On the polarizability and capacitance of the cubePolarizability Superellipse and Square The cube...

Polarizability Superellipse and Square The cube Numerical method Details To the movies

On the polarizability and capacitance of the cube

Johan Helsing

Lund University

Talk at FACM ’12, NJIT, May 19, 2012

Acknowledgement

The work presented has in part been carried out in cooperationwith or supported by:

Alexandru Aleman

Anders Karlsson

Ross McPhedran

Graeme Milton

Rikard Ojala

Karl-Mikael Perfekt

Daniel Sjoberg

the Swedish Research Council


A dielectric inclusion in a dielectric medium

νrS

e

VE

ǫ2 ǫ1


The electrostatic boundary value problem

U(r) continuous

limr→∞

∇U(r) = e

∆U(r) = 0

∆U(r) = 0

ǫ2∂

∂νr

U int(r) = ǫ1

∂

∂νr

Uext(r)


The polarizability α is given by

α = (ǫ2 − ǫ1)

∫

V

(e · ∇U(r))dr .


The polarizability α is given by

α = (ǫ2 − ǫ1)

∫

V

(e · ∇U(r))dr .

When does α exist? For what ǫ1, ǫ2 does the cube have an α?

Is it difficult to compute α?

Does a small rounding of a corner correspond to a small relativeperturbation of input data?

Is it important to compute α?

How high accuracy should one demand of a fast α-solver?

Are there many failed attempts to compute α for non-smooth S inthe literature?


Equivalent integral equation formulation

Ansatz which solves the PDE and takes care of r → ∞:

U(r) = e · r +

∫

S

G (r , r ′)ρ(r ′)dσr ′ ,

where ρ(r) is an unknown layer density.




U(r) = e · r +

∫

S

G (r , r ′)ρ(r ′)dσr ′ ,

where ρ(r) is an unknown layer density. Insertion of U(r) in theboundary condition at S :

(ǫ2 + ǫ1)

(ǫ2 − ǫ1)ρ(r) + 2

∫

S

∂

∂νrG (r , r ′)ρ(r ′)dσr ′ = −2 (e · νr ) , r ∈ S .

Abbreviated notation:

(−z + K ) ρ(r) = g(r) . (1)




U(r) = e · r +

∫

S

G (r , r ′)ρ(r ′)dσr ′ ,

where ρ(r) is an unknown layer density. Insertion of U(r) in theboundary condition at S :

(ǫ2 + ǫ1)

(ǫ2 − ǫ1)ρ(r) + 2

∫

S

∂

∂νrG (r , r ′)ρ(r ′)dσr ′ = −2 (e · νr ) , r ∈ S .

Abbreviated notation:

(−z + K ) ρ(r) = g(r) . (1)

The polarizability from ρ(r):

α = −ǫ1

∫

S

ρ(r) (e · r) dσr . (2)


When does(−z + K ) ρ(r) = g(r)

have a (unique) solution?



have a (unique) solution? Well, when z is not an eigenvalue of K .




What does the spectrum of K look like?




What does the spectrum of K look like? If S is smooth, then K iscompact and has a discrete spectrum accumulating at zero. Alleigenvalues zi are real and less than one in modulus.

The electrostatic equation is uniquely solvable for z /∈ [−1, 1], andoften otherwise as well.






Eigenvalues zi correspond to PLASMONS in α(z).






Eigenvalues zi correspond to PLASMONS in α(z).

Repetition: z = (ǫ1 + ǫ2)/(ǫ1 − ǫ2).


The Superellipse

The superellipse|r1|

k + |r2|k = 1

is a circle for k = 2 and “numerically” a square for k = 1016.

−1.5 −1 −0.5 0 0.5 1 1.5

−1

−0.5

0

0.5

1

Superellipse at k=2, 8, 32, 128


What does the spectrum of K look like for a superellipse?


What does the spectrum of K look like for a superellipse?

100

105

1010

1015

−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

superellipse |r1|k+|r

2|k=1

k

eige

nval

ues

z i

dark plasmonsbright plasmons


A zoom at low k reveals a fascinating pattern

2 3 4 5 6 7 8 9 1010

−8

10−6

10−4

10−2

100


2|k=1

k

posi

tive

eige

nval

ues

z i


Are there any phase transitions or critical points here?


The polarizability α(z), z ∈ [−1, 1], of a superellipse at k = 1012

(almost a square):

−1 −0.5 0 0.5 1

−10

−8

−6

−4

−2

0

2

4

6

8

10


2|k=1 with k=1012

x

α(x)

A myriad of PLASMONS.


Can we guess the spectrum of K for a square by studying thesuperellipse in the limit k → ∞?

100

105

1010

1015

−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5


2|k=1

k

eige

nval

ues

z i



Can we guess the spectrum of K for a square by studying thesuperellipse in the limit k → ∞?

100

105

1010

1015

−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5


2|k=1

k

eige

nval

ues

z i


The answer depends on what function space we consider.


What does the limit polarizability α+(x) = limy→0+ α(x + iy) looklike for a square? Does it exist at all? and if so, in what functionspaces lie the corresponding limit solutions ρ+(r) and U+(r)?


