absence of bound states for waveguides in 2d periodic ... · propagation of e.m. waves, maxwell...

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...

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