# The point scatterer approximation for wave dynamics

Andrea Mantile  and  Andrea Posilicano  Laboratoire de Mathématiques de Reims,
UMR9008 CNRS et Université de Reims Champagne-Ardenne, Moulin de la Housse BP
1039, 51687 Reims, France  DiSAT, Sezione di Matematica, Università
dell’Insubria, via Valleggio 11, I-22100 Como, Italy  andrea.mantile@univ-
reims.fr<EMAIL_ADDRESS>
###### Abstract.

Given an open, bounded and connected set $\Omega\subset\mathbb{R}^{3}$ and its
rescaling $\Omega_{\varepsilon}$ of size $\varepsilon\ll 1$, we consider the
solutions of the Cauchy problem for the inhomogeneous wave equation

$(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}})\partial_{tt}u=\Delta
u+f$

with initial data and source supported outside $\Omega_{\varepsilon}$; here,
$\chi_{S}$ denotes the characteristic function of a set $S$. We provide the
first-order $\varepsilon$-corrections with respect to the solutions of the
inhomogeneous free wave equation and give space-time estimates on the
remainders in the
$L^{\infty}((0,1/\varepsilon^{\tau}),L^{2}(\mathbb{R}^{3}))$-norm. Such
corrections are explicitly expressed in terms of the eigenvalues and
eigenfunctions of the Newton potential operator in $L^{2}(\Omega)$ and provide
an effective dynamics describing a legitimate point scatterer approximation in
the time domain.

##  1\. Introduction

Let $\Omega\subset\mathbb{R}^{3}$ be open, bounded and connected,
$y_{0}\in\Omega$ and let $\varepsilon\in\left[0,1\right)$ be a small-scale
parameter. Denoting with $\Omega_{\varepsilon}$ the rescaled domain

(1.1) $\Omega_{\varepsilon}:=\left\\{y=y_{0}+\varepsilon(x-y_{0})\,,\
x\in\Omega\right\\},$

we introduce the contrast function

$\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}\,,$

modeling a sharp discontinuity of a medium across the interface
$\partial\Omega_{\varepsilon}$. The perturbed wave equation

(1.2)
$(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}})\partial_{tt}u=\Delta
u\,,$

describes the interaction between a scalar wave and a small inhomogeneity
having high contrast with respect to an homogeneous background. Here, the
relative speed of propagation inside $\Omega_{\varepsilon}$ is given by the
number $\varepsilon$; hence, small values of $\varepsilon$ correspond to a
local regime of small relative speed of propagation. We are interested in the
asymptotic behavior of the solutions of the Cauchy problem for (1.2) as
$\varepsilon\searrow 0$.

Under the time-harmonic assumption $u(t,x)=e^{it\omega}u_{\omega}(x)$, the
corresponding stationary equation writes as

(1.3)
$\big{(}\Delta+(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}})\,\omega^{2}\big{)}u_{\omega}=0\,.$

It is known that, depending on the incident frequency $\omega$, the scattering
solutions $u^{sc}_{\omega}$ of (1.3) may concentrate around
$\Omega_{\varepsilon}$ as $\varepsilon\searrow 0$. Namely, specific values of
$\omega$, referred to as _resonant frequencies_ , are associated to a
scattering enhancement localized at $y_{0}$. At these specific frequencies,
the dominant part of the scattered field assumes the form

(1.4) $u_{\omega}^{sc}(x)\sim
u_{\omega}^{in}(y_{0})\,\Lambda_{\Omega}(\omega)\,\frac{e^{-i\omega|x-y_{0}|}}{4\pi|x-y_{0}|}\,,$

where $\Lambda_{\Omega}(\omega)$ is a scattering coefficient depending on the
physical parameters. Such asymptotic, provided in [3] under far-field
approximations, describes a diffusion from a _point-scatterer_ placed at
$y_{0}$.

The point scatterer model, introduced by Foldy in [10] and further developed
by Lax, [19], consists in using a (1.4)-like formula as an heuristic
approximation for the stationary scattering from a small obstacles placed at
$y_{0}$. Such an approximation is commonly adopted to describe scattering from
small Dirichlet obstacles. An interpretation, discussed in [14], shows that
the Foldy-Lax description of a point scatterer corresponds to modeling the
interaction between the wave and the scatterer in terms of a point
perturbation of the Laplacian, a class of singular perturbations studied in
[1], with a frequency-dependent scattering length. It is worth to remark that
such model has no direct counterpart in the time domain setting. However,
since the point-scatterer approximation provides an effective limit for the
asymptotic regime at specific frequencies, the asymptotic analysis of the wave
dynamics generated by solutions of (1.2) at small scale may suggests a
physically coherent definition of a point-scatterer in the time domain.

The analysis of the stationary problem (1.3) in the small scale limit
$\varepsilon\searrow 0$ enlightens the emergency of a discrete set of
_scattering resonances_ at resonant frequencies $\omega_{k}$ related to the
inverse spectral points of the _Newton potential operator_

$N_{0}:L^{2}(\Omega)\rightarrow L^{2}(\Omega)\,,\qquad
N_{0}u(x):=\frac{1}{4\pi}\int_{\Omega}\frac{u(y)\,dy}{|x-y|}$

by $\omega_{k}=\lambda_{k}^{-1/2}$, where
$\\{\lambda_{k}\\}_{k=1}^{+\infty}=\sigma_{d}(N_{0})$, see [3], [8]. This
suggests that the generator of the dynamics may have spectral resonances at
the energies $-\lambda_{n}^{-1}$. Such a spectral problem is considered in
[20], where this picture is validated and precise estimates for the imaginary
parts of resonances are provided. In particular, under suitable assumptions on
$\Omega$, it is shown that at each $\lambda_{k}\in\sigma_{d}(N_{0})$ there
corresponds a unique resonance of
$(\varepsilon^{2}\chi_{\Omega_{\varepsilon}}+\chi_{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}})\Delta$
converging to $-\lambda_{k}^{-1}$ with imaginary part of order $\varepsilon$.

According to the theory of second-order Cauchy problems with self-adjoint
generators, the solutions of (1.2) express in terms of time propagators which
are related to the inverse Laplace transform of the resolvent operator
$(-(\varepsilon^{2}\chi_{\Omega_{\varepsilon}}+\chi_{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}})\Delta+z^{2})^{-1}$,
see, e.g., [6, Sec I.3.14]. Hence, the relevant properties of the dynamical
system are encoded in the spectral profile of
$(\varepsilon^{2}\chi_{\Omega_{\varepsilon}}+\chi_{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}})\Delta$,
including eigenvalues and resonances. In view of the results from [20], we
expect that in the small-scale regime, the wave dynamics (1.2) may be governed
by a (possibly finite) number of resonant states whose survival time, defined
by the imaginary part of spectral resonances, is of order $1/\varepsilon$. In
this connection, the role of resonances on the asymptotic dynamics may be
relevant on a large time-scale.

In this work we focus on a direct asymptotic analysis of the dynamics
generated by solutions of the Cauchy problem for the wave equation (1.2). The
main results concern both the homogeneous case, considered in Theorem 5.1, and
the non-homogeneous case, in Theorem 5.2. In Theorem 5.3 a model for the
effective dynamics in the small-scale limit is provided in both cases. We next
present an overview of the outcomes for the homogeneous Cauchy problem; more
detailed results on such a case as well on the one in presence of a source
term are provided in the aforementioned theorems.

Let $u_{\varepsilon}$ and $u_{\text{{free}}}$ denote the classical solutions
of the Cauchy problems (here and in the following $t\geq 0$)

$\begin{cases}(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}})\partial_{tt}u_{\varepsilon}(t)=\Delta
u_{\varepsilon}(t)\,,\\\ u_{\varepsilon}(0)=\phi\,,\\\
\partial_{t}u_{\varepsilon}(0)=\psi\,,\end{cases}\qquad\begin{cases}\partial_{tt}u_{\text{{free}}}(t)=\Delta
u_{\text{{free}}}(t)\,,\\\ u_{\text{{free}}}(0)=\phi\,,\\\
\partial_{t}u_{\text{{free}}}(0)=\psi\,.\end{cases}$

Assume that $\phi,\psi\in H^{6}(\mathbb{R}^{3})$ are supported outside
$\Omega_{\varepsilon}$ and define $h:[0,+\infty)\rightarrow\mathbb{R}$
depending on these Cauchy data by

$h(t):=\int_{|y-y_{0}|<t}\frac{\Delta^{2}\phi(y)}{4\pi|y-y_{0}|}\,dy-\frac{1}{4\pi
t}\int_{|x-y_{0}|=t}\Delta\psi(x)\,d\sigma_{t}(y)\,.$

Notice that, by Kirchhoff’s formula, $h(t)=\Delta u_{\text{{free}}}(t,y_{0})$.
Then, for any $\tau\in(0,\frac{1}{11})$ there exists $\varepsilon_{0}>0$ small
enough such that the expansion

(1.5)
$u_{\varepsilon}(t,x)=\,u_{\text{{free}}}(t,x)+\varepsilon(\varepsilon^{2}-1)\,H(t-|x-y_{0}|)\,\frac{q(t-|x-y_{0}|)}{4\pi|x-y_{0}|}+{r}_{\varepsilon}(t,x)\,,$

holds for all $\varepsilon\in(0,\varepsilon_{0})$ and the remainder allows the
estimate

(1.6) $\sup_{0<t\leq
1/\varepsilon^{\tau}}\|{r}_{\varepsilon}(t,\cdot)\|_{L^{2}(\mathbb{R}^{3})}\lesssim\,o(\varepsilon)\big{(}\|\Delta^{3}\phi\|_{L^{2}(\mathbb{R}^{3}}+\|\Delta^{3}\psi\|_{L^{2}(\mathbb{R}^{3}}\big{)}\,.$

Here, $H$ is the Heaviside function and $q(t):=\sum_{k=1}^{+\infty}q_{k}(t)$,
with each $q_{k}(t)$ solving the Cauchy problem

(1.7)
$\begin{cases}\ddot{q}_{k}(t)=-\lambda_{k}^{-1}\,\big{(}q_{k}(t)-|\langle
e_{k},1\rangle|^{2}\,h(t)\big{)}\,,\\\
q_{k}(0)=\dot{q}_{k}(0)=0\,,\end{cases}$

where $e_{k}\in L^{2}(\Omega)$ are the eigenfunction corresponding to
$\lambda_{k}\in\sigma_{d}(N_{0})$.

The effective dynamics given by

(1.8)
$u_{\varepsilon,\text{{eff}}}(t,x)=\,u_{\text{{free}}}(t,x)+\varepsilon(\varepsilon^{2}-1)\,H(t-|x-y_{0}|)\,\frac{q(t-|x-y_{0}|)}{4\pi|x-y_{0}|}$

is the resulting point-scatterer approximation of the propagation, since it
solves the Cauchy problem

$\begin{cases}\partial_{tt}u_{\varepsilon,\text{{eff}}}(t)=\Delta
u_{\varepsilon,\text{{eff}}}(t)+\varepsilon(\varepsilon^{2}-1)\,q(t)\,\delta_{y_{0}}\\\
u_{\varepsilon,\text{{eff}}}(0)=\phi\\\
\partial_{t}u_{\varepsilon,\text{{eff}}}(0)=\psi\,,\end{cases}$

where $\delta_{y_{0}}$ denotes the Dirac delta distribution supported at the
point $y_{0}$. The spherical wave component of $u_{\varepsilon,\text{{eff}}}$
has the space profile of a Green function centered at $y_{0}$ modulated by a
function whose evolution in time is determined by the Cauchy data through the
solution of (1.7).

It is worth to remark that (1.5) holds for the specific class of
$H^{6}(\mathbb{R}^{3})$ Cauchy data with support away from $y_{0}$. This
condition is needed to control the high-energy contribution to the inverse
Laplace transform involved in the computation of the dynamics. It is unclear
if, for less regular data which may concentrate their $L^{2}$-mass close to
$y_{0}$, a different asymptotic may occur.

Since the solution $t\mapsto u_{\varepsilon}(t,\cdot)$ defines a continuous
flow in the Sobolev space
$H^{2}(\mathbb{R}^{3})\subset{\mathcal{C}}_{b}(\mathbb{R}^{3})$ and since
(1.8) is smooth outside $y_{0}$ but neither bounded nor continuous there, the
small remainder ${r}_{\varepsilon}(t,\cdot)$ must compensate such loss of
regularity close to $y_{0}$. It is worth to remark that this is not in
contradiction with the estimates (1.6), however, such circumstance prevents to
control the errors on a more regular Sobolev scale.

Although (1.5)-(1.7) does not explicitly involve resonant states, the resonant
energies $-\lambda_{k}^{-1}$ appears in the definition of the modulation
functions $q_{k}(t)$. This expansion holds on a large time scale which is,
however, shorter than the estimated resonances life-time. We expect that
(1.5)-(1.7) can be recasted in terms of quasi-bound states. Let us notice in
turns that, representing the quantum and the classical-waves dynamics for
point-perturbations of the Laplacian in terms of resonant states presents
well-known technical asperities (involving a careful estimate of the
contributions to the time-propagator from energies close to resonances) and
very few general results have been provided in this sense, see [2].

The small-scale expansions and the effective model presented in this work are
new. In particular, the asymptotic formulae resulting from our analysis are
given with global-in-space estimates of the errors on a large time scale. Such
formulae provide, in our opinion, a useful contribution in applications
involving time domain data. Let us recall, at this concern, that the
interaction of classical waves with _micro-resonators_ is attracting the
increasing interest of different branches of applied physics, where the
manipulation of the dynamics at micro-scales has several relevant
applications. For instance, these resonant-scattering phenomena have been used
for the design of acoustic and electromagnetic metamaterials, or in the
realization of contrast agents for various imaging strategies. Moreover, they
show to have a potential impact in classical and quantum information
processing. A vast literature has been devoted to these topics, see [11],
[13], [18], [21], [24] to cite a few.

While the small-$\varepsilon$ asymptotic of the stationary scattering problem
(1.3) has been well understood, only partial results have been provided so far
for the corresponding dynamics. An attempt to derive a Foldy-Lax model in the
time domain has been recently discussed in the two-dimensional case using
Dirichlet discs of small size, see [16]. As regards the model equation (1.3),
a possible approach to time-domain expansions was argued in [3] relying on the
previuos works [4] and [5, Appendix B]. This, however, would produce an
expansion only valid after truncating the high-frequency components of the
fields. To the best of our knowledge, none explicit formula has been produced
within such an approach, neither the relation with the asymptotic expansion of
the full time-domain solution has been discussed so far. In [23], the equation
(1.2), excited by a source term but with trivial Cauchy data, have been
considered in the perspective of using dielectric micro-resonators as contrast
agents for imaging in the time domain. The authors use known estimates in the
Laplace-transform domain to get the asymptotic expansion of the solution on a
finite time interval with a point-wise estimate of the error holding far from
the scatterer location, while, it deserves noting, none effective limit model
has been discussed. In the limit of their validity, the interesting asymptotic
expansions provided in [23] are coherent with the ones provided here,
enlightening the role of the Newton eigenvalues in the construction of the
dynamics.

######  Remark 1.1.

Without loss of generality, thanks to the translational invariance of the
equations, throughout the text we set for simplicity $y_{0}=0$ in the
definition (1.1) of the small domain $\Omega_{\varepsilon}$; the statements of
the Theorems 5.1, 5.2 and 5.3 easily adapt to the case of a generic point
$y_{0}$.

##  2\. Notation

$\bullet$ $\Omega_{\varepsilon}$ denotes the contracted set
$\Omega_{\varepsilon}:=\\{x\in\mathbb{R}^{3}:x=\varepsilon y\,,\
y\in\Omega\\}$, $0<\varepsilon<1$, where $\Omega\subset\mathbb{R}^{3}$ is a
given open, bounded and connected set containing the origin; $|\Omega|$
denotes the volume of $\Omega$.

$\bullet$ $\|\cdot\|$ denotes the norm in a space of square integrable
functions like $L^{2}(\Omega)$ and $L^{2}(\mathbb{R}^{3})$ with scalar product
$\langle\cdot,\cdot\rangle$; $\|\cdot\|$ also denotes the operator norm for
bounded linear operators acting between such spaces. Norms in different spaces
are specified with the appropriate subscripts.

$\bullet$ ${\mathscr{B}}(X,Y)$ denotes the Banach space of bounded operators
from the Banach space $X$ to the Banach space $Y$;
${\mathscr{B}}(X,X)\equiv{\mathscr{B}}(X)$.

$\bullet$ $\mathcal{L}$ denotes the Laplace transform.

$\bullet$ $H(t)$ denotes the Heaviside function.

$\bullet$ $\varrho(A)$ and $\sigma(A)$ denote the resolvent set and the
spectrum of the self-adjoint operator $A$; $\sigma_{d}(A)$ denotes its
discrete spectrum.

