Philippe Laurenc ¸ot

Christoph Walker
email: walker@ifam.uni-hannover.de

H 2 -REGULARITY FOR A TWO-DIMENSIONAL TRANSMISSION PROBLEM WITH GEOMETRIC CONSTRAINT

Keywords: Mathematics Subject Classification. 35B65 -35J25 -35J20. Key words and phrases. transmission problem, regularity, non-Lipschitz domain

The H 2 -regularity of variational solutions to a two-dimensional transmission problem with geometric constraint is investigated, in particular when part of the interface becomes part of the outer boundary of the domain due to the saturation of the geometric constraint. In such a situation, the domain includes some non-Lipschitz subdomains with cusp points, but it is shown that this feature does not lead to a regularity breakdown. Moreover, continuous dependence of the solutions with respect to the domain is established.

Introduction

The H 2 -regularity of variational solutions to a two-dimensional transmission problem with geometric constraint is investigated, in particular when part of the interface becomes part of the outer boundary of the domain due to the geometric constraint, a situation in which the domain includes some non-Lipschitz subdomains with cusp points.

To set up the geometric framework, let D := (-L, L) be a finite interval of R, L > 0, and let H > 0 and d > 0 be two positive parameters. Given a function u ∈ C( D, [-H, ∞)) with u(±L) = 0, we define the subdomain Ω(u) of D × (-H, ∞) by Ω(u) := {(x, z) ∈ D × R : -H < z < u(x) + d} = Ω 1 (u) ∪ Ω 2 (u) ∪ Σ(u) ,

where Ω 1 (u) := {(x, z) ∈ D × R : -H < z < u(x)} and Ω 2 (u) := {(x, z) ∈ D × R : u(x) < z < u(x) + d} are separated by the interface Σ(u) := {(x, z) ∈ D × R : z = u(x) > -H} .

Owing to the (geometric) constraint u ≥ -H, the lower boundary of Ω 2 (u), given by the graph of the function u, cannot go beyond the lower boundary D × {-H} of Ω 1 (u) but may coincide partly with it, along the so-called coincidence set

C(u) := {x ∈ D : u(x) = -H} , (1.1)
see Figures 1 and2. Clearly, the geometry of Ω(u), as well as the regularity of its boundary, heavily depends on whether min D {u} > -H or min D {u} = -H. Indeed, if min D {u} > -H (i.e. the graph of u is strictly separated from D × {-H} as in Figure 1), then the coincidence set C(u) is empty and Ω 1 (u) is connected. In contrast, if min

D {u} = -H so v Ω 1 (v) Ω 2 (v) D Σ(v) z -H 0 d -L L Figure 1.
Geometry of Ω(v) for a state v ∈ S with empty coincidence set.

w that the graph of u intersects D × {-H}, then C(u) = ∅ and Ω 1 (u) is disconnected with at least two (and possibly infinitely many) connected components, see Figures 2 and3.

Ω 1 (w) Ω 2 (w) D Σ(w) z -H 0 d -L L C(w)
For such a geometry, we study the regularity of variational solutions to the transmission problem div(σ∇ψ u ) = 0 in Ω(u) ,

(1.2a)

ψ u = σ∇ψ u • n Σ(u) = 0 on Σ(u) , (1.2b)
ψ u = h u on ∂Ω(u) , (1.2c)
where

σ := σ 1 1 Ω 1 (u) + σ 2 1 Ω 2 (u)
for some positive constants σ 1 = σ 2 , and n Σ(u) denotes the unit normal vector field to Σ(u) (pointing into Ω 2 (u)) given by n Σ(u) := (-∂ x u, 1)

1 + (∂ x u) 2 .

In (1.2c), h u is a suitable function reflecting the boundary behavior of ψ u , see Section 2 for details. In addition, • denotes the (possible) jump across the interface Σ(u); that is,

f (x, u(x)) := f | Ω 1 (u) (x, u(x)) -f | Ω 2 (u) (x, u(x)) , x ∈ D ,
whenever meaningful for a function f : Ω(u) → R.

Let us already mention that there are several features of the specific geometry of Ω(u) which may hinder the H 2 -regularity of the solution ψ u to (1.2). Indeed, on the one hand, the interface Σ(u) always intersects with the boundary ∂Ω(u) of Ω(u) and it follows from [START_REF] Lemrabet | Régularité de la solution d'un problème de transmission[END_REF] that this sole property prevents the H 2 -regularity of ψ u , unless σ and the angles between Σ(u) and ∂Ω(u) at the intersection points satisfy some additional conditions. On the other hand, Ω(u) and Ω 2 (u) are at best Lipschitz domains, while Ω 1 (u) may consist of non-Lipschitz domains with cusp points.

The particular geometry Ω(u) = Ω 1 (u) ∪ Ω 2 (u) ∪ Σ(u), in which the boundary value problem (1.2) is set, is encountered in the investigation of an idealized electrostatically actuated microelectromechanical system (MEMS) as described in detail in [START_REF] Laurenc | Heterogeneous dielectric properties in models for microelectromechanical systems[END_REF]. Such a device consists of an elastic plate of thickness d which is fixed at its boundary {±L}×(0, d) and suspended above a rigid conducting ground plate located at z = -H. The elastic plate is made up of a dielectric material and deformed by a Coulomb force induced by holding the ground plate and the top of the elastic plate at different electrostatic potentials. In this context, u represents the vertical deflection of the bottom of the elastic plate, so that the elastic plate is given by Ω 2 (u), while Ω 1 (u) denotes the free space between the elastic plate and the ground plate. An important feature of the model is that the elastic plate cannot penetrate the ground plate, resulting on the geometric constraint u ≥ -H. Still, a contact between the elastic plate and the ground plate -corresponding to a non-empty coincidence set C(u) -is explicitly allowed. The dielectric properties of Ω 1 (u) and Ω 2 (u) are characterized by positive constants σ 1 and σ 2 , respectively. The electrostatic potential ψ u is then supposed to satisfy (1.2) and is completely determined by the deflection u. The state of the MEMS device is thus described by the deflection u, and equilibrium configurations of the device are obtained as critical points of the total energy which is the sum of the mechanical and electrostatic energies, the former being a functional of u while the latter is the Dirichlet integral of ψ u . Owing to the nonlocal dependence of ψ u on u, minimizing the total energy and deriving the associated Euler-Lagrange equation demand quite precise information on the regularity of the electrostatic potential ψ u for an arbitrary, but fixed function u and its continuous dependence thereon. This first step of provisioning the required information is the main purpose of the present research, and we refer to the forthcoming paper [START_REF]Stationary solutions to a two-dimensional free boundary transmission problem arising in the modeling of an electrostatically actuated plate[END_REF] where the minimizing problem leading to the determination of u is analyzed.

Since the regularity of the variational solution ψ u to (1.2) is intimately connected with the regularity of the boundaries of Ω(u), Ω 1 (u), and Ω 2 (u), let us first mention that Ω(u) and Ω 2 (u) are always Lipschitz domains and that the measures of the angles at their vertices do not exceed π, a feature which complies with the H 2 -regularity of ψ u away from the interface Σ(u) [START_REF] Grisvard | Elliptic problems in nonsmooth domains[END_REF]. This property is shared by Ω 1 (u) when the coincidence set C(u) is empty, see Figure 1, so that it is expected that ψ| Ω i (u) belongs to H 2 (Ω i (u)), i = 1, 2, in that case. However, when C(u) is non-empty, the open set Ω 1 (u) is no longer connected and the boundary of its connected components is no longer Lipschitz, but features cusp points. Moreover, there is an interplay between the transmission conditions (1.2b) and the boundary condition (1.2c) when C(u) = ∅. Whether ψ| Ω i (u) still belongs to H 2 (Ω i (u)), i = 1, 2, in this situation is thus an interesting question, that we answer positively in our first result. For the precise statement, we introduce the functional setting we shall work with in the sequel. Specifically, we set S :

= {v ∈ H 2 (D) ∩ H 1 0 (D) : v ≥ -H in D and ± σ ∂ x v(±L) ≤ 0} , and S := {v ∈ H 2 (D) ∩ H 1 0 (D) : v > -H in D and ± σ ∂ x v(±L) ≤ 0} .
Clearly, the coincidence set C(u) is empty if and only if u ∈ S. In addition, the situation already alluded to, where C(u) is non-empty and Ω 1 (u) is a disconnected open set in R 2 with a non-Lipschitz boundary, corresponds to functions u ∈ S \ S. Also, we include the constraint ± σ ∂ x u(±L) ≤ 0 in the definition of S and S to guarantee that the way Σ(u) and ∂Ω(u) intersect does not prevent the H 2 -regularity of ψ u in smooth situations (i.e. u ∈ S ∩ W 2 ∞ (D)), see [START_REF] Lemrabet | Régularité de la solution d'un problème de transmission[END_REF]. Theorem 1.1. Suppose (2.1) below.

(a) For each u ∈ S, there is a unique variational solution

ψ u ∈ h u +H 1 0 (Ω(u)) to (1.2). Moreover, ψ u,1 := ψ u | Ω 1 (u) ∈ H 2 (Ω 1 (u)) and ψ u,2 := ψ u | Ω 2 (u) ∈ H 2 (Ω 2 (u)), and ψ u is a strong solution to the transmission problem (1.2). (b) Given κ > 0, there is c(κ) > 0 such that, for every u ∈ S satisfying u H 2 (D) ≤ κ, ψ u H 1 (Ω(u)) + ψ u,1 H 2 (Ω 1 (u)) + ψ u,2 H 2 (Ω 2 (u)) ≤ c(κ) .
It is worth emphasizing that, for i ∈ {1, 2}, the restriction of ψ u to Ω i (u) belongs to H 2 (Ω i (u)) for all u ∈ S. In particular, there is no regularity breakdown when the coincidence set C(u) is non-empty. A similar observation is made in [START_REF]Shape derivative of the Dirichlet energy for a transmission problem[END_REF] for a different geometric setting when one of the two subsets does not depend on the function u. Theorem 1.1 is an immediate consequence of Proposition 4.9 below. Its proof begins with quantitative H 2 -estimates on ψ u depending only on u H 2 (D) for sufficiently smooth functions in S, the H 2 -regularity of ψ u being guaranteed by [START_REF] Lemrabet | Régularité de la solution d'un problème de transmission[END_REF] in that case. Since the class of functions for which these estimates are valid is dense in S, we complete the proof with a compactness argument, the main difficulty to be faced being the dependence of Ω(u) on u. More precisely, we begin with a variational approach to (1.2) and first show in Section 3 by classical arguments that, given u ∈ S, the variational solution ψ u to (1.2) corresponds to the minimizer on h u + H 1 0 (Ω(u)) of the associated Dirichlet energy

J (u)[θ] := 1 2 Ω(u) σ|∇θ| 2 d(x, z) , θ ∈ h u + H 1 0 (Ω(u)) .
Thanks to this characterization, we use Γ-convergence tools to show the H 1 -stability of ψ u with respect to u in Section 3.2. Section 4 is devoted to the study of the H 2 -regularity of ψ u which we first establish in Section 4.1 for smooth functions u ∈ S ∩ W 2 ∞ (D) (thus having an empty coincidence set), relying on the analysis performed in [START_REF] Lemrabet | Régularité de la solution d'un problème de transmission[END_REF]. It is worth mentioning that the constraint involving σ in the definition of S comes into play here. For u ∈ S ∩W [START_REF] Grisvard | Elliptic problems in nonsmooth domains[END_REF]Lemma 4.3.1.2] allowing us to interchange derivatives with respect to x and z in some integrals involving second-order derivatives, its proof being provided in Appendix A. We then combine these estimates with the already proved H 1 -stability of variational solutions to (1.2) and use a compactness argument to extend the H 2 -regularity of ψ u to arbitrary functions u ∈ S in Section 4.3. In this step, special care is required to cope with the variation of the functional spaces with u. In fact, as a side product of the proof of Theorem 1.1, we obtain qualitative information on the continuous dependence of ψ u with respect to u, which we collect in the next result.

Theorem 1.3. Suppose (2.1) below. Let κ > 0, u ∈ S, and consider a sequence

(u n ) n≥1 in S such that u n H 2 (D) ≤ κ , n ≥ 1 , lim n→∞ u n -u H 1 (D) = 0 . (1.
3)

Setting M := d + max u L∞(D) , sup n≥1 { u n L∞(D) } , lim n→∞ (ψ un -h un ) -(ψ u -h u ) H 1 (Ω M ) = 0 . (1.4a) In addition, if i ∈ {1, 2} and U i is an open subset of Ω i (u) such that Ūi is a compact subset of Ω i (u), then ψ un,i ⇀ ψ u,i in H 2 (U i ) . (1.4b) Also, for any p ∈ [1, ∞), lim n→∞ ∇ψ un,2 (•, u n ) -∇ψ u,2 (•, u) Lp(D,R 2 ) = 0 , lim n→∞ ∇ψ un,2 (•, u n + d) -∇ψ u,2 (•, u + d) Lp(D,R 2 ) = 0 . (1.4c)
Clearly, the quantity M introduced in Theorem 1.3 is finite due to (1.3) and the continuous embedding of H 1 (D) in C( D).

