On Duclos–Exner’s conjecture about waveguides in strong uniform magnetic fields

Abstract

We consider the Dirichlet Laplacian with uniform magnetic field on a curved strip in two dimensions. We give a sufficient condition on the width and the curvature of the strip ensuring the existence of the discrete spectrum in the strong magnetic field limit, answering (negatively) a conjecture made by Duclos and Exner.

1 Introduction and statement of the main results

In this article, we address the question of existence of the discrete spectrum for a magnetic Laplacian with Dirichlet boundary condition on a two-dimensional curved waveguide.

1.1 What is a waveguide?

Let ${\boldsymbol \gamma } : \mathbb {R} \to \mathbb {R}^2$ be a smooth and injective curve with $|{\boldsymbol \gamma }'|= 1$ . We set $\mathbf N = ({\boldsymbol \gamma }')^\bot $ , where for $(a,b) \in \mathbb {R}^2$ we write $(a,b)^\bot $ for $(-b,a)$ . We denote by $\kappa $ the algebraic curvature of ${\boldsymbol \gamma }$ . It is defined by

$$ \begin{align*}{\boldsymbol \gamma}" = \kappa \mathbf N.\end{align*} $$

In this article, we work under the assumption that $\kappa $ is compactly supported. The function

$$ \begin{align*} \Theta : \left\{ \begin{array}{ccc} \mathbb{R} \times (-\delta,\delta) & \to & \mathbb{R}^2 \\ (s,t) & \mapsto & {\boldsymbol \gamma}(s) + t \mathbf N(s) \end{array} \right. \end{align*} $$

is smooth with bounded derivatives, and it is injective for $\delta> 0$ small enough. In this case, we set

$$ \begin{align*}\Omega = \Omega_{{\boldsymbol \gamma},\delta} = \Theta (\Omega_0)\,,\quad \mbox{ with }\quad\Omega_0 = \Omega_{0,\delta} =\mathbb{R} \times (-\delta,\delta).\end{align*} $$

The open set $\Omega $ is what we call a waveguide in this work.

1.2 The magnetic Laplacian with Dirichlet boundary conditions

The waveguide $\Omega $ is subject to a perpendicular uniform magnetic field with intensity B. That is why we consider a vector potential $\mathbf A = (A_1,A_2)$ that is smooth on $\overline {\Omega }$ and such that

(1.1) $$ \begin{align} \partial_{x_1} A_2 - \partial_{x_2} A_1 = 1. \end{align} $$

A fundamental property related to magnetic problems on simply connected domains is the gauge invariance. It is nothing but the fact that equation (1.1) only defines $\mathbf {A}$ up to adding a gradient vector field. Of course, it is trivial that there is a smooth solution to equation (1.1) since it is sufficient to consider $\mathbf {A}=(0,x_1)$ . Actually, one will see that there is a natural choice of vector potential in our setting. Finding a gauge that is adapted to the structure of the waveguide is in fact part of our problem, and it has been tackled in the past; see, for instance, [7] where a curvature-dependent gauge is introduced. We now assume that $\mathbf {A}$ can be chosen smooth on $\overline {\Omega }$ and bounded with bounded derivatives (at any order). It will be explained in Proposition 1.2 that we may indeed assume this.

For $B> 0$ , we consider on $\Omega $ the magnetic Laplacian corresponding to the uniform field equal to B:

(1.2) $$ \begin{align} (-i\nabla - B \mathbf A)^2 - B, \end{align} $$

subject to Dirichlet boundary conditions. The subtraction of B is made for the convenience of the analysis and does not change the presence or absence of the discrete spectrum (it is based on relating the Schrödinger operator to the square of a Dirac operator). In order to use semiclassical analysis, we also introduce the positive parameter $h = B^{-1}$ and set

$$ \begin{align*} \mathscr{P}_{h}=(-ih\nabla-\mathbf{A})^2-h. \end{align*} $$

The operator $\mathscr {P}_{h}$ is well defined and selfadjoint on the domain

$$ \begin{align*} \mathsf{Dom}(\mathscr{P}_{h}) = H^1_0(\Omega) \cap H^2(\Omega). \end{align*} $$

1.3 A subtle question and a conjecture by P. Duclos and P. Exner

Our aim is to study the existence of the discrete spectrum of $\mathscr {P}_h$ in the semiclassical limit $h\to 0$ (equivalent to the large magnetic field limit; see formula (1.2)). This question of existence is actually subtle since, when h goes to $0$ , not only the bottom of the spectrum moves but also the bottom of the essential spectrum. In this limit, it is natural to wonder if the bottom of the spectrum stays away from the threshold of the essential spectrum or collides with it. This question is all the more appealing that, when the magnetic field is zero, that is when considering the Dirichlet Laplacian on a strip, one knows that the discrete spectrum always exists as soon as the strip is not straight (see, for instance, [4] or the book [9, Chapter 1]). It is also known that (variable) magnetic fields can play against the existence of the discrete spectrum. Such considerations can be found in [14, Theorem 2.8 & Proposition 2.11] where a magnetic Hardy inequality is proved when the magnetic field has compact support and used to establish that the discrete spectrum is empty when the magnetic field is strong enough (see also the original work [5])¹.

In the mid nineties, buoyed by the momentum of their work [4], Pierre Duclos and Pavel Exner conjectured that the discrete spectrum of operator (1.2) is empty when the magnetic field is strong enough (and uniform). This conjecture was explicitly formulated 10 years ago during an “Open Problems” session in Barcelona; see [8]. Our main result disproves the conjecture when the waveguide has a fixed width $\delta $ assumed to be small enough but independent of B.

1.4 Main result

Our main result is the following.

Theorem 1.1. Assume that $\kappa ^2$ has a unique maximum, that is nondegenerate. There exist $\delta _0> 0$ and $h_0> 0$ such that for all $\delta \in (0,\delta _0)$ and all $h \in (0,h_0)$ we have

$$ \begin{align*} \inf \mathrm{sp}(\mathscr{P}_{h}) < \inf \mathrm{sp}_{\mathsf{ess}} (\mathscr{P}_{h}).\\[-15pt] \end{align*} $$

In particular, $\mathscr {P}_{h}$ has nonempty discrete spectrum.

We can be more precise and provide some bounds for the bottoms of spectrum and essential spectrum. For this, we compare the spectral properties of the magnetic Laplacian on $\Omega $ to those on $\Omega _0$ . On $\Omega _0$ , we set $\mathbf A_0(s,t) = (-t,0)$ and we consider in $L^2(\Omega _0)$ the operator $\mathscr {P}_{h,0} = (-ih\nabla - \mathbf A_0)^2 - h$ , with Dirichlet boundary conditions. For $h> 0$ , we set

$$ \begin{align*} \lambda_{\mathrm{ess}}(h) = \inf \mathrm{sp}(\mathscr{P}_{h,0}).\\[-15pt] \end{align*} $$

The following proposition gives a rather naive lower bound of the infimum of the essential spectrum. It is likely not optimal (due to the presence of the $h^2$ factor), but it will be sufficient for our analysis.

