# Index concepts for linear differential-algebraic equations in finite and
infinite dimensions

Mehmet Erbay, Birgit Jacob, Kirsten Morris, Timo Reis and Caren Tischendorf

###### Abstract

Different index concepts for linear differential-algebraic equations are
defined in the general Banach space setting, and compared. For regular finite-
dimensional linear differential-algebraic equations, all these indices exist
and are equivalent. For infinite-dimensional systems, the situation is more
complex. It is proven that although some indices imply others, in general they
are not equivalent. The situation is illustrated with a number of examples.

##  1 Introduction

In this article we take a closer look at the index terms for infinite-
dimensional differential-algebraic systems (DAE) of the form

$\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)+f(t),\quad t\geq 0,$ (1)

where $E\colon\mathcal{X}\to\mathcal{Z}$ is a bounded linear operator (denoted
by $E\in\mathcal{L}(\mathcal{X},\mathcal{Z})$), $(A,\mathcal{D}(A))$ is a
closed and densely defined linear operator from $\mathcal{X}$ to $\mathcal{Z}$
and $f\colon[0,\infty)\to\mathcal{Z}$. Throughout this article, $\mathcal{X}$
and $\mathcal{Z}$ are Banach spaces and the DAE (1) is assumed to be regular.
That is,

$\rho(E,A)\coloneqq\left\\{\lambda\in\mathbb{C}\,\middle|\,(sE-A)^{-1}\in\mathcal{L}(\mathcal{Z},\mathcal{X})\right\\}\not=\emptyset.$

By a solution of (1) we mean a classical solution, that is, a function
$x\colon[0,\infty)\rightarrow\mathcal{D}(A)$ such that $Ex(\cdot)$ is
continuously differentiable as a function with values in $\mathcal{Z}$, and
(1) is satisfied for every $t\geq 0$.

The index of a DAE can be defined in a number of various ways. Examples
include the differentiation index, the nilpotency index, the resolvent index
and the radiality index [4, 10, 18, 19]. Not all indices are defined in the
infinite-dimensional case. For instance, the nilpotency index of a DAE demands
a Weierstraß form (defined formally below), which is not always available.

Our aim in writing this paper is to ’collect’ all the index terms that are
applicable in the infinite-dimensional case, and to characterize and compare
them to each other. In particular, we investigate the resolvent index $p_{\rm
res\vphantom{d}}^{(E,A)}$, the chain index $p_{\rm chain}^{(E,A)}$, the
radiality index $p_{\rm rad}^{(E,A)}$, the nilpotency index $p_{\rm
nilp}^{(E,A)}$, the differentiation index $p_{\rm diff}^{(E,A)}$ and the
perturbation index $p_{\rm pert}^{(E,A)}$. Several of these indices have not
previously been defined for infinite-dimensional systems.

One of our main results is that if all the indices mentioned in the previous
paragraph exist then

$\displaystyle p_{\rm rad}^{(E,A)}+1\geq p_{\rm res\vphantom{d}}^{(E,A)}\geq
p_{\rm nilp}^{(E,A)}=p_{\rm diff}^{(E,A)}=p_{\rm chain}^{(E,A)}.$

If in addition the operator $A_{1}$ in the Weierstraß form generates a
$C_{0}$-semigroup, then

$\displaystyle p_{\rm rad}^{(E,A)}+1\geq p_{\rm res\vphantom{d}}^{(E,A)}\geq
p_{\rm nilp}^{(E,A)}=p_{\rm diff}^{(E,A)}=p_{\rm chain}^{(E,A)}=p_{\rm
pert}^{(E,A)}.$

Furthermore, Proposition 6.9 implies that in the finite-dimensional case,
equality holds in all these bounds.

We conclude the introduction with some notation. For a $\lambda\in\rho(E,A)$
we call $(\lambda E-A)^{-1}$, $R^{E}(\lambda,A)\coloneqq(\lambda E-A)^{-1}E$
and $L^{E}(\lambda,A)\coloneqq E(\lambda E-A)^{-1}$ the resolvent, right-$E$
resolvent and left-$E$ resolvent of $A$ respectively.

##  2 Weierstraß form

Consider differential-algebraic systems of the form (1). Let also
$\tilde{\mathcal{X}},\tilde{\mathcal{Z}}$ be Banach spaces,
$\tilde{E}\in\mathcal{L}(\tilde{\mathcal{X}},\tilde{\mathcal{Z}})$ and
$\tilde{A}\colon\mathcal{D}(\tilde{A})\subseteq\tilde{\mathcal{X}}\to\tilde{\mathcal{Z}}$
closed and densely defined.

######  Definition 2.1.

Two differential-algebraic systems $\frac{{\rm d}}{{\rm d}t}Ex=Ax$ and
$\frac{{\rm d}}{{\rm d}t}\tilde{E}x=\tilde{A}x$ are equivalent, denoted by
$(E,A)\sim(\tilde{E},\tilde{A})$, if there are two bounded isomorphisms
$P\colon\mathcal{X}\to\tilde{\mathcal{X}}$,
$Q\colon\mathcal{Z}\to\tilde{\mathcal{Z}}$, such that $E=Q^{-1}\tilde{E}P$ and
$A=Q^{-1}\tilde{A}P$.

######  Definition 2.2.

A bounded operator $N\in\mathcal{L}(\mathcal{X})$ is called nilpotent, if
there exists a $p\in\mathbb{N}$, such that $N^{l}\neq 0$ for all $l<p$ and
$N^{p}=0$. $p$ is called the degree of nilpotency.

This definition may be slightly different in other references. For example, in
[18] the degree of nilpotency is $p-1$ and not $p$.

######  Definition 2.3.

The DAE (1) has a Weierstraß form, if there exists a Hilbert space
$\mathcal{Y}=\mathcal{Y}^{1}\oplus\mathcal{Y}^{2}$, such that

$(E,A)\sim\left(\begin{bmatrix}I_{\mathcal{Y}^{1}}&0\\\
0&N\end{bmatrix},\begin{bmatrix}A_{1}&0\\\
0&I_{\mathcal{Y}^{2}}\end{bmatrix}\right),$ (2)

where $N\colon\mathcal{Y}^{2}\to\mathcal{Y}^{2}$ is a bounded linear nilpotent
operator, $A_{1}\colon D(A_{1})\subseteq\mathcal{Y}^{1}\to\mathcal{Y}^{1}$ is
a linear operator and $I_{\mathcal{Y}^{i}}$ indicates the identity operator on
the associated subspace $\mathcal{Y}^{i}$, $i=1,2$.

This form is also known variously as the quasi-Weierstraß form [2] or
Weierstraß canonical form [10]. In finite dimensions the operator $A_{1}$ is
generally a Jordan matrix $J$. In this case, the Weierstraß form is unique up
to isomorphisms and therefore the nilpotency degree of $N$ is uniquely
determined. To be more precise, assume that $(E,A)$ has two different
Weierstraß forms $\left(\big{[}\begin{smallmatrix}I&0\\\
0&N_{i}\end{smallmatrix}\big{]},\big{[}\begin{smallmatrix}J_{i}&0\\\
0&I\end{smallmatrix}\big{]}\right)$, $i=1,2$. Then the sizes of the Jordan
blocks $J_{1},J_{2}$ and of the nilpotent operators $N_{1},N_{2}$ coincide, as
well as the degree of nilpotency of these nilpotent operators [10, Lem. 2.10].

In the next few sections we will go through a variety of different index term
and generalise/adapt them for the infinite-dimensional case. Most of the terms
are already known in the finite-dimensional case.

##  3 Resolvent index

The resolvent index has already been defined in [4, p. 5], [19, p. 8] and [5,
ch. 6.1]. It has the advantage that it does not require a Weierstraß form.
Thus, this definition can be easily extended to the infinite-dimensional case.
The only difficulty encountered in calculating this index is the calculation
of the resolvent and its growth rate, which is a greater hurdle in the
infinite-dimensional case.

######  Definition 3.1 (resolvent index).

The resolvent index of $(E,A)$ is the smallest integer $p_{\rm
res\vphantom{d}}^{(E,A)}\in\mathbb{N}_{0}$, such that there exists a
$\omega\in\mathbb{R}$, $C>0$ with $(\omega,\infty)\subseteq\rho(E,A)$ and

$\left\|(\lambda E-A)^{-1}\right\|\leq C\left|\lambda\right|^{p_{\rm
res\vphantom{d}}^{(E,A)}-1}$ (3)

for all $\lambda\in(\omega,\infty)$. The resolvent index is called a complex
resolvent index, denoted by $p_{\rm c,res}^{(E,A)}\in\mathbb{N}_{0}$, if
$\mathbb{C}_{{\rm Re}\,>\omega}\subseteq\rho(E,A)$ and (3) holds for $p_{\rm
c,res}^{(E,A)}$.

Note that the resolvent index can also defined in a weaker form as seen in [5,
ch. 5& 6]. Clearly, $p_{\rm res\vphantom{d}}^{(E,A)}\leq p_{\rm
c,res}^{(E,A)}$. The next proposition shows that this index is uniquely
defined.

######  Proposition 3.2.

The resolvent index, given that it exists, is uniquely defined. To be more
precise, let $(E,A)\sim(\tilde{E},\tilde{A})$. Then
$p_{\mathrm{res}}^{(E,A)}=p_{\mathrm{res}}^{(\tilde{E},\tilde{A})}$ and
$p_{\mathrm{c,res}}^{(E,A)}=p_{\mathrm{c,res}}^{(\tilde{E},\tilde{A})}$.

###### Proof.

Since $(E,A)\sim(\tilde{E},\tilde{A})$ there exists two isomorphisms
$P\colon\mathcal{X}\to\tilde{\mathcal{X}}$,
$Q\colon\mathcal{Z}\to\tilde{\mathcal{Z}}$ with

$E=Q^{-1}\tilde{E}P\quad\text{and}\quad A=Q^{-1}\tilde{A}P.$

Assume that $(E,A)$ has resolvent index $p_{\rm res\vphantom{d}}^{(E,A)}$.
That is, there exists a $C>0$, $\omega\in\mathbb{R}$, such that
$(\omega,\infty)\subseteq\rho(E,A)$ and

$\|(\lambda E-A)^{-1}\|\leq C\left|\lambda\right|^{p_{\rm
res\vphantom{d}}^{(E,A)}-1},\quad\lambda>\omega.$

Since

$(\lambda\tilde{E}-\tilde{A})^{-1}\tilde{E}=P(\lambda
E-A)^{-1}EQ^{-1},\quad\lambda>\omega$

it follows that
$(\omega,\infty)\subseteq\rho(E,A)\subseteq\rho(\tilde{E},\tilde{A})$ and

$\|(\lambda\tilde{E}-\tilde{A})^{-1}\|\leq\left\|P\right\|\left\|(\lambda
E-A)^{-1}\right\|\left\|Q^{-1}\right\|\leq\tilde{C}\left|\lambda\right|^{p_{\rm
res\vphantom{d}}^{(E,A)}-1},\quad\lambda>\omega,$

for $\tilde{C}\coloneqq C\left\|P\right\|\left\|Q^{-1}\right\|$. Thus, the
resolvent index of $(\tilde{E},\tilde{A})$ is at most $p_{\rm
res\vphantom{d}}^{(E,A)}$. The other estimate follows from an equivalent
argument and switching $(E,A)$ and $(\tilde{E},\tilde{A})$. The statement
concerning the complex resolvent index follows similarly. ∎

Next, we will show the existence of the (complex) resolvent index for a
special class of systems, namely $\mathcal{X}=\mathcal{Z}$, $E$ is non-
negative and $A$ is dissipative. Note that we call $E$ non-negative, denoted
by $E\geq 0$, if $\langle Ex,x\rangle\geq 0$ for all $x\in\mathcal{X}$ and we
call $A$ dissipative, if ${\rm Re}\,\langle Ax,x\rangle\leq 0$ for all
$x\in\mathcal{D}(A)$. Such systems, are known as port-Hamiltonian DAEs or
abstract dissipative DAEs (see [5, ch. 7] and [12]).

######  Theorem 3.3.

Let $\mathcal{X}=\mathcal{Z}$, $E\in\mathcal{L}(\mathcal{X})$ be non-negative
self-adjoint and $A\colon\mathcal{D}(A)\subseteq\mathcal{X}\to\mathcal{X}$ be
dissipative. If there exists a $\omega>0$, such that
$(\omega,\infty)\subseteq\rho(E,A)$, then $p_{\mathrm{res}}^{(E,A)}\leq 2$. If
also $\mathbb{C}_{{\rm Re}\,>\omega}\subseteq\rho(E,A)$, then
$p_{\mathrm{c,res}}^{(E,A)}\leq 3$.

