Essential spectrum for Maxwell's equations

We study the essential spectrum of operator pencils associated with anisotropic Maxwell equations, with permittivity $\varepsilon$, permeability $\mu$ and conductivity $\sigma$, on finitely connected unbounded domains. The main result is that the essential spectrum of the Maxwell pencil is the union of two sets: namely, the spectrum of the pencil $\mathrm{div}((\omega\varepsilon + i \sigma) \nabla\,\cdot\,)$, and the essential spectrum of the Maxwell pencil with constant coefficients. We expect the analysis to be of more general interest and to open avenues to investigation of other questions concerning Maxwell's and related systems.


Introduction
In this paper we consider the essential spectrum of linear operator pencils arising from the Maxwell system where Ω ⊆ R 3 is a finitely connected domain, with boundary condition if Ω has a boundary. In these equations ω is the pencil spectral parameter, ε the electric permittivity, μ the magnetic permeability and σ is the conductivity; ν is the unit normal to the boundary. Lassas [15] already studied this problem on a bounded domain with C 1,1 boundary, so in this article our primary concern is to treat unbounded domains which provides additional sources for essential spectrum. However, even for bounded domains, we are able to relax the required boundary regularity to Lipschitz continuity. Like Lassas we allow the permittivity, permeability and conductivity to be tensor valued (i.e., we allow anisotropy); however, we make Λ −1 |η| 2 ≤ η · εη ≤ Λ|η| 2 , Λ −1 |η| 2 ≤ η · μη ≤ Λ|η| 2 , 0 ≤ η · ση ≤ Λ|η| 2 (2) almost everywhere in Ω.
As already mentioned, the case of bounded domains was treated by Lassas [15] under slightly stronger regularity assumptions; for infinite domains we assume that all the coefficients have a 'value at infinity' in the precise sense that for some scalar values μ 0 > 0, ε 0 > 0 and σ 0 ≥ 0. To allow a unified treatment of unbounded and bounded domains, it is convenient to assign values to ε 0 , μ 0 and σ 0 when Ω is bounded, and we choose ε 0 := 1, μ 0 := 1; σ 0 := 0, (Ω bounded).
We start by considering, in the Hilbert space the operator pencil ω → V ω defined from (1) in the space H 1 by Our aim is to study the essential spectrum of the pencil V ω .

Definition 1.
Let H 1 and H 2 be two Hilbert spaces. For each ω ∈ C, let L ω : H 1 → H 2 be a bounded linear operator. Adapting the definitions in [9, Ch. I, §4], we say that ω ∈ C lies in the 1. essential spectrum σ e,1 of the pencil ω → L ω if L ω is not semi-Fredholm (an operator is semi-Fredholm if its range is closed and its kernel or its cokernel is finite-dimensional); 2. essential spectrum σ e,2 of the pencil ω → L ω if L ω is not in the class F + of semi-Fredholm operators with finite-dimensional kernel; 3. essential spectrum σ e,3 of the pencil ω → L ω if L ω is not in the class F of Fredholm operators with finite-dimensional kernel and cokernel; 4. essential spectrum σ e,4 of the pencil ω → L ω if L ω is not Fredholm with index zero, where ind L ω = dim ker L ω − dim coker L ω . When these essential spectra coincide, we shall use the notation σ ess . With an abuse of terminology, we shall refer to the essential spectrum of L ω and write ω ∈ σ e,k (L ω ).
Take ω / ∈ σ e,1 (V ω ). Thus V ω is semi-Fredholm, namely the range of V ω is closed and its kernel or its cokernel is finite-dimensional. A direct calculation using integration by parts yields, thanks to the symmetry of the coefficients ε, μ and σ, Hence, ker V ω = coker V ω , and so implying that V ω is a Fredholm operator with index zero, namely ω / ∈ σ e,4 (V ω ). Finally, we introduce some homogeneous Sobolev spaces which are required for the Helmholtz decomposition for unbounded domains. For bounded domains, these coincide with the usual Sobolev spaces.