$\bullet$ $H^{2k}(\mathbb{R}^{3})$, $k\in{\mathbb{N}}$, denotes the Hilbert-
Sobolev space of order $2k$, i.e. the space of square integrale functions
$f:\mathbb{R}^{3}\to{\mathbb{C}}$ such that $\Delta^{k}f$ is square
integrable, $\Delta^{k}$ denoting the $k$-th power of the Laplacian.

$\bullet$ $\mathcal{G}_{z}$, $z\in{\mathbb{C}}\backslash(-\infty,0]$, denotes
the kernel function of $(-\Delta+z^{2})^{-1}$, i.e.,
$\mathcal{G}_{z}(x)=\frac{e^{-z|x|}}{4\pi|x|}$.

$\bullet$ $f(t,\varepsilon)\lesssim g(t,\varepsilon)$ means that there exists
$\kappa\geq 0$, independent of $t$ and $\varepsilon$ such that
$f(t,\varepsilon)\leq\kappa\,g(t,\varepsilon)$.

##  3\. The model operator and the corresponding wave equation

The wave equation (1.2) re-writes as

$\partial_{tt}u=A(\varepsilon)u\,,$

where the linear operator in $L^{2}(\mathbb{R}^{3})$ is defined by

$A(\varepsilon):H^{2}(\mathbb{R}^{3})\subset L^{2}(\mathbb{R}^{3})\to
L^{2}(\mathbb{R}^{3})\,,\qquad
A(\varepsilon):=(\varepsilon^{2}\chi_{\Omega_{\varepsilon}}+\chi_{{\mathbb{R}^{3}\backslash\Omega}})\Delta\,.$

It corresponds to an additive perturbation of the Laplacian:

(3.1)
$\begin{array}[c]{ccc}A(\varepsilon)=\Delta-T(\varepsilon)\,,&&T(\varepsilon):=(1-\varepsilon^{2})\chi_{\Omega_{\varepsilon}}\Delta\,.\end{array}$

Let us set

$R_{z}:=(-\Delta+z)^{-1}\,,\qquad z\in{\mathbb{C}}\backslash(-\infty,0]\,.$

We introduce the Hilbert space

$L_{\varepsilon}^{2}(\mathbb{R}^{3}):=L^{2}(\mathbb{R}^{3};(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}})dx)\,;$

the corresponding scalar product and norm are denoted by
$\langle\cdot,\cdot\rangle_{\varepsilon}$ and $\|\cdot\|_{\varepsilon}$
respectively; By the inequalities

(3.2) $\|u\|\leq\|u\|_{\varepsilon}\leq\varepsilon^{-1}\|u\|\,,$

one gets the equivalence, holding in the Banach space sense,

(3.3) $L_{\varepsilon}^{2}(\mathbb{R}^{3})\simeq L^{2}(\mathbb{R}^{3})\,.$

######  Remark 3.1.

Notice that, if $u\in L^{2}(\mathbb{R}^{3})$ is supported outside
$\Omega_{\varepsilon}$, then $\|u\|_{\varepsilon}=\|u\|$. Therefore, using
$\|\cdot\|_{\varepsilon}$ to also denote the norm in
${\mathscr{B}}(L_{\varepsilon}^{2}(\mathbb{R}^{3}))$, one has, for any
$L\in{\mathscr{B}}(L_{\varepsilon}^{2}(\mathbb{R}^{3}))$ and for any $u\in
L^{2}(\mathbb{R}^{3})$ supported outside $\Omega_{\varepsilon}$,

$\|Lu\|\leq\|L\|_{\varepsilon}\|u\|\,.$

######  Theorem 3.2.

$A(\varepsilon)$ is a closed operator in $L^{2}(\mathbb{R}^{3})$ and a non-
positive self-adjoint operator in $L_{\varepsilon}^{2}(\mathbb{R}^{3})$; it
has resolvent

(3.4)
$R_{z}(\varepsilon):=(-A(\varepsilon)+z)^{-1}=R_{z}-K_{z}(\varepsilon)\,,\qquad
z\in{\mathbb{C}}\backslash(-\infty,0]\,,$

where

$K_{z}(\varepsilon):=R_{z}(1+T(\varepsilon)R_{z})^{-1}T(\varepsilon)R_{z}\,,\qquad
z\in{\mathbb{C}}\backslash(-\infty,0]\,.$

###### Proof.

Since both
$(\varepsilon^{2}\chi_{\Omega_{\varepsilon}}+\chi_{{\mathbb{R}^{3}\backslash\Omega}})$
and
$(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{{\mathbb{R}^{3}\backslash\Omega}})$
are bounded and $\Delta:H^{2}(\mathbb{R}^{3})\subset L^{2}(\mathbb{R}^{3})\to
L^{2}(\mathbb{R}^{3})$ is closed,
$A(\varepsilon)=(\varepsilon^{2}\chi_{\Omega_{\varepsilon}}+\chi_{{\mathbb{R}^{3}\backslash\Omega}})\Delta$
is closed as well; furthermore, by

$\langle u,A(\varepsilon)u\rangle_{\varepsilon}=\langle u,\Delta u\rangle\leq
0\,,$

one gets $A(\varepsilon)\leq 0$ and
$\sigma(A(\varepsilon))\subseteq(\infty,0]$.

Let $z\in{\mathbb{C}}\backslash(-\infty,0]$ such that $\text{Re}(z)\geq 0$;
the identity

$\displaystyle\left\|u\right\|^{2}=\,$
$\displaystyle\left\|(-\Delta+z)R_{z}u\right\|^{2}=\left\|-\Delta
R_{z}u\right\|^{2}+|z|^{2}\left\|R_{z}u\right\|^{2}+2\text{Re}(z)\langle-\Delta
R_{z}u,R_{z}u\rangle$

implies

$\left\|-\Delta R_{z}\right\|\leq 1\,.$

By (3.1), this gives

(3.5) $\left\|T(\varepsilon)R_{z}\right\|\leq(1-\varepsilon^{2})\left\|\Delta
R_{z}\right\|\leq(1-\varepsilon^{2})<1\,.$

Hence,

(3.6)
$\sum_{n=0}^{+\infty}(-1)^{n}(T(\varepsilon)R_{z})^{n}=(1+T(\varepsilon)R_{z})^{-1}\,,$

where the Neumann series on the left converges in
${{\mathscr{B}}}(L^{2}(\mathbb{R}^{3}))$ for any
$z\in{\mathbb{C}}\backslash(-\infty,0]$ such that $\text{Re}(z)\geq 0$.
Furthermore, for such a $z$, by simple algebraic manipulations, one gets

(3.7)
$R_{z}(1+T(\varepsilon)R_{z})^{-1}=R_{z}-R_{z}(1+T(\varepsilon)R_{z})^{-1}T(\varepsilon)R_{z}\,,$

and

$(-A(\varepsilon)+z)R_{z}(1+T(\varepsilon)R_{z})^{-1}={\mathbb{1}}_{L^{2}(\mathbb{R}^{3})}\,,$
$R_{z}(1+T(\varepsilon)R_{z})^{-1}(-A(\varepsilon)+z)={\mathbb{1}}_{H^{2}(\mathbb{R}^{3})}\,.$

By the equivalence (3.3), $A(\varepsilon)$ is closed as an operator in
$L_{\varepsilon}^{2}(\mathbb{R}^{3})$; furthermore, $A(\varepsilon)$ is
symmetric in $L_{\varepsilon}^{2}(\mathbb{R}^{3})$. Hence, by

$\text{\rm ran}(-A(\varepsilon)\pm i)=\text{\rm
dom}(R_{z}(1+T(\varepsilon)R_{\pm
i})^{-1})=L_{\varepsilon}^{2}(\mathbb{R}^{3})\,,$

$A(\varepsilon)$ is self-adjoint as an operator in
$L_{\varepsilon}^{2}(\mathbb{R}^{3})$.

Finally, by [7, Theorem 2.19 and Remark 2.20], the resolvent formula (3.7)
extends to
$\varrho(A(\varepsilon))\cap\varrho(\Delta)={\mathbb{C}}\backslash(-\infty,0]$.
∎

######  Remark 3.3.

One can show that
$\sigma(A(\varepsilon))=\sigma_{ac}(A(\varepsilon))=(-\infty,0]$. However, for
the purposes of this work, it suffices to know, by $A(\varepsilon)\leq 0$,
that $\sigma(A(\varepsilon))\subseteq(-\infty,0]$.

######  Corollary 3.4.

For any $z\in{\mathbb{C}}\backslash(-\infty,0]$, for any integer $n\geq 1$ and
for any $u\in H^{2n}(\mathbb{R}^{3})$ supported outside $\Omega_{\varepsilon}$
one has

$K_{z}(\varepsilon)u=\frac{1}{z^{n}}\,K_{z}(\varepsilon)\Delta^{n}u\,.$

###### Proof.

For any self-adjoint operator $A$ and for any
$z\in\varrho(A)\cap{\mathbb{C}}\backslash\\{0\\}$, one has

$(-A+z)^{-1}=\frac{1}{z}+\frac{1}{z}\,(-A+z)^{-1}A\,.$

Iterating, one gets

(3.8)
$(-A+z)^{-1}=\sum_{k=1}^{n}\frac{1}{z^{k}}\,A^{k-1}+\frac{1}{z^{n}}\,(-A+z)^{-1}A^{n}\,.$

By $A(\varepsilon)u=\Delta u$ for any $u$ supported outside
$\Omega_{\varepsilon}$, by (3.4) and by applying (3.8) to both $A=\Delta$ and
$A=A(\varepsilon)$, one gets, whenever $u\in H^{2n}(\mathbb{R}^{3})$,

$\displaystyle R_{z}(\varepsilon)u=$
$\displaystyle\,\sum_{k=1}^{n}\frac{1}{z^{k}}\,A(\varepsilon)^{k-1}u+\frac{1}{z^{n}}\,R_{z}(\varepsilon)A(\varepsilon)^{n}u$
$\displaystyle=$
$\displaystyle\,\sum_{k=1}^{n}\frac{1}{z^{k}}\,\Delta^{k-1}u+\frac{1}{z^{n}}\,R_{z}\Delta^{n}u-\frac{1}{z^{n}}\,K_{z}(\varepsilon)\Delta^{n}u$
$\displaystyle=$
$\displaystyle\,R_{z}u-\frac{1}{z^{n}}\,K_{z}(\varepsilon)\Delta^{n}u$
$\displaystyle=$
$\displaystyle\,R_{z}(\varepsilon)u+K_{z}(\varepsilon)u-\frac{1}{z^{n}}\,K_{z}(\varepsilon)\Delta^{n}u\,.$

∎

Given a non-positive self-adjoint operator $A$ in a Hilbert space
${\mathcal{H}}$, we consider the Cauchy problem for the corresponding wave
equation, i.e.,

(3.9) $\begin{cases}\partial_{tt}u(t)=Au(t)\\\ u(0)=\phi\\\
\partial_{t}u(0)=\psi\,,\end{cases}$

where both the data $\phi$ and $\psi$ are in ${\mathcal{H}}$. We say that
$u\in{\mathcal{C}}(\mathbb{R}_{+};{\mathcal{H}})$ is a mild solution of (3.9)
whenever

$\int_{0}^{t}\int_{0}^{s}u(r)\,dr\,ds\equiv\int_{0}^{t}(t-s)u(s)\,ds\,\in\text{\rm
dom}(A)$

and

$u(t)=\phi+t\psi+A\int_{0}^{t}(t-s)u(s)\,ds$

for any $t\geq 0$. By [6, Proposition 3.14.4, Corollary 3.14.8 and Example
3.14.16], the unique mild solution of (3.9) is given by

(3.10) $u(t)=\cos(t(-A)^{1/2})\,\phi+(-A)^{-1/2}\sin(t(-A)^{1/2})\,\psi\,,$

where the sine and cosine operator functions are defined through the
functional calculus for the self-adjoint $A$. By [12, Theorem 7.4 in Chapter
2, Section 8], if $\phi\in\text{\rm dom}(A)$ and $\psi\in\text{\rm
dom}((-A)^{1/2})$, then $u$ in (3.10) is a classical solution, i.e.,

$u\in{\mathcal{C}}^{2}(\mathbb{R}_{+},{\mathcal{H}})\cap{\mathcal{C}}^{1}(\mathbb{R}_{+},\text{\rm
dom}((-A)^{1/2})\cap{\mathcal{C}}(\mathbb{R}_{+},\text{\rm dom}(A))\,.$

The Laplace transforms of the sine and the cosine functions give the relations

$\int_{0}^{\infty}e^{-\lambda\,t}\sin(tx)\,dt=x(x^{2}+\lambda^{2})^{-1}\,,\qquad\lambda>0\,,$
$\int_{0}^{\infty}e^{-\lambda\,t}\cos(tx))\,dt=\lambda(x^{2}+\lambda^{2})^{-1}\,,\qquad\lambda>0\,.$

Thus, by functional calculus, by the inversion of the Laplace-Stieltjes
transform of Lipschitz, ${\mathscr{B}}(L^{2}(\mathbb{R}^{n}))$-valued,
functions $F$ such that $F(0)=0$ (see [6, Theorem 2.3.4]), one gets

$\displaystyle(-A)^{-1/2}\sin(t(-A)^{1/2})\psi$
$\displaystyle\,=\lim_{\gamma\nearrow\infty}\frac{1}{2\pi
i}\int_{c-i\gamma}^{c+i\gamma}e^{tz}(-A+z^{2})^{-1}\psi\,dz\,,\qquad\psi\in
L^{2}(\mathbb{R}^{3})$

and

$\displaystyle(\cos(t(-A)^{1/2})-1)\phi$
$\displaystyle\,=\lim_{\gamma\nearrow\infty}\frac{1}{2\pi
i}\int_{c^{\prime}-i\gamma}^{c^{\prime}+i\gamma}e^{tz}(-A+z^{2})^{-1}A\phi\
\frac{dz}{z}\,,\qquad\phi\in\text{\rm dom}(A)\,,$

where $c>0$ and $c^{\prime}>0$ are arbitrary.

By applying these results to both $A=A(\varepsilon)$ and $A=\Delta$ one gets
the following

######  Lemma 3.5.

Let $\phi\in H^{2(m+1)}(\mathbb{R}^{3})$ and $\psi\in H^{2n}(\mathbb{R}^{3})$,
$m\geq 0$ and $n>0$, both having support disjoint from $\Omega_{\varepsilon}$;
let $u_{\varepsilon}$ and $u_{\text{\rm free}}$ be the classical solutions of
the Cauchy problems

$\begin{cases}(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}})\partial_{tt}u_{\varepsilon}(t)=\Delta
u_{\varepsilon}(t)\\\ u_{\varepsilon}(0)=\phi\\\
\partial_{t}u_{\varepsilon}(0)=\psi\,.\end{cases}\qquad\begin{cases}\partial_{tt}u_{\text{\rm
free}}(t)=\Delta u_{\text{\rm free}}(t)\\\ u_{\text{\rm free}}(0)=\phi\\\
u_{\text{\rm free}}(0)=\psi\,.\end{cases}$

Then,

$u_{\varepsilon}(t)=u_{\text{\rm free}}(t)-\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\\!\\!e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n}\psi\
\frac{dz}{z^{2n}}-\frac{1}{2\pi
i}\int_{c^{\prime}-i\infty}^{c^{\prime}+i\infty}\\!\\!e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m+1}\phi\
\frac{dz}{z^{2m+1}}\,,$

where $c>0$ and $c^{\prime}>0$ are arbitrary.

