# Asymptotic analysis of subwavelength halide perovskite resonators††thanks:
The work of KA was supported by the project ETH-34 20-2 and the work of BD was
supported by H2020 FETOpen project BOHEME under grant agreement No. 863179.

Konstantinos Alexopoulos Department of Mathematics, ETH Zurich, Rämistrasse
101, CH-8092 Zurich, Switzerland<EMAIL_ADDRESS>.
Bryn Davies Department of Mathematics, Imperial College London, 180 Queen’s
Gate, London SW7 2AZ, United Kingdom<EMAIL_ADDRESS>.

###### Abstract

Halide perovskites are promising materials with many significant applications
in photovoltaics and optoelectronics. In this paper, we use integral methods
to quantify the resonant properties of halide perovskite nano-particles. We
prove that, for arbitrarily small particles, the subwavelength resonant
frequencies can be expressed in terms of the eigenvalues of the Newtonian
potential associated with its shape. We also characterize the hybridized
subwavelength resonant frequencies of a dimer of two halide perovskite
particles. Finally, we examine the specific case of spherical resonators and
demonstrate that our new results are consistent with previous works.

###### Contents

  1. 1 Introduction
  2. 2 Single Resonators
    1. 2.1 Problem Setting
    2. 2.2 Integral Formulation
    3. 2.3 Reformulation as a Subwavelength Resonance Problem
    4. 2.4 Three Dimensions
      1. 2.4.1 Eigenvalue Calculation
      2. 2.4.2 Frequency and Wavenumber
      3. 2.4.3 Asymptotic Analysis
    5. 2.5 Two Dimensions
      1. 2.5.1 Eigenvalue Calculation
      2. 2.5.2 Frequency and Wavenumber
      3. 2.5.3 Asymptotic Analysis
  3. 3 Hybridization of Two Resonators
    1. 3.1 Three Dimensions
    2. 3.2 Two Dimensions
  4. 4 Example: Circular Resonators
  5. 5 Conclusion
  6. A Appendix
    1. A.1 Calculation of Three-dimensional Constants
    2. A.2 Calculation of Two-dimensional Constants

##  1 Introduction

Halide perovskites are set to be at the center of the next generation of
electromagnetic devices [11, 14, 15]. They are composed of crystalline
lattices which have octohedral shapes and contain atoms of heavier halides,
such as chlorine, bromine and iodine [1]. Their excellent optical and
electronic properties, combined with being cheap and easy to manufacture, have
paved the way for a perovskite revolution. A particular benefit of halide
perovskites is that their high absorption coefficient enables microscopic
devices (measuring only a few hundred nanometres) to absorb the complete
visible spectrum. Thus, we are able to design very small devices that are
lightweight and compact while also being low cost and efficient. Research is
ongoing to develop perovskites capable of fulfilling their theoretical
capabilities for use in applications such as optical sensors [8], solar cells
[15] and light-emitting diodes [17].

The dielectric permittivity of a halide perovskite $\varepsilon$ is given, in
terms of the frequency $\omega$ and wavenumber $k$ by [14]

$\varepsilon(\omega,k)=\varepsilon_{0}+\frac{\omega_{p}^{2}}{\omega^{2}_{exc}-\omega^{2}+\hslash\omega_{exc}k^{2}\mu_{exc}^{-1}-i\gamma\omega},$

where $\omega_{exc}$ is the frequency of the excitonic transition,
$\omega_{p}$ is the strength of the dipole oscillator, $\gamma$ is the damping
factor, $\mu_{exc}$ is related to the non-local response and $\varepsilon_{0}$
is the background dielectric constant. We refer to [14] for the values of
these constants for different halide perovskites at room temperature.
Meanwhile, the dispersion relation for the halide perovskites is observed [10]
to take the simplified form

$n^{2}=1+\frac{S_{0}\lambda_{0}^{2}}{1-\lambda_{0}^{2}k^{2}},$

where $n$ is the refractive index, $k$ is the wavenumber and $S_{0}$ and
$\lambda_{0}$ are positive constants that describe the average oscillator
strength, see [10] for the values for some different halide perovsksites.

Dielectric nano-particles, and other electromagnetic metamaterials, have been
studied using various techniques. In the case of extreme material parameters,
such as small particle size or large material contrast, asymptotic methods can
be used [3, 2, 7]. Likewise, homogenisation has been used to characterise
materials with periodic micro-structures [6, 9]. Multiple scattering
formulations are popular, particularly when a convenient choice of geometry
(e.g. cylindrical or spherical resonators) facilitates explicit formulas [16].
In this paper, we will use integral methods to study a general class of
geometries and will extend the previous theory, e.g. [2], to the case of
dispersive materials.

