# Characterisation for
Exponential Stability of port-Hamiltonian Systems

Sascha Trostorff Mathematisches Seminar, CAU Kiel, Germany Marcus Waurick
Institut für Angewandte Analysis, TU Bergakademie Freiberg, Germany

###### Abstract

Given an energy-dissipating port-Hamiltonian system, we characterise the
exponential decay of the energy via the model ingredients under mild
conditions on the Hamiltonian density $\mathcal{H}$. In passing, we obtain
generalisations for sufficient criteria in the literature by making regularity
requirements for the Hamiltonian density largely obsolete. The key assumption
for the characterisation (and thus the sufficient critera) to work is a
uniform bound for a family of fundamental solutions for some non-autonomous,
finite-dimensional ODEs. Regularity conditions on $\mathcal{H}$ for previously
known criteria such as bounded variation are shown to imply the key
assumption. Exponentially stable port-Hamiltonian systems with irregular
$L_{\infty}$-densities are provided.

Keywords port-Hamiltonian systems, Exponential stability, $C_{0}$-semi-groups,
Infinite-dimensional systems theory
MSC2020 93D23, 37K40, 47D06, 34G10

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 The Main Result
  4. 4 The Strategy of the Proof
  5. 5 A Sufficient Criterion for Exponential Stability
  6. 6 Characterisation for the Reference Case
  7. 7 Proof of Main Result – Theorem 3.1
  8. 8 The Condition (B)
    1. 8.1 A compatibility condition of $\mathcal{H}$ and $P_{1}$
    2. 8.2 A regularity condition on $\mathcal{H}$
  9. 9 Conclusion

##  1 Introduction

In [20] Arjen van der Schaft described the framework of port-Hamiltonian
systems. It has since then triggered manifold research and ideas. For this we
refer to [9, 21, 2, 18, 23] and the references therein, see also [8] for a
survey. The basic idea is to describe a physical – mostly energy conserving or
at least energy dissipating – phenomenon in terms of a partial differential
equation in the underlying physical domain together with suitable boundary
conditions. These boundary conditions now are what is thought of as so-called
ports. At these ports one can steer and measure data. Thus, the basic system
and particularly the one-plus-one-dimensional port-Hamiltonian system serves
as a prototype boundary control system. The emphasis is on hyperbolic type
partial differential equations. Quite naturally, it is of interest to
understand those boundary conditions leading to an evolution of the state
variable to have exponentially decaying energy. More precisely, we consider
the operator

$\mathcal{\mathcal{A}}\coloneqq P_{1}\partial_{x}\mathcal{H}+P_{0}\mathcal{H}$
(A1)

as an operator in $H\coloneqq L_{2}(a,b)^{d}$ (as sets) endowed with the
scalar-product $(u,v)\mapsto\langle\mathcal{H}u,v\rangle_{L_{2}}$. Here,
$(a,b)\subseteq\mathbb{R}$ is a bounded interval,
$P_{1}=P_{1}^{*}\in\mathbb{R}^{d\times d}$ is an invertible $d\times
d$-matrix, $P_{0}=-P_{0}^{*}\in\mathbb{R}^{d\times d}$ and $\partial_{x}$ is
the usual weak derivative operator. The mapping
$\mathcal{H}\colon(a,b)\to\mathbb{R}^{d\times d}$ is measurable, attains
values in the non-negative self-adjoint matrices and is strictly bounded away
from $0$ and bounded above. The port-Hamiltonian operator $\mathcal{A}$ is now
accompanied with suitable boundary conditions encoded in a full rank matrix
$W_{B}\in\mathbb{R}^{d\times 2d}$ satisfying

$W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\left(\begin{array}[]{cc}0&1_{d}\\\
1_{d}&0\end{array}\right)\left(W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\right)^{\ast}\geq 0$ (WB)

in the way that

$\operatorname{dom}(\mathcal{A})=\left\\{u\in H\,;\,\mathcal{H}u\in
H^{1}(a,b)^{d},\,W_{B}\left(\begin{array}[]{c}\left(\mathcal{H}u\right)(b)\\\
\left(\mathcal{H}u\right)(a)\end{array}\right)=0\right\\}.$ (A2)

The above conditions render $-\mathcal{A}$ to be the generator of a
$C_{0}$-semi-group of contractions (see e.g. [9, Theorem 7.2.4] or [7, Theorem
1.1], note also [7, Theorem 1.5] for conditions for $-\mathcal{A}$ being the
generator of a general $C_{0}$-semigroup). For the theory of $C_{0}$-semi-
groups we refer to the monographs [6, 11]. It is the aim of this article to
characterise all such settings for which this semi-group admits exponential
decay. The details of the definitions are given in Section 2. For the time
being we comment on approaches available in the literature and difficulties as
well as elements of our strategy to obtain our characterisation result. We
remark that stability and stabilisation of port-Hamiltonian systems is an
important topic in control theory, see e.g. [22, 14, 15, 3, 16].

By the celebrated Lumer–Phillips theorem (see e.g. [6, 3.15 Theorem]) for
$-\mathcal{A}$ to generate a semi-group of contractions it is equivalent that
$\mathcal{A}$ is $m$-accretive. This property is independent of the
Hamiltonian density $\mathcal{H}$ encoding the material coefficients in actual
physical systems. Hence, well-posedness and energy dissipation is not hinging
on the actual measurements of the material parameters. Thus, one might think
that, too, exponential stability is independent of the Hamiltonian. This is,
however, not the case as the example in [5, Section 5] demonstrates. This is
even more so surprising as the up to now only111If $\mathcal{H}$ is smooth, a
strict inequality in (WB) leads to exponential stability for the port-
Hamiltonian semi-group as well. However, by [9, Lemma 9.1.4] and its proof, a
strict inequality in (WB) implies the validity of the real part condition in
Theorem 1.1. available condition yielding an exponentially stable semi-group
is (almost) independent of the actual Hamiltonian.

######  Theorem 1.1 ([16, Theorem 3.5]).

Let $\mathcal{H}$ be of bounded variation and $\mathcal{A}$ given by (A1) and
(A2) is $m$-accretive. If there exist $c>0$ and $\eta\in\\{a,b\\}$ such that
for all $u\in\operatorname{dom}(\mathcal{A})$

$\operatorname{Re}\langle u,\mathcal{A}u\rangle_{H}\geq
c\|\mathcal{H}u(\eta)\|^{2},$

then $-\mathcal{A}$ generates an exponentially stable $C_{0}$-semi-group.

Note that the example in [5] assuring the dependence of $\mathcal{H}$ whether
or not the $C_{0}$-semi-group is exponentially stable, uses constant
$\mathcal{H}$. Hence, even in the class of constant Hamiltonians, the above
condition is not a characterisaion. Also, as a second drawback of the results
avaliable in the literature, the Hamiltonian always needs to satisfy certain
regularity requirements. Apart from the more recent advancement in Theorem
1.1, the results in [9, Theorem 9.1.3] require continuous differentiability or
Lipschitz continuity, see [22, Theorem III.2]. We refer to the results in [2]
for non-autnomous set ups.

The present article aims at replacing the regularity conditions altogether.
For this define $\Phi_{t}$ to be the fundamental solution associated with

$u^{\prime}(x)=\mathrm{i}tP_{1}^{-1}\mathcal{H}(x)u(x)\quad(x\in(a,b))$

with
$\Phi_{t}(a)=1_{d}\coloneqq\operatorname{diag}(1,\ldots,1)\in\mathbb{R}^{d\times
d}.$ The key assumption we shall impose here is

$\sup_{t\in\mathbb{R}}\|\Phi_{t}\|_{\infty}=\sup_{t\in\mathbb{R}}\sup_{x\in(a,b)}\|\Phi_{t}(x)\|<\infty$
(B)

We shall see below that (B) is satisfied, if $\mathcal{H}$ is of bounded
variation. Also, we will see that (B) is satisfied for scalar $\mathcal{H}$
not necessarily of bounded variation. The main theorem of the present article
is a characterisation result of exponential stability in case (B) is
satisfied.

######  Theorem 1.2.

Assume condition (B) and that $\mathcal{A}$ given by (A1) and (A2) is
$m$-accretive; i.e. $-\mathcal{A}$ generates a contraction semi-group. Then
the following conditions are equivalent:

  1. (i)

$-\mathcal{A}$ generates an exponentially stable $C_{0}$-semi-group.

  2. (ii)

For all $t\in\mathbb{R}$

$T_{t}\coloneqq W_{B}\left(\begin{array}[]{c}\Phi_{t}(b)\\\
1_{d}\end{array}\right)$

is invertible with $\sup_{t\in\mathbb{R}}\|T_{t}^{-1}\|<\infty.$

To the best of our knowledge this is the first characterisation result for
exponential stability of port-Hamiltonian systems. Together with a set of
examples warranting condition (B), this theorem contains all the available
sufficient criteria for exponential stability of port-Hamiltonian systems as
respective special cases as we will demonstrate below. In particular, we shall
provide a condition on the connection of $\mathcal{H}$ and $P_{1}$
guaranteeing the satisfaction of (B) – independently of any regularity
requirements for $\mathcal{H}$. A special case for this setting is that of
_scalar-valued_ Hamiltonian densities $\mathcal{H},$ that is, when we find a
bounded scalar function $h\colon(a,b)\to\mathbb{R}$ such that
$\mathcal{H}(x)=h(x)1_{d}$ for almost every $x$. The corresponding theorem
characterising exponential stability is then, in fact, fairly independent of
$h$ in the following sense:

######  Theorem 1.3.

Assume $\mathcal{H}$ to be scalar-valued and that $\mathcal{A}$ given by (A1)
and (A2) is $m$-accretive. Then the following conditions are equivalent

  1. (i)

$-\mathcal{A}$ generates an exponentially stable $C_{0}$-semi-group.

  2. (ii)

For all $t\in\mathbb{R}$

$\tau_{t}\coloneqq
W_{B}\left(\begin{array}[]{c}\mathrm{e}^{\mathrm{i}tP_{1}^{-1}}\\\
1_{d}\end{array}\right)$

is invertible with $\sup_{t\in\mathbb{R}}\|\tau_{t}^{-1}\|<\infty.$

The sufficient criterion Theorem 1.1 for scalar-valued $\mathcal{H}$ reads as
follows.

######  Theorem 1.4.

Let $\mathcal{H}$ be scalar-valued and $\mathcal{A}$ given by (A1) and (A2) is
$m$-accretive. If there exist $c>0$ and $\eta\in\\{a,b\\}$ such that for all
$u\in\operatorname{dom}(\mathcal{A})$

$\operatorname{Re}\langle u,\mathcal{A}u\rangle_{H}\geq
c\|(\mathcal{H}u)(\eta)\|^{2},$

then $-\mathcal{A}$ generates an exponentially stable $C_{0}$-semi-group.

We will demonstrate below that the real part condition in [16, Theorem 3.5]
implies (ii) from Theorem 1.2, thus providing an independent proof of [16,
Theorem 3.5] in a more general situation. By means of counterexamples, we will
show that the invertibility of $T_{t}$ is not sufficient for the uniform
boundedness of $T_{t}^{-1}.$ It will rely on future research to assess whether
condition (B) is needed at all in the present analysis. This leads us to our
two main open problems:

######  Problem 1.5.

Characterise all $P_{1}$ and $\mathcal{H}$ such that (B) holds.

It may well be that (B) is not needed altogether:

######  Problem 1.6.

Is (B) necessary for exponential stability of the port-Hamiltonian system at
hand?

We will provide a short outline of the manuscript next. We recall some of the
basic results related to port-Hamiltonian systems relevant to this article in
Section 2. The main result together wit a proof of the constant Hamiltonian-
case is elaborated on in Section 3. In this section we also apply our
characterisation to [5, Section 5]. In Section 4 we outline the strategy of
the proof of our main contribution. In particular, we provide a
characterisation for a certain subclass of port-Hamiltonian operators
congruent to the general case. Equipped with the accompanying reformulation of
general port-Hamiltonian systems, we shall show that the real part criterion
from Theorem 1.1 can be derived from our characterisation in Section 5. In
Section 6 we concentrate on the above mentioned subclass and provide a proof
for the respective characterisation. Section 8 is devoted to a discussion of
(B). We sum up our findings in the concluding Section 9.

##  2 Preliminaries

In this section, we recall the functional analytic setting of port-Hamiltonian
systems and detail some results from the literature characterising the
generation property of $\mathcal{A}$. To start out with we fix
$d\in\mathbb{N}$ and let $P_{1},P_{0}\in\mathbb{R}^{d\times d}$ be matrices
such that $P_{1}=P_{1}^{\ast}$ is invertible and $P_{0}=-P_{0}^{\ast}$.
Moreover, let $\mathcal{H}:(a,b)\to\mathbb{R}^{d\times d}$ be a measurable
function such that

$\exists m,M>0\>\forall
x\in(a,b):\>m\leq\mathcal{H}(x)=\mathcal{H}(x)^{\ast}\leq M,$

where the inequalities are meant in the sense of positive definiteness. We use
$\mathcal{H}$ to define a new inner product on $L_{2}(a,b)^{d}$ by setting

$\langle
u,v\rangle_{H}\coloneqq\langle\mathcal{H}u,v\rangle_{L_{2}(a,b)^{d}}\quad(u,v\in
L_{2}(a,b)^{d})$

and denote the Hilbert space $L_{2}(a,b)^{d}$ equipped with this new inner
product by $H$. Finally, we define the _port-Hamiltonian_ (operator)

$\displaystyle\mathcal{A}:\operatorname{dom}(\mathcal{A})\subseteq H$
$\displaystyle\to H,$ $\displaystyle u$ $\displaystyle\mapsto
P_{1}(\mathcal{H}u)^{\prime}+P_{0}\mathcal{H}u,$ (1)

with a suitable domain $\operatorname{dom}(\mathcal{A}).$ We recall the
following characterisation for $m$-accretivity of port-Hamiltonians:

######  Theorem 2.1 ([9, Theorem 7.2.4]).

Let $\mathcal{A}$ be as in (1). Then $\mathcal{A}$ is $m$-accretive (and
hence, $-\mathcal{A}$ generates a contraction semi-group) if and only if there
exists a matrix $W_{B}\in\mathbb{R}^{d\times 2d}$ with
$\operatorname{rk}W_{B}=d$ and

$W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\left(\begin{array}[]{cc}0&1_{d}\\\
1_{d}&0\end{array}\right)\left(W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\right)^{\ast}\geq 0$ (2)

such that

$\operatorname{dom}(\mathcal{A})=\left\\{u\in H\,;\,\mathcal{H}u\in
H^{1}(a,b)^{d},\,W_{B}\left(\begin{array}[]{c}\left(\mathcal{H}u\right)(b)\\\
\left(\mathcal{H}u\right)(a)\end{array}\right)=0\right\\}.$

We will also employ this generation result for a more particular situation,
where $\mathcal{H}$ is constant and equals the identity on $\mathbb{R}^{d}$
and $P_{1}$ is diagonal with only $1$s and/or $-1$s on the diagonal. For later
use, we briefly recall the rationale that relates these two results.

######  Lemma 2.2.

Let $a,b\in\mathbb{R}$ with $a<b$ and $d\in\mathbb{N}$. Moreover, let
$\mathcal{H}\in L_{\infty}(a,b;\mathbb{R}^{d\times d})$ with
$0<m\leq\mathcal{H}(x)=\mathcal{H}(x)^{\top}$ for almost every $x\in(a,b)$ and
let $H$ be the Hilbert space $L_{2}(a,b;\mathbb{R}^{d})$ endowed with the
inner product $\langle
u,v\rangle_{H}\coloneqq\langle\mathcal{H}u,v\rangle_{L_{2}}.$

  1. (a)

Define

$S:L_{2}(a,b)^{d}\to H,\quad u\mapsto\mathcal{H}^{-1}u.$

Then $S$ is a Banach space isomorphism with $S^{\ast}v=v$ for each $v\in
L_{2}(a,b)^{d}.$

  2. (b)

Define

$\Psi:L_{2}(a,b)^{d}\to L_{2}(-\tfrac{1}{2},\tfrac{1}{2})^{d},\quad
u\mapsto\left(s\mapsto\left(b-a\right)^{1/2}u\left((b-a)s+\frac{a+b}{2}\right)\right).$

Then $\Psi$ is unitary with
$\left(\Psi^{\ast}v\right)(x)=\sqrt{\frac{1}{b-a}}v\left(\frac{1}{b-a}x-\frac{1}{b-a}a-\frac{1}{2}\right)$
for $x\in(a,b)$ and $v\in L_{2}(-\frac{1}{2},\frac{1}{2})^{d}.$

###### Proof.

(a) $S$ clearly is bounded, one-to-one and onto; i.e., a Banach space
isomorphism. Moreover, for $u\in L_{2}(a,b)^{d},v\in H$ we compute

$\langle
Su,v\rangle_{H}=\langle\mathcal{H}\mathcal{H}^{-1}u,v\rangle_{L_{2}(a,b)^{d}}=\langle
u,v\rangle_{L_{2}(a,b)^{d}},$

which shows $S^{\ast}v=v.$
(b) The result is an elementary consequence of the substitution formula for
integrals. ∎

Applying the congruences induced by $S$ and $\Psi$ to the operator
$\mathcal{A},$ we define

$\displaystyle\mathcal{B}$ $\displaystyle\coloneqq\Psi
S^{\ast}\mathcal{A}S\Psi^{\ast}$
$\displaystyle\subseteq\Psi\left(P_{1}\partial_{x}+P_{0}\right)\Psi^{\ast}$
$\displaystyle=P_{1}\Psi\partial_{x}\Psi^{\ast}+P_{0}$
$\displaystyle=\frac{1}{b-a}P_{1}\partial_{s}+P_{0}$

on $L_{2}(-\frac{1}{2},\frac{1}{2})^{d}$. We emphasise that the congruence
singles out a particular domain for $\mathcal{B}$ inherited by that of
$\mathcal{A}$, that is,

$\operatorname{dom}(\mathcal{B})=\\{u\in
H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d};S\Psi^{*}u\in\operatorname{dom}(\mathcal{A})\\}.$

Moreover, we find a unitary matrix $U\in\mathbb{R}^{d\times d}$ such that

$U^{\ast}\frac{1}{b-a}P_{1}U=\operatorname{diag}(\lambda_{i})_{i\in\\{1,\ldots,d\\}}$
(3)

with $\lambda_{i}\neq 0$ for each $i\in\\{1,\ldots,d\\}$ due to the
invertibility of $P_{1}.$ By possibly changing coordinates, we may assume
$\lambda_{1},\dots,\lambda_{k}>0$ and $\lambda_{k+1},\ldots,\lambda_{d}<0.$ We
define

$Q\coloneqq\operatorname{diag}\left(\frac{1}{\sqrt{|\lambda_{i}|}}\right)_{i\in\\{1,\ldots,d\\}}$

and obtain

$D=Q\operatorname{diag}(\lambda_{i})_{i\in\\{1,\ldots,d\\}}Q^{\ast},$ (4)

where

$D\coloneqq\left(\begin{array}[]{cc}1_{k}&0\\\ 0&-1_{d-k}\end{array}\right)$

with $1_{\ell}=\operatorname{diag}(1,\ldots,1)\in\mathbb{R}^{\ell\times\ell}$,
$\ell\in\mathbb{N}$. In consequence,

$\mathcal{C}\coloneqq QU^{\ast}\mathcal{B}UQ^{\ast}\subseteq
D\partial_{s}+QU^{\ast}P_{0}UQ\eqqcolon D\partial_{s}+\widetilde{P_{0}}$ (5)

with

$\displaystyle\operatorname{dom}(\mathcal{C})$ $\displaystyle=\\{v\in
H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d};UQ^{*}v\in\operatorname{dom}(\mathcal{B})\\}$
$\displaystyle=\\{v\in
H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d};S\Psi^{*}UQ^{*}v\in\operatorname{dom}(\mathcal{A})\\}.$
(6)

We note that $\widetilde{P_{0}}$ is still skew-selfadjoint. Since
$\mathcal{C}$ is congruent to $\mathcal{A},$ we infer that $\mathcal{C}$ is
$m$-accretive if and only if $\mathcal{A}$ is $m$-accretive. In this case the
domain of $\mathcal{C}$ admits a handy description, which can be derived from
the general statement Theorem 2.1, as well.