(2) (Ω bounded) In this case we define the homogeneous Sobolev spaces to coincide with the usual Sobolev spaces: Remark 5. (a) Note that this definition does not coincide with Definition 1.31 in [4], which uses Fourier transforms to defineḢ s (R d ) and results in spaces which are not complete if s ≥ d/2. Our definition follows Dautray and Lions [8]. For clarity, we use our definition directly in "Appendix A". (b) If K is any compact subset of Ω with non-empty interior and Ω is bounded, then the usual H 1 -norm is equivalent to see Maz'ya [16]. In the case when Ω is unbounded, the norms onḢ 1 anḋ H 1 0 may be shown to be equivalent to the norm defined in (7), for any compact K ⊂ Ω with non-empty interior. Thus an equivalent definition ofḢ 1 (Ω), valid for bounded and unbounded Ω, is the closure of D(Ω) in the norm (7). However, for unbounded Ω, this is no longer equivalent to the H 1 -norm; e.g., the function given in polar coordinates by u(r) = 1/(r + 1) 3/2 does not lie in H 1 (R 3 ) but lies inḢ 1 (R 3 ).
We are now ready to state our main result.
Thanks to this result, the essential spectrum of the Maxwell pencil is decomposed into two parts.
• The essential spectrum of the operator div((ωε+ iσ)∇ · ): this component depends on the coefficients ε and σ directly. In particular, in the case when the coefficients ε and σ are continuous, it consists of the closure of the set of ω = iν, ν ∈ R, for which νε + σ is indefinite at some point in Ω: see Proposition 27. As in Lemma 2, a direct calculation shows that for this operator all essential spectra coincide, and so the notation σ ess is unambiguous. • The essential spectrum of the constant coefficient Maxwell pencil, is determined by the geometry of Ω. This can be computed explicitly in many cases of interest: we provide several examples below. It is worth observing that 0 always belongs to σ ess (V 0 ω ), since {0}⊕∇Ḣ 1 (Ω) ⊆ ker V 0 0 . In the next examples, we will calculate the essential spectrum of V 0 ω , for different choices of domains Ω.
Example 8. We consider here the case of the full space Ω = R 3 . We can make use of the Fourier transform to obtain a simple expression of this operator. Writing E(x) = R 3Ê (ξ)e ix·ξ dξ, the expression of the operator curl E in the Fourier domain is given by the multiplication operator iC(ξ)Ê(ξ), where Writing curl H in a similar way, we immediately see that V 0 ω is represented, in the Fourier domain, by the multiplication by the matrix By a standard argument, σ ess (V 0 ω )={ω ∈ C : det(A ω (ξ)) = 0 for some ξ ∈ R 3 }, so that In the particular case, when the conductivity at infinity is zero, i.e., σ 0 = 0, we simply have σ ess (V 0 ω ) = R. Example 9. Let us look at the case of the slab Ω = {x = (x , x 3 ) ∈ R 3 : 0 < x 3 < L}, for some L > 0. The derivation is very similar to the one presented above for the full space, the only difference being that the continuous Fourier transform in the third variable becomes a Fourier series. As a consequence, the continuous variable ξ 3 is replaced by a discrete variable n = 0, 1, . . .. More precisely, and, analogously, the range of n in each summation has been determined by the boundary conditions on x 3 = 0 and x 3 = L. Compared to the full space in Example 8, the continuous frequency variable ξ ∈ R 3 has become ξ := (ξ , nπ L ) ∈ R 2 × ( π L N). By calculations similar to those for the full space, we see that the essential spectrum is the set of ω ∈ C such that for some and it is easy to see that this coincides with the essential spectrum for the full space problem.
A direct calculation gives that the operators E → curl E and H → curl H may be written in Fourier coordinates as the multiplication operators by the matrices C iξ, π L1 n 1 , π L2 n 2 and C iξ, − π L1 n 1 , − π L2 n 2 , respectively, where the matrix C is defined in (8). As a consequence, in the Fourier domain, V 0 ω is a multiplication operator represented by the matrix If ω is such that det (A ω (n, ξ)) = 0 for every n ∈ N 2 and ξ ∈ R, then ω does not belong to the essential spectrum of V 0 ω . On the other hand, suppose that ω is such that det (A ω (n, ξ)) = 0 for some n ∈ N and ξ ∈ R. If n 1 = n 2 = 0, it is easy to see that there are no nonzero elements of ker A ω (n, ξ) satisfying (9). On the other hand, the vector (0, ωμ 0 L 2 , 0, πi, 0, ξL 2 ) belongs to ker A ω (0, 1, ξ) and satisfies (9) (and similarly if n 1 = 1 and n 2 = 0). As a consequence, we have that . In the particular case when σ 0 = 0, this set takes the simpler form Note that this set approaches the essential spectrum for the slab as L → +∞. This is expected: as L increases the cylinder becomes larger and larger in one direction.

