An efficient Multigrid calculation of the Far field
and Near Field map for Helmholtz and
Schrödinger problems
Department of Mathematics and Computer Science,
University of Antwerp, Belgium
Siegfried Cools∗ , Bram Reps† , Wim Vanroose‡
November 15, 2013
∗
BOF U. Antwerpen
FWO Flanders
‡
BOF U. Antwerpen
†
Far and Near field maps
Cross sections in chemical reactions
Near Field Scanning Microscope
Helmholtz equation: far- and near field.
Solvers and deforming the contour of volume integral.
Schrödinger equations are Helmholtz equations
No free lunch?
Discussion and Conclusions
Helmholtz equation
The total wave solution utot satisfies the homogeneous Helmholtz
equation
−∆ − k 2 (x) utot (x) = 0 on Ω ⊂ Rd , d ≥ 1,
with no wave source present within the domain. Space-dependent
wave number function k defined as
k 2 (x)−k 2
0
χ(x)=
k2
0
Ω
(
k(x),
k(x) =
k0 ,
for x ∈ O,
for x ∈ Ω \ O,
where k0 ∈ R, O ⊂ Ω ⊂ Rd .
O
χ(x) 6= 0
χ(x) = 0
Helmholtz equation
Decomposition (uin = e ik0 η·x = incoming wave, u = scattered wave)
utot = uin + u,
implies
⇒
⇒
−∆ − k 2 (x) utot (x) = 0
−∆ − k 2 (x) u(x) = (∆ + k 2 (x))uin (x)
−∆ − k 2 (x) u(x) = (k 2 (x) − k02 )uin (x),
yielding the inhomogeneous scattered wave equation
−∆ − k 2 (x) u(x) = f (x)
on
Ω ⊂ Rd ,
.
where f (x) = (k 2 (x) − k02 )uin (x) = k02 χ(x)uin (x).
(1)
Helmholtz equation
Scattered wave equation (1)
−∆ − k 2 (x) u(x) = f (x)
on
Ω ⊂ Rd ,
with f (x) = k02 χ(x)uin (x).
Solved for u on discretized subset ΩN ⊂ Ω (‘numerical box’)
with outgoing wave boundary conditions, e.g. PML, ECS.
Far- and near field map
Assume
−∆ − k 2 (x) u N (x) = k02 χ(x)uin (x)
−∆ − k02 u(x) = k02 χ(x)(uin (x) + u N (x))
{z
}
|
for x ∈ Ω,
for x ∈ Rd ,
g (x)
Analytic solution using Helmholtz Green’s function:
Z
u(x) =
G (x, x0 ) g (x0 ) dx0
d
ZR
=
G (x, x0 ) k02 χ(x0 ) uin (x0 ) + u N (x0 ) dx0 ,
Ω
x ∈ Rd .
. Calculate u in any point x ∈ Rd outside the numerical box, using
only the information inside the numerical box.
Far field mapping
Let x = (α, ρ) be given by a unit vector α ∈ Rd and length ρ ∈ R.
Asymptotic form of G (x, x0 ) (separable) yields the far field
(ρ → ∞) wave pattern for u
lim u(ρ, α) = D(ρ)F (α),
ρ→∞
where
Z
F (α) =
0
e −ik0 x ·α g (x0 ) dx0 .
Ω
is the far field amplitude map.
2D schematic:
F (α)
uin
O
α ∈ Rd ,
α
(2)
Far field mapping
Note that the far field integral can be split into a sum of two
contributions: F (α) = I1 + I2 , with
Z
Z
−ik0 x·α
I1 =
e
χ(x)uin (x)dx and I2 =
e −ik0 x·α χ(x)u N (x)dx
Ω
Ω
{z
}
{z
}
|
|
all factors known explicitly
requires u N (x) for x ∈ Ω !
Far field mapping
Note that the far field integral can be split into a sum of two
contributions: F (α) = I1 + I2 , with
Z
Z
−ik0 x·α
I1 =
e
χ(x)uin (x)dx and I2 =
e −ik0 x·α χ(x)u N (x)dx
Ω
Ω
{z
}
{z
}
|
|
all factors known explicitly
requires u N (x) for x ∈ Ω !
