Semi-uniform stabilization of anisotropic Maxwell's equations via boundary feedback on split boundary

We regard anisotropic Maxwell's equations as a boundary control and observation system on a bounded Lipschitz domain. The boundary is split into two parts: one part with perfect conductor boundary conditions and the other where the control and observation takes place. We apply a feedback control law that stabilizes the system in a semi-uniform manner without any further geometric assumption on the domain. This will be achieved by separating the equilibriums from the system and show that the remaining system is described by an operator with compact resolvent. Furthermore, we will apply a unique continuation principle on the resolvent equation to show that there are no eigenvalues on the imaginary axis.


Introduction
Let Ω ⊆ R 3 be a bounded and connected strongly Lipschitz domain, which boundary is split into Γ 0 and Γ 1 ̸ = ∅.Then we regard Maxwell's equations as a boundary control and observation system π τ E(t, ζ) = 0, (perfect conductor) t ≥ 0, ζ ∈ Γ 0 , (1h) with feedback law The traces π τ E, γ τ H and γ ν B are, roughly speaking, (ν × E ∂Ω ) × ν, ν × H ∂Ω and ν • B ∂Ω , respectively, for details see Appendix B. The permittivity ϵ and the permeability µ are Lipschitz continuous matrix-valued functions (i.e., we allow anisotropic and inhomogeneous materials) such that c −1 ≤ ϵ ≤ c and c −1 ≤ µ ≤ c for a c > 0 (in the sense of positive definitness).(The Lipschitz continuity of ϵ and µ is necessary to apply a unique continuation principle.)The feedback operator k is also matrix-valued and satisfies c −1 ≤ k ≤ c for a c > 0 (w.l.o.g. the same c)-we do not ask for any further regularity but measurability.The charge density ρ can be any L 2 (Ω) function.
The boundary feedback law illustrated in Figure 1 results in a Silver-Müller or Leontovich or impedance boundary condition 1 Our goal is to show that the system (1) is semi-uniformly stable, i.e., there exists a uniform decay rate such that every solution converges to an equilibrium state with this rate.More precisely this means that the corresponding semigroup is semi-uniformly stable, see Definition 4.1.Semi-uniform stability is a concept that was introduced in [2].It is a stability concept that is between strong stability and exponential stability, i.e., exponential stability implies semi-uniform stability and semi-uniform stability implies strong stability.
Stability of Maxwell's equations with Silver-Müller boundary conditions have been studied in several works.The goal was always to prove exponential stability and therefore additional assumptions were added.
The damping acts on the entire boundary.
• In [15] Ω has a C ∞ boundary and Γ 1 satisfies the geometric control condition.
The author worked with constant and scalar ϵ and µ.
• In [7] the authors regard a domain Ω with C ∞ boundary and ϵ and µ are scalar C ∞ functions.The damping acts on the entire boundary and they additionally assume the existence of a ζ 0 ∈ Ω such that However, they allow k to be in a class that also contains certain nonlinear operators.
• In [12] Ω has a C 2 boundary and the damping acts on the entire boundary ∂Ω.
The functions ϵ and µ are scalar C 1 and may be non-autonomous.The function k can be in a certain nonlinear class.
• In [1] Ω has a C ∞ boundary and the damping acts on the entire boundary.In that work there is an additional time delay in the boundary condition and a certain nonlinearity is allowed for k.
• In [17] Ω is strongly star-shaped and has a C 5 boundary and the damping acts on the entire boundary.They allow ϵ and µ to be even state-dependent, i.e., quasilinear Maxwell's equations.
Note that the entire boundary ∂Ω always satisfies the geometric control condition.Hence, all of the above references work, at least implicitly, with this condition.Apart from [15] none of these references work with split/mixed boundary conditions.
There are also other effects that can stabilize Maxwell's equations like distributed damping or memory terms, see e.g., [13,6].
