Summer course: Plasmonic resonances

Transcription

1 Summer course: Plasmonic resonances Faouzi TRIKI Grenoble-Alpes University, France Journées d été des mathématiciens tunisiens à l étranger Mediterranean Institute For The Mathematical Sciences July 20, 2016

2 The results presented in this course, have been done in collaboration with Habib AMMARI (ETH, Zurich) Eric BONNETIER (Grenoble-Alpes University) Charles DAPOGNY (Grenoble-Alpes University) Michael VOGELIUS (NSF & Rutgers University)

3 Outline of Part I Nanoparticles? Spectral problem Quasistatic approximation Biosensing problem Neumann Poincaré operator Poincaré variational problem

4 Nanoparticles? [A. Moores and F. Goettmann, New J. Chem., 2006, 30, ] the first syntheses of metallic small particles date back to the 4th or 5th century BC where gold specimen were reported in China and Egypt. their optical properties were used for coloration of glass, ceramics, china and pottery

5 Nanoparticles? the interesting diffractive properties of these particles are linked to resonances phenomena plasmon resonances may occur in metallic particles if the dielectric permittivity inside the particle is negative the wavelength of the incident excitation is much larger than the dimension of the particle for nanoscale metallic particles, these resonances occur in the optical frequency range and they result in an extremely large enhancement of the electromagnetic field near the boundary of the particles this phenomena has applications in many areas such as nanophotonics, nanolithography, near field microscopy, biosensors and medicine (for example gold nanoparticles allow heat from infrared lasers to be targeted on cancer tumors

6 Nanoparticles? The desired resonance frequencies as well as the local fields enhancement can be achieved by controlling the geometry of the metallic nanostructure. they are filled with a real metal: the electric permittivity ε depends on the frequency of the excitation ω. their size δ is very small compare to the incident wavelength λ = 2π/ω, that is λ = δω/2π 1. In practice δ is between 10 and 100 nm and λ 650 nm. the corresponding plasmonic resonances are in the the optical frequency range.

7 The spectral problem Let Ω be a bounded C 1,α -domain in R 2, with normalized size, that is Ω = 1. Assume that the metallic nanoparticle occupies Ω δ = z + δω, where the position z is fixed in R 2. The complex number ω is said to be a resonant frequency of the nanoparticle Ω δ if there exists a non-trivial solution E to the system (TE polarization): E + ω 2 ε(ω, x)µ 0 E = 0 in R 2 \ Ω δ Ω δ, εe = 0 on Ω δ, = 0 on Ω δ, E ν where the electric permittivity ε is given by ε(ω, x) = { ε0 for x R 2 \ Ω δ, ε 0ˆɛ(ω) for x Ω δ,

8 The spectral problem Drude model The metal that fills the nanoparticle is assumed to be real : Its dielectric constant is described by the Drude model: ˆɛ(ω) = ε ω 2 P ω 2 + iωγ, where ε > 0, ω P > 0 and Γ > 0 are the metal parameters that are usually fitted using experiment data [A. Moores and F. Goettmann, The plasmon band in noble nanoparticles: an introduction to theory and applications, New J. Chem., 2006, 30, ] For example the function ˆε(ω) with effective parameters: ε = 3.7 ev, ω P = 8.9 ev, Γ = ev for silver reproduce quite well the experimental values of the dielectric constant in the frequency range 0.8 ev to 4 ev [P.B. Johnson and R. W. Christy, Optical constants of the noble metals, Phys. Rev. B, 6, (1972).]

9 Quasi-static approximation Since δ is too small an asymptotic analysis needs to be performed to avoid costly and complex computations. Making the change of variables x = z + δx in the system, we obtain Ẽ + δ2 ω 2 ε(ω, x)µ 0 Ẽ = 0 in R 2 \ Ω Ω, εẽ = 0 on Ω, = 0 on Ω, Ẽ ν where the electric permittivity ε is given by ε(ω, x) = { ε0 for x R 2 \ Ω, ε 0ˆɛ(ω) for x Ω,

10 Consequently the condition R (ε) = R(ε 0ˆɛ(ω 0 )) < 0 is necessary to have a solution. Quasi-static approximation When δ 0, we obtain the spectral problem Ẽ = 0 in R2 \ Ω Ω, εẽ = 0 on Ω, = 0 on Ω. Ẽ ν This was observed by [ID Mayergoyz, DR Fredkin, Z Zhang - Physical Review B, 2005]. ω j (δ) ω 0 δ 0, where ω 0 is a complex value for which there exists a non-trivial solution Ẽ to the above problem. Assume that V = ε(ω 0, x)e, then. 1 ε V = 0 in R2.

