lecture 23 - umiacsramani/cmsc878r/11_25_03_lecture23.pdf · 2003. 11. 27. · Ł at the finest...
TRANSCRIPT
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Lecture 23
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Hierarchical Spatial Domains
Ε1
Ε3 Ε4
Ε2Ε1
Ε3 Ε4
Ε2
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Parameter s is determined by the box with most particles. Most boxes will be empty (ok) or have much fewer than s particles
Reducing number of levels is good Alternate strategy
Continue subdivision so that each parent box has > ℓ points and each leaf-box has ≤ ℓpoints
Consider an algorithm due to Cheng et al (J. Comput. Phys. 1999).
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At the finest level consider evaluation at box b. It is surrounded by boxes
Coarser, Same or Finer level Already considered, in a
coarser S|R translation Divide surrounding boxes not
already considered in toL1, L2, L3, L4 domains Boxes L1 share a boundary
with box b, or they lie within the sphere of box b and need to be evaluated directly.
Boxes L2 are separated by at least one box of size of bS|R translate these
Boxes in L3 contain sourcesthat lie outside the sphere of b and are closer than boxes L2R-preFMM these
Box b lies outside the sphere of boxes in L4. but is too close to S|R. S-preFMM these
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x*
xiR
S
R*
r*
r* < R*S-expansion: |y - x*| > R* > |xi - x*|R-expansion: |y - x*| < r* < |xi - x*|
For xi in L3 we can build R expansions in b
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Algorithm parameters & optimization Suggest that P and l should be balanced but do not provide
an optimal choice A good project
to implement this and find out optimal choices. compare this scheme with our scheme and check efficiency Perhaps develop a combined scheme.
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Outline
Review Vector analysis (Divergence & Gradient of potential) 3-D Cartesian coordinates & Spherical coordinates Laplaces equation and Helmholtz equation Green's function & Green's theorem Boundary element method FMM
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Gauss Divergence theorem
In practice we can write
Ω
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Integral Definitions of div, grad and curl
Elemental volume δτwith surface ∆S
τ 0∆S
τ 0∆S
τ 0∆S
1lim dSτ
1lim . dSτ1lim dSτ
δ
δ
δ
φ φδ
δ
δ
→
→
→
∇ ≡
∇⋅ ≡
∇× ≡ − ×
∫
∫
∫
n
D D n
D D n
!
!
!
n
dS
D=D(r), φ= φ(r)
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Greens formula
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Laplaces equation
Let G satisfy
∇2G (x) = δ (x)
Solution is
G (x) = − 1
4πrMore generally
G (x,y) = − 1
4π |x− y|
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Helmholtz equation
Let G satisfy
∇2G (x) + k2G= δ (x)
Solution is
G (x) = −exp (ikr)4πr
More generally
G(x, y) = −exp(ik |x− y|)4π |x− y|
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Discretize surface S into triangles Discretize`
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Greens formula Recall that the impulse-response is sufficient to characterize
a linear system Solution to arbitrary forcing constructed via convolution For a linear boundary value problem we can likewise use
the solution to a delta-function forcing to solve it. Fluid flow, steady-state heat transfer, gravitational
potential, etc. can be expressed in terms of Laplacesequation
Solution to delta function forcing, without boundaries, is called free-space Greens function
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Boundary Element Methods
Boundary conditions provide value of φj or qj
Becomes a linear system to solve for the other
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Accelerate via FMM
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