Notation. Given v ∈ S, f ∈ L 2 (Ω(v)), and i ∈ {1, 2}, we denote the restriction of f to Ω i (v) by f i ; that is,

f i := f | Ω i (v) .
Throughout the paper, c and (c k ) k≥1 denote positive constants depending only on L, H, d, V , σ 1 , and σ 2 . The dependence upon additional parameters will be indicated explicitly.

The Boundary Values

We state the precise assumptions on the function h v occurring in (1.2c). Roughly speaking, we assume that it is the trace on ∂Ω(v) of a function h v ∈ H 1 (Ω(u)) which is such that h| Ω i (v) belongs to H 2 (Ω i (v)) for i = 1, 2 and satisfies the transmission conditions (1.2b), as well as suitable boundedness and continuity properties with respect to u.

Specifically, for every v ∈ S, let

h v : D × (-H, ∞) → R be such that h v ∈ H 1 (Ω(v)) , h v,i := h v | Ω i (v) ∈ H 2 Ω i (v) , i = 1, 2 , (2.1a)
and suppose that h v satisfies the transmission conditions

h v = σ∇h v • n Σ(v) = 0 on Σ(v) . (2.1b) For κ > 0 given, there is c(κ) > 0 such that, for all v ∈ S satisfying v H 2 (D) ≤ κ, h v,i H 2 (Ω i (v)) ≤ c(κ) , i = 1, 2 . (2.1c)
Moreover, given v ∈ S and a sequence (v n ) n≥1 in S satisfying

lim n→∞ v n -v H 1 (D) = 0 ,
we assume that lim

n→∞ h vn -h v H 1 (D×(-H,M )) = 0 (2.1d)
and lim

n→∞ h vn (•, v n + d) -h v (•, v + d) C( D) = 0 , (2.1e)
where

M := d + max v L∞(D) , sup n≥1 { v n L∞(D) } < ∞ .
Observe that the convergence of (v n ) n≥1 , the continuous embedding of H

σ|∇h vn | 2 d(x, z) = Ω(v) σ|∇h v | 2 d(x, z) . (2.2)
From now on, we impose the conditions (2.1) throughout.

We finish this short section by providing an example of h v satisfying the imposed conditions (2.1).

Example 2.1. Let ζ ∈ C 2 (R) be such that ζ| (-∞,1] ≡ 0 and ζ| [1+d,∞) ≡ V for some V > 0. Given v ∈ S, put h v (x, z) := ζ(z -v(x) + 1) , -H ≤ z , x ∈ D . (2.3)
Then (2.1a)-(2.1e) are satisfied. In addition,

h v (x, -H) = 0 , h v (x, v(x) + d) = V , x ∈ D .
In the context of a MEMS device alluded to in the introduction, these additional properties mean that the ground plate and the top of the elastic plate are kept at constant potential. For instance, ζ(r) := V min{1, (r -1) 2 /d 2 } for r > 1 and ζ ≡ 0 on (-∞, 1] will do.

Variational Solution to (1.2)

In this section we investigate the properties of the variational solution ψ v to (1.2) for v ∈ S and, in particular, its H 1 -stability.

A Variational Approach to (1.2). Given v ∈ S we introduce the set of admissible potentials

A(v) := h v + H 1 0 (Ω(v)) , on which we define the functional

J (v)[θ] := 1 2 Ω(v) σ|∇θ| 2 d(x, z) , θ ∈ A(v) . (3.1)
The variational solution ψ v to the transmission problem (1.2) is then the minimizer of the functional J (v) on the set A(v):

Lemma 3.1. For each v ∈ S there is a unique minimizer ψ v ∈ A(v) of J (v) on A(v); that is, J (v)[ψ v ] = min θ∈A(v) J (v)[θ] . (3.2)
In addition,

Ω(v) σ|∇ψ v | 2 d(x, z) ≤ Ω(v) σ|∇h v | 2 d(x, z) . (3.3)
Proof. Let v ∈ S and recall that h v ∈ H 1 (Ω(v)) according to (2.1a). Thus, the existence of a minimizer ψ v of J (v) on A(v) readily follows from the direct method of calculus of variations due to the lower semicontinuity and coercivity of J (v) on A(v), the latter being ensured by the assumption σ ≥ min{σ 1 , σ 2 } > 0 and Poincaré's inequality. The uniqueness of ψ v is guaranteed by the strict convexity of J (v). Next, since obviously h v ∈ A(v), the inequality (3.3) is an immediate consequence of the minimizing property (3.2) of ψ v .

For further use, we report the following version of Poincaré's inequality for functions in

H 1 0 (Ω(v)) with a constant depending mildly on v ∈ S. Lemma 3.2. Let v ∈ S and θ ∈ H 1 0 (Ω(v)). Then θ L 2 (Ω(v)) ≤ 2 H + d + v L∞(D) ∂ z θ L 2 (Ω(v)) . Proof. For x ∈ D and z ∈ (-H, v(x) + d), θ(x, z) 2 = 2 z -H θ(x, y)∂ z θ(x, y) dy .
Hence, after integration with respect to (x, z) over Ω(v),

θ 2 L 2 (Ω(v)) = Ω(v) θ(x, z) 2 d(x, z) ≤ 2 H + d + v L∞(D) Ω(v) |θ(x, y)||∂ z θ(x, y)| d(x, z) ≤ 2 H + d + v L∞(D) θ L 2 (Ω(v)) ∂ z θ L 2 (Ω(v)) ,
from which we deduce the stated inequality.

3.2. H 1 -Stability of ψ v . The purpose of this section is to study the continuity properties of the solution ψ v to (3.2) with respect to v. More precisely, we aim at establishing the following result.

Proposition 3.3. Consider v ∈ S and a sequence (v n ) n≥1 in S such that v n → v in H 1 0 (D) , (3.4)
and set

M := d + max v L∞(D) , sup n≥1 { v n L∞(D) } , (3.5)
which is finite by (3.4) and the continuous embedding of

H 1 (D) in C( D). Then lim n→∞ (ψ vn -h vn ) -(ψ v -h v ) H 1 0 (D×(-H,M )) = 0 and lim n→∞ J (v n )[ψ vn ] = J (v)[ψ v ] .
To prove Proposition 3.3, we make use of a Γ-convergence approach and argue as in [7, Section 3.2] with minor changes. For the sake of completeness we provide a complete proof in Appendix C.

H 2 -Regularity

In the previous section we introduced the variational solution ψ v ∈ H 1 (Ω(v) to (1.2) for arbitrary v ∈ S and noticed its continuous dependence in H 1 (Ω(v) with respect to v. We now aim at improving the H 1 -regularity of

ψ v | Ω i (v) to H 2 (Ω i (v)) for i = 1, 2.
To this end we first consider the case of smooth functions v ∈ S ∩ W 2 ∞ (D) with empty coincidence sets and provide in Section 4.1 and Section 4.2 the corresponding H 2 -estimates that depend only on the norm of v in H 2 (D) (but not on its W 2 ∞ (D)-norm). In Section 4.3 we extend these estimates to the general case v ∈ S by means of a compactness argument.

4.1. H 2 -Regularity for v ∈ S ∩ W 2 ∞ (D).
Assuming that v is smoother with an empty coincidence set, see Figure 1, the existence of a strong solution ψ v to (1.2) is a consequence of the analysis performed in [START_REF] Lemrabet | Régularité de la solution d'un problème de transmission[END_REF].

Proposition 4.1. If v ∈ S ∩ W 2 ∞ (D), then the variational solution ψ v to (3.2) satisfies ψ v,i := ψ v | Ω i (v) ∈ H 2 (Ω i (v)) , i = 1, 2 ,
and the transmission problem

div(σ∇ψ v ) = 0 in Ω(v) , (4.1a
)

ψ v = σ∇ψ v • n Σ(v) = 0 on Σ(v) , (4.1b
)

ψ v = h v on ∂Ω(v) . (4.1c) Moreover, ∂ x ψ v + ∂ x v∂ z ψ v and -σ∂ x v∂ x ψ v + σ∂ z ψ v both belong to H 1 (Ω(v)).
Besides [START_REF] Lemrabet | Régularité de la solution d'un problème de transmission[END_REF], the proof of Proposition 4.1 requires the following auxiliary result.

Lemma 4.2. Let v ∈ S and consider φ ∈ L 2 (Ω(v)) such that

φ i := φ| Ω i (v) ∈ H 1 (Ω i (v)) , i = 1, 2 ,
and φ = 0 on Σ(v). Then φ ∈ H 1 (Ω(v)) and φ H 1 (Ω(v)) ≤ φ 1 H 1 (Ω 1 (v)) + φ 2 H 1 (Ω 2 (v)) . (4.2)
Proof. We set e x = (1, 0) and e z = (0, 1). Given θ ∈ C ∞ c Ω(v) and j ∈ {x, z} we note that

Ω(v) φ∂ j θ d(x, z) = Ω(v) div(φθe j ) d(x, z) - 2 i=1 Ω i (v) θ∂ j φ i d(x, z) = Σ(v) φ θe j • n Σ(v) dσ Σ(v) - 2 i=1 Ω i (v) θ∂ j φ i d(x, z) ,
due to Gauß' theorem. Thus, since φ = 0 on Σ(v),

Ω(v) φ∂ j θ d(x, z) ≤ φ 1 H 1 (Ω 1 (v)) + φ 2 H 1 (Ω 2 (v)) θ L 2 (Ω(v)) , for j = x, z and θ ∈ C ∞ c Ω(v) . Consequently, φ ∈ H 1 (Ω(v)).
Proof of Proposition 4.1. We check that the transmission problem (4.1) fits into the framework of [START_REF] Lemrabet | Régularité de la solution d'un problème de transmission[END_REF]. Since v ∈ S ∩ W 2 ∞ (D) and v(±L) = 0, the boundaries of Ω 1 (v) and Ω 2 (v) are W 2 ∞ -smooth curvilinear polygons and the interface Σ(v) meets the boundary ∂Ω(v) of Ω(v) at the vertices A ± := (±L, 0). Moreover, at the vertex A ± , the measures ω ±,1 and ω ±,2 of the angles between -e z and (1, ∓∂ x v(±L)) and between (1, ∓∂ x v(±L)) and e z , respectively, satisfy ω ±,1 + ω ±,2 = π, as well as

ω ±,2 ≥ π 2 if σ < 0 , ω ±,2 ≤ π 2 if σ > 0 ,
by definition of S. According to the analysis performed in [START_REF] Lemrabet | Régularité de la solution d'un problème de transmission[END_REF], these conditions guarantee that the variational solution ψ v to (3.2) provided by Lemma 3.1 satisfies

ψ v,i = ψ v | Ω i (v) ∈ H 2 (Ω i (v)
) for i = 1, 2 and solves the transmission problem (1.2) in a strong sense.

Next, owing to the just established H 2 -regularity of ψ v,1 and ψ v,2 , we may differentiate with respect to x the transmission condition ψ v (x, v(x)) = 0, x ∈ D, and find that

∂ x ψ v + ∂ x v∂ z ψ v = 0 on Σ(v) .
The stated H 1 -regularity of ∂ x ψ v + ∂ x v∂ z ψ v then follows from Lemma 4.2 and the boundedness of ∂ x v and ∂ 2 x v. In the same vein, due to (1.2b), the regularity of v, and the identity -σ∂

x v∂ x ψ v + σ∂ z ψ v 1 + (∂ x v) 2 = σ∇ψ v • n Σ(v) = 0 , the claimed H 1 -regularity of -σ∂ x v∂ x ψ v + σ∂ z ψ v is again a consequence of Lemma 4.2 and the boundedness of ∂ x v and ∂ 2 x v. 4.2. H 2 -Estimates on ψ v for v ∈ S ∩W 2 ∞ (D). The H 2 -regularity of ψ v being guaranteed by Proposition 4.1 for v ∈ S ∩ W 2 ∞ (D)
, the next step is to show that this property extends to any v ∈ S. To this end, we shall now derive quantitative H 2 -estimates on ψ v , paying special attention to their dependence upon the regularity of v. As in [START_REF]Shape derivative of the Dirichlet energy for a transmission problem[END_REF], it turns out to be more convenient to study a non-homogeneous transmission problem with homogeneous Dirichlet boundary conditions instead of (4.1). Specifically,

for v ∈ S ∩ W 2 ∞ (D), we define χ = χ v := ψ v -h v ∈ H 1 0 (Ω(v)) , (4.3)
where

ψ v ∈ H 1 (Ω(v)
) is the unique solution to (4.1) provided by Proposition 4.1. Since

ψ v,i = ψ v | Ω i (v) belongs to H 2 (Ω i (v)) for i = 1,
i := χ v | Ω i (v) ∈ H 2 (Ω i (v)) , i = 1, 2 . (4.4)
We omit in the following the dependence of χ on v for ease of notation.

