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Maxwell, eigenvalues, spectrum, periodic, crystal, Schroedinger
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\markright{A. Morame: Absolute Continuity of the Spectrum of Maxwell Operator} 

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\title{The Absolute Continuity of the Spectrum of Maxwell Operator in a Periodic Media}

\author{Abderemane MORAME} 

\date{}
 
 


\begin{document} 

\maketitle 

%adress 
\begin{center} University of Nantes,
  Dpt. Math\'ematiques, 
UMR 6629 du CNRS,\\ 
 B.P. 99208, 44322 Nantes Cedex 3, (FRANCE), \\ 
E.Mail: morame@math.univ-nantes.fr 
\end{center}
 
\begin{abstract} 
We prove that the spectrum of Maxwell operator, \\ 
$(B,E)\mapsto (-rot(E),a(x)rot(b(x)B),$ with non negative and periodic functions $a(x)$ and $b(x),$ 
is absolutely continuous.  
\end{abstract}

{\bf {Key words:}} Maxwell, eigenvalues, periodic, spectrum, crystal, Schr\"odinger 

{\bf {Mathematics Subject Classifications 1991:}} 35P05, 35Q60, 78A25, 78A50, 81Q10



\section{Introduction and main result } \label{intro}

The propagation of waves in  photonic crystals is subject to many interests in physics, applied 
mathematics and in mathematical physics, see for example the introduction and the bibliography  
of  \cite{FiKl} and \cite{FiKu}.  These waves are governed by 
the Maxwell equations in a periodic media, (see \cite{FiKl}), which can be written as follows 
\begin{equation}\label{maxeq} 
{\partial \over {\partial t}}B(t,x)=-rot(E(t,x)), \ \ \ \ 
{\partial \over {\partial t}}E(t,x)=a(x)rot(b(x)B(t,x)), 
\end{equation} 
$$ div(B(t,x))=0,\ \ \ \ div({1\over {a(x)}}E(t,x))=0.$$ 
if $rot(u(x))=\displaystyle ({\partial \over {\partial x_2}}u_3-{\partial \over {\partial x_3}}u_2, 
{\partial \over {\partial x_3}}u_1-{\partial \over {\partial x_1}}u_3, 
{\partial \over {\partial x_1}}u_2-{\partial \over {\partial x_2}}u_1)$ 
is the rotational of the vector field $\ u$ of $\ \bR ^3,$ and 
$ \ div(u)=\displaystyle \sum _{j=1}^{3}{\partial \over {\partial x_j}}u_j$  its divergent.

$\ a$ and $\ b$ are non negative functions which are $\ \Gamma $-periodic, 
\begin{equation}\label{hyp1} 
 a(x-c)=a(x),\ b(x-c)=b(x)\  \forall  \ c\ \in \Gamma ,\ 
 \forall  \ x\ \in \bR ^3; 
\end{equation} 
$$a(x)>0\ \ {\rm {and}}\ \  b(x)>0,$$
$\Gamma $ is a lattice in $\ \bR ^3,\ \Gamma =\{ c=\displaystyle \sum _{j=1}^{3}n_je_j;\ 
n=(n_1,n_2,n_3)\in \bZ ^3\} $ \\ 
with $\ \{ e_1,e_2,e_3\} $ a basis of 
$\ \bR ^3.$ 

 
$\ E$ denotes the electrical field, $\ B$ the magnetic field, 
$\ 1/a$ the dielectric constant and $\ 1/b$ the magnetic permeability.

The Maxwell equations can be written in the following form 
\begin{equation}\label{maxopdef}
{\partial \over {\partial t}}\Psi =iM(x,D)\Psi , \ \ \ D.\Psi ^{+}=D.({1\over {a(x)}}\Psi ^{-})=0, 
\end{equation} 
if $\ \Psi =(\Psi ^{+},\Psi ^{-})=(B,E)$, (we denote  $\ D.u=-idiv(u)$ for 
$\ u\in L^2(\bR ^3;\bC ^3)).$ \\ 
 $\ M(x,D)$ is the Maxwell operator, which is 
a $\ 6\times 6$ system of differential operators of order one, 
\begin{equation}\label{maxop}
M(x,D)=\pmatrix{ 0 & iR_{ot}(D) \cr 
 -ia(x)R_{ot}(D)b(x) & 0 } 
 \end{equation} 
$${\rm {with}}\ 
 R_{ot}(\xi )=i\pmatrix{0&-\xi _3&\xi _2\cr 
 \xi _3&0&-\xi _1\cr 
 -\xi _2&\xi _1&0} ;\ \ 
 D=-i\displaystyle ({{\partial }\over {\partial x_1}},{{\partial }\over {\partial x_2}},
{{\partial }\over {\partial x_3}}).$$ 
\ \ When $\ a(x),\ b(x)\ \in C^1(\bR ^3)$, it is easy to see that the Maxwell operator $\ M(x,D)$ 
 is essentially self-adjoint on 
\begin{equation}\label{defbH} 
\bH =L^{2}_{b}(\bR ^3;\bC ^3)\oplus L^{2}_{1/a}(\bR ^3;\bC ^3). 
\end{equation}  
($\ L^{2}_{w}(\bR ^3)=L^2(\bR ^3;w(x)dx)$, if $\ w(x)$ a density, $\ w>0$ and 
 $\ dx$ is the usual the Lebesgue measure). 
 
If we denote by $\ M_1(x,D)$ the unique self-adjoint operator on $\ \bH $ which is an 
extension of $\ M(x,D)$ from  the regular vector fields 
with compact support $\ C^{\infty }_{0}(\bR ^3;\bC ^3)\oplus C^{\infty }_{0}(\bR ^3;\bC ^3),$ \\ 
 then $\ M_1(x,D)$ has  zero as an eigenvalue of infinite multiplicity, (this comes 
from the non-ellipticity of the $\ M(x,D)$ as a differential operator).    
  