For simplicity, we consider the Helmholtz equation as a model of the
propagation of time-harmonic waves, and for the permittivity relation, we
consider the form

$\displaystyle\varepsilon(\omega,k)=\varepsilon_{0}+\frac{\alpha}{\beta-\omega^{2}+\eta
k^{2}-i\gamma\omega},$ (1.1)

where $\alpha,\beta,\gamma,\eta$ are positive constants. We will use an
approach based on representing the scattered solution using a Lippmann-
Schwinger integral formulation and then using asymptotic methods to
characterise the resonant frequencies in terms of the eigenvalues of the
resulting integral operator (which turns out to be the Newtonian potential).
This approach can handle a very general class of resonator shapes and can be
adapted to study solutions to the Maxwell’s equations [5]. A similar method
was used in [2] for a simpler, non-dispersive setting. This paper shows that
the asymptotic theory developed in [2] and elsewhere can be developed to model
real-world settings and can be used to influence high-impact design problems.

In section 2 of this paper we will introduce the problem setting and retrieve
its integral formulation. Then, we will study the one particle case for three
and two dimensions, using integral techniques to formulate the subwavelength
resonant problem and asymptotic approximations to study the resonant
frequencies. In section 3, we will use these methods to describe the
hybridization of two halide perovskite resonators. Again, we treat the three-
and two-dimensional cases separately. Passing from the integral to a matrix
formulation of the problem, we obtain the hybridized subwavelength resonant
frequencies. Finally, we examine the case of spherical resonators, making use
of the fact that eigenvalues and eigenfunctions of the Newtonian potential can
be computed explicitly in this case (see also [12]). We show that our findings
are qualitatively consistent with the ones of [2]. Hence, we show that the
asymptotic techniques used in [2] can be generalized to less straightforward
and more impactful physical settings.

##  2 Single Resonators

###  2.1 Problem Setting

Consider a single resonator occupying a bounded domain
$\Omega\subset\mathbb{R}^{d}$, for $d\in\\{2,3\\}$. We assume that the
particle is non-magnetic, so that the magnetic permeability $\mu_{0}$ is
constant on all of $\mathbb{R}^{d}$. We will consider a time-harmonic wave
with frequency $\omega\in\mathbb{C}$ (which we assume to have positive real
part). The wavenumber in the background
$\mathbb{R}^{d}\setminus\overline{\Omega}$ is given by
$k_{0}:=\omega\varepsilon_{0}\mu_{0}$ and we will use $k$ to denote the
wavenumber within $\Omega$. We, then, consider the following Helmholtz model
for light propagation:

$\displaystyle\begin{cases}\Delta u+\omega^{2}\varepsilon(\omega,k)\mu_{0}u=0\
\ \ &\text{ in }\Omega,\\\ \Delta u+k_{0}^{2}u=0&\text{ in
}\mathbb{R}^{d}\setminus\overline{\Omega},\\\ u|_{+}-u|_{-}=0&\text{ on
}\partial\Omega,\\\ \frac{\partial u}{\partial\nu}|_{+}-\frac{\partial
u}{\partial\nu}|_{-}=0&\text{ on }\partial\Omega,\\\ u(x)-u_{in}(x)&\text{
satisfies the outgoing radiation condition as }|x|\to\infty,\\\ \end{cases}$
(2.1)

where $u_{in}$ is the incident wave, assumed to satisfy

$\displaystyle(\Delta+k_{0}^{2})u_{in}=0\ \ \text{ in }\mathbb{R}^{d},$

and the appropriate outgoing radiation condition is the Sommerfeld radiation
condition, which requires that

$\lim_{|x|\to\infty}|x|^{\frac{d-1}{2}}\left(\frac{\partial}{\partial|x|}-ik_{0}\right)(u(x)-u_{in}(x))=0.$
(2.2)

In particular, we are interested in the case of small resonators. Thus, we
will assume that there exists some fixed domain $D$ such that $\Omega$ is
given by

$\Omega=\delta D+z,$ (2.3)

for some position $z\in\mathbb{R}^{d}$ and characteristic size $0<\delta\ll
1$. Then, making a change of variables, the Helmholtz problem (2.1) becomes

$\displaystyle\begin{cases}\Delta
u+\delta^{2}\omega^{2}\varepsilon(\omega,k)\mu_{0}u=0\ \ \ &\text{ in }D,\\\
\Delta u+\delta^{2}k_{0}^{2}u=0&\text{ in
}\mathbb{R}^{d}\setminus\overline{D},\\\ \end{cases}$ (2.4)

along with the same transmission conditions on $\partial D$ and far-field
behaviour. The behaviour of resonators which is of interest is when $\delta\ll
k_{0}^{-1}$, meaning the system can be described as being _subwavelength_. We
will study this by performing asymptotics in the regime that the frequency
$\omega$ is fixed while the size $\delta\to 0$.
We will characterise solutions to (2.1) in terms of the system’s resonant
frequencies. For a given wavenumber $k$, we define $\omega=\omega(k)$ to be a
_resonant frequency_ if it is such that there exists a non-trivial solution
$u$ to (2.1) in the case that $u_{in}=0$.

