Eigenvalues for Maxwell’s equations with dissipative boundary conditions

Abstract

Let $V (t) = e^{t G_{b}}$ , $t ⩾ 0$ , be the semigroup generated by Maxwell’s equations in an exterior domain $Ω \subset R^{3}$ with dissipative boundary condition $E_{\tan} - γ (x) (ν \land B_{\tan}) = 0$ , $γ (x) > 0$ , $\forall x \in Γ = \partial Ω$ . We prove that if $γ (x)$ is nowhere equal to 1, then for every $0 < ϵ ≪ 1$ and every $N \in N$ the eigenvalues of $G_{b}$ lie in the region $Λ_{ϵ} \cup R_{N}$ , where $Λ_{ϵ} = {z \in C : | Re z | ⩽ C_{ϵ} (| Im z |^{\frac{1}{2} + ϵ} + 1), Re z < 0}$ , $R_{N} = {z \in C : | Im z | ⩽ C_{N} {(| Re z | + 1)}^{- N}, Re z < 0}$ .

Keywords

Maxwell’s equations asymptotically disappearing solutions location of the eigenvalues

1. Introduction

Suppose that $K \subset {x \in R^{3} : | x | ⩽ a}$ is an open connected domain and $Ω : = R^{3} ∖ \bar{K}$ is an open connected domain with $C^{\infty}$ smooth boundary Γ. Consider the boundary problem $\begin{matrix} (1.1) & \begin{matrix} \partial_{t} E = curl B, \partial_{t} B = - curl E in R_{t}^{+} \times Ω, \\ E_{\tan} - γ (x) (ν \land B_{\tan}) = 0 on R_{t}^{+} \times Γ, \\ E (0, x) = e_{0} (x), B (0, x) = b_{0} (x) \end{matrix} \end{matrix}$ with initial data $f = (e_{0}, b_{0}) \in {(L^{2} (Ω))}^{6} = H$ . Here $ν (x)$ denotes the unit outward normal to $\partial Ω$ at $x \in Γ$ pointing into Ω, $⟨, ⟩$ denotes the scalar product in $C^{3}$ , $u_{\tan} : = u - ⟨ u, ν ⟩ ν$ , and $γ (x) \in C^{\infty} (Γ)$ satisfies $γ (x) > 0$ for all $x \in Γ$ . The solution of the problem (1.1) is given by a contraction semigroup $(E, B) = V (t) f = e^{t G_{b}} f$ , $t ⩾ 0$ , where the generator $G_{b}$ has domain $D (G_{b})$ that is the closure in the graph norm of functions $u = (v, w) \in {(C_{(0)}^{\infty} (R^{3}))}^{3} \times {(C_{(0)}^{\infty} (R^{3}))}^{3}$ satisfying the boundary condition $v_{\tan} - γ (ν \land w_{\tan}) = 0$ on Γ.

In an earlier paper [2] we proved that the spectrum of $G_{b}$ in $Re z < 0$ consists of isolated eigenvalues with finite multiplicity. If $G_{b} f = λ f$ with $Re λ < 0$ , the solution $u (t, x) = V (t) f = e^{λ t} f (x)$ of (1.1) has exponentially decreasing global energy. Such solutions are called asymptotically disappearing and they are invisible for inverse scattering problems. It was proved [2] that if there is at least one eigenvalue λ of $G_{b}$ with $Re λ < 0$ , then the wave operators $W_{\pm}$ are not complete, that is $Ran W_{-} \neq Ran W_{+}$ . Hence we cannot define the scattering operator S related to the Cauchy problem for the Maxwell system and (1.1) by the product $W_{+}^{- 1} W_{-}$ . For the perfect conductor boundary conditions for Maxwell’s equations, the energy is conserved in time and the unperturbed and perturbed problems are associated to unitary groups. The corresponding scattering operator $S (z) : {(L^{2} (S^{2}))}^{2} \to {(L^{2} (S^{2}))}^{2}$ satisfies the identity $\begin{matrix} (1.2) & S^{- 1} (z) = S^{*} (\bar{z}), z \in C \end{matrix}$ if $S (z)$ is invertible at z. The scattering operator $S (z)$ defined in [5] is such that $S (z)$ and $S^{*} (z)$ are analytic in the “physical” half plane ${z \in C : Im z < 0}$ and the above relation for conservative boundary conditions implies that $S (z)$ is invertible for $Im z > 0$ . For dissipative boundary conditions the relation (1.2) in general is not true and $S (z_{0})$ may have a non trivial kernel for some $z_{0}, Im z_{0} > 0$ . Lax and Phillips [5] proved that this implies that $i z_{0}$ is an eigenvalue of $G_{b}$ . The analysis of the location of the eigenvalues of $G_{b}$ is important for the location of the points where the kernel of $S (z)$ is not trivial.

Fig. 1.

Eigenvalues of $G_{b}$ .

The main result of this paper is the following (see Fig. 1).

Theorem 1.1.

Assume that for all $x \in Γ$ , $γ (x) \neq 1$ . Then for every $0 < ϵ ≪ 1$ and every $N \in N$ there are constants $C_{ϵ} > 0$ and $C_{N} > 0$ such that the eigenvalues of $G_{b}$ lie in the region $Λ_{ϵ} \cup R_{N}$ , where $\begin{matrix} \begin{matrix} Λ_{ϵ} = {z \in C : | Re z | ⩽ C_{ϵ} (| Im z |^{1 / 2 + ϵ} + 1), Re z < 0}, \\ R_{N} = {z \in C : | Im z | ⩽ C_{N} {(| Re z | + 1)}^{- N}, Re z < 0} . \end{matrix} \end{matrix}$

If $Re λ < 0$ and $G_{b} (E, B) = λ (E, B) \neq 0$ , then $\begin{matrix} (1.3) & \begin{matrix} λ E = curl B & on & Ω, \\ λ B = - curl E & on & Ω, \\ div E = div B = 0 & on & Ω, \\ E_{\tan} - γ (ν \land B_{\tan}) = 0 & on & Γ . \end{matrix} \end{matrix}$ This implies that $u : = (E, B)$ satisfies $\begin{matrix} Δ u - λ^{2} u = 0 on Ω . \end{matrix}$

The eigenvalues of $G_{b}$ are symmetric with respect to the real axis, so it is sufficient to examine the location of the eigenvalues whose imaginary part is nonnegative. The mapping $z \mapsto z^{2}$ maps the positive quadrant ${z \in C : Re z > 0, Im z > 0}$ bijectively to the upper half space. Denote by $\sqrt{z}$ the inverse map. The part of the spectral domain ${λ \in C : Re λ < 0, Im λ > 0}$ is mapped by $λ = i \sqrt{z}$ to the upper half plane ${z \in C : Im z > 0}$ . In ${z \in C : Im z ⩾ 0}$ introduce the sets $\begin{array}{l} Z_{1} : = {z \in C : Re z = 1, h^{δ} ⩽ Im z ⩽ 1}, 0 < h ≪ 1, 0 < δ < 1 / 2, \\ Z_{2} : = {z \in C : Re z = - 1, 0 ⩽ Im z ⩽ 1}, \\ Z_{3} : = {z \in C : | Re z | ⩽ 1, Im z = 1} . \end{array}$

Fig. 2.

Contours $Z_{1}$ , $Z_{2}$ , $Z_{3}$ , $δ = 1 / 2 - ϵ$ .

Set $λ = i \sqrt{z} / h$ , $z \in Z_{1} \cup Z_{2} \cup Z_{3}$ . To study the eigenvalues λ, $| λ | > R_{0}$ , it is sufficient to consider $0 < h ≪ 1$ . As z runs over the rectangle in Fig. 2, with $0 < h ≪ 1$ , λ sweeps out the large values in the intersection of left and upper half planes. The values of $z \in Z_{2}$ near the lower left hand corner, $z = - 1$ , of the rectangle go the spectral values near the negative real axis. The spectral analysis near these values in $Z_{2}$ for dissipative Maxwell’s equations does not have clear analogue with the spectral problems for the wave equation with dissipative boundary conditions. In fact, for the wave equation if $0 < γ (x) < 1$ , $\forall x \in Γ$ , the eigenvalues of the generator of the corresponding semigroup are located in the domain $Λ_{ϵ}$ (see Section 3, [8] and [6]). For Maxwell’s equations the eigenvalues of $G_{b}$ lie in the domain $Λ_{ϵ} \cup R_{N}$ and for $0 < γ (x) < 1$ and $γ (x) > 1$ we have the same location (see Appendix for the case $K = {x \in R^{3} : | x | ⩽ 1}$ ).