We deal with the Maxwell operator, we still denote by $\ M(x,D)$, which 
is   the restriction of $\ M_1(x,D)$ on the closure of its 
range $\ \bH _1={\overline {R(M_1(x,D))}}$. \\ 
$\ \bH _1 $ is a closed subspace of $\ \bH $ and its orthogonal 
subspace is the kernel of \\ 
$\ M_1(x,D),\ \bH _2=Ker(M_1(x,D)):$ 
$$\bH =L^{2}_{b}(\bR ^3;\bC ^3)\oplus L^{2}_{1/a}(\bR ^3;\bC ^3)=\bH _1\oplus \bH _2,$$ 
\begin{equation}\label{HrHd}
\bH _1=\{ (u,v)\in L^{2}_{b}(\bR ^3;\bC ^3)\oplus L^{2}_{1/a}(\bR ^3;\bC ^3);
\ D.u=D.({1\over a}v)=0\}
\end{equation} 
$$\bH _2=\{ (u,v)\in L^{2}_{b}(\bR ^3;\bC ^3)\oplus L^{2}_{1/a}(\bR ^3;\bC ^3);
\ R_{ot}(D)(bu)=R_{ot}(D)(v)=0\} .$$
The domain of $\ M(x,D)$  is the intersection of $\ \bH _1$ and  the Sobolev space of order one 
\begin{equation}\label{domMax} 
\cD (M(x,D))=\bH _1\bigcap (W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3;\bC ^3))
\end{equation} 
$$=\{ (u,v)\in W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3;\bC ^3);\ D.u=D.({1\over a}v)=0\} ,$$
and $\ M(x,D)$ is a self-adjoint operator on $\ \bH _1.$ More over, if 
$\ a$ an $\ b$ are in $\ C^{\infty }(\bR ^3),$ then 
$\ M(x,D)$ can be seen as a $\ 4\times 4$  elliptic  pseudo-differential 
system, in the meaning of \cite{Hor}. 

The solution of the Maxwell equations is given by 
$$\Psi _t=(B_t,E_t)=e^{itM(x,D)}\Psi _{0},$$
 if $\ \Psi _{0} \in \ \bH _1$.

Our main result is 
\begin{theorem}\label{th1} 
Under the assumptions (\ref{hyp1}) and if 
\begin{equation}\label{hyp2}  
a,\ b\ \in C^{\infty }(\bR ^3),
\end{equation} 
then the Maxwell operator $\ M(x,D)$ with domain (\ref{domMax}) is a self-adjoint operator 
on the Hilbert space $\ \bH _1$ defined in (\ref{HrHd}), and its spectrum 
is purely absolutely continuous. 
\end{theorem} 

To prove Theorem (\ref{th1}) we need to consider the Maxwell operator as a component of 
an elliptic differential operator of Dirac type. 

Let us consider the Hilbert space 
\begin{equation}\label{Hil2} 
\cH =(L^{2}_{b}(\bR ^3;\bC ^3)\oplus L^{2}_{b}(\bR ^3))\oplus 
(L^{2}_{1/a}(\bR ^3;\bC ^3)\oplus L^{2}_{a}(\bR ^3))
\end{equation} 
and the isometrical  injection 
\begin{equation}\label{inj} 
J:\ \bH \ \longmapsto \ \cH ,\ \ J(u,v)=(u,0,v,0), 
\end{equation} 
$\ \forall \ (u,v)\in \bH 
=L^{2}_{b}(\bR ^3;\bC ^3)\oplus L^{2}_{1/a}(\bR ^3;\bC ^3).$

Let us define the $\ 8\times 8$ systems of differential operators 
\begin{equation}\label{dir2} 
P_1(x,D)=J^{\star }M(x,D)J^{\star }, \ \ \ \ P_2(x,D)(u,f,v,g)= 
\end{equation} 
$$-i({1\over {b(x)}}\bigtriangledown (a(x)b(x)g),a(x)div({1\over {a(x)}}v),
\bigtriangledown (a(x)b(x)f),b(x)div(u)), $$
$\ \forall \ (u,f,v,g)\in \  
W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3)\oplus W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3),$\\ 
and $\ \ \ P(x,D)=P_1(x,D)+P_2(x,D),$ then 
\begin{equation}\label{dir1} 
P(x,D)(u,f,v,g)= 
\end{equation} 
$$-i(-rot(v)+{1\over {b(x)}}\bigtriangledown (a(x)b(x)g), 
a(x)div({1\over {a(x)}}v),a(x)rot(b(x)u)+\bigtriangledown (a(x)b(x)f),b(x)div(u)), $$ 
$\forall (u,f,v,g)\ \in \  
W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3)\oplus W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3).$ 
 
$\ P(x,D)$ is an elliptic differential $\ 8\times 8$ system, which is self-adjoint on 
$\ \cH $, (with domain the Sobolev space $\ W^1(\bR ^3;\bC ^8)).$  
$\ P(x,D)$ is of Dirac type:  
\begin{equation}\label{carD} 
P^2(x,D)(u,f,v,g)=(Q_1(x,D)u,q_1(x,D)f,Q_2(x,D)v,q_2(x,D)g),  
\end{equation} 
$\forall (u,f,v,g) \in \ 
W^2(\bR ^3;\bC ^3)\oplus W^2(\bR ^3)\oplus W^2(\bR ^3;\bC ^3)\oplus W^2(\bR ^3).$ 

$Q_1(x,D)$ and $\ Q_2(x,D)$ are $\ 3\times 3$ elliptic differential systems of order $\ 2$, which are 
 self-adjoint operators, on repectively $\ L^{2}_{b}(\bR ^3;\bC ^3)$ 
 and $\ L^{2}_{1/a}(\bR ^3;\bC ^3),$ (with domain $\ W^2(\bR ^3;\bC ^3)):$ 
\begin{equation}\label{defQ} 
Q_1(x,D)u=R_{ot}(D)[a(x)R_{ot}(D)b(x)u]+{1\over {b(x)}}D[a(x)b^2(x)D.u], 
\end{equation} 
$$Q_2(x,D)u=a(x)R_{ot}[b(x)R_{ot}u]+D[a^2(x)b(x)D.({1\over {a(x)}}u)], $$
$\ \ \forall \ u\in W^1(\bR ^3;\bC ^3);$ 
 $\ \ (D=-i\bigtriangledown )$. 