11 Quasi-static approximation The Quasi-static approximation is rigorously justified and the complete asymptotic expansion of the plasmonic resonances is derived. Theorem (BT 15) Let ω 0 be a the unique quasistatic resonance in B ρ (ω 0 ) C \ R, and denote by m its multiplicity. Then, there exists δ 0 > 0 such that for 0 < δ δ 0, the set of values of the ω 0 -group, satisfy 1 m m ω j (δ) = ω 0 + j=1 + holds for δ (0, δ 0 ). ω (2) p,0 p=1 ω p,n (2) n=1 p=0 1 (log(δ)) p + δ 2n (log(δ)) p n, ω p,n (1) n=0 p=0 δ 2n+1 (log(δ)) p n 1

12 Quasi-static approximation Why metallic nanoparticles? Ω ε(ω) Ω κ size δ << λ. permittivity ε(ω). plasmonic resonances 2π ω j (δ) 2π ω 0 as δ 0. the wavelength 2π ω 0 is within the optical range. size δ << λ. permittivity is a constant κ. scattering resonances 2π ω j (δ) 0 as δ 0. the resonances can not be measured.

13 Biosensing problem nanoparticle: Ω δ = z + δω thin layer: L δ = z + δl Denote Ω e δ = R2 \ (Ω δ L δ ) ε p (ω, x) = We consider the spectral problem: ε 0 for x Ω e δ, ε 0 κ for x L δ, ε 0ˆɛ(ω) for x Ω δ, ε(ω) L Ω κ E + ω 2 ε p (ω, x)µ 0 E = 0 in Ω e δ L δ Ω δ, ε p E = 0 on L δ Ω δ, = 0 on L δ Ω δ, E ν Let ω p,j (δ) be the plasmonic resonances of the perturbed nanoparticule.

14 Biosensing problem Biosensing problem: to determine κ(x) from the knowledge of the schift in plasmonic resonances: (ω p,j (δ) ω j (δ)). Making again the change of variables x = z + δx in the system, we obtain Ẽ + δ2 ω 2 ε p (ω, x)µ 0 Ẽ = 0 in Ω e L Ω, ε p Ẽ = 0 on L Ω, = 0 on L Ω, Ẽ ν where the electric permittivity ε p is given by ε p (ω, x) = ε 0 for x Ω e, ε 0 κ for x L, ε 0ˆɛ(ω) for x Ω,

15 Biosensing problem Quasi-static approximation The quasi-static approximation gives Ẽ = 0 in Ωe L Ω, ε p Ẽ = 0 on L Ω, = 0 on L Ω, Ẽ ν Let ω L,j (κ) be the quasi-static plasmonic resonances of the perturbed nanoparticule. We neglect in the rest of the talk the error in the quasi-static approximation. Biosensing problem: to determine κ from the knowledge of the schift in plasmonic resonances: (ω L,j (κ) ω 0,j ).

16 Neumann-Poincaré operator Recall that ( ) 1 x log( x y ) = δ y (x). 2π We now define, the single and double layer potentials of the functions ϕ H 1 2 ( Ω) and ψ H 1 2 ( Ω) respectively, by S 0 [ϕ](x) = 1 log( x y )ϕ(y)dσ(y), 2π Ω D 0 [ψ](x) = 1 (x y) ν(y) 2π x y 2 ψ(y)dσ(y), Ω for x R 2 \ Ω. We also define the boundary integral operator K 0 on H 1 2 ( Ω) by K 0 [ψ](x) = 1 (x y) ν(y) 2π x y 2 ψ(y)dσ(y), for x Ω. Ω

17 Neumann-Poincaré operator Let K0 be the adjoint of K 0, that is K0[ϕ](x) = 1 (x y) ν(x) 2π Ω x y 2 ϕ(y)dσ(y), for x Ω, ϕ H 1 2 ( Ω). K0 is called the Neumann-Poincaré operator (appears in many applications with different names). for x Ω. S 0 [ϕ](x) + = S 0 [ϕ](x), ( (D 0 [ψ](x)) ± = 1 ) 2 I + K 0 [ψ](x), ( ) ( ν S 0[ϕ](x) ± = ± 1 ) 2 I + K 0 [ϕ](x),

18 Neumann-Poincaré operator Lemma K 0 : H 1 2 ( Ω) H 1 2 ( Ω) is compact. Proof. Since Ω is a C 1,α domain, we have (x y) ν(x) x y 2 C(Ω) 1 for x, y Ω, x y. x y 1 α Lemma The operator A : H 1 2 ( Ω) R H 1 2 ( Ω) R, defined by ( ) A(ϕ, a) = S 0 [ϕ] + a, ϕdσ, Ω is invertible.