###### Proof.

For any $x\in\mathcal{X}$ define $z=(\lambda E-A)^{-1}x.$ By using the
dissipativity of $A$ and the positivity of $E$ we deduce that

$\displaystyle{\rm Re}\,\langle(\lambda E-A)^{-1}x,x\rangle=\langle
z,Ez\rangle{\rm Re}\,\lambda-{\rm Re}\,\langle z,Az\rangle\geq 0$

for all $\lambda\in\rho(E,A)\cap\mathbb{C}_{{\rm Re}\,>0}$, . Thus, for every
$x\in\mathcal{X}$ the function

$f_{x}\colon\lambda\mapsto\langle(\lambda
E-A)^{-1}x,x\rangle+\left\|x\right\|^{2}$

is positive real. Let $\mu>\omega$. Using [6, Thm. 3] implies

$\displaystyle\left|f_{x}(\lambda)\right|\leq\left|f_{x}(\mu)\right|\frac{\left|\lambda\right|^{2}+\left|\mu\right|^{2}+3\left|\lambda\mu\right|}{\mu{\rm
Re}\,\lambda}\leq\left|f_{x}(\mu)\right|\frac{5}{\mu}\frac{\left|\lambda\right|^{2}}{{\rm
Re}\,\lambda},\quad\lambda\in\mathbb{C}_{{\rm Re}\,>\mu}.$

Hence, together with the Riesz representation theorem we derive

$\displaystyle\left\|(\lambda
E-A)^{-1}\right\|=\sup_{\begin{subarray}{c}x,y\in\mathcal{X}\\\
\|x\|=\|y\|=1\end{subarray}}\left|\langle(\lambda
E-A)^{-1}x,y\rangle\right|\leq\sup_{\begin{subarray}{c}x\in\mathcal{X}\\\
\left\|x\right\|=1\end{subarray}}2\left|f_{x}(\mu)\right|\leq
K\frac{\left|\lambda\right|^{2}}{{\rm Re}\,\lambda},$

for all $\lambda\in\mathbb{C}_{{\rm Re}\,>\omega}$ with $K=(\left\|(\mu
E-A)^{-1}\right\|+1)\frac{10}{\mu}$. Thus, the complex resolvent index is at
most $3$ and if $\lambda>\omega$ is real, then
$\frac{\left|\lambda\right|^{2}}{{\rm Re}\,\lambda}=\lambda$ and the resolvent
index is at most $2$. ∎

######  Example 3.4.

($p_{\rm res\vphantom{d}}^{(E,A)}=2$ and
$p_{\mathrm{c,res}\vphantom{d}}^{(E,A)}=3$.) Define $A={\rm
diag\,}(A_{0},A_{1},A_{2},\ldots)$ with

$\displaystyle A_{0}=\begin{bmatrix}0&-1\\\ 1&0\end{bmatrix},\quad
A_{k}=\begin{bmatrix}0&\sqrt{k^{4}+1}\\\ -\sqrt{k^{4}+1}&-2\end{bmatrix},\quad
k\in\mathbb{N},$

and $\mathcal{D}(A)\coloneqq\\{x\in\ell^{2}\mid Ax\in\ell^{2}\\}$. Then $A$
can be extended to $\ell^{2}$, which will denoted by $A_{-1}$. Define
$E\in\mathcal{L}(\ell^{2})$,
$B\colon\mathbb{R}\to\mathcal{D}(A^{\ast})^{\prime}$ and
$C\colon\mathcal{D}(A)\to\mathbb{R}$ with $E={\rm
diag\,}(E_{0},E_{1},E_{2},\ldots)$ and
$B=(B_{0},B_{1},B_{2},\ldots)^{T}=C^{\ast}$, whereby

$\displaystyle E_{0}$ $\displaystyle=\begin{bmatrix}1&0\\\ 0&0\end{bmatrix},$
$\displaystyle E_{k}$ $\displaystyle=\begin{bmatrix}1&0\\\
0&1\end{bmatrix},\quad k\in\mathbb{N},$ $\displaystyle B_{0}$
$\displaystyle=\begin{bmatrix}0\\\ 1\end{bmatrix},$ $\displaystyle B_{k}$
$\displaystyle=\begin{bmatrix}0\\\ k^{\frac{5}{4}}\end{bmatrix},\quad
k\in\mathbb{N}.$

Consider the following system

$\displaystyle\frac{{\rm d}}{{\rm d}t}\underbrace{\begin{bmatrix}E&0&0\\\
0&0&0\\\ 0&0&0\end{bmatrix}}_{\mathcal{E}}\begin{bmatrix}x_{1}\\\ x_{2}\\\
x_{3}\end{bmatrix}=\underbrace{\begin{bmatrix}A_{-1}&B&0\\\ -C&0&I\\\
0&-I&0\end{bmatrix}}_{\mathcal{A}}\begin{bmatrix}x_{1}\\\ x_{2}\\\
x_{3}\end{bmatrix}$

on $\mathcal{X}=\ell^{2}\times\mathbb{R}\times\mathbb{R}$. Obviously,
$\mathcal{E}$ is non-negative and self-adjoint and, by its construction,
$\mathcal{A}$ with maximal domain is dissipative. Thus,
$(\mathcal{E},\mathcal{A})$ satisfies the conditions of Theorem 3.3.

It will now be shown that $p_{\rm
res\vphantom{d}}^{(\mathcal{E},\mathcal{A})}=2$ and $p_{\rm
c,res}^{(\mathcal{E},\mathcal{A})}=3$. For $s\in\rho(\mathcal{E},\mathcal{A})$

$(s\mathcal{E}-\mathcal{A})^{-1}=\begin{bmatrix}(sE-A)^{-1}&0&(sE-A)^{-1}B\\\
0&0&I\\\ -C(sE-A)^{-1}&-I&C(sE-A)^{-1}B\end{bmatrix}.$

It can be shown that

$G(s)\coloneqq
C(sE-A)^{-1}B=s+\sum_{k=1}^{\infty}\frac{k^{\frac{5}{2}}s}{s^{2}+2s+k^{4}+1},$

a function that George Weiss (Tel Aviv) scribbled on paper for one of the
authors. Thanks for that!
Since $G(s)\geq s$ for $s\geq 0$, the resolvent
$(s\mathcal{E}-\mathcal{A})^{-1}$ grows at least linearly along the real axis
and together with Theorem 3.3 this implies $p_{\rm
res\vphantom{d}}^{(\mathcal{E},\mathcal{A})}=2$.

Now consider growth in the entire right-hand-plane. Letting $\sigma>0$ and
$s_{n}\coloneqq\sigma+in^{2}$ for $n\in\mathbb{N}$,

$\displaystyle{\rm Re}\,G(s_{n})$
$\displaystyle=\sigma+\sum_{k=1}^{\infty}\frac{k^{\frac{5}{2}}(\sigma((1+\sigma)^{2}+k^{4})+(2+\sigma)n^{4})}{((1+\sigma)^{2}+(k^{4}-n^{4}))^{2}+4(1+\sigma)^{2}n^{4}}$
$\displaystyle\geq\sigma+\frac{n^{\frac{5}{2}}(\sigma((1+\sigma)^{2}+n^{4})+(2+\sigma)n^{4})}{(1+\sigma)^{4}+4(1+\sigma)^{4}n^{4}}$
$\displaystyle\geq\frac{2n^{\frac{5}{2}}n^{4}}{(1+\sigma)^{4}(1+4n^{4})}=\frac{2n^{\frac{5}{2}}}{5(1+\sigma)^{4}}\frac{5n^{4}}{1+4n^{4}}$
$\displaystyle\geq\frac{2n^{\frac{5}{2}}}{5(1+\sigma)^{4}},\quad
n\in\mathbb{N}.$

By choosing an $c>0$ and $N\in\mathbb{N}$ such that $|s_{n}|^{\frac{5}{4}}\leq
c\frac{2n^{\frac{5}{2}}}{5(1+\sigma)^{4}}$ for every $n\geq N$, one obtains

$\left|G(s_{n})\right|\geq{\rm
Re}\,G(s_{n})\geq\frac{1}{c}\left|s_{n}\right|^{\frac{5}{4}},\quad n\geq N.$

Hence, the growth of $G(s_{n})$ is more than quadratic along lines ${\rm
Re}\,s=\sigma$ parallel to the imaginary axis for every $\sigma>0$. Therefore,
using Theorem 3.3 it follows that $p_{\mathrm{c,res}}^{(E,A)}=3$.

######  Example 3.5.

(A class of well-posed systems with $p_{\rm res\vphantom{d}}^{(E,A)}\leq 2$)
Let $A_{o}$ with $\mathcal{D}(A_{o})\subset\mathcal{W}$ generate a
$C_{0}$-semigroup on $\mathcal{W}$, where $\mathcal{W}$ is a Hilbert space. We
indicate the growth bound of the semigroup by $\omega$. For $b\in\mathcal{W}$
and $c\in\mathcal{D}(A_{o}^{*})$ with $\langle b,c\rangle\not=0$ define the
operators $Bu=bu$ where $u\in\mathbb{C}$ and $Cz=\langle z,c\rangle$ for any
$z\in\mathcal{W}$. We define the DAE on
$\mathcal{Z}=\mathcal{W}\times\mathbb{C}$ by

$\frac{{\rm d}}{{\rm d}t}\underbrace{\begin{bmatrix}I&0\\\
0&0\end{bmatrix}}_{\eqqcolon E}x(t)=\underbrace{\begin{bmatrix}A_{o}&B\\\
C&0\end{bmatrix}}_{\eqqcolon A}x(t)\quad t\geq 0.$ (4)

Defining for $s\in\rho(A_{o}),$

$G(s)=\langle(sI-A_{o})^{-1}b,c\rangle$

we obtain

$(sE-A)^{-1}=\begin{bmatrix}(sI-A_{o})^{-1}-(sI-A_{o})^{-1}BG(s)^{-1}C(sI-
A_{o})^{-1}&(sI-A_{o})^{-1}BG(s)^{-1}\\\ G(s)^{-1}C(sI-
A_{o})^{-1}&-G(s)^{-1}\end{bmatrix}.$

Note that the condition $\langle b,c\rangle\not=0$ implies $G(s)\not=0$ for
$s\in\rho(A_{o})$. Thus, the resolvent of $(E,A)$ is non-empty and the system
is regular. Since

$\lim_{s\to\infty}sG(s)=\langle c,b\rangle,$

for large $s$, $G(s)^{-1}\leq Ms$ and so the resolvent index is at most $2$.

##  4 Chain index

The concept of Wong sequences [20] are needed to define the chain index. For
$\mathcal{X}=\mathcal{Z}=\mathbb{K}^{n}$ ($n\in\mathbb{N}$) we define

$\displaystyle\mathcal{V}_{0}$ $\displaystyle\coloneqq\mathbb{K}^{n},$
$\displaystyle\mathcal{V}_{i+1}$ $\displaystyle\coloneqq A^{-1}(E(V_{i})),$
$\displaystyle\quad i\in\mathbb{N},$ $\displaystyle\mathcal{W}_{0}$
$\displaystyle\coloneqq\left\\{0\right\\},$ $\displaystyle\mathcal{W}_{i+1}$
$\displaystyle\coloneqq E^{-1}(A(W_{i})),$ $\displaystyle\quad
i\in\mathbb{N}.$

Thus, there exists $k,l\in\mathbb{N}$ such that
$\mathcal{V}_{i+1}\subsetneq\mathcal{V}_{i}$, $0\leq i<l$,
$\mathcal{W}_{j}\subsetneq\mathcal{W}_{j+1}$, $0\leq j<k$ and
$\mathcal{V}_{l}=\mathcal{V}_{l+r}$, $\mathcal{W}_{k}=\mathcal{W}_{l+r}$ for
all $r\in\mathbb{N}$ [2, p. 3]. Let $x_{k}\in\mathcal{W}_{k}$. Then, there
exists a $x_{k-1}\in\mathcal{W}_{k-1}$ with $Ex_{k}=Ax_{k-1}$. Inductively,
one obtains

$\displaystyle\begin{split}Ex_{1}&=0,\\\ Ex_{2}&=Ax_{1},\\\ &\;\;\vdots\\\
Ex_{k}&=Ax_{k-1},\end{split}$ (5)

$x_{r}\in\mathcal{W}_{r}$, $0\leq r\leq k$, which is called a chain of length
$k$. The concept of such a chain is not new and can be found in [2, p. 15] for
the finite-dimensional case or [18, p. 18] for arbitrary Banach spaces. It
should be mentioned that the latter reference does not call (5) a chain.

######  Definition 4.1 (chain index).

We call $(x_{1},\ldots,x_{p})\in\mathcal{D}(A)^{p-1}\times\mathcal{X}$,
$p\in\mathbb{N}$, a chain of $(E,A)$ of length $p$, if $x_{1}\in{\rm
ker}\,E\setminus\\{0\\}$, $x_{k}\notin{\rm ker}\,A$, $k=1,\ldots,p-1$, and

$Ex_{k+1}=Ax_{k},\quad k=1,\ldots,p-1.$ (6)

The chain index $p_{\rm chain}^{(E,A)}\in\mathbb{N}_{0}$ of $(E,A)$ is the
supremum over all chain lengths of $(E,A)$.

With this definition we exclude the case where the chain index can be
infinite. Such an example is obtained when $(E,A)$ has a form as in (2),
whereby $N$ is quasi-nilpotent. For such systems the chain index is not
defined.

######  Proposition 4.2.

The chain index is, given that it exists, uniquely defined. To be more
specific, let $(E,A)\sim(\tilde{E},\tilde{A})$. Then
$p_{\mathrm{chain}}^{(E,A)}=p_{\mathrm{chain}}^{(\tilde{E},\tilde{A})}$.