$q_1(x,D)$ and $\ q_2(x,D)$ are elliptic differential operators of order $\ 2,$ and they are 
self-adjoint on repectively, $\ L^{2}_{b}(\bR ^3)$ 
 and $\ L^{2}_{a}(\bR ^3),$ (with domain $\ W^2(\bR ^3)):$ 
 \begin{equation}\label{defq} 
 q_1(x,D)f=a(x)D.[{1\over {a(x)}}D(a(x)b(x)f)],\ \ {\rm {and}}  
\end{equation} 
$$ q_2(x,D)f=b(x)D.[{1\over {b(x)}} D(a(x)b(x)f)],\ \  \forall\ f\in W^2(\bR ^3). $$
 

It comes from the the isometry of the injection $\ J$ that 
$\ P_1(x,D)$ is a self-adjoint operator on $\ \cH _1$, if its domain is 
\begin{equation}\label{domP1} 
\cD (P_1(x,D))= 
\end{equation} 
$$\{ (u,0,v,0)\in 
W^1(\bR ^3;\bC ^3)\oplus W^1(\bR^3)\oplus W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3); 
\ D.u=D.({1\over {a(x)}}v)=0\} ,$$ 
 $\ \cH _1$ is the closed subspace of $\ \cH ,$ 
\begin{equation}\label{def2H1} 
\cH _1=J(\bH _1)= 
\end{equation} 
$$\{ (u,0,v,0)\in 
L^{2}_{b}(\bR ^3;\bC ^3)\oplus L^{2}_{b}(\bR ^3)\oplus 
L^{2}_{1/a}(\bR ^3;\bC ^3)\oplus L^{2}_{a}(\bR ^3);\  
D.u=D.({1\over {a(x)}}v)=0\} . $$


In the same way, one can see easily that $\ P_2(x,D)$ is a self-adjoint operator on 
the orthogonal componet of $\ \cH _1,\ \cH _2,$ with domain 
\begin{equation}\label{domP2} 
\cD (P_2(x,D))= 
\end{equation} 
$$\{ (u,f,v,g)\in 
W^1(\bR ^3;\bC ^3)\oplus W^1(\bR^3)\oplus W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3); 
\ R_{ot}(D)(b(x)u)=R_{ot}(D)v=0\} .$$  
$\ \cH _2$  can be written 
\begin{equation}\label{def2H2} 
\cH _2=\cH _{1}^{\perp }= 
\end{equation} 
$$\{ (u,f,v,g)\in 
L^{2}_{b}(\bR ^3;\bC ^3)\oplus L^{2}_{b}(\bR ^3)\oplus 
L^{2}_{1/a}(\bR ^3;\bC ^3)\oplus L^{2}_{a}(\bR ^3); 
\ R_{ot}(D)(b(x)u)=R_{ot}(D)v=0\} .$$ 
 So $\ P(x,D)=P_1(x,D)\oplus P_2(x,D)$ and $\ P_1(x,D)$ is unitarly equivalent 
to the Maxwell operator $\ M(x,D).$ Then we have the proposition

\begin{proposition}\label{prop1}  Under the assumptions of Theorem (\ref{th1}), 
the spectrum of \\ 
Maxwell operator 
$\ M(x,D)$  is included in 
the spectrum of the Dirac type operator $\ P(x,D)$ defined by (\ref{dir1}). 
More precisely 
\begin{equation}\label{incl} 
\sigma _{sc}(M(x,D))\subset \sigma _{sc}(P(x,D))\ \ {\rm {and}}\ \ 
\sigma _{p}(M(x,D))\subset \sigma _{p}(P(x,D)). 
\end{equation} 
\end{proposition} 
For an operator $\ Q,\ \sigma (Q),\ \sigma _d(Q),\ 
\sigma _{ac}(Q),\ 
\sigma _{sc}(Q)\ {\rm {and}}\ 
\sigma _{p}(Q)$ denote its spectrum, discret spectrum, absolutely continuous spectrum, 
singular continuous spectrum and the point spectrum. 

Theorem (\ref{th1}) comes from Proposition (\ref{prop1}) and 
\begin{theorem}\label{th2} 
Under the assumptions of Theorem (\ref{th1}), 
the spectrum of the Dirac type operator $\ P(x,D)$ defined by (\ref{dir1}) 
is absolutely continuous. 

Consequently the   operators 
$\ Q_1(x,D),\ Q_2(x,D),\ q_1(x,D)$ and $\ q_2(x,D)$ defined in (\ref{defQ}) and 
(\ref{defq}) 
have only absolutely continuous spectrum. 
\end{theorem} 

\begin{remark}\label{rm1} 
The assumption (\ref{hyp2}) can be weakened to $\ a,\ b\ \in C^3(\bR ^3),$ 
following our proof. 
\end{remark} 

\vskip 0.5cm 

\begin{remark}\label{rm2}  
For any dimension $\ n,$ if $\ \Gamma $ is a lattice of 
$\ \bR ^n$ and if $\ a(x),\ b(x)$ and $\ k(x)$ are $\ \Gamma $-periodic 
functions on $\ \bR ^n$, such that $\ a(x),\ b(x)\ \in C^{\infty }(\bR ^n;]0,+\infty [)$ 
and $\ k(x)$ is a real and bounded potential, $\ k(x)\in L^{\infty }(\bR ^n)$, then 
the Schr\"odinger operator $\ q(x,D)$ defined by \\  
 $\ q(x,D)f=D.[a(x)D(b(x)f)]+k(x)f,\ \forall \ f\in W^2(\bR ^n),$ 
is self-adjoint on $\ L^{2}_{b}(\bR ^n).$ 

It comes easily from our proof that $\ q(x,D)$ has only absolutely continuous spectrum. 
\end{remark} 

 

\vskip 0.5cm 

\begin{remark}\label{rm3} In dimension two, we have just to take care of the modifications 
of the operators:  
$$R_{ot}(D)u=({{\partial }\over {\partial x_2}}u_3,-{{\partial }\over {\partial x_1}}u_3, 
{{\partial }\over {\partial x_1}}u_2-{{\partial }\over {\partial x_2}}u_1),$$ 
$$D.u=-i[{{\partial }\over {\partial x_1}}u_1+{{\partial }\over {\partial x_2}}u_2],\ \ 
Df=-i({{\partial }\over {\partial x_1}}f,{{\partial }\over {\partial x_2}}f,0), 
\ \ \forall u\in W^1(\bR ^2;\bC ^3),\ \forall f\in W^1(\bR ^2).$$ 

The square of the equivalent of Maxwell operator, defined as in (\ref{maxop}) 
(with $\ \xi _3=0$), becomes 
of diagonal type:  
$$M ^2(x,D)\Psi =(L_1(x,D){\widetilde {u}},q_{+}(x,D)u_3,L_2(x,D){\widetilde {v}}, 
q_{-}(x,D)v_3),$$ 
$\forall \Psi =(u,v)\in W^1(\bR ^2;\bC ^3)\oplus W^1(\bR ^2;\bC ^3),$ if 
$\ {\widetilde {u}}=(u_1,u_2)$ and $\ {\widetilde {v}}=(v_1,v_2)$. 

