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Scattering of transient waves AR & FJS & TQ by piecewise - - PowerPoint PPT Presentation

A piecewise homogeneous world of waves Scattering of transient waves AR & FJS & TQ by piecewise homogeneous obstacles Notation Problem Approximation Alexander Rieder & FranciscoJavier Sayas Theory & Tianyu Qiu


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A piecewise homogeneous world of waves AR & FJS & TQ Notation Problem Approximation Theory Estimates TDBIE Full discretization

Scattering of transient waves by piecewise homogeneous obstacles

Alexander Rieder & Francisco–Javier Sayas & Tianyu Qiu

Technical University of Vienna Department of Mathematical Sciences, University of Delaware Rice University

RICAM Workshop November 9, 2016

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The team (different stages of 2016)

Work with Alex Rieder (visiting from TUW), strongly based

  • n Tim Qiu’s with add-ons by Tom Brown
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Contents, disclaimers, warnings

Work in progress Acoustic waves on piecewise homogeneous materials — Galerkin BEM in space — RKCQ in time Don’t bother with the elastic case (trivial, trivial) Maxwell (I’m pretty, pretty, pretty sure it’s quite similar) Our numerical experiments are just convergence lines, showing not entirely expected results I’ll talk about analysis mainly Apologies for the very mathy slides

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Analysis? Really?

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Literature

Single trace formulations (FD): Von Petersdorff, 1989 Single and multi-trace formulations (FD): Claeys & Hiptmair, 2013. Additional work by them and Jerez-Hanckes Time domain tools from theory of evolutionary equations (say you own a copy of Pazy’s book and you understand what’s in it) Results on RK for abstract evolutionary equations: Alonso-Mallo & Palencia, 2003. With sequels RKCQ theory by Banjai, Lubich, & Melenk, 2011 The formulation can be used with other discretization methods (full Galerkin, etc)

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NOTATION AND SCATTERING PROBLEM

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Geometric setting

Ωℓ (ℓ = 1, . . . , L) mutually disjoint bounded Lipschitz domains (with possibly intersecting boundaries) Surrounding domain (air or something) Ω0 := Rd \

L

  • ℓ=1

Ωℓ Skeleton Γ :=

L

  • ℓ=1

∂Ωℓ Piecewise constant physical parameters: κ, c : Rd → (0, ∞) κ|Ωℓ ≡ κℓ > 0, c|Ωℓ ≡ cℓ > 0, ℓ = 0, . . . , L

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Scattering problem (scattered wave)

Incident wave. uinc : R → H1

loc(Ω0) (Extended by zero to

the interior domains as uinc.) c−2

0 uinc(t) = κ0∆uinc(t)

∀t ∈ R, supp uinc(t) ⊂ Ω0 ∀t ≤ 0. Scattered wave-field u : [0, ∞) → H1(Rd \ Γ). For alll t ≥ 0 c−2¨ u(t) = ∇ · (κ∇u)(t) in Rd \ Γ u(t) − uinc(t) ∈ H1(B) κ∇(u(t) − uinc(t)) ∈ H(div, B) u(0) = 0, ˙ u(0) = 0

(What’s B)

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Traces on one boundary

Interior, exterior, jump, average... γint

ℓ , γext ℓ , γℓ ·, {

{γℓ ·} } : H1(Rd \ ∂Ωℓ) → H1/2(∂Ωℓ) γℓu := γint

ℓ u − γext ℓ u,

{ {γℓu} } := 1

2(γint ℓ u + γext ℓ u).

Normal traces... γint

ν,ℓ, γext ν,ℓ, γν,ℓ ·, {

{γν,ℓ ·} } : H(div, Rd \ ∂Ωℓ) → H−1/2(∂Ωℓ)

All normals point out

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Packing up spaces and traces

Collecting fields in product spaces Hdiv :=

L

  • ℓ=0

H(div, Rd \ ∂Ωℓ), H :=

L

  • ℓ=0

H1(Rd \ ∂Ωℓ), H−1/2 :=

L

  • ℓ=0

H−1/2 (∂Ωℓ), H1/2 :=

L

  • ℓ=0

H1/2 (∂Ωℓ), γint, γext, γ ·, { {γ ·} } : H → H1/2, γint

ν , γext ν , γν ·, {

{γν ·} } : Hdiv → H−1/2

The field inside Ωℓ will be extended by zero outside. Why?