We will regard a domain Ω with Lipschitz boundary ∂Ω that is split into Γ 0 and Γ 1 .The boundary damping acts only on one part, namely Γ 1 -note that Γ 1 does not need to be connected.Moreover, ϵ and µ are Lipschitz continuous positive matrix-valued functions that are uniformly bounded from above and below.However, we do not show exponential stability, but semi-uniform stability and we do not prove an explicit decay rate.Most likely such a decay rate will depend on the geometry of Ω and Γ 1 .
However, in contrast to the listed literature we can, for example, deal with anisotropic Maxwell's equations on a cube that is damped on one face of the cube.
We will follow a similar approach as in [9], where semi-uniform stability for the wave equations was shown.That is splitting the equations in a time independent (for equilibriums) and a time dependent (for the "actual" dynamic) part and showing that the differential operator that corresponds to the time dependent part has no spectrum on the imaginary axis.This will be done by showing that the operator has a compact resolvent and employing a unique continuation principle.
The main theorem of this work reads as follows.
be an initial value that satisfies the boundary conditions and Gauß laws div ϵE 0 = ρ and div µH 0 = 0. Then the corresponding solution [ E H ](t, ζ) of (1) converges to an equilibrium state Ee He (ζ) for t → ∞.More precisely there exists a monotone decreasing f : [0, +∞) → [0, +∞) with lim t→+∞ f (t) = 0, which is independent of the initial values such that Remark 1.2.Clearly, we can replace the L 2 norm in the previous theorem by a weighted L 2 norm, e.g., the energy norm that is induced by ϵ and µ.
Remark 1.3.Note that the connectedness of Ω is not really necessary as long as Γ 1 has parts on every connected component.
Remark 1.4.We can actually also allow inhomogeneous boundary conditions for π τ E and γ ν H on Γ 0 , because they will disappear in the equilibrium.We only have to make sure that the inhomogeneity satisfies certain compatibility conditions (not every L 2 function is in the range of π τ ).
This work is structured in the following way: We will start by recalling the Sobolev spaces that correspond to our differential operators in Section 2. Then we will split the system into a static and a dynamic part in Section 3. In Section 4 we show that the dynamic part is semi-uniformly stable by a compact embedding and a unique continuation principle, which finally implies the main result Theorem 1.1.

Preliminary
For Ω ⊆ R d open and Γ ⊆ ∂Ω open we use the following notation (as in [3]) We will regard an open, bounded and connected Ω ⊆ R 3 with strongly Lipschitz boundary.For g ∈ C∞ (R 3 ) and f ∈ C∞ (R 3 ) 3 we define the differential operators These operators can be regarded as operators from L 2 (Ω) k1 to L 2 (Ω) k2 (with In the further we will omit the exponent k at L 2 (Ω) k , H 1 (Ω) k , C∞ (Ω) k , etc., if they are clear from the context.We introduce the maximal domain of these operators on L 2 (Ω): For g ∈ C∞ (R 3 ) and f ∈ C∞ (R 3 ) 3 there is the well-known integration by parts formula where ν denotes the outward pointing unit normal vector on the boundary.This formula can be extended to the maximal domain of the respective differential operator, such that we have for g ∈ H 1 (Ω) and f ∈ H(div, Ω) where γ 0 , the boundary trace, is the extension of g → g ∂Ω to H 1 (Ω) and γ ν , the normal trace, is the extension of f → ν • f ∂Ω to H(div, Ω).These mappings map onto H 1 /2 (∂Ω) and H − 1 /2 (∂Ω) respectively, where H 1 /2 (∂Ω) is range of γ 0 endowed with the range norm of H 1 (Ω), and H − 1 /2 (∂Ω) is its dual space.Basically, for curl there is a similar approach to extend the integration by parts formula that is valid for f, g ∈ C∞ (R 3 ) to the maximal domain of curl.We present this in Appendix B.
Note that every w ∈ C 3 can be represented as for a.e.ζ ∈ ∂Ω.
