≈ V (x, t) =

n

• j=0

Nj(x)Vj(t), (11) we get a finite dimensional linear system of equations; M ¨ U + KU − BU = 0. (12)

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 22 / 60

Absorbing Boundary Conditions Finite Element Approximation

Imposing ABC

we now take a step back in the above calculations and only consider the boundary term; 1 c

• ∂Ω

∂u ∂nv dx = 1 c

• Γr

∂u ∂x v dx

• Neumann=g

+1 c

• Γa

∂u ∂x v dx Where Γr denotes the reflecting boundary x = 0, and Γa denotes the absorbing at x = L. From (6) we have the following on the absorbing boundary ∂u ∂t + 1 c ∂u ∂x = 0 ⇒ 1 c ∂u ∂x = −∂u ∂t . Therefore the boundary term becomes 1 c

• Γa

∂u ∂x v dx = −1 c

• Γa

∂u ∂t v dx

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 23 / 60

Absorbing Boundary Conditions Finite Element Approximation

Imposing ABC

This expression 1 c

• Γa

∂u ∂x v dx = −1 c

• Γa

∂u ∂t v dx becomes, with our solution ansatz; −

• Γa

∂u ∂t v dx =

• i,j
• Γa

NiNj dx

• Ba

˙ uivi Note: our boundary in the one-dimensional case is simply the end point, x = L.

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 24 / 60

Absorbing Boundary Conditions Finite Element Approximation

Imposing ABC - Summary

Our final linear system of looks like this: M ¨ U + KU − g − Ba ˙ U = 0, (13) Ba

n,n = β and Ba i,j = 0 else. For the time stepping any suitable scheme

(such as Newmark) can be used.

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 25 / 60

Perfectly Matched Layer

Perfectly Matched Layer

Perfectly Matched Layer (PML) works by creating an absorbing layer in shape of analytical continuation around the region of interest.

Figure 4: [?]

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 26 / 60

Perfectly Matched Layer

The Wave Equation

One-dimensional ∂2u ∂t2 − ∂2u ∂x2 = 0, x > 0, t > 0 u(x, 0) = U(x) ∂ ∂t u(x, 0) = V (x) u(0, t) = 0, 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 27 / 60

Perfectly Matched Layer

But...

We need to come up with a layer where waves decay sufficiently fast. Hence, we need to define a domain in which the solution decays: x → z, for x > L. The goal is now to find such a transformation such that The solution u(z) → 0 sufficiently fast The solution is unharmed for x < L The whole process is computationally feasible

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 28 / 60

Perfectly Matched Layer The Approach

The Approach

Auxiliary function, v(x, t), that satisfies ∂u ∂t = ∂v ∂x , ∂v ∂t = ∂u ∂x (14) Then for the wave equation can be written as ∂ ∂t

• u

v

• =
• ∂

∂x ∂ ∂x

u v

• Nikolaj Takata Mücke (Technical University of Munich)

Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 29 / 60

Perfectly Matched Layer The Approach

Solution

This linear system of PDE’s has the solution in infinite space

• u

v

• =
• w1

w2

• ei(kx−ωt),

with w1 and w2 the constant amplitudes, ω the angular frquency and k the wavenumber.

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 30 / 60

Perfectly Matched Layer The Approach

Analytical Continuation

The solution

• u

v

• =
• w1

w2

• ei(kx−ωt),

is analytic in x. Hence we can freely analytically continue it and evaluate in the complex plane! Then we have ei(kx−ωt) = eik(Rex+iImx)e−iωt = eikRexe−iωt

• Oscillating

e−kImx

Decaying in x

Exponentially decaying in space for k > 0!

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 31 / 60

Perfectly Matched Layer The Approach

Analytical Continuation

Figure 5: [?]

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ABC vs. PML

Pros and Cons

ABC Relatively Easy to derive. Implementation (often) comes down to well-known techniques. Boundary conditions can be very complicated. Becomes very complicated in higher dimensions. Can be unstable at inhomogeneous boundaries. PML Does not reflect at the interface. Usually better for numerics than ABC. Only reflectionless for exact wave equation. Determining the complex curve can be very deficult. Change of coordinate can be hard to derive. Fails in inhomogeneous media.