According to (2.1a), (2.1b), and Proposition 4.1, χ solves the transmission problem div(σ∇χ) = -div(σ∇h v ) in Ω(v) , (4.5a)

χ = σ∇χ • n Σ(v) = 0 on Σ(v) , (4.5b
)

χ = 0 on ∂Ω(v) , (4.5c)
and it follows from (2.1a) that it is equivalent to derive H 2 -estimates on (ψ v,1 , ψ v,2 ) or (χ 1 , χ 2 ).

For that purpose, we transform (4.5) to a transmission problem on the rectangle R := D × (0, 1 + d). More precisely, we introduce the transformation

T 1 (x, z) := x, z + H v(x) + H , (x, z) ∈ Ω 1 (v) , (4.6)
mapping Ω 1 (v) onto the rectangle R 1 := D × (0, 1), and the transformation

T 2 (x, z) := (x, z -v(x) + 1) , (x, z) ∈ Ω 2 (v) , (4.7)
mapping Ω 2 (v) onto the rectangle

R 2 := D × (1, 1 + d). The interface separating R 1 and R 2 is Σ 0 := D × {1} , so that R = D × (0, 1 + d) = R 1 ∪ R 2 ∪ Σ 0 . It is worth pointing out here that T 1 is well-defined due to v ∈ S. Let (x, η) denote the new variables in R; that is, (x, η) = T 1 (x, z) for (x, z) ∈ R 1 and (x, η) = T 2 (x, z) for (x, z) ∈ R 2 . Then, (4.4) implies Φ := Φ 1 1 R 1 + Φ 2 1 R 2 ∈ H 1 0 (R) , Φ i := χ i • (T i ) -1 ∈ H 2 (R i ) , i = 1, 2 . (4.8)
For further use, we also introduce

σ(x, η) :=    σ 1 v(x) + H , (x, η) ∈ R 1 , σ 2 , (x, η) ∈ R 2 ,
and derive the following fundamental identity for Φ, which provides a connection between some integrals involving products of second-order derivatives of Φ and is in the spirit of [4,

Given v ∈ S ∩ W 2 ∞ (D), the function Φ defined in (4.8) satisfies 2 i=1 R i σ∂ 2 x Φ i ∂ 2 η Φ i d(x, η) = 2 i=1 R i σ|∂ x ∂ η Φ i | 2 d(x, η) -σ 1 R 1 ∂ x v (v + H) 2 ∂ η Φ 1 ∂ x ∂ η Φ 1 d(x, η) + 1 2 D ∂ 2 x v (∂ x v) 2 -1 (1 + (∂ x v) 2 ) 2 σ(∂ x Φ) 2 (x, 1) dx .
Proof. We adapt the proof of [START_REF]Shape derivative of the Dirichlet energy for a transmission problem[END_REF]Lemma 3.4] and [9, Lemme II.2.2]. Note that (4.5b), (4.6), (4.7), and (4.8) imply Φ = 0 on Σ 0 , so that

∂ x Φ = 0 on Σ 0 . (4.9) Consequently, since (∂ x Φ 1 , ∂ x Φ 2 ) lies in H 1 (R 1 ) × H 1 (R 2
) by (4.8), we may argue as in the proof of Lemma 4.2 and deduce from (4.9) that

F := ∂ x Φ ∈ H 1 (R) .
Moreover, by (4.8),

F (x, 0) = F (x, 1 + d) = 0 , x ∈ D . (4.10)
Similarly, setting

G := -σ ∂ x v 1 + (∂ x v) 2 ∂ x Φ + σ∂ η Φ we derive from (4.8) that G i := G| R i ∈ H 1 (R i ) for i = 1,
G 1 (x, 1) = σ 1 1 + (∂ x v(x)) 2 [-∂ x v(x)∂ x χ 1 (x, v(x)) + ∂ z χ 1 (x, v(x))] = σ 2 1 + (∂ x v(x)) 2 [-∂ x v(x)∂ x χ 2 (x, v(x)) + ∂ z χ 2 (x, v(x))] = G 2 (x, 1) ;
that is, G = 0 on Σ 0 , and we argue as in the proof of Lemma 4.2 to conclude that

G ∈ H 1 (R) .
In addition, by (4.8),

G(±L, η) = -σ(±L, η) ∂ x v 1 + (∂ x v) 2 (±L)∂ x Φ(±L, η) + σ(±L, η)∂ η Φ(±L, η) = -σ(±L, η) ∂ x v 1 + (∂ x v) 2 (±L)∂ x Φ(±L, η) for η ∈ (0, 1 + d). Hence, G(±L, η) + σ(±L, η) ∂ x v 1 + (∂ x v) 2 (±L)F (±L, η) = 0 , η ∈ (0, 1 + d) . (4.11)
Owing to (4.10), (4.11), and the H 1 -regularity of F and G, we are in a position to apply Lemma A.1 (see Appendix A) with

(V, W ) = (F, G) and τ ± = σ ∂ x v 1 + (∂ x v) 2 (±L) , to obtain the identity R ∂ x F ∂ η G d(x, η) = R ∂ η F ∂ x G d(x, η) .
(4.12)

Using the definitions of F and G, the identity (4.12) reads

2 i=1 R i ∂ 2 x Φ i -σ ∂ x v 1 + (∂ x v) 2 ∂ x ∂ η Φ i + σ∂ 2 η Φ i d(x, η) = 2 i=1 R i ∂ x ∂ η Φ i -σ ∂ x v 1 + (∂ x v) 2 ∂ 2 x Φ i -σ ∂ 2 x v[1 -(∂ x v) 2 ] [1 + (∂ x v) 2 ] 2 ∂ x Φ i d(x, η) + 2 i=1 R i ∂ x ∂ η Φ i ∂ x σ∂ η Φ i + σ∂ x ∂ η Φ i d(x, η) .
Noticing that the first terms on both sides of the above identity are the same and that

∂ x Φ i ∂ x ∂ η Φ i = 1 2 ∂ η (∂ x Φ i ) 2 implies that 2 i=1 R i σ ∂ 2 x v (∂ x v) 2 ) -1 [1 + (∂ x v) 2 ] 2 ∂ x Φ i ∂ x ∂ η Φ i d(x, η) = 1 2 D ∂ 2 x v (∂ x v) 2 -1 [1 + (∂ x v) 2 ] 2 σ(∂ x Φ) 2 (x, 1) dx ,
the assertion follows, recalling that We now translate the outcome of Lemma 4.3 in terms of the solution χ to (4.5).

∂ x σ = 0 in R 2 .
Lemma 4.5. Let v ∈ S ∩ W 2 ∞ (D). The solution χ = ψ v -h v to (4.5) satisfies 2 i=1 Ω i (v) σ ∂ 2 x χ i ∂ 2 z χ i d(x, z) = 2 i=1 Ω i (v) σ|∂ x ∂ z χ i | 2 d(x, z) - σ 2 2 D ∂ 2 x v(x) ∂ z χ 2 (x, v(x) + d) 2 dx - 1 2 D ∂ 2 x v(x) 1 + (∂ x v(x)) 2 σ|∇χ| 2 x, v(x) dx .
Proof. Let us first recall the regularity of Φ stated in (4.8) which validates the subsequent computations. Using the transformations T 1 and T 2 introduced in (4.6) and (4.7), respectively, we obtain

2 i=1 Ω i (v) σ ∂ 2 x χ i ∂ 2 z χ i d(x, z) = R 1 σ 1 v + H ∂ 2 x Φ 1 + η 2 ∂ x v v + H 2 - ∂ 2 x v v + H ∂ η Φ 1 -2η ∂ x v v + H ∂ x ∂ η Φ 1 + η 2 ∂ x v v + H 2 ∂ 2 η Φ 1 ∂ 2 η Φ 1 d(x, η) + R 2 σ 2 ∂ 2 x Φ 2 -2∂ x v∂ x ∂ η Φ 2 -∂ 2 x v∂ η Φ 2 + (∂ x v) 2 ∂ 2 η Φ 2 ∂ 2 η Φ 2 d(x, η) = 2 i=1 R i σ∂ 2 x Φ i ∂ 2 η Φ i d(x, η) + R 1 σ 1 v + H η 2 ∂ x v v + H 2 - ∂ 2 x v v + H ∂ η Φ 1 -2η ∂ x v v + H ∂ x ∂ η Φ 1 + η 2 ∂ x v v + H 2 ∂ 2 η Φ 1 ∂ 2 η Φ 1 d(x, η) + R 2 σ 2 -2∂ x v∂ x ∂ η Φ 2 -∂ 2 x v∂ η Φ 2 + (∂ x v) 2 ∂ 2 η Φ 2 ∂ 2 η Φ 2 d(x, η) .
We use Lemma 4.3 to express the first integral on the right-hand side and get

2 i=1 Ω i (v) σ ∂ 2 x χ i ∂ 2 z χ i d(x, z) = R 1 σ|∂ x ∂ η Φ 1 | 2 d(x, η) + R 2 σ|∂ x ∂ η Φ 2 | 2 d(x, η) + R 1 σ 1 v + H - ∂ x v v + H ∂ η Φ 1 ∂ x ∂ η Φ 1 -2η ∂ x v v + H ∂ x ∂ η Φ 1 ∂ 2 η Φ 1 + η 2 ∂ x v v + H 2 ∂ 2 η Φ 1 2 + 2η ∂ x v v + H 2 ∂ η Φ 1 ∂ 2 η Φ 1 -η ∂ 2 x v v + H ∂ η Φ 1 ∂ 2 η Φ 1 d(x, η) + R 2 σ 2 -2∂ x v∂ x ∂ η Φ 2 ∂ 2 η Φ 2 -∂ 2 x v∂ η Φ 2 ∂ 2 η Φ 2 + (∂ x v) 2 ∂ 2 η Φ 2 2 d(x, η) + 1 2 D ∂ 2 x v (∂ x v) 2 -1 (1 + (∂ x v) 2 ) 2 σ(∂ x Φ) 2 (x, 1) dx . (4.13)
We then compute separately the integrals over R i , i = 1, 2, and begin with the contribution of R 1 . We complete the square to get

I 1 := R 1 σ 1 v + H |∂ x ∂ η Φ 1 | 2 - ∂ x v v + H ∂ η Φ 1 ∂ x ∂ η Φ 1 -2η ∂ x v v + H ∂ x ∂ η Φ 1 ∂ 2 η Φ 1 + η 2 ∂ x v v + H 2 ∂ 2 η Φ 1 2 + 2η ∂ x v v + H 2 ∂ η Φ 1 ∂ 2 η Φ 1 -η ∂ 2 x v v + H ∂ η Φ 1 ∂ 2 η Φ 1 d(x, η) = R 1 σ 1 v + H ∂ x ∂ η Φ 1 2 + ∂ x v v + H 2 ∂ η Φ 1 2 + η 2 ∂ x v v + H 2 ∂ 2 η Φ 1 2 -2η ∂ x v v + H ∂ x ∂ η Φ 1 ∂ 2 η Φ 1 + 2η ∂ x v v + H 2 ∂ η Φ 1 ∂ 2 η Φ 1 -2 ∂ x v v + H ∂ η Φ 1 ∂ x ∂ η Φ 1 d(x, η) + R 1 σ 1 v + H - ∂ x v v + H 2 ∂ η Φ 1 2 + ∂ x v v + H ∂ η Φ 1 ∂ x ∂ η Φ 1 -η ∂ 2 x v v + H ∂ η Φ 1 ∂ 2 η Φ 1 d(x, η) = R 1 σ 1 (v + H) ∂ x ∂ η Φ 1 v + H - ∂ x v (v + H) 2 ∂ η Φ 1 -η ∂ x v (v + H) 2 ∂ 2 η Φ 1 2 d(x, η) + R 1 σ 1 ∂ x v 1 (v + H) 2 ∂ η Φ 1 ∂ x ∂ η Φ 1 - ∂ x v (v + H) 3 ∂ η Φ 1 2 d(x, η) - R 1 σ 1 ∂ 2 x v (v + H) 2 η ∂ η Φ 1 ∂ 2 η Φ 1 d(x, η) .
Thanks to the identities