Hence, we call π τ , the extension of g → (ν × g ∂Ω ) × ν, the tangential trace and γ τ , the extension of f → ν × g ∂Ω , the twisted tangential trace.If we want to emphasize that we are only interested on the part Γ 1 ⊆ ∂Ω of the tangential trace we denote this by π τ Γ1 and γ τ Γ1 for details see Appendix B.
For two Hilbert space X and Y we will use the notation For a strictly positive and bounded operator P on L 2 (Ω) we introduce ⟨x, y⟩ P := ⟨x, P y⟩ which establishes an equivalent inner product on L 2 (Ω).Corresponding to this new inner product we have ⊥ P and ⊕ P .For a strongly Lipschitz domain Ω and Γ 0 ⊆ ∂Ω we say that the pair (Ω, Γ 0 ) is a strongly Lipschitz pair, roughly speaking, if also ∂Γ 0 is strongly Lipschitz.For details see [3,22].

Split the system
In this section we will split our system in a time invariant part and the remaining dynamic part.This will simplify the analysis of the spectrum of the (remaining) dynamic part.From a certain point of view we factor out the eigenvectors corresponding to the eigenvalue 0 of the differential operator that describes the dynamic.
We can write the system (1b)-(1c) as From a semigroup perspective the first version is in a better form.The second version is for instance favored by the approach in [18].In order to analyze the stability of the system we have to separate the equilibrium states from the rest of the system.We call the remaining system the "true dynamic".
By setting all time derivatives to zero in (1), we obtain the equations for the equilibrium states.
where h is determined by the traces of the initial values.This static system is solvable by [3,Thm. 5.6].Note that if Ω is not simply connected, then the cohomology spaces contain more than the zero vector.Hence, the solutions of (3) are not unique, because for a solution Ee He also Ee He HΓ 1 ,Γ 0 ,µ(Ω) .For the "true" dynamic we regard the operator with the boundary conditions Note that in general π τ Γ1 and γ τ Γ1 map into different spaces.Hence, in order to meaningfully regard the second boundary condition we restrict ourselves to all those elements that are mapped into the pivot space L2 τ (Γ 1 ) under π τ Γ1 , γ τ Γ1 . 2 Combined with the boundary condition π τ Γ0 ϵ −1 D = 0 we can say that ϵ −1 D has an L 2 tangential trace on the entire boundary.Hence, in order to satisfy our boundary conditions we require/assume that Summarized we regard the following operator and domain The operator A 0 is a generator of a contraction semigroup as we will explain in Appendix B.4.Note that H : L 2 (Ω) → L 2 (Ω) is a strictly positive and bounded operator, by the assumptions on ϵ and µ.It comes very natural to use ⟨•, •⟩ H as an inner product, since the corresponding norm is the energy norm in the state space.
space (that contains Vτ (Γ 1 ) and V × τ (Γ 1 )), but in the end this only implies that all vector fields D, B whose tangential traces satisfy this condition in the larger space are already in the pivot space L 2 τ (Γ 1 ).
In order to satisfy the remaining conditions we define the state space The intersection with Ω), because these are static solutions and already included in the static solutions (as difference of two solutions of ( 3)).
Lemma 3.1.Let H be a Hilbert space that can be decomposed into where U 1 and U 2 are closed subspaces of H. Then for every strictly positive operator where ⊕ Q denotes the orthogonal sum w.r.t. the inner product ⟨x, y⟩ Q := ⟨x, Qy⟩.