Equation (1.3) implies that on Ω each eigenfunction $u = (E, B)$ of $G_{b}$ satisfies $\begin{matrix} (1.4) & \sqrt{z} E = \frac{h}{i} curl B, \sqrt{z} B = - \frac{h}{i} curl E, \end{matrix}$ and therefore $(- h^{2} Δ - z) E = (- h^{2} Δ - z) B = 0$ . For eigenfunctions $(E, B) \neq 0$ , we derive a pseudodifferential system on the boundary Γ involving $E_{\tan} = E - ⟨ E, ν ⟩ ν$ and $E_{nor} = ⟨ E, ν ⟩$ . A semi-classical analysis shows that for $z \in Z_{1} \cup Z_{3}$ this system implies that for h small enough we have $E |_{Γ} = 0$ which yields $E = B = 0$ . By scaling one concludes that the eigenvalues $λ = \frac{i \sqrt{z}}{h}$ of $G_{b}$ lie in the region $Λ_{ϵ} \cup M$ , where $\begin{matrix} M = {z \in C : | arg z - π | ⩽ π / 4, | z | ⩾ R_{0} > 0, Re z < 0} . \end{matrix}$ The strategy for the analysis of the case $z \in Z_{1} \cup Z_{3}$ is similar to that exploited in [9] and [8]. In these papers the semi-classical Dirichlet-to-Neumann map $N (z, h)$ plays a crucial role and the problem is reduced to the proof that some h-pseudodifferential operators is elliptic in a suitable class. For the Maxwell system the pseudodifferential equation on the boundary is more complicated. Using the equation $div E = 0$ , yields a pseudodifferential system for $E_{\tan}$ and $E_{nor}$ . We show that if $(E, B) \neq 0$ is an eigenfunction of $G_{b}$ , then $‖ E_{nor} ‖_{H_{h}^{1} (Γ)}$ is bounded by $C h ‖ E_{\tan} ‖_{H_{h}^{1} (Γ)}$ . The term involving $E_{nor}$ then plays the role of a negligible perturbation in the pseudodifferential system on the boundary and this reduces the analysis to one involving only $E_{\tan}$ . The system concerning $E_{\tan}$ has a diagonal leading term and we may apply the same arguments as those of [8] to conclude that $E_{\tan} = 0$ and hence $E_{nor} = 0$ .

The analysis of the case $z \in Z_{2}$ is more difficult since the principal symbol g of the pseudodifferential system for $E_{\tan}$ need not be elliptic at some points (see Section 3). Even where g is elliptic, if $| Im z | ⩽ h^{1 / 2}$ it is difficult to estimate the norm of the difference ${Op}_{h} (g) {Op}_{h} (g^{- 1}) - I$ . To show that the eigenvalues of $G_{b}$ lying in $M$ are in fact confined to the region $R_{N}$ for every $N \in N$ , we analyze the real part of the following scalar product in $L^{2} (Γ)$ $\begin{matrix} Q (z, E_{0}) : = Re {⟨ (N (z, h) - \sqrt{z} γ) E_{0}, E_{0} ⟩}_{L^{2} (Γ)}, E_{0} : = E |_{Γ} . \end{matrix}$ We follow the approach in [8,9] based on a Taylor expansion of $Q (z, E_{0})$ at $z = - 1$ and the fact that for $z = - 1$ we have $Q (- 1, E_{0}) = O (h^{N}), \forall N \in N$ . In the Appendix we treat the case when $K = {x \in R^{3} : | x | ⩽ 1}$ is a ball and $γ = const$ . We prove that for $γ \equiv 1$ the operator G has no eigenvalues in ${Re z < 0}$ , while for every $γ \in R^{+} ∖ {1}$ we have infinite number of real eigenvalues.

2. Pseudodifferential equation on the boundary

Introduce geodesic normal coordinates $(y_{1}, y^{'}) \in R^{3}$ on a neighborhood of a point $x_{0} \in Γ$ as follows. For a point x, $y^{'} (x)$ is the closest point in Γ and $y_{1} = dist (x, Γ)$ . Define $ν (x)$ to be the unit normal in the direction of increasing $y_{1}$ to the surface $y_{1} = constant$ through x. Thus $ν (x)$ is an extension of the unit normal vector to a unit vector field. The boundary Γ is mapped to $y_{1} = 0$ and $\begin{matrix} x = α (y_{1}, y^{'}) = β (y^{'}) + y_{1} ν (y^{'}) . \end{matrix}$ We have $\begin{matrix} \frac{\partial}{\partial x_{k}} = ν_{k} (y^{'}) \frac{\partial}{\partial y_{1}} + \sum_{j = 2}^{3} \frac{\partial y_{j}}{\partial x_{k}} \frac{\partial}{\partial y_{j}}, k = 1, 2, 3 . \end{matrix}$ Moreover, $\begin{matrix} \begin{matrix} \sum_{k = 1}^{3} ν_{k} (y^{'}) \frac{\partial y_{j}}{\partial x_{k}} (y_{1}, y^{'}) = ⟨ ν, \frac{\partial y_{j}}{\partial x} ⟩ = 0, j = 1, 2, 3, and \\ \sum_{k = 1}^{3} ν_{k} (x) \partial_{x_{k}} f (x) = \partial_{y_{1}} (f (α (y_{1}, y^{'}))) . \end{matrix} \end{matrix}$ Since $‖ ν (x) ‖ = 1$ , $⟨ ν, \partial_{x_{j}} ν ⟩ = 0$ , $j = 1, 2, 3$ .

A straight forward computation yields $\begin{matrix} \begin{matrix} ν (x) \land \frac{h}{i} curl u (x) & = i h \partial_{ν} u_{\tan} + (⟨ D_{x_{1}} u, ν ⟩, ⟨ D_{x_{2}} u, ν ⟩, ⟨ D_{x_{3}} u, ν ⟩) |_{\tan} \\ = i h \partial_{ν} u_{\tan} + ({grad}_{h} ⟨ u, ν ⟩) |_{\tan} - i h g_{0} (u_{\tan}), x \in Γ, \end{matrix} \end{matrix}$ where $\begin{matrix} \begin{matrix} D_{x_{j}} = - i h \partial_{x_{j}}, j = 1, 2, 3, {grad}_{h} f = {D_{x_{j}} f}_{j = 1, 2, 3}, \\ g_{0} (u_{\tan}) = {⟨ u_{\tan}, \partial_{x_{j}} ν ⟩}_{j = 1, 2, 3} . \end{matrix} \end{matrix}$ Setting $E_{nor} = ⟨ E, ν ⟩$ , from (1.3) one deduces $\begin{matrix} ν \land B = - \frac{1}{\sqrt{z}} ν \land \frac{h}{i} curl E = \frac{1}{\sqrt{z}} D_{ν} E_{\tan} - \frac{1}{\sqrt{z}} [({grad}_{h} E_{nor}) |_{\tan} - i h g_{0} (E_{\tan})], \end{matrix}$ where $D_{ν} = - i h \partial_{ν}$ and the boundary condition in (1.3) becomes $\begin{matrix} (2.1) & (D_{ν} - \frac{1}{γ} \sqrt{z}) E_{\tan} - ({grad}_{h} E_{nor}) |_{\tan} + i h g_{0} (E_{\tan}) = 0, x \in Γ . \end{matrix}$

