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\begin{document}





\title[]{\Large Factorization of Degenerate Cauchy Problems: The Linear Case}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\author[]{\large B. Thaller and S. Thaller ${}^*$}



\address{Inst. f. Mathematik, Universit\"at Graz, Heinrichstr. 36, A-8010 Graz, Austria}



\email{bernd.thaller@kfunigraz.ac.at, sigrid.thaller@kfunigraz.ac.at}



\maketitle



\bigskip\bigskip

\begin{center}

{\sc Institut f\"ur Mathematik, Universit\"at Graz, Heinrichstr. 36, A-8010 Graz, Austria}\\

\end{center}



\bigskip\bigskip



\begin{abstract}

We investigate an abstract degenerate Cauchy problem with a

non-invertible operator $M$ at the derivative.

The problem is formulated in a Hilbert space $\fH$ which can

be written as an orthogonal direct sum of $\Ker M$ and

$\Ran M^*$. Under certain conditions it is possible to reduce the problem

to an equivalent non-degenerate Cauchy problem in the

factor space $\fH/\Ker M$. The explicit form of the generator

for the restricted problem is investigated.

As an example we discuss the Dirac equation,

where our theory leads to a new interpretation of the nonrelativistic limit.

We show that this limit can be understood in terms of a degenerate Cauchy problem

where the generator of the restricted problem is a Schr\"odinger operator.

Finally, we describe some consequences for the treatment of

degenerate control systems.

In particular, we introduce dual methods of factorization in order to

investigate the compatibility of

factorization and transition to the dual system, which is essential for the

definition of basic notions like observability and detectability.

\end{abstract}



\vfill

${}^*$ Supported by SFB F003 ``Optimierung und Kontrolle''

%%%%%%%%%%%%%%%%%%%%%%%%%-------------> Erzwungener Seitenwechsel

\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%





%%%%%%%%%%%%%%%%%%%%%%%%%-------SECTION-------%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Introduction}\label{sec0}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%\setcounter{equation}{0}

\setcounter{prop}{0}

\bigskip

%

\noindent

We are going to study the abstract system

%

\begin{equation}\label{DEG CP 1}

\frac{d}{dt}\,M\,z(t) = A\,z(t) + f(t),

\end{equation}

%

where $M$ and $A$ are linear operators acting between Hilbert spaces

$\fH$ and $\fK$.

These systems are called ``degenerate'' or ``singular'', if the operator

$M$ is not invertible.

Under certain assumptions it is possible to restrict the whole problem

to the orthogonal complement $(\Ker M)^\bot$ of the kernel of $M$.

Here we investigate the structure of the restricted problem without

assuming that $M$ or $A$ be self-adjoint. Our approach is particularly

useful for systems of differential equations, as will be demonstrated

in Section~\ref{sec2}.



Degenerate Cauchy problems have received considerable interest during

the past few decades.

For finite dimensional linear systems the problem is understood completely,

because it is possible to transform the matrices $M$ and $A$ to

a common normal form which allows to classify the cases in which

Eq.~(\ref{DEG CP 1}) has a unique solution for each initial value.

This is discussed in the book by L. Dai \cite{2}, where

one can find many examples, applications to control theory,

and references to the earlier literature.



In the infinite dimensional case we mention the book by Carroll and

Showalter \cite{1} which also treats the nonlinear and

nonautonomous case. Showalter further considers the linear problem in

Refs.~\cite{16},  \cite{17}. In his setting

the operator $M$ has to be self-adjoint and non-negative.

The operators $M$ and $A$ are initially given as quadratic forms on some

vector space $E$.

The Hilbert space of the system is then defined to be

the completion $\fH_M$ of the factor space $E/\Ker M$ in the seminorm defined by $M$.

In this Hilbert space $M$ becomes the identity operator (Riesz map).

In order to obtain in $\fH_M$ a single-valued operator $A_0$ which

corresponds to $A$ one has to assume, e.g., that $A$ defines

a sectorial quadratic form (\cite{17}, cf. also Remark~\ref{Showrem}).

This assumption also implies that $A_0$ generates an analytic semigroup

for the factorized system.

Being very elegant, this approach has nevertheless the disadvantage

that the whole setup depends on the operator $M$, and it becomes difficult

to investigate what happens when $M$ is perturbed or approximated.

Our approach uses different assumptions allowing to express the

factorization in terms of orthogonal projection operators whose behavior

under perturbations is easier to describe. Moreover, we do not

require $M$ to be non-negative or self-adjoint. Likewise the assumption of

parabolicity, although convenient because it leads to a bounded

evolution operator, is not essential for the process of factorization.



In an interesting series of papers A.~Favini investigates degenerate Cauchy problems

of parabolic type in Banach spaces. The linear system is considered in

\cite{3} using Laplace transforms of the resolvent. In

\cite{5} the problem is treated also under the assumption that

the Banach space of the system can be written as a direct sum of

suitable subspaces (this is the main reason for our use of Hilbert

spaces in the present paper).

Most recently, in \cite{11} the authors discuss in particular

regularity properties of solutions. The paper \cite{4}

treats the controllability of degenerate systems, based on results

in \cite{3}. Results concerning existence, uniqueness and

regularity of solutions have also been obtained in more general

situations. \cite{6}, \cite{7} treat the case of

time dependent operators $M(t)$ and $A(t)$, nonlinear operators

are considered in \cite{8} and \cite{11}.

See \cite{9} and \cite{11} for applications and

\cite{10} for results concerning multivalued operators.



Among the older literature we further mention

\cite{13},  \cite{14},  \cite{15}.



In the present paper there is, of course, some overlap with the previously mentioned work.

But the emphasis of this paper is on the possibility of factorization and the relation

of the factorized problem with the original degenerate system -- without assuming parabolicity:

Making use of the decomposition of the

Hilbert space into a direct sum of $\Ker M$ and $\Ran M^*$ 

(resp. $\Ker M^*$ and $\Ran M$)

we formulate the conditions

which allow us to obtain an equivalent but nondegenerate

Cauchy problem in the factor space $\fH/(\Ker M) = \Ran M^*$ (resp. in $\Ran M$)

and give the explicit form of generators of the factorized problems.

The crucial assumption is that the restriction of $A$ to a mapping

from $\Ker M$ to $\Ker M^*$ is well defined and invertible

(Assumption~\ref{ASS2}). This allows to define a factorization

operator $Z_A$ which maps the solutions of the factorized

system to solutions of the original degenerate system.

It is interesting that this operator is generally unbounded.



We treat the basic concepts and the homogeneous equation ($f(t)=0$)

in Section~\ref{sec1}, the inhomogeneous equation is discussed

in Section~\ref{sec4}. As an example illustrating the application

of our method to systems of differential equations we consider

the (time dependent) Dirac equation of relativistic quantum mechanics which in

the nonrelativistic limit can be written as a degenerate

Cauchy problem. This degenerate Cauchy problem is by our

method of factorization easily seen to be equivalent (in a certain sense) to

the Schr\"odinger equation.



In Section \ref{sec3} we make some remarks concerning the setting of

control theory under simplifying assumptions. In particular

the conceptionally important relation between a control system and its

dual is compared to the corresponding relations of the factorized systems,

which leads to the notion of dual methods of factorization.

\medskip



%%%%%%%%%%%%%%%%%%%%%%%%%-------SECTION-------%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{The homogeneous equation}\label{sec1}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%\setcounter{equation}{0}

\setcounter{prop}{0}

\bigskip

%

%%%%%%%%%%%   NOTATION   %%%%%%%%%%%%%%%

\begin{notation}

For a linear operator $T$ we denote by $\Dom(T)$, $\Ker T$,

and $\Ran T$ its domain, kernel, and range, respectively.

If $\Dom(T)$ is dense, the adjoint operator is well defined and will be denoted by $T^*$.

The restriction to $\Dom$ of an operator $T$ defined on some larger domain

will be written as $T \restriction \Dom$.

\end{notation}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

\smallskip

\noindent

Let us consider the degenerate Cauchy problem

%

\begin{equation}\label{DCP}

\frac{d}{dt}\,M\,z(t) = A\,z(t),\qquad z(0) = z_0,

\end{equation}

%

where throughout this paper we make the following assumption:

\smallskip

%%

%%%%%%%%   ASSUMPTION 1   %%%%%%%%%%%%%%%

\begin{assumption}\label{ASS1}

Let $A$ and $M$ be closed linear operators which are

densely defined in some Hilbert space $\fH$ and map into a Hilbert space $\fK$.

\end{assumption}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

\smallskip

\noindent

Here $M$ need not be invertible.

Since $M$ is closed, $\Ker M$ is a closed subspace of $\fH$.

The orthogonal projection onto $\Ker M$ will

be denoted by $P$, and $P^\bot = \mathbf{1}-P$ projects onto the orthogonal complement

of the kernel.

Since the adjoint $M^*$ of a closed operator is again densely defined in $\fK$,

we have the relations

%

\begin{equation*}

P\fH = \Ker M,\qquad

P^\bot\fH = (\Ran M^*)^c.

\end{equation*}

%

(The symbol $^c$ denotes the closure).

For the projection $Q$ onto $\Ker M^*$ we obtain the analogous

relations

%

\begin{equation*}

Q\fK = \Ker M^*,\qquad

Q^\bot\fK = (\Ran M)^c.

\end{equation*}

%

We start with an obvious definition and a few observations.

\smallskip

%

%%%%%%%%%%%%% DEFINITION 1 %%%%%%%%%%%%%%

\begin{definition}\label{DEFINITION 1}

A strict solution of the degenerate Cauchy problem is a continuous function

$z:[0,\infty)\longrightarrow \fH$ such that $z(t)\in\Dom(A)\cap\Dom(M)$ for all $t\ge 0$,

$Mz$ is continuously differentiable,

and Eq.~(\ref{DCP}) holds.

\end{definition}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

\smallskip\noindent

Any strict solution $z$ of the degenerate Cauchy problem  Eq.~(\ref{DCP}) clearly satisfies

$z(t)\in \DA$ for all $t \ge 0$, where

%

\begin{equation*}

\DA = \{ z\in\Dom(A)\mid Az \in (\Ran M)^c \}.