and therefore h = u 2 .For w 1 = 0 and The next theorem is [3, Thm.5.3], we have just made some minor modifications to better suit our setting.Theorem 3.2 (Helmholtz decomposition).Let (Ω, Γ a ) be a strongly Lipschitz pair and Γ b = ∂Ω \ Γ a , and δ ∈ L b (L 2 (Ω)) a strictly positive operator.Then where the closedness of ∇ H1 Γa (Ω) is a consequence of Poincaré's inequality.Since ∇ H1 Γa (Ω) ⊆ HΓa (curl 0, Ω), the intersection of ( 5) and δ HΓa (curl 0, Ω) leads to By the same argument for D := curl with dom D = HΓ b (curl, Ω) we have where the closedness of curl HΓ b (curl, Ω) follows from [3,Lem. 5.2].Combining ( 7) and ( 6) leads to Proof.By Theorem 3.2 and ( 5) we have The space X H is a Hilbert space with the inner product ⟨x, y⟩ The closedness of the operators div and γ ν Γ0 implies the closedness of ker div and ker γ ν Γ0 .Hence, X H is the intersection of closed subspaces and therefore closed.By the definition of X H and Corollary 3.3 we have Finally, we introduce the differential operator and its domain that describes the dynamic of our system.Definition 3.5.We define A := A 0 X H , which we regard as an operator A Note that A is a generator of a contraction semigroup, since A 0 is a generator of a contraction semigroup and X H is closed and invariant under the semigroup T 0 generated by A 0 .In particular, the semigroup T that is generated by A is given by , see e.g., [8, ch.II sec.2.3].

Semi-uniform stability
We will regard the semigroup T that is generated by the operator A from the previous section, defined in Definition 3.5.Our goal is to show that this semigroup is semi-uniformly stable.Definition 4.1.Let A be the generator of the strongly continuous and bounded semigroup (T (t)) t≥0 on a Hilbert space X.Then we say that (T (t)) t≥0 is semiuniformly stable, if there exists a continuous and decreasing function f : [0, +∞) → [0, +∞) with lim t→+∞ f (t) = 0 and for every x ∈ dom A, where ∥•∥ dom A denotes the graph norm of A.
The difference between uniform stability and semi-uniform stability is that on the right-hand side of ( 8) there is the graph norm of the generator A instead of just the norm of the Hilbert space X.
It is very common to define semi-uniform stability by an equivalent characterization, namely item (iii) of the next theorem.

Theorem 4.2 ([2]
).Let T be a strongly continuous and bounded semigroup generated by A. Then the following assertions are equivalent.
Hence, by the previous theorem, the question about semi-uniform stability of A reduces to the following problem.A compact resolvent simplifies the task of finding resolvent points.Hence, the following theorem from [14,Thm. 4.1] will reduce the problem such that we only need make sure that iω is not an eigenvalue, see Theorem 4.6.
In particular, as a consequence of this compact embedding we obtain the following proposition.A special case for ϵ = µ = 1 has been regarded in [14,Thm. 5.6].
Hence, A has only point spectrum.So we only need to check, whether ker(A − iω) = {0} for all ω ∈ R. For convenience we provide a proof for this conclusion.Theorem 4.6.Let X be a Hilbert space and A : dom A ⊆ X → X be a closed and densely defined operator with compact resolvent.Then σ(A) = σ p (A), i.e., the spectrum of A contains only eigenvalues.Moreover, σ(A) is countable and has no accumulation points (or at most ∞ as accumulation point, if we regard the spectrum as subset of C ∪ {∞}).
Proof.Let λ ∈ ρ(A).Then (A − λ) −1 is a compact operator.Hence, the spectrum of (A − λ) −1 contains only countable eigenvalues that can only accumulate at 0. By the spectral mapping theorem (for unbounded operators or linear relations) we have Hence, the claim is true for A − λ.By shifting this operator by λ and again apply the spectral mapping theorem, we conclude the claim also for A. ❑ Now in order to show ker(A − iω) = {0} we make the following observation for supposed eigenvalues.
