#  On an inverse problem of nonlinear imaging with fractional damping

Barbara Kaltenbacher111 Department of Mathematics, Alpen-Adria-Universität
Klagenfurt<EMAIL_ADDRESS>William Rundell222 Department of
Mathematics, Texas A&M University, Texas 77843<EMAIL_ADDRESS>
###### Abstract

This paper considers the attenuated Westervelt equation in pressure
formulation. The attenuation is by various models proposed in the literature
and characterised by the inclusion of non-local operators that give power law
damping as opposed to the exponential of classical models. The goal is the
inverse problem of recovering a spatially dependent coefficient in the
equation, the parameter of nonlinearity $\kappa(x)$, in what becomes a
nonlinear hyperbolic equation with nonlocal terms. The overposed measured data
is a time trace taken on a subset of the domain or its boundary. We shall show
injectivity of the linearised map from $\kappa$ to the overposed data used to
recover it and from this basis develop and analyse Newton-type schemes for its
effective recovery.

Keywords: Inverse problems, damped nonlinear wave equation, ultrasound.

ams classification: 35R30, 35R11, 35K55, 78A46

##  1 Introduction

The problem of nonlinearity parameter imaging with ultrasound in lossy media
amounts to identification of the space dependent coefficient $\kappa(x)$ for
the attenuated Westervelt equation in pressure formulation

$\displaystyle\bigl{(}1-\kappa(x)v^{2}\bigr{)}_{tt}-c_{0}^{2}\triangle
v+Dv=r\mbox{ in }\Omega\times(0,T)$ (1) $\displaystyle v$
$\displaystyle=0\mbox{ on }\partial\Omega\times(0,T)\;;\qquad v(0)=0,\quad
v_{t}(0)$ $\displaystyle=0\;\mbox{ in }\;\Omega$

from observations Here $c_{0}>0$ is the wave speed (possibly space dependent
as well), and $Dv$ a damping term that will be specified below. For simplicity
we impose homogeneous Dirichlet boundary conditions here, but the ideas in
this paper extend to more realistic boundary conditions, such as absorbing
boundary conditions for avoiding spurious reflections and/or inhomogeneous
Neumann boundary conditions for modelling excitation via, e.g., some
transducer array. Note that the excitation here is modelled by an interior
source $r$, and we refer to a discussion on this in [16].

By letting our $v$ equation satisfy boundary (or possibly interior)
observations we obtain an inverse problem for the recovery of $\kappa$. These
measurements will be taken to be

$g(t)=v(x_{0},t)\quad\mbox{ or }\quad g(x,t)=v(x,t)\,,\
x\in\Sigma\,,t\in(0,T).$ (2)

at some point $x_{0}$ or on some surface $\Sigma$ contained in
$\overline{\Omega}$.

The inverse problem represented by equations (1) and (2) is challenging on at
least three counts. First, the underlying model equation is nonlinear and in
fact the nonlinearity occurs in the highest order term. Second, the unknown
coefficient $\kappa(x)$ is directly coupled to this term and third, is
spatially varying whereas the data $g(t)$ is in the “orthogonal” time
direction and this is well known to lead to several ill-conditioning of the
inversion of the map from data to unknown.

The use of ultrasound is a well-established protocol in the imaging of human
tissue and, besides the classical sonography methodology, there exist several
novel imaging principle, such as harmonic imaging or nonlinearity imaging. The
latter [2, 3, 5, 13, 24, 26, 27], relies on tissue-dependence, hence spatial
variation of the parameter of nonlinearity $B/A$ that is contained in
$\kappa$. See, e.g., It thus inherently needs a nonlinear acoustic model as an
underlying PDE and we refer to e.g., the review [15] and the references
therein for a brief derivation of the fundamental nonlinear acoustic
equations. The quantity of interest from an imaging perspective is the
coefficient $\kappa$ and its recovery in the case when the damping term was
$Du=b\triangle u_{t}$ was the subject of [16]. This is the classical
formulation of damping being proportional to velocity but there are may
alternative models that are prominent in the literature. We mention some of
these in the next section but the main change is the incorporation of non-
local terms involving either fractional derivatives in time or modifying the
operator $(-\triangle)$ to have the Laplacian raised to a fractional power
$(-\triangle)^{\beta}$. These have the effect of ameliorating the exponential
decay of the solution, by a fractional exponent in the frequencies in the case
of $(-\triangle)^{\beta}$ and by a power law decay in the case of a fractional
time derivative. The use of such operators in inverse problems is now well
documented in the literature (see, [14] and in particular, for the wave
equation in [17]).

In this paper we will provide analysis for the forwards problem and in
particular regularity and well-posedness for the coefficient-to-state map
$G:\kappa\mapsto v$ where $v$ solves (1). The forwards map is defined by
$F(\kappa)=\mbox{tr}_{\Sigma}v$, where $\mbox{tr}_{\Sigma}v$ denotes the time
trace of the space–and–time dependent function $v:(0,T)\times\Omega$ at the
observation surface $\Sigma$ (which may also just be a single point
$\Sigma=\\{x_{0}\\}$). Its linearization at $\kappa=0$ is $F^{\prime}(0)$ and
we will prove an injectivity result which will both show local uniqueness and
pave the way for the use of Newton’s method which we will analyse in section 4
then show reconstructions based on this approach in section 6.

##  2 The Imaging Problem

As already mentioned in the introduction, the inverse problem under
consideration is to recover the space dependent coefficient $\kappa(x)$ in the
attenuated Westervelt equation which can also be written in the form

$\displaystyle u_{tt}+c^{2}\mathcal{A}u+Du$
$\displaystyle=\kappa(x)(u^{2})_{tt}+r\mbox{ in }\Omega\times(0,T)$ (3)
$\displaystyle u(0)=0,\quad u_{t}(0)$ $\displaystyle=0\mbox{ in }\Omega$

from observations $g(x,t)\\!=\\!u(x,t)$, $x\in\Sigma,\;t\in\\!(0,T)$ where
$\Sigma$ may be a single point, typically located on $\partial\Omega$. Here
$c>0$ is the constant mean wave speed, and
$\mathcal{A}=-(c_{0}(x)^{2}\\!/c^{2})\triangle$ contains the possibly
spatially varying coefficient $c_{0}(x)>0$ and is equipped – for simplicity –
with homogeneous boundary conditions. Moreover $r=r(x,t)$ is a known source
term modelling excitation of the acoustic wave by a transducer array, see
[17].

The damping operator $D$ appearing in (3) is a differential operator
containing space and/or time derivatives. Classically, $D$ will consist of
integer derivatives, typical examples being $D=\mathcal{A}\partial_{t}$ or
$D=\partial_{t}$ often referred to as strong and weak damping, respectively.
We here list some of the (due to experimentally found power law frequency
dependence) practically relevant fractional damping models, that we have
already discussed in [17] in a different imaging context, namely for the
inverse pat/tat problem:

Time fractional models:

  * •

Caputo-Wismer model [6], [25, eq. (5)], called Kelvin wave equation in [4, eq.
(19)]

$D=b\mathcal{A}\partial_{t}^{\alpha}$ (4)

where typically $\alpha\in[0,1]$.

  * •

(Modified) Szabo model [22], [4, eq. (42)]

$D=b\partial_{t}^{\alpha+2}$ (5)

where $\alpha\in[-1,1]$, $b\geq 0$.

  * •

Fractional Zener (combined Caputo-Wismer-Szabo) model [12, 18], [4, eq. (30)]

$D=b_{1}\mathcal{A}\partial_{t}^{\alpha_{1}}+b_{2}\partial_{t}^{\alpha_{2}+2}$
(6)

where $\alpha_{1}\geq\alpha_{2}\in[0,1]$, $b_{1}\geq b_{2}c^{2}$, cf. [12,
Section III.B].

In these models $\partial_{t}^{\alpha}$ denotes the Djrbashian-Caputo
fractional time derivative, which here, due to the homogeneous initial
conditions, coincides with the Riemann-Liouville one.

Space fractional models:

  * •

Chen-Holm model [7, eq. (21)]

$D=b\mathcal{A}^{\beta}$ (7)

typically with $\beta\in[0,1]$, where Kelvin-Voigt damping is recovered when
$\beta=1$.

  * •

Treeby-Cox model [23, eq. (28)]

$D=b_{1}\mathcal{A}^{\beta}\partial_{t}+b_{2}\mathcal{A}^{\beta+1/2}$ (8)

typically $\beta\in[0,1]$, which is an extension of the former.

Here we use the spectral definition of the Laplacian which coincides with the
Riesz version on $\mathbb{R}^{d}$; however, they differ in case of bounded
$\Omega$.

In this paper, we will focus on two damping models namely (a) a combination of
(4) and (7), since we find it interesting to investigate the interplay of
space- and time-fractional derivatives and its influence on the ill-posedness
of the inverse problem; (b) (6) as it contains higher than second order time
derivatives which are in case $\alpha_{2}=1$ known to make the equation behave
wave-like (finite speed of propagation) in spite of the damping, which is
expected to influence the degree of ill-posedness of the inverse problem as
well. Thus we here focus on the two damping models

$\displaystyle D$
$\displaystyle=b\mathcal{A}^{\beta}\partial_{t}^{\alpha}\quad\mbox{
(combination of Caputo-Wismer and Chen-Holm model -- CWCH)}$ $\displaystyle D$
$\displaystyle=b_{1}\mathcal{A}\partial_{t}^{\alpha_{1}}+b_{2}\partial_{t}^{\alpha_{2}+2}\quad\mbox{
(fractional Zener -- FZ)}\,.$

##  3 Analysis of the forward problem

In this section, we consider well-posedness of the initial value problem for
the parameter-to-state map $G:\kappa\mapsto u$ where $u$ solves (3) and its
linearisation $z=G^{\prime}(\kappa)\underline{d\kappa}$

$\displaystyle(1-2\kappa u)z_{tt}+c^{2}\mathcal{A}z+Dz-4\kappa
u_{t}\,z_{t}-2\kappa u_{tt}\,z$
$\displaystyle=2\underline{\delta\kappa}(u\,u_{tt}+u_{t}^{2})\mbox{ in
}\Omega\times(0,T)$ (9) $\displaystyle z(0)=0,\quad z_{t}(0)$
$\displaystyle=0\mbox{ in }\Omega$

for given $\kappa$ and $\underline{\delta\kappa}$, respectively, in the
context of several damping models $D$. In order to prove Fréchet
differentiability, we will also have to consider the difference
$v=G(\tilde{\kappa})-G(\kappa)=\tilde{u}-u$, which solves

$\displaystyle(1-2\kappa u)v_{tt}$
$\displaystyle+c^{2}\mathcal{A}v+Dv-2\kappa(\tilde{u}_{t}+u_{t})\,v_{t}-2\kappa\tilde{u}_{tt}\,v$
(10)
$\displaystyle=2(\tilde{\kappa}-\kappa)(\tilde{u}\tilde{u}_{tt}+\tilde{u}_{t}^{2})\mbox{
in }\Omega\times(0,T)$ $\displaystyle v(0)=0,\quad v_{t}(0)$
$\displaystyle=0\mbox{ in }\Omega$

as well as the first order Taylor remainder
$w=G(\tilde{\kappa})-G(\kappa)-G^{\prime}(\kappa)(\tilde{\kappa}-\kappa)$
which satisfies

$\displaystyle(1-2\kappa u)w_{tt}$ $\displaystyle+c^{2}\mathcal{A}w+Dw-4\kappa
u_{t}\,w_{t}-2\kappa u_{tt}\,w$ (11)
$\displaystyle=2\underline{d\kappa}(v\tilde{u}_{tt}+uv_{tt}+(\tilde{u}_{t}+u_{t})v_{t})+2\kappa(vv_{tt}+v_{t}^{2})\mbox{
in }\Omega\times(0,T)$ $\displaystyle w(0)=0,\quad w_{t}(0)$
$\displaystyle=0\mbox{ in }\Omega\,,$

Here we can allow for spatially varying sound speed $c_{0}(x)$ for which we
only require

$c_{0}\in L^{\infty}(\Omega)\mbox{ and }c_{0}(x)\geq c>0$ (12)

unless otherwise stated, by setting

$\mathcal{A}={{\small-\frac{c_{0}^{2}}{c^{2}}}}\,\triangle$ (13)

where $-\triangle$ is the Laplace operator equipped with homogeneous Dirichlet
boundary conditions. We denote by $(\phi_{j},\lambda_{j})$ an eigensystem of
the operator $\mathcal{A}$ with domain
$\dot{H}^{2}(\Omega):=\mathcal{D}(\mathcal{A})$ which is selfadjoint and
positive definite with respect to the weighted $L^{2}$ space
$\dot{L}^{2}(\Omega):=L^{2}_{c^{2}/c_{0}^{2}}(\Omega)$. Note that we assume
$\mathcal{A}^{-1}:\dot{L}^{2}(\Omega)\to\dot{L}^{2}(\Omega)$ to be compact
(which is e.g., satisfied if $\Omega$ is bounded; for some comments on more
general domain and boundary settings we point to [17]) , so that the
eigensystem exists and is complete with $\lambda_{j}\to\infty$ as
$j\to\infty$. Moreover, this defines a scale of Hilbert spaces
$\dot{H}^{s}(\Omega):=\mathcal{D}(\mathcal{A}^{s/2})$, $s\in\mathbb{R}$, whose
norm can be defined via the eigensystem as
$\|v\|_{\dot{H}^{s}(\Omega)}=\left(\sum_{j=1}^{\infty}\lambda_{j}^{s}|\langle
v,\phi_{j}\rangle|^{2}\right)^{1/2}$. We will denote by
$\langle\cdot,\cdot\rangle$ the $\dot{L}^{2}$ inner product (that is, the
weighted one) on $\Omega$ whereas the use of the ordinary $L^{2}$ inner
product will be indicated by a subscript $\langle\cdot,\cdot\rangle_{L^{2}}$
Moreover, we use the abbreviations
$\|u\|_{L^{p}_{t}(L^{q})}=\|u\|_{L^{p}(0,t;L^{q}(\Omega))}$,
$\|u\|_{L^{p}(L^{q})}=\|u\|_{L^{p}(0,T;L^{q}(\Omega))}$ for space-time norms.

Throughout this paper, we denote by $\partial_{t}^{\alpha}$ the (partial)
Caputo-Djrbashian fractional time derivative of order $\alpha\in(n-1,n)$ with
$n\in\mathbb{N}$ by
$\partial_{t}^{\alpha}u=I_{t}^{n-\alpha}[\partial^{n}_{t}u]\;$, where
$\partial^{n}_{t}$ denotes the $n$-th integer order partial time derivative
and for $\gamma\in(0,1)$, and $I_{t}^{\gamma}$ is the Abel integral operator
defined by

$I_{t}^{\gamma}[v](t)=\frac{1}{\Gamma(\gamma)}\int_{0}^{t}\frac{v(s)}{(t-s)^{1-\gamma}}ds\,.$

For details on fractional differentiation and subdiffusion equations, we refer
to, e.g., [8, 9, 19, 20, 21]. See also the tutorial paper on inverse problems
for anomalous diffusion processes [14]. Whenever we use the Riemann-Liouville
fractional derivative $\partial^{n}_{t}I_{t}^{n-\alpha}$, this will be denoted
by ${}^{RL}\partial_{t}^{\alpha}$. These two versions of the fractional
derivative coincide when applied to a function whose Cauchy data up to order
$n-1$ at $t=0$ vanish.