19 Neumann-Poincaré operator Lemma Let (ϕ e, a e ) in H 1 2 ( Ω) R, be the unique solution of ( ) A(ϕ e, a e ) = S 0 [ϕ e ] + a e, ϕ e dσ = (0, 1). Ω Then i S 0 : H 1 2 ( Ω) H 1 2 ( Ω) is invertible if and only if a e 0. Define H ( Ω) = {ϕ H 1 2 ( Ω), ϕ, 1 1 2, 1 2 H 1 2 e ( Ω) = {ψ H 1 2 ( Ω), ϕe, ψ 1 2, 1 2 = 0}, = 0}. ii S 0 : H ( Ω) H 1 2 e ( Ω) is negative invertible operator.

20 Neumann-Poincaré operator Lemma Let λ be a real number. The operator λi + K0 is one to one on H ( Ω) if λ 1 2, and for λ (, 1 2 ] ( 1 2, + ), λi + K 0 is one to one on H 1 2 ( Ω). Proof. The argument is by contradiction. Assume that λ (, 1 2 ] ( 1 2, + ), and (λi + K 0 ) [ϕ] = 0 for ϕ 0 in H ( Ω). By Green formula we have K 0 [1] = 1 2. We deduce that (λ 1 2 ) ϕ, 1 1 2, 1 2 = 0. Hence ϕ, 1 1 2, 1 2 = 0, and 0 < A := S 0 [ϕ] 2 dx, B := S 0 [ϕ] 2 dx <. Ω R 2 \Ω

21 Neumann-Poincaré operator Proof. On the other hand, using the jump relations, we have ( A = 1 ) Ω 2 I + K 0 [ϕ]s 0 [ϕ]dσ, ( ) 1 B = 2 I + K 0 [ϕ]s 0 [ϕ]dσ. Consequently Ω λ = B A 2(B + A). Thus, λ < 1 2, which is in contradiction with the assumption. If λ = 1 2, then A = 0 and hence S 0[ϕ] = c on Ω. Since ϕ H ( Ω), the constant c H 1 2 e ( Ω). Hence c = 0, and finally ϕ = 0.

22 Neumann-Poincaré operator K 0 : H 1 2 ( Ω) H 1 2 ( Ω), is compact, has a real discrete spectrum in ( 1/2, 1/2], that is symmetric with respect to 0, with only 0 as an accumulation point (only in dimension two). We denote λ = 0 < λ + 1 λ 0 = 1 2, the eigenvalues in [0, 1 2 ], repeated according to their multiplicity, and λ 1 = λ+ 1 < λ 2 = λ+ 2 < < 0, the eigenvalues in ( 1 2, 0), repeated according to their multiplicity. σ p (K 0) = 0, 1/2, λ ± j, j 1. Finally, we have the following symmetrization S 0 K 0 = K 0 S 0.

23 Neumann-Poincaré operator One can introduce a new scalar product on H ( Ω): ϕ, ψ S := ϕ, S 0 [ψ] 1 2, 1, 2 for which K0 becomes selfadjoint [Khavinson-Putinar-Shapiro 09]. We denote ϕ 0, ϕ ± j the normalized eigenfunctions of K 0 associated respectively to 1/2 and λ ± j, j 1, that is { K 0 [ϕ ± j ] = λ ± j ϕ ± j ϕ ± j S = 1. We also have the following spectral decomposition on H ( Ω): (λi K0) 1 = 1 λ Q, ϕ ± j S ϕ ± j 0 + λ λ ±. j j=1 We now define the variational eigenvalues k 0 =, k = 1, k ± j := 2λ± j + 1 2λ ± for j 1. j 1

24 Neumann-Poincaré operator Back now to the quasi-static approximation. V (x) = S 0 [ϕ](x), is a generalized eigenfunction if and only if ϕ H 1/2 ( Ω) satisfies where (λ(ω)i K 0) [ϕ](x) = 0 x Ω, λ(ω) = ˆε(ω) + 1 2(ˆε(ω) 1). The quasi-static resonances (ω 0,j ) j are solutions to the dispersion relations ˆɛ(ω) = k ± j, 0 for j { } N, and are explicitly given by ω 2 p ± ε k ± 4 i Γ 2 j ) i ( Γ2 ± Γ 2 4 ωp 2 ε k ± j Γ2 and the value iγ corresponding to ˆɛ(ω) = 0. if k ± j ε 4 ω2 P Γ 2, if k ± j < ε 4 ω2 P Γ 2,