1 (v + H) 2 ∂ η Φ 1 ∂ x ∂ η Φ 1 - ∂ x v (v + H) 3 ∂ η Φ 1 2 = 1 2 ∂ x ∂ η Φ 1 v + H 2 , ∂ η Φ 1 ∂ 2 η Φ 1 = 1 2 ∂ η ∂ η Φ 1 2 ,
and the property ∂ η Φ 1 (±L, η) = 0 for η ∈ (0, 1) stemming from (4.8), we may perform integration by parts in the last two integrals on the right-hand side of the previous identity and obtain

I 1 = R 1 σ 1 (v + H) ∂ x ∂ η Φ 1 v + H - ∂ x v (v + H) 2 ∂ η Φ 1 -η ∂ x v (v + H) 2 ∂ 2 η Φ 1 2 d(x, η) - σ 1 2 D ∂ 2 x v (v + H) 2 ∂ η Φ 1 (x, 1) 2 dx .
Transforming the above identity back to Ω 1 (v) yields

I 1 = Ω 1 (v) σ 1 ∂ x ∂ z χ 1 2 d(x, z) - σ 1 2 D ∂ 2 x v(x) ∂ z χ 1 (x, v(x)) 2 dx . (4.14)
Next, arguing in a similar way,

I 2 := σ 2 R 2 ∂ x ∂ η Φ 2 2 -2∂ x v∂ x ∂ η Φ 2 ∂ 2 η Φ 2 -∂ 2 x v∂ η Φ 2 ∂ 2 η Φ 2 + (∂ x v) 2 ∂ 2 η Φ 2 2 d(x, η) = σ 2 R 2 ∂ x ∂ η Φ 2 2 -2∂ x v∂ x ∂ η Φ 2 ∂ 2 η Φ 2 + (∂ x v) 2 ∂ 2 η Φ 2 2 d(x, η) - σ 2 2 R 2 ∂ 2 x v∂ η (∂ η Φ 2 ) 2 d(x, η) = σ 2 R 2 ∂ x ∂ η Φ 2 -∂ x v∂ 2 η Φ 2 2 d(x, η) - σ 2 2 D ∂ 2 x v(x) ∂ η Φ 2 (x, 1 + d) 2 dx + σ 2 2 D ∂ 2 x v(x) ∂ η Φ 2 (x, 1) 2 dx .
Transforming this formula back to Ω 2 (v) yields

I 2 = σ 2 Ω 2 (v) ∂ x ∂ z χ 2 2 d(x, z) - σ 2 2 D ∂ 2 x v(x) ∂ z χ 2 (x, v(x) + d) 2 dx + σ 2 2 D ∂ 2 x v(x) ∂ z χ 2 (x, v(x)) 2 dx . (4.15) Finally, D ∂ 2 x v (∂ x v) 2 -1 [1 + (∂ x v) 2 ] 2 σ(∂ x Φ) 2 (x, 1) dx = D ∂ 2 x v (∂ x v) 2 -1 [1 + (∂ x v) 2 ] 2 σ(∂ x χ + ∂ x v∂ z χ) 2 (x, 1) dx ,
and we deduce from (4.13), (4.14), (4.15), and the above identity that

2 i=1 Ω i (v) σ ∂ 2 x χ i ∂ 2 z χ i d(x, z) = 2 i=1 Ω i (v) σ ∂ x ∂ z χ i 2 d(x, z) - σ 2 2 D ∂ 2 x v(x) ∂ z χ 2 (x, v(x) + d) 2 dx - 1 2 D ∂ 2 x v(x) σ ∂ z χ 2 2 (x, v(x)) dx + 1 2 D ∂ 2 x v (∂ x v) 2 -1 [1 + (∂ x v) 2 ] 2 σ ∂ x χ + ∂ x v∂ z χ 2 (x, v(x)) dx . (4.16)
It remains to simplify the last two integrals on the right-hand side of (4.16). To this end, we first recall that the regularity of χ allows us to differentiate with respect to x the transmission condition χ = 0 on Σ(v) to deduce that

∂ x χ + ∂ x v∂ z χ = 0 on Σ(v) , (4.17)
while the second transmission condition in (4.5b) reads

σ ∂ x v∂ x χ -∂ z χ = 0 on Σ(v) . (4.18)
In particular, (4.17) and (4.18) imply that, on Σ(v),

σ ∂ x v∂ x χ -∂ z χ ∂ x χ + ∂ x v∂ z χ = ∂ x χ 1 + ∂ x v∂ z χ 1 σ ∂ x v∂ x χ -∂ z χ + σ 2 ∂ x v∂ x χ 2 -∂ z χ 2 ∂ x χ + ∂ x v∂ z χ = 0 . Therefore, J := (∂ x v) 2 -1 σ ∂ x χ + ∂ x v∂ z χ 2 -[1 + (∂ x v) 2 ] 2 σ ∂ z χ 2 = (∂ x v) 2 -1 σ ∂ x χ + ∂ x v∂ z χ 2 -[1 + (∂ x v) 2 ] 2 σ ∂ z χ 2 -2∂ x v σ ∂ x v∂ x χ -∂ z χ ∂ x χ + ∂ x v∂ z χ = σ (∂ x v) 2 -1 -2(∂ x v) 2 ∂ x χ 2 + σ 2∂ x v (∂ x v) 2 -1 -2(∂ x v) 3 + 2∂ x v ∂ x χ∂ z χ + σ (∂ x v) 2 (∂ x v) 2 -1 + 2(∂ x v) 2 -[1 + (∂ x v) 2 ] 2 ∂ z χ 2 = -1 + (∂ x v) 2 σ ∂ x χ 2 + σ ∂ z χ 2 = -1 + (∂ x v) 2 σ|∇χ| 2 .
Hence,

(∂ x v) 2 -1 [1 + (∂ x v) 2 ] 2 σ ∂ x χ + ∂ x v∂ z χ 2 -σ ∂ z χ 2 = - 1 1 + (∂ x v) 2 σ|∇χ| 2 . ( 4.19)
Consequently, (4.16) and (4.19) entail

2 i=1 Ω i (v) σ ∂ 2 x χ i ∂ 2 z χ i d(x, z) = 2 i=1 Ω i (v) σ|∂ x ∂ z χ i | 2 d(x, z) - 1 2 D σ 2 ∂ 2 x v(x) ∂ z χ 2 (x, v(x) + d) 2 dx - 1 2 D ∂ 2 x v(x) 1 + (∂ x v(x)) 2 σ|∇χ| 2 x, v(x) dx , as claimed.
In order to estimate the boundary and the transmission terms in Lemma 4.5, we first report the following trace estimates. Lemma 4.6. Given κ > 0 and α ∈ (0, 1/2], there is c(α, κ) > 0 such that, for any

v ∈ S satisfying v H 2 (D) ≤ κ and θ ∈ H 1 (Ω 2 (v)), θ 2 (•, v) H α (D) + θ 2 (•, v + d) H α (D) ≤ c(α, κ) θ 2 (1-2α)/2 L 2 (Ω 2 (v)) θ 2 (2α+1)/2 H 1 (Ω 2 (v)) . Proof. Let θ ∈ H 1 (Ω 2 (v)). Using the transformation T 2 defined in (4.7) which maps Ω 2 (v) onto the rectangle R 2 = D × (1, 1 + d), we note that φ := θ • T -1 2 belongs to H 1 (R 2 ) with φ L 2 (R 2 ) = θ L 2 (Ω 2 (v)) (4.20) and ∇φ 2 L 2 (R 2 ) = ∂ x θ + ∂ x v∂ z θ 2 L 2 (Ω 2 (v)) + ∂ z θ 2 L 2 (Ω 2 (v))
, so that the continuous embedding of H 2 (D) in W 1 ∞ (D) and the assumed bound on v readily imply that

φ H 1 (R 2 ) ≤ c(κ) θ H 1 (Ω 2 (v)) . (4.21) By complex interpolation, [L 2 (R 2 ), H 1 (R 2 )] α+1/2 . = H α+1/2 (R 2 ) ,
from which we deduce that

φ H α+1/2 (R 2 ) ≤ c(α) φ (1-2α)/2 L 2 (R 2 ) φ (2α+1)/2 H 1 (R 2 ) .
Since α > 0, the trace maps H α+1/2 (R 2 ) continuously on H α (D × {1}), and we thus infer from (4.20) and (4.21) that

θ(•, v) H α (D) = φ(•, 1) H α (D) ≤ c(α) φ H α+1/2 (R 2 ) ≤ c(α) φ (1-2α)/2 L 2 (R 2 ) φ (2α+1)/2 H 1 (R 2 ) ≤ c(α, κ) θ (1-2α)/2 L 2 (Ω 2 (v)) θ (2α+1)/2 H 1 (Ω 2 (v)) .
The estimate for θ(•, v + d) H α (D) is proved in a similar way.

Based on Lemma 4.6 we are in a position to estimate the boundary and transmission terms in the identity provided by Lemma 4.5.

(ζ, κ) > 0 such that, if v ∈ S ∩ W 2 ∞ (D) satisfies v H 2 (D) ≤ κ, then the solution χ = χ v to (4.5) satisfies σ 2 2 D ∂ 2 x v(x) ∂ z χ 2 (x, v(x) + d) 2 dx ≤ c(ζ, κ) ∂ z χ 2 2(1-ζ) L 2 (Ω 2 (v)) ∂ z χ 2 2ζ H 1 (Ω 2 (v)) (4.22) and 1 2 D ∂ 2 x v(x) 1 + (∂ x v(x)) 2 σ|∇χ| 2 x, v(x) dx ≤ c(ζ, κ) ∇χ 2 2(1-ζ) L 2 (Ω 2 (v)) ∇χ 2 2ζ H 1 (Ω 2 (v)) .
σ 2 2 D ∂ 2 x v(x) ∂ z χ 2 (x, v(x) + d) 2 dx ≤ σ 2 2 ∂ 2 x v L 2 (D) ∂ z χ 2 (•, v + d) 2 L 4 (D) ≤ c(κ) ∂ z χ 2 (•, v + d) 2 H ζ-1/2 (D) ≤ c(ζ, κ) ∂ z χ 2 2(1-ζ) L 2 (Ω 2 (v)) ∂ z χ 2 2ζ H 1 (Ω 2 (v)) .
As for (4.23) we obtain analogously

σ 2 2 D ∂ 2 x v(x) 1 + (∂ x v(x)) 2 ∂ x χ 2 (x, v(x)) 2 + ∂ z χ 2 (x, v(x)) 2 dx ≤ σ 2 2 ∂ 2 x v L 2 (D) ∇χ 2 (•, v) 2 L 4 (D) ≤ c(ζ, κ) ∇χ 2 2(1-ζ) L 2 (Ω 2 (v)) ∇χ 2 2ζ H 1 (Ω 2 (v)) (4.24) and σ 1 2 D ∂ 2 x v(x) 1 + (∂ x v(x)) 2 ∂ x χ 1 (x, v(x)) 2 + ∂ z χ 1 (x, v(x)) 2 dx ≤ σ 1 2 ∂ 2 x v L 2 (D) ∇χ 1 (•, v) 2 L 4 (D) . (4.25)
At this point, we use (4.17) and (4.18) to show that

∂ x χ 1 = σ 1 + σ 2 (∂ x v) 2 σ 1 1 + (∂ x v) 2 ∂ x χ 2 + σ ∂ x v σ 1 1 + (∂ x v) 2 ∂ z χ 2 on Σ(v) , ∂ z χ 1 = σ ∂ x v σ 1 1 + (∂ x v) 2 ∂ x χ 2 + σ 1 + σ 2 (∂ x v) 2 σ 1 1 + (∂ x v) 2 ∂ z χ 2 on Σ(v) . Consequently, |∂ x χ 1 | ≤ max{σ 1 , σ 2 } σ 1 (|∂ x χ 2 | + |∂ z χ 2 |) on Σ(v) , |∂ z χ 1 | ≤ max{σ 1 , σ 2 } σ 1 (|∂ x χ 2 | + |∂ z χ 2 |) on Σ(v) , so that ∇χ 1 (•, v) L 4 (D) ≤ c ∇χ 2 (•, v) L 4 (D) .
Owing to (4.25) and the above inequality, we may then argue as in the proof of (4.24) to conclude that

