Spectrum of the Laplacian with mixed boundary conditions in a chamfered quarter of layer

We investigate the spectrum of a Laplace operator with mixed boundary conditions in an unbounded chamfered quarter of layer. This problem arises in the study of the spectrum of the Dirichlet Laplacian in thick polyhedral domains having some symmetries such as the so-called Fichera layer. The geometry we consider depends on two parameters gathered in some vector $\kappa=(\kappa_1,\kappa_2)$ which characterizes the domain at the edges. By exchanging the axes and/or modifying their orientations if necessary, it is sufficient to restrict the analysis to the cases $\kappa_1\ge0$ and $\kappa_2\in[-\kappa_1,\kappa_1]$. We identify the essential spectrum and establish different results concerning the discrete spectrum with respect to $\kappa$. In particular, we show that for a given $\kappa_1>0$, there is some $h(\kappa_1)>0$ such that discrete spectrum exists for $\kappa_2\in[-\kappa_1,0)\cup(h(\kappa_1),\kappa_1]$ whereas it is empty for $\kappa_2\in[0,h(\kappa_1)]$. The proofs rely on classical arguments of spectral theory such as the max-min principle. The main originality lies rather in the delicate use of the features of the geometry.


Formulation of the problem
The original motivation for this work comes from the study of the spectrum of the Laplace operator with Dirichlet boundary conditions in thick polyhedral domains having some symmetries.The archetype of such geometries is the so-called Fichera layer represented in Figure 1 right (see [10] for the original article that gave rise to the name).Exploiting symmetries, in certain cases one can reduce the analysis to the one of the spectrum of the Laplace operator with mixed boundary conditions in chamfered quarters of layers.More precisely, the geometries that we consider in this article are characterized by two parameters κ 1 , κ 2 ∈ R that we gather in some vector κ = (κ 1 , κ 2 ).Referring to carpentry and locksmith tools, we first define the "blade" (see Figure 1 left).Then we introduce the "incisor" (see Figure 1 centre).Let us give names to the different components of the boundary ∂Ω κ of Ω κ .First, denote by Σ κ the union of the two "horizontal" quadrants: Then consider the laterals sides of the incisor.Set We study the spectral problem with mixed boundary conditions where ∂ ν is the outward normal derivative on ∂Ω κ .Observe that by exchanging the axes and/or modifying their orientations, there is no loss of generality to restrict the analysis to the cases Denote by H 1 0 (Ω κ ; Σ κ ) the Sobolev space of functions of H 1 (Ω κ ) vanishing on Σ κ .Classically (see e.g.[14,15]), the variational formulation of Problem (5) writes where for a domain Ξ, (•, •) Ξ stands for the inner product of the Lebesgue spaces L 2 (Ξ) or (L 2 (Ξ)) 3  according to the case.Integrating first with respect to the x 3 variable and using the homogeneous Dirichlet condition on Σ κ for the functions in H 1 0 (Ω κ ; Σ κ ), one can prove that there holds the Friedrichs inequality where c κ > 0 is a constant which depends only on κ.As known e.g. from [4, §10.1] or [20, Ch.VIII.6], the variational problem (6) gives rise to the unbounded operator The operator A κ is positive definite and selfadjoint.Since Ω κ is unbounded, the embedding H 1 0 (Ω κ ; Σ κ ) ⊂ L 2 (Ω κ ) is not compact and A κ has a non-empty essential component σ ess (A κ ) ( [4,Thm. 10.1.5]).Note that the case plays a particular role.Indeed in this situation, if u is an eigenfunction associated with an eigenvalue of A κ , by extending u via even reflections with respect to the faces Γ κ 1 , Γ κ 2 , one gets an eigenvalue of the Dirichlet Laplacian in the Fichera layer F defined in (1).This latter problem has been studied in [6,2].More precisely, in [6] the authors give a characterization of the essential spectrum of the Dirichlet Laplacian and show that the discrete spectrum has at most a finite number of eigenvalues.The existence of discrete spectrum is proved in [2,Thm. 2].