25 Poincaré variational problem The space W 1, 1 0 (R 2 ) = { } u(x) (1 + x 2 ) 1/2 ln(2 + x 2 ) L2, u L 2, u(x) 0, x, equipped with the scalar product (u, v) W := u vdx, R 2 is a Hilbert space. Define T : W 1, 1 0 (R 2 ) W 1, 1 0 (R 2 ) by v W 1, 1 0 (R 2 ), Tu vdx = R 2 Ω u vdx.

26 Poincaré variational problem The operator T : W 1, 1 0 (R 2 ) W 1, 1 0 (R 2 ) is self-adjoint and bounded with with norm T 1. The spectral Poincaré variational problem is to find (w, β) W 1, 1 0 (R 2 ) \ {0} R, solution to β w vdx = w vdx, v W 1, 1 0 (R 2 ). R 2 Ω Integrating by parts, one immediately obtains that any eigenfunction w is harmonic in Ω and in Ω, and satisfies the transmission conditions on Ω: w + = w, w ν + = (1 1 β ) w ν

27 Poincaré variational problem Taking k = 1 1 β, and γ(x) = { 1 for x R 2 \ Ω, k for x Ω, we obtain that (w, k) is a solution to the system { div(γ(x) w) = 0 in R 2 w 0 as x, 1 Here γ(x) can be seen as the function approximation, defined by ε(ω 0,x) in the quasistatic ε(ω, x) = { ε0 for x R 2 \ Ω, ε 0ˆɛ(ω) for x Ω.

28 Poincaré variational problem H S = {S 0 [ϕ], ϕ H 1/2 ( Ω), Ω ϕdσ = 0} = {u W 1, 1 0 (R 2 ), u = 0 in Ω Ω, [u] D = 0} H S is the subspace of single layer potentials in W 1, 1 0 (R 2 ). Lemma The following assertions hold. i The eigenspace of T associated to the eigenvalue β = 1 is Ker(I T ) = {v Ω = 0, v Ω H 1 0 (Ω)}. This eigenspace does not contains any element of H S except v = 0. ii The eigenspace of T associated to the eigenvalue β = 0 is Ker(T ) = {v Ω W 1, 1 0 (Ω ), v Ω = 0} R.

29 Poincaré variational problem We can easily verify T H S H S. We further keep the notation T for the restriction of T to H S. Lemma We have T = 1 2 I + R, with R : H S H S is compact. Proof. 2 Ru(X ) v(x )dx = R 2 Ω ( u + ν + u ) v(x)dx. ν Since u H S, we can write u = S 0 [ϕ], with ϕ H ( Ω). On the other hand, we have u ν + + u u = 2K 0 [φ] = 2K 0 [ i H 1/2oS 1 0 [(u Ω, 0)] ] (X ), where i H 1/2 : H 1 2 ( Ω) R H 1 2 ( Ω) is defined by i H 1/2 (f, a) = f.

30 Poincaré variational problem Then T is a Fredholm operator of index zero and its spectrum is real, discrete, contained in (0, 1) and symmetric with respect to 1 2, with only 1 2 as an accumulation point. We denote (β ± n ) n 1 the eigenvalues of T, ordered as follows: 0 < β + 1 β , the eigenvalues in (0, 1/2] and, similarly, 1 > β 1 β 2 1 2, the eigenvalues in [1/2, 1). The eigenvalue 1/2 is the unique accumulation point of the spectrum.

31 Poincaré variational problem Lemma Let (w n ± ) n 1 be the eigenfunctions associated to (β n ± ) n 1. Then β n + Ω = max u(x) 2 dx u H S, w + 1,,w n + R u(x) 2 dx 2 and similarly = min F n H S dim(f n ) = n β n = min u H S, w 1,,w n = max F n H S dim(f n ) = n max u F n Ω u(x) 2 dx R u(x) 2 dx 2 Ω u(x) 2 dx R u(x) 2 dx, 2 min u F n Ω u(x) 2 dx R u(x) 2 dx. 2

32 Poincaré variational problem Lemma Let u be in H S. Then u(x) = n 1 u ± n w ± n (x), x R 2, where u ± n = R u(x) w ± 2 n (x)dx w ± R 2 n (x) 2. dx The series is convergent with respect to the norm W.