\end{equation*}

%

Obviously, we have

%

\begin{equation}\label{KER A}

\Ker A \subset \DA.

\end{equation}

%

%%

%%%%%%%%%%%%   LEMMA 1   %%%%%%%%%%%%%%%%

\begin{lemma}\label{LEMMA 1}

Under the Assumption \ref{ASS1} the operator $A\restriction\DA$ is closed.

\end{lemma}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{proof}

If $z_n$ is a sequence in $\DA\subset\Dom(A)$ such that $u=\lim z_n$ and $v=\lim Az_n$

both exist, then (since $A$ is closed) $u \in \Dom(A)$ and $Au=v$.

But each $Az_n$ is in the closed set $(\Ran M)^c$,

hence the limit $v$ is again in $(\Ran M)^c$. Thus, $u$ is in $\DA$.

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

\smallskip\noindent

In order to restrict the degenerate Cauchy problem Eq.~(\ref{DCP}) to the subspace

$P^\bot{\fH}=(\Ker M)^\bot$ we define the operator

%

\begin{equation}\label{RESTR M}

M^\bot=M\restriction P^\bot\Dom(M)

\end{equation}

%

which is invertible from $P^\bot\Dom(M)=(\Ker M)^\bot\cap\Dom(M)$ into $(\Ker M^*)^\bot$.

The operator $A$, however, will become a multi-valued operator $A_0$ on $(\Ker M)^\bot$

which can be defined as

%

\begin{align}\label{DEF A0}

&\Dom(A_0) = \{x\in (\Ker M)^\bot \mid (P^\bot)^{-1}\{x\}\cap \DA\ \not = \emptyset\},\\

&A_0x = A\{ (P^\bot)^{-1}\{x\}\cap \DA\} \subset (\Ran M)^c,

\quad \mbox{for all $x\in \Dom(A_0)$.}

\end{align}

%

Here $(P^\bot)^{-1}\{x\} = \{x+y\mid y\in\Ker M\}$, $x\in (\Ker M)^\bot$, is the inverse image of

$x$ under the projection $P^\bot$.

Instead of the degenerate Cauchy problem Eq.~(\ref{DCP}) we could investigate

%

\begin{equation}\label{MULTIVALUED DCP}

\frac{d}{dt}\,M^\bot\,x(t) \in A_0\,x(t),\qquad x(0) = P^\bot z_0,

\end{equation}

%

because for any family of vectors $x(t)$ satisfying this relation there exists a

family $z(t)$ in $\DA$ satisfying (\ref{DCP}). In general, the family $z(t)$

is neither unique nor continuous:

If $k(t)$ is an arbitrary family of vectors in $\Ker M\cap\Ker A$ then

$z(t)+k(t)$ also satisfies (\ref{DCP}).

In the following typical (non exclusive) cases, the operator $A_0$ will be single-valued:

\medskip



$(\mbox{A}):\quad \Ker A \supset \Ker M\cap \DA,$



\smallskip

$(\mbox{B}):\quad (P^\bot)^{-1}\{x\}\cap \DA \quad\mbox{consists of precisely one element.}$

\medskip

%

%%%%%%%%%%%%%%% REMARKS %%%%%%%%%%%%%%%%%%%%%%%%

%

\begin{rem}

Let $\fH=\fK$, assume that $A$, $M$ are self-adjoint, and $P\Dom(A)\subset\Dom(A)$.

Then $\Ker A \supset \Ker M \cap \Dom(A)$ is equivalent to $\DA$ dense in $\fH$.

If this is true, then $\DA = \Dom(A)$.

\end{rem}

%

\begin{proof}

Assume that $\DA$ is dense. $M$ self-adjoint means $\Ker M = (\Ran M)^\bot$.

Hence, for $y\in\Ker M\cap \Dom(A)$

%

\begin{equation*}

0 = (Ax,y) = (x,Ay),\quad\mbox{all $x\in\DA$.}

\end{equation*}

%

This implies $y\in \Ker A$.



Conversely, assume $\Ker A \supset B\equiv\Ker M \cap \Dom(A)$.

For all $y\in B$ we have $Ay=0$ and hence $(x,Ay) = (Ax,y)=0$ for all $x\in\Dom(A)$.

This implies $\Ran A \subset B^\bot$.

But $B$ is dense in $\Ker M$, because $P\Dom(A)$ is dense in $\Ker M$ and a subset of $B$.

Hence $B^\bot = (\Ker M)^\bot=(\Ran M)^c$.

Finally, $\Ran A \subset (\Ran M)^c$ is equivalent to $\Dom(A)=\DA$.

Hence $\DA$ is dense by Assumption~\ref{ASS1}.

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\smallskip\noindent

\begin{rem}\label{Showrem}

(cf. \cite{17}). Let $M$ be self-adjoint and $A$ be sectorial

(with vertex $0$ and semi-angle $\theta$, cf. \cite{12}, Sect.~VI.1.2).

Then Case~(A) holds.

\end{rem}

%

\begin{proof}

Let $x\in\Ker M\cap\DA$. Since $M$ is self-adjoint, we have $A \DA \bot \Ker M$, hence

$(x,Ax)=0$. But $A$ is sectorial, and therefore

%

\begin{equation*}

|(y,Ax)|^2\le (1+\tan\theta)^2\, {\mbox{Re}}\,(y,Ay)\,{\mbox{Re}}\,(x,Ax) = 0

\end{equation*}

%

for all $y\in\Dom(A)$ (see \cite{12}, loc.cit.).

Since $\Dom(A)$ is dense by Assumption~\ref{ASS1}, we have $Ax=0$, i.e., $x\in\Ker A$.

\end{proof}

Showalter uses this result as an important ingredient in Ref.~\cite{17}.

In his setup $M$ and $A$ are defined in the sense of quadratic forms.

$M$ must be non-negative in order to define a semi-scalar product.

(See also \cite{1}, \cite{15} and \cite{16}). 

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\medskip

\noindent

The degenerate Cauchy problem Eq.~(\ref{DCP}) is expected to have unique solutions only

if ${\Ker M \cap \Ker A = \{0\}}$,

and it is this case which we want to investigate further.

\smallskip

%%

%%%%%%%%%%%%% EXAMPLE 1 %%%%%%%%%%%%%%%%%

\begin{example}\label{EXAMPLE 1}

Let $\fH = \fK = L^2({\mathbb R})$, $A = d^2/dx^2$ on its natural domain

%

\begin{equation*}

\Dom(A) = \{z\in L^2\mid z, z' \,\,\mbox{absolutely continuous}, z''\in L^2\} \equiv W^{2,2}({\mathbb R}).

\end{equation*}

%

Let $I$ be the open interval $(-1,1)$ and $M$ be multiplication by the

characteristic function of ${\mathbb R}\setminus I$.

Clearly, $\Ran M$ is closed and consists of all $L^2$-functions vanishing on $I$.

Hence $\DA$ consists of those functions $z\in\Dom(A)$ which are linear on $I$

and thus satisfy

%

\begin{equation}\label{EX.BOUNDARY}

z'(+1) - z'(-1) = 0,\qquad z(+1) - z(-1) = 2z'(-1).

\end{equation}

%

The set $P^\bot\DA\subset \Ran M$ consists of all functions $u\in L^2$ vanishing on $I$,

which together with their derivative $u'$ are absolutely continuous on ${\mathbb R}\setminus I$,

such that $u''$ is in $L^2({\mathbb R}\setminus I)$, and such that the boundary conditions

(\ref{EX.BOUNDARY}) are satisfied for $u$.

On this domain we define the operator

%

\begin{equation}\label{EX.A0}

                         (A_0 u)(x) =

                             \begin{cases}

                                0 & \text{if $x\in I$,}\\

                                u''(x) & \text{if $x\not\in I$,}

                             \end{cases}

      \qquad

      \mbox{for all $u\in\Dom(A_0) \equiv P^\bot\DA$.}

\end{equation}

%

$\Dom(A_0)$ is the set of all $u\in \Ran M$

for which there exists a unique $v\in \Ker M$ such that $z=u+v\in \DA$.

(The function $v$ interpolates linearly on $I$ between the boundary values of

$u$ and vanishes outside $I$).

Moreover, $A_0u=Az$. This is the situation described in case (B) above.

Hence the degenerate Cauchy problem Eq.~(\ref{DCP})

is equivalent to the ordinary Cauchy problem

%

\begin{equation}\label{EX.CP}

\frac{d}{dt}\,u(t) = A_0\,u(t),\qquad u(0)=P^\bot z_0

\end{equation}

%

in the Hilbert space $L^2({\mathbb R}\setminus I)$.

It is easy to see that the operator $A_0=d^2/dx^2$ with the above boundary conditions

is self-adjoint and negative.

Hence for each $z_0\in\DA$, Eq.~(\ref{EX.CP}) has a unique solution $u(t)\in\Dom(A_0)$ ($t\ge 0$)

from which it is easy to obtain the unique strict solution of Eq.~(\ref{DCP}).

\end{example}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

\smallskip\noindent

The restriction to an ordinary Cauchy problem in $(\Ker M)^\bot$ is sometimes even possible,

if the operator $A$ does not leave this subspace invariant.

In order to investigate this case further we make the following assumption.

\smallskip

%%

%%%%%%%%   ASSUMPTION 2   %%%%%%%%%%%%%%%

\begin{assumption}\label{ASS2}

$P\DA \subset \Dom(A)$ and

the operator $(QAP)\restriction P\DA$ has a bounded inverse (which is an operator

from $Q\fK$ to $P\DA$).

\end{assumption}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

\smallskip\noindent

%%

%%%%%%%%%%%%% EXAMPLE 1 %%%%%%%%%%%%%%%%%

\begin{example}\label{EXAMPLE 1A}

Let $\fH=L^2({\mathbb R})^2 = L^2({\mathbb R},{\mathbb C}^2)$, and let

%

\begin{equation*}

     M=\begin{pmatrix}

                            1 & 0\\

                            0 & 0

       \end{pmatrix},\qquad

     A=

       \begin{pmatrix}

                      0 & \frac{\partial}{\partial x} \\

                      \frac{\partial}{\partial x}  &  a

       \end{pmatrix}.