Next $\begin{matrix} {grad}_{h} f (x) |_{\tan} = {\sum_{j = 2}^{3} \frac{\partial y_{j}}{\partial_{x_{k}}} D_{y_{j}} f (α (y_{1}, y^{'}))}_{k = 1, 2, 3} \end{matrix}$ and for $u = (u_{1}, u_{2}, u_{3}) \in C^{3}$ , $\begin{matrix} \begin{matrix} \frac{h}{i} div u (α (y_{1}, y^{'})) & = ⟨ D_{y_{1}} u (α (y_{1}, y^{'})), ν (y^{'}) ⟩ + \sum_{k = 1}^{3} \sum_{j = 2}^{3} \frac{\partial y_{j}}{\partial_{x_{k}}} D_{y_{j}} u_{k} (α (y_{1}, y^{'})) \\ = D_{y_{1}} (u_{nor} (y_{1}, y^{'})) + \sum_{j = 2}^{3} D_{y_{j}} ⟨ u_{\tan} (α (y_{1}, y^{'})), \frac{\partial y_{j}}{\partial x} ⟩ + h ⟨ u_{\tan}, Z ⟩, \end{matrix} \end{matrix}$ where $⟨ u (α (y_{1}, y^{'})), ν (y^{'}) ⟩ : = u_{nor} (y_{1}, y^{'})$ and Z depends on the second derivatives of $y_{j}$ , $j = 2, 3$ . Applying the operator $D_{y_{1}} - \frac{\sqrt{z}}{γ (y^{'})}$ to $div E (α (y_{1}, y^{'})) = 0$ , we find $\begin{matrix} \begin{matrix} (D_{y_{1}}^{2} - \frac{\sqrt{z}}{γ (y^{'})} D_{y_{1}}) E_{nor} (y_{1}, y^{'}) + \sum_{j = 2}^{3} D_{y_{j}} ⟨ (D_{y_{1}} - \frac{\sqrt{z}}{γ (y^{'})}) E_{\tan} (α (y_{1}, y^{'})), \frac{\partial y_{j}}{\partial x} ⟩ \\ = h ⟨ (D_{y_{1}} - \frac{\sqrt{z}}{γ}) E_{\tan}, Z ⟩ + h ⟨ E_{\tan}, Z_{1} ⟩, \end{matrix} \end{matrix}$ where $γ (y^{'}) : = γ (β (y^{'}))$ .

Taking the trace $y_{1} = 0$ and applying the boundary condition (2.1), yields $\begin{array}{rcl} (2.2) & \begin{matrix} (D_{y_{1}}^{2} + \sum_{j, μ = 2}^{3} \sum_{k = 1}^{3} \frac{\partial y_{j}}{\partial x_{k}} \frac{\partial y_{μ}}{\partial x_{k}} D_{y_{j}, y_{μ}}^{2}) E_{nor} (0, y^{'}) - \frac{\sqrt{z}}{γ (y^{'})} D_{y_{1}} E_{nor} (0, y^{'}) \\ = h ⟨ ({grad}_{h} E_{nor}) |_{\tan} (0, y^{'}), Z ⟩ + h Q_{1} (E_{\tan} (0, y^{'})), \end{matrix} \end{array}$ with $\begin{matrix} {‖ Q_{1} (E_{\tan} (0, y^{'})) ‖}_{L^{2} (R^{2})} ⩽ C_{2} {‖ E_{\tan} (0, y^{'}) ‖}_{H_{h}^{1} (R^{2})} . \end{matrix}$ Here $H_{h}^{s} (Γ)$ , $s \in R$ , denotes the semi-classical Sobolev spaces with norm $‖ {⟨ h \partial_{x} ⟩}^{s} u ‖_{L^{2} (Γ)}$ , $⟨ h \partial_{x} ⟩ = {(1 + ‖ h \partial_{x} ‖^{2})}^{1 / 2}$ . In the exposition below we use the spaces ${(L^{2} (Γ))}^{3}$ and ${(H_{h}^{s} (Γ))}^{3}$ of vector-valued functions but we will omit this in the notations writing simply $L^{2} (Γ)$ and $H_{h}^{s} (Γ)$ .

The operator $- h^{2} Δ_{x} - z$ in the coordinates $(y_{1}, y^{'})$ has the form $\begin{matrix} P (z, h) = D_{y_{1}}^{2} + r (y, D_{y^{'}}) + q_{1} (y, D_{y}) + h^{2} \tilde{q} - z \end{matrix}$ with $r (y, η^{'}) = ⟨ R (y) η^{'}, η^{'} ⟩$ , $q_{1} (y, η) = ⟨ q_{1} (y), η ⟩$ . Here $\begin{matrix} R (y) = {\sum_{k = 1}^{3} \frac{\partial y_{j}}{\partial x_{k}} \frac{\partial y_{μ}}{\partial x_{k}}}_{j, μ = 2}^{3} = {⟨ \frac{\partial y_{j}}{\partial x}, \frac{\partial y_{μ}}{\partial x} ⟩}_{j, μ = 2}^{3} \end{matrix}$ is a symmetric $(2 \times 2)$ matrix and $r (0, y^{'}, η^{'}) = r_{0} (y^{'}, η^{'})$ , where $r_{0} (y^{'}, η^{'})$ is the principal symbol of the Laplace–Beltrami operator $- h^{2} Δ_{Γ}$ on Γ equipped with the Riemannian metric induced by the Euclidean one in $R^{3}$ . We have $\begin{matrix} (P (z, h) E_{nor}) (0, y^{'}) = ⟨ P (z, h) E, ν ⟩ (0, y^{'}) + h Q_{2} (E (0, y^{'})), \end{matrix}$ where $\begin{matrix} {‖ Q_{2} (E (0, y^{'})) ‖}_{L^{2} (R^{2})} ⩽ C_{2} {‖ E (0, y^{'}) ‖}_{H_{h}^{1} (R^{2})} . \end{matrix}$ Since $P (z, h) E = 0$ , this lets us replace the terms with all second derivatives of $E_{nor}$ in (2.2) by $z E_{nor} (0, y^{'})$ modulo terms having a factor h and containing first order derivatives of $E_{nor}$ . This follows from the form of the matrix $R (y)$ given above. After a multiplication by $- \frac{γ (y^{'})}{\sqrt{z}}$ Eq. (2.2) yields $\begin{matrix} (2.3) & (D_{y_{1}} - γ (y^{'}) \sqrt{z}) E_{nor} (0, y^{'}) = h Q_{3} (E (0, y^{'})), \end{matrix}$ where $Q_{3} (E (0, y^{'}))$ has the same properties as $Q_{2} (E (0, y^{'}))$ .

Let $ψ (x) \in C_{0}^{\infty} (R^{3})$ be a cut-off function with support in small neighborhood of $x_{0} \in Γ$ . Replace $E, B$ by $E_{ψ} = E ψ$ , $B_{ψ} = B ψ$ . The above analysis works for $E_{ψ}$ and $B_{ψ}$ with lower order terms depending on ψ. We obtain $\begin{matrix} ⟨ (D_{ν} - γ (x) \sqrt{z}) E |_{Γ} ψ (x), ν (x) ⟩ = h Q_{3, ψ} (E |_{Γ}) . \end{matrix}$ Taking a partition of unity in a neighborhood of Γ, yields $\begin{matrix} (2.4) & ⟨ (D_{ν} - γ (x) \sqrt{z}) E |_{Γ}, ν ⟩ = h Q_{4} (E |_{Γ}), {‖ Q_{4} (E |_{Γ}) ‖}_{L^{2} (Γ)} ⩽ C ‖ E |_{Γ} ‖_{H_{h}^{1} (Γ)} . \end{matrix}$

For $z \in Z_{1} \cup Z_{2} \cup Z_{3}$ let $ρ (x^{'}, ξ^{'}, z) = \sqrt{z - r_{0} (x^{'}, ξ^{'})} \in C^{\infty} (T^{*} Γ)$ be the root of the equation $\begin{matrix} ρ^{2} + r_{0} (x^{'}, ξ^{'}) - z = 0 \end{matrix}$ with $Im ρ (x^{'}, ξ^{'}, z) > 0$ . For large $| ξ^{'} |$ , $\begin{matrix} ρ (x^{'}, ξ^{'}, z) \sim | ξ^{'} |, Im ρ (x^{'}, ξ^{'}, z) \sim | ξ^{'} |, \end{matrix}$ while for bounded $| ξ^{'} |$ , $\begin{matrix} Im ρ (x^{'}, ξ^{'}, z) ⩾ \frac{h^{δ}}{C} . \end{matrix}$