σ 1 2 D ∂ 2 x v(x) 1 + (∂ x v(x)) 2 ∂ x χ 1 (x, v(x)) 2 + ∂ z χ 1 (x, v(x)) 2 dx ≤ c(ζ, κ) ∇χ 2 2(1-ζ) L 2 (Ω 2 (v)) ∇χ 2 2ζ H 1 (Ω 2 (v)) , as claimed in (4.23).
We now gather the previous findings to deduce the following crucial H 2 -estimate on the solution ψ v to (4.1) for v ∈ S ∩ W 2 ∞ (D), which only depends on the

H 2 (D)-norm of v (but not on its W 2 ∞ (D)-norm). Proposition 4.8. Let κ > 0 and v ∈ S ∩ W 2 ∞ (D) be such that v H 2 (D) ≤ κ.
There is a constant c 0 (κ) > 0 such that the solution ψ v to (4.1) satisfies

χ H 1 (Ω(v)) + χ 1 H 2 (Ω 1 (v)) + χ 2 H 2 (Ω 2 (v)) ≤ c 0 (κ) , (4.26a
)

and ψ v H 1 (Ω(v)) + ψ v,1 H 2 (Ω 1 (v)) + ψ v,2 H 2 (Ω 2 (v)) ≤ c 0 (κ) , (4.26b
)

recalling that χ = ψ v -h v and χ i = χ| Ω i (v) , i = 1, 2. Proof. Let v ∈ S ∩ W 2 ∞ (D) with v H 2 (D) ≤ κ. Since σ is constant on Ω 1 (v) and on Ω 2 (v), it readily follows from (4.5a) that 2 i=1 Ω i (v) σ|∆χ i | 2 d(x, z) = 2 i=1 Ω i (v) σ|∆h v,i | 2 d(x, z) . Since |∆χ i | 2 = |∂ 2 x χ i | 2 + |∂ 2 z χ i | 2 + 2∂ 2 x χ i ∂ 2 z χ i , i = 1, 2 ,
we infer from Lemma 4.5 and the above two formulas that

2 i=1 Ω i (v) σ |∂ 2 x χ i | 2 + 2|∂ x ∂ z χ i | 2 + |∂ 2 z χ i | 2 d(x, z) = 2 i=1 Ω i (v) σ|∆h v,i | 2 d(x, z) + 2 2 i=1 Ω i (v) σ |∂ x ∂ z χ i | 2 -∂ 2 x χ i ∂ 2 z χ i d(x, z) ≤ 2 i=1 Ω i (v) σ|∆h v,i | 2 d(x, z) + σ 2 D ∂ 2 x v(x) ∂ z χ 2 (x, v(x) + d) 2 dx + D ∂ 2 x v(x) 1 + (∂ x v(x)) 2 σ|∇χ| 2 x, v(x) dx .
Using Lemma 4.7 with ζ = 7/8, along with the identity

2 i=1 Ω i (v) σ |∂ 2 x χ i | 2 + 2|∂ x ∂ z χ i | 2 + |∂ 2 z χ i | 2 d(x, z) = σ 1 ∇χ 1 2 H 1 (Ω 1 (v)) + σ 2 ∇χ 2 2 H 1 (Ω 2 (v))
, we further obtain

σ 1 ∇χ 1 2 H 1 (Ω 1 (v)) + σ 2 ∇χ 2 2 H 1 (Ω 2 (v)) ≤ 2 i=1 Ω i (v) σ|∆h v,i | 2 d(x, z) + c(κ) ∇χ 2 1/4 L 2 (Ω 2 (v)) ∇χ 2 7/4 H 1 (Ω 2 (v)) .
Hence, thanks to Young's inequality,

σ 1 ∇χ 1 2 H 1 (Ω 1 (v)) + σ 2 ∇χ 2 2 H 1 (Ω 2 (v)) ≤ 2 i=1 Ω i (v) σ|∆h v,i | 2 d(x, z) + σ 2 2 ∇χ 2 2 H 1 (Ω 2 (v)) + c(κ) ∇χ 2 2 L 2 (Ω 2 (v)) .
Recalling that

∇χ 2 2 L 2 (Ω 2 (v)) ≤ 1 σ 2 Ω(v) σ|∇χ| 2 d(x, z) ≤ 1 σ 2 Ω(v) σ|∇h v | 2 d(x, z) ≤ max{σ 1 , σ 2 } σ 2 ∇h v 2 L 2 (Ω(v))
w by (4.5) and that min{σ 1 , σ 2 } > 0, we conclude that

Ω 1 (w) Ω 2 (w) D Σ(w) z -H 0 d -L L C(w)
∇χ 1 2 H 1 (Ω 1 (v)) + ∇χ 2 2 H 1 (Ω 2 (v)) ≤ c(κ) ∆h v,1 2 L 2 (Ω 1 (v)) + ∆h v,2 2 L 2 (Ω 2 (v)) + ∇h v 2 L 2 (Ω(v)) .
χ H 1 (Ω(v)) + χ 1 H 2 (Ω 1 (v)) + χ 2 H 2 (Ω 2 (v)) ≤ c(κ) ∇h v L 2 (Ω(v)) + ∆h v,1 2
L 2 (Ω 1 (v)) + ∆h v,2 2
L 2 (Ω 2 (v))
. The bound (4.26a) then readily follows from the assumptions (2.1a) and (2.1c). Finally, (4.26a), together with (2.1a) and (2.1c), yields (4.26b). ∞ -regularity of v assumed in the previous sections and also allow for a non-empty coincidence set. Proposition 4.9. Let κ > 0 and v ∈ S be such that v H 2 (D) ≤ κ.

(a) The unique minimizer ψ v ∈ A(v) of J (v) on A(v) provided by Lemma 3.1 satisfies

ψ v,i = ψ v | Ω i (v) ∈ H 2 (Ω i (v)) , i = 1, 2 ,
and is a strong solution to the transmission problem (4.1). Moreover, there is c 1 (κ) > 0 such that

ψ v H 1 (Ω(v)) + ψ v,1 H 2 (Ω 1 (v)) + ψ v,2 H 2 (Ω 2 (v)) ≤ c 1 (κ) . (4.28) (b) Consider a sequence (v n ) n≥1 in S satisfying v n H 2 (D) ≤ κ , n ≥ 1 , and lim n→∞ v n -v H 1 (D) = 0 . (4.29) If i ∈ {1, 2} and U i is an open subset of Ω i (v) such that Ūi is a compact subset of Ω i (v), then ψ vn,i ⇀ ψ v,i in H 2 (U i ) ,
We supplement the H 2 -weak continuity of ψ v with respect to v reported in Proposition 4.9 with the continuity of the traces of ∇ψ v,2 on the upper and lower boundaries of Ω 2 (v). Proposition 4.11. Let κ > 0 and v ∈ S be such that v H 2 (D) ≤ κ and consider a sequence

(v n ) n≥1 in S satisfying (4.29). Then, for p ∈ [1, ∞), ∇ψ vn,2 (•, v n ) → ∇ψ v,2 (•, v) in L p (D, R 2 ) , (4.33) ∇ψ vn,2 (•, v n + d) → ∇ψ v,2 (•, v + d) in L p (D, R 2 ) , (4.34
)

and ∇ψ v,2 (•, v) Lp (D,R 2 ) + ∇ψ v,2 (•, v + d) Lp(D,R 2 ) ≤ c(p, κ) . ( 4

.35)

Proof. Recall first from (4.28) that

ψ vn,2 H 2 (Ω 2 (vn)) ≤ c 1 (κ) , n ≥ 1 . (4.36)
As in the proof of Lemma 4.6 we map Ω 2 (v) onto the rectangle R 2 = D × (1, 1 + d) and define, for (x, η) ∈ R 2 and n ≥ 1,

φ n (x, η) := ψ vn,2 (x, η + v n (x) -1) , φ(x, η) := ψ v,2 (x, η + v(x) -1) .
Let q ∈ (1, 2). Since

∇φ n (x, η) = ∂ x ψ vn + ∂ x v n ∂ z ψ vn , ∂ z ψ vn (x, η + v n (x) -1) , ∂ 2 x φ n (x, η) = ∂ 2 x ψ vn + 2∂ x v n ∂ x ∂ z ψ vn + (∂ x v n ) 2 ∂ 2 z ψ vn + ∂ 2 x v n ∂ z ψ vn (x, η + v n (x) -1) , ∂ x ∂ η φ n (x, η) = ∂ x ∂ z ψ vn + ∂ x v n ∂ 2 z ψ vn (x, η + v n (x) -1) , ∂ 2 η φ n (x, η) = ∂ 2 z ψ vn (x, η + v n (x) -1)
, it follows from (4.29), (4.36), the continuous embedding of H 2 (D) in C 1 ( D), and that of

H 1 (R 2 ) in L 2q/(2-q) (R 2 ) that φ n ∈ W 2 q (R 2 ) with φ n W 2 q (R 2 ) ≤ c(q, κ) , n ≥ 1 . (4.37)
Now, given p ∈ [1, ∞), we choose q ∈ (1, min{2, p}) satisfying 1 < 2/q < 1 + 1/p and s ∈ (2/q -1/p, 1). Since φ n ⇀ φ in W 2 q (R 2 ) by (2.1d), (4.37), and Proposition 4.9, the continuity of the trace as a mapping from

W 1 q (R 2 ) to W 1-1/q q (D × {1}
) and the compactness of the embedding of W

1-1/q q (D) in L p (D) imply that ∇φ n (•, 1) → ∇φ(•, 1) in W s-1/q q (D) (4.38) and ∇φ(•, 1) Lp (D) ≤ c(p, κ) . (4.39) That is, ∂ z ψ vn,2 (•, v n ) = ∂ η φ n (•, 1) → ∂ η φ(•, 1) = ∂ z ψ v,2 (•, v) in L p (D)
and, recalling (4.29) and the continuous embedding of

H 2 (D) in C 1 ( D), ∂ x ψ vn,2 (•, v n ) = ∂ x φ n (•, 1) -∂ x v n ∂ η φ n (•, 1) → ∂ x φ(•, 1) -∂ x v∂ η φ(•, 1) = ∂ x ψ v,2 (•, v) in L p (D) .
Next, after integrating by parts,

J(V, W ) := R ∂ x V ∂ η W -∂ η V ∂ x W d(x, η) = 1+d 0 (V ∂ η W )(x, η) x=L x=-L dη - R V ∂ x ∂ η W d(x, η) - D (V ∂ x W )(x, η) η=1+d η=0 + R V ∂ x ∂ η W d(x, η) .
Since V (x, 0) = V (x, 1 + d) = 0 for x ∈ D by (A.1a) and the second and fourth terms cancel each other out, we obtain

J(V, W ) = 1+d 0 V (L, η)∂ η W (L, η) dη - 1+d 0 V (-L, η)∂ η W (-L, η) dη .
Now, according to (A.1b) and the regularity of V and W ,

∂ η W (±L, η) = -τ ± 1 ∂ η V (±L, η) , η ∈ (0, 1) , ∂ η W (±L, η) = -τ ± 2 ∂ η V (±L, η) , η ∈ (1, 1 + d) , so that, since [η → V (±L, η)] ∈ C([0, 1 + d]) by (A.3), J(V, W ) = -τ + 1 1 0 (V ∂ η V )(L, η) dη -τ + 2 1+d 1 (V ∂ η V )(L, η) dη + τ - 1 1 0 (V ∂ η V )(-L, η) dη + τ - 2 1+d 1 (V ∂ η V )(-L, η) dη = -τ + 1 V (L, 1) 2 -V (L, 0) 2 2 -τ + 2 V (L, 1 + d) 2 -V (L, 1) 2 2 + τ - 1 V (-L, 1) 2 -V (-L, 0) 2 2 + τ - 2 V (-L, 1 + d) 2 -V (-L, 1) 2 2 = τ + 1 2 V (L, 0) 2 - τ - 1 2 V (-L, 0) 2 - τ + 2 2 V (L, 1 + d) 2 + τ - 2 2 V (-L, 1 + d) 2 (A.4) - τ + 1 -τ + 2 2 V (L, 1) 2 + τ - 1 -τ - 2 2 V (-L, 1) 2 .
On the one hand, it follows from (A.1) and the continuity (A.3) of V that

V (±L, 0) = lim x→±L V (x, 0) = 0 , V (±L, 1 + d) = lim x→±L V (x, 1 + d) = 0 . (A.5)
On the other hand, using (A.1b) along with the continuity (A.3) gives Combining (A.4), (A.5), and (A.6) leads us to J(V, W ) = 0 and we have proved that

τ ± 1 V (±L, 1) = lim ηր1 τ ± (η)V (±L, η) = -lim ηր1 W (±L, η) = -W (±L, 1) = -lim ηց1 W (±L, η) = lim ηց1 τ ± (η)V (±L, η) = τ ± 2 V (±L, 1) . Consequently, τ ± 1 -τ ± 2 V (±L, 1) = 0 . (A.6) -L L R x η S 4 S 1 S 2 S 3 Γ 4 Γ 3 Γ 2 Γ 1 0 1 1 + d
J(V, W ) = 0 , (V, W ) ∈ G 2 (R) . (A.7)
In other words, the identity stated in Lemma A.1 is valid for (V, W ) ∈ G 2 (R).