The crucial tool we need in obtaining the required estimates is the following
consequence of Alikhanov’s Lemma [1, Lemma 1]

$\partial_{t}^{\gamma}[w](t)w(t)\geq\tfrac{1}{2}(\partial_{t}^{\gamma}w^{2})(t)$
(14)

for any absolutely continuous function $w$. We apply it to
$w=\partial_{t}^{\alpha}v$ with $\gamma=1-\alpha$, using the identities

$\displaystyle\qquad\partial_{t}^{1-\alpha}w=\partial_{t}^{1-\alpha}\partial_{t}^{\alpha}v=\partial_{t}^{1-\alpha}I_{t}^{1-\alpha}v_{t}=v_{t}\,,$
$\displaystyle\qquad\int_{0}^{t}(\partial_{t}^{1-\alpha}w^{2})(s)\,ds=\int_{0}^{t}{}^{RL}\partial_{t}^{1-\alpha}[w^{2}](s)\,ds=\int_{0}^{t}\partial_{t}I_{t}^{\alpha}[w^{2}](s)\,ds=I_{t}^{\alpha}[w^{2}](t)$

that hold for $v_{t}\in L^{\infty}(0,T)$ (Theorem 2.3 in fde-lect.pdf) and for
$w^{2}\in W^{1,1}(0,T)$ with $w(0)=0$ (Theorem 2.2 in fde-lect.pdf). Note that
for $w=\partial_{t}^{\alpha}v$ we automatically have $w(0)=0$ and
$I_{t}^{\alpha}[w^{2}](t)=0$. After integration with respect to time this
implies the following result.

######  Lemma 3.1.

For $v\in W^{1,\infty}(0,T)$ with $(\partial_{t}^{\alpha}v)^{2}\in
W^{1,1}(0,T)$, and $t\in(0,T)$, the following estimate holds.

$\int_{0}^{t}\partial_{t}^{\alpha}[v](s)v_{t}(s)\,ds\geq
I_{t}^{\alpha}\left[(\partial_{t}^{\alpha}v)^{2}\right]\geq\frac{1}{2\Gamma(\alpha)t^{1-\alpha}}\|\partial_{t}^{\alpha}v\|_{L^{2}(0,t)}^{2}\,.$
(15)

###  3.1 Caputo-Wismer-Kelvin damping

We start with the Caputo-Wismer-Kelvin model

$D=b\mathcal{A}^{\beta}\partial_{t}^{\alpha}\quad\mbox{ with }\beta\in[0,1],\
b\geq 0$

and first of all consider the initial boundary value problem for the general
linear pde

$(1-\sigma)u_{tt}+c^{2}\mathcal{A}u+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}u+\mu
u_{t}+\rho u=h$ (16) $u(0)=u_{0}\,,\quad u_{t}(0)=u_{1}$ (17)

with constants $b$, $c>0$ and given space- and time dependent functions
$\sigma$, $\mu$, $\rho$, $h$ where $\sigma$ satisfies the non-degeneracy
condition

$\sigma(x,t)\leq\overline{\sigma}<1\mbox{ for all }x\in\Omega\,\quad
t\in(0,T)\,.$ (18)

In order to prove existence and uniqueness of solutions to (16), (17), we
apply the usual Faedo-Galerkin approach of discretisation in space with
eigenfunctions of $\mathcal{A}$, $u(x,t)\approx
u^{n}(x,t)=\sum_{i=1}^{n}u^{n}_{i}(t)\phi_{i}(x)$ and testing with $\phi_{j}$,
that is,

$\langle(1-\sigma)u^{n}_{tt}+c^{2}\mathcal{A}u^{n}+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}u^{n}+\mu
u^{n}_{t}+\rho u^{n}-h,v\rangle=0\,\quad
v\in\mbox{span}(\phi_{1},\ldots,\phi_{n})\,.$ (19)

This leads to the ode system

$(I-S^{n}(t)){\underline{u}^{n}}^{\prime\prime}(t)+b\,(\Lambda^{n})^{\beta}(\partial_{t}^{\alpha}\underline{u}^{n})(t)+M^{n}(t){\underline{u}^{n}}^{\prime}(t)+\Bigl{(}c^{2}\,\Lambda^{n}+R^{n}(t)\Bigr{)}\underline{u}^{n}(t)=\underline{h}^{n}$
(20)

with matrices and vectors defined by

$\displaystyle\underline{u}^{n}(t)=(u^{n}_{i}(t))_{i=1,\ldots
n}\,,\quad\underline{h}^{n}(t)=(\langle h(t),\phi_{i}\rangle)_{i=1,\ldots
n}\,,\quad\Lambda^{n}=\mbox{diag}(\lambda_{1},\ldots,\lambda_{n})\,,$ (21)
$\displaystyle
S^{n}(t)=(\langle\sigma(t)\phi_{i},\phi_{j}\rangle)_{i,j=1,\ldots n}\,,\quad
M^{n}(t)=(\langle\mu(t)\phi_{i},\phi_{j}\rangle)_{i,j=1,\ldots n}\,,$
$\displaystyle R^{n}(t)=(\langle\rho(t)\phi_{i},\phi_{j}\rangle)_{i,j=1,\ldots
n}\,.$

Existence of a unique solution $\underline{u}^{n}\in
C^{2}(0,T;\mathbb{R}^{n})$ to (20) follows from standard ode theory (Picard-
Lindelöf Theorem and Gronwall’s Inequality), as long as $\sigma$, $\mu$ and
$\rho$ are in $C(0,T;\dot{H}^{s}(\Omega))$ for some $s\in\mathbb{R}$ (noting
that the eigenfunctions $\phi_{j}$ are contained in $\dot{H}^{k}(\Omega)$ for
any $k\in\mathbb{N}$ and therefore the vector and matrix functions
$\underline{h}^{n}$, $S^{n}$, $M^{n}$, $R^{n}$ in (21) are well-defined and
contained in $C(0,T;\mathbb{R}^{n})$ and $C(0,T;\mathbb{R}^{n\times n})$,
respectively. Moreover due to (18), the symmetric matrix $S^{n}(t)$ is
positive definite with smallest eigenvalue bounded away from zero by
$1-\overline{\sigma}$ cf (18).

We multiply (20) with $(\Lambda^{n})^{2}{\underline{u}^{n}}^{\prime}(t)$ and
integrate with respect to time, using the identity

$\displaystyle(S^{n}(t){\underline{u}^{n}}^{\prime\prime}(t))^{T}(\Lambda^{n})^{2}{\underline{u}^{n}}^{\prime}(t)=\sum_{i=1}^{n}\sum_{j=1}^{n}\langle\sigma(t)\phi_{i},\phi_{j}\rangle\lambda_{j}^{2}{u^{n}_{i}}^{\prime\prime}(t){u^{n}_{j}}^{\prime}(t)$
$\displaystyle=\sum_{i=1}^{n}\sum_{j=1}^{n}\langle\sigma(t)\phi_{i},\phi_{j}\rangle\lambda_{i}{u^{n}_{i}}^{\prime\prime}(t)\lambda_{j}{u^{n}_{j}}^{\prime}(t)+\sum_{i=1}^{n}\sum_{j=1}^{n}\langle\sigma(t)\phi_{i},\phi_{j}\rangle(\lambda_{j}-\lambda_{i}){u^{n}_{i}}^{\prime\prime}(t)\lambda_{j}{u^{n}_{j}}^{\prime}(t)$
$\displaystyle=\frac{1}{2}\frac{d}{dt}\sum_{i=1}^{n}\sum_{j=1}^{n}\langle\sigma(t)\phi_{i},\phi_{j}\rangle\lambda_{i}{u^{n}_{i}}^{\prime}(t)\lambda_{j}{u^{n}_{j}}^{\prime}(t)-\frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n}\langle\sigma_{t}(t)\phi_{i},\phi_{j}\rangle\lambda_{i}{u^{n}_{i}}^{\prime}(t)\lambda_{j}{u^{n}_{j}}^{\prime}(t)$
$\displaystyle\qquad+\sum_{i=1}^{n}\sum_{j=1}^{n}\Bigl{(}\langle\sigma(t)\phi_{i},\mathcal{A}\phi_{j}\rangle-\langle\sigma(t)\phi_{j},\mathcal{A}\phi_{i}\rangle\Bigr{)}{u^{n}_{i}}^{\prime\prime}(t)\lambda_{j}{u^{n}_{j}}^{\prime}(t)$

where

$\displaystyle\langle\sigma(t)\phi_{i},\mathcal{A}\phi_{j}\rangle-\langle\sigma(t)\phi_{j},\mathcal{A}\phi_{i}\rangle=\langle\mathcal{A}[\sigma(t)\phi_{i}]-\sigma(t)\mathcal{A}\phi_{i},\phi_{j}\rangle$
$\displaystyle=\langle-\triangle[\sigma(t)\phi_{i}]+\sigma(t)\triangle\phi_{i},\phi_{j}\rangle_{L^{2}}=-\langle(c_{0}^{2}/c^{2})(\triangle\sigma(t)\,\phi_{i}+\nabla\sigma(t)\cdot\nabla\phi_{i}),\phi_{j}\rangle$

provided $\sigma(t)\phi_{i}\in\dot{H}^{s}(\Omega))$ for some $s\in\mathbb{R}$.
(Note that the latter identity also holds true in case of spatially varying
$c_{0}$ since we use the $c^{2}/c_{0}^{2}$ weighted $L^{2}$ inner product
then). Thus we have

$\displaystyle\int_{0}^{t}((I-S^{n}(t)){\underline{u}^{n}}^{\prime\prime}(s))^{T}(\Lambda^{n})^{2}{\underline{u}^{n}}^{\prime}(s)\,ds$
$\displaystyle=\int_{0}^{t}\Bigl{(}\frac{1}{2}\frac{d}{dt}\|\sqrt{1-\sigma}\mathcal{A}u^{n}_{t}\|_{\dot{L}^{2}(\Omega)}^{2}(t)+\frac{1}{2}\langle\sigma_{t}(t)\mathcal{A}u^{n}_{t}(t),\mathcal{A}u^{n}_{t}(t)\rangle$
$\displaystyle\qquad+\langle(c_{0}^{2}/c^{2})(\triangle\sigma(t)\,u^{n}_{tt}(t)+\nabla\sigma(t)\cdot\nabla
u^{n}_{tt}(t)),\mathcal{A}u^{n}_{t}(t)\rangle\Bigr{)}$
$\displaystyle=\frac{1}{2}\|\sqrt{1-\sigma(t)}\mathcal{A}u^{n}_{t}(t)\|_{\dot{L}^{2}(\Omega)}^{2}-\frac{1}{2}\|\sqrt{1-\sigma(0)}\mathcal{A}u^{n}_{t}(0)\|_{\dot{L}^{2}(\Omega)}^{2}$
$\displaystyle\quad+\int_{0}^{t}\langle\frac{1}{2}\sigma_{t}(s)\mathcal{A}u^{n}_{t}(s)+(c_{0}^{2}/c^{2})(\triangle\sigma(s)\,u^{n}_{tt}(s)+\nabla\sigma(s)\cdot\nabla
u^{n}_{tt}(s)),\mathcal{A}u^{n}_{t}(s)\rangle\,ds\Bigr{)}\,.$

Similarly, we have

$\displaystyle(M^{n}(t){\underline{u}^{n}}^{\prime}(t))^{T}(\Lambda^{n})^{2}{\underline{u}^{n}}^{\prime}(t)$
$\displaystyle=\langle\mu(t)\mathcal{A}u^{n}_{t}(t),\mathcal{A}u^{n}_{t}(t)\rangle-\langle(c_{0}^{2}/c^{2})(\triangle\mu(t)\,u^{n}_{t}(t)+\nabla\mu(t)\cdot\nabla
u^{n}_{t}(t)),\mathcal{A}u^{n}_{t}(t)\rangle$

and

$\displaystyle(R^{n}(t){\underline{u}^{n}}(t))^{T}(\Lambda^{n})^{2}{\underline{u}^{n}}^{\prime}(t)$
$\displaystyle=\langle\rho(t)\mathcal{A}u^{n}(t),\mathcal{A}u^{n}_{t}(t)\rangle-\langle(c_{0}^{2}/c^{2})(\triangle\rho(t)\,u^{n}(t)+\nabla\rho(t)\cdot\nabla
u^{n}(t)),\mathcal{A}u^{n}_{t}(t)\rangle\,.$

Finally,

$\displaystyle\int_{0}^{t}((\Lambda^{n})^{\beta}(\partial_{t}^{\alpha}\underline{u}^{n})(t))^{T}(\Lambda^{n})^{2}{\underline{u}^{n}}^{\prime}(t)=\int_{0}^{t}\sum_{j=1}^{n}\lambda_{j}^{2+\beta}\int_{0}^{t}(\partial_{t}^{\alpha}u^{n}_{j})(s){u^{n}_{j}}^{\prime}(s)\,ds$
$\displaystyle\geq\frac{1}{2\Gamma(\alpha)t^{1-\alpha}}\sum_{j=1}^{n}\lambda_{j}^{2+\beta}\int_{0}^{t}\Bigl{(}\partial_{t}^{\alpha}u^{n}_{j}(s)\Bigr{)}^{2}\,ds=\frac{1}{2\Gamma(\alpha)t^{1-\alpha}}\|\mathcal{A}^{1+\beta/2}\partial_{t}^{\alpha}u^{n}\|_{L^{2}_{t}(\dot{L}^{2})}^{2}\,.$

and

$\displaystyle\int_{0}^{t}(\Lambda^{n}\underline{u}^{n}(t))^{T}(\Lambda^{n})^{2}{\underline{u}^{n}}^{\prime}(t)=\sum_{j=1}^{n}\lambda_{j}^{3}\int_{0}^{t}u^{n}_{j}(s){u^{n}_{j}}^{\prime}(s)\,ds$
$\displaystyle=\frac{1}{2}\sum_{j=1}^{n}\lambda_{j}^{3}\int_{0}^{t}\frac{d}{dt}\Bigl{(}u^{n}_{j}\Bigr{)}^{2}(s)\,ds=\frac{1}{2}\|\mathcal{A}^{3/2}u^{n}(t)\|_{\dot{L}^{2}(\Omega)}^{2}-\frac{1}{2}\|\mathcal{A}^{3/2}u^{n}(0)\|_{\dot{L}^{2}(\Omega)}^{2}\,.$