33 Outline of Part II Motivation The Neumann-Poincaré operator The case of disks Inclusions with C 1+α boundaries Numerical illustration Conclusion

34 Motivation Let D 1, D 2 R 2 be two bounded smooth inclusions separated by a distance δ > 0, and { k X D1 D γ(x ) = 2 1 X R 2 \ D 1 D 2, where k 1 is a given real constant. Define the potential u solution to the PDE { div(γ(x ) u(x )) = 0 in R 2 u(x ) H(X ) 0 as X, where H is a given harmonic function ( H(X ) = 0 in R 2 ).

35 Motivation High contrast composite material (k [0, 1) (1, + ]) Are the gradients u uniformly bounded as δ 0? Do the bounds depend on the contrast k? Does the presence of narrow regions in between in favor stress concentration that could lead to fracture? Plasmonic resonances (k < 0 and H = 0) Plasmonic resonances of metallic nano-particles are the solutions to ε(ω) = k < 0; How do the resonances behave when δ 0? How the electric fields are confined inbetween the nanoparticles?

36 A model problem Assume that D 1 and D 2 are the translates of 2 reference touching inclusions D1 0 and D0 2, that is D 1 = D (0, δ/2), D 1 = D (0, δ/2). δ D 2 O δ e 2 δ D 1 - D1 0 and D0 2 are strictly convex and only meet at 0 - Γ 1 and Γ 2 have regularity C 1,α, for some 0 < α < 1 - Around 0, Γ i is parametrized by a curve x (x, ( 1) i [ψ i (x) + δ/2]) - ψ 1 (x) + ψ 2 (x) C x m as x 0

37 Related results Case of a non degenerate contrast (k 0, ) 2 disks in 2D for a conduction problem [Bonnetier-Vogelius 01] u W 1, (Ω) C Piecewise Hölder conductivities scalar equations and strongly elliptic systems [Li-Vogelius 01, Li-Nirenberg 03] ) Ω = D 0 ( N i=1d i Each D i has C 1,α boundary 0 < Λ γ Λ 1 and γ Di has regularity C 0,µ N u C 1,α (D i Ω ε) C ( u Ω ) i=0 where α α = inf(µ, 2(1+α) ) and C = C(Ω, ε, Λ, N, α, µ) is independent of δ

38 Related results Case of a possibly degenerate contrast (k 0) 2 disks in 2D [Ammari-Kang-Lim 05, Ammari-Kang-Lee-Lee-Lim 07, Ammari-Kang-Lee-Lim-Zribi 2010] 2 disks at a distance δ, of same radius r and conductivity k, meeting at X = 0 tangentially to the direction T. Then, u(x ) = u r (X ) + u s (X ) with u r C where C > 0 is independent of δ, and k. if k < 1 if k > 1 u s C 1 H(0) T 2k + δ/r u s (0) C 2 H(0) T 2k+ δ/r u s C 1 H(0) N 2k 1 + δ/r u s (0) C 2 H(0) N 2k 1 + δ/r

39 The Neumann-Poincaré operator Let G(X, Y ) = 1 ln X Y 2π be the Green function of Laplace operator, and S i f (X ) = G(X, Y )f (Y ) ds Γ i The potential u can be expressed as u(x ) = S 1 ϕ 1 (X ) + S 2 ϕ 2 (X ) + H(X ) Let λ = k+1 2(k 1) R \ [ 1/2, 1/2] operator defined on H 1/2 (Γ i ) by,δ and let Ki denote the K,δ i f (X ) = 1 (X Y ) ν i (X ) 2π Γ i X Y 2 f (Y ) ds Y.

40 The Neumann-Poincaré operator The layer potentials solve where K δ = (λi K δ ) ( ( ϕ1 ϕ 2 K,δ 1 L δ 2 L δ 1 K,δ 2 ) ) = ( ν H Γ1 ν H Γ2 ) is the Neumann-Poincaré operator and where for (ϕ 1, ϕ 2 ) C 0,α (Γ 1 ) C 0,α (Γ 2 ), L δ 2 ϕ 2(X ) = S 2 ϕ 2 (X ) X Γ 1 ν 1 L δ 1 ϕ 1(X ) = ν 2 S 1 ϕ 1 (X ) X Γ 2 Classical potential theory: When δ > 0, λi K,δ is a continuous linear mapping on C 0,α (Γ 1 ) C 0,α (Γ 2 ), invertible with bounded inverse, for any 0 < α < 1 and for λ > 1/2 In the integral equation, the two parameters δ and k are separated

