Absence of bound states for waveguides in 2D periodic ... · Propagation of e.m. waves, Maxwell equations Two-dimensional situation: Helmholtz equation (polarized waves) 3/36. Photonic

Absence of bound states for waveguides in 2D periodicstructures

Maria RadoszRice University

(Joint work with Vu Hoang)

Mathematical and Numerical Modeling in OpticsMinneapolis, December 13 2016

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Introduction: Photonic Crystals and Periodic media

Introduction: Floquet theory/Absolute continuity of spectra ofperiodic operators

The waveguide problem in 2D periodic structure and main result

Reformulation of the problem

Analytic continuation of resolvent operators

Proof of main result

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Photonic Crystals and Periodic media

Photonic crystal: artificial dielectric material.

Propagation of e.m. waves, Maxwell equations

Two-dimensional situation: Helmholtz equation (polarized waves)

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Photonic Crystals and Periodic media

Ω = (0,1)2

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Mathematical modeling

Start from Maxwell’s equations:

curlE =−µ0∂

∂t H

curlH = ε(x) ∂∂t E

divH = 0,divεE = 0

where ε(x) describes the material configuration.Assume E = (0,0,u). This leads to

ε(x)∂2u∂t2 −∆u = 0

Look for time-harmonic solutions u = eiωtv. This gives

−ε(x)ω2v−∆v = 0.5 / 36

Mathematical problem

In the following, let ε : R2→ R be a periodic function:

ε(x +m) = ε(x) (m ∈ Z2)

s.t. ε ∈ L∞(R2), 0< c ≤ ε.

Time-independent problem:

−∆u +ε(x)λu = 0, λ= ω2

Spectral problem: study the spectrum of Helmholtz operator

−1ε

∆

where D(−1ε∆) = H 2(R2). Self-adjoint in ε-weighted L2-space.

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Spectrum of periodic operators

Spectrum has band structure:

spec(−1ε

∆)

=⋃s∈N

λs((−π,π]d)

where λs(k) are band functions, k ∈ (−π,π]d is the quasimomentum (wave vector).

We expect that spec(−1ε∆)

has no point eigenvalues.

An eigenvalue implies existence of u ∈ L2(Rd)\0 (bound state)s.t.

(−∆−ω2ε)u = 0.

Hence, eiωtu solves the wave equation ⇒ “wave gets stuck”

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Absolute continuity of spectra of periodic operators

LetL =−q−1∇·A∇+ V

be an s.a. operator with periodic coefficients, D(L)⊂ L2(Rd). Want toprove: L has absolutely continuous spectrum.

Difficult: exclude point spectrum, i.e. prove that (L−λ)u = 0 hasno nontrivial solutions.

Schrodinger/Helmholtz [Thomas, ’73], Magnetic Schrodingeroperator [Birman-Suslina, ’00], [Sobolev, ’02], Divergence formoperator with symmetry [Friedlander ’03]

Problem in full generality still unsolved.

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Floquet-Bloch transformation

Recall constant coefficient operators are invariant

L[u(·+ s)] = L[u](·+ s) (s ∈ Rd)

w.r.t. arbitrary shifts.Periodic operators are invariant

L[u(·+n)] = L[u](·+n) (n ∈ Zd)

w.r.t. integer shifts.Can we construct an analogue of the Fourier transform?

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Gel’fand introduced the following transform (B = (−π,π)d = Brillouinzone)

(Vf )(x,k) := 1√|B|

∑n∈Zd

eik·(n−x)f (x−n)

TheoremV : L2(Rd)→ L2(Ω×B) = L2((0,1)d× (−π,π)d) is an isometricisomorphism

‖Vf ‖L2(Ω×B) = ‖f ‖L2(Rd).

Alternatively: V : L2(Rd)→ L2(B,L2(Ω)),(Vf )(k)(·) = Vf (·,k).

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It is important to understand the action of V on H s(Rd). Define

H sper(Ω) := u ∈H s(Ω) : Eu ∈H s

loc(Rd),

the space of periodic H s-functions. E is the periodic extension operator

(Eu)(x +n) = u(x) (x ∈ Rd ,n ∈ Zd).