######  Theorem 2.3 ([9, Theorem 7.2.4], [12, Theorem 2.8 (c)]).

Let $\mathcal{A}$ and $\mathcal{C}$ as in (A1),(A2) and (5),(6), respectively.
Then $\mathcal{A}$ is $m$-accretive if and only if there exists a matrix
$M\in\mathbb{R}^{d\times d}$ with $\|M\|\leq 1$ such that

$\operatorname{dom}(\mathcal{C})=\left\\{v\in
H^{1}\left(-\tfrac{1}{2},\tfrac{1}{2}\right){}^{d}\,;\,M\left(\begin{array}[]{c}\check{v}(1/2)\\\
\hat{v}(-1/2)\end{array}\right)=\left(\begin{array}[]{c}\check{v}(-1/2)\\\
\hat{v}(1/2)\end{array}\right)\right\\},$

where we decompose each vector $z\in\mathbb{R}^{d}$ into
$(\check{z},\hat{z})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k}$ according to the
block sizes of the matrix $D$.

######  Remark 2.4.

As we have outlined above, Theorem 2.3 can be derived from Theorem 2.1.
Indeed, note that the skew-selfadjointness of $P_{0}$ yields that
$\mathcal{A}$ is $m$-accretive if and only if $\mathcal{A}-P_{0}\mathcal{H}$
is. Thus, we can assume without loss of generality that $P_{0}=0$. Then
$\mathcal{C}\subseteq D\partial_{s}$ is itself a port-Hamiltonian operator.
Indeed, $P_{1}=D$ is selfadjoint and invertible. The boundary matrix $W_{B}$
in this special case can be identified using the description in Theorem 2.3 to
be

$\displaystyle W_{B}$
$\displaystyle=\left(\begin{array}[]{cc}M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)&M\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)-\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)D+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)D&-M\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)D-\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)D\end{array}\right)$

With this, using $\left(\begin{array}[]{cc}-D&D\\\
1_{d}&1_{d}\end{array}\right)^{-1}=\tfrac{1}{2}\left(\begin{array}[]{cc}-D&1_{d}\\\
D&1_{d}\end{array}\right)$ we compute

$\displaystyle W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\left(\begin{array}[]{cc}0&1_{d}\\\
1_{d}&0\end{array}\right)\left(W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\right)^{\ast}$
$\displaystyle=W_{B}\left(\begin{array}[]{cc}-D&D\\\
1_{d}&1_{d}\end{array}\right)^{-1}\left(\begin{array}[]{cc}0&1_{d}\\\
1_{d}&0\end{array}\right)\left(W_{B}\left(\begin{array}[]{cc}-D&D\\\
1_{d}&1_{d}\end{array}\right)^{-1}\right)^{\ast}$
$\displaystyle=\frac{1}{4}W_{B}\left(\begin{array}[]{cc}-D&1_{d}\\\
D&1_{d}\end{array}\right)\left(\begin{array}[]{cc}0&1_{d}\\\
1_{d}&0\end{array}\right)\left(\begin{array}[]{cc}-D&D\\\
1_{d}&1_{d}\end{array}\right)W_{B}^{*}$
$\displaystyle=\frac{1}{4}W_{B}\left(\begin{array}[]{cc}-2D&0\\\
0&2D\end{array}\right)W_{B}^{*}$
$\displaystyle=\frac{1}{2}\left(\begin{array}[]{cc}M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)D+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)D&-M\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)D-\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)D\end{array}\right)$
$\displaystyle\quad\quad\times\left(\begin{array}[]{cc}-D&0\\\
0&D\end{array}\right)\text{$\left(\begin{array}[]{c}D\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)M^{*}+D\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\\\ -D\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)M^{*}-D\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\end{array}\right)$}$
$\displaystyle=\frac{1}{2}\left(\begin{array}[]{cc}-M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)&-M\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)-\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\end{array}\right)$
$\times$$\left(\begin{array}[]{c}\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)M^{*}-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\\\ \left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)M^{*}-\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\end{array}\right)$
$\displaystyle=\frac{1}{2}\left(-M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)$
$\displaystyle\quad\quad\times\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)M^{*}-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)+\frac{1}{2}\left(-M\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)-\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\right)$
$\displaystyle\quad\quad\times\left(\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)M^{*}-\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\right)$
$\displaystyle=\frac{1}{2}\left(-M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)M^{*}+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)-\frac{1}{2}\left(M\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)M^{*}+\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\right)=\frac{1}{2}(1_{d}-MM^{\ast}).$

Thus, we read off that

$\displaystyle W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1\end{array}\right)^{-1}\left(\begin{array}[]{cc}0&1_{d}\\\
1_{d}&0\end{array}\right)\left(W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\right)^{\ast}\geq 0$ $\displaystyle\iff
1_{d}-MM^{*}\geq 0$ $\displaystyle\iff\|M\|^{2}=\|M^{*}\|^{2}=\|MM^{*}\|\leq
1.$

The norm condition in Theorem 2.3 is the corresponding positive definiteness
condition in Theorem 2.1.

##  3 The Main Result

In this section, we state the main result. For this, throughout this section,
we let

$\displaystyle\mathcal{A}\colon\operatorname{dom}(\mathcal{A})\subseteq H$
$\displaystyle\to H$ $\displaystyle u$ $\displaystyle\mapsto
P_{1}(\mathcal{H}u)^{\prime}+P_{0}\mathcal{H}u,$

be a port-Hamiltonian operator as in (1). Furthermore, we assume
$-\mathcal{A}$ to generate a $C_{0}$-semi-group of contractions. Hence, by
Theorem 2.1 we find a full rank matrix $W_{B}\in\mathbb{R}^{d\times 2d}$
satisfying (2) such that

$\operatorname{dom}(\mathcal{A})=\left\\{u\in H\,;\,\mathcal{H}u\in
H^{1}(a,b)^{d},\,W_{B}\left(\begin{array}[]{c}\left(\mathcal{H}u\right)(b)\\\
\left(\mathcal{H}u\right)(a)\end{array}\right)=0\right\\}.$

We recall $\Phi_{t}$, the fundamental solution associated with

$u^{\prime}(x)=\mathrm{i}tP_{1}^{-1}\mathcal{H}(x)u(x)\quad(x\in(a,b))$

such that $\Phi_{t}(a)=1_{d}$ and condition (B) stating
$\sup_{t\in\mathbb{R}}\|\Phi_{t}\|_{\infty}<\infty$.

######  Theorem 3.1 (Characterisation of Exponential Stability for port-
Hamiltonian Contraction Semi-groups).

Assume (B). Then the following conditions are equivalent:

  1. (i)

$-\mathcal{A}$ generates an exponentially stable $C_{0}$-semi-group.

  2. (ii)

For all $t\in\mathbb{R}$

$T_{t}\coloneqq W_{B}\left(\begin{array}[]{c}\Phi_{t}(b)\\\
1_{d}\end{array}\right)$

is invertible with $\sup_{t\in\mathbb{R}}\|T_{t}^{-1}\|<\infty.$

We specialise this result to the case when the Hamiltonian density
$\mathcal{H}$ is constant. For this we state a special case of Theorem 8.8,
which we prove in Subsection 8.2.

######  Proposition 3.2.

Let $\mathcal{H}_{0}=\mathcal{H}_{0}^{*}\in\mathbb{R}^{d\times d}$ be non-
negative and invertible. Then for all $Q_{1}=Q_{1}^{*}\in\mathbb{R}^{d\times
d}$ invertible

$\sup_{t\in\mathbb{R}}\|\mathrm{e}^{\mathrm{i}tQ_{1}\mathcal{H}_{0}}\|<\infty.$

######  Corollary 3.3.

Assume, additionally to the assumptions in this section,
$\mathcal{H}(x)=\mathcal{H}_{0}$ for some
$\mathcal{H}_{0}\in\mathbb{R}^{d\times d}$ and all $x\in(a,b)$. Then the
following conditions are equivalent:

  1. (i)

$-\mathcal{A}$ generates an exponentially stable $C_{0}$-semi-group.

  2. (ii)

For all $t\in\mathbb{R}$

$T_{t}\coloneqq
W_{B}\left(\begin{array}[]{c}\mathrm{e}^{\mathrm{i}tP_{1}^{-1}\mathcal{H}_{0}}\\\
1_{d}\end{array}\right)$

is invertible with $\sup_{t\in\mathbb{R}}\|T_{t}^{-1}\|<\infty.$

###### Proof.

Since $\mathcal{H}$ is constant equal $\mathcal{H}_{0}$,
$\Phi_{t}(b)=\mathrm{e}^{\mathrm{i}t(b-a)P_{1}^{-1}\mathcal{H}_{0}}$. Thus, by
Proposition 3.2, (B) holds, which in turn leads to the applicability of
Theorem 3.1. To obtain the claim it thus suffices to note that $t$ runs
through all reals, making the prefactor $(b-a)$ superfluous, and to read off
the equivalence stated in Theorem 3.1. ∎

Before we put this characterisation result into perspective of the results
available in the literature, we provide an example that confirms that
invertibility for $T_{t}$ alone is not sufficient to deduce the uniform bound
of the inverses.

######  Example 3.4.

Consider the matrices $M\coloneqq-\frac{1}{2}\left(\begin{array}[]{cc}1&1\\\
1&1\end{array}\right)$ (note that $\|M\|\leq 1$) and
$\mathcal{H}(x)\coloneqq\mathcal{H}_{0}\coloneqq\left(\begin{array}[]{cc}1&0\\\
0&\sqrt{2}\end{array}\right)$ for all $x\in(a,b)\coloneqq(0,1)$ as well as
$P_{1}=1_{2}=\left(\begin{array}[]{cc}1&0\\\ 0&1\end{array}\right)$. In Remark
2.4 we have seen that

$W_{B}=\left(\begin{array}[]{cc}M&-1_{2}\end{array}\right)$

leads to $-\mathcal{A}$ generate a contraction semi-group. We consider

$\displaystyle T_{t}$
$\displaystyle=W_{B}\left(\begin{array}[]{c}\mathrm{e}^{\mathrm{i}tP_{1}^{-1}\mathcal{H}_{0}}\\\
1_{2}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}M&-1_{2}\end{array}\right)\left(\begin{array}[]{c}\mathrm{e}^{\mathrm{i}t\mathcal{H}_{0}}\\\
1_{2}\end{array}\right)$
$\displaystyle=M\mathrm{e}^{\mathrm{i}t\mathcal{H}_{0}}-1_{2}$

and analyse the respective inverses for all $t\in\mathbb{R}$. We first show
that $T_{t}$ is invertible. For this, we compute

$\displaystyle\det(-T_{t})$
$\displaystyle=\det\left(\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right)+\frac{1}{2}\left(\begin{array}[]{cc}1&1\\\
1&1\end{array}\right)\left(\begin{array}[]{cc}\mathrm{e}^{\mathrm{i}t}&0\\\
0&\mathrm{e}^{\mathrm{i}\sqrt{2}t}\end{array}\right)\right)$
$\displaystyle=\det\left(\begin{array}[]{cc}1+\frac{1}{2}\mathrm{e}^{\mathrm{i}t}&\frac{1}{2}\mathrm{e}^{\mathrm{i}\sqrt{2}t}\\\
\frac{1}{2}\mathrm{e}^{\mathrm{i}t}&1+\frac{1}{2}\mathrm{e}^{\mathrm{i}\sqrt{2}t}\end{array}\right)$
$\displaystyle=1+\tfrac{1}{4}\mathrm{e}^{\mathrm{i}(1+\sqrt{2})t}+\tfrac{1}{2}\mathrm{e}^{\mathrm{i}t}+\tfrac{1}{2}\mathrm{e}^{\mathrm{i}\sqrt{2}t}-\tfrac{1}{4}\mathrm{e}^{\mathrm{i}(1+\sqrt{2})t}$
$\displaystyle=1+\tfrac{1}{2}\left(\mathrm{e}^{\mathrm{i}t}+\mathrm{e}^{\mathrm{i}\sqrt{2}t}\right).$

Now, as $2\operatorname{Re}\det(-T_{t})=2+(\cos(t)+\cos(\sqrt{2}t)),$ in order
for $\det T_{t}=0$ we require $\cos(t)+\cos(\sqrt{2}t)=-2$ which in turn
implies $t\in\left(2\mathbb{\mathbb{Z}}+1\right)\pi$ and
$\sqrt{2}t\in\left(2\mathbb{\mathbb{Z}}+1\right)\pi$, two conditions
contradicting one another. Hence, $T_{t}$ is invertible for all
$t\in\mathbb{R}$. In order to analyse the norm of $T_{t}^{-1}$ in
$t\in\mathbb{R}$, we compute

$\displaystyle-T_{t}^{-1}$
$\displaystyle=\left(1_{2}-M\mathrm{e}^{\mathrm{i}t\mathcal{H}_{0}}\right)^{-1}$
$\displaystyle=\frac{1}{\det(-T_{t})}\left(\begin{array}[]{cc}1+\frac{1}{2}\mathrm{e}^{\mathrm{i}\sqrt{2}t}&-\frac{1}{2}\mathrm{e}^{\mathrm{i}\sqrt{2}t}\\\
-\frac{1}{2}\mathrm{e}^{\mathrm{i}t}&1+\frac{1}{2}\mathrm{e}^{\mathrm{i}t}\end{array}\right).$

Next, we show that for the sequence $(t_{k})_{k}$ defined by

$t_{k}\coloneqq\frac{4}{1+\sqrt{2}}k\pi\quad(k\in\mathbb{N}),$

$(\det(-T_{t_{k}}))_{k}$ accumulates at $0$ confirming that
$\sup_{t\in\mathbb{R}}\|T_{t}^{-1}\|=\infty$ and that $-\mathcal{A}$ is not
exponentially stable. For this, for all $k\in\mathbb{N}$ we compute

$\displaystyle\operatorname{Im}\det(-T_{t_{k}})$
$\displaystyle=\frac{1}{2}\left(\sin\left(\frac{4}{1+\sqrt{2}}k\pi\right)+\sin\left(\sqrt{2}\frac{4}{1+\sqrt{2}}k\pi\right)\right)$
$\displaystyle=\sin\left(\frac{1}{2}\left(\frac{4}{1+\sqrt{2}}k\pi+\sqrt{2}\frac{4}{1+\sqrt{2}}k\pi\right)\right)\cos\left(\frac{1}{2}\left(\frac{4}{1+\sqrt{2}}k\pi-\sqrt{2}\frac{4}{1+\sqrt{2}}k\pi\right)\right)$
$\displaystyle=\sin(2k\pi)\cos\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}2k\pi\right)=0.$

Hence,

$\displaystyle\det\left(-T_{t_{k}}\right)$
$\displaystyle=1+\frac{1}{2}\left(\cos\left(\frac{4}{1+\sqrt{2}}k\pi\right)+\cos\left(\sqrt{2}\frac{4}{1+\sqrt{2}}k\pi\right)\right)$
$\displaystyle=1+\cos(2k\pi)\cos\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}2k\pi\right)$
$\displaystyle=1+\cos\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}2k\pi\right)\quad(k\in\mathbb{N}).$

Next, we recall that for $\alpha\in\mathbb{R}\setminus\mathbb{Q}$

$\left\\{\mathrm{e}^{\mathrm{i}\alpha k2\pi}\,;\,k\in\mathbb{N}\right\\}$

lies in dense in $S^{1}$. Thus, for all $\ell\in\mathbb{N}$, the set

$\mathrm{e}^{\mathrm{i}\alpha\ell 2\pi}\left\\{\mathrm{e}^{\mathrm{i}\alpha
k2\pi}\,;\,k\in\mathbb{N}\right\\}=\left\\{\mathrm{e}^{\mathrm{i}\alpha
k2\pi}\,;\,k\in\mathbb{N},k\geq\ell\right\\}$

is dense, too, in $S^{1}$. It follows that $S^{1}$ is the set of accumulation
values of $\left(\mathrm{e}^{\mathrm{i}\alpha k2\pi}\right)_{k\in\mathbb{N}}$.
Consequently, as
$\alpha\coloneqq\frac{1-\sqrt{2}}{1+\sqrt{2}}=-\left(1-\sqrt{2}\right)^{2}=-3+2\sqrt{2}\notin\mathbb{Q}$,
$-1$ is an accumulation value of
$\left(\cos\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}2k\pi\right)\right)_{k\in\mathbb{N}}$
and so $(\det(-T_{t_{k}})_{k})$ accumulates at $0$ implying
$\sup_{t\in\mathbb{R}}\|T_{t}^{-1}\|=\infty.$

Next, one could ask whether or not the exponential stability of the semi-group
generated by $-\mathcal{A}$ depends on the Hamiltonian density $\mathcal{H}$.
[5, Section 5] provided an example confirming dependence. We reprove this
result with the charactersation from above.

######  Example 3.5.

Let $\theta\in\mathbb{R}$, $\theta>-1$. Consider the matrices
$M\coloneqq\frac{1}{2}\left(\begin{array}[]{cc}1&1\\\ 1&-1\end{array}\right)$
(note that $\|M\|\leq 1$) and
$\mathcal{H}(x)\coloneqq\mathcal{H}_{\theta}\coloneqq\left(\begin{array}[]{cc}\frac{1}{1+\theta}&0\\\
0&1\end{array}\right)$ for all $x\in(a,b)\coloneqq(0,1)$ as well as
$P_{1}=-1_{2}$. In Remark 2.4 we have seen that

$W_{B}=\left(\begin{array}[]{cc}-1_{2}&M\end{array}\right)$

leads to $-\mathcal{A}_{\theta}\coloneqq-
P_{1}\partial_{x}\mathcal{H}_{\theta}=\partial_{x}\mathcal{H}_{\theta}$
generate a contraction semi-group. In order to assess for which $\theta$ this
semi-group is exponentially stable, using Corollary 3.3, we consider

$\displaystyle T_{t}^{\theta}$ $\displaystyle\coloneqq
W_{B}\left(\begin{array}[]{c}\mathrm{e}^{\mathrm{i}t(b-a)P_{1}^{-1}\mathcal{H}_{\theta}}\\\
1_{2}\end{array}\right)=\left(\begin{array}[]{cc}-1_{2}&M\end{array}\right)\left(\begin{array}[]{c}\mathrm{e}^{-\mathrm{i}t\mathcal{H}_{\theta}}\\\
1_{2}\end{array}\right)$
$\displaystyle=-\mathrm{e}^{-\mathrm{i}t\mathcal{H}_{\theta}}+M=-\left(\begin{array}[]{cc}\mathrm{e}^{-\mathrm{i}\frac{t}{1+\theta}}&0\\\
0&\mathrm{e}^{-\mathrm{i}t}\end{array}\right)+\frac{1}{2}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)$
$\displaystyle=-\left(\begin{array}[]{cc}\mathrm{e}^{-\mathrm{i}\frac{t}{1+\theta}}&0\\\
0&\mathrm{e}^{-\mathrm{i}t}\end{array}\right)\left(\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right)-\frac{1}{2}\left(\begin{array}[]{cc}\mathrm{e}^{\mathrm{i}\frac{t}{1+\theta}}&0\\\
0&\mathrm{e}^{\mathrm{i}t}\end{array}\right)\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\right)$
$\displaystyle=-\left(\begin{array}[]{cc}\mathrm{e}^{-\mathrm{i}\frac{t}{1+\theta}}&0\\\
0&\mathrm{e}^{-\mathrm{i}t}\end{array}\right)\left(\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right)-\frac{1}{2}\left(\begin{array}[]{cc}\mathrm{e}^{\mathrm{i}\frac{t}{1+\theta}}&\mathrm{e}^{\mathrm{i}\frac{t}{1+\theta}}\\\
\mathrm{e}^{\mathrm{i}t}&-\mathrm{e}^{\mathrm{i}t}\end{array}\right)\right).$

As $\theta$ is real, $T_{t}^{\theta}$ is invertible and the inverses are
bounded uniformly in $t$ if and only if, the respective properties hold for

$\displaystyle S_{t}^{\theta}$
$\displaystyle\coloneqq\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right)-\frac{1}{2}\left(\begin{array}[]{cc}\mathrm{e}^{\mathrm{i}\frac{t}{1+\theta}}&\mathrm{e}^{\mathrm{i}\frac{t}{1+\theta}}\\\
\mathrm{e}^{\mathrm{i}t}&-\mathrm{e}^{\mathrm{i}t}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1-\frac{1}{2}\mathrm{e}^{\mathrm{i}\frac{t}{1+\theta}}&-\frac{1}{2}\mathrm{e}^{\mathrm{i}\frac{t}{1+\theta}}\\\
-\frac{1}{2}\mathrm{e}^{\mathrm{i}t}&1+\frac{1}{2}\mathrm{e}^{\mathrm{i}t}\end{array}\right).$

With the help of Cramer’s rule – similar to Example 3.4 – it suffices to look
at

$\displaystyle\det S_{t}^{\theta}$
$\displaystyle=1-\frac{1}{2}\left(\mathrm{e}^{\mathrm{i}\frac{t}{1+\theta}}-\mathrm{e}^{\mathrm{i}t}\right)=\frac{1}{2}(2+\mathrm{e}^{\mathrm{i}t}-\mathrm{e}^{\mathrm{i}\frac{t}{1+\theta}})$
$\displaystyle=\frac{1}{2}(2+\mathrm{e}^{\mathrm{i}t}(1-\mathrm{e}^{-\mathrm{i}\frac{\theta}{1+\theta}t})).$

For $\theta=0$, we obtain that $S_{t}^{0}$ is invertible with bound for the
inverse uniform in $t$, leading to $-\mathcal{A}_{0}$ generating an
exponentially stable semi-group. If $t=3\pi$ and
$\frac{\theta}{1+\theta}=\frac{1}{3}$, that is, $\theta=1/2$, then
$S_{t}^{\theta}$ is not invertible and, hence, $-\mathcal{A}_{1/2}$ does not
generate an exponentially stable semi-group.