We recall some basic facts about h-pseudodifferential operators that the reader can find in [3]. Let X be a $C^{\infty}$ smooth compact manifold without boundary with dimension $d ⩾ 2$ . Let $(x, ξ)$ be the coordinates in $T^{*} (X)$ and let $a (x, ξ, h) \in C^{\infty} (T^{*} (X))$ . Given $m \in R$ , $l \in R$ , $δ > 0$ and a function $c (h) > 0$ , one denotes by $S_{δ}^{l, m} (c (h))$ the set of symbols so that $\begin{matrix} | \partial_{x}^{α} \partial_{ξ}^{β} a (x, ξ, h) | ⩽ C_{α, β} {(c (h))}^{- l - δ (| α | + | β |)} {(1 + | ξ |)}^{m - | β |}, \forall α, \forall β, (x, ξ) \in T^{*} (X) . \end{matrix}$ If $c (h) = h$ , we denote $S_{δ}^{l, m} (c (h))$ simply by $S_{δ}^{l, m}$ . Symbols restricted to a domain where $| ξ | ⩽ C$ will be denoted by $a \in S_{δ}^{l} (c (h))$ . The h-pseudodifferential operator with symbol $a (x, ξ, h)$ acts by $\begin{matrix} ({Op}_{h} (a) f) (x) : = {(2 π h)}^{- d + 1} \int_{T^{*} X} e^{- i ⟨ x - y, ξ ⟩ / h} a (x, ξ, h) f (y) d y d ξ . \end{matrix}$ For matrix valued symbols we use the same definition. This means that every element of a matrix symbol is in the class $S_{δ}^{l, m} (c (h))$ .

Now suppose that $a (x, ξ, h)$ satisfies the estimates $\begin{matrix} (2.5) & | \partial_{x}^{α} a (x, ξ, h) | ⩽ c_{0} (h) h^{- | α | / 2}, (x, ξ) \in T^{*} (X) \end{matrix}$ for $| α | ⩽ d - 1$ , where $c_{0} (h) > 0$ is a parameter. Then there exists a constant $C > 0$ independent of h such that $\begin{matrix} (2.6) & {‖ {Op}_{h} (a) ‖}_{L^{2} (X) \to L^{2} (X)} ⩽ C c_{0} (h) . \end{matrix}$ For $0 ⩽ δ < 1 / 2$ products of h-pseudodifferential operators are well behaved. If $a \in S_{δ}^{l_{1}, m_{1}}$ , $b \in S_{δ}^{l_{2}, m_{2}}$ and $s \in R$ , then $\begin{matrix} (2.7) & {‖ {Op}_{h} (a) {Op}_{h} (b) - {Op}_{h} (a b) ‖}_{H^{s} (X) \to H^{s - m_{1} - m_{2} + 1} (X)} ⩽ C h^{- l_{1} - l_{2} - 2 δ + 1} . \end{matrix}$

Let $u \in C^{3}$ be the solution of the Dirichlet problem $\begin{matrix} (2.8) & (- h^{2} Δ - z) u = 0 on Ω, u = F on Γ . \end{matrix}$ Introduce the semi-classical Dirichlet-to-Neumann map $\begin{matrix} N (z, h) : H_{h}^{s} (Γ) ∋ F ⟶ D_{ν} u |_{Γ} \in H_{h}^{s - 1} (Γ) . \end{matrix}$

G. Vodev [9] established for bounded domains $K \subset R^{d}$ , $d ⩾ 2$ , with $C^{\infty}$ boundary the following approximation of the interior Dirichlet-to-Neumann map $N_{int} (z, h)$ related to (2.8), where the equation $(- h^{2} Δ - z) u = 0$ is satisfied in K.

Theorem 2.1 ([9]).

For every $0 < ϵ ≪ 1$ there exists $0 < h_{0} (ϵ) ≪ 1$ such that for $z \in Z_{1, ϵ} : = {z \in Z_{1}, | Im z | ⩾ h^{\frac{1}{2} - ϵ}}$ and $0 < h ⩽ h_{0} (ϵ)$ we have $\begin{matrix} (2.9) & {‖ N_{int} (z, h) (F) - {Op}_{h} (ρ + h b) F ‖}_{H_{h}^{1} (Γ)} ⩽ \frac{C h}{\sqrt{| Im z |}} ‖ F ‖_{L^{2} (Γ)}, \end{matrix}$ where $b \in S_{0, 1}^{0} (Γ)$ does not depend on h and z. Moreover, ( 2.9 ) holds for $z \in Z_{2} \cup Z_{3}$ with $| Im z |$ replaced by 1.

With small modifications (2.9) holds for the Dirichlet-to-Neumann map $N (z, h)$ related to (2.8) (see [8]). Applying (2.9) with $N (z, h)$ and $F = E_{0} = E |_{Γ}$ , we obtain $\begin{matrix} (2.10) & {‖ ⟨ N (z, h) E_{0}, ν ⟩ - ⟨ {Op}_{h} (ρ) E_{0}, ν ⟩ ‖}_{L^{2} (Γ)} ⩽ \frac{C h}{\sqrt{| Im z |}} ‖ E_{0} ‖_{L^{2} (Γ)} . \end{matrix}$ Therefore (2.4) yields $\begin{matrix} (2.11) & ‖ ⟨ {Op}_{h} (ρ) - γ \sqrt{z}) E_{0}, ν ⟩ - h Q_{4} (E_{0}) ‖_{L^{2} (Γ)} ⩽ \frac{C h}{\sqrt{| Im z |}} ‖ E_{0} ‖_{L^{2} (Γ)} . \end{matrix}$ The commutator $[{Op}_{h} (ρ), ν (x)]$ is a pseudodifferential operator with symbol in $h^{1 - δ} S_{δ}^{0, 0}$ and so $\begin{matrix} {‖ [{Op}_{h} (ρ), ν_{k} (x)] E_{nor} ‖}_{H_{h}^{j} (Γ)} ⩽ C_{2} h^{1 - δ} ‖ E_{nor} ‖_{H_{h}^{j} (Γ)}, k = 1, 2, 3, j = 0, 1 . \end{matrix}$ The last estimate combined with (2.11) implies $\begin{matrix} (2.12) & {‖ ({Op}_{h} (ρ) - γ \sqrt{z}) E_{nor} - h Q_{4} (E_{0}) ‖}_{L^{2} (Γ)} ⩽ C_{3} (\frac{h}{| \sqrt{Im z} |} + h^{1 - δ}) ‖ E_{0} ‖_{L^{2} (Γ)} . \end{matrix}$

3. Eigenvalues-free regions

For $z \in Z_{1, ϵ}$ we have $ρ \in S_{δ}^{0, 1}$ with $0 < δ = 1 / 2 - ϵ < 1 / 2$ , while for $z \in Z_{2} \cup Z_{3}$ we have $ρ \in S_{0}^{0, 1}$ (see [9]). Since Γ is connected one has either $γ (x) > 1$ or $0 < γ (z) < 1$ . We present the analysis in the case where $0 < γ (x) < 1$ , $\forall x \in Γ$ . The case $1 < γ (x)$ is reduced to this case at the end of the section. Clearly, there exists $ϵ_{0} > 0$ such that $\begin{matrix} ϵ_{0} ⩽ γ (x) ⩽ 1 - ϵ_{0}, \forall x \in Γ . \end{matrix}$

Combing (2.4) and (2.9), yields $\begin{matrix} {‖ ⟨ ({Op}_{h} (ρ) - γ (x) \sqrt{z}) E_{0}, ν (x) ⟩ ‖}_{L^{2} (Γ)} ⩽ C \frac{h}{\sqrt{| Im z |}} ‖ E_{0} ‖_{L^{2} (Γ)} + C_{1} h ‖ E_{0} ‖_{H_{h}^{1} (Γ)}, \end{matrix}$ where for $z \in Z_{2} \cup Z_{3}$ we can replace $| Im z |$ by 1. This estimate for $E_{0}$ and the estimate for the commutator $[{Op}_{h} (ρ), ν_{k} (x)]$ imply $\begin{matrix} (3.1) & {‖ ({Op}_{h} (ρ) - γ (x) \sqrt{z}) E_{nor} ‖}_{L^{2} (Γ)} ⩽ \frac{C_{3} h}{\sqrt{| Im z |}} ‖ E_{0} ‖_{L^{2} (Γ)} + C_{4} h^{1 - δ} ‖ E_{0} ‖_{H_{h}^{1} (Γ)} . \end{matrix}$