We provide here a proof of the density of

G 2 (R) in G 1 (R) as claimed in Lemma A.2. It is adapted from that of [4, Lemma 4.3.1.3].
Proof of Lemma A.2. To cast the problem under consideration in a form which is as close as possible to that used in [4, Section 4.3.1], we recall that R is a polygon with vertices

Y 1 (s) := (L, s) , s ∈ (0, 1 + d) , (L + s, 0) , s ∈ (-2L, 0) , Y 2 (s) := (L -s, 1 + d) , s ∈ (0, 2L) , (L, 1 + d + s) , s ∈ (-1 -d, 0) , Y 3 (s) := (-L, 1 + d -s) , s ∈ (0, 1 + d) , (-L -s, 1 + d) , s ∈ (-2L, 0) , Y 4 (s) := (-L + s, 0) , s ∈ (0, 2L) , (-L, -s) , s ∈ (-1 -d, 0) .
We also set

(λ 1 , λ 2 , λ 3 , λ 4 ) := (1, τ + , 1, τ -) , (µ 1 , µ 2 , µ 3 , µ 4 ) := (0, 1, 0, 1) , (A.8)
and observe that the main difference to the situation studied in [4, Section 4.3.1] is that λ 2 and λ 4 are not constants, but piecewise constant and possibly discontinuous functions.

A preliminary step is the analogue of Lemma B.1 when (0, 1) is replaced by (0, ∞).

Lemma B.2. The space C ∞ c ((0, ∞)) is dense in H 1/2 (0, ∞) ∩ L 2 ((0, ∞), dx/x). Proof. Set X := H 1/2 (0, ∞) ∩ L 2 ((0, ∞), dx/x).
The proof is divided into two steps: we first show that C ∞ c ([0, ∞)) ∩ X is dense in X with an argument from [10]. We then use a method of truncation as in [START_REF] Triebel | Interpolation theory, function spaces, differential operators[END_REF] to complete the proof.

Step 1: Density of C ∞ c ([0, ∞)) ∩ X in X. Consider f ∈ X and δ ∈ (0, 1). As in [10], we define I j := (1/j, j) for j ≥ 1 and note that

D j := I j+1 \ Īj-1 = 1 j + 1 , 1 j -1 ∪ (j -1, j + 1) , j ≥ 2 .
Since (0, ∞) = I 2 ∪ j≥2 D j , there exists a partition of unity (ψ j ) j≥1 consisting of non-negative functions in C ∞ c ((0, ∞)) such that supp ψ 1 ⊂ I 2 and supp ψ j ⊂ D j for j ≥ 2 , (B.1a)

∞ j=1 ψ j (x) = 1 for x ∈ (0, ∞) , (B.1b)
and, for every compact subset K of (0, ∞), there exist an integer ℓ K ≥ 1 and an open subset O K of (0, ∞) such that

K ⊂ O K and ℓ K j=1 ψ j (x) = 1 for x ∈ O K . (B.1c)
Observe that, if x ∈ (1/2, 1) ∪ (1, 2), then x ∈ I 2 ∩ D 2 and x ∈ D k for k ≥ 3, while, if x ∈ (1/(j + 1), 1/j) ∪ (j, j + 1) for some j ≥ 2, then x ∈ D j ∩ D j+1 and x ∈ D k for 1 ≤ k ≤ j -1 and k ≥ j + 2. In addition, 1 ∈ I 2 but 1 ∈ D k for k ≥ 2. Consequently, given x ∈ (0, ∞), the series in (B.1b) has at most two non-vanishing terms.

Next, let (̺ ε ) ε∈(0,1) be a family of C ∞ -smooth mollifiers satisfying

supp ̺ ε ⊂ (-ε, ε) and R ̺ ε (x) dx = 1 for ε ∈ (0, 1) . (B.2)
Let j ≥ 3 and ε ∈ (0, 1). Owing to the properties of the convolution,

supp ̺ ε * (ψ j f ) ⊂ 1 j + 1 -ε, 1 j -1 + ε ∪ (j -1 -ε, j + 1 + ε) , so that, if ε ∈ (0, 1/(j + 1)(j + 2)), then supp ̺ ε * (ψ j f ) ⊂ 1 j + 2 , j + 2 . (B.3a) Similarly, for ε ∈ (0, 1/12), supp ̺ ε * (ψ 2 f ) ⊂ 1 4 , 4 , (B.3b)
and, for ε ∈ (0, 1/6),

supp ̺ ε * (ψ 1 f ) ⊂ 1 3 , 3 . (B.3c)
Since ̺ ε * (ψ j f ) ε∈(0,1) converges to ψ j f in H 1/2 (0, ∞) and in L 2 (0, ∞) for each j ≥ 1 as ε → 0, we may pick ε j ∈ (0, 1) such that

ε j ∈ 0, 1 (j + 1)(j + 2) , j ≥ 1 , (B.4a) ̺ ε j * (ψ j f ) -ψ j f H 1/2 (0,∞) ≤ δ2 -j , j ≥ 1 , (B.4b) ̺ ε j * (ψ j f ) -ψ j f L 2 (0,∞) ≤ δ √ j + 2 2 -j , j ≥ 1 . (B.4c)
A first consequence of (B.3), (B.4a), and (B.4c) is that, for j ≥ 1,

̺ ε j * (ψ j f ) -ψ j f L 2 ((0,∞),dx/x) = j+2 1/(j+2) ̺ ε j * (ψ j f ) -ψ j f (x) 2 dx x 1/2 ≤ j + 2 j+2 1/(j+2) ̺ ε j * (ψ j f ) -ψ j f (x) 2 dx 1/2 ≤ j + 2 ̺ ε j * (ψ j f ) -ψ j f L 2 (0,∞) ≤ δ2 -j . (B.5)
Still following the argument in [10], we now define

F := ∞ j=1 ̺ ε j * (ψ j f ) and F k := k j=1 ̺ ε j * (ψ j f ) , k ≥ 1 . (B.6) By (B.3), given x ∈ (0, ∞), ̺ ε j * (ψ j f ) (x) = 0 for j ≥ 2 + max x, 1
x , so that F (x) is actually a finite sum. As a consequence, F ∈ C ∞ ([0, ∞)) and, for k ≥ 2 and x ∈ I k , we infer from (B.1) and (B.6) that

F (x) = k+2 j=1 ̺ ε j * (ψ j f ) (x) = F k+2 (x) (B.7a) and f (x) = ∞ j=1 (ψ j f )(x) = k+2 j=1 (ψ j f )(x) . (B.7b) Let k ≥ 2.
Owing to (B.4b) and (B.7),

f -F H 1/2 (I k ) = k+2 j=1 ̺ ε j * (ψ j f ) - k+2 j=1 ψ j f H 1/2 (I k ) ≤ k+2 j=1 ̺ ε j * (ψ j f ) -ψ j f H 1/2 (I k ) ≤ δ . (B.8)
In particular,

F H 1/2 (I k ) ≤ δ + f H 1/2 (I k ) ≤ δ + f H 1/2 (0,∞) . (B.9)
We use Fatou's lemma first to deduce from (B.9) that F ∈ H 1/2 (0, ∞) with

F H 1/2 (0,∞) ≤ δ + f H 1/2 (0,∞) ,
and then from (B.8) that

f -F H 1/2 (0,∞) ≤ δ . (B.10)
Similarly, for k ≥ 2, it follows from (B.5) and (B.7) that

f -F L 2 (I k ,dx/x) = k+2 j=1 ̺ ε j * (ψ j f ) - k+2 j=1 ψ j f L 2 (I k ,dx/x) ≤ k+2 j=1 ̺ ε j * (ψ j f ) -ψ j f L 2 (I k ,dx/x) ≤ δ . (B.11)
In particular,

F L 2 (I k ,dx/x) ≤ δ + f L 2 (I k ,dx/x) , (B.12)
and we invoke again Fatou's lemma to derive first from (B.12) that F ∈ L 2 ((0, ∞), dx/x), and then from (B.11) that

f -F L 2 ((0,∞),dx/x) ≤ δ . (B.13)
According to (B.10) and (B.13), we have constructed a function F ∈ C ∞ ([0, ∞)) ∩ X lying in a δ-neighborhood of f in X. This result being valid whatever the value of δ ∈ (0, 1), we have established the density of C ∞ ([0, ∞))∩X in X. Finally, we use a standard truncation argument to deduce that

C ∞ c ([0, ∞)) ∩ X is dense in X. Step 2: Density of C ∞ c ((0, ∞)) in X.
We argue as in the proofs of [12, Theorems 2.9.2 (c) & 2.9. 3 (d)]. We fix χ ∈ C ∞ (R) such that χ(x) = 0 for x ∈ (-∞, 1], χ(x) = 1 for x ∈ [2, ∞), and χ(x) ∈ [0, 1] for x ∈ [START_REF] Amann | Analysis III[END_REF][START_REF] Brezis | Functional analysis, Sobolev spaces and partial differential equations[END_REF], and set χ λ (x) := χ(x/λ) for x ∈ R and λ ∈ (0, 1).

Let f ∈ X and δ ∈ (0, 1). According to the previous step, there is

F ∈ C ∞ c ([0, ∞)) ∩ X such that f -F H 1/2 (0,∞) + f -F L 2 ((0,∞),dx/x) ≤ δ 2 . (B.14) Since F ∈ C ∞ c ([0, ∞)
) ∩ X, we observe that, for ε ∈ (0, 1),

F L 2 ((0,∞),dx/x) ≥ 1 0 |F (x)| 2 dx x + ε = 1 0 F (0) + x 0 F ′ (y) dy 2 dx x + ε ≥ 1 0 |F (0)| 2 2 - x 0 F ′ (y) dy 2 dx x + ε ≥ |F (0)| 2 2 ln 1 + 1 ε -F ′ 2
L∞(0,1)

1 0 x 2 x + ε dx ≥ |F (0)| 2 2 ln 1 + 1 ε - F ′ 2 L∞(0,1)
2 .

Letting ε → 0 in the above inequality implies that F (0) = 0, from which we deduce that

|F (x)| ≤ A 1 x , x ∈ [0, 2] , (B.15)
with A 1 := F ′ L∞(0,2) . Now, for λ ∈ (0, 1), it follows from (B.15) that

F -χ λ F L 2 (0,∞) = 2λ 0 1 -χ λ (x) 2 |F (x)| 2 dx 1/2 ≤ A 1 2λ 0 x 2 dx 1/2 ≤ √ 3A 1 λ 3/2 (B.16) and F ′ -(χ λ F ) ′ L 2 (0,∞) = (1 -χ λ )F ′ + F χ ′ λ L 2 (0,∞) ≤ (1 -χ λ )F ′ L 2 (0,∞) + F χ ′ λ L 2 (0,∞) ≤ 2λ 0 |F ′ (x)| 2 dx 1/2 + A 1 λ 2λ λ x 2 χ ′ x λ 2 dx 1/2 ≤ √ 2λ F ′ L∞(0,2) + 7λ 3 A 1 χ ′ L∞(R) ≤ A 2 √ λ , (B.17)
where

A 2 := 2A 1 (1 + χ ′ L∞(R) )
. By interpolation, we deduce from (B.16) and (B.17) that

F -χ λ F H 1/2 (0,∞) ≤ A F -χ λ F 1/2 H 1 (0,∞) F -χ λ F 1/2 L 2 (0,∞) ≤ Aλ (B.18)
for some constant A > 0 depending only on F and χ. Similarly, by (B.15),

F -χ λ F L 2 ((0,∞),dx/x) = 2λ 0 1 -χ λ (x) 2 |F (x)| 2 dx x 1/2 ≤ A 1 2λ 0 x dx 1/2 = √ 2A 1 λ . (B.19)
Thanks to (B.18) and (B.19), there is λ δ ∈ (0, 1) such that