##  4 The Strategy of the Proof

We outline the strategy of the proof of our main result Theorem 1.2 (and
Theorem 3.1). Again, we assume the following setting. Throughout this section,
we let

$\displaystyle\mathcal{A}\colon\operatorname{dom}(\mathcal{A})\subseteq H$
$\displaystyle\to H$ $\displaystyle u$ $\displaystyle\mapsto
P_{1}(\mathcal{H}u)^{\prime}+P_{0}\mathcal{H}u,$

be a port-Hamiltonian operator as in (1), where $H$ is the
$\mathcal{H}$-weighted $L_{2}$-space, so that $-\mathcal{A}$ generates a
$C_{0}$-semi-group of contractions. Furthermore, let
$W_{B}\in\mathbb{R}^{d\times 2d}$ be a full rank matrix satisfying (2) such
that

$\operatorname{dom}(\mathcal{A})=\left\\{u\in H\,;\,\mathcal{H}u\in
H^{1}(a,b)^{d},\,W_{B}\left(\begin{array}[]{c}\left(\mathcal{H}u\right)(b)\\\
\left(\mathcal{H}u\right)(a)\end{array}\right)=0\right\\}.$

Let $\Phi_{t}$ be the fundamental solution associated with

$u^{\prime}(x)=\mathrm{i}tP_{1}^{-1}\mathcal{H}(x)u(x)\quad(x\in(a,b))$

such that $\Phi_{t}(a)=1_{d}$ and condition (B) stating
$\sup_{t\in\mathbb{R}}\|\Phi_{t}\|_{\infty}<\infty$. As a first
simplification, we argue, why we may assume $P_{0}=0$ without any restriction
of generality.

######  Theorem 4.1 (Trotter–Kato [19], [6, 5.8 Corollary]).

Let $X$ be a Hilbert space and $\mathcal{B}$ a generator of a $C_{0}$-semi-
group $U$ on $X$. Let $P\in L(X)$. Then $P+\mathcal{B}$ is a generator of the
semi-group given by the strong operator topology limit of

$U_{P}(t)\coloneqq\lim_{n\to\infty}\left(\mathrm{e}^{\frac{t}{n}P}\mathrm{e}^{\frac{t}{n}\mathcal{B}}\right)^{n}\quad(t\geq
0).$

In particular, if $P$ is skew-selfadjoint and $U$ exponentially stable, then
$U_{P}$ is exponentially stable.

Since, in the weighted Hilbert space $H$, $P_{0}\mathcal{H}$ is skew-
selfadjoint and bounded, the semi-group generated by $P_{0}\mathcal{H}$ is
unitary. Thus, $-\mathcal{A}+P_{0}\mathcal{H}$ generates an exponentially
semi-group, if and only if so does $-\mathcal{A}$. Hence, we shall assume

$P_{0}=0$ (7)

in the following.

One core argument for the characterisation of exponential stability is the
celebrated Gearhart–Prüß theorem.

######  Theorem 4.2 (Gearhart–Prüß, see [13]).

$-\mathcal{A}$ generates an exponentially stable $C_{0}$-semi-group, if, and
only if,

$\sup_{z\in\mathbb{C}_{\operatorname{Re}>0}}\|(z+\mathcal{A})^{-1}\|_{L(H)}<\infty.$
(8)

The second main ingredient for the desired characterisation uses the
reformulation of port-Hamiltonian systems in the reference case as outlined in
Section 2. For this, we briefly recall (keeping in mind our assumption
$P_{0}=0$) that

$\mathcal{C}=QU^{\ast}\Psi S^{\ast}\mathcal{A}S\Psi^{\ast}UQ^{\ast}\subseteq
D\partial_{s}$ (9)

as an operator on $L_{2}(-\tfrac{1}{2},\tfrac{1}{2})^{d}$ with
$D=\left(\begin{array}[]{cc}1_{k}&0\\\ 0&-1_{d-k}\end{array}\right)$ for some
$k\in\\{0,\ldots,d\\}$ and

$\operatorname{dom}(\mathcal{C})=\\{v\in
H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d};S\Psi^{*}UQ^{*}v\in\operatorname{dom}(\mathcal{A})\\},$
(10)

where the unitaries $S$, the transformation used to transfer the setting into
the unweighted Hilbert space $L_{2}(a,b)^{d}$, and $\Psi$, the transformation
transforming from $L_{2}(a,b)^{d}$ to $L_{2}(-\tfrac{1}{2},\tfrac{1}{2})^{d}$,
are given in Lemma 2.2 and $U$ as much as $Q=Q^{*}$ are constant, invertible
$d\times d$-matrices that arise as certain transformation matrices from a
particular diagonalisation procedure for $P_{1}$, see again Section 2 for the
details.

######  Proposition 4.3.

Assume that $-\mathcal{A}$ generates a $C_{0}$-semi-group of contractions.
Then the following conditions are equivalent:

  1. (i)

$-\mathcal{A}$ generates an exponentially stable semi-group.

  2. (ii)

For all $t\in\mathbb{R}$, $\mathrm{i}t+\mathcal{A}$ is boundedly invertible
and

$\sup_{t\in\mathbb{R}}\|(\mathrm{i}t+\mathcal{A})^{-1}\|<\infty.$

  3. (iii)

For $\mathcal{O}(s)\coloneqq
QU^{\ast}\mathcal{H}\left((b-a)s+\frac{a+b}{2}\right)UQ$,
$s\in(-\tfrac{1}{2},\tfrac{1}{2})$, and $\mathcal{C}$ from (9) and (10), we
have that $\mathrm{i}t\mathcal{O}+\mathcal{C}$ is boundedly invertible and

$\sup_{t\in\mathbb{R}}\|\left(\mathrm{i}t\mathcal{O}+\mathcal{C}\right)^{-1}\|<\infty.$

###### Proof.

For the equivalence of (i) and (ii), note that, since $-\mathcal{A}$ generates
a contraction semi-group, by the Lumer–Philips theorem, $\mathcal{A}$ is
$m$-accretive. Hence, the resolvent condition in Theorem 4.2 is equivalent to
(ii). For the equivalence of (ii) and (iii) we use (9) and compute for
$t\in\mathbb{R}$

$\displaystyle\mathrm{i}t+\mathcal{A}$
$\displaystyle=\mathrm{i}t+S\Psi^{\ast}\mathcal{B}\Psi S^{\ast}$
$\displaystyle=\mathrm{i}t+S\Psi^{\ast}UQ^{-1}\mathcal{C}Q^{-1}U^{\ast}\Psi
S^{\ast}$ $\displaystyle=S\Psi^{\ast}UQ^{-1}\left(\mathrm{i}tQU^{\ast}\Psi
S^{\ast}S\Psi^{\ast}UQ+\mathcal{C}\right)Q^{-1}U^{\ast}\Psi S^{\ast}.$

Let us analyse the operator $QU^{\ast}\Psi S^{\ast}S\Psi^{\ast}UQ.$ For $u\in
L_{2}(a,b)^{d}$ and $s\in(-\tfrac{1}{2},\tfrac{1}{2})$ we obtain

$\displaystyle\left(QU^{\ast}\Psi S^{\ast}S\Psi^{\ast}UQu\right)(s)$
$\displaystyle=\sqrt{b-a}QU^{\ast}\left(S^{\ast}S\Psi^{\ast}UQu\right)\left((b-a)s+\frac{a+b}{2}\right)$
$\displaystyle=\sqrt{b-a}QU^{\ast}\mathcal{H}\left((b-a)s+\frac{a+b}{2}\right)\left(\Psi^{\ast}UQu\right)\left((b-a)s+\frac{a+b}{2}\right)$
$\displaystyle=QU^{\ast}\mathcal{H}\left((b-a)s+\frac{a+b}{2}\right)UQu(x).$

Thus, $QU^{\ast}\Psi S^{\ast}S\Psi^{\ast}UQ$ is a multiplication operator
induced by the function $s\mapsto\mathcal{O}(s)$ provided in (iii) and hence,
the equivalence of (ii) and (iii) follows. ∎

######  Remark 4.4.

Note that $\mathcal{O}$ satisfies the same assumptions as $\mathcal{H},$ that
is, we find $c,C>0$ such that

$\forall
s\in(-\tfrac{1}{2},\tfrac{1}{2}):\,c\leq\mathcal{O}(s)=\mathcal{O}(s)^{\ast}\leq
C.$

There are more properties inherited by $\mathcal{O}$ from $\mathcal{H}$. This
will be important, when we translate condition (iii) in Proposition 4.3 to
conditions for the model ingredients of $\mathcal{A}$.

Given Proposition 4.3, we need to characterise the boundedness of
$t\mapsto\left\|\left(\mathrm{i}t\mathcal{O}+\mathcal{C}\right)^{-1}\right\|_{L(L_{2})}$.
For this, we introduce the fundamental solution

$\Gamma_{t}\colon(-\tfrac{1}{2},\tfrac{1}{2})\to\mathbb{C}^{d\times d}$ (11)

associated to the ordinary differential equation

$\displaystyle v^{\prime}(s)$
$\displaystyle=\mathrm{i}D\mathcal{O}(s)v(s)\in\mathbb{C}^{d}\quad(s\in(-\tfrac{1}{2},\tfrac{1}{2}))$
(12)

uniquely determined by

$\Gamma_{t}(-\tfrac{1}{2})=1_{d}.$ (13)

For the desired characterisation, we recall that by Theorem 2.3,
$-\mathcal{A}$ generates a contraction semi-group, if and only if we find
$M\in\mathbb{R}^{d\times d}$ with $\|M\|\leq 1$ such that

$\operatorname{dom}(\mathcal{C})=\left\\{v\in
H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d}\,;\,M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)=\left(\begin{array}[]{c}\check{v}(-\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)\right\\},$ (14)

where we decomposed any $y\in\mathbb{R}^{d}$ into the first $k$ and the last
$d-k$ coordinates; i.e.,
$y=(\check{y},\hat{y})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k}$ according to
the block structure of $D=\left(\begin{array}[]{cc}1_{k}&0\\\
0&-1_{d-k}\end{array}\right).$ The main new ingredient for the
characterisation of exponential stability now reads as follows.

######  Theorem 4.5.

Let $M\in\mathbb{R}^{d\times d}$ with $\|M\|\leq 1$ such that (14) holds. For
$t\in\mathbb{R}$ let $\Gamma_{t}\in
C([-\tfrac{1}{2},\tfrac{1}{2}];\mathbb{C}^{d\times d})$ be given by (11),
(12), and (13) and assume

$\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}<\infty.$

Then the following statements are equivalent.

  1. (i)

For all $t\in\mathbb{R}$, $(\mathrm{i}t\mathcal{O}+\mathcal{C})$ is boundedly
invertible on $L_{2}(-\frac{1}{2},\frac{1}{2})^{d}$ with
$\sup_{t\in\mathbb{R}}\|(\mathrm{i}t\mathcal{O}+\mathcal{C})^{-1}\|<\infty.$

  2. (ii)

For all $t\in\mathbb{R}$,

$S_{t}\coloneqq M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)$

is invertible and satisfies $\sup_{t\in\mathbb{R}}\|S_{t}^{-1}\|<\infty$.

A first almost immediate consequence of the rationale above is the following
statement.

######  Corollary 4.6.

Let $M\in\mathbb{R}^{d\times d}$ with $\|M\|\leq 1$ such that (14) holds and
recall that $\mathcal{A}$ and $\mathcal{C}$ are related by (9). For
$t\in\mathbb{R}$ let $\Gamma_{t}\in
C([-\tfrac{1}{2},\tfrac{1}{2}];\mathbb{C}^{d\times d})$ be given by (11),
(12), and (13) and assume

$\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}<\infty.$

Then the following conditions are equivalent:

  1. (i)

$-\mathcal{A}$ generates an exponentially stable contraction semi-group.

  2. (ii)

For all $t\in\mathbb{R}$,

$S_{t}\coloneqq M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)$

is invertible and satisfies $\sup_{t\in\mathbb{R}}\|S_{t}^{-1}\|<\infty$.

###### Proof.

By Theorem 2.3, $-\mathcal{A}$ generates a contraction semi-group. Hence, by
Proposition 4.3, this semi-group is exponentially stable, if and only if, for
all $t\in\mathbb{R}$, $(\mathrm{i}t\mathcal{O}+\mathcal{C})$ is boundedly
invertible on $L_{2}([-\frac{1}{2},\frac{1}{2}])^{d}$ with
$\sup_{t\in\mathbb{R}}\|(\mathrm{i}t\mathcal{O}+\mathcal{C})^{-1}\|<\infty.$
Under assumption $\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}<\infty$, the
boundedness of the inverses $(\mathrm{i}t\mathcal{O}+\mathcal{C})$ is
characterised by (ii) in Theorem 4.5. ∎

We prove Theorem 4.5 in Section 6. Next, we apply the result in order
generalise results from the literature.

##  5 A Sufficient Criterion for Exponential Stability

It is the aim of this section to put the characteristion into perspective of
the literature. For this, throughout this section, we let

$\displaystyle\mathcal{A}\colon\operatorname{dom}(\mathcal{A})\subseteq H$
$\displaystyle\to H$ $\displaystyle u$ $\displaystyle\mapsto
P_{1}(\mathcal{H}u)^{\prime}+P_{0}\mathcal{H}u,$

be a port-Hamiltonian operator as in (1). We recall $\Phi_{t}$, the
fundamental solution associated with

$u^{\prime}(x)=\mathrm{i}tP_{1}^{-1}\mathcal{H}(x)u(x)\quad(x\in(a,b))$

such that $\Phi_{t}(a)=1_{d}$ and condition (B) stating
$\sup_{t\in\mathbb{R}}\|\Phi_{t}\|_{\infty}<\infty$.

######  Theorem 5.1.

Assume (B) and that $-\mathcal{A}$ generates a contraction semi-group. If
there exist $c>0$ and $\eta\in\\{a,b\\}$ such that for all
$u\in\operatorname{dom}(\mathcal{A})$

$\operatorname{Re}\langle u,\mathcal{A}u\rangle_{H}\geq
c\|\mathcal{H}u(\eta)\|^{2},$ (15)

then $-\mathcal{A}$ generates an exponentially stable $C_{0}$-semi-group.

######  Remark 5.2.

In Subsection 8.2, we confirm that Theorem 5.1 implies Theorem 1.1. In fact,
we shall provide criteria warranting (B) to be satisfied. One of them being
$\mathcal{H}$ to be of bounded variation, see also Section 8. Another
criterion requires a structural hypothesis on the interplay of $\mathcal{H}$
and $P_{1}$, which is independent of regularity, thus, showing that the
statements in Theorem 5.1 and Theorem 1.1 are not equivalent, eventually
proving that Theorem 5.1 is a proper generalisation of Theorem 1.1.

For the proof of Theorem 5.1, we aim to apply Corollary 4.6. For this, we deal
with (B) first. We recall the definition of $\Gamma_{t}$ in (11), (12), and
(13) and the matrices $U$ and $Q$ from Section 2. Note that (3) and (4) yield

$U^{\ast}\frac{1}{b-a}P_{1}U=Q^{-1}DQ^{-1}$

and, thus,

$P_{1}^{-1}=\left((b-a)UQ^{-1}DQ^{-1}U^{*}\right)^{-1}=\frac{1}{b-a}UQDQU^{\ast}.$
(16)

######  Proposition 5.3.

For all $t\in\mathbb{R}$,

$\Phi_{t}(x)=UQ\Gamma_{t}\left(\frac{1}{b-a}(x-a)-\frac{1}{2}\right)Q^{-1}U^{\ast}\quad(x\in(a,b)).$

In particular $\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}<\infty$ if and
only if $\sup_{t\in\mathbb{R}}\|\Phi_{t}\|_{\infty}<\infty$.

###### Proof.

We set $\Delta_{t}(x)\coloneqq
UQ\Gamma_{t}\left(\frac{1}{b-a}(x-a)-\frac{1}{2}\right)Q^{-1}U^{\ast}.$ Note
that

$\Delta_{t}(a)=UQ\Gamma_{t}\left(-\frac{1}{2}\right)Q^{-1}U^{\ast}=1_{d}\quad(t\in\mathbb{R}).$

Next, we compute, using
$\mathcal{O}(s)=QU^{\ast}\mathcal{H}\left((b-a)s+\frac{a+b}{2}\right)UQ$,
$s\in(-\tfrac{1}{2},\tfrac{1}{2})$, from (iii) in Proposition 4.3, for all
$x\in(a,b$)

$\displaystyle\Delta_{t}^{\prime}(x)$
$\displaystyle=UQ\Gamma_{t}^{\prime}\left(\frac{1}{b-a}(x-a)-\frac{1}{2}\right)Q^{-1}U^{\ast}$
$\displaystyle=\frac{1}{b-a}\mathrm{i}tUQD\mathcal{O}(\frac{1}{b-a}(x-a)-\frac{1}{2})\Gamma_{t}\left(\frac{1}{b-a}(x-a)-\frac{1}{2}\right)Q^{-1}U^{\ast}$
$\displaystyle=\frac{1}{b-a}\mathrm{i}tUQD\mathcal{O}(\frac{1}{b-a}(x-a)-\frac{1}{2})Q^{-1}U^{*}UQ\Gamma_{t}\left(\frac{1}{b-a}(x-a)-\frac{1}{2}\right)Q^{-1}U^{\ast}$
$\displaystyle=\frac{1}{b-a}\mathrm{i}tUQDQU^{\ast}\mathcal{H}\left(x\right)UQQ^{-1}U^{*}UQ\Gamma_{t}\left(\frac{1}{b-a}(x-a)-\frac{1}{2}\right)Q^{-1}U^{\ast}$
$\displaystyle=\frac{1}{b-a}\mathrm{i}tUQDQU^{\ast}\mathcal{H}\left(x\right)\Delta_{t}(x)$
$\displaystyle=\mathrm{i}tP_{1}^{-1}\mathcal{H}(x)\Delta_{t}(x),$

where in the last equality we used (16). Thus, by uniqueness of the
fundamental matrix, $\Delta_{t}(x)=\Phi_{t}(x)$ for all $x\in[a,b]$. ∎

For the application of Corollary 4.6, we reformulate the estimate provided in
Theorem 5.1 next.

