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An adaptive PML technique for time-harmonic scattering problems - - PowerPoint PPT Presentation

0.5 setgray0 0.5 setgray1 An adaptive PML technique for time-harmonic scattering problems Following a paper by Zhiming Chen and Xuezhe Liu Manuel Largo An adaptive PML technique for time-harmonic scattering problems p. 1/51 Overview


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An adaptive PML technique for time-harmonic scattering problems

Following a paper by Zhiming Chen and Xuezhe Liu

Manuel Largo

An adaptive PML technique for time-harmonic scattering problems – p. 1/51

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Overview

Introduction, Hankel functions

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Overview

Introduction, Hankel functions PML formulation

An adaptive PML technique for time-harmonic scattering problems – p. 2/51

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Overview

Introduction, Hankel functions PML formulation Finite Elements and the Main Theorem

An adaptive PML technique for time-harmonic scattering problems – p. 2/51

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Overview

Introduction, Hankel functions PML formulation Finite Elements and the Main Theorem Implementation and Examples

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First Part

INTRODUCTION AND HANKEL FUNCTIONS

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Task

We want to show how we can adapt finite element mesh size.

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Task

We want to show how we can adapt finite element mesh size. To do so, we need an a posteriori error estimate to control the error we make when discretizing space.

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Task

We want to show how we can adapt finite element mesh size. To do so, we need an a posteriori error estimate to control the error we make when discretizing space. We extend the idea of using a posteriori error estimates to determine the PML parameters and propose an adaptive PML technique for solving the Helmholtz-type scattering problem.

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Task

We want to show how we can adapt finite element mesh size. To do so, we need an a posteriori error estimate to control the error we make when discretizing space. We extend the idea of using a posteriori error estimates to determine the PML parameters and propose an adaptive PML technique for solving the Helmholtz-type scattering problem. We will first introduce and prove some error estimates, later construct an algorithm to adapt mesh size with a posteriori error control.

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Scattering problem

So, lets derive a PML technique for solving Helmholtz-type scattering problems with perfectly conducting boundary.

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Scattering problem

So, lets derive a PML technique for solving Helmholtz-type scattering problems with perfectly conducting boundary. Let D ∈ R2 denote the bounded domain (scatterer) with boundary ΓD, g ∈ H−1/2(ΓD) determined by the incoming wave, n the unit outer normal to ΓD.

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Scattering problem

So, lets derive a PML technique for solving Helmholtz-type scattering problems with perfectly conducting boundary. Let D ∈ R2 denote the bounded domain (scatterer) with boundary ΓD, g ∈ H−1/2(ΓD) determined by the incoming wave, n the unit outer normal to ΓD. Helmholtz-type scattering problem (constant k): ∆u + k2u =

in R2\ ¯

D ∂u ∂n = −g

  • n ΓD

√r ∂u ∂r − iku

0 as r = |x| → ∞

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Hankel functions

First, consider the Bessel equation for functions of order ν: z2 d2y dz2 + z dy dz + (z2 − ν2)y = 0, ν ∈ C.

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Hankel functions

First, consider the Bessel equation for functions of order ν: z2 d2y dz2 + z dy dz + (z2 − ν2)y = 0, ν ∈ C. The so called Bessel function of the first kind Jν(z) is defined as the solution to the Bessel differential equation with non singular values at the

  • rigin.

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Hankel functions

First, consider the Bessel equation for functions of order ν: z2 d2y dz2 + z dy dz + (z2 − ν2)y = 0, ν ∈ C. The so called Bessel function of the first kind Jν(z) is defined as the solution to the Bessel differential equation with non singular values at the

  • rigin.

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Hankel functions (cont)

The so called Bessel function of the second kind Yν(z) is defined as the solution to the Bessel differential equation with singular values at the

  • rigin.

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Hankel functions (cont)

The so called Bessel function of the second kind Yν(z) is defined as the solution to the Bessel differential equation with singular values at the

  • rigin.

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Hankel functions (cont)

We introduce now the Hankel function of the first kind and order ν H(1)

ν (z), z ∈ C, and the Hankel function of the second kind and order ν

H(2)

ν (z), z ∈ C, are defined by

H(1)

ν (z)

≡ Jν(z) + iYν(z), H(2)

ν (z)

≡ Jν(z) − iYν(z).