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 33 / 60

Non-Linear Absorbing Boundary Condition

Non-Linear Absorbing Boundary Condition

Why is non-linearity a problem? General problems when solving numerically Difficult to identify in- and outgoing waves

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Non-Linear Absorbing Boundary Condition

Westervelt Equation

c−2utt − ∆u − β∆ut = γ(u2)tt (15) Strategy

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Non-Linear Absorbing Boundary Condition

Strategy

1 Linearise 2 Express linearised equation in terms of Pseudo-differential operators 3 Derive ABC’s Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 36 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Operators

But first... A (very!) Brief Introduction to Ps.d.o’s

An extension of differential operators Gives us a way to derive generalised inverse operators Makes us able to derive general properties of solutions to PDE’s

Existence Regularity

We can derive many of these properties by looking at polynomials

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 37 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Operators

Example - Helmholtz

P(x, D)u = (∆ + k2)u = 0, k > 0, x ∈ Rn Fourier transformation - Differentiation corresponds to multiplication! F(P(x, D)u) = (k2 − ξ2)ˆ u(ξ) = 0 Inverse Fourier transformation F−1 [F(P(x, D)u)] = P(x, D)u =

• (k2 − ξ2)ˆ

u(ξ)eiξx dξ = 0

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 38 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Operators

Example - Helmholtz

P(x, D)u =

• (k2 − ξ2)ˆ

u(ξ)eiξx dξ = 0 Let’s denote p(x, ξ) = (k2 − ξ2) and call it the symbol of the operator. This means that our ps.d.o is defined by P(x, D)u =

• p(x, ξ)ˆ

u(ξ)eiξx dξ In other words; the Pseudo-differential Operator is the Fourier Integral Representation of the differential operator.

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 39 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Operators

Composition of Ps.d.o’s

The composition (product) of two ps.d.o’s is a new ps.d.o; P1 ◦ P2 = P (16) with symbol satisfying p(x, ξ) ∼

• α≥0

i|α| α! Dα

ξ p1(x, ξ)Dα x p2(x, ξ)

(17) For our purposes, we can switch the ” ∼ ” with ” = ”.

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 40 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Operators

Back to Westervelt Equation!

c−2utt − ∆u − β∆ut = γ(u2)tt (18) Strategy

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 41 / 60

Non-Linear Absorbing Boundary Condition Linearisation

Linearisation

c−2utt − ∆u − β∆ut = γ(u2)tt (19) We linearise by inserting a reference solution, u(0)(x), and add a small perturbation ǫw(x, t); c−2 u(0) + ǫw(x, t)

• tt − ∆
• u(0) + ǫw(x, t)
• −β∆
• u(0) + ǫw(x, t)
• t = γ(
• u(0) + ǫw(x, t)

2)tt.

Doing some reductions and ignore ǫ2 terms...

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 42 / 60

Non-Linear Absorbing Boundary Condition Linearisation

Linearisation

Finally, by replacing w by u, we get

• c−2 − 2γu(0)

u − ∆u − β∆ut = 2γu(0)

t ut

⇒ν2utt − uxx − βutxx − 2γu(0)

t ut = 0

with ν2 = ν2(u(0)), where ν2(v) = c−2 − 2γv. The operator form is given by D1u = 0, D1 = ν2∂2

t − ∂2 x − β∂txx − 2γu(0) t ∂t

(20) c−2 − 2γu = ν2(u) > 0, ν ∈ C∞

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Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

Pseudo-Differential Representation

We now want to identify in- and outgoing waves as we did in the linear wave equation. But our problem is much more complicated!

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 44 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

Nirenberg Factorisation

We can factorise D1; D1 = −(∂x − A)(∂x − B) + R (21) where A = A(x, t, Dt), B = B(x, t, Dt) are ps.d.o’s with symbols a(x, t, τ) =

k

• j=0

aj(x)τ j, b(x, t, τ) =

k

• j=0

bj(x)τ j and Dt = −i∂t. Furthermore, R is a smoothing ps.d.o; which means that we can "neglect" it.

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Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

ν2∂2

t − ∂2 x − β∂txx − 2γu(0) t ∂t = −(∂x − A)(∂x − B) + R

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 46 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

Computing the Symbols

How does A and B look like? - or more specificly: how does their symbols, a and b, look like? Since we want to compute explicit boundary conditions we need to compute the symbols to gain anything from the ps.d.o representation!

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 47 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

Computing the Symbols

We start by developing the factorisation D1 = −(∂x − A)(∂x − B) + R = −∂2

x + (A + B)∂x + Bx − AB + R

Which at the symbolic level is d1(x, t, τ) = −∂2

x + (a + b)∂x + bx − ab + R

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Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

... and the symbolic representation of the non-ps.d.o. representation is given by ν2(iτ)2 − ∂2

x − β(iτ)∂xx − 2γu(0) t (iτ)

• btained by Fourier transformation in time.

Combing the two symbol representations: ν2(iτ)2 − ∂2

x − β(iτ)∂xx − 2γu(0) t (iτ) = −∂2 x + (a + b)∂x + bx − ab + R

⇒ ν2(iτ)2 − β(iτ)∂xx − 2γu(0)

t (iτ) = (a + b)∂x + bx − ab + R

We can use this equality to compute a and b!

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Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

But first... What is ab?