Let $(x^{'}, ξ^{'})$ be coordinates on $T^{*} (Γ)$ . Consider the symbol $\begin{matrix} c (x^{'}, ξ^{'}, z) : = ρ (x^{'}, ξ^{'}, z) - γ (x^{'}) \sqrt{z}, x^{'} \in Γ . \end{matrix}$ Following the analysis in Section 3, [8], we know that c is elliptic in the case $0 < γ (x^{'}) < 1$ and if $z \in Z_{1}$ we have $c \in S_{δ}^{0, 1}$ , $| Im z | c^{- 1} \in S_{δ}^{0, - 1}$ , while if $z \in Z_{2} \cup Z_{3}$ one gets $c \in S_{0}^{0, 1}$ , $c^{- 1} \in S_{0}^{0, - 1}$ . This implies $\begin{matrix} {‖ {Op}_{h} (c^{- 1}) {Op}_{h} (c) E_{nor} ‖}_{H_{h}^{1} (Γ)} ⩽ \frac{C}{| Im z |} {‖ {Op}_{h} (c) E_{nor} ‖}_{L^{2} (Γ)} . \end{matrix}$ On the other hand, according to Section 7 in [3], the symbol of the operator ${Op}_{h} (c^{- 1}) {Op}_{h} (c) - I$ is given by $\begin{matrix} \begin{matrix} \sum_{j = 1}^{N} \frac{{(i h)}^{j}}{j!} \sum_{| α | = j} D_{ξ^{'}}^{α} (c^{- 1}) (x^{'}, ξ^{'}) D_{y^{'}}^{α} c (y^{'}, η^{'}) |_{x^{'} = y^{'}, ξ^{'} = η^{'}} + {\tilde{b}}_{N} (x^{'}, ξ^{'}) \\ : = b_{N} (x^{'}, ξ^{'}) + {\tilde{b}}_{N} (x^{'}, ξ^{'}), \end{matrix} \end{matrix}$ where $\begin{matrix} | \partial_{x^{'}}^{α} {\tilde{b}}_{N} (x^{'}, ξ^{'}) | ⩽ C_{α} h^{N (1 - 2 δ) - s_{d} - | α | / 2} . \end{matrix}$ Taking into account the estimates for $c^{- 1}$ and c, and applying (2.5), and (2.6) yields $\begin{matrix} {‖ ({Op}_{h} (c^{- 1}) {Op}_{h} (c) - I) E_{nor} ‖}_{H_{h}^{j} (Γ)} ⩽ C_{5} \frac{h}{| Im z |^{2}} ‖ E_{nor} ‖_{H_{h}^{j} (Γ)}, j = 0, 1 . \end{matrix}$ Repeating the argument in Section 3 in [8] concerning the case $0 < γ (x^{'}) < 1$ , for $z \in Z_{1}$ and $0 < δ < 1 / 2$ , one finds $\begin{matrix} (3.2) & \begin{matrix} ‖ E_{nor} ‖_{H_{h}^{1} (Γ)} & ⩽ {‖ ({Op}_{h} (c^{- 1}) {Op}_{h} (c) - I) E_{nor} ‖}_{H_{h}^{1} (Γ)} + {‖ {Op}_{h} (c^{- 1}) {Op}_{h} (c) E_{nor} ‖}_{H_{h}^{1} (Γ)} \\ ⩽ C_{6} h^{1 - 2 δ} ‖ E_{0} ‖_{L^{2} (Γ)} + C_{5} h^{1 - 2 δ} ‖ E_{nor} ‖_{H_{h}^{1} (Γ)} + C_{7} h^{1 - δ} ‖ E_{0} ‖_{H_{h}^{1} (Γ)} . \end{matrix} \end{matrix}$ Clearly, $\begin{matrix} ‖ E_{0} ‖_{H_{h}^{k} (Γ)} ⩽ ‖ E_{\tan} ‖_{H_{h}^{k} (Γ)} + B_{k} ‖ E_{nor} ‖_{H_{h}^{k} (Γ)}, k \in N \end{matrix}$ with $B_{k}$ independent of h. Hence we can absorb the terms involving the norms of $E_{nor}$ in the right hand side of (3.2) choosing h small enough, and we get $\begin{matrix} (3.3) & ‖ E_{nor} ‖_{H_{h}^{1} (Γ)} ⩽ C h^{1 - 2 δ} ‖ E_{\tan} ‖_{H_{h}^{1} (Γ)} . \end{matrix}$ The analysis of the case $z \in Z_{2} \cup Z_{3}$ is simpler since in the estimates above we have no coefficient $| Im z |^{- 1}$ and we obtain the same result with a factor h on the right hand side of (3.3).

With a similar argument it is easy to show that $\begin{matrix} (3.4) & ‖ E_{nor} ‖_{L^{2} (Γ)} ⩽ C^{'} h^{1 - 2 δ} ‖ E_{\tan} ‖_{L^{2} (Γ)} . \end{matrix}$ In fact from (2.12) one obtains $\begin{matrix} {‖ {Op}_{h} (c^{- 1}) [({Op}_{h} (ρ) - γ \sqrt{z}) E_{nor} - h Q_{4} (E_{0})] ‖}_{L^{2} (Γ)} ⩽ \frac{C_{8}}{| Im z |} (\frac{h}{\sqrt{| Im z |}} + h^{1 - δ}) ‖ E_{0} ‖_{L^{2} (Γ)} \end{matrix}$ and $\begin{matrix} {‖ {Op}_{h} (c^{- 1}) Q_{4} (E_{0}) ‖}_{L^{2} (Γ)} ⩽ \frac{C_{9}}{| Im z |} ‖ E_{0} ‖_{L^{2} (Γ)} . \end{matrix}$ Combining these estimates with the estimate of $‖ {Op}_{h} (c^{- 1}) {Op}_{h} (c) - I ‖_{L^{2} (Γ) \to L^{2} (Γ)}$ yields (3.4).

Going back to Eq. (2.1), we have $\begin{array}{rcl} (D_{ν} - \frac{1}{γ} \sqrt{z}) E & = & (D_{ν} - γ \sqrt{z}) E_{nor} ν - (\frac{1}{γ} - γ) \sqrt{z} E_{nor} ν \\ (3.5) & + i h g_{0} (E_{\tan}) + ({grad}_{h} (E_{nor})) |_{tan}, x \in Γ . \end{array}$ Notice that for the first term on the right hand side of (3.5) we can apply the equality (2.4), while for $E_{nor}$ and $({grad}_{h} (E_{nor})) |_{tan}$ we have a control by the estimate (3.3). Consequently, setting $E_{0} = E |_{Γ}$ , the right hand side of (3.5) is bounded by $C h^{1 - 2 δ} ‖ E_{0} ‖_{H_{h}^{1} (Γ)}$ . Next $\begin{matrix} 1 < \frac{1}{1 - ϵ_{0}} ⩽ \frac{1}{γ (x)} ⩽ \frac{1}{ϵ_{0}}, \forall x \in Γ . \end{matrix}$ This corresponds to the case (B) examined in Section 4 of [8]. The approximation of the operator $N (z, h)$ given by (2.9) yields the estimate $\begin{matrix} (3.6) & {‖ ({Op}_{h} (ρ) - \frac{1}{γ} \sqrt{z}) E_{0} ‖}_{L^{2} (Γ)} ⩽ C (\frac{h}{\sqrt{| Im z |}} ‖ E_{0} ‖_{L^{2} (Γ)} + h^{1 - 2 δ} ‖ E_{0} ‖_{H_{h}^{1} (Γ)}) . \end{matrix}$ For $z \in Z_{1} \cup Z_{3}$ the symbol $\begin{matrix} d (x^{'}, ξ^{'}, z) : = ρ (x^{'}, ξ^{'}, z) - \frac{1}{γ (x^{'})} \sqrt{z} \end{matrix}$ is elliptic (see Section 4, [8]) and $d \in S_{δ}^{0, 1}$ , $d^{- 1} \in S_{δ}^{0, - 1}$ . Then from (3.6) we estimate $‖ E_{0} ‖_{H_{h}^{1} (Γ)}$ and obtain $E_{0} = 0$ for h small enough. This implies $E = B = 0$ .