Example: u ∈H 1per(Ω) implies e.g. that u|x1=0 = u|x1=1 in the trace

sense.

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TheoremV : H s(Rd)→ L2(B,H s

per(Ω)) is an topological isomorphism. Theinverse transform is

(V−1g)(x) = 1√|B|

∫B

g(x,k)eik·xdk

Let a family of cell operators be defined by

L(k) =−1ε

(∇+ ik) · (∇+ ik) =−1ε

∆k

acting on H 2per(Ω).

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The key fact is: the operator L is decomposed into operators L(k)under the transform V :

V (Lu)(x,k) = L(k)[Vu(x,k)] (u ∈D(L)).

orLu = V−1[L(k)Vu(x,k)].

Sometimes this is written symbolically as

L =

⊕∫B

L(k)dk

(direct integral of operators).13 / 36


Further consequences:

((L−λ)−1f )(x) = 1√(2π)d

∫B

(L(k)−λ)−1Vf (·,k)dk

Write spec(L(k)) as

λ1(k)≤ λ2(k)≤ . . .≤ λs(k)≤ . . . .

Thenspec(L) =

⋃k∈B

spec(L(k)) =⋃

n∈Nλs(B).

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Band structure

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Thomas argument

Consider the problem of excluding point spectrum for −ε−1∆,ε, 1ε ∈ L∞(Rd):

Existence of a u ∈H 2(Rd) solving (−∆−λε)u = 0 implies:

(−∆k−λε)v = 0 has a nontrivial solution v ∈H 1per((0,1)d)

for a positive measure set of k ∈ [−π,π]d .

Extension into the complex plane: analytic Fredholm theory((0,1)d bounded) implies:

(−∆k−λε)v = 0 has nontrivial solution for all k of the form

k = (k,0, . . . ,0), k ∈ C.

Study −∆k using Fourier series.16 / 36

Thomas argument (continued)

Expand v ∈H 2((0,1)d) into Fourier series v =∑

m∈2πZd cmeimx ,then

−∆kv =∑m

(m +k) · (m +k)cmeimx .

Let k = (π+ iτ,0, . . . ,0) and compute

| Im(m +k)2| = | Im[(m +πe1)2 + 2i(m1 +π)τ − τ2]|

= 2|m1 +π||τ | ≥ 2c0|τ |

since |m1 +π| ≥ c0 > 0 for all m1 ∈ 2πZ.

⇒ ‖−∆−1(π+iτ,0,...,0)‖ ≤ C/|τ |

(−∆(π+iτ,0,...,0)−λε)−1 exists for τ sufficiently large (Neumannseries). Contradiction!

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The waveguide problem in 2D periodic structure

strip S := R× (0,1), unit cell Ω = (0,1)2, ε= ε0 +ε1

ε0,ε1 ∈ L∞(R2,R), ε0 periodic with respect to Z2,

ε1(x1,x2 + m) = ε1(x1,x2) (m ∈ Z)

suppε1 ⊂ (0,1)×R, infM |ε1|> 0 on some open set M18 / 36

Guided modes vs Bound states

Perturbations create extra spectrum.

guided modes ψ: Exist and corresponding new spectrum iscontinuous

(−∆−λε)ψ = 0, 0 6= ψ ∈H 2(S)

But ψ(x1,x2 + m) = eiβmψ(x1,x2) (not decaying in x2 direction).

bound states: localized standing waves, corresponding spectrum ispoint spectrum

(−∆−λε)ψ = 0, 0 6= ψ ∈H 2(R2).

We show that this is impossible.

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Guided modes and Leaky waveguides

Existence of guided mode spectrum:

strong defects: Kuchment-Ong (’03,’10)

weak defects: Parzygnat-Avniel-Lee-Johnson (’10),Brown-Hoang-Plum-(R.)-Wood (’15,’16)

Absence of bound states:

“hard-wall”waveguides: Sobolev-Walthoe (’02), Friedlander (’03),Suslina-Shterenberg (’03)

“soft-wall”waveguides with asymptotically constant background:Filonov-Klopp (’04,’05), Exner-Frank (’07)

“soft-wall”waveguides with periodic background: Hoang-Radosz(’14)

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Main result

Consider a Helmholtz-type spectral problem on R2 of the form

−∆u = λεu. (∗)

Definitionσ(−1

ε∆)∩ (R\σ(− 1ε0

∆)) is called guided mode spectrum.