41 The Neumann-Poincaré operator When δ > 0, the operators K,δ are compact, but not self-adjoint However, due to the Calderón identity SK δ = K,δ S where S ( ϕ1 ϕ 2 ) = ( (S1 ϕ 1 + S 2 ϕ 2 ) Γ1 (S 1 ϕ 1 + S 2 ϕ 2 ) Γ2 ) one can introduce a new scalar product in an appropriate subspace of L 2 < ϕ, ψ > S = < Sϕ, ψ > L 2 = S 1 [ϕ 1 ]ψ 1 Γ 1 S 2 [ϕ 2 ]ψ 2 Γ 2 for which K δ is also self-adjoint. [Carleman, Krein, Khavinson-Putinar-Shapiro]

42 The Neumann-Poincaré operator Consequently, if λ δ n ( 1/2, 1/2], ϕ δ n denote the eigenelements of K,δ and if the data decomposes as ( ) ν H Γ1 ν H Γ2 = n 1 α n ϕ δ n then one can write the solution of the integral equation (λi K δ )ϕ = ν H as ϕ = ( ϕ1 ϕ 2 ) α n = λ λ δ n 1 n ( ϕ δ 1,n ϕ δ 2,n with convergence in the sense of the scalar product <, > S, and u = S 1 ϕ 1 + S 2 ϕ 2 + H The singularities of ϕ are likely to depend on how λ λ δ n becomes small as δ 0 and λ ±1/2. )

43 Uniform bounds for a non-degenerate contrast Theorem (ABTV) Let 0 < k 0, be fixed and Ω be a bounded open domain containing D 1 D 2. The solution u is bounded in C 1,α 0 (Ω \ (D 1 D 2 )) C 1,α 0 (D 1 ) C 1,α 0 (D 2 ), uniformly with respect to δ, for any 0 < α 0 < α. H. Ammari, E. Bonnetier, F, Triki and M. Vogelius. Elliptic estimates in composite media with smooth inclusions: an integral equation approach, Ann. Sci. Éc. Norm. Supér. volume 48, no. 2, p50, (2015).

44 The case of 2 disks centered at ( 1 δ, 0) and (1 + δ, 0) [see also Ammari-Ciraolo-Kang-Lee-Milton, McPhedran] One can Transform into 2 concentric disks of radii ρ and 1/ρ with the conformal mapping ξ = z a z + a a = δ(2 + δ) ρ = a δ a + δ and look for u(x, y) = Re(f (ξ))

45 One can explicitly compute the eigenelements of K,δ : ( ) 1 + ρ kn 2n = 1 ρ 2n λ λ n = λ (1/2 n 2 δ + O(δ)) ( ) 1 ρ k n + 2n = 1 + ρ 2n λ λ + n = λ ( 1/2 + n 2 δ + O(δ)) δ(2 + δ) δ where ρ = δ(2 + δ) + δ and the expansion ( ϕ1 ϕ 2 ) = n 1 α n λ λ δ n ( ϕ δ 1,n ϕ δ 2,n ) converges pointwise (so that one recovers the pointwise estimates on u δ shown before)

46 Inclusions with C 1+α boundaries A min-max principle Let D = D 1 D 2, D = R 2 \ D W 1, 1 0 (R 2 ) = { } u(x ) (1 + X 2 ) 1/2 ln(2 + X 2 ) L2, u L 2, u(x ) 0, X H S = {S 1 ϕ 1 + S 2 ϕ 2, ϕ i H 1/2 (Γ i ), ϕ 1 + Γ 1 ϕ 2 = 0} Γ 2 = {u W 1, 1 0 (R 2 ), u = 0 in D D, [u] D = 0} H S is the subspace of single layer potentials in W 1, 1 0 (R 2 ). The operator T δ defined by v W 1, 1 0 (R 2 ), R 2 T δ u v = D u v is self-adjoint and satisfies T δ 1. It is also Fredholm of index 0

47 If (w, β) H S R is an eigenpair of T δ, then v W 1, 1 0 (R 2 ), T δ w v R 2 = β w v = R 2 so that w = 0 in D D and D w v [w] Γ1 Γ 2 = 0 n w + = (1 1/β) n w w is thus solution to div(a δ (X ) w(x )) = 0 in R 2 w 0 as X with a δ = 1 D + (1 1/β)1 D. In other words, w = S 1 ϕ 1 + S 2 ϕ 2 with ( ) (λi K,δ ϕ1 ) = 0 and λ = k + 1 2(k 1) ϕ 2 = 1/2 β is an eigenvalue of K,δ

