Immersed Interface Methods for Elliptic Boundary Value ...€¦ · Graduiertenkolleg Mathematik und...
Transcript of 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
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Problem Statement
complex problemsfrom
real life modelling
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Problem Statement
iterative approach
many linear problemshave to be solved
complex problemsfrom
real life modelling
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Problem Statement
iterative approach
many linear problemshave to be solved
for the linear case
need fast & accurate
solvers
complex problemsfrom
real life modelling
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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x
y
von Mises stress
‖~u‖2Von Mises stress(“color zoom”)
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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?
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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?
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YES!at least — if seen asymptotically when h −→ 0
on realistic meshes. . . not always. . .
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YES!at least — if seen asymptotically when h −→ 0
on realistic meshes. . . not always. . .
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YES!at least — if seen asymptotically when h −→ 0
on realistic meshes. . . not always. . .
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YES!at least — if seen asymptotically when h −→ 0
on realistic meshes. . . not always. . .
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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
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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
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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
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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
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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
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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
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RedEJIIM
EJIIM system:
AU + ΨJ = F
DU + J = F, J = ([~u ] , [∂~u ] , [∂2~u ])
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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 ])
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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 ])
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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
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RedEJIIM Results: Textiles
Example: temperature distribution
EJIIM solution RedEJIIM solution
difference:‖uEJIIM− uRedEJIIM‖∞ ≈ 0.16, ‖u‖∞ = 5
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May be you wonder . . .
Answer:
For the moment the linear elasticityworks in “easy” geometries.
For more complicated cases
refer to the future research!
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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!
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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!
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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!
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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!
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Algorithms
Convergence Analysis
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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!
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“Fact”:
AU = F
‖U‖∞ ≤ C‖F‖∞
A: a matrix of discrete Laplace
C: some constant
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a little bit of
Experimental Numerics
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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)
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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)
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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)
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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)
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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)
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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)
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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!
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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) . . .
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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