###  2.2 Integral Formulation

Let $G(x,k)$ be the outgoing Helmholtz Green’s function in $\mathbb{R}^{d}$,
defined as the unique solution to

$\displaystyle(\Delta+k^{2})G(x,k)=\delta_{0}(x)\ \ \text{ in
}\mathbb{R}^{d},$

along with the outgoing radiation condition (2.2). It is well known that $G$
is given by

$\displaystyle G(x,k)=\begin{cases}-\frac{i}{4}H^{(1)}_{0}(k|x|),\ \ \
&d=2,\\\ -\frac{e^{ik|x|}}{4\pi|x|},&d=3,\\\ \end{cases}$ (2.5)

where $H_{0}^{(1)}$ is the Hankel function of first kind and order zero.

######  Theorem 2.1 (Lippmann-Schwinger Integral Representation Formula).

The solution to the Helmholtz problem (2.1) is given by

$\displaystyle
u(x)-u_{in}(x)=-\delta^{2}\omega^{2}\xi(\omega,k)\int_{D}G(x-y,\delta
k_{0})u(y)\mathrm{d}y,\ \ x\in\mathbb{R}^{d},$ (2.6)

where the function $\xi:\mathbb{C}\to\mathbb{C}$ describes the permittivity
contrast between $D$ and the background and is given by

$\xi(\omega,k)=\mu_{0}(\varepsilon(\omega,k)-\varepsilon_{0}).$

###### Proof.

We see from (2.4) that $\Delta
u+\delta^{2}\omega^{2}\varepsilon(\omega,k)\mu_{0}u=0$ in $D$, so it holds
that

$\displaystyle\Delta
u+\delta^{2}k_{0}^{2}u=-\delta^{2}\omega^{2}\xi(\omega,k)u,$

where $\xi(\omega,k)=\mu_{0}(\varepsilon(\omega,k)-\varepsilon_{0})$.
Therefore, the Helmholtz problem $(\ref{2.4})$ becomes

$\displaystyle(\Delta+\delta^{2}k_{0}^{2})(u(y)-u_{in}(y))=-\delta^{2}\omega^{2}\xi(\omega,k)u(y)\chi_{D}(y),\
\ y\in\mathbb{R}^{d}\setminus\partial D,$

with $\chi_{D}$ being the indicator function of the set $D$. Then, we know
that for $x\in\mathbb{R}^{d}$ and $y\in\mathbb{R}^{d}\setminus\partial D$, the
following identity holds:

$\displaystyle\nabla_{y}\cdot\Big{[}G(x-y,\delta k_{0})\nabla_{y}\Big{(}u(y)$
$\displaystyle-
u_{in}(y)\Big{)}-\Big{(}u(y)-u_{in}(y)\Big{)}\nabla_{y}G(x-y,\delta
k_{0})\Big{]}$
$\displaystyle=-\delta^{2}\omega^{2}\xi(\omega,k)u(y)G(x-y,\delta
k_{0})\chi_{D}(y)-\Big{(}u(y)-u_{in}(y)\Big{)}\delta_{0}(x-y).$

Let $S_{R}$ be a large sphere with radius $R$ large enough so that $D\subset
S_{R}$. Integrating the above identity over $y\in S_{r}\setminus\partial D$
and letting $R\to+\infty$, we can use the radiation condition (2.2) in the far
field and the transmission conditions on $\partial D$ to obtain the desired
integral representation formula. ∎

We are interested in understanding how the formula from Theorem 2.1 behaves in
the case that $\delta$ is small. In particular, we wish to understand the
operator $K^{\delta k_{0}}_{D}:L^{2}(D)\to L^{2}(D)$ given by

$\displaystyle K^{\delta k_{0}}_{D}[u](x)=-\int_{\Omega}G(x-y,\delta
k_{0})u(y)\mathrm{d}y,\ \ \ \ x\in D.$ (2.7)

The Helmholtz Green’s function has helpful asymptotic expansions which
facilitate this. However, the behaviour is quite different in two and three
dimensions, so we must now consider these two settings separately. We will
first work on the three-dimensional case, and then treat the two-dimensional
setting as the asymptotic expansions are more complicated.

###  2.3 Reformulation as a Subwavelength Resonance Problem

In order to reveal the behaviour of the system of nano-particles, we will
characterise the properties of the operator $K^{\delta k_{0}}_{D}$. We observe
that the Lippmann-Schwinger formulation (2.6) of the problem is equivalent to

$\displaystyle(u-u_{in})(x)=\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D}[u](x).$

This is equivalent to

$\displaystyle(I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D})[u](x)=u_{in}(x),$

which gives

$\displaystyle u(x)=(I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D})^{-1}[u_{in}](x),$

for all $x\in D$, where $I$ denotes the identity operator. Then, the
subwavelength resonance problem is to find $\omega\in\mathbb{C}$ close to 0,
such that the operator $(I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D})^{-1}$ is singular, or equivalently, such that there exists $u\in
L^{2}(D)$, $u\neq 0$ with

