a continuous approach for distributing points on the
TRANSCRIPT
A continuous approach for distributing pointson the sphere
using fast Fourier transforms
13th International Conference Approximation Theory7 -10 March 2010, San Antonio
Manuel Graf in joint work with Daniel Potts 1
andRainer Backofen, Simon Praetorius, Axel Voigt, Thomas Witkowski 2
1Chemnitz University of Technology, Germany, supported by DFG2Dresden University of Technology, Germany, supported by DFG
Outline
1 Preliminaries
2 Fast Algorithms
3 Model of Crystal Growth
4 Numerical Approach
5 Numerical Examples
Preliminaries
sphere S2 := {x ∈ R3 : ‖x‖2 = 1}• spherical coordinates (θ, ϕ) ∈ [0, π]× [0, 2π)
x(θ, ϕ) := (sin θ cosϕ, sin θ sinϕ, cos θ)>
• metricdS2(x1, x2) := arccos x>1 x2
• measure∫S2
f (x)dµS2(x) =1
4π
∫ 2π
0
∫ π
0f (x(θ, ϕ)) sin θdθdϕ
Preliminaries
Hilbert space
L2(S2) :={
f : S2 → C :
∫S2
|f (x)|2 dµS2(x) <∞}
spherical harmonics Y kn of degree n and order k
Y kn (x(θ, ϕ)) := e−ikϕP
|k|n (θ)
polynomial space of degree N
ΠN(S2) := span{Y kn : n = 0, . . . ,N; k = −n, . . . , n}
dN := (N + 1)2
Preliminaries
scattered sampling set X = {xi : i = 0, . . . ,M − 1} ⊂ S2
• mesh norm
δ(X ) := 2 maxx∈S2
mini=0,...,M−1
dS2(x, xi )
• associated quadrature rule with weights wi ≥ 0
Q(X ) := {(xi ,wi ) : i = 0, . . . ,M − 1}
degree of exactness N if∫S2
f (x)dµS2(x) =M−1∑i=0
wi f (xi ) for all f ∈ ΠN(S2)
Preliminaries
Theorem (sufficient condition - Mhaskar, Narcowich, Ward 01;Filbir,Themistoclakis 08)
A sampling set X (S2) supports nonnegative quadrature weights fordegree of exactness N, if the mesh norm satisfies
δ(X (S2)) ≤ 1
306N.
Theorem (necessary condition - Yudin 95; Reimer 00)
If a sampling set X (S2) supports nonnegative quadrature weightsfor degree of exactness N = 2L− 1, then the mesh norm satisfies
δ(X (S2)) ≤ 2 arccos zL ≤4π
N + 2,
where zL is the greatest zero of the L-th Legendre polynomial.
Fast Algorithms
• nonequispaced spherical Fourier matrix
Y := (Y kn (xi )) ∈ CM×dN
for sampling set X = {xi : i = 0, . . . ,M − 1} ⊂ S2
• fast algorithms1 for evaluation of
f = Yf, fi := f (xi ) =N∑
n=0
n∑k=−n
f kn Y k
n (xi ) (nfsft)
and
f = Y∗diag(w)f, f kn =
M∑i=0
wi fiY kn (xi ) (adjoint nfsft)
in O(N2log2N + M) arithmetic operations
1NFFT Library (Keiner,Kunis,Potts): http//www.tu-chemnitz.de/∼potts/nfft
Fast Algorithms
HEALPix grid XHS of size M = 12S2 ∈ N
• used in cosmic microwavebackground data analysis
• lacks exact integration scheme
numerical computation of the quadrature weights wi
(G, Kunis, Potts (Appl. Comput. Harm. Anal. 2009))
S M N rel. residual time
175 367500 512 3.242945e-13 9h185 410700 512 1.918440e-14 35min375 1687500 1024 1.088695e-14 97min
Motivation: Grystal Growth on the Sphere
• particle ordering on the sphere(Thomson problem)
minimizeM−1∑i=0
M−1∑i 6=j=0
‖xi−xj‖−12
(Backofen, Voigt, Witkowski (Phys. Rev. E 2010))
• crystal defects, grain boundaryscars
• ...
Model of Crystal Growth
phase field crystal model
• density function on the sphere of radius r > 0
ρ : rS2 → R
• minimizing free energy functional
F(ρ) :=
∫rS2
−|∇rS2ρ|2 +1
2|∆rS2ρ|+
1
2(1− ε)ρ2 +
1
4ρ4dµrS2
• L2 gradient flow ∂tρ = − δFδρ (Swift, Hohenberg (Phys. Rev. A 1977))
• H−1 gradient flow ∂tρ = ∆ δFδρ (Elder et al. (Phys. Rev. Lett. 2002))
∂tρ(rx, t) = ∆rS2
[((∆rS2 + 1)2 − ε
)ρ(rx, t) + ρ(rx, t)3
]
Numerical Approach
spectral method
• approximation by a spherical polynomial of degree N
ρ(rx, t) :=N∑
n=0
n∑k=−n
ρkn(t)Y k
n (x), x ∈ S2, t ∈ [0,∞)
• in spectral domain the PDE becomes a system of ODEs
dρkn
dt= −n(n + 1)
r 2
[((1− n(n + 1)
r 2
)2
− ε
)ρkn + (ρ3)k
n
]
with n = 0, . . . ,N, k = −n, . . . , n
Numerical Approach
semi-implicit time-integration scheme
ρt+1 − ρt
4t= −n(n + 1)
r 2
[((1− n(n + 1)
r 2
)2
− ε
)ρt+1 + (ρ3
t )
]
computation of (ρ3t ), t ≥ 0
• given ρt ∈ CdN
• evaluate ρt on a given sampling set X ⊂ rS2 (nfsft)
• compute pointwise ρ3t
• get spherical Fourier coefficients (ρ3t )
(adjoint nfsft & quadrature)
Numerical Examples
numerical solution for an icosahedral symmetric initial density ρ0
on a sphere rS2 with radius r = 100 using a HEALPix grid of sizeS = 100 and a polynomial degree N = 114
t = 0 t = 150 t = 300
t = 900 t = 2400 t = 4500
Numerical Examples
the solution ρt after t = 4500 time steps in the Fourier domain
n
k
0 10 20 30 40 50 60 70 80 90 100 110
−110
−90
−70
−50
−30
−10
0
10
30
50
70
90
1100
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
Numerical Examples
distributing many points on the sphere
• r = 600, ε = 0.4, ρ = −0.3︸ ︷︷ ︸≈ 100.000 particels
• Coulomb energy is only0.0004% above lower bound
1
2M2+cM
32 , c = −0.553051...
(Kuijlaars, Saff (Trans. Amer. Math. Soc. 1998))