###### Proof.

Since $(E,A)\sim(\tilde{E},\tilde{A})$ there exists two isomorphisms
$P\colon\mathcal{X}\to\tilde{\mathcal{X}}$,
$Q\colon\mathcal{Z}\to\tilde{\mathcal{Z}}$, such that

$E=Q^{-1}\tilde{E}P\quad\text{and}\quad A=Q^{-1}\tilde{A}P.$

Let $p_{\rm chain}^{(E,A)}\in\mathbb{N}$. Thus, there exists a chain
$(x_{1},\ldots,x_{p_{\rm chain}^{(E,A)}})$, such that

$Ex_{1}=0,Ex_{2}=Ax_{1},\ldots,Ex_{p_{\rm chain}^{(E,A)}}=Ax_{p_{\rm
chain}^{(E,A)}-1}.$

Since $E=Q^{-1}\tilde{E}P$, $A=Q^{-1}\tilde{A}P$ and $Q$ is an isomorphism,

$\tilde{E}Px_{1}=0,\tilde{E}Px_{2}=\tilde{A}Px_{1},\ldots,\tilde{E}Px_{p_{\rm
chain}^{(E,A)}}=\tilde{A}Px_{p_{\rm chain}^{(E,A)}-1}.$

Because $P$ is an isomorphism, $Px_{1}\neq 0$ and thus $Px_{1}\in{\rm
ker}\,\tilde{E}\backslash\\{0\\}$. If we assume that $Px_{k}\in{\rm
ker}\,\tilde{A}$ for any $1\leq k\leq p_{\rm chain}^{(E,A)}-1$, then
$0=Q^{-1}\tilde{A}Px_{k}=Ax_{k}$, which would be a contradiction to
$x_{k}\in\mathcal{X}\backslash{\rm ker}\,A$. Therefore, $Px_{k}\notin{\rm
ker}\,\tilde{A}$ for all $1\leq k\leq p_{\rm chain}^{(E,A)}-1$. Hence,
$(Px_{1},\ldots,Px_{p_{\rm chain}^{(E,A)}})$ denotes a chain of
$(\tilde{E},\tilde{A})$ of length $p_{\rm chain}^{(E,A)}$ . Therefore, the
chain index of $(\tilde{E},\tilde{A})$ is bounded from below by $p_{\rm
chain}^{(E,A)}$ and, equivalently, the chain index of $(E,A)$ is bounded from
below by $p_{\rm chain}^{(\tilde{E},\tilde{A})}$. ∎

##  5 Radiality index

In this section we introduce a less well-known index, the radiality index.

######  Definition 5.1 (radiality index).

The radiality index of $(E,A)$ is the smallest number $p_{\rm
rad}^{(E,A)}\in\mathbb{N}_{0}$, such that there exists a
$\omega\in\mathbb{R}$, $C>0$ with $(\omega,\infty)\subseteq\rho(E,A)$ and

$\displaystyle\begin{split}\left\|(\lambda_{0}E-A)^{-1}E\cdot\ldots\cdot(\lambda_{p_{\rm
rad}^{(E,A)}}E-A)^{-1}E\right\|&\leq C\prod_{k=0}^{p_{\rm
rad}^{(E,A)}}\frac{1}{\left|\lambda_{k}-\omega\right|},\\\
\left\|E(\lambda_{0}E-A)^{-1}\cdot\ldots\cdot E(\lambda_{p_{\rm
rad}^{(E,A)}}E-A)^{-1}\right\|&\leq C\prod_{k=0}^{p_{\rm
rad}^{(E,A)}}\frac{1}{\left|\lambda_{k}-\omega\right|}\end{split}$ (7)

for all $\lambda_{0},\ldots,\lambda_{p_{\rm{rad}}^{(E,A)}}>\omega$. The
radiality index is called a complex radiality index, denoted by $p_{\rm
c,rad}^{(E,A)}$, if $\mathbb{C}_{{\rm Re}\,>\omega}\subseteq\rho(E,A)$ and (7)
holds for $p_{\rm{rad}}^{(E,A)}$.

Clearly, $p_{\rm rad}^{(E,A)}\leq p_{\rm c,rad}^{(E,A)}$. The radiality index
$p=p_{\rm rad}^{(E,A)}\in\mathbb{N}_{0}$, originally known as weak
$(E,p)$-radiality [18, p. 21], is not a commonly used index in finite-
dimensional problems. However, the radiality index is one of the most useful
index terms in infinite dimensions (see [18, 8]).

To get a better understanding for this definition one can start by looking at
the radiality index with $p_{\rm rad}^{(E,A)}=0$, $\mathcal{X}=\mathcal{Z}$
and $E=I_{\mathcal{X}}$. Then, (7) translates into the well known Hille-Yosida
type estimate. This suggests that the radiality index implies further well-
posedness results. In fact, as soon as slightly stronger assumptions, namely
strong $(E,p)$-radiality, are considered, one obtains not only the well-
posedness of the system, but also the existence of a Weierstraß-form [18, Sec.
2.5 & 2.6].

######  Proposition 5.2.

The radiality index is unique. In fact, let $(E,A)\sim(\tilde{E},\tilde{A})$.
Then $p_{\mathrm{rad}}^{(E,A)}=p_{\mathrm{rad}}^{(\tilde{E},\tilde{A})}$ and
$p_{\mathrm{c,rad}}^{(E,A)}=p_{\mathrm{c,rad}}^{(\tilde{E},\tilde{A})}$.

###### Proof.

Since $(E,A)\sim(\tilde{E},\tilde{A})$ there exists two isomorphisms
$P\colon\mathcal{X}\to\tilde{\mathcal{X}}$,
$Q\colon\mathcal{Z}\to\tilde{\mathcal{Z}}$, such that

$E=Q^{-1}\tilde{E}P\quad\text{and}\quad A=Q^{-1}\tilde{A}P.$

Assume, that $(E,A)$ has radiality index $p_{\rm rad}^{(E,A)}$. Thus, there
exists positive constants $C>0$, $\omega>0$, such that
$(\omega,\infty)\subseteq\rho(E,A)$ and

$\displaystyle\left\|(\lambda_{0}E-A)^{-1}E\cdot\ldots\cdot(\lambda_{p_{\rm
rad}^{(E,A)}}E-A)^{-1}E\right\|$
$\displaystyle\leq\frac{C}{(\lambda_{0}-\omega)\cdot\ldots\cdot(\lambda_{p_{\rm
rad}^{(E,A)}}-\omega)},$
$\displaystyle\left\|E(\lambda_{0}E-A)^{-1}\cdot\ldots\cdot E(\lambda_{p_{\rm
rad}^{(E,A)}}E-A)^{-1}\right\|$
$\displaystyle\leq\frac{C}{(\lambda_{0}-\omega)\cdot\ldots\cdot(\lambda_{p_{\rm
rad}^{(E,A)}}-\omega)},$

for all $\lambda_{0},\ldots,\lambda_{p_{\rm rad}^{(E,A)}}\in\rho(E,A)$. Since

$P(\lambda E-A)^{-1}Q^{-1}=(\lambda\tilde{E}-\tilde{A})^{-1}$

for all $\lambda\in\rho(E,A)$ we derive
$(\omega,\infty)\subseteq\rho(E,A)\subseteq\rho(\tilde{E},\tilde{A})$ and
therefore

$\displaystyle\|(\lambda_{0}\tilde{E}-\tilde{A})^{-1}\tilde{E}\cdot\ldots$
$\displaystyle\cdot(\lambda_{p_{\rm
rad}^{(E,A)}}\tilde{E}-\tilde{A})^{-1}\tilde{E}\|$
$\displaystyle\leq\left\|P\right\|\|(\lambda_{0}E-A)^{-1}E\cdot\ldots\cdot(\lambda_{p_{\rm
rad}^{(E,A)}}E-A)^{-1}E\|\|P^{-1}\|$
$\displaystyle\leq\frac{\tilde{C}}{(\lambda_{0}-\omega)\cdot\ldots\cdot(\lambda_{p_{\rm
rad}^{(E,A)}}-\omega)},$
$\displaystyle\|\tilde{E}(\lambda_{0}\tilde{E}-\tilde{A})^{-1}\cdot\ldots$
$\displaystyle\cdot\tilde{E}(\lambda_{p_{\rm
rad}^{(E,A)}}\tilde{E}-\tilde{A})^{-1}\|$
$\displaystyle\leq\left\|Q\right\|\|E(\lambda_{0}E-A)^{-1}\cdot\ldots\cdot
E(\lambda_{p_{\rm rad}^{(E,A)}}E-A)^{-1}\|\left\|Q^{-1}\right\|$
$\displaystyle\leq\frac{\tilde{C}}{(\lambda_{0}-\omega)\cdot\ldots\cdot(\lambda_{p_{\rm
rad}^{(E,A)}}-\omega)}$

for all $\lambda_{0},\ldots,\lambda_{\rm rad}^{(E,A)}>\omega$, for
$\tilde{C}\coloneqq
2C\max\\{\left\|P\right\|,\left\|P^{-1}\right\|,\left\|Q\right\|,\left\|Q^{-1}\right\|\\}$.
Consequently, $(\tilde{E},\tilde{A})$ has at most radiality index $p_{\rm
rad}^{(E,A)}$ . By an identical argument, $(E,A)$ has at most radiality index
$p_{\rm rad}^{(\tilde{E},\tilde{A})}$. The equality
$p_{\mathrm{c,rad}}^{(E,A)}=p_{\mathrm{c,rad}}^{(\tilde{E},\tilde{A})}$ is
shown in the same way. ∎

######  Example 5.3.

(A system with radiality index 0.)  We consider the dynamics on an interval
$[a,b]$ of an undamped beam fixed at one end and free at the other are given
by the partial differential equations

$\displaystyle\rho\frac{\partial^{2}v(\xi,t)}{\partial t^{2}}$
$\displaystyle=\alpha\frac{\partial}{\partial\xi}\frac{\partial
v(\xi,t)}{\partial\xi}-\gamma\beta\frac{\partial}{\partial\xi}\frac{\partial
p(\xi,t)}{\partial\xi}$ (8a)
$\displaystyle\mu\frac{\partial^{2}p(\xi,t)}{\partial t^{2}}$
$\displaystyle=\beta\frac{\partial}{\partial\xi}\frac{\partial
p(\xi,t)}{\partial\xi}-\gamma\beta\frac{\partial}{\partial\xi}\frac{\partial
v(\xi,t)}{\partial\xi},$ (8b)

where $\xi\in[a,b[$, $t>0$, $v(\xi,t)$ is the longitudinal displacement and
$p(\xi,t)$ electric charge. The material parameters are material density
$\rho>0,$ magnetic permeability $\mu\geq 0,$ elastic stiffness $\alpha_{1}>0$,
impermittivity $\beta>0$, piezoelectric coefficients $\gamma>0$ and we define

$\alpha=\alpha_{1}+\gamma^{2}\beta,$

One choice of boundary conditions is subject to the boundary conditions

$\displaystyle v(a,t)$ $\displaystyle=0,$ (9a) $\displaystyle p(a,t)$
$\displaystyle=0,$ (9b) $\displaystyle\beta\frac{\partial
p(b,t)}{\partial\xi}-\gamma\beta\frac{\partial v(b,t)}{\partial\xi}$
$\displaystyle=0,$ (9c) $\displaystyle\alpha\frac{\partial
v(b,t)}{\partial\xi}-\gamma\beta\frac{\partial p(b,t)}{\partial\xi}$
$\displaystyle=0.$ (9d)

The total energy is

$\displaystyle\mathcal{H}(t)=\frac{1}{2}\int\limits_{a}^{b}\rho\left(\frac{\partial
v(\xi,t)}{\partial t}\right)^{2}+\alpha_{1}\left(\frac{\partial
v(\xi,t)}{\partial\xi}\right)^{2}+\mu\left(\frac{\partial p(\xi,t)}{\partial
t}\right)^{2}+\beta\left(\frac{\partial
p(\xi,t)}{\partial\xi}-\gamma\frac{\partial
v(\xi,t)}{\partial\xi}\right)^{2}d\xi.$ (10)

This model was shown to be a well-posed port-Hamiltonian system associated
with a contraction semigroup [14]. However, there are quite a few choices for
the state variables. We use here a different choice of state variable,
suggested by Hans Zwart (personal communication):

$\boldsymbol{z}(\xi,t)\coloneqq\begin{bmatrix}z_{1}(\xi,t)\\\ z_{2}(\xi,t)\\\
z_{3}(\xi,t)\\\ z_{4}(\xi,t)\end{bmatrix}=\begin{bmatrix}\frac{\partial
v(\xi,t)}{\partial\xi}\\\ \sqrt{\rho}\frac{\partial v(\xi,t)}{\partial t}\\\
\frac{\partial p(\xi,t)}{\partial\xi}\\\ \sqrt{\mu}\frac{\partial
p(\xi,t)}{\partial t}\\\ \end{bmatrix}\,.$

Defining

$Q\coloneqq\begin{bmatrix}\alpha&0&-\gamma\beta&0\\\ 0&I&0&0\\\
-\gamma\beta&0&\beta&0\\\ 0&0&0&I\end{bmatrix},\quad
E\coloneqq\begin{bmatrix}\sqrt{\rho}&0&0&0\\\ 0&\sqrt{\rho}&0&0\\\
0&0&\sqrt{\mu}&0\\\ 0&0&0&\sqrt{\mu}\end{bmatrix},\quad
P_{1}\coloneqq\begin{bmatrix}0&I&0&0\\\ I&0&0&0\\\ 0&0&0&I\\\
0&0&I&0\end{bmatrix},$

the PDAE can be written

$\frac{\partial}{\partial t}E\boldsymbol{z}(\xi,t)=A\boldsymbol{z}(\xi,t),$

where $A=P_{1}\frac{\partial}{\partial\xi}Q$. This choice of state variables
yields equations in the standard pH form in [9].

