Immersed Interface Methods for Elliptic Boundary Value ...€¦ · Graduiertenkolleg Mathematik und...

Immersed Interface Methods for EllipticBoundary Value Problems

M.Sc. Vita Rutka

Supervisor: Prof. Dr. H. Neunzert

25.02.2005

Financial support:

Fraunhofer ITWM Kaiserslautern, Department Flows and Complex Structures

Graduiertenkolleg Mathematik und Praxis

Technical University of Kaiserslautern, Department of Mathematics

0-0

Problem Statement

complex problemsfrom

real life modelling

0-1

Problem Statement

iterative approach

many linear problemshave to be solved

complex problemsfrom

real life modelling

0-2

Problem Statement

iterative approach

many linear problemshave to be solved

for the linear case

need fast & accurate

solvers

complex problemsfrom

real life modelling

0-3

Problem Statement

iterative approach

many linear problemshave to be solved

for the linear case

need fast & accurate

solvers

complex problemsfrom

real life modelling

Concrete problem: complex domains

0-4

Equations to solve

Linear elasticity

µ∆~u + (µ + λ)grad div ~u = ~f in Ω

~u = ~uD on ∂ΩD , σ~n = ~τ on ∂ΩN

~u : displacementsσ : stress tensorε : strain tensor

σkm = 2µεkm + λδkmεqq

εjk =1

2

“

∂kuj + ∂ju

k”

Poisson as “the” elliptic PDE

∆u = f in Ω

u = uD on ∂ΩD , ∂nu = τ on ∂ΩN

0-5

IIM Road Map

RedEJIIM

methodIB

GhostFluid

IIM

EJIIM

REGULAR GRID

2nd order methodsPreserved jumpsSharp interface

explicit variablesC

artesian jumps as

"Preserved" jumps"Sharp" interface

less additionalvariables

0-6

IIM Road Map

RedEJIIM

methodIB

GhostFluid

IIM

EJIIM

REGULAR GRID

"Preserved" jumps"Sharp" interface

2nd order methodsPreserved jumpsSharp interface

explicit variablesC

artesian jumps as

less additionalvariables

0-7

EJIIM

PSfrag replacements

L~u = ~f

b.c.

Ω

Embed the original domain in a“box”. B.C.⇒ Jumps

Standard FD at the regular points

Irregular points:

ΛU + correction = F

correction =X

s

2X

‖~α‖1=0

ψm,s[∂mu

i]αs

Extrapolation + B.C. ⇒h

∂~αu

ii

αs

= Fs −X

i∈grid

ds,jUij

0-8

EJIIM

PSfrag replacements

L~u = ~f

L~u = ~0

Ω−

Ω+

jumps

b.c

.

1. Embed the original domain in a“box”. B.C.⇒ Jumps

Standard FD at the regular points

Irregular points:

ΛU + correction = F

correction =X

s

2X

‖~α‖1=0

ψm,s[∂mu]αs

Extrapolation + B.C. ⇒h

∂~αu

ii

αs

= Fs −X

i∈grid

ds,iUi

0-9

EJIIM

PSfrag replacements

ΛU = F

ΛU = 0

α1

α2

α3

α4

jumps

b.c

.

1. Embed the original domain in a“box”. B.C.⇒ Jumps

2. Standard FD at the regular points

Irregular points:

ΛU + correction = F

correction =X

s

2X

‖~α‖1=0

ψm,s[∂~αu

i]αs

Extrapolation + B.C. ⇒h

∂~αu

i

αs

= Fs −X

i∈grid

ds,iUi

0-10

EJIIM

PSfrag replacements

ΛU = F

ΛU = 0

α1

α2

α3

α4

jumps

b.c

.

1. Embed the original domain in a“box”. B.C.⇒ Jumps

2. Standard FD at the regular points

Irregular points:

ΛU + c.t. = F

c.t. =X

s

2X

‖~α‖1=0

ψ~α,s[∂mu

i]αs

Extrapolation + B.C. ⇒h

∂~αu

i

αs

= Fs −X

i∈grid

ds,iUi

0-11

EJIIM

PSfrag replacements

ΛU = F

ΛU = 0

~y1~y2

~y3

~y4

jumps

b.c

.