Hence, π τ Γ1 ϵ −1 D = 0 and by the boundary condition we also have γ τ Γ1 µ −1 B = 0. ❑ Since the unique continuation principle from Appendix A only works for ω ̸ = 0 we need to check 0 ∈ ρ(A) separately.Proof.Note that by Theorem 4.6 the spectrum σ(A) contains only eigenvalues.Hence, if we assume that 0 ̸ ∈ ρ(A), then 0 is an eigenvalue.Then by Lemma 4.7 the corresponding eigenvector Combined with the remaining boundary condition for D we obtain π τ ϵ −1 D = 0 on ∂Ω.Thus, for ϕ ∈ H(curl, Ω), using Ax = 0, we obtain Similarly, for ψ ∈ HΓ0 (curl, Ω) we have which implies that B ⊥ µ −1 ran HΓ0 (curl, Ω).Since B ∈ curl HΓ0 (curl, Ω) we conclude B = 0, which leads to x = 0. Therefore, 0 is not an eigenvalue and 0 ∈ ρ(A).❑ Now we have collected all the tools to prove that the dynamic part of Maxwell's equation is semi-uniformly stable.
Proof.By Theorem 4.2 we have already reduced the question to Problem 4.3, i.e., we have to show iR ⊆ ρ(A).By Proposition 4.5, A has a compact resolvent, which implies that the spectrum σ(A) contains only eigenvalues (Theorem 4.6).
If iw ∈ iR \ {0} is an eigenvalue of A, then by Lemma 4.7 the corresponding eigenvector has vanishing tangential trace on Γ 1 .From the principle of unique continuation (Theorem A.2) follows that this eigenvector can only be 0. Hence, there are no non-zero imaginary eigenvalues and consequently iR \ {0} ⊆ ρ(A).
Finally, Proposition 4.8 yields 0 ∈ ρ(A).Thus iR ⊆ ρ(A) and the semigroup that is generated by A is semi-uniformly stable. ❑ Since every solution of the system (1) can be decomposed into an equilibrium and a dynamic part (that solves the abstract Cauchy problem corresponding to A), we conclude the main theorem.
be an initial value that satisfies the boundary conditions and Gauß laws div ϵE 0 = ρ and div µH 0 = 0. Then the corresponding solution [ E H ](t, ζ) of (1) converges to an equilibrium state Ee He (ζ) for t → ∞.More precisely there exists a monotone decreasing f : [0, +∞) → [0, +∞) with lim t→+∞ f (t) = 0, which is independent of the initial values such that Recall that Proof.Let Ee He be a solution of (3) with h = γ τ H, i.e., an equilibrium state.Then we define the initial value for the "dynamic" part by

Note that Ee
He is in general-depending on the cohomology groups-not unique.Moreover, D0 B0 satisfies div D 0 = div B 0 = 0 and γ ν Γ0 B 0 = 0, but is not necessarily in . Hence, we shift Ee He by the and consequently (the updated) . Hence, D0 B0 ∈ X H and even in dom A. We denote the semi-uniformly stable semigroup that is generated by A by (T (t)) t≥0 and the decay function by f (Theorem 4.9).Hence, the solution The solution of ( 1) is then given by Note that H = ϵ −1 0 0 µ −1 is a bounded and boundedly invertible operator, which implies that c Redefining f as c(c + 1)f finishes the proof.❑

Conclusion
We have shown that Maxwell's equation without damping terms can be semiuniformly stabilized by a simple boundary feedback.This was done without any geometric assumptions on the boundary like the geometric control condition, convexity, star-shapedness, etc. and with only mild conditions on the matrix-valued functions ϵ and µ, i.e., strict positivity and Lipschitz continuity.
The key ingredients were a compact resolvent and a unique continuation principle.The other arguments do not really depend on our particular differential operator, but can also be done for an entire class of systems.In particular the port-Hamiltonian systems that are discussed in [20,19].In fact, if there were a generalization of the compact resolvent and the unique continuation principle for those port-Hamiltonian systems, we could conclude the same semi-uniform stability result.