######  Proposition 5.4.

Assume that $-\mathcal{A}$ generates a semi-group of contractions and recall
(9) and (10). Then the following conditions are equivalent:

  1. (i)

There exist $c>0$ and $\eta\in\\{a,b\\}$ such that for all
$u\in\operatorname{dom}(\mathcal{A})$

$\operatorname{Re}\langle u,\mathcal{A}u\rangle_{H}\geq
c\|\mathcal{H}u(\eta)\|^{2}.$

  2. (ii)

There exist $c>0$ and $\eta\in\\{-\tfrac{1}{2},\tfrac{1}{2}\\}$ such that for
all $v\in\operatorname{dom}(\mathcal{C})$

$\operatorname{Re}\langle v,\mathcal{C}v\rangle_{L_{2}}\geq c\|v(\eta)\|^{2}.$

###### Proof.

(i)$\Rightarrow$(ii): Let $v\in\operatorname{dom}(\mathcal{C})$. Then, by
(10), $S\Psi^{*}UQ^{*}v\in\operatorname{dom}(\mathcal{A})$. We compute with
the help of Lemma 2.2

$\displaystyle(\mathcal{H}S\Psi^{*}UQ^{*}v)(\eta)$
$\displaystyle=(\Psi^{*}UQ^{*}v)(\eta)$
$\displaystyle=\sqrt{\frac{1}{b-a}}UQ^{*}v\left(\frac{1}{b-a}\eta-\frac{1}{b-a}a-\frac{1}{2}\right).$

Hence, using (9), that is, $\mathcal{C}=QU^{\ast}\Psi
S^{\ast}\mathcal{A}S\Psi^{\ast}UQ^{\ast}$

$\displaystyle\operatorname{Re}\langle v,\mathcal{C}v\rangle_{L_{2}}$
$\displaystyle=\operatorname{Re}\langle
S\Psi^{*}UQ^{*}v,\mathcal{A}S\Psi^{\ast}UQ^{\ast}v\rangle_{H}$
$\displaystyle\geq c\left\|(\mathcal{H}S\Psi^{*}UQ^{*}v)(\eta)\right\|^{2}$
$\displaystyle=\frac{c}{b-a}\left\|UQ^{*}v\left(\frac{1}{b-a}\eta-\frac{1}{b-a}a-\frac{1}{2}\right)\right\|^{2}$
$\displaystyle\geq\widetilde{c}\left\|v\left(\widetilde{\eta}\right)\right\|^{2},$

where $\widetilde{\eta}\in\\{-\tfrac{1}{2},\tfrac{1}{2}\\}$ and some
$\widetilde{c}>0$. Note that in the last estimate we used that $Q$ is an
invertible matrix.

(ii)$\Rightarrow$(i): Using (9) and (10), we can use almost the exact same
argument as above. ∎

Next, we provide an integration by parts formula valid for the operator
$D\partial_{s}$, which is a straightforward consequence of the classical
result for $H^{1}$-functions. For this, we recall

$D=\left(\begin{array}[]{cc}1_{k}&0\\\ 0&-1_{d-k}\end{array}\right)$

and that we decompose any vector $y\in\mathbb{R}^{d}$ according to
$y=(\check{y},\hat{y})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k}$.

######  Proposition 5.5.

(a) For all $v\in H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d}$,

$\operatorname{Re}\langle
v,D\partial_{s}v\rangle_{L_{2}}=\frac{1}{2}\left(\left\|\check{v}(\tfrac{1}{2})\right\|^{2}-\left\|\check{v}(-\tfrac{1}{2})\right\|^{2}+\left\|\hat{v}(-\tfrac{1}{2})\right\|^{2}-\left\|\hat{v}(\tfrac{1}{2})\right\|^{2}\right)$

(b) Let $M\in\mathbb{R}^{d\times d}$ with $\|M\|\leq 1$ and assume that
$\mathcal{C}$ satisfies (14). Then for all
$v\in\operatorname{dom}(\mathcal{C})$

$\displaystyle\operatorname{Re}\langle v,\mathcal{C}v\rangle_{L_{2}}$
$\displaystyle=\frac{1}{2}\left(\left\|\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)\right\|^{2}-\left\|M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)\right\|^{2}\right)$

###### Proof.

(a) By the classical intgeration by parts formula, we compute for
$v=(\check{v},\hat{v})\in H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{k}\times
H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d-k}$

$\displaystyle 2\operatorname{Re}\langle v,D\partial_{s}v\rangle_{L_{2}}$
$\displaystyle=\left\langle\left(\begin{array}[]{c}\check{v}\\\
\hat{v}\end{array}\right),\left(\begin{array}[]{c}\check{v}^{\prime}\\\
-\hat{v}^{\prime}\end{array}\right)\right\rangle+\left\langle\left(\begin{array}[]{c}\check{v}^{\prime}\\\
-\hat{v}^{\prime}\end{array}\right),\left(\begin{array}[]{c}\check{v}\\\
\hat{v}\end{array}\right)\right\rangle$
$\displaystyle=\left\|\check{v}(\tfrac{1}{2})\right\|^{2}-\left\|\check{v}(-\tfrac{1}{2})\right\|^{2}+\left\|\hat{v}(-\tfrac{1}{2})\right\|^{2}-\left\|\hat{v}(\tfrac{1}{2})\right\|^{2}$

(b) This is a straightforward consequence of the formula in (a) and the domain
constraint of $\mathcal{C}$. For this recall (5) and that $\widetilde{P_{0}}$
is skew-selfadjoint. ∎

The next results is easily obtained using affine functions.

######  Lemma 5.6.

Let $M\in\mathbb{R}^{d\times d}$ with $\|M\|\leq 1$ and assume that
$\mathcal{C}$ satisfies (14). Then the mapping

$\mathrm{tr}\colon\operatorname{dom}(\mathcal{C})\ni
v\mapsto\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)\in\mathbb{R}^{d}$

is onto.

Equipped with the latter auxiliary results, one can reformulate the estimate
derived for $\mathcal{C}$ in Proposition 5.4 into an estimate for $M$.

######  Proposition 5.7.

Let $M\in\mathbb{R}^{d\times d}$ with $\|M\|\leq 1$ and assume that
$\mathcal{C}$ satisfies (14). Let $c>0$. Then the following statements are
equivalent:

  1. (i)

There exists $\eta\in\\{-\tfrac{1}{2},\tfrac{1}{2}\\}$ such that for all
$v\in\operatorname{dom}(\mathcal{C})$

$\operatorname{Re}\langle v,\mathcal{C}v\rangle_{L_{2}}\geq c\|v(\eta)\|^{2}.$

  2. (ii)

Either

$\left\|y\right\|^{2}-\left\|My\right\|^{2}\geq
2c\left(\left\|(My)^{\lor}\right\|^{2}+\|\hat{y}\|^{2}\right)\quad(y\in\mathbb{R}^{d})$

or

$\left\|y\right\|^{2}-\left\|My\right\|^{2}\geq
2c\left(\|\check{y}\|^{2}+\left\|(My)^{\land}\right\|^{2}\right)\quad(y\in\mathbb{R}^{d}).$

###### Proof.

By Proposition 5.5, for all $v\in\operatorname{dom}(\mathcal{C}),$

$\operatorname{Re}\langle
v,\mathcal{C}v\rangle_{L_{2}}=\frac{1}{2}\left(\left\|\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)\right\|^{2}-\left\|M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)\right\|^{2}\right).$

Moreover, for all $v\in\operatorname{dom}(\mathcal{C})$ we have
$M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)=\left(\begin{array}[]{c}\check{v}(-\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)$. Hence,

$\left(\begin{array}[]{c}\check{v}(-\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)=\left(\begin{array}[]{c}\left(M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)\right)^{\lor}\\\
\hat{v}(-\frac{1}{2})\end{array}\right),\quad\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)=\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\left(M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)\right)^{\land}\end{array}\right).$

Thus, employing $\mathrm{tr}$ from Lemma 5.6, we deduce for all
$v\in\operatorname{dom}(\mathcal{C})$

$\operatorname{Re}\langle
v,\mathcal{C}v\rangle_{L_{2}}=\frac{1}{2}\left(\left\|\mathrm{tr}\,v\right\|^{2}-\left\|M\mathrm{tr}\,v\right\|^{2}\right)$

and for $\eta\in\\{-\tfrac{1}{2},\tfrac{1}{2}\\}$

$v(\eta)=\begin{cases}\left(\begin{array}[]{c}\left(M\left(\mathrm{tr}\,v\right)\right)^{\lor}\\\
\left(\mathrm{tr}\,v\right)^{\land}\end{array}\right),&\eta=-\tfrac{1}{2},\\\
\left(\begin{array}[]{c}\left(\mathrm{tr}\,v\right)^{\lor}\\\
\left(M\left(\mathrm{tr}\,v\right)\right)^{\land}\end{array}\right),&\eta=\tfrac{1}{2}.\end{cases}$

Since, by Lemma 5.6, $v\mapsto\mathrm{tr}\,v$ maps onto $\mathbb{R}^{d}$, the
assertion follows. ∎

The proof of Theorem 5.1 requires some more insight into the fundmantal
solution $\Gamma_{t}$.

######  Proposition 5.8.

Let $\mathcal{U}\in L_{\infty}(-\tfrac{1}{2},\tfrac{1}{2};\mathbb{C}^{d\times
d})$ taking values in the self-adjoint matrices. Denote by
$\Pi\colon\left[-\tfrac{1}{2},\tfrac{1}{2}\right]\to\mathbb{C}^{d\times d}$
the fundamental solution satisfying $\Pi(-\tfrac{1}{2})=1_{d}$ associated with
the ordinary differential equation

$v^{\prime}(s)=\mathrm{i}D\mathcal{U}(s)v(s)\in\mathbb{C}^{d}\quad(s\in(-\tfrac{1}{2},\tfrac{1}{2})),$
(17)

where $D=\left(\begin{array}[]{cc}1_{k}&0\\\ 0&-1_{d-k}\end{array}\right)$ for
some $k\in\\{0,\ldots,d\\}$.

Then the following statements hold:

  1. (a)

$\Pi(s)^{-1}=D\Pi(s)^{*}D$ for all
$s\in\left[-\tfrac{1}{2},\tfrac{1}{2}\right]$.

  2. (b)

The matrices

$V_{1}\coloneqq\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Pi(\tfrac{1}{2}),\quad
V_{2}\coloneqq\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Pi(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)$

are invertible and $\|V_{j}^{-1}\|\leq 1+\|\Pi(\frac{1}{2})\|$,
$j\in\\{1,2\\}$.

  3. (c)

The matrix $W\coloneqq V_{2}V_{1}^{-1}$ is unitary.

  4. (d)

For all
$y\coloneqq(\check{y},\hat{y})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k}$,

$\|y\|^{2}\leq(1+\|\Pi(\tfrac{1}{2})\|)^{2}\left(\|\check{y}\|^{2}+\|(Wy)^{\land}\|^{2}\right).$

  5. (e)

For all
$y\coloneqq(\check{y},\hat{y})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k}$,

$\|y\|^{2}\leq\left(1+\|\Pi(\tfrac{1}{2})\|\right)^{2}\|\Pi(\tfrac{1}{2})\|^{2}\left(\|\hat{y}\|^{2}+\|(Wy)^{\lor}\|^{2}\right).$

###### Proof.

(a) We compute the derivative of $\Psi\coloneqq\Pi(\cdot)^{-1}.$ For
$s\in\left[-\tfrac{1}{2},\tfrac{1}{2}\right]$ we have

$\displaystyle\Psi^{\prime}(s)$ $\displaystyle=-\Psi(s)\Pi^{\prime}(s)\Psi(s)$
$\displaystyle=-\mathrm{i}\Psi(s)D\mathcal{U}(s)\Pi(s)\Psi(s)$
$\displaystyle=-\mathrm{i}\Psi(s)D\mathcal{U}(s).$

Consequently,

$\Psi^{\prime}(s)^{\ast}=\mathrm{i}\mathcal{U}(s)D\Psi(s)^{\ast}.$

Multiplying the latter equation with $D$ from the left and from the right, we
infer

$D\Psi^{\prime}(s)^{\ast}D=\mathrm{i}D\mathcal{U}(s)D\Psi(s)^{\ast}D$

and thus, $s\mapsto D\Psi(s)^{\ast}D$ is the fundamental solution for (17)
with $D\Psi(-\frac{1}{2})^{\ast}D=D^{2}=1_{d}.$ Hence,
$\Psi(s)=D\Pi(s)^{\ast}D.$

(b) We write $\Pi(\tfrac{1}{2})=\left(\begin{array}[]{cc}\pi_{11}&\pi_{12}\\\
\pi_{21}&\pi_{22}\end{array}\right)$ according to the block structure of $D$.
Using (a), we obtain

$\left(\begin{array}[]{cc}\pi_{11}&-\pi_{12}\\\
-\pi_{21}&\pi_{22}\end{array}\right)=D\Pi(\tfrac{1}{2})D=\left(\Pi(\tfrac{1}{2})^{*}\right)^{-1}=\left(\begin{array}[]{cc}\pi_{11}^{*}&\pi_{21}^{*}\\\
\pi_{12}^{*}&\pi_{22}^{*}\end{array}\right)^{-1}.$

It follows that

$\left(\begin{array}[]{cc}1_{k}&0\\\
0&1_{d-k}\end{array}\right)=\left(\begin{array}[]{cc}\pi_{11}^{*}&\pi_{21}^{*}\\\
\pi_{12}^{*}&\pi_{22}^{*}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}&-\pi_{12}\\\
-\pi_{21}&\pi_{22}\end{array}\right)=\left(\begin{array}[]{cc}\pi_{11}^{*}\pi_{11}-\pi_{21}^{*}\pi_{21}&-\pi_{11}^{*}\pi_{12}+\pi_{21}^{*}\pi_{22}\\\
\pi_{12}^{*}\pi_{11}-\pi_{22}^{*}\pi_{21}&\pi_{22}^{*}\pi_{22}-\pi_{12}^{*}\pi_{12}\end{array}\right).$
(18)

In the following we prove (b) only for $V\coloneqq V_{1}$. The corresponding
statement for $V_{2}$ follows the same way interchanging indices
$1\leftrightarrow 2$ and $k\leftrightarrow d-k$.

For $V=V_{1}$ we deduce from (18)

$\|\pi_{22}\check{y}\|^{2}=\langle\pi_{22}^{*}\pi_{22}\check{y},\check{y}\rangle\geq\langle\check{y},\check{y}\rangle\quad(\check{y}\in\mathbb{R}^{k})$

confirming that $\pi_{22}$ is one-to-one, hence invertible. The inverse is
bounded by $1$. From

$V=\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Pi(\tfrac{1}{2})=\left(\begin{array}[]{cc}1_{k}&0\\\
\pi_{21}&\pi_{22}\end{array}\right)$

we thus obtain the invertibility of $V$. The formula

$V^{-1}=\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{22}^{-1}\pi_{21}&\pi_{22}^{-1}\end{array}\right)=\left(\begin{array}[]{cc}1_{k}&0\\\
0&\pi_{22}^{-1}\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
-\pi_{22}^{-1}\pi_{21}&0\end{array}\right)$ (19)

yields the norm estimate as $\|\pi_{22}^{-1}\|\leq 1$ and
$\|\pi_{21}\|\leq\|\Pi(\tfrac{1}{2})\|$.

(c) We recall $\Pi(\tfrac{1}{2})=\left(\begin{array}[]{cc}\pi_{11}&\pi_{12}\\\
\pi_{21}&\pi_{22}\end{array}\right)$, (18) and (19) from (b). In particular,
note that $\pi_{11}$ is invertible and $\|\pi_{11}^{-1}\|\leq 1$. Then we
compute

$\displaystyle W^{-1}$
$\displaystyle=\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Pi(\tfrac{1}{2})\right)\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Pi(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)^{-1}$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&0\\\
\pi_{21}&\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}^{-1}&-\pi_{11}^{-1}\pi_{12}\\\
0&1_{d-k}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}\pi_{11}^{-1}&-\pi_{11}^{-1}\pi_{12}\\\
\pi_{21}\pi_{11}^{-1}&\pi_{22}-\pi_{21}\pi_{11}^{-1}\pi_{12}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&0\\\
\pi_{21}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}^{-1}&0\\\
0&\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&-\pi_{12}\\\
0&1_{d-k}\end{array}\right).$

In order to confirm that $W$ (or, equivalently, $\widetilde{W}\coloneqq
W^{-1})$ is unitary, we appeal to (18) and obtain

$\displaystyle\widetilde{W}^{*}\widetilde{W}$
$\displaystyle=\left(\left(\begin{array}[]{cc}1_{k}&0\\\
\pi_{21}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}^{-1}&0\\\
0&\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&-\pi_{12}\\\
0&1_{d-k}\end{array}\right)\right)^{*}\left(\begin{array}[]{cc}1_{k}&0\\\
\pi_{21}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}^{-1}&0\\\
0&\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&-\pi_{12}\\\
0&1_{d-k}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{12}^{*}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\left(\pi_{11}^{-1}\right)^{*}&0\\\
0&\pi_{22}^{*}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&\pi_{21}^{*}\\\
0&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&0\\\
\pi_{21}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}^{-1}&0\\\
0&\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&-\pi_{12}\\\
0&1_{d-k}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{12}^{*}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\left(\pi_{11}^{-1}\right)^{*}&0\\\
0&\pi_{22}^{*}\end{array}\right)\left(\begin{array}[]{cc}1_{k}+\pi_{21}^{*}\pi_{21}&\pi_{21}^{*}\\\
\pi_{21}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}^{-1}&0\\\
0&\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&-\pi_{12}\\\
0&1_{d-k}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{12}^{*}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\left(\pi_{11}^{-1}\right)^{*}&0\\\
0&\pi_{22}^{*}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}^{*}\pi_{11}&\pi_{21}^{*}\\\
\pi_{21}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}^{-1}&0\\\
0&\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&-\pi_{12}\\\
0&1_{d-k}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{12}^{*}&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&\left(\pi_{11}^{-1}\right)^{*}\pi_{21}^{*}\pi_{22}\\\
\pi_{22}^{*}\pi_{21}\pi_{11}^{-1}&\pi_{22}^{*}\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&-\pi_{12}\\\
0&1_{d-k}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&\left(\pi_{11}^{-1}\right)^{*}\pi_{21}^{*}\pi_{22}\\\
\pi_{22}^{*}\pi_{21}\pi_{11}^{-1}-\pi_{12}^{*}&\pi_{22}^{*}\pi_{22}-\pi_{12}^{*}\left(\pi_{11}^{-1}\right)^{*}\pi_{21}^{*}\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&-\pi_{12}\\\
0&1_{d-k}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&\left(\pi_{11}^{-1}\right)^{*}\pi_{21}^{*}\pi_{22}-\pi_{12}\\\
\pi_{22}^{*}\pi_{21}\pi_{11}^{-1}-\pi_{12}^{*}&\pi_{22}^{*}\pi_{22}-\pi_{12}^{*}\left(\pi_{11}^{-1}\right)^{*}\pi_{21}^{*}\pi_{22}-\pi_{22}^{*}\pi_{21}\pi_{11}^{-1}\pi_{12}+\pi_{12}^{*}\pi_{12}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&\left(\pi_{11}^{-1}\right)^{*}\left(\pi_{21}^{*}\pi_{22}-\pi_{11}^{*}\pi_{12}\right)\\\
\left(\pi_{22}^{*}\pi_{21}-\pi_{12}^{*}\pi_{11}\right)\pi_{11}^{-1}&\pi_{22}^{*}\pi_{22}-\pi_{12}^{*}\left(\pi_{11}^{-1}\right)^{*}\pi_{21}^{*}\pi_{22}-\pi_{22}^{*}\pi_{21}\pi_{11}^{-1}\pi_{12}+\pi_{12}^{*}\pi_{12}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}1_{k}&0\\\
0&\pi_{22}^{*}\pi_{22}-\pi_{12}^{*}\left(\pi_{11}^{-1}\right)^{*}\pi_{21}^{*}\pi_{22}-\pi_{22}^{*}\pi_{21}\pi_{11}^{-1}\pi_{12}+\pi_{12}^{*}\pi_{12}\end{array}\right).$

Finally, since $\pi_{11}^{*}\pi_{12}=\pi_{21}^{*}\pi_{22}$ (see the top right
corner in (18)) we conclude

$\displaystyle-\pi_{12}^{*}\left(\pi_{11}^{-1}\right)^{*}\pi_{21}^{*}\pi_{22}-\pi_{22}^{*}\pi_{21}\pi_{11}^{-1}\pi_{12}$
$\displaystyle=-\pi_{12}^{*}\left(\pi_{11}^{-1}\right)^{*}\pi_{11}^{*}\pi_{12}-\left(\pi_{11}^{*}\pi_{12}\right)^{*}\pi_{11}^{-1}\pi_{12}$
$\displaystyle=-\pi_{12}^{*}\pi_{12}-\pi_{12}^{*}\pi_{12}=-2\pi_{12}^{\ast}\pi_{12},$

and hence,

$\displaystyle\pi_{22}^{*}\pi_{22}-\pi_{12}^{*}\left(\pi_{11}^{-1}\right)^{*}\pi_{21}^{*}\pi_{22}-\pi_{22}^{*}\pi_{21}\pi_{11}^{-1}\pi_{12}+\pi_{12}^{*}\pi_{12}$
$\displaystyle=\pi_{22}^{\ast}\pi_{22}-\pi_{12}^{\ast}\pi_{12}=1_{d-k}.$

which yields the assertion in (c).