The goal of this paper is to get similar information for the operator A κ with respect to the parameter κ.In the present work, we will also show that the spectrum of the Dirichlet Laplacian in Ω κ , i.e. with homogeneous Dirichlet boundary conditions everywhere on ∂Ω κ , has a rather simple structure with only essential spectrum and no discrete spectrum.This note is organized as follows.In Section 2, we describe the essential spectrum of A κ (Theorem 2.1).Then in Section 3, we state the results for the discrete spectrum of A κ (the main outcome of the present work is Theorem 3.2).The next four sections contain the proof of the different items of Theorem 3.2.In Section 8, we illustrate the theory with some numericals results.Finally we establish the above mentioned result related to the Dirichlet Laplacian in Ω κ in the Appendix (Proposition 8.1).Introducing the angle α 1 ∈ [0, π/2) such that κ 1 = tan α 1 , the blade (2) can also be defined as
The continuous spectrum of Problem (8) coincides with the ray [π 2 , +∞).When α 1 = 0 (straight end), working with the decomposition in Fourier series in the vertical direction, one can prove that the discrete spectrum is empty.On the other hand, for all α 1 ∈ (0, π/2), it has been shown in [12] that there is at least one eigenvalue below the continuous spectrum (see also [17] for more general shapes).Notice that by extending Π α 1 by reflection with respect to γ α 1 , we obtain a broken strip that we can also call a V-shaped domain.This allows us to exploit all the results from [9,19] (see also [8] as well as the amendments in [18]) to get information on µ α 1 1 , the smallest eigenvalue of (8).In particular, the function α 1 → µ α 1  1 is smooth and strictly decreasing on (0, π/2).Additionally, we have lim (see Figure 8 for a numerical approximation of α 1 → µ α 1 1 ).By adapting the approach proposed in [6, §3.1], one establishes the next assertion.The only point to be commented here is that there holds

Discrete spectrum
For the discrete spectrum σ d (A κ ), our main results are as follows: 2) There exists h(κ 1 ) ∈ (0, κ 1 ) such that: 1 is strictly increasing on [−κ 1 , 0) and strictly decreasing on (h(κ 1 ), The items 1)-3) of Theorem 3.2 are illustrated by Figure 3.Note in particular that we have the following mechanism for positive κ 2 : diminishing κ 2 from the value κ 1 makes the eigenvalue λ κ . Theorem 3.1 can be established quite straightforwardly by working with symmetries.The rest of the present note is dedicated to the proof of the statements of Theorem 3.2.

Absence of eigenvalues for small positive κ 2
The goal of this section is to prove an intermediate result to establish the item 2) of Theorem 3.2.Therefore we assume that κ 2 ≥ 0. In that situation, the integral I Γ κ 1 in ( 19) is positive because α 2 ∈ [0, π/2) and our argument of the previous section does not work for showing the existence of discrete spectrum.Of course, this does not yet guarantee that σ d (A κ ) is empty.Actually we will see in Section 6 that σ d (A κ ) is non-empty for certain κ with κ 2 > 0. For the moment, combining the calculations of Section 4 with the approach of [16], we show the following result.Proposition 5.1.For all κ 1 > 0, there exists δ(κ 1 ) > 0 such that σ d (A κ ) is empty for Proof.
In the remaining part of the proof, we establish (23).Consider the mixed boundary-value problem As pictured in Figure 6 left, the domain Ω κ 1− in ( 21) is a semi-infinite prism with a trapezoidal cross-section and a skewed end.When κ 2 = 0, the trapezoid is simply the unit square and the continuous spectrum of the problem (24) coincides with the ray [π 2 , +∞).In that situation, the problem (24) admits an eigenvalue at µ α 1  1 ∈ (0, π 2 ) with α 1 = arctan κ 1 (see the text above ( 9) for the definition of that quantity), a corresponding eigenfunction being w such that where v is an eigenfunction of (8) associated with µ α 1  1 .Now let us consider the situation κ 2 > 0 small.Then the map is a diffeomorphism whose Jacobian matrix is close to the identity and whose Hessian matrix is small.Using these properties, we deduce that the discrete spectrum of the problem (24) is still non-empty for κ 2 small enough.This comes from the fact that the cut-off point of the essential spectrum satisfies the estimate and the first (smallest) eigenvalue of the discrete spectrum, which is simple 1 , admits the expansion Here C > 0 is a constant independent of κ 2 .These properties can be justified using classical results of the perturbation theory for linear operators, see e.g.[13,Ch. 7], [4, Ch. 10], [20, Ch.XII].From the minimum principle, to establish (23), we see that it suffices to show that Let w κ 2 1 be an eigenfunction of Problem (24) associated with τ κ 2 1 .Together with (29), consider the asymptotic ansatz 1 Since Ω κ 1− is unbounded, we cannot directly apply the classical Krein-Rutman theorem to prove that the first eigenvalue τ κ 2 1 of (24) is simple.However, this can be established for example by exploiting that the eigenfunctions associated with τ κ 2  1 are exponentially decaying at infinity and by approximating them by eigenfunctions of operators set in bounded domains (where we can apply the Krein-Rutman theorem).With this, we show that each eigenfunction associated with τ κ 2  1 is either non-negative or non-positive in Ω κ 1− , which is possible only if τ κ 2 1 is a simple eigenvalue.In the proof, one needs to use the fact that eigenfunctions cannot vanish on sets of positive area owing to the theorem on unique continuation.