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ANOTHER WAY OF WRITING THE PROBLEM

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Enforcing interface continuity

Y : = {(γint

ℓ u)L ℓ=0 : u ∈ H1(Rd)}

= {φ ∈ H1/2 : ∃u ∈ H1(Rd), φ = γintu}, X : = {(γint

ν,ℓv)L ℓ=0 : v ∈ H(div, Rd)}

= {λ ∈ H−1/2 : ∃v ∈ H(div, Rd), λ = γint

ν v}.

Lemma (Restricting and gluing — odd conditions) Let U ∈ H satisfy γU ∈ Y and γextU ∈ Y and let u : Rd → R be defined by u|Ωℓ := uℓ|Ωℓ. Then u ∈ H1(Rd). Similarly, let V ∈ Hdiv satisfy γνV ∈ X and γext

ν V ∈ X and

let v : Rd → Rd be defined by v|Ωℓ := vℓ|Ωℓ. Then v ∈ H(div, Rd).

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A trivial example: disjoint homogeneous

  • bstacles

H±1/2(∂Ω0) ≡

L

  • ℓ=1

H±1/2(∂Ωℓ)

L

  • ℓ=1

H1/2(∂Ωℓ) ∋ (φ1, . . . , φL) ← → ((φ1, . . . , φL), φ1, . . . , φL) ∈ Y

L

  • ℓ=1

H−1/2(∂Ωℓ) ∋ (λ1, . . . , λL) ← → ((−λ1, . . . , −λL), λ1, . . . , λL) ∈ X

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An observation and more notation

The spaces X and Y are respective polar to each other, i.e., Y◦ = {µ ∈ H−1/2 : φ, µΓ = 0 ∀φ ∈ Y} = X and X ◦ = {φ ∈ H1/2 : φ, µΓ = 0 ∀µ ∈ X} = Y

I call’em polar sets because I cannot pronounce annihilator

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The reformulation: with pieces

Data β0 := (γint

0 uinc, 0, . . . , 0),

β1 := (κ0γint

ν,0∇∂−1 t

uinc, 0, . . . , 0) Unknowns U = (uℓ)L

ℓ=0 : [0, ∞) → H,

V = (vℓ)L

ℓ=0 : [0, ∞) → Hdiv

First order system (ℓ = 0, . . . , L) ˙ uℓ(t) = c2

ℓ ∇ · vℓ(t),

˙ vℓ(t) = κℓ∇uℓ(t), t ≥ 0 Transmission conditions (t ≥ 0) γU(t) + β0(t) ∈ Y γνV(t) + β1(t) ∈ X γextU(t) ∈ X ◦ = Y γext

ν V(t) ∈ Y◦

Vanishing initial conditions

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The formulation: all together and why it works

Data: Uinc := ( uinc, 0, . . . , 0) Unknowns U : [0, ∞) → H, V : [0, ∞) → Hdiv First order system ˙ U(t) = T c2∇ · V(t), ˙ V(t) = T κ∇U(t), t ≥ 0 Transmission conditions γU + Uinc(t) ∈ Y γext(U + Uinc)(t) ∈ Y γνT κ∇(U + Uinc)(t) ∈ X γext

ν T κ∇(U + Uinc)(t) ∈ X

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AN APPROXIMATION TO THE PROBLEM

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Discrete spaces

Choose two closed (finite dimensional) subspaces Xh ⊂ X Yh ⊂ Y Look at their polar sets X ◦

h := {φ ∈ H1/2 : φ, µΓ = 0

∀µ ∈ Xh}, Y◦

h := {λ ∈ H−1/2 : ψ, λΓ = 0

∀ψ ∈ Yh}. Note that Xh ⊂ X = Y◦ ⊂ Y◦

h

Yh ⊂ Y = X ◦ ⊂ X 0

h .

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Continuous vs semidiscrete (1)

Unknowns U : [0, ∞) → H, V : [0, ∞) → Hdiv First order system ˙ U(t) = T c2∇ · V(t), ˙ V(t) = T κ∇U(t), ∀t, t ≥ 0 Transmission conditions γU(t) + β0(t) ∈ Y γνV(t) + β1(t) ∈ X γextU(t) ∈ X ◦ γext

ν V(t) ∈ Y◦

Vanishing initial conditions

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Continuous vs semidiscrete (2)