(d) We recall $\Pi(\tfrac{1}{2})=\left(\begin{array}[]{cc}\pi_{11}&\pi_{12}\\\
\pi_{21}&\pi_{22}\end{array}\right)$, (18) and (19) from (b). Then we compute
for $y=(\check{y},\hat{y})\in\mathbb{R}^{d}$

$\displaystyle\left(\begin{array}[]{c}0\\\
\left(Wy\right)^{\land}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}0&0\\\ 0&1_{d-k}\end{array}\right)Wy$
$\displaystyle=\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)V_{2}V_{1}^{-1}y$
$\displaystyle=\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}\pi_{11}&\pi_{12}\\\
0&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{22}^{-1}\pi_{21}&\pi_{22}^{-1}\end{array}\right)y$
$\displaystyle=\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{22}^{-1}\pi_{21}&\pi_{22}^{-1}\end{array}\right)y=\left(\begin{array}[]{cc}0&0\\\
-\pi_{22}^{-1}\pi_{21}&\pi_{22}^{-1}\end{array}\right)y.$

Hence,

$\left(\begin{array}[]{c}\check{y}\\\
\left(Wy\right)^{\land}\end{array}\right)=\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{22}^{-1}\pi_{21}&\pi_{22}^{-1}\end{array}\right)y$

and, thus,

$y=\left(\begin{array}[]{cc}1_{k}&0\\\
\pi_{21}&\pi_{22}\end{array}\right)\left(\begin{array}[]{c}\check{y}\\\
\left(Wy\right)^{\land}\end{array}\right).$

Hence,

$\|y\|^{2}\leq(1+\|\Pi(\tfrac{1}{2})\|)^{2}\left(\|\check{y}\|^{2}+\|(Wy)^{\land}\|^{2}\right).$

(e) We recall $\Pi(\tfrac{1}{2})=\left(\begin{array}[]{cc}\pi_{11}&\pi_{12}\\\
\pi_{21}&\pi_{22}\end{array}\right)$, (18) and (19) from (b). Then we compute
for $y=(\check{y},\hat{y})\in\mathbb{R}^{d}$

$\displaystyle\left(\begin{array}[]{c}\left(Wy\right)^{\lor}\\\
0\end{array}\right)$ $\displaystyle=\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)Wy$ $\displaystyle=\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)V_{2}V_{1}^{-1}y=\left(\begin{array}[]{cc}\pi_{11}&\pi_{12}\\\
0&0\end{array}\right)\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{22}^{-1}\pi_{21}&\pi_{22}^{-1}\end{array}\right)y$
$\displaystyle=\left(\begin{array}[]{cc}\pi_{11}-\pi_{12}\pi_{22}^{-1}\pi_{21}&\pi_{12}\pi_{22}^{-1}\\\
0&0\end{array}\right)y.$

Hence,

$\displaystyle\left(\begin{array}[]{c}\left(Wy\right)^{\lor}\\\
\hat{y}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}\pi_{11}-\pi_{12}\pi_{22}^{-1}\pi_{21}&\pi_{12}\pi_{22}^{-1}\\\
0&1_{d-k}\end{array}\right)y$
$\displaystyle=\left(\begin{array}[]{cc}\pi_{11}&\pi_{12}\\\
\pi_{21}&\pi_{22}\end{array}\right)\left(\begin{array}[]{cc}1_{k}&0\\\
-\pi_{22}^{-1}\pi_{21}&\pi_{22}^{-1}\end{array}\right)y=\Pi(\tfrac{1}{2})V_{1}^{-1}y.$

and, thus,

$\displaystyle\|y\|$
$\displaystyle=\left\|\left(\Pi(\tfrac{1}{2})V_{1}^{-1}\right)^{-1}\left(\begin{array}[]{c}\left(Wy\right)^{\lor}\\\
\hat{y}\end{array}\right)\right\|$
$\displaystyle=\|V_{1}\|\|\Pi(\tfrac{1}{2})^{-1}\|\left\|\left(\begin{array}[]{c}\left(Wy\right)^{\lor}\\\
\hat{y}\end{array}\right)\right\|\leq\left(1+\|\Pi(\tfrac{1}{2})\|\right)\|D\Pi(\tfrac{1}{2})D\|\left\|\left(\begin{array}[]{c}\left(Wy\right)^{\lor}\\\
\hat{y}\end{array}\right)\right\|,$

where we used (a). The assertion thus follows from the unitarty of $D$. ∎

###### Proof of Theorem 5.1.

Condition (B) implies due to Proposition 5.3 that
$\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}<\infty$. Thus, we are in the
position to apply Corollary 4.6. Hence, in order to have that $-\mathcal{A}$
generates a $C_{0}$-semi-group, we need to show that

$S_{t}\coloneqq M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)$

is invertible and satisfies $\sup_{t\in\mathbb{R}}\|S_{t}^{-1}\|<\infty$,
where $M\in\mathbb{R}^{d\times d}$ satisfies $\|M\|\leq 1$ and (14) and
$\mathcal{C}$ satisfies (9). Since $\mathcal{A}$ satisfies (15), we obtain
using Proposition 5.4 and Proposition 5.7 that there exists $c>0$ such that
either

$\left\|y\right\|^{2}-\left\|My\right\|^{2}\geq
c\left(\left\|(My)^{\lor}\right\|^{2}+\|\hat{y}\|^{2}\right)\quad(y\in\mathbb{R}^{d})$
(20)

or

$\left\|y\right\|^{2}-\left\|My\right\|^{2}\geq
c\left(\|\check{y}\|^{2}+\left\|(My)^{\land}\right\|^{2}\right)\quad(y\in\mathbb{R}^{d}).$
(21)

Next, in order to prove that $S_{t}$ is invertible with uniform (in $t$) bound
for the inverses, we let $y\in\mathbb{R}^{d}$ and define

$z\coloneqq S_{t}y.$

Our aim is to show that there exists $\kappa\geq 0$ independently of
$t\in\mathbb{R}$ such that

$\|y\|\leq\kappa\|z\|.$ (22)

This then confirms (ii) in Corollary 4.6. For this, we apply Proposition 5.8
to $\Pi=\Gamma_{t}$ (thus $\mathcal{U}=t\mathcal{O})$ and infer

$\displaystyle U_{t}$
$\displaystyle\coloneqq\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)^{-1}$

is unitary for all $t\in\mathbb{R}.$ Moreover,

$V_{t}\coloneqq\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})$

is invertible and

$\|V_{t}^{-1}\|\leq
1+\|\Gamma_{t}(\tfrac{1}{2})\|\leq\left(1+\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}\right)\eqqcolon\Lambda.$

Next, we compute

$\displaystyle z$ $\displaystyle=S_{t}y$
$\displaystyle=M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)y-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)y$
$\displaystyle=MU_{t}V_{t}y-V_{t}y$
$\displaystyle=\left(MU_{t}-1_{d}\right)y_{t},$

where $y_{t}\coloneqq V_{t}y.$ Since

$\|y\|=\|V_{t}^{-1}\|\|y_{t}\|\leq\Lambda\|y_{t}\|$

in order to show (22) in view of assumption (B) and Proposition 5.3, it thus
suffices to find $\kappa\geq 0$ independently of $t\in\mathbb{R}$, $y_{t}$ and
$z$ such that

$\|y_{t}\|\leq\kappa\|z\|.$ (23)

The proof of the previous estimate requires the distinction between two cases
according to which inequality – (20) or (21) – is satisfied.

To begin with, we assume that (20) holds. Applying this inequality to
$U_{t}y_{t}$, we compute for $y_{t}=MU_{t}y_{t}-z$

$\displaystyle\|y_{t}\|^{2}$
$\displaystyle=\|z\|^{2}-2\operatorname{Re}\langle
z,MU_{t}y_{t}\rangle+\|MU_{t}y_{t}\|^{2}$
$\displaystyle\leq\|z\|^{2}-2\operatorname{Re}\langle
z,MU_{t}y_{t}\rangle+\|U_{t}y_{t}\|^{2}-c\left(\|(MU_{t}y_{t})^{\lor}\|^{2}+\|(U_{t}y_{t})^{\land}\|^{2}\right)$
$\displaystyle=\|z\|^{2}-2\operatorname{Re}\langle
z,y_{t}+z\rangle+\|y_{t}\|^{2}-c\left(\|\check{y}_{t}+\check{z}\|^{2}+\|(U_{t}y_{t})^{\land}\|^{2}\right),$

where in the last equality we used the definition of $z$ and that $U_{t}$ is
unitary. Hence,

$\displaystyle 0$ $\displaystyle\leq\|z\|^{2}-2\operatorname{Re}\langle
z,y_{t}+z\rangle-c\left(\|\check{y}_{t}+\check{z}\|^{2}+\|(U_{t}y_{t})^{\land}\|^{2}\right)$
$\displaystyle=-\|z\|^{2}-2\operatorname{Re}\langle
z,y_{t}\rangle-c\left(\|\check{y}_{t}\|^{2}+2\operatorname{Re}\langle\check{y}_{t},\check{z}\rangle+\|\check{z}\|^{2}+\|(U_{t}y_{t})^{\land}\|^{2}\right)$
$\displaystyle\leq-2\operatorname{Re}\langle
z,y_{t}\rangle-c\left(\|\check{y}_{t}\|^{2}+2\operatorname{Re}\langle\check{y}_{t},\check{z}\rangle+\|(U_{t}y_{t})^{\land}\|^{2}\right).$

With Young’s inequality ($2ab\leq\frac{1}{\varepsilon}a^{2}+\varepsilon
b^{2},a,b\in\mathbb{R},\varepsilon>0)$, for all $\gamma>0$ we find
$\widetilde{\kappa}\geq 0$ such that

$\|\check{y}_{t}\|^{2}+\|(U_{t}y_{t})^{\land}\|^{2}-\gamma\|y_{t}\|^{2}\leq\widetilde{\kappa}\left(\|z\|^{2}+\|\check{z}\|^{2}\right)\leq
2\widetilde{\kappa}\|z\|^{2}.$ (24)

In a similar manner, assuming (21) holds, we obtain for all $\gamma>0$ some
$\widetilde{\kappa}\geq 0$ such that

$\|\hat{y}_{t}\|^{2}+\|(U_{t}y_{t})^{\lor}\|^{2}-\gamma\|y_{t}\|^{2}\leq
2\widetilde{\kappa}\|z\|^{2}.$ (25)

With the help of the inequalities in Proposition 5.8 (d) and (e) (applied to
$W=U_{t}$) in conjunction with (24) and (25), respectively, we obtain (23) by
choosing $\gamma>0$ small enough and thus conclude the proof. ∎

The observations in Proposition 5.8 yield another criterion for exponential
stability.

######  Theorem 5.9.

Assume (B) and that $-\mathcal{A}$ generates a contraction semi-group. Let
$M\in\mathbb{R}^{d\times d},$ $\|M\|\leq 1$ such that $\mathcal{C}$ given by
(9) satisfies (14). If $\|M\|<1$, then $-\mathcal{A}$ generates an
exponentially stable $C_{0}$-semi-group.

###### Proof.

We apply Corollary 4.6. Note that (B) together with Proposition 5.3 implies
that $t\mapsto\|\Gamma_{t}\|$ is uniformly bounded in $t$; denote
$\Lambda\coloneqq\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}$. In order to
obtain that $-\mathcal{A}$ is exponentially stable, we need to show that

$S_{t}\coloneqq M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)$

is invertible, with $\|S_{t}^{-1}\|$ uniformly bounded in $t$. Proposition 5.8
applied to $\Pi=\Gamma_{t}$ yields that

$V_{t}\coloneqq\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)$

is invertible with $\|V_{t}^{-1}\|\leq 1+\|\Gamma_{t}(\tfrac{1}{2})\|\leq
1+\Lambda$ and that

$U_{t}\coloneqq\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)V_{t}^{-1}$

is unitary. Hence,

$S_{t}=MU_{t}V_{t}-V_{t}=\left(MU_{t}-1_{d}\right)V_{t}.$

Since $\|MU_{t}\|=\|M\|<1$, by the unitarity of $U_{t}$, a Neumann series
argument leads to invertibility of $S_{t}$ and the estimate

$\|S_{t}^{-1}\|\leq\|V_{t}^{-1}\|\sum_{k=0}^{\infty}\|M\|^{k}\leq\frac{1+\Lambda}{1-\|M\|}.$

Thus, $-\mathcal{A}$ generates and exponentially stable semi-group, by
Corollary 4.6. ∎

######  Remark 5.10.

This criterion can be seen as the version of [9, Lemma 9.1.4] with the weaker
prerequisite (B) instead of continuous differentiability of $\mathcal{H}$, see
also Section 8 below. Indeed, for the particular case of a port-Hamiltonian
system given by $\mathcal{C}$ with appropriate boundary condition encoded by
$M$; the additional requirement in [9, Lemma 9.1.4], that is, ((WB)) with
strict inequality is equivalent to $\|M\|<1$, see also Remark 2.4. The general
statement of Theorem 5.9 remains true, if one replaces the condition $\|M\|<1$
by ((WB)) with strict inequality. This, in turn, can be shown using Theorem
5.1 in conjunction with (the proof of) [9, Lemma 9.1.4].

##  6 Characterisation for the Reference Case

In this section, we aim to prove Theorem 4.5. For this, throughout this
section, we let

$\mathcal{O}\in L_{\infty}(-\tfrac{1}{2},\tfrac{1}{2};\mathbb{R}^{d\times
d})\text{ with }\mathcal{O}(s)=\mathcal{O}(s)^{\ast}\geq m\text{ for a.e.
}s\in(-\tfrac{1}{2},\tfrac{1}{2})$

for some $m>0$. Next, let $k\in\\{0,\ldots,d\\}$ and

$D=\left(\begin{array}[]{cc}1_{k}&0\\\
0&-1_{d-k}\end{array}\right)\in\mathbb{R}^{d\times d}.$

As usual, we decompose
$y=(\check{y},\hat{y})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k}$ and let for
some $M\in\mathbb{R}^{d\times d}$ with $\|M\|\leq 1$ the operator
$\mathcal{C}\subseteq D\partial_{s}$ on
$L_{2}(-\tfrac{1}{2},\tfrac{1}{2})^{d}$ be such that

$\operatorname{dom}(\mathcal{C})=\left\\{v\in
H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d};M\left(\begin{array}[]{c}\check{v}(\tfrac{1}{2})\\\
\hat{v}(-\tfrac{1}{2})\end{array}\right)=\left(\begin{array}[]{c}\check{v}(-\tfrac{1}{2})\\\
\hat{v}(\tfrac{1}{2})\end{array}\right)\right\\}.$ (26)

In order to estimate the resolvent
$\left(\mathrm{i}t\mathcal{O}+\mathcal{C}\right)^{-1}$ uniformly in
$t\in\mathbb{R}$ we need to solve the ordinary differential equation

$\mathrm{i}t\mathcal{O}(s)v(s)+Dv^{\prime}(s)=\widetilde{f}(s)\quad(s\in(-\tfrac{1}{2},\tfrac{1}{2})),$

for any given $\widetilde{f}\in L_{2}(-\tfrac{1}{2},\tfrac{1}{2})^{d}$ whilst
respecting the boundary condition formulated in (26). Equivalently, however,
by multipliying with $D$ and using that $D^{2}=1_{n}$ (and switching signs in
$t$, which is insignificant as $t$ runs through $\mathbb{R}$ anyway) we may
consider

$-\mathrm{i}tD\mathcal{O}(s)v(s)+v^{\prime}(s)=f(s)\coloneqq
D\widetilde{f}(s)\quad(s\in(-\tfrac{1}{2},\tfrac{1}{2}))$ (27)

instead. Of course, this equation is solvable and posseses an $n$-dimensional
affine space of solutions in $H^{1}(-\frac{1}{2},\frac{1}{2})^{d}\subseteq
C[-\frac{1}{2},\frac{1}{2}]{}^{d}$ by the Picard–Lindelöf Theorem. Throughout
this section, we therefore fix the fundamental matrix $\Gamma_{t}\in
C([-\frac{1}{2},\frac{1}{2}];\mathbb{C}^{d\times d})$ associated to (27), that
is, associated to

$v^{\prime}(s)=\mathrm{i}tD\mathcal{O}(s)v(s)\in\mathbb{C}^{d}\quad(s\in(-\tfrac{1}{2},\tfrac{1}{2}))$

and subject to $\Gamma_{t}(-\frac{1}{2})=1_{d}.$ With these preparations, we
may state the main ingredient for our characterisation of exponential
stability.

######  Theorem 6.1.

Assume

$\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}<\infty.$

Then the following statements are equivalent.

  1. (i)

For all $t\in\mathbb{R}$, $(\mathrm{i}t\mathcal{O}+\mathcal{C})$ is boundedly
invertible on $L_{2}(-\frac{1}{2},\frac{1}{2})^{d}$ with
$\sup_{t\in\mathbb{R}}\|(\mathrm{i}t\mathcal{O}+\mathcal{C})^{-1}\|<\infty.$

  2. (ii)

For all $t\in\mathbb{R}$, the matrix

$S_{t}\coloneqq M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)$

is invertible and satisfies $\sup_{t\in\mathbb{R}}\|S_{t}^{-1}\|<\infty$.