Now recall that we have $\begin{matrix} Re λ = - \frac{Im \sqrt{z}}{h}, Im λ = \frac{Re \sqrt{z}}{h} . \end{matrix}$ Suppose that $z \in Z_{1}$ . Then $\begin{matrix} | Re λ | ⩾ C {(h^{- 1})}^{1 - δ}, | Im λ | ⩽ C_{1} h^{- 1} ⩽ C_{2} | Re λ |^{\frac{1}{1 - δ}} . \end{matrix}$ So if $\begin{matrix} | Re λ | ⩾ C_{3} | Im λ |^{1 - δ}, Re λ ⩽ - C_{4} < 0, \end{matrix}$ there are no eigenvalues $λ = \frac{i \sqrt{z}}{h}$ of $G_{b}$ . In the same way we handle the case $z \in Z_{3}$ and we conclude that if $z \in Z_{1} \cup Z_{3}$ for every $ϵ > 0$ the eigenvalues $λ = \frac{i \sqrt{z}}{h}$ of $G_{b}$ lie in the domain $Λ_{ϵ} \cup M$ , where $\begin{matrix} M = {z \in C : | arg z - π | ⩽ π / 4, | z | ⩾ R_{0} > 0, Re z < 0}, \end{matrix}$ $Λ_{ϵ}$ being the domain introduced in Theorem 1.1. Of course, if we consider the domain $\begin{matrix} Z_{3, δ_{0}} = {z \in C : | Re z | ⩽ 1, Im z = δ_{0} > 0}, \end{matrix}$ instead of $Z_{3}$ , we obtain an eigenvalue-free region with $M$ replaced by $\begin{matrix} M_{δ_{0}} = {z \in C : | arg z - π | ⩽ arctg δ_{0}, | z | ⩾ R_{0} (δ_{0}) > 0, Re z < 0} . \end{matrix}$

The investigation of the case $z \in Z_{2}$ is more complicated since the symbol d may vanish for $Im z = 0$ and $(x_{0}^{'}, ξ_{0}^{'}) \in T^{*} (Γ)$ satisfying the equation $\begin{matrix} \sqrt{1 + r_{0} (x_{0}^{'}, ξ_{0}^{'})} - \frac{1}{γ (x_{0}^{'})} = 0 . \end{matrix}$

To cover this case and to prove that the eigenvalues $λ = \frac{i \sqrt{z}}{h}$ with $z \in Z_{2}$ are confined in the domain $R_{N}$ , $\forall N \in N$ , we follow the arguments in [9] and [8]. For $z \in Z_{2}$ we introduce an operator $T (z, h)$ that yields a better approximation of $N (z, h)$ . In fact, $T (z, h)$ is defined by the construction of the semi-classical parametrix in Section 3, [9] for the problem (2.8) with $F = E_{0}$ . We refer to [9] for the precise definition of $T (z, h)$ and more details. For our exposition we need the next proposition. Since $(Δ - z) E = 0$ , as in [9], we obtain

Proposition 3.1.

For $z \in Z_{2}$ and every $N \in N$ we have the estimate $\begin{matrix} (3.7) & {‖ N (z, h) E_{0} - T (z, h) E_{0} ‖}_{H_{h}^{1} (Γ)} ⩽ C_{N} h^{- s_{0}} h^{N} ‖ E_{0} ‖_{L^{2} (Γ)} \end{matrix}$ with constants $C_{N}, s_{0} > 0$ , independent of $E_{0}, h$ and z, and $s_{0}$ independent of N.

Proof of Theorem 1.1 in the case

z \in Z_{2}

Consider the system $\begin{matrix} (3.8) & \{\begin{matrix} (D_{ν} - \frac{1}{γ} \sqrt{z}) E_{\tan} - ({grad}_{h} E_{nor}) |_{\tan} + i h g_{0} (E_{\tan}) = 0, & x \in Γ, \\ {div}_{h} E_{\tan} + {div}_{h} (E_{nor} ν) = 0, & x \in Γ, \end{matrix} \end{matrix}$ where ${div}_{h} F = \sum_{k = 1}^{3} D_{x_{k}} F_{k}$ .

Take the scalar product ${⟨, ⟩}_{L^{2} (Γ)}$ in $L^{2} (Γ)$ of the first equation of (3.8) and $E_{\tan}$ . Applying Green formula, it easy to see that $\begin{matrix} (3.9) & - Re {⟨ {grad}_{h} E_{nor} |_{\tan}, E_{\tan} ⟩}_{L^{2} (Γ)} = - Re {⟨ {div}_{h} E_{\tan}, E_{nor} ⟩}_{L^{2} (Γ)} . \end{matrix}$ We claim that $\begin{matrix} (3.10) & Im {⟨ g_{0} (E_{\tan}), E_{\tan} ⟩}_{L^{2} (Γ)} = 0 . \end{matrix}$ Let $E_{\tan} = (w_{1}, w_{2}, w_{3})$ . Then $\begin{matrix} {⟨ g_{0} (E_{\tan}), E_{\tan} ⟩}_{C^{3}} = \sum_{k, j = 1}^{3} w_{k} \frac{\partial ν_{k}}{\partial x_{j}} \overline{w_{j}} = \frac{1}{q} \sum_{k, j = 1}^{3} w_{k} \frac{\partial V_{k}}{\partial x_{j}} \overline{w_{j}} = \frac{1}{q} {⟨ S w, w ⟩}_{C^{3}}, \end{matrix}$ where $S : = {\frac{\partial V_{k}}{\partial x_{j}}}_{k, j = 1}^{3}$ with $V (x) = q (x) ν (x)$ , $q (x) > 0$ because $\sum_{k = 1}^{3} (\partial_{x_{j}} q) w_{k} ν_{k} = 0$ . Thus if the boundary is given locally by $x_{3} = G (x_{1}, x_{2})$ , we choose $V (x) = (- \partial_{x_{1}} G, - \partial_{x_{2}} G, 1)$ and it is obvious that S is symmetric. Therefore $Im {⟨ S w, w ⟩}_{C^{3}} = 0$ and this proves the claim. Hence (3.10) implies $\begin{matrix} (3.11) & Re [i h {⟨ g_{0} (E_{\tan}), E_{\tan} ⟩}_{L^{2} (Γ)}] = 0 . \end{matrix}$ From the $L^{2} (Γ)$ scalar product of the second equation in (3.8) with $E_{nor}$ , we obtain $\begin{matrix} (3.12) & Re {⟨ {div}_{h} E_{\tan}, E_{nor} ⟩}_{L^{2} (Γ)} + Re {⟨ D_{ν} E_{nor}, E_{nor} ⟩}_{L^{2} (Γ)} = 0 . \end{matrix}$ In fact, $\begin{matrix} {div}_{h} (E_{nor} ν) = D_{ν} E_{nor} - i h E_{nor} div ν \end{matrix}$ and $Im (div ν | E_{nor} |^{2}) = 0$ .