Appendix A. Unique continuation
For ϵ and µ as in the beginning and ω ̸ = 0 we regard the following stationary system, Recall that ϵ and µ are strictly positive matrix-valued functions.In particular, there exists a c > 0 such that where W 1,∞ (Ω) denotes the Sobolev space of Lipschitz continuous functions.

Appendix B. Background on the differential and trace operators
In this section we want to give a little background on the operator A 0 defined in (4) and ultimately justify that it generates a contraction semigroup.We still have the standing assumption that (Ω, Γ 1 ) is a strongly Lipschitz pair.In order to do that we have to explain the boundary spaces and traces that correspond to curl.We recall the operator A 0 from (4) In order to better understand the domain of A 0 we have to take a closer look at the traces π τ Γ0 , π τ Γ1 and γ τ Γ1 .First of all note that the differential operator curl can be written as Therefore, it has the form curl = L ∂ := 3 i=1 ∂ i L i .Hence, the operator matches the form of the differential operators in [20].That work presents a general approach to boundary traces and boundary spaces that correspond to such differential operators.These boundary spaces and traces can be derived by an integration by parts formula.Nevertheless we want to present here a sketch of this construction that is adjusted just for the curl operator.
For the curl operator we have for f, g ∈ C∞ (R 3 ) the following integration by parts formula We can restrict ourselves to the boundary space L 2 τ (∂Ω) := {ϕ ∈ L 2 (∂Ω) | ν • ϕ = 0}, as both arguments of the L 2 (∂Ω) inner product in (11) belong to that space anyway.
In the following we want to show that similar to the integration by parts formula for div-∇ we can extend the integration parts formula by continuity on the maximal domain of the differential operator.The price to pay is that the L 2 inner product on the boundary has to be replaced by a dual pairing.For div-∇ this would be the dual pairing of (H 1 /2 (∂Ω), H − 1 /2 (∂Ω)), which forms a Gelfand triple with the pivot space L 2 (∂Ω).Unfortunately unlike in the div-∇ case the boundary spaces that correspond to curl do not establish a Gelfand triple, at least not in the usual sense where we have continuous embeddings.However, we get something that is almost a Gelfand triple, what we call quasi Gelfand triple, a notion that was introduced in [20] or more detailed in [19,21].
In particular we will give a sketch of the construction of the boundary spaces and traces that correspond to curl.B.1.Boundary spaces.Note that the integration by parts formula (11) can easily be extended to H 1 (Ω).Hence, we want to take the step to extend ( 11) from H 1 (Ω) to H(curl, Ω).In order to do this we introduce the spaces Note that every element of M (Γ 1 ) can be extended to M (∂Ω) by setting it to zero on Γ 0 .We define the spaces V τ (Γ 1 ) and Vτ (Γ 1 ) as the completions of M (Γ 1 ) and M (Γ 1 ), respectively, with respect to the range norms ∥ϕ∥ Vτ (Γ1) := inf (Ω) (ν×γ0g|Γ 1 )×ν=ϕ ∥g∥ H(curl,Ω) for ϕ ∈ M (Γ 1 ), respectively.These two norms are really norms by [20,Lem. 6.3].By construction we have that and πτ Γ1 : , see e.g., [3].Now we can extend π τ Γ1 and πτ Γ1 by density and continuity to H(curl, Ω) and HΓ0 (curl, Ω), respectively.We will use the same symbols for these extensions and call both of them tangential trace.If Γ 1 = ∂Ω, then clearly V τ (∂Ω) = Vτ (∂Ω) and we will just write π τ instead of π τ ∂Ω or πτ ∂Ω .Sometimes it is convenient to also leave out the circle on πτ Γ1 even if we work with elements of HΓ0 (curl, Ω) that are mapped into Vτ (Γ 0 ).Moreover, if it is clear from the context, we will also just use π τ instead of π τ Γ1 and πτ Γ1 .
Remark B.1.For a short moment we want to distinguish between π τ Γ1 and its continuous extension on H(curl, Ω), by denoting the extension by π τ Γ1 .Then it can be shown that π τ Γ1 : H(curl, Ω) → V τ (Γ 1 ) is surjective and .