This together with Young’s inequality yields the energy estimate

$\displaystyle\frac{1}{2}\|\sqrt{1-\sigma(t)}\mathcal{A}u^{n}_{t}(t)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{b}{2\Gamma(\alpha)t^{1-\alpha}}\|\mathcal{A}^{1+\beta/2}\partial_{t}^{\alpha}u^{n}\|_{L^{2}_{t}(\dot{L}^{2})}^{2}+\frac{c^{2}}{2}\|\mathcal{A}^{3/2}u^{n}(t)\|_{\dot{L}^{2}(\Omega)}^{2}$
(22)
$\displaystyle\leq\frac{1}{2}\|\sqrt{1-\sigma(0)}\mathcal{A}u^{n}_{t}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\mathcal{A}^{3/2}u^{n}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{1}{2\epsilon}\|\mathcal{A}u^{n}_{t}\|_{L^{2}_{t}(\dot{L}^{2})}^{2}$
$\displaystyle\
+\frac{\epsilon}{2}\|-\frac{1}{2}\sigma_{t}\mathcal{A}u^{n}_{t}-\mu\mathcal{A}u^{n}_{t}-\rho\mathcal{A}u^{n}+\mathcal{A}h$
$\displaystyle\quad\
+(c_{0}^{2}/c^{2})(-\triangle\sigma\,u^{n}_{tt}-\nabla\sigma\cdot\nabla
u^{n}_{tt}+\triangle\mu\,u^{n}_{t}+\nabla\mu\cdot\nabla
u^{n}_{t}+\triangle\rho\,u^{n}+\nabla\rho\cdot\nabla
u^{n})\|_{L^{2}_{t}(\dot{L}^{2})}^{2}$
$\displaystyle\leq\frac{1}{2}\|\sqrt{1-\sigma(0)}\mathcal{A}u^{n}_{t}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\mathcal{A}^{3/2}u^{n}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{1}{2\epsilon}\|\mathcal{A}u^{n}_{t}\|_{L^{2}_{t}(\dot{L}^{2})}^{2}$
$\displaystyle\quad+\frac{\epsilon}{2}\Bigl{(}\frac{1}{2}\|\sigma_{t}\|_{L^{2}_{t}(L^{\infty})}\|\mathcal{A}u^{n}_{t}\|_{L^{\infty}_{t}(L^{2})}+\|\mu\|_{L^{2}_{t}(L^{\infty})}\|\mathcal{A}u^{n}_{t}\|_{L^{\infty}_{t}(L^{2})}$
$\displaystyle\qquad\qquad+\|\rho\|_{L^{2}_{t}(L^{\infty})}\|\mathcal{A}u^{n}\|_{L^{\infty}_{t}(L^{2})}+\|\mathcal{A}h\|_{L^{2}_{t}(L^{2})}$
$\displaystyle\qquad\quad+\frac{\|c_{0}\|_{L^{\infty}(\Omega)}}{c}\bigl{(}\|\triangle\sigma\|_{L^{\infty}(L^{4})}\|u^{n}_{tt}\|_{L^{2}(L^{4})}+\|\nabla\sigma\|_{L^{\infty}_{t}(L^{\infty})}\|\nabla
u^{n}_{tt}\|_{L^{2}_{t}(L^{2})}$
$\displaystyle\qquad\qquad+\|\triangle\mu\|_{L^{2}_{t}(L^{2})}\|u^{n}_{t}\|_{L^{\infty}_{t}(L^{\infty})}+\|\nabla\mu\|_{L^{2}(L^{4})}\|\nabla
u^{n}_{t}\|_{L^{\infty}(L^{4})}$
$\displaystyle\qquad\qquad+\|\triangle\rho\|_{L^{2}_{t}(L^{2})}\|u^{n}\|_{L^{\infty}_{t}(L^{\infty})}+\|\nabla\rho\|_{L^{2}(L^{4})}\|\nabla
u^{n}\|_{L^{\infty}(L^{4})}\bigl{)}\Bigr{)}^{2}$

Here we can make use of the fact that
$\|u^{n}\|_{L^{\infty}(0,t;Z)}\leq\|u^{n}(0)\|_{Z}+\sqrt{T}\|u^{n}_{t}\|_{L^{2}(0,t;Z)}$
and the embedding estimates

$\displaystyle\|v\|_{L^{4}(\Omega)}\leq C_{H^{1},L^{4}}\|\nabla v\|\,,\quad
v\in H_{0}^{1}(\Omega)\,,$ (23) $\displaystyle\|v\|_{L^{\infty}(\Omega)}\leq
C_{H^{2},L^{\infty}}\|\mathcal{A}v\|_{\dot{L}^{2}(\Omega)}\,,\quad v\in
H_{0}^{1}(\Omega)\cap H^{2}(\Omega)\,,$

in order to further estimate

$\displaystyle\|u^{n}_{tt}\|_{L^{2}(L^{4})}$ $\displaystyle\leq
C_{H^{1},L^{4}}\|\nabla u^{n}_{tt}\|_{L^{2}_{t}(L^{2})}$
$\displaystyle\|u^{n}_{t}\|_{L^{\infty}_{t}(L^{\infty})}$ $\displaystyle\leq
C_{H^{2},L^{\infty}}\|\mathcal{A}u^{n}_{t}\|_{L^{\infty}_{t}(L^{2})}$
$\displaystyle\|u^{n}\|_{L^{\infty}_{t}(L^{\infty})}$ $\displaystyle\leq
C_{H^{2},L^{\infty}}\|\mathcal{A}u^{n}\|_{L^{\infty}_{t}(L^{2})}$
$\displaystyle\|\nabla u^{n}\|_{L^{\infty}_{t}(L^{\infty})}$
$\displaystyle\leq
C_{H^{2},L^{\infty}}\|\mathcal{A}^{3/2}u^{n}\|_{L^{\infty}_{t}(L^{2})}$

Now we proceed with estimating $\|\nabla u^{n}_{tt}\|_{L^{2}}^{2}$ by
multiplying the ode (20) with $\Lambda^{n}{\underline{u}^{n}}^{\prime\prime}$,
that is testing (19) with $v=\mathcal{A}u^{n}_{tt}(t)$, and using integration
by parts (note that all terms in
$(1-\sigma)u^{n}_{tt}+c^{2}\mathcal{A}u^{n}+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}u^{n}+\mu
u^{n}_{t}+\rho u^{n}-h$ vanish on $\partial\Omega$) as well as Young’s
inequality

$\displaystyle
0=\langle(1-\sigma)u^{n}_{tt}+c^{2}\mathcal{A}u^{n}+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}u^{n}+\mu
u^{n}_{t}+\rho u^{n}-h,\mathcal{A}u^{n}_{tt}\rangle$
$\displaystyle=\|\sqrt{1\\!-\\!\sigma}\nabla
u^{n}_{tt}\|_{L^{2}(\Omega)}^{2}+\langle-\nabla\sigma\,u^{n}_{tt}+\nabla\Bigl{(}c^{2}\mathcal{A}u^{n}+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}u^{n}+\mu
u^{n}_{t}+\rho u^{n}\\!-\\!h\Bigr{)},\nabla u^{n}_{tt}\rangle_{L^{2}(\Omega)}$
$\displaystyle\geq\|\sqrt{1-\sigma}\nabla
u^{n}_{tt}\|_{L^{2}(\Omega)}^{2}-\frac{1}{2}(1-\overline{\sigma})\|\nabla
u^{n}_{tt}\|_{L^{2}(\Omega)}^{2}$
$\displaystyle\quad-\frac{1}{2(1-\overline{\sigma})}\|-\nabla\sigma\,u^{n}_{tt}+c^{2}\nabla\mathcal{A}u^{n}+b\nabla\mathcal{A}^{\beta}\partial_{t}^{\alpha}u^{n}$
$\displaystyle\hskip 85.35826pt+\nabla\mu\,u^{n}_{t}+\mu\nabla
u^{n}_{t}+\nabla\rho\,u^{n}+\rho\nabla u^{n}-\nabla h\|_{L^{2}(\Omega)}^{2}$

which yields

$\displaystyle\|\nabla
u^{n}_{tt}\|_{L^{2}_{t}(L^{2})}\leq\frac{1}{1-\overline{\sigma}}$
$\displaystyle\cdot\Bigl{(}\|\nabla\sigma\|_{L^{\infty}(L^{4})}\|u^{n}_{tt}\|_{L^{2}(L^{4})}^{2}+c^{2}\|\mathcal{A}^{3/2}u^{n}\|_{L^{2}_{t}(L^{2})}+b\|\mathcal{A}^{1/2+\beta}\partial_{t}^{\alpha}u^{n}\|_{L^{2}_{t}(L^{2})}$
$\displaystyle\qquad+\|\nabla\mu\|_{L^{2}(L^{4})}\|u^{n}_{t}\|_{L^{\infty}(L^{4})}+\|\mu\|_{L^{2}_{t}(L^{\infty})}\|\nabla
u^{n}_{t}\|_{L^{\infty}_{t}(L^{2})}$
$\displaystyle\qquad+\|\nabla\rho\|_{L^{2}_{t}(L^{2})}\|u^{n}\|_{L^{\infty}_{t}(L^{\infty})}+\|\rho\|_{L^{2}_{t}(L^{\infty})}\|\nabla
u^{n}\|_{L^{\infty}_{t}(L^{2})}\Bigr{)}$

where we can again employ the embedding estimates (23) and assume

$\|\nabla\sigma\|_{L^{\infty}(L^{4})}<\frac{1-\overline{\sigma}}{C_{H^{1}\to
L^{4}}}$ (24)

in order to extract an estimate of the form

$\displaystyle\|\nabla u^{n}_{tt}\|_{L^{2}_{t}(L^{2})}$ (25)
$\displaystyle\leq
C\Bigl{(}\|\mathcal{A}^{3/2}u^{n}\|_{L^{2}_{t}(L^{2})}+\|\mathcal{A}^{1/2+\beta}\partial_{t}^{\alpha}u^{n}\|_{L^{2}_{t}(L^{2})}+\|\nabla
u^{n}_{t}\|_{L^{\infty}_{t}(L^{2})}+\|\nabla
u^{n}\|_{L^{\infty}_{t}(L^{2})}\Bigr{)}$

Adding a multiple (factor
$\epsilon(\|\triangle\sigma\|_{L^{\infty}(L^{4})}^{2}(C_{H^{1},L^{4}})^{2}+\|\nabla\sigma\|_{L^{\infty}_{t}(L^{\infty})}^{2})$)
of the square of (25) to (22), making $\epsilon$ small enough (so that
$C^{2}\epsilon<\frac{b}{2\Gamma(\alpha)t^{1-\alpha}})$ and all terms
containing $L^{\infty}(0,T)$ norms of $u^{n}$ on the right hand side of (22)
can be dominated by left hand side terms) using the fact that $1/2+\beta\leq
1+\beta/2$ for $\beta\in[0,1]$ and Gronwall’s inequality we end up with an
estimate of the form

$\displaystyle\|\nabla
u^{n}_{tt}\|_{L^{2}(L^{2})}^{2}+\|\mathcal{A}u^{n}_{t}\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}+\|\mathcal{A}^{1+\beta/2}\partial_{t}^{\alpha}u^{n}\|_{L^{2}_{t}(\dot{L}^{2})}^{2}+\|\mathcal{A}^{3/2}u^{n}\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}$
(26) $\displaystyle\leq
C(T)\Bigl{(}\|\mathcal{A}u^{n}_{t}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\|\mathcal{A}^{3/2}u^{n}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\|\mathcal{A}h\|_{L^{2}_{t}(\dot{L}^{2})}^{2}\Bigr{)}$

which via weak limits shows existence of a solution to the homogeneous initial
boundary value problem for (16) and transfers to $u$ as

$\displaystyle\|u\|_{U}^{2}:=\|\nabla
u_{tt}\|_{L^{2}(L^{2})}^{2}+\|\mathcal{A}u_{t}\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}+\|\mathcal{A}^{1+\beta/2}\partial_{t}^{\alpha}u\|_{L^{2}_{t}(\dot{L}^{2})}^{2}+\|\mathcal{A}^{3/2}u\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}$
(27) $\displaystyle\qquad\leq
C(T)\Bigl{(}\|\mathcal{A}u_{t}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\|\mathcal{A}^{3/2}u(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\|\mathcal{A}h\|_{L^{2}_{t}(\dot{L}^{2})}^{2}\Bigr{)}$

The required regularity on $\sigma$, $\mu$, $\rho$, $h$, $u_{0}$, $u_{1}$ is,
besides (18) and (24)

$\displaystyle\sigma\in H^{1}(0,T;L^{\infty}(\Omega))\cap
L^{\infty}(0,T;W^{2,4}(\Omega)\cap W^{1,\infty}(\Omega))$ (28)
$\displaystyle\mu\in L^{2}(0,T;H^{2}(\Omega))\,,\quad\rho\in
L^{2}(0,T;H^{2}(\Omega))$ $\displaystyle h\in
L^{2}(0,T;\dot{H}^{2}(\Omega))\,,\quad u_{0}\in\dot{H}^{3}(\Omega)\,,\quad
u_{1}\in\dot{H}^{2}(\Omega)$

######  Proposition 3.1.

Under conditions (18), (24), (28), there exists a unique solution

$u\in U:=H^{2}(0,T;\dot{L}^{2}(\Omega))\cap
W^{1,\infty}(0,T;\dot{H}^{2}(\Omega))\cap L^{\infty}(0,T;\dot{H}^{3}(\Omega))$
(29)

to the initial boundary value problem (16), (17). This solution satisfies the
estimate (27).

The energy estimate leading to this result has been obtained by basically
“multiplying (16) with $\mathcal{A}^{2}u_{t}$”, that is, taking the
$\dot{L}^{2}(\Omega)$ inner product of the pde with $\mathcal{A}^{2}u_{t}$ and
using selfadjointness of $\mathcal{A}$ in $\dot{L}^{2}(\Omega)$.

Later on, we will also need less regular solutions along with estimates on
them. Since the proofs are actually somewhat simpler then, we skip the details
on Galerkin approximation and only provide the energy estimates.

Multiplying (16) with $\mathcal{A}u_{t}$ we obtain

$\displaystyle\frac{1}{2}\|\sqrt{1-\sigma(t)}\nabla
u_{t}(t)\|_{L^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\mathcal{A}u(t)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{1}{2\Gamma(\alpha)t^{1-\alpha}}\|\mathcal{A}^{(1+\beta)/2}\partial_{t}^{\alpha}u\|_{L^{2}_{t}(\dot{L}^{2})}^{2}$
(30) $\displaystyle\leq\frac{1}{2}\|\sqrt{1-\sigma(0)}\nabla
u_{t}(0)\|_{L^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\mathcal{A}u(0)\|_{\dot{L}^{2}(\Omega)}^{2}$
$\displaystyle\quad-\int_{0}^{t}\langle\frac{1}{2}\sigma_{t}(s)\nabla
u_{t}(s)+\nabla\sigma(s)\,u_{tt}(s)+\nabla\mu\,u_{t}+\mu\nabla
u_{t}+\nabla\rho\,u+\rho\nabla u\\!-\\!\nabla h,\nabla
u_{t}(s)\rangle_{L^{2}(\Omega)}\,ds$
$\displaystyle\leq\frac{1}{2}\|\sqrt{1-\sigma(0)}\nabla
u_{t}(0)\|_{L^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\mathcal{A}u(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{1}{2\epsilon}\|\nabla
u_{t}\|_{L^{2}_{t}(L^{2})}^{2}$
$\displaystyle\qquad+\frac{\epsilon}{2}\Bigl{(}\frac{1}{2}\|\sigma_{t}\|_{L^{2}_{t}(L^{\infty})}\|\nabla
u_{t}\|_{L^{\infty}_{t}(L^{2})}+\|\nabla\sigma\|_{L^{\infty}_{t}(L^{\infty})}\|u_{tt}\|_{L^{2}_{t}(L^{2})}$
$\displaystyle\qquad\qquad+\|\nabla\mu\|_{L^{2}(L^{4})}\|u_{t}\|_{L^{\infty}(L^{4})}+\|\mu\|_{L^{2}_{t}(L^{\infty})}\|\nabla
u_{t}\|_{L^{\infty}_{t}(L^{2})}$
$\displaystyle\qquad\qquad+\|\nabla\rho\|_{L^{2}_{t}(L^{2})}\|u\|_{L^{\infty}_{t}(L^{\infty})}+\|\rho\|_{L^{2}(L^{4})}\|\nabla
u\|_{L^{\infty}(L^{4})}+\|\nabla h\|_{L^{2}_{t}(L^{2})}\Bigr{)}^{2}\,.$

