On the polarizability and capacitance of the cube

Transcription

1 On the polarizability and capacitance of the cube Johan Helsing Lund University Double Seminar at CSC, May 10, 2012

2 Acknowledgement The work presented has in part been carried out in cooperation with or supported by: Alexandru Aleman Anders Karlsson Ross McPhedran Graeme Milton Rikard Ojala Karl-Mikael Perfekt Daniel Sjöberg the Swedish Research Council

3 A dielectric inclusion in a dielectric medium e ǫ 2 ǫ 1 V E S ν r

4 The electrostatic boundary value problem lim U(r) = e r U(r) continuous ǫ 2 ν r U int (r) = ǫ 1 ν r U ext (r) U(r) = 0 U(r) = 0

5 The polarizability α is given by α = (ǫ 2 ǫ 1 ) (e U(r))dr. V

6 The polarizability α is given by α = (ǫ 2 ǫ 1 ) When does α exist? V (e U(r))dr.

7 The polarizability α is given by α = (ǫ 2 ǫ 1 ) V (e U(r))dr. When does α exist? For what ǫ 1, ǫ 2 does the cube have an α?

8 The polarizability α is given by α = (ǫ 2 ǫ 1 ) V (e U(r))dr. When does α exist? For what ǫ 1, ǫ 2 does the cube have an α? Is it difficult to compute α?

9 The polarizability α is given by α = (ǫ 2 ǫ 1 ) V (e U(r))dr. When does α exist? For what ǫ 1, ǫ 2 does the cube have an α? Is it difficult to compute α? Does a small rounding of a corner correspond to a small relative perturbation of input data?

10 The polarizability α is given by α = (ǫ 2 ǫ 1 ) V (e U(r))dr. When does α exist? For what ǫ 1, ǫ 2 does the cube have an α? Is it difficult to compute α? Does a small rounding of a corner correspond to a small relative perturbation of input data? Is it important to compute α?

11 The polarizability α is given by α = (ǫ 2 ǫ 1 ) V (e U(r))dr. When does α exist? For what ǫ 1, ǫ 2 does the cube have an α? Is it difficult to compute α? Does a small rounding of a corner correspond to a small relative perturbation of input data? Is it important to compute α? How high accuracy should one demand of a fast α-solver?

12 The polarizability α is given by α = (ǫ 2 ǫ 1 ) V (e U(r))dr. When does α exist? For what ǫ 1, ǫ 2 does the cube have an α? Is it difficult to compute α? Does a small rounding of a corner correspond to a small relative perturbation 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 in the literature?

13 Equivalent integral equation formulation Ansatz which solves the PDE and takes care of r : U(r) = e r + G(r,r )ρ(r )dσ r, where ρ(r) is an unknown layer density. S

14 Equivalent integral equation formulation Ansatz which solves the PDE and takes care of r : U(r) = e r + G(r,r )ρ(r )dσ r, S where ρ(r) is an unknown layer density. Insertion of U(r) in the boundary condition at S: (ǫ 2 + ǫ 1 ) (ǫ 2 ǫ 1 ) ρ(r) + 2 G(r,r )ρ(r )dσ r = 2(e ν r ), r S. ν r S

15 Equivalent integral equation formulation Ansatz which solves the PDE and takes care of r : U(r) = e r + G(r,r )ρ(r )dσ r, S where ρ(r) is an unknown layer density. Insertion of U(r) in the boundary condition at S: (ǫ 2 + ǫ 1 ) (ǫ 2 ǫ 1 ) ρ(r) + 2 G(r,r )ρ(r )dσ r = 2(e ν r ), r S. ν r Abbreviated notation: S ( z + K) ρ(r) = g(r). (1)

16 Equivalent integral equation formulation Ansatz which solves the PDE and takes care of r : U(r) = e r + G(r,r )ρ(r )dσ r, S where ρ(r) is an unknown layer density. Insertion of U(r) in the boundary condition at S: (ǫ 2 + ǫ 1 ) (ǫ 2 ǫ 1 ) ρ(r) + 2 G(r,r )ρ(r )dσ r = 2(e ν r ), r S. ν r Abbreviated notation: The polarizability from ρ(r): α = ǫ 1 S ( z + K) ρ(r) = g(r). (1) S ρ(r)(e r) dσ r. (2)

17 When does have a (unique) solution? ( z + K) ρ(r) = g(r)

18 When does ( z + K) ρ(r) = g(r) have a (unique) solution? Well, when z is not an eigenvalue of K.

19 When does ( z + K) ρ(r) = g(r) have a (unique) solution? Well, when z is not an eigenvalue of K. What does the spectrum of K look like?