\end{equation*}

%

The matrix differential operator $A$ is defined on its natural domain, the Sobolev space

$W^{1,2}({\mathbb R})^2$. Assumption~\ref{ASS2} implies $a\ne 0$. We find that

%

\begin{equation*}

\DA = \left\{    \bigl(

\begin{smallmatrix}

  f \\

  g

\end{smallmatrix}

                 \bigr) \in W^{1,2}({\mathbb R})^2                

      \Bigm|

g(x) = -{\textstyle \frac{1}{a}}\,f'(x)

      \right\}.

\end{equation*}

%

We see that $z=\bigl(

\begin{smallmatrix}

  f \\

  g

\end{smallmatrix}

                 \bigr) \in \DA$ implies $f\in W^{2,2}({\mathbb R})$.

Moreover, $z\in\DA$ is uniquely determined by its part $f=P^\bot z$

in $(\Ker M)^\bot$. This is also the content of the next lemma.

\end{example}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

\smallskip\noindent

%

%%

%%%%%%%%%%%%   LEMMA 2   %%%%%%%%%%%%%%%%

\begin{lemma}\label{LEMMA 2}

Under the Assumptions \ref{ASS1} and \ref{ASS2} a vector $z\in\fH$ is in the subspace $\DA$

if and only if

%

\begin{equation*}

z\in\Dom(A)\,\,\,{\mbox{and}}\,\,\,Pz = -(QAP)^{-1} QAP^\bot z.

\end{equation*}

%

\end{lemma}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{proof}

By definition, $z\in\DA$ iff $z\in\Dom(A)$ and $QAz=0$.

By Assumption~\ref{ASS2} the vectors $Pz$ and $P^\bot z$ are in $\Dom(A)$.

Writing $z=Pz+P^\bot z$ gives

$QAPz+QAP^\bot z=0$ from which the result follows immediately, since $QAP$ is invertible

on the range of the projection $Q$.

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

%

%%%%%%%%%%%%%%%%%%%% REMARKS %%%%%%%%%%%%

\begin{rem}\label{REMARK 12}

According to Lemma~\ref{LEMMA 2}, any $x\in P^\bot\DA\subset (\Ker M)^\bot$ uniquely determines

$z\in\DA$ such that $x=P^\bot z$:

%

\begin{equation}\label{15}

z = (\id -(QAP)^{-1}QA)x.

\end{equation}

%

Hence the set $(P^\bot)^{-1}\{x\}\cap \DA$ contains precisely one element,

which is just the case (B) described above.

\end{rem}

%

\begin{rem}

From Lemma~\ref{LEMMA 2} above we conclude 

for $z\in \Ker M\cap \DA$ that $x = P^\bot z = 0$ and hence $z=0$ by Eq.~(\ref{15}).

Therefore $\Ker M\cap \DA = \{0\}$, and the condition (A) described above

is satisfied in a trivial way.

In particular, because of Eq.~(\ref{KER A}), we have

%

\begin{equation*}

\Ker M \cap \Ker A = \{0\}.

\end{equation*}

%

\end{rem}

%

\begin{rem}

Example~\ref{EXAMPLE 1} is a counter-example for Assumption~\ref{ASS2}, because

in this case $P\DA\not\subset \Dom(A)$. (Note, however, that $M$ and $-A$ are non-negative

and thus Remark~\ref{Showrem} applies).

%One could also redefine $P$ and $P^\bot$ as appropriate ``identification

%operators'', such that, e.g., $P^\bot:L^2({\mathbb R})\to L^2({\mathbb R}\setminus I)$, $P^\bot f(x) = f(x)$

%for all $x\in {\mathbb R}\setminus I$.

\end{rem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\medskip\noindent

According to the Remark~\ref{REMARK 12} we define the operator ${Z_A}$ by

%

\begin{equation}\label{DEFINE Z}

{Z_A} = P^\bot -P(QAP)^{-1}QAP^\bot.

\end{equation}

%

This operator is defined on $\Dom(Z_A)\supset P^\bot\DA$.

The restriction $Z_A\restriction P^\bot\DA$,

given by $\id -(QAP)^{-1}QA$ on $P^\bot\DA$,

is the inverse of the projection $P^\bot\restriction \DA$:

%

\begin{equation}\label{ZP}

{Z_A}P^\bot = \id \quad \mbox{on $\DA$}, \qquad P^\bot {Z_A} = \id  \quad \mbox{on $P^\bot\DA$}.

\end{equation}

%

Now we simply set

%

\begin{equation}\label{FACTOR A0 1}

A_0 = A{Z_A} \quad\mbox{on $\Dom(A_0) = P^\bot\DA$}

\end{equation}

%

and for all $z\in\DA$ we find, writing $x=P^\bot z$,

%

\begin{equation}\label{A0x}

A_0x = Az.

\end{equation}

%

Since $Az=Q^\bot Az$ for all $z\in\DA$, the operator $A_0$ can be written in the more symmetric form

%

\begin{equation}\label{A0sym}

A_0 = Q^\bot \! A P^\bot - Q^\bot \! A P \,(QAP)^{-1} \,QAP^\bot \quad\mbox{on $\Dom(A_0)$}.

\end{equation}

%

Obviously, we can also factorize $A_0$ with the help of the operator

%

\begin{equation}\label{DEFINE Y}

{Y_A} = Q^\bot - Q^\bot \! A P \,(QAP)^{-1} \,Q.

\end{equation}

%

From ${Y_A} AP = 0$ we find ${Y_A} A P^\bot =  {Y_A} A $ and

%

\begin{equation}\label{FACTOR A0 2}

A_0 = {Y_A} A \quad\mbox{on $\Dom(A_0) = P^\bot\DA$.}

\end{equation}

%

%%%%%%%%%%%%%%% REMARKS %%%%%%%%%%%%%%%%%%



\begin{rem}

For the Example~\ref{EXAMPLE 1A} above

$A_0$ can be identified with $-\frac{1}{a}\partial^2/\partial x^2$ on $W^{2,2}({\mathbb R})$

and $M^\bot=\id$ on $L^2({\mathbb R})$.

Hence for $a<0$ the factorized system defines a well-posed (non-degenerate) Cauchy problem 

(in the Hilbert space $L^2({\mathbb R})$).

\end{rem}

%

\begin{rem}

In the finite dimensional case, it is well known that Eq.~(\ref{DCP}) has only

unique solutions if the matrix pencil $(A,M)$ is similar to $(A_1\oplus \id,\id\oplus N)$,

where $N$ is nilpotent (see \cite{2}).

This is equivalent to $(A,M)$ being a regular pencil,

which means that there exists a $\lambda_0\in {\mathbb C}$ such that $A-\lambda_0 M$ is invertible.

We are going to assume this type of regularity.

\end{rem}

%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



%%%%%%%%   ASSUMPTION 3   %%%%%%%%%%%%%%%

\begin{assumption}\label{ASS3}

The operator $A$ has a bounded inverse.

\end{assumption}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

%%

%%%%%%%%%%%%%   REMARKS   %%%%%%%%%%%%%%%

%

\begin{rem}

Under Assumption~\ref{ASS1} this is equivalent to $A$ injective with $\Ran A=\fK$.

By Eq.~(\ref{KER A}), this implies that $A\restriction\DA$ has a

bounded inverse as an operator from $\DA$ to $Q^\bot\fK$.

In particular, $A\DA = Q^\bot\fK$.

In this case the operator $A_0$ is also invertible, the inverse operator

%

\begin{equation}\label{INVERSE A0}

A_0^{-1}= (A{Z_A})^{-1} = P^\bot A^{-1}\restriction Q^\bot\fK

\end{equation}

%

is bounded and defined on all of $Q^\bot\fK$.

\end{rem}

%

\begin{rem}\label{WLG}

Whenever $z(t)$ is a solution of Eq.~(\ref{DCP}), then

$y(t) = \exp(-\lambda_0t)z(t)$ is a solution of Eq.~(\ref{DCP}) with $A$ replaced by $A-\lambda_0 M$.

Hence if $(A,M)$ is a regular operator pencil, then Assumption~\ref{ASS3} means no further

restriction.

\end{rem}

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

%%%%%%%%%%%%   LEMMA 3   %%%%%%%%%%%%%%%%

\begin{lemma}\label{LEMMA 3}

Under the Assumptions \ref{ASS1}---\ref{ASS3}

the operator $A_0$ defined in (\ref{FACTOR A0 1})

is closed on $\Dom(A_0)=P^\bot\DA$.

%

\end{lemma}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{proof}

Let $(x_n)$ be a sequence in $\Dom(A_0)$ which converges to $x$ in $P^\bot\fH$,

such that $A_0x_n=A{Z_A}x_n$ is also convergent.

By continuity of $A^{-1}$ (Assumption~\ref{ASS3})

the sequence $z_n \equiv A^{-1}A{Z_A}x_n ={Z_A}x_n$ is convergent,

and for each $n$, $z_n\in\DA$.

Since $Az_n$ converges and since $A$ is closed on $\DA$ we find $z\in\DA$ and

%

\begin{equation}\label{pf1}

Az=\lim Az_n = \lim A{Z_A}x_n.

\end{equation}

%

The projection $P^\bot$ is continuous, hence

%

\begin{equation*}

P^\bot z = \lim P^\bot z_n = \lim P^\bot {Z_A}x_n = \lim x_n = x.

\end{equation*}

%

From $z\in\DA$ we therefore conclude that $x = P^\bot z\in P^\bot\DA=\Dom(A_0)$.

Moreover, by (\ref{ZP}), $z={Z_A}P^\bot z$, which implies $Az=A{Z_A}x$.

Together with (\ref{pf1}) this implies the result.

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

\smallskip\noindent

By construction, for all $z\in\DA$, we have $Az=A_0x$, where

$x=P^\bot z$.

Moreover, for $z\in\Dom(M)$, $Mz=M^\bot x$ with the invertible operator $M^\bot$

defined in Eq.~(\ref{RESTR M}).

Hence we can replace the degenerate Cauchy problem Eq.~(\ref{DCP})

by

%

\begin{equation}\label{DCP3}

\frac{d}{dt}\,M^\bot x(t) = A_0\, x(t),\qquad x(0) = P^\bot z_0.