Proposition 1.1. We have

$$ \begin{align*} \mathrm{sp}_{\mathrm{ess}}(\mathscr{P}_h)=\mathrm{sp}_{\mathrm{ess}}(\mathscr{P}_{h,0})=\mathrm{sp}(\mathscr{P}_{h,0})=[\lambda_{\mathrm{ess}}(h),+\infty)\\[-15pt] \end{align*} $$

and

$$ \begin{align*} \lambda_{\mathrm{ess}}(h)\geqslant \frac{(\pi h)^2}{4\delta^2} e^{-\delta^2/h}.\\[-15pt] \end{align*} $$

To prove an upper bound on the bottom of the spectrum, we first introduce on $\Omega _0$ the function $\phi _0$ defined by

$$ \begin{align*} \phi_0(s,t)=\frac{t^2-\delta^2}2.\\[-15pt] \end{align*} $$

Then we define $\hat \phi _0 = \phi _0 \circ \Theta ^{-1} \in \mathscr C^\infty (\overline \Omega )$ . In particular, $\hat \phi _0$ vanishes on $\partial \Omega $ . In order to perform the analysis of the bottom of the spectrum, we will use a function $\phi $ , looking like $\hat \phi _0$ at infinity, defined thanks to the following proposition. We will use the following notation for the Schwartz space

$$ \begin{align*}\mathscr{S}(\overline{\Omega})=\{\psi\in\mathscr{C}^\infty(\overline{\Omega}) : \forall (\alpha,\beta)\in\mathbb{N}^2\,,\quad \exists C_{\alpha,\beta}>0 : \|x^\alpha\partial^\beta\psi\|_{\infty}\leqslant C_{\alpha,\beta}\}.\end{align*} $$

Proposition 1.2. There exists a unique $\phi \in \mathscr {C}^\infty (\overline {\Omega })$ such that $\Delta \phi =1$ , $\phi _{|\partial \Omega }=0$ , and $\phi -\hat \phi _0\in \mathscr {S}(\overline {\Omega })$ . Moreover, there exists $c_0>0$ such that $\partial _\nu \phi \geqslant c_0$ on $\partial \Omega $ , $\nu $ being the outward pointing normal to the boundary.

Then, by gauge invariance, we can choose $\mathbf A = (\nabla \phi )^\bot $ in the definition of $\mathscr P_h$ . In particular, we may assume that $\mathbf A$ is bounded on $\Omega $ , as announced in Section 1.2. Here comes our result ensuring the existence of the discrete spectrum.

Theorem 1.2. Assume that $\phi $ given by Proposition 1.2 has a unique minimum $\phi _{\min }$ (reached at $x_{\min } \in \Omega $ ) that is nondegenerate and smaller than $\min \phi _0 = - \delta ^2 / 2$ . Then, as $h \to 0$ , we have

$$ \begin{align*}\inf\mathrm{sp}(\mathscr{P}_h)\leqslant \frac{J}{\pi}\sqrt{\det\mathrm{Hess}_{x_{\min}}\phi}e^{2\phi_{\min}/h} \big(1+o(1)\big),\end{align*} $$

with

$$ \begin{align*}J=2\underset{f\in\mathscr{E}}{\inf}\|(\partial_\nu \phi) ^{\frac12}f\|^2_{\partial\Omega},\end{align*} $$

and

$$ \begin{align*}\mathscr{E}=\{f\in\mathscr{O}(\Omega)\cap H^1(\Omega) : f(x_{\min})=1\},\end{align*} $$

where $\mathscr {O}(\Omega )$ is the set of holomorphic functions on $\Omega $ .

Remark 1.3.

1. 1. The set $\mathscr {E}$ is not empty as we can see by considering a function of the form $f\colon z\mapsto c(z-z_1)^{-2}$ with $z_1\notin \overline {\Omega }$ and c such that $f(x_{\min })=1$ .

2. 2. Due to a classical trace theorem and the fact that $\partial _\nu \phi $ is bounded, J is finite.

3. 3. The fact that $\phi $ has a unique minimum (which is nondegenerate) can be ensured under explicit assumptions on the curvature $\kappa $ and on the width of the waveguide; see Proposition 1.3 below.

4. 4. By using Proposition 1.1 and under the assumption on $\phi $ in Theorem 1.2, we have $\inf \mathrm {sp}(\mathscr P_h) < \inf \mathrm {sp}_{\mathsf {ess}}(\mathscr P_h)$ .

Our proof of Theorem 1.2 is based on extensions of strategies used in [1]², where the asymptotic simplicity of the low-lying eigenvalues is established, under generic assumptions on $\Omega $ . Let us emphasize that, in [1], $\Omega $ is assumed to be bounded and that the assumption on $\phi $ can be ensured, in the uniform magnetic field case, when $\Omega $ is strictly convex (thanks to the works by Kawohl [12, 13]). In the present setting, $\Omega $ is neither bounded, nor convex. Moreover, in our unbounded setting, one needs to be very careful since the functional spaces (such as the Hardy spaces) involved in [1] are no more obviously well defined. The study of such spaces on strips³ has an interest of its own, and their use to deduce precise spectral asymptotics will be the object of a future work. Fortunately, we do not need them to disprove Duclos–Exner’s conjecture.

To complete our analysis, it remains to give a sufficient condition under which the assumption of Theorem 1.2 is satisfied.

Proposition 1.3. Assume that $\kappa \in \mathscr {C}^\infty _0(\mathbb {R})$ and that $\kappa ^2$ has a unique maximum, which is nondegenerate. There exists $\delta _0>0$ such that, for all $\delta \in (0,\delta _0)$ , $\phi $ has a unique minimum $\phi _{\min }$ in $\Omega $ that is nondegenerate. Moreover, $\phi _{\min }<(\phi _0)_{\min }$ .

Theorem 1.1 follows from Proposition 1.1, Proposition 1.2, Theorem 1.2 and Proposition 1.3. Due to our motivation to disprove a conjecture from the nineties, we provide the reader with rather self-contained proofs (and sometimes recall basic arguments). In Section 2, we analyze the essential spectrum and we prove Proposition 1.1. In Section 3, the existence of the function $\phi $ is established and we prove Propositions 1.2 and 1.3. In Section 4, we prove Theorem 1.2.

2 The essential spectrum

In this section, we prove Proposition 1.1, which follows from Propositions 2.1 and 2.2. We first recall a classical result.

Lemma 2.1. Let $\phi \in \mathscr C^\infty (\overline \Omega )$ be bounded with bounded derivatives and $\mathbf {A} = (\nabla \phi )^\bot $ . For all $\psi \in H^1_0(\Omega )$ , we have

$$ \begin{align*}\|(-ih\nabla-\mathbf{A})\psi\|_{L^2(\Omega)}^2-h\|\psi\|_{L^2(\Omega)}^2=4h^2\int_{\Omega}e^{-2\phi/h}|\partial_{\bar{z}} u|^2\mathrm{d}x,\end{align*} $$

where $u:=e^{\phi /h}\psi \in H^1_0(\Omega )$ and $z = x_1 + i x_2$ , $x=(x_1,x_2)$ .