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Hankel functions (cont)

We introduce now the Hankel function of the first kind and order ν H(1)

ν (z), z ∈ C, and the Hankel function of the second kind and order ν

H(2)

ν (z), z ∈ C, are defined by

H(1)

ν (z)

≡ Jν(z) + iYν(z), H(2)

ν (z)

≡ Jν(z) − iYν(z). Asymptotic behaviour: H(1)

ν (z)

  • 2

πz ei(z− 1

2 νπ− 1 4 π),

H(2)

ν (z)

  • 2

πz e−i(z− 1

2 νπ− 1 4 π).

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Hankel functions, H(1)

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Hankel functions, H(1)

1

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Hankel functions, H(1)

−1

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Lemma 1

Lemma 1: For any ν ∈ R, z ∈ C++ = {z ∈ C : ℑ(z) ≥ 0, ℜ(z) ≥ 0}, and Θ ∈ R such that 0 < Θ ≤ |z|, we have |H(1)

ν (z)| ≤ e −ℑ(z) r 1− Θ2

|z|2 |H(1)

ν (Θ)|

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Lemma 1

Lemma 1: For any ν ∈ R, z ∈ C++ = {z ∈ C : ℑ(z) ≥ 0, ℜ(z) ≥ 0}, and Θ ∈ R such that 0 < Θ ≤ |z|, we have |H(1)

ν (z)| ≤ e −ℑ(z) r 1− Θ2

|z|2 |H(1)

ν (Θ)|

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Lemma 1 (cont.)

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Second Part

PML FORMULATION

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Setup

Let the scatterer D be contained in the interior of the circle BR = {x ∈ R2 : |x| < R}, and ΩR = BR\ ¯ D. We now surround the domain ΩR with a PML layer ΩPML = {x ∈ R2 : R < |x| < ρ}.

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The PML formulation

Look at the domain R2\ ¯

  • BR. The solution u of the scattering problem can

be written under the polar coordinates as follows: u(r, θ) =

  • n∈Z

H(1)

n (kr)

H(1)

n (kR)

ˆ uneinθ, ˆ un = 1 2π 2π u(R, θ)e−inθdθ. H(1)

n

denotes the just discussed Hankel function of the first kind and order

  • n. It can be shown that this series converges uniformly for r > R.

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Dirichlet-to-Neumann operator

We now introduce the so called Dirichlet-to-Neumann operator T : H1/2(ΓR) → H−1/2(ΓR), where ΓR = ∂BR. It is definied as follows: for any f ∈ H1/2(ΓR), Tf =

  • n∈Z

k H(1)′

n

(kR) H(1)

n (kR)

ˆ fneinθ, ˆ fn = 1 2π 2π fe−inθdθ.

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Dirichlet-to-Neumann operator

We now introduce the so called Dirichlet-to-Neumann operator T : H1/2(ΓR) → H−1/2(ΓR), where ΓR = ∂BR. It is definied as follows: for any f ∈ H1/2(ΓR), Tf =

  • n∈Z

k H(1)′

n

(kR) H(1)

n (kR)

ˆ fneinθ, ˆ fn = 1 2π 2π fe−inθdθ. Looking at the representation of the solution u in polar coordinates: u(r, θ) =

  • n∈Z

H(1)

n (kr)

H(1)

n (kR)

ˆ uneinθ, ˆ un = 1 2π 2π u(R, θ)e−inθdθ, it is obvious that it satisfies ∂u ∂n

ΓR = Tu.

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Reformulation

Let a : H1(ΩR) × H1(ΩR) → C be the sesquilinear form a(ϕ, ψ) =

  • ΩR
  • ∇ϕ · ∇ ¯

ψ − k2ϕ ¯ ψ

  • dx − Tϕ, ψΓR.

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Reformulation

Let a : H1(ΩR) × H1(ΩR) → C be the sesquilinear form a(ϕ, ψ) =

  • ΩR
  • ∇ϕ · ∇ ¯

ψ − k2ϕ ¯ ψ

  • dx − Tϕ, ψΓR.

Given g ∈ H−1/2(ΓR), find u ∈ H1(ΓR) such that a(u, ψ) = g, ψΓD ∀ψ ∈ H1(ΩR), µ > 0.

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Reformulation

Let a : H1(ΩR) × H1(ΩR) → C be the sesquilinear form a(ϕ, ψ) =

  • ΩR
  • ∇ϕ · ∇ ¯

ψ − k2ϕ ¯ ψ

  • dx − Tϕ, ψΓR.

