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Schroedinger, semiclassical, magnetic field, Neumann problem,  
eigenvalue, superconductivity,  
Schroedinger, semiclassical, magnetic field, Neumann problem
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\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 $ (<X|X>)^{\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<j\leq 3}B_{ij}(x)dx_i\wedge dx_j,\ \ 
B_{ij}(x)=\frac{\pa A_j}{\pa x_i}(x)-\frac{\pa A_i }{\pa x_j} (x)\;. 
\end{equation} 
If needed $B_{ij}$ is extended as an antisymmetric matrix. 
The intensity of the magnetic field is the non negative function 
\begin{equation}\label{magint} 
|B|(x)=\sup \{ |B_x(X,Y)|;\ X,\; Y\; \in \; T_x{\overline {\Omega }},\ 
|X|=|Y|=1\} , \ \ \forall \; x\; \in \; {\overline {\Omega }}. 
\end{equation} 
One associates to the magnetic field $B$,  a vector field 
$ \vec{H}:=V(B)\; \in \; T{\overline {\Omega }}$ by 
\begin{equation}\label{defvecmg} 
B_x(X,Y)= \omega_3 (X,Y,V_x(B)),\ \ \forall \; X,\ Y\; \in \; T_x{\overline {\Omega }}\;. 
\end{equation} 
In the initial  coordinates, we have~: 
\begin{equation}\label{defvecmg2}  
V(B)=\sum _{j=1}^{3}b_j(x)\frac{\pa }{\pa x_j},\ \with \ 
b_1=B_{23},\ b_2=B_{31} \ {\rm {and}} \ b_3=B_{12}\;. 
\end{equation} 
Then the  magnetic intensity satisfies~:
\begin{equation}\label{magin}
 |B|=|V(B)|\;.
\end{equation}
We associate  to the magnetic potential $A$  the 
sesquilinear form on $H^1(\Omega)$~:
\begin{equation}\label{defqA} 
u\mapsto q^{h}_{A}(u)=\int _{\Omega }|{\sqrt {-1}}hdu+uA|^2\, dx^3\;. 
\end{equation} 
Here $ h\; \in \; ]0,+\infty [$ is a parameter. 
The associated differential operator is the magnetic  Laplacian 
which is  given in the standard coordinates by  
\begin{equation}\label{defP} 
P^{h}_{A}=(hD-A)^2=\sum _{j=1}^{3}(hD_{x_j}-A_j(x))^2\;. 
\end{equation} 


\subsection{Local coordinates}\label{corgener}  

If $y=(y_1,y_2,y_3)$ are new local coordinates, that is,  if 
$\Vg _{x_0}$ is a neighborhood of some point 
$ x_0\; \in \; \Omega$ and if $x\mapsto y=\Phi(x)$ is a diffeomorphism of $
\Vg _{x_0}$ onto $\Og \subset \rz^3$,  then  
in the new basis
$
(\frac{\pa }{\pa y_1}, \frac{\pa }{\pa y_2}, \frac{\pa }{\pa y_3})$ 
of $ T\Vg _{x_0}$, the standard metric $ g_0$ and 
the volume $\omega_3$ becomes 
\begin{equation}\label{metric} 
g_0=\sum _{1\leq i,j\leq 3}g_{ij}dy_i\otimes dy_j\;,
\end{equation}
and
\begin{equation}\label{volum} 
\omega_3 =|g|^{\frac 1 2 }dy_1\wedge dy_2\wedge dy_3, 
\end{equation}
with  
$$
\begin{array}{ll}
g_{ij} & =
<\frac{\pa x}{\pa y_i}|\frac{\pa x}{\pa y_j}> \;,\\
  <X|Y>& =\sum _{1\leq i,j\leq 3}g_{ij}
{\widetilde {X}}_i{\widetilde {Y}}_j\;, \\
|g| &= \det( g_{ij})\;,
\end{array}
$$ 
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 i<j\leq 3}{\widetilde {B}}_{ij}dy_i\wedge dy_j\ \ 
\with \ {\widetilde {B}}_{ij}=\frac{\pa {\widetilde {A}}_j}{\pa y_i} 
-\frac{\pa {\widetilde {A}}_i}{\pa y_j}. 
\end{equation} 
So the vector magnetic field becomes  
\begin{equation}\label{newvecmag} 
V(B)=\sum _{j=1}^{3}{\widetilde {b}}_j\frac{\pa }{\pa y_j}\;,
\end{equation}
with 
\begin{equation}\label{newvecmaga}
 {\widetilde {b}}_1=|g|^{-\frac 1 2 }{\widetilde {B}}_{23}\;,\; 
{\widetilde {b}}_2  =|g|^{-\frac 1 2 } {\widetilde {B}}_{31}\;,\; {\widetilde {b}}_3=|g|^{-\frac 
1 2 }  
{\widetilde {B}}_{12}\;,
\end{equation} 
and the intensity of the magnetic field  becomes~:
\begin{equation}\label{newvecmmagb}
 |B|=( \sum _{i,j}g_{ij}{\widetilde {b}}_i{\widetilde {b}}_j)^{\frac 1 2 }\;.
\end{equation}
The sesquilinear form takes the following form, for $u$ supported in $\Vg_{x_0}$,  
\begin{equation}\label{qAb} 
q^{h}_{A}(u)=\int _{\Vg _{x_0}}|g|^{\frac 1 2 }\sum _{1\leq i,j\leq 3}g^{ij}
[hD_{ y_i}u-{\widetilde {A}}_iu]\times 
{\overline {[hD_{y_j}u-{\widetilde {A}}_ju]}}\; dy^3\;, 
\end{equation} 
 and  the associated differential 
operator is 
\begin{equation}\label{Pb} 
P^{h}_A= |g|^{-\frac 1 2 }\sum _{1\leq i,j\leq 3}(hD_{y_j}-{\widetilde {A}}_j)
\cdot  |g|^{\frac 1 2 } g^{ij} (hD_{ y_i} - {\widetilde {A}}_i )\;. 
\end{equation} 
Here $g^{ij}$ is the inverse matrix of the matrix $g_{ij}$.