Question: does the guided mode spectrum really correspond to trulyguided modes? Are there possibly localized standing waves?

Theorem (V.H., M.R.)

Let λ ∈ R. In H 2(R2), the equation (∗) has only the trivial solution.JMP 55, 033506-2014

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Reformulation of the problem, Floquet-Bloch reductionto S

Recall the partial Floquet-Bloch transform in x2 direction

(Vf )(x1,x2,k2) := 1√2π∑n∈Z

eik2(n−x2)f (x1,x2−n).

V : L2(R2)→ L2(S × (−π,π)) is isometry

−1ε

∆ =∫ ⊕

[−π,π)−1ε

∆k2 dk2

σ

(−1ε

∆)

=⋃

k2∈[−π,π]σ

(−1ε

∆k2

).

where−∆k2 :=−(∇+ i(0,k2)) · (∇+ i(0,k2)) with domain H 2

per(S)22 / 36

Reformulation of the problem, Floquet-Bloch reductionto S

The problem −∆u = λεu has a nontrivial solution in H 2(R2)

Floquet-Bloch reduction in x2-direction.⇐⇒

The problem (−∆k2−λε)u = 0, u ∈H 2per(S), has a nontrivial

solution for k2 in a set of positive measure P.

However, standard Thomas approach not possible, since S notbounded!

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Derivation of a Fredholm-type problem on Ω

Fix λ /∈ σ(− 1ε0

∆), let k2 ∈ P. Define G(k2) : L2(Ω)→ L2(Ω) by

G(k2)v := ε1(−∆k2−λε0)−1v

where v = v on Ω and v = 0 outside.Lemma

If k2 ∈ R and u ∈H 2per(S), u 6= 0 solves

(−∆k2−λε)u = 0,

then v ∈ L2(Ω) defined by v = ε1u solves

v +λG(k2)v = 0 on Ω

and v 6= 0.24 / 36

Derivation of a Fredholm-type problem on Ω

Proofv ≡ 0 gives a contradiction to u 6= 0 by a unique continuationprinciple.

λ /∈ σ(− 1ε0

∆)⇒ (−∆k2−λε0)−1 exists as a bounded operator in

L2(S).

0 = (−∆k2−λ(ε0 +ε1))u =⇒0 = u +λ(−∆k2−λε0)−1ε1u.

=⇒ 0 = ε1u +λε1(−∆k2−λε0)−1ε1u.

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For k2 close to the real axis, by Floquet-Bloch reduction in x1-direction

G(k2)f (x) = ε1((−∆k2−λε0)−1f )(x)

= ε1

∫ π

−πeik1x1(T (k1,k2)e−ik1 ·f )(x) dk1

= ε1

∫ π

−π(H (k1,k2)f )(x) dk1 (x ∈ Ω)

where

T (k) = T (k1,k2) := 12π (−∆k−λε0)−1 (k ∈ C2)

T (k) : L2(Ω)→ L2(Ω)

H (k1,k2)r := eik1x1T (k1,k2)[e−ik1·r ] k1,k2 ∈ C

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G(k2)f (x) = ε1

∫ π

−π(H (k1,k2)f )(x) dk1.

Properties of H (e.g. Steinberg [’68], Kato [’76]):

k1 7→H (k1,k2) is meromorphic

H (k1 + 2πm,k2) = H (k1,k2)

isolated poles of H (·,k2) are analytic in k2

In general, the poles qj(k2) of are algebraic functions of k2, i.e. ∃analytic g

qj(k2)= g

(p√

k2− k02

)(multivalued complex function)

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G(k2)f (x) = ε1

∫ π

−π(H (k1,k2)f )(x) dk1.