$q_{+}(x,D)f=D.[a(x)D(b(x)f)]$ and $\ q_{-}(x,D)f=a(x)D.[b(x)Df]$,\\ 
 so by Remark 
(\ref{rm2}) $\ q_{+}(x,D)$ and $\ q_{-}(x,D)$  have only absolutely continuous spectrum, 
(as self-adjoint operator on $\ L^{2}_{b}(\bR ^2) $ and respectively on 
$\ L^{2}_{1/a}(\bR ^2)).$ 

The operator $\ L_1(x,D)=(L_{1}^{+}(x,D),L_{1}^{-}(x,D))$ is defined as follow: 

$L_{1}^{+}(x,D)w=-{{\partial }\over {\partial x_2}}[a(x)({{\partial }\over {\partial x_2}}(b(x)w_1)
-{{\partial }\over {\partial x_1}}(b(x)w_2))],$ 

$L_{1}^{-}(x,D)w= 
{{\partial }\over {\partial x_1}}[a(x)({{\partial }\over {\partial x_2}}(b(x)w_1)
-{{\partial }\over {\partial x_1}}(b(x)w_2))])$,\\ 
 $ \ \forall \ w\in  \cD (L_1(x,D))=W^2(\bR ^2;\bC ^2)\bigcap \bH ^{+}_{1},$

$\ \ \bH ^{+}_{1}=\{ w\in L^2(\bR ^2;\bC ^2);\ div(w)=0\} .$ 



In the same way $\ L_2(x,D)=(L_{2}^{+}(x,D),L_{2}^{-}(x,D))$, if 

$L_{2}^{+}(x,D)w=-a(x){{\partial }\over {\partial x_2}}[b(x)({{\partial }\over {\partial x_2}}w_1
-{{\partial }\over {\partial x_1}}w_2)]$\\ 
and  $\ L_{2}^{-}(x,D)w=a(x)
{{\partial }\over {\partial x_1}}[b(x)({{\partial }\over {\partial x_2}}w_1
-{{\partial }\over {\partial x_1}}w_2)])$,\\ 
 $\ \forall \ w\in  \cD (L_{2}(x,D))=W^2(\bR ^2;\bC ^2)\bigcap \bH ^{-}_{1},$ 
 
$\ \ \bH ^{-}_{1}=\{ w\in L^2(\bR ^2;\bC ^2);\ div(a^{-1}(x)w)=0\} .$ 

The Theorem (\ref{th1}) is valid for $\ M(x,D)$, so 
$\ L_1(x,D)$ and $\ L_{2}(x,D)$  are self-adjoint operators,  on respectively 
 the Hilbert space $\ \bH ^{+}_{1}$ and $\ \bH ^{-}_{1}$,\\  
 and their spectrums are absolutely continuous. 
\end{remark}



 
 

\section{Proofs } \label{proof} 



\centerline  {\underline {Proof of the essentially self-adjointness }} 

\vskip 0.5cm 

As $\ M(x,D)$ with domain $\ C^{\infty }_{0}(\bR ^3)\oplus C^{\infty }_{0}(\bR ^3)$ is a 
symetric operator on the Hilbert space $\ \bH ,$ we have just to check that 
$\ M(x,D)\Psi =\pm i\Psi $ has no solution in $\ \bH .$ 

If $\ \Psi =(u,v)\in \bH $ is such that $\ M(x,D)\Psi =\pm i\Psi ,$\\ 
then $\ div(u)=div({1\over {a(x)}}v)=0$ in $L^2(\bR ^3).$ 

 But $\ v\in L^3(\bR ^3;\bC ^3),\ \ rot(v)=\pm u\ \in L^2(\bR ^3;\bC ^3)$ \\ 
 and $\ div(v)={1\over {a(x)}}(\bigtriangledown a(x)).v\ \in L^2(\bR ^3),$ so 
 $\ v\in W^1(\bR ^3;\bC ^3).$ 
 
 In the same way 
 $\ u\in L^3(\bR ^3;\bC ^3),\ \ rot(u)=\mp {1\over {a(x)b(x)}}v-
 {1\over {b(x)}}(\bigtriangledown b(x))\times  u\ \in L^2(\bR ^3;\bC ^3)$ and
 $\ div(u)=0\ \in L^2(\bR ^3),$ so 
 $\ u\in W^1(\bR ^3;\bC ^3).$ 
 
But $\ M(x,D)$ with domain $\ W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3;\bC ^3)$ is a 
symetric operator on $\ \bH ,$ 
so $\ M(x,D)\Phi =\pm i\Phi $ has only $\ \Phi =0$ as 
a solution in $\ W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3;\bC ^3),$ 
then $\ \Psi =(u,v)=(0,0)\ \bullet $ 

The previous proof shows easily that, if $\ \Psi =(u,v)\ \in \bH $ is such 
$\ M(x,D)\Psi \ \in \bH $ and  $\ div(u)=div({1\over {a(x)}}v)=0,$ \\ 
 then $\ \Psi \in W^1(\bR ^3;\bC ^3)\oplus W^1(\bR ^3;\bC ^3).$ 

 

As for the proof of Maxwell operator, it is easy to show that 
$\ P_2(x,D)$ is essentially self-adjoint on $\ \cH $ 
and, \\ 
if 
$\ \Phi =(u,f,v,g)\ \in \cH $ is such that $\ P_2(x,D)\Phi \ \in \cH $ \\ 
and $\ rot(b(x)u)=rot(v)=0$ then $\ \Phi \ \in W^1(\bR ^3;\bC ^8).$ 

So $\ P_2(x,D)$ with domain given by (\ref{domP2}) is a self-adjoint operator 
on $\ \cH _2.$ 