Given g ∈ H−1/2(ΓR), find u ∈ H1(ΓR) such that a(u, ψ) = g, ψΓD ∀ψ ∈ H1(ΩR), µ > 0. sup

0=ψ∈H1(ΩR)

|a(ϕ, ψ)| ψH1(ΩR) ≥ µϕH1(ΩR) ∀ϕ ∈ H1(ΩR).

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PML formulation

Let α(r) = 1 + iσ(r) be the PML model medium property with σ ∈ C(R), σ ≥ 0,

and σ = 0 for r ≤ R.

We denote by ˜ r the complex radius defined by ˜ r = ˜ r(r) =    r

if r ≤ R,

r

0 α(t)dt = rβ(r)

if r ≥ R.

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PML formulation

Let α(r) = 1 + iσ(r) be the PML model medium property with σ ∈ C(R), σ ≥ 0,

and σ = 0 for r ≤ R.

We denote by ˜ r the complex radius defined by ˜ r = ˜ r(r) =    r

if r ≤ R,

r

0 α(t)dt = rβ(r)

if r ≥ R.

Lets introduce now the PML equation: ∇ · (A∇w) + αβk2w = 0

in ΩPML,

where A = A(x) is a matrix which satisfies, in polar coordinates, ∇ · (A∇) = 1 r ∂ ∂r βr α ∂ ∂r

  • + α

βr2 ∂2 ∂θ2 .

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PML formulation (cont)

Now, the PML solution ˆ u in Ωρ = Bρ\ ¯ D is defined as the solution of the system ∇ · (A∇ˆ u) + αβk2ˆ u =

in Ωρ,

∂ˆ u ∂n = −g

  • n ΓD,

ˆ u =

  • n Γρ.

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PML formulation (cont)

Now, the PML solution ˆ u in Ωρ = Bρ\ ¯ D is defined as the solution of the system ∇ · (A∇ˆ u) + αβk2ˆ u =

in Ωρ,

∂ˆ u ∂n = −g

  • n ΓD,

ˆ u =

  • n Γρ.

Again, we introduce the sesquilinear form ˆ a : H1(ΩR) × H1(ΩR) → C by ˆ a(ϕ, ψ) =

  • ΩR

(A∇ϕ · ∇ ¯ ψ − k2αβϕ ¯ ψ)dx − ˆ Tϕ, ψΓR, and ˆ a(ˆ u, ψ) = g, ψΓD ∀ψ ∈ H1(ΩR).

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PML formulation (cont)

Similar to the previous problem, we can reformulate the problem in the bounded domain ΩR by imposing the boundary condition ∂ˆ u ∂n

ΓR = ˆ

T ˆ u, where ˆ T : H1/2(ΓR) → H−1/2(ΓR) is defined as follows: given f ∈ H1/2(ΓR), ˆ Tf = ∂ζ ∂n

ΓR,

where ζ ∈ H1(ΩPML) satisfies ∇ · (A∇ζ) + αβk2ζ =

in ΩPML,

ζ = f

  • n ΓR,

ζ =

  • n Γρ.

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The PML equation in the layer

Lets look now at the Dirichlet problem in the PML layer ΩPML only: The solution w solves ∇ · (A∇w) + αβk2w =

in ΩPML,

w =

  • n ΓR,

w = q

  • n Γρ.

where q ∈ H1/2(Γρ).

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The PML equation in the layer

Lets look now at the Dirichlet problem in the PML layer ΩPML only: The solution w solves ∇ · (A∇w) + αβk2w =

in ΩPML,

w =

  • n ΓR,

w = q

  • n Γρ.

where q ∈ H1/2(Γρ). With ˆ b : H1(ΩPML) × H1(ΩPML) → C defined to be ˆ b(ϕ, ψ) = ρ

R

2π βr α ∂ϕ ∂r ∂ ¯ ψ ∂r + α βr ∂ϕ ∂θ ∂ ¯ ψ ∂θ − αβk2rϕ ¯ ψ

  • drdθ,

we can write down the weak formulation for this problem: given q ∈ H1/2(Γρ), find w ∈ H1(ΩPML) such that w = 0 on ΓR, w = q on Γρ, and ˆ b(w, ϕ) = 0 ∀ϕ ∈ H1

0(ΩPML).

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Medium property

We make the following assumption for the fictitious medium property σ: (H1): σ = σ0

  • r−R

ρ−R

m for some σ0 > 0 and m ∈ N.

