Content-Type: multipart/mixed; boundary="-------------0110100814765" This is a multi-part message in MIME format. ---------------0110100814765 Content-Type: text/plain; name="01-362.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="01-362.keywords" Schroedinger, semiclassical, magnetic field, Neumann problem, eigenvalue, superconductivity, Schroedinger, semiclassical, magnetic field, Neumann problem ---------------0110100814765 Content-Type: application/x-tex; name="magnet.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="magnet.tex" %% Version travaill\'ee le 17 f\'evrier 2001, le 18 f\'evrier, %% le 6 Avril. Le 7 Juin. Le 8 juin. Le 20 juin. Le 27 Juin. %% Le 1 Juillet, le 2 Juillet 2001 le 19 juillet le 20 juillet le 23 Juillet %% Le 24 Juillet. Le 30 juillet.Le 2 Aout. Le 7 Aout. Le 13 Aout. Le %% 16 Aout, le 18 Aout, le 20 Aout, le 21 Aout, le 22 Aout, le 3 septembre, %% le 4 septembre, le 9 septembre, le 10 septembre, le 16 septembre, %% Le 18 Septembre, le 20 septembre, le 24 Septembre, le 25 septembre %% Le 27 Septembre %% Document de travail %% realise en partant de l'article soumis au journal indien. %% Synthese du fichier Held6.tex (BH) et du fichier morameperso1 (AM) %This document is written in LATEX . Texte principal. %%\documentstyle[12pt]{article} \documentclass{article} % \usepackage{showkeys} % \def\nz{\ifmmode {I\hskip -3pt N} \else {\hbox {$I\hskip -3pt N$}}\fi} \def\zz{\ifmmode {Z\hskip -4.8pt Z} \else {\hbox {$Z\hskip -4.8pt Z$}}\fi} \def\qz{\ifmmode {Q\hskip -5.0pt\vrule height6.0pt depth 0pt \hskip 6pt} \else {\hbox {$Q\hskip -5.0pt\vrule height6.0pt depth 0pt\hskip 6pt$}}\fi} \def\rz{\ifmmode {I\hskip -3pt R} \else {\hbox {$I\hskip -3pt R$}}\fi } \def\cz{\ifmmode {C\hskip -4.8pt\vrule height5.8pt \hskip 6.3pt} \else {\hbox {$C\hskip -4.8pt\vrule height5.8pt \hskip 6.3pt$}}\fi} %Macros \def\const{{\rm const.\,}} % constante \def\div{{\rm div}\,}% divergence \def\det{{\rm det}\,} \def\grad{{\rm grad}\, }% gradient \def\rot{{\rm curl}\,}% rotationnel \def\curl{{\rm curl}\,}% rotationnel \def\rotadj{{\rm rot^* }}% rotationnel \def\Hess{{\rm Hess\,}}% Hessian \def\sp{{\rm Sp\,}} \def\supp{\mathop{\rm supp} \nolimits} % Support \def\minB {{b}} %inf B \def\minBb {{b'}} %inf B au bord % \def\qed{\hbox {\hskip 1pt \vrule width 4pt height 6pt depth 1.5pt \hskip 1pt}\\} % cqfd % \def\Ag {{\cal A}} %A gothique \def\Bg {{\cal B}} %B gothique \def\Cg {{\cal C}} %C gothique \def\Fg {{\cal F}} \def\Hg {{\cal H}} %H gothique \def\Hb {{\bf H}} %H \def\hb {{\bf h}} %H \def\Ig {{\cal I}} %I gothique \def\Jg {{\cal J}} %J gothique \def\Eg {{\cal E}} %E gothique \def\Kg {{\cal K}} %K gothique \def\Lg {{\cal L}} %L gothique \def\Ng {{\cal N}} %N gothique \def\Og {{\cal O}} % O gothique ou grand O \def\Pg {{\cal P}} %P gothique \def\Qg {{\cal Q}} %Q gothique \def\Rg {{\cal R}} %R gothique \def\Rb {{\bf R}} %R bold \def\Nb {{\bf N}} %N bold \def\Sg {{\cal S}} %S gothique \def\Vg {{\cal V}} %V gothique \def\Wg {{\cal W}} %V gothique \def\Ug {{\cal U}} %U gothique \def \Sp{{\rm \, Sp\;}} \def \Tr{{\rm \, Tr\;}} \def\For {{\rm \;for\;}} \def\for {{\rm \;for\;}} \def\Or {{\rm \;or\;}} \def\If {{\rm \;if \;}} \def\and {{\rm \; and \;}} \def\dist {{\rm \; dist \;}} \def\then {{\rm \; then \;}} \def\with {{\rm \; with \;}} \def\Im {{\rm \; Im\;}} \def\Re {{\rm \;Re\;}} % \newcommand {\pa}{\partial} \newcommand {\ar}{\rightarrow} \newcommand {\Ar}{\Rightarrow} \newcommand {\ot}{\otimes} \newcommand{\tnp}[1]{#1\index{#1}} % changement de la numerotation \makeatletter \@addtoreset{equation}{section} \renewcommand{\theequation}{\thesection.\arabic{equation}} \makeatother % %\def\baselinestretch{2} % Environments \newtheorem{theorem}{Theorem}[section] \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \title{Magnetic bottles for the Neumann problem: curvature effects in the case of dimension 3} \author{ Bernard Helffer$^{1,3}$ and Abderemane Morame $^{2}$} %\makeindex \begin{document} \bibliographystyle{plain} \maketitle %address \noindent $^1${\small D\'epartement de Math\'ematiques, UMR CNRS 8628, B\^at. 425, Universit\'e Paris-Sud,\\ F-91405 Orsay Cedex, France. \\ E.Mail: Bernard.Helffer@math.u-psud.fr}\\ $^2${\small D\'epartement de Math\'ematiques, UMR CNRS 6629 , Universit\'e de Nantes,\\ 2, rue de la Houssini\`ere, B.P. 92208, 44322 Nantes Cedex 3, France.\\ E.Mail: morame@math.univ-nantes.fr}\\ \noindent $^3${\small This work has been partially supported by the European Union under the TMR Program EBRFXT 960001.}\\ % %RESUME % \begin{abstract} In \cite{HeMo3}, we have analyzed the recent results obtained on the Neumann realization of the Schr\"odinger operator in the case of dimension~$3$ by Lu and Pan. After presenting a short treatment of their spectral analysis of key-models, we show briefly how to implement the techniques of Helffer-Morame in order to give some localization of the ground state. This leaves open the question of the localization by curvature effect which was conjectured by Bernoff-Sternberg in the case of dimension $2$ and solved by \cite{LuPa2}, \cite{PiFeSt} and \cite{HeMo2}. We shall now analyze the curvature effects in the case of dimension~$3$. \end{abstract} \newpage \tableofcontents \newpage \section{Introduction}\label{Section1} Motivated by the superconductivity\footnote{ One can find in Chapter 4 of \cite{S-JSaTh} a physical presentation of the problem we are considering. We particularly emphasize on their subsection 4.3 where they analyze (with partially heuristic arguments) the angular dependence of the nucleation field. We will give a mathematical proof of, what they describe for example p.~87~: `` For type II superconductors\footnote{The authors mean $\kappa \geq \frac {1}{ \sqrt{2}}$, but we treat only $\kappa$ large} the above calculation shows that superconductivity is not entirely destroyed for $H_{c_2} < H < H_{c_3}$. A superconducting sheath remains close to the surface parallel to the applied field. Conversely, when the field is decreased below $H_{c_3}$, a superconducting sheath appears at the surface before superconductivity is restored in the bulk at $H=H_{c_2}$. If the sample is a long cylinder with the applied field parallel to the axis, the sheath will cover all the surface of the cylinder. If it is a sphere, the sheath will be restricted to a small zone near the equatorial plane when $H$ is close to $H_{c_3}$. When the field is decreased towards $H_{c_2}$ the sheath will progressively extend up to the poles.'' } (see \cite{He4} or \cite{LuPa4} for a review on this aspect), we would like to analyze the groundstate energy (that is the lowest eigenvalue) of the Neumann realization of the Schr\"odinger operator with magnetic potential in an open set $\Omega$ in $\rz^3$, which will be assumed in all the paper to be smooth up to the boundary~: \begin{equation} P_{A}^h = \sum_{j=1}^3 (h D_{x_j} - A_j (x))^2 \;, \end{equation} where $A = (A_1,A_2,A_3)$ is a magnetic potential. This realization will be denoted by $P_{A,\Omega}^{h,N}$ but we will sometimes use shorter notations when there is no ambiguity. If $\lambda (h)$ is the lowest eigenvalue and $u^h(x)$ is a corresponding normalized eigenstate, we would like to discuss the asymptotic of $\lambda (h)$ as $h\ar 0$, with the hope to analyze after the localization of $u^h$. This problem was treated in the case of dimension $2$ by Bernoff-Sternberg \cite{BeSt}, Lu-Pan \cite{LuPa1, LuPa2, LuPa3}, Del Pino-Fellmer-Sternberg \cite{PiFeSt} and Helffer-Morame \cite{HeMo2} in a rather complete way. Later a recent work by Lu-Pan \cite{LuPa5} was starting the analysis of the dimension $3$ and this was completed slightly in \cite{HeMo3}. So this paper is the continuation of the programme, with a particular emphasis on the constant magnetic field case.\\ We start by establishing rough estimates, but sufficiently accurate for determining the effective order of the curvature effect. \begin{theorem}\label{theorough}.\\ Let $\Omega $ be a bounded open set of $\rz ^3$ with smooth boundary $\pa \Omega$. Let $P^{h,N}_{A,\Omega}$ be the Neumann operator on $L^2(\Omega )$ associated to the Schr\"odinger operator with constant magnetic field $(hD-A)^2$ and let us assume that the vector magnetic field $ H = \rot (A)$ is constant and of intensity equal to $b$.\\ If $\lambda (h)=\inf \sp (P^{h,N}_{A,\Omega})$ is the first eigenvalue of $P^{h,N}_{A,\Omega}$, then there exists a constant $C_0$ such that \begin{equation}\label{estgros1all} |\lambda (h) - b\Theta_0 h|\leq C_0 h^{\frac 43} \; . \end{equation} \end{theorem} We observe that in the case of dimension $2$, the corresponding error was in $\Og(h^\frac 32)$. We then continue with the analysis of the curvature effects, which are the analog in dimension $3$ of the results first conjectured by Bernoff-Sternberg (see \cite{BeSt}) in the case of dimension $2$. We establish the corresponding conjecture in the case of dimension $3$. \begin{conjecture}\label{BNbisa}.\\ Let $P_{A,\Omega}^{h,N}$ be the Neumann realization on $L^2(\Omega )$ of the magnetic Laplacian \break $(hD-A)^2$, where $h \in ]0,1[$ is a small parameter and $A$ corresponds to constant magnetic field. Under a natural generic assumption (satisfied when $\Omega$ is a strictly convex open set in $\rz^3$), there exists $ \eta >0 $ and ${\hat \gamma}_0 $ such that~: \begin{equation}\label{BN1bisa} \inf \; \Sp (P_{A,\Omega}^{h,N}) = b\Theta_0 h + {\hat \gamma}_0 b^{\frac 2 3 } h^{\frac 4 3 } + \Og(h^{\eta + \frac 4 3 })\;, \end{equation} where $ {\hat \gamma}_0$ is a complicate explicitely computable magnetic invariant. \end{conjecture} The more precise statements are given in Section \ref{Section9}, once all the terms appearing in the computation of ${\hat \gamma}_0$ have been defined. We will prove in this paper the upper bound part of this conjecture and establish the lower bound in an unfortunately non generic, but geometrically natural case, that we now briefly describe. It is observed by \cite{LuPa5} and this will be recalled in Section \ref{Section4}, that the ground state $u^h$ is localized (as $h\ar 0$) near the boundary $\pa \Omega$ but more precisely on the set~: \begin{equation}\label{Gamma} \Gamma = \{ x\in \pa \Omega\;|\; H \cdot N(x) =0\;\}\;, \end{equation} where $N(x)$ is the normal at $x$ to $\pa \Omega$.\\ It is natural to assume that~: \begin{equation} \label{Gammareg} \Gamma \mbox{ is a regular submanifold of } \pa \Omega\;. \end{equation} At each point $x$ of $\Gamma$, we will associate to $H$ (see Definition \ref{defnormalcurvature}) a notion of normal curvature along the magnetic field $\kappa_{n,B} (x)$. It is natural to assume that \begin{equation} \kappa_{n,B} \neq 0\;,\;\mbox{ on }\Gamma\;. \end{equation} The case, where we can prove the full conjecture is the case when $H$ is assumed to be orthogonal to $\Gamma$~: \begin{equation}\label{add} H \cdot T(x) = 0\;,\; \forall x\in \Gamma\;, \end{equation} where $T(x)$ is a unit tangent vector at $x$ to $\Gamma$.\\ This occurs for example for an ellipsoid when $H$ is parallel to one of the principal axis. In this case, we shall show that~: \begin{equation} {\hat \gamma}_0 = c_0 \inf_{x\in \Gamma} |\kappa_{n,B}(x) |^\frac 23\;, \end{equation} where $c_0$ is a universal strictly positive constant attached to spectral invariants related to two model Hamiltonians respectively defined on $\rz$ and $\rz^+$ and which will be analyzed in Subsections \ref{Subsection2.2} and \ref{Subsection2.4}. Although the methods of proof can also lead to localization results for the ground state (see \cite{HeMo2}, \cite{HeMo3}) or more generally for minimizers of the Ginzburg-Landau functional (see \cite{HePa}), this will not be discussed here. In Sections \ref{Section2} and \ref{Section3}, we recall previous results extracted of \cite{DaHe}, \cite {HeMo2}, \cite{LuPa5}, \cite{HeMo3}, which will play an important role in the analysis. In Section \ref{Section4}, we recall the results of \cite{LuPa5} devoted to the case when the magnetic field is not constant. It is also shown there that the problem is reduced in the constant magnetic case to a neighborhood of the boundary. In Section \ref{Section5}, we make explicit our choice of coordinates near the boundary. Section \ref{Section6} gives rough upper bounds for the ground state energy by constructing quasimodes. In Section \ref{Section7}, we present our first lower bounds which are sufficiently accurate for giving the right order for the remainder. In Section \ref{Section8}, we go further in the choice of adapted coordinates, taking in particular account of the fact that the magnetic field is constant. This permits to introduce our magnetic invariants attached to $\Gamma$ and to present in Section \ref{Section9} our main results in a more precise form. Section \ref{Section10} is devoted to the research of simpler models for the magnetic potentials obtained in the adapted coordinates by suitable gauge transformations and by neglecting ``small'' terms. Section \ref{Section11} is devoted to the estimate of the errors done in the use of the previous models. In Section \ref{Section12}, we start by some heuristics which lead to the spectral analysis of a simplified model which will be used in the general case. Section \ref{Section13} is devoted to the proof of the accurate upper bounds. We start with the analysis of the particular case, which is simpler, for helping the reader. Section \ref{Section14} is devoted to the proof of accurate lower bounds, but in a particular case. \section{ The case of dimension $2$}\label{Section2} \subsection{Main results for the models with constant magnetic fields}\label{Subsection2.1} We first consider the operator~: $$ P_{A}^h:=(h D_{x_1} - A_1)^2 + (h D_{x_2} - A_2)^2\;, $$ with $A=(- \frac b 2 x_2, \frac b 2 x_1)$, $h>0$ and $b>0 $, and analyze the spectrum of its realization in $\rz^2$ or of its Neumann realization in $\rz^2_+$. We observe that by homogeneity, one can reduce the analysis to $h=1$ and $b=1$. It is well known that in the case of $\rz^2$ the spectrum is a point spectrum and that the eigenvalues are given by $(2n+1)$ with ($n\in \nz$), each eigenvalue being of infinite multiplicity. One way, to see this is to show the unitary equivalence (via a gauge transformation, a partial Fourier transform and a translation) with the harmonic oscillator $(- \pa_t^2 + t^2)$ but seen as an unbounded operator on $L^2(\rz^2_{t,s})$. The case of the Neumann realization in $\rz^2_+$ is a little more delicate. A unitary transformation leads this time to the analysis of the Neumann realization of the operator of \begin{equation}\label{defQ} Q(t,D_t;s):= D_t^2 + (t-s)^2 \mbox{ on } L^2(\rz_+\times \rz) \end{equation} which is reduced to the analysis of the family ($s\in \rz$) of operators~: \begin{equation} Q(s):=D_t^2 + (t-s)^2\;, \end{equation} defined on $L^2(\rz_+)$. If $\mu(s)$ is defined as the lowest eigenvalue of the Neumann realization of $Q(s)$ in $\rz^+$, one easily shows (cf \cite{ReSi}) \begin{proposition}\label{Proposition2.3}.\\ The infimum of the spectrum of the Neumann realization of $P^h_{ A}$ is given by $ b h \Theta_0$ with~: \begin{equation}\label{defThetazero} \Theta_0:= \inf_{s\in \rz} \mu(s)\;. \end{equation} \end{proposition} \subsection{An important one-dimensional family of operators on $\rz$}\label{Subsection2.2} We recall the main properties of the groundstate energy of the Neumann realization on $\rz^+$ of \begin{equation}\label{defQs} Q(s):=D_t^2 + (t-s)^2\;, \end{equation} in function of the parameter $s\in \rz$. Let us just list some properties of $\mu(s)$ and of the correspoing $\varphi^s$ and refer to \cite{DaHe}, \cite{HeMo2} for proofs or details. The Neumann eigenvalue $\mu(s)$ satisfies the following properties~: \begin{enumerate} \item $\lim_{s \ar -\infty} \mu(s) = + \infty$~; \item $\mu$ is decreasing for $s<0$~; \item $\mu(0) =1$~; \item $\lim_{s\ar +\infty} \mu(s) =1$~; \item $\mu$ admits in $]0,+\infty[$ a unique minimum $\Theta_0 <1$ at some $\xi_0>0$. So \begin{equation}\label{proptheta0} \Theta_0:= \inf_{s\in \rz} \mu(s)= \mu(\xi_0) < 1\;. \end{equation} This minimum is non degenerate and more precisely~: \begin{equation}\label{musecond} 0 < \mu''(\xi_0) < 2\;. \end{equation} \item We introduce $\varphi_0=\varphi^{\xi_0}$. We have~: \begin{equation}\label{ortha} \int _{\rz _{+}} (t-\xi _0)|\varphi_0(t)|^2\, dt = 0\;, \end{equation} which just corresponds to the condition $\mu'(\xi_0) =0$. \item $\varphi_0$ is rapidly decreasing at $\infty$. \end{enumerate} Let us just mention that the proof of some of these properties is based on the following identity, relating the first eigenfunction $\varphi^s$ and $\mu(s)$~: \begin{equation}\label{Dau} ||\varphi^s||^2 \mu'(s) = (s^2 - \mu(s)) (\varphi^s(0))^2\;. \end{equation} \subsection{Applications: main results in dimension $2$} As an application of the analysis of the models, one gets (cf \cite{LuPa3}) via a partition of unity for the lower bound and a suitable construction of quasimodes for the upper bound, the following general result (in the two-dimensional case)~: \begin{theorem} \label{Theoremdim2}.\\ If $\lambda (h)$ is the lowest eigenvalue of the Neumann realization of $P_{A}^h$ in $\Omega$, then we have~: \begin{equation}\label{lupa} \lim_{h\ar 0} \frac{\lambda (h)}{h} = \min \left( \inf_{x\in \Omega} |B(x)|, \Theta_0 \inf_{x\in \pa \Omega} |B(x)|\right)\;, \end{equation} where $B =\curl A$. \end{theorem} We emphasize that there is no assumption that the magnetic field is constant. In the case of the constant magnetic field, one can actually have more precise results for $\lambda (h)$. When $B$ is constant, the minimum in (\ref{lupa}) is obtained by the second term and we show in \cite{BeSt}, \cite{PiFeSt} and \cite{HeMo2} the \begin{theorem}\label{courbure}.\\ If the magnetic field is constant of intensity $b$, then~: \begin{equation} \lambda(h) = b \Theta_0 h - c_1 b^\frac 12 \left(\sup_{x\in \pa \Omega} \kappa (x)\right) h^\frac 32 + o(h^\frac 32)\;, \end{equation} where $c_1$ is a universal strictly positive universal constant and $\kappa(x)$ is the curvature of $\pa \Omega$ at $x\in \pa \Omega$. \end{theorem} \subsection{The Montgomery's model and a second important family of operators on $\rz$.}\label{Subsection2.4} We just discuss a model that we shall meet indirectly later and which is interesting. It first appears in \cite{Mon} but see also \cite{HeMo1}. We consider in $\rz^2_{x,y}$, and for some parameter $\gamma >0$ the operator~: \begin{equation} P:= h^2 D_x^2 + (hD_y - \frac \gamma 2 x^2)^2\;. \end{equation} The magnetic potential is $\vec A = (0,\frac \gamma 2 x^2)$ and we have~: $$ \curl A = \gamma x\;. $$ So the magnetic field vanishes along a line. Let us briefly describe the spectral analysis. After a Fourier transform in the $y$-variable, we first get~: $$ \hat P = h^2 D_x^2 + ( h\eta - \frac \gamma 2 x^2)^2\;, $$ and this leads to the analysis of the family, parametrized by $\eta\in \rz$, of selfadjoint operators on $L^2(\rz)$~: $$ {\hat P}(\eta) = h^2 D_x^2 + ( h\eta - \frac \gamma 2 x^2)^2\;. $$ Using a simple dilation, we get~: \begin{equation} \inf \Sp(\hat P)=\inf_\eta \inf \Sp ({\hat P}(\eta)) = h^\frac 43 |\frac \gamma 2 |^\frac 23 \inf_\rho \inf \sp (D_r^2 + (r^2 -\rho)^2)\;. \end{equation} Let us recall some properties of the family of operators \begin{equation} S(\rho) = D_r^2 + (r^2 - \rho)^2\;, \end{equation} and of the corresponding ground state $\psi^\rho$, which were established in \cite{Mon}, \cite{HeMo1} and \cite{PaKw}. \begin{enumerate} \item There exists a unique $\rho = \rho_{min}$ such that~: \begin{equation}\label{nuzero} \hat \nu _0 := \inf_\rho \inf \Sp (D_r^2 + (r^2-\rho)^2) = \inf \Sp (D_r^2 + (r^2-\rho_{min})^2)\;. \end{equation} \item $\psi^\rho$ belongs to $\Sg(\rz)$ and is even. \end{enumerate} We shall later use the notation~: \begin{equation}\label{psizero} \psi_0 = \psi^{\rho_{min}}\;. \end{equation} \section{Constant magnetic field: models in $\rz^3$ and in $\rz^3_+$.}\label{Section3} As in the case of dimension $2$, where the first thing was to understand the model with constant magnetic feld in $\rz^2$ and $\rz^2_+$, we shall discuss the case of $\rz^3$ and $\rz^3_+$. \subsection{Model in $\rz^3$.}\label{Subsection3.1} We start (after some gauge transformation) of the Schr\"odinger operator with constant magnetic field in dimension $3$. \begin{equation}\label{moddim3} P^h(\vec {H}):= h^2 D_{x_1}^2 + (hD_{x_2} - H_3 x_1)^2 + (h D_{x_3} + H_2 x_1 - H_1 x_2)^2\;. \end{equation} The following lemma has the age of quantum mechanics~: \begin{lemma}.