Deformation of the integral in k1-plane: (D=the enclosed region)

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Thus for k2 close to the real axis,

G(k2)f = ε1

∫[−π,π]+iτ1

H (k1,k2)f dk1 + 2πiN+∑j=1

ε1 res(H (·,k2)f ,q+j (k2)) (∗∗)

q+j (k2) are those poles of H (·,k2) which lie in the upper half-plane

when k2 ∈ [π− δ2,π+ δ2] for some small δ2 > 0

Idea: to construct analytic continuation of G, use the rhs of (∗∗)as a definition!

Problem: since q+j (k2) are algebraic in k2 (root-like singularities),

there exists no direct continuation of the rhs for all k2, Imk2 > 0.

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LemmaThere exist a continuous path Γ : [0,∞)→ C satisfying

(i) Γ(0) ∈ R, (−∆k2 −λε)u = 0 has a nontrivial solution for k2 in a smallball around Γ(0).

(ii) t 7→ ImΓ(t) is nondecreasing,

(iii) ImΓ(t)→+∞ for t→∞,

with the property that there exists a neighborhood

N (Γ) :=N (Γ([0,∞)))

of the path Γ and a N ∈ N such that the number of poles of T (·,k2) in D isequal to N for all k2 ∈N (Γ).

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The path Γ and a neighborhood N (Γ) on which there exists ananalytic continuation of G(k2). Picture in k2 plane:

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DefinitionFor k2 ∈N (Γ) let

q+j (k2) (j = 1, . . . ,N +)

denote the poles of H (·,k2) in D with the property that Imq+j (Γ(0))> 0, i.e.

those poles which initially lie in the upper half-plane. For any k2 ∈N (Γ) define

A(k2)r := ε1

∫[−π,π]+iτ1

H (k1,k2)r dk1 + 2πiN+∑j=1

ε1 res(H (·,k2)r ,q+j (k2))

for all r ∈ L2(Ω).

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Properties of A(k2):

k2 7→A(k2) is analytic on N (Γ)

for all k2 ∈N (Γ), A(k2) is compact

A careful study of the poles q+j (k2) as Imk2→∞ reveals (hard!)

Theorem

There exist constants C = C (δ,τ1,λ)> 0,M = M (δ,τ1,λ)> 0 such thatfor k2 ∈N (Γ) of the form k2 = Rek2 + i(π2 + `) with ` ∈ 2πN, ` >M ,

‖A(k2)‖ ≤ C`−1.

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The key estimate

The technical heart of the whole construction is (need 2D)

Theorem

For s(m,k) = (m +k)2, ξ = (ξ1, ξ2), η = (η1,η2) ∈ R2, the followingestimates hold:

|s(m,ξ + iη)|2 ≥ [(m2 + ξ2)2−η21]2 + [(m1 + ξ1)2−η2

2]2

|s(m,ξ + iη)|2 ≥ 2[(m1 + ξ1)η1 + (m2 + ξ2)η2]2

Proof:

|s(m,ξ + iη)|2 = [(m2 + ξ2)2−η21]2 + [(m1 + ξ1)2−η2

2]2

+2[(m1 + ξ1)η1 + (m2 + ξ2)η2]2

+2[(m2 + ξ2)(m1 + ξ1) +η1η2]2.34 / 36

Proof of the main result

‖A(Rek2 + i(π2 + `))‖ ≤ C`−1 −∆u = λεu = 0for some u ∈H 2(R2)\0

⇓ ⇓v +λA(k2)v = 0 only has (−∆k2 −λε)w = 0 has

the trivial solution for ` large nontrivial solutionfor k2 ∈ P

⇓ ⇓v +λA(k2)v = 0 has nontrivial v +λG(k2)v = 0 has

solution only for a discrete nontrivial solutionset of k2 ∈ [−π,π] ! for almost all k2 ∈ P

↑Contradiction!

since A(k2) = G(k2)for k2 ∈ P

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Thank you for your attention!

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Absence of bound states for waveguides in 2D periodic ... · Propagation of e.m. waves, Maxwell equations Two-dimensional situation: Helmholtz equation (polarized waves) 3/36. Photonic