Helmholtz Decomposition and Related Operators
We shall treat both bounded and unbounded Lipschitz domains Ω ⊆ R 3 . The latter are our primary interest, as the bounded case has already been studied by Lassas [15], albeit under slightly stronger assumptions on the boundary regularity. However, in the definitions which follow, we deal with both cases. The first decomposition result which we require is true without restrictions on the topology of Ω. Although it is standard, we present a proof since it shows how the homogeneous Sobolev spaces arise in a natural way.
1. The space L 2 (Ω; C 3 ) admits the following orthogonal decompositions: in which
To decompose the Maxwell pencil we need to decompose the spaces H(div 0, Ω) and H 0 (div 0, Ω) further, by using vector potentials in some suitable spaces, which we now introduce.
• The spaceẊ T (Ω) is the completion of H(curl, Ω) ∩ H 0 (div 0, Ω) with respect to the seminorm u is the kernel of the curl operator restricted toẊ T (Ω), namely The spaces K T (Ω) and K N (Ω) are closed inẊ T (Ω) and inẊ N (Ω), respectively, and so we can consider the quotient spaceṡ The curl operator is well-defined and injective on these spaces. To avoid cumbersome notation, we will in the following identify curl ψ for ψ ∈Ẋ T (Ω)/K T (Ω) or ψ ∈Ẋ N (Ω)/K N (Ω) with the vector in L 2 (Ω; C 3 ) given by curl acting on any representative of the equivalence class ψ. The curl operator maps these quotient spaces into the space of divergence-free fields, with appropriate boundary conditions.
We make the following assumption.
Assumption 14. The spaces K T (Ω) and K N (Ω) are finite-dimensional and This assumption is verified in many cases of theoretical and practical interest.
Proposition 15. Assumption 14 is verified in any of the following cases: Remark 16. We have decided not to provide the details of the assumptions of parts (2) and (3), since they are rather lengthy and are not needed for the rest of the paper. In simple words, these assumptions require ∂Ω to be a finite union of connected surfaces and that there exist a finite number of cuts within Ω which divide it into multiple simply connected domains. The number of cuts is given by dim K T (Ω), and the number of connected components of ∂Ω by dim K N (Ω) + 1. Thus, for simply connected domains with connected boundaries, the decomposition is even simpler: K T (Ω) and K N (Ω) are trivial and can be omitted. Proof.
(2) This part is proved in [4] (see also [8, Chapter IXA] and [10,ter I, §3] for the smooth case). The construction of the spaces K T (Ω) and K N (Ω) is described explicitly.  6) The arguments are standard and explicit, but it is not easy to find precise statements in the literature. We detail the derivation in "Appendix A", which contains a general construction for a larger class of cylinders.

Lemma 17. Let Ω ⊆ R 3 be a Lipschitz domain satisfying Assumption 14.
There exist bounded operators T N : Proof. In view of (13b), every F ∈ L 2 (Ω, C 3 ) admits a unique decomposition into three orthogonal vectors, . By the closed graph theorem, T N is bounded. The definition of T T follows similarly by using the other Helmholtz decomposition (13a).

Proof of the Main Result
In a first part, we introduce a series of equivalent reformulations of our problem to obtain a form where the two contributions to the essential spectrum in our main result can easily be separated.