\subsection{Adapted local coordinates near the boundary}\label{corbord2} 

Let $ \phi (x)=(y_1,y_2)$ be local coordinates on the boundary and $G$ the induced metric by 
$g_0$ on $\pa \Omega$ in these coordinates.. 
Then for $ \epsilon >0$ 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-(<X,Y>)^2\;,\\
 <X|X\wedge Y>=<Y| X\wedge Y>=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} =<V(B)|\frac{\pa }{\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} = <V(B)|\frac{\pa }{\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} = <V(B)|\frac{\pa }{\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
$$ 
\frac{\pa {\widetilde {b}}_3}{\pa r}
=\frac{\pa }{\pa r}<V(B)|N>
=<V(B)|\frac{\pa N}{\pa r}>\;,
$$
 when $ t=0$. \\
\noindent Then, using also (\ref{BN.l2.1}), (\ref{BN.l2.5bis}) and (\ref{secondfa}), we obtain~:
$$
<V(B)|\frac{\pa N}{\pa r}> = b \cos \theta(s) <\frac{\pa}{\pa r}\;|\;\frac{\pa N}
{\pa r}> + b \sin \theta (s)  <\frac{\pa}{\pa s}\;|\;\frac{\pa N}
{\pa r}>\;.
$$
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}} (|<T (x_0) | \frac H b>| )\;,\\
\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)=
<H |\frac{\pa N}{\pa r} (x_0)>= - b K_{11}(x_0),
\ \frac{\pa {\widetilde b}_3}{\pa s} (0)=
<H |\frac{\pa N }{\pa s} (x_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. 
 What remains is consequently to prove the lower bound in the conjecture in the general case. As 
in \cite{HeMo2}, 
 we need also to describe the applications to the localization of the ground states which was 
mentioned in the introduction as our main
 motivation. We hope to come back to these points in a near future \cite{HeMo4}.
\vskip 1.cm



{\bf Acknowledgements.}\\
The first author  would like to thank X-B.~Pan for transmitting his preprint 
\cite{PaKw}  before publication.  He is also 
greatful
 to the organizers of the conferences or workshops 
 in Goa (December 2000),  Cardiff (July 2001) and  Stockholm
 (August 2001)  where some of these results were presented.


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\newblock 
Semi-classical analysis for the Schr\"odinger operator with magnetic wells,
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\newblock Magnetic bottles in connection with superconductivity.
\newblock To appear in Journal of Functional Analysis (2001).
 
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\newblock Magnetic bottles in connection with superconductivity:
Case of dimension 3.
\newblock April 2001 (submitted)

 
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\newblock Magnetic bottles  for the Neumann problem: Curvature effects in the case of dimension 
$3$.
 Part II. Lower bounds and localization.
\newblock In preparation.


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\newblock Upper critical field and location of surface nucleation of superconductivity.
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 Henri Poincar\'e (Section Analyse Non Lin\'eaire).


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Multiple wells in the semiclassical limit I.\\ 
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\newblock Eigenvalue problems of Ginzburg-Landau operator in bounded
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\bibitem[{LuPa3}]{LuPa3} K.~Lu, X-B.~Pan~:
\newblock Gauge invariant eigenvalue problems on $\rz^2$ and $\rz^2_+$.
\newblock Trans. AMS, 352 (3), 2000, p.~1247-1276.

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\newblock Ginzburg-Landau system and surface nucleation of superconductivity.
\newblock Proceeding of the IMS Workshop on Reaction-Diffusion systems.
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\bibitem[{LuPa5}]{LuPa5} K.~Lu, X-B.~Pan~:
\newblock Surface nucleation of superconductivity in $3$-dimension.
\newblock J. of Differential Equations 168(2)  (2000), p. 386-452.

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\newblock Upper critical for superconductors with edges and corners.
\newblock Preprint 2000. Submitted.


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\newblock Schr\"odinger operators with
non-degenerately vanishing magnetic fields in bounded domain.
\newblock Preprint. To appear in Transactions of the AMS.

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\end{thebibliography}


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