$\displaystyle u(x)-\delta^{2}\omega^{2}\xi(\omega,k)\int_{D}G(x-y,\delta
k_{0})u(y)\mathrm{d}y=0,\ \ \ \text{ for }x\in D.$ (2.8)

###  2.4 Three Dimensions

We consider the three-dimensional case, $d=3$. Let us define the Newtonian
potential on $D$ to be $K^{(0)}_{D}:L^{2}(D)\to L^{2}(D)$ such that

$K^{(0)}_{D}[u](x):=-\int_{D}u(y)G(x-y,0)\mathrm{d}y=-\frac{1}{4\pi}\int_{D}\frac{1}{|x-y|}u(y)\mathrm{d}y.$
(2.9)

Similarly, we define the operators $K^{(n)}_{D}:L^{2}(D)\to L^{2}(D)$, for
$n=1,2,\dots,$ as

$K^{(n)}_{D}[u](x)=-\frac{i}{4\pi}\int_{D}\frac{(i|x-y|)^{n-1}}{n!}u(y)\mathrm{d}y.$
(2.10)

Then, in order to capture the behaviour of $K^{\delta k_{0}}_{D}$ for small
$\delta$, we can use an asymptotic expansion in terms of small $\delta$.

######  Lemma 2.2.

Suppose that $d=3$. The operator $K^{\delta k_{0}}_{D}$ can be rewritten as

$\displaystyle K^{\delta k_{0}}_{D}=\sum_{n=0}^{\infty}(\delta
k_{0})^{n}K^{(n)}_{D},$

where the series converges in the $L^{2}(D)\to L^{2}(D)$ operator norm if
$\delta k_{0}$ is small enough.

###### Proof.

This follows from an application of the Taylor expansion on the operator
$K^{\delta k_{0}}_{D}$. Indeed, using Taylor expansion on the second variable
of $G$, we can see that

$G(x-y,\delta k_{0})=\sum_{n=0}^{\infty}(\delta
k_{0})^{n}\frac{\partial^{n}G}{\partial k^{n}}(x-y,k)\Big{|}_{k=0}.$

Since we are working in three dimensions,

$G(x,k)=-\frac{e^{ik|x|}}{4\pi|x|}$

and, thus,

$G(x-y,0)=-\frac{1}{4\pi|x-y|}\quad\text{and}\quad\frac{\partial^{n}G}{\partial
k^{n}}(x-y,k)=-\frac{i}{4\pi}\frac{(i|x-y|)^{n-1}}{n!}\quad\text{for }n>0.$

Hence,

$\displaystyle G(x-y,\delta k_{0})$ $\displaystyle=\sum_{n=0}^{\infty}(\delta
k_{0})^{n}\frac{\partial^{n}G}{\partial
k^{n}}(x-y,k)\Big{|}_{k=0}=\sum_{n=0}^{\infty}(\delta
k_{0})^{n}\left(-\frac{i}{4\pi}\right)\frac{(i|x-y|)^{n-1}}{n!},$

and then, multiplying by $u$ and integrating over $D$, gives

$\displaystyle K^{\delta k_{0}}_{D}[u](y)$
$\displaystyle=\sum_{n=0}^{\infty}(\delta k_{0})^{n}K^{(n)}_{D}[u](y),$

which is the desired result. ∎

####  2.4.1 Eigenvalue Calculation

In order to study the operator $K^{\delta k_{0}}_{D}$, we need to find its
eigenvalues. From Lemma 2.2, if $k_{0}$ is fixed then we can write our
operator as

$\displaystyle K^{\delta k_{0}}_{D}=K^{(0)}_{D}+\delta
k_{0}K^{(1)}_{D}+O(\delta^{2})\ \ \ \text{ as }\delta\to 0.$ (2.11)

Since we know that $K^{(0)}_{D}$ is self-adjoint, we know that it admits
eigenvalues. Let us denote such an eigenvalue by $\lambda_{0}$, and by $u_{0}$
the associated eigenvector. Let us now consider the problem

$K^{\delta k_{0}}_{D}u_{\delta}=\lambda_{\delta}u_{\delta},$ (2.12)

where $\lambda_{\delta}$ denotes an eigenvalue of the operator $K^{\delta
k_{0}}_{D}$ and $u_{\delta}$ denotes the associated eigenvector. Using this
eigenvalue problem, we will find the resonant frequency $\omega_{s}$ and the
associated wavenumber $k_{s}$ of the halide perovskite nano-particle.