\end{equation}

%

How to proceed now depends on the precise assumptions on $M$.

%%

%%%%%%%%   ASSUMPTION 4   %%%%%%%%%%%%%%%

\begin{assumption}\label{ASS4}

Let $\Dom_A\subset\Dom(M)$ and assume that at least one of the following statements is true:



\smallskip

(Case a): The operator $M$ has a closed range.



\smallskip

(Case b): The operator $M$ has a closed domain.

\end{assumption}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



%%

%%%%%%%%%%%%%   REMARKS   %%%%%%%%%%%%%%%

%

\begin{rem}

If M is closed and densely defined (Assumption \ref{ASS1}),

we can give the equivalent formulation (using the closed graph theorem):



\smallskip

{\it (Case a): $(M^\bot)^{-1}$ is bounded and

defined on all of $Q^\bot\fK$.}



\smallskip

{\it (Case b): $M^\bot$ is bounded and defined on all of $P^\bot\fH$.}

\end{rem}

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\medskip\noindent

If Assumption~\ref{ASS4} (Case a) is fulfilled, we may define the operator

%

\begin{equation}\label{OpA1}

A_1 = A_0(M^\bot)^{-1}

\end{equation}

%

on the natural domain

%

\begin{equation}\label{DA1}

\Dom(A_1)=\{y\in Q^\bot\fK\mid (M^\bot)^{-1}y\in\Dom(A_0)\} =

M^\bot P^\bot\DA =

M\DA,

\end{equation}

%

where it is closed, because it is the product of a closed operator $A_0$

and a bounded operator $(M^\bot)^{-1}$.

The operator $A_1$ is densely defined in the Hilbert space

%

\begin{equation}\label{KNULL}

\fK_0 = (M \DA)^c.

\end{equation}

%

If Assumption~\ref{ASS4} (Case b) is fulfilled, we define instead

%

\begin{equation}\label{OpA2}

A_2 = (M^\bot)^{-1}A_0.

\end{equation}

%

This operator is closed on

%

\begin{equation}\label{DA2}

\Dom(A_2)=\{x\in P^\bot\DA\mid A_0x\in\Ran M\} =

A_0^{-1}\Ran M,

\end{equation}

%

because it is the product of a boundedly invertible operator $(M^\bot)^{-1}$  with a closed operator $A_0$.

The operator $A_2$ is densely defined in the Hilbert space

%

\begin{equation}\label{HNULL}

\fH_0 = (P^\bot\DA)^c.

\end{equation}

%

%

%%%%%%%%   ASSUMPTION 5   %%%%%%%%%%%%%%%

\begin{assumption}\label{ASS5}

$A_1$ (resp. $A_2$) generates a strongly continuous semigroup in $\fK_0$

(resp. $\fH_0$).

\end{assumption}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%%

%%%%%%%%%%%   THEOREM 1   %%%%%%%%%%%%%%%

\begin{theorem}\label{THEOREM 1}

The Assumptions \ref{ASS1}---\ref{ASS5} imply the following statements



\noindent (Case a):

For each initial value $z_0 \in \DA$

the degenerate Cauchy problem (\ref{DCP}) has the unique strict solution

%

\begin{equation}\label{SOLUTION A}

z(t) = {Z_A}(M^\bot)^{-1}e^{A_1 t}Mz_0.

\end{equation}

%



\noindent (Case b):

For each initial value $z_0 \in A^{-1}\Ran M$ the unique strict solution

of Eq.~(\ref{DCP}) is given by

%

\begin{equation}\label{SOLUTION B}

z(t) = {Z_A}e^{A_2t}P^\bot z_0.

\end{equation}

%

\end{theorem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{proof}

{\it (Case a):} Let $z_0\in\DA$.

Then $Mz_0\in\Dom(A_1)$, cf. Eq.~(\ref{DA1}).

By Assumption \ref{ASS5} the Cauchy problem

%

\begin{equation}\label{OCP1}

\frac{d}{dt}\,y(t) = A_1\,y(t),     \qquad y(0) = Mz_0

\end{equation}

%

has a unique solution $y(t) = \exp(A_1t)\,y(0)$

which is continuously differentiable and is in $\Dom(A_1)$ for all $t\ge 0$.

Since $(M^\bot)^{-1}$ is bounded,

$x = (M^\bot)^{-1}y$ is again continuously differentiable with $x(t)\in P^\bot\DA$.

Define $z(t) = {Z_A}x(t)$. Then $Mz = M^\bot x = y$ is continuously differentiable and

%

\begin{equation*}

Az(t) = A_0x(t) = A_1y(t) = \frac{d}{dt}\,y(t) = \frac{d}{dt}\,Mz(t).

\end{equation*}

%

This shows that $Az$ is continuous, and since $A^{-1}$ is bounded,

$z$ is also continuous.

Hence Eq.~(\ref{SOLUTION A})

is the unique strict solution of the degenerate Cauchy problem Eq.~(\ref{DCP})

in the sense of Definition \ref{DEFINITION 1}.



{\it (Case b):} For $z_0\in\DA$ with $Az_0\in\Ran M$ we have $P^\bot z_0\in\Dom(A_2)$,

cf. {Eq}. (\ref{DA2}).

Hence the Cauchy problem

%

\begin{equation}\label{OCP2}

\frac{d}{dt}\,x(t) = A_2\,x(t),     \qquad x(0) = P^\bot z_0

\end{equation}

%

has the unique solution $x(t)=\exp(A_2t)\,x(0)$, which is continuously differentiable

and in $\Dom(A_2)$ for all $t\ge 0$.

Hence $z(t)={Z_A}x(t)\in\DA$, and $Mz(t)=M^\bot x(t)$ is continuously differentiable,

because $M^\bot$ is bounded. Hence

%

\begin{equation*}

\frac{d}{dt} \,M z(t) = M^\bot A_2 x(t) = A_0 x(t) = A z(t)

\end{equation*}

%

is continuous.

This implies (by the boundedness of $A^{-1}$) that $z$ is continuous.

Therefore  Eq.~(\ref{SOLUTION B})

is the unique strict solution of Eq.~(\ref{DCP}).

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

%%

%%%%%%%%%%%%%   REMARKS   %%%%%%%%%%%%%%%

%

\begin{rem}\label{SIMPLIFY}

If $M^\bot$ is bounded and has a bounded inverse, then the cases a and b are equivalent

and

%

\begin{align*}

&\Dom(A_2) = (M^\bot)^{-1}\Dom(A_1),       \quad

A_2 = (M^\bot)^{-1} A_1 M^\bot,           \\

&e^{A_2 t} = (M^\bot)^{-1} e^{A_1 t} M^\bot.

\end{align*}

%

\end{rem}

%

\begin{rem}

If, in addition to the assumptions of the theorem, $A$ is self-adjoint in $\fH=\fK$,

and $M$ is self-adjoint and positive, then it is

useful to factorize the degenerate Cauchy problem in a more symmetric way.

If $M^\bot$ is bounded and has a bounded inverse, then the same is true for

the positive square root $(M^\bot)^{1/2}$. Write $u(t) = (M^\bot)^{1/2} x(t)$

to obtain the ordinary Cauchy problem $\dot u(t) = A_3 u(t)$ with

%

\begin{equation*}

A_3 = (M^\bot)^{-1/2} \,A_0 \,(M^\bot)^{-1/2}.

\end{equation*}

%

Here the generator $A_3$ is self-adjoint, if it is densely defined.

\end{rem}

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

\medskip

\noindent

The conditions necessary for Assumption \ref{ASS5} to hold are usually given

in terms of the resolvent of the generator (Hille-Yoshida theorem).

In the next lemma the resolvents of $A_1$ and $A_2$ are expressed in terms of

the operators $A$ and $M$.

%%

%%%%%%%%%%%%   LEMMA 4   %%%%%%%%%%%%%%%%

\begin{lemma}\label{LEMMA 4}

Under the Assumptions \ref{ASS1}---\ref{ASS4} we find

for $\lambda$ in the resolvent set $\rho(A_1)$ of $A_1$ (Case a)

%

\begin{equation*}

(A_1 - \lambda)^{-1}Q^\bot = M(A-\lambda M)^{-1}Q^\bot,

\end{equation*}

%

and for $\lambda \in \rho(A_2)$ (Case b)

%

\begin{equation*}

(A_2-\lambda)^{-1}P^\bot = P^\bot (A-\lambda M)^{-1}M.

\end{equation*}

%

\end{lemma}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{proof}

We note that $\Dom(M)\supset\DA$ and hence

$\Dom(M^\bot)\supset \Dom(A_0)$. Thus we calculate (in Case a):

%

\begin{align*}

(A_1-\lambda)^{-1}Q^\bot

  &= \bigl(A_0(M^\bot)^{-1}-\lambda M^\bot(M^\bot)^{-1}\bigr)^{-1}Q^\bot\\

  &= M^\bot(A_0-\lambda M^\bot)^{-1}Q^\bot\\

  &= M^\bot\bigl((A-\lambda M^\bot P^\bot){Z_A}\bigr)^{-1}Q^\bot\\

  &= M(A-\lambda M)^{-1}Q^\bot,

\end{align*}

%

and (in Case b):

%

\begin{alignat}{1}

(A_2-\lambda)^{-1}P^\bot

  &= \bigl((M^\bot)^{-1}A_0-\lambda (M^\bot)^{-1}M^\bot\bigr)^{-1}P^\bot\nonumber\\

  &= (A_0-\lambda M^\bot)^{-1}M^\bot P^\bot\nonumber\\

  &= \bigl((A-\lambda M^\bot P^\bot){Z_A}\bigr)^{-1}M\nonumber\\

  &= P^\bot (A-\lambda M)^{-1}M.\label{147}

\end{alignat}

%

In these calculations we used Eq.~(\ref{ZP}).

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

\medskip\noindent

Hence the assumptions on the resolvents of $A_1$ (or $A_2$) can be expressed

in terms of the operators

%

\begin{equation*}

S(\lambda) = M(A-\lambda M)^{-1},\qquad\mbox{or}\quad

R(\lambda) = (A-\lambda M)^{-1}M.