Proof. We have

$$ \begin{align*}\begin{split} 4h^2\int_{\Omega}e^{-2\phi/h}|\partial_{\bar{z}} u|^2\mathrm{d}x&=\int_{\Omega}|e^{-\phi/h}(h\partial_{1}+ih\partial_2) u|^2\mathrm{d}x\\ &=\int_{\Omega}|(h\partial_{1}+ih\partial_2) e^{-\phi/h}u-[h\partial_1+ih\partial_2,e^{-\phi/h}]u|^2\mathrm{d}x\\ &=\int_{\Omega}|(h\partial_{1}+i\partial_2\phi+ih\partial_2+\partial_1\phi) \psi|^2\mathrm{d}x\\ &=\int_{\Omega}|(h\partial_{1}-iA_1+ih\partial_2+A_2) \psi|^2\mathrm{d}x\\ &=\int_{\Omega}|(L_1+iL_2) \psi|^2\mathrm{d}x\,,\quad L_j=-ih\partial_j-A_j.\\ \end{split}\end{align*} $$

Then, we get

$$ \begin{align*} \begin{split} 4h^2\int_{\Omega}e^{-2\phi/h}|\partial_{\bar{z}} u|^2\mathrm{d}x&=\|(-ih\nabla-\mathbf{A})\psi\|^2+2\mathrm{Re}\,\langle L_1\psi,iL_2\psi\rangle\\ &=\|(-ih\nabla-\mathbf{A})\psi\|^2+2\mathrm{Im}\,\langle L_1\psi,L_2\psi\rangle.\\ \end{split} \end{align*} $$

Note that

$$ \begin{align*} 2\mathrm{Im}\,\langle L_1\psi,L_2\psi\rangle & =2\mathrm{Im}\,\langle \psi,L_1L_2\psi\rangle\\ & =2\mathrm{Im}\,\langle \psi,L_2L_1\psi+[L_1,L_2]\psi\rangle\\ & =2\mathrm{Im}\,\langle L_2\psi,L_1\psi\rangle-2h\|\psi\|^2. \end{align*} $$

The conclusion follows.

Proposition 2.1. For all $h>0$ , we have

$$ \begin{align*}\mathrm{sp}(\mathscr{P}_{h,0})=[\lambda_{\mathrm{ess}}(h),+\infty),\end{align*} $$

and

$$ \begin{align*}\lambda_{\mathrm{ess}}(h)\geqslant \frac{(\pi h)^2}{4\delta^2} e^{-\delta^2/h}. \end{align*} $$

Proof. By using the Fourier transform with respect to s, we have

$$ \begin{align*}\mathscr{P}_{h,0}=\int^\oplus \mathscr{P}_{h,0,\xi}\mathrm{d}\xi,\end{align*} $$

where the operator

$$ \begin{align*}\mathscr{P}_{h,0,\xi}=-h^2\partial_t^2+(\xi+t)^2-h\end{align*} $$

is equipped with the Dirichlet conditions at $t=\pm \delta $ . Let us denote by $(\gamma _n(\xi ,h))_{n\geqslant 1}$ the increasing sequence of its eigenvalues. A straightforward application of the min-max theorem shows that, for all $h>0$ ,

$$ \begin{align*}\lim_{\xi\to\pm\infty}\gamma_n(\xi,h)=+\infty.\end{align*} $$

We get

$$ \begin{align*}\mathrm{sp}(\mathscr{P}_{h,0})=[\min_{\xi\in\mathbb{R}}\gamma_1(\xi,h),+\infty)=\mathrm{sp}_{\mathrm{ess}}(\mathscr{P}_{h,0}).\end{align*} $$

By the min-max principle, we have

$$ \begin{align*}\inf \mathrm{sp}(\mathscr{P}_{h,0})=\inf_{\psi\in H^1_0(\Omega_0)\setminus\{0\}}\frac{\|(-ih\nabla-\mathbf{A}_0)\psi\|^2-h\|\psi\|^2}{\|\psi\|^2},\end{align*} $$

and, by letting $\psi =e^{-\phi _0/h}u$ , we get

$$ \begin{align*}\inf \mathrm{sp}(\mathscr{P}_{h,0})=\inf_{u\in H^1_0(\Omega_0)\setminus\{0\}}\frac{4h^2\|e^{-\phi_0/h}\partial_{\bar{z}}u\|^2}{\|e^{-\phi_0/h}u\|^2}.\end{align*} $$

This allows to get the rough lower bound

$$ \begin{align*} \inf \mathrm{sp}(\mathscr{P}_{h,0}) & \geqslant e^{-\delta^2/h}\inf_{u\in H^1_0(\Omega_0)\setminus\{0\}}\frac{4h^2\|\partial_{\bar{z}}u\|^2}{\|u\|^2}\\ & \geqslant h^2e^{-\delta^2/h}\lambda^{\mathrm{Dir}}_1((-\delta,\delta))\\ & \geqslant \frac{(\pi h)^2}{4\delta^2} e^{-\delta^2/h} . \end{align*} $$

This last argument already appeared in [11, Theorem 3.1].

Let us recall the following classical result.

Lemma 2.2. Consider $(T_1,\mathrm {Dom}(T_1))$ and $(T_2,\mathrm {Dom}(T_2))$ two closed operators on a Banach space E and having the same domain. Assume that there exists $z_0\in \rho (T_1)\cap \rho (T_2)$ such that the operator $K : (T_1-z_0)^{-1}-(T_2-z_0)^{-1} : E\to E$ is compact. Then,

$$ \begin{align*}\mathrm{sp}_{\mathrm{ess}}(T_1)=\mathrm{sp}_{\mathrm{ess}}(T_2).\end{align*} $$

Proof. Let us recall the proof and note that it does not require the selfadjointness of $T_1$ or $T_2$ . We recall that $\lambda \in \mathrm {sp}_{\mathrm {ess}}(T_1)$ if and only if $T_1-\lambda $ is not a Fredholm operator with index $0$ .

Consider $\lambda \notin \mathrm {sp}_{\mathrm {ess}}(T_1)$ , and write

$$ \begin{align*}\begin{split} T_2-\lambda=T_2-z_0+(z_0-\lambda)&=\left(\mathrm{Id}+(z_0-\lambda)(T_2-z_0)^{-1}\right)(T_2-z_0)\\ &=\left(\mathrm{Id}+(\lambda-z_0)K+(z_0-\lambda)(T_1-z_0)^{-1}\right)(T_2-z_0)\\ &=\left((\lambda-z_0)K+(T_1-\lambda)(T_1-z_0)^{-1}\right)(T_2-z_0). \end{split} \end{align*} $$

Now, notice that $T_2-z_0 : \mathrm {Dom}(T_2)\to E$ is Fredholm with index $0$ (since it is bijective). The operator $(T_1-z_0)^{-1} : E\to \mathrm {Dom}(T_1)$ is also bijective and thus Fredholm with index $0$ . Therefore, $(T_1-\lambda )(T_1-z_0)^{-1} : E\to E$ is also Fredholm with index $0$ (see [3, Corollary 5.7]). Since K is compact,

$$ \begin{align*}(\lambda-z_0)K+(T_1-\lambda)(T_1-z_0)^{-1}\end{align*} $$

is still Fredholm with index $0$ (see [3, Corollary 5.9]). Thus, $T_2-\lambda $ is Fredholm with index $0$ (again by [3, Corollary 5.7]).

Thanks to Lemma 2.2, it is rather easy to get the following.

Proposition 2.2. For all $h>0$ , we have $\mathrm {sp}_{\mathrm {ess}}(\mathscr {P}_h)=\mathrm {sp}_{\mathrm {ess}}(\mathscr {P}_{h,0})$ .

Proof. The operator $\mathscr {P}_h$ is unitarily equivalent to the selfadjoint operator $\widetilde {\mathscr {P}}_h$ (on $L^2(\Omega _0,\mathrm {d}s\mathrm {d}t)$ with domain $\mathsf {Dom}(\widetilde {\mathscr P}_h) = H^2(\Omega ) \cap H^1_0(\Omega ) = \mathsf {Dom}(\mathscr P_{h,0})$ ) given by

$$ \begin{align*}\widetilde{\mathscr{P}}_h=-\partial^2_t+(a^{-\frac12}(D_s-\tilde A(s,t)) a^{-\frac12})^2-\frac{\kappa^2}{4a^2}-h\,,\quad a(s,t)=1-t\kappa(s),\end{align*} $$

where $\tilde A(s,t)=t-\kappa (s)\frac {t^2}{2}$ . The unitary transformation is made of a changing of coordinates via $\Theta $ , a flattening of the metric induced by the Jacobian of $\Theta $ and a change of magnetic gauge. Since $\kappa $ is compactly supported, we see that $\widetilde {\mathscr {P}}_h$ acts as $\mathscr {P}_{h,0}$ away from a compact set.