\\ The bottom of the spectrum of the selfadjoint realization of $ P^h( \vec {H})$ in $\rz^3$ is \begin{equation} \inf \sp \left( P^{h,N}_{\rz^3}( \vec {H})\right) = b h\;, \end{equation} where $$ b =||\,H\, ||= \sqrt{H_1^2 + H_2^2 + H_3^2} $$ is the intensity of $H$. \end{lemma} \subsection{Models in halfspaces.}\label{Subsection3.2} We refer to \cite{LuPa5} and \cite{HeMo3} for the proof of the results presented in this subsection. If $N$ is a unit vector in $\rz^3$, we now consider the Neumann realization in $\Omega:=\{ x\in \rz^3\;|\; x\cdot N >0\}$. After a rotation, we can assume in the proofs that $N=(1,0,0)$, so $\Omega$ is $\rz^3_+:=\{ x_1 >0\}$.\\ After scaling, we can assume that $h=1$ and $|| H || =1$. After some rotation in the $(x_2,x_3)$ variables, we can assume that the new magnetic field is $(\beta_1,\beta_2, 0)$ and we are reduced to the problem of analyzing~: $$ P(\beta_1, \beta_2):=D_{x_1}^2 + D_{x_2}^2 + (D_{x_3} + \beta_2 x_1 - \beta_1 x_2)^2\;, $$ in $\{ x_1 >0\}$, where~: $$ \beta_1^2 + \beta_2^2 =1\;.$$ We introduce~: $$ \beta_2 = \cos \vartheta\;,\; \beta_1 = \sin \vartheta\;, $$ and we observe that, if $N$ is the external normal to $x_1=0$, we have~: \begin{equation}\label{orth} \vec{H}\cdot N = - \sin \vartheta\;. \end{equation} By partial Fourier transform, we arrive to~: \begin{equation} L(\vartheta, \tau) = D_{x_1}^2 + D_{x_2}^2 + (\tau + \cos \vartheta \; x_1 - \sin \vartheta \; x_2 )^2\;, \end{equation} in $x_1 >0$ and with Neumann condition on $x_1 =0$. The bottom of the spectrum of $L(\vartheta, \tau)$ is given by~: \begin{equation}\label{bott} \sigma(\vartheta):=\inf \Sp\left( L(\vartheta, D_t)\right) = \inf_{\tau} \left(\inf \Sp ( L(\vartheta,\tau) )\right)\;. \end{equation} \begin{proposition}\label{Propvartheta}.\\ The bottom of the spectrum of the Neumann realization of $P^h( \vec{H})$ in \break $\Omega:=\{x\in \rz^3\;|\; x\cdot N > 0\}$ is~: \begin{equation} \inf \; \sp P^{h,N}_\Omega( \vec{H}) = \sigma (\vartheta) \; b \; h\;, \end{equation} where $\vartheta \in [-\frac \pi 2, \frac \pi 2]$ is defined by (\ref{orth}). \end{proposition} By symmetry considerations, we observe also that~: \begin{equation}\label{sym} \sigma(\vartheta) = \sigma (- \vartheta) = \sigma ( \pi - \vartheta)\;. \end{equation} It is consequently enough to look at the restriction to $[0,\frac \pi 2]$. \subsection{Properties of $\vartheta \mapsto \sigma(\vartheta)$.}\label{Subsection3.3} Let us now list the main properties of the function $\vartheta \mapsto \sigma(\vartheta)$ on $[0,\frac \pi 2 ]\;$ . Most of them are established in \cite{LuPa5} but see also \cite{HeMo3}. \begin{enumerate} \item $\sigma$ is continuous on $[0, \frac \pi 2]$. \item \begin{equation} \sigma(0) = \Theta_0 < 1\;. \end{equation} \item \begin{equation} \sigma(\frac \pi 2 )=1\;. \end{equation} \item \begin{equation}\label{roughlow} \sigma(\vartheta) \geq \Theta_0 (\cos \vartheta)^2 + (\sin \vartheta)^2\;. \end{equation} \item If $\vartheta \in ]0,\frac \pi 2[$, the spectrum of $L(\vartheta,\tau)$ is independent of $\tau$ and its essential spectrum is contained in $[1, +\infty[$. \item For $\vartheta \in ]0,\frac \pi 2 [$, $\sigma(\vartheta)$ is an isolated eigenvalue of $L(\vartheta,\tau)$, with finite multiplicity. \item The function $\sigma $ is strictly increasing on $[0, \frac \pi 2[$. \item The function $\sigma $ has the following expansion for $\vartheta$ small~: \begin{equation}\label{asympexp} \sigma(\vartheta) \sim \Theta_0 + \sum_{n\geq 1} \alpha_n |\vartheta|^n\;, \end{equation} with $\alpha_1 = \sqrt{\frac{\mu''(\xi_0)}{2}}>0$. \end{enumerate} A first consequence of this analysis is \begin{proposition} \label{Proposition3.5}.\\ When $b=|| H ||$ is fixed the bottom of the spectrum of $P_\Omega^{h,N}(\vec{H})$ in $\Omega :=\{ x\cdot N >0 \}$ is minimal when $\vartheta=0$ that is, according to (\ref{orth}), that is when the magnetic field vector satisfies $\vec{H}\cdot N =0$. \end{proposition} \section{First results for general magnetic fields} \label{Section4} In the case of dimension $2$, under the assumption that $\curl A:= B(x) >0$, the basic estimate at the interior was the inequality~: \begin{equation} \label{basic} h \int B(x) |u(x)|^2 dx \leq \int |(h\nabla - i A)u|^2 dx \;,\; \forall u \in C_0^\infty (\Omega)\;. \end{equation} This was not enough for understanding the Neumann problem. One should more carefully analyze the case of $\rz^2_+$ and, in the case of the constant magnetic field, one should also analyze more complicate models (like for example the case of the disk). The most spectacular result was~: \begin{theorem}\label{Theorem4.1}.\\ When the magnetic field is constant, the groundstate of the Neumann realization of $P^h_{A}$ in an open regular bounded set $\Omega\subset \rz^2$ is localized in the neighborhood of the points of the boundary of maximal curvature. \end{theorem} In the case of dimension $3$, estimate (\ref{basic}) should be replaced by the weaker estimate established in \cite{HeMo1} (Theorem 3.1) \begin{theorem}\label{Theorem4.2}.\\ There exist $C$ and $h_0 >0$ such that, for all $h\in ]0,h_0]$, we have~: \begin{equation} h \int_\Omega (||H (x)|| - C h^\frac 14) |u(x)|^2 dx \leq \int_\Omega |(h\nabla - i A)u|^2 dx \;,\; \forall u \in C_0^\infty (\Omega) \end{equation} \end{theorem} If this result is essentially sufficient for analyzing the Dirichlet problem in $\Omega$, it is necessary to implement the analysis given in the first part in order to treat the Neumann problem. Near each point of the boundary $x$, we have to use the lower bound obtained for the model with constant magnetic field $H = H(x)$. Following for example the proof in \cite{HeMo2} and using the same partition of unity, we get~: \begin{theorem}\label{Theorem4.3}.\\ \begin{equation} h \int_\Omega W_h (x) |u(x)|^2 dx \leq \int |(h\nabla - i A)u|^2 dx \;,\; \forall u \in H^1(\Omega) \end{equation} where \begin{equation} \begin{array}{lll} W_h (x) &= || H(x) || - C h^\frac 14\;, \; &\mbox{ if } d(x, \pa \Omega) \geq 2 h^\frac 38\;,\\ &= ||H(s(x))|| \; \sigma(\vartheta(x) - C h^\frac 14 &\mbox{ if } d(x, \pa \Omega) \leq 2 h^\frac 38\;. \end{array} \end{equation} \end{theorem} Here we recall that $\vartheta (x)$ satisfies~: \begin{equation}\label{defis} || H (s(x))|| \cdot \sin \vartheta (x) = - H (s(x))\cdot N (s(x))\;, \end{equation} where $s(x)$ is, for $x$ near $\pa \Omega$, the point in $\pa \Omega$ such that~: $$ d(x,\pa \Omega) = d(x, s(x))\;, $$ and we observe that, due to (\ref{sym}), $\sigma ( \vartheta(x))$ is well defined by (\ref{defis}).\\ The first consequence (compare with Theorem \ref{Theoremdim2}) is~: \begin{theorem}\label{Theorem4.4}.\\ \begin{equation}\label{minim} \lim_{h\ar 0} \left(\lambda (h)/h\right) = \min \left(\inf_{x\in \Omega} ||H (x)||\;,\; \inf_{x\in \pa \Omega} ||H (x)|| \; \sigma(\vartheta(x))\right)\;. \end{equation} \end{theorem} The lower bound is a direct consequence of Theorem \ref{Theorem4.3} and the proof of the upper bound is sketched in \cite{LuPa5}. When $H(x)$ is constant of intensity $b$ , then the minimum in (\ref{minim}) is obtained for the second term and we have~: \begin{equation}\label{minim1} \lim_{h\ar 0} \left(\lambda (h)/h\right) = b \left(\inf_{x\in \pa \Omega} \; \sigma(\vartheta(x))\right)\;. \end{equation} But it should be (see (\ref{latera})~) one point such that $\vartheta(x) =0$, so we get~: \begin{equation}\label{minim2} \lim_{h\ar 0} \left(\lambda (h)/h\right) = \Theta_0 b\;. \end{equation} \begin{remark}.\\ As in \cite{HeMo2}, this leads to localization theorems ( \cite{LuPa5}, \cite{HeMo3} but see also \cite{PiFeSt}). In particular, if the magnetic field is constant, then the ground state is localized near the points of the boundary where the magnetic field is parallel to the tangent space. \end{remark} {\bf Application}.\\ An interesting case is the case when $H$ is constant and when $\Omega$ is the ellipsoid~: $$ a_1 x_1^2 + a_2 x_2^2 + a_3 x_3^2 \leq 1\;.$$ The set of points where $\vec{H}= (H_1,H_2,H_3)$ is parallel is obtained by intersecting the boundary of the ellipsoid with the plane~: $$ a_1 x_1\, H_1 + a_2 x_2 H_2 + a_3 x_3 H_3 =0\;. $$ More generally, if the surface is strictly convex and if the magnetic field is constant, it is possible to show that the set of points of the boundary where $H$ is parallel to the tangent space is a $C^\infty$ curve. We emphasize that Theorem \ref{Theorem4.4} does not explain all the situation. In the case with constant magnetic field it would be nice to show the role of some curvature in the localization as in the case of dimension $2$. {\bf This is actually our goal to give an answer to this question.} \section{Adapted coordinates} \label{Section5} \subsection{Magnetic geometrical invariants}\label{maginv} The standard coordinates on $ \rz^3$ will be denoted by $ x=(x_1,x_2,x_3)$ and the standard flat metric by $ g_0$. We will also use $ <\; .\; |\; .\; >$ for $ g_0(\; .\; ,\; .\; )$ or for $ g_{0}^{\star }(\; .\; ,\; .\; )$, and $ | X | $ for $ ()^{\frac 1 2 }$, (if $ X$ is a vector field or a one-form). The standard volume on $\rz^3$ will be denoted by $\omega_3 = dx^3 =dx_1\wedge dx_2\wedge dx_3 $ and will fix also the orientation of $\rz^3$. Let $A$ be a smooth magnetic potential one-form on $ {\overline {\Omega }}$, : $$ \displaystyle A=\sum _{j=1}^{3}A_j(x)dx_j,\ \ A_j\; \in \; C^{\infty }({\overline {\Omega }};\rz ).$$ The magnetic field $ B$ is the two-form: \begin{equation}\label{defmag} B=dA=\sum _{1\leq i \;,\\ & =\sum _{1\leq i,j\leq 3}g_{ij} {\widetilde {X}}_i{\widetilde {Y}}_j\;, \\ |g| &= \det( g_{ij})\;, \end{array} $$ where $\displaystyle \ X=\sum _j{\widetilde {X}}_j\frac{\pa }{\pa y_j}$ and $\displaystyle \ Y=\sum _j{\widetilde {Y}}_j\frac{\pa }{\pa y_j}$.\\ The magnetic potential is given in the new coordinates by \begin{equation}\label{newmagpot} A=\sum _{j=1}^{3}{\widetilde {A}}_jdy_j\ \ \with \ {\widetilde {A}}_j=\sum _{k=1}^{3}A_k\frac{\pa x_k}{\pa y_j}. \end{equation} The magnetic field is also given by \begin{equation}\label{newmagfi} B=\sum _{1\leq i0$ small enough (and modifying a little $ \Vg_{x_0}$ if necessary), we can define local coordinates on $ \Vg_{x_0}$, $$ \Phi \; :\; \Vg_{x_0} \; \to \; \Sg \times ]0,\epsilon [, \ \ \Phi (x)=(y_1,y_2,y_3) $$ such that \begin{equation}\label{adapcoo} y_3(x)=\dist (x,\phi^{-1}(y_1,y_2))=\dist (x,\pa \Omega ); \end{equation} so \begin{equation}\label{adapcoo2} x=\phi^{-1}(y_1,y_2)+y_3N(\phi^{-1}(y_1,y_2)),\ \ \forall \ x\; \in \; \Vg_{x_0}, \end{equation} where $N(x)$ is the interior normal unit to $ \pa \Omega $ at the point $ x \; \in \; \pa \Omega$.\\ Then we get by simple computation the form of the standard flat metric $ g_0$ in these new coordinates, \begin{equation}\label{adapmetra} \begin{array}{ll} g_0 &=\sum_{1\leq i,j\leq 3}g_{ij}\; dy_i\otimes dy_j \\ & =dy_{3}\otimes dy_3 + G \\ &\quad + 2 y_3 \sum_{1\leq i,j\leq 2} <\frac{\pa N}{\pa y_i}|\frac{\pa x}{\pa y_j}> dy_i\otimes dy_j \\ & \quad + y_{3}^{2}\sum_{1\leq i,j\leq 2} <\frac{\pa N}{\pa y_i}|\frac{\pa N}{\pa y_j}> dy_i\otimes dy_j\;. \end{array} \end{equation} In this case we have, using (\ref{newmagpot}) and (\ref{adapcoo2}), $$ {\widetilde {A}}_3=\sum_{j=1}^{3}A_j N_j\;. $$ \begin{remark}\label{Remconv}.\\ In the following, we choose a finite family of open sets $\Vg_{x_j}$ ($x_j\in \pa \Omega$) covering a tubular neighborhood of $\pa \Omega$. When we shall speak later of local coordinates, we will mean that we take some element belonging to the family. \end{remark} \section{Rough upper bounds}\label{Section6} % Voici une proposition intermediaire To prove our results, we need some rough preliminary estimates (with some control of the remainder) of the ground state energy. \begin{proposition}\label{estgros}.\\ Let $\Omega $ be a bounded open set of $\ \rz ^3$ with smooth boundary $\pa \Omega$. Let $P^{h,N}_{A,\Omega}$ be the Neumann operator on $L^2(\Omega )$ associated to the Schr\"odinger operator with constant magnetic field $(hD-A)^2$ and let us assume that the vector magnetic field $ H = \rot (A)$ is constant.\\ If $\lambda (h) =\inf \sp (P^{h,N}_{A,\Omega})$ is the first eigenvalue of $P^{h,N}_{A,\Omega}$, then there exists a constant $C_0$ such that \begin{equation}\label{estgros1up} \lambda (h)\leq b\Theta_0 h + C_0 h^{\frac 43} \; . \end{equation} \end{proposition} \begin{remark}.\\ It is not necessary to assume that the magnetic field is constant as the proof of this proposition will show. If one choose a point $x_0$ such that $H(x_0)$ is tangent to $\pa \Omega$, one can get the same result with $b= ||\, H(x_0)\,||$. One can then optimize by choosing a point of this kind such that $|B(x_0)|$ is minimal. If $\Gamma$ is the set introduced in (\ref{Gamma}) we will prove~: \begin{equation}\label{estgros1upa} \lambda (h)\leq \Theta_0 \inf_{x\in \Gamma}|H(x)|\, h + C_0 h^{\frac 43} \; . \end{equation} \end{remark} {\bf {Proof of Proposition \ref{estgros}}}.\\ As the flux of the magnetic field through the boundary is zero, \begin{equation}\label{latera} \int_{\pa \Omega }\rot(A(x))\cdot N(x)ds =\int_{\Omega }\div \rot (A) dx=0\;, \end{equation} if $ N $ is the interior normal unit of $\pa \Omega $, there exists $x_0\; \in \; \pa \Omega$ such that the vector magnetic field is tangent to $\pa \Omega $ at $\ x_0$: $$ \rot(A(x_0))\cdot N(x_0)=0. $$ We take local coordinates $(y_1,y_2)$ in a neighbourhood of $W_0$ of $x_0$ in $\pa \Omega $ such that the metric is $\delta_{ij}$ at $ x_0$ and $\frac{\pa }{\pa y_1}$ is parallel to the vector magnetic field. \\ We consider the adapted coordinates $\ y=(y_1,y_2,y_3)$, with $y_3(x)=d(x,\pa \Omega ),$ and then $\ g_{ij}(x_0)=\delta _{ij} $. If ${\widetilde {A}}$ is the magnetic potential in the new coordinates, $A\cdot dx={\widetilde {A}}\cdot dy$, then \begin{equation}\label{est2} \begin{array}{l} \frac{\pa {\widetilde A}_3}{\pa y_2}(x_0)- \frac{\pa {\widetilde A}_2}{\pa y_3}(x_0)=b\\ \frac{\pa {\widetilde A}_1}{\pa y_3}(x_0) -\frac{\pa {\widetilde A}_3}{\pa y_1}(x_0)=0\\ \frac{\pa {\widetilde A}_2}{\pa y_1}(x_0)- \frac{\pa {\widetilde A}_1}{\pa y_2}(x_0))=0\;. \end{array} \end{equation} We can find a a gauge transform $\exp i \frac{\varphi}{h}$, with $\varphi$ real, such that, in (\ref{qAc}), $$ {\widetilde A}_1(x_0) =0\;,\; {\widetilde A}_2(x_0)=0\;,\; {\widetilde A}_3=0\;. $$ We first estimate the error occuring when eliminating the terms vanishing to order $3$.\\ If $u\in H^1(\Omega )$ is such that \begin{equation}\label{est3} {\rm {supp}}(u) \; \subset \; \{ x;\ |x-x_0|\leq h^{\delta }\} \end{equation} with $\delta >0$, and if $\ h$ is small enough, then, for some $C>0$, \begin{equation}\label{est4} q^{h}_{A}(e^{i \frac{\varphi}{h} }u)\leq (1+h^{\delta }C)q^{h}_{A^0}(u)+ C [h^{3\delta }(q^{h}_{A^0}(u))^{\frac 1 2 }\cdot \| u\| + h^{6\delta }\| u\| ^2]\; , \end{equation} where $$ A^{0}_{1}=R_1(y), \ A^{0}_{2}=-b y_3 + R_2 (y) ,\ A^{0}_{3}=0\;, $$ the $R_j(y)$ are polynomial functions, homogeneous of order two, $$ R_j(y)=\sum _{|\alpha |=2}a_{j,\alpha }y^{\alpha }, $$ and \begin{equation}\label{est4b} q^{h}_{A^0}(u)=\int _{|y|\leq C h^{\delta },\ y_3>0} |(hD_yu-A^0(y)u|^2\; dy\; . \end{equation} We take the function $ u$ in the form $$ u(y_1,y_2,y_3):=e^{-ib^{\frac 1 2 }y_2 \xi _0/h^{\frac 1 2 }}v(y) $$ with \begin{equation}\label{est5} v(y_1,y_2,y_3)= h^{- \frac 14 - \delta}\varphi _0(b^{\frac 1 2 }h^{-\frac 1 2 } y_3 )\chi (4h^{-\delta } y_3 ) \chi (4h^{-\delta }(y_{1}^{2}+y_{2}^{2})^{\frac 1 2 })\; . \end{equation} Here $ \chi $ is a cut off function equal to one on $[-\frac 1 2 ,\frac 1 2 ]$ and supported in $[-1,1]$, $\varphi_0$ is introduced in Subsection \ref{Subsection2.2} and $\delta \in ]0,\frac 12[$. \\ Observing that~: $$ \int _{\rz _{+}}(|D_t\varphi _0(t)|^2+|(t-\xi _0)\varphi _0(t)|^2)dt =\Theta_0\int _{\rz _{+}} |\varphi _0(t)|^2dt, $$ and that $ t \mapsto \varphi _0(t)$ is an exponentially decreasing function at infinity,\\ we get when $\ \delta = \frac 1 3 \; ,$ \begin{equation}\label{est6} q^{h}_{\hat A^0}(u)\leq (b\Theta_0 h + C\, h^{\frac 43 })\| u\|^2 \; , \end{equation} where $$ {\hat A}^0_{1}=0, \ {\hat A}^0_{2}=-b y_3 ,\ {\hat A}^0_{3}=0\;.$$ We then have to compare $q^{h}_{\hat A^0}(u)$ and $q^{h}_{A^0}(u)$. The first try could be to use~: \begin{equation}\label{est4a} q^{h}_{A^0}(e^{i \frac{\varphi}{h} }u)\leq (1+h^{\delta }C)q^{h}_{\hat A^0}(u)+ C [h^{2\delta }(q^{h}_{\hat A^0}(u))^{\frac 1 2 }\cdot \| u\| + h^{4\delta }\| u\| ^2]\; , \end{equation} This leads to error terms of size $h^{1+ \delta }$, $ h^{\frac 12 + 2 \delta}$, $h^{4 \delta}$ and $h^{2- 2 \delta}$. But this leads only, using (\ref{est6}) and (\ref{est4a}) with $\delta =\frac 13$, to (\ref{estgros1up}) with $\Og(h^\frac 76 )$ instead of $\Og(h^\frac 43)$ as expected. In order to get effectively (\ref{estgros1up}), we need to use (\ref{ortha}). We have indeed to analyze more carefully the term which was bounded from above in (\ref{est4a}) by $h^{2\delta }(q^{h}_{\hat A^0}(u))^{\frac 1 2 }\cdot \| u\|$. The terms which were estimated by $h^{\frac 12 + 2 \delta}$ are of the form $$ h^{-\frac 1 2 - 2 \delta}\int y_j y_k (\xi_0 - \frac{b^\frac 12 y_3}{h^{\frac 12} }) \varphi_0^2(b^\frac 12 h^{-\frac 12} y_3) \chi ^2(4 h^{-\delta} (y_1^2 + y_2^2)^{\frac 12}) dy\;, $$ with $j$ and $k$ equal to $1$ or $2$. But they actually vanish due to (\ref{ortha}). The other terms have actually better upper bounds. We observe indeed that for example~: $$ h^{-\frac 1 2 - 2 \delta}\int y_2 y_3 (\xi_0 - \frac{b^\frac 12 y_3}{h^\frac 12}) \varphi_0^2(b^\frac 12 h^{-\frac 12} y_3) \chi ( h^{-\delta} y_3)\chi ^2(4 h^{-\delta} (y_1^2 + y_2^2)^\frac 12) dy\;, $$ can be estimated by $\Og(h^{1+ \delta})$.\\ This suggests strongly that the next term in the expansion of $\lambda (h)$ is $\Og(h^\frac 43)$, but to go further we need to analyze the model more carefully in the neighborhood of the points of the boundary where $H(x)\cdot N(x)$ vanishes. For this we need to enter more deeply in the geometry of the boundary in connection with the magnetic vector field, and this will be done in Section \ref{Section8}. \section{Rough lower bounds}\label{Section7} We assume that the magnetic field is constant. Let $u^h$ be an eigenfunction associated to the eigenvalue $\lambda (h)=\inf \; \Sp (P_{A,\Omega}^{h,N})$. \subsection{A priori estimates} As proved in \cite{HeMo2}, we have, without the assumption (\ref{Gammareg}), the following behavior of $u^h$. \begin{proposition}\label{pBN}.\\ If $t=t(x)=d(x,\pa \Omega )$, then, for any $k\in \nz $, there exists a constant $C_k$ depending only on $k$, such that \begin{equation}\label{pBN1} \| t^{\frac k 2 }u^h\| \leq C_kh^{\frac k 4 }\| u^h\| \end{equation} and \begin{equation}\label{pBN2} \| t^{\frac k 2 }(hD-A)u^h\| \leq C_k h^{(k+2)/4} \| u^h\| \;. \end{equation} \end{proposition} \noindent {\bf {Proof of Proposition \ref{pBN}}} .\\ Proposition \ref{estgros} and the fact that $\Theta_0<1$, give the existence of $h_0>0$ and of a constant $ C_0>1$ such that \begin{equation}\label{pBN3} C_{0}^{-1} b h \leq b h - \lambda (h) \leq C_0 b h\; . \end{equation} Of course this a very rough estimate.\\ Let us remark that (\ref{pBN1}) and (\ref{pBN2}) are valid when $k=0$. As for the 2-dimensional case in \cite{HeMo1}, we proceed by recursion. \\ By changing $t$ away from the boundary, we can assume that $x \mapsto t(x)=d(x,\pa \Omega )$ is extended as a $ C^1({\overline {\Omega }})$ function. By choosing suitable coordinates and after a gauge transform, we can assume that $$ A_1(x)=0,\ \ A_2(x)=-\frac{b}{2}x_3,\ \ A_3(x)=\frac{b}{2}x_2\; . $$ As $\ t/\pa \Omega =0$, we have by integrating by parts and for $k>0$, \begin{equation}\label{pBN4} \begin{array}{l} ihb\int_{\Omega }t^k|u^h|^2dx = \int_{\Omega }t^k[(hD_2-A_2),(hD_3-A_3)]u^h\;\cdot\;{\overline u}^hdx \\ =\int_{\Omega }t^k\{ (hD_3-A_3)u^h{\overline {(hD_2-A_2)u^h}} -(hD_2-A_2)u^h{\overline {(hD_3-A_3)u^h}}\} dx\\ +ihk\int_{\Omega }t^{k-1}\{ \frac{\pa t}{\pa x_2}(hD_3-A_3)u^h -\frac{\pa t}{\pa x_3}(hD_2-A_2)u^h\} {\overline u}^hdx \end{array} \end{equation} and \begin{equation}\label{pBN5} \int_{\Omega }t^k|(hD-A)u^h|^2dx= \int_{\Omega }\{ \lambda (h)t^k|u^h|^2- ihkt^{k-1}( (\nabla_x t)\cdot (hD-A)u^h){\overline u}^h\} dx\; . \end{equation} If $k=1$ we get from (\ref{pBN4}) \begin{equation}\label{pBN6} hb\int_{\Omega }t|u^h|^2dx \leq \int_{\Omega }t|(hD-A)u^h|^2dx + C h \| (hD-A)u^h\| \cdot \| u^h\| \; , \end{equation} and then, we use (\ref{pBN5}) to get \begin{equation}\label{pBN7} hb\int_{\Omega }t|u^h|^2dx \leq \lambda (h)\int_{\Omega }t|u^h|^2dx + C h \| (hD-A)u^h\| \cdot \| u^h\| \; . \end{equation} So \begin{equation}\label{pBN8} (hb-\lambda (h))\int_{\Omega }t|u^h|^2dx \leq C h{\sqrt {\lambda (h)}} \| u^h\| ^2\; . \end{equation} We obtain (\ref{pBN1}) with $k=1$ from (\ref{pBN3}) and (\ref{pBN8}), and then it is easy to get (\ref{pBN2}) for $k=1$ from (\ref{pBN5}) using (\ref{pBN3}), (\ref{pBN1}) with $k=1$ and (\ref{pBN2}) with $k=0.