F -χ λ δ F H 1/2 (0,∞) + F -χ λ δ F L 2 ((0,∞),dx/x) ≤ δ 2 .
Together with (B.14), the above estimate ensures that

f -χ λ δ F H 1/2 (0,∞) + f -χ λ δ F L 2 ((0,∞),dx/x) ≤ δ ,
E ∈ L L 2 (0, 1), L 2 (0, ∞) ∩ L H 1 (0, 1), H 1 (0, ∞) , E ∈ L L 2 ((0, 1), dx/x), L 2 ((0, ∞), dx/x) ,
and satisfies Ef = f a.e. on (0, 1). By interpolation, E ∈ L H 1/2 (0, 1), H 1/2 (0, ∞) . Now, let f ∈ H 1/2 (0, 1)∩L 2 ((0, 1), dx/x) and δ ∈ (0, 1). Since Ef belongs to

H 1/2 (0, ∞)∩ L 2 ((0, ∞), dx/x), we infer from Lemma B.2 that there is ξ ∈ C ∞ c ((0, ∞)) such that Ef -ξ H 1/2 (0,∞) + Ef -ξ L 2 ((0,∞),dx/x) ≤ δ 2 . (B.20) We again fix χ ∈ C ∞ (R) such that χ(x) = 0 for x ∈ (-∞, 1], χ(x) = 1 for x ∈ [2, ∞), and χ(x) ∈ [0, 1] for x ∈ [1, 2]. For λ ∈ (0, 1/4) and x ∈ (0, ∞), we set ξ λ (x) := ξ(x)χ 1 -x λ ,
and observe that

ξ λ (x) = 0 for x ∈ [1 -λ, ∞) and ξ λ (x) = ξ(x) for x ∈ (0, 1 -2λ] . Since ξ ∈ C ∞ c ((0, ∞)) and χ ∈ C ∞ (R), we conclude that ξ λ ∈ C ∞ c ((0, 1
)) for all λ ∈ (0, 1/4). Furthermore,

ξ -ξ λ L 2 (0,1) ≤ √ 2λ ξ L∞(0,1) , ξ -ξ λ L 2 ((0,1),dx/x) ≤ 2 √ λ ξ L∞(0,1) , ξ ′ -ξ ′ λ L 2 (0,1) ≤ 2 √ λ ξ C 1 ([0,1]) . (B.21)
By interpolation, we infer from (B.21) that

ξ -ξ λ H 1/2 (0,1) ≤ b ξ -ξ λ 1/2 H 1 (0,1) ξ -ξ λ 1/2 L 2 (0,1) ≤ b ξ C 1 ([0, 1
]) for some positive constant b depending only on χ. The above estimate ensures that (ξ λ ) λ∈(0,1/4) is bounded in H 1/2 (0, 1), while it follows from (B.21) that (ξ λ ) λ∈(0,1/4) converges to ξ in L 2 ((0, 1), (1 + 1/x)dx). Consequently, there is a sequence (λ j ) j≥1 in (0, 1/4),

λ j → 0 as j → ∞, such that lim j→∞ 1 0 ξ λ j (x) -ξ(x) 2 1 + 1 x dx = 0 , (B.22) ξ λ j ⇀ ξ in H 1/2 (0, 1) . (B.23) Introducing θ := θ1 Ω(v) , we compute θn -θ 2 L 2 (Ω M ) = Ω(vn)∩Ω(v) |θ n -θ| 2 d(x, z) + Ω(vn)∩(Ω M \Ω(v)) |θ n | 2 d(x, z) + (Ω M \Ω(vn))∩Ω(v) |θ| 2 d(x, z) ≤ Ω(vn)∩Ω(v) |θ n -θ| 2 d(x, z) + 2 Ω(vn)∩(Ω M \Ω(v)) |θ n -θ| 2 d(x, z) + 2 Ω(vn)∩(Ω M \Ω(v)) |θ| 2 d(x, z) + (Ω M \Ω(vn))∩Ω(v) |θ| 2 d(x, z) ≤ 2 θ n -θ 2 L 2 (Ω M ) + 2 Ω(vn)∩(Ω M \Ω(v)) |θ| 2 d(x, z) + (Ω M \Ω(vn))∩Ω(v) |θ| 2 d(x, z) .
Owing to (C.1),

lim n→∞ |Ω(v n ) ∩ (Ω M \ Ω(v))| = lim n→∞ |(Ω M \ Ω(v n )) ∩ Ω(v)| = 0 ,
a property which, together with (C.2) and Lebesgue's theorem, entails that the right-hand side of the above inequality converges to zero as n → ∞. Consequently, ( θn ) n≥1 converges to θ in L 2 (Ω M ). This property, along with (C.4), implies that θ ∈ H 1 0 (Ω M ) and, bearing in mind that Ω(v) ⊂ Ω M and the compactness of the embedding of H

1 (Ω(v)) in H 3/4 (Ω(v)), θn ⇀ θ in H 1 (Ω M ) , θn → θ in H 3/4 (Ω(v)) . (C.5)
Invoking (2.1d) and the continuity of the trace, we deduce

θn + h vn → θ + h v in L 2 (∂Ω(v)) . (C.6)
We now claim that (C.5) and (C.6) imply θ ∈ H 1 0 (Ω(v)). Indeed, on the one hand, θ = θ vanishes on D × {-H} and on {±L} × (-H, d). On the other hand, we infer from Hölder's inequality that

h vn (x, v n (x) + d) -( θn + h vn )(x, v(x) + d) = ( θn + h vn )(x, v n (x) + d) -( θn + h vn )(x, v(x) + d) = vn(x)+d v(x)+d ∂ z ( θn + h vn )(x, z) dz ≤ |v n (x) -v(x)| 1/2 M -H |∂ z ( θn + h vn )(x, z)| 2 dz 1/2
for a.e. x ∈ D, and thus

D |h vn (x, v n (x) + d) -( θn + h vn )(x, v(x) + d)| 2 dx ≤ D |v n (x) -v(x)| M -H |∂ z ( θn + h vn )(x, z)| 2 dzdx ≤ v n -v L∞(D) min{σ 1 , σ 2 } Ω M σ|∇( θn + h vn )(x, z)| 2 d(x, z) = 2 v n -v L∞(D) min{σ 1 , σ 2 } G(v n )[θ n ] + Ω M \Ω(vn) σ|∇h vn (x, z)| 2 d(x, z) .
Note that the sum embraced with curly brackets is bounded due to the boundedness of (G(v n )[θ n ]) n≥1 and (2.1d). Consequently, the uniform convergence (C.1) guarantees that the right-hand side of the above inequality converges to zero as n → ∞. Hence, due to (2.1e) and (C.6), we conclude that θ = 0 on {(x, v(x)+d) : x ∈ D} and thus θ ∈ H 1 0 (Ω(v)). Now, by (2.1d) and (C.5), θn + h vn ⇀ θ + h v in H 1 0 (Ω M ) , so that Step 2: Existence of a Recovery Sequence. We only need to provide a recovery sequence for θ ∈ H 1 0 (Ω(v)), in view of the definition of the functional G(v). Note that θ ∈ H 1 0 (Ω M ) and that f := -∆θ ∈ H -1 (Ω M ) can be considered also as an element of H -1 (Ω(v n )) by restriction. Let now θ n ∈ H 1 0 (Ω(v n )) denote the unique weak solution to -∆θ n = f in Ω(v n ) , θ n = 0 on ∂Ω(v n ) . Since θ and θ both belong to H 1 0 (Ω M ), we readily deduce from the above equation that θ = θ, so that θ n → θ in H 1 0 (Ω M ). Since θ n ∈ H 1 0 (Ω(v n )) and θ ∈ H 1 0 (Ω(v)), this convergence yields, along with (2.1d) and (2. and recall that χ n is a minimizer of G(v n ) in H 1 0 (Ω(v n )) by Lemma 3.1 and Ω M = D × (-H, M ). Since (v n ) n≥1 is bounded in H 1 (D), it follows from Lemma 3.1, Lemma 3.2, and (2.2) that (χ n ) n≥1 is bounded in H 1 0 (Ω M ). Hence, there are a subsequence (n j ) j≥1 and χ ∈ H 1 0 (Ω M ) such that χ n j → χ in L 2 (Ω M ) and χ n j ⇀ χ in H 1 0 (Ω M ). By Lemma C.1 and the Fundamental Theorem of Γ-Convergence, see [START_REF] Maso | An introduction to Γ-convergence[END_REF]Corollary 7.20], χ is a minimizer of the functional G(v) on L 2 (Ω M ). Clearly, from the definition of G(v), we see that χ + h v ∈ A(v) minimizes the functional J (v) on A(v), hence ψ v = χ + h v by Lemma 3.1. Consequently, χ = ψ vh v is the unique cluster point of the sequence (χ n ) n≥1 in L 2 (Ω M ) and this sequence is compact in that space and weakly compact in H 1 0 (Ω M ). Combining these properties leads us to conclude that χ n → χ in L 2 (Ω M ) and χ n ⇀ χ in H 1 0 (Ω M ). Moreover, the Fundamental Theorem of Γ-Convergence, see [START_REF] Maso | An introduction to Γ-convergence[END_REF]Corollary 7.20], also ensures that

Ω M σ|∇( θ + h v )| 2 d(x, z) ≤ lim inf
lim n→∞ J (v n )[ψ vn ] = lim n→∞ G(v n )[χ n ] = G(v)[χ] = J (v)[ψ v ] .
In particular, this property, along with (2.1d), implies that lim n→∞ χ n H 1 0 (Ω M ) = χ H 1 0 (Ω M ) .

Since (χ n ) n≥1 converges weakly to χ in H 1 0 (Ω M ), this gives the strong convergence of (χ n ) n≥1 in H 1 0 (Ω M ).

We now check that ψ v is a strong solution to (4.1). As a minimizer of J (v) on A(v) according to Lemma 3.1, the function ψ v satisfies Proof of Proposition 4.9 (b). Proposition 4.9 (b) is now a straightforward consequence of Proposition 4.9 (a) and Lemma 4.10.

Figure 2 .

 2 Figure 2. Geometry of Ω(w) for a state w ∈ S with non-empty coincidence set.

Remark 1 . 2 .

 12 When the upper part Ω 2 (v) is clamped at its lateral boundaries in the sense that u ∈ H 2 0 (D) := {v ∈ H 2 (D) ∩ H 1 0 (D) : ∂ x v(±L) = 0} , Theorem 1.1 applies whatever the values of σ 1 and σ 2 .

Remark 4 . 4 .

 44 If ∂ x v(±L) = 0, then(4.11) reduces to G(±L, η) = 0 for η ∈ (0, 1 + d) and the crucial identity (4.12) used in the proof of Lemma 4.3 directly follows from[START_REF] Grisvard | Elliptic problems in nonsmooth domains[END_REF] Lemma 4.3.1.2]. For the general case v ∈ S, we require the extension given in Lemma A.1.

Lemma 4 . 7 .

 47 Let ζ ∈ (3/4, 1) and κ > 0. There is c

  To prove (4.22), let us first note that H ζ-1/2 (D) embeds continuously into L 4 (D). We use the Cauchy-Schwarz inequality and Lemma 4.6 with α = ζ -1/2 and deduce

Figure 3 .

 3 Figure 3. Geometry of Ω(w) for a state w ∈ S with non-empty and disconnected coincidence set.

( 4

 4 .27) Owing to the continuous embedding of H 2 (D) in C( D), combining (4.27) and Lemma 3.2 leads us to the estimate

4. 3 .

 3 H 2 -Regularity and H 2 -Estimates on ψ v for v ∈ S. Finally, we extend Proposition 4.1 and Proposition 4.8 by showing the H 2 -regularity of ψ v and the corresponding H 2 -estimates for an arbitrary v ∈ S; that is, we drop the additional W 2

Figure 4 .

 4 Figure 4. The rectangle R.

S 1 :

 1 = (L, 0) , S 2 := (L, 1 + d) , S 3 := (-L, 1 + d) , S 4 := (-L, 0) , and edges Γ 1 := D × {0} = (S 4 , S 1 ) , Γ 2 := {L} × (0, 1 + d) = (S 1 , S 2 ) , Γ 3 := D × {1 + d} = (S 2 , S 3 ) , Γ 4 := {-L} × (0, 1 + d) = (S 3 , S 4 ) , see Figure 4. We next introduce local charts (Y i ) 1≤i≤4 in the neighborhood of the vertices (S i ) 1≤i≤4 defined by

  and completes the proof, sinceχ λ δ F ∈ C ∞ c ((0, ∞)). Proof of Lemma B.1. The derivation of Lemma B.1 from Lemma B.2 is also adapted from the proofs of [12, Theorems 2.9.2 (c) & 2.9.3 (d)]. First, arguing as in the proof of [2, Theorem 8.6], we construct an extension operator

  n→∞Ω M σ|∇( θn + h vn )| 2 d(x, z) . (C.7) Since θn ∈ H 1 0 (Ω(v n )),Ω M \Ω(vn) σ|∇( θn + h vn )| 2 d(x, z) = Ω M \Ω(vn) σ|∇h vn | 2 d(x, z) ,and we deduce from (2.1d) thatlim n→∞ Ω M \Ω(vn) σ|∇( θn + h vn )| 2 d(x, z) = Ω M \Ω(v) σ|∇h v | 2 d(x, z) = Ω M \Ω(v) σ|∇( θ + h v )| 2 d(x, z) , (C.8)the last equality being due to θ ∈ H 1 0 (Ω(v)). Combining (C.7) and (C.8) gives (C.3).