The pde provides us with

$\displaystyle\|u_{tt}\|_{L^{2}_{t}(L^{2})}=\|-\frac{1}{1-\sigma}\Bigl{(}c^{2}\mathcal{A}u+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}u+\mu
u_{t}+\rho u-h\Bigr{)}\|_{L^{2}(L^{2})}$ $\displaystyle\leq
C(\|\mu\|_{L^{2}_{t}(L^{\infty})}+\|\rho\|_{L^{2}(L^{4})})\times$
$\displaystyle\qquad\sup_{t\in(0,T)}\Bigl{(}\|\sqrt{1\\!-\\!\sigma(t)}\nabla
u_{t}(t)\|_{L^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\mathcal{A}u(t)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{1}{2\Gamma(\alpha)t^{1-\alpha}}\|\mathcal{A}^{\beta}\partial_{t}^{\alpha}u\|_{L^{2}_{t}(\dot{L}^{2})}^{2}\Bigr{)}$

which, due to the condition $\beta\leq(1+\beta)/2$, allows us to dominate the
$\epsilon u_{tt}$ term on the right hand side of (30). Thus, using Gronwall’s
inequality, we get an estimate of the form

$\displaystyle\|u\|_{U_{lo}}^{2}:=\|u_{tt}\|_{L^{2}_{t}(\dot{L}^{2})}^{2}+\|\nabla
u_{t}\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}+\|\mathcal{A}^{(1+\beta)/2}\partial_{t}^{\alpha}u\|_{L^{2}_{t}(\dot{L}^{2})}^{2}+\|\mathcal{A}u\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}$
(31) $\displaystyle\leq C(T)\Bigl{(}\|\nabla
u_{t}(0)\|_{L^{2}(\Omega)}^{2}+\|\mathcal{A}u(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\|\nabla
h\|_{L^{2}(L^{2})}^{2}\Bigr{)}$

provided

$\displaystyle\sigma\in H^{1}(0,T;L^{\infty}(\Omega))\cap
L^{\infty}(0,T;W^{1,\infty}(\Omega))\,,$ (32) $\displaystyle\mu\in
L^{2}(0,T;L^{\infty}(\Omega)\cap W^{1,4}(\Omega))\,,\quad\rho\in
L^{2}(0,T;H^{1}(\Omega))$ $\displaystyle h\in L^{2}(0,T;H^{1}(\Omega))\,,\quad
u_{0}\in\dot{H}^{2}(\Omega)\,,\quad u_{1}\in\dot{H}^{1}(\Omega)$

######  Proposition 3.2.

Under the conditions (18), (32), there exists a unique solution $u$, where
$u\in H^{2}(0,T;\dot{L}^{2}(\Omega))\cap
W^{1,\infty}(0,T;\dot{H}^{1}(\Omega))\cap
L^{\infty}(0,T;\dot{H}^{2}(\Omega))$, to the initial boundary value problem
(16), (17), and this solution satisfies the estimate (31).

An even lower regularity estimate can be obtained by multiplication of (16)
with $u_{t}$, which yields

$\displaystyle\frac{1}{2}\|\sqrt{1-\sigma(t)}u_{t}(t)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\nabla
u(t)\|_{L^{2}(\Omega)}^{2}+\frac{1}{2\Gamma(\alpha)t^{1-\alpha}}\|\mathcal{A}^{\beta/2}\partial_{t}^{\alpha}u\|_{L^{2}_{t}(\dot{L}^{2})}^{2}$
$\displaystyle\leq\frac{1}{2}\|\sqrt{1-\sigma(0)}u_{t}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\nabla
u(0)\|_{L^{2}(\Omega)}^{2}$
$\displaystyle\qquad-\int_{0}^{t}\langle\frac{1}{2}\sigma_{t}(s)u_{t}(s)+\mu\,u_{t}+\rho
u-h,u_{t}(s)\rangle\,ds$
$\displaystyle\leq\frac{1}{2}\|\sqrt{1-\sigma(0)}u_{t}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\nabla
u(0)\|_{L^{2}(\Omega)}^{2}+\frac{1}{2\epsilon}\|u_{t}\|_{L^{2}_{t}(\dot{L}^{2})}^{2}$
$\displaystyle\qquad+\frac{\epsilon}{2}\Bigl{(}\frac{1}{2}\|\sigma_{t}\|_{L^{2}_{t}(L^{\infty})}\|u_{t}\|_{L^{\infty}_{t}(\dot{L}^{2})}+\|\mu\|_{L^{2}_{t}(L^{\infty})}\|u_{t}\|_{L^{\infty}_{t}(\dot{L}^{2})}$
$\displaystyle\qquad\qquad+\|\rho\|_{L^{2}(L^{4})}\|u\|_{L^{\infty}(L^{4})}+\|h\|_{L^{2}_{t}(\dot{L}^{2})}\Bigr{)}^{2}\,,$

(where we have used the fact that
$\|\frac{c}{c_{0}}\|_{L^{\infty}(\Omega)}\leq 1$), hence an estimate of the
form

$\displaystyle\|u\|_{U_{vl}}^{2}:=\
\|u_{t}\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}+\|\mathcal{A}^{\beta/2}\partial_{t}^{\alpha}u\|_{L^{2}(L^{2})}^{2}+\|\nabla
u\|_{L^{\infty}_{t}(L^{2})}^{2}$ (33) $\displaystyle\leq
C(T)\Bigl{(}\|u_{t}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\|\nabla
u(0)\|_{L^{2}(\Omega)}^{2}+\|h\|_{L^{2}_{t}(\dot{L}^{2})}^{2}\Bigr{)}$

Here it obviously suffices to assume

$\displaystyle\sigma\in H^{1}(0,T;L^{\infty}(\Omega))\,,\quad\mu\in
L^{2}_{t}(L^{\infty})\,,\quad\rho\in L^{2}(L^{4})$ (34) $\displaystyle h\in
L^{2}_{t}(\dot{L}^{2})\,,\quad u_{0}\in\dot{H}^{1}(\Omega)\,,\quad
u_{1}\in\dot{L}^{2}(\Omega)$

######  Proposition 3.3.

Under conditions (18) and (34), there exists a unique solution $u$ lying in
the space $u\in H^{2}(0,T;\dot{H}^{-1}(\Omega))\cap
W^{1,\infty}(0,T;\dot{L}^{2}(\Omega))\cap L^{\infty}(0,T;\dot{H}^{1}(\Omega))$
and satisfying the initial boundary value problem (16), (17). This solution
satisfies the estimate (33).

We proceed to proving well-posedness of the nonlinear problem (3) with
$D=b\mathcal{A}^{\beta}\partial_{t}^{\alpha}$ by applying a fixed point
argument to the operator $\mathcal{T}$ mapping $v$ to the solution of

$\displaystyle(1-2\kappa
v)u_{tt}+c^{2}\mathcal{A}u+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}u-2\kappa
v_{t}u_{t}$ $\displaystyle=r\mbox{ in }\Omega\times(0,T)$ (35) $\displaystyle
u(0)=0,\quad u_{t}(0)$ $\displaystyle=0\mbox{ in }\Omega$

that is, of (16), (17) with $\sigma(x,t)=-2\kappa(x)v(x,t)$,
$\mu(x,t)=-2\kappa(x)v_{t}(x,t)$, $\rho(x,t)=0$, $h(x,t)=r(x,t)$. We first
consider self-mapping of $\mathcal{T}$. Even in case of constant $\kappa$, the
regularity requirements on $\sigma$, $\mu$, $\rho$ force us into the high
regularity scenario of Proposition 3.1. For spatially variable $\kappa$, due
to the estimates

$\displaystyle\|\triangle$
$\displaystyle\mu\|_{L^{2}(L^{2})}=2\|\triangle(\kappa
v_{t})\|_{L^{2}(L^{2})}$ $\displaystyle\leq
2\Bigl{(}\|\triangle\kappa\|_{L^{2}(\Omega)}\|v_{t}\|_{L^{2}(0,T;L^{\infty}(\Omega)}+2\|\nabla\kappa\|_{L^{4}(\Omega)}\|\nabla
v_{t}\|_{L^{2}(L^{4})}+\|\kappa\|_{L^{\infty}(\Omega)}\|\triangle
v_{t}\|_{L^{2}(L^{2})}\Bigr{)}$ $\displaystyle\|\sigma_{t}$
$\displaystyle\|_{L^{2}_{t}(L^{\infty})}\leq
2C_{H^{2},L^{\infty}}\|\triangle(\kappa v_{t})\|_{L^{2}(L^{2})}$
$\displaystyle\|\triangle\sigma$
$\displaystyle\|_{L^{\infty}(L^{4})}=2\|\triangle(\kappa
v)\|_{L^{\infty}(L^{4})}$ $\displaystyle\quad\leq
2\Bigl{(}\|\triangle\kappa\|_{L^{4}(\Omega)}\|v\|_{L^{\infty}(L^{\infty})}+2\|\nabla\kappa\|_{L^{4}(\Omega)}\|\nabla
v\|_{L^{\infty}(L^{\infty})}+\|\kappa\|_{L^{\infty}(\Omega)}\|\triangle
v\|_{L^{\infty}(L^{4})}\Bigr{)}$ $\displaystyle\|\nabla\sigma$
$\displaystyle\|_{L^{\infty}_{t}(L^{\infty})}=2\|\nabla(\kappa
v)\|_{L^{\infty}_{t}(L^{\infty})}\leq
2\Bigl{(}\|\nabla\kappa\|_{L^{\infty}(\Omega)}\|v\|_{L^{\infty}(L^{\infty})}+\|\kappa\|_{L^{\infty}(\Omega)}\|\nabla
v\|_{L^{\infty}(L^{\infty})}\Bigr{)}$

the regularity

$\kappa\in W^{2,4}(\Omega)\cap W^{1,\infty}(\Omega)$ (36)

is sufficient for obtaining the regularity (28) for any for any $v\in U$. To
achieve the nondegeneracy and smallness conditions (18), (24), we use the
estimates

$\displaystyle\|\sigma\|_{L^{\infty}_{t}(L^{\infty})}\leq
2\|\kappa\|_{L^{\infty}(\Omega)}\|v\|_{L^{\infty}_{t}(L^{\infty})}$ (37)
$\displaystyle\|\nabla\sigma\|_{L^{\infty}(L^{4})}\leq
2\|\nabla\kappa\|_{L^{4}(\Omega)}\|v\|_{L^{\infty}(L^{\infty})}+2\|\kappa\|_{L^{\infty}(\Omega)}\|\nabla
v\|_{L^{\infty}(L^{4})}$

and additionally to (36) require smallness of $v$.

Proposition 3.1 yields that $\mathcal{T}$ is a self-mapping on

$B_{R}=\\{v\in U\ :\ \|v\|_{U}\leq R\\}$

provided the initial and right hand side data are sufficiently small so that

$C(T)\Bigl{(}\|\mathcal{A}u_{1}\|_{\dot{L}^{2}(\Omega)}^{2}+\|\mathcal{A}^{3/2}u_{0}\|_{\dot{L}^{2}(\Omega)}^{2}+\|\mathcal{A}r\|_{L^{2}_{t}(\dot{L}^{2})}^{2}\Bigr{)}\leq
R^{2}$ (38)

with $C(T)$ as in Proposition 3.1. In view of (18), (24), (37), we choose $R$
such that

$\Bigl{(}C_{H^{1}\to
L^{4}}\Bigl{(}\|\nabla\kappa\|_{L^{4}(\Omega)}C_{H^{2},L^{\infty}}+\|\kappa\|_{L^{\infty}(\Omega)}C_{H^{1},L^{4}}\Bigr{)}+\|\kappa\|_{L^{\infty}(\Omega)}C_{H^{2},L^{\infty}}\Bigr{)}R<\frac{1}{2}\,.$
(39)

Contractivity of $\mathcal{T}$ can be shown by taking $v^{(1)}$ and $v^{(2)}$
in $B_{R}$ and considering $u^{(1)}=\mathcal{T}v^{(1)}$ and
$u^{(2)}=\mathcal{T}v^{(2)}$, whose differences $\overline{u}=u^{(1)}-u^{(2)}$
and $\overline{v}=v^{(1)}-v^{(2)}$ solve

$(1-2\kappa
v^{(1)})\overline{u}_{tt}+c^{2}\mathcal{A}\overline{u}+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}\overline{u}_{t}-2\kappa
v_{t}^{(1)}\overline{u}_{t}=2\kappa\overline{v}_{t}u^{(2)}_{t}+2\kappa\overline{v}u^{(2)}_{tt}$
(40)

with homogeneous initial conditions. Similarly to above, with $\sigma=-2\kappa
v^{(1)}$, $\mu=-2\kappa v^{(1)}_{t}$, $\rho(x,t)=0$,
$h=2\kappa\overline{v}_{t}u^{(2)}_{t}+2\kappa\overline{v}u^{(2)}_{tt}$, since
$v^{(1)}$, $u^{(2)}\in B_{R}$ (the latter due to the already shown self-
mapping property of $\mathcal{T}$) we satisfy the conditions (18), (24), (28)
on $\sigma$ and $\mu$. However $h$ in general fails to be contained in
$L^{2}(0,T;\dot{H}^{2}(\Omega))$ (in particular the term
$2\kappa\overline{v}u^{(2)}_{tt}$), hence we move to the lower order
regularity regime from Proposition 3.2. To this end, we estimate

$\|\nabla h\|_{L^{2}(L^{2})}\leq
2\|\kappa\|_{L^{4}(\Omega)}\Bigl{(}\|\overline{v}_{t}\|_{L^{2}(L^{4})}\|u^{(2)}_{t}\|_{L^{\infty}_{t}(L^{\infty})}+\|\overline{v}\|_{L^{\infty}_{t}(L^{\infty})}\|u^{(2)}_{tt}\|_{L^{2}(L^{4})}\Bigr{)}\,.$

Thus imposing the additional smallness condition

$\theta:=2\|\kappa\|_{L^{4}(\Omega)}\sqrt{C}C_{H^{1},L^{4}}C_{H^{2},L^{\infty}}(\sqrt{T}+1)R<1$

on $R$ and employing from Proposition 3.2, we obtain contractivity

$\|\mathcal{T}v^{(1)}-\mathcal{T}v^{(2)}\|_{U_{lo}}=\|\overline{u}\|_{U_{lo}}\leq\theta\|\overline{v}\|_{U_{lo}}=\theta\|v^{(1)}-v^{(2)}\|_{U_{lo}}$

######  Theorem 3.1.

For any $\alpha\in(0,1)$, $T>0$, $\kappa\in W^{2,4}(\Omega)\cap
W^{1,\infty}(\Omega)$ there exists $R_{0}>0$ such that for any data
$u_{0}\in\dot{H}^{3}(\Omega)$, $u_{1}\in\dot{H}^{2}(\Omega)$, $r\in
L^{2}(0,T;\dot{H}^{2}(\Omega))$ satisfying

$\|\mathcal{A}u_{1}\|_{\dot{L}^{2}(\Omega)}^{2}+\|\mathcal{A}^{3/2}u_{0}\|_{\dot{L}^{2}(\Omega)}^{2}+\|\mathcal{A}r\|_{L^{2}(\dot{L}^{2})}^{2}\leq
R_{0}^{2}$ (41)

there exists a unique solution $u\in U$ of

$\displaystyle(1-2\kappa
u)u_{tt}+c^{2}\mathcal{A}u+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}u$
$\displaystyle=2\kappa(u_{t})^{2}+r\mbox{ in }\Omega\times(0,T)$ (42)
$\displaystyle u(0)=u_{0},\quad u_{t}(0)$ $\displaystyle=u_{1}\mbox{ in
}\Omega\,.$

Existence of the linearisation of $G$ requires well-posedness of (9) with
$D=b\mathcal{A}\partial_{t}^{\alpha}$, that is, (16), (17) with
$\sigma=-2\kappa u$, $\mu=-4\kappa u_{t}$, $\rho=-2\kappa u_{tt}$,
$h=2\underline{\delta\kappa}(u\,u_{tt}+u_{t}^{2})$. Due to the appearance of a
$u_{tt}$ term we are in a similar situation to the contractivity proof above
and therefore the lower regularity Proposition 3.2 is the right framework for
analysing the linearisation of the forward problem.

