Boundary integral methods for implicitly defined dynamic interfaces - - PowerPoint PPT Presentation

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Boundary integral methods for implicitly defined dynamic interfaces - - PowerPoint PPT Presentation

Boundary integral methods for implicitly defined dynamic interfaces Celebrating Prof. Yoshikazu Gigas 60th Birthday Richard Tsai with C. Chen, C. Kublik, Y. Wu KTH Royal Institute of Technology, Sweden and The University of Texas


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SLIDE 1

Boundary integral methods for implicitly defined dynamic interfaces

Richard Tsai

with 


  • C. Chen, C. Kublik, Y. Wu

KTH Royal Institute of Technology, Sweden
 and
 The University of Texas at Austin, USA

Celebrating Prof. Yoshikazu Giga’s 60th Birthday

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SLIDE 2

Among the important things

  • The existence of a second stomach for sweets
  • The importance of “unagi” for survival of summer
  • The completion of a meal by soba-yu
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The L-solutions of Giga-Sato

Embed the solution graph as a level curve of a higher dimensional function, and look at convergence in Hausdorff distance of the subgraphs:

u(x) x y x phi>0 phi<0

ut +H(t,x,u,ux) = 0 ⇐ ⇒ φt −φyH(t,x,y,−φx φy ) = 0.

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SLIDE 4

Multivalued solutions

−0.4 −0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 −0.4 −0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 x y

v(y)

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SLIDE 5

Very singular diffusion of Giga

φt + ˜ H(t,x,y,φx,φy) = M|∇φ| ∂ ∂y( φy |φy|).

Numerical solution verifies the entropy condition. (Equal area rule).

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −0.4 −0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −0.4 −0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

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SLIDE 6

Bunching and anisotropy

−2.5 −2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 −2.5 −2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 −2.5 −2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5 −2.5 −2 −1.5 −1 −0.5 0.5 1 1.5 2 2.5

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SLIDE 7

[Ohtsuka-Giga-T:2013] Computation by T. Ohtsuka

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SLIDE 8

Multivalued solutions for high frequency waves

  • tS þ jrxSj2

2 þ V ðxÞ ¼ 0;

[Jin,Liu, Osher-T:2005]

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SLIDE 9

Geometric motion of an interface

Mullins-Sekerka dynamics: Ω(t) Ω(t) vn = ∂u ∂n

  • normal velocity of a moving interface

∆u = 0 x / ∈ ∂Ω(t) u = κ, x ∈ ∂Ω(t)

+ far field condition

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SLIDE 10

Solution of linear PDEs by boundary integrals

  • Solving for :

γ

  • Evaluating the solution:

f(x) = λγ(y) + Z

∂Ω

K(x, y)γ(y)dSy u(x) = Z

∂Ω

˜ K(x, y)γ(y)dSy

∆u = 0, x ∈ Ω u(x) = f(x), x ∈ ∂Ω

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SLIDE 11

Breaking of convexity

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SLIDE 12

Volume preservation

−4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4

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SLIDE 13

Merging

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SLIDE 14

Nontrivial surface area-volume relation

Isothermal at infinity.

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SLIDE 15

Closed simple curve Γ : γ(s) = (X(s),Y(s)), parametrized by arclength s. I = Z f ◦γ(s)ds = Z

R2 f(x,y)δ(Γ,x,y)dxdy.

φ<0 φ>0

Explicit curves:

δ(Γ,x,y) =

R δ(x − X(s))δ(y − Y(s))ds. Implicit curves:

δ(Γ,x,y) = δ(φ(x,y))|∇φ(x,y)|.

Surface integral: Z

Rd f(x)δ(φ)|∇φ|dx.

[Peskin, Lai]

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SLIDE 16

Eε,h =

  • ∑i,j δε(Γ,xi,j)f(xi,j)h2 −

R

Γ f(γ(s))ds

  • Typically Eε,h −

→ 0 if ε = hα,0 ≤ α < 1. (wide support!)

Non-convergence if e.g. δcos

Ch

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SLIDE 17

Relative error E = |Sh − S|/S. Radii: r× > r◦ > r.