Let us now apply Lemma 2.2 with $T_1=\mathscr {P}_{h,0}$ , $T_2=\widetilde {\mathscr {P}}_h$ and $z_0=i$ . The resolvent formula gives

$$ \begin{align*}K=(T_1-z_0)^{-1}(T_2-T_1)(T_2-z_0)^{-1}.\end{align*} $$

In our case, we have

$$ \begin{align*}T_2-T_1=a^{-\frac12}\left[(D_s-\tilde A) a^{-1}(D_s-\tilde A)\right]a^{-\frac12}-(D_s-t)^2-\frac{\kappa^2}{4a^2}.\end{align*} $$

Computing some commutators shows that we can find three smooth functions $W_1$ , $W_2$ and $W_3$ on $\overline {\Omega _0}$ , compactly supported with respect to s, such that

$$ \begin{align*}T_2-T_1=W_1(s,t)D^2_s+W_2(s,t)D_s+W_3(s,t).\end{align*} $$

Then, by elliptic regularity and the Kolmogorov–Riesz theorem (see [3, Theorem 4.14 & Remark 4.15]), we notice that $W(\widetilde {\mathscr {P}}_h-i)^{-1} : L^2(\Omega _0)\to H^1(\Omega _0)$ is compact for all $W\in \mathscr {C}^\infty _0(\overline {\Omega _0})$ . This shows that the terms involving $W_2$ and $W_3$ in K are compact operators on $L^2(\Omega _0)$ (by using that the set of compact operators forms an ideal). Concerning the term involving $W_1$ , we notice, on the one hand, that $D^2_s(\widetilde {\mathscr {P}}_h-i)^{-1}$ is bounded on $L^2(\Omega _0)$ and, on the other hand, that $(\mathscr {P}_{h,0}-i)^{-1}W_1 : L^2(\Omega _0)\to L^2(\Omega _0)$ is compact since the operators

$$ \begin{align*}[(\mathscr{P}_{h,0}-i)^{-1},W_1]=-(\mathscr{P}_{h,0}-i)^{-1}[\mathscr{P}_{h,0},W_1](\mathscr{P}_{h,0}-i)^{-1}\end{align*} $$

and $W_1(\mathscr {P}_{h,0}-i)^{-1} : L^2(\Omega _0)\to L^2(\Omega _0)$ are compact.

Applying Lemma 2.2, the conclusion follows.

3 On the function $\phi $

In this section, we prove Propositions 1.2 and 1.3. We recall that $\phi _0$ and $\hat \phi _0$ were defined before Proposition 1.2.

3.1 Proof of Proposition 1.2

Assume that two functions $\phi _1$ and $\phi _2$ satisfy the conclusions of the proposition. Then $\phi _1 - \phi _2$ is harmonic in $\Omega $ and belongs to $H^1_0(\Omega )$ . This implies that $\phi _1 = \phi _2$ and gives uniqueness.

Since the tube $\Omega $ is straight at infinity, we have $\Delta \hat \phi _0=1$ outside a compact set. In particular, $1-\Delta \hat \phi _0\in L^2(\Omega )$ . By the Poincaré inequality (see, for instance, [4] for the case of a waveguide) and the Riesz representation theorem, there exists a unique $f_0\in H^1_0(\Omega )$ such that

$$ \begin{align*} \forall \varphi\in H^1_0(\Omega), \quad \int_{\Omega}\nabla f_0\cdot\nabla\varphi\,\mathrm{d}x=\int_\Omega (1-\Delta\hat \phi_0) \varphi\,\mathrm{d}x. \end{align*} $$

Then $-\Delta f_0 =1-\Delta \hat \phi _0$ in the sense of distributions, and $f_0$ belongs to $\mathscr C^\infty (\overline \Omega )$ by elliptic regularity (notice that elliptic regularity on the straight waveguide $\Omega _0$ is proved as in the classical case of the half-space and that we can then deduce elliptic regularity on $\Omega $ via the diffeomorphism $\Theta $ with the same proof as for bounded domains; see, for instance, [2, §9.6]).

Let $V=1-\Delta \hat \phi _0$ , and consider a bounded Lipschitzian function $\Phi $ on $\Omega $ . We have

$$ \begin{align*}\langle -\Delta f_0,e^{2\Phi}f_0\rangle=\int_{\Omega}Ve^{2\Phi}f_0\mathrm{d}x.\end{align*} $$

Taking the real part and integrating by parts in the left-hand side, we get the “Agmon formula”

$$ \begin{align*} \|\nabla(e^{\Phi}f_0)\|_{L^2(\Omega)}^2-\|f_0e^{\Phi}\nabla\Phi\|_{L^2(\Omega)}^2=\mathrm{Re}\,\int_{\Omega}Ve^{2\Phi}f_0\mathrm{d}x. \end{align*} $$

Since V has compact support, it follows that

$$ \begin{align*} \|\nabla(e^{\Phi}f_0)\|_{L^2(\Omega)}^2-\|f_0e^{\Phi}\nabla\Phi\|_{L^2(\Omega)}^2\leqslant \|V\|_{L^2(\Omega)}\|f_0\|_{L^2(\Omega)}\underset{\mathrm{supp V}}{\max} e^{2\Phi}. \end{align*} $$

By the Poincaré inequality, we have

$$ \begin{align*} \|\nabla(e^{\Phi}f_0)\|^2\geqslant\lambda_1(\Omega)\|e^{\Phi}f_0\|^2,\end{align*} $$

where $\lambda _1(\Omega )>0$ is the infimum of the spectrum of the Dirichlet Laplacian on $\Omega $ . This shows that

$$ \begin{align*} \big( \lambda_1(\Omega)-\|\nabla\Phi\|^2_\infty \big) \int_{\Omega}e^{2\Phi}|f_0|^2\mathrm{d}x\leqslant \|V\|_{L^2(\Omega)}\|f_0\|_{L^2(\Omega)}\underset{\mathrm{supp V}}{\max} e^{2\Phi}. \end{align*} $$

Choosing $\Phi (x)=\Phi _m(x)=\alpha \min (\langle x\rangle ,m)$ (with $\alpha>0$ fixed small enough) and letting $m\to +\infty $ , we see by the Fatou lemma that there exists $C> 0$ such that

$$ \begin{align*} \int_{\Omega}e^{2\alpha\langle x\rangle}|f_0|^2\mathrm{d}x \leqslant C \|f_0\|_{L^2(\Omega)}. \end{align*} $$

Coming back to the Agmon formula, we deduce that

$$ \begin{align*} \int_{\Omega}e^{2\alpha\langle x\rangle}|\nabla f_0|^2\mathrm{d}x < + \infty . \end{align*} $$

Then we adapt the classical strategy used for elliptic regularity. Let $R> 0$ be so large that $\kappa $ is supported in $(-R,R)$ . Let $\chi _0 \in C^\infty (\Omega _0;[0,1])$ equal to 0 on $(-R,R) \times (-\delta ,\delta )$ and equal to 1 on $\Omega \setminus (-R-1,R+1) \times (-\delta ,\delta )$ . Let $\chi = \chi _0 \circ \Theta ^{-1}$ . Then the above arguments apply with $f_0$ replaced by $\partial _s (\chi f_0) = {\boldsymbol \gamma }' \cdot \nabla (\chi f_0)$ . Since $\Delta (\chi f_0)$ is compactly supported, we deduce that for any $\beta \in \mathbb N^2$ with $|\beta | = 2$ we have

$$ \begin{align*} \int_{\Omega}e^{2\alpha\langle x\rangle}|\partial^\beta f_0|^2\mathrm{d}x < + \infty . \end{align*} $$

We proceed by induction on $|\beta |$ to get the same estimate for any $\beta \in \mathbb N^2$ , and we deduce in particular that $f_0$ belongs to the Schwartz class $\mathscr {S}(\overline {\Omega })$ .