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Medium property

We make the following assumption for the fictitious medium property σ: (H1): σ = σ0

  • r−R

ρ−R

m for some σ0 > 0 and m ∈ N. We know that β(r) = r−1 r

0 α(t)dt, and therefore β(r) = 1 + iˆ

σ(r), where ˆ σ(r) = 1 r r

R

σ(t)dt = σ0 m + 1 r − R r r − R ρ − R m . Therefore, ˆ σ ≤ σ ∀r ≥ R.

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Medium property

We make the following assumption for the fictitious medium property σ: (H1): σ = σ0

  • r−R

ρ−R

m for some σ0 > 0 and m ∈ N. We know that β(r) = r−1 r

0 α(t)dt, and therefore β(r) = 1 + iˆ

σ(r), where ˆ σ(r) = 1 r r

R

σ(t)dt = σ0 m + 1 r − R r r − R ρ − R m . Therefore, ˆ σ ≤ σ ∀r ≥ R. (H2) There exists a unique solution to the Dirichlet PML problem in the PML layer ΩPML.

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Theorem 1

We give the following theorem (without proof) as the main objective of this subsection: Theorem 1 Let (H1)-(H2) be satisfied. There exists a constant C > 0 independent of k, R, ρ, and σ0 such that the following estimates hold: |α|−1∇wL2(ΩPML) ≤ C ˆ C−1(1 + kR)|α0|qH1/2(Γρ),

  • ∂w

∂n

  • H−1/2(ΓR)

≤ C ˆ C−1(1 + kR)2|α0|2qH1/2(Γρ). where α0 = 1 + iσ0.

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Theorem 1

We give the following theorem (without proof) as the main objective of this subsection: Theorem 1 Let (H1)-(H2) be satisfied. There exists a constant C > 0 independent of k, R, ρ, and σ0 such that the following estimates hold: |α|−1∇wL2(ΩPML) ≤ C ˆ C−1(1 + kR)|α0|qH1/2(Γρ),

  • ∂w

∂n

  • H−1/2(ΓR)

≤ C ˆ C−1(1 + kR)2|α0|2qH1/2(Γρ). where α0 = 1 + iσ0. We will need these estimates later to prove the main theorem of this talk . . .

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Propagation operator

To prove the convergence of the just considered PML problem to the

  • riginal scattering problem, we need to introduce the propagation operator

P : H1/2(ΓR) → H1/2(Γρ) defined as (Lassas and Somersalo): P(f) =

  • n∈Z

H(1)

n (k˜

ρ) H(1)

n (kR)

ˆ fneinθ, ˆ fn = 1 2π 2π fe−inθdθ.

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Propagation operator

To prove the convergence of the just considered PML problem to the

  • riginal scattering problem, we need to introduce the propagation operator

P : H1/2(ΓR) → H1/2(Γρ) defined as (Lassas and Somersalo): P(f) =

  • n∈Z

H(1)

n (k˜

ρ) H(1)

n (kR)

ˆ fneinθ, ˆ fn = 1 2π 2π fe−inθdθ. One can also show that P(f)H1/2(Γρ) ≤ e

−kℑ(˜ ρ) r 1− R2

|˜ ρ|2 fH1/2(ΓR)

∀r ≥ R.

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D2N mapping

Lemma 2: Let (H1)-(H2) be satisfied. Then, we have Tf − ˆ TfH−1/2(ΓR) ≤ C ˆ C−1(1 + kR)2|α0|2e

−kℑ(˜ ρ) r 1− R2

|˜ ρ|2 fH1/2(ΓR).

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D2N mapping

Lemma 2: Let (H1)-(H2) be satisfied. Then, we have Tf − ˆ TfH−1/2(ΓR) ≤ C ˆ C−1(1 + kR)2|α0|2e

−kℑ(˜ ρ) r 1− R2

|˜ ρ|2 fH1/2(ΓR).

Theorem 2: Let again (H1)-(H2) be satisfied. Then, for sufficiently large σ0 > 0, the PML problem has a unique solution ˆ u ∈ H1(Ωρ). Moreover, we have the following estimate: u − ˆ uH1(ΩR) ≤ C ˆ C−1(1 + kR)2|α0|2e

−kℑ(˜ ρ) r 1− R2

|˜ ρ|2 ˆ

uH1/2(ΓR).

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Third Part

FINITE ELEMENTS AND THE MAIN THEOREM

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The Finite Element Method (FEM)

Task: By discretization, transform a variational boundary value problem to a system of finite number of equations for real unknowns. I.e. transform the linear variational problem u ∈ V : a(u, v) = f(v) ∀v ∈ V to uN ∈ Vh : a(uN, vN) = f(vN) ∀vN ∈ Vh.