  Since the Hausdorff distance dH in Ω M (see [5, Section 2.2.3]) satisfies d H (Ω(v n ), Ω(v)) ≤ v nv L∞(D) → 0 by (C.1) and since Ω M \ Ω(v n ) has a single connected component for every n ≥ 1 as v n ≥ -H, it follows from [13, Theorem 4.1] and [5, Theorem 3.2.5] that θ n → θ in H 1 0 (Ω M ), where θ ∈ H 1 0 (Ω M ) is the unique weak solution to -∆ θ = f = -∆θ in Ω M , θ = 0 on ∂Ω M .

  [START_REF] Brezis | Functional analysis, Sobolev spaces and partial differential equations[END_REF],Ω(v) σ|∇(θ + h v )| 2 d(x, z) = Ω(v) σ |∇θ| 2 + 2∇θ • ∇h v + |∇h v | 2 d(x, z) = lim n→∞ Ω M σ |∇θ n | 2 + 2∇θ n • ∇h vn d(x, z) + lim n→∞ Ω(vn) σ|∇h vn | 2 d(x, z) = lim n→∞ Ω(vn) σ|∇(θ n + h vn )| 2 d(x, z) ; that is, G(v)[θ] = lim n→∞ G(v n )[θ n ] .Hence, (θ n ) n≥1 is a recovery sequence for θ.Thanks to the just established two properties, we have proved theΓ-convergence of (G(v n )) n≥1 to G(v) in L 2 (Ω M ).Proposition 3.3 is now an almost immediate consequence of Lemma C.1. Proof of Proposition 3.3. For n ≥ 1, set χ n := ψ vnh vn ∈ H 1 0 (Ω(v n )) ⊂ H 1 0 (Ω M ) ,

Ω

  (v) σ∇ψ v • ∇θ d(x, z) = 0 , θ ∈ H 1 0 (Ω(v)) . (D.5)Thus, since (ψ v,1 , ψ v,2 ) ∈ H 2 (Ω 1 (v)) × H 2 (Ω 2 (v)) it readily follows that div(σ∇ψ v ) = 0 in Ω(v)as claimed in (4.1a). Moreover, owing to v ∈ C( D), we can write the open set {x ∈ D : v(x) > -H} as a countable union of open intervals ((a i , b i )) i∈I , see [1, IX.Proposition 1.8]. Let i ∈ I and setO i (v) := {(x, z) ∈ (a i , b i ) × R : -H < z < v(x) + d} ⊂ Ω(v) .For each θ ∈ D(O i (v)) we infer from (D.5) and Gauß' theorem that0 = O i (v) σ∇ψ v • ∇θ d(x, z) = b i a i σ∇ψ v • n Σ(v) θ (x, v(x)) dx , hence σ∇ψ v • n Σ(v) (•, v(•)) = 0 a.e. in (a i , b i ). Therefore, σ∇ψ v • n Σ(v) = 0 on Σ(v)as stated in (4.1b). Finally, since ψ v ∈ H 1 (Ω(v)) we have ψ v = 0 on Σ(v), while (4.1c) is due to ψ v ∈ A(v).

recalling that ψ vn,i = ψ vn | Ω i (vn) .

The proof involves three steps: we first establish Proposition 4.9 (b) under the additional assumption sup n≥1 ψ vn,1 H 2 (Ω 1 (vn)) + ψ vn,2 H 2 (Ω 2 (vn)) < ∞ .

Building upon this result, we take advantage of the density of S ∩ W 2 ∞ (D) in S and of the estimates derived in Proposition 4.8 to verify Proposition 4.9 (a) by a compactness argument. Combining the previous steps leads us finally to a complete proof of Proposition 4.9 (b). We thus start with the proof of Proposition 4.9 (b) when the solutions (ψ vn ) n≥1 to (4.1) associated with the sequence (v n ) n≥1 satisfies the above additional bound. We state this result as a separate lemma for definiteness. Lemma 4.10. Let κ > 0 and v ∈ S be such that v H 2 (D) ≤ κ and consider a sequence (v n ) n≥1 in S satisfying (4.29). Assume further that, for each n ≥ 1, (ψ vn,1 , ψ vn,2 ) belongs to

The proof of Lemma 4.10 is very close to that of [7, Proposition 3.13 & Corollary 3.14]. For the sake of completeness we provide a detailed proof in Appendix D.

Proof of Proposition 4.9 (a). Let v ∈ S be such that v H 2 (D) ≤ κ. We may choose a sequence

Owing to (4.32) and the regularity property

and (ψ vn ) n≥1 satisfies (4.30) with µ = c 0 (2κ). We then infer from Lemma 4.10 that (ψ v,1 , ψ v,2 ) belongs to

Combining the above bound with (2.1d) and Lemma 4.10 gives (4.28). It remains to check that ψ v is a strong solution to (4.1) which can be done as in [START_REF]Shape derivative of the Dirichlet energy for a transmission problem[END_REF]Corollary 3.14], see Appendix D for details.

Proof of Proposition 4.9 (b). Proposition 4.9 (b) is now a straightforward consequence of Proposition 4.9 (a) and Lemma 4.10.

Proof of Theorem 1.1. The proof of Theorem 1.1 readily follows from Proposition 4.9 (a). This appendix is devoted to the proof of the identity (4.12), which can be seen as a variant of [START_REF] Grisvard | Elliptic problems in nonsmooth domains[END_REF]Lemma 4.3.1.2] with piecewise constants linear constraints on the boundaries instead of constant ones.

where τ ± are piecewise constants functions of the form

As in the proof of [4, Lemma 4.3.1.3], the core of the proof of Lemma A.2 is to establish the density of the space Z 2 (∂R) of traces of functions in G 2 (R) in the space Z 1 (∂R) of traces of functions in G 1 (R), after identifying these two trace spaces. The proof is almost identical to that of [4, Lemma 4.3.1.3] and we postpone it to the end of this appendix.

Proof of Lemma A.1. Due to Lemma A.2, it suffices to prove the identity in Lemma A.1 when (V, W ) belongs to G 2 (R). This additional regularity allows us to use integration by parts to interchange the derivatives and guarantees the continuity of both V and W on R. Indeed, H 2 (R) embeds continuously in C α ( R) for all α ∈ (0, 1) by [11, Chapter 2, Theorem 3.8] and we deduce that

Now, according to [4, Theorem 1.5.2.3] and the definition of G 1 (R), the space Z 1 (∂R) is the subspace of

with δ := min{1 + d, 2L} and (v 5 , w 5 ) := (v 1 , w 1 ). Owing to (A.8) and (A.9), the integrability properties listed in (A.9) simplify and Z 1 (∂R) is isomorphic to

,

We next turn to Z 2 (∂R) and deduce from [4, Theorem 1.5.2.8] that it is the subspace of

with (v 5 , w 5 ) = (v 1 , w 1 ). As above, the continuity requirements in (A.11) simplify due to (A.8) and we conclude that Z 2 (∂R) is isomorphic to P 2 (∂R, R 2 ) with

,

where P 2 (∂R) is the subspace of 4 i=1 H 3/2 (Γ i ) defined by

by Lemma B.1 below and obviously included in P 2 (∂R), the density of P 2 (∂R) in P 1 (∂R) follows, and that of Z 2 (∂R) in Z 1 (∂R) as well. The remainder of the proof is then the same as in [START_REF] Grisvard | Elliptic problems in nonsmooth domains[END_REF]Lemma 4.3.1.3], to which we refer.

Appendix B. A Density Result

In this appendix, we recall a density result which is stated without proof in [START_REF] Grisvard | Elliptic problems in nonsmooth domains[END_REF]Lemma 4.3.1.3] and used in the proof of Lemma A.2. We provide a proof for the sake of completeness.

Lemma B.1. The space C ∞ c ((0, 1)) is dense in H 1/2 (0, 1) ∩ L 2 ((0, 1), dx/x).

According to (B.23) and Mazur's lemma, there is a sequence ( ξk ) k≥1 made up of convex combinations of (ξ λ j ) j≥1 such that ( ξk ) k≥1 converges strongly to ξ in H 1/2 (0, 1). Each ξk being a convex combination of (ξ λ j ) j≥1 , it is obvious that ξk ∈ C ∞ c ((0, 1)) for each k ≥ 1 and that (B.22) entails that lim

Therefore, there is

Recalling that Ef = f a.e. in (0, 1), we combine (B.20) and (B.24) to conclude that ξk δ lies in a δ-neighborhood of f in H 1/2 (0, 1) ∩ L 2 ((0, 1), dx/x), thereby completing the proof.

Appendix C. Proof of Proposition 3.3

For M > 0, we set Ω M := D × (-H, M ) and define, for v ∈ S,

. The next result is devoted to the stability of G(v) with respect to v which we express in terms of Γ-convergence of functionals.

Lemma C.1. Consider v ∈ S and a sequence (v n ) n≥1 in S satisfying (3.4). Then

where M is defined in (3.5) and

Proof. The proof follows the lines of that of [START_REF]Shape derivative of the Dirichlet energy for a transmission problem[END_REF]Proposition 3.11]. A first consequence of (3.4) and the continuous embedding of

Step 1: Asymptotic Lower Semicontinuity. Let (θ n ) n≥1 be an arbitrary sequence in L 2 (Ω M ) and θ ∈ L 2 (Ω M ) such that

In order to prove that

we may assume without loss of generality that θ n ∈ H 1 0 (Ω(v n )) for all n ≥ 1 and that (G(v n )[θ n ]) n≥1 is bounded, since (C.3) is clearly satisfied otherwise owing to the definition of G. Therefore, denoting the extension of θ n by zero in Ω M \ Ω(v n ) by θn , it follows from assumption (2.2) that ( θn ) n≥1 is bounded in H 1 0 (Ω M ), so that ( θn ) n≥1 is weakly relatively compact in H 1 0 (Ω M ) .

(C.4)

Appendix D. Proof of Proposition 4.9

We first establish Lemma 4.10 which is the building block of the proof of Proposition 4.9.

Proof of Lemma 4.10. It first follows from (4.29) and the continuous embedding of H 2 (D) in C 1 ( D) that

We may then apply Proposition 3.3 to obtain that

Now, fix i ∈ {1, 2} and let U i be any open subset of Ω i (v) such that Ūi is a compact subset of Ω i (v). Owing to (D.1), there is an integer N ≥ 1 such that U i ⊂ Ω i (v n ) for n ≥ N . We then infer from (2.1d), (4.30), and (D.2) that (ψ vn,i ) n≥N is bounded in H 2 (U i ) and ψ vn,i → ψ v,i in H 1 (U i ) as n → ∞. Thus, (ψ vn,i ) n≥N is a weakly compact sequence in H 2 (U i ) which has a unique cluster point ψ v,i for that topology, so that ψ v,i ∈ H 2 (U i ) and

In particular, we deduce from (4.30) and (D.3) that

for (l, k) ∈ {(2, 0), (1, 1), (0, 2)}. We then use Fatou's lemma to conclude that ∂ l x ∂ k z ψ v,i belongs to L 2 (Ω i (v)) for (l, k) ∈ {(2, 0), (1, 1), (0, 2)} with

Therefore, ψ v,i ∈ H 2 (Ω i (v)) and (4.31) readily follows from (4.30) and the above estimate.

Proof of Proposition 4.9 (a). Let v ∈ S be such that v H 2 (D) ≤ κ. We may choose a sequence

Owing to (D.4) and the regularity property v n ∈ S ∩ W 2 ∞ (D), n ≥ 1, Proposition 3.3 guarantees that (ψ vn,1 , ψ vn,2 ) belongs to H 2 (Ω 1 (v n )) × H 2 (Ω 2 (v n )) and (ψ vn ) n≥1 satisfies (4.30) with µ = c 0 (2κ). We then infer from Lemma 4.10 that (ψ v,1 , ψ v,2 ) belongs to H 2 (Ω 1 (v)) × H 2 (Ω 2 (v)) and satisfies

Combining the above bound with (2.1d) and Lemma 4.10 gives (4.28).