###### Proof.

One gets, by (3.4) and by Corollary 3.4, for any $\phi\in
H^{2(m+1)}(\mathbb{R}^{3})$ and $\psi\in H^{2n}(\mathbb{R}^{3})$ both having
support disjoint from $\Omega_{\varepsilon}$,

$\displaystyle(-A(\varepsilon))^{-1/2}\sin(t(-A(\varepsilon))^{1/2})\psi$
$\displaystyle=$ $\displaystyle\,\lim_{\gamma\nearrow\infty}\frac{1}{2\pi
i}\int_{c-i\gamma}^{c+i\gamma}e^{tz}R_{z^{2}}\psi\,dz-\lim_{\gamma\nearrow\infty}\frac{1}{2\pi
i}\int_{c-i\gamma}^{c+i\gamma}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n}\psi\
\frac{dz}{z^{2n}}$ $\displaystyle=$
$\displaystyle\,(-\Delta)^{-1/2}\sin(t(-\Delta)^{1/2})\psi-\lim_{\gamma\nearrow\infty}\frac{1}{2\pi
i}\int_{c-i\gamma}^{c+i\gamma}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n}\psi\
\frac{dz}{z^{2n}}$

and

$\displaystyle(\cos(t(-A(\varepsilon))^{1/2})-1)\phi$ $\displaystyle=$
$\displaystyle\,\lim_{\gamma\nearrow\infty}\frac{1}{2\pi
i}\int_{c-i\gamma}^{c+i\gamma}e^{tz}R_{z^{2}}\Delta\phi\
\frac{dz}{z}-\lim_{\gamma\nearrow\infty}\frac{1}{2\pi
i}\int_{c^{\prime}-i\gamma}^{c^{\prime}+i\gamma}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m+1}\phi\
\frac{dz}{z^{2m+1}}$ $\displaystyle=$
$\displaystyle\,(\cos(t(-\Delta)^{1/2})-1)\phi-\lim_{\gamma\nearrow\infty}\frac{1}{2\pi
i}\int_{c^{\prime}-i\gamma}^{c^{\prime}+i\gamma}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m+1}\phi\
\frac{dz}{z^{2m+1}}\,.$

Writing $z=c+i\gamma$, $c>0$, $\gamma\in\mathbb{R}$, one has

$\|R_{z^{2}}\|\leq\begin{cases}{|z|^{-2}}&|\gamma|<c\\\
{(2c|\gamma|)^{-1}}&|\gamma|\geq
c\end{cases}\,,\quad\|R_{z^{2}}(\varepsilon)\|_{\varepsilon}\leq\begin{cases}{|z|^{-2}}&|\gamma|<c\\\
{(2c|\gamma|)^{-1}}&|\gamma|\geq c\end{cases}\,,$

and so, by Remark 3.1, whenever $\varphi\in L^{2}(\mathbb{R}^{3})$ is
supported outside $\Omega_{\varepsilon}$,

(3.11)
$\|K_{z^{2}}(\varepsilon)\varphi\|\leq\|R_{z^{2}}\varphi\|+\|R_{z^{2}}(\varepsilon)\varphi\|\leq\|\varphi\|\begin{cases}{2|z|^{-2}}&|\gamma|<c\\\
{(c|\gamma|)^{-1}}&|\gamma|\geq c\,.\end{cases}$

Hence, the above improper operator-valued integrals are absolutely convergent
and one gets,

$(-A(\varepsilon))^{-1/2}\sin(t(-A(\varepsilon))^{1/2})\psi=(-\Delta)^{-1/2}\sin(t(-\Delta)^{1/2})\psi-\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\\!\\!e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n}\psi\
\frac{dz}{z^{2n}}$

and

$\cos(t(-A(\varepsilon))^{1/2})\phi=\cos(t(-\Delta)^{1/2})\phi-\frac{1}{2\pi
i}\int_{c^{\prime}-i\infty}^{c^{\prime}+i\infty}\\!\\!e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m+1}\phi\
\frac{dz}{z^{2m+1}}\,,$

where $c>0$ and $c^{\prime}>0$ are arbitrary. The proof is then concluded by
(3.10). ∎

######  Remark 3.6.

By $|\Omega_{\varepsilon}|\searrow 0$ as $\varepsilon\searrow 0$, for any
$u\in H^{2}(\mathbb{R}^{3})$ one has

$\lim_{\varepsilon\searrow 0}\|A(\varepsilon)u-\Delta
u\|^{2}=\lim_{\varepsilon\searrow
0}\,(1-\varepsilon^{2})^{2}\\!\int_{\Omega_{\varepsilon}}|\Delta
u(x)|^{2}\,dx=0\,.$

Therefore, by [17, Theorems 2.25 and 2.29], $A(\varepsilon)$ converges to the
free Laplacian in norm resolvent sense, i.e., for any
$z\in{\mathbb{C}}\backslash(-\infty,0]$, there holds

(3.12) $\lim_{\varepsilon\searrow
0}\left\|(-A(\varepsilon)+z)^{-1}-(-\Delta+z)^{-1}\right\|=0\,.$

If the sine and the cosine operator function of $A(\varepsilon)$ had norms in
$\mathscr{B}(L^{2}(\mathbb{R}^{3}))$ uniformly bounded with respect to
$\varepsilon$, then (3.12) would imply the convergence of the solutions of the
Cauchy problem for the wave equation for $A(\varepsilon)$ to the ones for
$\Delta$, and that would be true for generic initial data (see [12, Theorem
8.6 in Section 8, Chapter 2]). However, by (3.2), these operator norms behave
like $1/\varepsilon$, and so, the convergence of the dynamics generated by
$A(\varepsilon)$ to the free one is not guaranteed. As we will prove in the
following sections, thanks to Lemma 3.5, convergence to the free case surely
holds whenever the initial data are supported outside $\Omega_{\varepsilon}$.

##  4\. Operator estimates

In the following, we use the identification

$L^{2}(\mathbb{R}^{3})\equiv
L^{2}(\mathbb{R}^{3}\backslash\Omega_{\varepsilon})\oplus
L^{2}(\Omega_{\varepsilon})\,,$

provided by the unitary map $u\mapsto
1_{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}u\oplus
1_{\Omega_{\varepsilon}}$. Here and below, given the measurable domain
$D\subset\mathbb{R}^{3}$ we denote by $1_{D}:L^{2}(\mathbb{R}^{3})\to
L^{2}(D)$ the bounded linear operator given by the restriction to $D$; then,
the adjoint $1^{*}_{D}:L^{2}(D)\to L^{2}(\mathbb{R}^{3})$ corresponds to the
extension by zero. Notice that $1^{*}_{D}1_{D}u=\chi_{D}u$.

In such a framework, the free resolvent rewrites as the operator block matrix

(4.1)
$R_{z}=\begin{bmatrix}1_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}}R_{z}1^{*}_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}}&1_{\Omega_{\varepsilon}^{c}}R_{z}1^{*}_{\Omega_{\varepsilon}}\\\
1_{\Omega_{\varepsilon}}R_{z}1^{*}_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}}&1_{\Omega_{\varepsilon}}R_{z}1^{*}_{\Omega_{\varepsilon}}\end{bmatrix}:L^{2}(\mathbb{R}^{3}\backslash\Omega_{\varepsilon})\oplus
L^{2}(\Omega_{\varepsilon})\to
L^{2}(\mathbb{R}^{3}\backslash\Omega_{\varepsilon})\oplus
L^{2}(\Omega_{\varepsilon})\,.$

By the identity

$T(\varepsilon)R_{z}=(1-\varepsilon^{2})\chi_{\Omega_{\varepsilon}}\Delta
R_{z}=-(1-\varepsilon^{2})1_{\Omega_{\varepsilon}}(1-zR_{z})\,,$

making use of (4.1), we get

$\displaystyle 1+T(\varepsilon)R_{z}$ $\displaystyle=\begin{bmatrix}1&0\\\
0&1\end{bmatrix}-(1-\varepsilon^{2})\left(\begin{bmatrix}0&0\\\
0&1\end{bmatrix}-z\begin{bmatrix}0&0\\\
1_{\Omega_{\varepsilon}}R_{z}1^{*}_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}}&1_{\Omega_{\varepsilon}}R_{z}1^{*}_{\Omega_{\varepsilon}}\end{bmatrix}\right)$
$\displaystyle\left.=\begin{bmatrix}1&0\\\
(1-\varepsilon^{2})z1_{\Omega_{\varepsilon}}R_{z}1^{*}_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}}&\varepsilon^{2}+(1-\varepsilon^{2})z1_{\Omega_{\varepsilon}}R_{z}1^{*}_{\Omega_{\varepsilon}}\end{bmatrix}\,.\right.$

Hence, for any $z\in{\mathbb{C}}$ such that
$z^{2}\in{\mathbb{C}}\backslash(-\infty,0]$, introducing the notations

$S_{z}(\varepsilon):=\varepsilon^{2}+(1-\varepsilon^{2})z^{2}1_{\Omega_{\varepsilon}}R_{z^{2}}1^{*}_{\Omega_{\varepsilon}}\,,$
(4.2)
$P_{z}(\varepsilon):=(1-\varepsilon^{2})S_{z}(\varepsilon)^{-1}z^{2}1_{\Omega_{\varepsilon}}R_{z^{2}}1^{*}_{\Omega_{\varepsilon}^{c}}\,,$

we get

(4.3) $(1+T(\varepsilon)R_{z^{2}})^{-1}=\begin{bmatrix}1&0\\\
P_{z}(\varepsilon)&S_{z}(\varepsilon)^{-1}\end{bmatrix}$

and so

$\displaystyle(1+T(\varepsilon)R_{z^{2}})^{-1}T(\varepsilon)=\,$
$\displaystyle(1-\varepsilon^{2})\begin{bmatrix}1&0\\\
P_{z}(\varepsilon)&S_{z}(\varepsilon)^{-1}\end{bmatrix}\begin{bmatrix}0&0\\\
1_{\Omega_{\varepsilon}}\Delta
1^{*}_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}}&1_{\Omega_{\varepsilon}}\Delta
1^{*}_{\Omega_{\varepsilon}}\end{bmatrix}$ $\displaystyle=\,$
$\displaystyle(1-\varepsilon^{2})\begin{bmatrix}0&0\\\
S_{z}(\varepsilon)^{-1}1_{\Omega_{\varepsilon}}\Delta
1^{*}_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}}&S_{z}(\varepsilon)^{-1}1_{\Omega_{\varepsilon}}\Delta
1^{*}_{\Omega_{\varepsilon}}\end{bmatrix}$ $\displaystyle=\,$
$\displaystyle(1-\varepsilon^{2})1^{*}_{\Omega_{\varepsilon}}S_{z}(\varepsilon)^{-1}1_{\Omega_{\varepsilon}}\Delta\,.$

This gives

$\displaystyle
K_{z^{2}}(\varepsilon)=R_{z^{2}}(1+T(\varepsilon)R_{z^{2}})^{-1}T(\varepsilon)R_{z^{2}}=\,$
$\displaystyle(1-\varepsilon^{2})R_{z^{2}}1^{*}_{\Omega_{\varepsilon}}S_{z}(\varepsilon)^{-1}1_{\Omega_{\varepsilon}}\Delta{R}_{z^{2}}\,.$

Introducing the unitary dilation operator

$U(\varepsilon):L^{2}(\mathbb{R}^{3})\to L^{2}(\mathbb{R}^{3})\,,\qquad
U(\varepsilon)u(x):=\varepsilon^{3/2}u(\varepsilon x)\,,$

one has

$U(\varepsilon)\chi_{\Omega_{\varepsilon}}=\chi_{\Omega}U(\varepsilon)\,,\qquad
U(\varepsilon)\chi_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}}=\chi_{{\mathbb{R}^{3}\backslash\Omega}}U(\varepsilon)$

and

$U(\varepsilon)R_{z^{2}}=\varepsilon^{2}R_{(\varepsilon
z)^{2}}U(\varepsilon)\,.$

Hence, defining

(4.4) $N_{z}:=1_{\Omega}R_{z^{2}}1^{*}_{\Omega}:L^{2}(\Omega)\to
L^{2}(\Omega)\,,$

one has

$\displaystyle
1_{\Omega_{\varepsilon}}R_{z^{2}}1^{*}_{\Omega_{\varepsilon}}=\varepsilon^{2}1_{\Omega_{\varepsilon}}U(\varepsilon)^{*}R_{(\varepsilon
z)^{2}}U(\varepsilon)1^{*}_{\Omega_{\varepsilon}}$ $\displaystyle=\,$
$\displaystyle\varepsilon^{2}1_{\Omega_{\varepsilon}}U(\varepsilon)^{*}(\chi_{\Omega}+\chi_{{\mathbb{R}^{3}\backslash\Omega}})R_{(\varepsilon
z)^{2}}(\chi_{\Omega}+\chi_{{\mathbb{R}^{3}\backslash\Omega}})U(\varepsilon)1^{*}_{\Omega_{\varepsilon}}$
$\displaystyle=\,$
$\displaystyle\varepsilon^{2}1_{\Omega_{\varepsilon}}(U(\varepsilon)^{*}\chi_{\Omega}+\chi_{{\mathbb{R}^{3}\backslash\Omega}}U(\varepsilon)^{*})R_{(\varepsilon
z)^{2}}(\chi_{\Omega}U(\varepsilon)+U(\varepsilon)\chi_{{\mathbb{R}^{3}\backslash\Omega}})1^{*}_{\Omega_{\varepsilon}}$
$\displaystyle=\,$
$\displaystyle\varepsilon^{2}1_{\Omega_{\varepsilon}}U(\varepsilon)^{*}\chi_{\Omega}R_{(\varepsilon
z)^{2}}\chi_{\Omega}U(\varepsilon)1^{*}_{\Omega_{\varepsilon}}$
$\displaystyle=\,$
$\displaystyle\varepsilon^{2}1_{\Omega_{\varepsilon}}U(\varepsilon)^{*}1^{*}_{\Omega}1_{\Omega}R_{(\varepsilon
z)^{2}}1^{*}_{\Omega}1_{\Omega}U(\varepsilon)1^{*}_{\Omega_{\varepsilon}}$
$\displaystyle=\,$
$\displaystyle\varepsilon^{2}1_{\Omega_{\varepsilon}}U(\varepsilon)^{*}1^{*}_{\Omega}N_{\varepsilon
z}1_{\Omega}U(\varepsilon)1^{*}_{\Omega_{\varepsilon}}\,.$

Defining

${M}_{z}(\varepsilon):=1+(1-\varepsilon^{2})z^{2}{N}_{\varepsilon z}\,,$

there follows

(4.5)
$S_{z}(\varepsilon)=\varepsilon^{2}1_{\Omega_{\varepsilon}}U^{*}_{\varepsilon}1^{*}_{\Omega}{M}_{z}(\varepsilon)1_{\Omega}U_{\varepsilon}1^{*}_{\Omega_{\varepsilon}}\qquad$

and then, defining

(4.6)
$G_{z}(\varepsilon):=\varepsilon^{-3/2}R_{z^{2}}\chi_{\Omega_{\varepsilon}}U_{\varepsilon}^{*}1^{*}_{\Omega}\equiv\varepsilon^{-3/2}R_{z^{2}}U_{\varepsilon}^{*}1^{*}_{\Omega}$

one finally gets the following

######  Lemma 4.1.

Let $K_{z}(\varepsilon)$ be as in Theorem 3.2. Then, for any $u\in
H^{2}(\mathbb{R}^{3})$ and for any $z\in{\mathbb{C}}$ such that
$z^{2}\in{\mathbb{C}}\backslash(-\infty,0]$,

(4.7)
$K_{z^{2}}(\varepsilon)u=(1-\varepsilon^{2})\varepsilon\,G_{z}(\varepsilon){M}_{z}(\varepsilon)^{-1}G_{\bar{z}}(\varepsilon)^{*}\Delta
u\,.$

###### Proof.

$\displaystyle K_{z^{2}}(\varepsilon)u=\,$ $\displaystyle
R_{z^{2}}(1+T(\varepsilon)R_{z^{2}})^{-1}T(\varepsilon)R_{z^{2}}u$
$\displaystyle=\,$
$\displaystyle(1-\varepsilon^{2})\varepsilon^{-2}R_{z^{2}}1^{*}_{\Omega_{\varepsilon}}1_{\Omega_{\varepsilon}}U_{\varepsilon}^{*}1_{\Omega}{M}_{z}(\varepsilon)^{-1}1^{*}_{\Omega}U_{\varepsilon}1_{\Omega_{\varepsilon}}^{*}1_{\Omega_{\varepsilon}}\Delta{R}_{z^{2}}u$
$\displaystyle=\,$
$\displaystyle(1-\varepsilon^{2})\varepsilon^{-2}R_{z^{2}}\chi_{\Omega_{\varepsilon}}U_{\varepsilon}^{*}1^{*}_{\Omega}{M}_{z}(\varepsilon)^{-1}1_{\Omega}U_{\varepsilon}\chi_{\Omega_{\varepsilon}}{R}_{z^{2}}\Delta
u$ $\displaystyle=\,$ $\displaystyle(1-\varepsilon^{2})\varepsilon
G_{z}(\varepsilon){M}_{z}(\varepsilon)^{-1}G_{\bar{z}}(\varepsilon)^{*}\Delta
u\,.$

∎

###  4.1. The Newton potential operator of $\Omega$

Now, we introduce the not negative, symmetric operator $N_{0}$ in
$L^{2}(\Omega)$ defined by

$N_{0}:L^{2}(\Omega)\to L^{2}(\Omega)\,,\qquad
N_{0}u(x):=\frac{1}{4\pi}\int_{\Omega}\frac{u(y)\,dy}{|x-y|}\,.$

By Sobolev’s inequality (see, e.g., [22, Appendix 2 to I.1])