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The Finite Element Method (FEM)

Task: By discretization, transform a variational boundary value problem to a system of finite number of equations for real unknowns. I.e. transform the linear variational problem u ∈ V : a(u, v) = f(v) ∀v ∈ V to uN ∈ Vh : a(uN, vN) = f(vN) ∀vN ∈ Vh. Do it by triangulation of space Ω:

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FEM basis functions

Basis functions φ1, . . . , φN for a finite element space Vh built on a mesh Mh satisfy: each φi associated with a single cell/edge/face/vertex of Mh, supp(φi) = { ¯ K : K ∈ Mh, p ⊂ ¯ K}, if φi associated with cell/edge/face/vertex p.

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FEM basis functions

Basis functions φ1, . . . , φN for a finite element space Vh built on a mesh Mh satisfy: each φi associated with a single cell/edge/face/vertex of Mh, supp(φi) = { ¯ K : K ∈ Mh, p ⊂ ¯ K}, if φi associated with cell/edge/face/vertex p.

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FEM nodal basis

Let Vh(Mh) = Nh := set of nodes of Mh. Then, the nodal basis is defined as: If Nh = {a1, . . . , aN}, nodal basis Φh := {φ1, . . . , φN} defined by φi(aj) = δij.

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FEM nodal basis

Let Vh(Mh) = Nh := set of nodes of Mh. Then, the nodal basis is defined as: If Nh = {a1, . . . , aN}, nodal basis Φh := {φ1, . . . , φN} defined by φi(aj) = δij.

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Finite element approximation

Now, we introduce the finite element approximation of the PML problem. From now on, we assume g ∈ L2(ΓD). Let b : H1(Ωρ) × H1(Ωρ) → C be the sesquilinear form given by b(ϕ, ψ) =

  • Ωρ

(A∇ϕ · ∇ ¯ ψ − αβk2ϕ ¯ ψ)dx.

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Finite element approximation

Now, we introduce the finite element approximation of the PML problem. From now on, we assume g ∈ L2(ΓD). Let b : H1(Ωρ) × H1(Ωρ) → C be the sesquilinear form given by b(ϕ, ψ) =

  • Ωρ

(A∇ϕ · ∇ ¯ ψ − αβk2ϕ ¯ ψ)dx. Furthermore, denote by H1

(0)(Ωρ) = {v ∈ H1(Ωρ) : v = 0 on Γρ}. Then, we

can write down the weak formulation for the PML problem: given g ∈ L2(ΓD), find ˆ u ∈ H1

(0)(Ωρ) such that

b(ˆ u, ψ) =

  • ΓD

g ¯ ψds ∀ψ ∈ H1

(0)(Ωρ).

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Finite element notation

Let Γh

ρ, which consists of piecewise segments whose vertices lie on

Γρ, be an approximation of Γρ.

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Finite element notation

Let Γh

ρ, which consists of piecewise segments whose vertices lie on

Γρ, be an approximation of Γρ. Let Mh be a regular triangulation of the domain Ωh

ρ.

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Finite element notation

Let Γh

ρ, which consists of piecewise segments whose vertices lie on

Γρ, be an approximation of Γρ. Let Mh be a regular triangulation of the domain Ωh

ρ.

Assume the elements K ∈ Mh may have one curved edge align with ΓD, such that Ωh

ρ = K∈Mh K.

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SLIDE 63

Finite element notation

Let Γh

ρ, which consists of piecewise segments whose vertices lie on

Γρ, be an approximation of Γρ. Let Mh be a regular triangulation of the domain Ωh

ρ.

Assume the elements K ∈ Mh may have one curved edge align with ΓD, such that Ωh

ρ = K∈Mh K.

Let Vh ⊂ H1(Ωh

ρ) be the conforming linear finite element space over

Ωh

ρ, and V 0 h = {vh ∈ Vh : vh = 0 on Γh ρ}.

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SLIDE 64

Finite elements

Now, we can formulate the finite element approximation to the variational PML problem: find uh ∈ V 0

h such that

b(uh, ψh) =

  • ΓD

g ¯ ψhds ∀ψh ∈ V 0

h .

and the discrete inf-sup condition sup

0=ψh∈V 0

h

|b(ϕh, ψh)| ψhH1(Ωρ) ≥ ˆ µϕhH1(Ωρ) ∀ϕh ∈ V 0

h , ˆ

µ > 0.