######  Proposition 2.3.

Let $\lambda_{\delta}$ denote an eigenvalue of the operator $K^{\delta
k_{0}}_{D}$ in dimension three. Then, if $\delta$ is small, it is
approximately given by

$\displaystyle\lambda_{\delta}\approx\lambda_{0}+\delta k_{0}\langle
K^{(1)}_{D}u_{0},u_{0}\rangle.$ (2.13)

###### Proof.

We start by truncating the $O(\delta^{2})$ term in the expansion (2.11). Then,
we have that

$\displaystyle K^{\delta k_{0}}_{D}u_{\delta}=\lambda_{\delta}u_{\delta}\ $
$\displaystyle\Leftrightarrow\ (K^{(0)}_{D}+\delta
k_{0}K^{(1)}_{D})u_{\delta}=\lambda_{\delta}u_{\delta}\ \Leftrightarrow\
\langle(K^{(0)}_{D}+\delta
k_{0}K^{(1)}_{D})u_{\delta},u_{0}\rangle=\langle\lambda_{\delta}u_{\delta},u_{0}\rangle.$

Since $K^{(0)}_{D}$ is self-adjoint, we have that

$\displaystyle\ \lambda_{0}\langle u_{\delta},u_{0}\rangle+\delta k_{0}\langle
K^{(1)}_{D}u_{\delta},u_{0}\rangle=\lambda_{\delta}\langle
u_{\delta},u_{0}\rangle.$

Finally, since $u_{\delta}\approx u_{0}$ we find that

$\displaystyle\lambda_{\delta}=\lambda_{0}+\delta k_{0}\frac{\langle
K^{(1)}_{D}u_{\delta},u_{0}\rangle}{\langle
u_{\delta},u_{0}\rangle}\approx\lambda_{0}+\delta k_{0}\langle
K^{(1)}_{D}u_{0},u_{0}\rangle,$

which is the desired result. ∎

The following corollary is a direct consequence of Proposition 2.13:

######  Corollary 2.3.1.

Let $\lambda_{\delta}$ denote an eigenvalue of the operator $K^{\delta
k_{0}}_{D}$ in three dimensions. Then, if $\delta$ is small, it is
approximately given by

$\displaystyle\lambda_{\delta}\approx\lambda_{0}-\frac{i}{4\pi}\delta
k_{0}\mathbb{B},$ (2.14)

where $\mathbb{B}:=(\int_{D}u_{0}(y)\mathrm{d}y)^{2}$ is a constant.

###### Proof.

From (2.10), we get that

$\displaystyle
K^{(1)}_{D}[u](x)=-\frac{i}{4\pi}\int_{D}\frac{(i|x-y|)^{1-1}}{1!}u(y)\mathrm{d}y=-\frac{i}{4\pi}\int_{D}u(y)\mathrm{d}y.$

Let us define the constant $\mathbb{B}:=(\int_{D}u_{0}(y)\mathrm{d}y)^{2}$.
Then, we observe that

$\displaystyle\langle
K^{(1)}_{D}u_{0},u_{0}\rangle=\int_{D}\left(-\frac{i}{4\pi}\right)u_{0}(y)\int_{D}\overline{u_{0}}(x)\mathrm{d}x\mathrm{d}y=-\frac{i}{4\pi}\left(\int_{D}u_{0}(y)\mathrm{d}y\right)^{2}=-\frac{i}{4\pi}\mathbb{B}.$

Substituting into (2.13) gives the desired result. ∎

####  2.4.2 Frequency and Wavenumber

Let us now find the resonant frequency $\omega_{\delta}$ and wavenumber
$k_{\delta}$ associated to this eigenvalue, which will also constitute the
basis of our analysis of the operator $K^{\delta k_{0}}_{D}$. From (2.8), we
see that if $u=u_{\varepsilon}$, then
$1=\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}$ so, using Corollary 2.14
we have that

$\displaystyle\varepsilon(\omega,k)=\frac{1}{\mu_{0}\delta^{2}\omega^{2}\left(\lambda_{0}-\frac{i}{4\pi}\delta
k_{0}\mathbb{B}\right)}+\varepsilon_{0}.$ (2.15)

In order to study halide perovskite particles, we want the permittivity
$\varepsilon(\omega,k)$ to be given by (1.1), that is,

$\varepsilon(\omega,k)=\varepsilon_{0}+\frac{\alpha}{\beta-\omega^{2}+\eta
k^{2}-i\gamma\omega},$ (2.16)

where $\alpha,\beta,\gamma,\eta$ are positive constants. Comparing the two
expressions (2.15) and (2.16) we see that

$\displaystyle\begin{cases}\alpha\mu_{0}\delta^{2}\omega^{2}\lambda_{0}-\beta+\omega^{2}-\eta
k^{2}=0,\\\
\gamma\omega-\mu_{0}\frac{1}{4\pi}\alpha\delta^{3}\omega^{2}k_{0}\mathbb{B}=0.\end{cases}$
(2.17)

We study these two equations separately. First, we look at the second equation
of (2.17). We have that

$\displaystyle\omega\left(\gamma-\mu_{0}\frac{1}{4\pi}\alpha\delta^{3}\omega
k_{0}\mathbb{B}\right)=0,$

meaning that

$\displaystyle\omega=0\quad\text{ or
}\quad\omega=\frac{4\pi\gamma}{\alpha\mu_{0}\delta^{3}k_{0}\mathbb{B}}.$