Basically we can repeat the previous construction for the twisted tangential trace f → ν × γ 0 f .Note that for smooth functions g → (ν × g ∂Ω ) × ν is really the projection of g on its tangential component.Therefore, the name tangential trace is justified for π τ .Furthermore, for smooth functions we have ν × f ∂Ω = ν × π τ f , which tells us that ν × f ∂Ω is a 90 degree (or π 2 ) rotated version of the tangential component of f ∂Ω .The axis of rotation is the normal vector ν.
Even though it is basically a repetition we will execute the construction also for the twisted tangential trace.Hence, we define the twisted version of M (Γ 1 ) and M (Γ 1 ) as respectively.Furthermore we define V × τ (Γ 1 ) and V× τ (Γ 1 ) as the completion of M × (Γ 1 ) and M × (Γ 1 ), respectively with respect to the range norms respectively.The cross symbol × in the superscript indicates that we deal with a twisted version of M (Γ 1 ) and V τ (Γ 1 ), respectively.In particular, the operation ν × • can be extended to a unitary mapping between V τ (Γ 1 ) and V × τ (Γ 1 ), and a unitary operator between Vτ (Γ 1 ) and V× τ (Γ 1 ).By construction the mapping and γτ Γ1 : ).Hence, we can extend these mappings by continuity and density to H(curl, Ω) and HΓ0 (curl, Ω), respectively.We still denote these extension by γ τ Γ1 and γτ Γ1 .If Γ 1 = ∂Ω we will just write γ τ .Also if it is clear from the context that we only want to regard the trace on Γ 1 , we will just write γ τ instead of γ τ Γ1 .We will sometimes also omit the circle at γτ , if is is clear that we regard elements in HΓ0 (curl, Ω).
Hence, there exists a g ∈ H1 Γ0 (Ω) and an f ∈ H 1 (Ω) such that π τ Γ1 g = ϕ and γ τ Γ1 f = ψ.By γ 0 g Γ0 = 0, the integration by parts formula (11) and Cauchy-Schwarz's inequality we have Since this is true for all g ∈ H1 Γ0 (Ω) and f ∈ H 1 (Ω) such that π τ Γ1 g = ϕ and γ τ Γ1 f = ψ, we can apply an infimum on the right-hand side and obtain . ❑ Now we can define the following dual pairing by a limit.
Hence, for the orthogonal complements (in H(curl, Ω)) we have the reverse inclusion.
The next lemmas also show similarities to the notion of boundary data spaces from [16,Sec. 5.2].
Finally we define what it means for an f ∈ H(curl, Ω) to have an L 2 (Γ 1 ) tangential trace.There are two a priori different approaches: a weak approach by representation in an inner product and a strong approach by convergence.
) is a unitary mapping.Therefore, we can modify (f n ) n∈N such that we also have ∥f n − f ∥ H(curl,Ω) → 0. 3   Since we have a proper definition for π τ f ∈ L 2 τ (Γ 1 ) we can finally properly define the space ĤΓ1 (curl, Ω).
If we want to characterize L 2 tangential traces on Γ 1 for functions that additionally have a homogeneous tangential trace on Γ 0 , then it turns out to be more convenient to regard this in a combined way.More precisely, we do not look at the space ĤΓ1 (curl, Ω) ∩ HΓ0 (curl, Ω), but at Ĥ∂Ω (curl, Ω) ∩ HΓ0 (curl, Ω).This means that we regard functions that have an L 2 tangential trace on the entire boundary and on one part of the boundary this tangential trace vanishes.One might hope that these spaces coincide, but it is even unclear to us, whether HΓ0 (curl, Ω) ∩ HΓ1 (curl, Ω) = H∂Ω (curl, Ω) ( 3 If we really want to prove this modification of fn such that the other properties are preserved, we would also use ker γτ Γ 1 = ker γτ Γ 1 ∩ H 1 (Ω)

H(curl,Ω)
. holds true.Hence, these questions can be seen as open problems.