48 As a consequence, the eigenvalues of K,δ are given by the min-max principle βn δ, = max min D u 2 F n H u F n\{0} S u R 2 2 dim(f n ) = n βn δ,+ = min F n H S dim(f n ) = n max u F n\{0} D u 2 R 2 u 2 so that, the nondegenerate eigenvalues of K,δ satisfy 0 < β + 1 β+ 2 1/2 β 2 β 1 1 < 1 or in other words 1/2 < λ 1 λ 2 0 λ+ 2 λ+ 1 < 1

49 Inclusions with C 1+α boundaries Step 1: Let ε > 0 small enough with respect to the sizes of D 1, D 2 and w + = w Γ1 B(0,ε) w = w Γ2 B(0,ε) Theorem the quantity [w] 1/2 = ε w 2 Ḣ 1/2 (Γ j ) + j=1,2 ε w + (x) w (x) 2 ψ 1 (x) + ψ 2 (x) + δ ( defines a norm, which is equivalent to the norm u 2 R 2 on H S uniformly in δ. 1/2 ) 1/2

50 Recall the variational principle for the eigenvalues of K,δ We obtain estimates βn δ,+ = min F n H S dim(f n ) = n 1 C bδ,+ n β δ,+ n max u F n\{0} D u 2 R 2 u 2 C b δ,+ n n 1, with a constant C independent of δ, where bn δ,+ := min F n Ḣ 1/2 dim(f n) = n max w F n\{0} 2 j=1 w 2 Ḣ 1/2 ( D j ) 2 j=1 w 2 Ḣ 1/2 ( D j ) + ε ε w + (x) w (x) 2 x m +δ dx

51 Step 2: Comparison with disks Let X 1 = (0, 1 + δ 2 m /2), X 2 = (0, 1 δ 2 m /2), B 1 = B 1 (X 1 ), B 2 = B 1 (X 2 ) Let D1 0 = D 1 δ/2e 2, D2 0 = D 2 + δ/2e 2 Let Π : D1 0 D0 2 be a diffeomorphism such that Π(x 1, ψ(x 1 )) = (x 1, ψ 2 (x 1 )), for x 1 ε Let G : B 1 (0, 1) D1 0 be a diffeomorphism such that G(0, 0) = (0, 0), and consider G 1 = τ( δ 2 ) G τ( δ 2 m 2 ) G 2 = τ( δ 2 ) Π G τ( δ 2 m 2 ) Θ where τ(α) is the translation by αe 2 and Θ the symmetry wrt the x 1 - axis.

52 Then G 1 : B 1 D 1 and for x 1 < ε, G 2 : B 2 D 2 G 1 1 (x 1, ψ(x 1 ) + δ/2) = G 1 2 (x 1, ψ 2 (x 1 ) δ/2) This parametrization of D 1, D 2 leaves the H 1/2 norm essentially unchanged : for some C independent of δ 1 C v Ḣ 1/2 ( D 1 ) v G 1 Ḣ1/2 ( B 1 (0,1+ δ2/m 2 )) C v Ḣ 1/2 ( D 1 )

53 Noticing that 1 x m + δ 2δ2/m 1 x 2 shows that + δ2/m b + n (D 1, D 2 ) 2 j=1 w 2 Ḣ 1/2 ( D j ) 2 j=1 w 2 Ḣ 1/2 ( D j ) + ε ε Cδ 1 2 m w + (x) w (x) 2 x m +δ dx 2 j=1 w 2 Ḣ 1/2 ( D j ) 2 j=1 w 2 Ḣ 1/2 ( D j ) + ε ε Cδ 1 2 m w + (x) w (x) 2 x 2 +δ 2 m dx 2 j=1 w G j 2 Ḣ 1/2 ( D j ) 2 j=1 w G j 2 Ḣ 1/2 ( D j ) + ε ε w G + j (x) w G x 2 +δ 2 m j (x) 2 dx Cδ 1 2 m b + n (B 1, B 2 ) Cδ 1 1 m

54 Theorem - D 1, D 2 are convex, with smooth boundary, and meet at X = 0 tangentially to the x axis when δ = 0 - their boundary are parametrized by curves (x, ψ j (x)) around X = 0 with ψ 1 (x) + ψ 2 (x) C x m for some m 2 Then the non-degenerate eigenvalues of the Neumann-Poincaré operator form 2 families λ δ n,+ = 1/2 cn,+ δ δ m 1 m λ δ n, = 1/2 + cn, δ δ m 1 m + o(δ m 1 m ) + o(δ m 1 m ) where the c δ n,± form an increasing sequence of positive numbers, that only depend on the shapes of the inclusions and satisfy c δ n,± n as n