Generally, $\mu$ is very small and it is often taken to be zero, which yields
the quasi-static piezo-electric beam. If $\mu=0$ the operator $E$ becomes
singular and the fourth state variable becomes identically zero. The PDAE
becomes

$\frac{\partial}{\partial t}\begin{bmatrix}\sqrt{\rho}&0&0&0\\\
0&\sqrt{\rho}&0&0\\\ 0&0&0&0\\\
0&0&0&0\end{bmatrix}\begin{pmatrix}\frac{\partial v(\xi,t)}{\partial\xi}\\\
\sqrt{\rho}\frac{\partial v(\xi,t)}{\partial t}\\\ \frac{\partial
p(\xi,t)}{\partial\xi}\\\
0\end{pmatrix}=P_{1}\frac{\partial}{\partial\xi}Q\begin{pmatrix}\frac{\partial
v(\xi,t)}{\partial\xi}\\\ \sqrt{\rho}\frac{\partial v(\xi,t)}{\partial t}\\\
\frac{\partial p(\xi,t)}{\partial\xi}\\\ 0\end{pmatrix}.$

Removing the fourth column of $E,Q$ since the 4th variable is zero, we obtain

$\frac{\partial}{\partial t}\begin{bmatrix}\sqrt{\rho}&0&0\\\
0&\sqrt{\rho}&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}\begin{pmatrix}\frac{\partial
v(\xi,t)}{\partial\xi}\\\ \sqrt{\rho}\frac{\partial v(\xi,t)}{\partial t}\\\
\frac{\partial p(\xi,t)}{\partial\xi}\\\
\end{pmatrix}=\begin{bmatrix}0&I&0&0\\\ I&0&0&0\\\ 0&0&0&I\\\
0&0&I&0\end{bmatrix}\frac{\partial}{\partial\xi}{\begin{bmatrix}\alpha&0&-\gamma\beta\\\
0&I&0\\\ -\gamma\beta&0&\beta\\\
0&0&0\end{bmatrix}}\begin{pmatrix}\frac{\partial v(\xi,t)}{\partial\xi}\\\
\sqrt{\rho}\frac{\partial v(\xi,t)}{\partial t}\\\ \frac{\partial
p(\xi,t)}{\partial\xi}\end{pmatrix}.$

This PDAE is radial of degree zero; see [8, sect. IV].

Next, we compare the radiality index with the resolvent index.

######  Proposition 5.4.

If the (complex) radiality index $p_{\rm rad}^{(E,A)}$ exists, then the
(complex) resolvent index also exists . Furthermore, $p_{\rm
rad}^{(E,A)}+1\geq p_{\rm res\vphantom{d}}^{(E,A)}$ ($p_{\rm
c,rad}^{(E,A)}+1\geq p_{\rm c,res\vphantom{d}}^{(E,A)}$).

###### Proof.

We have to show that there exists a $C>0$, $\omega>0$, such that
$(\omega,\infty)\subseteq\rho(E,A)$ ($\mathbb{C}_{{\rm
Re}\,>\omega}\subseteq\rho(E,A)$) and (3) holds for all
$\lambda\in(\omega,\infty)$ ($\lambda\in\mathbb{C}_{{\rm Re}\,>\omega}$) and a
$p\leq p_{\rm{rad}}+1$ ($p\leq p_{\rm c,rad}+1$). The first part of this
statement is implied by existence of the radiality index. Hence, we only have
to prove (3). This has already been shown in [18, Lem. 3.1.1] for an
$(E,p)$-sectorial operator pair $(E,A)$. However, since the definition of
$(E,p)$-sectorial introduced in [18, sec. 3.1] coincides with the radiality
index except that (7) holds for a sector
$\left\\{\mu\in\mathbb{C}\,\middle|\,\left|\arg(\mu-\omega)\right|<\theta,\mu\neq\omega\right\\}$
for a given $\omega\in\mathbb{R}$ and $\theta\in(\frac{\pi}{2},\pi)$, one can
simply follow the proof of [18, Lem. 3.1.1] for positive $\lambda>\omega$ (for
$\lambda\in\mathbb{C}_{{\rm Re}\,>\omega}$). ∎

The following example shows that in general existence of the resolvent index
does not imply that the radiality index exists.

######  Example 5.5.

(A system where the resolvent index exists, but the radiality index does not.)
Let $\mathcal{X}=L^{2}(0,\infty)\times\mathbb{R}$ and define

$\displaystyle E=\begin{bmatrix}I&0\\\ 0&0\end{bmatrix},\quad
A=\begin{bmatrix}\partial_{x}&0\\\ \delta_{0}&1\end{bmatrix}$

with $\mathcal{D}(A)\coloneqq H^{1}_{0}(0,\infty)\times\mathbb{R}$, where
$\delta_{0}\colon L^{2}(0,\infty)\to\mathbb{R},x\mapsto x(0)$. Let $\lambda>0$
and $\Big{(}\begin{smallmatrix}f\\\ g\end{smallmatrix}\Big{)}\in\mathcal{X}$.
Then

$\displaystyle(\lambda E-A)\begin{pmatrix}x\\\
y\end{pmatrix}=\begin{pmatrix}f\\\ g\end{pmatrix}$

if and only if

$\displaystyle\begin{pmatrix}x\\\
y\end{pmatrix}=\begin{pmatrix}\int_{\cdot}^{\infty}\mathrm{e}^{\lambda(\cdot-s)}f(s)\,\mathrm{d}s\\\
-g-\int_{0}^{\infty}\mathrm{e}^{-\lambda s}f(s)\,\mathrm{d}s\end{pmatrix}.$

Hence,

$\displaystyle(\lambda E-A)^{-1}\begin{pmatrix}f\\\
g\end{pmatrix}=\begin{pmatrix}\int_{\cdot}^{\infty}\mathrm{e}^{\lambda(\cdot-s)}f(s)\,\mathrm{d}s\\\
-g-\int_{0}^{\infty}\mathrm{e}^{\lambda(0-s)}f(s)\,\mathrm{d}s\end{pmatrix}$

and $\left\|(\lambda E-A)^{-1}\right\|\leq M=M\lambda^{0}$ for a $M>0$ for all
$\lambda\geq 0$. Thus, $(E,A)$ has resolvent index $p_{\rm res}^{(E,A)}=1$.
Furthermore, we have

$\displaystyle\left((\lambda E-A)^{-1}E\right)^{p+1}\begin{pmatrix}f\\\
g\end{pmatrix}=\begin{pmatrix}\int_{\cdot}^{\infty}\left(\int_{s_{p+1}}^{\infty}\left(\ldots\left(\int_{s_{2}}^{\infty}\mathrm{e}^{\lambda(\cdot-
s_{1})}f(s_{1})\,\mathrm{d}s_{1}\right)\ldots\right)\,\mathrm{d}s_{p}\right)\,\mathrm{d}s_{p+1}\\\
-\int_{0}^{\infty}\left(\int_{s_{p+1}}^{\infty}\left(\ldots\left(\int_{s_{2}}^{\infty}\mathrm{e}^{\lambda(0-s_{1})}f(s_{1})\,\mathrm{d}s_{1}\right)\ldots\right)\,\mathrm{d}s_{p}\right)\,\mathrm{d}s_{p+1}\end{pmatrix}.$

Let $f(\tau)=\tau\mathrm{e}^{-\lambda\tau}$. Then,
$\left\|f\right\|_{L^{2}}=\frac{1}{\lambda^{\frac{3}{2}}}$ and

$\displaystyle\begin{split}\left\|\left((\lambda
E-A)^{-1}E\right)^{p+1}\right\|&\geq\left\|\left((\lambda
E-A)^{-1}E\right)^{p+1}\begin{pmatrix}\lambda^{\frac{3}{2}}f\\\
0\end{pmatrix}\right\|\\\
&\geq\left|\int_{0}^{\infty}\left(\int_{s_{p+1}}^{\infty}\left(\cdots\left(\int_{s_{2}}^{\infty}\mathrm{e}^{-2\lambda
s_{1}}s_{1}\,\mathrm{d}s_{1}\right)\right)\,\mathrm{d}s_{p}\right)\,\mathrm{d}s_{p+1}\right|\\\
&=\frac{p+1}{2^{p+1}}\frac{1}{\lambda^{p+1-\frac{3}{2}}}\end{split}$ (11)

for every $p\in\mathbb{N}$. Consequently, the radiality index does not exist.
(Because if it exists, then there would exist $C>0$, $\omega>0$, such that

$\displaystyle\left\|\left((\lambda E-A)^{-1}E\right)^{p+1}\right\|\leq
C\frac{1}{(\lambda-\omega)^{p+1}}$

for all $\lambda>\omega$, which would contradict (11).)

##  6 Nilpotency index

Next, we are going to look at what is probably the best known index, the
nilpotency index (also known as the Weierstraß-index). As the alternative term
suggests, the most important part of the definition is the existence of the
Weierstraß form.

######  Definition 6.1 (nilpotency index).

Assume, that the DAE (1) has a Weierstraß form given by (2). Then, the
nilpotency index of $(E,A)$, denoted by $p_{\rm
nilp}^{(E,A)}\in\mathbb{N}_{0}$, is the nilpotency degree of $N$ if it is
present and $0$ if $N$ is absent. In the latter case one has
$(E,A)\sim(I_{\mathcal{Y}^{1}},A_{1})$ in (2).

######  Proposition 6.2.

Assume that the DAE (1) has a Weierstraß form. Let $\lambda\in\mathbb{C}$ such
that $\lambda E-A$ is bijective (which exists by our regularity assumption on
(1)). Then the nilpotency index of $(E,A)$ is is the smallest number
$k\in\mathbb{N}$, such that

${\rm ker}\,\big{(}(\lambda E-A)^{-1}E\big{)}^{k}={\rm ker}\,\big{(}(\lambda
E-A)^{-1}E\big{)}^{k+1}.$

In particular, the nilpotency index is well-defined.

###### Proof.

Assume that $P\colon\mathcal{X}\to\mathcal{Y}^{1}\times\mathcal{Y}^{2}$,
$Q\colon\mathcal{Z}\to\mathcal{Y}^{1}\times\mathcal{Y}^{2}$, such that

$E=Q^{-1}\begin{bmatrix}I_{\mathcal{Y}^{1}}&0\\\ 0&N\end{bmatrix}P,\quad
A=Q^{-1}\begin{bmatrix}A_{1}&0\\\ 0&I_{\mathcal{Y}^{2}}\end{bmatrix}P.$

Then the result follows, since for all $k\in\mathbb{N}$,

$\big{(}(\lambda E-A)^{-1}E\big{)}^{k}=Q\begin{bmatrix}(\lambda
I_{\mathcal{Y}^{1}}-A_{1})^{-k}&0\\\ 0&(\lambda
N-I_{\mathcal{Y}^{2}})^{-1}N^{k}\end{bmatrix}Q^{-1}.\qed$

According to this definition, the nilpotency index is always a natural number
(including $0$). This means that the nilpotency index can never be $\infty$,
since we require a nilpotent operator $N$ in (2). To handle this more general
situation, one would have to replace the nilpotency of $N$ with quasi-
nilpotency, namely with $\sigma(N)=\\{0\\}$. For example,
$N\colon\ell^{2}\to\ell^{2}$,
$(x_{1},x_{2},,x_{3},\ldots)\mapsto(0,\frac{x_{1}}{2^{1}},\frac{x_{2}}{2^{2}},\frac{x_{3}}{2^{3}},\ldots)$.

A disadvantage of the nilpotency index is that it requires the Weierstraß
form. In finite dimensions one only needs the system to be regular to obtain
such a form, see, for instance, [10, Def. 2.9] for a constructive procedure.
There is no standard procedure to obtain the Weierstraß form for infinite-
dimensional systems, and, in fact, it has not been proven that such a form
always exists.

######  Proposition 6.3.

  1. (a)

If the nilpotency index exists, then the chain index also exists with $p_{\rm
nilp}^{(E,A)}=p_{\rm chain}^{(E,A)}$.

  2. (b)

If the nilpotency index and resolvent index exist, then $p_{\rm
res\vphantom{d}}^{(E,A)}\geq p_{\rm nilp}^{(E,A)}$.

  3. (c)

If the nilpotency index and radiality index exist, then $p_{\rm
rad}^{(E,A)}+1\geq p_{\rm nilp}^{(E,A)}$.

###### Proof.

We start with the proof of Part (a). Since
$(E,A)\sim(\tilde{E},\tilde{A})\coloneqq\left(\Big{[}\begin{smallmatrix}I_{\mathcal{Y}^{1}}&0\\\
0&N\end{smallmatrix}\Big{]},\Big{[}\begin{smallmatrix}A_{1}&0\\\
0&I_{\mathcal{Y}^{2}}\end{smallmatrix}\Big{]}\right)$ for a Hilbert space
$\mathcal{Y}=\mathcal{Y}^{1}\times\mathcal{Y}^{2}$, where $N$ has nilpotency
degree $p_{\rm nilp}^{(E,A)}$, there exists two isomorphisms
$P\colon\mathcal{X}\to\mathcal{Y}^{1}\times\mathcal{Y}^{2}$,
$x\mapsto(P_{1}x,P_{2}x)$,
$Q\colon\mathcal{Z}\to\mathcal{Y}^{1}\times\mathcal{Y}^{2}$,
$z\mapsto(Q_{1}z,Q_{2}z)$, with

$E=Q^{-1}\begin{bmatrix}I_{\mathcal{Y}^{1}}&0\\\
0&N\end{bmatrix}P\quad\text{and}\quad A=Q^{-1}\begin{bmatrix}A_{1}&0\\\
0&I_{\mathcal{Y}^{2}}\end{bmatrix}P.$ (12)

For simplicity we will show, that $p_{\rm nilp}^{(\tilde{E},\tilde{A})}=p_{\rm
chain}^{(\tilde{E},\tilde{A})}$ (because then the rest follows from
Proposition 4.2 and 6.2). Let $x_{k}=\left(\begin{smallmatrix}x_{k,1}\\\
x_{k,2}\end{smallmatrix}\right)$, $k=1,\ldots,p$ denote an arbitrary chain of
length $p-1$. Then $x_{k,1}=0$ for all $k=1,\ldots,p$ and

$\tilde{E}x_{1}=0,\tilde{E}x_{2}=\tilde{A}x_{1},\quad\ldots\quad,\tilde{E}x_{p}=\tilde{A}x_{p-1}.$

Hence $Nx_{l,2}=x_{l-1,2}$ for all $2\leq l$ and
$x_{1,2}=N^{1}x_{2,2}=\ldots=N^{p-1}x_{p,2}$ . Since $x_{1,2}\neq 0$,
$p>p_{\rm nilp}^{(E,A)}$ is not possible. Thus, the existence of a Weierstraß
form implies that the chain index also exists and that it is at most $p_{\rm
nilp}^{(E,A)}$. In order to prove that these two index-terms are equal, we
need to show that there does exists a chain of length $p_{\rm nilp}^{(E,A)}$.
This follows directly by choosing
$x_{k}\coloneqq\left(\begin{smallmatrix}x_{k,1}\\\
x_{k,2}\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\\ N^{p_{\rm
nilp}^{(\tilde{E},\tilde{A})}-k}z\end{smallmatrix}\right)$, $k=1,\ldots,p_{\rm
nilp}^{(\tilde{E},\tilde{A})}$, for a $z\in\mathcal{Y}^{2}$ such that
$N^{p_{\rm nilp}^{(\tilde{E},\tilde{A})}}z=0$ and $N^{p_{\rm
nilp}^{(\tilde{E},\tilde{A})}-1}z\neq 0$.