\vskip 0.5cm 

\centerline  {{\underline {Proof of  Proposition }}(\ref{prop1})} 

\vskip 0.5cm 

The ellipticity of $\ P(x,D)$ ensure that $\ P(x,D)$ with domain $\ W^1(\bR ^3;\bC ^8)$ 
is self-adjoint on $\ \cH $, which is the closure of $\ P(x,D)$ with initial 
domain $\ C^{\infty }_{0}(\bR ^3;\bC ^8).$ 
 It is easy to see that zero is not an eigenvalue of $\ P(x,D),$ 
so $\ \cH ={\overline {R(P(x,D))}}.$ More precisely, for every $\ \Phi \ \in \cH ,$ there 
exists a sequence $\ (\Phi _j)$ in $\ C^{\infty }_{0}(\bR ^3;\bC ^8),$ such that 
$\ (P(x,D)\Phi _j)$ converges in 
$\ \cH $ to $\ \Phi .$ But $\ P_1(x,D)\Phi _j\ \in \cH _1,\ 
\ P_2(x,D)\Phi _j\ \in \cH _2$ and \\  
$P_1(x,D)P_2(x,D)\Psi =P_2(x,D)P_1(x,D)\Psi =0,\ \forall \Psi \in  W^2(\bR ^3;\bC ^8).$ 

So  $\ P_1(x,D)\Phi _j$ and $\ P_2(x,D)\Phi _j$ are orthogonal in 
$\ \cH $ for every integer $\ j,$  
then  $\ \Phi =\Phi ^{+}+\Phi ^{-},$ with $\ \Phi ^{+}$ the limit in $\ \cH $ of 
$\ (P_1(x,D)\Phi _j)$ and $\ \Phi ^{-}$ the one of $\ (P_2(x,D)\Phi _j)$. 
It is easy to see that the decomposition 
\begin{equation}\label{decomp} 
\Phi =\Phi ^{+}+\Phi ^{-},\ \Phi ^{+}\in \cH _1={\overline {P_1(x,D)(W^1(\bR ^3;\bC ^8))}}, 
\end{equation} 
$$\Phi ^{-}\in \cH _2={\overline {P_2(x,D)(W^1(\bR ^3;\bC ^8))}},$$
is unique, so the sum $\ \cH =\cH _1\oplus \cH _2 $ is orthogonal. 

More over, if $\ \Phi \in W^1(\bR ^3;\bC ^8),$ 
then   $\ \Phi ^{+},\ \Phi ^{-}\in W^1(\bR ^3;\bC^8).$ \\ 
To be convinced, consider the equivalent norm on 
$\ W^1(\bR ^3;\bC ^8),$\\ 
$\ \| \Psi \| ^{2}=\| P(x,D)\Psi \| ^{2}_{\cH }+\| \Psi \| ^2_{\cH }$, 
for any $\ \Psi \in  W^1(\bR ^3;\bC ^8),$\\ 
and use that $\ P^3(x,D)(C^{\infty }_{0}(\bR ^3;\bC ^8))$ is dense in 
$\ \cH $ to see that $\ P(x,D)(C^{\infty }_{0}(\bR ^3;\bC ^8))$ is dense in 
$\ W^1(\bR ^3;\bC ^8).$ 

We have proved that 
\begin{equation}\label{decompP} 
P(x,D)=P_1(x,D)\oplus P_2(x,D):
\end{equation} 
$$[W^{1}(\bR ^3;\bC ^8)\bigcap \cH _1]
\oplus [W^{1}(\bR ^3;\bC ^8)\bigcap \cH _2]\ \longmapsto \ \cH _1\oplus \cH _2,$$
the last sum is orthogonal, the $\ P_k(x,D)$ are self-adjoint on $\ \cH _k$, 
 so Proposition (\ref{prop1}) follows from (\ref{decompP}) and the unitary equivalence 
between $\ M(x,D)$ and $\ P_1(x,D).$  





\vskip 0.5cm 

\centerline  {{\underline {Proof of  Theorem }}(\ref{th2})} 

\vskip 0.5cm 
 
For any differential $\ \ n\times n$ system $\ S(x,D)=\displaystyle 
\sum _{|\alpha |\leq m}A_{\alpha }(x)D^{\alpha }$\\ 
on $\ \bR ^3$, with $\ \Gamma $-periodic $\ n\times n$ 
matrices $\ A_{\alpha }(x),$ and for any 
$\ \Theta \in \bC ^3,$ we will consider the differential $\ \ n\times n$ 
system on the torus $\ \bT ^3=\bR ^3/\Gamma ,$

$\ \ \  S(x,D-\Theta )=\displaystyle 
\sum _{|\alpha |\leq m}A_{\alpha }(x)(D-\Theta )^{\alpha }.$ 

Then the  ellipticity of $\ P(x,D-\Theta ),$ ensure that, with domain 
$\ W^1(\bT ^3;\bC ^8),$\\ 
$\ \ P(x,D-\Theta )$ is a closed operator on the Hilbert 
space $\ \cH (\bT ^3),$ 
\begin{equation}\label{2.1} 
\cH (\bT ^3)=L^{2}_{b}(\bT ^3;\bC ^3)\oplus L^{2}_{b}(\bT ^3)\oplus 
L^{2}_{1/a}(\bT ^3;\bC ^3)\oplus L^{2}_{a}(\bT ^3). 
\end{equation} 
Its adjoint is $\ P(x,D-{\overline {\Theta }}),$ so $\ P(x,D-\theta )$ is self-adjoint 
for any $\ \theta \in \bR ^3.$ 

But $\ P(x,D-\Theta )$ is of compact resolvent, so its spectrum is discrete, (for any 
$\ \Theta \in \bC ^3).$ 

The Floquet theory, 
(see \cite{Sim} or \cite{Kuch}),  enables to see that the spectrum of $\ P(x,D)$ 
is composed  of the bands formed by the eigenvalues of 
$\ P(x,D-\theta )$ when $\ \theta $ varies in $\ \bR ^3.$ 

For any fixed $\ \theta \in \bR ^3$ and $\ \omega \in \bS ^2, 
\ (\ \bS ^2$ denotes  the two-dimensional unit sphere), 
for $\ z$ varying in $\ \bC,$  the eigenvalues of 
$\ P(x,D-\theta -z\omega )$ can be labelled such that they become analytical in 
$\ z$ in a neighbourhood of the real line, (see \cite{Sim} for the 
particular case of Laplace operator).  