55 Numerical illustration Pick R > 2 large enough, so that D 1 D 2 B R/2 and let B δ : H0 1(B R) H0 1(B R) defined by v H0 1 (B R ), B δ u(x ) v(x ) B R = u(x ) v(x ). D 1 D 2 B δ is self adjoint and of Fredholm type, thus has a spectral decomposition. We denote bn δ,± its eigenvalues. Theorem (BTT) Let n 1. There exists a constant C independent of δ and n such that 1 C bδ,+ n β δ,+ n Cb δ,+ n.

56 Numerical illustration We estimate numerically the rate of convergence to 0 of the first non-degenerate eigenvalue b δ,+ 1. To this end, we use the min-max principle to approximate b δ,+ b δ,+ 1,N = min u V N D 1 D 2 u(x ) 2 dx B R u(x ) 2 dx 1 by where V N is a finite dimensional subspace of H0 1(B R). Let X 1 = (x 1 + iy 1 ) D 1, X 2 = (x 2 + iy 2 ) D 2 and n N. Define φ ± n,1, φ± n,2 : R2 C by φ n,1 (z) = (z X 1 ) n, φ n,2 (z) = (z X 2 ) n, where z = x + iy. Let w m, m 1 be the H0 1 (D) functions which are harmonic in B R \ D and such that { { w4n 3 = Re(φ n,1 ) in D 1 w4n 2 = Im(φ n,1 ) in D 1 w 4n 3 = 0 in D 2, { w4n 1 = 0 in D 1 w 4n 1 = Re(φ n,2 ) in D 2, w 4n 2 = 0 in D 2, { w4n = 0 in D 1 w 4n = Im(φ n,2 ) in D 2.

57 Numerical illustration We consider a conformal triangulation T of B R, which is refined in the neck between the 2 inclusions. Figure: Mesh for δ = 1/16.

58 Numerical illustration Let ŵ m, m 1 denote the H 1 projection of w m on the space of functions which are piecewise linear on T. We define V N as the vector space generated by the functions ŵ m, m 4N. We note that the functions w m, m 1 are linearly independent. Together with the functions w 0,1, w 0,2 in H 1 0 (B R) defined by w 0,i = 0 in B R \ D, and { w0,1 = 1 in D 1 w 0,1 = 0 in D 2, { w0,2 = 0 in D 1 w 0,2 = 1 in D 2, they from a basis of H 1 0 (B R).

59 Numerical illustration To compute the eigenvalues b δ,+ 1,N, we form the matrices A and B with entries A i,j = ŵ i ŵ j, D 1 D 2 B i,j = ŵ i ŵ j, B R and then compute the generalized eigenvalues of the system AU = βbu. We have used the software Freefem++ to compute the vectors ŵ m, and Scilab to solve the above matrix eigenvalue problem.

60 Numerical illustration Figure: Domains D 1 and D 2 The points X 1 and X 2 in the construction of the space V N, are the centers of the perturbed discs.

61 Numerical illustration We deduce from the asymptotic analysis that as δ tends to 0. log b δ,+ 1,N log c+ 1 + m 1 m log δ The following table provide the numerical results for δ between 1/2 and 1/2 7, and N = 39. m Equation of the line Theory Error 1 m = 2 t = s 2 = m = 6 t = s m = 9 t = s We remark that the computed slopes are in a good agreement with the expected theoretical values.

62 Numerical illustration Figure: eigenfunction for m=9

63 Conclusion We obtain the asymptotic of the eigenvalues of K,δ as δ 0 which may be used for design purposes in optimizing plasmon resonances In the case of disks, one can read off the blow up rate of u δ on the eigenvalues of the Neumann-Poincaré operator How do we derive the pointwise convergence of the spectral decomposition? In the case of disks, [Kang-Lim-Yun] obtained an asymptotic expansion of the solution, isolating a dipole-like singular part. It would be nice to extend this to more general geometries of contact Using the same approach can we derive estimates of the resonances and the gradients in the 3d case? Inverse problems related to surface plasmonic resonances (Biosensing) (Phd thesis of A. Banani). Asymptotic expansion of plasmonic resonances of periodic distributed nano-particles, nano-holes (Phd thesis of L. Salesses)+[Bonnetier-Dapogny-Triki 16].