Taking together (3.9), (3.11) and (3.12), we conclude that $\begin{matrix} \begin{matrix} Re [{⟨ (D_{ν} - \frac{\sqrt{z}}{γ}) E_{\tan}, E_{\tan} ⟩}_{L^{2} (Γ)} + {⟨ D_{ν} E_{nor} ν, E_{nor} ν ⟩}_{L^{2} (Γ)}] \\ = Re {⟨ D_{ν} E, E ⟩}_{L^{2} (Γ)} - Re {⟨ \frac{\sqrt{z}}{γ} E_{\tan}, E_{\tan} ⟩}_{L^{2} (Γ)} = 0 . \end{matrix} \end{matrix}$

Here we have used the fact that $\begin{matrix} {⟨ D_{ν} E_{\tan}, E_{nor} ν ⟩}_{C^{3}} = D_{ν} ({⟨ E_{\tan}, E_{nor} ν ⟩}_{C^{3}}) = 0 . \end{matrix}$ Applying Proposition 3.1 with $E |_{Γ} = E_{0}$ , yields $\begin{matrix} (3.13) & | Re {⟨ T (z, h) E_{0}, E_{0} ⟩}_{L^{2} (Γ)} - Re {⟨ \frac{\sqrt{z}}{γ} E_{\tan}, E_{\tan} ⟩}_{L^{2} (Γ)} | ⩽ C_{N} h^{- s_{0}} h^{N} ‖ E_{0} ‖_{L^{2} (Γ)} . \end{matrix}$ For $z = - 1$ , as in Lemma 3.9 in [9] and Lemma 4.1 in [8], we have $\begin{matrix} | Re {⟨ T (- 1, h) E_{0}, E_{0} ⟩}_{L^{2} (Γ)} | ⩽ C_{N} h^{- s_{0} + N} ‖ E_{0} ‖_{L^{2} (Γ)}^{2} = 0 . \end{matrix}$ Consequently, by using Taylor formula for the real-valued function $\begin{matrix} Re [{⟨ T (z, h) E_{0}, E_{0} ⟩}_{L^{2} (Γ)} - {⟨ \frac{\sqrt{z}}{γ} E_{\tan}, E_{\tan} ⟩}_{L^{2} (Γ)}], \end{matrix}$ we get for every $N \in N$ the estimate $\begin{array}{rcl} (3.14) & \begin{matrix} | Im [{⟨ (\frac{\partial T}{\partial z} (z_{t}, h)) E_{0}, E_{0} ⟩}_{L^{2} (Γ)} - {⟨ \frac{γ_{1}}{2 \sqrt{z_{t}}} E_{\tan}, E_{\tan} ⟩}_{L^{2} (Γ)}] | \\ ⩽ C_{N} \frac{h^{- s_{0} + N}}{| Im z |} ‖ E_{0} ‖_{L^{2} (Γ)}^{2}, \end{matrix} \end{array}$ where $z_{t} = - 1 + i t Im z$ , $0 < t < 1$ , $γ_{1} = γ^{- 1}$ .

According to Lemma 3.9 in [9], in (3.14) we can replace $\frac{\partial T}{\partial z} (z_{t}, h)$ by ${Op}_{h} (\frac{\partial ρ}{\partial z} (z_{t}))$ and this yields an error term bounded by $C h ‖ E_{0} ‖_{H_{h}^{- 1} (Γ)}^{2}$ . On the other hand, $\begin{matrix} \begin{matrix} | {⟨ {Op}_{h} (\frac{\partial ρ}{\partial z} (z_{t})) E_{\tan}, E_{nor} ν ⟩}_{L^{2} (Γ)} + {⟨ {Op}_{h} (\frac{\partial ρ}{\partial z} (z_{t})) E_{nor}, E_{\tan} ν ⟩}_{L^{2} (Γ)} | \\ ⩽ C h ‖ E_{0} ‖_{L^{2} (Γ)}^{2} \end{matrix} \end{matrix}$ since the estimate (3.4) holds for $z \in Z_{2}$ with factor h and $\frac{\partial ρ}{\partial z} (z_{t}) \in S_{0}^{0, - 1}$ .

Thus the problem is reduced to a lower bound of $\begin{matrix} \begin{matrix} J & : = | Im [{⟨ ({Op}_{h} (\frac{\partial ρ}{\partial z} (z_{t})) - \frac{γ_{1}}{2 \sqrt{z}}) E_{\tan}, E_{\tan} ⟩}_{L^{2} (Γ)} + {⟨ {Op}_{h} (\frac{\partial ρ}{\partial z} (z_{t})) E_{nor} ν, E_{nor} ν ⟩}_{L^{2} (Γ)}] | \\ ⩾ | Im {⟨ ({Op}_{h} (\frac{\partial ρ}{\partial z} (z_{t})) - \frac{γ_{1}}{2 \sqrt{z}}) E_{\tan}, E_{\tan} ⟩}_{L^{2} (Γ)} | - C_{1} ‖ E_{nor} ‖_{L^{2} (Γ)}^{2} . \end{matrix} \end{matrix}$ Since $γ_{1} (x) > 1$ , $\forall x \in Γ$ , applying the analysis of Section 4 in [8] for the scalar product involving $E_{\tan}$ , one deduces $\begin{matrix} | Im {⟨ ({Op}_{h} (\frac{\partial ρ}{\partial z} (z_{t})) - \frac{γ_{1}}{2 \sqrt{z}}) E_{\tan}, E_{\tan} ⟩}_{L^{2} (Γ)} | ⩾ η_{1} ‖ E_{\tan} ‖_{L^{2} (Γ)}^{2}, η_{1} > 0 . \end{matrix}$ By using once more the estimate (3.4), for h small enough we obtain $\begin{matrix} J ⩾ η_{1} (‖ E_{\tan} ‖_{L^{2} (Γ)}^{2} + ‖ E_{nor} ‖_{L^{2} (Γ)}^{2}) - B_{0} h ‖ E_{\tan} ‖_{L^{2} (Γ)}^{2} ⩾ η_{2} ‖ E_{0} ‖_{L^{2} (Γ)}^{2}, 0 < η_{2} < η_{1} . \end{matrix}$ Consequently, (3.14) yields $\begin{matrix} (η_{2} - B_{1} h) ‖ E_{0} ‖_{L^{2} (Γ)}^{2} ⩽ C_{N} \frac{h^{- s_{0} + N}}{| Im z |} ‖ E_{0} ‖_{L^{2} (Γ)}^{2} \end{matrix}$ and for small h we conclude that for $z \in Z_{2}$ the eigenvalues $λ = \frac{i \sqrt{z}}{h}$ of $G_{b}$ lie in the region $R_{N}$ . This completes the analysis of the case $0 < γ (x) < 1$ , $\forall x \in Γ$ .

To study the case $γ (x) > 1$ , $\forall x \in Γ$ , we write the boundary condition in (1.1) as $\begin{matrix} \frac{1}{γ (x)} (ν \land E_{\tan}) - (ν \land (ν \land B_{\tan})) = \frac{1}{γ (x)} (ν \land E_{\tan}) + B_{tan} = 0 . \end{matrix}$ Next $\begin{matrix} ν \land E = \frac{1}{\sqrt{z}} ν \land \frac{h}{i} curl B = - \frac{1}{\sqrt{z}} D_{ν} B_{\tan} + \frac{1}{\sqrt{z}} [({grad}_{h} B_{nor}) |_{\tan} - i h g_{0} (B_{\tan})] \end{matrix}$ and one obtains $\begin{matrix} (3.15) & (D_{ν} - γ (x) \sqrt{z}) B_{\tan} - ({grad}_{h} B_{nor}) |_{\tan} + i h g_{0} (B_{\tan}) = 0, x \in Γ, \end{matrix}$ which is the same as (2.1) with $E_{\tan}$ , $E_{nor}$ replaced respectively by $B_{\tan}$ , $B_{nor}$ and $\frac{1}{γ (x)}$ replaced by $γ (x) > 1$ . We apply the operator $D_{y_{1}} - γ \sqrt{z}$ to the equation $div B = 0$ and repeat without any change the above analysis concerning $E_{\tan}$ , $E_{nor}$ . Thus the proof of Theorem 1.1 is complete. □

Remark 3.2.

The result of Theorem 1.1 holds for obstacles $K = ⋃_{j = 1}^{J} K_{j}$ , where $K_{j}$ , $j = 1, \dots, J$ are open connected domains with $C^{\infty}$ boundary and $K_{i} \cap K_{j} = \emptyset$ , $i \neq j$ . Let $Γ_{j} = \partial K_{j}$ , $j = 1, \dots, J$ . In this case we may have $γ (x) < 1$ for some obstacles $Γ_{j}$ and $γ (x) > 1$ for other ones. The proof extends with only minor modifications. The construction of the semi-classical parametrix in [9] is local and for the Dirichlet-to-Neumann map $N_{j} (z, h)$ related to $Γ_{j}$ we get the estimate $\begin{matrix} {‖ N_{j} (z, h) (F) - {Op}_{h} (ρ + h b) F ‖}_{H_{h}^{1} (Γ_{j})} ⩽ \frac{C h}{\sqrt{| Im z |}} ‖ F ‖_{L^{2} (Γ_{j})} . \end{matrix}$ The boundary condition in (1.1) is local and we can reduce the analysis to a fixed obstacle $K_{j}$ . If $(E, B) \neq 0$ is an eigenfunction of $G_{b}$ , our argument implies $E_{\tan} = 0$ for $x \in Γ_{j}$ if $0 < γ (x) < 1$ on $Γ_{j}$ and $B_{\tan} = 0$ for $x \in Γ_{j}$ in the case $γ (x) > 1$ on $Γ_{j}$ . By the boundary condition we get $E_{\tan} = 0$ on Γ and this yields $E = B = 0$ since the Maxwell system with boundary condition $E_{\tan} = 0$ has no eigenvalues in ${z \in C : Re z < 0}$ .