1. Embed the original domain in a“box”. B.C.⇒ Jumps

2. Standard FD at the regular points

Irregular points:

ΛU + c.t. = F

c.t. =X

s

X

i

2X

‖~α‖1=0

ψi,~α(~ys)[∂~αu

i]~ys

Extrapolation + B.C. ⇒h

∂~αu

ii

~ys

= Fs −X

j∈grid

ds,jUij

0-12

EJIIM

PSfrag replacements

ΛU = F

ΛU = 0

~y1~y2

~y3

~y4

jumps

b.c

.

1. Embed the original domain in a“box”. B.C.⇒ Jumps

2. Standard FD at the regular points

Irregular points:

ΛU + c.t. = F

c.t. =X

s

X

i

2X

‖~α‖1=0

ψi,~α(~ys)[∂~αu

i]~ys

Extrapolation + B.C. ⇒h

∂~αu

ii

~ys

= Fs −X

j∈grid

ds,jUij

EJIIM system

„A Ψ

D I

« „U

J

«

=

„F

F

«

U : grid function J : jumpsA : standard FD matrix F : extended rhsΨ : corrections I : identity

D, F : extrapolation

0-13

EJIIM Results

Computations done for 3D Poisson and elasticity equations

2-nd order convergence in max norm wrt the grid refinement confirmed

“Toy” for real situations: cut torus

Example: compression, mild steel

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

x 10−3

0 0.02 0.04 0.06 0.08 0.10

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

x

ynorm of displacements

1

2

3

4

5

6

7

0 0.02 0.04 0.06 0.08 0.10

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

x

y

von Mises stress

‖~u‖2Von Mises stress(“color zoom”)

0-14

EJIIM Results: Textiles

Example: compression, mild steel

‖~u ‖2von Misesstress

von Mises stressalong three cuts

5 10 15

0 20 40 60

0

20

40

60

x3 = 17

x1

x2

2 4 6 8 10

0 20 40 60

0

20

40

60

x3 = 29

x1

x2

2 4 6 8 10 12

0 20 40 60

0

20

40

60

x3 = 41

x1

x20-15

Fact:EJIIM system actually is solved by reducing it to the Schur complement forjumps

(I−DA−1Ψ)J = F −DA−1F

Observation:unknown variables are living only at the boundary

Conclusion:method is cheap

INDEED?

0-16

Fact:EJIIM system actually is solved by reducing it to the Schur complement forjumps

(I−DA−1Ψ)J = F −DA−1F

Observation:unknown variables are living only at the boundary

Conclusion:method is cheap

INDEED?

0-17

YES!at least — if seen asymptotically when h −→ 0

on realistic meshes. . . not always. . .

0-18

YES!at least — if seen asymptotically when h −→ 0

on realistic meshes. . . not always. . .

0-19

YES!at least — if seen asymptotically when h −→ 0

on realistic meshes. . . not always. . .

0-20

YES!at least — if seen asymptotically when h −→ 0

on realistic meshes. . . not always. . .

0-21

YES!at least — if seen asymptotically when h −→ 0

on realistic meshes. . . not always. . .

EJIIM (linear elasticity): at each intersection point

jumps in up to (at least) 2-nd order derivatives as unknowns ⇒ 10×3 = 30variables at each intersection point!

Without “smoothing property”: 57 variablesint. point would be needed!

each jump approximated using ≈ 50 grid points(volume approximation by least squares). Coefficients are stored.

Simple calculation — technical textile:

603 grid points ⇒ 603 × 3 = 648000 volume variables30584 intersection points ⇒ 30584 × 30 = 917520 > 648000 jump variables!

Take approximation into account: 30584 × 30 × 50 = 45876000

0-22

YES!at least — if seen asymptotically when h −→ 0

on realistic meshes. . . not always. . .

EJIIM (linear elasticity): at each intersection point

jumps in up to (at least) 2-nd order derivatives as unknowns ⇒ 10×3 = 30variables at each intersection point!

Without “smoothing property”: 57 variablesint. point would be needed!

each jump approximated using ≈ 50 grid points(volume approximation by least squares). Coefficients are stored.

Simple calculation — technical textile:

603 grid points ⇒ 603 × 3 = 648000 volume variables30584 intersection points ⇒ 30584 × 30 = 917520 > 648000 jump variables!