$\int_{\Omega\times\Omega}\frac{dx\,dy}{|x-y|^{2}}=\int_{\mathbb{R}^{3}\times\mathbb{R}^{3}}\chi_{\Omega}(x)\chi_{\Omega}(y)\,\frac{dx\,dy}{|x-y|^{2}}\lesssim\|\chi_{\Omega}\|^{2}_{L^{3/2}(\mathbb{R}^{3})}$

and so $N_{0}$ is Hilbert-Schmidt and hence compact (see, e.g., [22, Theorem
A.28]). Since $-\Delta N_{0}u=u$, one gets $\ker(N_{0})=\\{0\\}$. By the
spectral theory of compact symmetric operators (see, e.g., [15, Section 6]),
$\sigma_{d}(N_{0})=\sigma(N_{0})\backslash\\{0\\}$ and the orthonormal
sequence $\\{e_{k}\\}_{1}^{+\infty}$ of eigenvectors corresponding to the
discrete spectrum is an orthonormal base of
$\ker(N_{0})^{\perp}=L^{2}(\Omega)$. We denote by
$\\{\lambda_{k}\\}_{1}^{+\infty}=\sigma_{d}(N_{0})$, the set of eigenvalues
(counting multiplicities) indexed in decreasing order. One has
$\lambda_{1}=\|N_{0}\|$ and $\lambda_{k}\searrow 0$. Furthermore, for any
$u\in L^{2}(\Omega)$ and for any bounded measurable function
$f:[\,0,\|N_{0}\|\,]\to{\mathbb{C}}$,

(4.8) $f(N_{0})u=\sum_{k=1}^{+\infty}f(\lambda_{k})\langle e_{k},u\rangle
e_{k}\,.$

######  Lemma 4.2.

Let $z=c+i\gamma$, $c>0$, $\gamma\in\mathbb{R}$. Then, $1+z^{2}N_{0}$ has a
bounded inverse and

$\|(1+z^{2}N_{0})^{-1}\|\leq\begin{cases}1&|\gamma|\leq c\\\
{|z|^{2}}{(2c|\gamma|\,)^{-1}}&|\gamma|>c\end{cases}\quad.$

###### Proof.

By
$\bar{z}^{2}=(c^{2}-\gamma^{2})-i2c\gamma\in{\mathbb{C}}\backslash(-\infty,0]$
and by $\sigma(-N_{0})\subset\left[-\|N_{0}\|,0\,\right]$, one gets
$z^{-2}=|z|^{-4}\bar{z}^{2}\in\varrho(-N_{0})$. Hence,

$(1+z^{2}N_{0})={z^{2}}\,(N_{0}+z^{-2})$

has a bounded inverse and, by the functional calculus for $N_{0}$,

$\|(1+z^{2}N_{0})^{-1}\|=\sup_{\lambda\in\sigma(N_{0})}\,\frac{1}{|1+z^{2}\lambda|}\leq\sup_{0\leq\lambda\leq\|N_{0}\|}\,\frac{1}{|1+z^{2}\lambda|}=\left(\inf_{0\leq\lambda\leq\|N_{0}\|}\,|1+z^{2}\lambda|\right)^{-1}\,.$

The quadratic polynomial

$p(\lambda):=|1+z^{2}\lambda|^{2}=\big{(}1+(c^{2}-\gamma^{2})\lambda\big{)}^{2}+4c^{2}\gamma^{2}\lambda^{2}=(\gamma^{2}+c^{2})^{2}\lambda^{2}-2(\gamma^{2}-c^{2})\lambda+1$

has minimum at $\lambda=\lambda_{\circ}$,

$\lambda_{\circ}:=\frac{\gamma^{2}-c^{2}}{(\gamma^{2}+c^{2})^{2}}\,,$

where it assumes the value

$p(\lambda_{\circ})=1-\frac{(\gamma^{2}-c^{2})^{2}}{(\gamma^{2}+c^{2})^{2}}=\frac{4c^{2}\gamma^{2}}{|z|^{4}}\,.$

Therefore,

$\inf_{0\leq\lambda\leq\|N_{0}\|}\,p(\lambda)=\begin{cases}p(0)&\lambda_{\circ}<0\\\
p(\lambda_{\circ})&0\leq\lambda_{\circ}\leq\|N_{0}\|\\\
p(\|N_{0}\|)&\lambda_{\circ}>\|N_{0}\|\end{cases}\,.$

However, since $\lambda_{\circ}\searrow 0$ as $|\gamma|\nearrow+\infty$, for
our later purposes it suffices to use the rougher estimate

$\inf_{0\leq\lambda\leq\|N_{0}\|}\,p(\lambda)\geq\inf_{\lambda\geq
0}\,p(\lambda)=\begin{cases}p(0)&\lambda_{\circ}<0\\\
p(\lambda_{\circ})&\lambda_{\circ}\geq
0\end{cases}=\begin{cases}1&|\gamma|\leq c\\\
4c^{2}\gamma^{2}|z|^{-4}&|\gamma|>c\ .\end{cases}$

∎

######  Lemma 4.3.

Let $z\in{\mathbb{C}}\backslash(-\infty,0]$. Then

${M}_{z}(\varepsilon)=(1+z^{2}N_{0})+{M}^{(1)}_{z}(\varepsilon)\,,$

where

$\|{M}^{(1)}_{z}(\varepsilon)\|\leq\varepsilon\,|z|^{2}\left({\frac{|\Omega|}{4\pi}}^{1/2}|z|+\|N_{0}\|\right)\,.$

###### Proof.

One has

$\displaystyle{M}^{(1)}_{z}(\varepsilon)=\,$
$\displaystyle{M}_{z}(\varepsilon)-(1+z^{2}N_{0})=z^{2}\big{(}(1-\varepsilon^{2})({N}_{\varepsilon
z}-N_{0})-\varepsilon^{2}N_{0}\big{)}$

and so, by $\varepsilon<1$,

$\|{M}^{(1)}_{z}(\varepsilon)\|\leq|z|^{2}\left(\|N_{\varepsilon
z}-N_{0}\|+\varepsilon^{2}\|N_{0}\|\right)\leq|z|^{2}\left(\|N_{\varepsilon
z}-N_{0}\|+\varepsilon\,\|N_{0}\|\right)\,.$

Furthermore,

$N_{\varepsilon z}-N_{0}=z\int_{0}^{\varepsilon}N_{sz}^{(1)}ds\,,$

where

$N^{(1)}_{z}:L^{2}(\Omega)\to L^{2}(\Omega)\,,\qquad
N^{(1)}_{z}u(x):=-\frac{1}{4\pi}\int_{\Omega}\,e^{-z|x-y|}\,u(y)\,dy\,.$

Then, by

$\int_{\Omega}|N^{(1)}_{z}u(x)|^{2}dx\leq\frac{e^{-2\text{Re}(z)\,d_{\Omega}}}{(4\pi)^{2}}\left(\int_{\Omega}|u(y)|\,dy\right)^{\\!\\!2}\leq\frac{|\Omega|}{(4\pi)^{2}}\,\|u\|^{2}_{L^{2}(\Omega)}\,,$

one obtains

$\displaystyle\|N_{\varepsilon
z}-N_{0}\|\leq|z|\int_{0}^{\varepsilon}\|N^{(1)}_{sz}\|\,ds\leq\varepsilon\,\frac{|z|}{4\pi}\,|\Omega|^{1/2}\,.$

∎

######  Lemma 4.4.

Let $\varepsilon\in(0,1)$ and $z=c+i\gamma$, $c>0$, $\gamma\in\mathbb{R}$,
such that

(4.9)
$\|(1+z^{2}N_{0})^{-1}\|\,\|M^{(1)}_{z}(\varepsilon)\|\leq\frac{1}{2}\,.$

Then

${M}_{z}(\varepsilon)^{-1}=(1+z^{2}N_{0})^{-1}+\Lambda_{z}(\varepsilon)$

and

$\|\Lambda_{z}(\varepsilon)\|\leq
2\varepsilon\,|z|^{2}\left({\frac{|\Omega|}{4\pi}}^{1/2}\\!\\!|z|+\|N_{0}\|\right)\,\|(1+z^{2}N_{0})^{-1}\|^{2}\,.$

###### Proof.

By

${M}_{z}(\varepsilon)=(1+z^{2}N_{0})(1+(1+z^{2}N_{0})^{-1}{M}^{(1)}_{z}(\varepsilon))$

and by (4.9), one gets

${M}_{z}(\varepsilon)^{-1}=\left(\sum_{n=0}^{+\infty}((1+z^{2}N_{0})^{-1}{M}^{(1)}_{z}(\varepsilon))^{n}\right)(1+z^{2}N_{0})^{-1}$

and

$\|{M}_{z}(\varepsilon)^{-1}\|\leq 2\,\|(1+z^{2}N_{0})^{-1}\|\,.$

Hence, by Lemma 4.3 and by

(4.10)
$\Lambda_{z}(\varepsilon)=-{M}_{z}(\varepsilon)^{-1}M^{(1)}_{z}(\varepsilon)(1+z^{2}N_{0})^{-1}\,,$

one gets

$\displaystyle\|\Lambda_{z}(\varepsilon)\|\leq\|{M}_{z}(\varepsilon)^{-1}\|\,\|M^{(1)}_{z}(\varepsilon)\|\,\|(1+z^{2}N_{0})^{-1}\|$
$\displaystyle\,\leq
2\varepsilon\,|z|^{2}\left({\frac{|\Omega|}{4\pi}}^{1/2}\\!\\!|z|+\|N_{0}\|\right)\,\|(1+z^{2}N_{0})^{-1}\|^{2}\,.$

∎

######  Lemma 4.5.

Let $z=c+i\gamma$, $c>0$, $\gamma\in\mathbb{R}$ such that $|\gamma|\leq c$.
Then,

$\|\Lambda_{z}(\varepsilon)\|\leq\varepsilon\,2c^{2}\left({\frac{|\Omega|^{1/2}c}{2\sqrt{2}\pi}}+\|N_{0}\|\right)\,.$

###### Proof.

By $\gamma^{2}\leq c^{2}$ one gets Re$(z^{2})\geq 0$. Then, by (3.5), (3.6),
and (4.3), one obtains

$\|S_{z}(\varepsilon)^{-1}\|\leq\|(1+T(\varepsilon)R_{z^{2}})^{-1}\|\leq\frac{1}{\varepsilon^{2}}\,.$

Hence, by (4.5) and by

$\|1_{\Omega}U_{\varepsilon}1^{*}_{\Omega_{\varepsilon}}\|=\|1_{\Omega_{\varepsilon}}U^{*}_{\varepsilon}1^{*}_{\Omega}\|=1\,,$

one has

$\|M_{z}(\varepsilon)^{-1}\|\leq 1\,.$

The proof is then concluded by (4.10), by Lemmata 4.2, 4.3 and by
$|z|\leq\sqrt{2}\,c$. ∎

######  Lemma 4.6.

Suppose that $z=c+i\gamma$, $|\gamma|>c>0$. Let

$c_{1}:=\frac{1}{4}\left(\frac{|\Omega|^{1/2}}{2\sqrt{2}\pi}+\|N_{0}\|\right)^{\\!\\!-1}\,.$

If

(4.11) $\frac{\varepsilon}{c}\leq
c_{1}\,,\qquad|\gamma|^{4}\leq\frac{c}{\varepsilon}\,c_{1}\,,$

then (4.9) holds true.

###### Proof.

By $|\gamma|>c$, by Lemmata 4.3 and 4.2,

$\displaystyle\|(1+z^{2}N_{0})^{-1}\|\,\|{M}^{(1)}_{z}(\varepsilon)\|\leq$
$\displaystyle\,\varepsilon\,\frac{(c^{2}+\gamma^{2})^{2}}{2c|\gamma|}\left(\frac{|\Omega|^{1/2}}{4\pi}\,(c^{2}+\gamma^{2})^{1/2}+\|N_{0}\|\right)$
$\displaystyle\leq$
$\displaystyle\,\varepsilon\,\frac{2|\gamma|^{3}}{c}\left(\frac{|\Omega|^{1/2}}{2\sqrt{2}\pi}\,|\gamma|+\|N_{0}\|\right).$

If $|\gamma|\leq 1$, then

$\|(1+z^{2}N_{0})^{-1}\|\,\|{M}^{(1)}_{z}(\varepsilon)\|\leq\varepsilon\,\frac{2}{c}\left(\frac{|\Omega|^{1/2}}{2\sqrt{2}\pi}+\|N_{0}\|\right).$

If $|\gamma|\geq 1$, then

$\|(1+z^{2}N_{0})^{-1}\|\,\|{M}^{(1)}_{z}(\varepsilon)\|\leq\varepsilon\,|\gamma|^{4}\,\frac{2}{c}\left(\frac{|\Omega|^{1/2}}{2\sqrt{2}\pi}+\|N_{0}\|\right).$

Hence, (4.11) implies (4.9). ∎

Given

$G_{z}(\varepsilon):L^{2}(\Omega)\to L^{2}(\mathbb{R}^{3})\,,$

as in (4.6), and

(4.12) $G_{z}:L^{2}(\Omega)\to L^{2}(\mathbb{R}^{3})\,,\qquad G_{z}u:=\langle
1,u\rangle_{L^{2}(\Omega)}\,{\mathcal{G}}_{z}\,,\qquad{\mathcal{G}}_{z}(x):=\frac{e^{-z|x|}}{4\pi|x|}\,,$

let

$G^{(1)}_{z}(\varepsilon):L^{2}(\Omega)\to L^{2}(\mathbb{R}^{3})\,,$

be such that

$G_{z}(\varepsilon)=G_{z}+G^{(1)}_{z}(\varepsilon)\,.$

The next result gives bounds on the norms of such three bounded operators:

######  Lemma 4.7.

Let $z=c+i\gamma$, $c>0$, $\gamma\in\mathbb{R}$. Then,

1)

$\|G_{z}\|\leq\left(\frac{|\Omega|}{8\pi c}\right)^{\\!\\!1/2}\,,$

2)

$\|G^{(1)}_{z}(\varepsilon)\|\leq\varepsilon^{1/2}\,\left(\int_{\Omega}|x|\,dx\right)^{1/2}\,\begin{cases}2+|z|^{-2}&|\gamma|\leq
c\\\ 1+(2c|\gamma|\,)^{-1}(1+|z|^{2})&|\gamma|>c\end{cases}\,,$

3)

$\|G_{z}(\varepsilon)\|\leq\,n_{s}\,|\Omega|\,\varepsilon^{s-3/2}\,\begin{cases}\left(2+|z|^{-2}\right)^{s/2}|z|^{s-2}&|\gamma|\leq
c\\\
\big{(}1+(c+|\gamma|\,)^{2})\big{)}^{s/2}\left(2c|\gamma|\right)^{-1}&|\gamma|>c\end{cases}\,.$

Here, $0\leq s<\frac{3}{2}$ and $n_{s}$ denotes the norm of the Sobolev
embedding $H^{s}(\mathbb{R}^{3})\hookrightarrow L^{q}(\mathbb{R}^{3})$,
$q=\frac{6}{3-2s}$.

