SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
SMA-HPC ©2002 MIT
First Kind Issues
3-D Laplace’s
Equation
The singular Kernel Saves
the day
z
Spike Function=1 on a disk
x2 + y 2 > R
σ0 =1
x2 + y2 ≤ R
R 2π
1
σ 0 ( x′)dS ′ = ∫
xci − x′
0
∫
Singular Kernel
disk
Case
Smooth Kernel
Case
σ0 = 0
∫
∫
0
1
rdrdθ = 2π R
r
1 σ 0 ( x′)dS ′ = π R 2
disk
Smooth kernel → 0 faster as R → 0, more singular
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Convergence Analysis
3-D Laplace’s
Equation
Quick review of FEM
Convergence for Laplace
Partial Differential Equation form
∇ 2u = f in Ω
u = 0 on Γ
Ω is the volume domain
Γ is the problem surface
“Nearly” Equivalent weak form
1
∇
u
∇
v
dx
=
f
v
dx
for
a
ll
v
∈
H
(Ω)
∫Ω
∫Ω
14243 1
424
3
a (u ,v)
l (v )
Introduced an abstract notation for the equation u must satisfy
a (u , v) = l (v) for all v ∈ H 1 ( Ω )
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Convergence Analysis
3-D Laplace’s
Equation
Quick review of FEM
Convergence for Laplace
n
Introduce an approximate solution u n = ∑ α iϕ i
i =1
⇒ u is a weighted sum of basis functions
n
The basis functions define a space
n


X n = v ∈ X n | v = ∑ β iϕ i for some β i 's 
i =1


Example
“Hat” basis functions
ϕ2
ϕ1
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ϕ4 ϕ6
ϕ3 ϕ5
Piecewise linear Space
3-D Laplace’s
Equation
Convergence Analysis
Quick review of FEM
Convergence for Laplace
Key Idea
a(u , u ) defines a norm on H 01 ( Ω )
a(u, u ) ≡ u
U is restricted to be 0 at 0 and1!!
Using the norm properties, it is possible to show
If a (u , ϕ i ) = l (ϕ i ) for all ϕ i ∈ {ϕ1 , ϕ 2 ,..., ϕ n }
n
Then u − u n = min wn ∈X n u − wn
1
424
3
1
424
3
Pr ojection
Solution
Error
Error
SMA-HPC ©2002 MIT
Convergence Analysis
3-D Laplace’s
Equation
Quick review of FEM
Convergence for Laplace
The question is only
u
un
How well can you fit u with a member of Xn
But you must measure the error in the ||| ||| norm
1
u −u ≤ u −Π u = O 
For piecewise linear:
1
424
3
n
n
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error
a
n
Convergence Analysis
3-D Laplace’s
Equation
Applying FEM approach to
first kind integral equations
“Weak” Form for the integral equation
1
v
x
∫∫Γ ( ) x − x′ σ ( x′) dS ′dS = ∫Γ v ( x ) Ψ ( x ) dS for all v ∈ H ( Γ )
14444
4244444
3 1442443
l( v )
a(σ , v )
The difficulty is defining H ( Γ ) with right properties
Must exclude σ ( x ) ' s where ∫
1
σ ( x′) dS ′ = 0
x − x′
H ( Γ ) is a fractional Sobolev Space
We won’t
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.
say
more about this
3-D Laplace’s
Equation
Convergence Analysis
Applying FEM approach to
first kind integral equations
Use FEM key Idea
a (σ , σ ) defines a norm on H ( Γ ) a (σ , σ ) = σ
n
σ = ∑αi ϕ i ( x )
n
i =1
{
Basis Functions
n


X n = v ∈ X n | v = ∑ β iϕ i for some β i 's 
i =1


Using the norm properties, it is possible to show
If a (σ , ϕ i ) = l (ϕ i ) for all ϕ i ∈ {ϕ1 , ϕ 2 ,..., ϕ n }
n
Then σ − σ n = min wn∈X n
1424
3
Solution
Error
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σ − wn
1424
3
Pr ojection
Error
MEMS Performance Depends on Air Damping of
Complicated 3-D Structures
Bosch angular rate sensor
ADXL76 accelerometer
TI 3x3 mirror array
Resonator
Lucent micromirror
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Drag In MEMS is Incompressible Stokes
Velocity integral equation for Stokes flow
1
uj (xo ) = − ∫ fi (x)Gij (x − xo )ds
8πµ s
where
G
ij
=
δ
ij
R
+
xˆ i xˆ
R
3
j
;
R = xo − x ;
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xˆi = xoi − xi
Null Space of the Stokes Equation
Constant pressure a singular mode, generates zero velocity.
Differential Form of Stokes,
independent of absolute
pressure
Integral Form of Stokes,
constant pressure must not
change velocity
r
0 = −∇P + µ∇ u

r
∇•u = 0

2
r
r
r r
1
uj (xo ) =−
fi (x)Gij (x − xo )ds
∫
8πµ s
If P = constant, u i = 0; fi =−Pni
r r
r
⇒∫Gij (x − xo )ni (x)ds = 0
s
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Null Space of the Singular BEM Operators
• Stokes Integral Operator has a null space
– The solution is not uniquely defined.
– A pressure boundary condition is needed.
• Null space must be removed
– so as to avoid numerical error.
F = F correct solution + ΧN + ε
• Two-step method:
1. Modify GMRES to calculate a null-space-free solution.
2. Use pressure condition to adjust solution
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Krylov Subspace Iterative methods
Linear System
Start with Ax = b
Determine the Krylov Subspace r 0 = b − Ax 0
{
Krylov Subspace ≡ span r 0 , Ar 0 ,..., Ak r 0
}
Select Solution from the Krylov Subspace
k +1
0
k
k
0
0
k 0
x = x + y , y ∈ span r , Ar ,..., A r
{
k
GMRES picks a residual-minimizing y .
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}
Modify Krylov-Subspace Method to Calculate
Null-Space-Free Solution
• The discretized Stokes equation
GF =U
• The Krylov subspace is
κ = span {U , GU , G U , G U , G U ,LL}
2
• If
κ ⊥ Null (G)
3
4
then F = F ⊥ ⊥ Null (G )
Remove Null(G) from every Krylov subspace vector
SMA-HPC ©2002 MIT
FastStokes Simulation Result
In-plane motion, 3-D steady incompressible Stokes, 16k panels
Total
Couette Model
1-D Stokes
FastStokes
123.7
137.1
223.7
Drag Force (nN)
Q
Bottom Top Inter- End and
finger others
108.9
14.8
50.1
108.9 13.5 14.8
45.20
123.0 26.8
73.8
27.7
(55%)
(12%)
(33%)
Measurement
Computation finished in 10 minutes
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27
Micromirror Q-factor
Gimbal
Mirror
Rotate
Mode
Mirror Mirror+Gimbal
1
Mirror
Mirrror Mirror+Gimbal
2
Mirror
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Q
Simulated
2.36
3.14
4.69
10.16
Error
Measured (%)
2.31
2.16
3.45
8.99
4.27
9.84
10.63
4.42
Summary
Reminder about 2nd Kind theory
Convergence Theory
Fredholm Alternative for 2nd Kind
Finite Dimensional Null Space
First Kind Convergence Theory, sort of
Connection to the FEM results
MEMS Drag Example