\end{equation*}

%

If, for some $\lambda\in{\mathbb C}$, the operator $S(\lambda)$ is 

bounded and everywhere defined, we say that $\lambda$ belongs to

the generalized resolvent set $\rho_S(A)$. If the same holds

for $R(\lambda)$, we say $\lambda \in \rho_R(A)$.

The operators $S(\lambda)$ and $R(\lambda)$ are

pseudoresolvents, i.e., they satisfy the resolvent equation

%

\begin{equation*}

S(\lambda)-S(\mu) = (\lambda-\mu)S(\lambda)S(\mu),

\quad\mbox{all $\lambda$ and $\mu$ in $\rho_S(A)$,}

\end{equation*}

%

and similarly for $R$. Thus the generalized resolvent sets are

open subsets of ${\mathbb C}$.

If Assumptions \ref{ASS1}---\ref{ASS4} hold, then

%

\begin{equation*}

\rho_S(A) = \rho(A_1),\qquad\mbox{or}\quad

\rho_R(A) = \rho(A_2)

\end{equation*}

%

(in Case a or b, respectively).

From general semigroup theory we find, e.g., the following sufficient condition replacing

Assumption \ref{ASS5}:

%

%

%%%%%%%%   ASSUMPTION 6   %%%%%%%%%%%%%%%

\begin{assumption}\label{ASS6}

Let $\{\lambda\in{\mathbb C}\mid \mbox{Re}\,\lambda>\omega\} \subset \rho_S(A)$

for some real constant $\omega$, and assume that there exists a constant $1\ge K>0$ such that

%

\begin{equation}\label{RESOLVENT COND}

\|M(A-\lambda M)^{-1}Q^\bot\|\le \frac{K}{\mbox{Re}\,\lambda-\omega},

\quad\mbox{for all $\lambda$ with $\mbox{Re}\,\lambda>\omega$,}

\end{equation}

%

(or assume a similar condition on $P^\bot(A-\lambda M)^{-1}M$ in Case b).

\end{assumption}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%%

%%%%%%%%%%%%%   REMARKS   %%%%%%%%%%%%%%%

%

%

\begin{rem}

In view of Remark \ref{WLG} we can assume without loss of generality

that the constant $\omega$ in Eq.~(\ref{RESOLVENT COND}) is negative.

\end{rem}

%

\begin{rem}

The operator $A_1$ is invertible, its inverse is given by

%

\begin{equation*}

(A_1)^{-1}= M^\bot (A_0)^{-1} = MA^{-1}\restriction Q^\bot\fK.

\end{equation*}

%

Since $(A_1)^{-1}$ is closed and defined on all of $Q^\bot\fK$, it is bounded.

The operator $T=MA^{-1}$ is the object of investigation in Favini's paper

\cite{5} and, e.g., in \cite{7},  \cite{11}.

Instead of (\ref{DCP}), Favini considers the

equation

%

\begin{equation*}

\frac{d}{dt}\,T\,z(t) = z(t)

\end{equation*}

%

under suitable assumptions on $T$. These assumptions are formulated in terms

of the operator $A(A-\lambda M)^{-1}$ and insure analyticity of the semigroup

generated by the inverse of $T\restriction Q^\bot\fK$ (which is just $A_1$).

\end{rem}

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



In the general case we expect no continuous dependence on the initial conditions,

because the operator ${Z_A}$ in Eqs.~(\ref{SOLUTION A}) and (\ref{SOLUTION B}) is not continuous.

In special cases, however, the time evolution defined in Theorem~\ref{THEOREM 1}

extends to a bounded operator.

%

%%%%%%%%%%%%% EXAMPLE %%%%%%%%%%%%%%%%%%%

%

\begin{example}\label{EXAMPLE 2}

For simplicity, we assume that $M^\bot$ is bounded and has a bounded inverse.

Hence $\rho_S(A) = \rho_R(A)$.

Let $A_2$ be the generator of a bounded holomorphic semigroup

(see \cite{12}, Section IX.1.6).

This is the case, if $\rho_R(A)$ contains a sector

$|\mbox{arg}\,z|<\frac{\pi}{2} + \delta$, $\delta>0$, and if

%

\begin{equation}\label{ANALYTICITY}

\|(A-\lambda M)^{-1}M\|\le \frac{K}{|\lambda|}

\quad\mbox{for $|\mbox{arg}\,\lambda|<\frac{\pi}{2} + \delta - \epsilon$}

\end{equation}

%

holds for any $\epsilon>0$, with a suitable constant $K>0$.

Thus $\exp{A_2 t}$ is given by a Dunford-Taylor integral (where $\Gamma$ is a suitable

path within the sector mentioned above, see [12], p.~490)

%

\begin{equation}\label{DUNFORD}

e^{A_2 t} = -\frac{1}{2\pi\rmi}

\int_{\Gamma} e^{\lambda t}\,(A_2-\lambda)^{-1}\,d\lambda.

\end{equation}

%

Inserting this into Eq.~(\ref{SOLUTION B}) and using

%

\begin{equation*}

{Z_A}\,(A_2-\lambda)^{-1}\,P^\bot = (A-\lambda M)^{-1}M = R(\lambda)

\end{equation*}

%

(cf. Eq.~(\ref{147})) we obtain immediately

%

\begin{equation}\label{ANALYTIC SOLUTION}

z(t) = -\frac{1}{2\pi\rmi}

\Bigl(\int_\Gamma e^{\lambda t}

\,R(\lambda)\,d\lambda\Bigr)z_0.

\end{equation}

%

It is clear that this defines a bounded evolution operator

on all of $\fH$. In this case $z(t)$ depends continuously on the initial value.

As $t\to 0$, $z$ might be discontinuous (in $t$), but $P^\bot z(t)\to P^\bot z_0$.

For $z_0\in\DA$, however, $z$ is a strict solution and is therefore continuous in $t$.

\end{example}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%



\smallskip

\noindent



%%%%%%%%%%%%%%%%%%%%%%%%%-------SECTION-------%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Example: The Dirac equation in the nonrelativistic limit}\label{sec2}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%

%\setcounter{equation}{0}

\setcounter{prop}{0}


\bigskip\noindent

%

%

In this section we show that the Dirac equation of relativistic quantum mechanics

has a singularity in the nonrelativistic limit which can be understood

in the framework of degenerate Cauchy problems.

The Dirac equation is a system of linear

partial differential equations which has the following abstract structure

(after subtraction of the ``rest-energy'' $mc^2$,

see \cite{18}, Chapters~5 and~6 for details)

%

\begin{equation}\label{D1}

\frac{d}{dt}\,\psi(t) = -\rmi \bigl(cD + mc^2(\tau - \id) + V\bigr)\,\psi(t).

\end{equation}

%

Here $m>0$ denotes a mass, $c$ the speed of light.

$\tau$ is a unitary involution (which means $\tau^2=\tau^*\tau=\tau\tau^*=\id$) in the

given Hilbert space $\fH$,

the operator $D$ is self-adjoint and anticommutes with $\tau$,

%

\begin{equation*}

\tau\Dom(D)\subset\Dom(D),\qquad D\tau+\tau D=0 \quad\mbox{on $\Dom(D)$}.

\end{equation*}

%

The operator $V$ is symmetric, commutes with $\tau$,



%

\begin{equation*}

\tau\Dom(V)\subset\Dom(V),\qquad V \tau-\tau V=0 \quad\mbox{on $\Dom(V)$},

\end{equation*}

%

and is relatively bounded with respect to $D$ (but not necessarily with relative bound less than $1$).



\begin{rem}\label{REMARK DIR1}

Here we work in a setting which is much more general than relativistic quantum mechanics.

In \cite{18} it is shown how the concrete special cases fit into the abstract approach.

Here we only mention that 

the Dirac equation in its usual form is obtained in the special case $\fH=\C^4 \otimes L^2(\R^3)$,

$D=-\rmi\alpha\cdot\nabla$,

$\tau=\beta$, where $\alpha_1$, $\alpha_2$, $\alpha_3$, and $\beta$ are the famous Dirac matrices.

In this case $V$ is multiplication by a diagonal matrix-valued function (electric or

scalar potential).

\end{rem}



The Hilbert space decomposes into a direct sum of the eigenspaces belonging to the

eigenvalues $\pm 1$ of $\tau$, $\fH = \fH_+\oplus\fH_-$.

We write $\fH_\pm = P_\pm\fH$, where $P_\pm$ are the orthogonal projections

%

\begin{equation}\label{P+-}

P_\pm = {\textstyle{\frac{1}{2}}}\,\bigl(\id \pm \tau\bigr).

\end{equation}

%

In order to investigate the nonrelativistic limit ($c=\infty$) we

make for $0<c<\infty$ the transformation $\psi(t)\to z(t) = (P_+ + cP_-)\psi(t)$

and multiply (\ref{D1}) with $P_++P_-/c$ from the left.

This gives the following equation for $z$:

%

\begin{equation*}

\frac{d}{dt}\,\bigl(P_+ + P_-/c^2\bigr)\,z(t) =

-\rmi \bigl(D + VP_+ - 2mP_- + VP_-/c^2\bigr)\,z(t).

\end{equation*}

%

In the nonrelativistic limit $c=\infty$ this becomes formally

%

\begin{equation}\label{nrlim}

\frac{d}{dt}\,P_+\,z(t) =

-\rmi \bigl(D + VP_+ - 2mP_-\bigr)\,z(t).

\end{equation}

%

This is a degenerate Cauchy problem of the form (\ref{DCP}), where $M=P_+$ is a self-adjoint projection operator

(hence it is bounded and has closed range and both cases (a) and (b) in Assumption~\ref{ASS4} apply here).

The theory described in Section~\ref{sec1} gives the tools for the factorization of the degenerate equation~(\ref{nrlim}).



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%THEOREM

\begin{theorem}\label{THEOREM3.1}

Under the above assumptions $A = -\rmi(D + VP_+ - 2mP_-)$ is closed on the domain

%

\begin{equation}\label{DEFDA}

\DA = \bigl\{ z \in \fH \bigm| P_+z \in \Dom(D^2), P_-z=\frac{1}{2m}\,D\,P_+z \bigr\}.