We set $\phi =\hat \phi _0-f_0$ . It is smooth, it satisfies the Dirichlet condition, $\phi - \hat \phi _0$ belongs to $\mathscr S(\overline \Omega )$ and $\Delta \phi =1$ . It remains to discuss the uniform positivity of the normal derivative. By the Hopf lemma, we already know that $\partial _\nu \phi> 0$ on $\partial \Omega $ , so it is enough to show that this estimate is uniform at infinity.

We have

$$ \begin{align*} \partial_\nu \phi= \partial_\nu \hat \phi_0 - \partial_\nu f_0. \end{align*} $$

Since $\Theta $ is a rotation at infinity, we see by the explicit expression of $\phi _0$ that there exists $c_1>0$ such that, for all $x\in \partial \Omega $ with a sufficiently large curvilinear abscissa,

$$ \begin{align*} \partial_\nu \hat \phi_0 \geqslant 2 c_1. \end{align*} $$

On the other hand, since $f_0 \in \mathscr S(\overline \Omega )$ we have

$$ \begin{align*}\lim_{\substack{|x|\to+\infty\\ x \in \partial \Omega}} \partial_\nu f_0(x)=0.\end{align*} $$

Then $\partial _\nu \phi (x) \geqslant c_1$ for $x \in \partial \Omega $ large enough, and we deduce the uniform positivity of $\partial _\nu \phi $ on $\partial \Omega $ .

3.2 Proof of Proposition 1.3

For $(s,t) \in \Omega _0$ , we set

$$ \begin{align*} a(s,t) = \det \big( \mathrm{Jac} (\Theta)(s,t) \big) = 1 - t \kappa(s)\, \end{align*} $$

and notice that $a(s,t)\geqslant \frac 12$ as soon as $\delta $ is small enough.

Let $\tilde \phi = \phi \circ \Theta $ . For $s \in \mathbb {R}$ and $\tau \in (-1,1)$ , we set

$$ \begin{align*} a_\delta(s,\tau) = a(s,\delta \tau) \quad \text{and} \quad \psi(s,\tau)=\delta^{-2} a_\delta(s,\tau)^{\frac 12} \tilde\phi(s,\delta\tau). \end{align*} $$

Finally, we define on $\mathbb {R} \times (-1,1)$ the differential operator

$$ \begin{align*} \mathscr{M}_\delta=\partial_\tau^2+\delta^2 \big( a^{-\frac12}_\delta \partial_s a_\delta^{-\frac12} \big)^2+\frac{\delta^2\kappa^2}{4 a_\delta^2}, \end{align*} $$

where $a^{-\frac 12}_\delta \partial _s a_\delta ^{-\frac 12} : u \mapsto a^{-\frac 12}_\delta \partial _s (a_\delta ^{-\frac 12} u )$ .

Lemma 3.1. We have $\mathscr {M}_\delta \psi =a_{\delta }^{\frac 12}$ and $\psi ( \cdot ,\pm 1)=0$ .

Proof. Since $\tilde \phi (s,\pm \delta )=0$ we have $\psi ( \cdot ,\pm 1)=0$ for all $s \in \mathbb {R}$ . In the tubular coordinates, the equality $\Delta \phi =1$ reads

$$ \begin{align*} \big( a^{-1}\partial_s a^{-1} \partial_s+ a^{-1}\partial_t a \partial_t \big) \tilde\phi=1. \end{align*} $$

Setting $\check \phi =a^{\frac 12}\tilde \phi $ , we get

$$ \begin{align*}\left[\big(a^{-\frac12}\partial_sa^{-\frac12}\big)^2+\big(a^{-\frac12}\partial_ta^{\frac12}\big)\big(a^{\frac12}\partial_ta^{-\frac12}\big)\right]\check\phi=a^{\frac12},\end{align*} $$

or

$$ \begin{align*}\left[\big(a^{-\frac12}\partial_sa^{-\frac12}\big)^2+ \Big(\partial_t-\frac{\kappa}{2a}\Big) \Big(\partial_t + \frac{\kappa}{2a}\Big)\right]\check\phi=a^{\frac12},\end{align*} $$

which gives

$$ \begin{align*}\left[\big(a^{-\frac12}\partial_sa^{-\frac12}\big)^2+\partial_t^2+\frac{\kappa^2}{4 a^2}\right]\check\phi=a^{\frac12}.\end{align*} $$

Since $\psi (s,\tau )=\delta ^{-2}\check \phi (s,\delta \tau )$ , the conclusion follows.

Proof of Proposition 1.3

We look for an approximation $\Psi _5$ of $\psi $ , in the sense that

(3.1) $$ \begin{align} \mathscr{M}_\delta(\psi-\Psi_5)=\mathscr{O}_{H^2(\mathbb{R}\times (-1,1))}(\delta^5)\,,\quad \psi-\Psi_5\in H^2\cap H^1_0(\mathbb{R}\times[-1,1]). \end{align} $$

By elliptic regularity, this will give

$$ \begin{align*} \|\psi-\Psi_5\|_{H^4(\mathbb{R}\times[-1,1])}=\mathscr{O}(\delta^3), \end{align*} $$

and then, by Sobolev embeddings,

(3.2) $$ \begin{align} \|\psi-\Psi_5\|_{\mathscr{C}^2(\mathbb{R}\times[-1,1])}=\mathscr{O}(\delta^3). \end{align} $$

We look for $\Psi _5$ of the form $\psi _0 + \delta \psi _1 + \delta ^2 \psi _2 + \delta ^3 \psi _3 + \delta ^4 \psi _4$ . Note that we could proceed similarly to get a rest of order $\mathscr O(\delta ^N)$ in $\mathscr {C}^k(\mathbb {R}\times [-1,1])$ for any N and k.