Decomposing H 1 using (11) and (12) allows us to transform the Maxwell operator V ω . More precisely, for E ∈ H 0 (curl, Ω) and H ∈ H(curl, Ω) consider the decompositions where q E ∈Ḣ 1 0 (Ω; C), q H ∈Ḣ 1 (Ω; C), Ψ E ∈ H 0 (curl, Ω)∩curl(Ẋ T (Ω)/K T (Ω)), Ψ H ∈ H(curl, Ω) ∩ curl(Ẋ N (Ω)/K N (Ω)), h T ∈ K T (Ω) and h N ∈ K N (Ω). We now wish to discard the contribution coming from K T (Ω) and K N (Ω). To this end, we introduce the space equipped with the canonical product norm Define the projection map where E, H are given by (15), and its right inverse W −1 : Since the decompositions in (11) and (12) are orthogonal, for any (E, H) ∈ H 1 we have Instead of the operator V ω , we consider This does not change the essential spectrum, as the following lemma shows.
Let ω ∈ σ e,2 (V ω ). By Remark 3, there exists a sequence of functions u n = (∇q E,n +Ψ E,n +h N,n , ∇q H,n +Ψ H,n +h T,n ) in H 1 , u n H1 = 1, u n 0 in H 1 such that V ω u n L 2 → 0. Then there exists c > 0 such that ||W u n || H2 ≥ c for all sufficiently large n. This follows from the fact that otherwise by (17) we would have that, at least on a subsequence, P NT u n := (h N,n , h T,n ) satisfies ||P NT u n || H1 → 1. However, the range of P NT is the finite-dimensional space Setũ n = W u n /||W u n || H2 . Then, ||ũ n || H2 = 1 and and hence ω is in the σ e,2 essential spectrum ofṼ ω .
By definition ofṼ ω and (6), we obtaiñ where M ω F = (ωε + iσ)F and M μ F = μF . In order to simplify this operator even further, we need the following elementary result.

Lemma 19.
Let P H denote the orthogonal projection onto the space H.

The map ζ
given by is an isomorphism.
Proof. We use the characterization of the essential spectrum by Weyl singular sequences given in Remark 3. Sinceζ is continuous, the inclusion σ e,2 (Ṽ ω ) ⊆ σ e,2 (Ṽ ω ) is immediate. Let us now show the reverse inclusion.
Since the index is invariant under compact perturbations and the operator (x, y) → (P Z T (0, y), 0) + (0, P Y T (x, y)) is finite rank (dim Y < +∞), we have that T is Fredholm with index 0 if and only if Since dim ker T = dim ker T + dim Y and dim coker T = dim coker T we have that T is Fredholm with index 0 if and only if T is Fredholm with Finally, observe that T = P Z • T . Since the range of T is contained in Z ⊕ {0} we have dim ker T = dim ker T and dim coker T = dim coker T + dim Y . Hence, T is Fredholm with ind(T ) = − dim Y if and only if T is Fredholm with index 0. This concludes the proof.

Proof. Recall that
Now, recalling that Ψ H ∈Ẋ T (Ω) and Ψ E ∈Ẋ N (Ω), by (14) and (18) we have that in which [·] denotes the equivalence class in the appropriate quotient space.
In order to compute the essential spectrum ofṼ ω we now decompose the coefficients in the Maxwell system. As a consequence of our Hypotheses (3,4), whether Ω be bounded or unbounded, for each δ > 0 the Maxwell coefficients admit a decomposition in which the terms μ 0 , ε 0 and σ 0 are constant and do not depend on δ, the terms μ c , ε c and σ c are compactly supported, and the terms μ δ , ε δ , σ δ are essentially bounded, with here the norms are defined by ||a|| L ∞ (Ω) := ess sup x∈Ω ||a(x)|| 2 for a ∈ L ∞ (Ω; R 3×3 ), where ||A|| 2 denotes the induced norm sup v∈R 3 \{0} |Av| |v| for A ∈ R 3×3 .
In the expression forṼ ω appearing in (20) the Maxwell coefficients appear linearly in the multiplication operators M μ (multiplication by μ) and M ω (multiplication by ωε + iσ). The decomposition (21) of the coefficients is partially reflected in the following decomposition ofṼ ω : Note that the operatorṼ ω,0 is independent of δ. Further, the operatorṼ ω,c is compact and the operatorṼ ω,δ is O(δ)-small in a suitable norm, as we show in the following two lemmata.