Summary: Far field map calculation in a nutshell:
I
Step 1. Solve the Helmholtz eqn. (1) for u to obtain a
solution u N on a numerical domain.
I
Step 2. Calculate/approximate the Fourier integral (2) on the
given numerical domain.
Far field mapping
Note that the far field integral can be split into a sum of two
contributions: F (α) = I1 + I2 , with
Z
Z
−ik0 x·α
I1 =
e
χ(x)uin (x)dx and I2 =
e −ik0 x·α χ(x)u N (x)dx
Ω
Ω
{z
}
{z
}
|
|
all factors known explicitly
requires u N (x) for x ∈ Ω !
Summary: Far field map calculation in a nutshell:
I
Step 1. Solve the Helmholtz eqn. (1) for u to obtain a
solution u N on a numerical domain.
main computational bottleneck
I
Step 2. Calculate/approximate the Fourier integral (2) on the
given numerical domain.
Iterative Solution
I
Helmholtz equation is hard to solve with iterative methods.
−∆ − k 2 (x) u(x) = f (x)
(3)
Elman, Ernst and O’Leary, 2002
Ernst and Gander, 2012
I
Complex Shifted Laplacian as a preconditioner.
−∆ − (1 + iβ)k 2 (x) u(x) = f (x)
Erlangga, Oosterlee and Vuik, 2006, Cools and Vanroose, 2013
(4)
Multigrid
Two-grid correction scheme
• Relax ν1 times on Ah v h = f h
(e.g. ω-Jacobi, Gauss-Seidel, . . . ).
• Compute r h = f h − Ah v h , restrict
r 2h = Ih2h r h , and solve
A2h e 2h = r 2h .
h 2h
• Interpolate e h = I2h
e and correct
h
h
h
v ←v +e .
• Relax ν2 times on Ah v h = f h .
Multigrid V-cycle = recursion.
Convergence
Erlangga, Oosterlee, Vuik, SISC, 2006
Complex contour method
={z}
For u and χ analytic the far
field integral
Z
I2 =
e
−ik0 x·α
Ω
Z1
Z2
γ
<{z}
N
χ(x) u (x) dx
| {z }
difficulty
can be calculated over a complex contour Z = Z1 + Z2 , rather
than over the real domain Ω, i.e.
Z
I2 =
e
Z1
|
−ik0 z·α
N
χ(z)u (z)dz +
{z
}
requires u N (z) for z ∈ Z1 !
Z
Z2
e −ik0 z·α χ(z)u N (z)dz.
Complex contour method
={z}
For u and χ analytic the far
field integral
Z
I2 =
e
−ik0 x·α
Ω
Z1
Z2
γ
<{z}
N
χ(x) u (x) dx
| {z }
difficulty
can be calculated over a complex contour Z = Z1 + Z2 , rather
than over the real domain Ω, i.e.
Z
I2 =
e
Z1
|
−ik0 z·α
N
χ(z)u (z)dz +
{z
}
requires u N (z) for z ∈ Z1 !
Z
Z2
e −ik0 z·α χ(z)u N (z)dz.
Solving Helmholtz eqn. on complex contour
Complex Shifted Laplacian (CSL) system with β ∈ R
−∆ − (1 + iβ)k 2 (x) u(x) = f (x)
is efficiently solvable using multigrid. Erlangga Oosterlee Vuik (2004)
Discretized:
1
2
−
L + (1 + iβ)k uh = bh ,
h2
with L = Laplacian stencil matrix. Division by (1 + iβ) yields
1
bh
2
−
L + k uh =
2
(1 + iβ)h
1 + iβ
= the
system discretized with
√ original Helmholtz
.
iγ
h̃ = 1 + iβ h = ρe h.
Reps, Vanroose, bin Zubair (2010)
(5)
Solving Helmholtz eqn. on complex contour
Rule-of-thumb for CSL Helmholtz system (5) can be efficiently
solved using multigrid for β > 0.5.