Take approximation into account: 30584 × 30 × 50 = 45876000

0-23

situation improves with grid refinement

Technical textile after refinement:1203 × 3 = 5184000 volume variables

120652 intersection points⇒ 3619560 < 5184000 jump variables

number of grid pointsper direction

costs

computational cubic

quadratic

0-24

situation improves with grid refinement

Technical textile after refinement:1203 × 3 = 5184000 volume variables

120652 intersection points⇒ 3619560 < 5184000 jump variables

number of grid pointsper direction

costs

computational cubic

quadratic

availableresources

But:

often a refinement is not possible!

an algorithm improvement is needed

0-25

situation improves with grid refinement

Technical textile after refinement:1203 × 3 = 5184000 volume variables

120652 intersection points⇒ 3619560 < 5184000 jump variables

number of grid pointsper direction

costs

computational cubic

quadratic

quadratic

availableresources

But:

often a refinement is not possible!

an algorithm improvement is needed

0-26

Consider equations in divergence form

div(L~u ) = ~f in Ω , ~u = ~uD on ∂ΩD , L~u~n := D~u = ~τ on ∂ΩN

L: a first order differential operatorD: co-normal derivative operator

Poisson: L = grad ~u

Linear elasticity: L = σ

Fact:solution can be determined from the so called Cauchy data (~u and D~u) onthe boundary:

~u =

∫

∂Ω

g(~x, ~z )D~u(~z ) ds~z −∫

∂Ω

D~z g(~x, ~z )~u(~z ) ds~z +

∫

Ω

g(~x, ~z )f(~z ) d~z

seek an algorithm which would use only [~u ] and [D~u ] as additionalvariables

0-27

RedEJIIM

EJIIM system:

AU + ΨJ = F

DU + J = F, J = ([~u ] , [∂~u ] , [∂2~u ])

0-28

RedEJIIM

EJIIM system:

AU + ΨJ = F

DU + J = F, J = ([~u ] , [∂~u ] , [∂2~u ])

Construct an operator

J = extension(Jred) = EJred

︸ ︷︷ ︸

linear part

+ ext︸︷︷︸

constant part

Tools: differentiation along the interface + least squares fit

RedEJIIM system

AU + ΨEJred = F − Ψext

DredU + Jred = 0, Jred = ([~u ] , [D~u ])

0-29

RedEJIIM

EJIIM system:

AU + ΨJ = F

DU + J = F, J = ([~u ] , [∂~u ] , [∂2~u ])

Construct an operator

J = extension(Jred) = EJred

︸ ︷︷ ︸

linear part

+ ext︸︷︷︸

constant part

Tools: differentiation along the interface + least squares fit

RedEJIIM system

AU + ΨEJred = F − Ψext

DredU + Jred = 0, Jred = ([~u ] , [D~u ])

0-30

RedEJIIM Results in 3D

second order convergence confirmed in simple geometries (spheres)

“Toy” for real situations: cut torus

Example: temperature distributionGiven: temperature at the cuts, isolation otherwise

2 4 6 8 10

0 0.02 0.04 0.06 0.08 0.10

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

x1

x2

x3 = 0.049

0.5 1 1.5 2

x 10−3

0 0.02 0.04 0.06 0.08 0.10

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

x1

x2

x3 = 0.049

EJIIM solution |uEJIIM − uRedEJIIM|

‖uEJIIM − uRedEJIIM‖∞ ≈ 2.8e-3 , ‖u‖∞ = 10 , 1013 grid points

0-31

RedEJIIM Results: Textiles

Example: temperature distribution

EJIIM solution RedEJIIM solution

difference:‖uEJIIM− uRedEJIIM‖∞ ≈ 0.16, ‖u‖∞ = 5

0-32

May be you wonder . . .

Answer:

For the moment the linear elasticityworks in “easy” geometries.

For more complicated cases

refer to the future research!

0-33

May be you wonder . . .

Poisson solutions?show only

Why does she

Answer:

For the moment the linear elasticityworks in “easy” geometries.

For more complicated cases

refer to the future research!

0-34

May be you wonder . . .

Poisson solutions?show only

Why does she

Answer:

For the moment the linear elasticityworks in “easy” geometries.

For more complicated cases

refer to the future research!

0-35

May be you wonder . . .

Poisson solutions?show only

Why does she

Answer:

For the moment the linear elasticityworks in “easy” geometries.

For more complicated cases

refer to the future research!

0-36

May be you wonder . . .

Poisson solutions?show only

Why does she

Answer:

For the moment the linear elasticityworks in “easy” geometries.

For more complicated cases

refer to the future research!

0-37

Algorithms

Convergence Analysis

0-38

EJIIM system:(A ΨD I

)(UJ

)

=

(F

F

)

solve:

1. (I −DA−1Ψ)J = F − DA−1F

2. U = A−1(F −ΨJ)

in the following:

analyse A−1

important not only for EJIIM but also many other methods!