###### Proof.

1)

$\|G_{z}\|^{2}\leq|\Omega|\,\|{\mathcal{G}}_{z}\|^{2}=\frac{|\Omega|}{4\pi}\,\int_{0}^{\infty}e^{-2\text{Re}(z)r}\,dr=\,\frac{|\Omega|}{8\pi\text{Re}(z)}$

2) By the continuous embedding of $H^{2}(\Omega)$ into the space of Hölder-
continuous functions of order $\frac{1}{2}$, one gets

$\displaystyle\|(G^{*}_{\bar{z}}(\varepsilon)-G^{*}_{\bar{z}})u\|^{2}=$
$\displaystyle\,\int_{\Omega}\left|\int_{\mathbb{R}^{3}}({\mathcal{G}}_{z}(\varepsilon
x-y)-{\mathcal{G}}_{z}(y))u(y)\,dy\right|^{2}dx$ $\displaystyle=$
$\displaystyle\,\int_{\Omega}\big{|}R_{z^{2}}u(\varepsilon
x)-R_{z^{2}}u(0)\big{|}^{2}dx\leq\varepsilon\int_{\Omega}|x|\,dx\,\|R_{z^{2}}u\|^{2}_{H^{2}(\Omega)}$
$\displaystyle=$ $\displaystyle\,\varepsilon\int_{\Omega}|x|\,dx\
\|(-\Delta+z^{2}+1-z^{2})R_{z^{2}}u\|^{2}$ $\displaystyle\leq$
$\displaystyle\,\varepsilon\int_{\Omega}|x|\,dx\
\left(\|u\|+(1+|z|^{2})\|R_{z^{2}}u\|\right)^{2}$ $\displaystyle\leq$
$\displaystyle\,\varepsilon\int_{\Omega}|x|\,dx\
\begin{cases}\left(2+|z|^{-2}\right)^{2}\|u\|^{2}&|\gamma|\leq c\\\
\left(1+(2c|\gamma|\,)^{-1}(1+|z|^{2})\right)^{2}\|u\|^{2}&|\gamma|>c\end{cases}\,.$

3) By the inequalities,

$\displaystyle\|R_{z^{2}}\|\leq\begin{cases}|z|^{-2}&|\gamma|\leq c\\\
\left(2c|\gamma|\right)^{-1}&|\gamma|>c\end{cases}\,,\qquad\|(-\Delta+1)R_{z^{2}}\|\leq\begin{cases}2+|z|^{-2}&|\gamma|\leq
c\\\ 1+(2c|\gamma|\,)^{-1}(1+|z|^{2})&|\gamma|>c\end{cases}\,,$

and by interpolation, one obtains

$\|R_{z^{2}}\|_{L^{2}(\mathbb{R}^{3}),H^{s}(\mathbb{R}^{3})}\leq\begin{cases}\left(2+|z|^{-2}\right)^{s/2}|z|^{s-2}&|\gamma|\leq
c\\\
\left(1+(2c|\gamma|\,)^{-1}(1+|z|^{2})\right)^{s/2}\left(2c|\gamma|\right)^{s/2-1}&|\gamma|>c\end{cases}\,,\qquad
0\leq s\leq 2\,.$

Then, by the Sobolev embedding

$H^{s}(\mathbb{R}^{3})\hookrightarrow L^{q}(\mathbb{R}^{3})\,,\qquad
q=\frac{6}{3-2s}\,,\quad\quad 0\leq s<\frac{3}{2}\,,$

and by Hölder’s inequality with $\frac{1}{p}=\frac{s}{3}$, so that
$\frac{1}{p}+\frac{1}{q}=\frac{1}{2}$, one gets

$\displaystyle\|G_{z}(\varepsilon)u\|=\varepsilon^{-3/2}\left(\int_{\mathbb{R}^{3}}|\chi_{\Omega_{\varepsilon}}(x)R_{z^{2}}u(x)|^{2}dx\right)^{1/2}$
$\displaystyle\leq\,$
$\displaystyle\varepsilon^{-3/2}|\Omega_{\varepsilon}|^{1/p}\|R_{z^{2}}u\|_{L^{q}(\mathbb{R}^{3})}$
$\displaystyle\leq\,$ $\displaystyle
n_{s}\,|\Omega|\,\varepsilon^{s-3/2}\|R_{z^{2}}u\|_{H^{s}(\mathbb{R}^{3})}$
$\displaystyle\leq\,$ $\displaystyle
n_{s}\,|\Omega|\,\varepsilon^{s-3/2}\,\|u\|\begin{cases}\left(2+|z|^{-2}\right)^{s/2}|z|^{s-2}&|\gamma|\leq
c\\\
\left(1+(2c|\gamma|\,)^{-1}(1+|z|^{2})\right)^{s/2}\left(2c|\gamma|\right)^{s/2-1}&|\gamma|>c\,.\end{cases}$

∎

##  5\. Asymptotic dynamics

By (3.4), (4.7) and by the results in the previous section, one has

$R_{z^{2}}(\varepsilon)=R_{z^{2}}-K_{z^{2}}(\varepsilon)=R_{z^{2}}-(1-\varepsilon^{2})\,\varepsilon\,G_{z}(\varepsilon){M}_{z}(\varepsilon)^{-1}G_{\bar{z}}(\varepsilon)^{*}\Delta$

and

$\displaystyle
G_{z}(\varepsilon){M}_{z}(\varepsilon)^{-1}G_{\bar{z}}(\varepsilon)^{*}$
$\displaystyle=\,$ $\displaystyle
G_{z}(\varepsilon)(1+z^{2}N_{0})^{-1}G_{\bar{z}}(\varepsilon)^{*}+G_{z}(\varepsilon)\Lambda_{z}(\varepsilon)G_{\bar{z}}(\varepsilon)^{*}$
(5.1) $\displaystyle=\,$ $\displaystyle
G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}+G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{(1)}(\varepsilon)^{*}$
$\displaystyle+\,$ $\displaystyle
G_{z}^{(1)}(\varepsilon)(1+z^{2}N_{0})^{-1}G_{\bar{z}}(\varepsilon)^{*}+G_{z}(\varepsilon)\Lambda_{z}(\varepsilon)G_{\bar{z}}(\varepsilon)^{*}\,.$

Then, combining the estimates provided in Lemmata 4.2-4.7 with Lemma 3.5, one
gets

######  Theorem 5.1.

Let $\phi$ and $\psi$ be in $H^{6}(\mathbb{R}^{3})$ and supported outside
$\Omega_{\varepsilon}$; let $u_{\varepsilon}$ and $u_{\text{\rm free}}$ be the
classical solutions of the Cauchy problems

$\begin{cases}(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}})\partial_{tt}u_{\varepsilon}(t)=\Delta
u_{\varepsilon}(t)\\\ u_{\varepsilon}(0)=\phi\\\
\partial_{t}u_{\varepsilon}(0)=\psi\,.\end{cases}\qquad\begin{cases}\partial_{tt}u_{\text{\rm
free}}(t)=\Delta u_{\text{\rm free}}(t)\\\ u_{\text{\rm free}}(0)=\phi\\\
u_{\text{\rm free}}(0)=\psi\,.\end{cases}$

Then, for any $\tau\in(0,\frac{1}{11})$, $\tau^{\prime}\in(0,\frac{1}{9})$ and
for any $\varepsilon>0$ such that

$\varepsilon\leq{\rm
min}\left\\{1,c_{1}^{4/(3\tau+1)},c_{1}^{4/(3\tau^{\prime}+1)}\right\\}\,,$

one has

$u_{\varepsilon}(t)=\,u_{\text{\rm
free}}(t)+\varepsilon\,(\varepsilon^{2}-1)\,v(t)+r_{\varepsilon}(t)+r^{\prime}_{\varepsilon}(t)\,,\qquad
v(t,x)=H(t-|x|)\,\frac{q(t-|x|)}{4\pi|x|}\,,$

where

$\displaystyle q(t):=\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\int_{|y|<t}\bigg{(}\bigg{(}1-\cos\bigg{(}\frac{t-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\bigg{)}\frac{\Delta^{2}\phi(y)}{4\pi|y|}+\frac{1}{\lambda_{k}^{1/2}}\,\sin\bigg{(}\frac{t-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta\psi(y)}{4\pi|y|}$
$\displaystyle\,\bigg{)}\,dy$

and

$\sup_{0<t\leq
1/\varepsilon^{\tau}}\|r_{\varepsilon}(t)\|\lesssim\,\varepsilon^{1+(1-11\tau)/2}\,\|\Delta^{3}\psi\|\,,\qquad\sup_{0<t\leq
1/\varepsilon^{\tau^{\prime}}}\|r^{\prime}_{\varepsilon}(t)\|\lesssim\,\varepsilon^{1+(1-9\tau^{\prime})/4}\,\|\Delta^{3}\phi\|\,.$

###### Proof.

By Lemma 3.5, there holds, for any $\phi\in
H^{2(m^{\prime}+1)}(\mathbb{R}^{3})$, $m^{\prime}\geq 0$, and $\psi\in
H^{2n^{\prime}}(\mathbb{R}^{3})$, $n^{\prime}\geq 1$, supported outside
$\Omega_{\varepsilon}$,

$u_{\varepsilon}(t)=u_{\text{\rm free}}(t)-\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n^{\prime}}\psi\
\frac{dz}{z^{2n^{\prime}}}-\frac{1}{2\pi
i}\int_{c^{\prime}-i\infty}^{c^{\prime}+i\infty}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m^{\prime}+1}\phi\
\frac{dz}{z^{2m^{\prime}+1}}\,,$

where $c>0$ and $c^{\prime}>0$ are arbitrary. Using Corollary 3.4, one has,
for any $1\leq n<n^{\prime}$,

$\displaystyle\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n^{\prime}}\psi\,\frac{dz}{z^{2n^{\prime}}}=\frac{1}{2\pi
i}\int_{c-ic_{1}(c/\varepsilon)^{1/4}}^{c+ic_{1}(c/\varepsilon)^{1/4}}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n}\psi\,\frac{dz}{z^{2n}}+I(t,\varepsilon)$

and, for any $0\leq m<m^{\prime}$,

$\displaystyle\frac{1}{2\pi
i}\int_{c^{\prime}-i\infty}^{c^{\prime}+i\infty}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m^{\prime}+1}\phi\,\frac{dz}{z^{2m^{\prime}+1}}=\frac{1}{2\pi
i}\int_{c^{\prime}-ic_{1}(c^{\prime}/\varepsilon)^{1/4}}^{c^{\prime}+ic_{1}(c^{\prime}/\varepsilon)^{1/4}}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m+1}\phi\,\frac{dz}{z^{2m+1}}+I^{\prime}(t,\varepsilon)\,,$

where

$I(t,\varepsilon):=\frac{1}{2\pi
i}\int_{c-i\infty}^{c-ic_{1}(c/\varepsilon)^{1/4}}\\!\\!\\!e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n^{\prime}}\psi\,\frac{dz}{z^{2n^{\prime}}}+\frac{1}{2\pi
i}\int_{c+ic_{1}(c/\varepsilon)^{1/4}}^{c+i\infty}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n^{\prime}}\psi\,\frac{dz}{z^{2n^{\prime}}}\,,$
$I^{\prime}(t,\varepsilon):=\frac{1}{2\pi
i}\int_{c^{\prime}-i\infty}^{c^{\prime}-ic_{1}(c^{\prime}/\varepsilon)^{1/4}}\\!\\!\\!e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m^{\prime}+1}\phi\,\frac{dz}{z^{2m^{\prime}+1}}+\frac{1}{2\pi
i}\int_{c^{\prime}+ic_{1}(c^{\prime}/\varepsilon)^{1/4}}^{c^{\prime}+i\infty}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m^{\prime}+1}\phi\,\frac{dz}{z^{2m^{\prime}+1}}\,.$

By (3.11), whenever

(5.2) $c_{1}\left(\frac{c}{\varepsilon}\right)^{1/4}\geq c\,,\qquad
c_{1}\left(\frac{c^{\prime}}{\varepsilon}\right)^{1/4}\geq c^{\prime}\,,$

one gets

$\|I(t,\varepsilon)\|\lesssim\frac{e^{ct}}{c}\,\int_{c_{1}(c/\varepsilon)^{1/4}}^{+\infty}\frac{d\gamma}{\gamma^{2n^{\prime}+1}}\
\|\Delta^{n^{\prime}}\psi\|\lesssim\left(\frac{\varepsilon}{c}\right)^{n^{\prime}/2}\,\frac{e^{ct}}{c}\
\|\Delta^{n^{\prime}}\psi\|\,.$
$\|I^{\prime}(t,\varepsilon)\|\lesssim\frac{e^{c^{\prime}t}}{c^{\prime}}\,\int_{c_{1}(c^{\prime}/\varepsilon)^{1/4}}^{+\infty}\frac{d\gamma}{\gamma^{2m^{\prime}+2}}\
\|\Delta^{m^{\prime}+1}\phi\|\lesssim\left(\frac{\varepsilon}{c^{\prime}}\right)^{m^{\prime}/2+1/4}\,\frac{e^{c^{\prime}t}}{c^{\prime}}\
\|\Delta^{m^{\prime}+1}\phi\|\,,$

Now, we re-write $K_{z^{2}}(\varepsilon)$ by using (5) and provide estimates
on the various terms. We begin by considering the evolution of the initial
datum $\psi$.

One has

$\displaystyle\,\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{n}\psi\,\frac{dz}{z^{2n}}$
$\displaystyle=$
$\displaystyle\,(1-\varepsilon^{2})\varepsilon\big{(}v_{1}(t)+I_{0}(t,\varepsilon)+I_{1}(t,\varepsilon)+I_{1}(t,\varepsilon)^{*}+{I_{2}}(t\varepsilon)\big{)}+I(t,\varepsilon)\,,$

where

(5.3) $v_{0}(t):=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{tz}G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}\Delta^{n+1}\psi\,\frac{dz}{z^{2n}}$
$\displaystyle I_{0}(t,\varepsilon):=$ $\displaystyle\,-\frac{1}{2\pi
i}\int_{c-i\infty}^{c-ic_{1}(c/\varepsilon)^{1/4}}e^{tz}G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}\Delta^{n+1}\psi\,\frac{dz}{z^{2n}}$
$\displaystyle-\frac{1}{2\pi
i}\int_{c+ic_{1}(c/\varepsilon)^{1/4}}^{c+i\infty}e^{tz}G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}\Delta^{n+1}\psi\,\frac{dz}{z^{2n}}\,,$
$I_{1}(t,\varepsilon):=\frac{1}{2\pi
i}\int_{c-ic_{1}(c/\varepsilon)^{1/4}}^{c+ic_{1}(c/\varepsilon)^{1/4}}e^{tz}G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{(1)}(\varepsilon)^{*}\Delta^{n+1}\psi\,\frac{dz}{z^{2n}}\,,$
$I_{1}(t,\varepsilon)^{*}:=\frac{1}{2\pi
i}\int_{c-ic_{1}(c/\varepsilon)^{1/4}}^{c+ic_{1}(c/\varepsilon)^{1/4}}e^{tz}G_{z}^{(1)}(\varepsilon)(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}\Delta^{n+1}\psi\,\frac{dz}{z^{2n}}\,,$
$I_{2}(t,\varepsilon):=\frac{1}{2\pi
i}\int_{c-ic_{1}(c/\varepsilon)^{1/4}}^{c+ic_{1}(c/\varepsilon)^{1/4}}e^{tz}G_{z}(\varepsilon)\Lambda_{z}(\varepsilon)G_{z}(\varepsilon)^{*}\Delta^{n+1}\psi\,\frac{dz}{z^{2n}}\,.$

By Lemmata 4.2 and 4.7, whenever $|\gamma|>c$,

$\|G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}\|\leq\,\|{G}_{z}\|\,\|{G}_{\bar{z}}\|\,\|(1+z^{2}N_{0})^{-1}\|\lesssim\frac{1}{c^{2}}\,\frac{|z|^{2}}{|\gamma|\
}$

and so,

$\|I_{0}(t,\varepsilon)\|\lesssim\frac{e^{ct}}{c^{2}}\,\int_{c_{1}(c/\varepsilon)^{1/4}}^{+\infty}\frac{d\gamma}{\gamma^{2n-1}}\
\|\Delta^{n+1}\psi\|\lesssim\left(\frac{\varepsilon}{c}\right)^{(n-1)/2}\,\frac{e^{ct}}{c^{2}}\,\|\Delta^{n+1}\psi\|\,.$

By Lemmata 6.1 and 6.2 in the Appendix,

$\|I_{1}(t,\varepsilon)\|\lesssim\varepsilon^{1/2}\,\frac{e^{tc}}{c^{2n+3/2}}\
\|\Delta^{n+1}\psi\|\,,\qquad\|I_{2}(t,\varepsilon)\|\lesssim\varepsilon^{1/2}\,\frac{e^{tc}}{c^{2n+1}}\
\|\Delta^{n+1}\psi\|\,.$

Then, the estimate for $\|I_{1}(t,\varepsilon)^{*}\|$ and
$\|I_{2}(t,\varepsilon)^{*}\|$ are the same as the ones for
$\|I_{1}(t,\varepsilon)\|$ and $\|I_{2}(t,\varepsilon)\|$.