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SLIDE 65

Finite elements

Now, we can formulate the finite element approximation to the variational PML problem: find uh ∈ V 0

h such that

b(uh, ψh) =

  • ΓD

g ¯ ψhds ∀ψh ∈ V 0

h .

and the discrete inf-sup condition sup

0=ψh∈V 0

h

|b(ϕh, ψh)| ψhH1(Ωρ) ≥ ˆ µϕhH1(Ωρ) ∀ϕh ∈ V 0

h , ˆ

µ > 0. Since we are interested in a posterior error estimates and the associated adaptive algorithm, we simply assume that the discrete problem has a unique solution uh ∈ V 0

h .

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SLIDE 66

Finite elements, definitions

For any K ∈ Mh, denote by hK its diameter.

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SLIDE 67

Finite elements, definitions

For any K ∈ Mh, denote by hK its diameter. Let Bh denote the set of all sides that do not lie on ΓD and Γh

ρ.

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SLIDE 68

Finite elements, definitions

For any K ∈ Mh, denote by hK its diameter. Let Bh denote the set of all sides that do not lie on ΓD and Γh

ρ.

For any e ∈ Bh, he stands for its length.

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SLIDE 69

Finite elements, definitions

For any K ∈ Mh, denote by hK its diameter. Let Bh denote the set of all sides that do not lie on ΓD and Γh

ρ.

For any e ∈ Bh, he stands for its length. For any K ∈ Mh, introduce the residual Rh := ∇ · (A∇uh|K) + αβk2uh|K.

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SLIDE 70

Finite elements, definitions

For any K ∈ Mh, denote by hK its diameter. Let Bh denote the set of all sides that do not lie on ΓD and Γh

ρ.

For any e ∈ Bh, he stands for its length. For any K ∈ Mh, introduce the residual Rh := ∇ · (A∇uh|K) + αβk2uh|K. For any interior side e ∈ Bh, which is the common side of K1 and K2 ∈ Mh, define the jump residual across e: Je := (A∇uh|K1 − A∇uh|K2) · νe, where the unit normal vector νe to e points from K2 to K1.

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SLIDE 71

Finite elements, definitions

For any K ∈ Mh, denote by hK its diameter. Let Bh denote the set of all sides that do not lie on ΓD and Γh

ρ.

For any e ∈ Bh, he stands for its length. For any K ∈ Mh, introduce the residual Rh := ∇ · (A∇uh|K) + αβk2uh|K. For any interior side e ∈ Bh, which is the common side of K1 and K2 ∈ Mh, define the jump residual across e: Je := (A∇uh|K1 − A∇uh|K2) · νe, where the unit normal vector νe to e points from K2 to K1. If e = ΓD ∩ ∂K for some element K ∈ Mh, then we define the jump residual to be: Je := 2(∇uh|K · n + g).

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SLIDE 72

Finite elements, definitions (cont)

For any K ∈ Mh, denote by ηK the local error estimator which is defined by ηK = max

x∈ ˜ K

w(x) ·

  • hKRh2

L2(K) + 1

2

  • e⊂∂K

heJe2

L2(e)

1/2 , where ˜ K is the union of all elements having nonempty intersection with K, and w(x) =      1

if x ∈ ¯

ΩR, |α0α|e

−kℑ(˜ r) r 1− r2

|˜ r|2

if x ∈ ΩPML.

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SLIDE 73

Main Theorem

Theorem 3: There exists a constant C depending only on the minimum angle of the mesh Mh such that the following a posterior error estimate is valid: u − uhH1(ΩR) ≤ C ˆ C−1 Λ(kR)(1 + kR)

K∈Mh

η2

K

1/2 + C ˆ C−1(1 + kR)2|α0|2e

−kℑ(˜ ρ) r 1− R2

|˜ ρ|2 uhH1/2(ΓR),

where Λ(kR) = max

  • 1, |H(1)′

(kR)| |H(1) (kR)|

  • .

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SLIDE 74

Main Theorem

Theorem 3: There exists a constant C depending only on the minimum angle of the mesh Mh such that the following a posterior error estimate is valid: u − uhH1(ΩR) ≤ C ˆ C−1 Λ(kR)(1 + kR)

K∈Mh

η2

K

1/2 + C ˆ C−1(1 + kR)2|α0|2e

−kℑ(˜ ρ) r 1− R2

|˜ ρ|2 uhH1/2(ΓR),

where Λ(kR) = max

  • 1, |H(1)′

(kR)| |H(1) (kR)|

  • .