0-39

“Fact”:

AU = F

‖U‖∞ ≤ C‖F‖∞

A: a matrix of discrete Laplace

C: some constant

0-40

a little bit of

Experimental Numerics

0-41

A numerically observed property of inverse finite difference Laplaceand linear elasticity operators:

in max norm under grid refinements

truncation error: solution error:

PSfrag replacementsO(h2)

O(h2)

O(h)

applyA−1

PSfrag replacements

O(h2)O(h)

O(h2)

O(h2)

O(h 2)

0-42

A Small Experiment

Λu =1

h2(ui+1 − 2ui + ui−1) =

−1 at x = α

0 otherwise, u(0) = u(1) = 0

α : some grid point

rhs: solution:

1

−1

−0.5

0

x

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 1

0

0.002

0.004

0.006

0.008

0.01

x

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

Observe:‖u‖∞ decreases as O(h).

Indeed:

uL = h(1 − α)x

uR = αh(1 − x)= O(h)

0-43

A Small Experiment

Λu =1

h2(ui+1 − 2ui + ui−1) =

−1 at x = α

0 otherwise, u(0) = u(1) = 0

α : some grid point

rhs: solution:

1

−1

−0.5

0

x

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 1

0

0.002

0.004

0.006

0.008

0.01

x

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

Observe:‖u‖∞ decreases as O(h).

Indeed:

uL = h(1 − α)x

uR = αh(1 − x)= O(h)

0-44

A Small Experiment

Λu =1

h2(ui+1 − 2ui + ui−1) =

−1 at x = α

0 otherwise, u(0) = u(1) = 0

α : some grid point

rhs: solution:

1

−1

−0.5

0

x

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 1

0

0.002

0.004

0.006

0.008

0.01

x

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

Observe:‖u‖∞ decreases as O(h).

Indeed:

uL = h(1 − α)x

uR = αh(1 − x)= O(h)

0-45

A Small Experiment

Λu =1

h2(ui+1 − 2ui + ui−1) =

−1 at x = α

0 otherwise, u(0) = u(1) = 0

α : some grid point

rhs: solution:

1

−1

−0.5

0

x

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 1

0

0.002

0.004

0.006

0.008

0.01

x

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

Observe:‖u‖∞ decreases as O(h).

Indeed:

uL = h(1 − α)x

uR = αh(1 − x)= O(h)

0-46

A Small Experiment

Λu =1

h2(ui+1 − 2ui + ui−1) =

−1 at x = α

0 otherwise, u(0) = u(1) = 0

α : some grid point

rhs: solution:

1

−1

−0.5

0

x

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 1

0

0.002

0.004

0.006

0.008

0.01

x

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

Observe:‖u‖∞ decreases as O(h).

Indeed:

uL = h(1 − α)x

uR = αh(1 − x)= O(h)

0-47

One More Experiment

Discrete 5-point Laplace in 2D

Λu =1

h2(ui+1,j + ui−1,j + ui,j+1 + ui,j−1 − 4uij)

Random O(1) right hand side near some curve, homogeneous Dirichlet b.c.

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

1−1

−0.5

0

xy

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 0

0.20.4

0.60.8

1

0

0.2

0.4

0.6

0.8

10

0.005

0.01

0.015

xy

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

n ‖u‖∞ ratio

11 9.16e-3 —21 6.94e-3 1.3241 3.24e-3 2.1481 2.09e-3 1.54161 9.01e-4 2.32321 4.47e-4 2.01

Observe: solution decreases like O(h)

3D: a similar situation!

0-48

One More Experiment

Discrete 5-point Laplace in 2D

Λu =1

h2(ui+1,j + ui−1,j + ui,j+1 + ui,j−1 − 4uij)

Random O(1) right hand side near some curve, homogeneous Dirichlet b.c.

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

1−1

−0.5

0

xy

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 0

0.20.4

0.60.8

1

0

0.2

0.4

0.6

0.8

10

0.005

0.01

0.015

xy

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

n ‖u‖∞ ratio

11 9.16e-3 —21 6.94e-3 1.3241 3.24e-3 2.1481 2.09e-3 1.54161 9.01e-4 2.32321 4.47e-4 2.01

Observe: solution decreases like O(h)

3D: a similar situation!