Putting the previous estimates together, one gets

$\displaystyle\,{\mathcal{I}}(t,\varepsilon):=\left\|(1-\varepsilon^{2})\varepsilon\big{(}I_{0}(t,\varepsilon)+I_{1}(t,\varepsilon)+I_{1}(t,\varepsilon)^{*}+{I_{2}}(t,\varepsilon)\big{)}+I(t,\varepsilon)\right\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon\bigg{(}\left(\frac{\varepsilon}{c}\right)^{n/2-1/2}\,\frac{e^{ct}}{c^{2}}+\varepsilon^{1/2}\left(\frac{e^{tc}}{c^{2n+3/2}}+\frac{e^{tc}}{c^{2n+1}}\right)\bigg{)}\|\Delta^{n+1}\psi\|+\left(\frac{\varepsilon}{c}\right)^{n^{\prime}/2}\,\frac{e^{ct}}{c}\,\|\Delta^{n^{\prime}}\psi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon\bigg{(}\left(\frac{\varepsilon}{c}\right)^{n/2-1/2}\,\frac{e^{ct}}{c^{2}}+\varepsilon^{1/2}\,\frac{e^{tc}}{c^{2n+3/2}}\bigg{)}\,\|\Delta^{n+1}\psi\|+\left(\frac{\varepsilon}{c}\right)^{n^{\prime}/2}\,\frac{e^{ct}}{c}\,\|\Delta^{n^{\prime}}\psi\|\,.$

Then, taking $c=\varepsilon^{\tau}$, $\tau>0$, and $t\leq\varepsilon^{-\tau}$,

$\displaystyle{\mathcal{I}}(t,\varepsilon)\lesssim$
$\displaystyle\,\varepsilon\left(\varepsilon^{(1-\tau)(n/2-1/2)-2\tau}+\varepsilon^{1/2-(2n+3/2)\tau}\right)\|\Delta^{n+1}\psi\|+\varepsilon^{(1-\tau)n^{\prime}/2-\tau}\,\|\Delta^{n^{\prime}}\psi\|$

and, by (5.2), $\varepsilon\leq c_{1}^{4/(3\tau+1)}$. We choose $\tau$ such
that

$(1-\tau)\left(\frac{n}{2}-\frac{1}{2}\right)-2\tau>0\,,\quad\frac{1}{2}-\left(2n+\frac{3}{2}\right)\tau>0\,,\quad(1-\tau)\,\frac{n^{\prime}}{2}-\tau>1\,,$

equivalently,

$\tau<\frac{n-1}{n+3}\,,\quad\tau<\frac{1}{4n+3}\,,\quad\tau<\frac{n^{\prime}-2}{n^{\prime}+2}$

Taking $n=2$, $n^{\prime}=3$, one obtains the bound $\tau<\frac{1}{11}$ and

$\displaystyle{\mathcal{I}}(t,\varepsilon)\lesssim$
$\displaystyle\,\varepsilon\left(\varepsilon^{(1-3\tau)/2}+\varepsilon^{(1-11\tau)/2}\right)\|\Delta^{3}\psi\|+\varepsilon^{(3-5\tau)/2}\,\|\Delta^{3}\psi\|\lesssim\varepsilon^{1+(1-11\tau)/2}\|\Delta^{3}\psi\|\,.$

Proceeding as above, one has, as regards the evolution of the initial datum
$\phi$,

$\displaystyle\,\frac{1}{2\pi
i}\int_{c^{\prime}-i\infty}^{c^{\prime}+i\infty}e^{tz}K_{z^{2}}(\varepsilon)\Delta^{m+1}\phi\,\frac{dz}{z^{2m+1}}$
$\displaystyle\,=(1-\varepsilon^{2})\varepsilon\big{(}v^{\prime}_{0}(t)+I^{\prime}_{0}(t,\varepsilon)+I^{\prime}_{1}(t,\varepsilon)+I^{\prime}_{1}(t,\varepsilon)^{*}+{I^{\prime}_{2}}(t,\varepsilon)\big{)}+I^{\prime}(t,\varepsilon)\,,$

where

(5.4) $v^{\prime}_{0}(t):=\frac{1}{2\pi
i}\int_{c^{\prime}-i\infty}^{c^{\prime}+i\infty}e^{tz}G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}\Delta^{m+2}\phi\,\frac{dz}{z^{2m+1}}\,,$
$\displaystyle I^{\prime}_{0}(t,\varepsilon):=$ $\displaystyle\,-\frac{1}{2\pi
i}\int_{c^{\prime}-i\infty}^{c^{\prime}-ic_{1}(c^{\prime}/\varepsilon)^{1/4}}e^{tz}G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}\Delta^{m+2}\phi\,\frac{dz}{z^{2m+1}}$
$\displaystyle-\frac{1}{2\pi
i}\int_{c^{\prime}+ic_{1}(c^{\prime}/\varepsilon)^{1/4}}^{c^{\prime}+i\infty}e^{tz}G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}\Delta^{m+2}\phi\,\frac{dz}{z^{2m+1}}\,,$
$I^{\prime}_{1}(t,\varepsilon):=\frac{1}{2\pi
i}\int_{c^{\prime}-ic_{1}(c^{\prime}/\varepsilon)^{1/4}}^{c^{\prime}+ic_{1}(c^{\prime}/\varepsilon)^{1/4}}e^{tz}G_{z}(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{(1)}(\varepsilon)^{*}\Delta^{m+2}\phi\,\frac{dz}{z^{2m+1}}\,,$
$I^{\prime}_{1}(t,\varepsilon)^{*}:=\frac{1}{2\pi
i}\int_{c^{\prime}-ic_{1}(c^{\prime}/\varepsilon)^{1/4}}^{c^{\prime}+ic_{1}(c^{\prime}/\varepsilon)^{1/4}}e^{tz}G_{z}^{(1)}(\varepsilon)(1+z^{2}N_{0})^{-1}G_{\bar{z}}^{*}\Delta^{m+2}\phi\,\frac{dz}{z^{2m+1}}\,,$
$I^{\prime}_{2}(t,\varepsilon):=\frac{1}{2\pi
i}\int_{c^{\prime}-ic_{1}(c^{\prime}/\varepsilon)^{1/4}}^{c^{\prime}+ic_{1}(c^{\prime}/\varepsilon)^{1/4}}e^{tz}G_{z}(\varepsilon)\Lambda_{z}(\varepsilon)G_{z}(\varepsilon)^{*}\Delta^{m+2}\phi\,\frac{dz}{z^{2m+1}}\,.$

One has

$\|I^{\prime}_{0}(t,\varepsilon)\|\lesssim\frac{e^{c^{\prime}t}}{c^{\prime
2}}\,\int_{c_{1}(c^{\prime}/\varepsilon)^{1/4}}^{+\infty}\frac{d\gamma}{\gamma^{2m}}\
\|\Delta^{m+2}\phi\|\lesssim\left(\frac{\varepsilon}{c^{\prime}}\right)^{m/2-1/4}\,\frac{e^{c^{\prime}t}}{c^{\prime
2}}\,\|\Delta^{m+2}\phi\|\,.$

As regard the estimates for $\|I^{\prime}_{1}(t,\varepsilon)\|$ and
$\|I^{\prime}_{2}(t,\varepsilon)\|$, it suffices to proceed as in the proofs
of Lemmata 6.1 and 6.2, with the foresight to take into account the different
scaling properties of $(c^{\prime 2}+\gamma^{2})^{-(m+1/2)}$ with respect to
$(c^{2}+\gamma^{2})^{-n}$; this means that in the estimates, after the scaling
$\gamma\mapsto\gamma/c^{\prime}$, the exponent $2n$ has to be replaced by
$2m+1$. Therefore, one gets

$\|I^{\prime}_{1}(t,\varepsilon)\|\lesssim\,\varepsilon^{1/2}\,\frac{e^{tc}}{c^{2m+5/2}}\
\|\Delta^{m+2}\phi\|\,,\qquad\|I^{\prime}_{2}(t,\varepsilon)\|\lesssim\,\varepsilon^{1/2}\,\frac{e^{tc}}{c^{2(m+1)}}\
\|\Delta^{m+2}\phi\|$

and the estimates for $\|I^{\prime}_{1}(t,\varepsilon)^{*}\|$ and
$\|I^{\prime}_{2}(t,\varepsilon)^{*}\|$ are the same as the ones for
$\|I^{\prime}_{1}(t,\varepsilon)\|$ and $\|I^{\prime}_{2}(t,\varepsilon)\|$.

Putting such estimates together, one gets

$\displaystyle\,{\mathcal{I}^{\prime}}(t,\varepsilon):=\left\|(1-\varepsilon^{2})\varepsilon\big{(}I^{\prime}_{0}(t,\varepsilon)+I^{\prime}_{1}(t,\varepsilon)+I^{\prime}_{1}(t,\varepsilon)^{*}+{I^{\prime}_{2}}(t,\varepsilon)\big{)}+I^{\prime}(t,\varepsilon)\right\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon\bigg{(}\left(\frac{\varepsilon}{c^{\prime}}\right)^{m/2-1/4}\,\frac{e^{c^{\prime}t}}{c^{\prime
2}}+\varepsilon^{1/2}\left(\frac{e^{tc^{\prime}}}{c^{\prime
2m+5/2}}+\frac{e^{tc^{\prime}}}{c^{\prime
2m+2}}\right)\bigg{)}\|\Delta^{m+2}\phi\|+\left(\frac{\varepsilon}{c^{\prime}}\right)^{m^{\prime}/2+1/4}\,\frac{e^{c^{\prime}t}}{c^{\prime}}\,\|\Delta^{m^{\prime}+1}\phi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon\bigg{(}\left(\frac{\varepsilon}{c^{\prime}}\right)^{m/2-1/4}\,\frac{e^{c^{\prime}t}}{c^{\prime
2}}+\varepsilon^{1/2}\,\frac{e^{tc^{\prime}}}{c^{\prime
2m+5/2}}\bigg{)}\|\Delta^{m+2}\phi\|+\left(\frac{\varepsilon}{c^{\prime}}\right)^{m^{\prime}/2+1/4}\,\frac{e^{c^{\prime}t}}{c^{\prime}}\,\|\Delta^{m^{\prime}+1}\phi\|\,.$

Then, taking $c^{\prime}=\varepsilon^{\tau^{\prime}}$, $\tau^{\prime}>0$, and
$t\leq\varepsilon^{-\tau^{\prime}}$,

$\displaystyle{\mathcal{I}^{\prime}}(t,\varepsilon)\lesssim$
$\displaystyle\,\varepsilon\left(\varepsilon^{(1-\tau^{\prime})(m/2-1/4)-2\tau^{\prime}}+\varepsilon^{1/2-(2m+5/2)\tau^{\prime}}\right)\|\Delta^{m+2}\phi\|$
$\displaystyle+\varepsilon^{(1-\tau^{\prime})(m^{\prime}/2+1/4)-\tau^{\prime}}\,\|\Delta^{m^{\prime}+1}\phi\|$

and, by (5.2), $\varepsilon\leq c_{1}^{4/(3\tau^{\prime}+1)}$. We choose
$\tau^{\prime}$ such that

$(1-\tau^{\prime})\left(\frac{m}{2}-\frac{1}{4}\right)-2\tau^{\prime}>0\,,\quad\frac{1}{2}-\left(2m+\frac{5}{2}\right)\tau>0\,,\quad(1-\tau^{\prime})\left(\frac{m^{\prime}}{2}+\frac{1}{4}\right)-\tau^{\prime}>1\,,$

equivalently,

$\tau^{\prime}<\frac{2m-1}{2m+7}\,,\quad\tau^{\prime}<\frac{1}{4m+5}\,,\quad\tau^{\prime}<\frac{2m^{\prime}-3}{2m^{\prime}+5}\,.$

Taking $m=1$ and $m^{\prime}=2$, one obtains the bound
$\tau^{\prime}<\frac{1}{9}$ and

$\displaystyle{\mathcal{I}^{\prime}}(t,\varepsilon)\lesssim$
$\displaystyle\,\varepsilon\left(\varepsilon^{(1-9\tau^{\prime})/4}+\varepsilon^{(1-9\tau^{\prime})/2}\right)\|\Delta^{3}\phi\|+\varepsilon^{(5-9\tau^{\prime})/4}\|\Delta^{3}\phi\|\lesssim\varepsilon^{1+(1-9\tau^{\prime})/4}\|\Delta^{3}\phi\|\,.$

Thus,

$u(t)=u_{\rm
free}(t)+\varepsilon(\varepsilon^{2}-1)v(t)+r_{\varepsilon}(t)+r^{\prime}_{\varepsilon}(t)\,,$
$v(t):=v_{0}(t)+v^{\prime}_{0}(t)\,,\quad\|r_{\varepsilon}(t)\|\leq{\mathcal{I}}(t,\varepsilon)\,,\quad\|r^{\prime}_{\varepsilon}(t)\|\leq{\mathcal{I}^{\prime}}(t,\varepsilon)\,.$

By (5.3), with $n=2$, (5.4), with $m=1$, and by (4.12),

$\displaystyle v(t)=$ $\displaystyle\,\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{tz}{\mathcal{G}}_{z}\langle
1,(1+z^{2}N_{0})^{-1}1\rangle\langle{\mathcal{G}}_{\bar{z}},\Delta^{3}\psi\rangle\,\frac{dz}{z^{4}}$
$\displaystyle+$ $\displaystyle\,\frac{1}{2\pi
i}\int_{c^{\prime}-i\infty}^{c^{\prime}+i\infty}e^{tz}\langle
1,(1+z^{2}N_{0})^{-1}1\rangle\langle{\mathcal{G}}_{\bar{z}},\Delta^{3}\phi\rangle\,\frac{dz}{z^{3}}\,.$

By the spectral resolution of $N_{0}$ (see (4.8)), the linear operator
${\mathcal{G}}_{z}\langle
1,(1+z^{2}N_{0})^{-1}1\rangle\langle{\mathcal{G}}_{\bar{z}},\cdot\rangle$ has
the kernel

$K_{z}(x,y)=\sum_{k=1}^{+\infty}\frac{|\langle
e_{k},1\rangle|^{2}}{1+z^{2}\lambda_{k}}\
{\mathcal{G}}_{z}(x){\mathcal{G}}_{z}(y)=\sum_{k=1}^{+\infty}\frac{|\langle
e_{k},1\rangle|^{2}}{1+z^{2}\lambda_{k}}\
\frac{e^{-z(|x|+|y|)}}{(4\pi)^{2}|x|\,|y|}\,.$

Let

$K^{(\ell)}_{k,x,y}(z):=\frac{1}{z^{\ell}}\,\frac{e^{-z(|x|+|y|)}}{1+z^{2}\lambda_{k}}\
,\qquad\ell\geq 0\,.$

Then,

$[{\mathcal{L}}^{-1}K^{(0)}_{k,x,y}](t)=H(t-|x|-|y|)\,\frac{1}{{\lambda_{k}^{1/2}}}\,\sin\bigg{(}\frac{t-|x|-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}$

and, by the Laplace transform properties,

$[{\mathcal{L}}^{-1}K^{(\ell+1)}_{k,x,y}](t)=\int_{0}^{t}[{\mathcal{L}}^{-1}K^{(\ell)}_{k,x,y}](s)\,ds\,.$

Hence,

$[{\mathcal{L}}^{-1}K^{(\ell)}_{k,x,y}](t)=H(t-|x|-|y|)\,\lambda_{k}^{(\ell-1)/2}f^{(m)}_{k,x,y}\left(\frac{t-|x|-|y|}{{\lambda_{k}^{1/2}}}\right)\,,$
$f^{(\ell+1)}_{k,x,y}(t)=\int_{0}^{t}f^{(\ell)}_{k,x,y}(s)\,ds\,,\qquad
f^{(0)}_{k,x,y}(t)=\sin t\,.$

The latter recursive relations give

$f^{(3)}_{k,x,y}(t)=\frac{t^{2}}{2}-1+\cos t\,,\qquad
f^{(4)}_{k,x,y}(t)=\frac{t^{3}}{6}-t+\sin t\,.$

Therefore,

$\displaystyle\,[{\mathcal{L}}^{-1}K^{(4)}_{k,x,y}](t)$ $\displaystyle=$
$\displaystyle\,H(t-|x|-|y|)\,\lambda_{k}^{3/2}\left(\frac{(t-|x|-|y|)^{3}}{6\lambda_{k}^{3/2}}-\frac{t-|x|-|y|}{\lambda_{k}^{1/2}}+\sin\bigg{(}\frac{t-|x|-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\right)\,,$
$\displaystyle[{\mathcal{L}}^{-1}K^{(3)}_{k,x,y}](t)=H(t-|x|-|y|)\,\lambda_{k}\left(\frac{(t-|x|-|y|)^{2}}{2\lambda_{k}}-1+\cos\bigg{(}\frac{t-|x|-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\right)\
.$

Let either $\ell=3$ or $\ell=4$. Then, for any $\varphi\in
H^{6}(\mathbb{R}^{3})$,

$\displaystyle\int_{c-i\infty}^{c+i\infty}\int_{\mathbb{R}^{3}}\sum_{k=1}^{+\infty}\left|e^{tz}\frac{|\langle
e_{k},1\rangle|^{2}}{1+z^{2}\lambda_{k}}\
{\mathcal{G}}_{z}(x){\mathcal{G}}_{z}(y)\,\Delta^{3}\varphi(y)\,\right|\,dy\,\frac{dz}{z^{\ell}}$
$\displaystyle\lesssim$
$\displaystyle\,\frac{e^{c(t-|x|)}}{|x|}\int_{-\infty}^{+\infty}\sum_{k=1}^{+\infty}\frac{|\langle
e_{k},1\rangle|^{2}}{|1+(c+i\gamma)^{2}\lambda_{k}|}\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{\ell/2}}\int_{\mathbb{R}^{3}}e^{-c|y|}\,|\Delta^{3}\varphi(y)|\,\frac{dy}{|y|}$
$\displaystyle\lesssim$
$\displaystyle\,\frac{e^{c(t-|x|)}}{|x|}\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\int_{-\infty}^{+\infty}\sup_{0\leq\lambda\leq\|N_{0}\|}\frac{1}{|1+(c+i\gamma)^{2}\lambda|}\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{\ell/2}}\,\times$
$\displaystyle\times\int_{\mathbb{R}^{3}}e^{-c|y|}\,|\Delta^{3}\varphi(y)|\,\frac{dy}{|y|}$
$\displaystyle\lesssim$
$\displaystyle\,\frac{e^{c(t-|x|)}}{|x|}\,|\Omega|\int_{-\infty}^{+\infty}{\rm
max}\left\\{1,\frac{c^{2}+\gamma^{2}}{2c|\gamma|}\right\\}\frac{d\gamma}{(c^{2}+\gamma^{2})^{\ell/2}}\int_{\mathbb{R}^{3}}e^{-c|y|}\,|\Delta^{3}\varphi(y)|\,\frac{dy}{|y|}<+\infty\,.$