###### Proof of Theorem 4.5.

The proof is an immediate consequence of the potentially more general
situation in Theorem 6.1. For this particularly note that the assumption on
$\mathcal{H}$ to be bounded taking values in the self-adjoint, strictly
positive $d\times d$-matrices with strictly positive lower bound yields the
same properties for $\mathcal{O}$ in Theorem 4.5. Hence, Theorem 6.1 is
applicable. ∎

We start with some observations for the matrix $S_{t}$ and the fundamental
matrix $\Gamma_{t}$.

######  Lemma 6.2.

Let $\Gamma_{t}$ be the fundamental matrix associated with (27) such that
$\Gamma_{t}(-\frac{1}{2})=1_{d}$ and let

$S_{t}\coloneqq M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right).$

  1. (a)

For all $t\in\mathbb{R}$,
$\left(\Gamma_{t}(s)\right)^{-1}=D\Gamma_{t}(s)^{\ast}D$ and, hence,
$\|\Gamma_{t}(s)^{-1}\|=\|\Gamma_{t}(s)\|,$ since $D$ is unitary.

  2. (b)

$\mathbb{R}\ni t\mapsto\Gamma_{t}\in
C([-\frac{1}{2},\frac{1}{2}];\mathbb{R}^{d\times d})$ is continuous and hence,
$\mathbb{R}\ni t\mapsto S_{t}\in\mathbb{R}^{d\times d}$ is continuous.

  3. (c)

If $S_{t}$ is invertible for some $t\in\mathbb{R}$, then

$\frac{\kappa_{t}}{\|M\|+1}\leq\|S_{t}^{-1}\|\leq\kappa_{t}+\kappa_{t}\|\Gamma_{t}(\tfrac{1}{2})\|+1$

where

$\kappa_{t}\coloneqq\left\|S_{t}^{-1}\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)\right\|.$

###### Proof.

(a): The statement follows from Proposition 5.8 (a) applied to
$\Pi=\Gamma_{t}$.

(b): Let $r,t\in\mathbb{R}$ and $u_{r},u_{t}$ be solutions of

$\displaystyle u_{r}^{\prime}(s)+\mathrm{i}rD\mathcal{O}(s)u_{r}(s)$
$\displaystyle=0,$ $\displaystyle
u_{t}^{\prime}(s)+\mathrm{i}tD\mathcal{O}(s)u_{t}(s)$ $\displaystyle=0$

with $u_{r}(-\tfrac{1}{2})=u_{t}(-\tfrac{1}{2})=e_{i}$ für some
$i\in\\{1,\ldots,d\\}.$ Then for all
$s\in\left[-\tfrac{1}{2},\tfrac{1}{2}\right]$

$\displaystyle\partial_{s}\|u_{t}(s)-u_{r}(s)\|^{2}$
$\displaystyle=2\operatorname{Re}\langle
u_{t}^{\prime}(s)-u_{r}^{\prime}(s),u_{t}(s)-u_{r}(s)\rangle$
$\displaystyle=2\operatorname{Re}\langle\mathrm{i}tD\mathcal{O}(s)u_{t}(s)-\mathrm{i}rD\mathcal{O}(s)u_{r}(s),u_{t}(s)-u_{r}(s)\rangle$
$\displaystyle\leq\|\mathcal{O}\|_{\infty}\|tu_{t}(s)-ru_{r}(s)\|\|u_{t}(s)-u_{r}(s)\|$
$\displaystyle\leq\|\mathcal{O}\|_{\infty}\left(|t-r|\|u_{t}(s)\|+|r|\|u_{t}(s)-u_{r}(s)\|\right)\|u_{t}(s)-u_{r}(s)\|$

and thus,

$\|u_{t}(s)-u_{r}(s)\|^{2}\leq\|\mathcal{O}\|_{\infty}|t-r|\int_{-\frac{1}{2}}^{s}\|u_{t}(\sigma)\|\|u_{t}(\sigma)-u_{r}(\sigma)\|\,\mathrm{d}\sigma+\|\mathcal{O}\|_{\infty}|r|\int_{-\frac{1}{2}}^{s}\|u_{t}(\sigma)-u_{r}(\sigma)\|^{2}\,\mathrm{d}\sigma.$

Employing Gronwall’s inequality (see also Theorem 8.14 below), we infer

$\displaystyle\|u_{t}(s)-u_{r}(s)\|^{2}$
$\displaystyle\leq\|\mathcal{O}\|_{\infty}\mathrm{e}^{\|\mathcal{O}\|_{\infty}|r|(s+\frac{1}{2})}|t-r|\int_{-\frac{1}{2}}^{s}\|u_{t}(\sigma)\|\|u_{t}(\sigma)-u_{r}(\sigma)\|\,\mathrm{d}r$
$\displaystyle\leq\|\mathcal{O}\|_{\infty}\mathrm{e}^{\|\mathcal{O}\|_{\infty}|r|(s+\frac{1}{2})}|t-r|\int_{-\frac{1}{2}}^{s}\|u_{t}(\sigma)\|\,\mathrm{d}r\|u_{t}-u_{r}\|_{\infty}.$

Supremum over $s$ yields

$\|u_{t}-u_{r}\|_{\infty}\leq\|\mathcal{O}\|_{\infty}\mathrm{e}^{\|\mathcal{O}\|_{\infty}|r|}|t-r|\int_{-\frac{1}{2}}^{\frac{1}{2}}\|u_{t}(\sigma)\|\,\mathrm{d}\sigma\to
0\quad(r\to t),$

which shows the continuity of $t\mapsto\Gamma_{t}.$

(c) Since clearly $\kappa_{t}\leq\|S_{t}^{-1}\|(\|M\|+1)$ the first inequality
follows. For the second inequality, let $y\in\mathbb{R}^{d}$ and $w\coloneqq
S_{t}^{-1}y.$ We decompose
$y=(\check{y},\hat{y})\in\mathbb{R}^{k}\times\mathbb{R}^{d-k}$ and set
$w_{1}\coloneqq S_{t}^{-1}(\check{y},0)$ and $w_{2}\coloneqq
S_{t}^{-1}(0,\hat{y}).$ Then $w=w_{1}+w_{2}.$ Moreover, we compute

$S_{t}w_{2}=\left(\begin{array}[]{c}0\\\
\hat{y}\end{array}\right)=-\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)\left(\begin{array}[]{c}0\\\
\hat{y}\end{array}\right)$

and hence, $\|w_{2}\|\leq\kappa_{t}\|(0,\hat{y})\|\leq\kappa_{t}\|y\|.$
Moreover,

$\displaystyle S_{t}\left(\begin{array}[]{c}\check{y}\\\ 0\end{array}\right)$
$\displaystyle=M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)\left(\begin{array}[]{c}\check{y}\\\
0\end{array}\right)$ $\displaystyle=\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)\Gamma_{t}(\tfrac{1}{2})\left(\begin{array}[]{c}\check{y}\\\
0\end{array}\right)-\left(\begin{array}[]{c}\check{y}\\\ 0\end{array}\right)$
$\displaystyle=\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)\Gamma_{t}(\tfrac{1}{2})\left(\begin{array}[]{c}\check{y}\\\
0\end{array}\right)-S_{t}w_{1}.$

Hence,

$\|w_{1}\|\leq\|w_{1}+(\check{y},0)\|+\|(\check{y},0)\|\leq\left(\kappa_{t}\|\Gamma_{t}(\tfrac{1}{2})\|+1\right)\|y\|$

and thus

$\|w\|\leq\|w_{1}\|+\|w_{2}\|\leq\left(\kappa_{t}+\kappa_{t}\|\Gamma_{t}(\tfrac{1}{2})\|+1\right)\|y\|,$

which shows the asserted estimate. ∎

We are now in the position to prove the main result of the present section.

###### Proof of Theorem 6.1.

(i) $\Rightarrow$ (ii): Let $y\in\mathbb{R}^{d}$ and $t\in\mathbb{R}$ such
that $S_{t}y=0.$ We define for $s\in\left[-\frac{1}{2},\frac{1}{2}\right]$

$v(s)\coloneqq\Gamma_{t}(s)y.$

Then $v^{\prime}(s)=\mathrm{i}tD\mathcal{O}(s)v(s)$ and $v(-\frac{1}{2})=y.$
Moreover, using $S_{t}y=0$, we infer

$\displaystyle M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)$
$\displaystyle=M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)y$
$\displaystyle=\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)y$
$\displaystyle=\left(\begin{array}[]{c}\check{v}(-\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)$

and, thus, $v\in\operatorname{dom}(\mathcal{C}).$ Hence, we infer
$0=\left(-\mathrm{i}t\mathcal{O}+\mathcal{C}\right)v$ and thus, $v=0$ by
invertibility of $\mathrm{i}(-t)\mathcal{O}+\mathcal{C}.$ Hence,
$y=v(-\frac{1}{2})=0$ and, thus, $S_{t}$ is invertible. Assume now that
$\sup_{t\in\mathbb{R}}\|S_{t}^{-1}\|=\infty$. Then we find a sequence
$(t_{n})_{n\in\mathbb{N}}$ with $|t_{n}|\to\infty$ and
$\|S_{t_{n}}^{-1}\|\to\infty$ (note that $\sup_{t\in K}\|S_{t}^{-1}\|<\infty$
for each compact set $K\subseteq\mathbb{R}$ by continuity of $t\mapsto S_{t}$,
see Lemma 6.2(b)). By Lemma 6.2(c) it follows that

$\left\|S_{t_{n}}^{-1}\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)\right\|\to\infty$

Thus, by the finite-dimensional analogue of the uniform boundedness principle,
we find $y\in\mathbb{R}^{d}$ with

$\left\|S_{t_{n}}^{-1}\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)y\right\|\to\infty.$

We set $\widetilde{y}_{n}\coloneqq
S_{t_{n}}^{-1}\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)y$ as well as
$y_{n}\coloneqq\frac{1}{\|\widetilde{y}_{n}\|}\widetilde{y}_{n}$ and
$z_{n}\coloneqq\frac{1}{\|\widetilde{y}_{n}\|}y$ for each $n\in\mathbb{N}.$
Then $S_{t_{n}}y_{n}=\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)z_{n}$ and $z_{n}\to 0.$ We define $f_{n}\in
L_{2}(-\frac{1}{2},\frac{1}{2})^{d}$ by

$f_{n}(s)\coloneqq-\Gamma_{t_{n}}(s)\Gamma_{t_{n}}(\tfrac{1}{2})^{-1}z_{n}$

and $v_{n}\in L_{2}(-\frac{1}{2},\frac{1}{2})^{d}$ by

$v_{n}(s)\coloneqq\Gamma_{t_{n}}(s)y_{n}+\Gamma_{t_{n}}(s)\int_{-\frac{1}{2}}^{s}\Gamma_{t_{n}}(\sigma)^{-1}f_{n}(\sigma)\,\mathrm{d}\sigma=\Gamma_{t_{n}}(s)y_{n}-\Gamma_{t_{n}}(s)\Gamma_{t_{n}}(\tfrac{1}{2})^{-1}(s+\tfrac{1}{2})z_{n}.$

Then $v_{n}^{\prime}(s)+\mathrm{i}(-t_{n})D\mathcal{O}(s)v_{n}(s)=f_{n}(s)$
and $v_{n}(-\frac{1}{2})=y_{n}.$ Moreover,

$v_{n}(\tfrac{1}{2})=\Gamma_{t_{n}}(\tfrac{1}{2})y_{n}-z_{n}$

and thus,

$\displaystyle M\left(\begin{array}[]{c}\check{v}_{n}(\frac{1}{2})\\\
\hat{v}_{n}(-\frac{1}{2})\end{array}\right)-\left(\begin{array}[]{c}\check{v}_{n}(-\frac{1}{2})\\\
\hat{v}_{n}(\frac{1}{2})\end{array}\right)$
$\displaystyle=M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\left(\Gamma_{t_{n}}(\tfrac{1}{2})y_{n}-z_{n}\right)+M\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)y_{n}$
$\displaystyle\quad-\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)y_{n}-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\left(\Gamma_{t_{n}}(\tfrac{1}{2})y_{n}-z_{n}\right)$
$\displaystyle=M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t_{n}}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)y_{n}-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t_{n}}(\tfrac{1}{2})\right)y_{n}$
$\displaystyle\quad-\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)z_{n}$
$\displaystyle=S_{t_{n}}y_{n}-\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)z_{n}=0.$

We get $v_{n}\in\operatorname{dom}(\mathcal{C})$ and hence,
$\left(\mathcal{C}+\mathrm{i}(-t_{n})\mathcal{O}\right)v_{n}=Df_{n},$ which
yields
$v_{n}=(\mathrm{i}\left(-t_{n}\right)\mathcal{O}+\mathcal{C})^{-1}Df_{n}.$
However, since $\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|<\infty$,
$\|\Gamma_{t}(\tfrac{1}{2})^{-1}\|=\|\Gamma_{t}(\tfrac{1}{2})\|,$ by Lemma 6.2
(a), and $z_{n}\to 0$, we infer that $f_{n}\to 0$. Thus, since
$\sup_{t\in\mathbb{R}}\|(\mathrm{i}\mathcal{O}+\mathcal{C})^{-1}\|<\infty,$ by
(i), $v_{n}\to 0$. In particular, for a suitable subsequence (which we do not
relabel), we obtain $v_{n}(s)\to 0$ for almost every
$s\in[-\frac{1}{2},\frac{1}{2}].$ For these $s$ we obatin

$\|y_{n}\|\leq\|\Gamma_{t_{n}}(s)^{-1}v_{n}(s)\|+\|\Gamma_{t_{n}}(\tfrac{1}{2})^{-1}\|\|z_{n}\|.$

Now, since $\Gamma_{t}$ is uniformly bounded, so is $\Gamma_{t}(\cdot)^{-1}$
by Lemma 6.2(a), and hence, $y_{n}\to 0,$ contradicting $\|y_{n}\|=1.$ Hence
$\|S_{t}^{-1}\|$ is uniformly bounded in $t$.

(ii) $\Rightarrow$ (i): Let $t\in\mathbb{R}$ and $f\in
L_{2}(-\frac{1}{2},\frac{1}{2})^{d}.$ We define

$z\coloneqq-\Gamma_{t}(\tfrac{1}{2})\int_{-\frac{1}{2}}^{\frac{1}{2}}\Gamma_{t}(\sigma)^{-1}f(\sigma)\,\mathrm{d}\sigma$

and $y\coloneqq S_{t}^{-1}\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)z.$ Moreover, we set

$v(s)\coloneqq\Gamma_{t}(s)y+\Gamma_{t}(s)\int_{-\frac{1}{2}}^{s}\Gamma_{t}(\sigma)^{-1}f(\sigma)\,\mathrm{d}\sigma\quad(s\in(-\tfrac{1}{2},\tfrac{1}{2})).$

Then clearly, $v^{\prime}(s)+\mathrm{i}(-t)D\mathcal{O}(s)v(s)=f(s)$ and
$v(-\frac{1}{2})=y.$ Moreover, $v(\frac{1}{2})=\Gamma_{t}(\frac{1}{2})y-z$ and
thus,

$\displaystyle M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)-\left(\begin{array}[]{c}\check{v}(-\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)$
$\displaystyle=M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)y-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)y$
$\displaystyle\quad-\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)z$
$\displaystyle=S_{t}y-\left(M\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)z=0.$

Hence, $v\in\operatorname{dom}(\mathcal{C}).$ Consequently,
$(\mathrm{i}(-t)\mathcal{O}+\mathcal{C})v=Df.$ Moreover, with
$\Lambda_{0}\coloneqq\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}$ and
$\Lambda_{1}\coloneqq\sup_{t\in\mathbb{R}}\|S_{t}^{-1}\|$ ($<\infty$, by (ii)
) we estimate

$\displaystyle\|v\|_{L_{2}(-\frac{1}{2},\frac{1}{2})^{d}}$
$\displaystyle\leq\|\Gamma_{t}\|_{\infty}\left(\|y\|+\|\Gamma_{t}^{-1}\|_{\infty}\|f\|\right)$
$\displaystyle\leq\|\Gamma_{t}\|_{\infty}\|S_{t}^{-1}\|(\|M\|+1)\|z\|+\|\Gamma_{t}\|_{\infty}^{2}\|f\|_{L_{2}}$
$\displaystyle\leq 2\Lambda_{0}\Lambda_{1}\|z\|+\Lambda_{0}^{2}\|f\|_{L_{2}}$

where we have used Lemma 6.2(a). Now, since

$\|z\|\leq\|\Gamma_{t}\|_{\infty}^{2}\|f\|_{L_{2}}\leq\Lambda_{0}^{2}\|f\|_{L_{2}},$

uniform boundedness of $(\mathrm{i}t\mathcal{O}+\mathcal{C})^{-1}$ follows, if
$(\mathrm{i}t\mathcal{O}+\mathcal{C})$ is one-to-one for all $t\in\mathbb{R}$.
For this, let $t\in\mathbb{R}$ and $v\in\operatorname{dom}(\mathcal{C})$ with
$\left(\mathrm{i}(-t)\mathcal{O}+\mathcal{C}\right)v=0.$ Then

$v(s)=\Gamma_{t}(s)v(-\tfrac{1}{2})$

and hence,

$\displaystyle 0$
$\displaystyle=M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)-\left(\begin{array}[]{c}\check{v}(-\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)$
$\displaystyle=M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)v(-\tfrac{1}{2})$
$\displaystyle\quad-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)v(-\tfrac{1}{2})$
$\displaystyle=S_{t}v(-\tfrac{1}{2}).$

Since, by (ii), $S_{t}$ is invertible, $v(-\frac{1}{2})=0$ and, hence, $v=0.$
∎

##  7 Proof of Main Result – Theorem 3.1

We aim to transfer the characterisation result to the original setting. For
this we briefly recall that

$\displaystyle\mathcal{A}\colon\operatorname{dom}(\mathcal{A})\subseteq H$
$\displaystyle\to H$ $\displaystyle u$ $\displaystyle\mapsto
P_{1}(\mathcal{H}u)^{\prime}+P_{0}\mathcal{H}u,$

is a port-Hamiltonian operator as in (1) such that $-\mathcal{A}$ generates a
$C_{0}$-semi-group of contractions. By Theorem 2.1 we find a full rank matrix
$W_{B}\in\mathbb{R}^{d\times 2d}$ with (2); that is

$W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\left(\begin{array}[]{cc}0&1_{d}\\\
1_{d}&0\end{array}\right)\left(W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\right)^{\ast}\geq 0$

and

$\operatorname{dom}(\mathcal{A})=\left\\{u\in H\,;\,\mathcal{H}u\in
H^{1}(a,b)^{d},\,W_{B}\left(\begin{array}[]{c}\left(\mathcal{H}u\right)(b)\\\
\left(\mathcal{H}u\right)(a)\end{array}\right)=0\right\\}.$

Let $\Phi_{t}$ be the fundamental solution associated with

$u^{\prime}(x)=\mathrm{i}tP_{1}^{-1}\mathcal{H}(x)u(x)\quad(x\in(a,b))$

such that $\Phi_{t}(a)=1_{d}$; condition (B) states
$\sup_{t\in\mathbb{R}}\|\Phi_{t}\|_{\infty}<\infty$. Before we prove Theorem
3.1 with the help of Corollary 4.6, we relate the matrix
$M\in\mathbb{R}^{d\times d}$ with $\|M\|\leq 1$ in the formulation of the
boundary conditions for $\mathcal{C}$ given by (9) in (14) to the matrix
$W_{B}$ stated to exist here. This is done next.