\end{equation}

%

For all initial values $z_0\in\DA$ the degenerate abstract Dirac equation (\ref{nrlim}) has a unique

strict solution $z(t)$ which

is in $\DA$ for all $t$. Moreover, $z(t)$ is given by

%

\begin{equation}\label{DirRes}

z(t) = \Bigl(1+\frac{1}{2m}\,D\Bigr)\, \exp(A_0 t) \,P_+\,z_0,

\end{equation}

%

where

%

\begin{equation}\label{DEFA0}

A_0 = -\rmi \Bigl(\frac{D^2}{2m} + V\Bigr) P_+.

\end{equation}

%

The operator $\rmi A_0$ is self-adjoint on $\Dom(A_0)=P_+\DA = \Dom(D^2)P_+ \subset \fH_+$.

\end{theorem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\begin{proof}

First we note that $V$ is bounded relative to $D^2$ with infinitesimal relative bound,

hence $D^2/2m+V$ is self-adjoint on $\Dom(D^2)$. The self-adjointness of $\rmi A_0$ follows immediately.

In the notation of Section~\ref{sec1} we have $Q=P=P_-$, because $M=P_+$ is a self-adjoint projection.

Note that $D$ maps $\Ker M\cap \Dom(D)$ into $\Ran M$ and $\Ran M\cap\Dom(D)$ into $\Ker M$. 

We find

%

\begin{align*}

&Q A P^\bot = P_-AP_+ = -\rmi D P_+ = -\rmi P_- D, \\

&P^\bot A Q = P_+AP_- = -\rmi D P_- = -\rmi P_+ D, \\

&Q A P      = P_-AP_- = 2m\rmi P_-,                \\

&Q^\bot A P^\bot = P_+ A P_+ = -\rmi VP_+.

\end{align*}

%

The domain $\DA=\{z\in\Dom(A)\mid QAz=0\}$ is according to Lemma~\ref{LEMMA 2} characterized as the set

of all $z$ with $P_+z$ and $P_-z$ in $\Dom(A)=\Dom(D)$ and

%

\begin{equation*}

P_-z=\frac{1}{2m}\,D\,P_+z.

\end{equation*}

%

From this we may conclude that $P_+z\in\Dom(D^2)$. Hence $\DA$ can be described as in Eq.~(\ref{DEFDA}), $P_+\DA=P_+\Dom(D^2)$,

and we can define the operator $A_0$ as in Eq.~(\ref{A0sym}) to end up with Eq.~(\ref{DEFA0}).

For all $z\in\DA$ we have $Az=A_0P_+z$ and the closedness of $A$ on $\DA$ follows easily

from the self-adjointness (and hence closedness) of $A_0$ on $P_+\DA$.

We have $A_0=A_1=A_2$ because $M^\bot=(M^\bot)^{-1} = 1\restriction \fH_+$.

Let $x(t)=\exp(A_0 t) P_+z_0$ be a solution of the factorized problem

with an initial value $P_+z_0$ in $\Dom(A_0)$.

Let ${Z_A}$ be defined as in Eq.~(\ref{DEFINE Z}) as the operator which

maps $P^\bot\Dom(D^2)$ onto $\DA$,

%

\begin{equation*}

{Z_A}= \Bigl(1+\frac{1}{2m}\,D\Bigr)\,P_+.

\end{equation*}

%

We note that

$Z_A$ is well defined (but generally unbounded) on $\Dom(D)\supset \Dom(A_0)$.

In order to show the continuity of the solution $z(t) = Z_A x(t)$ of the degenerate problem,

we had to assume that $A^{-1}$ is bounded (see Assumption~\ref{ASS3}).

This is not true here, but

a somewhat tedious calculation shows that

%

\begin{align*}

(\rmi A-\lambda M)^{-1} &= \bigl(D + (V-\lambda)P_+ - 2mP_-\bigr)^{-1}\\

& = \Bigl(Z_A + \frac{D + \lambda}{2m} P_-\Bigr)\,R(\lambda)

+Z_A\, \bigl(R(\lambda)-R_0(\lambda)\bigr)\,\frac{D}{2m}\,P_-,

\end{align*}

%%

where

%

\begin{equation*}

R(\lambda) = \Bigl(\frac{D^2}{2m} + VP_+ - \lambda\Bigr)^{-1},\qquad

R_0(\lambda) = \Bigl(\frac{D^2}{2m} - \lambda\Bigr)^{-1}.

\end{equation*}

%%

Since $D^2/2m$ is nonnegative and $V$ is infinitesimally relatively bounded, $R(\lambda)$ and

$R_0(\lambda)$ are bounded for $\lambda$ sufficiently negative or for $\lambda$ with

$\mbox{Im}\,\lambda\neq 0$. This implies also the boundedness of $(\rmi A-\lambda M)^{-1}$, because $DR(\lambda)D$ and

$DR_0(\lambda)D$ are bounded.

Hence there exists a $\lambda_0\in\C$ such that $(A-\lambda_0 M)^{-1}$ is bounded and we can apply

Theorem~\ref{THEOREM 1} with $A$ replaced by $A-\lambda_0 M$ (cf. Remark~\ref{WLG}).

Note that $Z_{A-\lambda_0 M}=Z_A$ and $A_0$ has to be replaced

by $A_0-\lambda_0$. Theorem~\ref{THEOREM 1} now implies that $y(t)=Z_A\exp(A_0 t-\lambda_0 t) P_+z_0$

is a strict solution of the degenerate problem with $A-\lambda_0 M$.

The solution $z(t)$ of the original problem is, according to Remark~\ref{WLG}, finally given by $\exp(\lambda_0 t)y(t)$.

%

\end{proof}

%

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

\medskip\noindent

%

\begin{rem}

For the applications in quantum mechanics the reduced Cauchy problem in

$\fH_+ = (\Ker M)^\bot$ is the Schr\"odinger equation:

If $D$ is defined as in Remark~\ref{REMARK DIR1} above,

then the properties of Dirac matrices imply that $D^2 = -\Delta$ is the Laplace operator.

\end{rem}



\smallskip\noindent

\begin{rem}

The time evolution $z(t) = {Z_A}\,\exp(A_0 t)\,P_+z_0$ can be defined for all 

initial values $z_0\in\Dom(D^2)$

and $z(t)$ is even a strict solution if $z_0\in\DA$.

Nevertheless the time evolution operator is unbounded and cannot be

extended to all of $\fH$.



For example, if $V=0$, ${Z_A}$ commutes with $\exp(A_0 t)$ and

%

\begin{equation*}

z(t) = \exp\left({-\rmi\frac{D^2}{2m} t}\right) \Bigl(1+\frac{1}{2m}\,D\Bigr)P_+z_0

\end{equation*}

%

implies that

%

\begin{equation}\label{z(t)}

\|z(t)\|^2 = \|P_+z_0\|^2 + \frac{1}{2m}\|DP_+z_0\|^2.

\end{equation}

%

Hence the domain of the evolution operator cannot be extended to initial values $z_0\not\in\Dom(D)$.

Eq.~(\ref{z(t)}) implies that (in the case $V=0$) ${Z_A}\,\exp(A_0 t)\,P_+$ is continuous

from $\Dom(D)$, equipped with the graph-norm, into the Hilbert space $\fH$.

A more detailed investigation the time evolution of the Dirac equation in the nonrelativistic limit, scattering

theory and relativistic corrections in the framework of degenerate Cauchy problems

will be published elsewhere.

\end{rem}

\smallskip

\noindent



%%%%%%%%%%%%%%%%%%%%%%%%%-------SECTION-------%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{The inhomogeneous equation}\label{sec4}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%

%\setcounter{equation}{0}

\setcounter{prop}{0}

\bigskip\noindent

%

%

In this section we briefly discuss the modifications necessary in order to treat the

inhomogeneous equation

%

\begin{equation}\label{INHOMOGEN 1}

\frac{d}{dt}\,M\,z(t) = A\,z(t) + f(t).

\end{equation}

%

We make the following simplifying assumption.

%%

%%%%%%%%   ASSUMPTION 7   %%%%%%%%%%%%%%%

\begin{assumption}\label{ASS7}

In addition to Assumptions~\ref{ASS1}--\ref{ASS5}

let $M^\bot = M\restriction P^\bot\fH$ be bounded and have a

bounded inverse.

Furthermore, let $A_0$ be densely defined in $P^\bot\fH$.

\end{assumption}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%

%

Some consequences of this assumption are described in the Remark~\ref{SIMPLIFY}.

We also have $\fH_0=P^\bot\fH$, and $\fK_0 = Q^\bot\fK$,

cf. Eqs.~(\ref{KNULL}) and (\ref{HNULL}).



A necessary condition for $z$ to be a solution of Eq.~(\ref{INHOMOGEN 1}) is now given by

%

\begin{equation}\label{INHOMOGEN 2}

z(t) = {Z_A}P^\bot z(t) - (QAP)^{-1}Qf(t), \quad\mbox{for all $t\ge 0$}

\end{equation}

%

where ${Z_A}$ is defined as in Eq.~(\ref{DEFINE Z}).

This is an immediate consequence of the requirement that $Az(t) + f(t)$ be in $\Ran M$.

We can easily restrict (\ref{INHOMOGEN 1}) to

%

\begin{align}

\frac{d}{dt}\,M^\bot\,x(t) &= A_0\,x(t) + \bigl(Q^\bot - Q^\bot A P(QAP)^{-1}Q\bigr) f(t)\nonumber\\

                           &= A_0 \,x(t) + Y_A\,f(t),          \label{INHOMOGEN 3}

\end{align}

%

where $A_0 = A Z_A = Y_A A$ is given as in Eqs.~(\ref{FACTOR A0 1}) and (\ref{FACTOR A0 2}).