We use the theorem from earlier to represent the symbol of the composition of of two ps.d.o’s: ab(x, t, τ) ∼

• k,l,n≥0

(−i)n n! ∂n

τ a1−k(x, t, τ)∂n τ b1−l(x, t, τ)

=

• j≥0, k+l+n=j

(−i)n n! ∂n

τ a1−k(x, t, τ)∂n τ b1−l(x, t, τ)

This is a polynomial of order O(τ 2−j).

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 50 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

... which means

(a + b)∂x + bx − ab + R =

• j≥0

(a1−j + b1−j)∂x +

• j≥0

∂xb1−j) −

• j≥0, k+l+n=j

(−i)n n! ∂n

τ a1−k∂n τ b1−l

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 51 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

ν2(iτ)2 − β(iτ)∂xx − 2γu(0)

t (iτ) =

• j≥0

(a1−j + b1−j)∂x +

• j≥0

∂xb1−j −

• j≥0, k+l+n=j

(−i)n n! ∂n

τ a1−k∂n τ b1−l

• O(τ 2−j)

To compute ai and bi we match coeffcients of τ j. The more coefficients we take the higer accuracy of ABC! - but also higher complexity.

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 52 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

O(τ 2), j = 0 :

• a1 + b1 = 0

ν2(iτ)2 = −a1b1 O(τ), j = 1 :

• a0 + b0 = 0

β(iτ)∂2

x + 2γu(0) t (iτ) = a1b0 + a0b1 − ia1τb1t − b1x

O(τ 0), j = 2 :

    

a−1 + b−1 = 0 −a1b−1 − a0b0 − a−1b1 + i(a1τb0t + a0τb1t) −i2

2 a1ττb1tt + b0x = 0

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 53 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

a1 = −ν(iτ), a0 = 1 2ν

• Gν + 2γu(0)

t

• ,

a−1 = γµ 2ν(iτ) b1 = ν(iτ), b0 = − 1 2ν

• Gν + 2γu(0)

t

• ,

b−1 = − γµ 2ν(iτ) where µ = G

1

2ν

• Gν + 2γu(0)

t

• −

1

2ν

• Gν + 2γu(0)

t

2

and G = (∂x + ν∂t). Note: ai = −bi

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 54 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

And therefore D1 = −(∂x − A)(∂x − B) + R = ν2∂2

t − ∂2 x − β∂txx − 2γu(0) t ∂t

has the symbol − (∂x − a) (∂x − b) + R = − (∂x − a) (∂x + a) + R = −

 ∂x −

2

• j=0

a1−j

   ∂x +

2

• j=0

a1−j

  + R

with a1 = −ν(iτ), a0 = 1 2ν

• G(ν) + 2γu(0)

t

• ,

a−1 = γµ 2ν(iτ)

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 55 / 60

Non-Linear Absorbing Boundary Condition Pseudo-Differential Representation

Summary

What have we done so far? Linearised our PDE D1 = ν2∂2

t − ∂2 x − β∂txx − 2γu(0) t ∂t

Formulated the PDE in terms of ps.d.o’s D1 = −(∂x − A)(∂x − B) + R, Computed the symbols of A and B and realised D1 = −(∂x − A)(∂x + A) + R,

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 56 / 60

Non-Linear Absorbing Boundary Condition Absorbing Boundary Conditions

Absorbing Boundary Conditions

We have now done the hard work! The ps.d.o. representation, D1 = −(∂x − A)(∂x + A) + R, makes it possible for us to identify in- and outgoing waves. According to [?], the operator ∂x + a(x, t, Dt) represents left-propagating waves!

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 57 / 60

Non-Linear Absorbing Boundary Condition Absorbing Boundary Conditions

Absorbing Boundary Conditions

... Which means ∂x + a(x, t, Dt) = 0 is the operator we need at the boundary! This gives us

 ∂x +

k

• j=0

a1−j(x, t, Dt)

  u

• x=L

= 0 k denotes the order of our ABC.

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 58 / 60

Non-Linear Absorbing Boundary Condition Absorbing Boundary Conditions

0th order ABC

k = 0. We use only a1:

 ∂x +

• j=0

a1−j(x, t, Dt)

  u

• x=L

= (∂x + ν∂t) u|x=0 = L This almost the same as for the linear wave equation! Note: Remember, a1(x, t, τ) = ν(iτ) corresponds to a1(x, t, Dt) = −ν∂t.

Nikolaj Takata Mücke (Technical University of Munich) Absorbing Boundary Conditions and Perfectly Matched Layers for Acoustics 19/1 - 2018 59 / 60

Non-Linear Absorbing Boundary Condition Absorbing Boundary Conditions

1st order ABC

k = 1. We use a1 and a0:

 ∂x +

1

• j=0

a1−j(x, t, Dt)

  u

• x=L

=

• ux + νut − 1

2ν

• (νx + ννt)u + 2γu(0)

t u

• x=L

= 0

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