$ \\ If $k\geq 2,$ we get in the same way \begin{equation}\label{pBN9} (hb-\lambda (h))\int_{\Omega }t^k|u^h|^2dx \leq hkC\| t^{-1+\frac k 2 }(hD-A)u^h\| \cdot \| t^{\frac k 2 } u^h\| \; , \end{equation} which gives, using (\ref{pBN3}), \begin{equation}\label{pBN10} \| t^{\frac k 2 } u^h\| \leq C k \| t^{-1+\frac k 2 }(hD-A)u^h\| \; . \end{equation} Using (\ref{pBN10}) and (\ref{pBN5}) we get \begin{equation}\label{pBN11} \| t^k | (hD-A)u^h \| ^2 \leq C k h [ \| t^{\frac k 2 } u^h\| ^2 + \| t^{-1+\frac k 2 } | (hD-A)u^h \| ^2 ] \; . \end{equation} Then we can proceed by recursion. (\ref{pBN2}) for $k=j-2$ and (\ref{pBN10}) for $k=j$ give (\ref{pBN1}) for $k=j$. Formulas (\ref{pBN11}) with $k=j$ and (\ref{pBN2}) for $k=j-2$ give (\ref{pBN2}) with $k=j$. \subsection{A partition of unity} \noindent Let $\ (\chi_{\gamma }(z))_{\gamma \in \zz^3 }$ be a partition of unity of $\ \rz^3.$ For example we can take $$ \chi_{\gamma }\in C^{\infty }(\rz^3;\rz )\ {\rm {and}}\ \supp (\chi_{\gamma }) \subset \gamma +[-1,1]^3,\ \forall \gamma \in \zz^3 , $$ \begin{equation}\label{low1} \sum_{\gamma }\chi_{\gamma }^{2}(z)=1 \ {\rm {and}}\ \sum_{\gamma } |\bigtriangledown \chi_{\gamma }(z)|^2\; < \; \infty . \end{equation} If $\ \tau (h)$ is a function of $h$ such that $\tau (h)\in ]0,\epsilon(\Omega)[$, where $\epsilon (\Omega)$ is the geometric constant which is the maximal $\epsilon$ for the property that $\{ d(x,\pa \Omega) < \epsilon \}$ is a regular tubular neighborhood in $\overline \Omega$ of $\pa \Omega$, we will define the functions \begin{equation}\label{low2} \chi_{\gamma ,\tau (h)}(z)=\chi_{\gamma }(z/\tau (h)),\ \ \forall \; \gamma \; \in \; \zz^3 \;. \end{equation} So we get a new partition of unity such that \begin{equation}\label{low3} \sum_{\gamma }\chi_{\gamma ,\tau (h)}^{2}(z)=1, \ \ \ \sum_{\gamma} |\nabla \chi_{\gamma ,\tau (h)}(z)|^2 \leq C\, \tau (h)^{-2}\;, \end{equation} and $$ \supp (\chi_{\gamma ,\tau (h)})\subset \tau (h)\gamma +[-\tau (h),\tau (h)]^3 . $$ Then, for any $\ u\; \in \; H^1(\Omega )$, we have~: \begin{equation}\label{low4} q^{h}_{A}(u)=\sum_{\gamma } \left[q^{h}_{A}(\chi_{\gamma ,\tau (h)}u) -h^2\|\; |\nabla \chi_{\gamma ,\tau (h)}| \,u \; \|^{2}\right]\;. \end{equation} Let us define \begin{equation}\label{low5} \begin{array} {l} \Gamma_{\tau (h)}(\Omega )= \{ \gamma \in \zz^3 ;\ \supp (\chi_{\gamma ,\tau (h)})\cap \Omega \neq \emptyset \} \\ \Gamma^{0}_{\tau (h)}(\Omega )= \{ \gamma \in \Gamma_{\tau (h)}(\Omega );\ \dist (\supp (\chi_{\gamma ,\tau (h)}),\partial \Omega )>\tau (h) \} \\ \Gamma^{1}_{\tau (h)}(\Omega )= \{ \gamma \in \Gamma_{\tau (h)}(\Omega );\ \dist (\supp (\chi_{\gamma ,\tau (h)}),\partial \Omega )\leq \tau (h) \} \end{array} \end{equation} \subsection{Proof of Theorem \ref{theorough}: Lower bounds} \label{Subsection7.3} For the moment, we give a rough lower bound of the ground state energy in the case of constant magnetic field. \begin{proposition}\label{Lowb}.\\ Under the assumptions of Proposition \ref{estgros}, there exists $C_0$ such that~: \begin{equation}\label{estgros1low} b\Theta_0 h - C_0 h^{\frac 43} \leq \lambda (h) \; . \end{equation} \end{proposition} {\bf {Proof of Proposition \ref{Lowb} }}.\\ We proceed as in the proof of Proposition 9.2 of \cite{HeMo2} and use the partition of unity introduced in the previous subsection. \paragraph{Far from the boundary}.\\ As $\rot A $ is constant, the standard estimate gives~: $$ \| (hD - A)w\|^{2}_{L^2(\rz^3)}\geq b h \| w\|^{2}_{L^2(\rz^3)} \; ,\ \forall \; w \; \in \; C^{\infty }_{0}(\rz^3)\; . $$ So \begin{equation}\label{3est1} q^{h}_{A}(\chi_{\gamma ,\tau (h)}u) \geq b h \| \chi_{\gamma ,\tau (h)}u \|^2 \; , \ \forall \; \gamma \; \in \; \Gamma^{0}_{\tau (h)}(\Omega ) \; . \end{equation} \paragraph{Near the boundary}.\\ Suppose $\gamma \in \Gamma^{1}_{\tau (h)}(\Omega )$; then there exists local coordinates\footnote{belonging to the family introduced in Subsection \ref{corbord2}} adapted to the boundary $y=(y_1,y_2,y_3)$ such that $\ y_3(x)=d(x,\pa \Omega )$. Then, for $\tau (h)<1$ and for some integer $\ k\in \nz ^{\star },$ \begin{equation}\label{3est2} \begin{array}{ll} q^{h}_{A}(\chi_{\gamma ,\tau (h)}u) & \geq (1 - C\tau (h))q^{h}_{{\widetilde A}^{(k)}}(\chi_{\gamma ,\tau (h)}u) \\ & \quad - C (\tau (h))^{k+1} \| \chi_{\gamma ,\tau (h)}u\| \cdot (q^{h}_{A}(\chi_{\gamma ,\tau (h)}u))^{1/2}\\ & \quad - C (\tau (h))^{2k +2} \| \chi_{\gamma ,\tau (h)}u\|^2 \; , \end{array} \end{equation} with, for some $ y_0 \in \pa \Omega \cap \{ \tau (h) + {\rm {supp}}(\chi_{\gamma ,\tau (h)})\} $, $$ {\widetilde A}^{(k)}={\widetilde A}(y_0) + \sum_{1\leq |\alpha |\leq k}(y-y_0)^{\alpha } \frac{\pa ^{\alpha }\widetilde A}{\pa y^{\alpha }} (y_0) $$ and $$ \begin{array}{l} q^{h}_{{\widetilde A}^{(k)}}(w) \\ = |g|^{1/2}(y_0) \int_{\rz^2\times \rz_{+}} [\;|(hD_{y_3} -{\widetilde A}^{(k)}_{3})w|^2 \\ \quad + \sum_{1\leq i,j\leq 2} g^{ij}(y_0)(hD_{y_i} - {\widetilde A}^{(k)}_{i})w \cdot {\overline {(hD_{y_j} - {\widetilde A}^{(k)}_{j})w}}] dy \; . \end{array} $$ Let us remark that \begin{equation}\label{3est3} \| \chi_{\gamma ,\tau (h)} u \|^2 \leq (1 + C \tau (h)) |g|^{1/2}(y_0) \int_{\rz^2\times \rz_{+}} | \chi_{\gamma ,\tau (h)} u |^2 dy\; , \end{equation} and by (\ref{newvecmag}) \begin{equation}\label{3est4} b^2= |g|^{-1}(y_0)[({\widetilde H}^{0}_{3})^2 + \sum_{1\leq j,k\leq 2} g^{jk}(y_0){\widetilde H}^{0}_{j} {\widetilde H}^{0}_{k} ] \end{equation} if \begin{equation}\label{3est5} \begin{array}{l} {\widetilde H}^{0}_{1} = \frac{\pa {\widetilde A}_3 }{\pa y_2} (y_0) - \frac{\pa {\widetilde A}_2 }{\pa y_3} (y_0),\\ {\widetilde H}^{0}_{2} = \frac{\pa {\widetilde A}_1 }{\pa y_3}(y_0) - \frac{\pa {\widetilde A}_3 }{\pa y_1} (y_0),\\ {\widetilde H}^{0}_{3} = \frac{\pa {\widetilde A}_2}{\pa y_1} (y_0) - \frac{\pa {\widetilde A}_1 }{\pa y_2}(y_0)\; . \end{array} \end{equation} We observe that by (\ref{orth}) $$ \sin \vartheta(y_0) = - |g|^{-\frac 12}(y_0) {\tilde H}_3^0 / b\;. $$ So, the study of the constant magnetic field in a half-space done in Section \ref{Section2}, (more precisely (\ref{roughlow}), property $7.$ in Subsection \ref{Subsection3.3} and Theorem \ref{asympexp}), (\ref{3est4}) and (\ref{3est5}) show that there exists $\epsilon_1>0,$ such for all $w \in C^{\infty }({\overline {\rz_+}^3}) $ with compact support, \begin{equation}\label{3est6} q^{h}_{{\widetilde A}^{(1)}}(w)\geq b h \left(\Theta_0 + \epsilon_1 |g|^{-1/2}(y_0)\frac{|{\widetilde H}^{0}_{3}|}{b}\right) |g|^{1/2}(y_0) \int_{\rz^2\times \rz_{+}} | w |^2dy\;, \end{equation} We will consider two cases. Let $C_1>1$ to be chosen later. For more simplicity, we can assume, after a rotation and a dilation in $(y_1,y_2)$, that the metric $g$ is standard at $y_0$~: \begin{equation}\label{gflat} g(y_0)=(\delta_{ij}). \end{equation} {\bf {First case:}} $\displaystyle \frac{|{\widetilde H}^{0}_{3}|}{b}\geq \frac{C_1 }{\epsilon_1} h^{-1/2} \tau(h)^2$. \\ We take $k=1$. Then (\ref{3est2}),(\ref{3est3}) and (\ref{3est6}) prove that \begin{equation}\label{3est7} \begin{array}{l} q^{h}_{A}(\chi_{\gamma ,\tau (h)}u) + C (\tau (h))^2 \| \chi_{\gamma ,\tau (h)}u\| \cdot (q^{h}_{ A}(\chi_{\gamma ,\tau (h)}u))^{1/2} \\ \geq [ b h (\Theta_0 + C_1h^{-1/2} \tau(h)^2) - C (h\tau (h) + (\tau (h))^4)]\cdot \| \chi_{\gamma ,\tau (h)}u\| ^2 \; . \end{array} \end{equation} This implies~: \begin{equation} \begin{array}{l} \left((q^{h}_{ A}(\chi_{\gamma ,\tau (h)}u))^{1/2} + C \tau(h)^2 \| \chi_{\gamma ,\tau (h)}u\|\right)^2\\ \quad \geq b h (\Theta_0 + C_1h^{-1/2} \tau(h)^2)\cdot (1 - C \tau (h))\| \chi_{\gamma ,\tau (h)}u\|^2\;. \end{array} \end{equation} Taking the square root, we get~: \begin{equation} \begin{array}{l} (q^{h}_{ A}(\chi_{\gamma ,\tau (h)}u))^{1/2} \\ \quad \geq \left( ( b h)^\frac 12 (\Theta_0 + C_1h^{-1/2} \tau(h)^2)^\frac 12 \cdot (1 - C \tau (h))\quad - C \tau (h)^2\right)\| \chi_{\gamma ,\tau (h)}u\|\;. \end{array} \end{equation} This finally gives, if $C_1$ is large enough and if, for some $\epsilon_0 >0$, the weight $\tau(h)$ satisfies the condition $ h^{\frac 14 + \epsilon_0}\geq \tau (h)\geq h^{1/2}$, \begin{equation}\label{labellis} q^{h}_{ A}(\chi_{\gamma ,\tau (h)}u) \geq h\; \left( b (\Theta_0 + \frac 12 C_1 h^{-1/2} \tau(h)^2) - C \tau (h) \right) \; \| \chi_{\gamma ,\tau (h)}u\|^2\;. \end{equation} So if $ h^{\frac 14 + \epsilon_0} \geq \tau (h)\geq h^{1/2}$ and $C_1$ is large enough, there exists $ h_0>0$ such that \begin{equation}\label{cas1} q^{h}_{ A}(\chi_{\gamma ,\tau (h)}u) \geq b h \Theta_0 \| \chi_{\gamma ,\tau (h)}u\|^2,\ \forall h\in ]0,h_0]\; . \end{equation} We now keep $C_1$ fixed such that (\ref{cas1}) is satisfied. {\bf {Second case:}} $\displaystyle \frac{|{\widetilde H}^{0}_{3}|}{b}< \frac{C_1}{\epsilon_1}\; h^{-1/2} \tau(h)^2$. \\ We proceed by steps~:\\ {\bf Step1}: We use (\ref{3est4}) but we take $k=2$. This leads to~: \begin{equation}\label{3est22} \begin{array}{ll} q^{h}_{A}(\chi_{\gamma ,\tau (h)}u) & \geq (1 - C\tau (h))q^{h}_{{\widetilde A}^{(k)}}(\chi_{\gamma ,\tau (h)}u) \\ & \quad - C (\tau (h))^{3} \| \chi_{\gamma ,\tau (h)}u\| \cdot (q^{h}_{A}(\chi_{\gamma ,\tau (h)}u))^{1/2}\\ & \quad - C (\tau (h))^{4} \| \chi_{\gamma ,\tau (h)}u\|^2 \; , \end{array} \end{equation}\ {\bf Step 2}: Let us now remark that \begin{equation}\label{3est2bbb} \begin{array}{ll} q^{h}_{{\widetilde A}^{(2)}}(\chi_{\gamma ,\tau (h)}u) & \geq q^{h}_{{\widetilde A}^{(2,1)}}(\chi_{\gamma ,\tau (h)}u) \\ & \quad - C [\tau (h) \| y_3\chi_{\gamma ,\tau (h)}u\| + \| y^2_3\chi_{\gamma ,\tau (h)}u\| ] \cdot (q^{h}_{A}(\chi_{\gamma ,\tau (h)}u))^{1/2}\\ & \quad - C [\tau (h)^{2} \| y_3 \chi_{\gamma ,\tau (h)}u\|^2 + \| y^2_3\chi_{\gamma ,\tau (h)}u\| ^2]\; , \end{array} \end{equation} with \begin{equation}\label{defa21} \begin{array}{l} {\widetilde A}^{(2,1)}={\widetilde A}(y_0) + \sum_{j=1}^{3}(y_j-y_{0j})\frac{\pa \widetilde A}{\pa y_j} (y_0)\\ + \sum_{j=1}^{2}(y_j-y_{0j})^2\frac{\pa ^2\widetilde A}{\pa y_j^2} (y_0) + 2 (y_1-y_{01})(y_2-y_{02})\frac{\pa ^2\widetilde A}{\pa y_1 \pa y_2} (y_0) \; . \end{array} \end{equation} {\bf Step 3}: But it is easy to find a real polynomial function $p(y)$ such that \begin{equation}\label{3est9bbb} q^{h}_{{\widetilde A}^{(2,1)}}(\chi_{\gamma ,\tau (h)}u) = q^{h}_{{\widetilde A}^{(2,1,0)}}(e^{i \frac p h}\chi_{\gamma ,\tau (h)}u) \end{equation} with \begin{equation}\label{defa210} \begin{array}{l} {\widetilde A}^{(2,1,0)}_{1}={\widetilde H}^0_2 y_3 + a_1 y^2_2 + b_1(y),\\ {\widetilde A}^{(2,1,0)}_{2}= - {\widetilde H}^0_1 y_3 + {\widetilde H}^0_3 y_1 + a_2 y^2_1 + b_2(y),\\ {\widetilde A}^{(2,1,0)}_{3}=0 \; , \end{array} \end{equation} where~: $$ b_1(y) = c_{11}\; y_1 y_3 + c_{12}\; y_2 y_3\;,\; b_2(y) = c_{21} \; y_1 y_3 + c_{22} \; y_2 y_3\;. $$ {\bf Step 4}: Modulo an error like in the right hand side of (\ref{3est2bbb}), we are reduced to the analysis of \begin{equation}\label{defa211} \begin{array}{ll} {\widetilde A}^{(2,1,1)}_{1} & ={\widetilde H}^0_2 y_3 + a_1 y^2_2,\\ {\widetilde A}^{(2,1,1)}_{2} & = - {\widetilde H}^0_1 y_3 + {\widetilde H}^0_3 y_1 + a_2 y^2_1 ,\\ {\widetilde A}^{(2,1,1)}_{3} & =0 \;. \end{array} \end{equation} We write \begin{equation}\label{3est10bbb} \begin{array}{ll} q^{h}_{{\widetilde A}^{(2,1,1)}}(w) & = h^2 \| D_{y_3} w \| ^2 \\ &\quad + {\tilde b}^2 \| \left(y_3-{\tilde b}^{-1}(\sin \theta L_1 - \cos \theta L_2)\right) w\| ^2 \\ &\quad + \|(\cos \theta L_1 + \sin \theta L_2) w \| ^2 \; , \end{array} \end{equation} with $$ \begin{array}{ll} {\tilde b} & = [ ({\widetilde H}^0_1)^2 + ({\widetilde H}^0_2)^2 ]^{1/2}\;,\\ {\widetilde H}^0_1 & = {\tilde b} \cos \theta ,\\ {\widetilde H}^0_2& = {\tilde b} \sin \theta ,\\ \end{array} $$ and \begin{equation}\label{3est11bbb} L_1= hD_{y_1} - a_1 y^2_2,\ \ L_2 = hD_{y_2} - {\widetilde H}^0_3 y_1 - a_2 y^2_1 \; . \end{equation} As the operator $\sin \theta L_1 - \cos \theta L_2$, a priori defined on $\Sg(\rz^2)$, has a self-adjoint realization on $L^2(\rz ^2)$. This is indeed, after rotation, an operator of the form $\frac 1 i \pa_t + \phi(t,y)$ which is unitary equivalent to $\frac 1 i \pa_t$. Hence, we can consider the spectral representation of $\sin \theta L_1 - \cos \theta L_2 $ and the analysis of Subsection \ref{Subsection2.1} to see that \begin{equation}\label{3est12bbb} h^2 \| D_{y_3}w\| ^2 + {\tilde b}^2 \| (y_3- {\tilde b}^{-1} (\sin \theta L_1 - \cos \theta L_2) ) w\| ^2 \geq h {\tilde b} \Theta_0 \| w\| ^2 \; . \end{equation} {\bf Step 5}: We now compare $b$ and ${\tilde b}$. We are in the case when $ \frac{|{\widetilde H}^{0}_{3}|}{b}< C_1 h^{-1/2} \tau(h)^2 .$ \\ So \begin{equation}\label{3est13bbb} |b - {\tilde b}|\leq C \; C^2_1 h^{-1} \tau(h)^4 \; . \end{equation} {\bf Step 6}: Then (\ref{3est22}), (\ref{gflat}) and (\ref{3est2bbb}) - - (\ref{3est13bbb}) prove that, if $u^h$ is the eigenfunction associated to the ground state energy of $P_{A,\Omega}^{h,N}$ and if $\tau (h)= h^{1/3}$, then \begin{equation}\label{3est14bbba} \begin{array}{ll} q^h_A(\chi_{\gamma ,\tau (h)} u^h)& \geq [h b \Theta_0 - C h^{4/3}] \| \chi_{\gamma ,\tau (h)} u^h\| ^2 \\ &\quad - C h^\frac 23 || t \chi_{\gamma ,\tau (h)} u^h||^2 - C || t^2 \chi_{\gamma ,\tau (h)} u^h||^2\\ &\quad - C h^\frac 13 || t \chi_{\gamma ,\tau (h)} u^h||\left[q^h_A(\chi_{\gamma ,\tau (h)} u^h)\right]^\frac 12\\ &\quad - C \;|| t^2 \chi_{\gamma ,\tau (h)} u^h||\left[q^h_A(\chi_{\gamma ,\tau (h)} u^h)\right]^\frac 12 \;. \end{array} \end{equation} We now eliminate the two last lines at the price of a worst error term. We get first the existence of $C$ such that~: \begin{equation} \begin{array}{ll} (1 + C h^\frac 13) \; q^h_A(\chi_{\gamma ,\tau (h)} u^h)& \geq \left( h b \Theta_0 - C h^{4/3}\right) \| \chi_{\gamma ,\tau (h)} u^h\| ^2 \\ &\quad - C h^\frac 13 || t \chi_{\gamma ,\tau (h)} u^h||^2 - C h^{-\frac 13} || t^2 \chi_{\gamma ,\tau (h)} u^h||^2 \;, \end{array} \end{equation} which leads to the existence of $C$ such that~: \begin{equation}\label{3est14bbb} \begin{array}{ll} q^h_A(\chi_{\gamma ,\tau (h)} u^h)& \geq \left( h b \Theta_0 - C h^{4/3} \right) \| \chi_{\gamma ,\tau (h)} u^h\| ^2 \\ &\quad - C h^\frac 13 || t \chi_{\gamma ,\tau (h)} u^h||^2 - C h^{-\frac 13} || t^2 \chi_{\gamma ,\tau (h)} u^h||^2 \;. \end{array} \end{equation} The two last terms will be controlled by summation using Proposition \ref{pBN}. \vskip 0.5cm \paragraph{End of the proof}.\\ With $\tau (h)=h^{1/3}$, we get from (\ref{cas1}) and (\ref{3est14bbb}) that \begin{equation}\label{3est9a} \begin{array}{ll} q^{h}_{A}(\chi_{\gamma ,\tau (h)}u^h) & \geq [ b\Theta_0 h - C \, h^{4/3}]\cdot \| \chi_{\gamma ,\tau (h)} u^h\| ^2 \\ &\quad - C h^\frac 13 || t \chi_{\gamma ,\tau (h)} u^h||^2 - C h^{-\frac 13} || t^2 \chi_{\gamma ,\tau (h)} u^h||^2 \; , \end{array} \end{equation} if $\gamma \; \in \; \Gamma^{1}_{\tau (h)}(\Omega )$.\\ Using Proposition \ref{pBN}, we verify that~: \begin{equation} \sum_{\gamma \in \Gamma_{\tau(h)}^1} \left(h^\frac 13 || t \chi_{\gamma ,\tau (h)} u^h||^2 + h^{-\frac 13} || t^2 \chi_{\gamma ,\tau (h)} u^h||^2 \right) \leq C h^\frac 43\;. \end{equation} So we get \begin{equation}\label{3est9} \sum_{\gamma \in \Gamma_{\tau(h)}^1} q^{h}_{A}(\chi_{\gamma ,\tau (h)}u^h) \geq \left( b\Theta_0 h \right)\cdot \sum_{\gamma \in \Gamma_{\tau(h)}^1} \| \chi_{\gamma ,\tau (h)} u^h\| ^2 \; - \; C \, h^{4/3}||u_h||^2\;. \end{equation} Then (\ref{estgros1low}) is deduced from (\ref{low2}) - - (\ref{low5}) with $\tau (h)=h^{1/3}$, (\ref{3est1}) and from (\ref{3est9}). \section{Refined adapted coordinates on the boundary}\label{Section8} \subsection{Curvatures}\label{ssCurv} For the following geometric properties of embedded surface in $\rz^3$, we refer mainly to the first chapter of \cite{DHKW} (see also the volume two of Spivak's book \cite{Sp}). Let us suppose more specifically that $ x_0$ is a point of the boundary $ \pa \Omega $. The neighborhood of $x_0$ $\Vg_{x_0}$ can be chosen such that, in $W_{x_0}:= \pa \Omega \cap \Vg_{x_0}$, there exist local coordinates $(y_1,y_2)$, i.e. there exists an open subset $\Sg $ of $ \rz^2$ and a diffeomorphism \begin{equation}\label{diffbor} \phi \; :\; \pa \Omega \cap \Vg_{x_0}\; \to \; \Sg ,\ \ \phi (x)=(y_1,y_2). \end{equation} We denote by $y\mapsto x(y)$ its inverse. Then $\{ \frac{\pa x}{\pa y_1},\; \frac{\pa x}{\pa y_2}\} $ is a basis of $T\Wg_{x_0}$. \\ The first fundamental form of $ \pa \Omega $ is the restriction of $g_0$ to $\pa \Omega$ and it is denoted by $G$. In the local coordinates $(y_1,y_2)$, $G$ is given by~: \begin{equation}\label{firstf} G=\sum_{1\leq i,j\leq 2}G_{ij}dy_i\otimes dy_j,\ \ G_{ij}=\sum_{k=1}^{3}\frac{\pa x_k}{\pa y_i}\cdot \frac{\pa x_k}{\pa y_j}. \end{equation} The element of area is given by $$ ds^2=|G|^{\frac 1 2 }dy_1\wedge dy_2,$$ with $$ |G|=\det(G_{ij})_{1\leq i,j\leq 2}),$$ The unit normal vector to $ \pa \Omega $ is defined by \begin{equation}\label{normal} N=\frac{{\frac{\pa x}{\pa y_1}}\wedge {\frac{\pa x}{\pa y_2}}} {|{\frac{\pa x}{\pa y_1}}\wedge {\frac{\pa x}{\pa y_2}}|}\;. \end{equation} For any vector fields $X$ and $Y$, $X\wedge Y$ is the vector field defined by \begin{equation}\label{rela} \begin{array}{l} \omega_3 (X,Y,X\wedge Y)=|X|^2 \cdot |Y|^2-()^2\;,\\ ==0 \;. \end{array} \end{equation} When $\pa \Omega$ is the boundary of a bounded regular domain, we can choose moreover these coordinates such that $N$ is the interior normal unit vector, and we observe, using (\ref{rela}), that~: $$ \omega_3 (\frac{\pa x }{\pa y_1},\frac{\pa x}{\pa y_2}, N)>0\;. $$ Remembering that $|N|^2 =1$, we get that, for $ j=1,\ 2$, $$ \frac{\pa N}{\pa y_j}\; \in \; T\pa \Omega\; .$$ This permits us to introduce the Weingarten map $k$ from $T\pa \Omega$ into $T\pa \Omega$ , by~: $$ k(\sum_{j=1}^{2}\lambda_j\frac{\pa x}{\pa y_j})= -\sum_{j=1}^{2}\lambda_j\frac{\pa N}{\pa y_j} \;. $$ This map is independent of the choice of the coordinates respecting (\ref{normal}) with $N$ inner normal.\\ The second fundamental form $K$ on $T\pa \Omega $ is defined, for $X$ and $Y$ in $ T\pa \Omega $, by \begin{equation}\label{secondf} K(X,Y)=G(X,k(Y))=\sum_{1\leq i,j\leq j}K_{ij}\; X_i\,Y_j\;, \end{equation} where \begin{equation}\label{secondfa} K_{ij}= -<\frac{\pa x}{\pa y_i}|\frac{\pa N}{\pa y_j}>, \end{equation} and $X_j$ and $Y_j$ are the components of $X$ and $Y$ in the local basis $(\frac{\pa }{\pa y_j})$.\\ Observing that $ <\frac{\pa x}{\pa y_i}|N>=0$, it is easy to see, by differentiating this equality, that \begin{equation}\label{Kdef} K_{ij}=<\frac{\pa^2x}{\pa y_i\pa y_j}|N>\;. \end{equation} In particular $K$ is symmetric. We recall that the Gauss curvature is \begin{equation}\label{gauss} \kappa^G =\det(k)=(K_{11}K_{22}-K^{2}_{12})/(G_{11}G_{22}-G^{2}_{12}), \end{equation} and that the mean curvature is \begin{equation}\label{mincurv} \kappa^M =\frac{1}{2}{\rm {tr}}(k)= \frac{1}{2}(k_{11}+k_{22}), \end{equation} where $(k_{ij})_{ij}$ is the matrix $G^{-1} K$.\\ Let $a >0$. If $[-a,+a]\ni t \mapsto \gamma (t) \in \pa \Omega $ defines a parametrized curve, we denote its image by $\Gamma$. We assume moreover that it is parametrized by arc length, that is~: \begin{equation} |\gamma'(\tau )|=1\;. \end{equation} The curvature vector $(\kappa_g, \kappa_n)$ of $\Gamma $ is defined\footnote{ The definition of the geodesic curvature is not uniform in the literature. Some authors (\cite{DHKW}) use the opposite sign.} by \begin{equation} \label{curvec} \frac{d^2 \gamma}{d\tau^2} (\tau )= - \kappa_g(\tau )\, N(\gamma (\tau )) \wedge \gamma'(\tau ) \; +\; \kappa_n(\tau )\, N(\gamma (\tau ))\;. \end{equation} The function $\kappa_g$ is called the geodesic curvature and $ \kappa_n$ the normal curvature of $\Gamma$, at the point $\gamma(\tau)$. \\ As shown in \cite{DHKW} (Formula (18)~), the normal curvature is given by the equation \begin{equation} \label{normcurv} \kappa_n(\tau )=K(\gamma'(\tau ), \gamma'(\tau ))\;. \end{equation} \begin{remark}\label{geode}.