######  Proposition 3.4.

Under the assumptions of Theorem 3.1, for any $\underline{\delta\kappa}\in
W^{1,\infty}(\Omega)$ there exists a unique solution $z\in U_{lo}$ of

$\displaystyle(1-2\kappa
u)z_{tt}+c^{2}\mathcal{A}z+b\mathcal{A}^{\beta}\partial_{t}^{\alpha}z-4\kappa
u_{t}\,z_{t}-2\kappa u_{tt}\,z$
$\displaystyle=2\underline{\delta\kappa}(u\,u_{tt}+u_{t}^{2})\mbox{ in
}\Omega\times(0,T)$ (43) $\displaystyle z(0)=0,\quad z_{t}(0)$
$\displaystyle=0\mbox{ in }\Omega\,,$

where $u\in U$ solves (42).

In order to prove Fréchet differentiability we also need to bound the solution
of (11) that is, (16), (17) with $\sigma=-2\kappa u$, $\mu=-4\kappa u_{t}$,
$\rho=-2\kappa u_{tt}$,
$h=2\underline{d\kappa}(v\tilde{u}_{tt}+uv_{tt}+(\tilde{u}_{t}+u_{t})v_{t})+2\kappa(vv_{tt}+v_{t}^{2})$,
where $v$ can be bounded analogously to $z$ by Proposition 3.2; in particular
we can only expect to have $v_{tt}\in L^{2}(0,T;L^{2}(\Omega))$, so $h\in
L^{2}(0,T;H^{1}(\Omega))$ is out of reach and we show Fréchet
differentiability in the very low regularity regime of Proposition 3.3.

######  Theorem 3.2.

For any $\alpha\in(0,1)$, $T>0$, $\bar{R}>0$ there exists $R_{0}>0$ such that
for any data $u_{0}\in\dot{H}^{3}(\Omega)$, $u_{1}\in\dot{H}^{2}(\Omega)$,
$r\in L^{2}(0,T;\dot{H}^{2}(\Omega))$ satisfying (41), the parameter-to-state
map $G:B_{\bar{R}}(0)\to U$ is well-defined according to Theorem 3.1.
Moreover, it is Fréchet differentiable as an operator $G:B_{\bar{R}}(0)\to
U_{vl}$. Here $B_{\bar{R}}(0)=\\{\kappa\in W^{2,4}(\Omega)\cap
W^{1,\infty}(\Omega)\,:\,\|\kappa\|_{W^{2,4}(\Omega)\cap
W^{1,\infty}(\Omega)}\leq\bar{R}\\}$.

###  3.2 Fractional Zener damping

Consider

$D=b_{1}\mathcal{A}\partial_{t}^{\alpha_{1}}+b_{2}\partial_{t}^{\alpha_{2}+2}\quad\mbox{
with }b_{2}>0,\ b_{1}\geq b_{2}c^{2},\ 1\geq\alpha_{1}\geq\alpha_{2}>0,$

where based on the analysis in [17] we expect to get well-posedness of the
nonlinear forward problem only in case $\alpha_{1}=1$, so we first of all
focus on this case. Later on we will also prove a well-posedness result on the
equation linearized at $\kappa=0$ in the practically relevant case
$\alpha_{1}=\alpha_{2}=:\alpha$.

Again we first of all consider the initial boundary value problem for the
general linear pde

$(1-\sigma)u_{tt}+c^{2}\mathcal{A}u+b_{1}\mathcal{A}u_{t}+b_{2}\partial_{t}^{\alpha_{2}+2}u+\mu
u_{t}+\rho u=h$ (44) $u(0)=u_{0}\,,\quad
u_{t}(0)=u_{1}\,,\quad(u_{tt}(0)=u_{2}\mbox{ in case
$\alpha_{2}>\frac{1}{2}$)}$ (45)

with given space- and time dependent functions $\sigma$, $\mu$, $\rho$, $h$.

Again we skip the details about the Faedo-Galerkin approach and the
discretisation index $n$ and only provide the crucial energy estimate. We
multiply (44) with $\mathcal{A}u_{tt}$ and integrate with respect to time,
using the inequalities and identities

$\int_{0}^{t}\langle\mathcal{A}u(s),\mathcal{A}u_{tt}(s)\rangle\,ds=\langle\mathcal{A}u(t),\mathcal{A}u_{t}(t)\rangle-\langle\mathcal{A}u(0),\mathcal{A}u_{t}(0)\rangle-\int_{0}^{t}\|\mathcal{A}u_{t}(s)\|_{\dot{L}^{2}(\Omega)}^{2}$

and

$\displaystyle\int_{0}^{t}\langle\partial_{t}^{\alpha_{2}+2}[u](s),\mathcal{A}u_{tt}(s)\rangle\,ds=\int_{0}^{t}\langle\partial_{t}^{\alpha_{2}}[\nabla
u_{tt}](s),\nabla u_{tt}(s)\rangle\,ds$
$\displaystyle\geq\frac{1}{2}\int_{0}^{t}\partial_{t}^{\alpha_{2}}\left[\|\nabla
u_{tt}\|_{\dot{L}^{2}(\Omega)}^{2}\right](s)\,ds=\frac{1}{2}I_{t}^{1-\alpha}\left[\|\nabla
u_{tt}\|_{L^{2}(\Omega)}^{2}\right](t)\geq\frac{1}{2\Gamma(1-\alpha)t^{\alpha}}\|\nabla
u_{tt}\|_{L^{2}_{t}(L^{2})}^{2}\,,$

where the latter equality holds provided $u_{tt}(0)=0$ and we have applied
(14) (actually to the Fourier components of the Galerkin discretisation
$w=\lambda_{j}^{1/2}{u^{n}_{j}}^{\prime\prime}$) with $\gamma=\alpha_{2}$.

This yields the energy estimate

$\displaystyle\Bigl{(}\frac{b_{2}}{2\Gamma(1-\alpha)t^{\alpha}}+1-\overline{\sigma}\Bigr{)}\|\nabla
u_{tt}\|_{L^{2}_{t}(L^{2})}^{2}+\frac{b_{1}}{2}\|\mathcal{A}u_{t}(t)\|_{\dot{L}^{2}(\Omega)}^{2}$
$\displaystyle\leq\frac{b_{1}}{2}\|\mathcal{A}u_{t}(0)\|_{L^{2}_{t}(\dot{L}^{2})}^{2}+c^{2}\langle\mathcal{A}u(0),\mathcal{A}u_{t}(0)\rangle$
$\displaystyle\quad-c^{2}\langle\mathcal{A}u(t),\mathcal{A}u_{t}(t)\rangle+c^{2}\|\mathcal{A}u_{t}\|_{L^{2}_{t}(\dot{L}^{2})}^{2}+\int_{0}^{t}\langle\nabla\sigma
u_{tt}+\nabla(\mu u_{t}+\rho u-h)(s),\nabla u_{tt}(s)\rangle\,ds$
$\displaystyle\leq\frac{b_{1}}{2}\|\mathcal{A}u_{t}(0)\|_{L^{2}_{t}(\dot{L}^{2})}^{2}+c^{2}\langle\mathcal{A}u(0),\mathcal{A}u_{t}(0)\rangle$
$\displaystyle\qquad+\frac{b_{1}}{4}\|\mathcal{A}u_{t}(t)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{c^{4}}{b_{1}}\|\mathcal{A}u(t)\|_{\dot{L}^{2}(\Omega)}^{2}+c^{2}\|\mathcal{A}u_{t}\|_{L^{2}_{t}(\dot{L}^{2})}^{2}$
$\displaystyle\qquad+\|\nabla\sigma\|_{L^{\infty}(L^{4})}\|u_{tt}\|_{L^{2}(L^{4})}\|\nabla
u_{tt}\|_{L^{2}_{t}(L^{2})}+\frac{\epsilon}{2}\|\nabla
u_{tt}\|_{L^{2}_{t}(L^{2})}^{2}$
$\displaystyle\qquad+\frac{1}{2\epsilon}\Bigl{(}\|\nabla\mu\|_{L^{\infty}_{t}(L^{2})}\|u_{t}\|_{L^{2}(0,t;L^{\infty}(\Omega)}+\|\mu\|_{L^{\infty}(L^{4})}\|\nabla
u_{t}\|_{L^{2}(L^{4})}$
$\displaystyle\qquad\qquad+\|\nabla\rho\|_{L^{2}_{t}(L^{2})}\|u\|_{L^{\infty}_{t}(L^{\infty})}+\|\rho\|_{L^{2}(L^{4})}\|\nabla
u\|_{L^{\infty}(L^{4})}+\|\nabla h\|_{L^{2}_{t}(L^{2})}\Bigr{)}^{2}$

Here we assume nondegeneracy

$\sigma(x,t)\leq\overline{\sigma}<\frac{b_{2}}{\Gamma(1-\alpha)T^{\alpha}}+1\mbox{
for all }x\in\Omega\,\quad t\in(0,T)\,.$ (46)

and smallness of $\nabla\sigma$

$\|\nabla\sigma\|_{L^{\infty}(L^{4})}<\frac{\frac{b_{2}}{\Gamma(1-\alpha)T^{\alpha}}+1-\overline{\sigma}}{C_{H^{1}\to
L^{4}}}$ (47)

and choose
$\epsilon<\frac{b_{2}}{\Gamma(1-\alpha)T^{\alpha}}+1\\!-\\!\underline{\sigma}-C_{H^{1}\\!\to
L^{4}}\|\nabla\sigma\|_{L^{\infty}(L^{4})}$ to obtain, using Gronwall’s Lemma,

$\displaystyle\|\nabla
u_{tt}\|_{L^{2}(L^{2})}^{2}+\|\mathcal{A}u_{t}\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}$
$\displaystyle\leq
C\Bigl{(}\|\mathcal{A}u_{t}(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\|\nabla
h\|_{L^{2}_{t}(\dot{L}^{2})}^{2}\Bigr{)}$ (48)

The required regularity on $\sigma$, $\mu$, $\rho$, $h$, $u_{0}$, $u_{1}$, is,
besides (46), (47)

$\displaystyle\mu\in L^{\infty}(0,T;H^{1}(\Omega))\,,\quad\rho\in
L^{2}(0,T;H^{1}(\Omega))\,,$ (49) $\displaystyle h\in
L^{2}(0,T;\dot{H}^{1}(\Omega))\,,\quad u_{0},u_{1}\in\dot{H}^{2}(\Omega),\quad
u_{2}=0\,.$

######  Proposition 3.5.

Under conditions (46), (47), (49), there exists a unique solution

$u\in U:=H^{2}(0,T;\dot{H}^{1}(\Omega))\cap
W^{1,\infty}(0,T;\dot{H}^{2}(\Omega))$ (50)

to the initial boundary value problem (44), (45), and this solution satisfies
the estimate (48).

######  Theorem 3.3.

For any $\alpha_{2}\in(0,1)$, $T>0$, $\kappa\in W^{1,4}(\Omega)$ there exists
$R_{0}>0$ such that for any data $u_{0},u_{1}\in\dot{H}^{2}(\Omega)$, $r\in
L^{2}(0,T;\dot{H}^{1}(\Omega))$ satisfying

$\|\mathcal{A}u_{1}\|_{\dot{L}^{2}(\Omega)}^{2}+\|\mathcal{A}u_{0}\|_{\dot{L}^{2}(\Omega)}^{2}+\|\nabla
r\|_{L^{2}(L^{2})}^{2}\leq R_{0}^{2}$ (51)

there exists a unique solution $u\in U$ of

$\displaystyle b_{2}\partial_{t}^{\alpha_{2}+2}u+(1-2\kappa
u)u_{tt}+c^{2}\mathcal{A}u+b_{1}\mathcal{A}u_{t}$
$\displaystyle=2\kappa(u_{t})^{2}+r\mbox{ in }\Omega\times(0,T)$ (52)
$\displaystyle u(0)=u_{0},\quad u_{t}(0)=u_{1},\quad u_{tt}$
$\displaystyle=0\mbox{ in }\Omega\,.$

######  Proposition 3.6.

Under the assumptions of Theorem 3.3, for any $\underline{\delta\kappa}\in
L^{3}(\Omega)$ there exists a unique solution $z\in U$ of

$\displaystyle b_{2}\partial_{t}^{\alpha_{2}+2}z+(1\\!-\\!2\kappa
u)z_{tt}\\!+\\!c^{2}\mathcal{A}z\\!+\\!b_{1}\mathcal{A}z_{t}\\!-\\!4\kappa
u_{t}\,z_{t}\\!-\\!2\kappa u_{tt}\,z$
$\displaystyle=2\underline{\delta\kappa}(u\,u_{tt}+u_{t}^{2})\mbox{ in
}\Omega\times(0,T)$ (53) $\displaystyle z(0)=0,\quad z_{t}(0),\quad z_{tt}=0$
$\displaystyle=0\mbox{ in }\Omega\,.$

where $u\in U$ solves (52).

######  Theorem 3.4.

For any $\alpha_{2}\in(0,1)$, $T>0$, $\bar{R}>0$ there exists $R_{0}>0$ such
that for any data $u_{0},u_{1}\in\dot{H}^{2}(\Omega)$, $r\in
L^{2}(0,T;\dot{H}^{2}(\Omega))$ satisfying (51), the parameter-to-state map
$G:B_{\bar{R}}(0)\to U$ is well-defined according to Theorem 3.3. Moreover, it
is Fréchet differentiable as an operator $G:B_{\bar{R}}(0)\to U$. Here
$B_{\bar{R}}(0)=\\{\kappa\in
W^{1,4}(\Omega)\,:\,\|\kappa\|_{W^{1,4}(\Omega)}\leq\bar{R}\\}$.