where wκ 2 1 is a small remainder.Insert (29), (28) into (24) and collect the terms of order κ 2 .We obtain As for the Neumann boundary condition of (24), using in particular that on Γ κ 1 , at order κ 2 , we find Since the smallest eigenvalue µ α 1 1 is simple, there exists only one compatibility condition to satisfy to ensure that the problem (29)-(30) has a non trivial solution.It can be written as 6 Existence of eigenvalues for κ 2 close to κ 1 > 0 We start this section by proving that the discrete spectrum σ d (A κ ) of the operator A κ can also be non-empty for certain positive κ 2 .This happens for example in the case κ 1 = κ 2 , which we now assume.We adapt the proof of [2, Thm.2] and exhibit a function φ ∈ H 1 0 (Ω κ ; Σ κ ) satisfying (11).First, note that for κ 1 = κ 2 , the domain Ω κ is symmetric with respect to the "bisector" cross-section Let us divide Ω κ into the two congruent domains Accordingly, we set where v is as in (13).Since ψ ε is continuous on Υ κ and decays exponentially at infinity, it belongs to H 1 0 (Ω κ ; Σ κ ).Moreover, we have Using that v solves (8), we get Remark that Υ κ is included in the plane (O, ê1 , ê3 ).Let (x 1 , x2 , x3 ) denote the coordinates in the basis (34).We have Exploiting the exponential decay of v(ξ) as ξ 1 → +∞, (see Section 2), one finds that the first integral of the right hand side above is O(ε).For the second one, using that v is independent of x 1 , we can write Remarking also that x2 = 0 on Υ κ , this gives where } and where ν stands for the outward unit normal vector to Υ κ (in the plane (O, ê1 , ê3 )).Using that κ 1 = κ 2 > 0, we find ν • ê1 < 0 on L κ .Since there holds v ̸ ≡ 0 on L κ , gathering (35) and (36), we deduce that we have I ε Υ κ < 0 for ε small enough.From (33), we deduce for ε small enough.Then by symmetry, we obtain We conclude that the inequality ( 11) is satisfied by the function (32) which proves the following statement.
Since the eigenvalues of the discrete spectrum are stable with respect to small perturbations of the operator, Theorem 6.1 and diffeomorphisms similar to (25) imply that σ d (A κ ) is not empty for κ 2 in a neighbourhood of κ 1 .With Proposition 5.1, this allows us to introduce h(κ 1 ) ∈ (0, κ 1 ) as the infimum of the numbers δ such that σ d (A κ ) is non-empty for all κ 2 ∈ (δ, κ 1 ]. On the other hand, we have the following monotonicity result: Lemma 6.2.Consider some κ = (κ 1 , κ 2 ) with κ 1 > 0 and κ 2 ∈ (0, κ 1 ] such that A κ has a nonempty discrete spectrum.Let λ κ 1 denote the first (smallest) eigenvalue of σ d (A κ ).For ε > 0 small, set κ ε := (κ 1 , κ 2 + ε) and denote by λ κ ε 1 the first eigenvalue of σ d (A κ ε ).Then, we have Proof.Using again the minimum principle, we can write Now define the function ψ ε such that where u κ ∈ H 1 0 (Ω κ ; Σ κ ) is an eigenfunction associated with the first eigenvalue of σ d (A κ ).Clearly ψ ε is a non zero element of H 1 0 (Ω κ ε ).Besides, we find According to (38), these identities imply The strict inequality in (37) follows from the fact that the derivative ∂u κ /∂x 2 cannot be null in the whole domain Ω κ .This completes the proof of the lemma.