20 When does ( z + K) ρ(r) = g(r) have a (unique) solution? Well, when z is not an eigenvalue of K. What does the spectrum of K look like? If S is smooth, then K is compact and has a discrete spectrum accumulating at zero. All eigenvalues are real and less than one in modulus.

21 When does ( z + K) ρ(r) = g(r) have a (unique) solution? Well, when z is not an eigenvalue of K. What does the spectrum of K look like? If S is smooth, then K is compact and has a discrete spectrum accumulating at zero. All eigenvalues are real and less than one in modulus. The electrostatic equation is uniquely solvable for z / [ 1,1], and often otherwise as well.

22 When does ( z + K) ρ(r) = g(r) have a (unique) solution? Well, when z is not an eigenvalue of K. What does the spectrum of K look like? If S is smooth, then K is compact and has a discrete spectrum accumulating at zero. All eigenvalues are real and less than one in modulus. The electrostatic equation is uniquely solvable for z / [ 1,1], and often otherwise as well. Eigenvalues z correspond to PLASMONS in α(z).

23 The Superellipse The superellipse r 1 k + r 2 k = 1 is a circle for k = 2 and numerically a square for k = Superellipse at k=2, 8, 32,

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

25 What does the spectrum of K look like for a superellipse? superellipse r 1 k + r 2 k = eigenvalues z i dark plasmons bright plasmons k

26 A zoom at low k reveals a fascinating pattern superellipse r k + r k = dark plasmons bright plasmons positive eigenvalues z i k Are there any phase transitions or critical points here?

27 What does the polarizability α(z), z [ 1,1], look like for a superellipse (almost square)?

28 What does the polarizability α(z), z [ 1,1], look like for a superellipse (almost square)? superellipse r 1 k + r 2 k =1 with k= α(x) x A myriad of PLASMONS.

29 Can we guess the spectrum of K for a square by studying the superellipse in the limit k? superellipse r 1 k + r 2 k = eigenvalues z i dark plasmons bright plasmons k

30 Can we guess the spectrum of K for a square by studying the superellipse in the limit k? superellipse r 1 k + r 2 k = eigenvalues z i dark plasmons bright plasmons k The answer depends on what function space we consider.

31 What does the limit polarizability α + (x) = lim y 0 + α(x + iy) look like for a square? Does it exist at all? and if so, in what function spaces lie the corresponding limit solutions ρ + (r) and U + (r)?

32 What does the limit polarizability α + (x) = lim y 0 + α(x + iy) look like for a square? Does it exist at all? and if so, in what function spaces lie the corresponding limit solutions ρ + (r) and U + (r)? The square: polarizability α + (x) real{α + (x)} 8 imag{α + (x)} x Yes, α + (x) exists, but do we see any PLASMONS?

33 Comparison: superellipse r 1 k + r 2 k =1 with k=10 12 The square: polarizability α(x) x α + (x) real{α + (x)} 8 imag{α + (x)} x

34 Theory for Lipschitz sets The polarizability α + (x) exists. The polarizability α(z) admits an integral representation dµ(x) α(z) = x z = µ (x)dx, σ µ x z R where σ µ = {x : µ (x) > 0} and the measure dµ(x) relate to α + (x) as µ (x) = I{α + (x)}/π, x R. The spectral measure is of great interest in theoretical material science. One can show that for z L, L : σ µ L [ 1,1], the electrostatic equation has no finite-energy solution U(r). For all other z there are solutions U(r). This has consequences.

35 What about the cube?

36 What about the cube? The cube: polarizability real{α + (x)} imag{α + (x)} 10 α + (x) x No one has succeeded in computing this (in its entirety) before. The reason being, perhaps, the behavior of U + (r) and ρ + (r).

37 Some cube records Table: Reference values, estimated relative errors, and best previous estimates of the polarizability α + (x) and the capacitance C of the cube. present reference values relerr previous results relerr α + ( 1) α + ( 0.6) i 10 8 α + (0.25) i 10 7 α + (1) C

38 10 0 The cube: convergence Estimated relative error in α + (x) and C α + ( 1) α + ( 0.6) α + (0.25) α + (1) C Number of discretization points n 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), and cube with zero permittivity (x = 1).