\\ Note (cf \cite{DHKW}) that the geodesic curvature vanishes when $\gamma$ is a geodesic. Let us also recall that the scalar curvature of $ \Gamma $ is $$ \kappa =\mid \frac{d^2 \gamma}{d\tau^2} \mid = (\kappa_{g}^{2}+\kappa_{n}^{2})^{\frac 1 2 }\;. $$ \end{remark} \begin{example}{The ellipsoid}\label{exellipsoid}.\\ Let us consider the case of the ellipsoid~: $$ a_1 x_1^2 + a_2 x_2^2 + a_3 x_3^2 =1\;. $$ Let us compute $K$ in the case when one can parametrize by the two first coordinates and let us consider the case when $x_3 > 0$. So we have~: $$ x_1 = y_1\;,\; x_2 = y_2\;,\; x_3= \frac{1}{\sqrt a_3} \sqrt{1- a_1 y_1^2 - a_2 y_2^2}\;. $$ The interior normal is given by~: $$ N = - \frac 1 \nu \left( \begin{array}{c} a_1 y_1\\ a_2 y_2 \\ a_3 x_3 (y_1,y_2) \end{array} \right)\;, $$ where~: $$ \nu = \sqrt{ a_1^2 y_1^2 + a_2 ^2 y_2^2 + a_3^2 x_3^2}\;. $$ We can now compute $K$ in these coordinates. We first observe that~: $$ \pa N/\pa {y_1} = c_1 N - \frac 1 \nu \left( \begin{array}{c} a_1\\ 0 \\ a_3 \frac{\pa x_3}{\pa y_1} \end{array} \right)\;, $$ and $$ \pa N/\pa {y_2} = c_2 N - \frac 1 \nu \left( \begin{array}{c} 0\\ a_2\\ a_3 \frac{\pa x_3}{\pa y_2} \end{array} \right)\;. $$ We use the notation $\pa_j = \frac{\pa}{\pa y_j}$. This leads to~: $$ \begin{array}{lll} K_{11} &= - \langle \pa_1 N \,,\, \pa_1 x \rangle &= \frac{1}{\nu} a_1 + \frac{1}{\nu} a_3 |\pa_1 x_3|^2\\ K_{22} &= - \langle \pa_2 N \,,\, \pa_2 x \rangle &= \frac{1}{\nu} a_2 + \frac{1}{\nu} a_3 |\pa_2 x_3|^2\\ K_{12} &= - \langle \pa_1 N \,,\, \pa_2 x \rangle &= \frac{1}{\nu} a_3 (\pa_1 x_3)(\pa_2 x_3)\;. \end{array} $$ We immediately verify the strict positivity of $K$. We note indeed that~: $$ \begin{array}{ll} K (u\,,\, u ) & = \frac{1}{\nu} \left( a_1 u_1^2 + a_2 u_2^2 + a_3 ( (\pa_1 x_3) \, u_1 + (\pa_2 x_3) \, u_2)^2 \right)\\ & \geq a_1 u_1^2 + a_2 u_2^2 \;. \end{array} $$ Similarly, we have $$ \begin{array}{ll} G (u \,,\, u ) & = \left( u_1^2 + u_2^2 + ( (\pa_1 x_3) \, u_1 + (\pa_2 x_3) \, u_2)^2 \right) \;. \end{array} $$ This leads to the following expression for the Gauss curvature~: $$ \kappa^G=\det k = \frac{a_1 a_2 a_3}{\nu^4}=\frac{a_1 a_2 a_3}{(a_1^2 x_1^2 + a_2^2 x_2^2 + a_3 ^2 x_3^2)^2} \;. $$ \end{example} \subsection{Local coordinates near a curve inside the boundary} \label{sslccb} Let $\Gamma $ be a curve in $\pa \Omega $ parametrized by arc lengths on some interval $I$ ($I=[-a,+a]$)~: $\Gamma =\{ \gamma (s);\ s\in I \}$. So we have $ |\gamma'(s)|=1 $. Then, there exists a neighborhood $\Wg_{x_0}$ of $x_0=\gamma (0)$ in $\pa \Omega$, such that, for any $z\; \in \; \Wg_{x_0}\cap \Gamma$, there exists a unique geodesic $\Lambda_z$ through $z$ and normal to $\Gamma $. The neighborhood $ \Wg_{x_0}$ of $ x_0$ can also be choosen such that \begin{equation}\label{cu1} \forall \; x\; \in \; \Wg_{x_0},\ \exists !\; z=z(x)\; \in \; \Gamma \cap \Wg_{x_0} \ s.t.\ d_{\pa \Omega }(x,z)=d_{\pa \Omega }(x,\Gamma )\;, \end{equation} where $d_{\pa \Omega }(\; .\; ,\; .\; )$ denotes the distance on $\pa \Omega$. Then, there exists an open set $S$ of $\rz^2$ and a regular diffeomorphism \begin{equation}\label{cu2} \phi :\; \Wg_{x_0}\; \to \; S,\ \phi (x)=(r,s)\ {\rm {with}}\ \pm r=d_{\pa \Omega }(x,\Gamma )=d_{\pa \Omega }(x,\gamma (s))\;. \end{equation} We observe that~: $$ x(0,s)=\gamma (s)\;. $$ We choose a positive orientation (and this determines the choice of the sign of $r$), by imposing~: \begin{equation}\label{cu3} \frac{\pa x}{\pa r}(0,s)\wedge \frac{\pa x }{\pa s} (0,s)=N(\gamma (s)), \end{equation} where $N(x)$ is the interior normal of $\pa \Omega $ at the point $x \in \pa \Omega$. Then $(r,s)$ are local coordinates in $ \Wg_{x_0}$ and observing that, for any fixed $s$, $r \mapsto \; x(r,s)$ is a parametrization by arc lengths of the geodesic $\Lambda_{\gamma (s)}$, we have \begin{equation}\label{cu4a} |\frac{\pa x}{\pa r}(r,s)|=1\;, \end{equation} and \begin{equation}\label{cu4b} < \frac{\pa x}{\pa r}(0,s)|\frac{\pa x}{\pa s}(0,s)>=0\;. \end{equation} More precisely we have the following Lemma. \begin{lemma}\label{curve}.\\ In the above local coordinates, the metric $G$ on $\pa \Omega $ is diagonal~: \begin{equation}\label{cu5} G = dr\otimes dr +\alpha (r,s)\, ds\otimes ds. \end{equation} On the curve $\Gamma$, we have \begin{equation}\label{cu6} \alpha (0,s)=1,\ \frac{\pa \alpha}{\pa r} (0,s)=-2\kappa_g(s)\ {\rm {and}}\ \frac{\pa \alpha}{\pa s} (0,s)=0\;. \end{equation} where $\kappa_g(s)$ denotes geodesic curvature of the curve $\Gamma $ at $\gamma (s)$. \end{lemma} {\bf {Proof}}.\\ As for any fixed $s$, the map $ r \mapsto x(r,s)$ is a parametrization by arc lengths of the geodesic $\Lambda_{\gamma (s)}$, the curvature of $\Lambda_{\gamma (s)}$ is given, using (\ref{curvec}), (\ref{normcurv}) and Remark \ref{geode}, by \begin{equation}\label{cu7} \frac{\pa^2x }{\pa r^2}(r,s)= K(\frac{\pa x}{\pa r} (r,s),\frac{\pa x}{\pa r}(r,s))\, N(x(r,s))\;. \end{equation} Then, we get from (\ref{cu7}) that \\ \begin{equation}\label{cu7a} \frac{\pa }{\pa r}<\frac{\pa x }{\pa s}|\frac{\pa x}{\pa r}>= <\frac{\pa^2 x}{\pa s\pa r}|\frac{\pa x}{\pa r}>= \frac{1}{2}\frac{\pa }{\pa s}<\frac{\pa x}{\pa r} |\frac{\pa x}{\pa r}>\;. \end{equation} So, using in addition (\ref{cu4a}), we have \begin{equation}\label{cu8} \frac{\pa }{\pa r}<\frac{\pa x}{\pa s}|\frac{\pa x}{\pa r}> =0\;. \end{equation} But $\Lambda_{\gamma (s)}$ is normal to $\Gamma$; so $<\frac{\pa x}{\pa s} (0,s)|\frac{\pa x}{\pa r}(0,s)>=0$. Then, using (\ref{cu8}), we get~: \begin{equation}\label{cu9} <\frac{\pa x}{\pa s}(r,s)|\frac{\pa x }{\pa r} (r,s)>=0\;. \end{equation} This shows that the metric is diagonal as announced in (\ref{cu5}) with $$ \alpha (r,s)= <\frac{\pa x }{\pa s} (r,s)|\frac{\pa x}{\pa s}(r,s)>\;. $$ We get also from the orthogonality of $\gamma'$ and $\gamma''$ that $$ \frac{\pa \alpha}{\pa s} (0,s)=0\;. $$ But, differentiating the identity (\ref{cu9}) with respect to $s$, we have \begin{equation}\label{cu10} \frac{\pa \alpha}{\pa r} (r,s)=2<\frac{\pa^2 x}{\pa s\pa r}| \frac{\pa x}{\pa s}>=-2<\frac{\pa x}{\pa r}|\frac{\pa^2 x}{\pa s^2}>. \end{equation} But, by (\ref{curvec}), we have~: $$ \frac{\pa^2 x}{\pa s^2}(0,s)= \kappa_n(\gamma (s))N(\gamma (s))+\kappa_g(\gamma (s)) \frac{\pa x}{\pa s} (0,s)\wedge N(\gamma (s))\;, $$ so (\ref{cu3}) and (\ref{cu4b}) lead to \begin{equation}\label{cu11} \frac{\pa^2 x}{\pa s^2} (0,s)= \kappa_n(\gamma (s))N(\gamma (s))+\kappa_g(\gamma (s)) \frac{\pa x}{\pa r}(0,s). \end{equation} The formulas (\ref{cu10}) and (\ref{cu11}) give the non-obvious part of (\ref{cu6}) $$ \frac{\pa \alpha}{\pa r} (0,s) = - 2\kappa_g(\gamma (s))\;.$$ \begin{remark}\label{2f}.\\ In the coordinates $(r,s)$ introduced in Lemma \ref{curve}, the second fondamental form is given (see (\ref{Kdef})~) $$ K=K_{11}\; dr\otimes dr\; +\; K_{12}dr\otimes ds \; +\; K_{21}\; ds\otimes dr\;+\; K_{22}\; ds\otimes ds\;,$$ is given by $$ \begin{array}{ll} K_{11}(r,s) & =<\frac{\pa^2 x}{\pa r^2}(r,s)|N(x(r,s))>, \\ K_{22}(r,s) & =<\frac{\pa^2 x}{\pa s^2}(r,s)|N(x(r,s))>,\\ K_{12}(r,s)&= <\frac{\pa^2 x}{\pa r\pa s}(r,s)|N(x(r,s))>\\ K_{21}(r,s) & = K_{12}(r,s)\;. \end{array} $$ The function $K_{11}(r,s)$ is the normal curvature of the geodesic $\Lambda_{\gamma (s)}$ at $x(r,s)$ and the function $K_{22}(0,s)=\kappa_n(\gamma (s))$ is the normal curvature of the curve $\Gamma$ at $x(0,s)=\gamma (s)$. %% Note also that %% \begin{equation} %% K_{12}(0,s) = \kappa_g(\gamma(s))\;. %% \end{equation} \end{remark} \subsection{Local coordinates near a curve in the boundary} \label{sslccbo} We come back to previous computations and relate them to the curvatures. Let $ \phi (x)=(y_1,y_2)$ be local coordinates of the boundary as defined in (\ref{diffbor}). We have observed in (\ref{adapmetra}), that \begin{equation}\label{adapmetr} g_0 = dy_{3}\otimes dy_3 + \sum_{1\leq i,j\leq 2} [G_{ij}(y_1,y_2)-2 y_3 K_{ij}(y_1,y_2) + y_{3}^{2} L_{ij}] dy_i\otimes dy_j\;, \end{equation} where \begin{itemize} \item $ G=\sum_{1\leq i,j\leq 2} G_{ij}\, dy_i\otimes dy_j$, \item $ K=\sum_{1\leq i,j\leq 2}K_{ij}\, dy_i\otimes dy_j$, \item $ L=\sum_{1\leq i,j\leq 2}L_{ij}\, dy_i\otimes dy_j =\sum_{1\leq i,j\leq 2} <\frac{\pa N}{\pa y_i}| \frac{\pa N}{\pa y_j}>dy_i\otimes dy_j$. \end{itemize} $G$, $K$ and $L$ are respectively called the first, second and third fundamental forms on $\pa \Omega$. If we take local coordinates $(y_1,y_2)=(r,s)$ on the boundary given by Lemma~\ref{curve}, the sesquilinear form introduced in (\ref{qAb}) becomes \begin{equation}\label{qAc} \begin{array}{l} q^{h}_{A}(u)\\ =\int_{\Vg_{x_0}}|g|^{\frac 1 2 }[|hD_{y_3}-{\widetilde {A}}_3 u|^2 \; + \sum_{1\leq i,j\leq 2}g^{ij} (hD_{ y_i}u-{\widetilde {A}}_iu)\cdot {\overline {(hD_{y_j} u - {\widetilde {A}}_ju)}}]\; dy^3, \end{array} \end{equation} for $ u $ supported in $ \Vg_{x_0}$, the associated differential operator is \begin{equation}\label{Pc} \begin{array}{ll} P^{h}_A& = (hD_{y_3}-{\widetilde {A}}_3)^2 + \frac {h}{2i} |g|^{-1} (\frac{\pa }{\pa y_3}|g|)(hD_{y_3}-{\widetilde {A}}_3)\;\\ &\quad + |g|^{-\frac 1 2 }\sum_{1\leq i,j\leq 2}(hD_{y_j}-{\widetilde {A}}_j) ( |g|^{\frac 1 2 } g^{ij} (hD_{ y_i} -{\widetilde {A}}_i))\;. \end{array} \end{equation} If we now consider the coordinates introduced in Subsection \ref{sslccb}, that is $(y_1,y_2)=(s,t)$ and complete by $t=y_3$ introduced in Subsection \ref{corbord2}, then \begin{equation}\label{0y2} |g|=\alpha (r,s)-2t[\alpha (r,s)K_{11}(r,s) + K_{22}(r,s)] +t^2 \epsilon_3 (r,s,t), \end{equation} and, for $ 1\leq i,j\leq 2$, \begin{equation}\label{Oy3} (g^{ij})_{1\leq i,j\leq 2}=\pmatrix{ 1 & 0 \cr 0 & \alpha^{-1}} +2t\pmatrix{ K_{11} & \alpha^{-1} K_{12} \cr \alpha^{-1}K_{21} & \alpha^{-2}K_{22}} + t^2 R\;, \end{equation} where $ \epsilon_3$ and $ R_{ij} $ are smooth functions. \subsection{More magnetic geometry}\label{ssmmg} We assume that the magnetic field $H=\rot A$ is constant and we can assume, without loss of generality, that~: \begin{equation}\label{constf} A(x)=\frac{b}{2}(0,-x_3,x_2) \end{equation} for some fixed $b>0$.\\ Let $\Omega $ be bounded open set of $\rz^3$ with regular boundary $\pa \Omega$. We now assume that (\ref{Gammareg}) is satisfied. \begin{remark}\label{remconv}.\\ We observe that this assumption is satisfied when $\Omega$ is strictly convex. \end{remark} In this situation, we can introduce the following definition~: \begin{definition}\label{defnormalcurvature}.\\ At each point $x$ of $\Gamma$, we introduce the normal curvature along the magnetic field $H = V(B)$ by~: \begin{equation}\label{BN0b} \kappa_{n,B}(x):=K_x(\gamma' \wedge N \;,\; \frac{H}{|H|})\;, \end{equation} where $K$ denotes the second fundamental form on the surface $\pa \Omega $. \end{definition} We have already introduced $\kappa_{n,B}$ in (\ref{BN0b}). Similarly, we can define~: \begin{equation}\label{kappat} \kappa_{t,B}(s) =K(\gamma'(s),\frac{V(B)}{|V(B)|})\;. \end{equation} We observe that we have~: \begin{equation}\label{defkappas1} \begin{array}{ll} \kappa_{n,B}(s)&=K(\frac{\pa }{\pa r},\frac{V(B)}{|V(B)|}) =\cos \theta (s) K_{11}(0,s)+\sin \theta (s) K_{12}(0,s)\;;\\ \kappa_{t,B}(s) &=K(\frac{\pa }{\pa s},\frac{V(B)}{|V(B)|}) =\cos \theta (s) K_{12}(0,s) +\sin \theta (s) K_{22}(0,s) \; . \end{array} \end{equation} Let us observe that the angle $\theta(s)$ is not ``free'' in our picture. We can indeed, by computing $V(B)$ in the new coordinates (and remembering (\ref{magin}) and Lemma \ref{curve}), get the relation~: $$ (b \cos \theta(s))^2 + (1- 2 \kappa_g(s)r) [ b \sin \theta(s) + r b (2 \kappa_g(s) - \theta'(s)) \sin \theta(s)]^2 + \Og (r^2) = b^2\;. $$ Looking at the coefficient of $r$, we get \begin{equation} \kappa_g(s) = \theta'(s)\;, \end{equation} or \begin{equation} \theta(s)=0\;. \end{equation} The case $\theta$ constant corresponds to the case when $\gamma$ is a geodesic. In fact we have the geometrical fact: \begin{proposition}\label{geodesic}.\\ The assumption $V(B)$ is constant (of norm equal to one) and tangent to the surface $\pa \Omega $ along the curve $\ \Gamma $ implies that \begin{equation}\label{zerocurva} \kappa_{t,B}(x)=0,\ \ \forall \; x\; \in \; \Gamma \;. \end{equation} If $ s \mapsto \gamma (s)$ is a parametrization of $\Gamma $ by arc length, with $$ \theta (s)=\mbox{ \rm Arcsin }\left(({\gamma '}(s),V(B))\right) $$ then \begin{equation}\label{deriveangle} {\theta'}(s)=\kappa_g(\gamma (s)),\ \ \forall \; s\; . \end{equation} \end{proposition} {\bf {Proof of Proposition \ref{geodesic}}}.\\ Let us write~: \begin{equation}\label{defangle} V(B)=\sin(\theta (s)){\gamma' }(s)+ \cos (\theta (s))[{\gamma' }(s)\wedge N(\gamma (s)]\;, \end{equation} As $V(B)$ is constant, we can differentiate with respect to $s$ (\ref{defangle}) and we get~: $$ \begin{array}{l} \theta'(s) \left[\cos (\theta (s) )\gamma'(s) - \sin(\theta (s) )\gamma'(s) \wedge N(\gamma (s) )\right] \\ + \sin (\theta(s) )\gamma''(s)\\ + \cos (\theta (s) ) \left[\gamma''(s)\wedge N(\gamma (s) )+ \gamma'(s) \wedge (N\circ \gamma)'(s)\right ]\\ =0\; . \end{array} $$ But using the coordinates of Lemma \ref{curve}, the formula (\ref{secondfa}) proves $$ (N\circ \gamma )'= - K_{12}(\gamma )\gamma' \wedge N(\gamma) -K_{22}(\gamma )\gamma'\; , $$ and (\ref{cu11}) becomes $$ \gamma''=\kappa_g (\gamma ) \gamma'\wedge N(\gamma ) +\kappa_n(\gamma) N(\gamma )\; $$ So $$ \begin{array}{l} \theta' [\cos \theta \,\gamma' - \sin \theta \, \gamma' \wedge N(\gamma )]\\ + \sin \theta \, [\kappa_g (\gamma)\gamma' \wedge N(\gamma) +\kappa_n (\gamma) N(\gamma) ]\\ + \cos \theta\, [-\kappa_g (\gamma) \gamma' +K_{12}(\gamma) N(\gamma) ] \\ =0\; , \end{array} $$ and then, expressing the previous equality on the basis $\gamma'\wedge N$, $\gamma'$, $N$, \begin{equation}\label{VdeB} \begin{array}{ll} \cos \theta \, [\theta' -\kappa_g (\gamma)]& =0\\ \sin \theta\, [-\theta' +\kappa_g (\gamma)] & = 0 \\ \sin \theta\, \kappa_n(\gamma )+\cos \theta\, K_{12}(\gamma) &=0 \; . \end{array} \end{equation} We get (\ref{deriveangle}), and using (\ref{normcurv}) we get also (\ref{zerocurva}) from $\kappa_n (\gamma ) =K_{22} (\gamma)$ and from (\ref{defkappas1}).\\ \begin{remark}\label{remgeodesic}.\\ In the case when $\pa \Omega $ is strictly convex, ($K>0)$, then (\ref{zerocurva}) implies that \begin{equation}\label{nonzerocurv} \kappa_{n,B} \neq \; 0,\ \forall \; x\; \in \; \Gamma \; . \end{equation} One can also meet degenerate cases where $K$ is not invertible (locally cylindric domains).\\ When $\theta (s) =0$, we deduce from (\ref{defkappas1}) and (\ref{zerocurva}) that $K_{12}(x(s)) =0$. So the curvature matrix $K$ becomes diagonal. \end{remark} {\bf Proof of Remark \ref{remgeodesic}}.\\ We observe that (\ref{defkappas1}) can be rewritten in the form~: \begin{equation}\label{relut} \left( \begin{array}{c} \kappa_{n,B}\\ \kappa_{t,B} \end{array} \right) = K \left( \begin{array}{c} \cos \theta \\ \sin \theta \end{array} \right) \end{equation} Observing that $K$ is inversible when $\Omega$ is strictly convex ($K$ is actually strictly positive), we immediately see that $|\kappa_{n,B}| + |\kappa_{t,B}|\neq 0$. \begin{remark} \label{addrem}.\\ The condition that $\theta' = \kappa_g$ is obtained also in the case $\theta \equiv 0$. So in this case we have effectively $\kappa_g =0$. So $\Gamma$ should be a geodesic. Conversely, when $\Gamma$ is a geodesic, we should have $\theta \equiv 0$ if $V(B)$ is constant. We observe indeed that by computing the circulation of $V(B)$ along $\Gamma$ and using Stokes Lemma (or more simply that $V(B)\cdot \gamma' = \frac{d}{dt} (V(B) \cdot \gamma)$), we should have~: $$ \int_{\Gamma} \sin \theta(s) ds = 0\;. $$ When $\theta$ is constant, this implies $\theta =0$. \end{remark} \begin{example}.\\ In the case of the ellipsoid considered in Example \ref{exellipsoid}, it is interesting to compute our invariants. Take for simplification, the case when $B =(0,0,1)$. Then $\Gamma$ is the intersection of the ellipsoid with $x_3 =0$. So we get an ellipse in this plane. We can now observe that the condition (\ref{BN0bb}) is satisfied. The vector field $H$ is orthogonal to $\Gamma$. We observe that~: $$ \langle H \;,\; N\rangle = - |B| \frac{a_3 x_3}{\nu}\;. $$ This leads to~: $$ |\kappa_{n,B} (y_1,y_2)| = \frac{a_3 }{\nu}\;. $$ The minimum of $\kappa_{n,B}$ (which appears in Formula (\ref{BN2})) is then obtained at the point where $\nu$ is maximal. If we assume for example that $a_1 > a_2$, we get that this maximum is obtained at $x_2=x_3 =0$ and equal to $a_1$. This is not conformed to the intuition we got from the two-dimensional case. \end{example} \section{Curvature effects on the ground state energy in the case of constant magnetic field: Main statements}\label{Section9} \begin{theorem}\label{BN}.\\ Let $P^{h,N}_{A,\Omega}$ be the Neumann realization of the magnetic Laplacian $(hD-A)^2$ on $L^2(\Omega )$ , where $h \in ]0,1[$ is a small parameter. We assume that (\ref{Gammareg}) is satisfied and that~: \begin{equation}\label{BN0ba} \kappa_{n,B}\neq 0, \forall x \in \; \Gamma \;. \end{equation} We assume moreover that~: \begin{equation}\label{BN0bb} \mbox{ The magnetic field $H$ is orthogonal to $\Gamma$.} \end{equation} Then there exists $ \eta >0 $ such that~: \begin{equation}\label{BN1} \inf \; \Sp (P^{h,N}_{A,\Omega}) = b\Theta_0 h + {\hat \gamma}_0 b^{\frac 2 3 } h^{\frac 4 3 } + \Og(h^{\eta + \frac 4 3 })\;, \end{equation} where \begin{equation}\label{BN2} {\hat \gamma}_0=\frac{1}{2}{\hat \nu}_0 [\mu''(\xi_0)]^{\frac 1 3 } \inf_{x\in \Gamma }|\kappa_{n,B}(x)|^{\frac 2 3 }. \end{equation} \end{theorem} Here the constants $\Theta_0$, $\xi_0$ and the function $ \mu (t)$ are those defined in subsections \ref{Subsection2.1} and \ref{Subsection2.2}. The constant $ {\hat \nu}_0$ is defined in Subsection \ref{Subsection2.4}. The assumption (\ref{BN0b}) says that the constant magnetic field is tangent exactly at order one on $\pa \Omega$, for any $x\in \Gamma $. We observe indeed, using (\ref{secondfa}) and (\ref{BN0b}), that~: \begin{equation}\label{varthetaa} \kappa_{n,B} = - \frac{\pa}{\pa r}( \langle N\;,\; \frac{H}{|H|}\rangle)\;. \end{equation} Without the assumption (\ref{BN0bb}), we shall only prove~: \begin{theorem}\label{BNbis}.\\ Let $P^{h,N}$ be the Neumann realization on $L^2(\Omega )$ of the magnetic Laplacian \break $(hD-A)^2$, where $h \in ]0,1[$ is a small parameter. We assume that (\ref{Gammareg}) is satisfied. Then there exists $ \eta >0 $ such that~: \begin{equation}\label{BN1bis} \inf \; \Sp (P^{h,N}) \leq b\Theta_0 h + {\hat \gamma}_0 b^{\frac 2 3 } h^{\frac 4 3 } + \Og(h^{\eta + \frac 4 3 })\;, \end{equation} where \begin{equation}\label{BN2bis} {\hat \gamma}_0 = (\frac 12) \hat \nu_0 ( \mu''(\xi_0))^\frac 13 \inf_{x\in\Gamma} \left( (\kappa_{n,B}(x)^\frac 2 3 ( \frac 12 \mu''(\xi_0) \sin^2 \theta (x) + \cos^2 \theta (x))^\frac 13\right)\;. \end{equation} \end{theorem} When $\theta \equiv 0$, we recover the statement of Theorem \ref{BN}. We conjecture that we have equality in the general case. \begin{remark}\label{Remintrin}.\\ The expression in (\ref{BN2bis}) can be related more directly to the curvature by using the following formula~: \begin{equation} \begin{array}{l} \kappa_{n,B}(x)^2 ( \frac 12 \mu''(\xi_0) \sin^2 \theta (x) + \cos^2 \theta (x))\\ \quad = \kappa^G(x)^2\, \left( \kappa_n(x)^2 + \frac 12 \mu''(\xi_0) K_{x} ( T(x), T(x)\wedge N(x))\right)\times \\ \quad\quad\quad \times \left( \kappa_n(x)^2 + K_{x} ( T(x), T(x)\wedge N(x)) \right)^{-2}\;. \end{array} \end{equation} Here, we recall that, for any $x\in \pa \Omega$, $N(x)$ is the unitary interior nornal vector, $\kappa^G(x)$ is the Gauss curvature and $K_x(\cdot ,\cdot )$ is the curvature form, (the second fondamental form). Moreover, if $x\in \Gamma$, $T(x)$ denotes the unit tangent vector with a given orientation on each connected component of $\Gamma$ and $\kappa_n(x)=K_x( T(x), T(x))$ the normal curvature of $\Gamma $ at $x$. This formula is established using the results of Section \ref{Section8}. \end{remark} \section{Towards the model: new normal forms}\label{Section10} In this section, the vector magnetic field $V(B)$ is assumed to be constant, more precisely $$ V(B)=b\frac{\pa }{\pa x_1},\ (b=|V(B)|>0).