Lecture 23

Hierarchical Spatial Domains

Ε1

Ε3 Ε4

Ε2Ε1

Ε3 Ε4

Ε2

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).

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

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

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.

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

Gauss Divergence theorem

In practice we can write

Ω

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)

Greens formula

Laplaces equation

Let G satisfy

∇2G (x) = δ (x)

Solution is

G (x) = − 1

4πrMore generally

G (x,y) = − 1

4π |x− y|

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|

Discretize surface S into triangles Discretize`

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

Boundary Element Methods

Boundary conditions provide value of φj or qj

Becomes a linear system to solve for the other

Accelerate via FMM