Proof. By a direct calculation it is easy to see that div((ωε 0 + iσ 0 )Ψ E ) = 0 and h(μ 0 Ψ H ) = 0, using that ε 0 , σ 0 and μ 0 are scalar. Since the operators div : are bounded, it is enough to show that the operators are compact. We now prove that F T is compact, the other proof is completely analogous. Let R > 0 be big enough so that K := supp(ωε c +iσ c ) ⊆ B(0, R)∩Ω and χ ∈ C ∞ (Ω) be a cutoff function such that χ ≡ 1 in K and supp χ ⊆ B(0, R) ∩ Ω. Setting Ω R = B(0, R) ∩ Ω, the operator F T may be expressed via the following compositions where the third operator is the multiplication by ωε c + iσ c and the fourth operator is simply the extension by zero. Therefore, since the embedding [19] (see also [4, Theorem 2.8]), the operator F T is compact.

Lemma 24.
There exists a constant C > 0 depending only on Ω and on the coefficients μ, ε and σ, such that for each δ > 0 we have Proof. Note that by (16) we have Thus, since the four operators in (24) are bounded, there exists a constant C > 0 depending only on Ω and on the coefficients μ, ε and σ, such that where the second inequality follows from (22). This concludes the proof.
It is helpful to recall thatṼ ω : H 2 → H 3 , where and Proposition 25. The σ e,2 essential spectrum ofṼ ω is the union of the σ e,2 essential spectra of the two block operator pencils Here the operators I N and I T are the canonical mappings fromẊ N (Ω) anḋ X T (Ω) to the quotient spacesẊ N (Ω)/K N (Ω) andẊ T (Ω)/K T (Ω), respectively.
Proof. By inspection of (23), the operator pencilṼ ω,0 may be written as the block lower triangular operator matrix pencil in which A ω and D ω are as in Eq. (25) and the off-diagonal component is given by The proof is divided into several steps.
Remark 26. The text [18] contains many interesting results on essential spectra of block-operator matrices and pencils; Theorem 2.4.1 is very close to what we would need, but our pencilṼ ω,0 is lower triangular rather than diagonally dominant.
We are now ready to prove our main result.
Proof of Theorem 6. We commence the proof by observing the following identity: This is an immediate consequence of Lemmas 18 and 20 and of Proposition 25. We now consider σ e,2 (A ω ) and σ e,2 (D ω ) in more detail. The essential spectrum of A ω consists of the point {0}, arising from the (2, 2) diagonal entry of A ω , which has ω = 0 as an eigenvalue of infinite multiplicity and is otherwise invertible; and of the essential spectrum of the pencil in the (1, 1) entry, which is as stated in the theorem, namely In order to deal with the essential spectrum of D ω we observe that if we replace V ω by a new pencil V 0 ω in which the coefficients have the constant values ε 0 , μ 0 and σ 0 , then D ω will be unchanged while A ω will be replaced by a pencil A ω,0 in which all the coefficients are constant. For the constant coefficient pencil A ω,0 we see that 0 lies in the σ e,2 essential spectrum as we reasoned before, while the (1, 1) term is invertible and Fredholm precisely when ωε 0 + iσ 0 = 0, by the Babuška-Lax-Milgram theorem; hence σ e,2 (A ω,0 ) = {0, −iσ 0 /ε 0 }. Using (27) for the constant coefficient pencil, we now have We now prove that the σ e,2 essential spectrum of A ω already contains the set {0, −iσ 0 /ε 0 }. The (2, 2) component has 0 as an eigenvalue of infinite multiplicity. If Ω is bounded, we have σ 0 = 0 and so the claim is proven. Otherwise, for the point −iσ 0 /ε 0 we observe that by the Hypothesis (3), given n > 0 there exists R n > 0 such that if ω 0 := −iσ 0 /ε 0 then Choosing any function φ n ∈ C ∞ 0 (Ω) with support in {x ∈ Ω : |x| > R n }, with ∇φ n L 2 (Ω) = 1, we see that Since the supports of the sequence (∇φ n ) n∈N move off to infinity, the sequence converges weakly to zero; it is therefore a singular sequence in ∇Ḣ 1 0 (Ω) for the (1, 1) element of A ω0 . Thus ω 0 lies in the σ e,2 essential spectrum of A ω . Combining the observations (27), (28) and (29) with the fact that σ e,2 (A ω ) ⊇ {0, −iσ 0 /ε 0 } completes the proof.