$$ The set $\Gamma =\{ x\in \pa \Omega ;\ V(B)\in T_x(\pa \Omega ) \} $ is assumed to be a regular curve. \\ The angle between the normal to $\Gamma $ in $T_x\pa \Omega$ and $V(B)$ will be denoted by $\theta (x)$. \\ We will work near a point of $\Gamma$ $x_0$ to be determined and will look after a good approximation of the operator in a small box around $x_0$. \\ We consider local coordinates $(r,s,t)$ in a neighborhood ${\cal V}_{x_0}$ of $x_0$, such that $(r,s)$ are the coordinates of Lemma \ref{curve}, and such that $t=t(x)={\rm {distance}} (x,\Gamma ))$. \subsection{Normal form for the magnetic potential} In this subsection we shall show how after a suitable gauge transform, we can arrive to a more tractable model. We assume that the magnetic field is constant. \begin{lemma}\label{BN.l2}.\\ Let ${\widetilde A}$ be the magnetic potential defining ${\widetilde B}$, in the coordinates $(r, s, t)$ defined near $x_0 = (0,s_0,0)$. Let ${\widetilde A}^{(2)}$ the Taylor expansion to order $2$ of $\widetilde A$. Then there exists a polynomial function of $ y=(r,s-s_0,t)$, $p^0(y)$ such that, \begin{equation}\label{potmod} {\tilde A}^{(2)} = A^0 + \grad_y p^0\;, \end{equation} with \begin{equation} \begin{array}{ll} { { A}}^0_1 & =bt [\sin \theta_0 + \kappa_g (x_0)\cos \theta_0 \, (s-s_0) -\sin \theta_0 \, \kappa^M (x_0)t] \;,\\ { { A}}^0_2 & =-bt[\cos \theta_0 \, - \kappa_g (x_0)\cos \theta_0 \, r -\kappa_g (x_0)\sin \theta_0 \, (s-s_0) -\cos \theta_0 \, \kappa^M(x_0) t] \\ & \quad - \frac{1}{2}b\kappa_{n,B}(x_0)r^2\\ {{A}}^0_3 & = 0\;. \end{array} \end{equation} \end{lemma} {\bf {Proof}}.\\ \noindent To prove this lemma, we have just to determine the Taylor expansion up to order~$1$ of the magnetic field in the coordinates $(r,s-s_0,t)$ at the point $(0,s_0,0)$. Let us first recall that we have assumed that $A$ was given by (\ref{constf}). So the corresponding vector $V(B)$ introduced in (\ref{defvecmg}) is~: $$ V(B)=b\frac{\pa }{\pa x_1}\;. $$ Let us analyze its expression in the new coordinates~: \begin{equation}\label{BN.l2.1} V(B)=b\frac{\pa }{\pa x_1}= {\widetilde {b}}_1\frac{\pa }{\pa r}+ {\widetilde {b}}_2\frac{\pa }{\pa s}+{\widetilde {b}}_3\frac{\pa }{\pa t}, \end{equation} and, by (\ref{adapmetr}), (\ref{cu5}) and (\ref{BN.l2.1}), we get \begin{equation}\label{BN.l2.2} b \frac{\pa x_1 }{\pa r} == {\widetilde {b}}_1(1-2tK_{11})-2t{\widetilde {b}}_2K_{12} +\Og(t^2), \end{equation} \begin{equation}\label{BN.l2.3} b \frac{\pa x_1}{\pa s} = = {\widetilde {b}}_2(\alpha -2tK_{22}) -2t{\widetilde {b}}_1K_{12} +\Og(t^2), \end{equation} and \begin{equation}\label{BN.l2.4} b \frac{\pa x_1}{\pa t} = ={\widetilde {b}}_3. \end{equation} Using (\ref{cu6}), (\ref{BN.l2.2}), (\ref{BN.l2.3}) and Schwarz Lemma for $x_1$, we get~: \begin{equation}\label{BN.12.4a} \frac{\pa {\widetilde {b}}_1}{\pa s} =\frac{\pa }{\pa r}(\alpha {\widetilde {b}}_2) = - 2\kappa_g{\widetilde {b}}_2+\frac{\pa {\widetilde {b}}_2}{\pa r}\;, \end{equation} when $t=r=0$.\\ We have~: \begin{equation}\label{BN.l2.5bis} {\widetilde {b}}_1=b \cos \theta (s),\ \ {\widetilde {b}}_2= b \sin \theta(s), \\{\widetilde {b}}_3=0\;, \end{equation} when $t=r=0$.\\ This leads by differentiation with respect to $s$ to~: \begin{equation}\label{BN.l2.6bis} \begin{array}{ll} \frac{\pa {\tilde {b}}_1}{\pa s}&= - b \sin \theta(s) \, \theta'(s)\;,\\ \frac{\pa {\tilde {b}}_2}{\pa s}&= b \cos \theta(s) \, \theta'(s)\;,\\ \frac{\pa {\tilde {b}}_3}{\pa s} &= 0\;, \end{array} \end{equation} when $t=r=0$. Then we get by coming back to (\ref{BN.12.4a}) that (\ref{BN.l2.5bis}) leads to~: \begin{equation}\label{BNl2.6abis} \frac{\pa {\widetilde {b}}_2}{\pa r} = - b \sin \theta \, \theta' + 2 \kappa_g b \, \sin \theta\;. \end{equation} Differentiating (\ref{BN.l2.2}) with respect to $r$, we get for $ t=0$, \begin{equation} \frac{\pa {\widetilde {b}}_1}{\pa r} =b\frac{\pa^2 x_1 }{\pa r^2}\;, \end{equation} and by (\ref{cu6}) $$ <\frac{\pa x}{\pa r}|\frac{\pa x}{\pa r}>=1\;. $$ So differentiating the last equality, we get~: $$ <\frac{\pa x }{\pa r} |\frac{\pa^2 x}{\pa r^2}>=0 $$ But when $t=r=0$, $\frac{\pa x}{ \pa r}$ is normal to $\Gamma$, so $$ <\frac{\pa x_1 }{\pa r} |\frac{\pa^2 x_1}{\pa r^2}>=0\;. $$ This gives finally~: \begin{equation}\label{BN.l2.7} \frac{\pa {\widetilde {b}}_1}{\pa r} = b<\frac{\pa x }{\pa r} |\frac{\pa^2 x}{\pa r^2}>=0\;, \end{equation} when $ t=r=0$. \noindent Let us now use (\ref{BN.l2.4}) to write that $$ \frac{\pa {\widetilde {b}}_3}{\pa r} =\frac{\pa }{\pa r} =\;, $$ when $ t=0$. \\ \noindent Then, using also (\ref{BN.l2.1}), (\ref{BN.l2.5bis}) and (\ref{secondfa}), we obtain~: $$ = b \cos \theta(s) <\frac{\pa}{\pa r}\;|\;\frac{\pa N} {\pa r}> + b \sin \theta (s) <\frac{\pa}{\pa s}\;|\;\frac{\pa N} {\pa r}>\;. $$ This leads to \begin{equation}\label{BN.l2.8bis} \frac{\pa {\widetilde {b}}_3}{\pa r} = -b\, \cos \theta (s) \, K_{11} - b \sin \theta(s)\, K_{12}\;, \end{equation} when $ t=r=0$. \noindent But by (\ref{BN.l2.2}), (\ref{BN.l2.4}) and Schwarz Lemma, we have \begin{equation} \frac{\pa {\widetilde {b}}_3}{\pa r} = \frac{\pa {\widetilde {b}}_1}{\pa t} - 2{\widetilde {b}}_1 K_{11} -2{\widetilde {b}}_2 K_{12} \end{equation} when $ t=r=0$.\\ \noindent So, from (\ref{BN.l2.8bis}), we get~: \begin{equation}\label{BN.l2.9bis} \frac{\pa {\widetilde {b}}_1}{\pa t} = b \cos \theta (s) K_{11} + b \sin \theta (s) K_{12}\;, \end{equation} when $t=r=0$. Once again, Schwarz Lemma, (\ref{BN.l2.3}), and (\ref{BN.l2.4}) lead to \begin{equation} \frac{\pa {\widetilde {b}}_3}{\pa s} = \alpha \frac{\pa {\widetilde {b}}_2}{\pa t} -2 {\widetilde {b}}_2 K_{22} - 2 {\widetilde {b}}_1 K_{12} \end{equation} when $ t=0$. \\ Then, (\ref{cu6}), (\ref{BN.l2.5bis}) and (\ref{BN.l2.6bis}) lead to \begin{equation}\label{BN.l2.10bis} \frac{\pa {\widetilde {b}}_2}{\pa t} = 2 b \cos \theta K_{12} + 2 b \sin \theta K_{22}\;, \end{equation} when $t=r=0$. \noindent It remains to determine $ \frac{\pa{\widetilde {b}}_3 }{\pa t}$ when $t=r=0$. Let us denote the magnetic field $$ {\widetilde B}={\widetilde B}_{23}\, ds\wedge dt+ {\widetilde B}_{31}\, dt\wedge dr+{\widetilde B}_{12}\, dr\wedge ds\;. $$ So by (\ref{newmagfi}) and using the property that~: $$ \frac{\pa {\widetilde B}_{23}}{\pa r} + \frac{\pa {\widetilde B}_{31}}{\pa s} + \frac{\pa {\widetilde B}_{12}}{\pa t}=0\;, $$ we get from (\ref{newvecmaga}) that \begin{equation} |g|^{\frac 1 2 }[ \frac{\pa {\widetilde b}_{1}}{\pa r} + \frac{\pa {\widetilde b}_{2}}{\pa s} + \frac{\pa {\widetilde b}_{3}}{\pa t}] + \frac 12 |g|^{-\frac 12} [(\pa_t |g|) {\widetilde b_3} + (\pa_r |g|) {\widetilde b}_1 + (\pa_s|g|){\widetilde b}_2 ]=0\;, \end{equation} Let us now use this formula on $t=r=0$. \noindent Using (\ref{BN.l2.7}), (\ref{BN.l2.6bis}), we get first~: $$ \frac{\pa {\widetilde b}_{3}}{\pa t} + b \cos \theta(s) \theta'(s) = - \frac 12 |g|^{-1} ((\pa_t |g|) {\widetilde b}_3 + (\pa_r |g|) {\widetilde b}_1 + (\pa_s |g|) {\widetilde b}_2)\;, $$ on $t=r=0$.\\ \noindent We now use (\ref{0y2}) and Lemma \ref{curve} (more precisely (\ref{cu6})), we obtain, using (\ref{BN.l2.5bis})~: \begin{equation} (\pa_r |g|) = - 2 \kappa_g (s) \;,\; \mbox{ when } t=r=0\;. \end{equation} This leads to~: \begin{equation}\label{BN.l2.14bis} \frac{\pa {\widetilde {b}}_3}{\pa t} = b\, (\kappa_g - \theta')\, \cos \theta \ {\rm {when}}\ \ t=r=0. \end{equation} We actually need a more complete expansion of $|g|$. We have~: \begin{equation} |g| = 1 - 2 \kappa_g(s_0) r - 2 (K_{11}(0,s_0)+ K_{22}(0,s_0))t + \Og ((t,r,s-s_0)^2)\;. \end{equation} We can now establish the Taylor expansion of the magnetic field in the coordinates $(r,s,t)$. We can now establish the formulas in full generality. For $t=r=0$, we have, with $\theta =\theta(s_0)$~: \begin{equation}\label{taylorH1} \begin{array}{ll} {\tilde b}_1 &= b \cos \theta\,,\\ \frac{\pa {\tilde b}_1}{\pa r}& =0,\\ \frac{\pa {\tilde b}_1}{\pa s} &= - b \sin \theta \,,\\ \frac{\pa {\tilde b}_1}{\pa t} & = b \cos \theta K_{11} + b \sin \theta K_{12}\;; \end{array} \end{equation} \begin{equation}\label{taylorH2} \begin{array}{ll} {\tilde b}_2 &= b \sin \theta(s)\;,\\ \frac{\pa {\tilde b}_2}{\pa r} &= b(2\kappa_g-\theta') \sin \theta\;,\\ \frac{\pa {\tilde b}_2}{\pa s} & = b \cos \theta \, \theta'\;,\\ \frac{\pa {\tilde b}_2}{\pa t} & = 2 b \cos \theta K_{12} + 2 b \sin \theta K_{22}\;; \end{array} \end{equation} \begin{equation}\label{taylorH3} \begin{array}{ll} {\tilde b}_3 &= 0\;,\\ \frac{\pa {\tilde b}_3}{\pa r} &= - b \cos \theta K_{11} - b \sin \theta K_{12},\\ \frac{\pa {\tilde b}_3}{\pa s} &=0,\\ \frac{\pa {\tilde b}_3}{\pa t} &= b \cos \theta (\kappa_g - \theta')\;. \end{array} \end{equation} Using the Taylor extension (to order $1$) of $|g|^\frac 12$, $$ |g|^\frac 12 = 1 - \kappa_g (s_0) r - (K_{11}(0,s_0) + K_{22}(0,s_0)) t + \Og (|(r,s-s_0,t)|)\;, $$ we obtain the model~: \begin{equation} \begin{array}{ll} {\tilde b}_1^0 & = b \cos \theta(s_0) - b \sin \theta(s_0) \theta'(s_0) (s-s_0) + (b \cos \theta(s_0) K_{11}(0,s_0)\\ &\quad + b \sin \theta(s_0) K_{22}(0,s_0)) t\;,\\ {\tilde b}_2^0& = b \sin \theta(s_0) - b \cos \theta (s_0) \theta'(s_0) (s-s_0) + b \sin \theta (2 \kappa_g - \theta'(s_0))r\\ & \quad + (2 b \cos \theta K_{12} + 2 b \sin \theta(s_0) K_{22} ) t\;,\\ {\tilde b}_3^0 & = - (b \cos \theta(s_0) K_{11} + b \sin \theta(s_0) K_{12}) r\\ &\quad + b \cos \theta(s_0) (\kappa_g - \theta'(s_0)) t\;. \end{array} \end{equation} These formulas lead to the introduction of the ``model'' corresponding magnetic field~: \begin{equation} \begin{array}{ll} {\widetilde B}_{23}^0 & = b \cos \theta_0 - b \sin \theta_0\, \theta'(s_0) (s-s_0) - b \cos \theta_0 \kappa_g r\\ &\quad + b ((\cos \theta_0 -1) K_{11} + ( \sin \theta_0 -1) K_{22}) t\;,\\ {\widetilde B}_{31}^0& = b \sin \theta_0 - b \cos \theta_0 \,\theta'(s_0) (s-s_0) + b \sin \theta_0 (\kappa_g - \theta')r\\ &\quad + b(2 \cos \theta_0 K_{12} + \sin \theta_0 K_{22} ) t\;,\\ {\widetilde B}_{12}^0 & = - b ( \cos \theta_0 K_{11} + \sin \theta_0 K_{12}) r + b \cos \theta_0 (\kappa_g - \theta'(s_0)) t\;, \end{array} \end{equation} with $\theta_0 = \theta (s_0)$, $K_{ij}= K_{ij}(0,s_0)$.\\ This corresponds to the Taylor expansion up to order $1$ of $\widetilde B$ at the point $(0,s_0,0)$. \begin{equation}\label{bjGen} \begin{array}{ll} {\widetilde {b}}_1 & =b[\cos (\theta (s_0)) -\theta ^{\prime }(s_0)\sin (\theta (s_0))(s-s_0) +\kappa_{n,B}(x_0)t]\\ & \quad +{\Og}{(|y|^2)}\\ {\widetilde {b}}_2 & =b[\sin (\theta (s_0))+ \sin (\theta (s_0))(2\kappa_g(x_0)-\theta'(s_0))r\\ & \qquad + \theta'(s_0)\cos (\theta (s_0))(s-s_0)+ 2\kappa_{t,B}(x_0)t]\\ & \quad +{\Og}{(|y|^2)}\\ {\widetilde {b}}_3 & =b[-\kappa_{n,B}(x_0)r+ (\kappa_g(x_0)-\theta ^{\prime }(s_0))\cos (\theta (s_0))t] +{\Og}{(|y|^2)} \end{array} \end{equation} As from (\ref{0y2}) $$ |g|^{\frac 1 2 }=1-\kappa_gr-2\kappa ^Mt+{\Og}{(|y|^2)}\; ,$$ we get using the relation between the magnetic field and the vector magnetic field given by (\ref{newmagfi}) and (\ref{newvecmag}) \begin{equation}\label{Bsimp1} \begin{array}{ll} {\widetilde {B}}_{23}& =b[\cos \theta_0 - \kappa_g (x_0)\cos\theta_0 \, r -\theta ^{\prime }(s_0)\sin \theta_0\, (s-s_0)] \\ & \qquad +b (\kappa_{n,B}(x_0)-2\cos \theta_0 \, \kappa ^M(x_0))t\\ & \quad +{\Og}{(|y|^2)}\\ {\widetilde {B}}_{31}&=b[\sin \theta_0\, + \sin \theta_0\, (\kappa_g(x_0)-\theta ^{\prime }(s_0))r+ \theta^{\prime }(s_0)\cos (\theta_0)(s-s_0)]\\ & \quad + 2bt [\kappa_{t,B}(x_0)-\sin \theta_0\, \kappa ^M(x_0)]\\ & \quad +{\Og}{(|y|^2)}\\ {\widetilde {B}}_{12}& = b[-\kappa_{n,B}(x_0)r + (\kappa_g(x_0)-\theta ^{\prime }(s_0))\cos \theta_0 \, t]\\ & \quad +{\Og}{(|y|^2)} \end{array} \end{equation} with $ \theta_0:=\theta (s_0).$\\ If we write that $|V(B)|=b$ , using the relation $[\theta ^{\prime }(s_0)-\kappa_g(x_0)]=0$, this leads first to~: \begin{equation}\label{bjBis} \begin{array}{ll} {\widetilde {b}}_1& =b[\cos (\theta_0) -\theta^{\prime }(s_0)\sin (\theta_0)(s-s_0) +\kappa_{n,B}(x_0)t]+ \Og {(|y|^2)}\\ {\widetilde {b}}_2 & =b[\sin (\theta_0)+ \sin (\theta_0)\kappa_g(x_0)r + \kappa_g(x_0)\cos (\theta_0)(s-s_0)]+ \Og {(|y|^2)}\\ {\widetilde {b}}_3 & =-b\kappa_{n,B}(x_0)r + \Og {(|y|^2)} \end{array} \end{equation} and then to~: \begin{equation}\label{BsimpBis} \begin{array}{ll} {\widetilde {B}}_{23}& = b[\cos \theta_0 - \kappa_g (x_0)\cos \theta_0 \, r -\kappa_g (x_0)\sin \theta_0 \, (s-s_0)\\ & \qquad +(\kappa_{n,B}(x_0)-2\cos \theta_0 \, \kappa^M(x_0))t]\\ &\quad + \Og {(|y|^2)}\\ {\widetilde {B}}_{31}& =b[\sin \theta_0 \, + \kappa_g (x_0)\cos \theta_0 \, (s-s_0) -2\sin \theta_0 \, \kappa^M(x_0)t]\\ & \quad +{\Og}{(|y|^2)}\\ {\widetilde {B}}_{12} & =-b \kappa_{n,B}(x_0)r +{\Og}{(|y|^2)} \end{array} \end{equation} By comparison of the curls on the left and right hand side of (\ref{potmod}) we get the lemma. \subsection{Towards simplified models} If we neglect in addition the terms of order $t^2$, we get the potential $A^{00}$ whose components are given by~: \begin{equation}\label{A00a} A^{00}_{1}=bt [\sin \theta_0 \, + \kappa_g(x_0)\cos \theta_0 \, (s-s_0)]\;, \end{equation} \begin{equation}\label{A00b} A^{00}_{2} = - bt [\cos \theta_0 \, -\kappa_g (x_0)\cos \theta_0 \, r -\kappa_g \sin \theta_0 \, (s-s_0)] -b \frac{\kappa_{n,B}}{2}r^2\;, \end{equation} and \begin{equation}\label{A00c} A_3^{00}=0\;. \end{equation} The corresponding magnetic field is~: \begin{equation} \begin{array}{ll} {B}_{23}^{00}& = b \cos \theta_0 - b \cos \theta_0 \kappa_g r - b \sin \theta_0\, \kappa_g (s-s_0)\;,\\ {B}_{31}^{00}& = b \sin \theta_0 - b \cos \theta_0 \,\kappa_g (s-s_0)\;,\\ {B}_{12}^{00}& = - b \kappa_{n,B} r\;. \end{array} \end{equation} If we neglect in addition the terms corresponding to $t (s-s_0)$, we get the magnetic potential \begin{equation}\label{A000} \begin{array}{ll} A^{01}_{1} & = bt \sin \theta_0 \, \\ A^{01}_{2} & = - bt [\cos \theta_0 \, -\kappa_g (x_0)\cos \theta_0 \, r] -b \frac{\kappa_{n,B}}{2}r^2\;,\\ A^{01}_3 & =0\; . \end{array} \end{equation} The corresponding magnetic field is~: \begin{equation}\label{B000} \begin{array}{ll} {B}_{23}^{01}& = b \cos \theta_0 - b \cos \theta_0 \kappa_g r \;,\\ {B}_{31}^{01}& = b \sin \theta_0 \;,\\ {B}_{12}^{01}& = - b \kappa_{n,B} r\;. \end{array} \end{equation} But we should probably first start with the model, where we neglect also the terms in $\Og(|r t|)$~: \begin{equation} \begin{array}{ll} {\hat A}_1 &= b \sin \theta_0 \, t \;, \\ {\hat A}_2 &= - b \cos \theta_0 \, t - \frac b2 \kappa_{n,B}(s_0) r^2 \;, \\ {\hat A}_3 &= 0\;. \end{array} \end{equation} Neglecting also the geodesic curvature, this leads to the model~: \begin{equation}\label{modle} (hD_r-{\hat A}_{1})^2 + (hD_s-{\hat A}_{2})^2 + h^2 D^{2}_{t}\;. \end{equation} \section{Comparison lemmas}\label{Section11} Although the problem is easier in the case of upper bounds where we work with explicit quasimodes, we need in the two cases (proof of upper bounds and proof of lower bounds) comparison lemmas permitting to control the error made when considering the simplified models. \subsection{A first comparison lemma} We are interested in the energy $ q^{h}_{A}(u)$, introduced in (\ref{qAc}), of some function $ u$ in $H^2(\Omega )$ such that \begin{equation}\label{BN.1} \supp (u)\; \subset \; Q(x_0) \;, \end{equation} where \begin{equation}\label{BN.1a} Q(x_0)=\{ x\in \Omega ;\ (r(x),s(x))\in (r_0,s_0)+[-h^{\delta },h^{\delta }]^2,\ t(x)\in [0,\epsilon_2]\}\; , \end{equation} for some $ \delta \in ]0,1[$. From now on, we assume in this section that $$ (r(x_0),s(x_0))=(0,s_0)\;. $$ \begin{lemma}\label{BN.l1}.\\ If (\ref{BN.1}) is satisfied, then \begin{equation}\label{BN.2} \begin{array}{l} (1 - C \, h^{2\delta} )q^{h}_{{\tilde A}^{(2)}}(u) - C\, \| t^{\frac 1 2 }(hD_x-A)u\|^2 \\ -C\, [q^{h}_{{\tilde A}^{(2)}}(u)]^{\frac 1 2 }\cdot \|(h^{3\delta }+h^{2\delta }t+h^{\delta }t^2+t^3) u\| \ -C \, \|(h^{3\delta }+h^{2\delta }t+h^{\delta }t^2+t^3) u\|^2 \\ \leq q^{h}_{A}(u)\\ \leq (1+ C h^{2\delta })\, q^{h}_{{\tilde A}^{(2)}}(u) + C\, \| t^{\frac 1 2 }(hD_x-A)u\|^2 \\ + \, C\, [q^{h}_{{\tilde A}^{(2)}}(u)]^{\frac 1 2 }\cdot \|(h^{3\delta }+h^{2\delta }t+h^{\delta }t^2+t^3) u\|\\ \ + C \, \|(h^{3\delta }+h^{2\delta }t+h^{\delta }t^2+t^3) u\|^2 \;, \end{array} \end{equation} where~: \begin{equation}\label{BN.3} \begin{array}{ll} q^{h}_{{\tilde A}^{(2)}}(u)& =\int_{Q(x_0)}(1-r\kappa_g(x_0)) [|(hD_t-{\tilde A}^{(2)}_{3})u|^2\\ &\quad + ((1+2 r \kappa_g(x_0))|(hD_s-{\tilde A}^{(2)}_{2})u|^2 + |(hD_r-{\tilde A}^{(2)}_{1})u|^2]\; drdsdt\;, \end{array} \end{equation} $ {\tilde A}^{(2)}$ is the second order polynomial part of the Taylor expansion at $x_0$ of $ {\widetilde {A}}$ defined in (\ref{qAc})~: $$ {\tilde A}^{(2)}(y)=\sum_{|\beta |\leq 2} \frac{\pa^{\beta }{\widetilde {A}}}{\pa y^{\beta }}(0,s_0,0)\frac{y^{\beta }}{\beta !}\;, $$ with $ y=(y_1,y_2,y_3)=(r,s-s_0,t)$.\\ The function $ \kappa_g(x_0)$ is the geodesic curvature at $ x_0$ introduced in (\ref{curvec}). \end{lemma} We have actually two types of errors in order to control the comparison of $q_A^h (u)$ and $q_{{\tilde A}^{(2)}}^h (u)$.\\ {\bf The first error} is that we replace the initial metric in the coordinates $(r,s,t)$ by the new metric (see Lemma \ref{curve})~: $$ g_0^{new}= dt\otimes dt + dr\otimes dr + (1 - 2 r \kappa_g (r_0,s_0)) ds \otimes ds\;, $$ with the corresponding $$ |g^{new}|^\frac 12 := (1- r \kappa_g(r_0,s_0))\;, $$ and we consider a similar linearization (with respect to $r$ and $t$) for $g^{ij}$.\\ We then observe that, on the support of $u$, we have~: $$ |g|^\frac 12 - |g^{new}|^\frac 12 = \Og (t) + \Og (h^{2 \delta})\;, $$ $$ g^{ij}_{new} - g^{ij}= \Og (t) + \Og (h^{2 \delta})\;. $$ {\bf The second error} occurs when replacing $A$ by ${\tilde A}^{(2)}$ and this leads to an error in~: $$ A - {\tilde A}^{(2)} = \Og (|y|^3) = \Og ( (|t| + h^\delta)^3)\;. $$ Once these estimates are satisfied, the lemma follows easily using the Cauchy-Schwarz inequality, after having writen~: $$(h D_y -A) = (h D_y- {\tilde A}^{(2)}) + (A-{\tilde A}^{(2)})\;. $$ \subsection{A second comparison lemma (case $\theta\equiv 0$)} We now perform a gauge transform and a change of unknown in order to go back to the Lebesgue measure. We also decide to consider as remainder the term in $\Og(t^2)$ in $A$. We recall from (\ref{deriveangle}) that, in this case $$ \kappa_g(x_0) =0\;, $$ for $ x_0\in \Gamma$. \begin{lemma}\label{LemmaBN.l3}.\\ If (\ref{BN.1}) and (\ref{BN0bb}) are satisfied, then \begin{equation}\label{BN.4} \begin{array}{l} q^{h}_{A^{00}}(v)-C\| t^{\frac 1 2 }(hD_x-A)u\|^2\\ -C[q^{h}_{A^{00}}(v) ]^{\frac 1 2 }\cdot \|(h^{3\delta }+h+h^{2\delta }t+t^2) u\| \\ -C\|(h^{3\delta }+h+h^{2\delta }t+t^2) u\|^2 \\ \leq q^{h}_{A}(u)\\ \leq q^{h}_{A^{00}}(v)+C\| t^{\frac 1 2 }(hD_x-A)u\|^2 \\ +C[q^{h}_{A^{00}}(v) ]^{\frac 1 2 }\cdot \|(h^{3\delta }+h+h^{2\delta }t+t^2) u\| \\ +C\|(h^{3\delta }+h+h^{2\delta }t+t^2) u\|^2 \end{array} \end{equation} where \begin{equation}\label{quasim} v= e^{-ip^0/h} u\;, \end{equation} \begin{equation}\label{BN.5} q^{h}_{A^{00}}(v)=\int_{Q(x_0)} [|hD_tv|^2+ |(hD_s-A^{00}_{2})v|^2+ |hD_rv|^2]\;dr ds dt\;, \end{equation} \begin{equation} A^{00}_{2}(r,t;s_0):=-bt -\frac{b}{2}{\kappa}_{n,B}(x_0)r^2 \;, \end{equation} and where $ p^0$ is the polynomial function introduced in Lemma \ref{BN.l2}. \end{lemma} Lemma \ref{LemmaBN.l3} comes easily from Lemmas \ref{BN.l1} and \ref{BN.l2} which give that $$ \begin{array}{l} q^{h}_{A^{00}}(v) - C[q^{h}_{A^{00}}(v) ]^{\frac 1 2 }\cdot [h\| u\| +\|t^2 u\| ] -C[h^2\| u\|^2+\|t^2 u\|^2] \\ \leq q^{h}_{{\widetilde A}^{(2)}}( u)\\ \leq q^{h}_{A^{00}}(v)+ C[q^{h}_{A^{00}}(v) ]^{\frac 1 2 }\cdot [h\| u\| +\|t^2 u\| ] +C[h^2\| u\|^2+\|t^2 u\|^2]\;. \end{array} $$ \section{Spectral theory for a simple model}\label{Section12} \subsection{Heuristics}\label{Subsection12.1} We would like to understand the following simplified model considered in (\ref{modle})~: \begin{equation} P_0:= (h D_r - \sin \theta \, t)^2 + (h D_s + \cos \theta \, t + \gamma \frac{r^2}{2})^2 + h^2 D_t^2 \end{equation} on $\rz^2 \times \rz^+$. Here $\theta $ is assumed to be fixed and $\gamma$ is a real parameter. We have moreover assumed for simplification assumed that $b=1$.\\ We will take later \begin{equation}\label{later} \gamma = \kappa_{n,B}(s_0)\;,\; \theta=\theta (s_0)\;, \end{equation} but it is better to keep them as an independent parameter for the first part of the analysis. We are interested in the analysis of the bottom of the spectrum but will concentrate on the research of $L^2$ normalized solutions $u^h$ such that~: $\langle P_0 u^h \;,\; u^h \rangle$ is minimal. .