We now consider the linear problem in case $\alpha_{1}=\alpha_{2}=:\alpha$,
$b_{1}=b_{2}c^{2}+\delta$ with $\delta\geq 0$

$(1-\sigma)u_{tt}+c^{2}\mathcal{A}u+b_{1}\mathcal{A}\partial_{t}^{\alpha}u+b_{2}\partial_{t}^{\alpha+2}u+\mu
u_{t}+\rho u=h$ (54)

in which the differential operator can partially be factorised as

$\displaystyle(1-\sigma)\partial_{tt}+c^{2}\mathcal{A}+(b_{2}c^{2}+\delta)\mathcal{A}\partial_{t}^{\alpha}+b_{2}\partial_{t}^{\alpha+2}+\mu\partial_{t}+\rho\mbox{id}$
$\displaystyle=\Bigl{(}\partial_{tt}+c^{2}\mathcal{A}\Bigr{)}\Bigl{(}b_{2}\partial_{t}^{\alpha}+\mbox{id}\Bigr{)}-\sigma\partial_{tt}+\delta\mathcal{A}\partial_{t}^{\alpha}+\mu\partial_{t}+\rho\mbox{id}$

Thus, up to the “perturbation” terms containing $\sigma$, $\delta$, $\mu$, and
$\rho$, the auxiliary function $\tilde{u}=b_{2}\partial_{t}^{\alpha}u+u$
satisfies a wave equation $\tilde{u}_{tt}+c^{2}\mathcal{A}\tilde{u}=h$.
Motivated by this fact, we multiply (54) with $\mathcal{A}\tilde{u}_{t}$ to
obtain the energy identity

$\displaystyle\frac{1}{2}\|\nabla(b_{2}\partial_{t}^{\alpha}u+u)_{t}(t)\|_{L^{2}(\Omega)}^{2}$
$\displaystyle+\frac{c^{2}}{2}\|\mathcal{A}(b_{2}\partial_{t}^{\alpha}u+u)(t)\|_{\dot{L}^{2}(\Omega)}^{2}$
(55)
$\displaystyle\qquad+\delta\int_{0}^{t}\langle\nabla\partial_{t}^{\alpha}u(s),\nabla(b_{2}\partial_{t}^{\alpha}u+u)_{t}(s)\rangle\,ds$
$\displaystyle=\frac{1}{2}\|\nabla(b_{2}\partial_{t}^{\alpha}u+u)_{t}(0)$
$\displaystyle\|_{L^{2}(\Omega)}^{2}+\frac{c^{2}}{2}\|\mathcal{A}(b_{2}\partial_{t}^{\alpha}u+u)(0)\|_{\dot{L}^{2}(\Omega)}^{2}$
$\displaystyle\quad+\int_{0}^{t}\langle\nabla(h+\sigma u_{t}t-\mu u_{t}-\rho
u)(s),\nabla(b_{2}\partial_{t}^{\alpha}u+u)_{t}(s)\rangle\,ds\,.$

The term containing $\delta$ can be nicely tackled by means of Lemma 3.1

$\displaystyle\int_{0}^{t}\langle\nabla\partial_{t}^{\alpha}u(s)$
$\displaystyle,\nabla(b_{2}\partial_{t}^{\alpha}u+u)_{t}(s)\rangle\,ds$
$\displaystyle=\frac{b_{2}}{2}\|\nabla\partial_{t}^{\alpha}u(t)\|_{L^{2}(\Omega)}^{2}-\frac{b_{2}}{2}\|\nabla\partial_{t}^{\alpha}u(0)\|_{L^{2}(\Omega)}^{2}+\int_{0}^{t}\langle\nabla\partial_{t}^{\alpha}u(s),\nabla
u_{t}(s)\rangle\,ds$
$\displaystyle\geq\frac{b_{2}}{2}\|\nabla\partial_{t}^{\alpha}u(t)\|_{L^{2}(\Omega)}^{2}+\frac{1}{2\Gamma(\alpha)t^{1-\alpha}}\|\nabla\partial_{t}^{\alpha}u\|_{L^{2}_{t}(L^{2})}^{2}\,,$

(which reflects the physical fact that $\delta$ is the diffusivity of sound
and therefore the corresponding term models damping). However, in the term
containing $\sigma$ this is inhibited by the time-dependence of $\sigma$. Thus
in case $\alpha_{1}=\alpha_{2}=:\alpha<1$ we have to restrict ourselves to the
linearisation of the forward problem at $\kappa=0$ (where also $\mu=0$,
$\rho=0$), where the above together with Young’s inequality yields the energy
estimate

$\displaystyle\frac{1}{2}\|\nabla$
$\displaystyle(b_{2}\partial_{t}^{\alpha}u+u)_{t}\|_{L^{\infty}_{t}(L^{2})}^{2}+c^{2}\|\mathcal{A}(b_{2}\partial_{t}^{\alpha}u+u)\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}$
(56) $\displaystyle+\delta
b_{2}\|\nabla\partial_{t}^{\alpha}u(t)\|_{L^{\infty}_{t}(L^{2})}^{2}+\frac{1}{2\Gamma(\alpha)t^{1-\alpha}}\|\nabla\partial_{t}^{\alpha}u\|_{L^{2}_{t}(L^{2})}^{2}$
$\displaystyle\leq\|\nabla(b_{2}\partial_{t}^{\alpha}u+u)_{t}(0)\|_{L^{2}(\Omega)}^{2}+c^{2}\|\mathcal{A}(b_{2}\partial_{t}^{\alpha}u+u)(0)\|_{\dot{L}^{2}(\Omega)}^{2}+\frac{1}{2}\|\nabla
h\|_{L^{1}_{t}(L^{2})}^{2}$

The pde yields an estimate of $\tilde{u}_{tt}$ as follows

$\displaystyle\|(b_{2}\partial_{t}^{\alpha}u+u)_{tt}\|_{L^{\infty}_{t}(L^{2})}=\|h-c^{2}\mathcal{A}(b_{2}\partial_{t}^{\alpha}u+u)\|_{L^{\infty}_{t}(\dot{L}^{2})}^{2}$
(57)

To extract temporal reguarity of $u$ from regularity of
$b_{2}\partial_{t}^{\alpha}u+u$ for $b_{2}>0$, we make use of known regularity
results of time fractional ODEs: $b_{2}\partial_{t}^{\alpha}u+u=\tilde{u}\in
W^{k,\infty}(0,T;Z)$ implies $u\in W^{k+\alpha,\infty}(0,T;Z)$.

In order to formulate a result on this linearisation, we will postulate the
regularity of $u$ that is to be “expected” from (56), (57) (but apparently not
provable):

$u\in U:=W^{2+\alpha,\infty}(0,T;\dot{L}^{2}(\Omega))\cap
W^{1+\alpha,\infty}(0,T;\dot{H}^{1}(\Omega))\cap
W^{\alpha,\infty}(0,T;\dot{H}^{2}(\Omega))$ (58)

The required regularity of $\underline{\delta\kappa}$ to guarantee
$h=2\underline{\delta\kappa}(u\,u_{tt}+u_{t}^{2})\in L^{1}(0,t;L^{2}(\Omega))$
is therefore $\underline{\delta\kappa}\in L^{\infty}(\Omega)$.

We thus obtain the following result on the linearisation of the forward
problem at $\kappa=0$:

######  Proposition 3.7.

For any $u\in U$ and $\underline{\delta\kappa}\in L^{\infty}(\Omega)$ there
exists a unique solution $z\in U$ of

$\displaystyle
b_{2}\partial_{t}^{\alpha+2}z+z_{tt}+c^{2}\mathcal{A}z+b_{1}\mathcal{A}\partial_{t}^{\alpha}z$
$\displaystyle=2\underline{\delta\kappa}(u\,u_{tt}+u_{t}^{2})\mbox{ in
}\Omega\times(0,T)$ $\displaystyle z(0)=0,\quad z_{t}(0),\quad z_{tt}=0$
$\displaystyle=0\mbox{ in }\Omega\,.$

##  4 Injectivity of the linearised forward operator

The forwards map is defined by $F(\kappa)=\mbox{tr}_{\Sigma}u$, where
$\mbox{tr}_{\Sigma}v$ denotes the time trace of the space and time dependent
function $v:(0,T)\times\Omega$ at the observation surface $\Sigma$ (which may
also just be a single point $\Sigma=\\{x_{0}\\}$) and $u$ solves

$\displaystyle u_{tt}+c^{2}\mathcal{A}u+Du$
$\displaystyle=\kappa(x)(u^{2})_{tt}+r\mbox{ in }\Omega\times(0,T)$ (59)
$\displaystyle u(0)=0,\quad u_{t}(0)$ $\displaystyle=0\mbox{ in }\Omega\,.$

Its linearisation at $\kappa=0$ in direction $\underline{\delta\kappa}$ is
$F^{\prime}(0)\underline{\delta\kappa}=\mbox{tr}_{\Sigma}z_{0}$, where $z_{0}$
solves

$z_{tt}+c^{2}\mathcal{A}z+Dz=\underline{\delta\kappa}(u_{0}^{2})_{tt}$ (60)

where

$u_{0,tt}+c^{2}\mathcal{A}u_{0}+Du_{0}=r.$ (61)

Both pdes (60), (61) come with homogeneous initial conditions.

As in the previous section, we consider the two damping models

$D=b\mathcal{A}^{\beta}\partial_{t}^{\alpha}\quad\mbox{ ({\sc cwch})}$ (62)

and

$D=b_{1}\mathcal{A}\partial_{t}^{\alpha_{1}}+b_{2}\partial_{t}^{\alpha_{2}+2}\quad\mbox{
({\sc fz})}\,.$ (63)

The Laplace transformed solutions to the corresponding resolvent equation

$\hat{w}(\lambda,s)=\frac{1}{\omega(\lambda,s)}\mbox{ with
}\omega(\lambda,s)=\begin{cases}s^{2}+b\lambda^{\beta}s^{\alpha}+c^{2}\lambda\mbox{
for {\sc cwch}}\\\ b_{2}s^{2+\alpha_{2}}+s^{2}+b_{1}\lambda
s^{\alpha_{1}}+c^{2}\lambda\mbox{ for {\sc fz}}\end{cases}$

will play a crucial role in the proofs below.

From [17, Lemma 4.1] we conclude that in case of cwch, the function
$\omega^{{\sc cwch}}(\lambda,\cdot)$ has precisely two complex-conjugate zeros
$p^{{\sc cwch}}_{+}(\lambda)$, $p^{{\sc cwch}}_{-}(\lambda)$, which lie in the
left hand complex plane.

For fz, we first consider the particular parameter configuration
(corresponding to vanishing viscosity of sound)

$b_{1}=b_{2}c^{2}\mbox{ and }\alpha_{1}=\alpha_{2}$ (64)

in which we can factorise
$\omega^{FZ}(\lambda,s)=(b_{2}s^{\alpha_{2}}+1)(s^{2}+c^{2}\lambda)$ and get
the roots

$p^{FZ}_{0}=-\frac{1}{b_{2}}\mbox{ (only in case
$\alpha_{1}=\alpha_{2}=1$)}\,,\qquad p^{FZ}_{\pm}(\lambda)=\pm
ic\sqrt{\lambda}$

Note that $p^{FZ}_{0}$ is independent of $\lambda$, but
$p^{FZ}_{\pm}(\lambda)$ obviously allows to distinguish between different
$\lambda$’s. This distinction is possible in general, a fact that has already
been shown for the cwch case with $\beta=1$ in [17, Remark 4.1]. As an
additional result, that is not needed for the uniqueness proof but might be
convenient for the computation of poles and residues, we state that the poles
are single in certain cases.

######  Lemma 4.1.

The poles of $\hat{w}^{{\sc cwch}}$ and of $\hat{w}^{FZ}$ (except for
$p^{FZ}_{0}$ in case $\alpha_{1}=\alpha_{2}=1$) differ for different
$\lambda$. Moreover, in the case cwch and in the case fz with (64) the poles
are single.

###### Proof.

For cwch, let $f(z)=z^{2}+c^{2}\lambda$, $g(z)=b\lambda^{\beta}z^{\alpha}$.
Then for a sufficiently large $R>c^{2}\lambda$ let $C_{R}$ be the circle
radius $R$, centre at the origin. Then $|g(z)|<|f(z)|$ on $C_{R}$ and so
Rouché’s theorem shows that $f(z)$ and $(f+g)(z)$ have the same number of
roots, counted with multiplicity, within $C_{R}$. For $f$ these are only at
$z=\pm i\sqrt{\lambda}c$ so the same must be true of $f+g$ and so
$\omega^{{\sc cwch}}$ has precisely one single root in the third and in the
fourth quadrant, respectively.

Suppose now that $\hat{w}^{{\sc cwch}}$ has a pole at $re^{i\theta}$, where
$\pi/2<\theta<\pi$, for both $\lambda_{1}$ and $\lambda_{2}$. Then for
$s=re^{i\theta}$

$s^{2}+b\lambda_{1}^{\beta}s^{\alpha}+c^{2}\lambda_{1}=0\qquad
s^{2}+b\lambda_{2}^{\beta}s^{\alpha}+c^{2}\lambda_{2}=0$

so that

$\frac{b}{c^{2}}\frac{(\lambda_{1}^{\beta}-\lambda_{2}^{\beta})}{(\lambda_{1}-\lambda_{2})}=-s^{\alpha}$

Now if $\lambda_{1}\not=\lambda_{2}$ then the left hand side is positive and
real and so $\alpha\theta=\pi$. This means that $\theta>\pi$, a contradiction.

In case of fz, assuming that $p$ is a pole of both
$\hat{w}^{FZ}(\lambda_{1},\cdot)$ and $\hat{w}^{FZ}(\lambda_{2},\cdot)$ we
have

$0=\omega(\lambda_{1},p)-\omega(\lambda_{2},p)=(\lambda_{1}-\lambda_{2})[b_{1}p^{\alpha_{1}}+c^{2}]\,,$

where due $\omega(\lambda_{1},p)=0$, the term in brackets
$\,b_{1}p^{\alpha_{1}}+c^{2}=-\frac{p^{2}}{\lambda_{1}}(b_{2}p^{\alpha_{2}}+1)\not=0\,$,
hence $\lambda_{1}=\lambda_{2}$. In the factorisable case (64) of fz,
obviously all roots are single. ∎

As in [16] (where we used the classical damping term
$D=b\mathcal{A}\partial_{t}$), we assume that $r$ has the form

$r(x,t)=f(x)\chi^{\prime\prime}(t)+c^{2}\mathcal{A}f(x)\chi(t)+D[f(x)\chi(t)]$
(65)

with some function $f$ in the domain of $\mathcal{A}$ vanishing only on a set
of measure zero and some twice differentiable function $\chi$ of time such
that $(\chi^{2})^{\prime\prime}(t_{0})\not=0$ for some $t_{0}>0$. With (65),
the solution $u_{0}$ of equation (61) is clearly given by
$u_{0}(x,t)=f(x)\chi(t)$, so that $\underline{d\kappa}(u_{0}^{2})_{tt}$ can be
written in the form

$\bigl{(}\underline{d\kappa}(x)u_{0}^{2}(x,t)\bigr{)}_{tt}=\sum_{j=1}^{\infty}a_{j}\phi_{j}(x)\psi_{j}(t)$
(66)

where $a_{j}$ are the coefficients of $\underline{d\kappa}\cdot f$ with
respect to the eigenfunction basis $(\phi_{j})_{j\in\mathbb{N}}$, and
$\psi_{j}=(\chi^{2})^{\prime\prime}$.

We can rewrite equation (60) as

$z_{j}^{\prime\prime}(t)+c^{2}\lambda_{j}z_{j}(t)+D_{j}z_{j}=a_{j}\psi_{j}(t)\,,\quad
t>0\,,\qquad z_{j}(0)=0\,,\ z_{j}^{\prime}(0)=0$ (67)

for all $j\in\mathbb{N}$, where

$z_{0}(x,t)=\sum_{j=1}^{\infty}z_{j}(t)\phi_{j}(x)\,,\qquad
D_{j}=\begin{cases}b\lambda_{j}^{\beta}\partial_{t}^{\alpha}\mbox{ for {\sc
cwch}}\\\
b_{2}\partial_{t}^{2+\alpha_{2}}+b_{1}\lambda_{j}\partial_{t}^{\alpha_{1}}\mbox{
for {\sc fz}}\end{cases}$

Applying the Laplace transform to both sides of (67) yields

$\hat{z}_{j}(s)=\hat{w}_{j}(s)a_{j}\widehat{\psi}_{j}(s),\qquad\mbox{where }\
\hat{w}_{j}(s)=\frac{1}{\omega(\lambda_{j},s)}\,,\qquad s\in\mathbb{C}\,,$
(68)

and we have used homogeneity of the initial conditions.