What does the limit polarizability α+(x) = limy→0+ α(x + iy) looklike for a square? Does it exist at all? and if so, in what functionspaces lie the corresponding limit solutions ρ+(r) and U+(r)?

−1 −0.5 0 0.5 1

−10

−8

−6

−4

−2

0

2

4

6

8

10

The square: polarizability

x

α+(x

)

real{α+(x)}

imag{α+(x)}

Yes, α+(x) exists, but do we see any PLASMONS?


Comparison:

−1 −0.5 0 0.5 1

−10

−8

−6

−4

−2

0

2

4

6

8

10


2|k=1 with k=1012

x

α(x)

−1 −0.5 0 0.5 1

−10

−8

−6

−4

−2

0

2

4

6

8

10

The square: polarizability

x

α+(x

)

real{α+(x)}

imag{α+(x)}


Theory for Lipschitz sets

The polarizability α+(x) exists.

The polarizability α(z) admits an integral representation

α(z) =

∫

R

dµ(x)

x − z=

∫

σµ

µ′(x)dx

x − z,

where σµ = {x : µ′(x) > 0} and the measure dµ(x) relate toα+(x) as

µ′(x) = ℑ{α+(x)}/π , x ∈ R .

The spectral measure is of great interest in theoretical materialscience.

One can show that for z ∈ L, L : σµ ⊂ L ⊂ [−1, 1], theelectrostatic equation has no finite-energy solution U(r). For allother z there are solutions U(r). This has consequences.


What about the cube?


What about the cube?

−1 −0.5 0 0.5 1−10

−5

0

5

10

15

20

The cube: polarizability

x

α+(x

)

real{α+(x)}

imag{α+(x)}

No one has succeeded in computing this (in its entirety) before.The reason being, perhaps, the behavior of U+(r) and ρ+(r).


Some cube records

Table: Reference values, estimated relative errors, and best previousestimates of the polarizability α+(x) and the capacitance C of the cube.

present reference values relerr previous results relerr

α+(−1) 3.644305190268 10−11 3.6442 3 · 10−5

α+(−0.6) 5.85574775 + 16.64205643i 10−8

α+(0.25) −2.76289925 + 3.08034035i 10−7

α+(1) −1.638415712936517 10−14

−1.6383 6 · 10−5

C 0.66067815409957 10−13 0.66067813 10−7


103

104

105

10−15

10−10

10−5

100

The cube: convergence

Number of discretization points n

Est

imat

ed r

elat

ive

erro

r in

α+(x

) an

d C

α+(−1)

α+(−0.6)

α+(0.25)

α+(1)

C

Figure: Convergence of α+(x) and the capacitance C of a unit cube.The values of x correspond to: cube with infinite permittivity (x = −1),resonance in corners (x = −0.6), resonance along edges (x = 0.25), andcube with zero permittivity (x = 1).


Table: Numerical progress for the capacitance of the unit cube.

value year Author0.6555 1957 Reitan–Higgins0.661 1961 Greenspan–Silverman0.6596 1967 Cochran0.661 1990 Brown0.66067475 1992 Goto–Shi–Yoshida0.6632 1994 Douglas–Zho–Hubbard0.6606751 1997 Given–Hubbard–Douglas0.6606785 1997 Read0.660692 2001 Mansfield–Douglas–Garboczi0.6601 2002 Bai–Lonngren0.6606835 2003 Hwang–Mascagni0.6606782 2004 Hwang–Mascagni0.6606 2004 Wintle0.6606780 2004 Mascagni–Simonov0.66067674 2004 Read0.6606749 2005 Ong–Lim0.66067786 2008 Lazic–Stefancic–Abraham0.6606746 2009 Mukhopadhyay–Majumdar0.66067813 2010 Hwang–Mascagni–Won0.678 2011 Bontzios–Dimopoulos–Hatzopoulos0.66067815409957 2012 Helsing–Perfekt


Numerical method

The computation of α+(x) relies on the solution ρ+(r) to

(I + λK ) ρ+(r) = λg(r) ,

but how is it possible to obtain accurate numerical approximationsto ρ+(r), which barely lies in W−1,∞?


Numerical method

The computation of α+(x) relies on the solution ρ+(r) to

(I + λK ) ρ+(r) = λg(r) ,

but how is it possible to obtain accurate numerical approximationsto ρ+(r), which barely lies in W−1,∞?

Well, thanks toK = K ⋆ + K ◦ ,

ρ(r) = (I + λK ⋆)ρ(r) ,(

I + λK ◦(I + λK ⋆)−1)

ρ(r) = λg(r) ,

which gives a solution ρ+(r) in L∞.


Discretization

Discretization of(I + λK ) ρ(r) = λg(r)

gives(Icoa + λK◦

coaR) ρcoa = λgcoa , (3)

where the small block diagonal matrix

R = PTW (Ifin + λK⋆

fin)−1P

is obtained via a fast recursion.


12

3

4

56

7

8

γ1

Figure: A coarse mesh with eight quadrature panels on a closed surface.A fine mesh with 14 panels is created by refinement in a direction towardthe corner γ1.