The important exponentially decaying factor e

−kℑ(˜ r) r 1− r2

|˜ r|2 in the PML

region ΩPML allows us to take thicker PML layers without introducing unnecessary fine meshes away from the fixed domain ΩR.

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SLIDE 75

Symmetry in ˆ T

For any ϕ ∈ H1(ΩR), let ˜ ϕ be its extension in ΩPML such that ∇ · ( ¯ A∇ ˜ ϕ) + ¯ α ¯ βk2 ˜ ϕ =

in ΩPML,

˜ ϕ = ϕ

  • n ΓR,

˜ ϕ =

  • n Γρ.

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SLIDE 76

Symmetry in ˆ T

For any ϕ ∈ H1(ΩR), let ˜ ϕ be its extension in ΩPML such that ∇ · ( ¯ A∇ ˜ ϕ) + ¯ α ¯ βk2 ˜ ϕ =

in ΩPML,

˜ ϕ = ϕ

  • n ΓR,

˜ ϕ =

  • n Γρ.

Lemma 3: Let (H2) be satisfied. For any ϕ, ψ ∈ H1(ΩPML), we have ˆ Tϕ, ψΓR = ˆ T ¯ ψ, ¯ ϕΓR.

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SLIDE 77

Symmetry in ˆ T

For any ϕ ∈ H1(ΩR), let ˜ ϕ be its extension in ΩPML such that ∇ · ( ¯ A∇ ˜ ϕ) + ¯ α ¯ βk2 ˜ ϕ =

in ΩPML,

˜ ϕ = ϕ

  • n ΓR,

˜ ϕ =

  • n Γρ.

Lemma 3: Let (H2) be satisfied. For any ϕ, ψ ∈ H1(ΩPML), we have ˆ Tϕ, ψΓR = ˆ T ¯ ψ, ¯ ϕΓR. Whenever no confusion of the notation incurred, we shall write in the following ˜ ϕ as ϕ in ΩPML.

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SLIDE 78

Error representation formula

Lemma 4: For any ϕ ∈ H1(ΩR), which is extended to be a function in H1(Ωρ), and ϕh ∈ V 0

h , we have

a(u − uh, ϕ) =

  • ΓD

g(ϕ − ϕh) − b(uh, ϕ − ϕh) + Tuh − ˆ Tuh, ϕΓR.

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SLIDE 79

Error representation formula

Lemma 4: For any ϕ ∈ H1(ΩR), which is extended to be a function in H1(Ωρ), and ϕh ∈ V 0

h , we have

a(u − uh, ϕ) =

  • ΓD

g(ϕ − ϕh) − b(uh, ϕ − ϕh) + Tuh − ˆ Tuh, ϕΓR. Lets now prove this important Lemma!

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SLIDE 80

Interpolation Operator

Since we are going to interpolate nonsmooth functions satisfying boundary conditions, we resort to an interpolation operator Πh : H1

(0)(Ωh ρ) → V 0 h of Scott-Zhang.

Notation: Let Nh = {ai}N

i=1 be the set of all nodes of Mh.

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SLIDE 81

Interpolation Operator

Since we are going to interpolate nonsmooth functions satisfying boundary conditions, we resort to an interpolation operator Πh : H1

(0)(Ωh ρ) → V 0 h of Scott-Zhang.

Notation: Let Nh = {ai}N

i=1 be the set of all nodes of Mh.

Let {φi}N

i=1 be the corresponding nodal basis of Vh.

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SLIDE 82

Interpolation Operator

Since we are going to interpolate nonsmooth functions satisfying boundary conditions, we resort to an interpolation operator Πh : H1

(0)(Ωh ρ) → V 0 h of Scott-Zhang.

Notation: Let Nh = {ai}N

i=1 be the set of all nodes of Mh.

Let {φi}N

i=1 be the corresponding nodal basis of Vh.

For any node ai which is interior to Ωh

ρ or on the boundary ΓR, we take

σi = e, any side in Bh having ai as one of its vertex.

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SLIDE 83

Interpolation Operator

Since we are going to interpolate nonsmooth functions satisfying boundary conditions, we resort to an interpolation operator Πh : H1

(0)(Ωh ρ) → V 0 h of Scott-Zhang.

Notation: Let Nh = {ai}N

i=1 be the set of all nodes of Mh.

Let {φi}N

i=1 be the corresponding nodal basis of Vh.