(a) Γ

200 400 0.06 0.08 0.1 0.12

(b) δL

h

200 400 0.01 0.012 0.014 0.016

(c) δL

2h

10

2

10

3

10

−5

10

−4

10

−3

10

−2

(d)

δcos

2h (dΓ)

10

2

10

3

10

−2

10

−1

(e)

δcos

2h (phi)

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SLIDE 18

Theorem

∑i,j∈Ip,q δL

ε(dΓ(xi,j))h2 exact,

if (p,q) = ∇dΓ are relative prime and ε = ε(p,q) =

|p|+|q|

p2+q2 h.

ε (p,q) Γ

−1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5 Polar plot of ε(θn) θn ε(θn)

[Engquist-Tornberg-T:2005]

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SLIDE 19

Defining a “fat” integral equation

  • Proper definition of the equation for computation using

implicit surfaces: 
 
 
 
 Need extension of and the unknown from to

  • Parameterization of the interface using nearby level sets
  • Averaging over different parameterizations

f

γ

∂Ω {−✏ < < ✏}

f(x) = λρ(y) + Z

∂Ω

Φ(x, y)ρ(y)dSy

fext(x) ⇡ λρext(y) + Z

Rd Φ(x, y)ρext(y) δ✏(φ)|rφ|dy ?

∂Ω = {x : φ(x) = 0}

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SLIDE 20

Advantages of implicit interface boundary integral method

  • Inherit certain advantages of boundary integral methods: 


solution on exterior domain problem, easy for both Dirichlet, Neumann, Robin, transmission problems.

  • Inherit certain benefits from implicit interface formulation: 

  • ne grid for complicated, topologically changing

dynamic interface

  • Flexible on the grid geometry used to embed the
  • interface. Multi-resolution grid possible.
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SLIDE 21

Inverse problem and shape optimization

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SLIDE 22

Analysis of cancellous bone from micro-CT scan

Compute effective material properties.

Segmented by an MBO type scheme [Esedoglu-T]

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SLIDE 23

LIDARs and 3D scanners

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SLIDE 24

New formulas for integration over implicit surfaces

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SLIDE 25

∂Ω ∂Ωη

y∗ z∗ = y∗ + J ~ ds z = y + ~ ds y

Parametrizing by a nearby level set

y∗ = y ηrd(y) Z

∂Ω

ρ(γ(s))Φ(x, γ(s))ds = Z

∂Ωη

ρ(y∗(sη))Φ(x, y∗(sη))Jηdsη

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SLIDE 26

θ1 θ2 ⌘~ n ds dτ

R2 R1

The Jacobian

A0 = (R1 − η)θ1 · (R2 − η)θ2 Aη = ds dτ = R1θ1 · R2θ2 Aη A0 = R1R2 (R1 − η)(R2 − η) = 1 + (κ1 + κ2)η + O(η2) Jη = 1 + 2Hηη + Gηη2 In fact, we have:

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SLIDE 27

Average the identities

h z∗ = z d(z)rd(z) I0 = Z

∂Ω

ρ(γ(s))Φ(x, γ(s))ds = Z

∂Ωη

ρ(y∗(sη))Φ(x, y∗(sη))Jηdsη = Iη I0 = Z

@Ω

ρ(γ(s))Φ(x, γ(s))ds = Z ✏

−✏

δ✏(η)I⌘dη = Z

Rn ρ(z∗)Φ(x, z∗)δ✏(dΩ(z))J(z; dΩ)dz

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SLIDE 28

The singular values of

Proposition 3. The following identity holds for sufficiently small ✏: ˆ

Σ

g(x)dSx = ˆ

R3 g(PΣ(y)) 12K✏((y))dy,

where K✏ is any symmetric kernel supported in [−✏, ✏] having unit mass.

λ1λ2 = 0 at the boundary of Σ

DPΣ

PΣy =y∗ =y φ(y)rφ(y)

φ(y) = min

x∈Σ |y − x|

[Kublik-T:2015]

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SLIDE 29

The Jacobian and singular values of

Proposition 1. Let 1(x) be the largest singular value of DPΓ(x). The following identity holds for ✏ < ∞, where ∞ is the maximal unsigned curvature of Γ. ˆ

Γ

gds = 1 2⇡ ˆ

R3 g(PΓ(x))1(x)K✏()

  • dx.