We now prove Part (b). From the definition of the resolvent index, there
exists $C>0,\omega>0$, such that $(\omega,\infty)\subseteq\rho(E,A)$ and

$\left\|(\lambda E-A)^{-1}\right\|\leq C\left|\lambda\right|^{p_{\rm
res\vphantom{d}}^{(E,A)}-1}.$

By Proposition 3.2 , for some $\tilde{C}>0$

$\displaystyle\left\|\begin{bmatrix}(\lambda
I_{\mathcal{Y}^{1}}-A_{1})^{-1}&0\\\ 0&(\lambda
N-I_{\mathcal{Y}^{2}})^{-1}\end{bmatrix}\right\|\leq\tilde{C}\left|\lambda\right|^{p_{\rm
res\vphantom{d}}^{(E,A)}-1}$ (13)

for all $\lambda>\omega$. Since $N$ is nilpotent with nilpotency degree
$p_{\rm nilp}^{(E,A)}$ , $(\lambda
N-I_{\mathcal{Y}^{2}})^{-1}=-\sum_{i=0}^{p_{\rm nilp}^{(E,A)}-1}(\lambda
N)^{i}$ and together with (13) we derive

$\displaystyle\left\|(\lambda N-I_{\mathcal{Y}^{2}})^{-1}\right\|$
$\displaystyle=\lambda^{p_{\rm
nilp}^{(E,A)}-1}\left\|I_{\mathcal{X}^{2}}\frac{1}{\lambda^{p_{\rm
nilp}^{(E,A)}-1}}+N\frac{1}{\lambda^{p_{\rm nilp}^{(E,A)}-2}}+\ldots+N^{p_{\rm
nilp}^{(E,A)}-2}\frac{1}{\lambda}+N^{p_{\rm nilp}^{(E,A)}-1}\right\|$
$\displaystyle\leq\tilde{C}\lambda^{p_{\rm res\vphantom{d}}^{(E,A)}-1}$

for all $\lambda>\omega$ and therefore $p_{\rm nilp}^{(E,A)}\leq p_{\rm
res\vphantom{d}}^{(E,A)}$.

Finally, in order to prove Part (c) we assume that the radiality index $p_{\rm
rad}^{(E,A)}$ exists. That is, there exists $C>0$ and $\omega>0$ such that

$\displaystyle\left\|((\lambda N-I_{Y^{2}})^{-1}N)^{p_{\rm
rad}^{(E,A)}+1}\right\|$
$\displaystyle\leq\left\|((\lambda\tilde{E}-\tilde{A})^{-1}\tilde{E})^{p_{\rm
rad}^{(E,A)}+1}\right\|$
$\displaystyle\leq\left\|P^{-1}\right\|\left\|((\lambda E-A)^{-1}E)^{p_{\rm
rad}^{(E,A)}+1}\right\|\left\|P\right\|$
$\displaystyle\leq\frac{C}{(\lambda-\omega)^{p_{\rm rad}^{(E,A)}+1}}$

for all $\lambda>\omega$. Now, $((\lambda
N-I_{\mathcal{Y}^{2}})^{-1}N)^{p_{\rm rad}^{(E,A)}+1}=\frac{1}{\lambda^{p_{\rm
rad}^{(E,A)}+1}}\left(-\sum_{k=1}^{p_{\rm nilp}^{(E,A)}-1}(\lambda
N)^{k}\right)^{p_{\rm rad}^{(E,A)}+1}$ is a polynomial of degree
$1\leq\rm{deg}\leq(p_{\rm nilp}^{(E,A)}-1)$. Thus,

$\left\|\left(-\sum_{k=1}^{p_{\rm nilp}^{(E,A)}-1}(\lambda
N)^{k}\right)^{p_{\rm rad}^{(E,A)}+1}\right\|\leq C\frac{\lambda^{p_{\rm
rad}^{(E,A)}+1}}{(\lambda-\omega)^{p_{\rm rad}^{(E,A)}+1}}.$ (14)

The left-hand-side contains terms of the form $(\lambda N)^{n}$ where $n\geq
p_{\rm rad}^{(E,A)}+1.$ Since it is bounded by the right-hand-side of (14)
which is bounded as $\lambda\to\infty,$ $p_{\rm nilp}^{(E,A)}\leq p_{\rm
rad}^{(E,A)}+1$, as was to be proven. ∎

######  Remark 6.4.

For non-negative and self-adjoint $E\in\mathcal{L}(\mathcal{X})$ and
dissipative $A\colon\mathcal{D}(A)\subseteq\mathcal{X}\to\mathcal{X}$ the
nilpotency index and the chain index is at most $2$. This follows directly
from Theorem 3.3 and Proposition 6.3.

######  Example 6.5.

(A system with $p_{\rm rad}^{(E,A)}=1$ and $p_{\rm nilp}^{(E,A)}=2$) Let us
recall Example 3.5. Defining

$\displaystyle R_{1,s}$ $\displaystyle=(sI-A_{o})^{-1}(I-BG(s)^{-1}C(sI-
A_{o})^{-1}),$ $\displaystyle R_{2,s}$ $\displaystyle=G(s)^{-1}C(sI-
A_{o})^{-1},$ $\displaystyle L_{2,s}$ $\displaystyle=(sI-
A_{o})^{-1}BG(s)^{-1}$

one has

$\displaystyle(sE-A)^{-1}E$ $\displaystyle=\begin{bmatrix}R_{1,s}&0\\\
R_{2,s}&0\end{bmatrix},$ $\displaystyle E(sE-A)^{-1}$
$\displaystyle=\begin{bmatrix}R_{1,s}&L_{2,s}\\\ 0&0\end{bmatrix}.$

As a particular example, let $\mathcal{W}=L^{2}(0,1)$,

$A_{o}z=z^{\prime\prime},\quad D(A_{o})=\\{w\in H^{2}(0,1)\mid
z^{\prime}(0)=z^{\prime}(1)=0\\},\quad b=c=\mathds{1}.$ (15)

With these definitions of $A_{o}$, $B$ and $C$,

$\displaystyle(sI-A_{o})^{-1}Bu$ $\displaystyle=\frac{1}{s}\mathds{1}u,$
$\displaystyle C(sI-A_{o})^{-1}z$ $\displaystyle=\frac{1}{s}\langle
z,\mathds{1}\rangle,$ $\displaystyle G(s)$ $\displaystyle=C(sI-
A_{o})^{-1}B=\frac{1}{s}$

and so $R_{1,s}z=(sI-A_{o})^{-1}z-\mathds{1}\langle z,1\rangle\frac{1}{s}$,
$R_{2,s}z=\langle z,\mathds{1}\rangle$, $L_{2,s}u=\mathds{1}u$. Since
$R_{2,s}$ is independent of $s$, the radiality degree must be larger than $0$.

Define the projection onto $\mathcal{W}_{1}\coloneqq{\rm
ker}\,C\subset\mathcal{W}$,

$\displaystyle Q_{c}z$ $\displaystyle\coloneqq z-\frac{\langle
z,c\rangle}{\langle c,c\rangle}\,c.$

Then, with $\mathcal{W}_{2}\coloneqq{\rm span\,}c$, $Q_{c}$ splits
$\mathcal{W}$ into $\mathcal{W}_{1}\oplus\mathcal{W}_{2}$. It is easy to see
that for $\alpha\mathds{1}\in{\rm span\,}c$ $R_{1,s}\alpha\mathds{1}=0$ and
for $z\in{\rm ker}\,C$ $R_{1,s}z=(sI-A_{o})^{-1}z\in{\rm ker}\,C$. Thus, for
$w=\alpha z\mathds{1}\in W_{1}\oplus W_{2}$

$\displaystyle(sE-A)^{-1}E(\mu E-A)^{-1}E\begin{pmatrix}z+\alpha\mathds{1}\\\
u\end{pmatrix}$ $\displaystyle=\begin{bmatrix}R_{1,s}R_{1,\mu}&0\\\
R_{2,s}R_{1,\mu}&0\end{bmatrix}\begin{pmatrix}z+\alpha\mathds{1}\\\
u\end{pmatrix}=\begin{pmatrix}R_{1,s}R_{1,\mu}z\\\ 0\end{pmatrix},$
$\displaystyle E(sE-A)^{-1}E(\mu E-A)^{-1}\begin{pmatrix}z+\alpha\mathds{1}\\\
u\end{pmatrix}$
$\displaystyle=\begin{bmatrix}R_{1,s}R_{1,\mu}&R_{2,s}L_{2,\mu}\\\
0&0\end{bmatrix}\begin{pmatrix}z+\alpha\mathds{1}\\\
u\end{pmatrix}=\begin{pmatrix}R_{1,s}R_{1,\mu}z\\\ 0\end{pmatrix}.$

Since $A_{o}$ is the generator of a contraction semigroup,

$\displaystyle\left\|R_{1,s}R_{1,\mu}\right\|\leq\frac{1}{s\mu},\quad
s,\mu>0.$

Hence, the radiality index $(E,A)$ is $1$.

Furthermore, it is possible to rewrite $(E,A)$ into a Weierstraß form with
$p_{\rm nilp}^{(E,A)}=2$. For more concise notation, define
$\tilde{C}u\coloneqq\frac{1}{\langle c,c\rangle}cu.$

We can define the isomorphisms
$U\colon\mathcal{W}\times\mathbb{C}\to\mathcal{W}_{1}\times\mathcal{W}_{2}\times\mathbb{C}$
and
$V\colon\mathcal{W}_{1}\times\mathcal{W}_{2}\times\mathbb{C}\to\mathcal{W}\times\mathbb{C}$

$U\coloneqq\begin{bmatrix}Q_{c}&0\\\\[6.45831pt] 0&\tilde{C}\\\\[6.45831pt]
C&0\end{bmatrix},\qquad V\coloneqq\begin{bmatrix}I&I&0\\\ 0&0&I\end{bmatrix}$

which have inverses

$U^{-1}\coloneqq\begin{bmatrix}I&A_{o}&B\\\\[6.45831pt]
0&C&0\end{bmatrix},\qquad V^{-1}\coloneqq\begin{bmatrix}Q_{c}&0\\\
I-Q_{c}&0\\\ 0&I\end{bmatrix}\,.$

These mappings will be applied to (4) to obtain a splitting of the system into
equations on $\mathcal{W}_{1}\times\mathcal{W}_{2}\times\mathbb{C}$. Noting
that

  * •

if $z\in\mathcal{W}_{1},$ $A_{o}z\in\mathcal{W}_{1},$

  * •

$\tilde{C}C=I-Q_{c},$ $=0$ on ${\rm ker}\,C$, $I$ on span $c$

  * •

$CB=1,$

the isomorphisms $U$ and $V$ lead to

$\displaystyle\tilde{E}$ $\displaystyle\coloneqq U\begin{bmatrix}I&0\\\
0&0\end{bmatrix}V$ $\displaystyle\tilde{A}$ $\displaystyle\coloneqq
U\begin{bmatrix}A_{o}&B\\\ C&0\end{bmatrix}V$
$\displaystyle=\begin{bmatrix}Q_{c}&Q_{c}&0\\\ 0&0&0\\\
\tilde{C}&\tilde{C}&0\end{bmatrix}\,,\;$
$\displaystyle=\begin{bmatrix}Q_{c}A_{o}-Q_{c}A_{o}\tilde{C}C&Q_{c}A_{o}-Q_{c}A_{o}\tilde{C}C&0\\\
\tilde{C}C&\tilde{C}C&0\\\ \tilde{C}(A_{o})&\tilde{C}A_{o}&1\end{bmatrix}\,$
$\displaystyle=\begin{bmatrix}I&0&0\\\ 0&0&0\\\ 0&\tilde{C}&0\end{bmatrix}\,$
$\displaystyle=\begin{bmatrix}A_{o}&0&0\\\ 0&I&0\\\ 0&0&1\end{bmatrix}.$

Define

$N=\begin{bmatrix}0&0\\\
\tilde{C}&0\end{bmatrix},\qquad\tilde{A}_{o}z=z^{\prime\prime}$

with domain $D(A_{o})\cap{\rm ker}\,C,$ which is dense in ${\rm ker}\,C$ [13].
The operator $A_{o}$ will generate a $C_{0}$-semigroup on
$\mathcal{W}_{1}={\rm ker}\,C.$ Via the isomorphisms $U$ and $V$ the system
(4) is equivalent to the Weierstraß form

$\frac{{\rm d}}{{\rm d}t}\begin{bmatrix}I_{\mathcal{W}_{1}}&0\\\
0&N\end{bmatrix}\begin{pmatrix}z_{1}(t)\\\
z_{2}(t)\end{pmatrix}=\begin{bmatrix}\tilde{A}_{o}&0\\\
0&I_{\mathcal{W}_{2}\times\mathbb{C}}\end{bmatrix}\begin{pmatrix}z_{1}(t)\\\
z_{2}(t)\end{pmatrix}.$ (16)

A simple calculation shows that $N^{2}=0$, and thus, the nilpotency index of
the system is $2$.