\\ \paragraph{Scaling}.\\ We first introduce the following scaling~: $t = h^\frac 12 {\tilde t}$, $r = h^\frac 13 {\tilde r}$ and the coefficients of the operator being independent of $s$ we take a Fourier transform in $s$. Dividing by $ h$, we get that $P_0 /h$ is unitary equivalent to~: \begin{equation} P_1:= (h^\frac 16 D_{\tilde r} - \sin \theta {\tilde t})^2 + (h^\frac 12\sigma + \cos \theta {\tilde t} + \gamma h^\frac 16 \frac{{\tilde r}^2}{2})^2 + D_{\tilde t}^2 \end{equation} on $\rz^2 \times \rz^+$. We can rewrite the operator in the form~: \begin{equation}\label{Heur2} \begin{array}{ll} P_1 & = D_{\tilde t}^2 + \left({\tilde t}- h^\frac 16 \left[ \sin \theta D_{\tilde r} + \cos \theta (h^\frac 13 \sigma + \frac \gamma 2 {\tilde r}^2 )\right]\right)^2 \\ & \quad + h^\frac 16 \left( \cos \theta D_{\tilde r} - \sin \theta (h^\frac 13 \sigma + \frac \gamma 2 {\tilde r}^2)\right)^2\;. \end{array} \end{equation} We note that we have ``formally '' a lower bound of the type~: \begin{equation}\label{Heur3} \begin{array}{ll} P_1 & \geq \mu ( h^\frac 16 \sin \theta D_{\tilde r} + \cos \theta (h^\frac 13 \sigma + \frac \gamma 2 {\tilde r}^2 )) \\ & \quad + \left( h^\frac 16 \cos \theta D_{\tilde r} - \sin \theta (h^\frac 13 \sigma + \frac \gamma 2 {\tilde r}^2)\right)^2\;, \end{array} \end{equation} which leads naturally to first consider the operator~: \begin{equation}\label{Heur4} \begin{array}{ll} P_2 &: = \mu ( h^\frac 16 \sin \theta D_{\tilde r} + \cos \theta (h^\frac 13 \sigma + \frac \gamma 2 {\tilde r}^2 )) \\ & \quad + \left( h^\frac 16 \cos \theta D_{\tilde r} - \sin \theta (h^\frac 13 \sigma + \frac \gamma 2 {\tilde r}^2)\right)^2\;, \end{array} \end{equation} now considered as an operator on $L^2(\rz^2_{{\tilde r}, \sigma})$. In order to find a suitable quasimode, it is better to first decompose it as an Hilbertian integral of operators $P_2(\sigma)$, this time defined on $L^2(\rz)$, and to look for a minimization on $\sigma$. In this context, it is natural to replace $\mu$ by its approximation at the bottom. We recover a differential model, which will give a good understanding of the problem, modulo an error term which has to be controlled. We consequently analyze the family (depending on $\sigma$) \begin{equation} \begin{array}{ll} P_3 (\sigma) &:= \Theta_0 + \delta_0 ( h^\frac 16 (\sin \theta D_{\tilde r} + \cos \theta (h^\frac 13 \sigma + \frac \gamma 2 {\tilde r}^2 ))-\xi_0 )^2 \\ & \quad + h^\frac 13 \left( \cos \theta D_{\tilde r} - \sin \theta (h^\frac 13 \sigma + \frac \gamma 2 {\tilde r}^2)\right)^2\;. \end{array} \end{equation} with $\delta_0 = \frac 12 \mu''(\xi_0)$. It is now better to introduce, \begin{equation}\label{sigmatheta} \sigma(\theta) = - \xi_0 \cos \theta\;,\; \rho(\theta) = + \xi_0 \sin \theta\;, \end{equation} \begin{equation}\label{newsigma} \sigma = \sigma(\theta) h^{-\frac 12} + h^{-\frac 13} {\hat \sigma}\;, \end{equation} which permits to rewrite $P_3(\sigma)$ in the form~: $$ \begin{array}{ll} P_3 (\sigma) &:= \Theta_0 + \delta_0 ( h^\frac 16 (\sin \theta (D_{\tilde r} + \rho(\theta) h^{-\frac 16}) + \cos \theta ({\hat \sigma} + \frac \gamma 2 {\tilde r}^2 )) )^2 \\ & \quad + h^\frac 13 \left( \cos \theta (D_{\tilde r} + \rho(\theta) h^{-\frac 16}) - \sin \theta ({\hat \sigma} + \frac \gamma 2 {\tilde r}^2)\right)^2\;. \end{array} $$ A translation in $\hat \sigma$ and a gauge transform by $\exp i h^{-\frac 16} \rho(\theta) r$ leads then to the analysis of the family~: \begin{equation}\label{Pe4} \begin{array}{ll} P_4 ({\hat \sigma}) &:= \Theta_0 + h^\frac 13 \delta_0 ( (\sin \theta D_{\tilde r} + \cos \theta ({\hat \sigma} + \frac \gamma 2 {\tilde r}^2 )) )^2 \\ & \quad + h^\frac 13 \left( \cos \theta D_{\tilde r} - \sin \theta ({\hat \sigma} + \frac \gamma 2 {\tilde r}^2)\right)^2\;. \end{array} \end{equation} \subsection{Analysis of the simplified model}\label{sectionsupersimplifiedmodel} It is then natural to introduce~: \begin{equation}\label{Pe5a} P_5(\hat \sigma):= h^{-\frac 13} (P_4(\hat \sigma) - \Theta_0)\;, \end{equation} which becomes independent of $h$~: \begin{equation}\label{Pe5b} \begin{array}{ll} P_5( \hat \sigma) &= \delta_0 \left(\cos \theta (\frac \gamma 2 r^2 + \hat \sigma) + \sin \theta D_r \right)^2\\ &\qquad + \left(\cos \theta D_r - \sin \theta (\frac \gamma 2 r^2 + \hat \sigma) \right)^2\;. \end{array} \end{equation} Here we have omitted the tilde's for the next computations.\\ Our aim is to minimize over $\hat \sigma$ (and then to minimize over the points of $\Gamma$, remembering that $\gamma = \kappa_{n,B}(x)$ and $\theta = \theta(x)$ with $x$ on $\Gamma$).\\ Let us now show, that by a gauge transform, we can rewrite $P_5(\hat \sigma)$ in the form~: \begin{equation} P_6(\hat \sigma)= c D_r^2 + d (r^2-\rho)^2 \;. \end{equation} We look for a gauge transformation of the form~: \begin{equation}\label{tthetar} t(\theta,r) = \alpha(\theta) (\frac \gamma 6 r^3 + {\hat \sigma r}) \;. \end{equation} We consider \begin{equation} P_6 (\hat \sigma):= \exp - i t(\theta,r) \cdot P_5 (\hat \sigma) \cdot \exp i t(\theta,r)\;. \end{equation} The function $\alpha(\theta)$ in (\ref{tthetar}) is chosen such that the coefficients of the operator \break $(\frac \gamma 2 r^2 + \hat \sigma) D_r + D_r (\frac \gamma 2 r^2 + \hat \sigma)$ vanishes. This leads to~: \begin{equation} \alpha (\theta) = \frac{\sin \theta \, \cos \theta \, (1 - \delta_0)}{\delta_0 \sin^2 \theta + \cos^2 \theta}\;. \end{equation} Of course, we have \begin{equation} c = \cos^2 \theta + \delta_0 \sin^2 \theta\;, \end{equation} and \begin{equation} \rho = 2 \hat \sigma / \gamma\;. \end{equation} But for this value of $\alpha(\theta)$, we get~: $$ d= (\frac \gamma 2)^2 \left(\delta_0 ( \cos \theta + \alpha(\theta)\sin \theta)^2 + (-\sin \theta + \alpha(\theta) \cos \theta )^2\right)\;. $$ After computation, this gives~: \begin{equation} d = (\delta_0 (\frac \gamma 2)^2) / ( \delta_0 \sin^2 \theta + \cos^2 \theta)\;, \end{equation} We now rescale the operator $c D_r^2 + d (r^2 - \rho)^2$. We recall, that we deleted the tilde's for simplicity. But we come back to the former writing\footnote{This is better, if we want to follow all the scalings we have done in the construction of quasimodes.} and consider~: $$ P_6(\hat \sigma) = c D_{\tilde r}^2 + d ({\tilde r}^2 - \rho)^2\;. $$ This means that we perform a new scaling~: $$ \tilde r = (\frac c d) ^\frac 16 r'\;, $$ such that $P_6(\hat \sigma)$ becomes in the new coordinates~: \begin{equation} P_7(\hat \sigma) = d^\frac 13 c^\frac 23 \left( (D_{r'}^2 + ((r')^2 - \rho')^2\right)\;, \end{equation} with $$ \rho ' = (\frac c d) ^{-\frac 13} \rho\;. $$ We observe that $c$ and $d$ are independent of $\hat \sigma$. So in order to minimize over $\sigma$ the bottom of the spectrum of the initial operator, we will have to minimize over $\rho'$, the bottom of the spectrum of $\left( D_{r'}^2 + ((r')^2 - \rho')^2\right)$ which is obtained for $\rho'=\rho_{min} $ and take the value $\hat \nu _0$ introduced in (\ref{nuzero}). This corresponds to \begin{equation} \label{hatsigma} \hat \sigma = \frac \gamma 2 \; (\frac cd )^\frac 13 \; \rho_{min} \;, \end{equation} with \begin{equation}\label{csurd} \frac cd = (\cos^2 \theta + \delta_0 \sin^2 \theta)^2 (\delta_0 (\frac \gamma 2)^2)^{-1}\;. \end{equation} So, the bottom of the spectrum of $P_7$, is given for this value of $\hat \sigma$ by~: \begin{equation}\label{botto} d^\frac 13 c^\frac 23 {\hat \nu_0} = (\frac 12)^\frac 23 \delta_0^\frac 13 |\gamma|^\frac 2 3 ( \delta_0 \sin^2 \theta + \cos^2 \theta)^\frac 13 {\hat \nu_0}\;. \end{equation} If we now remember the values of $\gamma$ and $\theta$ in our application, this leads to~: \begin{proposition}.\\ The ground state energy of the model operator $P_4$ is given by~: \begin{equation}\label{formmod} \inf_{\hat \sigma}\inf \sp P_4(\hat \sigma) = \Theta_0 + h^\frac 13 {\hat \nu_0} (\frac 12)^\frac 23 \delta_0^\frac 13 |\gamma |^\frac 2 3 ( \delta_0 \sin^2 \theta + \cos^2 \theta)^\frac 13 \;. \end{equation} Moreover the minimum is obtained for ${\hat \sigma}$ defined in (\ref{hatsigma}). \end{proposition} In our application $\gamma$ and $\theta$ are not independent but satisfy (\ref{later}) for some $ x $ in $\Gamma$. This suggests that we have to look for a minimum over $\Gamma$ of the expression~: $$\gamma^2 \left( \delta_0 \sin^2 \theta(x) + \cos^2 \theta (x)\right)\;, $$ that is~: \begin{equation}\label{minimiz} \ell^2:= \inf_{x\in \Gamma} \left((\kappa_{n,B} (x))^2 ( \delta_0 \sin^2 \theta(x) + \cos^2 \theta (x)) \right). \end{equation} \section{Proof of Theorem \ref{BNbis}: upper bounds in general}\label{Section13} \subsection{Analysis of the case when $\theta \equiv 0$}\label{Subsection13.1} In order to help the reader, we first treat the particular case when $\theta \equiv 0$. Let us show the upper bound in the estimate~(\ref{BN1}) of Theorem~\ref{BN}. \paragraph{Analysis of the model}.\\ Taking into account the assumtion (\ref{BN0bb}) and (\ref{deriveangle}), the model operator is, (see Lemma \ref{LemmaBN.l3} ), \begin{equation}\label{BN.6} P^{h,N,0}= h^2D_{r}^{2}+ (hD_s+bt +\frac{b}{2} {\kappa }_{n,B}(x_0) r^2 )^2 + h^2 D_{t}^{2} \end{equation} associated to the Neumann problem in the $t$-variable for $ t\in ]0,+\infty [$, and the Dirichlet problem for the variables $(r,s) \in I \times J$. \\ \noindent Here $I$ and $ J$ are intervals in $ \rz $, which may depend on $h$. For the moment, we take $ I=J=]-h^{\delta },h^{\delta }[$ and we look for a test function $ u $ satisfying (\ref{BN.1}) in the form (cf (\ref{quasim})) \begin{equation}\label{quasim1a} u =e^{ip^0/h} {\tilde u}\;, \end{equation} with \begin{equation}\label{quasim2a} {\tilde u}(r,s, t)=c h^{- \frac \delta 2} e^{ - i \rho \kappa ^{0}_{n,B} s /(2h^{\frac 13}) } e^{ - i b^{\frac 12}\xi_0 s /h^{\frac 12} } \chi ( h^{ - \delta } C^{ -1} s) v(r,t) \end{equation} where $c$ is a normalization constant, $\ \rho $ is a fixed constant which will be defined later on, $\chi $ is an even function on $\rz $ supported on $[ - \frac{1}{2} , \frac{1}{2} ]$ and equal to $1$ on $ [ - \frac{1}{4} , \frac{1}{4} ]$.\\ As $s \mapsto \chi^h(s) := c h^{-\frac \delta 2} \chi ( h^{ - \delta } C^{ -1} s)$ is an even function and $$ \| hD_s \chi^h \|_{L^2(\rz_s)}^2 \leq C h^{2 - 2 \delta } \ , $$ we have just to search for a normalized function $v(r,t)$ satisfying \begin{equation}\label{BN1.o2} q^{h}_{M^{00}}(v)\leq [ h b \Theta_0 + h^{\frac 43} b^{\frac 23} {\hat \nu}_0 (\mu''(\xi_0))^{\frac 13}{\gamma}_0 + C\, h^{\eta +\frac 43} ] \end{equation} with \begin{equation}\label{BN1.oo3} q^{h}_{M^{00}}(v):=\int_{h^{\delta } Q_2} [ |hD_r v |^2 + | M^{00}_{2} v |^2 + h^2 |D_t v |^2 ] \; drdt \;, \end{equation} $Q_2=]- C,C[ \times [0 , C [$, and \begin{equation} M^{00}_{2}(r,t)= b^{\frac 12} ( b^{\frac 12} t - h^{\frac 12} \xi_0) + \frac{b}{2} \kappa^{0}_{n,B} ( r^2 - h^{\frac 23} b^{-1} \rho ) \; . \end{equation} We introduce the scaling $ b^{\frac 12}( h^{-\frac 13} r , h^{-\frac 12} t) = ( {\hat r} , {\hat t})$ and take $v$ in the form~: $$ v (r,t) = b^\frac 12 h^{-\frac {5}{12}}\; v_0 ({\hat r}, {\hat t})\;. $$ We then delete the hats and we get easily that \begin{equation}\label{BN.oo6} q^{h}_{M^{00}}(v) = b h \; {\tilde q}^{h} (v_0) \end{equation} with \begin{equation}\label{BN.oo7} {\tilde q}^{h}(v_0):= \int_{ Q^h } [ | D_t v_0 |^2 + |[ t - \xi_0 + h^{\frac 16} \frac{b^{-\frac 12}}{2} \kappa^{0}_{n,B} ( r^2 - \rho ) v_0 |^2 + h^{\frac 13} | D_r v_0 |^2 ] \; drdt \; , \end{equation} $$ Q^h=]-h^{\delta -\frac 13} b^{\frac 12} C ,h^{\delta -\frac 13} b^{\frac 12} C[ \times [0 , h^{\delta -\frac 12} b^{\frac 12} [\; . $$ \noindent Let $\varphi_0(t)$ the normalized eigenfunction of $ Q(\xi_0)$ associated to the groundstate energy $ \Theta_0$ (cf Subsection \ref{Subsection2.2}). Then one can check easily as in \cite{HeMo2} that \begin{equation}\label{BN1.o3} \| Q(\xi )\phi^{\xi }-[\Theta_0+\frac {1}{2}(\xi -\xi_0)^2 \mu''(\xi_0)]\phi^{\xi }\| \leq C |\xi -\xi_0|^3\;, \end{equation} if $ |\xi -\xi_0|\leq c_0$, for some sufficiently small constant $c_0$, where \begin{equation} \label{BN1.o3bis} \phi^{\xi }(t)=\varphi_0(t)+(\xi -\xi_0)\varphi_1(t) +(\xi -\xi_0)^2\varphi_2(t), \end{equation} $$ \varphi_1=2[Q(\xi_0)-\Theta_0]^{-1, r}[(t-\xi_0)\varphi_0]$$ and $$ \varphi_2=2[Q(\xi_0)-\Theta_0]^{-1,r}[(t-\xi_0)\varphi_1- <(t-\xi_0)\varphi_1|\varphi_0>\varphi_0]\;. $$ Here $ (Q(\xi_0) -\Theta_0)^{-1,r}$ is the regularized resolvent defined by~: $$ \begin{array}{lll} (Q(\xi_0) -\Theta_0)^{-1,r} \varphi_0 & = 0\;,&\\ (Q(\xi_0) -\Theta_0)^{-1,r} u&= (Q(\xi_0) -\Theta_0)^{-1}u &\mbox{ if } \langle u | \varphi_0 \rangle =0\;. \end{array} $$ Let $ \psi_0$ be the normalized eigenfunction (cf Subsection \ref{Subsection2.4}) associated to the groundstate energy of the Hamiltonian $ D^{2}_{r}+(r^2 -\rho_{min})^2$ on $ L^2(\rz )$ and \begin{equation}\label{BN1.o6} \begin{array}{l} \int_{\rz } [|D_r\psi_0|^2 + (r^2 -\rho_{min})^2|\psi_0|^2]dr\\ ={\hat \nu}_0\\ = \inf_{\eta \in \rz } \inf\{ \int_{\rz }[|D_rf|^2+(r^2-\eta )^2|f|^2]dr;\ \| f\| =1\}\; . \end{array} \end{equation} We recall that $ r \mapsto \psi_0(r)$ is an even function. \\ We take \begin{equation}\label{BN1.oo7} \rho = \alpha ^{-2}(b) \rho_{min} \ {\rm {with}}\ \alpha (b)=[\frac{1}{8}\mu''(\xi_0) b^{-1}(\kappa_{n,B}(x_0))^2]^{\frac 16}\; ,. \end{equation} $$ v_0(r,t)= \psi_0(\alpha (b) r) \phi^{\xi (h,r)} (t) \chi (C h^{-\delta + \frac 13} r) \chi (C h^{-\delta + \frac 12} t)\; , $$ with $$ \xi (h,r)=\xi_0 - h^{\frac 16} \, \frac{b^{-\frac 12}}{2} \kappa^{0}_{n,B} (r^2 - \rho )\; . $$ As $ r\mapsto v_0(r,t)$ is an even function, we get easily that $$ {\tilde q}^h(v_0) \leq [\Theta_0 + {\hat \nu}_0\alpha^2(b) h^{\frac 13} + C\, h^{\frac 12} ] \| v_0 \|^2 \;. $$ \paragraph {Conclusion of the proof}.\\ The upper bound in (\ref{BN1}) follows with $ \delta $ equal to $\frac{5}{16}$ and $ \eta = \frac {1}{24}$, in (\ref{BN1}). Here we take $x_0$ in $\Gamma $ such that \begin{equation}\label{choix} |\kappa_{n,B}(x_0)|= \inf_{x\in \Gamma }|\kappa_{n,B}(x)|\;. \end{equation} \subsection{Upper bounds in the generic case}\label{Subsection13.2} We have seen in the previous subsection how one can heuristically get a candidate for the main term of the ground state. We give now the rigorous proof of the upper bound. \begin{proposition}\label{over} .\\ Let $P^{h,N}$ be the Neumann magnetic Laplace operator on $L^2(\Omega )$ $(hD-A)^2$, where $\Omega $ is a bounded open set of $\rz^3$ with smooth boundary $\pa \Omega $. We assume that the magnetic field $ H=\rot A $ is constant, that \begin{equation}\label{over1} \Gamma =\{ x\in \pa \Omega ;\ H\in T_x(\pa \Omega )\} \ {\rm {is \ a\ finite\ collection\ of\ smooth\ closed\ curves.}} \end{equation} and that~: \begin{equation} \forall \; x\; \in \; \Gamma , \ \ \kappa^G (x) \neq 0 \;. \end{equation} Then there exist $\eta >0$ and $C_0 > 0$ such that, for all $ h\in ]0,1] ,$ \begin{equation}\label{over3} \inf \sp ( P^{h,N} ) \leq h b \Theta_0 + h^{\frac 4 3 } b^{\frac 2 3 }{\hat \nu}_0 (\mu''(\xi_0))^{\frac 1 3 }{ \gamma}_0 + C_0 \, h^{\eta +\frac 4 3 } ; \end{equation} where $b = |H|$, $ {\hat \nu}_0$ is introduced in (\ref{nuzero}) and \begin{equation}\label{gammazero} { \gamma}_0 = \inf_{x\in \Gamma } {\tilde \gamma} (x) , \end{equation} \begin{equation}\label{hatgammax} \begin{array}{ll} {\tilde \gamma (x)}^3 & = \frac{1}{8} (\kappa^G(x))^{2} \left [ \kappa^{2}_{n}(x) + \frac{1}{2} \mu''(\xi_0) K^{2}_{x}(T(x) , T(x) \wedge N(x) )\right ] \times \\ &\quad\quad \times \left[ \kappa^{2}_{n}(x) + K^{2}_{x}(T(x) , T(x) \wedge N(x) ) \right]^{-2} \;. \end{array} \end{equation} \end{proposition} \begin{remark}.\\ Let us remark that when the assumptions of Theorem \ref{BN} are satisfied, $$ K_x(T(x) , T(x) \wedge N(x) ) =0 \mbox{ on }\Gamma\;,$$ so (\ref{over3}) is the same upperbound given in (\ref{BN1}) of the same theorem. If $\kappa^G \neq 0$, then $\kappa_{n,B}$ is different of $0$. Note that if we assume in addition that $$ \kappa_n (x) \neq 0 \;,$$ then $H$ can not be tangent to $\Gamma$. This would mean indeed that $\theta = \frac \pi 2$, and by (\ref{defkappas1}) and (\ref{zerocurva}) that $K_{22} =0$. \end{remark} \noindent {\bf Proof of Proposition \ref{over}} .\\ Any constant independent of $ h$ will be denoted by $ C $. We still use the notation $ b = |H|$. \\ As for the proof of the upper bound in (\ref{BN1}) of Theorem \ref{BN}, we have just to find an $L^2$ normalized function $ u $, supported in $h^{\delta }Q$, with $Q=]- C,C[^2 \times [0 , C [$, such that \begin{equation}\label{over4} q^{h}_{A^{00}}(u):=\int_{h^{\delta } Q} [ |(h D_r - A^{00}_{1})u|^2 + (1 + 2\kappa^{0}_{g} r) |(h D_s - A^{00}_{2})u|^2 + h^2 |D_t u|^2 ]\; drdsdt \end{equation} satisfies \begin{equation}\label{over5} q^{h}_{A^{00}}(u)\leq \Theta_0 b \, h + b^{\frac 2 3 } {\hat \nu_0}(\mu''(\xi_0))^{\frac 1 3 }{ \gamma}_0 \, h^\frac 43 + C \, h^{\eta +\frac 4 3 } \end{equation} for some $\delta \in ] \frac{5}{18} , \frac{1}{3} [\;$ and some $ \eta > 0$,\\ and \begin{equation}\label{over6} h^{-1} \| (1+h^{-\frac 1 4 }t^{\frac 1 2 }) (hD_y-{\tilde A}) u \|^2 + \| (1 + h^{-\frac 1 2 }t)^2 u \|^2 \leq C \; ; \end{equation} with \begin{equation}\label{def0} \begin{array}{ll} A^{00}_{1}(r,s,t)&= b t [\sin \theta_0 \; + \; \kappa^{0}_{g} \cos \theta_0\; s]\;, \\ A^{00}_{2}(r,s,t)& = b t [ -\cos \theta_0 \; + \; \kappa^{0}_{g} \cos \theta_0 \; r \;+\; \kappa^{0}_{g} \sin \theta_0\; s] - \frac{b}{2} \kappa^{0}_{n,B} r^2\;, \\ \theta_0 &= \theta (x_0) = {\rm {arcsin}} (|| )\;,\\ \kappa^{0}_{g}&=\kappa_g(x_0)\;,\\ \kappa^{0}_{n,B} & = \kappa_{n,B}(x_0) \;, \end{array} \end{equation} and where $ x_0 \in \Gamma $ is chosen such that $$ {\tilde \gamma }(x_0)={ \gamma}_0 \;. $$ We have chosen a system of coordinates $(r,s,t)$ such that~: $x_0= (0,0,0)$. We recall that $A^{00}$ was introduced in (\ref{A00a})-(\ref{A00c}).\\ We search for a function of the form \begin{equation}\label{over6a} \begin{array}{l} u(r,s,t;h)=\\ \quad c h^{- \frac \delta 2} \exp( - \frac {i}{h^\frac 13} \rho \kappa^0_{n,B} s)\, \cdot\, \exp (\frac {i}{h^{\frac 1 2 }} ( \sin \theta_0 r - \cos \theta_0 s )b^{\frac 1 2 } \xi_0 ) \, \cdot \, \chi ( h^{ - \delta } C^{ -1} s) \; v(r,t) \;, \end{array} \end{equation} where $c$ is a normalization constant, $\rho$ is a constant to be determined later, $\chi $ is an even function on $\rz $ supported on $[ - \frac{1}{2} , \frac{1}{2} ]$ and equal to $1$ on $ [ - \frac{1}{4} , \frac{1}{4} ]$.\\ As $s \mapsto \chi^h(s) := c h^{-\frac \delta 2} \chi ( h^{ - \delta } C^{ -1} s)$ is an even function and $$ \| (hD_s - b \kappa^{0}_{g} \sin \theta_0 s t) \chi^h \|_{L^2(\rz_s)}^2 \leq C (h^{2 - 2 \delta } + h^{2\delta} t^2) \ , $$ we have just to search for a normalized function $v(r,t)$ satisfying \begin{equation}\label{over7} q^{h}_{M^{00}}(v)\leq h b \Theta_0 + h^{\frac 4 3 } b^{\frac 2 3 } {\hat \nu}_0 (\mu''(\xi_0))^{\frac 1 3 }{ \gamma}_0 + C\, h^{\eta +\frac 4 3 } \;, \end{equation} with \begin{equation}\label{over8} q^{h}_{M^{00}}(v):=\int_{h^{\delta } Q_2} [ |(hD_r - M^{00}_{1}) v |^2 + | M^{00}_{2} v |^2 + h^2 |D_t v |^2 ] \; drdt \;, \end{equation} $Q_2=]- C,C[ \times [0 , C [$, and \begin{equation} \begin{array}{ll} M^{00}_{1}(r,t)&= b^{\frac 1 2 } \sin \theta_0 ( b^{\frac 1 2 } t - h^{\frac 1 2 } \xi_0) \\ M^{00}_{2}(r,t)&= (1 + 2\kappa^{0}_{g}r) [ - b^{\frac 1 2 } \cos \theta_0 ( b^{\frac 1 2 } t - h^{\frac 1 2 } \xi_0) + \kappa^{0}_{g} \cos \theta_0 b r t \\ & \qquad \quad - \frac{b}{2} \kappa^{0}_{n,B} (r^2- h^{\frac 23} b^{-1}\rho)] \; . \end{array} \end{equation} Moreover $v $ must satisfy \begin{equation}\label{over9} h^{-1} \int_{h^{\delta } Q_2} (1+h^{-\frac 1 2 }t) [ |(hD_r - M^{00}_{1})v|^2 + | M^{00}_{2}v|^2 +h^2|D_t v|^2 ]\; drdt \leq C \;, \end{equation} and \begin{equation}\label{over9a} \| (1 + h^{-\frac 1 2 }t)^2 v \|^2 \leq C \; . \end{equation} So we can neglect the $ t r^2$ and $ r^3$ terms in $ M^{00}_{2}$ which will lead to error terms in (\ref{over7}), and we get the new \begin{equation}\label{over10} \begin{array}{ll} M^{01}_{1}(r,t) & := b^{\frac 1 2 } \sin \theta_0 ( b^{\frac 1 2 } t - h^{\frac 1 2 } \xi_0) \\ M^{01}_{2}(r,t) & := - b^{\frac 1 2 } \cos \theta_0 ( b^{\frac 1 2 } t - h^{\frac 1 2 } \xi_0) + h^{\frac 1 2 } b^{\frac 1 2 } \xi_0 \kappa^{0}_{g} \cos \theta_0 r - \frac{b}{2} \kappa^{0}_{n,B} (r^2- h^{\frac 23} b^{-1}\rho)\; . \end{array} \end{equation} We introduce the scaling $ b^{\frac 1 2 }( h^{-\frac 1 3 } r , h^{-\frac 1 2 } t) = ( {\hat r} , {\hat t})$ and take $v$ in the form~: $$ v (r,t) = b^\frac 12 h^{-\frac {5}{12}}\; v_0 ({\hat r}, {\hat t})\;. $$ We then delete the hats and we get easily that \begin{equation}\label{over11} q^{h}_{M^{01}}(v) = b h \; {\tilde q}^{h} (v_0) \end{equation} with \begin{equation}\label{over12} {\tilde q}^{h}(v_0):= \int_{ Q^h } [ | D_t v_0 |^2 + |( t - \xi_0 - h^{\frac 1 6 } L^h_1(r,D_r)) v_0 |^2 + h^{\frac 1 3 } | L^h_2(r,D_r)) v_0 |^2 ] \; drdt \; , \end{equation} $$ Q^h=]- C b^{\frac 1 2 } h^{\delta -\frac 1 3 }, C b^{\frac 1 2 } h^{\delta -\frac 1 3 } [ \times [0 , h^{\delta -\frac 1 2 } b^{\frac 1 2 } [\;, $$ and with \begin{equation}\label{over13} \begin{array}{l} L^h_1(r,D_r) = \sin \theta_0\, D_r - \cos \theta_0\, \frac{b^{-\frac 1 2 }}{2} [ \kappa^{0}_{n,B} (r^2 -\rho) - 2 h^{\frac 1 6 }\xi_0 \kappa^{0}_{g} \cos \theta_0\, r ]\;;\\ L^h_2(r,D_r) = \cos \theta_0\, D_r + \sin \theta_0\, \frac{b^{-\frac 1 2 }}{2} [ \kappa^{0}_{n,B} (r^2-\rho) - 2 h^{\frac 1 6 }\xi_0 \kappa^{0}_{g} \cos \theta_0\, r ] \; . \end{array} \end{equation} We search for a function $ v_0$ such that $$ \| ( 1 + r^2) v_0 \| \leq C \| v_0 \| .