For $\omega=\frac{4\pi\gamma}{\alpha\mu_{0}\delta^{3}k_{0}\mathbb{B}}$, we
obtain that

$\displaystyle\eta
k^{2}=(1+\alpha\mu_{0}\delta^{2}\lambda_{0})\omega^{2}-\beta,$

which has solutions

$\displaystyle
k=\pm\sqrt{\frac{16\pi^{2}\gamma^{2}(1+\alpha\mu_{0}\delta^{2}\lambda_{0})}{\alpha^{2}\mu_{0}^{2}\delta^{6}k_{0}^{2}\mathbb{B}^{2}\eta}-\frac{\beta}{\eta}}.$
(2.18)

The case of $\omega=0$ is not of physical interest here. Thus, denoting this
specific frequency by $\omega_{\delta}$ and the associated wavenumber by
$k_{\delta}$, we will work with

$\displaystyle\omega_{\delta}=\frac{4\pi\gamma}{\alpha\mu_{0}\delta^{3}k_{0}\mathbb{B}}\quad\text{
and }\quad
k_{\delta}=\sqrt{\frac{16\pi^{2}\gamma^{2}(1+\alpha\mu_{0}\delta^{2}\lambda_{0})}{\alpha^{2}\mu_{0}^{2}\delta^{6}k_{0}^{2}\mathbb{B}^{2}\eta}-\frac{\beta}{\eta}},$
(2.19)

where we have chosen the wavenumber $k_{\delta}$ to be positive.

####  2.4.3 Asymptotic Analysis

Let us now return to the problem of studying the singularities of the operator
$(I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta k_{0}}_{D})^{-1}$. We have the
following equivalence:

$\displaystyle(I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D})^{-1}=0\ \Leftrightarrow\
\left(I-\delta^{2}\omega^{2}\xi(\omega,k)\sum_{n=0}^{\infty}(\delta
k_{0})^{n}K^{(n)}_{D}\right)^{-1}=0.$

We define

$A_{n}:=\delta^{2}\omega^{2}\xi(\omega,k)K^{(n)}_{D}=-\delta^{2}\omega^{2}\xi(\omega,k)\frac{i}{4\pi}\int_{D}\frac{(i|x-y|)^{n-1}}{n!}u(y)\mathrm{d}y.$

Then, it holds that

$\displaystyle\left(I-\delta^{2}\omega^{2}\xi(\omega,k)\sum_{n=0}^{\infty}(\delta
k_{0})^{n}K^{(n)}_{D}\right)^{-1}$
$\displaystyle=\left(I-\sum_{n=0}^{\infty}(\delta k_{0})^{n}A_{n}\right)^{-1}$
$\displaystyle=\sum_{i=0}^{\infty}\left((I-A_{0}-\delta
k_{0}A_{1})^{-1}\sum_{n=2}^{\infty}(\delta
k_{0})^{n}A_{n}\right)^{i}(I-A_{0}-\delta k_{0}A_{1})^{-1}$
$\displaystyle=(I-A_{0}-\delta k_{0}A_{1})^{-1}$ $\displaystyle\ \ \
+(I-A_{0}-\delta k_{0}A_{1})^{-1}(\delta k_{0})^{2}A_{2}(I-A_{0}-\delta
k_{0}A_{1})^{-1}+O(\delta^{4}).$

Thus, the above equivalence yields

$\displaystyle(I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D})^{-1}=0\ \Leftrightarrow\
\left(I-\delta^{2}\omega^{2}\xi(\omega,k)\sum_{n=0}^{\infty}(\delta
k_{0})^{n}K^{(n)}_{D}\right)^{-1}=0\ \Leftrightarrow$
$\displaystyle(I-A_{0}-\delta k_{0}A_{1})^{-1}+(I-A_{0}-\delta
k_{0}A_{1})^{-1}(\delta k_{0})^{2}A_{2}(I-A_{0}-\delta
k_{0}A_{1})^{-1}+O(\delta^{4})=0.$

Using this expression, we obtain the following proposition.

######  Proposition 2.4.

Let $d=3$ and let $\omega_{\delta}$ be defined by (2.19). Then, as $\delta\to
0$, the $O(\delta^{4})$ approximation of the subwavelength resonant
frequencies $\omega_{s}$ and the associated wavenumbers $k_{s}$ of the single
halide perovskite resonator $\Omega=\delta D+z$ satisfy

$1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}=-\delta^{4}k_{0}^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle.$ (2.20)

###### Proof.

For $\psi\in L^{2}(D)$, and dropping the $O(\delta^{4})$ term, we have

$\Big{[}(I-A_{0}-\delta k_{0}A_{1})^{-1}+(I-A_{0}-\delta
k_{0}A_{1})^{-1}(\delta k_{0})^{2}A_{2}(I-A_{0}-\delta
k_{0}A_{1})^{-1}\Big{]}[\psi]=0.$ (2.21)

We apply a pole-pencil decomposition, as it is defined in [4], to the term
$(I-A_{0}-\delta k_{0}A_{1})^{-1}[\psi]$ and obtain