Unknowns Uh : [0, ∞) → H, V h : [0, ∞) → Hdiv First order system ˙ Uh(t) = T c2∇ · V h(t), ˙ V

h(t) = T κ∇Uh(t),

t ≥ 0 Transmission conditions γUh(t) + β0(t) ∈ Yh γνVh(t) + β1(t) ∈ Xh γextUh(t) ∈ X ◦

h

γext

ν V h(t) ∈ Y◦ h

Vanishing initial conditions

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How to build Xh and Yh (1)

Decompose skeleton into flat faces Γ = Γ1 ∪ . . . ∪ ΓM so that ∂Ωℓ = ∪{Γi : i ∈ I(ℓ)} ∀ℓ Triangulate Γ respecting the faces: Γh Spaces on the skeleton: Ph := {λh : Γ → R : λh|e ∈ Pk(e) ∀e ∈ Γh}, Qh := {φh ∈ C(Γ) : φh|e ∈ Pk+1(e) ∀e ∈ Γh}, where Pk(e) is the space of polynomials of degree up to k defined on (tangential coordinates of) e

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How to build Xh and Yh (2)

Construction of Yh ≡ Qh Qh := {φh ∈ C(Γ) : φh|e ∈ Pk+1(e) ∀e ∈ Γh} Yh := {(φh|∂Ωℓ)L

ℓ=0 : φh ∈ Qh}.

Construction of Xh ≡ Ph Ph := {λh : Γ → R : λh|e ∈ Pk(e) ∀e ∈ Γh}, Xh := {(sℓλh|∂Ωℓ)L

ℓ=0 : λh ∈ Ph},

where sℓ : Γ → {−1, 0, 1} is an orientation function.

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THE FUNCTIONAL MACHINERY (= YOUR EDUCATION) AT WORK

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More product spaces

Kinetic energy space (U, V) ∈ H := L2(Rd)L+1 × (L2(Rd)d)L+1 Kinetic energy inner product: (U, V)2

H := L

  • ℓ=0

c−2

ℓ uℓ2 Rd + L

  • ℓ=0

κℓvℓ2

Rd.

Yes, (d + 1)(L + 1) fields. All

  • f this is happening to you
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The (kind of) skewsymmetric differential

  • perator

Potential energy space V := H × Hdiv. Operator: A⋆ : V → H A⋆(U, V) := (T c2∇ · V, T κ∇U). (including material parameter scaling) (U, V)V ≡ (U, V)H + A⋆(U, V)H

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The T.C. operator (using the hammer big time)

Space for TCs M := (Y◦

h)′ × (X ◦ h )′ × X ′ h × YY ′ h

Operator for TCs: B : V → M B(U, V) := (γU|Y◦

h , γνV|X ◦ h , γextU|Xh, γext

ν V|Yh).

This might require an explanation.

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It fits, it fits...

  • Data. Ξ : [0, ∞) → M

Ξ(t) := −(β0(t)|Y◦

h , β1(t)|X ◦ h , 0, 0)

Unknowns (Uh, V h) : [0, ∞) → V = H × Hdiv Equations ( ˙ Uh(t), ˙ V

h(t)) = A⋆(Uh(t), V h(t))

∀t ≥ 0, B(Uh(t), V h(t)) = Ξ(t) ∀t ≥ 0, (Uh(0), V h(0)) = 0.

And now what?

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STABILITY AND CONVERGENCE: A THEORETICAL CHECKLIST

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The kernel of B

B(U, V) := (γU|Y◦

h , γνV|X ◦ h , γextU|Xh, γext

ν V|Yh).

The following are equivalent:

1

(U, V) ∈ ker B

2

(note: (X ◦

h )◦ = Xh and (Y◦ h)◦ = Yh)

γU ∈ Yh γνV ∈ Xh γextU ∈ X ◦

h

γext

ν V ∈ Y◦ h

3

(note: Xh ⊂ Y◦

h and Yh ⊂ X ◦ h )

γU ∈ Yh γνV ∈ Xh γintU ∈ X ◦

h

γint

ν V ∈ Y◦ h

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First one out

Conservation (double dissipativity) (A(U, V), (U, V))H = 0 ∀(U, V) ∈ ker B = D(A) Proof. (A(U, V), (U, V))H =

L

  • ℓ=0

γint

ℓ uℓ, γint ν,ℓvℓ∂Ωℓ − γext ℓ uℓ, γext ν,ℓvℓ∂Ωℓ

= γintU, γνVΓ + γU, γext

ν VΓ

= 0

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Second one out

Sign flipping = time reversal The operator Φ(U, V) = (U, −V) ... is an isometric involution in H preserves ker B and satisfies ΦA⋆ = −A⋆Φ.