######  Proposition 7.1.

There exists an invertible matrix $L\in\mathbb{R}^{n\times n}$ such that

$L\left(\begin{array}[]{cc}M&-1_{d}\end{array}\right)=W_{B}\left(\begin{array}[]{cc}UQ&0\\\
0&UQ\end{array}\right)\left(\begin{array}[]{cccc}1_{k}&0&0&0\\\
0&0&0&1_{d-k}\\\ 0&0&1_{k}&0\\\ 0&1_{d-k}&0&0\end{array}\right).$

###### Proof.

We recall from (9) and (10)

$\mathcal{C}=QU^{\ast}\Psi S^{\ast}\mathcal{A}S\Psi^{\ast}UQ,$

and

$\operatorname{dom}(\mathcal{C})=\operatorname{dom}(\mathcal{A}S\Psi^{\ast}UQ).$

Using the domain characterisations in (A2) and (14) of
$\operatorname{dom}(\mathcal{A})$ and $\operatorname{dom}(\mathcal{C})$,
respectively, we know that for $v\in H^{1}(-\frac{1}{2},\frac{1}{2})^{d}$ we
have

$\displaystyle M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)-\left(\begin{array}[]{c}\check{v}(-\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)=0$ $\displaystyle\Leftrightarrow
v\in\operatorname{dom}(\mathcal{C})$ $\displaystyle\Leftrightarrow
S\Psi^{\ast}UQv\in\operatorname{dom}(\mathcal{A})$
$\displaystyle\Leftrightarrow
W_{B}\left(\begin{array}[]{c}(\mathcal{H}S\Psi^{\ast}UQv)(b)\\\
(\mathcal{H}S\Psi^{\ast}UQv)(a)\end{array}\right)=0.$

Now, we observe that

$\left(\mathcal{H}S\Psi^{\ast}UQv\right)(x)=\sqrt{\frac{1}{b-a}}\mathcal{H}(x)\mathcal{H}(x)^{-1}UQv\left(\frac{1}{b-a}(x-a)-\frac{1}{2}\right).$

Hence,

$M\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\end{array}\right)-\left(\begin{array}[]{c}\check{v}(-\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)=0\Leftrightarrow
W_{B}\left(\begin{array}[]{c}UQv(\frac{1}{2})\\\
UQv(-\frac{1}{2})\end{array}\right)=0,$

which can be reformulated as

$\left(\begin{array}[]{cc}M&-1_{d}\end{array}\right)\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\\\ \check{v}(-\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)=0\Leftrightarrow
W_{B}\left(\begin{array}[]{cc}UQ&0\\\
0&UQ\end{array}\right)\left(\begin{array}[]{cccc}1_{k}&0&0&0\\\
0&0&0&1_{d-k}\\\ 0&0&1_{k}&0\\\
0&1_{d-k}&0&0\end{array}\right)\left(\begin{array}[]{c}\check{v}(\frac{1}{2})\\\
\hat{v}(-\frac{1}{2})\\\ \check{v}(-\frac{1}{2})\\\
\hat{v}(\frac{1}{2})\end{array}\right)=0.$

Thus, the matrices $\left(\begin{array}[]{cc}M&-1_{d}\end{array}\right)$ and
$W_{B}\left(\begin{array}[]{cc}UQ&0\\\
0&UQ\end{array}\right)\left(\begin{array}[]{cccc}1_{k}&0&0&0\\\
0&0&0&1_{d-k}\\\ 0&0&1_{k}&0\\\ 0&1_{d-k}&0&0\end{array}\right)$ have the same
kernels and thus, we find an invertible matrix $L\in\mathbb{R}^{d\times d}$ as
desired. ∎

######  Proposition 7.2.

Let $S_{t}$ be as in Corollary 4.6 and $L\in\mathbb{R}^{d\times d}$ as in
Proposition 7.1. Then

$LS_{t}Q^{-1}U^{\ast}=W_{B}\left(\begin{array}[]{c}\Phi_{t}(b)\\\
1_{d}\end{array}\right).$

###### Proof.

We use Proposition 5.3 and compute

$\displaystyle S_{t}$
$\displaystyle=M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)\Gamma_{t}(\tfrac{1}{2})+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\Gamma_{t}(\tfrac{1}{2})\right)$
$\displaystyle=M\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)Q^{-1}U^{\ast}\Phi_{t}(b)UQ+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\right)$
$\displaystyle\quad\quad-\left(\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)Q^{-1}U^{\ast}\Phi_{t}(b)UQ\right)$
$\displaystyle=\left(\begin{array}[]{cc}M&-1_{d}\end{array}\right)\left(\begin{array}[]{c}\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)Q^{-1}U^{\ast}\Phi_{t}(b)UQ+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\\\ \left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)Q^{-1}U^{\ast}\Phi_{t}(b)UQ\end{array}\right).$

Hence,

$\displaystyle LS_{t}$ $\displaystyle=W_{B}\left(\begin{array}[]{cc}UQ&0\\\
0&UQ\end{array}\right)\left(\begin{array}[]{cccc}1_{k}&0&0&0\\\
0&0&0&1_{d-k}\\\ 0&0&1_{k}&0\\\
0&1_{d-k}&0&0\end{array}\right)\left(\begin{array}[]{c}\left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)Q^{-1}U^{\ast}\Phi_{t}(b)UQ+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)\\\ \left(\begin{array}[]{cc}1_{k}&0\\\
0&0\end{array}\right)+\left(\begin{array}[]{cc}0&0\\\
0&1_{d-k}\end{array}\right)Q^{-1}U^{\ast}\Phi_{t}(b)UQ\end{array}\right)$
$\displaystyle=W_{B}\left(\begin{array}[]{cc}UQ&0\\\
0&UQ\end{array}\right)\left(\begin{array}[]{c}Q^{-1}U^{\ast}\Phi_{t}(b)UQ\\\
1_{d}\end{array}\right)$
$\displaystyle=W_{B}\left(\begin{array}[]{c}\Phi_{t}(b)\\\
1_{d}\end{array}\right)UQ,$

from which the asserted formula follows. ∎

Using this representation of $S_{t},$ we can formulate our main theorem; which
is a compressed formulation of Theorem 3.1.

######  Theorem 7.3.

Let $P_{1},P_{0}\in\mathbb{R}^{d\times d}$ such that $P_{1}=P_{1}^{\ast}$ is
invertible, $P_{0}^{*}=-P_{0}$, and $\mathcal{H}:[a,b]\to\mathbb{R}^{d\times
d}$ be a measurable function such that

$\exists m,M>0\>\forall
x\in[a,b]:\>m\leq\mathcal{H}(x)=\mathcal{H}(x)^{\ast}\leq M.$

We consider $\mathcal{A}\subseteq
P_{1}\mathcal{H}\partial_{x}+P_{0}\mathcal{H}$ on the Hilbert space
$H\coloneqq(L_{2}(a,b)^{d},\langle\mathcal{H}\cdot,\cdot\rangle_{L_{2}})$ with
domain

$\operatorname{dom}(\mathcal{A})=\left\\{u\in H\,;\,\mathcal{H}u\in
H^{1}([a,b])^{d},\,W_{B}\left(\begin{array}[]{c}\left(\mathcal{H}u\right)(b)\\\
\left(\mathcal{H}u\right)(a)\end{array}\right)=0\right\\}$

for a matrix $W_{B}\in\mathbb{R}^{d\times 2d}$ satisfying
$\operatorname{rk}W_{B}=d$ and

$W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\left(\begin{array}[]{cc}0&1_{d}\\\
1_{d}&0\end{array}\right)\left(W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\right)^{\ast}\geq 0.$

Moreover, assume that the fundamental matrix $\Phi_{t}$ associated to the ODE-
system

$u^{\prime}(x)=\mathrm{i}tP_{1}^{-1}\mathcal{H}(x)u(x)\quad(x\in[a,b])$

with $\Phi_{t}(a)=1_{d}$ for each $t\in\mathbb{R}$ satisfies
$\sup_{t\in\mathbb{R}}\|\Phi_{t}\|<\infty.$ Then the following statements are
equivalent:

  1. (i)

the (contraction-)semi-group $\left(\mathrm{e}^{-t\mathcal{A}}\right)_{t\geq
0}$ is exponentially stable,

  2. (ii)

for each $t\in\mathbb{R}$ the matrix

$T_{t}\coloneqq W_{B}\left(\begin{array}[]{c}\Phi_{t}(b)\\\
I_{d}\end{array}\right)$

is invertible with $\sup_{t\in\mathbb{R}}\|T_{t}^{-1}\|<\infty.$

###### Proof of Theorem 7.3 and Theorem 3.1.

Recall that by Theorem 4.1, without loss of generality $P_{0}=0.$ Now, the
proof follows from Corollary 4.6 by using Proposition 5.3, and Proposition 7.1
and Proposition 7.2. ∎

##  8 The Condition (B)

This section is devoted to a discussion of condition (B). For this, throughout
this section, we let $a,b\in\mathbb{R},$ $a<b$ and

$\mathcal{H}\colon\left[a,b\right]\to\mathbb{R}^{d\times d}\in
L_{\infty}(a,b;\mathbb{R}^{d\times d})$

satisfying $\mathcal{H}(x)=\mathcal{H}(x)^{*}\geq m$ for some $m>0$.
Furthermore, let $P_{1}=P_{1}^{*}\in\mathbb{R}^{d\times d}$ invertible. For
$t\in\mathbb{R}$ we define the fundamental matrix $\Phi_{t}\in
C([-\tfrac{1}{2},\tfrac{1}{2}];\mathbb{C}^{d\times d})$ associated to

$u^{\prime}(x)=\mathrm{i}tP_{1}^{-1}\mathcal{H}(x)u(x)\in\mathbb{C}^{d}\quad(x\in(a,b))$

subject to $\Phi_{t}(a)=1_{d}$. Then, as in Section 2, we introduce invertible
matrices $U,Q\in\mathbb{C}^{d\times d}$, $Q$ diagonal and real and $U$ unitary
such that (16) holds, that is,

$\frac{1}{b-a}P_{1}=UQ^{-1}DQ^{-1}U^{*}$

where

$D=\left(\begin{array}[]{cc}1_{k}&0\\\ 0&-1_{d-k}\end{array}\right)$

for some $k\in\\{0,\ldots,d\\}.$ With

$\mathcal{O}(s)\coloneqq
QU^{\ast}\mathcal{H}\left((b-a)s+\frac{a+b}{2}\right)UQ,$ (28)

$s\in(-\tfrac{1}{2},\tfrac{1}{2})$, for $t\in\mathbb{R}$, we define the
fundamental matrix $\Gamma_{t}\in
C([-\tfrac{1}{2},\tfrac{1}{2}];\mathbb{C}^{d\times d})$ associated to

$v^{\prime}(s)=\mathrm{i}tD\mathcal{O}(s)v(s)\in\mathbb{C}^{d}\quad(s\in(-\tfrac{1}{2},\tfrac{1}{2}))$

subject to $\Gamma_{t}(-\tfrac{1}{2})=1_{d}$. In this section, we focus on the
conditions

$\sup_{t\in\mathbb{R}}\|\Phi_{t}\|_{\infty}<\infty$ (29)

and

$\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}<\infty.$ (30)

Condition (B) is the same as (29) and the two conditions (29) and (30) are
equivalent by Proposition 5.3. Whilst we do not yet know of any
counterexamples, we only managed to provide sufficient conditions on $D$ and
$\mathcal{O}$, or $P_{1}$ and $\mathcal{H}$ warranting (30) and (29). These
conditions either require some compatibility properties of $D$ and
$\mathcal{O}$ ($P_{1}$ and $\mathcal{H}$) or regularity properties for
$\mathcal{O}$ (or $\mathcal{H}$).

###  8.1 A compatibility condition of $\mathcal{H}$ and $P_{1}$

We start off with a condition irrespective of any regularity of $\mathcal{H}$.
In terms of $D$ and $\mathcal{O}$ the condition reads as follows.

######  Theorem 8.1.

Assume that, for almost every $s\in(-\tfrac{1}{2},\tfrac{1}{2}),$

$D\mathcal{O}(s)=\mathcal{O}(s)D.$

Then (30) holds.

###### Proof.

Let $v\in H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d}$ be a solution of

$v^{\prime}(s)=\mathrm{i}tD\mathcal{O}(s)v(s).$

Since $D\mathcal{O}(s)$ is selfadjoint as product of two selfadjoint commuting
matrices, we obtain for $s\in(-\tfrac{1}{2},\tfrac{1}{2})$

$\partial_{s}\|v(s)\|^{2}=2\langle
v(s),v^{\prime}(s)\rangle=2\mathrm{i}t\langle
v(s),D\mathcal{O}(s)v(s)\rangle\in\mathrm{i}\mathbb{R}.$

Hence, $s\mapsto\|v(s)\|$ is constant and so, for all $y\in\mathbb{R}^{d}$ and
$s\in(-\tfrac{1}{2},\tfrac{1}{2})$

$\|\Gamma_{t}(s)y\|=\|\Gamma_{t}(-\tfrac{1}{2})y\|=\|y\|$

so that $\sup_{t\in\mathbb{R}}\|\Gamma_{t}\|_{\infty}=1<\infty$. ∎

For the formulation of Theorem 8.1 in terms of $P_{1}$ and $\mathcal{H}$ we
introduce the following spectral subspaces of $P_{1}$. We set

$\displaystyle E_{+}$
$\displaystyle\coloneqq\operatorname{lin}\\{x\in\mathbb{R}^{d}\,;\,\exists\lambda>0:P_{1}x=\lambda
x\\},$ $\displaystyle E_{-}$
$\displaystyle\coloneqq\operatorname{lin}\\{x\in\mathbb{R}^{d}\,;\,\exists\lambda<0:P_{1}x=\lambda
x\\}.$

Then $E_{+}\oplus E_{-}=\mathbb{R}^{d}$ in the sense of an orthogonal sum,
since $P_{1}$ is selfadjoint and invertible.

Theorem 8.1 in the language $P_{1}$ and $\mathcal{H}$ is the following.

######  Theorem 8.2.

Assume that, for almost every $x\in(a,b)$,

$\mathcal{H}(x)[E_{+}]\subseteq E_{+}.$

Then (29) holds.

The proof of Theorem 8.2 is an immediate consequence of Theorem 8.1,
Proposition 5.3 and part (b) of the following lemma.

######  Lemma 8.3.

(a) Let
$\mathcal{O}_{0}=\mathcal{O}_{0}^{*},\mathcal{H}_{0}=\mathcal{H}_{0}^{*}\in\mathbb{R}^{d\times
d}$ and assume that

$\mathcal{H}_{0}=UQ^{-1}\mathcal{O}_{0}Q^{-1}U^{*}.$

Then

$D\mathcal{O}_{0}=\mathcal{O}_{0}D\iff\mathcal{H}_{0}[E_{+}]\subseteq E_{+}.$

(b) The following conditions are equivalent.

  1. (i)

For almost every $s\in(-\tfrac{1}{2},\tfrac{1}{2}),$
$D\mathcal{O}(s)=\mathcal{O}(s)D.$

  2. (ii)

For almost every $x\in\left[a,b\right]$, $\mathcal{H}(x)[E_{+}]\subseteq
E_{+}.$

###### Proof.

(a) We recall that
$Q=\operatorname{diag}(\frac{1}{\sqrt{|\lambda_{i}|}})_{i\in\\{1,\ldots,d\\}},$
where $\lambda_{1},\ldots,\lambda_{k}>0$ and
$\lambda_{k+1},\ldots,\lambda_{d}<0$ are the eigenvalues of $P_{1}$ counted
with multiplicity. Moreover, $U\in\mathbb{R}^{d\times d}$ is the unitary
matrix, where the rows $u_{j}$ of $U$ are orthonormal vectors with
$P_{1}u_{j}=\lambda_{j}u_{j}.$

Assume first that $D\mathcal{O}_{0}=\mathcal{O}_{0}D$; that is
$\mathcal{O}_{0}=\left(\begin{array}[]{cc}\mathcal{O}_{1}&0\\\
0&\mathcal{O}_{2}\end{array}\right)$ for some
$\mathcal{\mathcal{O}}_{1}\in\mathbb{R}^{k\times k}$ and
$\mathcal{O}_{2}\in\mathbb{R}^{(d-k)\times(d-k)}$, and let
$y\in\mathbb{R}^{d}$ with $P_{1}y=\lambda y$ for some $\lambda>0.$ Since $y\in
E_{+},$ we find $\mu_{1},\ldots,\mu_{k}\in\mathbb{R}$ such that
$y=\sum_{j=1}^{k}\mu_{j}u_{j}.$ Hence, it suffices to show that
$\mathcal{H}_{0}u_{j}\in E_{+}$ for $j\in\\{1,\ldots,k\\}.$ We have that

$\displaystyle\mathcal{H}_{0}u_{j}$
$\displaystyle=UQ^{-1}\mathcal{O}_{0}Q^{-1}U^{\ast}u_{j}$
$\displaystyle=UQ^{-1}\mathcal{O}_{0}\sqrt{|\lambda_{j}|}e_{j}$
$\displaystyle\in UQ^{-1}[\mathbb{R}^{k}\times\\{0\\}]\subseteq E_{+}.$

Assume conversely, that $\mathcal{H}_{0}[E_{+}]\subseteq E_{+}$. Due to the
selfadjointness of $\mathcal{H}_{0}$ we infer that
$\mathcal{H}_{0}[E_{-}]\subseteq E_{-}$ as $E_{-}=E_{+}^{\bot}.$ We have to
show that $\mathcal{O}_{0}e_{j}\in\mathbb{R}^{k}\times\\{0\\}$ and
$\mathcal{O}_{0}e_{\ell}\in\\{0\\}\times\mathbb{R}^{d-k}$ for all
$j\in\\{1,\ldots,k\\}$ and $\ell\in\\{k+1,\ldots,d\\}.$ This, however follows
from

$\displaystyle\mathcal{O}_{0}e_{j}$
$\displaystyle=QU^{\ast}\mathcal{H}_{0}UQe_{j}$
$\displaystyle=\frac{1}{\sqrt{|\lambda_{j}|}}QU^{\ast}\mathcal{H}_{0}u_{j}$
$\displaystyle\in QU^{\ast}[E_{+}]\subseteq\mathbb{R}^{k}\times\\{0\\}$

and analogously $\mathcal{O}_{0}e_{\ell}\in\\{0\\}\times\mathbb{R}^{d-k}$.

(b) The statement follows from (a) using
$\mathcal{H}(x)=UQ^{-1}\mathcal{O}\left(\frac{1}{b-a}x-\frac{1}{b-a}a-\frac{1}{2}\right)Q^{-1}U^{\ast}$
for almost every $x\in(a,b)$. ∎

######  Example 8.4.

Let $\mathcal{H}$ be _scalar_ -_valued_ ; i.e., there exists bounded scalar
function $h\colon\left[a,b\right]\to\mathbb{R}$ such that
$\inf_{x\in\left[a,b\right]}h(x)>0$ with $\mathcal{H}(x)=h(x)1_{d}$ for almost
every $\mathcal{H}(x)=h(x)1_{d}$. Then condition (ii) in Lemma 8.3 (b) is
satisfied and hence, (B) holds for the corresponding $(\Phi_{t})_{t}$.