0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 2.5 x K(x)=(1−cos(2 π x))/x

δ✏(x; Γ) λ1 = 0 near the end points of Γ

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SLIDE 30

High order approximation

n Relative Error Order 32 6.2030 ⇥ 10−3

  • 64

1.8073 ⇥ 10−4 5.10 128 6.6838 ⇥ 10−6 4.76 256 4.1530 ⇥ 10−7 4.01 512 5.0379 ⇥ 10−8 3.04

Integration over a torus:

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SLIDE 31

Surfaces with boundaries

n Relative Error Order 32 1.1726 ⇥ 10−2

  • 64

1.1733 ⇥ 10−3 3.32 128 9.1325 ⇥ 10−4 0.36 256 3.8238 ⇥ 10−4 1.26 512 7.8308 ⇥ 10−5 2.29

3/4 sphere and third-order one-sided finite differencing:

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SLIDE 32

n Relative Error Order 60 5.5078 ⇥ 10−3

  • 120

1.1476 ⇥ 10−3 2.63 240 2.3409 ⇥ 10−4 2.29 480 3.7166 ⇥ 10−5 2.66

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SLIDE 33

Summation over point clouds

−1 1 −1 1 −1 −0.5 0.5 1

  • 30×30 uniformly distributed point clouds sampling in spherical coordinate the quarter

sphere patch.

  • 50 × 50 × 50 uniform Cartesian grid discretizing [−1, 1]3.
  • Relative error using = 0.05 = dx: −0.56.
  • Relative error using = 0.2 = 4dx: −0.061.

has improved regularity

{dΓN = η} P : x 2 {dΓN = η} 7! x ηrdΓN (x) “interpolates”

ΓN ⊂ Σ

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SLIDE 34

Implicit interface boundary integral equation

Discretized by simple Riemann sum over the grid nodes. Solve the resulting linear system by an iterative solver.

x, y ∈ {−✏ ≤ dΩ ≤ ✏} δ✏(x; dΩ) := δ✏(dΩ(x))J(x; dΩ) x∗ = x dΩ(x)rdΩ(x)

[Kublik-Tanushev-T:2013]

f(x∗) = λρ(x∗) + Z

Rd Φ(x∗, y∗)ρ(y∗)δ✏(y; dΩ)dy

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SLIDE 35

Regularization

  • Need regularization of near singularity


  • Approximate surfaces by paraboloids
  • Approximate weakly in a neighborhood of x on the

paraboloid (the 3D case):

x Ω y

∂Φ ∂n ∂Φ ∂n ∂Φ ∂n (x, y) ' A✏(x, y) for ||x y|| < ε Z

U(x;")

∂Φ ∂ny (x, y)α(y)dS(y) ' Z

˜ U(x;")

A✏(x, y)α(y)dS(y)

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SLIDE 36

Regularization

  • sculating paraboloid

Z

U(x;h)

∂Φ ∂ny (x, y)dS(y) ' 1 8πh(κx + κy)|U(x; h)|

O P 0

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SLIDE 37

Condition numbers

h

  • Cond. number with the tangent regularization
  • Cond. number with the paraboloid regularization

4dx 6.2113 7.8322 2dx 7.3343 6.9390 dx 7.0467 7.4197

dx 2

6.8948 6.7572 h

  • Cond. number with the tangent regularization
  • Cond. number with the paraboloid regularization

4dx 8.0859 7.8024 2dx 8.2791 7.7919 dx 7.4127 8.1231

dx 2

7.8830 8.1192

N = 503 N = 803

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SLIDE 38

Exterior Neumann Problem for the Helmholtz Equation

dx Re(ue) Re(uKT T ) ErrKT T Re(unew) Errnew

4 256 4 512

dx Im(ue) Im(uKT T ) ErrKT T Im(unew) Errnew

4 256 4 512

k = 2.4048255577 . . . ✏ = 0.145 0 = 0.005 ✏0 = 0.15

     ∆u(x) + k2u(x) = 0 x ∈ ¯ Ωc

∂u ∂n(x) = g(x)

x ∈ ∂Ω lim|x|→∞ |x|

1 2 (

∂ ∂|x| − ik)u(x) = 0

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SLIDE 39

Solutions to Helmholtz equation

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SLIDE 40

Happy Birthday!!