0 50 100 150 200

0

50

100

150

200

nz = 501760 50 100 150 200

0

50

100

150

200

nz = 163840 50 100 150 200

0

50

100

150

200

nz = 33792

Figure: Kfin = K⋆

fin+ K◦

fin.


Details

(I + λK ) ρ(r) = λg(r) .

Assumption: K is compact some distance away from the corners γj

and g(r) is piecewise smooth.

Let K (r , r ′) be the kernel of K . Split K (r , r ′) into two functions

K (r , r ′) = K ⋆(r , r ′) + K ◦(r , r ′) ,

where K ⋆(r , r ′) is zero except for when r and r ′ simultaneously lieclose to the same γj . In this latter case K ◦(r , r ′) is zero.

The kernel split corresponds to an operator split K = K ⋆ + K ◦,where K ◦ is compact. Discretization on a fine mesh gives

(Ifin + λK⋆

fin + λK◦

fin) ρfin = λgfin .


Details

The transformρfin = (Ifin + λK⋆

fin)−1ρfin

leads to(

Ifin + λK◦

fin (Ifin + λK⋆

fin)−1)

ρfin = λgfin .

Note:

K◦

fin(Ifin + λK⋆

fin)−1 is, in practice, a compact operator

ρfin is piecewise smooth

neither gfin, ρfin, nor K◦

finneeds a fine mesh for resolution.


Details

Let P be a prolongation operator from the coarse mesh to the finemesh and let Q be a restriction operator in the opposite direction

QP = I .

Then

gfin = Pgcoa

ρfin = Pρcoa

K◦

fin = PK◦

coaPTW .

Compare reduced SVD. Here PW = WfinPW−1coa and W has

quadrature weights on the diagonal. Taken together

(

Icoa + λK◦

coaPTW (Ifin + λK⋆

fin)−1P

)

ρcoa = λgcoa . (4)


Details

WithR = PT

W (Ifin + λK⋆

fin)−1

P

(4) becomes(Icoa + λK◦

coaR) ρcoa = λgcoa .

We only need the fine mesh for the construction of the small blockdiagonal matrix R. One block per corner.

The blocks of R can be obtained via a fast recursion, where step i

inverts and compresses contributions to R involving the outermostpanels on level i of a locally n-ply refined mesh.

Newton’s method can accelerate the recursion in situations thatdemand extreme refinement (very many recursion steps). Thiscorresponds to inverting a giant full matrix in sublinear time.Boundary singularities do no longer pose any practical problems.


The fast recursion

Ri = PTWbc

(

F{R−1(i−1)} + I◦b + λK◦

b

)

−1Pbc , i = 1, . . . , n .

part ofcoarse mesh

refined n = 4

i = 1

i = 2

i = 3

Figure: Recursion on a refined mesh surrounding a corner.


Matlab

Kmat=Kinit(zloc,wloc,nzloc,96);

MAT=eye(96)+lambda*Kmat;

starL=[17:80];

R=inv(MAT(starL,starL));

myerr=1;

while myerr>eps

Rold=R;

MAT(starL,starL)=inv(R);

R=Pwbc’*inv(MAT)*Pbc;

myerr=norm(R-Rold,’fro’)/norm(R,’fro’);

end


An extreme example – 20 years too late?

Figure: A unit cell of a random checkerboard with a million squares. Thepermittivity ratio is ǫ2/ǫ1 = 106 and the effective permittivity is obtainedwith a relative error of 10−9.


References

J. Helsing and K.-M. Perfekt (2012) “On the polarizability andcapacitance of the cube”, arXiv:1203.5997v2.

J. Helsing, R.C. McPhedran, and G.W. Milton (2011) “Spectralsuper-resolution in metamaterial composites”, New J. Phys., 13(11),115005.

J. Helsing (2011) “The effective conductivity of arrays of squares: largerandom unit cells and extreme contrast ratios”, J. Comput. Phys.,230(20), 7533–7547.

J. Helsing (2011) “A fast and stable solver for singular integral equationson piecewise smooth curves”, SIAM J. Sci. Comput., 33(1), 153–174.

J. Helsing (2009) “Integral equation methods for elliptic problems withboundary conditions of mixed type”, J. Comput. Phys., 228(23), pp.8892–8907.

J. Helsing and R. Ojala (2008) “Corner singularities for elliptic problems:integral equations, graded meshes, quadrature, and compressed inversepreconditioning”, J. Comput. Phys., 227(20), 8820–8840.


−6 −5 −4 −3 −2 −1 0 1−5

−4

−3

−2

−1

0

1

2

3

4

5

Staggered array of square cylinders at p1=0.49999999995

permittivity ratio σ

effe

ctiv

e pe

rmitt

ivity

εef

f(σ)

real{εeff}

imag{εeff}

−6 −5 −4 −3 −2 −1 0 1

10−15

10−10

10−5

100

Staggered array of square cylinders at p1=0.49999999995

|εef

f(σ)ε

eff

∗(1

/σ∗ )−

1|

permittivity ratio σ

Figure: Effective permittivity (polarizability) and error estimates.

On the polarizability and capacitance of the cubePolarizability Superellipse and Square The cube...

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