Numerics and discussion
In this section, we illustrate some of the results above.In Figure 8, we represent an approximation of the first eigenvalue of the 2D problem (8) set in the pointed strip Π α 1 with respect to α 1 ∈ (0, 9π/20).We use a rather crude method which consists in truncating the domain at ξ 2 = 12 (see the picture of Figure 7 left) and imposing homogeneous Dirichlet boundary condition on the artificial boundary.Then we compute the spectrum in this bounded geometry by using a classical P2 finite element method.To proceed, we use the library Freefem++ [11] and display the results with Matlab2 and Paraview3 .The values we get are coherent with the ones recalled in (9).For more details concerning the numerical analysis of this problem, we refer the reader to [7].In this work, one can also find an interesting study concerning the case α → (π/2) − .
In Figure 9-11, we fix κ 1 = 1 (equivalently α 1 = π/4) and compute the first eigenvalue of σ d (A κ ) for κ 2 ∈ {−1, −0.1, 1}.For κ 1 = 1, the bound of the essential spectrum of A κ is λ κ † ≈ 0.929π 2 (see Figure 8 as well as [6]).For each of the three κ 2 , in agreement with Theorem 3.2, we find an eigenvalue below the essential spectrum.Actually, in each situation our numerical experiments seem to indicate that there is only one eigenvalue in the discrete spectrum, which is a result that we have not proved.Interestingly, for κ 2 = −0.1 (Figure 10), the eigenfunction is not particularly localized at the intersection of the obliques sides.This is related to the so-called Agmon estimates which guarantee that the decay rate coincides with the square root of the difference between the lower bound of the essential spectrum λ κ † and the eigenvalue λ κ 1 (see [1,8]).For κ 2 = −0.1 the quantity λ κ † − λ κ 1 is rather small.We emphasize that here we simply compute the spectrum of the Laplace operator with mixed boundary conditions in the bounded domain {x ∈ Ω κ |x 1 < 6 and x 2 < 6}.At x 1 = 6 and x 2 = 6, we impose homogeneous Neumann boundary condition.Admittedly, this is a very naive approximation, especially in the case of Figure 10.Our approximation lacks precision in that situation, this is the reason why we do not give the value of the corresponding eigenvalue.
In Figure 12, we represent eigenfunctions associated with two different eigenvalues of σ d (A κ ) for κ = (3, −3).For the 2D problem (8) in the pointed strip, the cardinal of the discrete spectrum can be made as large as desired by considering sufficiently sharp angles.We imagine that a similar phenomenon occurs in our geometry Ω κ .However to prove such a result is an open problem.At least the numerics of Figure 12 suggest that we can have more than one eigenvalue in σ d (A κ ).
As mentioned in the introduction, the fact that the discrete spectrum of A κ for κ = (1, 1) is not empty ensures that the Dirichlet Laplacian in the so-called Fichera layer F of Figure 1 right admits an eigenvalue below the essential spectrum.This can be proved by playing with symmetries and reconstructing F from three versions of Ω (1,1) .Now, gluing six domains Ω (1,−1) , we can create the cubical structure pictured in Figure 13 (note that its boundary is not Lipschitz).Then, from Theorem 3.2 which guarantees that σ d (A κ ) contains at least one eigenvalue, we deduce that the Dirichlet Laplacian in this geometry has at least one eigenvalue.Using this in (47) gives ∂u/∂ν = 0 on Γ κ 2 .Since we also have u = 0 on Γ κ 2 and ∆ x u + λu = 0 in Ω κ , the theorem of unique continuation (see e.g.[3], [5,Thm. 8.6]) guarantees that u ≡ 0 in Ω κ .

Figure 2 :
Figure 2: Domain Π α 1 corresponding with a cut of the blade B κ 1 in the plane x 2 = 0.