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The tough cookie

Uniform coercivity For all (F, F) ∈ H and Ξ ∈ M, there exists a unique (U, V) ∈ V such that (U, V) = A⋆(U, V) + (F, F), B(U, V) = Ξ, and there exists a constant C > 0, depending only on the geometry and the physical parameters, and thus independent of the choice of Xh and Yh, such that (U, V)H ≤ C((F, F)H + ΞM).

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Idea of the proof

Write the equations U = T c2∇ · V + F V = T κ∇U + F in terms of U only T c−2U = ∇ · (T κ∇U) + ∇ · F + T c−2F Keep TC for U as essential, use conditions for V (written as conditions for U) as natural, and find your variational

  • formulation. Reconstruct V = T κ∇U + F
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Easier said than done?

a(U, W) :=

L

  • ℓ=0

c−2

ℓ (uℓ, wℓ)Rd + L

  • ℓ=0

κℓ(∇uℓ, ∇wℓ)Rd\∂Ωℓ, b(W) :=

L

  • ℓ=0

c−2

ℓ (fℓ, wℓ)Rd − L

  • ℓ=0

(f ℓ, ∇wℓ)Rd\∂Ωℓ + γintW, ξ2X ◦

h ,(X ◦ h )′ + γW, ξ4Yh,Y′ h.

U ∈ H, γU|Y◦

h = ξ1,

γextU|Xh = ξ3, a(U, W) = b(W) ∀W ∈ W. W :={W ∈ H : γW ∈ Yh, γextW ∈ X ◦

h }

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Consequences (1)

1

The operators ±A are dissipative ±(A(U, V), (U, V))H ≤ 0 ∀(U, V) ∈ D(A) = ker B

2

I ± A : D(A) → H are surjective

3

±A are maximal dissipative

4

A is the infinitesimal generator of a group of isometries in H

5

Strong and mild solutions for IVP ( ˙ U(t), ˙ V(t)) = A(U(t), V(t)) t ∈ R (U(0), V(0)) = (U0, V 0)

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Consequences (2) — wordless slide

( ˙ Uh(t), ˙ V

h(t)) = A⋆(Uh(t), V h(t))

∀t ≥ 0, B(Uh(t), V h(t)) = Ξ(t) ∀t ≥ 0, (Uh(0), V h(0)) = 0. (Uh(t), V h(t)H ≤Ct max

0≤τ≤t(Ξ(τ)M + ˙

Ξ(τ)M), ( ˙ Uh(t), ˙ V

h(t)H ≤Ct max 0≤τ≤t(Ξ(τ)M + ˙

Ξ(τ)M + ¨ Ξ(τ)M). Ξ(t)M ≤ β0(t)H1/2 + β1(t)H−1/2 = γint

0 uinc(t)H1/2(∂Ω0) + κ0∂int ν,0∂−1 t

∇uinc(t)H−1/2(∂Ω0)

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RETARDED POTENTIAL FORMULATION

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Calderón Calculus for the overlapped problem

Kirchhoff’s representation operator G(∂t) :=

  • S(∂t)

−D(∂t)

  • Jump relations

∂ν · γ ·

  • G(∂t) = I

TDBIE C(∂t) := { {∂ν ·} } { {γ ·} }

  • G(∂t) =

K′(∂t) W(∂t) V(∂t) −K(∂t)

  • Exterior traces
  • ∂ext

ν

γext

  • G(∂t) = C(∂t) − 1

2I

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Associated system of TDBIE

Boundary unknowns and data Λh := (γν ˙ V

h + ˙

β

1, γUh + β0)⊤,

Θ := ( ˙ β

1, β0)⊤.

Semidiscrete equations Λh ∈ Xh × Yh, QκC(∂t)Q−1

κ Λh, Ψ = Qκ(C(∂t) − 1 2I)Q−1 κ Θ, Ψ

∀Ψ ∈ Yh × Xh, Reconstruction Uh = G(∂t)Q−1

κ (Λh − Θ),

V h = T κ∇∂−1

t

Uh,

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Equivalent form

Transmission conditions γUh(t) + β0(t) ∈ Yh γνVh(t) + β1(t) ∈ Xh γextUh(t) ∈ X ◦

h

γext

ν V h(t) ∈ Y◦ h

Boundary integral conditions Λh = (γν ˙ V

h + ˙

β

1, γUh + β0)⊤ ∈ Xh × Yh

Λh − Θ = (γν ˙ V

h, γUh)⊤

(γextUh(t), γext

ν

˙ V

h(t))⊤ = Qκ(C(∂t)− 1 2I)Q−1 κ (Λh−Θ) ∈ Y◦ h×X ◦ h

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Exact vs semidiscrete

Λ ∈ X × Y, Qκ(C(∂t) − 1

2I)Q−1 κ (Λ − Θ), Ψ = 0

∀Ψ ∈ Y × X, U = G(∂t)Q−1

κ (Λ − Θ)