0-49

One More Experiment

Discrete 5-point Laplace in 2D

Λu =1

h2(ui+1,j + ui−1,j + ui,j+1 + ui,j−1 − 4uij)

Random O(1) right hand side near some curve, homogeneous Dirichlet b.c.

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

1−1

−0.5

0

xy

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 0

0.20.4

0.60.8

1

0

0.2

0.4

0.6

0.8

10

0.005

0.01

0.015

xy

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

n ‖u‖∞ ratio

11 9.16e-3 —21 6.94e-3 1.3241 3.24e-3 2.1481 2.09e-3 1.54161 9.01e-4 2.32321 4.47e-4 2.01

Observe: solution decreases like O(h)

3D: a similar situation!

0-50

One More Experiment

Discrete 5-point Laplace in 2D

Λu =1

h2(ui+1,j + ui−1,j + ui,j+1 + ui,j−1 − 4uij)

Random O(1) right hand side near some curve, homogeneous Dirichlet b.c.

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

1−1

−0.5

0

xy

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 0

0.20.4

0.60.8

1

0

0.2

0.4

0.6

0.8

10

0.005

0.01

0.015

xy

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

n ‖u‖∞ ratio

11 9.16e-3 —21 6.94e-3 1.3241 3.24e-3 2.1481 2.09e-3 1.54

161 9.01e-4 2.32321 4.47e-4 2.01

Observe: solution decreases like O(h)

3D: a similar situation!

0-51

One More Experiment

Discrete 5-point Laplace in 2D

Λu =1

h2(ui+1,j + ui−1,j + ui,j+1 + ui,j−1 − 4uij)

Random O(1) right hand side near some curve, homogeneous Dirichlet b.c.

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

1−1

−0.5

0

xy

f

PSfrag replacements

O(h2)O(h)

O(h2)O(h2) 0

0.20.4

0.60.8

1

0

0.2

0.4

0.6

0.8

10

0.005

0.01

0.015

xy

u

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

n ‖u‖∞ ratio

11 9.16e-3 —21 6.94e-3 1.3241 3.24e-3 2.1481 2.09e-3 1.54

161 9.01e-4 2.32321 4.47e-4 2.01

Observe: solution decreases like O(h)

3D: a similar situation!

0-52

Smoothing Property

Theorem:

Λ: (an elliptic) finite difference operator, Gh: its Green’s function in Ω∗ :=]0, 1√

2[2⊂ R

2 with

|Gh(~x, ~z )| ≤ C1 + C2

∣∣log(‖~x − ~z ‖2

2 + h2)∣∣ , h = 1/(

√2m) .

Solution of Λu = f in Ω∗, u = 0 on ∂Ω∗ can be estimated by

‖u‖∞ ≤ ρ(supp(f))‖f‖∞ ,

where with M := #(supp(f)) we have

(A) M = O(m2) ⇒ ρ = O(1)

(B) M = O(m) and all points of supp(f) are distributed along some fixedcurve ⇒ ρ = O(h)

(C) M = O(1) ⇒ ρ = O(h2| log h|).

(D) . . .

0-53

Smoothing Property

Theorem:

Λ: (an elliptic) finite difference operator, Gh: its Green’s function in Ω∗ :=]0, 1√

2[2⊂ R

2 with

|Gh(~x, ~z )| ≤ C1 + C2

∣∣log(‖~x − ~z ‖2

2 + h2)∣∣ , h = 1/(

√2m) .

Solution of Λu = f in Ω∗, u = 0 on ∂Ω∗ can be estimated by

‖u‖∞ ≤ ρ(supp(f))‖f‖∞ ,

where with M := #(supp(f)) we have

(A) M = O(m2) ⇒ ρ = O(1)

(B) M = O(m) and all points of supp(f) are distributed along some fixedcurve ⇒ ρ = O(h)

(C) M = O(1) ⇒ ρ = O(h2| log h|).

(B) . . .

0-54

Smoothing Property

Theorem:

Λ: (an elliptic) finite difference operator, Gh: its Green’s function in Ω∗ :=]0, 1√

2[2⊂ R

2 with

|Gh(~x, ~z )| ≤ C1 + C2

∣∣log(‖~x − ~z ‖2

2 + h2)∣∣ , h = 1/(

√2m) .

Solution of Λu = f in Ω∗, u = 0 on ∂Ω∗ can be estimated by

‖u‖∞ ≤ ρ(supp(f))‖f‖∞ ,

where with M := #(supp(f)) we have

(A) M = O(m2) ⇒ ρ = O(1)

(B) M = O(m) and all points of supp(f) are distributed along some fixedcurve ⇒ ρ = O(h)

(C) M = O(1) ⇒ ρ = O(h2| log h|)

(D) . . .