Hence, by Fubini’s theorem, one has

$\displaystyle\,\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{tz}{\mathcal{G}}_{z}(x)\langle
1,(1+z^{2}N_{0})^{-1}1\rangle\langle{\mathcal{G}}_{\bar{z}},\Delta^{3}\psi\rangle\,\frac{dz}{z^{4}}$
$\displaystyle=$
$\displaystyle\,\frac{1}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\,\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{tz}\left(\int_{\mathbb{R}^{3}}K^{(4)}_{k,x,y}(z)\,\frac{\Delta^{3}\psi(y)}{4\pi|y|}\,dy\right)dz$
$\displaystyle=$
$\displaystyle\,\frac{1}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\int_{\mathbb{R}^{3}}[{\mathcal{L}}^{-1}K^{(4)}_{k,x,y}](t)\,\frac{\Delta^{3}\psi(y)}{4\pi|y|}\,dy$
$\displaystyle=$
$\displaystyle\,\frac{1}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\int_{\mathbb{R}^{3}}H(t-|x|-|y|)\,\lambda_{k}^{3/2}\bigg{(}\,\frac{(t-|x|-|y|)^{3}}{6\lambda_{k}^{3/2}}$
$\displaystyle-\frac{t-|x|-|y|}{\lambda_{k}^{1/2}}+\sin\bigg{(}\frac{t-|x|-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\bigg{)}\frac{\Delta^{3}\psi(y)}{4\pi|y|}\,dy$
$\displaystyle=$
$\displaystyle\,\frac{H(t-|x|)}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\lambda_{k}^{3/2}\int_{|y|<t-|x|}\bigg{(}\,\frac{(t-|x|-|y|)^{3}}{6\lambda_{k}^{3/2}}$
$\displaystyle-\frac{t-|x|-|y|}{\lambda_{k}^{1/2}}+\sin\bigg{(}\frac{t-|x|-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\bigg{)}\frac{\Delta^{3}\psi(y)}{4\pi|y|}\,dy\,.$

Analogously,

$\displaystyle\,\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{tz}{\mathcal{G}}_{z}(x)\langle
1,(1+z^{2}N_{0})^{-1}1\rangle\langle{\mathcal{G}}_{\bar{z}},\Delta^{3}\phi\rangle\,\frac{dz}{z^{3}}$
$\displaystyle=$
$\displaystyle\,\frac{1}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\,\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{tz}\left(\int_{\mathbb{R}^{3}}K^{(3)}_{k,x,y}(z)\,\frac{\Delta^{3}\phi(y)}{4\pi|y|}\,dy\right)dz$
$\displaystyle=$
$\displaystyle\,\frac{1}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\int_{\mathbb{R}^{3}}[{\mathcal{L}}^{-1}K^{(3)}_{k,x,y}](t)\,\frac{\Delta^{3}\phi(y)}{4\pi|y|}\,dy$
$\displaystyle=$
$\displaystyle\,\frac{1}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\int_{\mathbb{R}^{3}}H(t-|x|-|y|)\,\lambda_{k}\bigg{(}\,\frac{(t-|x|-|y|)^{2}}{2\lambda_{k}}$
$\displaystyle-1+\cos\bigg{(}\frac{t-|x|-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\bigg{)}\frac{\Delta^{3}\phi(y)}{4\pi|y|}\,dy$
$\displaystyle=$
$\displaystyle\,\frac{H(t-|x|)}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\lambda_{k}\int_{|y|<t-|x|}\bigg{(}\,\frac{(t-|x|-|y|)^{2}}{2\lambda_{k}}$
$\displaystyle-1+\cos\bigg{(}\frac{t-|x|-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\bigg{)}\frac{\Delta^{3}\phi(y)}{4\pi|y|}\,dy\,.$

The proof is then concluded by Lemma 6.3. ∎

As regards the inhomogeneous Cauchy problem, one has the following

######  Theorem 5.2.

Let $f\in L_{loc}^{1}(\mathbb{R}_{+},H^{6}(\mathbb{R}^{3}))$ be such that
$f(t)$ is supported outside $\Omega_{\varepsilon}$ for any
$t\in\mathbb{R}_{+}$; let $\widehat{u}_{\varepsilon}$ and
$\widehat{u}_{\text{\rm free}}$ be the mild (classical if furthermore
$f\in{\mathcal{C}}(\mathbb{R}_{+},H^{2}(\mathbb{R}^{3}))\,$) solutions of the
inhomogeneous Cauchy problems

$\begin{cases}(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}})\partial_{tt}\widehat{u}_{\varepsilon}(t)=\widehat{u}_{\varepsilon}(t)+f(t)\\\
\widehat{u}_{\varepsilon}(0)=0\\\
\partial_{t}\widehat{u}_{\varepsilon}(0)=0\,.\end{cases}\qquad\begin{cases}\partial_{tt}\widehat{u}_{\text{\rm
free}}(t)=\Delta\widehat{u}_{\text{\rm free}}(t)+f(t)\\\
\widehat{u}_{\text{\rm free}}(0)=0\\\ \widehat{u}_{\text{\rm
free}}(0)=0\,.\end{cases}$

If

$\|\Delta^{3}f(t)\|\lesssim t^{\alpha}\,,\quad t\gg 1\,,\qquad\alpha\geq-1\,,$

then, for any $\tau\in\left(0,\frac{1}{11+2(\alpha+1)}\right)$ and for any
$\varepsilon>0$ such that

$\varepsilon\leq{\rm min}\left\\{1,c_{1}^{4/(3\tau+1)}\right\\}\,,$

one has

$\widehat{u}_{\varepsilon}(t)=\widehat{u}_{\text{\rm
free}}(t)+\varepsilon\,(\varepsilon^{2}-1)\,\widehat{v}(t)+\widehat{r}_{\varepsilon}(t)\,,\qquad\widehat{v}(t,x)=H(t-|x|)\,\frac{\widehat{q}(t-|x|)}{4\pi|x|}\,,$

where

$\displaystyle\widehat{q}(t)\\!:=\\!\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\,\frac{1}{\lambda_{k}^{1/2}}\int_{0}^{t}\\!\\!$
$\displaystyle\int_{|y|<t-s}\\!\\!\sin\bigg{(}\frac{t-s-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta
f(s,y)}{4\pi|y|}\,dy\,ds$

and

$\sup_{0<t\leq
1/\varepsilon^{\tau}}\|\widehat{r}_{\varepsilon}(t)\|\lesssim\begin{cases}\varepsilon^{1+(1-11\tau)/2}\ln\varepsilon^{-1}&\alpha=-1\\\
\varepsilon^{1+(1-(11+2(\alpha+1)\tau)/2}&\alpha>-1\,.\end{cases}$

If $f\in L^{1}(\mathbb{R}_{+},H^{6}(\mathbb{R}^{3}))$, then, for any
$\tau\in(0,\frac{1}{11})$ and for any $\varepsilon>0$ as above,

$\sup_{0<t\leq
1/\varepsilon^{\tau}}\|\widehat{r}_{\varepsilon}(t)\|\lesssim\varepsilon^{1+(1-11\tau)/2}\,.$

###### Proof.

By [9, Theorem 4.1 and Lemma 4.2 in Section II.4] one gets the existence of
the solutions with the stated regularity and

$\widehat{u}_{\text{\rm
free}}(t)=\int_{0}^{t}(-\Delta)^{-1/2}\sin((t-s)(-\Delta)^{1/2})f(s)\,ds\,,$
$\widehat{u}_{\varepsilon}(t)=\int_{0}^{t}(-A(\varepsilon))^{-1/2}\sin((t-s)(-A(\varepsilon))^{1/2})f(s)\,ds\,.$

Therefore, by the same reasonings as in the proof of Theorem 5.1,

$\displaystyle\widehat{u}_{\varepsilon}(t)=\widehat{u}_{\text{\rm free}}(t)$
$\displaystyle\varepsilon\,(\varepsilon^{2}-1)\,\int_{0}^{t}\bigg{(}\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{(t-s)z}{\mathcal{G}}_{z}\langle
1,(1+z^{2}N_{0})^{-1}1\rangle\langle{\mathcal{G}}_{\bar{z}},\Delta^{3}f(s)\rangle\,\frac{dz}{z^{4}}\bigg{)}\,ds+\widehat{r}_{\varepsilon}(t)\,,$

where

$\displaystyle\sup_{0<t\leq
1/\varepsilon^{\tau}}\|\widehat{r}_{\varepsilon}(t)\|\lesssim$
$\displaystyle\,\varepsilon^{1+(1-11\tau)/2}\int_{0}^{1/\varepsilon^{\tau}}\|\Delta^{3}f(s)\|\,ds\,.$

This gives

$\sup_{0<t\leq
1/\varepsilon^{\tau}}\|\widehat{r}_{\varepsilon}(t)\|\lesssim\varepsilon^{1+(1-11\tau)/2}$

whenever $f\in L^{1}(\mathbb{R}_{+},H^{6}(\mathbb{R}^{3}))$. If
$\|\Delta^{3}f(t)\|\lesssim t^{\alpha}$, $t\gg 1$, then

$\int_{0}^{1/\varepsilon^{\tau}}\|\Delta^{3}f(s)\|\,ds\lesssim\begin{cases}1+\ln\varepsilon^{-1}&\alpha=-1\\\
1+\varepsilon^{-\tau(\alpha+1)}&\alpha>-1\,,\end{cases}$

and so in this case

$\sup_{0<t\leq
1/\varepsilon^{\tau}}\|\widehat{r}_{\varepsilon}(t)\|\lesssim\begin{cases}\varepsilon^{1+(1-11\tau)/2}\ln\varepsilon^{-1}&\alpha=-1\\\
\varepsilon^{1+(1-11\tau)/2-\tau(\alpha+1)}&\alpha>-1\,.\end{cases}$

Similarly to the proof of Theorem 5.1,

$\displaystyle\frac{1}{4\pi|x|}\int_{0}^{t}\int_{c-i\infty}^{c+i\infty}\int_{\mathbb{R}^{3}}\sum_{k=1}^{+\infty}\left|e^{(t-s)z}|\langle
e_{k},1\rangle|^{2}K^{(4)}_{k,x,y}(z)\,\frac{\Delta^{3}f(s,y)}{4\pi|y|}\,\right|\,dy\,dz\,ds$
$\displaystyle\lesssim$
$\displaystyle\,\frac{e^{-c|x|}}{|x|}\int_{-\infty}^{+\infty}\sum_{k=1}^{+\infty}\frac{|\langle
e_{k},1\rangle|^{2}}{|1+(c+i\gamma)^{2}\lambda_{k}|}\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{3}}\int_{\mathbb{R}^{3}}e^{-c|y|}\,\int_{0}^{t}e^{c(t-s)}|\Delta^{3}f(s,y)|\,ds\,\frac{dy}{|y|}$
$\displaystyle\lesssim$
$\displaystyle\,\frac{e^{-c|x|}}{|x|}\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\int_{-\infty}^{+\infty}{\rm
max}\left\\{1,\frac{c^{2}+\gamma^{2}}{2c\gamma}\right\\}\frac{d\gamma}{(c^{2}+\gamma^{2})^{3}}\,\times$
$\displaystyle\times\int_{\mathbb{R}^{3}}e^{-c|y|}\,\int_{0}^{t}|\Delta^{3}f(s,y)|\,ds\,\frac{dy}{|y|}<+\infty$

and so, by Fubini’s theorem,

$\displaystyle\,\int_{0}^{t}\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{(t-s)z}{\mathcal{G}}_{z}(x)\langle
1,(1+z^{2}N_{0})^{-1}1\rangle\langle{\mathcal{G}}_{\bar{z}},\Delta^{3}f(s)\rangle\,\frac{dz}{z^{4}}\,ds$
$\displaystyle=$
$\displaystyle\,\int_{0}^{t}\frac{1}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\,\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}e^{(t-s)z}\left(\int_{\mathbb{R}^{3}}K^{(4)}_{k,x,y}(z)\,\frac{\Delta^{3}f(s,y)}{4\pi|y|}\,dy\right)dz\,ds$
$\displaystyle=$
$\displaystyle\,\frac{1}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\int_{0}^{t}\int_{\mathbb{R}^{3}}[{\mathcal{L}}^{-1}K^{(4)}_{k,x,y}](t-s)\,\frac{\Delta^{3}f(s,y)}{4\pi|y|}\,dy\,ds$
$\displaystyle=$
$\displaystyle\,\frac{1}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\int_{0}^{t}\int_{\mathbb{R}^{3}}H(t-s-|x|-|y|)\,\lambda_{k}^{3/2}\bigg{(}\,\frac{(t-s-|x|-|y|)^{3}}{6\lambda_{k}^{3/2}}$
$\displaystyle-\frac{t-s-|x|-|y|}{\lambda_{k}^{1/2}}+\sin\bigg{(}\frac{t-s-|x|-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\bigg{)}\frac{\Delta^{3}f(s,y)}{4\pi|y|}\,dy\,ds$
$\displaystyle=$
$\displaystyle\,\frac{H(t-|x|)}{4\pi|x|}\,\sum_{k=1}^{+\infty}|\langle
e_{k},1\rangle|^{2}\lambda_{k}^{3/2}\int_{0}^{t-|x|}\int_{|y|<t-s-|x|}\bigg{(}\,\frac{(t-s-|x|-|y|)^{3}}{6\lambda_{k}^{3/2}}$
$\displaystyle-\frac{t-s-|x|-|y|}{\lambda_{k}^{1/2}}+\sin\bigg{(}\frac{t-s-|x|-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\bigg{)}\frac{\Delta^{3}f(s,y)}{4\pi|y|}\,dy\,ds\,.$

The proof is then concluded by Lemma 6.3. ∎

###  5.1. Effective dynamics

In this subsection, $\phi$, $\psi$ and $f(t)$ are as in Theorems 5.1 and 5.2,
$\delta_{0}$ denotes the Dirac delta distribution supported at the origin and
$Mu(t)$ denotes the spherical mean of the continuous function $u$ over the
sphere $\\{x\in\mathbb{R}^{3}:|x|=t\\}$, i.e. (here $\sigma_{t}$ denotes the
surface measure)

$Mu(t):=\frac{1}{4\pi t^{2}}\int_{|x|=t}u(x)\,d\sigma_{t}(x)\,.$

######  Theorem 5.3.

Let $u_{\varepsilon}$ and $u_{\varepsilon,\text{\rm eff}}$ be the solutions of
the inhomogeneous Cauchy problems

$\begin{cases}(\varepsilon^{-2}\chi_{\Omega_{\varepsilon}}+\chi_{{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}})\partial_{tt}u_{\varepsilon}(t)=\Delta
u_{\varepsilon}(t)+f(t)\\\ u_{\varepsilon}(0)=\phi\\\
\partial_{t}u_{\varepsilon}(0)=\psi\,,\end{cases}\\!\\!\\!\begin{cases}\partial_{tt}u_{\varepsilon,\text{\rm
eff}}(t)=\Delta u_{\varepsilon,\text{\rm
eff}}(t)+\varepsilon(\varepsilon^{2}-1)\big{(}\sum_{k=1}^{\infty}q_{k}(t)\big{)}\delta_{0}\\\
u_{\varepsilon,\text{\rm eff}}(0)=\phi\\\ \partial_{t}u_{\varepsilon,\text{\rm
eff}}(0)=\psi\,,\end{cases}$

where, for any $k\geq 1$, $q_{k}(t)$ solves the Cauchy problem

(5.5) $\begin{cases}\lambda_{k}\,\ddot{q}_{k}(t)=-q_{k}(t)+|\langle
e_{k},1\rangle|^{2}\,h(t)\\\ q_{k}(0)=0\\\ \dot{q}_{k}(0)=0\end{cases}$

with

$h(t):=\Delta(u_{\text{\rm free}}+\widehat{u}_{\text{\rm free}})(t,0)=\frac{d\
}{dt}\,\big{(}tM{\Delta\phi}(t)\big{)}+tM{\Delta\psi}(t)+\int_{0}^{t}(t-s)(M{\Delta
f(s)})(t-s)\,ds\,.$

Then, there exist $\tau\in(0,\frac{1}{11})$ and
$\tau^{\prime\prime}\in(0,\frac{1}{2})$ such that

$\sup_{0<t\leq
1/\varepsilon^{\tau}}\|u_{\varepsilon}(t)-u_{\varepsilon,\text{\rm
eff}}(t)\|\lesssim\varepsilon^{1+\tau^{\prime\prime}}\,,\qquad\varepsilon\ll
1\,.$