Then, as for Theorem 2.1 of \cite{Mo}, the Floquet theory  
 and  an argument due to Thomas \cite{Tho} and generalized by Reed and Simon \cite{Sim}, 
 lead to 

\begin{lemma}\label{lem1} The singular continuous spectrum of 
$\ P(x,D)$ is empty: 
$$\ \sigma _{sc}(P(x,D))=\emptyset .$$ 
$\ \ \ \mu  \in \bR $ is an eigenvalue of $\ P(x,D)$ iff  
 $\ \mu $ is an eigenvalue 
 of \\ 
 $\ P(x,D-\Theta ),\ \ \forall \ \Theta \in \bC ^3 .$ 
\end{lemma} 



It remains to prove that $\ P(x,D)$ has no eigenvalue, or equivalently 
 that $\ P(x,D-\Theta )$ has no $\ \Theta $-independent eigenvalue. 

By \cite{Tho} or \cite{Sim}, we know the existence of $\ \theta \in \bR ^3,\ \omega \in \bS ^2$   
  and a constant $\ C>0$ such that 
\begin{equation}\label{2.2} 
\| (D-\theta -i\lambda \omega )^2f\| _{L^2(\bT ^3)}\geq 
\lambda {1\over C}\| f\| _{L^2(\bT ^3)},\ \ \forall \ f\in W^1(\bT ^3),\ \lambda >C. 
\end{equation} 

Suppose now $\ \mu $ is an eigenvalue of $\ P(x,D)$, then $\ \mu $ is an eigenvalue of 
$\ P(x,D-\theta -i\lambda \omega )$ for any real $\ \lambda .$ Let $\ \Phi ^{\lambda }$ be 
an associated eigenvector of unit norm: 
\begin{equation}\label{2.3} 
P(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }=\mu \Phi ^{\lambda }\ \ {\rm {and}}\ \ 
\| \Phi ^{\lambda }\| _{\cH (\bT ^3)}=1 . 
\end{equation} 
The ellipticity of $\ P(x,D-\theta -i\lambda \omega )$ ensure that 
$\ \Phi ^{\lambda }\in W^3(\bT ^3;\bC ^8),$ (in fact 
$\ \Phi ^{\lambda }\in C^{\infty }(\bT ^3;\bC ^8)$ with the hypothesis (\ref{hyp2}), 
see for example \cite{Hor}),
 so for $\ k=1$ or $\ k=2,$ 
$P_k(x,D-\theta -i\lambda \omega )(P(x,D-\theta -i\lambda \omega )\Phi ^{\lambda })=$

$P^{2}_{k}(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }=
\mu P_k(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }.$ 

Let us prove that either $\ P_1(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }\neq 0$\\ 
or $\ P_2(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }\neq 0.$ 


If $\ P_1(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }=
P_2(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }=0,$ \\ 
and if $\ \Phi ^{\lambda }=(u^{\lambda },f^{\lambda },v^{\lambda },g^{\lambda }),$ with 
$\ u^{\lambda },\ v^{\lambda }\in W^3(\bT ^3;\bC ^3),\ f^{\lambda },\ g^{\lambda }\ 
\in W^3(\bT ^3;\bC ),$ then  
$$ R_{ot}(D-\theta -i\lambda \omega )v^{\lambda }=
R_{ot}(D-\theta -i\lambda \omega )(b(x)u^{\lambda })$$
$$=
(D-\theta -i\lambda \omega )(a(x)b(x)g^{\lambda })=
(D-\theta -i\lambda \omega )(a(x)b(x)f^{\lambda })=0$$ 
and $\ (D-\theta -i\lambda \omega ).({1\over {a(x)}}v^{\lambda })=
(D-\theta -i\lambda \omega ).(u^{\lambda })=0;$\\ 
so if (\ref{2.2}) is satisfied $\ f^{\lambda }=g^{\lambda }=0.$ 

 
Let $\ {\bf 1}_3$ denotes the $\ 3\times 3$ identity matrix and let us remind the formulas,   
$\ \forall 
\ w\in W^2(\bT ^3;\bC ^3),$
\begin{equation}\label{2.0}  
 (D-\theta -i\lambda \omega )^2{\bf 1}_3w=
R^{2}_{ot}(D-\theta -i\lambda \omega )w+(D-\theta -i\lambda \omega )[(D-\theta -i\lambda \omega ).w], 
\end{equation} 
$$\ \ (D-\theta -i\lambda \omega ).R_{ot}(D-\theta -i\lambda \omega )w=0\ \ {\rm {and}}\ \ 
R_{ot}(D-\theta -i\lambda \omega )[(D-\theta -i\lambda \omega )w]=0.$$
 \ \ As  
   $\ R_{ot}(D-\theta -i\lambda \omega )(b(x)u^{\lambda })=0,$ if $\ \lambda $ is large enough,\\ 
   then, by (\ref{2.2}) and (\ref{2.0}),  we can write 
\begin{equation}\label{2.21} 
b(x)u^{\lambda }=(D-\theta -i\lambda \omega )h^{\lambda }, 
\end{equation} 
$h^{\lambda }=s(D-\theta -i\lambda \omega )
[(D-\theta -i\lambda \omega ).(b(x)u^{\lambda })],$ 
 if $\ s(D-\theta -i\lambda \omega )$ is the inverse of $\ (D-\theta -i\lambda \omega )^2.$


But  $\ b(x)u^{\lambda }\in W^3(\bT ^3;\bC ^3),$ then $ \ h^{\lambda }\ \in W^4(\bT ^3;\bC ).$ 

As  $\ (D-\theta -i\lambda \omega ).u^{\lambda }=0,$ then  (\ref{2.21}) entails  
$$(D-\theta -i\lambda \omega )^2h^{\lambda }+i{1\over {b(x)}}(\bigtriangledown b(x)).
[(D-\theta -i\lambda \omega )h^{\lambda }]=
$$ 
\begin{equation}\label{2.22} 
{\sqrt {b(x)}}[(D-\theta -i\lambda \omega )^2({1\over {\sqrt {b(x)}}}h^{\lambda })+
(\Delta  {1\over {\sqrt {b(x)}}})h^{\lambda }]=0. 
\end{equation} 
If $\ \lambda $ is large enough, (\ref{2.2}) and (\ref{2.22}) lead to 
$\ h^{\lambda }=0$ 
and then $\ \ \ u^{\lambda }=0.$ 