$(I-A_{0}-\delta k_{0}A_{1})^{-1}[\psi]=\frac{\langle u_{\delta},\psi\rangle
u_{\delta}}{1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}}+R(\delta)[\psi],$

where the remainder term $R(\delta)[\psi]$ can be dropped. Hence, (2.21) is,
at leading order, equivalent to

$\displaystyle\frac{\langle u_{\delta},\psi\rangle
u_{\delta}}{1-\delta^{2}\omega^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}}+\frac{\langle
u_{\delta},\psi\rangle
u_{\delta}}{1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}}(\delta
k_{0})^{2}A_{2}\frac{\langle u_{\delta},\psi\rangle
u_{\delta}}{1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}}=0,$

which reduces to

$\displaystyle\frac{u_{\delta}}{1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}}+\frac{u_{\delta}\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle}{(1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta})^{2}}\delta^{4}k_{0}^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})=0,$

which can be rearranged to give the desired result. ∎

######  Remark.

In the appendix, we will recover a formula for $\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle$.

Before we move on to the consequences of this proposition, we recall that

$\displaystyle
K^{(2)}_{D}[u](x)=-\frac{i}{4\pi}\int_{D}\frac{i|x-y|}{2}u(y)\mathrm{d}y=\frac{1}{8\pi}\int_{D}|x-y|u(y)\mathrm{d}y.$

Thus, we can define the constant $\mathbb{F}$ to be such that

$\displaystyle\langle K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle$
$\displaystyle=\int
K^{(2)}_{D}[u_{\delta}](x)\overline{u_{\delta}}(x)\mathrm{d}x=\frac{1}{8\pi}\int_{D}\int_{D}|x-y|u_{\delta}(y)\overline{u_{\delta}}(x)\mathrm{d}y\mathrm{d}x=:\frac{1}{8\pi}\mathbb{F}.$

So, from (2.20), we see that

$\displaystyle
1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}=-\frac{\delta^{4}k_{0}^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})}{8\pi}\mathbb{F}.$
(2.22)

Then the following two corollaries are immediate consequences of the
Proposition 2.4.

######  Corollary 2.4.1.

Let $d=3$. Then, as $\delta\to 0$, the $O(\delta^{8})$ approximation of the
subwavelength resonant frequencies of the halide perovskite resonator
$\Omega=\delta D+z$ are given by

$1-\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}\lambda_{\delta}(\varepsilon(\omega_{s},k_{s})-\varepsilon_{0})\frac{64\pi^{2}\gamma^{2}}{\alpha^{2}\mu_{0}\delta^{4}k_{0}^{2}\mathbb{B}^{2}}=-\frac{\delta^{4}\omega_{s}^{2}k_{0}^{2}\xi(\omega_{s},k_{s})}{8\pi}\mathbb{F}.$
(2.23)

###### Proof.

By a direct calculation and using (2.19), we have

$\displaystyle
1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}$
$\displaystyle\
=1-\delta^{2}\omega_{s}^{2}\mu_{0}(\varepsilon(\omega_{s},k_{s})-\varepsilon_{0})\lambda_{\delta}$
$\displaystyle\overset{(\ref{w_e k_e for
d=3})}{=}1-\delta^{2}\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}\mu_{0}(\varepsilon(\omega_{s},k_{s})-\varepsilon_{0})\lambda_{\delta}\frac{64\pi^{2}\gamma^{2}}{\alpha^{2}\mu_{0}^{2}\delta^{4}k_{0}^{2}\mathbb{B}^{2}}$
$\displaystyle\
=1-\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}\lambda_{\delta}(\varepsilon(\omega_{s},k_{s})-\varepsilon_{0})\frac{64\pi^{2}\gamma^{2}}{\alpha^{2}\mu_{0}\delta^{4}k_{0}^{2}\mathbb{B}^{2}}.$

Then, using (2.22), the result follows. ∎

######  Corollary 2.4.2.

Let $d=3$. Then, as $\delta\to 0$, the $O(\delta^{4})$ approximation of the
subwavelength resonant frequencies of the halide perovskite resonator
$\Omega=\delta D+z$ can be computed as

$1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}=-\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}(\varepsilon(\omega_{s},k_{s})-\varepsilon_{0})\frac{8\pi\gamma^{2}\mathbb{F}}{\alpha^{2}\mu_{0}\delta^{2}\mathbb{B}^{2}}.$
(2.24)

###### Proof.

Again, we can calculate this directly:

$\displaystyle\frac{\delta^{4}k_{0}^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})}{8\pi}\mathbb{F}$
$\displaystyle\
=\frac{1}{8\pi}\delta^{4}\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}k_{0}^{2}\mu_{0}(\varepsilon(\omega_{s},k_{s})-\varepsilon_{0})\omega_{\delta}^{2}$
$\displaystyle\overset{(\ref{w_e k_e for
d=3})}{=}\frac{1}{8\pi}\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}\delta^{4}k_{0}^{2}\mu_{0}(\varepsilon(\omega_{s},k_{s})-\varepsilon_{0})\frac{64\pi^{2}\gamma^{2}}{\alpha^{2}\mu_{0}^{2}\delta^{4}k_{0}^{2}\mathbb{B}^{2}}$
$\displaystyle=\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}(\varepsilon(\omega_{s},k_{s})-\varepsilon_{0})\frac{8\pi\gamma^{2}}{\alpha^{2}\mu_{0}\delta^{2}\mathbb{B}^{2}}.$

Then, using (2.22), the result follows. ∎

We finish our analysis of the three-dimensional case with the following
proposition.