###### Proof.

By Theorems 5.1 and 5.2, it suffices to show that, for any $k\geq 1$,

$q_{k}(t):=q_{1,k}(t)+q_{k,2}(t)+q_{k,3}(t)\,,$

solves (5.5), where

$q_{1,k}(t):=|\langle
e_{k},1\rangle|^{2}\int_{|y|<t}\bigg{(}1-\cos\bigg{(}\frac{t-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\bigg{)}\frac{\Delta^{2}\phi(y)}{4\pi|y|}\,dy$
$q_{k,2}(t):=|\langle
e_{k},1\rangle|^{2}\,\frac{1}{\lambda_{k}^{1/2}}\int_{|y|<t}\sin\bigg{(}\frac{t-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta\psi(y)}{4\pi|y|}\,dy\,,$
$q_{k,3}(t):=|\langle
e_{k},1\rangle|^{2}\,\frac{1}{\lambda_{k}^{1/2}}\int_{0}^{t}\\!\\!\int_{|y|<t-s}\\!\\!\sin\bigg{(}\frac{t-s-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta
f(s,y)}{4\pi|y|}\,dy\,ds\,.$

One has,

$\displaystyle\ddot{q}_{1,k}(t)$ $\displaystyle\,=|\langle
e_{k},1\rangle|^{2}\,\frac{1}{\lambda^{1/2}_{k}}\frac{d\
}{dt}\int_{|y|<t}\sin\bigg{(}\frac{t-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta^{2}\phi(y)}{4\pi|y|}\,dy$
$\displaystyle\,=|\langle
e_{k},1\rangle|^{2}\,\frac{1}{\lambda_{k}}\int_{|y|<t}\cos\bigg{(}\frac{t-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta^{2}\phi(y)}{4\pi|y|}\,dy$
$\displaystyle\,=-\frac{1}{\lambda_{k}}\left(q_{1,k}(t)-|\langle
e_{k},1\rangle|^{2}\int_{|y|<t}\frac{\Delta^{2}\phi(y)}{4\pi|y|}\,dy\right)\,,$

and

$\displaystyle\ddot{q}_{2,k}(t)$ $\displaystyle\,=|\langle
e_{k},1\rangle|^{2}\,\frac{1}{\lambda_{k}}\frac{d\
}{dt}\int_{|y|<t}\cos\bigg{(}\frac{t-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta\psi(y)}{4\pi|y|}\,dy$
$\displaystyle\,=|\langle
e_{k},1\rangle|^{2}\,\frac{1}{\lambda_{k}}\left(tM\Delta\psi(t)-\frac{1}{\lambda_{k}^{1/2}}\int_{|y|<t}\cos\bigg{(}\frac{t-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta^{2}\phi(y)}{4\pi|y|}\,dy\right)$
$\displaystyle\,=-\frac{1}{\lambda_{k}}\left(\,q_{2,k}(t)-|\langle
e_{k},1\rangle|^{2}tM{\Delta\psi}(t)\right)\,,$
$\displaystyle\,\ddot{q}_{3,k}(t)=|\langle
e_{k},1\rangle|^{2}\,\frac{1}{\lambda_{k}}\,\frac{d\
}{dt}\int_{0}^{t}\\!\\!\int_{|y|<t-s}\\!\\!\cos\bigg{(}\frac{t-s-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta
f(s,y)}{4\pi|y|}\,dy\,ds$ $\displaystyle\,=|\langle
e_{k},1\rangle|^{2}\,\frac{1}{\lambda_{k}}\int_{0}^{t}\left((t-s)(M\Delta
f(s))(t-s)-\frac{1}{\lambda_{k}^{1/2}}\int_{|y|<t-s}\sin\bigg{(}\frac{t-s-|y|}{{\lambda_{k}^{1/2}}}\bigg{)}\frac{\Delta^{2}f(s,y)}{4\pi|y|}\,dy\right)$
$\displaystyle\,=-\frac{1}{\lambda_{k}}\left(\,q_{2,k}(t)-|\langle
e_{k},1\rangle|^{2}\int_{0}^{t}(t-s)M({\Delta f(s)})(t-s)\,ds\right)\,.$

Furthermore, by Green’s identity and by
$0\notin\text{supp}(\phi)\subseteq{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}$,

$\displaystyle\int_{|y|<t}\frac{\Delta^{2}\phi(y)}{4\pi|y|}\,dy=\frac{1}{4\pi
t^{2}}\int_{|y|=t}x\\!\cdot\\!\nabla\Delta\phi(y)\,d\sigma_{t}(y)+\frac{1}{4\pi
t^{2}}\int_{|y|=t}\Delta\phi(y)\,d\sigma_{t}(y)=\frac{d\
}{dt}\,tM{\Delta\phi}(t)\,.$

Hence, by Kirchhoff’s formula, one gets

$\frac{d\
}{dt}\,\big{(}tM{\Delta\phi}(t)\big{)}+tM{\Delta\psi}(t)+\int_{0}^{t}(t-s)(M{\Delta
f(s)})(t-s)\,ds=u^{\Delta}_{\text{\rm free}}(t,0)\,,$

where $u^{\Delta}_{\text{\rm free}}$ denotes the solution of the Cauchy
problem for the free wave equation with initial data $\Delta\phi$,
$\Delta\psi$ and source $\Delta f(t)$; such a solution coincides with
$\Delta(u_{\text{\rm free}}+\widehat{u}_{\text{\rm free}})$. ∎

##  6\. Appendix. Auxiliary results.

######  Lemma 6.1.

$\|I_{1}(t,\varepsilon)\|\lesssim\varepsilon^{1/2}\,\frac{e^{tc}}{c^{2n+3/2}}\
\|\Delta^{n+1}\psi\|\,.$

###### Proof.

By Lemmata 4.2 and 4.7,

$\displaystyle\|I_{1}(t,\varepsilon)\|$ $\displaystyle\lesssim$
$\displaystyle\,e^{tc}\int_{0}^{c_{1}(c/\varepsilon)^{1/4}}\\!\\!\\!\|G_{c+i\gamma}\|\,\|(1+(c+i\gamma)^{2}N_{0})^{-1}\|\,\|G_{c+i\gamma}^{(1)}(\varepsilon)\|\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{n}}\
\|\Delta^{n+1}\psi\|$ $\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,\frac{e^{tc}}{c^{1/2}}\bigg{(}\int_{0}^{c}\left(1+\frac{1}{c^{2}+\gamma^{2}}\right)\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{n}}$
$\displaystyle+\int_{c}^{c_{1}(c/\varepsilon)^{1/4}}\frac{c^{2}+\gamma^{2}}{c\gamma}\left(1+(1+c^{2}+\gamma^{2})\,\frac{1}{c\gamma}\right)\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{n}}\bigg{)}\|\Delta^{n+1}\psi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,\frac{e^{tc}}{c^{1/2}}\bigg{(}\int_{0}^{c}\frac{d\gamma}{(c^{2}+\gamma^{2})^{n}}+\int_{0}^{c}\frac{d\gamma}{(c^{2}+\gamma^{2})^{n+1}}$
$\displaystyle+\frac{1}{c^{2}}\int_{c}^{c_{1}(c/\varepsilon)^{1/4}}\bigg{(}\frac{c}{\gamma}+\frac{1}{\gamma^{2}}+\frac{c^{2}}{\gamma^{2}}+1\bigg{)}\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{n-1}}\bigg{)}\|\Delta^{n+1}\psi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,\frac{e^{tc}}{c^{1/2}}\bigg{(}\frac{1}{c^{2n-1}}\int_{0}^{1}\frac{d\gamma}{(1+\gamma^{2})^{n}}+\frac{1}{c^{2n+1}}\int_{0}^{1}\frac{d\gamma}{(1+\gamma^{2})^{n+1}}$
$\displaystyle+\frac{1}{c^{2n-1}}\int_{1}^{+\infty}\bigg{(}\frac{1}{\gamma}+\frac{1}{c^{2}\gamma^{2}}+\frac{1}{\gamma^{2}}+1\bigg{)}\,\frac{d\gamma}{(1+\gamma^{2})^{2n-1}}\bigg{)}\|\Delta^{n+1}\psi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,\frac{e^{tc}}{c^{1/2}}\bigg{(}\frac{1}{c^{2n-1}}+\frac{1}{c^{2n+1}}\bigg{)}\|\Delta^{n+1}\psi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,\frac{e^{tc}}{c^{2n+3/2}}\
\|\Delta^{n+1}\psi\|$

∎

######  Lemma 6.2.

$\|I_{2}(t,\varepsilon)\|\lesssim\varepsilon^{1/2}\,\frac{e^{tc}}{c^{2n+1}}\
\|\Delta^{n+1}\psi\|\,.$

###### Proof.

By Lemmata 4.4, 4.5, 4.6 and 4.7 (with $s=5/4$),

$\displaystyle\|I_{2}(t,\varepsilon)\|$ $\displaystyle\lesssim$
$\displaystyle\,e^{tc}\int_{0}^{c_{1}(c/\varepsilon)^{1/4}}\\!\\!\\!\|G_{c+i\gamma}(\varepsilon)\|^{2}\|\Lambda_{c+i\gamma}(\varepsilon)\|\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{n}}\
\|\Delta^{n+1}\psi\|$ $\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,e^{tc}\bigg{(}\int_{0}^{c}\bigg{(}1+\frac{1}{c^{2}+\gamma^{2}}\bigg{)}^{\\!\\!5/4}\\!\\!\frac{c^{2}(1+c)}{(c^{2}+\gamma^{2})^{3/4}}\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{n}}$
$\displaystyle+\int_{c}^{c_{1}(c/\varepsilon)^{1/4}}\\!\\!\\!\frac{(1+(c+\gamma)^{2})^{5/4}}{(c\gamma)^{2}}\,(1+(c^{2}+\gamma^{2})^{1/2})\,\frac{(c^{2}+\gamma^{2})^{3}}{(c\gamma)^{2}}\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{n}}\bigg{)}\|\Delta^{n+1}\psi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,e^{tc}\bigg{(}\int_{0}^{c}\bigg{(}1+\frac{1}{c^{2}+\gamma^{2}}\bigg{)}^{\\!\\!5/4}\frac{c^{2\,}d\gamma}{(c^{2}+\gamma^{2})^{n+3/4}}$
$\displaystyle+\int_{c}^{c_{1}(c/\varepsilon)^{1/4}}(1+(c+\gamma)^{2})^{5/4}\,\frac{1+(c^{2}+\gamma^{2})^{1/2}}{(c\gamma)^{4}}\,\frac{d\gamma}{(c^{2}+\gamma^{2})^{n-3}}\bigg{)}\|\Delta^{n+1}\psi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,e^{tc}\bigg{(}\frac{1}{c^{2n-3/2}}\int_{0}^{1}\bigg{(}c^{2}+\frac{1}{1+\gamma^{2}}\bigg{)}^{\\!\\!5/4}\frac{d\gamma}{(1+\gamma^{2})^{n+3/4}}$
$\displaystyle+\frac{1}{c^{2n+1}}\int_{1}^{+\infty}(1+c^{2}(1+\gamma)^{2})^{5/4}\,\frac{1+c(1+\gamma^{2})^{1/2}}{\gamma^{4}}\,\frac{d\gamma}{(1+\gamma^{2})^{n-3}}\bigg{)}\|\Delta^{n+1}\psi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,e^{tc}\bigg{(}\frac{1}{c^{2n-3/2}}+\frac{1}{c^{2n+1}}\bigg{)}\|\Delta^{n+1}\psi\|$
$\displaystyle\lesssim$
$\displaystyle\,\varepsilon^{1/2}\,\frac{e^{tc}}{c^{2n+1}}\
\|\Delta^{n+1}\psi\|\,.$

∎

######  Lemma 6.3.

Let $\varphi\in H^{6}(\mathbb{R}^{3})$ be supported outside
$\Omega_{\varepsilon}$. Then

$\int_{|x|<t}\\!\\!\left(\frac{(t-|x|)^{2}}{2\lambda}-1+\cos\bigg{(}\frac{t-|x|}{\lambda^{1/2}}\bigg{)}\right){\Delta^{3}\varphi(x)}\,\frac{dx}{|x|}=\frac{1}{\lambda}\int_{|x|<t}\\!\\!\left(1-\cos\bigg{(}\frac{t-|x|}{\lambda^{1/2}}\bigg{)}\right)\Delta^{2}\varphi(x)\,\frac{dx}{|x|}$

and

$\displaystyle\int_{|x|<t}\left(\frac{(t-|x|)^{3}}{6\lambda^{3/2}}-\frac{t-|x|}{\lambda^{1/2}}+\sin\bigg{(}\frac{t-|x|}{\lambda^{1/2}}\right)\bigg{)}{\Delta^{3}\varphi(x)}\,\frac{dx}{|x|}=\frac{1}{\lambda^{2}}\int_{|x|<t}\sin\bigg{(}\frac{t-|x|}{\lambda^{1/2}}\bigg{)}\Delta\varphi(x)\,\frac{dx}{|x|}\,.$

###### Proof.

For any $g\in{\mathcal{C}}^{2}(\mathbb{R}_{+})$ and for any $x\not=0$, one has

$\nabla\left(\frac{g(|x|)}{|x|}\right)=\left(g^{\prime}(|x|)-\,\frac{g(|x|)}{|x|}\right)\frac{x}{|x|^{2}}\,,\qquad\Delta\left(\frac{g(|x|)}{|x|}\right)=\frac{g^{\prime\prime}(|x|)}{|x|}\,.$

Hence, by Green’s formula and by
$0\notin\text{supp}(\varphi)\subseteq{\mathbb{R}^{3}\backslash\Omega_{\varepsilon}}$,

$\displaystyle\int_{|x|<t}\frac{g(|x|)}{|x|}\,{\Delta^{n}\varphi(x)}\,dx=\int_{|x|<t}\Delta\left(\frac{g(|x|)}{|x|}\right)\,\Delta^{n-1}\varphi(x)\,dx$
$\displaystyle+\frac{g(t)}{t^{2}}\int_{|x|=t}x\\!\cdot\\!\nabla\Delta^{n-1}\varphi(x)\,d\sigma_{t}(x)-\frac{1}{t}\int_{|x|=t}x\\!\cdot\\!\nabla\left(\frac{g(|x|)}{|x|}\right)\Delta^{n-1}\varphi(x)\,d\sigma_{t}(x)$
(6.1)
$\displaystyle=\int_{|x|<t}\frac{g^{\prime\prime}(|x|)}{|x|}\,\Delta^{n-1}\varphi(x)\,dx+\frac{g(t)}{t^{2}}\int_{|x|=t}x\\!\cdot\\!\nabla\Delta^{n-1}\varphi(x)\,d\sigma_{t}(x)$
$\displaystyle-\frac{1}{t}\left(g^{\prime}(t)-\,\frac{g(t)}{t}\right)\int_{|x|=t}\Delta^{n-1}\varphi(x)\,d\sigma_{t}(x)\,.$

Taking
$g(s)=\frac{(t-s)^{2}}{2\lambda}-1+\cos\left(\frac{t-s}{\lambda^{1/2}}\right)$,
one gets, by (6) and by $g(t)=g^{\prime}(t)=0$

$\displaystyle\int_{|x|<t}\left(\frac{(t-|x|)^{2}}{2\lambda}-1+\cos\bigg{(}\frac{t-|x|}{\lambda^{1/2}}\bigg{)}\right){\Delta^{3}\varphi(x)}\,\frac{dx}{|x|}$
$\displaystyle=\frac{1}{\lambda}\int_{|x|<t}\left(1-\cos\bigg{(}\frac{t-|x|}{\lambda^{1/2}}\bigg{)}\right)\Delta^{2}\varphi(x)\,\frac{dx}{|x|}\,.$

Taking
$g(s)=\frac{(t-s)^{3}}{6\lambda^{3/2}}-\frac{t-s}{\lambda^{1/2}}+\sin\left(\frac{t-s}{\lambda^{1/2}}\right)$,
one gets, by using (6) twice and by
$g(t)=g^{\prime}(t)=g^{\prime\prime}(t)=g^{\prime\prime\prime}(t)=0$,

$\displaystyle\int_{|x|<t}\left(\frac{(t-|x|)^{3}}{6\lambda^{3/2}}-\frac{t-|x|}{\lambda^{1/2}}+\sin\bigg{(}\frac{t-|x|}{\lambda^{1/2}}\bigg{)}\right){\Delta^{3}\varphi(x)}\,\frac{dx}{|x|}$
$\displaystyle=\int_{|x|<t|}\frac{g^{\prime\prime}(|x|)}{|x|}\,\Delta^{2}\varphi(x)\,dx=\int_{|x|<t|}\frac{g^{\prime\prime\prime\prime}(|x|)}{|x|}\,\Delta\varphi(x)\,dx$
$\displaystyle=\frac{1}{\lambda^{2}}\int_{|x|<t}\sin\bigg{(}\frac{t-|x|}{\lambda^{1/2}}\bigg{)}\Delta\varphi(x)\,\frac{dx}{|x|}\,.$

∎

Acknowledgements. Part of this work was done while A.P. was visiting the
Laboratoire de Mathématiques CNRS de Reims; this stay was financially
supported by the Laboratoire Ypatia de Sciences Mathématiques.

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