In the same way $\ R_{ot}(D-\theta -i\lambda \omega )v^{\lambda }=0,$ leads to  
\begin{equation}\label{2.23} 
v^{\lambda }=(D-\theta -i\lambda \omega )k^{\lambda }, 
\end{equation} 
$k^{\lambda }=s(D-\theta -i\lambda \omega )
[(D-\theta -i\lambda \omega ).v^{\lambda }], $ 
(if $\ \lambda $ is large enough),\\ 
 and 
$\ (D-\theta -i\lambda \omega ).({1\over {a(x)}}v^{\lambda })=0$  entails  
$$(D-\theta -i\lambda \omega )^2k^{\lambda }+i{1\over {a(x)}}(\bigtriangledown a(x)).
[(D-\theta -i\lambda \omega )k^{\lambda }]=
$$ 
\begin{equation}\label{2.24} 
{\sqrt {a(x)}}[(D-\theta -i\lambda \omega )^2({1\over {\sqrt {a(x)}}}k^{\lambda })+
(\Delta  {1\over {\sqrt {a(x)}}})k^{\lambda }]=0, 
\end{equation} 
this last equation will give $\ k^{\lambda }=0$ and then $\ v^{\lambda }=0,$ 
(if $\ \lambda $ is large enough, thanks to (\ref{2.2})). 



We have proved that, if $\ \lambda $ is large enough, then 
either 

$\ P_1(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }\neq 0$\\ 
or $\ \ P_2(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }\neq 0.$ \\ 
So equivalently to (\ref{2.3}), we can suppose that, either 
\begin{equation}\label{2.4} 
P(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }=
P_1(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }=\mu \Phi ^{\lambda },\ \ 
\| \Phi ^{\lambda }\| _{\cH (\bT ^3)}=1  
\end{equation} 
$$\Phi ^{\lambda }=(u^{\lambda },0,v^{\lambda },0),\ u^{\lambda },\ v^{\lambda } \ \in 
W^2(\bT ^3;\bC ^3)\ \ {\rm {and}}\ \ 
(D-\theta -i\lambda \omega ).u^{\lambda }=(D-\theta -i\lambda \omega ).
({1\over {a(x)}}v^{\lambda })=0$$ 
or 
\begin{equation}\label{2.5} 
P(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }=
P_2(x,D-\theta -i\lambda \omega )\Phi ^{\lambda }=\mu \Phi ^{\lambda },\ \ 
\| \Phi ^{\lambda }\| _{\cH (\bT ^3)}=1  
\end{equation} 
$$\Phi ^{\lambda }=(u^{\lambda },f^{\lambda },v^{\lambda },g^{\lambda }),\ 
 u^{\lambda },\ v^{\lambda }\in 
W^2(\bT ^3;\bC ^3),\ f^{\lambda },\ g^{\lambda }\ \in W^2(\bT ^3)$$ 
and  
$\ \ \ \ R_{ot}(D-\theta -i\lambda \omega )(b(x)u^{\lambda })=
R_{ot}(D-\theta -i\lambda \omega )v^{\lambda }=0.$ \\ 
( It comes from this proof that, if $\ \lambda $ is large enough then 
 $\ \ \  \mu \neq 0). $ 

\vskip 0.5cm  

1) {\underline {Suppose}}  (\ref{2.4}). 

For any $\ w\ \in W^2(\bT ^3;\bC ^3),$ 
$$R_{ot}(D-\theta -i\lambda \omega )[a(x)R_{ot}(D-\theta -i\lambda \omega )b(x)w]=
a(x)b(x)R^{2}_{ot}(D-\theta -i\lambda \omega )w+$$ 
\begin{equation}\label{2.10} 
(\bigtriangledown (a(x)b(x)))\times 
R_{ot}(D-\theta -i\lambda \omega )w
 +ia(x)[(D-\theta -i\lambda \omega ).w](\bigtriangledown b(x))-
\end{equation} 
$$i[a(x)(\bigtriangledown b(x)).(D-\theta -i\lambda \omega )]w+V(x)w$$
with $\ V(x)$ the $\ 3\times 3$ matrix,\\ 
 $\ (V_{ij}(x))_{1\leq i,j\leq 3}=
-div(a(x)(\bigtriangledown b(x))){\bf 1}_3+({{\partial }\over {\partial x_j}}(a(x)
{{\partial }\over {\partial x_i}}b(x)))_{1\leq i,j\leq 3}.$ 

When using (\ref{2.10}), the equations (\ref{2.4}) lead to 
\begin{equation}\label{2.11} 
a(x)b(x)(D-\theta -i\lambda \omega )^2{\bf 1}_3u^{\lambda }=
a(x)b(x)R^{2}_{ot}(D-\theta -i\lambda \omega )u^{\lambda }= 
\end{equation} 
$$ \mu ^2u^{\lambda }
 -(\bigtriangledown (a(x)b(x)))\times 
R_{ot}(D-\theta -i\lambda \omega )u^{\lambda }+
i[a(x)(\bigtriangledown b(x)).(D-\theta -i\lambda \omega )]u^{\lambda }+V(x)u^{\lambda }$$ 
and then, using that\\ 
 $\ {\sqrt {b(x)}}[(D-\Theta )^2f-
i{1\over {b(x)}}(\bigtriangledown b(x)).(D-\Theta )f]= 
(D-\Theta )^2({\sqrt {b(x)}}f)+(\Delta ({\sqrt {b(x)}}))f,$\\ 
 we get 
 \begin{equation}\label{2.12} 
(D-\theta -i\lambda \omega )^2{\bf 1}_3({\sqrt {b(x)}}u^{\lambda })=
-(\Delta ({\sqrt {b(x)}}))u^{\lambda }+
\end{equation} 
$${1\over {a(x)}}{1\over {\sqrt {b(x)}}}\{ \mu ^2u^{\lambda }-
(\bigtriangledown (a(x)b(x)))\times 
R_{ot}(D-\theta -i\lambda \omega )u^{\lambda }+V(x)u^{\lambda }\} .$$ 
Using (\ref{2.4}) again we get 
 \begin{equation}\label{2.13} 
(D-\theta -i\lambda \omega )^2{\bf 1}_3({\sqrt {b(x)}}u^{\lambda })=-
(\Delta ({\sqrt {b(x)}}))u^{\lambda }+ 
\end{equation}  
$$
{1\over {a(x)}}{1\over {\sqrt {b(x)}}}\{ \mu ^2u^{\lambda }
-(\bigtriangledown (a(x)b(x)))\times 
[i\mu {1\over {a(x)b(x)}}v^{\lambda }-{1\over {b(x)}}(\bigtriangledown b(x))\times u^{\lambda }]
+V(x)u^{\lambda }\} . $$
The estimate (\ref{2.2}), the equations (\ref{2.4}) and the equality (\ref{2.13}) 
entail the existence of a constant $\ C_1>0$ such that 
\begin{equation}\label{2.14} 
\| u^{\lambda }\| _{L^2(\bT ^3)}\leq C_1{1\over {\lambda }},\ \ \forall \ 
\lambda >C_1(1+|\mu |). 
\end{equation} 