Systems of the form (4) are known as Hessenberg index 2 systems in the finite-
dimensional situation, see for example, [11].

This problem is identical to the question of establishing the zero dynamics of
a system with state-space realization $(A_{o},B,C)$. That is, finding the
largest space $\mathcal{W}_{1}$ on which for initial conditions in
$\mathcal{W}_{1}$ the dynamics remain in $\mathcal{W}_{1}$. Clearly
$\mathcal{W}_{1}\subseteq{\rm ker}\,C$. In this case, the largest space is
$\mathcal{W}_{1}={\rm ker}\,C$. A more general situation with $c\in
D(A_{o}^{*})$ and $b\neq c$ is described in [21]. The case where $c\notin
D(A_{o}^{*})$ is treated in [13]. If $c\notin D(A_{o}^{*})$, in general the
zero dynamics are only associated with an integrated semigroup [13]. A similar
type of construction was shown to hold for a class of boundary control systems
with collocated observation; that is, the unbounded generalization of $\langle
b,c\rangle\neq 0$ in [16] and for diffusion problems on an interval in [3].

######  Example 6.6.

(A system where the nilpotency index exists, but resolvent index does not.)
Let $A_{0}=-\partial^{2}_{x}+ix$ be the complex Airy operator on
$\mathcal{X}^{1}=L^{2}(\mathbb{R})$ with

$\mathcal{D}(A_{0})=\left\\{u\in H^{2}(\mathbb{R})\,\middle|\,xu\in
L^{2}(\mathbb{R})\right\\}.$

By [7, 1] we have $\sigma(A_{0})=\emptyset$ and there exists a constant $C>0$,
such that

$\left\|(\lambda I_{\mathcal{X}^{1}}-A_{0})^{-1}\right\|=C({\rm
Re}\,\lambda)^{-\frac{1}{4}}\mathrm{e}^{\frac{4}{3}({\rm
Re}\,\lambda)^{\frac{3}{2}}}$

for all ${\rm Re}\,\lambda>0$. Thus, $\left\|(\lambda
I_{\mathcal{X}^{1}}-A_{0})^{-1}\right\|\to\infty$ for ${\rm
Re}\,\lambda\to\infty$. Let $N\colon\mathcal{X}^{2}\to\mathcal{X}^{2}$ be a
nilpotent operator of degree $p\in\mathbb{N}$. Then

$\frac{{\rm d}}{{\rm d}t}\underbrace{\begin{bmatrix}I_{\mathcal{X}^{1}}&0\\\
0&N\end{bmatrix}}_{=E}\begin{pmatrix}x_{1}\\\
x_{2}\end{pmatrix}=\underbrace{\begin{bmatrix}A_{0}&0\\\
0&I_{\mathcal{X}^{2}}\end{bmatrix}}_{=A}\begin{pmatrix}x_{1}\\\
x_{2}\end{pmatrix}$

defines a DAE with nilpotency index $p=p_{\rm nilp}^{(E,A)}$, but since
$\left\|(\lambda I_{\mathcal{X}^{1}}-A_{0})^{-1}\right\|$ grows exponentially
the resolvent index does not exist.

######  Example 6.7.

(A system where the nilpotency index exists, but radiality index does not.)
Example 6.6 shows that in general the existence of the nilpotency index does
not imply the existence of the radiality index.

######  Example 6.8.

(A system where the resolvent index exists, but nilpotency index does not.)
Consider the system

$\displaystyle\tfrac{\partial}{\partial t}x_{1}(t,\xi)$
$\displaystyle=-\tfrac{\partial}{\partial\xi}x_{1}(t,\xi),$ (17)
$\displaystyle 0$ $\displaystyle=-x_{1}(t,0),$ (18) $\displaystyle 0$
$\displaystyle=-x_{1}(t,1)+x_{2}(t),$ (19) $\displaystyle\tfrac{\rm d}{{\rm
d}t}x_{2}(t)$ $\displaystyle=x_{3}(t),$ (20)

which corresponds to a differential-algebraic equation $\frac{{\rm d}}{{\rm
d}t}Ex(t)=Ax(t)$ with

$\displaystyle\mathcal{X}$
$\displaystyle=L^{2}([0,1])\times\mathbb{C}^{2},\quad\mathcal{D}(A)=H^{1}([0,1])\times\mathbb{C}^{2},$
(21) $\displaystyle\mathcal{Z}$
$\displaystyle=L^{2}([0,1])\times\mathbb{C}^{3},$

and

$E=\begin{bmatrix}I_{L^{2}}&0&0\\\ 0&0&0\\\ 0&0&0\\\ 0&1&0\end{bmatrix},\quad
A=\begin{bmatrix}-\frac{\partial}{\partial\xi}&0&0\\\ -\delta_{0}&0&0\\\
-\delta_{1}&1&0\\\ 0&0&1\end{bmatrix},$ (22)

where $\delta_{\xi}\in H^{1}([0,1])^{*}$ is the evaluation operator at
$\xi\in[0,1]$. The resolvent fulfills, for all
$x_{1},x_{2},x_{2}\in\mathbb{C}$, $x_{2}\in L^{2}([0,1])$,

$(\lambda E-A)^{-1}\begin{pmatrix}x_{1}\\\ x_{2}\\\ x_{3}\\\
x_{4}\end{pmatrix}=\begin{pmatrix}{\rm e}^{-s}x_{3}+\int_{0}^{1}{\rm
e}^{-s(1-\xi)}f(\xi)d\xi\\\ s{\rm e}^{-s}x_{3}+s\int_{0}^{1}{\rm
e}^{-s(1-\xi)}f(\xi)d\xi-x_{2}\\\ {\rm e}^{-s\cdot}x_{3}+\int_{0}^{\cdot}{\rm
e}^{-s(\cdot-\xi)}f(\xi)d\xi\end{pmatrix},$

which gives $p_{\rm res\vphantom{d}}^{(E,A)}=2$. It has however been shown in
[17] that the system has no nilpotency index.

As we have already seen through Example 5.5 and 6.6 the nilpotency index, the
resolvent index and the radiality index do not have to coincide. This is
different in finite dimensions as we will see next.

######  Proposition 6.9.

Let $\mathcal{X}$ and $\mathcal{Z}$ be finite-dimensional. Then $p_{\rm
nilp}^{(E,A)}=p_{\rm rad}^{(E,A)}+1=p_{\rm res\vphantom{d}}^{(E,A)}$.

###### Proof.

In [4, Lem. 2.1] it is shown that $p_{\rm nilp}^{(E,A)}=p_{\rm
res\vphantom{d}}^{(E,A)}$. Thus it remains to show $p_{\rm
nilp}^{(E,A)}=p_{\rm rad}^{(E,A)}+1$.

By Proposition 6.3 (c) we already have $p_{\rm rad}^{(E,A)}+1\geq p_{\rm
nilp}^{(E,A)}$. Let $\left(\big{[}\begin{smallmatrix}I&0\\\
0&N\end{smallmatrix}\big{]},\big{[}\begin{smallmatrix}J&0\\\
0&I\end{smallmatrix}\big{]}\right)$ be the Weierstraß form of $(E,A)$. For
$\lambda\in\rho(E,A)$ the right-resolvent of this form is

$R_{\lambda}\coloneqq\left(\lambda\begin{bmatrix}I&0\\\
0&N\end{bmatrix}-\begin{bmatrix}J&0\\\
0&I\end{bmatrix}\right)^{-1}\begin{bmatrix}I&0\\\
0&N\end{bmatrix}=\begin{bmatrix}(\lambda I-J)^{-1}&0\\\ 0&(\lambda
N-I)^{-1}N\end{bmatrix}$ (23)

and since $N$ commutes with $(\lambda N-I)^{-1}=-\sum_{k=0}^{p_{\rm
nilp}^{(E,A)}-1}(\lambda N)^{k}$ this is the same as the left-resolvent. Let
$\lambda_{0},\ldots,\lambda_{p_{\rm nilp}^{(E,A)}-1}\in\rho(E,A)$. Using the
nilpotency of $N$ we get

$(\lambda_{0}N-I)^{-1}N\cdot\ldots\cdot(\lambda_{p_{\rm
nilp}^{(E,A)}-1}N-I)^{-1}N=(\lambda_{0}N-I)^{-1}\cdot\ldots\cdot(\lambda_{p_{\rm
nilp}^{(E,A)}-1}N-I)^{-1}N^{p_{\rm nilp}^{(E,A)}}=0.$ (24)

Thus,

$R_{\lambda_{0}}\cdot\ldots\cdot R_{\lambda_{p_{\rm
nilp}^{(E,A)}-1}}=\begin{bmatrix}\displaystyle\prod_{k=0}^{p_{\rm
nilp}^{(E,A)}-1}(\lambda_{k}I-J)^{-1}&0\\\ 0&0\end{bmatrix}.$

Since $J$ is in Jordan form there exists a $C>0$, such that for all
$\lambda_{0},\ldots,\lambda_{p_{\rm nilp}^{(E,A)}-1}>\omega$ we have

$\left\|(\lambda_{0}I-J)^{-1}\cdot\ldots\cdot(\lambda_{p_{\rm
nilp}^{(E,A)}-1}I-J)^{-1}\right\|\leq
K\frac{1}{(\lambda_{0}-\omega)\cdot\ldots\cdot(\lambda_{p_{\rm
nilp}^{(E,A)}-1}-\omega)}.$ (25)

Thus, $(E,A)$ has at most radiality index $p_{\rm nilp}^{(E,A)}-1$ and
therefore $p_{\rm nilp}^{(E,A)}\geq p_{\rm rad}^{(E,A)}+1$. ∎

##  7 Differentiation index

The differentiation index is based on taking formal derivatives up to a
certain order which leads to an ordinary differential equation. For details,
see, for instance, [11, Sec. 3.10] and [10, Sec. 3.3]. Here we present a
generalization of this concept to abstract differential-algebraic equations of
the form (1) with the restriction that only the case where $f\equiv 0$ is
considered. An extension to nonzero $f$ is a subject for future research.
Using the fact that the operators $E$ and $A$ do not depend on time, the
$k$-th formal derivative of (1) with $f=0$ is

$\frac{{\rm d}}{{\rm d}t}E\frac{d^{k}}{dt^{k}}x(t)=A\frac{d^{k}}{dt^{k}}x(t).$

The collection of the first $\mu$ formal derivatives of (1) is

$\underbrace{\begin{bmatrix}A&-E\\\ &A&-E\\\ &&\ddots&\ddots\\\
&&&A&-E\end{bmatrix}}_{\eqqcolon M_{\mu}}\begin{pmatrix}x(t)\\\ \frac{{\rm
d}}{{\rm d}t}x(t)\\\ \vdots\\\ \frac{d^{\mu}}{dt^{\mu}}x(t)\\\
\frac{d^{\mu+1}}{dt^{\mu+1}}x(t)\end{pmatrix}=0$ (26)

which will be referred to as a derivative array of order $\mu$. The operator
$M_{\mu}$ maps from $\mathcal{D}(A)^{\mu+1}\times\mathcal{X}$ to
$\mathcal{Z}^{\mu+1}$.

######  Definition 7.1 (Differentiation index).

Let $\mathcal{X},\mathcal{Z}$ be Banach spaces, and let a linear differential-
algebraic equation $\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$ be given, where
$E\in\mathcal{L}(\mathcal{X},\mathcal{Z})$, $(A,\mathcal{D}(A))$ is a closed
and densely defined linear operator from $\mathcal{X}$ to $\mathcal{Z}$. The
DAE $\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$ has differentiation index $p_{\rm
diff}^{(E,A)}\in\mathbb{N}_{0}$, if the following holds:

  1. (a)

There exists a Banach space $\tilde{\mathcal{X}}$, such that
$\tilde{\mathcal{X}}$ is densely embedded into $\mathcal{X}$, an operator
$(S,\mathcal{D}(S))$ which is the generator of a strongly continuous semigroup
on $\tilde{\mathcal{X}}$, and the operator $M_{p_{\rm diff}^{(E,A)}}$ as
defined in (26) satisfies

$\begin{pmatrix}x_{0}\\\ x_{1}\\\ \vdots\\\ x_{p_{\rm
diff}+1}\end{pmatrix}\in{\rm ker}\,M_{p_{\rm
diff}^{(E,A)}}\quad\Longrightarrow\,x_{0}\in\mathcal{D}(S)\text{ with
}x_{1}=Sx_{0}.$

  2. (b)

The number $p_{\rm diff}^{(E,A)}$ is minimal with the properties in (a).

We call

$\frac{{\rm d}}{{\rm d}t}x(t)=Sx(t)$ (27)

an abstract completion ODE of (1).

Before establishing a connection between the solutions of the abstract
completion ODE and the DAE itself, we introduce a lemma that characterizes
solutions. This result is a direct consequence of the integration by parts
formula. It is therefore omitted.

######  Lemma 7.2.

Let $\mathcal{X},\mathcal{Z}$ be Banach spaces, and let a linear differential-
algebraic equation $\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$ be given, where
$E\in\mathcal{L}(\mathcal{X},\mathcal{Z})$, $(A,\mathcal{D}(A))$ is a closed
and densely defined linear operator from $\mathcal{X}$ to $\mathcal{Z}$.

  1. (a)

If $x\colon[0,\infty)\to X$ solves $\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$, then
for all test functions $\varphi\colon[0,\infty)\to\mathbb{R}$ (i.e., those
which are smooth and have compact support in $(0,\infty)$),

$-\int_{\mathbb{R}_{\geq}0}\frac{{\rm d}}{{\rm
d}t}\varphi(t)Ex(t)dt=\int_{\mathbb{R}_{\geq}0}\varphi(t)Ax(t)dt.$ (28)

  2. (b)

Conversely, if $x\colon[0,\infty)\to\mathbb{R}$ fulfills (28) for all test
functions $\varphi\colon[0,\infty)\to\mathbb{R}$, and it is continuously
differentiable with $x(t)\in\mathcal{D}(A)$ for all $t\geq 0$, then it is a
solution of $\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$.