0-55

Smoothing Property

Theorem:

Λ: (an elliptic) finite difference operator, Gh: its Green’s function in Ω∗ :=]0, 1√

2[2⊂ R

2 with

|Gh(~x, ~z )| ≤ C1 + C2

∣∣log(‖~x − ~z ‖2

2 + h2)∣∣ , h = 1/(

√2m) .

Solution of Λu = f in Ω∗, u = 0 on ∂Ω∗ can be estimated by

‖u‖∞ ≤ ρ(supp(f))‖f‖∞ ,

where with M := #(supp(f)) we have

(A) M = O(m2) ⇒ ρ = O(1)

(B) M = O(m) and all points of supp(f) are distributed along some fixedcurve ⇒ ρ = O(h)

(C) M = O(1) ⇒ ρ = O(h2| log h|).

(D) . . .

0-56

Proof

Take for concreteness the “interface” case

Solution of Λu = f in Ω∗ , u = 0 on ∂Ω∗:

u(~x ) = h2 P

~z∈Ω∗

h

Gh(~x, ~z )f(~z )

Assume, ~x ∈ supp(f) (worst case!)

Radially symmetric estimate of Gh (r := ‖~x−~z‖2):

|Gh(~x, ~z )| ≤ w(r) := C1 + C2

˛˛log(r2 + h

2)˛˛

Introduce layers L with thickness h

Count number of support points in each layer,#(Li ∩ supp(f))

Estimate

|u(~x)|(1)

≤ h2 P

~z∈Ω∗

h

w(r)f(~z)

≤ ‖f‖∞PR

i=0 w(ri)#(Li ∩ supp(f))| z

=O(1)

= h2O(1)

PR

i=1 w(ri)| z

‖f‖∞

=O(h)=:ρ

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-57

Proof

Take for concreteness the “interface” case

1. Solution of Λu = f in Ω∗ , u = 0 on ∂Ω∗:

u(~x ) = h2 P

~z∈Ω∗

h

Gh(~x, ~z )f(~z )

Assume, ~x ∈ supp(f) (worst case!)

Radially symmetric estimate of Gh (r := ‖~x−~z‖2):

|Gh(~x, ~z )| ≤ w(r) := C1 + C2

˛˛log(r2 + h

2)˛˛

Introduce layers L with thickness h

Count number of support points in each layer,#(Li ∩ supp(f))

Estimate

|u(~x)|(1)

≤ h2 P

~z∈Ω∗

h

w(r)f(~z)

≤ ‖f‖∞PR

i=0 w(ri)#(Li ∩ supp(f))| z

=O(1)

= h2O(1)

PR

i=1 w(ri)| z

‖f‖∞

=O(h)=:ρ

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-58

Proof

Take for concreteness the “interface” case

1. Solution of Λu = f in Ω∗ , u = 0 on ∂Ω∗:

u(~x ) = h2 P

~z∈Ω∗

h

Gh(~x, ~z )f(~z )

2. Assume, ~x ∈ supp(f) (worst case!)

Radially symmetric estimate of Gh (r := ‖~x−~z‖2):

|Gh(~x, ~z )| ≤ w(r) := C1 + C2

˛˛log(r2 + h

2)˛˛

Introduce layers L with thickness h

Count number of support points in each layer,#(Li ∩ supp(f))

Estimate

|u(~x)|(1)

≤ h2 P

~z∈Ω∗

h

w(r)f(~z)

≤ ‖f‖∞PR

i=0 w(ri)#(Li ∩ supp(f))| z

=O(1)

= h2O(1)

PR

i=1 w(ri)| z

‖f‖∞

=O(h)=:ρ

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-59

Proof

Take for concreteness the “interface” case

1. Solution of Λu = f in Ω∗ , u = 0 on ∂Ω∗:

u(~x ) = h2 P

~z∈Ω∗

h

Gh(~x, ~z )f(~z )

2. Assume, ~x ∈ supp(f) (worst case!)

3. Radially symmetric estimate of Gh (r := ‖~x−~z‖2):

|Gh(~x, ~z )| ≤ w(r) := C1 + C2

˛˛log(r2 + h

2)˛˛

Introduce layers L with thickness h

Count number of support points in each layer,#(Li ∩ supp(f))