Under the assumptions made above there are two equivalent methods of 

dealing with Eq.~(\ref{INHOMOGEN 3})

(cf. cases (a) and (b) described in Section \ref{sec1}),

either by transforming to

%

\begin{equation}\label{INHOMOGEN 4}

\frac{d}{dt}\,y(t) = A_1\,y(t) + Y_A \,f(t),

\end{equation}

%

(where $y(t) = M^\bot x(t)$) or to

%

\begin{equation}\label{INHOMOGEN 5}

\frac{d}{dt}\,x(t) = A_2\,x(t) + (M^\bot)^{-1}Y_A \,f(t),

\end{equation}

%

where (as in Section \ref{sec1})

%

\begin{equation*}

A_1 = A_0(M^\bot)^{-1},\quad A_2 = (M^\bot)^{-1}A_0.

\end{equation*}

%

Since we assume that $A^{-1}$ is bounded, we can define $g(t) = P^\bot A^{-1}f(t)$ and

write Eq.~(\ref{INHOMOGEN 5}) as

%

\begin{equation}\label{INHOMOGEN 7}

\frac{d}{dt}\,x(t) = A_2\,\bigl(x(t) + g(t)\bigr).

\end{equation}

%

Whenever $g(t)$ is in $\Dom(A_2)=P^\bot\DA$,

a solution of this equation is given by:

%

\begin{align*}

x(t) &= e^{A_2 t}P^\bot z_0 + \int_0^t e^{A_2 (t-s)}A_2 g(s)\,ds\\

     &= e^{A_2 t}P^\bot z_0 + A_2 \int_0^t e^{A_2 (t-s)}g(s)\,ds.

\end{align*}

%

It is well known (see \cite{12}, Section IX.1.5) that the expression

%

\begin{equation}\label{INHOMOGEN 8}

h(t) = A_2 \int_0^t e^{A_2 (t-s)}g(s)\,ds

\end{equation}

%

is well defined whenever $g(t)$ is a continuously differentiable function

of $t$ (which is the case if $f$ is continuously differentiable).

In order to obtain a solution of the original degenerate problem

we have to build the expression

%

\begin{equation}\label{INHOMOGEN 9}

z(t) = {Z_A} x(t)  - (QAP)^{-1}Qf(t)

\end{equation}

%

which is only possible if $x(t)\in\Dom(Z_A)$.



%

%%%%%%%%%%%%   REMARK   %%%%%%%%%%%%%%

%

\begin{rem}

In the important special case where

the factorization operator $Z_A$ is bounded, the solution

$z(t)$ is well defined for all initial values $z_0$ and

for all continuously differentiable functions $f$.

It is a strict solution in the sense of Definition \ref{DEFINITION 1}

only if $z(t)\in\DA$ and $z_0\in\DA$.

\end{rem}

% 

\begin{rem}

Another special case, where $z(t)$, given by Eq.~(\ref{INHOMOGEN 9}), is

well defined is obtained if $A$ and $M$ satisfy Eq.~(\ref{ANALYTICITY})

(see, e.g., \cite{11}).

In this case $A_2$ generates a bounded analytic semigroup and

$h(t)$ is in $\Dom(A_2)$, because $A_2^2\exp(A_2t)$ is bounded.

\end{rem}

%

\smallskip\noindent

Finally we discuss the inhomogeneous degenerate equation for functions

$f$ with values in $\Ran M$. The following theorem is a direct

analogue of Theorem \ref{THEOREM 1}. Under the simplifying assumptions of this

section (cf. also Remark \ref{SIMPLIFY}) it can be formulated as follows.



%

%%

%%%%%%%%%%%   THEOREM 2   %%%%%%%%%%%%%%%

\begin{theorem}\label{THEOREM 2}

Let $f$ be a continuously differentiable function with values in $\Ran M$.

Under Assumption~\ref{ASS7},

Eq.~(\ref{INHOMOGEN 1}) has a unique strict solution $z$ for each initial

value $z_0\in\DA$. It is given by

%

\begin{align*}

z(t) &= {Z_A}\,x(t) - (QAP)^{-1}Q\,f(t) = {Z_A}\,x(t),\\

x(t) &= e^{A_2 t}P^\bot z_0 + \int_0^t e^{A_2 (t-s)}\,(M^\bot)^{-1} f(s)\,ds.

\end{align*}

%

\end{theorem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{proof} 

For every initial value $P^\bot z_0\in\Dom(A_2) = P^\bot\DA$

the function $t\to x(t)$ is continuously differentiable, and

$x(t)\in\Dom(A_2)= P^\bot\DA$ for all $t\ge 0$

(see \cite{12}, Section IX.1.5).

Therefore $z(t) = {Z_A}x(t)$ is well defined.

Using the continuity of $M^\bot$ we find that $Mz(t)=M^\bot x(t)$

is continuously differentiable with

%

\begin{align*}

\frac{d}{dt}\,Mz(t) &= M^\bot\,\frac{d}{dt}\,x(t)                     \\

                    &= M^\bot\bigl(A_2\,x(t) + (M^\bot)^{-1} f(t)\bigr) \\

                    &= Az(t) + f(t).

\end{align*}

% 

This shows that $Az(t)$ and hence $z(t) = A^{-1}Az(t)$ is continuous.

Hence $z$ is a strict solution in the sense of Definition \ref{DEFINITION 1}.

This finishes the proof.

Let us finally

note that $x(t)$ can be defined for all $z_0$, and for each $t\ge 0$

%

\begin{equation*}

f(t) = \int_0^t e^{A_2 (t-s)}\,(M^\bot)^{-1} f(s)\,ds

\end{equation*}

%

is in $\Dom(A_2)$ and hence ${Z_A}f(t)$ is well defined.

But ${Z_A}\exp(A_2 t)P^\bot$ need not be defined for

initial values not in $\DA$, because ${Z_A}$ may be unbounded.

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\medskip



%%%%%%%%%%%%%%%%%%%%%%%%%-------SECTION-------%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Some remarks about control theory}\label{sec3}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%

%\setcounter{equation}{0}

\setcounter{prop}{0}

\bigskip\noindent

%

%

In this section we make some remarks concerning the basic concepts of

control theory under the simplifying assumptions of the previous section.

Leaving a more exhaustive treatment of degenerate control problems to further

investigation, our purpose here is to illustrate the possible usefulness of

our approach by discussing a special case to be described in the following.



In order to define a degenerate control problem,

we need an operator $B$ mapping a control space

$U$ into $\fK$, and an operator $C$ mapping $\fH$ into the output space $V$,

$U$ and $V$ being suitable Banach spaces.

The system of equations

%

\begin{align}

\frac{d}{dt}\,M\,z(t) &= A\,z(t) + B\,u(t),\label{DEGCONTROL 1}\\

v(t) &= C z(t).\label{DEGCONTROL 2}

\end{align}

%

will be called the control problem defined by the ordered quadrupel $(M,A,B,C)$.

Besides the Assumption \ref{ASS7} we assume that $B$ and $C$ are bounded and satisfy

%

\begin{equation}\label{CONSISTENCY}

\Ran B \subset (\Ran M)^c,\qquad \Ker C \supset \Ker M.

\end{equation}

%

Hence Theorem~\ref{THEOREM 2} applies to this situation and

Eq.~(\ref{DEGCONTROL 1}) has a unique strict solution $z(t,u,z_0)$ whenever

$Bu$ is continuously differentiable and $z_0\in\DA$.



%

%%%%%%%%%%%%   REMARK   %%%%%%%%%%%%%%

%

\begin{rem}

The output $v(t) = Cx(t)$ is well defined and continuous in $t$

for all initial values $z_0\in\fH$.

This is because by Eq.~(\ref{CONSISTENCY}) $Cz=Cx$ holds for any

$z$ and $x=P^\bot z$.

Hence $C{Z_A}=CP^\bot {Z_A} = C$ is bounded and

$C{Z_A}\exp(A_2 t)P^\bot$ extends to all of $\fH$.

\end{rem}

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

\smallskip\noindent

We will also consider the ``dual problem'' defined by $(-M^*,A^*,C^*,B^*)$.

Because of Eq.~(\ref{CONSISTENCY}) we find that

%

\begin{equation}\label{CONSISTENCY 2}

\Ran C^* \subset (\Ran M^*)^c,\qquad \Ker B^* \supset \Ker M^*.

\end{equation}

%

Under the Assumptions~\ref{ASS1}--\ref{ASS5} of Section~\ref{sec1}

we can easily restrict (\ref{DEGCONTROL 1}), (\ref{DEGCONTROL 2}) to

%

\begin{align}

\frac{d}{dt}\,M^\bot\,x(t) &= A_0\,x(t) + B\,u(t),\label{RESTR DCP}\\

v(t) &= C x(t)\nonumber.

\end{align}

%

where $A_0$ is given as in Eqs.~(\ref{A0x}) and (\ref{A0sym}).



Because of Eq.~(\ref{CONSISTENCY 2}), the dual problem

can be restricted in a similar way.

When defining the factorization operators $Z_{A^*}$ and $Y_{A^*}$ for the dual system

one has to take into account that the roles of $Q$ and $P$ are interchanged.

Hence

%

\begin{equation*}

Z_{A^*}= Q^\bot - Q(PA^*Q)^{-1}PA^*Q^\bot = (Y_A)^*,\qquad Y_{A^*} = (Z_A)^*

\end{equation*}

%

and we can define $(A^*)_0 = A^*Z_{A^*}=Y_{A^*}A^*$. We see immediately that

%

\begin{equation}\label{SYMMETRY}

(A^*)_0 = P^\bot A^* Q^\bot - P^\bot A^* Q (PA^*Q)^{-1}PA^*Q^\bot = (A_0)^*.

\end{equation}

%



With Assumption~\ref{ASS7} the two formulations Case~a and b of Section~\ref{sec1}

are equivalent,

and we have two equivalent methods of dealing with (\ref{RESTR DCP}):

Method (a) shows that the control problem

$(M,A,B,C)$ in $\fH$ can be reduced to $(\id,A_1,B_1,C_1)$ in

$Q^\bot\fK$, where

%

\begin{equation*}

A_1 = A_0(M^\bot)^{-1},\quad B_1 = B,\quad C_1 = C(M^\bot)^{-1}.

\end{equation*}

%

On the other hand, by method (b) we obtain an equivalent problem in $P^\bot\fH$

which is defined by the operators $(\id,A_2,B_2,C_2)$, where

%

\begin{equation*}

A_2 = (M^\bot)^{-1}A_0,\quad B_2 = (M^\bot)^{-1} B,\quad C_2 = C.