######  Theorem 7.3.

Let $\mathcal{X},\mathcal{Z}$ be Banach spaces, and let
$E\in\mathcal{L}(\mathcal{X},\mathcal{Z})$, and $(A,\mathcal{D}(A))$ be a
closed and densely defined linear operator from $\mathcal{X}$ to
$\mathcal{Z}$. Assume that the linear differential-algebraic equation
$\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$ has differentiation index $p_{\rm
diff}^{(E,A)}$. If $x$ solves $\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$, then it
fulfills the abstract completion ODE (27).

###### Proof.

Assume that $x$ solves $\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$. Lemma 7.2
implies that for all test functions $\varphi\colon[0,\infty)\to\mathbb{R}$

$-\int_{\mathbb{R}_{\geq}0}\frac{{\rm d}}{{\rm
d}t}\varphi(t)Ex(t)dt=\int_{\mathbb{R}_{\geq}0}\varphi(t)Ax(t)dt.$

Now testing with $(-1)^{k}\frac{d^{k}}{dt^{k}}\varphi(t)$, $k=0,\ldots,p_{\rm
diff}^{(E,A)}+1$, we obtain that the vectors

$\displaystyle x_{k}$
$\displaystyle=(-1)^{k}\int_{\mathbb{R}_{\geq}0}\frac{d^{k}}{dt^{k}}\varphi(t)x(t)dt,\quad
k=0,\ldots,p_{\rm diff}^{(E,A)}+1,$

satisfy

$\begin{bmatrix}A&-E\\\ &A&-E\\\ &&\ddots&\ddots\\\
&&&A&-E\end{bmatrix}\begin{pmatrix}x_{0}\\\ x_{1}\\\ \vdots\\\ x_{p_{\rm
diff}^{(E,A)}}\\\ x_{p_{\rm diff}^{(E,A)}+1}\end{pmatrix}=-0,$

and the definition of the differentiation index yields

$x_{1}=Sx_{0}.$

Thus, for all test functions $\varphi\colon[0,\infty)\to\mathbb{R}$

$-\int_{\mathbb{R}_{\geq}0}\tfrac{{\rm d}}{{\rm
d}t}\varphi(t)x(t)dt=\int_{\mathbb{R}_{\geq}0}\varphi(t)Sx(t)dt.$

Another application of Lemma 7.2 now shows that $x$ is a solution of the
abstract completion ODE of (1). ∎

######  Remark 7.4.

Let $\mathcal{X},\mathcal{Z}$ be Banach spaces, and let
$E\in\mathcal{L}(\mathcal{X},\mathcal{Z})$, and $(A,\mathcal{D}(A))$ be a
closed and densely defined linear operator from $\mathcal{X}$ to
$\mathcal{Z}$. Assume that the linear differential-algebraic equation
$\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$ has differentiation index $p_{\rm
diff}^{(E,A)}$, and let (27) be an abstract completion ODE. Let
$x_{0}\in\mathcal{X}$, such that the differential-algebraic initial value
problem $\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$, $Ex_{0}=Ex(0)$ has a solution
$x$. Since the solution of the abstract completion boundary control system
(27) with initial value $x(0)=x_{0}\in\mathcal{X}$ is unique, we can conclude
that $x$ is the unique solution of the differential-algebraic initial value
problem $\frac{{\rm d}}{{\rm d}t}Ex(t)=Ax(t)$.

######  Example 7.5.

(A system with $p_{\rm diff}^{(E,A)}=2$) Once again, consider the system
(17)–(20), which corresponds to a differential-algebraic equation $\frac{{\rm
d}}{{\rm d}t}Ex(t)=Ax(t)$ with spaces (21) and operators as in (22).
The system does not have differentiation index zero, since $E$ has a
nontrivial nullspace. To analyze whether it has differentiation index one,
assume that $x_{1}(0,\cdot)=x_{10}$ for some given $x_{10}\in L^{2}([0,1])$.
Then the solution of the initial-boundary value problem (17), (18) reads
$x_{1}\colon[0,\infty)\to L^{2}([0,1])$ with

$(x_{1}(t))(\xi)=x_{1}(t,\xi)=\begin{cases}x_{0}(\xi-t):&\xi>t,\\\
0:&\xi<t.\end{cases}$

By thereafter plugging this into (18) and (19), we obtain that

$x_{2}(t)=\begin{cases}x_{0}(1-t):&t\leq 1,\\\ 0:&t>1,\end{cases}$

and (20) yields that $x_{3}(t)=\frac{d}{dt}x_{2}(t)$. This requires
differentiability of $x_{2}$, which is fulfilled, if $x_{0}\in H^{3}([0,1])$
with $x_{0}(0)=x_{0}^{\prime}(0)=x_{0}^{\prime\prime}(0)=0$. In this case,

$x_{3}(t)=\begin{cases}-\frac{\partial}{\partial\xi}x_{10}(1-t):&t\leq 1,\\\
0:&t>1.\end{cases}$

Hence, a classical solution of (17)–(20) exists if

$x_{10}\in H^{3}([0,1])\text{ with
}x_{10}(0)=x_{10}^{\prime}(0)=x_{10}^{\prime\prime}(0)=0.$

Next we analyse the differentiation index $p_{\rm diff}^{(E,A)}\in\mathbb{N}$:
Since with

$x_{1}=\left(\begin{smallmatrix}0\\\ 0\\\
1\end{smallmatrix}\right),\,x_{2}=\left(\begin{smallmatrix}0\\\ 1\\\
0\end{smallmatrix}\right)\in\mathcal{X},$ $\left(\begin{smallmatrix}{0}\\\
x_{1}\\\ x_{2}\end{smallmatrix}\right)\in{\rm ker}\,M_{1}$

one has $p_{\rm diff}^{(E,A)}>1$.

It will now be shown that $p_{\rm diff}^{(E,A)}=2.$ To this end, let
$x_{k}\in\mathcal{X}$, $k=0,1,2,3$, partitioned as

$x_{k}=\left(\begin{smallmatrix}x_{k1}\\\ x_{k2}\\\
x_{k3}\end{smallmatrix}\right),\quad x_{k1}\in
L^{2}(0,1),\;x_{k2},x_{k3}\in\mathbb{C},$

and

$\left(\begin{smallmatrix}x_{0}\\\ x_{1}\\\ x_{2}\\\
x_{3}\end{smallmatrix}\right)\in{\rm ker}\,M_{2}.$ (29)

Then

$\tfrac{\partial}{\partial\xi}x_{11}=-x_{21},\quad\tfrac{\partial}{\partial\xi}x_{21}=-x_{21},$

and

$0=\delta_{0}x_{11},\quad
0=\delta_{0}x_{21}=-\delta_{0}\tfrac{\partial}{\partial\xi}x_{11},$

which implies that

$x_{11}\in\tilde{\mathcal{X}}_{1}\coloneqq\left\\{x\in
H^{2}(0,1)\left|\vphantom{x\in
H^{2}(0,1)}x(0)=x^{\prime}(0)=0\right.\right\\}.$

Analogously, we see that

$x_{01}\in\tilde{\mathcal{Y}}_{1}\coloneqq\left\\{x\in
H^{3}(0,1)\left|\vphantom{x\in
H^{3}(0,1)}x(0)=x^{\prime}(0)=x^{\prime\prime}(0)=0\right.\right\\}.$

Further, by

$\displaystyle x_{11}$ $\displaystyle=-\tfrac{\partial}{\partial\xi}x_{10},$
$\displaystyle x_{12}$ $\displaystyle=x_{03},$ $\displaystyle x_{13}$
$\displaystyle=x_{22}=\delta_{1}x_{21}=-\delta_{1}\tfrac{\partial}{\partial\xi}x_{11}=\delta_{1}\tfrac{\partial^{2}}{\partial\xi^{2}}x_{10},$

we have that (29) implies that, for

$\tilde{\mathcal{X}}\coloneqq\tilde{\mathcal{X}}_{1}\times\mathbb{C}^{2}$

and $(S,\mathcal{D}(S))$ with

$\mathcal{D}(S)=\tilde{\mathcal{Y}}_{1}\times\mathbb{C}^{2},\quad
S\begin{pmatrix}x_{01}\\\ x_{02}\\\
x_{03}\end{pmatrix}=\begin{pmatrix}-\frac{\partial}{\partial\xi}x_{01}\\\
x_{03}\\\ x_{10}^{\prime\prime}(1)\end{pmatrix}.$

The operator $S$ is indeed the generator of a strongly continuous semigroup on
$\tilde{\mathcal{X}}$, namely $T(\cdot)$ with

$T(t)\begin{pmatrix}x_{01}\\\ x_{02}\\\
x_{03}\end{pmatrix}=\begin{pmatrix}\mathbb{S}_{t}x_{01}\\\
x_{02}+tx_{03}+\int_{0}^{t}\int_{0}^{\tau}\big{(}\mathbb{S}_{s}x_{10}\big{)}(1)dsd\tau\\\
x_{03}+\int_{0}^{t}\big{(}\mathbb{S}_{\tau}x_{10}\big{)}(1)d\tau\end{pmatrix},$

where $\mathbb{S}_{t}$ denotes the right shift of length $t$ on functions
defined on the interval $[0,1]$.
In the case where the initial condition fulfills

$x_{02}=x_{01}(1),\quad x_{03}=-x_{01}^{\prime}(1),$

it can be seen that

$x(t)=T(t)\begin{pmatrix}x_{01}\\\ x_{02}\\\ x_{03}\end{pmatrix}$

is indeed a solution of (17)-(20).

######  Proposition 7.6.

  1. (a)

If the differentiation index $p_{\rm diff}^{(E,A)}$ exists, then the chain
index exists and $p_{\rm chain}^{(E,A)}\leq p_{\rm diff}^{(E,A)}$.

  2. (b)

If the nilpotency index $p_{\rm nilp}^{(E,A)}$ exists and the operator $A_{1}$
in the Weierstraß form generates a $C_{0}-$semigroup, then the differentiation
index exists and $p_{\rm nilp}^{(E,A)}=p_{\rm diff}^{(E,A)}$.

###### Proof.

  1. (a)

Assume that the differentiation index exists. Further, let
$(x_{1},\ldots,x_{p})$, $p\in\mathbb{N}$, be a chain of $(E,A)$. Then

$M_{p-1}\left(\begin{smallmatrix}0\\\ x_{1}\\\ \vdots\\\
x_{p}\end{smallmatrix}\right)=0.$

This implies that there does not exist a mapping
$S\colon\mathcal{D}(S)\to\mathcal{X}$ such that every $z\in{\rm ker}\,M_{p-1}$
satisfies

$z=\left(\begin{smallmatrix}x_{0}\\\ Sx_{0}\\\ x_{2}\\\ \vdots\\\
x_{p}\end{smallmatrix}\right)$

for some $x_{0}\in\mathcal{X}$, $x_{2},\ldots,x_{p}\in\mathcal{D}(A)$.
Consequently, $p\leq p_{\rm diff}^{(E,A)};$ that is, the chain index of
$(E,A)$ does not exceed the differentiation index of $(E,A)$.

  2. (b)

Assume that $(E,A)$ is in Weierstraß form (2) with $\nu=p_{\rm nilp}^{(E,A)}$
and $N^{\nu}=0$, $N^{\nu-1}\neq 0$. Let
$x_{k}\in\mathcal{X}=\mathcal{Y}^{1}\times\mathcal{Y}^{2}$, $k\in\mathbb{N}$,
be partitioned as

$x_{k}=\left(\begin{smallmatrix}x_{k,1}\\\
x_{k,2}\end{smallmatrix}\right),\quad
x_{k,1}\in\mathcal{Y}^{1},\;x_{k,2}\in\mathcal{Y}^{2}.$

Using the Weierstraß form, it follows that

$\left(\begin{smallmatrix}x_{0}\\\ \vdots\\\
x_{k+1}\end{smallmatrix}\right)\in{\rm ker}\,M_{k}$ (30)

if, and only if, for some $x_{1,0}\in\mathcal{D}(A_{1}^{k+1})$,
$x_{2,0}\in\mathcal{Y}^{2}$,

$x_{1,i}=A_{1}^{i}x_{1,0},\quad N^{i}x_{2,i}=x_{2,0},\quad i=1,\ldots,k+1.$
(31)

Let $k<\nu$. Then by choosing some $z\in\mathcal{Y}^{2}$ with $N^{\nu-1}z\neq
0$, we obtain (30) with

$x_{0}=\left(\begin{smallmatrix}0\\\
0\end{smallmatrix}\right),\,x_{1}=\left(\begin{smallmatrix}0\\\
N^{\nu-1}z\end{smallmatrix}\right),\ldots,\,x_{k+1}=\left(\begin{smallmatrix}0\\\
N^{\nu-k-1}z\end{smallmatrix}\right).$

Since $x_{0}=0$ and $x_{1}\neq 0$, an operator $S$ with the properties as
described in Definition 7.1 cannot exist.

On the other hand, the equivalence between (30) and (31) implies that if
$k=\nu$

$x_{1}=\left(\begin{smallmatrix}A_{1}x_{0,1}\\\ 0\end{smallmatrix}\right).$

Consequently, we can choose $\tilde{\mathcal{X}}=\mathcal{X}$ and
$S\colon\mathcal{D}(S)\subset\mathcal{X}\to\mathcal{X}$ with
$\mathcal{D}(S)=\mathcal{D}(A_{1})\times\mathcal{Y}^{2}$ and

$S=\begin{bmatrix}A_{1}&0\\\ 0&0\end{bmatrix},$

which is the generator of the strongly continuous semigroup

$T(\cdot)=\begin{bmatrix}T_{1}(\cdot)&0\\\
0&I_{\mathcal{Y}^{2}}\end{bmatrix},$

where $T_{1}(\cdot)$ is the semigroup generated by $A_{1}$. This shows that
the differentiation index coincides with the nilpotency index of $N$.∎

##  8 Perturbation index

Finally, we define the perturbation index for DAEs on possibly infinite-
dimensional spaces. As the name indicates, the perturbation index is a measure
of sensitivity of solutions with respect to perturbations of the problem.