39 Table: Numerical progress for the capacitance of the unit cube. value year Author Reitan Higgins Greenspan Silverman Cochran Brown Goto Shi Yoshida Douglas Zho Hubbard Given Hubbard Douglas Read Mansfield Douglas Garboczi Bai Lonngren Hwang Mascagni Hwang Mascagni Wintle Mascagni Simonov Read Ong Lim Lazić Štefančić Abraham Mukhopadhyay Majumdar Hwang Mascagni Won Bontzios Dimopoulos Hatzopoulos Helsing Perfekt

40 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 approximations to ρ + (r), which barely lies in W 1,?

41 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 approximations to ρ + (r), which barely lies in W 1,? Well, thanks to K = 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.

42 Discretization Discretization of (I + λk) ρ(r) = λg(r) gives (I coa + λk coa R) ρ coa = λg coa, (3) where the small block diagonal matrix R = P T W (I fin + λk fin ) 1 P is obtained via a fast recursion.

43 Polarizability Superellipse and Square The cube Numerical method Details To the movies γ 1 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 toward the corner γ nz = nz = Figure: Kfin = K fin + K fin nz =

44 Details (I + λk)ρ(r) = λg(r). Assumption: K is compact some distance away from the corners γ j 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 lie close to the same γ j. 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 (I fin + λk fin + λk fin ) ρ fin = λg fin.

45 Details The transform ρ fin = (I fin + λk fin ) 1 ρ fin leads to ( I fin + λk fin (I fin + λk fin ) 1) ρ fin = λg fin. Note: K fin (I fin + λk fin ) 1 is, in practice, a compact operator ρ fin is piecewise smooth neither g fin, ρ fin, nor K fin needs a fine mesh for resolution.

46 Details Let P be a prolongation operator from the coarse mesh to the fine mesh and let Q be a restriction operator in the opposite direction Then QP = I. g fin = Pg coa ρ fin = P ρ coa K fin = PK coa PT W. Compare reduced SVD. Here P W = W fin PWcoa 1 and W has quadrature weights on the diagonal. Taken together ( ) I coa + λk coa PT W (I fin + λk fin ) 1 P ρ coa = λg coa. (4)

47 Details With (4) becomes R = P T W (I fin + λk fin ) 1 P (I coa + λk coa R) ρ coa = λg coa. We only need the fine mesh for the construction of the small block diagonal 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 outermost panels on level i of a locally n-ply refined mesh. Newton s method can accelerate the recursion in situations that demand extreme refinement (very many recursion steps). This corresponds to inverting a giant full matrix in sublinear time. Boundary singularities do no longer pose any practical problems.

48 The fast recursion R i = P T Wbc ( F{R 1 (i 1) } + I b + λk b) 1 Pbc, i = 1,...,n. part of coarse mesh i = 1 i = 2 refined n = 4 i = 3 Figure: Recursion on a refined mesh surrounding a corner.

49 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

50 An extreme example 20 years too late? Figure: A unit cell of a random checkerboard with a million squares. The permittivity ratio is ǫ 2 /ǫ 1 = 10 6 and the effective permittivity is obtained with a relative error of 10 9.

51 References J. Helsing and K.-M. Perfekt (2012) On the polarizability and capacitance of the cube, arxiv: v2. J. Helsing, R.C. McPhedran, and G.W. Milton (2011) Spectral super-resolution in metamaterial composites, New J. Phys., 13(11), J. Helsing (2011) The effective conductivity of arrays of squares: large random unit cells and extreme contrast ratios, J. Comput. Phys., 230(20), J. Helsing (2011) A fast and stable solver for singular integral equations on piecewise smooth curves, SIAM J. Sci. Comput., 33(1), J. Helsing (2009) Integral equation methods for elliptic problems with boundary conditions of mixed type, J. Comput. Phys., 228(23), pp J. Helsing and R. Ojala (2008) Corner singularities for elliptic problems: integral equations, graded meshes, quadrature, and compressed inverse preconditioning, J. Comput. Phys., 227(20),

52 Staggered array of square cylinders at p = Staggered array of square cylinders at p = effective permittivity ε eff (σ) real{εeff} imag{εeff} permittivity ratio σ ε eff (σ)ε eff (1/σ ) permittivity ratio σ Figure: Effective permittivity (polarizability) and error estimates.