There exist $M_0,\dots ,M_4 \in \mathcal L(H^{4}(\mathbb {R}\times [-1,1]),H^2(\mathbb {R}\times [-1,1]))$ (we denote by $\mathcal L (\mathcal H_1,\mathcal H_2)$ the space of bounded operators from $\mathcal H_1$ to $\mathcal H_2$ ) such that in $\mathcal L(H^{4}(\mathbb {R}\times [-1,1]),H^2(\mathbb {R}\times [-1,1]))$ we have

$$ \begin{align*} \mathscr{M}_\delta = \sum_{k=0}^4 \delta^k M_k + \mathscr O(\delta^5). \end{align*} $$

In particular,

$$ \begin{align*} M_0=\partial^2_\tau\,,\quad M_1=0\,,\quad M_2=\partial^2_s+\frac{\kappa^2}{4}. \end{align*} $$

Similarly, in $H^2(\mathbb {R}\times [-1,1])$ we have by Lemma 3.1

$$ \begin{align*} \mathscr M_\delta \psi = \sum_{k=0}^4 \delta^k\alpha_k+ \mathscr O (\delta^5), \end{align*} $$

with

$$ \begin{align*} \alpha_0 = 1, \quad \alpha_1 = - \frac { \kappa \tau}2 , \quad \alpha_2 = - \frac { \tau^2 \kappa^2} 8 , \end{align*} $$

and $\alpha _3,\alpha _4 \in \mathscr C^\infty (\overline \Omega )$ . We compute $\psi _k$ by induction on k. It satisfies

$$ \begin{align*} M_0 \psi_0 = \alpha_0, \quad M_0 \psi_1 = -M_1 \psi_0 + \alpha_1,\quad M_0 \psi_k = - \sum_{j=2}^k M_j \psi_{k-j} + \alpha_k, \end{align*} $$

with the Dirichlet condition $\psi _k (\cdot ,\pm 1) = 0$ . This gives in particular

$$ \begin{align*}\psi_0(s,\tau)=\frac{\tau^2-1}{2}, \quad \psi_1(s,\tau)=\frac{\kappa(s)}{12}(\tau-\tau^3). \end{align*} $$

Then $\psi _2$ has to be a solution of

$$ \begin{align*}{M}_0 \psi_2=-{M}_2 \psi_0-\frac{\kappa^2\tau^2}{8}=\frac{\kappa^2}{4}\left(-\frac{\tau^2-1}{2}-\frac{\tau^2}{2}\right)=\frac{\kappa^2}{4}\left(\frac12-\tau^2\right).\end{align*} $$

This leads to take

$$ \begin{align*}\psi_2(s,\tau)=\frac{\kappa^2}{4}\left(\frac{\tau^2-1}{4}-\frac{\tau^4-1}{12}\right)=\frac{\kappa^2}{4}\left(\frac{\tau^2}{4}-\frac{\tau^4}{12}-\frac16\right).\end{align*} $$

Due to the asymptotic behavior of $\phi $ given in Proposition 1.2 and the fact that $a = 1$ at infinity, $a ^{\frac 12} \tilde \phi - \phi _0$ and hence $\psi -\psi _0$ belong to the Schwartz class. Thus, $\Psi _5$ satisfies equation (3.1) and hence equation (3.2). Now setting $\Psi = \psi _0 + \delta \psi _1 + \delta ^2 \psi _2$ we deduce

$$ \begin{align*} \|\psi-\Psi\|_{\mathscr{C}^2(\mathbb{R}\times[-1,1])}=\mathscr{O}(\delta^3). \end{align*} $$

This gives

$$ \begin{align*} \|\delta^{-2}\tilde\phi(s,\delta\tau)-a(s,\delta\tau)^{-\frac12} \Psi\|_{\mathscr{C}^2(\mathbb{R}\times[-1,1])}=\mathscr{O}(\delta^3) \end{align*} $$

or

$$ \begin{align*} \left \| \delta^{-2}\tilde\phi(s,\delta\tau)-\left(1+\delta\tau\frac\kappa2+\delta^2\frac{3}{8}\tau^2\kappa^2\right) \Psi \right \|_{\mathscr{C}^2(\mathbb{R}\times[-1,1])}=\mathscr{O}(\delta^3). \end{align*} $$

Then

(3.3) $$ \begin{align} \left\| \delta^{-2}\tilde\phi(s,\delta\tau)-f_\delta(s,\tau) \right\|_{\mathscr{C}^2(\mathbb{R}\times[-1,1])} =\mathscr{O}(\delta^3)\,, \end{align} $$

where we have set

$$ \begin{align*} f_{\delta}(s,\tau)=\psi_0+\delta\left(\psi_1+\frac{\tau\kappa} 2\psi_0\right)+\delta^2\left(\frac{3\tau^2\kappa^2}{8}\psi_0+\frac{\tau\kappa}{2} \psi_1+ \psi_2\right). \end{align*} $$

We have

$$ \begin{align*}\begin{split} f_{\delta}(s,\tau) =\frac{\tau^2-1}{2}-\frac{\delta\kappa(s)}{6}\left(\tau-\tau^3\right)+\delta^2\kappa^2P_2(\tau)\,, \end{split}\end{align*} $$

where

$$ \begin{align*}P_2(\tau)=\frac{3\tau^2(\tau^2-1)}{16}+\frac{\tau(\tau-\tau^3)}{24}+\frac{\tau^2}{16}-\frac{\tau^4}{48}-\frac{1}{24}.\end{align*} $$

Let us explain why $f_\delta $ has a unique minimum, nonattained at infinity and which is nondegenerate. Firstly, when $s\notin \mathrm {supp}\,\kappa $ , we have

$$ \begin{align*} f_\delta(s,\tau)=\frac{\tau^2-1}{2}\geqslant-\frac12=f_\delta(s,0). \end{align*} $$

This shows that $f_\delta $ has a minimum in $\Omega _0$ . This minimum is in fact strictly less than $-\frac 12$ and thus attained at points where the curvature is not $0$ . Indeed, consider $s_0$ the maximum of $\kappa ^2$ . We have $\kappa (s_0)\neq 0$ , $\kappa '(s_0)=0$ , and $\kappa (s_0)\kappa "(s_0)<0$ . Let us notice that

$$ \begin{align*} f_{\delta}\left(s_0,\frac{\delta\kappa(s_0)}{6}\right) & =-\frac{1}{2}+\delta^2\kappa^2(s_0)\left(\frac{1}{72}-\frac{1}{36}-\frac{1}{24}\right)+\mathscr{O}(\delta^3)\\ & =-\frac{1}{2}-\frac{\delta^2\kappa^2(s_0)}{18}+\mathscr{O}(\delta^3). \end{align*} $$

This shows that, for $\delta $ small enough,

$$ \begin{align*}\inf_{(s,\tau)\in\mathbb{R}\times(-1,1)} f_{\delta}(s,\tau)\leqslant -\frac{1}{2}-\frac{\delta^2\max \kappa^2}{18}+C\delta^3<-\frac12\,\end{align*} $$

and that the infimum is a minimum (which is not attained at infinity).

Now, we prove that for $\delta $ small enough all the possible minima are nondegenerate. Consider a minimum $(s_1,\tau _1)$ of $f_\delta $ . We have $\tau _1\in (-1,1)$ and $\kappa (s_1)\neq 0$ . Moreover, we must have

$$ \begin{align*}\partial_\tau f_\delta(s_1,\tau_1)=0,\end{align*} $$

which implies that

(3.4) $$ \begin{align} \tau_1=\frac{\delta\kappa(s_1)}{6}+\mathscr{O}(\delta^2). \end{align} $$

Then,

$$ \begin{align*}f_\delta(s_1,\tau_1)=-\frac{1}{2}-\frac{\delta^2\kappa^2(s_1)}{18}+\mathscr{O}(\delta^3).\end{align*} $$

With the upper bound on the minimum, we deduce that

$$ \begin{align*}0\leqslant \kappa^2(s_0)-\kappa^2(s_1)\leqslant C\delta.\end{align*} $$

By using the uniqueness and nondegeneracy of the minimum, this implies that

(3.5) $$ \begin{align} s_1=s_0+\mathscr{O}(\delta^{\frac12})\,,\quad\tau_1=\frac{\delta\kappa(s_0)}{6}+\mathscr{O}(\delta^2)\,, \end{align} $$

where we used equation (3.4) and that $\kappa '(s_0)=0$ .