######  Definition 8.1 (perturbation index).

Consider the DAE (1) on $[0,T]$ with input $\delta\in C^{p_{\rm
pert}^{(E,A)}-1}([0,T],\mathcal{Z})$ and solution $x:$

$\displaystyle\frac{\text{d}}{\text{d}t}Ex(t)=Ax(t)+\delta(t),\quad t\geq 0.$
(32)

The perturbation index $p_{\rm pert}^{(E,A)}$ is the smallest number
$p\in\mathbb{N}$ such that there exists a constant $c>0$ satisfying the
following for all $\delta:$

$\displaystyle\max_{0\leq t\leq T}\|x(t)\|_{\mathcal{X}}$ $\displaystyle\leq
c\left(\|x(0)\|_{\mathcal{X}}+\sum_{i=0}^{p-1}\max_{0\leq t\leq
T}\left\|\frac{\text{d}^{i}}{\text{d}t^{i}}\delta(t)\right\|_{\mathcal{Z}}\right),$
$\displaystyle\text{if }p>0,$ (33) $\displaystyle\max_{0\leq t\leq
T}\|x(t)\|_{\mathcal{X}}$ $\displaystyle\leq
c\left(\|x(0)\|_{\mathcal{X}}+\int_{0}^{T}\|\delta(t)\|_{\mathcal{Z}}{\rm
d}t\right),$ $\displaystyle\text{if }p=0.$ (34)

First we show that the perturbation index is identical for equivalent systems.

######  Proposition 8.2.

Consider two differential-algebraic systems $\frac{{\rm d}}{{\rm d}t}Ex=Ax$,
$\frac{{\rm d}}{{\rm d}t}\tilde{E}\tilde{x}=\tilde{A}\tilde{x}$ with
$(E,A)\sim(\tilde{E},\tilde{A})$. Then

$p_{\rm pert}^{(E,A)}=p_{\rm pert}^{(\tilde{E},\tilde{A})}.$

###### Proof.

Let $P\colon\mathcal{X}\to\tilde{\mathcal{X}}$,
$Q\colon\mathcal{Z}\to\tilde{\mathcal{Z}}$, be bounded isomorphisms as in
Definition 2.1. Then the result follows, since $x$ is a solution of (32), if,
and only if, $\tilde{x}=Px$ is a solution of $\frac{{\rm d}}{{\rm
d}t}\tilde{E}\tilde{x}=\tilde{A}\tilde{x}+\tilde{\delta}$ for
$\tilde{\delta}=Q^{-1}\delta$. ∎

######  Proposition 8.3.

The following holds:

  1. (a)

If the nilpotency index $p_{\rm nilp}^{(E,A)}$ and the perturbation index
$p_{\rm pert}^{(E,A)}$ exist, then $p_{\rm nilp}^{(E,A)}\leq p_{\rm
pert}^{(E,A)}$.

  2. (b)

If the nilpotency index $p_{\rm nilp}^{(E,A)}$ exists and $A_{1}$ in (2)
generates a $C_{0}$-semigroup, then the perturbation index also exists and
$p_{\rm nilp}^{(E,A)}=p_{\rm pert}^{(E,A)}$.

###### Proof.

By Proposition 8.2, it is no loss of generality to assume that the system is
in Weierstraß form, that is,

$(E,A)=\left(\begin{bmatrix}I_{\mathcal{Y}^{1}}&0\\\
0&N\end{bmatrix},\begin{bmatrix}A_{1}&0\\\
0&I_{\mathcal{Y}^{2}}\end{bmatrix}\right)$

and $p\coloneqq p_{\rm nilp}^{(E,A)}$.

First, we show Part (a). If $p_{\rm nilp}^{(E,A)}=0$, then nothing has to be
shown. If $p_{\rm nilp}^{(E,A)}=1$, then $N=0$ and thus, for the second part
of a solution $x$ of (32) holds $x_{2}(t)=-\delta_{2}(t)$. Consequently, the
maximum of $x$ cannot be estimated by the integral of $\|\delta_{2}\|$, and
thus also an estimate of the form (34) is not possible. Thus, we have $p_{\rm
pert}^{(E,A)}\leq 1$ in this case.
Finally, we assume that $p\coloneqq p_{\rm nilp}^{(E,A)}\geq 2$. Seeking for a
contradiction, we assume that $p_{\rm pert}^{(E,A)}<p$. Then there exists some
$c>0$, such that for all solutions $x$ of (32), it holds

$\max_{0\leq t\leq T}\|x(t)\|_{\mathcal{X}}\leq
c\left(\|x(0)\|_{\mathcal{X}}+\sum_{i=0}^{p-2}\max_{0\leq t\leq
T}\left\|\frac{\text{d}^{i}}{\text{d}t^{i}}\delta(t)\right\|_{\mathcal{Z}}\right)$
(35)

Let $v_{2}\in\mathcal{Y}_{2}$ with $N^{p-1}v_{2}\neq 0$. We choose
$\delta_{1}=0$ $x_{1}(0)=0$ and $\delta_{2,n}\colon[0,T]\to\mathcal{Y}_{2}$
with

$\delta_{2,n}(t)=\frac{{\rm Re}\big{(}\imath^{p}{\rm e}^{\imath
nt}\big{)}}{n^{p-1}}v_{2}.$

Then, by using that ${\rm Re}\big{(}\imath^{p}{\rm e}^{\imath nt}\big{)}$ is a
trigonometric function, we have

$\forall\,n\geq\frac{\pi}{T}:\quad\max_{0\leq t\leq
T}\|N^{p-1}\delta_{2,n}^{(p-1)}\|\geq\|N^{p-1}v_{2}\|>0,$ (36)

and, further

$\displaystyle\forall\,k=1,\ldots,p-2:\quad\lim_{n\to\infty}\max_{0\leq t\leq
T}\|\delta_{2,n}^{(k)}\|=0,$ (37) $\displaystyle\delta_{2,n}(0)={\rm
Re}\left.\big{(}\imath{\rm e}^{\imath nt}\big{)}\right|_{t=0}v_{2}=0.$ (38)

Let $\delta_{n}=\left(\begin{smallmatrix}0\\\
\delta_{2,n}\end{smallmatrix}\right)$. Since if $x_{1,n}=0$ the solution
$x_{n}=\left(\begin{smallmatrix}x_{1,n}\\\ x_{2,n}\end{smallmatrix}\right)$ of
$\frac{\text{d}}{\text{d}t}Ex_{n}(t)=Ax_{n}(t)+\delta_{n}(t)$ has
$x_{1,n}(0)=0$ and also

$x_{2,n}=-\sum_{i=0}^{p-1}N^{i}\frac{\text{d}^{i}}{\text{d}t^{i}}\delta_{2,n}(t),$

we obtain from (36) and (37) that

$\liminf_{n\to\infty}\max_{0\leq t\leq T}\|x_{n}(t)\|>0.$

On the other hand, by (36) and (38), we obtain

$\lim_{n\to\infty}\underbrace{\|x_{n}(0)\|_{\mathcal{X}}}_{=0}+\sum_{i=0}^{p-2}\max_{0\leq
t\leq
T}\left\|\frac{\text{d}^{i}}{\text{d}t^{i}}\delta_{n}(t)\right\|_{\mathcal{Z}}=0,$

which contradicts (35).

Next, we will show Part (b). Let $x$ be a solution of DAE (32), and partition

$x(t)=\begin{pmatrix}x_{1}(t)\\\
x_{2}(t)\end{pmatrix},\quad\delta(t)=\begin{pmatrix}\delta_{1}(t)\\\
\delta_{2}(t)\end{pmatrix}$

according to the structure of the Weierstraß form. Then

$\displaystyle\frac{\text{d}}{\text{d}t}x_{1}(t))$
$\displaystyle=A_{1}x_{1}(t)+\delta_{1}(t),$ $\displaystyle t\geq 0,$ (39)
$\displaystyle N\frac{\text{d}}{\text{d}t}x_{2}(t)$
$\displaystyle=x_{2}(t)+\delta_{2}(t),$ $\displaystyle t\geq 0.$ (40)

Let $S$ be the semigroup generated by $A_{1}$. Then (39) yields

$\displaystyle x_{1}(t)$
$\displaystyle=S(t)x_{1}(0)+\int_{0}^{t}S(t-\tau)\delta_{1}(\tau)\,\text{d}\tau.$

Since $S$ is bounded on $[0,T]$, we have
$c_{0}\coloneqq\sup_{t\in[0,T]}\|S(t)\|<\infty$,
$c_{1}=\max\\{c_{0},c_{0}T\\}$, and we can conclude

$\displaystyle\max_{t\in[0,T]}\|x_{1}(t)\|_{\mathcal{Y}_{1}}$
$\displaystyle\leq c_{0}\|x_{1}(0)\|_{\mathcal{Y}_{1}}+Tc_{0}\max_{0\leq t\leq
T}\|\delta_{1}(t)\|_{\mathcal{Y}_{1}}$ (41) $\displaystyle\leq
c_{1}\big{(}\|x_{1}(0)\|_{\mathcal{Y}_{1}}+\max_{0\leq t\leq
T}\|\delta_{1}(t)\|_{\mathcal{Y}_{1}}\big{)}.$ (42)

In the case where $p=0$, we have $\mathcal{X}=\mathcal{Y}_{1}$, $x=x_{1}$ and
$\delta=\delta_{1}$, and we obtain from (41) that $p_{\rm pert}^{(E,A)}=0$. If
this is not the case, we find from (40) that

$\displaystyle x_{2}(t)$
$\displaystyle=N\frac{\text{d}}{\text{d}t}x_{2}(t)-\delta_{2}(t)=N\frac{\text{d}}{\text{d}t}\left(N\frac{\text{d}}{\text{d}t}x_{2}(t)-\delta_{2}(t)\right)-\delta_{2}(t)$
$\displaystyle=N^{2}\frac{\text{d}^{2}}{\text{d}t^{2}}x_{2}(t))-N\frac{\text{d}}{\text{d}t}\delta_{2}(t)+\delta_{2}(t)=N^{2}\frac{\text{d}^{2}}{\text{d}t^{2}}\left(N\frac{\text{d}}{\text{d}t}x_{2}(t)+\delta_{2}(t)\right)-N\frac{\text{d}}{\text{d}t}\delta_{2}(t)-\delta_{2}(t)$
$\displaystyle=N^{3}\frac{\text{d}^{3}}{\text{d}t^{3}}x_{2}(t)-N^{2}\frac{\text{d}^{2}}{\text{d}t^{2}}\delta_{2}(t)-N\frac{\text{d}}{\text{d}t}\delta_{2}(t)+\delta_{2}(t)=\ldots=-\sum_{i=0}^{p-1}N^{i}\frac{\text{d}^{i}}{\text{d}t^{i}}\delta_{2}(t).$

Boundedness of $E$ yields that $N$ is bounded, and thus
$c_{2}\coloneqq\max_{i=0}^{p-1}\|N^{i}\|<\infty$. Consequently,

$\displaystyle\max_{0\leq t\leq T}\|x_{*2}(t)\|_{\mathcal{X}_{2}}\leq
c_{2}\left(\sum_{i=0}^{p-1}\max_{0\leq t\leq
T}\left\|\frac{\text{d}^{i}}{\text{d}t^{i}}\delta_{2}(t)\right\|_{\tilde{\mathcal{Z}}_{2}}\right).$
(43)

Collecting the bounds (41) and (43) implies that, setting
$c\coloneqq\max\\{c_{1},Tc_{1},c_{2}\\},$

$\displaystyle\max_{0\leq t\leq T}\|x(t)\|_{\mathcal{X}}\leq
c\left(\|x(0)\|_{\mathcal{X}}+\sum_{i=0}^{p-1}\max_{0\leq t\leq
T}\left\|\frac{\text{d}^{i}}{\text{d}t^{i}}\delta(t)\right\|_{\mathcal{Z}}\right)\,.$

Thus, $p_{\rm pert}^{(E,A)}\leq p_{\rm nilp}^{(E,A)}$. ∎

######  Remark 8.4.

In [15], a class of linear infinite-dimensional differential-algebraic systems
was examined, which do not necessarily have a nilpotency index. It was
revealed that a perturbation analysis necessitates the consideration of
stronger norms for the initial value. These norms may correspond to a type of
spatial perturbation index for systems governed by partial differential-
algebraic equations.

##  9 Summary

The purpose of this paper was, as explained in the introduction, to review the
common indices for DAEs and to extend them to infinite-dimensional DAEs where
this has not already been done. Rigorous definitions were given for the
nilpotency, resolvent, radiality, chain, differentiation and perturbation
indices. The definitions of the differentiation and perturbation indices are
as far as we know, new in the Banach space setting. Unlike finite-dimensional
DAEs, the existence of an index is not guaranteed, and furthermore the various
indices are not equivalent for general DAEs. Each index reveals different
information about the system. We were able to show that in some cases
existence of a particular index will imply existence of some other other
indices. One implication is that the chain index is the least restrictive
index since existence of any of the other indices implies existence of the
chain index, as well as a bound on the chain index.

This work has revealed a number of open questions for DAEs on Banach spaces.
Under what conditions does a particular index exist? If two indices exist,
when are they equal? Which indices are more useful for the study of DAEs? It
is hoped that this paper will inspire research on these and related questions.

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