Conversion to angle γ
h̃ =
p
1 + iβ h = ρ exp(iγ)h
⇔ γ=
arctan(β)
2
≈ 13.28◦
β=0.5
Note: softened to γ ≈ 9.5◦ with
GMRES
as smoother substitute.
p
.
1 + βi = ρexp(iγ)
={z}
h̃
h
γ
<{z}
Numerical validation (2D)
χ(x, y ) = −
1
5
e
−(x 2 +(y −4)2 )
+e
−(x 2 +(y +4)2 )
Object of interest |χ| (modulus)
Real domain with ECS |χ(x)|
(θECS = 45◦ )
Complex contour |χ(z)|
(γ = 14.6◦ )
Numerical validation (2D)
χ(x, y ) = −
1
5
e
−(x 2 +(y −4)2 )
+e
−(x 2 +(y +4)2 )
Scattered wave solution |u|
Real domain with ECS |u(x)|
LU factorization
Complex contour |u(z)|
V(1,1) cycles (tolres = 10−6 )
Numerical validation (2D)
χ(x, y ) = −
1
5
e
−(x 2 +(y −4)2 )
+e
−(x 2 +(y +4)2 )
Far field amplitude map
Real domain with ECS F (α)
Complex contour F (α)
kFre − Fex k2
= 9.37e-5
kFex k2
kFco − Fex k2
= 1.39e-4
kFex k2
Multigrid performance (3D)
Solving the Helmholtz system - complex contour γ = 10◦
GMRES(3)-smoothed V(1,1) cycles (tolres = 10−6 )
nx ×ny ×nz
k0 = 1/4
k0 = 1/2
k0 = 1
k0 = 2
k0 = 4
163
323
643
1283
2563
10 (0.79s.)
0.24
12 (0.92s.)
0.31
7 (0.62s.)
0.13
2 (0.28s.)
0.00
1 (0.20s.)
0.00
9 (4.65s.)
0.20
10 (4.96s.)
0.24
13 (6.59s.)
0.32
8 (4.24s.)
0.14
2 (1.35s.)
0.00
9 (44.2s.)
0.21
10 (48.3s.)
0.22
11 (54.6s.)
0.27
13 (63.9s.)
0.33
7 (36.1s.)
0.12
9 (352s.)
0.20
10 (390s.)
0.23
10 (387s.)
0.24
11 (428s.)
0.27
13 (503s.)
0.33
9 (2778s.)
0.20
9 (2797s.)
0.21
10 (3079s.)
0.24
10 (3006s.)
0.24
11 (3306s.)
0.26
Multigrid performance (3D)
Solving the Helmholtz system - complex contour γ = 10◦
k0 h = 0.625
GMRES(3)-smoothed V(1,1) cycles (tolres = 10−6 )
nx ×ny ×nz
k0 = 1/4
k0 = 1/2
k0 = 1
k0 = 2
k0 = 4
163
323
643
1283
2563
10 (0.79s.)
0.24
12 (0.92s.)
0.31
7 (0.62s.)
0.13
2 (0.28s.)
0.00
1 (0.20s.)
0.00
9 (4.65s.)
0.20
10 (4.96s.)
0.24
13 (6.59s.)
0.32
8 (4.24s.)
0.14
2 (1.35s.)
0.00
9 (44.2s.)
0.21
10 (48.3s.)
0.22
11 (54.6s.)
0.27
13 (63.9s.)
0.33
7 (36.1s.)
0.12
9 (352s.)
0.20
10 (390s.)
0.23
10 (387s.)
0.24
11 (428s.)
0.27
13 (503s.)
0.33
9 (2778s.)
0.20
9 (2797s.)
0.21
10 (3079s.)
0.24
10 (3006s.)
0.24
11 (3306s.)
0.26
Multigrid performance (3D)
Solving the Helmholtz system (k0 = 1) - complex contour
γ = 10◦
GMRES(3)-smoothed FMG(1,1) cycle
nx × ny × nz
163
323
643
1283
2563
CPU time
kr k2
0.20 s.
3.3e-5
0.78 s.
7.9e-5
6.24 s.