Footnotes

Acknowledgements

We thank Georgi Vodev for many useful discussions and remarks concerning an earlier version of the paper.

Vesselin Petkov was partially supported by the ANR project Nosevol BS01019 01.

In this appendix, assume that $γ > 0$ is constant. Our purpose is to study the eigenvalues of $G_{b}$ in case the obstacle is equal to the ball $B_{3} = {x \in R^{3} : | x | ⩽ 1}$ . Setting $λ = i μ$ , $Im μ > 0$ , an eigenfunction $(E, B) \neq 0$ of $G_{b}$ satisfies $\begin{matrix} (A.1) & curl E = - i μ B, curl B = i μ E . \end{matrix}$ Replacing B by $H = - B$ yields for $(E, H) \in {(H^{2} (| x | ⩽ 1))}^{6}$ , $\begin{matrix} (A.2) & \{\begin{matrix} curl E = i μ H, curl H = - i μ E, & for x \in B_{3}, \\ E_{\tan} + γ (ν \land H_{\tan}) = 0, & for x \in S^{2} . \end{matrix} \end{matrix}$ Expand $E (x)$ , $H (x)$ in the spherical functions $Y_{n}^{m} (ω)$ , $n = 0, 1, 2, \dots, | m | ⩽ n$ , $ω \in S^{2}$ and the modified Hankel functions $h_{n}^{(1)} (z)$ of first kind. An application of Theorem 2.50 in [4] (in the notation of [4] it is necessary to replace ω by $μ \in C ∖ {0}$ ) says that the solution of the system (A.2) for $| x | = r = 1$ has the form $\begin{matrix} \begin{matrix} E_{\tan} (ω) = \sum_{n = 1}^{\infty} \sum_{| m | ⩽ n} [α_{n}^{m} (h_{n}^{(1)} (μ) + \frac{d}{d r} h_{n}^{(1)} (μ r) |_{r = 1}) U_{n}^{m} (ω) + β_{n}^{m} h_{n}^{(1)} (μ) V_{n}^{m} (ω)], \\ H_{\tan} (ω) = - \frac{1}{i μ} \sum_{n = 1}^{\infty} \sum_{| m | ⩽ n} [β_{n}^{m} (h_{n}^{(1)} (μ) + \frac{d}{d r} h_{n}^{(1)} (μ r) |_{r = 1}) U_{n}^{m} (ω) + μ^{2} α_{n}^{m} h_{n}^{(1)} (μ) V_{n}^{m} (ω)] . \end{matrix} \end{matrix}$

Here $U_{n}^{m} (ω) = \frac{1}{\sqrt{n (n + 1)}} {grad}_{S^{2}} Y_{n}^{m} (ω)$ and $V_{n}^{m} (ω) = ν \land U_{n}^{m} (ω)$ for $n \in N, - n ⩽ m ⩽ n$ form a complete orthonormal basis in $\begin{matrix} L_{t}^{2} (S^{2}) = {u \in {(L^{2} (S^{2}))}^{3} : ⟨ ν, u ⟩ = 0 on S^{2}} . \end{matrix}$ To find a representation of $ν \land H_{\tan}$ , observe that $ν \land (ν \land U_{n}^{m}) = - U_{n}^{m}$ , so $\begin{matrix} (ν \land H_{\tan}) (ω) = - \frac{1}{i μ} \sum_{n = 1}^{\infty} \sum_{| m | ⩽ n} [β_{n}^{m} (h_{n}^{(1)} (μ) + \frac{d}{d r} h_{n}^{(1)} (μ r) |_{r = 1}) V_{n}^{m} (ω) - μ^{2} α_{n}^{m} h_{n}^{(1)} (μ) U_{n}^{m} (ω)] \end{matrix}$ and the boundary condition in (A.2) is satisfied if $\begin{array}{l} (A.3) & α_{n}^{m} [h_{n}^{(1)} (μ) + \frac{d}{d r} (h_{n}^{(1)} (μ r)) |_{r = 1} - γ i μ h_{n}^{(1)} (μ)] = 0, \forall n \in N, | m | ⩽ n, \\ (A.4) & - \frac{β_{n}^{m} γ}{i μ} [h_{n}^{(1)} (μ) + \frac{d}{d r} (h_{n}^{(1)} (μ r)) |_{r = 1} - \frac{i μ}{γ} h_{n}^{(1)} (μ)] = 0, \forall n \in N, | m | ⩽ n . \end{array}$ For $γ \equiv 1$ , there are no eigenvalues.

For the case $γ \neq 1$ , there are an infinite number of real eigenvalues.

It is easy to see that for $γ > 1$ the equation $g_{n} (w) = 0$ has no complex roots. Denote by $\begin{matrix} z_{j}, Re z_{j} < 0, j = 1, \dots, n, n ⩾ 1 \end{matrix}$ the roots of $R_{n} (w) = 0$ . Suppose that $g_{n} (w_{0}) = 0$ , $n ⩾ 1$ with $Re w_{0} > 0$ , $Im w_{0} \neq 0$ . Then $\begin{matrix} Im [\frac{1 - γ}{2 w_{0}} + w_{0} \sum_{j = 1}^{n} \frac{1}{w_{0} - z_{j}}] = 0 \end{matrix}$ and $\begin{array}{l} - \frac{(1 - γ) Im w_{0}}{2 | w_{0} |^{2}} + Re w_{0} [- \sum_{j = 1}^{n} \frac{Im w_{0}}{| w_{0} - z_{j} |^{2}} + \sum_{j = 1}^{n} \frac{Im z_{j}}{| w_{0} - z_{j} |^{2}}] \\ (A.10) & + Im w_{0} \sum_{j = 1}^{n} \frac{Re w_{0} - Re z_{j}}{| w_{0} - z_{j} |^{2}} = 0 . \end{array}$

On the other hand, if $z_{j}$ with $Im z_{j} \neq 0$ is a root of $R_{n} (w) = 0$ , then ${\bar{z}}_{j}$ is also a root and $\begin{array}{l} \frac{Im z_{j}}{| w_{0} - z_{j} |^{2}} - \frac{Im z_{j}}{| w_{0} - {\bar{z}}_{j} |^{2}} & = \frac{Im z_{j}}{| w_{0} - z_{j} |^{2} | w_{0} - {\bar{z}}_{j} |^{2}} (| w_{0} - {\bar{z}}_{j} |^{2} - | w_{0} - z_{j} |^{2}) \\ = \frac{4 Im w_{0} {(Im z_{j})}^{2}}{| w_{0} - z_{j} |^{2} | w_{0} - {\bar{z}}_{j} |^{2}} . \end{array}$

Equation (A.10) becomes $\begin{matrix} (A.11) & Im w_{0} [\frac{γ - 1}{2 | w_{0} |^{2}} - \sum_{j = 1}^{n} \frac{Re z_{j}}{| w_{0} - z_{j} |^{2}} + \sum_{Im z_{j} > 0} \frac{4 Re w_{0} {(Im z_{j})}^{2}}{| w_{0} - z_{j} |^{2} | w_{0} - {\bar{z}}_{j} |^{2}}] = 0 . \end{matrix}$ The term in the brackets $[\dots]$ is positive, and one concludes that $Im w_{0} = 0$ .

Repeating the argument of the Appendix in [8], one can show that for $0 < γ < 1$ the complex eigenvalues of $G_{b}$ lie in the region $\begin{matrix} {z \in C : | arg z - π | > π / 4, Re z < 0} . \end{matrix}$

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