Estimate

|u(~x)|(1)

≤ h2 P

~z∈Ω∗

h

w(r)f(~z)

≤ ‖f‖∞PR

i=0 w(ri)#(Li ∩ supp(f))| z

=O(1)

= h2O(1)

PR

i=1 w(ri)| z

‖f‖∞

=O(h)=:ρ

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-60

Proof

Take for concreteness the “interface” case

1. Solution of Λu = f in Ω∗ , u = 0 on ∂Ω∗:

u(~x ) = h2 P

~z∈Ω∗

h

Gh(~x, ~z )f(~z )

2. Assume, ~x ∈ supp(f) (worst case!)

3. Radially symmetric estimate of Gh (r := ‖~x−~z‖2):

|Gh(~x, ~z )| ≤ w(r) := C1 + C2

˛˛log(r2 + h

2)˛˛

4. Introduce layers L with thickness h

Count number of support points in each layer,#(Li ∩ supp(f))

Estimate

|u(~x)|(1)

≤ h2 P

~z∈Ω∗

h

w(r)f(~z)

≤ ‖f‖∞PR

i=0 w(ri)#(Li ∩ supp(f))| z

=O(1)

= h2O(1)

PR

i=1 w(ri)| z

‖f‖∞

=O(h)=:ρ

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-61

Proof

Take for concreteness the “interface” case

1. Solution of Λu = f in Ω∗ , u = 0 on ∂Ω∗:

u(~x ) = h2 P

~z∈Ω∗

h

Gh(~x, ~z )f(~z )

2. Assume, ~x ∈ supp(f) (worst case!)

3. Radially symmetric estimate of Gh (r := ‖~x−~z‖2):

|Gh(~x, ~z )| ≤ w(r) := C1 + C2

˛˛log(r2 + h

2)˛˛

4. Introduce layers L with thickness h

5. Count number of support points in each layer,#(Li ∩ supp(f))

Estimate

|u(~x)|(1)

≤ h2 P

~z∈Ω∗

h

w(r)f(~z)

≤ ‖f‖∞PR

i=0 w(ri)#(Li ∩ supp(f))| z

=O(1)

= h2O(1)

PR

i=1 w(ri)| z

‖f‖∞

=O(h)=:ρ

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-62

Proof

Take for concreteness the “interface” case

1. Solution of Λu = f in Ω∗ , u = 0 on ∂Ω∗:

u(~x ) = h2 P

~z∈Ω∗

h

Gh(~x, ~z )f(~z ) (1)

2. Assume, ~x ∈ supp(f) (worst case!)

3. Radially symmetric estimate of Gh (r := ‖~x−~z‖2):

|Gh(~x, ~z )| ≤ w(r) := C1 + C2

˛˛log(r2 + h

2)˛˛

4. Introduce layers L with thickness h

5. Count number of support points in each layer,#(Li ∩ supp(f))

6. Estimate

|u(~x)|(1)

≤ h2 P

~z∈Ω∗

h

w(r)f(~z)

≤ ‖f‖∞PR

i=0 w(ri)#(Li ∩ supp(f))| z

=O(1)

= h2O(1)

PR

i=1 w(ri)| z

‖f‖∞

=O(h)=:ρ

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-63

essential issue:

proof does not make

any use

of the

maximum principle!

0-64

The required estimates of the discrete Green’s function

hold for standard 5 point discrete Laplace

∆hu :=1

h2(ui+1,j + ui−1,j + ui,j+1 + ui,j−1 − 4ui,j) = grad±div∓u

(proofs in literature)

for discrete linear elasticity equations (standard central finite differenceapproximation):

Λ~u = µ∆h~u + (µ + λ)

(

D(1,0)+ D

(1,0)− u1 + D

(1,0)0 D

(0,1)0 u2

D(1,0)0 D

(0,1)0 u1 + D

(0,1)+ D

(0,1)− u2

)

. . . there are a lot of reasons to believe that the estimate holds

0-65

The required estimates of the discrete Green’s function

hold for standard 5 point discrete Laplace

∆hu :=1

h2(ui+1,j + ui−1,j + ui,j+1 + ui,j−1 − 4ui,j) = grad±div∓u

(proofs in literature)

for discrete linear elasticity equations (standard central finite differenceapproximation):

Λ~u = µ∆h~u + (µ + λ)

(

D(1,0)+ D

(1,0)− u1 + D

(1,0)0 D

(0,1)0 u2

D(1,0)0 D

(0,1)0 u1 + D

(0,1)+ D

(0,1)− u2

)