Let us now estimate the second derivative of $f_\delta $ at $(s_1,\tau _1)$ . We have

$$ \begin{align*}\partial^2_s f_\delta(s_1,\tau_1)=-\frac{\kappa(s_0)\kappa"(s_0)}{9}\delta^2+o(\delta^2)\,,\quad \partial_s\partial_{\tau}f_\delta(s_1,\tau_1)=\mathscr{O}(\delta^{\frac32}),\end{align*} $$

and

$$ \begin{align*}\partial^2_\tau f_\delta(s_1,\tau_1)=1+\mathscr{O}(\delta^2).\end{align*} $$

We infer that there exist $\delta _0,c>0$ such that for all $\delta \in (0,\delta _0)$ and all minimum $(s_1,\tau _1)$ ,

$$ \begin{align*}\mathrm{Hess}_{(s_1,\tau_1)}f_\delta\geqslant c\delta^2.\end{align*} $$

By definition, this means that the minima are nondegenerate.

Let us finally prove that there is only one minimum. Consider two minima $X_1=(s_1,\tau _1)$ and $X_2=(s_2,\tau _2)$ . From equation (3.5), we have, uniformly in $\lambda \in [0,1]$ ,

(3.6) $$ \begin{align} X_1+\lambda(X_2-X_1)=\left(s_0,\frac{\delta\kappa(s_0)}{6}\right)+(\mathscr{O}(\delta^{\frac12}),\mathscr{O}(\delta^2)). \end{align} $$

Since the differential of $f_\delta $ vanishes at $X_1$ , the Taylor formula gives

$$ \begin{align*}f_\delta(X_2)-f_\delta(X_1)=\int_0^1(1-\lambda) \mathrm{Hess}_{X_1+\lambda(X_2-X_1)}f_\delta(X_2-X_1,X_2-X_1)\mathrm{d}\lambda.\end{align*} $$

By using equation (3.6), we deduce as before that there exist $\delta _0,c>0$ such that for all $\delta \in (0,\delta _0)$ and all $\lambda \in [0,1]$ ,

$$ \begin{align*}\mathrm{Hess}_{X_1+\lambda(X_2-X_1)}f_\delta\geqslant c\delta^2.\end{align*} $$

This shows that

$$ \begin{align*}0=f_\delta(X_2)-f_\delta(X_1)\geqslant \frac{c\delta^2}{2}|X_1-X_2|^2.\end{align*} $$

Therefore, for $\delta $ small enough, $f_\delta $ has a unique minimum $X(\delta )$ , which is not attained at infinity and nondegenerate, and

$$ \begin{align*}\mathrm{Hess}_{X(\delta)}f_\delta\geqslant c\delta^2.\end{align*} $$

By a perturbative argument using equation (3.3), this shows that $\delta ^{-2}\tilde \phi (s,\delta \tau )$ has also a unique minimum, which is not attained at infinity and nondegenerate. The same conclusion follows for $\phi $ .

4 Upper bound for the bottom of the spectrum

This last section is devoted to the proof of Theorem 1.2. From the min-max principle, we have

$$ \begin{align*}\inf\mathrm{sp}(\mathscr{P}_h)=\inf_{\psi\in H^1_0(\Omega)\setminus\{0\}}\frac{\|(-ih\nabla-\mathbf{A})\psi\|^2-h\|\psi\|^2}{\|\psi\|^2}.\end{align*} $$

From Lemma 2.1, we have

(4.1) $$ \begin{align} \inf\mathrm{sp}(\mathscr{P}_h)=\inf_{u\in H^1_0(\Omega)\setminus\{0\}}\frac{4h^2\int_{\Omega}e^{-2\phi/h}|\partial_{\bar{z}} u|^2\mathrm{d}x}{\int_{\Omega}e^{-2\phi/h}|u|^2\mathrm{d}x}. \end{align} $$

Let us construct a convenient test function. It is natural to consider a test function in the form

$$ \begin{align*} u(x)=f(x)\chi(x),\end{align*} $$

where $f\in \mathscr {O}(\Omega )\cap H^1(\Omega )$ is such that $f(x_{\min })\neq 0$ and the cut-off function $\chi $ ensures that u satisfies the boundary condition. It is chosen of the form $\chi = \rho \circ \Theta ^{-1}$ with $\rho (s,\pm \delta )=0$ and $\rho (s,t)= 1$ for all $s \in \mathbb {R}$ and $t\in (-\delta + \epsilon ,\delta - \epsilon )$ . This function $\rho $ will be determined below to optimize an upper bound; see equation (4.5).

4.1 Estimate of the numerator

By using the change of variable $\Theta $ , we have

(4.2) $$ \begin{align} 4h^2\int_{\Omega}e^{-2\phi/h}|\partial_{\bar{z}} u|^2\mathrm{d}x =h^2\int_{\Omega_0}e^{-2\tilde\phi(s,t)/h}|\tilde f(s,t)|^2\left(a^{-2}|\partial_s\rho|^2+|\partial_t\rho|^2\right) a(s,t)\mathrm{d}s\mathrm{d}t\,, \end{align} $$

with $\tilde \phi =\phi \circ \Theta $ and $\tilde f=f\circ \Theta $ . Since $\rho $ is constant on $\mathbb {R} \times [-\delta + \epsilon , \delta - \epsilon ]$ , the right-hand side is actually an integral over $\mathbb {R} \times \big ( (-\delta , -\delta + \epsilon ) \cup (\delta -\epsilon , \delta ) \big )$ . We begin with the contribution of the integral over $\mathbb {R} \times (\delta -\epsilon , \delta )$ . We will choose $\epsilon $ smaller that h. Then, using the Taylor formula to expand $\tilde \phi (s,t)$ and $a(s,t)$ near $t=\delta $ , we have

$$ \begin{align*} \int_{\mathbb{R}}\int_{\delta-\epsilon}^\delta \,e^{-2\tilde\phi(s,t)/h}&|\partial_t\rho|^2 |\tilde f(s,t)|^2a(s,t)\mathrm{d}t\mathrm{d}s\\ &\leqslant(1+C\epsilon+C\epsilon^2/h)\int_{\mathbb{R}}\int_{\delta-\epsilon}^\delta \,e^{-2(t-\delta)\partial_t\tilde\phi(s,\delta)/h}|\partial_t\rho|^2 |\tilde f(s,t)|^2a(s,\delta)\mathrm{d}t\mathrm{d}s. \end{align*} $$

We also want to replace $ |\tilde f(s,t)|^2$ by $ |\tilde f(s,\delta )|^2$ . To do so, we remark that, for all $(s,t)\in \mathbb {R}\times (\delta -\epsilon ,\delta )$ ,

$$ \begin{align*} \left||\tilde f(s,t)|^2-|\tilde f(s,\delta)|^2\right|&\leqslant 2\int_{t}^\delta|\tilde f(s,\tau)||\partial_t\tilde f(s,\tau)|\mathrm{d}\tau\\ &\leqslant\left(\|\tilde f(s,\cdot)\|^2_{L^2([\delta-\epsilon,\delta])}+\|\partial_t\tilde f(s,\cdot)\|^2_{L^2([\delta-\epsilon,\delta])}\right)\, \end{align*} $$

so that

$$ \begin{align*} \int_{\mathbb{R}}\left(\int_{\delta-\epsilon}^\delta \,e^{-2(t-\delta)\partial_t\tilde\phi(s,\delta)/h}|\partial_t\rho|^2\mathrm{d}t\right) a(s,\delta)\left| |\tilde f(s,t)|^2- |\tilde f(s,\delta)|^2\right|\mathrm{d}s \leqslant \int_{\mathbb{R}}a(s,\delta) R(s,\epsilon,h)\mathrm{d}s,\end{align*} $$

with

(4.3) $$ \begin{align} R(s,\epsilon,h) =\left( \|\tilde f(s,\cdot)\|^2_{L^2([\delta-\epsilon,\delta])} +\|\partial_t\tilde f(s,\cdot)\|^2_{L^2([\delta-\epsilon,\delta])} \right)\int_{\delta-\epsilon}^\delta e^{-2(t-\delta)\partial_t\tilde\phi(s,\delta)/h}|\partial_t\rho|^2\mathrm{d}t. \end{align} $$