2.7e-5
53.3 s.
1.1e-5
462 s.
4.6e-6
R
Intel
CoreTM i7-2720QM 2.20GHz CPU, 6MB Cache, 8GB RAM.
Multigrid performance (3D)
Solving the Helmholtz system (k0 = 1) - complex contour
γ = 10◦
GMRES(3)-smoothed FMG(1,1) cycle
R
Intel
CoreTM i7-2720QM 2.20GHz CPU, 6MB Cache, 8GB RAM.
System with two electrons
r1 = (x, y , z)1
r2 = (x, y , z)2
(6)
Schrödinger equation is a Helmholtz equation
1
− (∆r1 + ∆r2 ) + V (r1 , r2 ) − E u(r1 , r2 ) = f (r1 , r2 )
2
(7)
Schrödinger equation is a Helmholtz equation
1
− (∆r1 + ∆r2 ) + V (r1 , r2 ) − E u(r1 , r2 ) = f (r1 , r2 )
2
(7)
Defining k 2 (r1 , r2 ) := 2(E − V (r1 , r2 )) leads to a 6D Helmholtz
equation
−∆6D − k 2 (r1 , r2 ) u(r1 , r2 ) = f (r1 , r2 )
(8)
Schrödinger equation is a Helmholtz equation
1
− (∆r1 + ∆r2 ) + V (r1 , r2 ) − E u(r1 , r2 ) = f (r1 , r2 )
2
(7)
Defining k 2 (r1 , r2 ) := 2(E − V (r1 , r2 )) leads to a 6D Helmholtz
equation
−∆6D − k 2 (r1 , r2 ) u(r1 , r2 ) = f (r1 , r2 )
(8)
Again the solution outside the numerical box can be found from
Z
u(r1 , r2 ) =
G (r1 , r2 , s1 , s2 ) f (s1 , s2 ) − k02 χ(s1 , s2 )u N (s1 , s2 ) ds1 ds2
V
(9)
Example in 2D. Spectral properties.
The Hamiltonian
1 d2 1 d2
−
−4.5 exp(−x 2 )−4.5 exp(−y 2 )+2 exp(−(x+y )2 )
2 dx 2 2 dy 2
(10)
It can be written as
H=−
H = H1d ⊗ I + I ⊗ H1d + 2 exp(−(x + y )2 )
(11)
where
1 d2
− 4.5 exp(−x 2 )
2 dx 2
The eigenvalues of 2D system are then approximately:
H1d = −
λ2D ≈ µ1d + µ1d
where H1D φi = µi φi
(12)
(13)
Eigenvalues of 1D Hamiltonian
The example
1 d2
− 4.5 exp(−x 2 )
2 dx 2
For the example the 1D Hamiltonian has eigenvalues
H1d = −
(14)
σ(H1D ) = {−1.0215} ∪ [0, ∞(
When we discretize the Lapacian with finite differences. h → he iθ
Real contour
Complex Contour
Eigenvalues of 2D Hamiltonian
The eigenvalues of
H = H1d ⊗ I + I ⊗ H1d
Real Contour
Complex Contour
Solution depends on the energy E
µ0 < E < 0
0<E
Solution depends on the energy E
single ionization
double ionization
={z}
Z1
γ
Z2
<{z}
2D model problem
1 d2 1 d2
−
−4.5 exp(−x 2 )−4.5 exp(−x 2 )+2 exp(−(x+y )2 )
2 dx 2 2 dy 2
(15)
We discretize with 2552 grid points.
E<
−1.0215 < E ¡
<0<E
no scattering
single ionization
single and double ionization
H=−
Validation of the resulting far field map
·10−3
Total Cross section
single ionization
CC single ionization
double ionization
CC double ionization
crosssection(au)
4
3
2
1
0
−1
0
1
E
2
3
V-cycle convergence rate as function of Energy.
2552 , V-cycle with GMRES(3) smoother.
3D model problem
H = −∆3D + V (x) + V (y ) + V (z) + V12 (x, y ) + V12 (y , z) + V12 (z, x)
(16)
We discretize with 2553 grid points.