Λh ∈ Xh × Yh, Qκ(C(∂t) − 1

2I)Q−1 κ (Λh − Θ), Ψ = 0

∀Ψ ∈ Yh × Xh, Uh = G(∂t)Q−1

κ (Λh − Θ)

We lucked out big time Use Θ = Λ − ΠΛ as data and the stability theorem

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Error estimate

U(t) − Uh(t)H ≤ Ct max

0≤τ≤t 2

  • j=0

Λ(j)(τ) − ΠΛ(j)(τ), φ(t) − φh(t)H1/2 ≤ Ct max

0≤τ≤t 2

  • j=0

Λ(j)(τ) − ΠΛ(j)(τ), λ(t) − λh(t)H−1/2 ≤ Ct max

0≤τ≤t 3

  • j=1

Λ(j)(τ) − ΠΛ(j)(τ).

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A CQRK=RK DISCRETIZATION

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What we discovered worked better

Instead of this Λh ∈ Xh × Yh, Qκ(C(∂t) − 1

2I)Q−1 κ (Λh − Θ), Ψ = 0

∀Ψ ∈ Yh × Xh, Uh = G(∂t)Q−1

κ (Λh − Θ)

(this uses γ0uinc and γν,0∇uinc)

Solve this (D(∂t) = diag(1, ∂−1

t

)) Λh ∈ Xh × Yh, Qκ(C(∂t) − 1

2I)Q−1 κ (Λh − D(∂t)Σ), Ψ = 0

∀Ψ ∈ Yh × Xh, Uh = G(∂t)Q−1

κ (Λh − D(∂t)Σ)

(this uses γ0 ˙ uinc and γν,0∇uinc)

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Equivalent odd PDE problem

Unknowns Uh : [0, ∞) → H, V h : [0, ∞) → Hdiv First order system ∂tUh = T c2∇ · V h(t), ∂tV h(t) = T κ∇Uh(t), t ≥ 0 Transmission conditions ∂tγUh(t) + σ0(t) ∈ Yh ∂tγνVh(t) + σ1(t) ∈ Xh γextUh(t) ∈ X ◦

h

γext

ν V h(t) ∈ Y◦ h

Vanishing initial conditions

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A piecewise homogeneous world of waves AR & FJS & TQ Notation Problem Approximation Theory Estimates TDBIE Full discretization

That Lubich magic

We use CQ for all occurrences of ∂t (algorithms by Sauter, Banjai, Schanz) Λh ∈ Xh × Yh, Qκ(C(∂t) − 1

2I)Q−1 κ (Λh − D(∂t)Σ), Ψ = 0

∀Ψ ∈ Yh × Xh, Uh = G(∂t)Q−1

κ (Λh − D(∂t)Σ)

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SLIDE 47

A piecewise homogeneous world of waves AR & FJS & TQ Notation Problem Approximation Theory Estimates TDBIE Full discretization

It’s equivalent to...

... aplying RK to the following strange ‘ODE’ ∂tUh = T c2∇ · V h(t) ∂tV h(t) = T κ∇Uh(t) ∂tγUh(t) + σ0(t) ∈ Yh ∂tγνVh(t) + σ1(t) ∈ Xh γextUh(t) ∈ X ◦

h

γext

ν V h(t) ∈ Y◦ h

Keyword Order reduction

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A piecewise homogeneous world of waves AR & FJS & TQ Notation Problem Approximation Theory Estimates TDBIE Full discretization

What we know (so far)

Full classical order for fields away from the interfaces (LD analysis following Banjai, Lubich, Melenk) If you don’t work hard you get stage order for boundary fields If you work harder (using Alonso-Mallo & Palencia), you can do better. But not optimal? (We wish we knew) The semidiscretization results apply for full space-time Galerkin Multi-trace? It’s second-kindy...

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SLIDE 49

A piecewise homogeneous world of waves AR & FJS & TQ Notation Problem Approximation Theory Estimates TDBIE Full discretization

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