Thus, assuming that
$F^{\prime}(0)\underline{d\kappa}=\mbox{tr}_{\Sigma}z_{0}=0$ implies that

$0=\hat{z}_{0}(x_{0},s)=\sum_{j=1}^{\infty}a_{j}\phi_{j}(x_{0})\hat{w}_{j}(s)\widehat{\psi}_{j}(s)\,,\qquad\mbox{
for all }s\in\mathbb{C}\,,\ x_{0}\in\Sigma\,.$

Considering the residues at some pole $p_{m}$ corresponding to the eigenvalue
$\lambda_{m}$ and using the fact that by Lemma 4.1,
$(s-p_{m})\hat{w}(\lambda_{j},s)=0$ for $j\not=m$ yields

$\displaystyle 0$
$\displaystyle=\mbox{Res}(\hat{z}_{0}(x_{0};p_{m}))=\sum_{j=1}^{\infty}a_{j}\phi_{j}(x_{0})\lim_{s\to
p_{m}}(s-p_{m})^{\ell_{m}}\hat{w}(\lambda_{j},s)\widehat{\psi}_{j}(s)$
$\displaystyle=\mbox{Res}(\hat{w}_{m};p_{m})\widehat{\psi}_{j}(p_{m})\sum_{k\in
K_{m}}a_{k}\phi_{k}(x_{0})\,.$

Here $\ell_{m}$ is the multiplicity of $p_{m}$ as a root of
$\omega(\lambda_{m},\cdot)$ and $K_{m}\subseteq\mathbb{N}$ is an enumeration
of the eigenspace basis $(\phi_{k})_{k\in K_{m}}$ corresponding to the
eigenvalue $\lambda_{m}$. Assuming now that

$\widehat{\psi}_{k}(p_{m})\not=0$ (69)

and there exists points $x_{0,m,1},\ldots x_{0,m,N_{m}}\in\Sigma$,
$N_{m}\geq\\#K_{m}$ such that

$\mbox{the matrix }\phi_{k}(x_{0,m,i})_{k\in
K_{m},i\in\\{1,\ldots,N_{m}\\}}\mbox{ has full rank }\\#K_{m}$ (70)

we can conclude that $a_{k}=0$ for all $k\in K_{m}$.

Now since $(u_{0}^{2})_{tt}(t_{0})=f(\chi^{2})^{\prime\prime}(t_{0})$ only
vanishes on a set of measure zero and (66), we can conclude that
$\underline{d\kappa}=0$ almost everywhere.

######  Theorem 4.1.

Under the above assumptions (66), (69), (70) for all $m\in\mathbb{N}$, $k\in
K_{m}$, the linearised derivative at $\kappa=0$, $F^{\prime}(0)$ is injective.

In particular, (70) is satisfied in the spatially 1-dimensional case
$\Sigma=\\{x_{0}\\}$, where all eigenvalues of $\mathcal{A}$ are single, i.e.,
$\\#K_{m}=1$ for all $m$, provided none of the eigenfunctions vanish at
$x_{0}$; this can be achieved by taking $x_{0}$ on the boundary and where
$\phi_{j}$ is subject to non-Dirichlet conditions.

##  5 Ill-posedness of the linearised inverse problem

As in the injectivity section, we consider the linear(ised at $\kappa=0$)
problem of recovering $\underline{\delta\kappa}(x)$ from time trace
observations

$h(t)=\mbox{tr}_{\Sigma}z_{0}=z_{0}(x_{0},t_{0}),\quad x_{0}\in\Sigma,$

where $z_{0}$ solves

$z_{tt}+c^{2}\mathcal{A}z+Dz=\underline{\delta\kappa}(u_{0}^{2})_{tt}$ (71)

with $u_{0}$ solving

$u_{0,tt}+c^{2}\mathcal{A}u_{0}+Du_{0}=r$ (72)

both with homogeneous initial conditions.

Again, we assume that the excitation $r$ has been chosen such that $u_{0}$
takes the form $u_{0}(x,t)=f(x)\chi(t)$ and employ the shorthand notation
$\tilde{\chi}=(\chi^{2})^{\prime\prime}$. Using the eigensystem of
$\mathcal{A}$ we can then write

$z_{0}(x,t)=\sum_{j=1}^{\infty}\phi(x)z_{j}(t)$

where

$\hat{z}_{j}(s)=\hat{w}_{j}(s)\langle\underline{\delta\kappa}\cdot
f,\phi_{j}\rangle\widehat{\tilde{\chi}}(s),\qquad\hat{w}_{j}(s)=\frac{1}{\omega(\lambda_{j},s)}\,,\qquad
s\in\mathbb{C}\,,$ (73)

and

$\omega(\lambda,s)=\begin{cases}s^{2}+b\lambda^{\beta}s^{\alpha}+c^{2}\lambda\mbox{
for {\sc cwch}}\\\ b_{2}s^{2+\alpha_{2}}+s^{2}+b_{1}\lambda
s^{\alpha_{1}}+c^{2}\lambda\mbox{ for {\sc fz}}\,.\end{cases}$

As in the injectivity section we obtain (for simplicity in the 1-d case where
all eigenvalues are single)

$\displaystyle\mbox{Res}(\hat{h}(p_{m}))=\mbox{Res}(\hat{w}_{m};p_{m})\widehat{\tilde{\chi}}(p_{m})\langle\underline{\delta\kappa}\cdot
f,\phi_{m}\rangle\phi_{m}(x_{0})\,.$

that is,

$\displaystyle\langle\underline{\delta\kappa}\cdot
f,\phi_{m}\rangle=\mbox{Res}(\hat{h}(p_{m}))\left(\mbox{Res}(\hat{w}_{m};p_{m})\widehat{\tilde{\chi}}(p_{m})\phi_{m}(x_{0})\right)^{-1}\,.$
(74)

(In our numerical example we had $\chi(t)=t$, hence
$\widehat{\tilde{\chi}}(p_{m})=2$.)

By l’Hospital’s rule we have

$\mbox{Res}(\hat{w}_{m};p_{m})=\lim_{s\to
p_{m}}\frac{s-p_{m}}{\omega(\lambda,s)}=\lim_{s\to
p_{m}}\frac{1}{\omega^{\prime}(\lambda,s)}=\begin{cases}\frac{1}{2p_{m}+\alpha
b\lambda_{j}^{\beta}p_{m}^{\alpha-1}}\mbox{ for {\sc cwch}}\\\
\frac{1}{(2+\alpha_{2})b_{2}p_{m}^{1+\alpha_{2}}+2p_{m}+\alpha_{1}b_{1}\lambda_{j}p_{m}^{\alpha_{1}-1}}\mbox{
for {\sc fz}}\,.\end{cases}$

Thus, the factor multiplied with $\langle\underline{\delta\kappa}\cdot
f,\phi_{m}\rangle$ in (74) only mildly grows with $p_{m}$.

The major ill-posedness seems to lie in the evaluation of the residue of the
observations $\mbox{Res}(\hat{h}(p_{m}))$ at the poles $p_{m}$, form knowledge
of $h(t)$ for $t>0$, that is, from
$\hat{h}(s)=\int_{0}^{\infty}e^{-st}h(t)\,dt$ for $s$ with nonnegative real
part (so that the integral defining the Laplace transform is well-defined). If
these poles lie on the imaginary axis (wave equation), this is still well
posed. The further left the poles lie, the more ill-posed this problem.

###  5.1 Location of the poles of the relaxation functions

Motivated by their role for the degree of ill-posedness of the inverse
problem, we develop some further results – beyond those stated in Lemma 4.1 as
essentials for our uniqueness proof – for each of the two models under
consideration and also provide some computational results with plots of these
poles for several parameter configurations.

The cwch model: Poles of the cwch model are the roots of the function

$\omega_{\text{\thinspace\sc
cwch}}(s):=s^{2}+b\lambda^{\beta}s^{\alpha}+c^{2}\lambda=0\,.$

We recall that according to [17, Lemma 4.1] which relies on Rouché’s theorem,
they lie in the left hand complex plane and this is easily shown by the
following alternative argument.

$x$$y$$\bullet$$s$................................................................................................................................................................................................................................................................$b\lambda^{\beta}s^{\alpha}$......................................................................................................................................................................................................................................................................................$s^{2}$....................................................................................................................................................................................................................................$c^{2}\lambda$

Suppose $s$ is a root in the first quadrant. Then ${\bf s}$, the line joining
the origin to the point $s$ can be split into a component in the direction of
the positive real axis and one in the direction of the positive imaginary
axis. Then since $\alpha\leq 1$, ${\bf{s}^{\alpha}}$ has components in the
same directions. Similarly, the vector ${\bf{s}^{2}}$ has a component parallel
to the real axis and again one in the direction of the positive imaginary
axis. Since $\lambda\geq 0$, the same is true of the vector
$b\lambda^{\beta}{\bf s}^{\alpha}$. The third vector representing
$c^{2}\lambda{\bf x}$ points along the real axis. However, the sum of these
three vectors cannot add to zero contradicting the claim the root $s$ lay in
the first quadrant. An identical argument shows $s$ cannot lie in the fourth
quadrant and hence cannot lie in the right half plane.

For $b=0$ the poles are along the imaginary axis and spaced exactly as the
eigenvalue sequence $\\{\lambda_{n}\\}$ stretched by the factor $c^{2}$. As
$b$ increases so does the (negative) real component of the poles which follow
a curve whose rough slope is determined by the ratio of $b$ and $c^{2}$. The
powers $\alpha$ and $\beta$ also are a factor that influence the skewness of
the curve along which the poles align. The magnitude of the real and imaginary
parts show the relative strengths of the damping and oscillation effects
respectively in the equation.

The roots of $\omega_{\text{\thinspace\sc cwch}}(s)$ are shown in Figure 1
with $b=0.1$, $\lambda_{n}=n^{2}\pi^{2}$, and for both $c=1$ and $c=5$, as
well as $\alpha=\frac{1}{2}$ and $\alpha=\frac{9}{10}$ and illustrate the
above point.

$-6$$-5$$-4$$-3$$-2$$-1$0
0510152025303540$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\alpha=\frac{1}{2}$,
$\;c=1$ $\star$ $\;\beta=1$$\diamond$ $\;\beta=\frac{3}{4}$$\circ$
$\;\beta=\frac{1}{2}$ $-25$$-20$$-15$$-10$$-5$0
0102030405060708090$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\alpha=\frac{1}{2}$,
$\;c=5$ $\star$ $\;\beta=1$$\diamond$ $\;\beta=\frac{3}{4}$$\circ$
$\;\beta=\frac{1}{2}$ $-25$$-20$$-15$$-10$$-5$0
051015202530$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\alpha=\frac{9}{10}$,
$\;c=1$ $\star$ $\;\beta=1$$\diamond$ $\;\beta=\frac{3}{4}$$\circ$
$\;\beta=\frac{1}{2}$ $-30$$-25$$-20$$-15$$-10$$-5$0
04080120160$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\alpha=\frac{9}{10}$,
$\;c=5$ $\star$ $\;\beta=1$$\diamond$ $\;\beta=\frac{3}{4}$$\circ$
$\;\beta=\frac{1}{2}$

Figure 1:  Roots of $\;\omega_{\text{\thinspace\sc cwch}}(s)\;$ for various
$\alpha$, $\beta$, $c$ values.

Some notes on how these poles were computed. For rational $\alpha=p/q$,
$\omega_{\text{\thinspace\sc cwch}}(s)$ can be written as $z^{2q}+Bz^{p}+C$
with $B=b\lambda^{\beta}$, $C=c^{2}\lambda$ and where $s=z^{p}$. Now the
$2q^{\rm th}$ degree polynomial can be represented as the characteristic
polynomial of a $2q\times 2q$ matrix. Then the roots of this polynomial are
calculated by computing the eigenvalues of the companion matrix. This gives a
good approximation even for reasonably large $q$ values but additional care
must be taken, see, for example, [10]. Given now the values of $\\{z_{n}\\}$
for $\lambda\in\\{\lambda_{n}\\}$, one can recover $\\{s_{n}\\}$ from
$s_{n}=z_{n}^{p}$. This is subject to considerable round-off error for even
modest values of $p$. However it is usually sufficient as an initial
approximation for Newton’s method to then compute a more exact value of the
roots of $\omega$ to desired accuracy. This is also successful for real
$\alpha$ by first taking a rational approximation $\alpha\approx p/q$ for the
initial approximation of the roots and then proceeding as above.

The fz model: Poles of the fractional Zener model are the roots of the
function

$\omega_{\,\text{\sc fz}}(s):=b_{2}s^{2+\alpha_{2}}+s^{2}+b_{1}\lambda
s^{\alpha_{1}}+c^{2}\lambda=0.$

There is a more complex relationship here and more constants whose value can
affect the outcome. In the case that $\alpha_{1}=\alpha_{2}=\alpha$, we can
re-write this as

$\omega_{\,\text{\sc
fz}}(s):=(b_{2}s^{\alpha}+1)(s^{2}+c^{2}\lambda)+\delta\lambda
s^{\alpha}=0,\qquad\mbox{where }\delta:=b_{1}-c^{2}b_{2}$

and $\delta$ needs to be nonnegative, cf. [12, Section iii.b]. If $\delta=0$
then $\omega_{\,\text{\sc fz}}(s)$ factors. There will be two roots at $\pm
i\,\sqrt{\lambda}\,c$ on the imaginary axis and a potential root coming from
$s^{\alpha}+1/b_{2}=0$. The latter only exists in case $\alpha=1$ for
otherwise writing $s=re^{i\theta}$ with $\theta\in(-\pi,\pi]$ we have that
$\alpha\theta\in(-\pi,\pi)$ and therefore
$\Im(s^{\alpha}+1)=r^{\alpha}\sin(\alpha\theta)=0$ implies $\alpha\theta=0$,
hence $\Re(s^{\alpha}+1/b_{2})=r^{\alpha}\cos(\alpha\theta)+1/b_{2}>0$. In
case $\alpha=1$ we obviously have a root at $-1/b_{2}$, whose modulus,
notably, does not increase with $\lambda$, as opposed to the two other complex
conjugate roots of $\omega_{\,\text{\sc fz}}$.

Clearly, physical reasoning leads us to the conclusion that in case of a
nonegative diffusivity of sound $\delta\geq 0$, all poles need to have
nonpositive real part. However, the complex analysis arguments from [17, Lemma
4.1] via Rouché’s Theorem, using as a bounding function the dominant power
part $f(z)=b_{2}z^{2+\alpha_{2}}+c^{2}\lambda$, does not seem to directly
carry over to the FZ case. This is basically due to the fact that we cannot
say anything about the number of roots of the non-polynomial function $f$.
Additionally, asymptotics in terms of powers of $s$ will be much less
effective here since, for small $b_{2}$ and/or $\alpha_{2}$, the term $s^{2}$
will be de facto dominant even for relatively large magnitudes of $s$.

Therefore we have to take a different path to conclude that also in the fz
case, the poles lie in the left hand complex plane. We do so by means of
energy estimates similar to those in section 3.2, which basically corresponds
to the mentioned physical argument. As a (partial) counterpart to [17, Lemma
4.1] we state the following.