\end{equation*}

%

The dual problem $(-M^*,A^*,C^*,B^*)$ can be factorized in a similar way.

This leads to an

ordinary control problem $(-\id,(A^*)_1,(C^*)_1,(B^*)_1)$ in $P^\bot\fH$

(=factorization by method a) 

and to

$(-\id,(A^*)_2,(C^*)_2,(B^*)_2)$ in $Q^\bot\fK$ (method b).

Using Eq.~(\ref{SYMMETRY}) and

%

\begin{equation*}

(M^*)^\bot = (M^\bot)^*,

\end{equation*}

%

we obtain the following relations

%

\begin{align*}

&(A^*)_1 = (A^*)_0(M^{*\bot})^{-1} = ((M^\bot)^{-1}A_0)^* = (A_2)^*,

\\

&(C^*)_1 = C^* = (C_2)^*,\\

&(B^*)_1 = B^*(M^{*\bot})^{-1} = ((M^\bot)^{-1} B)^* = (B_2)^*,

\end{align*}

%

and by similar calculations

%

\begin{equation*}

(A^*)_2 = (A_1)^*,\quad (C^*)_2 = (C_1)^*,\quad (B^*)_2 = (B_1)^*.

\end{equation*}

%

In the control theory dealing with nondegenerate systems

it is customary to define a dual system for each control problem.

According to this definition, the dual of the ordinary control problem

$(\id,A_1,B_1,C_1)$ obtained by method (a) is given by $(-\id,(A_1)^*,(C_1)^*,(B_1)^*)$.

From the result above this dual system is the same as $(-\id,(A^*)_2, (C^*)_2,(B^*)_2)$,

which is just the control problem obtained from $(-M^*$, $A^*$, $C^*$, $B^*)$

by factorization via method (b).



Hence we find the following diagram which exhibits the compatibility

of duality and factorization:

%

\begin{equation*}

\begin{array}{ccc}

(M,A,B,C) &\overset{\mbox{\scriptsize duality}}\longleftrightarrow &(-M^*,A^*,C^*,B^*)\\

& & \\

\!\!\!\mbox{\scriptsize (a)}\,\Big\updownarrow & &    \,\,\,\Big\updownarrow\,\mbox{\scriptsize (b)}\\

& & \\

(1,A_1,B_1,C_1)   &\overset{\mbox{\scriptsize duality}}\longleftrightarrow  &

{(-1,(A^*)_2,(C^*)_2,(B^*)_2)}

\end{array}

\end{equation*}

%

There is, of course, a corresponding diagram with (a) and (b) exchanged:

%

\begin{equation*}

\begin{array}{ccc}

(M,A,B,C) &\overset{\mbox{\scriptsize duality}}\longleftrightarrow &(-M^*,A^*,C^*,B^*)\\

& & \\

\!\!\!\mbox{\scriptsize (b)}\,\Big\updownarrow & &    \,\,\,\Big\updownarrow\,\mbox{\scriptsize (a)}\\

& & \\

(1,A_2,B_2,C_2)   &\overset{\mbox{\scriptsize duality}}\longleftrightarrow  &

{(-1,(A^*)_1,(C^*)_1,(B^*)_1)}

\end{array}

\end{equation*}

%

This shows that methods (a) and (b) are dual methods of factorization.



\smallskip

\noindent

In control theory one is also interested in the possibility of feedback control.

This is achieved by an operator $K$ defined on a domain $\Dom(K)\subset\fH$

and mapping into the control space $U$.

The feedback-control system is now defined by

%

\begin{equation}\label{FEEDBACK}

\frac{d}{dt}\,M\,z(t) = (A + BK)\, z(t).

\end{equation}

%

We assume that $\Dom(K)\supset\Dom(A)$, hence the operator $BK$ is relatively bounded with

respect to $A$. We can factorize (\ref{FEEDBACK}) as in Section~\ref{sec1}.

The range condition (\ref{CONSISTENCY}) implies that $QB=0$ and hence $\Dom_{A+BK} = \DA$.

We define $(A+BK)_0 = (A+BK)Z_{A+BK}$ on $P^\bot\DA$, where

%

\begin{align*}

Z_{A+BK} &= \id - \bigl(Q(A+BK)P\bigr)^{-1}Q(A+BK)\\

	        &= \id - (QAP)^{-1}QA = Z_A \quad\mbox{on $P^\bot\DA$.}

\end{align*}

%

For the factorized problem we therefore obtain (by method a) the generator $A_1 + B_1K_1$,

where

$K_1 = K{Z_A}(M^\bot)^{-1}$ and (by method b) $A_2+B_2K_2$ where $K_2 = K$

(here $A_j$ and $B_j$, $j=1,2$, are defined as above).

The solution of the degenerate feedback-system (\ref{FEEDBACK}) is therefore given by

%

\begin{equation*}

z(t) = S_{A+BK}(t)\, z_0

\end{equation*}

%

with the (possibly unbounded) evolution operator

%

\begin{equation*}

 S_{A+BK}(t) = Z_A\,e^{(A_2+B_2K_2)t}\,P^\bot = Z_A\,(M^\bot)^{-1}\,e^{(A_1+B_1K_1)t}\,M^\bot\,P^\bot.

\end{equation*}

%





\noindent

The standard notions of control theory can now be generalized to degenerate problems

as follows.

%

%%%%%%%%%%%%% DEFINITION %%%%%%%%%%%%%%%%%%%

%

\begin{definition}\label{DEFINITION 4 }

The degenerate control system $(M,A,B,C)$ is called {\em approximately controllable},

if for all $\epsilon>0$, all $T>0$ and all $z_0, z_1\in (\DA)^c$

there exists a control function $u$ such that $\|z(T,u,z_0)-z_1\|<\epsilon$.



It is called {\em observable}, if the dual system $(-M^*,A^*,C^*,B^*)$ is approximately

controllable.



The degenerate control system is called {\em stabilizable}, if there exists a bounded operator

$K$ such that $ S_{A+BK}(t)$ is bounded for $t\ge 0$ with $\|S_{A+BK}(t)\|\le \mu\exp(-\omega t)$,

$\mu\ge 1$,  $\omega > 0$.



The degenerate control system is called {\em detectable}, if the dual system is stabilizable.

\end{definition}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%

%%%%%%%%%%%%% THEOREM %%%%%%%%%%%%%%%%%%%

%

\begin{theorem}

Let $(M,A,B,C)$ satisfy the assumptions formulated in this section and assume that the

factorization operator $Z_A$ is bounded.

Then the degenerate system $(M,A,B,C)$ is controllable (observable, stabilizable, detectable)

if and only if the

ordinary control system $(1,A_j, B_j, C_j)$ ($j=1$ or $2$) is controllable  (observable, stabilizable, detectable).

\end{theorem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

\begin{proof}

The proof is immediate after the observations made in this section. If $Z_A$ is bounded,

then it extends to a bounded operator defined everywhere on $(P^\bot\DA)^c=P^\bot\fH$

and maps this subspace onto $(\DA)^c$.

A solution of the degenerate problem can thus be defined according to Theorem~\ref{THEOREM 2}

for all initial values $z_0$ and is given by

%

\begin{align*}

z(t,u,z_0) &= {Z_A}\,x(t,u,P^\bot z_0),\\

x(t,u,x_0) &= e^{A_2 t}\,x_0 + \int_0^t e^{A_2 (t-s)}\,B_2\,u(s)\,ds.

\end{align*}

%

Here $x(t,u,x_0)$ is a solution of the system factorized by method (b). 

For controllability, we just note that $\|z(T,u,z_0)-z_1\| \le \|Z_A\|\,\|x(T,u,x_0)-x_1\|$,

where $x_{0,1} = P^\bot z_{0,1}$.

Hence if $x_1$ can be approximated by a solution $x(T,u,x_0)$ of the

factorized system,

then $Z_A x(T,u,x_0)$ is a solution of the degenerate problem which approximates the given vector $z_1$. 



If the factorized system is stabilizable this means that there exists a bounded operator $K_2$ such that

$(A_2+B_2K_2)$ is generator of an exponentially stable semigroup. Hence

$\|S_{A+BK}(t)\|\le \|Z_A\|\,\mu\,\exp(-\omega t)$,

where $\mu\ge 1$ and $\omega>0$.

The result follows since $\|Z_A\|\ge 1$.



In the same way one shows that controllability and stabilizability of the system factorized

according to method (a) implies the corresponding properties of the degenerate system.

In a similar way controllability (stabilizability) of the degenerate system implies these

properties for the factorized systems.



Concerning the properties observability and detectability,

which are defined with the help of the dual system,

the commuting diagrams above together with the preceeding part of this proof show the equivalence of

the degenerate and the factorized systems.

\end{proof}

%

%

\smallskip

%%%%%%%%%%%%   REMARK   %%%%%%%%%%%%%%

%

\begin{rem}

Under the assumptions of the theorem and according to our observations in Section~\ref{sec1},

the degenerate control system is

stabilizable, if there exists an operator $K$

and constants $\mu\ge 1$ and $\omega > 0$ such that

\begin{equation*}

\|(A+BK-\lambda M)^{-1}M\| \le \frac{\mu}{\mbox{Re}\, \lambda -\omega}

\quad\mbox{for all $\lambda$ with $\mbox{Re}\, \lambda > \omega$.}

\end{equation*}

\end{rem}

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%



\medskip

\begin{ack}

We would like to thank W. Desch and F. Kappel for stimulating discussions and helpful suggestions.

\end{ack}





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%







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\bibitem{3} Favini, A., {\em Laplace transform method for a class of degenerate evolution problems}, Rend. Mat. {\bf 12}, 511--536 (1979). 



\bibitem{4} Favini, A., {\em Controllability conditions of linear degenerate evolution systems}, Appl. Math. Optim. {\bf 6}, 153--167 (1980). 



\bibitem{5} Favini, A., {\em Abstract potential operator and spectral methods for a class of degenerate evolution problems}, J. Diff. Equat. {\bf 39}, 212--225 (1981). 



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