In the same way the equations from (\ref{2.4}), \\ 
$\ \ a(x)R_{ot}(D-\theta -i\lambda \omega )(b(x)R_{ot}(D-\theta -i\lambda \omega )v^{\lambda })=
\mu ^2v^{\lambda },$\\ 
and $\ \ (D-\theta -i\lambda \omega ).({1\over {a(x)}}v^{\lambda })=0,$ \\ 
lead to the equivalent equation to (\ref{2.12}) 
\begin{equation}\label{2.15} 
(D-\theta -i\lambda \omega )^2{\bf 1}_3({1\over {\sqrt {a(x)}}}v^{\lambda })=-
(\Delta ({\sqrt {a(x)}})){{v^{\lambda }}\over {a(x)}}+ 
\end{equation} 
$$
{1\over {\sqrt {a(x)}}}{1\over {b(x)}}\{ \mu ^2{{v^{\lambda }}\over {a(x)}}-
(\bigtriangledown (a(x)b(x)))\times 
R_{ot}(D-\theta -i\lambda \omega ){{v^{\lambda }}\over {a(x)}}
+W(x){{v^{\lambda }}\over {a(x)}}\}  , 
$$ 
$\ W(x)$ is defined as $\ V(x)$ interchanging $\ a(x)$ and $\ b(x).$ 

(\ref{2.4}) and (\ref{2.15}) ensure the equivalent estimate to (\ref{2.14}), 
for some constant $ C_2>0,$  
\begin{equation}\label{2.16} 
\| v^{\lambda }\| _{L^2(\bT ^3)}\leq C_2{1\over {\lambda }},\ \ \forall \ \lambda >C_2(1+|\mu |). 
\end{equation} 

But (\ref{2.4}) gives the existence of a constant $\ C_3$ such that 
\begin{equation}\label{2.17} 
\| u^{\lambda }\| ^{2}_{L^2(\bT ^3)}+\| v^{\lambda }\| ^{2}_{L^2(\bT ^3)} 
\geq {1\over {C_{3}}},\ \ \forall \ \lambda >C_3, 
\end{equation} 
then, if $\ \lambda $ is large enough, (\ref{2.14}) coupled with  (\ref{2.16}), and 
(\ref{2.17}) are contradictory, so (\ref{2.4}) is not accurate. \\  

\vskip 0.5cm 

2) {\underline {Suppose}}  (\ref{2.5}). 

\vskip 0.5cm 

From (\ref{2.5}) we get 
\begin{equation}\label{2.31} 
a(x)(D-\theta -i\lambda \omega ).[{1\over {a(x)}}(D-\theta -i\lambda \omega )(a(x)b(x)f^{\lambda })]
=\mu ^2f^{\lambda }, 
\end{equation} 
or equivalently, (using the formula $\ (D-\Theta )^2f+i{1\over {a(x)}}(\bigtriangledown a(x)).
(D-\Theta )f=$\\ 
$\ {\sqrt {a(x)}}[(D-\Theta )^2({1\over {\sqrt {a(x)}}}f)+
(\Delta ({1\over {\sqrt {a(x)}}}))f]),$ that 
\begin{equation}\label{2.32} 
(D-\theta -i\lambda \omega )^2({\sqrt {a(x)}}b(x)f^{\lambda })=   
\end{equation} 
$${1\over {\sqrt {a(x)}}}\mu ^2f^{\lambda }-(\Delta ({1\over {\sqrt {a(x)}}}))a(x)b(x)f^{\lambda }.$$
We have also  
\begin{equation}\label{2.33} 
b(x)(D-\theta -i\lambda \omega ).[{1\over {b(x)}}(D-\theta -i\lambda \omega )(a(x)b(x)g^{\lambda })]
=\mu ^2g^{\lambda } 
\end{equation} 
which is equivalent to 
\begin{equation}\label{2.34} 
(D-\theta -i\lambda \omega )^2(a(x){\sqrt {b(x)}}g^{\lambda })=   
\end{equation} 
$$\mu ^2{1\over {\sqrt {b(x)}}}g^{\lambda }-
(\Delta ({1\over {\sqrt {b(x)}}}))a(x)b(x)g^{\lambda }].$$ 
Applying  again (\ref{2.2}), to the equalities (\ref{2.32}) and (\ref{2.34}),  
 leads to \\  
$f^{\lambda }=g^{\lambda }=0$, and then by (\ref{2.5}) $\ u^{\lambda }=v^{\lambda }=0,$ 
so $\ \Phi ^{\lambda }=0.$ 

We have proved that, for any fixed real $\ \mu,$  
if (\ref{2.2}) is satisfied, for any large $\ \lambda ,\ \ \mu $ cannot be 
an eigenvalue of $ \ P(x,D-\theta -i\lambda \omega ),$ then 
by Lemma (\ref{lem1}) $\ \mu $ is not an eigenvalue of $\ P(x,D)$,\\ 
so $\ \sigma _p(P(x,D))=\emptyset ,$ 
 using again the same Lemma, we get Theorem (\ref{th2}) $\ \ \bullet $ 








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\end{thebibliography}


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