######  Lemma 5.1.

The roots of $\omega_{\,\text{\sc fz}}(s)$ with $\alpha_{1}=\alpha_{2}=\alpha$
and $\delta:=b_{1}-c^{2}b_{2}\geq 0$ lie in the left hand complex plane.

###### Proof.

We consider the following initial value problem for the relaxation equation

$b_{2}\partial_{t}^{2+\alpha_{2}}w+w^{\prime\prime}+b_{1}\lambda\partial_{t}^{\alpha_{1}}c^{2}\lambda
w=0\,,\quad w(0)=0\,,\ w^{\prime}(0)=1\,,\quad w^{\prime\prime}(0)=0\,.$ (75)

The Laplace transform $\hat{w}$ of its solution satisfies

$(b_{2}s^{2+\alpha_{2}}+s^{2}+b_{1}\lambda
s^{\alpha_{1}}c^{2}\lambda)\hat{w}(s)=b_{2}s^{\alpha_{2}}+1$

and therefore $\hat{w}(s)=\frac{b_{2}s^{\alpha_{2}}+1}{\omega_{\,\text{\sc
fz}}(s)}$. Now if $\alpha_{1}=\alpha_{2}=\alpha$ and
$\delta:=b_{1}-c^{2}b_{2}\geq 0$, analogously to the proof of Proposition 3.7,
we obtain an energy estimate for $\tilde{w}:=b_{2}\partial_{t}^{\alpha}w+w$ by
multiplying (75) with $w^{\prime\prime}$ and integrating with respect to time

$\displaystyle\frac{1}{2}|\tilde{w}_{t}(t)|^{2}+\frac{c^{2}\lambda}{2}|\tilde{w}(t)|^{2}+\delta\lambda\Bigl{(}\frac{b_{2}}{2}|\partial_{t}^{\alpha}w(t)|^{2}+\frac{1}{2\Gamma(\alpha)t^{1-\alpha}}\int_{0}^{t}|\partial_{t}^{\alpha}w(\tau)|^{2}\,d\tau\Bigr{)}$
$\displaystyle\leq\frac{1}{2}|\tilde{w}_{t}(0)|^{2}+\frac{c^{2}\lambda}{2}|\tilde{w}(0)|^{2}$

for all $t\geq 0$. This implies uniform boundedness
$|\tilde{w}(t)|\leq\sqrt{(c^{2}\lambda_{\min})^{-1}|\tilde{w}_{t}(0)|^{2}+|\tilde{w}(0)|^{2}}=:C$
by a constant independent of $\lambda$. Taking Laplace transforms

$|\hat{\tilde{w}}(s)|=\left|\int_{0}^{\infty}e^{-st}\tilde{w}(t)\,dt\right|\leq\int_{0}^{\infty}e^{-\Re(s)t}\,dt\
C=\frac{C}{\Re(s)}$

for $\Re(s)>0$, we see that $\hat{\tilde{w}}(s)$ cannot have any poles in the
right half plane. Due to the identity
$\hat{\tilde{w}}(s)=(b_{2}s^{\alpha}+1)\hat{w}(s)-b_{2}s^{\alpha-1}w(0)=(b_{2}s^{\alpha}+1)\hat{w}(s)=\frac{(b_{2}s^{\alpha}+1)^{2}}{\omega_{\,\text{\sc
fz}}(s)}$ (where the numerator has no zeros in case $\alpha\in(0,1)$), the
assertion follows. ∎

The effect of $\delta$ on the poles in the fz model can also be assessed by
means of the implicit function theorem, applied to the function

$f(r,\theta;\delta)=\left(\begin{array}[]{l}b_{2}r^{2+\alpha_{2}}\cos((2+\alpha_{2})\theta)+r^{2}\cos(2\theta)+b_{1}\lambda
r^{\alpha_{1}}\cos(\alpha_{1}\theta)+c^{2}\lambda\\\
b_{2}r^{2+\alpha_{2}}\sin((2+\alpha_{2})\theta)+r^{2}\sin(2\theta)+b_{1}\lambda
r^{\alpha_{1}}\sin(\alpha_{1}\theta)\end{array}\right)$

whose zeros are the magnitudes and arguments of the roots $s=re^{i\theta}$ of
$\omega^{\,\text{\sc fz}}$. Now

$\frac{\partial f}{\partial(r,\theta)}=\left(\begin{array}[]{cc}A_{1}&A_{2}\\\
B_{1}&B_{2}\end{array}\right),\quad\frac{\partial
f}{\partial\delta}=\left(\begin{array}[]{c}C_{1}\\\ C_{2}\end{array}\right)$

and using Cramer’s rule this yields

$\frac{\partial
r}{\partial\delta}=\frac{B_{1}C_{2}-B_{2}C_{1}}{A_{1}B_{2}-A_{2}B_{1}}\,,\quad\frac{\partial\theta}{\partial\delta}=\frac{C_{1}A_{2}-C_{2}A_{1}}{A_{1}B_{2}-A_{2}B_{1}}\,,$

where $b_{1}=b_{2}c^{2}+\delta$ and for $(r,\theta)$ satisfying
$f(r,\theta;\delta)=0$. Then

$\displaystyle A_{1}$
$\displaystyle=(2+\alpha_{2})b_{2}r^{1+\alpha_{2}}\cos((2+\alpha_{2})\theta)+2r\cos(2\theta)+\alpha_{1}b_{1}\lambda
r^{\alpha_{1}-1}\cos(\alpha_{1}\theta)$
$\displaystyle=-\frac{1}{r}\Bigl{(}\alpha_{2}r^{2}\cos(2\theta)+(2+\alpha_{2}-\alpha_{1})b_{1}\lambda
r^{\alpha_{1}}\cos(\alpha_{1}\theta)+(2+\alpha_{2})c^{2}\lambda\Bigr{)}$
$\displaystyle A_{2}$
$\displaystyle=(2+\alpha_{2})b_{2}r^{1+\alpha_{2}}\sin((2+\alpha_{2})\theta)+2r\sin(2\theta)+\alpha_{1}b_{1}\lambda
r^{\alpha_{1}-1}\sin(\alpha_{1}\theta)$
$\displaystyle=-\frac{1}{r}\Bigl{(}\alpha_{2}r^{2}\sin(2\theta)+(2+\alpha_{2}-\alpha_{1})b_{1}\lambda
r^{\alpha_{1}}\sin(\alpha_{1}\theta)\Bigr{)}$ $\displaystyle B_{1}$
$\displaystyle=-rA,\qquad B_{2}=rA_{1},\qquad C_{1}=\lambda
r^{\alpha_{1}}\cos(\alpha_{1}\theta),\qquad C_{2}=\lambda
r^{\alpha_{1}}\sin(\alpha_{1}\theta)\,.$

This results in

$\displaystyle A_{1}B_{2}-A_{2}B_{1}$ $\displaystyle=r(A_{1}^{2}+A_{2}^{2})$
$\displaystyle B_{1}C_{2}-B_{2}C_{1}$ $\displaystyle=\lambda
r^{\alpha_{1}}\bigl{(}\alpha_{2}r^{2}\cos((2\\!-\\!\alpha_{1})\theta)+(2\\!+\\!\alpha_{2}\\!-\\!\alpha_{1})b_{1}\lambda
r^{\alpha_{1}}+(2\\!+\\!\alpha_{2})c^{2}\lambda\cos(\alpha_{1}\theta)\bigr{)}$
$\displaystyle C_{1}A_{2}-C_{2}A_{1}$ $\displaystyle=\lambda
r^{\alpha_{1}-1}\bigl{(}-\alpha_{2}r^{2}\sin((2-\alpha_{1})\theta)+(2+\alpha_{2})c^{2}\lambda\sin(\alpha_{1}\theta)\bigr{)}\,.$

and therefore $\,\frac{\partial r}{\partial\delta}>0\,$,
$\quad\frac{\partial\theta}{\partial\delta}>0\,$, for the case of the known
roots $r=c\sqrt{\lambda}$, $\theta=\pm\pi/2$ at $\delta=0$, for
$\alpha_{1}=\alpha_{2}$. That is, increasing $\delta$ tends to move the poles
into the left hand complex plane, which is intuitive in view of its physical
role as a diffusivity of sound.

Also here we have employed the method for numerically computing roots as
described above. In particular we use this in order to illustrate the
influence of $\delta>0$ on the behaviour of the roots, see Figure 2.

$-80$$-60$$-40$$-20$0
0102030405060$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$
$\star$ $\;\delta=0.1$$\diamond$ $\;\delta=0.5$$\circ$ $\;\delta=1.0$$\bullet$
$\;\delta=2.0$ $-7$$-6$$-5$$-4$$-3$$-2$$-1$0
020406080100120$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$

Figure 2: Roots of $\mathbf{\omega_{\,\text{\sc fz}}(s)}$ for various $\delta$
values with: left $\;\mathbf{\alpha=\frac{1}{2}}\,$, right
$\;\mathbf{\alpha=\frac{9}{10}}$

The location of the poles can thus be computed from knowledge of the constants
$b_{1}$, $b_{2}$, $c$, the exponents $\alpha$ and $\beta$ as well as the
values of $\lambda_{n}$. These poles are also obtainable from the time trace
measurements as the zeros of the relaxation function which is the Laplace
transform of this data $g$. Thus assuming the spectrum
$\\{\lambda_{n}\\}_{n=1}^{\infty}$ of $\mathcal{A}$ was known it is perfectly
reasonable that a least-squares fit could be made to determine the damping
constants contained in the term $D$ appearing in equation (3) and/or the wave
speed $c$. While an ill-conditioned problem, it would be particularly feasible
if the time trace data were measured at several points along an arc rather
than at a single point. It is further conceivable that spectral information on
the eigenvalues of $\mathcal{A}$ could be determined, in particular those of
the low frequency. This in turn might be used to obtain knowledge on either a
coefficient in $\mathcal{A}$ or on the domain $\Omega$ itself as there is
geometrical information contained in the lowest few eigenvalues. See, for
example, [11].

##  6 Reconstructions of $\mathbf{\kappa}$

The boundary conditions for our test cases were homogeneous Dirichlet at $x=0$
and homogeneous Neumann at $x=1$ with a nonhomogeneous driving term $r(x,t)$
with greater weight near $x=1$. Thus the solution was small in the region near
$x=0$ in comparison to near $x=1$ where the data $h(t)=u(1,t)$ was measured.
The consequence of this was that $\kappa(x)$ for $x$ small was multiplied by
terms that were also small in comparison to that at the rightmost endpoint and
resulting in greater ill-conditioning of the inversion near $x=0$. This will
be apparent in each of the reconstructions to be shown below.

The data $h(t)$ was computed by the direct solver at the endpoint $x=1$ and a
sample of $50$ points were taken to which uniformly distributed random noise
was added as representing the actual data measurements that formed the
overposed data. This was then filtered by a smoothing routine based on
penalising the $H^{2}$ norm with regularisation parameter based on the
estimate of the noise and then up-resolved to the working size for the inverse
solver.

Discretisation of $\kappa$ was done by means of a fixed set of $40$ chapeau
basis functions and we applied a regularised frozen Newton iteration, stopped
by the discrepancy principle, for numerically solving the discretised inverse
problem.

Figure 3 shows the reconstruction of a piecewise linear $\kappa$ for the
values $\alpha=1$, $\alpha=0.9$, $\alpha=0.5$ and $\alpha=0.25$. In each case
the damping coefficient $b$ was kept at $b=0.1$ and the wave speed at $c=1$.

0.00.20.40.60.81.0
0.000.050.100.15$\kappa(x)$$\alpha=1.0$....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
0.00.20.40.60.81.0
0.000.050.100.15$\kappa(x)$$\alpha=0.9$...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
0.00.20.40.60.81.0
0.000.050.100.15$\kappa(x)$$\alpha=0.5$........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
0.00.20.40.60.81.0
0.000.050.100.15$\kappa(x)$$\alpha=0.25$.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

Figure 3: Reconstructions of $\kappa(x)$ for various $\alpha$ values. Noise =
0.1%

The $(L^{\infty},\ L^{2})$ norm difference for the final versus the actual
reconstruction were: $(0.109,0.078)$, $(0.116,0.084)$, $(0.184,0.126)$,
$(0.315,0.191)$, respectively and show the increase in resolution possible
with a decrease in $\alpha$.

Note that the reconstructions of $\kappa$ are clearly superior at the right-
hand endpoint due to imposed conditions as the wave is essentially
transmitting information primarily from right to left but the amplitude is
damped as it travels. The smaller the fractional damping the lesser is this
effect which is also apparent from these figures.

The reconstructions naturally worsen with increasing noise levels as Figure 4
shows.

0.00.20.40.60.81.0
0.000.050.100.15$\kappa(x)$$\alpha=0.9$............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
0.00.20.40.60.81.0
0.000.050.100.15$\kappa(x)$$\alpha=0.25$.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

Figure 4: Reconstructions of $\kappa(x)$ for $\alpha=0.25,\;0.9$ Noise = 0.5%
(blue) and 1% (green)

Figure 5 shows the singular values of the Jacobian matrix used in the (frozen)
Newton method. Note that if the function $\kappa$ can be well represented by a
small number of basis functions then the dependence with respect to $\alpha$
will be fairly weak. On the other hand, if a large number of basis functions
are needed for $\kappa$ to be represented, then the dependence on $\alpha$
becomes much stronger although by this point the condition number of the
Jacobian is already extremely high for all $\alpha$ and relatively few
singular values are likely to be usable in any reconstruction with data
subject to extremely small noise levels.

051015202530 $-12$$-8$$-4$0 $\star$ $\alpha=0.25$$\diamond$
$\alpha=0.5$$\circ$ $\alpha=0.9$$\bullet$
$\alpha=1.0$$c=1$$\log_{10}(\sigma_{n})$$n$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$
051015202530 $-12$$-8$$-4$0 $\star$ $\alpha=0.25$$\diamond$
$\alpha=0.5$$\circ$ $\alpha=0.9$$\bullet$
$\alpha=1.0$$c=5$$\log_{10}(\sigma_{n})$$n$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\circ$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\diamond$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$$\star$

Figure 5: Singular values for various $\alpha$ values: left; $c=1$, right;
$c=5$

The effect of damping is to directly contribute to the ill-conditioning and
thus it is clear that for fixed $\alpha$ and $c$ this will increase as the
coefficient $b$ increases. The degree of ill-conditioning as a function of the
wave speed $c$ is less clear.

Figure 5 shows the singular values $\\{\sigma_{n}\\}$ of the Jacobian matrix
for both $c=1$ and $c=5$. This illustrates the decay of the singular values
and hence the level of ill-conditioning does depend on $c$ but certainly not
uniformly for all values of the fractional exponent $\alpha$. For $\alpha$
near unity, that is damping approaches or is at the classical paradigm, there
is a considerable increase in the smaller, high index singular values
indicating the problem is much less ill-posed. for larger wave speeds $c$. For
the smaller index $\sigma_{n}$ the ratio $\sigma_{n}/\sigma_{1}$ is almost the
same indicating at most a weak effect due to the wave speed. Thus for a
function $\kappa(x)$ requiring only a small number of basis functions the
effect of wave speed is relatively minimal but this changes quite dramatically
if a larger number of singular values are required. For $\alpha$ less than
about one half the condition number $\sigma_{n}/\sigma_{1}$ becomes relatively
independent of $c$ – at least in the range indicated.

### Acknowledgment

The work of the first author was supported by the Austrian Science Fund fwf
under the grants P30054 and doc78.

The work of the second author was supported in part by the National Science
Foundation through award dms-1620138.

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