$$ We can neglect terms with $ h^{\frac 1 6 }$ in factor, and the $L^{h}_{j}$ are replaced by~: \begin{equation}\label{over13n} \begin{array}{l} L^{0}_1(r,D_r) = \sin \theta_0\, D_r - \cos \theta_0\, \frac{b^{-\frac 1 2 }}{2} [ \kappa^{0}_{n,B} (r^2 -\rho)]\;;\\ L^{0}_2(r,D_r) = \cos \theta_0\, D_r + \sin \theta_0\, \frac{b^{-\frac 1 2 }}{2} [ \kappa^{0}_{n,B} (r^2-\rho) ] \; . \end{array} \end{equation} By (\ref{BN1.o3}) and using Fourier transform, we know that there exist functions $ \varphi_j(t),\ j =0,\ldots ,2$ exponentially decreasing at infinity on $[ 0 , + \infty [ ,$ with their derivatives, such that, for any $ \psi \in \Sg(\rz )$ , if \begin{equation}\label{over15} w^h(r,t) = \varphi_0(t) \psi (r) + \varphi_1(t) h^{\frac 1 6 }L_1^{0}(r,D_r) \psi (r) + \varphi_2(t) h^{\frac 1 3 }(L_1^{0}(r,D_r) )^2\psi (r) , \end{equation} then we get, \begin{equation}\label{over16} \begin{array}{l} \| [ D^{2}_{t} + (t - \xi_0 - h^{\frac 1 6 } L_1^{0}(r,D_r) )^2 - (\Theta_0 + \frac{1}{2} \mu''(\xi_0) h^\frac 13 (L_1^{0}(r,D_r))^2 ) ] w^h \| \\ \leq C_e h^{\frac 1 2 } \| L_1^{0}(r,D_r))^3 \,w^h \| . \end{array} \end{equation} This is indeed a consequence of the estimate, for $\tau \in \rz$ and $$ w(t) = \phi_0 + \tau \phi_1 + \tau^2 \phi_2\;. $$ \begin{equation}\label{over16bis} \| [ D^{2}_{t} + (t - \xi_0 - \tau )^2 - (\Theta_0 + \frac{1}{2} \mu''(\xi_0) \tau^2) ] w \| \leq C_e | \tau|^3\; , \end{equation} and of a suitable functional calculus. \\ The inequality (\ref{over16bis}) was already used for $|\tau| \leq \tau_0$ in (\ref{BN1.o3}) but let us just observe that it is also true for $|\tau| > \tau_0 >0$, if one realizes that the square of the left hand side of (\ref{over16bis}) is a polynomial of degree $4$ with respect to the $\tau$ variable.\\ For defining the functional calculus, we can, for example, when $\theta_0 \neq 0$, use a gauge transform which transforms $L_1^h$ into $\sin \theta_0 D_r$ and the proof can then be done by partial Fourier transform. In the case when $\theta_0=0$ a direct proof can be done. Note that the constant $C_e$ is independent of $\theta_0$. Let us define \begin{equation}\label{over17} M(r,D_r) := \frac{1}{2} \mu''(\xi_0) L_1^{0}(r,D_r)^2 + L_2^{0} (r,D_r) ^2\;. \end{equation} At this point, we have justified the heuristic part of the previous section and can use the results obtained on the model. Let \begin{equation} \psi^{0}(r) = (\frac cd)^{-\frac {1}{12} } \; \exp i \varphi (r) \; \psi_0 ((\frac cd)^{-\frac 16 } r) \end{equation} and \begin{equation}\label{over19} \psi = \psi^0\cdot \chi (C^{-1} h^{-\delta + \frac 13 } r)\;. \end{equation} We take \begin{equation} \rho = b^\frac 13 (\frac cd)^\frac 13 \rho_{min}\;, \end{equation} with $c$ and $d$ as in (\ref{csurd}), with $\gamma = \kappa^0_{n,B}$ and $\theta=\theta_0$. We then consider, with $w^h(r,t)$ defined in (\ref{over15}) with (\ref{over19}), \begin{equation}\label{over19b} v_0(r,t) = w^h(r,t) \chi (\frac 1C \, h^{-\delta +\frac 1 2 }t) \;. \end{equation} We recall that $\psi_0$ is exponentially deacreasing at infinity with all its derivatives. \\ Then we get easily from (\ref{over16}) and (\ref{formmod}) that \begin{equation}\label{over20} {\tilde q}^{h}_{A^{00}}(v_0) \leq [\Theta_0 + h^{\frac 1 3 } {\hat \nu}_0 b^{-\frac 13} {\hat \gamma}_1 + C\, h^{\frac 1 2 } ] \cdot \| v_0 \|^2 \;, \end{equation} with \begin{equation} \hat \gamma_1 = (\frac 12)^\frac 23 \delta_0^\frac 13 |\kappa^0_{n,B}|^\frac 23 (\delta_0 \sin^2 \theta_0 + \cos^2 \theta_0)^\frac 13\;. \end{equation} This gives (\ref{over7}) using Remark \ref{Remintrin}. \subsection{Upper bounds in a degenerate case} \begin{proposition}\label{Impr}.\\ Under the assumptions of Proposition \ref{estgros} and if moreover there exists $x_0 \in \pa \Omega $ such that the Gauss curvature at $x_0$ vanishes, $H$ is tangent to $\pa \Omega $ at $x_0$and the normal curvature at $x_0$ along the vector $H$ vanishes~: \begin{equation}\label{Hestgros} H \; \in \; T_{x_0} \pa \Omega \;, \; \kappa^G (x_0)=0 \;, \; K_{x_0}(H,H) = 0 \; , \end{equation} then the upper bound is of the form \begin{equation}\label{estgros2} \lambda (h)\leq b \Theta_0 h + C_0 \, h^{\frac 3 2 } \; . \end{equation} \end{proposition} {\bf {Proof of Proposition \ref{Impr}}}.\\ Let $\Gamma $ be the geodesic of $\pa \Omega $ through $ x_0$ and normal to $ H$. We consider the coordinates, in a neighbourhood of $ W_0$ of $ x_0$, given by Lemma \ref{curve}, and associated to $\Gamma $. Let us remark that the geodesic curvature of $\Gamma $ vanishes, because $\Gamma$ is a geodesic. \\ Then, if $y=(y_1,y_2,y_3)=(r,s,t)$ are the adapted coordinates in a neighbourhood of $ W_0$, and if ${\widetilde b}_j$ are the new components of the vector magnetic field~: $$ H= {\widetilde b}_1(y)\frac{\pa }{\pa r} + {\widetilde b}_2(y)\frac{\pa }{\pa s} + {\widetilde b}_3(y)\frac{\pa }{\pa t} \;, $$ then, assuming $y(x_0)=0$, \begin{equation}\label{2est1} {\widetilde b}_1(0)=b,\ \; {\widetilde b}_2(0)= {\widetilde b}_3(0)=0 \; . \end{equation} As $ {\widetilde b}_j=< H | \frac{\pa }{\pa y_j} >$ and as $ H$ is constant, using (\ref{secondfa}), we get \begin{equation}\label{2est2} \frac{\pa {\widetilde b}_3 }{\pa r} (0)= = - b K_{11}(x_0), \ \frac{\pa {\widetilde b}_3}{\pa s} (0)= = - b K_{12}(x_0) \; . \end{equation} But the assumption (\ref{Hestgros}) leads to \begin{equation}\label{2est3} K_{11}(x_0)=K_{12}(x_0)=0 \; , \end{equation} so (\ref{2est2}) and (\ref{2est3}) shows that \begin{equation}\label{2est4} \begin{array}{l} {\widetilde b}_1=b+b_{11}r+b_{12}s+b_{13}t + {\Og}(|y|^2)\\ {\widetilde b}_2=b_{21}r+b_{22}s+b_{23}t + {\Og}(|y|^2)\\ {\widetilde b}_3=b_{33}t + {\Og}(|y|^2)\; . \end{array} \end{equation} But the asymptotic behavior (\ref{0y2}) and (\ref{cu6}) (with $\kappa_g=0)$ prove that \begin{equation}\label{2est5} |g|=1-4t\kappa^M(x_0) + {\Og}(|y|^2) \; . \end{equation} Then, in the new cordinates, the magnetic potential ${\widetilde A}$ satisfies by (\ref{newmagfi}), (\ref{newvecmag}) $(H=V(B)),$ (\ref{2est4}) and (\ref{2est5}) \begin{equation}\label{2est6} \begin{array}{l} {\widetilde B}_{23} = b + \beta_{11}r + \beta_{12}s + \beta_{13} t + {\Og}(|y|^2)\\ {\widetilde B}_{31} = \beta_{21}r + \beta_{22}s + \beta_{23} t + {\Og}(|y|^2)\\ {\widetilde B}_{12} = \beta_{33} t + {\Og}(|y|^2) \; . \end{array} \end{equation} Then we can choose a gauge ${\widetilde A} $ such that \begin{equation}\label{2est7} {\tilde A}_1(y) = t S_1(y) + {\Og}(|y|^3),\ {\tilde A}_2(y) = - t b + t S_2(y) + {\Og}(|y|^3),\ {\tilde A}_3(y) = 0 \; , \end{equation} where $y \mapsto S_j(y)$ is the linear form $$ S_j(y)=S_{j1} r + S_{j2} s + S_{j3} t\;. $$ Using again (\ref{cu6}) with $\kappa_g=0$ we get that \begin{equation}\label{2est8} ||g|^{\frac 1 2 }(y)g^{\star }(X,X) - |X|^2| \leq C\; (t + |y|^2)|X|^2 \; ,\ \forall \; X\; \in \; \rz^3 \; . \end{equation} As in the proof of (\ref{estgros1up}, we get, if ${\rm {supp}}(u) \; \subset \; \{ x;\ |x-x_0|\leq h^{\delta }\} $ with $\delta >0$, that, for some constant $C = C(\delta ) >0$, independent of $h$ and $ u$, \begin{equation}\label{2est9} \begin{array}{ll} q^{h}_{A}(e^{i\varphi }u)& \leq (1+h^{2\delta }C)q^{h}_{A^0}(u) \\ &+ C \| t^{\frac 1 2 }(hD_y - {\widetilde A})u \|^2 \\ & + C[h^{3\delta }(q^{h}_{A^0}(u))^{\frac 1 2 }\cdot \| u\| + h^{6\delta }\| u\|^2]\; , \end{array} \end{equation} where $ \varphi (y)$ a real function, $$ A^{0}_{1}=tS_1(y), \ A^{0}_{2}=-bt+tS_2(y),\ A^{0}_{3}=0\;, $$ and $q^{h}_{A^0}(u)$ defined as in (\ref{est4b}) .\\ We take the function $ u$ of the form $$ u=e^{-ib^{\frac 1 2 }y_2\xi_0/h^{\frac 1 2 }}\; v(y)\;, $$ with $ v$ as in (\ref{est5}), and we get easily from (\ref{2est9}) with $ \delta =\frac 1 4$, that $$ q^{h}_{A}(e^{i\varphi }u) \leq (b\Theta_0 h + C \; h^{\frac 3 2 })\| u\|^2 $$ and (\ref{estgros2}) follows. \section{ Proof of Theorem \ref{BN}: accurate lower bounds in a particular case.}\label{Section14} \subsection{Introduction}\label{ssintrod} Before to enter into the technicalities, let us explain the ideas which lead to the partion of unity we shall consider. We cut $\overline{\Omega}$ in different zones. \begin{itemize} \item The first zone corresponds to the points which are far from the boundary. They are treated rather easily, exactly as in the case of dimension $2$. Here the model with constant magnetic field in $\rz^3$ is relevant. This corresponds to the introduction of $\Gamma_{\tau(h)}^{0}$ already done in (\ref{low5}). The other zones correspond to a finer decomposition of $\Gamma_{\tau(h)}^{1}$ also already introduced in (\ref{low5}). \item The second zone $\Gamma^{11}{\tau(h)}$ corresponds to the points which are near $\Gamma$ at a distance which is $\Og(h^\delta)$. The model which was analyzed in the previous sections is essentially relevant. \item The third zone $\Gamma_{\tau(h)}^{13}$ corresponds to the points which are near $\pa \Omega$ but far from $\Gamma$. Here the model analyzed in Section \ref{Section3} is relevant. \item Unfortunately, we need an intermediate zone This will correspond to the introduction of $\Gamma_{\tau(h)}^{12}$ corresponding to points near $\Gamma$ but not in the second zone. In some sense, we need to be able to interpolate between the second zone and the third zone. \end{itemize} Then, we have to glue together the estimates introduced in each zone and to control the errors related to the partition. \subsection{A partition of unity}\label{sspartition} We proceed as in the proof of Proposition \ref{Lowb} . In the partition of unity (\ref{low3}), we take \begin{equation}\label{ajout1} \tau (h) = h^{\delta } \ \ {\rm {with}} \ \delta \; \in \; ]\frac{5}{18} , \frac{1}{3} [\; . \end{equation} Let us denote by $u^h$ a normalized eigenfunction associated to the ground state energy $\lambda (h)$ of $P^{h,N} :$ \begin{equation}\label{defuh} u^h\; \in \; H^1(\Omega ),\ \ \| u^h\| =1,\ \lambda (h) \| u^h\| ^2 =q^{h}_{A}(u^h)\; \leq q^{h}_{A}(v)/\| v\| ^2, \ \forall v\in H^1(\Omega )\; . \end{equation} As announced in Subsection \ref{ssintrod}, we split $\Gamma^{1}_{\tau (h)}$, which was defined by (\ref{low5}), in the following way. For $C_1>1$ fixed large enough, let us define \begin{equation}\label{low5bb} \begin{array}{l} \Gamma^{11}_{\tau (h)}=\{ \gamma \in \Gamma^{1}_{\tau (h)};\ \dist ({\rm {supp}}(\chi _{\gamma ,\tau (h)}), \Gamma )< C_1 \tau (h)\} \\ \Gamma^{12}_{\tau (h)}=\{ \gamma \in \Gamma^{1}_{\tau (h)};\ C_1 \tau (h) \leq \dist ({\rm {supp}}(\chi _{\gamma ,\tau (h)}), \Gamma )< C_{1}^{-1} \} \\ \Gamma^{13}_{\tau (h)}=\{ \gamma \in \Gamma^{1}_{\tau (h)};\ C_{1}^{-1} \leq \dist ({\rm {supp}}(\chi _{\gamma ,\tau (h)}), \Gamma ) \} \; . \end{array} \end{equation} \subsection{ First zone~: $ \Gamma^{0}_{\tau (h)}$ } This was already analyzed in Subsection \ref{Subsection7.3}. We established in (\ref{3est1}), that~: \begin{lemma}\label{BN.LL4}.\\ \begin{equation}\label{ajout2} \sum_{\gamma \in\Gamma^{0}_{\tau(h)}} q^{h}_{A}(\chi_{\gamma ,\tau (h)}u^h)\geq hb \sum_{\gamma \in\Gamma^{0}_{\tau(h)}} \| \chi_{\gamma ,\tau (h)}u \| ^2\;. \end{equation} \end{lemma} \subsection{Second zone~: $ \Gamma^{11}_{\tau (h)}$ } We assume that ${\rm {supp}}(\chi _{\gamma ,\tau (h)}) \; \subset \; Q(x_0)$ with $x_0\in \Gamma$ and $Q(x_0)$ satisfying (\ref{BN.1a}). \\ We begin by establishing a lower bound of the ground state energy of the model operator. We consider the model operator (\ref{BN.6}), (coming from Lemma \ref{LemmaBN.l3}), on $L^2(I\times J\times\rz_{+})$ with $I=J=[- C\, h^{\delta }, C\, h^{\delta }]$, for some $\delta >0$ and $C>0$, and Dirichlet condition on $\pa I \times J\times\rz_{+}\cup I\times \pa J \times \rz_{+}$ and Neumann condition on $I\times J\times \{ 0\} .$ \\ We estimate from below its ground state energy by the ground state energy associated to the periodic conditions on a larger interval containing $J$, for example on $L^2(I\times \Gamma \times \rz_{+})$. We denote in the same way this operator by $P^{h,N,0}.$ Using Parseval formula we get \begin{equation}\label{BN1.l1} \sp (P^{h,N,0})=\cup_{k\in \zz }\sp (P^{h,N,0}_{k}) \end{equation} where, $$ P^{h,N,0}_{k}=h^2D_{t}^{2}+ [\frac{2k\pi }{|\Gamma |}h +tb +\frac{r^2}{2}b\kappa_{n,B}(x_0)]^2 + h^2 D_{r}^{2} $$ is defined on $L^2(I\times \rz_{+})$ with the Dirichlet condition on $\pa I \times \rz_{+}$ and the Neumann condition on $I\times \{ 0\}$ . For simplicity we assume that the length of the curve $\Gamma $ is $2\pi ,\ |\Gamma |= 2\pi $. So \begin{equation}\label{BN1.l2} P^{h,N,0}_{k}=h^2D_{t}^{2} + [kh + tb + \frac{r^2}{2}b\kappa_{n,B}(x_0)]^2+h^2D_{r}^{2} \end{equation} Then we have, for the suitable realizations \begin{equation}\label{BN1.l3} \inf \sp (P^{h,N,0}_{k})\geq \inf \sp (h^2D_{r}^{2}+ \mu^{h}_{k}(r)), \end{equation} where $\mu^{h}_{k}(r)$ is the ground state energy of the Neumann problem on $L^2(\rz_{+})$ \\ associated to $$ P^{h,N,0}_{r,k}=h^2D_{t}^{2}+ [kh + t b + \frac{r^2}{2}b\kappa_{n,B}(x_0)]^2\;. $$ By scaling, we have \begin{equation}\label{BN1.l4} \sp (P^{h,N,0}_{r,k})=hb \times \sp (Q(\xi_{h,k}(r))), \end{equation} with \begin{equation}\label{ksihkr} \xi_{h,k}(r)=-b^{-\frac 12}h^{-\frac 12} [kh+\frac{r^2}{2} b \kappa_{n,B}(x_0)], \end{equation} and $Q(\xi)$ is defined in (\ref{defQs}). Now, as $|r|\leq h^{\delta }$, with $\delta > 1/4$, $$ |\xi_{h,k}(r)-\xi_0|\leq C h^{ - \tau_0 +\frac 16} $$ when $$ k\; \in \; I_0(h) : = [- C h^{ - \tau_0 -\frac 13} - h^{-\frac 12}\xi_0 , -h^{-\frac 12}\xi_0 + C h^{ - \tau_0 -\frac 13}],$$ and $$ |\xi_{h,k}(r)-\xi_0|\geq \frac 1C h^{ - \tau_0 +\frac 16}\;,$$ when $$ |k-h^{-\frac 12}\xi_0|\geq C |h^{ - \tau_0 - \frac 13}\xi_0|\;. $$ Here $\ \tau_0 $ is non negative, choosen small enough, for example \begin{equation}\label{deftau0} \tau_0 = {\rm {min}} \{ \delta - \frac{1}{4} , \ \frac{1}{36} \} \; . \end{equation} So we can use (\ref{BN1.o3}) to get that, for some constant $\ C_0 >0,$ \begin{equation}\label{BN1.la5} Q(\xi_{h,k}(r)))\\ \geq \Theta_0 + \frac{1}{C_0} h^{-2 \tau_0 + \frac 13} , \ {\rm {if}} \ k \notin I_0(h)\; , \end{equation} so, if $ \ k \notin I_0(h)$, then \begin{equation}\label{BN1.lb5} \inf \sp (h^2D_{r}^{2}+ \mu^{h}_{k}(r)) \geq \ h b \Theta _0 + \frac{1}{C_0} h^{-2 \tau_0 + \frac 43}, \ \end{equation} Now, we have just to consider the case when $k \in I_0(h).$ \\ We use also (\ref{BN1.o3}) to get \begin{equation}\label{BN1.lc5} Q(\xi_{h,k}(r)))\\ \geq \Theta_0 + \frac{\mu''(\xi_0)}{2} (\xi_{h,k}(r) - \xi_0)^2 - C h^{- 3 \tau_0 + \frac 12} \; . \end{equation} By using min-max principle, the ground state energy of the Dirichlet problem on $ I $ is less then the one on $\ \rz ,$ and then by scaling we get that \begin{equation}\label{BN1.ld5} \begin{array}{l} \inf \sp (h^2D_{r}^{2}+ \mu^{h}_{k}(r)) \geq - C h^{- 3 \tau_0 + \frac 32} \\ \quad + h \inf \sp \{ b \Theta _0 + \frac{1}{2} h^{\frac 13} (b^2 \mu''(\xi_0) \kappa^{2}_{n,B}(x_0))^{\frac 13} [ D^{2}_{r} + (r^2 - \rho_{h,k})^2 ]\} \; , \end{array} \end{equation} with $$ - \rho_{h,k} = {\sqrt 2} h^{-\frac 16} (b^2 \mu''(\xi_0) \kappa^{2}_{n,B}(x_0))^{-\frac 16} [\xi _0 + h^{\frac 12} \frac{1}{\sqrt 2} k \kappa^{\frac 12}_{n,B}(x_0)] ), $$ and then, if $ k \in I_0(h)$, \begin{equation}\label{BN1.l5} \begin{array}{l} \inf \sp (h^2D_{r}^{2}+ \mu^{h}_{k}(r)) \geq \\ h b \Theta _0 + \frac{1}{2} h^{\frac 43}{\hat \nu}_0 (b^2 \mu''(\xi_0) \kappa^{2}_{n,B}(x_0))^{\frac 13} - C h^{- 3 \tau_0 + \frac 32} \; . \end{array} \end{equation} Then, using a modified version of Lemma \ref{BN.l1}, where $q^{h}_{{\tilde A}^{(2)}}(u)$ and $q^{h}_{A}(u)$ are interchanged, and also Lemma \ref{LemmaBN.l3} (with $q^{h}_{A^{00}}(u)$ and $q^{h}_{A}(u)$ interchanged), we get easily from (\ref{BN1.l1}), (\ref{BN1.l3}), (\ref{deftau0}), (\ref{BN1.lb5}) and (\ref{BN1.l5}), the following lemma \begin{lemma}\label{BN.p1}.\\ Let $\delta \in ]\frac 14, \frac 13[$, $C_0>0$. Then there exists ${\tilde C}_0$ such that, for any $x_0\; \in \; \Gamma$, for any $u \in H^1(\Omega )$ such that $$ {\rm {supp }} (u)\subset \{ x\in {\overline {\Omega }};\ |x-x_0|\leq C_0 h^{\delta }\} \;, $$ for any $h \in ]0,1]$, we have~: \begin{equation}\label{BN.p1.1} \begin{array}{l} q^{h}_{A}(u)\geq [ hb\Theta_0 + h^{\frac 43} \frac{{\hat \nu}_0}{2} [\mu''(\xi_0) \kappa^{2}_{n,B}(x_0)]^{\frac 13}]\| u\|^2\\ - {\tilde C}_0 \| t^{\frac 12 }(hD_x-A)u\|^2 - {\tilde C}_0 [q^{h}_{A}(u) ]^{\frac 12}\cdot \|(h^{3\delta }+h+h^{2\delta }t+t^2) u\| \\ - {\tilde C}_0 \|(h^{\delta + \frac 12}+h^{3\delta } + h + h^{2\delta }t + t^2) u\|^2 \end{array} \end{equation} \end{lemma} This lemma and (\ref{pBN1}) prove that for the ground state eigenfunction defined in (\ref{defuh}), we have, for any $\epsilon >0$, \begin{equation} \begin{array}{l} q^{h}_{A}(\chi_{\gamma ,\tau (h)} u^h)\geq [hb \Theta_0 + h^{\frac 43} {\hat \gamma}_0 b^{\frac 23 }] \| \chi_{\gamma ,\tau (h)} u^h\|^2 \\ - C(h^{1 + \epsilon } +h^{6\delta - \epsilon }) \| \chi_{\gamma ,\tau (h)} u^h\|^2 \\ \qquad - C\| t^{\frac 12} (hD-A) \chi_{\gamma ,\tau (h)} u^h\|^2\\ \qquad - C h^{-\epsilon} || t^2 \, \| \chi_{\gamma ,\tau (h)} u^h\|^2 \;. \end{array} \end{equation} So, choosing $\epsilon = 3 \delta - \frac 12$, we have proved that there exists $C >O$, such that, for any $\ \gamma \in \Gamma^{11}_{\tau (h)} $ defined by (\ref{low5bb}), we have \begin{equation}\label{ajout5z} \begin{array}{l} q^{h}_{A}(\chi_{\gamma ,\tau (h)} u^h)\geq [hb \Theta_0 + h^{\frac 43} {\hat \gamma}_0 b^{\frac 23} - C h^{\frac {17}{12}}]\| \chi_{\gamma ,\tau (h)} u^h\|^2 \\ - Ch^{3\delta + \frac 12 } \| \chi_{\gamma ,\tau (h)} u^h\|^2 - C\| t^{\frac 12} (hD-A) \chi_{\gamma ,\tau (h)} u^h\|^2 - C h^{\frac 12 - 3 \delta} || t^2 \, \chi_{\gamma ,\tau (h)} u^h\|^2 \;. \end{array} \end{equation} Summing over $\gamma \in \Gamma^{11}_{\tau(h)}$, we will obtain \begin{lemma}\label{BN.LL1} .\\ There exists $C >O$, such that, for $\delta \in ]\frac 14,\frac 13[$, there exists $C$ and $h_0$ such that, for all $h\in ]0,h_0]$, \begin{equation}\label{ajout5} \begin{array}{ll} \sum_{\gamma \in \Gamma^{11}_{\tau (h)}} q^{h}_{A}(\chi_{\gamma ,\tau (h)} u^h)& \geq [hb \Theta_0 + h^{\frac 43} {\hat \gamma}_0 b^{\frac 23}] \sum_{\gamma \in \Gamma^{11}_{\tau (h)}} \| \chi_{\gamma ,\tau (h)} u^h\|^2 \\ &\quad - C h^{3\delta + \frac 12} - C h^{\frac{17}{12}}\;. \end{array} \end{equation} \end{lemma} {\bf Proof}.