Next, we provide a proof for Theorem 1.3. For this, we provide the detailed
statement next,

######  Corollary 8.5.

Let $\mathcal{H}$ be scalar-valued. Let
$P_{0}^{*}=-P_{0}\in\mathbb{R}^{d\times d}$. Consider $\mathcal{A}\subseteq
P_{1}\mathcal{H}\partial_{x}+P_{0}\mathcal{H}$ on the Hilbert space
$H\coloneqq(L_{2}(a,b)^{d},\langle\mathcal{H}\cdot,\cdot\rangle_{L_{2}})$ with
domain

$\operatorname{dom}(\mathcal{A})=\left\\{u\in H\,;\,\mathcal{H}u\in
H^{1}([a,b])^{d},\,W_{B}\left(\begin{array}[]{c}\left(\mathcal{H}u\right)(b)\\\
\left(\mathcal{H}u\right)(a)\end{array}\right)=0\right\\}$

for a matrix $W_{B}\in\mathbb{R}^{d\times 2d}$ satisfying
$\operatorname{rk}W_{B}=d$ and

$W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\left(\begin{array}[]{cc}0&1_{d}\\\
1_{d}&0\end{array}\right)\left(W_{B}\left(\begin{array}[]{cc}-P_{1}&P_{1}\\\
1_{d}&1_{d}\end{array}\right)^{-1}\right)^{\ast}\geq 0.$

Then the following statements are equivalent:

  1. (i)

the (contraction-)semi-group $\left(\mathrm{e}^{-t\mathcal{A}}\right)_{t\geq
0}$ is exponentially stable,

  2. (ii)

for each $t\in\mathbb{R}$ the matrix

$\tau_{t}\coloneqq
W_{B}\left(\begin{array}[]{c}\mathrm{e}^{\mathrm{i}tP_{1}^{-1}}\\\
1_{d}\end{array}\right)$

is invertible with $\sup_{t\in\mathbb{R}}\|\tau_{t}^{-1}\|<\infty.$

###### Proof of Theorem 1.3 and Corollary 8.5.

By Theorem 7.3 we need to look at $\Phi_{t}$ for scalar-valued $\mathcal{H}$.
Thus, let $\mathcal{H}=h1_{d}$ for some scalar function $h$. Then
differentiation shows that

$\Phi_{t}(x)=\mathrm{e}^{\mathrm{i}t\int_{a}^{x}h(\sigma)\,\mathrm{d}\sigma
P_{1}^{-1}}.$

As $h$ is scalar, by Example 8.4, $t\mapsto\|\Phi_{t}\|_{\infty}$ is bounded.
Since
$\Phi_{t}(b)=\mathrm{e}^{\mathrm{i}t\int_{a}^{b}h(\sigma)\,\mathrm{d}\sigma
P_{1}^{-1}}$ and $\int_{a}^{b}h(\sigma)\,\mathrm{d}\sigma\neq 0$, the second
condition here is equivalent to the second one in Theorem 7.3. This shows the
assertion. ∎

With the results of this section, we can also prove a theorem from the
introduction.

###### Proof of Theorem 1.4.

The claim follows using Example 8.4 and Theorem 5.1. ∎

The next example provides a set-up for which the Hamiltonian density can be as
rough as $L_{\infty}$, but the corresponding port-Hamiltonian operator still
generates an exponentially stable semi-group.

######  Example 8.6.

Let

$\displaystyle M$
$\displaystyle=\left(\begin{array}[]{cc}\tfrac{1}{2}&-\tfrac{1}{2}\\\
\tfrac{1}{2}&\tfrac{1}{2}\end{array}\right),\quad
P_{1}=\left(\begin{array}[]{cc}1&0\\\ 0&-1\end{array}\right),\quad
W_{B}=\left(\begin{array}[]{cc}\left(\begin{array}[]{cc}\tfrac{1}{2}&0\\\
\tfrac{1}{2}&-1\end{array}\right)&\left(\begin{array}[]{cc}-1&-\tfrac{1}{2}\\\
0&\tfrac{1}{2}\end{array}\right)\end{array}\right)$

and $\mathcal{H}$ be scalar-valued. Then the corresponding port-Hamiltonian
$-\mathcal{A}$ generates a contraction semi-group (see Remark 2.4).
Furthermore, using the formula in (ii) in Corollary 8.5, we get

$\displaystyle\tau_{t}$
$\displaystyle=\left(\begin{array}[]{cc}\left(\begin{array}[]{cc}\tfrac{1}{2}&0\\\
\tfrac{1}{2}&-1\end{array}\right)&\left(\begin{array}[]{cc}-1&-\tfrac{1}{2}\\\
0&\tfrac{1}{2}\end{array}\right)\end{array}\right)\left(\begin{array}[]{c}\left(\begin{array}[]{cc}\mathrm{e}^{\mathrm{i}t}&0\\\
0&\mathrm{e}^{-\mathrm{i}t}\end{array}\right)\\\
\left(\begin{array}[]{cc}1&0\\\ 0&1\end{array}\right)\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}\tfrac{1}{2}&0\\\
\tfrac{1}{2}&-1\end{array}\right)\left(\begin{array}[]{cc}\mathrm{e}^{\mathrm{i}t}&0\\\
0&\mathrm{e}^{-\mathrm{i}t}\end{array}\right)+\left(\begin{array}[]{cc}-1&-\tfrac{1}{2}\\\
0&\tfrac{1}{2}\end{array}\right)\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}\tfrac{1}{2}\mathrm{e}^{\mathrm{i}t}-1&-\tfrac{1}{2}\\\
\tfrac{1}{2}\mathrm{e}^{\mathrm{i}t}&\tfrac{1}{2}-\mathrm{e}^{-\mathrm{i}t}\end{array}\right).$

Next,

$\displaystyle\det\tau_{t}$
$\displaystyle=(\tfrac{1}{2}\mathrm{e}^{\mathrm{i}t}-1)(\tfrac{1}{2}-\mathrm{e}^{-\mathrm{i}t})+\tfrac{1}{4}\mathrm{e}^{\mathrm{i}t}=\tfrac{1}{2}\mathrm{e}^{\mathrm{i}t}-1+\mathrm{e}^{-\mathrm{i}t}.$

This expression is $2\pi$-periodic. It thus, suffices to consider
$t\in[0,2\pi)$. Since $\operatorname{Im}\det\tau_{t}=-\frac{1}{2}\sin t=0$ if
and only if $t\in\left\\{0,\pi\right\\}$. For these values, however, we have
$\det\tau_{0}=\frac{1}{2}$ and $\det\tau_{\pi}=-\frac{5}{2}.$ By continuity,
$\min_{t\in\mathbb{R}}\left|\det\tau_{t}\right|=\min_{t\in[0,2\pi]}\left|\det\tau_{t}\right|>0.$
Thus, Cramer’s rule implies that $\tau_{t}$ is invertible with uniform bound
for the inverse. Hence, the corresponding port-Hamiltonian semi-group is
exponentially stable.

###  8.2 A regularity condition on $\mathcal{H}$

The aim of this subsection is to show that Theorem 5.1 contains all the cases
already contained in Theorem 1.1 next. For this recall the setting from the
beginning of Section 8; also we briefly define what it means for a function to
be of bounded variation.

######  Definition 8.7.

Let $I\subseteq\mathbb{R}$ be an open interval, $\alpha\in L_{1}(I)$. Then
$\alpha$ is of _bounded variation_ , if

$\sup\\{|\int_{I}\alpha(x)\phi^{\prime}(x)\,\mathrm{d}x|;\phi\in
C_{c}^{1}(I),\|\phi\|_{\infty}\leq 1\\}<\infty.$

By the Riesz–Markov representation theorem, there exists a unique signed Radon
measure, $\mathrm{D}_{x}\alpha$, on the Borel sets of $I$ such that

$-\int_{I}\alpha\phi^{\prime}=\int_{I}\phi\,\mathrm{d}\mathrm{D}_{x}\alpha\eqqcolon\langle\mathrm{D}_{x}\alpha,\phi\rangle,$

for each $\phi\in C_{c}^{1}(I)$. Moreover

$\|\mathrm{D}_{x}\alpha\|\coloneqq|\mathrm{D}_{x}\alpha|(I)=\sup\\{|\int_{I}\alpha(x)\phi^{\prime}(x)\,\mathrm{d}x|;\phi\in
C_{c}^{1}(I),\|\phi\|_{\infty}\leq 1\\},$

where $|\mathrm{D}_{x}\alpha|$ denotes the variation of the measure
$\mathrm{D}_{x}\alpha.$ The space

$BV(I)\coloneqq\\{\alpha\in L_{1}(I);\alpha\text{ of bounded variation}\\}$

becomes a Banach space, if endowed with the norm given by

$\|\alpha\|_{BV}\coloneqq\|\alpha\|_{L_{1}}+\|\mathrm{D}_{x}\alpha\|.$

A matrix-valued function $G\colon(a,b)\to\mathbb{R}^{d\times d}$ is called _of
bounded variation,_ if all its components are of bounded variation. In this
case we set

$\mathrm{D}_{x}G\coloneqq\left(\mathrm{D}_{x}G_{ij}\right)_{i,j\in\\{1,\ldots,n\\}}$

as a matrix-valued measure.

######  Theorem 8.8.

Assume that $\mathcal{H}$ is of bounded variation. Then (B) (i.e., (29)) is
satisfied.

This results immediately yields a proof for Proposition 3.2:

###### Proof of Proposition 3.2.

Any constant is of bounded variation. Thus, the result follows from Theorem
8.8. ∎

As in the previous subsection, we detour the proof of Theorem 8.8 via
$\mathcal{O}$. For this, some preliminaries are in order.

######  Proposition 8.9.

Let $\mathcal{O}$ be as in (28). Then

$\mathcal{H}\text{ is of bounded variation }\iff\mathcal{O}\text{ is of
bounded variation.}$

###### Proof.

The statement is immediate from the formula for $\mathcal{O}$, see (28). ∎

######  Theorem 8.10.

Assume that $\mathcal{O}$ is of bounded variation. Then (30) holds.

For the proof of this theorem, we require some classical results. The material
is widely known. We shall, however, summarise and prove some particular
findings needed in the present situation.

######  Theorem 8.11 ([10, Theorem 7.2 and Theorem 5.13]).

Let $I=(a,b)\subseteq\mathbb{R}$ be an interval and $\alpha\in BV(I)$. Then
there is a right-continuous representative, $\alpha_{r}$, of $\alpha$.
Moreover, we have for all $x,x^{\prime}\in(a,b)$, $x^{\prime}<x$

$\alpha_{r}(x)-\alpha_{r}(x^{\prime})=\mathrm{D}_{x}\alpha([x,x^{\prime})).$

######  Theorem 8.12.

Let $I\subseteq\mathbb{R}$ be an open and bounded interval, $\alpha\in
BV(I)\cap L_{\infty}(I).$

If $u\in H^{1}(I)$, then $\alpha u\in BV(I)\cap L_{\infty}(I)$ and

$\mathrm{D}_{x}(\alpha
u)=u\mathrm{D}_{x}\alpha+u^{\prime}\alpha\,\mathrm{d}\lambda,$

where $\,\mathrm{d}\lambda$ denotes the Lebesgue measure and
$u\mathrm{D}_{x}\alpha$ is the measure $\mathrm{D}_{x}\alpha$ with density
$u$. Moreover, we have

$\|\mathrm{D}_{x}(\alpha
u)\|\leq\|\mathrm{D}_{x}\alpha\|\|u\|_{H^{1}}+\|\alpha\|_{\infty}\|u\|_{H^{1}}\sqrt{b-a}.$

###### Proof.

Let $\phi\in C_{c}^{1}(I)$. Assume that $u$ is continuously differentiable.
Then we compute

$\displaystyle-\int_{I}\alpha u\phi^{\prime}d\lambda$
$\displaystyle=-\int_{I}\alpha((u\phi)^{\prime}-u^{\prime}\phi)\,\mathrm{d}\lambda$
$\displaystyle=\langle\mathrm{D}_{x}\alpha,u\phi\rangle+\int_{I}u^{\prime}\alpha\phi\,\mathrm{d}\lambda$
$\displaystyle=\langle
u\mathrm{D}_{x}\alpha+u^{\prime}\alpha\,\mathrm{d}\lambda,\phi\rangle$

We estimate

$\displaystyle|\int_{I}\alpha u\phi^{\prime}\,\mathrm{d}\lambda|$
$\displaystyle\leq\|\mathrm{D}_{x}\alpha\|\|u\phi\|_{\infty}+\|\alpha\|_{\infty}\|u^{\prime}\phi\|_{1}$
$\displaystyle\leq\|\mathrm{D}_{x}\alpha\|\|u\|_{\infty}\|\phi\|_{\infty}+\|\alpha\|_{\infty}\|u^{\prime}\|_{2}\|\phi\|_{2}$
$\displaystyle\leq\|\mathrm{D}_{x}\alpha\|\|u\|_{H^{1}}\|\phi\|_{\infty}+\|\alpha\|_{\infty}\|u\|_{H^{1}}\sqrt{b-a}\|\phi\|_{\infty}$
$\displaystyle=\left(\|\mathrm{D}_{x}\alpha\|\|u\|_{H^{1}}+\|\alpha\|_{\infty}\|u\|_{H^{1}}\sqrt{b-a}\right)\|\phi\|_{\infty}.$

Hence, $\alpha u\in BV(I)$ and

$\|\mathrm{D}_{x}(\alpha
u)\|\leq\|\mathrm{D}_{x}\alpha\|\|u\|_{H^{1}}+\|\alpha\|_{\infty}\|u\|_{H^{1}}\sqrt{b-a}.$

Moreover, we estimate

$\|\alpha
u\|_{1}\leq\|u\|_{\infty}\|\alpha\|_{1}\leq\|u\|_{H^{1}}\|\alpha\|_{1}.$

In particular, let $(u_{n})_{n}$ be a sequence of continuously differentiable
functions converging to $u$ in $H^{1}$. Then, by the estimates above,
$\mathrm{D}_{x}(\alpha u)\in BV(I)$. Moreover, using the product formula from
the beginning of the proof for $u_{n}$ and letting $n\to\infty$, we infer

$\mathrm{D}_{x}(\alpha
u)=u\mathrm{D}_{x}\alpha+u^{\prime}\alpha\,\mathrm{d}\lambda.$

Next, we recall Gronwall’s inequality for locally finite measures.

######  Definition 8.13.

Let $I\subseteq\mathbb{R}$ be an interval. We call a Borel measure $\mu$ on
$I$ _locally finite_ , if for all compact $K\subseteq I$, $\mu(K)<\infty$.

######  Theorem 8.14 ([17, Lemma A.1]).

Let $I\subseteq\mathbb{R}$ be an interval. Let $u\colon I\to\mathbb{R}$ and
$\alpha\colon I\to[0,\infty)$ measurable. Assume that for a locally finite
Borel measure $\mu$ on $I$, we have $u$ is locally integrable and there exists
$a\in I$ such that for all $t>a$

$u(t)\leq\alpha(t)+\int_{[a,t)}|u(s)|\,\mathrm{d}\mu(s).$

Then

$u(t)\leq\alpha(t)+\int_{[a,t)}\alpha(s)\exp(\mu((s,t)))\,\mathrm{d}\mu(s)\quad(t>a).$

Now, we are in the position to show the main result of this subsection.

###### Proof of Theorem 8.10.

Let $v\in H^{1}(-\tfrac{1}{2},\tfrac{1}{2})^{d}$ and $t\in\mathbb{R}$ satisfy

$v^{\prime}(s)=\mathrm{i}tD\mathcal{O}(s)v(s),\quad\left(s\in(-\tfrac{1}{2},\tfrac{1}{2})\right),$

where without loss of generality, we chose the right-continuous representative
of $\mathcal{O}$, which is possible by Theorem 8.11. Next, note that

$\displaystyle\langle v(s),\mathcal{O}(s)v(s)\rangle_{\mathbb{C}^{d}}$
$\displaystyle=\sum_{j=1}^{d}\sum_{k=1}^{d}v_{j}(s)\mathcal{O}_{jk}(s)v_{k}(s)$
$\displaystyle=\langle\mathcal{O}(s),v(s)v(s)^{\top}\rangle_{\mathbb{C}^{d\times
d}}$

for all $s\in(-\tfrac{1}{2},\tfrac{1}{2})$. Thus, by Theorem 8.12, we obtain

$\displaystyle\mathrm{D}_{s}\langle v,\mathcal{O}v\rangle_{\mathbb{C}^{d}}$
$\displaystyle=\langle\mathcal{O},\partial_{s}(vv^{\top})\rangle\,\mathrm{d}\lambda+\langle\textrm{D}_{s}\mathcal{O},vv^{\top}\rangle$
$\displaystyle=\left(\langle v^{\prime},\mathcal{O}v\rangle+\langle
v,\mathcal{O}v^{\prime}\rangle\right)\,\mathrm{d}\lambda+\langle\textrm{D}_{s}\mathcal{O},vv^{\top}\rangle$
$\displaystyle=\left(\langle\mathrm{i}tD\mathcal{O}v,\mathcal{O}v\rangle+\langle\mathcal{O}v,\mathrm{i}tD\mathcal{O}v\rangle\right)\,\mathrm{d}\lambda+\langle\textrm{D}_{s}\mathcal{O},vv^{\top}\rangle$
$\displaystyle=\langle\mathrm{D}_{s}\mathcal{O},vv^{\top}\rangle$

For $s,t\in(-\frac{1}{2},\frac{1}{2})$ with $s<t$ we therefore obtain by
Theorem 8.11

$\displaystyle\langle v(t),\mathcal{O}(t)v(t)\rangle_{\mathbb{C}^{d}}$
$\displaystyle=\langle
v(s),\mathcal{O}(s)v(s)\rangle+\int_{[s,t)}\,\mathrm{d}\langle
v(\sigma)v^{\top}(\sigma),\textrm{D}_{s}\mathcal{O}(\sigma)\rangle$
$\displaystyle=\langle
v(s),\mathcal{O}(s)v(s)\rangle+\sum_{j=1}^{d}\sum_{k=1}^{d}\int_{[s,t)}v_{j}(\sigma)v_{k}(\sigma)\mathrm{D}_{s}\mathcal{O}_{jk}(\sigma)$

Hence, using Remark 4.4, for some $c>0$

$\displaystyle c\|v(t)\|^{2}$ $\displaystyle\leq\langle
v(t),\mathcal{O}(t)v(t)\rangle$
$\displaystyle\leq\|v(s)\|^{2}\|\mathcal{O}\|_{\infty}+\frac{1}{2}\int_{[s,t)}\|v(\sigma)\|^{2}\,\mathrm{d}\mu(\sigma),$

where $\mu\coloneqq\sum_{j,k=1}^{d}|\mathrm{D}_{s}\mathcal{O}_{jk}|$. Note
that $\mu$ is a finite measure on $(-\tfrac{1}{2},\frac{1}{2})$. Using Theorem
8.14, we get

$\displaystyle\|v(t)\|^{2}$
$\displaystyle\leq\tfrac{1}{c}\|v(s)\|^{2}\|\mathcal{O}\|_{\infty}\left(1+\frac{1}{2c}\int_{[s,t)}\mathrm{e}^{\frac{1}{2c}\mu\left((\sigma,t)\right)}\,\mathrm{d}\mu(\sigma)\right)$
$\displaystyle\leq\tfrac{1}{c}\|v(s)\|^{2}\|\mathcal{O}\|_{\infty}\left(1+\tfrac{1}{2c}\mathrm{e}^{\frac{1}{2c}\mu((-\tfrac{1}{2},\tfrac{1}{2}))}\mu((-\tfrac{1}{2},\tfrac{1}{2}))\right),$

Letting now $s\to-\frac{1}{2}$ in the last inequality, we derive

$\|v(t)\|\leq
C\|v(-\tfrac{1}{2})\|^{2}\quad(t\in[-\tfrac{1}{2},\tfrac{1}{2}])$

for some constant $C>0$. The latter confirms (30). ∎

###### Proof of Theorem 8.8.

By Proposition 8.9, $\mathcal{O}$ is of bounded variation. Hence, by Theorem
8.10, (30) holds. This, in turn, yields that (29) is satifsied by Proposition
5.3. ∎

We finally obtain a proof of Theorem 1.1.

###### Proof of Theorem 1.1.

The statement is an immediate consequence of Theorem 5.1 and Theorem 8.8. ∎

##  9 Conclusion

We presented a new characterisation of exponential stability for port-
Hamiltonian systems. The strategy of proof is based on a reformulation that
goes back to [12] and several standard results in semi-group theory. The
characteristion works only if a certain family of fundamental solutions of a
non-autonomous ODE-system is uniformly bounded. Whether or not this
boundedness is needed lies beyond the scope of this article and can be
considered an open problem for the time-being. We emphasise that as the
strategy above uses existence and boundedness of the resolvent of the
generator on the imaginary axis it might be possible, invoking results such as
[1, 4], to show algebraic decay for certain set-ups. Whether at all these set-
ups exists and how they maybe characterised will be addressed in future work.

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