E<
−1.8 < E <
−1.0215¡ E
no scattering
single
single and double
<0<E
single, double and trip
3D Benchmark
Multigrid Convergence rate
2553 grid points, V-cycle with GMRES(3) smoother,
1
0.8
0.6
0.4
0.2
0
−4
−2
0
2
E
4
6
8
No free lunch?
The larger β, or γ (rotation angle), the larger the damping in the
Helmholtz equation, the better the multigrid convergence.
−∆ − (1 + iβ)k 2 u = f
(17)
In principle, the value of the integral
Z
u(x) =
G (x, x0 ) k02 χ(x0 ) uin (x0 ) + u N (x0 ) dx0 ,
Ω
should be independent of the complex shift β.
x ∈ Rd .
Example with Coulomb Potentials
Incoming wave is evanenescant wave
φn (x) sin(kn y )
wherep
φn (x) is an eigenstate with eigenvalue λn and
kn = 2(E − λn )
However, φn is calculated numerically
(18)
Incoming wave on the contour
φn (x) sin(kn y )
(19)
Contaminates the volume integral
The scattered wave is contaminated
After multiplication with exponentially growing function the volume integral is contaminated
Possible Solutions? WIP
I
I
Smaller rotation angles?
Shift and invert iteration increases the accuracy of the
eigenstate. But also object needs to be know very accurately.
Comments
1. For educational purposes we have used a volume integral to
calculate the far and near field map. Using Green’s second
theorem it can be rewritten as a surface integral. It is
cheaper and gives the same results.
2. The volume integral is a oscillatory integral. The method of
steepest descent allows to calculate the integral without
resolving all the oscillation. However, here we have product of
an exponentially increasing function and a decaying function.
3. A detailed analysis of the discretization of the complex
contour is required. For the integral only points around the
origin are required. However, we still need to solve the PDE
along the contour accurately, this might require more grid
points.
Conclusions
I
There are many applications that require an integral over
solution of the Helmholtz equations. This gives additional
freedom to develop a solver.
I
We have deformed the contour of the integral and this
requires us to solve a Complex Shifted Laplacian problem.
I
Complex shifted Laplacian problems can be solved scalable
with multigrid. But any solver becomes easier: Domain
Decomposition, ...
I
We have validated and bench-marked the method on 3D
Helmholtz and Schrodinger problems.
I
Extra care is required rounding errors might accumulate and
contaminate the result.
Outlook
I
I
We are working with computational chemists to develop
solvers for Schrödinger equations. Coupled electronic and
molecular motion: unsolved because of the very high number
of dimension.
Explore near field integral:
I
I
Phase Contrast Tomography, Near field Microscopy.
Inverse problems. We reconstruct the object on the contour.
References
[1] Y.A. Erlangga, C.W. Oosterlee and C. Vuik. A novel multigrid based
preconditioner for heterogeneous Helmholtz problems. SIAM Journal on
Scientific Computing 27(4):1471–1492, 2006.
[2] H.C Elman, O.G Ernst, D.P O’Leary, A Multigrid method enhanced by
Krylov subspace iteration for Discrete Helmholtz Equations, SIAM
Journal on Scientific Computing, 23(4):1291-1315 (2001).
[3] O.G Ernst and M.J. Gander, Why it is difficult to solve Helmholtz
problem with Classical Iterative methods, Numerical Analysis of
Multiscale Problems, LNCS, 83:325-263, 2012.
[4] W. Vanroose, D.A. Horner, F. Martin, T.N. Rescigno and C.W. McCurdy.
Double photoionization of aligned molecular hydrogen. Physical Review
A, 74(5):052702, 2006.
[5] B. Reps, W. Vanroose and H. bin Zubair. On the indefinite Helmholtz
equation: Complex stretched absorbing boundary layers, iterative analysis,
and preconditioning. Journal of Computational Physics
229(22):8384–8405, 2010.
[6] S. Cools, B. Reps and W. Vanroose. An efficient multigrid method
calculation of the far field map for Helmholtz problems. Submitted
SISC arXiv:1211.4461, 2013.