For any node ai which is interior to Ωh

ρ or on the boundary ΓR, we take

σi = e, any side in Bh having ai as one of its vertex. For any node ai which is on the boundary Γh

ρ, we take σi as any side

  • n Γh

ρ with one vertex ai.

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SLIDE 84

Interpolation Operator (cont)

Let ai,1 = ai, and {ai,j}2

j=1 the set of nodal points in σi with nodal

basis {φi,j}2

j=1.

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SLIDE 85

Interpolation Operator (cont)

Let ai,1 = ai, and {ai,j}2

j=1 the set of nodal points in σi with nodal

basis {φi,j}2

j=1.

Let {ψi,j}2

j=1 be the L2(σi) dual basis:

  • σi

ψi,j(x)φi,k(x)dx = δjk, j, k = 1, 2.

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SLIDE 86

Interpolation Operator (cont)

We now define the interpolation operator Πh : H1(Ωh

ρ) → Vh to be

Πhv(x) =

N

  • i=1

φi(x)

  • σi

ψi(x)v(x)dx. One can show the following properties of Πh: Πhv ∈ V 0

h if v ∈ H1 (0)(Ωh ρ).

v − ΠhvL2(K) ≤ Chk∇vL2( ˜

K),

v − ΠhvL2(e) ≤ Ch1/2

e

∇vL2(˜

e).

˜ K and ˜ e denote the union of all elements in Mh having non-empty intersection with K ∈ Mh and the side e, respectively.

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SLIDE 87

Fourth Part

IMPLEMENTATION AND EXAMPLES

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SLIDE 88

Implementation

We use the a posteriori error estimate in the main theorem to determine the PML parameters. Just as before, we choose the PML medium property to be a power function. So, only the thickness ρ − R of the layer and the medium parameter σ0 are left to be specified.

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SLIDE 89

Implementation

We use the a posteriori error estimate in the main theorem to determine the PML parameters. Just as before, we choose the PML medium property to be a power function. So, only the thickness ρ − R of the layer and the medium parameter σ0 are left to be specified. First, we choose the exponentially decaying factor to be small such that it becomes negligible compared with the finite element discretization errors. Now, we set up an algorithm to adapt mesh size according to the a posteriori error estimate.

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SLIDE 90

Algorithm

Let TOL > 0 be the tolerance for the error. Set m = 2. Now, the strategy is: Choose ρ and σ0 such that the exponentially decaying factor ˆ ω ≤ 10−8;

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SLIDE 91

Algorithm

Let TOL > 0 be the tolerance for the error. Set m = 2. Now, the strategy is: Choose ρ and σ0 such that the exponentially decaying factor ˆ ω ≤ 10−8; Set the computational domain Ωρ = Bρ\¯ ΓD and generate an initial mesh Mh over Ωρ;

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SLIDE 92

Algorithm

Let TOL > 0 be the tolerance for the error. Set m = 2. Now, the strategy is: Choose ρ and σ0 such that the exponentially decaying factor ˆ ω ≤ 10−8; Set the computational domain Ωρ = Bρ\¯ ΓD and generate an initial mesh Mh over Ωρ; While ERR > TOL do refine the mesh Mh: if ηK > 1

2 max ˆ K∈Mh η ˆ K, refine the element

K ∈ Mh; solve the discrete problem (3.3) on Mh; compute error estimators on Mh;

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SLIDE 93

Algorithm

Let TOL > 0 be the tolerance for the error. Set m = 2. Now, the strategy is: Choose ρ and σ0 such that the exponentially decaying factor ˆ ω ≤ 10−8; Set the computational domain Ωρ = Bρ\¯ ΓD and generate an initial mesh Mh over Ωρ; While ERR > TOL do refine the mesh Mh: if ηK > 1

2 max ˆ K∈Mh η ˆ K, refine the element

K ∈ Mh; solve the discrete problem (3.3) on Mh; compute error estimators on Mh; End While.

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SLIDE 94

Example 1: Unit circle

Let the scatterer D be the unit circle. Let the exact solution be u = H(1)

0 (kr), where r = |x|. Take R = 2, and k = 1. (ρ = 4R and σ0 = 10)

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SLIDE 95

Example 1: Unit circle (cont)

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SLIDE 96

Example 2

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SLIDE 97

Example 2 (cont)

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SLIDE 98

Example 2

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SLIDE 99

Example 2

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SLIDE 100

The End

Remarks / Questions

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