Anyway, for our purpose it suffices to work with Ĥ∂Ω (curl, Ω) ∩ HΓ0 (curl, Ω).A strong approach to that space would be to regard limits of H1 Γ0 (Ω) elements w.r.t.∥•∥ Ĥ∂Ω (curl,Ω) , which is the closure of H1 Γ0 (Ω) in Ĥ∂Ω (curl, Ω) ∩ HΓ0 (curl, Ω).Again the question that arises is: "Is H1 Γ0 (Ω) dense in Ĥ∂Ω (curl, Ω) ∩ HΓ0 (curl, Ω) (w.r.t.∥•∥ Ĥ∂Ω (curl,Ω) )?" (15) This is the actual formulation of the open problem of [23, eq. (5.20)] and at the end of section 5 in [23].Also this question was answered recently in [22].Hence, long story short, we do not have to distinguish between strongly and weakly in L 2 τ (Γ 1 ) in neither case.This is in particular of advantage, because we want to formulate our boundary conditions in L 2 τ (Γ 1 ) and the framework of quasi Gelfand triples would regard one of Vτ (Γ 1 ) ∩ L 2 τ (Γ 1 ) and V × τ (Γ 1 ) ∩ L 2 τ (Γ 1 ) strongly and the other one -for duality reasons -weakly.Therefore, now we avoid to have two different concepts of L , that if we define the domain of the upper right block strongly, then we have to define the domain of the (adjoint) lower left block weakly.The same goes for the corresponding traces.In this particular case, where both operators are basically the same, this leads to the strange situation where we have to introduce two a priori different definitions for the domain of the same operator.Luckily as we have discussed in this section both approaches coincide and consequently we just need one definition.
In order to justify that A 0 generates a contraction semigroup we want to use [19,Thm. 5.3.6] or originally [20,Thm. 7.6 & Ex. 7.8].For convenience we provide [19,Thm. 5.3.6] as Theorem B.12.In order to understand how this theorem is applicable, we have to translate our setting into their notation.In particular our setting is a special case of a general theory.As we have remarked at the beginning of this section our operator fits into the framework of [20], because we can decompose the curl operator into curl = 3 i=1 ∂ i L i = L ∂ , as we did in (10).The corresponding Hermitian transposed operator L H ∂ is in our case (by the skew-adjointness of L i ) Consequently, the block operator P ∂ that is regarded by the theory in [20,19] matches our block differential operator In our case the additional P 0 is just 0. For the boundary operators and boundary space we have the following translation π L = π τ , L ν = γ τ and L 2 π (Γ 1 ) = L 2 τ (Γ 1 ).However, since the boundary condition is formulated in L 2 τ (Γ 1 ), we can also integrate this directly into the space and obtain our original condition. 4Therefore, we finally conclude the following corollary.
Alternatively, it was also shown in [23] that A 0 generates a contraction semigroup.Note for elements in the domain of A 0 we can apply the extended integration by parts formula Lemma B.4 and replace the dual pairing by an L 2 τ (Γ 1 ) inner product as all elements in dom A 0 have L 2 tangential traces.Hence, applying this integration by parts formula twice for both rows of A 0 gives the following lemma.

Proposition 4 . 5 .
The operator A has a compact resolvent.Proof.By definition of A we have dom A = dom A 0 ∩ X H . Hence, by construction and Theorem 4.4 we have dom

Figure 3 .
Figure 3. Difference between ordinary and quasi Gelfand triples
).Note that the restriction bar Γ1 at π τ Γ1 is an abuse of notation.It indicates that we are only interested in 2 τ (Γ 1 ) tangential traces.B.4. Contraction semigroup.Finally, we want to consider the actual question of this section.It lies in the nature of block operators like