Therefore,

(4.4) $$ \begin{align} \begin{aligned} &\quad {\int_{\mathbb{R}}\int_{\delta-\epsilon}^\delta \,e^{-2\tilde\phi(s,t)/h}|\partial_t\rho|^2 |\tilde f(s,t)|^2a(s,t)\mathrm{d}t\mathrm{d}s}\\ &\leqslant(1+C\epsilon+C\epsilon^2/h)\int_{\mathbb{R}}\int_{\delta-\epsilon}^\delta \,e^{-2(t-\delta)\partial_t\tilde\phi(s,\delta)/h}|\partial_t\rho|^2 |\tilde f(s,t)|^2a(s,\delta)\mathrm{d}t\mathrm{d}s \\ &\leqslant (1+C\epsilon+C\epsilon^2/h)\Big(\int_{\mathbb{R}}\int_{\delta-\epsilon}^\delta \,e^{-2(t-\delta)\partial_t\tilde\phi(s,\delta)/h}|\partial_t\rho|^2 |\tilde f(s,\delta)|^2a(s,\delta)\mathrm{d}t\mathrm{d}s\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\int_{\mathbb{R}} a(s,\delta) R(s,\epsilon,h)\mathrm{d}s\Big). \end{aligned} \end{align} $$

Looking at the right-hand side suggests to consider a function $\rho $ that minimizes $\int _{\delta -\epsilon }^\delta e^{-2(t-\delta )\partial _t\tilde \phi (s,\delta )/h}|\partial _t\rho |^2\mathrm {d}t$ among the $H^1$ -functions equal to $1$ in $\delta -\epsilon $ and $0$ in $\delta $ . This leads to the explicit choice

(4.5) $$ \begin{align} \rho(s,t)= \frac{1 - e^{2(t-\delta)\partial_t\tilde\phi(s,\delta)/h}}{1 - e^{-2\epsilon\partial_t\tilde\phi(s,\delta)/h}}\,,\quad\forall (s,t)\in\mathbb{R}\times(\delta-\epsilon,\delta). \end{align} $$

The minimum satisfies

$$ \begin{align*} \int_{\delta-\epsilon}^\delta e^{-2(t-\delta)\partial_t\tilde\phi(s,\delta)/h}|\partial_t\rho|^2\mathrm{d}t = \frac{2\partial_t\tilde\phi(s,\delta)}{h(1-e^{-2\epsilon\partial_t\tilde\phi(s,\delta)/h})}. \end{align*} $$

We recall from Proposition 1.2 that $\partial _t\tilde \phi (s,\delta ) = \partial _\nu \phi (\Theta (s,\delta ))$ is uniformly positive. Choosing $\epsilon =h|\ln h|$ , we get, uniformly with respect to s,

(4.6) $$ \begin{align} \int_{\delta-\epsilon}^\delta \,e^{-2(t-\delta)\partial_t\tilde\phi(s,\delta)/h}|\partial_t\rho|^2\mathrm{d}t = \frac{2\partial_t\tilde\phi(s,\delta)}{h}+o(h^{-1})=\mathscr{O}(h^{-1})\,, \end{align} $$

where we used that $\Theta $ and $\Theta ^{-1}$ have uniformly bounded Jacobians.

Using that $f\in H^1(\Omega )$ , we get

$$ \begin{align*} \int_{\mathbb{R}}\left( \|\tilde f(s,\cdot)\|^2_{L^2([\delta-\epsilon,\delta])} +\|\partial_t\tilde f(s,\cdot)\|^2_{L^2([\delta-\epsilon,\delta])} \right)\mathrm{d}s\underset{\epsilon\to0}{\longrightarrow}0\, \end{align*} $$

so that, with equations (4.3) and (4.6), it follows that

$$ \begin{align*}\int_{\mathbb{R}} a(s,\delta) R(s,\epsilon,h)\mathrm{d}s=o_{h\to 0}(h^{-1}).\end{align*} $$

With equation (4.4), this gives

$$ \begin{align*} &{\int_{\mathbb{R}}\int_{\delta-\epsilon}^\delta e^{-2\tilde\phi(s,t)/h}|\partial_t\rho|^2 |\tilde f(s,t)|^2a(s,t)\mathrm{d}t\mathrm{d}s} \\ &\quad \leqslant 2h^{-1}\int_{\mathbb{R}}\partial_\nu \phi (\Theta(s,\delta))|\tilde f(s,\delta)|^2a(s,\delta)\mathrm{d}s +o_{h\to0}(h^{-1}). \end{align*} $$

Let us now come back to equation (4.2). Considering the term with the tangential derivative, we get with similar computations

$$ \begin{align*} \int_{\mathbb{R}}\int_{\delta-\epsilon}^\delta \,e^{-2\tilde\phi(s,t)/h}|\partial_s\rho|^2 |\tilde f(s,t)|^2 a(s,t)^{-1}\mathrm{d}t\mathrm{d}s =o_{h\to0}(h^{-1}). \end{align*} $$

We play the same game with the contribution of the integral over $\mathbb {R} \times (-\delta ,-\delta +\epsilon )$ in equation (4.2) (notice that $\partial _t\tilde \phi (s,-\delta ) = -\partial _\nu \phi (\Theta (s,-\delta ))$ is now uniformly negative). We get

$$ \begin{align*} 4h^2\int_{\Omega}e^{-2\phi/h}|\partial_{\bar{z}} u|^2\mathrm{d}x & \leqslant 2h\|(\partial_\nu \phi)^{\frac12}f\|^2_{\partial\Omega}+o_{h\to0}(h). \end{align*} $$

4.2 Estimate of the denominator and conclusion

We have

$$ \begin{align*} \int_{\Omega}e^{-2\phi/h}|u|^2\mathrm{d}x & = \int_{\Omega}e^{-2\phi/h}|f(x)\chi(x)|^2\mathrm{d}x \\ & = e^{-2\phi_{\min}/h}\int_{\Omega}e^{-2(\phi-\phi_{\min})/h}|f(x)\chi(x)|^2\mathrm{d}x. \end{align*} $$

The Laplace method yields (notice that $\chi (x_{\min }) = 1$ for h small enough)

$$ \begin{align*}\int_{\Omega}e^{-2\phi/h}|u|^2\mathrm{d}x= h e^{-2\phi_{\min}/h}\left( |f(x_{\min})|^2\frac{\pi}{\sqrt{\det\mathrm{Hess}_{x_{\min}}\phi}}+o_{h\to0}(1)\right).\end{align*} $$

With equation (4.1), this shows that

$$ \begin{align*}\inf\mathrm{sp}(\mathscr{P}_h)\leqslant 2\sqrt{\det\mathrm{Hess}_{x_{\min}}\phi}\frac{\|(\partial_\nu \phi)^{\frac12}f\|^2_{\partial\Omega}}{\pi|f(x_{\min})|^2}(1+o_{h\to0}(1))e^{2\phi_{\min}/h},\end{align*} $$

and Theorem 1.2 since this estimate holds for all the functions f in $\mathscr {E}$ .