We conclude this section with a more explicit description of the essential spectrum of the divergence form operator div((ωε + iσ)∇ · ) in the case of continuous coefficients.
Proof. If (ω) = 0 then the real part of ωε+iσ is definite, and the result follows by the Lax-Milgram theorem. If ω = iν is purely imaginary, this reasoning still works if νε + σ is uniformly definite in Ω. It remains only to show that if νε + σ is indefinite at some point x 0 ∈ Ω, then 0 lies in the essential spectrum of div((ωε + iσ)∇ · ).
We prove the result by constructing a Weyl singular sequence. Define a := νε + σ and a 0 := a(x 0 ). Let χ : [0, ∞) → [0, 1] be a smooth cutoff function such that χ(t) = 1 for 0 ≤ t ≤ 1 and χ(t) = 0 for all t ≥ 2. Let θ ∈ R 3 be a unit vector chosen such that θ T a(x 0 )θ = 0. For each sufficiently small δ > 0 and large r > 0 let A direct calculation shows that ∇u r,δ in sup-norm is O(r −1 δ −5/2 ) + O(δ −3/2 ). We suppose that rδ 5 2 1, so that the δ −3/2 term dominates; we have ∇u r,δ L ∞ (B 2δ (x0)) = O(δ −3/2 ) and u r,δ Ḣ1 0 (Ω) ≥ c for some c > 0 independent of r and δ. If v is any smooth test function then so that the u r,δ tend to zero weakly in H 1 0 (Ω) as r +∞ and δ 0, with r ≥ δ − 5 2 . To complete the proof that 0 lies in the essential spectrum of our operator we show that div(a ∇u r,δ ) H −1 (Ω) can be made arbitrarily small. It is easy to see that div(a ∇u r,δ ) We compute ∇u r,δ by direct differentiation of eqn. (30) and deduce that for in the last step we have used the fact that div a 0 ∇ r −1 exp(irθ · x) = 0, which follows immediately from θ T a 0 θ = 0. Integration by parts yields We estimate the final inner product by observing that χ δ is O(δ −3/2 ), its gradient is O(δ −5/2 ) and its second derivatives O(δ −7/2 ), while its support is a ball whose volume is O(δ 3 ): thus for some constant C > 0. Substituting this back into (31) we obtain div(a ∇u r,δ ) Letting r ∞ and then letting δ 0 we obtain the required result.
and Laplacian. For simplicity of notation, we shall write a b to mean a ≤ Cb for some positive constant C depending only on Ω . We assume that the crosssection Ω satisfies the following additional hypothesis.
by the Lax Milgram theorem, and so ψ 1 is uniquely determined. To obtain the remaining components of ψ we rewrite (33) and (34) as Again take the Fourier transform with respect to x 1 and obtain Using the second and third identities in (37) yieldŝ for a.e. ξ ∈ R.
It remains to check the first and fourth identities in (37) and the first component of (35). For the first identity in (37) we observe that Here we have used, for the second equality, the fact that div f = 0. For the fourth identity in (37), by (36) we have Finally, for the first component of (35), using F to denote the Fourier transform, where we have used the fact that ψ 1 = 0 and f · ν = 0 on ∂Ω in the last step.
Step (2) The only difference between this case and the one above lies in the boundary conditions. We no longer have f · ν = 0 on the boundary. This time f ∈ H(div 0, Ω) and we seek ψ such that ψ · ν = 0 on ∂Ω.
The calculations follow as above except that problem (36) is replaced by in which the reason for the slightly curious Neumann boundary condition will become clear shortly. As above, for almost every ξ ∈ R, this problem admits a unique solutionψ 1 (ξ) ∈Ḣ 1 (Ω ) by the Lax Milgram theorem. Having found ψ 1 , we constructψ 2 andψ 3 as before, and the verification of (38) and (39) is similar to the calculations for (33) and (34). This leaves the boundary condition (40): sinceψ · ν =ψ 2 ν 2 +ψ 3 ν 3 , we havê the equality at the last step coming from the boundary equation in (41).
(3) This follows by using the Fourier transform with respect to the variable x 2 and the Fourier series in the variable x 3 , as in cases (1) and (2) above; the calculations are completely analogous.