. . . there are a lot of reasons to believe that the estimate holds

0-66

The required estimates of the discrete Green’s function

hold for standard 5 point discrete Laplace

∆hu :=1

h2(ui+1,j + ui−1,j + ui,j+1 + ui,j−1 − 4ui,j) = grad±div∓u

(proofs in literature)

for discrete linear elasticity equations (standard central finite differenceapproximation):

Λ~u = µ∆h~u + (µ + λ)

(

D(1,0)+ D

(1,0)− u1 + D

(1,0)0 D

(0,1)0 u2

D(1,0)0 D

(0,1)0 u1 + D

(0,1)+ D

(0,1)− u2

)

. . . there are a lot of reasons to believe that the estimate holds

0-67

The required estimates of the discrete Green’s function

hold for standard 5 point discrete Laplace

∆hu :=1

h2(ui+1,j + ui−1,j + ui,j+1 + ui,j−1 − 4ui,j) = grad±div∓u

(proofs in literature)

for discrete linear elasticity equations (standard central finite differenceapproximation):

Λ~u = µ∆h~u + (µ + λ)

(

D(1,0)+ D

(1,0)− u1 + D

(1,0)0 D

(0,1)0 u2

D(1,0)0 D

(0,1)0 u1 + D

(0,1)+ D

(0,1)− u2

)

. . . there are a lot of reasons to believe that the estimate holds

0-68

similar results hold also in three dimensions

for the discrete Green’s function require an estimate

|Gh(~x, ~z )| ≤ C1 + C21

‖~x − ~z ‖2 + h

0-69

Application to EJIIM

Question:under which conditions is U returned with 2-nd order accuracy?

U = A−1(F −ΨJ)

0-70

Application to EJIIM

Question:under which conditions is U returned with 2-nd order accuracy?

U = A−1(F −ΨJ)

Answer:

O(h3) O(h2) O(h)

jumps a [~u ] [∂~α~u] , ‖~α‖1 = 1 [∂~α~u] ,‖~α‖1 = 2

geometry position normals and tangents

rhs at the regular points at the irregular points

b.c. Dirichlet Neumann, traction

athis means that for one sided approximations (operator D) quadratic polynomials suffice

0-71

Application to EJIIM

Question:under which conditions is U returned with 2-nd order accuracy?

U = A−1(F −ΨJ)

Answer:

O(h3) O(h2) O(h)

jumps a [~u ] [∂~α~u] , ‖~α‖1 = 1 [∂~α~u] ,‖~α‖1 = 2

geometry position normals and tangents

rhs at the regular points at the irregular points

b.c. Dirichlet Neumann, traction

athis means that for one sided approximations (operator D) quadratic polynomials suffice

0-72

Summary

EJIIM is extended to three dimensional boundary value problems (Poissonand linear elasticity)

0-73

Summary

EJIIM is extended to three dimensional boundary value problems (Poissonand linear elasticity)

Implemented algorithms are able to deal with highly complex problems

0-74

Summary

EJIIM is extended to three dimensional boundary value problems (Poissonand linear elasticity)

Implemented algorithms are able to deal with highly complex problems

A memory reduced algorithm (RedEJIIM) is proposed

0-75

Summary

EJIIM is extended to three dimensional boundary value problems (Poissonand linear elasticity)

Implemented algorithms are able to deal with highly complex problems

A memory reduced algorithm (RedEJIIM) is proposed

Smoothing property is proven

0-76

Summary

EJIIM is extended to three dimensional boundary value problems (Poissonand linear elasticity)

Implemented algorithms are able to deal with highly complex problems

A memory reduced algorithm (RedEJIIM) is proposed

Smoothing property is proven

0-77

Outlook

Algorithms:

preconditioner

local grid refinement

parallelisation FEM/FVM formulation

other 3D applications

other

composite materials

more fast solvers

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-78

Outlook

Algorithms:

preconditioner

local grid refinement

parallelisation FEM/FVM formulation

other 3D applications

other

composite materials

more fast solvers

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

Convergence proof:

systems space variable coefficients

unequal step sizesanalyse of full EJIIM system

error bounds on realistic grids

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-79

Immersed Interface Methodsfor

EllipticBoundary Value Problems

RedEJIIMEJIIM

in 3D

Smoothing

property

PSfrag replacements

O(h2)O(h)

O(h2)O(h2)

0-80