\\ Let $$ r_\gamma^{11}(h):= C h^{3\delta + \frac 12 } \| \chi_{\gamma ,\tau (h)} u^h\|^2 + C\| t^{\frac 12} (hD-A) \chi_{\gamma ,\tau (h)} u^h\|^2 + C h^{\frac 12 - 3 \delta} || t^2 \, \chi_{\gamma ,\tau (h)} u^h\|^2\;. $$ We obtain~: $$ \begin{array}{ll} \sum_{\gamma \in \Gamma^{11}_{\tau(h)} } r_\gamma^{11}(h)& \leq C h^{3\delta + \frac 12 } \sum_{\gamma \in \Gamma^{11}_{\tau(h)}}\| \chi_{\gamma ,\tau (h)} u^h\|^2 \\ &\quad + C h^{\frac 12 - 3 \delta} \sum_{\gamma \in \Gamma^{11}_{\tau(h)}} || t^2 \, \chi_{\gamma ,\tau (h)} u^h\|^2 \\ &\quad + C\sum_{\gamma \in \Gamma^{11}_{\tau(h)}}\| t^{\frac 12} (hD-A) \chi_{\gamma ,\tau (h)} u^h\|^2 \;. \end{array} $$ We use then Proposition \ref{pBN} for the two first terms of the right hand side. The last term is treated in the following way. As for (\ref{low4}), we have also \begin{equation} \| t^{\frac 12} (hD - A) u^h\| ^2 = \sum_{\gamma } [ \| t^{\frac 12} (hD - A)\chi_{\gamma ,\tau (h)} u^h\| ^2 - h^2 \| t^\frac 12\, |\nabla \chi_{\gamma ,\tau (h)}| u^h\| ^2 ] \; . \end{equation} So \begin{equation}\label{ajout8} \sum_{\gamma \in \Gamma^{11}_{\tau(h)} } \| t^{\frac 12} (hD - A)\chi_{\gamma ,\tau (h)} u^h\| ^2 \leq C (h^{ \frac 32} + h^{3 -2\delta})\;. \end{equation} \subsection{ Third zone~: $ \Gamma^{13}_{\tau (h)}$ } We use the proof of Proposition \ref{Lowb}. We observe that we are in the first case considered in the proof of this proposition. Formula (\ref{labellis}) gives that there exists $c_3 >0$ and $h_0 >0$ such that, for any $u\in H^1(\Omega )$ and any $h\in ]0,h_0]$, \begin{equation}\label{ajout3z} q^{h}_{A}(\chi_{\gamma ,\tau (h)}u)\geq hb(\Theta_0 + c_3) \| \chi_{\gamma ,\tau (h)}u \| ^2, \ \ \forall\gamma \in \Gamma^{13}_{\tau (h)} \; . \end{equation} Taking $u= u^h$ and summing over $\gamma \in \Gamma^{13}_{\tau(h)}$, we get \begin{lemma}\label{BN.LL3}.\\ There exists $c_3 >0$ and $h_0 >0$ such that for any $h\in ]0,h_0]$, \begin{equation}\label{ajout3} \sum_{\gamma \in\Gamma^{13}_{\tau(h)}} q^{h}_{A}(\chi_{\gamma ,\tau (h)}u^h)\geq hb(\Theta_0 + c_3) \sum_{\gamma \in\Gamma^{13}_{\tau(h)}} \| \chi_{\gamma ,\tau (h)}u \| ^2\;. \end{equation} \end{lemma} \subsection{Fourth zone~: $ \Gamma^{12}_{\tau (h)}$ } As for the second zone, we assume that~: $ {\rm {supp}}( \chi_{\gamma ,\tau (h)})\ \subset \; Q(x_0)$, but with $x_0=(r_0,0,s_0)$, $r_0\neq 0$. For simplicity of notation, we take $s_0 =0$. We have also to consider the case when $J = [- C\, h^{\delta }, C\, h^{\delta }]$, with $I=r_0+[-h^{\delta }, h^{\delta }]$, but instead of the third zone, with $|r_0|$ small enough. We will also impose $|r_0|\geq C h^\delta$, with $C$ large enough in order to have the property that the corresponding $I(h)$ is always sufficiently far from $0$. We will also estimate from below the ground state energy of this Dirichlet problem by the one associated to the periodic problem on a larger interval containing $J$. \paragraph{Adapted normal forms}.\\ We will make a Taylor expansion at the point $(r_0,0,0)$ instead of $(0,0,0)$.\\ It comes easily from (\ref{taylorH1}-\ref{taylorH3}) that \begin{equation}\label{BN1.lb1} V(B)(x_0)=(b + \beta_{10} r^{2}_{0})\frac{\pa }{\pa r} + \beta_{20} r^{2}_{0}\frac{\pa }{\pa s} + \beta_{30} r_0\frac{\pa }{\pa t} \end{equation} with \begin{equation}\label{BN1.lb2} |\beta_{10}|+ |\beta_{20}|\leq C\;,\; |\beta_{30}+ b \kappa_{n,B}(x_{00})|\leq C |r_{0}| \;, \; \end{equation} and $x_{00} \in \Gamma $, such $s(x_{00})=s(x_0):=0).$\\ Moreover, if we write \begin{equation}\label{BN1.lb3} V(B)(x)={\tilde b}_1(x)\frac{\pa }{\pa r} + {\tilde b}_2(x)\frac{\pa }{\pa s} + {\tilde b}_3(x)\frac{\pa }{\pa t} \; , \end{equation} then it comes also from (\ref{taylorH1}-\ref{taylorH3}) that \begin{equation}\label{BN1.lb4} \begin{array}{l} | \frac{\pa {\tilde b}_1}{\pa r}(x_0)| + | \frac{\pa {\tilde b}_2}{\pa r}(x_0)| + | \frac{\pa {\tilde b}_3}{\pa r}(x_0) + b \kappa_{n,B}(x_{00})| \leq C|r_0| \\ | \frac{\pa {\tilde b}_1}{\pa s}(x_0)| + | \frac{\pa {\tilde b}_2}{\pa s}(x_0)| + | \frac{\pa {\tilde b}_2}{\pa s}(x_0) | \leq C|r_0| \; . \end{array} \end{equation} In the same way, we get from (\ref{cu5}), (\ref{cu6}), the assumption (\ref{BN0bb}) and (\ref{deriveangle}) that \begin{equation}\label{BN1.lbb5} | \alpha (x_0) - 1 |\leq C |r_{0}| \; . \end{equation} So the magnetic field in the coordinates $y=(r-r_0,s,t)$ satisfies \begin{equation}\label{BN1.lbb6} \begin{array}{l} {\widetilde B}_{23}(y) = b + \beta_{10}r^{2}_{0} + \beta_{11}r_0(r-r_0) + \beta_{12}r_0 s + \beta_{13} t + \Og(|y|^2) \\ {\widetilde B}_{31}(y) = \beta_{20}r^{2}_{0} + \beta_{21}r_0 (r-r_0) + \beta_{22}r_0 s + \beta_{23} t + \Og(|y|^2) \\ {\widetilde B}_{12}(y) = \beta_{30}r_0 + \beta_{31}(r-r_0) + \beta_{32}r_0 s + \beta_{33} t + \Og(|y|^2)\; ; \end{array} \end{equation} with \begin{equation}\label{BN1.lbb7} \begin{array}{l} |\beta_{11}| + |\beta_{12}| + |\beta_{21}| + |\beta_{22}| + |\beta_{23}| + |\beta_{32}| + |\beta_{33}| \leq C\;\\ |\beta_{30} + b \kappa_{n,B}(x_{00})| + | \beta_{31} + b \kappa_{n,B}(x_{00})| \leq C | r_{0}| \; . \end{array} \end{equation} Then we can after a change of gauge get $ {\widetilde {A}}(y)$ such that \begin{equation} \label{BN1.lb6} \begin{array}{l} |{\widetilde {A}}_2 + bt - [\beta_{30}r_0(r-r_0) + \frac{1}{2} \beta_{31}(r-r_0)^2] |\\ + |{\widetilde {A}}_1|+|{\widetilde {A}}_3| \\ \leq C\, [t(r^{2}_{0} + |y|)\,+\, |r_0|\cdot |y|^2], \end{array} \end{equation} Here we observe that a term like $\Og(|y|^3)$ is controlled by $C r_0 |y|^2$.\\ We observe indeed that, with $$ { A}^{00} = ( 0, -bt + \beta_{30}r_0(r-r_0) + \frac{1}{2} \beta_{31}(r-r_0)^2, 0)\;,$$ we have $$\curl A^{00} = (b, 0, \beta_{30}r_0 + \beta_{31}(r-r_0))\;. $$ This leads to~: $$ {\tilde B} - \curl { A}^{00} = (\beta_{10}r^{2}_{0} , \beta_{20} r^2_0,0) + |r_0| \Og (y) + \Og (|y|^2)\;. $$ If we introduce ${\tilde A}^1 = { A}^{00} + (\beta_{20} r_0^2 t, - \beta_{10} r_0^2 t, 0)$, we get~: $$ {\tilde B} - \curl {\tilde A}^1 = \Og (|y|^2)\;. $$ We then define~: $$ \widetilde A (y)= {\tilde A}^1 (y) - \int_{0}^{1 }\;\left[ y \wedge ({\tilde B}(s y)-\curl {\tilde A}^1(sy ))\right]\; s \, ds\;, $$ which has the right property. We refer to \cite{Tha} (Formula (8.153)) for this standard formula which is called the transversal gauge (or Poincar\'e gauge). \paragraph{Comparison}.\\ As for Lemma \ref{LemmaBN.l3}, we get from (\ref{BN1.lb6}) the following lemma. \begin{lemma}\label{BN.l4}.\\ If (\ref{BN.1}) is satisfied, then \begin{equation}\label{BN1.l4.1} \begin{array}{l} q^{h}_{A}(u)-C[\|\;( |r_0| + t^{\frac 12})(hD_x-A)u\|^2+ h^{\delta }\|(hD_x-A)u\|^2] \\ -C[ q^{h}_{A^{00}}(v)]^{\frac 12}\cdot \|(h^{2\delta }|r_0|+h+(h^{\delta }+r^{2}_{0}) t+t^2) u\| \\ -C\|(h^{2\delta }|r_0|+h+(h^{\delta }+r^{2}_{0}) t+t^2) u\|^2 \\ \leq q^{h}_{A^{00}}(v)\\ \leq q^{h}_{A}(u)+ C[\| (|r_0| + t^{\frac 12})\, (hD_x-A)u\|^2+ +h^{\delta }\|(hD_x-A)u\|^2] \\ + C [q^{h}_{A}(u) ]^{\frac 12}\cdot \|(h^{2\delta }|r_0|+h+(h^{\delta }+r^{2}_{0}) t+t^2) u\| \\ + \|(h^{2\delta }|r_0| + h + (h^{\delta } + r^{2}_{0}) t+t^2) u\|^2 \end{array} \end{equation} where \begin{equation}\label{quasim.l4} v=(\alpha (r,s))^{\frac 12}e^{-ip^0/h} u\;, \end{equation} for some smooth real function $p^0$, \begin{equation}\label{BN1.l4.2} q^{h}_{A^{00}}(v)=\int_{Q(x_0)} [|hD_tv|^2+ |(hD_s-A^{00}_{2})v|^2+ |hD_rv|^2]\;dr ds dt\;, \end{equation} \begin{equation} A^{00}_{2} (r,t):= -b t + \beta_{30}r_0 (r -r_0) +\frac{1}{2} \beta_{31}(r-r_0)^2 \;, \end{equation} and $r_0=r(x_0)$. \end{lemma} We recall that the constants $\beta_{30}$ and $\beta_{31} $ satisfy (\ref{BN1.lbb7}), so by the assumption (\ref{BN0ba}), there exists $C_0>0$, such that, for $|r_0|\leq \frac{1}{C_0}$, \begin{equation}\label{BN1.lb6a} C_{0}^{-1}\leq |\beta_{30}| \;. \end{equation} \paragraph{Lower bound for the model}.\\ By translation, scaling and expanding in Fourier series, we get easily that in (\ref{BN1.l4.2}), we have \begin{equation}\label{BN1.lb7} q^{h}_{A^{00}}(v)\geq hb (\inf_{k\in \zz }\inf \sp (H^{h,k}))\times \| v\|^2, \end{equation} where $H^{h,k}$ is the selfadjoint operator on $L^2(I(h)\times \rz_{+})$ defined by imposing a Dirichlet condition on $\pa I(h)\times \rz_{+}$ and a Neumann condition on $I(h)\times \{ 0\} $, and by considering the differential operator \begin{equation}\label{BN1.lb8} H^{h,k} w = D^{2}_{r} w +D^{2}_{t} w + (t-\xi_{h,k}(r))^2 w, \ \forall v\in C^{\infty }_{0}(I(h)\times \rz_{+})\;. \end{equation} Here $I(h)=]-h^{\delta -\frac 12}b^{\frac 12},h^{\delta -\frac 12}b^{\frac 12}[$ and \begin{equation}\label{BN1.lb9} \xi_{h,k}(r)=a_0 h^{\frac 12} r^2 + 2 d_0 r - h^{\frac 12} b^{-\frac 12} k \; ; \end{equation} with $\ a_0 := - \frac{1}{2} b^{-3/2} \beta_{31}$ and $\ d_0 := - \frac{1}{2} b^{-1} \beta_{30} r_0 .$ \\ So , by (\ref{BN1.lbb7}) \begin{equation}\label{BN1.lbb9} C^{-1} \leq | a_0 | \leq C ,\ \ {\rm {and}} \ C^{-1} | r_0 |\leq | d_0 | \leq C | r_0 | \; . \end{equation} Then we get that, for the Dirichlet realizations in $I(h)$ \begin{equation}\label{BN1.lb10} \inf \sp (H^{h,k}) \geq \inf \sp \left(D^{2}_{r} + \mu (\xi_{h,k}(r))\right) \;. \end{equation} \begin{lemma}\label{BN.l5}.\\ Let $\delta \in ]0, \frac 13[$. There exist some constants $ C_0\geq 1$ and $c_1 >0$ such that, for all $ (h,k)\; \in ]0,1[\times \zz$, and $r_0$ s.t. \begin{equation}\label{BN.l5.1} |r_0|\in [C_0h^{\delta },\frac{1}{C_0}] \;, \end{equation} we have, for the realization in $I(h)$~: \begin{equation}\label{BN.l5.2} \inf \sp (D^{2}_{r}+\mu (\xi_{h,k}(r))) \; \geq \; \Theta_0 + c_1 |r_0|\,. \end{equation} \end{lemma} This lemma comes easily, after a translation in the $r$ variable, from the following one. \begin{lemma}\label{LemmaBN.l5bis}.\\ There exist strictly positive constants $c_0$ and $c_1$ such that, for any $\zeta\in \rz$, for any $\gamma \neq 0$, the Dirichlet realization of the operator~: $$ P_{\zeta,h}:= h D_r^2 + \mu ( \zeta + \frac{\gamma r^2}{h^\frac 12}) - \Theta_0\;, $$ in an open interval $I(h)$ such that~: $$ I(h) \subset ] 0, +\infty[ \cup ]- \infty, 0[\;, $$ we have~: $$ \langle P_{\zeta,h}u| u \rangle \geq c_0 (\inf_{r\in I(h)} \min(|\gamma r|, c_1)\;) || u ||^2\;,\; \forall u\in C_0^\infty(I(h))\;. $$ \end{lemma} {\bf Proof}.\\ The proof is based on a bracket argument. According to the properties of $\mu$ described in Subsection \ref{Subsection2.2}, we can find a regular increasing function $f$ such that~: $$ \mu (t) - \Theta_0 = \left(f (t-\xi_0)\right)^2\;, $$ with~: $$ f(0) =0\;,\; f'(t)> 0\;,\; \lim_{t\ar -\infty} f(t) =-\infty\;,\;\lim_{t\ar \infty} f(t) =1\;. $$ Using the identity~: $$ [ h^\frac 12 D_r\;,\; f(\frac{r^2}{h^\frac 12}+\zeta -\xi_0) ]= \frac 2 i \gamma r f'(\frac{r^2}{h^\frac 12}+\zeta - \xi_0)\;, $$ we observe\footnote{ This is the estimate~: $$ || L_1 u||^2 + || L_2 u|| ^2 \geq \pm i \langle [L_1,L_2] u\;, \; u\rangle\;. $$ } that, according to the assumption that $I(h)$ avoids $0$ $$ \langle P_{\zeta,h}u| u \rangle \geq 2 \langle |r| |\gamma|f'(\frac{r^2}{h^\frac 12}+\zeta -\xi_0) u\;,\; u \rangle \;. $$ and we add the trivial estimate~: $$ \langle P_{\zeta,h}u| u \rangle \geq \langle f^2(\frac{r^2}{h^\frac 12}+\zeta - \xi_0)u\;,\; u\rangle \;. $$ This leads to $$ \begin{array}{ll} \langle P_{\zeta,h}u| u \rangle & \geq \frac 12 \inf_{r\in I(h)} \left( 2 |r||\gamma| f'(\frac{r^2}{h^\frac 12}+\zeta - \xi_0) + f^2(\frac{r^2}{h^\frac 12}+\zeta - \xi_0) \right) || u||^2\\ &\geq \inf_{r\in I(h)} ( |\gamma r|, \frac 12) \inf_{s\in \rz} (f'(s) + f(s)^2) || u||^2 \;. \end{array} $$ We can take $c_1= \frac 12 $ and the lemma will consequently follow from~: $$ \inf_{s\in \rz} (f'(s) + f(s)^2) >0\;. $$ But the proof is easy. \vskip 0.5cm \noindent Using Lemma \ref{BN.l4} and (\ref{BN.l5.2}), we get the following lemma \begin{lemma}\label{BN.p2}.\\ Let $\delta \in \; ]0,\frac{1}{3}[$, $C_0 >0$. Then there exists $c_0 >0$, such that, for any $C_1 >0$, there exists $C >0$, such that, for any $x_0 \in \pa \Omega$ such that \begin{equation}\label{BN.p2.1} \rho_0:=d(x_0,\Gamma )\in [C_1h^{\delta },\frac{1}{C_1}]\;, \end{equation} for any $u$ in $H^1(\Omega )$ supported in $ \{ x\in {\overline {\Omega }};\ |x-x_0|\leq C_0 \, h^{\delta }\} $, for any $h\in ]0,1]$, we have \begin{equation}\label{BN.p2.2} \begin{array}{l} q^{h}_{A}(u)\geq [ hb\Theta_0 + h c_0\rho_0]\| u \|^2\\ -C [\| (\rho_0 + t^{\frac 12})(hD_x-A)u \|^2+h^{\delta }\| (hD_x-A)u \|^2]\\ -C [q^{h}_{A}(u) ]^{\frac 12}\cdot \|(h^{2\delta }\rho_0+h+(h^{\delta }+\rho^{2}_{0}) t+t^2) u\| \\ -C \|h^{2\delta }\rho_0 + h + (h^{\delta }+\rho^{2}_{0}) t + t^2) u \|^2 \; \end{array} \end{equation} \end{lemma} We first deduce~: \begin{equation}\label{BN.p2.2a} \begin{array}{ll} (1 + C \rho_0^2 + C h^{\delta}) q^{h}_{A}(u)& \geq [ hb\Theta_0 + h c_0\rho_0]\| u \|^2\\ &\quad -C \| t^{\frac 12})(hD_x-A)u \|^2 \\ &\quad -C [q^{h}_{A}(u) ]^{\frac 12}\cdot \|(h^{2\delta }\rho_0+h+(h^{\delta }+\rho^{2}_{0}) t+t^2) u\| \\ &\quad -C \|h^{2\delta }\rho_0 + h + (h^{\delta }+\rho^{2}_{0}) t + t^2) u \|^2 \; \end{array} \end{equation} \noindent The mixed term $[q^{h}_{A}(u) ]^{\frac 12}\cdot \|(h^{2\delta }\rho_0+h+(h^{\delta }+\rho^{2}_{0}) t+t^2) u\|$ is to analyze carefully and we analyze each of the terms appearing in its expression.\\ \noindent Let us first consider the term~: $$ T_1 = [q^{h}_{A}(u) ]^{\frac 12}\cdot \|(h^{2\delta }\rho_0 u\|\;. $$ We use the estimate~: $$ T_1 \leq D^{-1} \rho_0 q^{h}_{A}(u) + D \rho_0^{-1} \|h^{2\delta }\rho_0 u ||^2\;, $$ with the idea that we will have to choose $D$ large enough. We have indeed $$ D \rho_0^{-1} \|h^{2\delta }\rho_0 u ||^2 = \rho_0 h h^{4 \delta -1} D ||u||^2\;. $$ This is OK, for $h$ small enough, under the condition~: $\delta > \frac 14$. We can for example take $ D = h^{\frac 12 -2\delta}$. \noindent Let us now consider the second term~: $$ T_2 = h^{\delta } \; [q^{h}_{A}(u) ]^{\frac 12}\cdot \| t u\| \;. $$ Here we write~: $$ T_2 \leq D^{-1} \rho_0 q^{h}_{A}(u) + D \rho_0^{-1} h^{2 \delta} || tu ||^2\;, $$ where $D$ has to be large, and we observe that~: $$ D \rho_0 |\rho_0^{-1} h^{\delta}|^2 || t u||^2 \leq C D \rho_0 C_1^{-2} || tu||^2\;. $$ This is essentially OK, if $C_1$ is large enough because the term $|| tu||^2$ is hoped to be of order $\Og(h)$. But we have to be careful with this point, which will be treated later. The third term to consider is~: $$ T_3 = [q^{h}_{A}(u) ]^{\frac 12}\cdot || \rho^{2}_{0} t u\| \;. $$ We can use~: $$ T_3 \leq D^{-1} \rho_0 q^{h}_{A}(u) + (D \rho_0^{2}) \rho_0 || tu ||^2\;, $$ with again $D$ large. The other terms are smaller or controlled in the same way. . We can for example use that $\Og (t^2)$ is controlled by $\rho_0 \Og(|t|)$ on the support of $u$. This implies the following lemma: \begin{lemma}\label{BN.LL2} .\\ There exists $\hat c_0 > 0$, $C$ and $C_1^0$, such that, if $C_1 $ in (\ref{low5bb}) satisfies $ C_1\geq C_1^0$ then, for any $ u\; \in \; H^1(\Omega )$, \begin{equation}\label{ajout4} \begin{array}{ll} q^h_A(\chi_{\gamma ,\tau (h)}u)& \geq h (b\Theta_0 + \rho_0{\hat c}_0 )\, \| \chi_{\gamma ,\tau (h)}u \| ^2\\ &\quad - C \, \| t^{\frac 12} (hD_x-A) \chi_{\gamma ,\tau ( h)} u \|^2 \\ &\quad - \frac{C }{C_1} \rho_0 \, || t \chi_{\gamma ,\tau (h)} u \|^2 \;, \end{array} \end{equation} for all $\gamma \in \Gamma^{12}_{\tau (h)}$. \end{lemma} We plan to use this estimate with $u=u^h$ and to sum over the $\gamma$'s in $\Gamma^{12}_{\tau(h)}$. Here we meet a problem with the term $\rho_0 \sum_\gamma || t \chi_\gamma u^h||^2$. Proposition \ref{pBN} can indeed gives that~: $$ \frac {C}{C_1} \rho_0 \sum_\gamma || t \chi_\gamma u^h||^2 \leq \frac {C}{C_1} \rho_0 || t u^h||^2 \leq \frac {C'}{C_1} \rho_0 h\;, $$ but this error is much too large and the fact that we can increase $C_1$ is not enough. One could think that one could get instead the better~: $$ \frac {C}{C_1} \rho_0 \sum_{\gamma \in \Gamma^{12}_{\tau (h)}} || t \chi_\gamma u^h||^2 \leq \frac {C'}{C_1} \rho_0 h \sum_{\gamma\in \Gamma^{12}_{\tau (h)}} || \chi_\gamma u^h||^2\;, $$ but this is probably not true. We will actually show the weaker~: \begin{equation} \frac {C}{C_1} \rho_0 \sum_{\gamma \in \Gamma^{12}_{\tau (h)}} || t \chi_\gamma u^h||^2 \leq \frac {C'}{C_1} \rho_0 h \sum_{\gamma\in \Gamma^{12}_{\tau (h)}} || \chi_\gamma u^h||^2 + \Og (h^{2-2\delta})\;. \end{equation} In order to get this more accurate upper bound, we need to get a~: \paragraph{Local control of $|| t \chi u^h ||$}.\\ We follow the proof of Proposition \ref{pBN}, but we have to take account of the presence of $\chi$. One first obtains (see (\ref{pBN4}) ) that~: $$ \begin{array}{ll} hb\int t^2\chi ^2|u^h|^2 & \leq \int t^2 |(hD-A)\chi u^h|^2 \\ &\quad +C h \| (hD-A) \chi u^h\| \cdot \| t\chi u^h\|\;. \end{array} $$ Commuting $\chi$ and $(hD-A)$ in the second term of the right hand side, we get~: \begin{equation}\label{estloc1} \begin{array}{ll} hb\int t^2\chi ^2|u^h|^2 & \leq \int t^2 |(hD-A)\chi u^h|^2 \\ &\quad +C h \| \chi (hD-A)u^h\| \cdot \| t\chi u^h\| \\ & \quad + C h \| t \, |\nabla \chi |\, (hD-A)u^h\| \cdot \| t\chi u^h\|\;. \end{array} \end{equation} Let us now analyze $\int t^2 |(hD-A)\chi u^h|^2$. Starting from~: $$ \int t^2 |(hD-A)\chi u^h|^2 = \int t^2 \left( \chi (hD-A)u^h + h \nabla \chi u^h\right)\cdot \left(\overline{\chi (hD-A)u^h + h \nabla \chi u^h}\right)\;, $$ we get~: $$ \begin{array}{ll} \int t^2 |(hD-A)\chi u^h|^2 & \leq \int t^2 \chi^2 (hD-A)^2 u^h \overline{u^h}\\ &\quad + h^2 \int t^2 |\nabla \chi|^2 |u^h|^2\\ &\quad + 2 h || t \chi u^h||\; || t (\nabla \chi)\cdot (hD-A) u^h||\\ &\quad + C h || t \chi u^h||\; || \chi (hD-A) u^h|| \end{array} $$ This leads to $$ \begin{array}{ll} \int t^2 |(hD-A)\chi u^h|^2 & \leq \lambda(h) \int t^2 |\chi u^h|^2 \\ &\quad + h^2 \int t^2 |\nabla \chi|^2 |u^h|^2\\ &\quad + 2 h || t \chi u^h||\; || t (\nabla \chi)\cdot (hD-A) u^h||\\ &\quad + C h || t \chi u^h|| \; || t \chi (hD-A) u^h|| \end{array} $$ We then get~: $$ \begin{array}{ll} (hb - \lambda(h)) \int t^2 |\chi u^h|^2 & \leq h^2 \int t^2 |\nabla \chi|^2 |u^h|^2 \\ &\quad + C h || t \chi u^h||\; || t \chi (hD-A) u^h||\\ & \quad + C h \| t \, |\nabla \chi |\, (hD-A)u^h\| \cdot \| t\chi u^h\|\;. \end{array} $$ This leads to the estimate~: $$ \| t\chi u^h\|^2 \leq C [\, \| \chi (hD-A)u^h\| + \| t |\nabla \chi | (hD-A)u^h\|\,]\; \| t\chi u^h\| + h \int t^2 |\nabla \chi|^2 |u^h|^2 \; . $$ and consequently to \begin{equation}\label{est111} \| t\chi u^h\|^2 \leq C [\, \| \chi (hD-A)u^h\|^2 + \| t |\nabla \chi | (hD-A)u^h\|^2 + h \int t^2 |\nabla \chi|^2 |u^h|^2 \, ] \; . \end{equation} On the other hand, we have~: $$ \| \chi (hD-A)u^h\| ^2\leq \lambda (h) \| \chi u^h\| ^2 + 2 \| \chi (hD-A)u^h\| \cdot \| \, |\nabla \chi | \, u^h\| $$ and consequently to~: \begin{equation}\label{est112} \| \chi (hD-A)u^h\| ^2\leq C\left[ h \| \chi u^h\| ^2 + h^2 \| \, |\nabla \chi |\, u^h\| ^2\right]\; . \end{equation} Finally the two estimates (\ref{est111}) et (\ref{est112}) give \begin{equation}\label{est113} \| t\chi u^h\| ^2 \leq C \left[ h \| \chi u^h\| ^2 + h^2 \| \, |\nabla \chi | \, u^h\| ^2 +h\| t \, |\nabla \chi | \, u^h\| ^2 + \| t \,|\nabla \chi |\, (hD-A)u^h\| ^2\right] \; . \end{equation} We now take $\chi = \chi_\gamma$ and sum over $\gamma \in \Gamma^{12}_{\tau (h)}$. Using Proposition \ref{pBN}, we get, with \begin{equation} r_\gamma^1 := h^2 \| \, |\nabla \chi_\gamma | \, u^h\| ^2 +h \| t \, |\nabla \chi_\gamma | \, u^h\| ^2 + \| t \,|\nabla \chi_\gamma |\, (hD-A)u^h\| ^2 \end{equation} \begin{equation} \sum_{\gamma \in \Gamma^{12}_{\tau(h)} } r_\gamma^1 \leq C h^{2-2 \delta}\;. \end{equation} We can treat the second error term in (\ref{ajout4}) as in the second step. This implies the following lemma: \begin{lemma}\label{BN.LL2new} .\\ There exists $\tilde c_0 > 0$, $C$ and $C_1^0$, such that, if $C_1 $ in (\ref{low5bb}) satisfies $ C_1\geq C_1^0$ then \begin{equation}\label{ajout4a} \begin{array}{ll} q^h_A(\chi_{\gamma ,\tau (h)}u)& \geq h \left(b\Theta_0 + \rho_0{\tilde c}_0 \right) \| \chi_{\gamma ,\tau (h)}u^h \| ^2\\ &\quad - r^{12}_\gamma(h) \;, \end{array} \end{equation} for all $\gamma \in \Gamma^{12}_{\tau (h)}$, with $$ \sum_{\gamma \in \Gamma^{12}_{\tau (h)}} r^{12}_\gamma \leq C ( h^{2-2\delta} + h^\frac 32)\;.$$ \end{lemma} Because $\delta$ will be chosen in $]\frac 14,\frac 13[$. The last remainder is smaller. \subsection{End of the proof} Then (\ref{ajout1}), (\ref{ajout3}), (\ref{ajout2} and (\ref{low4}) prove that, for $\delta \in ]\frac{5}{18},\frac 13[$, there exists $C$ and $h_0$ such that, for all $h\in ]0,h_0]$, \begin{equation} \lambda (h) \geq [hb\Theta_0 + h^{\frac 43} {\hat \gamma }_0b^{\frac 23} ] - C (h^{2 - 2\delta } + h^{3\delta +\frac 12} ) - C h^{\frac {17}{12}}\; . \end{equation} Then, the lower bound in Theorem \ref{BN} is obtained by taking for example $\delta = \frac {3}{10}$. This leads to~: \begin{equation}\label{ajout9} \lambda (h) \geq [hb\Theta_0 + h^{\frac 43} {\hat \gamma }_0 b^{\frac 23} ] - C h^{\frac 75}\; . \end{equation} \section{Conclusion}\label{Section15} In this article, we have established a conjecture extending Bernoff-Sternberg conjecture in the case of dimension $3$. We have proved the upper bound part in full generality and established the lower bound part in a special case. 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