######  Proposition 2.5.

Let $d=3$. For $\omega$ close to the resonant frequency $\omega_{s}$, the
field scattered by the halide perovskite nano-particle $\Omega=\delta D+z$ can
be approximated by

$\displaystyle u(x)-u_{in}(x)\approx\frac{1+\delta^{2}\omega^{2}G(x-y,\delta
k_{0})\xi(\omega,k)}{\delta^{2}\omega^{2}\left(\lambda_{\delta}-\lambda_{0}+\frac{i}{4\pi}\delta
k_{0}\mathbb{B}\right)\xi(\omega,k)}\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}.$

###### Proof.

Our goal is to find $u\in L^{2}(D)$, $u\neq 0$ such that (2.8) is satisfied.
Using the pole-pencil decomposition, we can rewrite

$\displaystyle u(x)-u_{in}(x)\approx-\delta^{2}\omega^{2}\xi(\omega,k)$
$\displaystyle G(x-y,\delta k_{0})\frac{\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}$
$\displaystyle+\delta^{4}k_{0}^{2}\omega^{2}\xi(\omega,k)\frac{\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{(1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta})^{2}}.$

Using (2.20) and plugging it in the above expression, we obtain

$\displaystyle u(x)-u_{in}(x)$ $\displaystyle\approx\
-\delta^{2}\omega^{2}\xi(\omega,k)G(x-y,\delta k_{0})\frac{\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{-\delta^{4}k_{0}^{2}\omega^{2}\xi(\omega,k)\langle
K^{(2)}_{D}[u_{\delta}],\upsilon_{\delta}\rangle}$ $\displaystyle\ \ \ \ \ \ \
+\delta^{4}k_{0}^{2}\omega^{2}\xi(\omega,k)\frac{\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{\delta^{8}k_{0}^{4}\omega^{4}\xi(\omega,k)^{2}\langle
K^{(2)}_{D}[u_{\delta}],\upsilon_{\delta}\rangle^{2}}$ $\displaystyle=\
\frac{G(x-y,\delta k_{0})\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{\delta^{2}k_{0}^{2}\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle}+\frac{\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{\delta^{4}k_{0}^{2}\omega^{2}\xi(\omega,k)\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle}.$

Thus, defining the constant $\mathbb{F}:=8\pi\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle$, we obtain

$\displaystyle u(x)-u_{in}(x)\approx\
\frac{8\pi\Big{(}1+\delta^{2}\omega^{2}G(x-y,\delta
k_{0})\xi(\omega,k)\Big{)}}{\delta^{4}k_{0}^{2}\omega^{2}\xi(\omega,k)\mathbb{F}}\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}.$ (2.25)

From the Appendix A.1, we have that

$\mathbb{F}=\frac{8\pi}{\delta^{2}k_{0}^{2}}\left(\lambda_{\delta}-\lambda_{0}+\frac{i}{4\pi}\delta
k_{0}\mathbb{B}\right).$

Plugging it into (2.25), we obtain

$u(x)-u_{in}(x)\approx\frac{8\pi\Big{(}1+\delta^{2}\omega^{2}G(x-y,\delta
k_{0})\xi(\omega,k)\Big{)}}{\delta^{4}k_{0}^{2}\omega^{2}\xi(\omega,k)\mathbb{F}}\frac{\delta^{2}k_{0}^{2}}{8\pi}\frac{1}{\lambda_{\delta}-\lambda_{0}+\frac{i}{4\pi}\delta
k_{0}\mathbb{B}}\langle u_{in},u_{\delta}\rangle\int_{D}u_{\delta},$

from which the result follows. ∎

###  2.5 Two Dimensions

We now turn our attention to the two-dimensional setting. We define the
Newtonian potential on $D$, $K^{(0)}_{D}:L^{2}(D)\to L^{2}(D)$, by

$\displaystyle
K^{(0)}_{D}[u](x):=-\int_{D}u(y)G(x-y,0)\mathrm{d}y=-\frac{1}{2\pi}\int_{D}\log|x-y|u(y)\mathrm{d}y.$

We also define the operators $K^{(-1)}_{D}:L^{2}(D)\to L^{2}(D)$ and
$K^{(1)}_{D}:L^{2}(D)\to L^{2}(D)$ by

$\displaystyle
K^{(-1)}_{D}[u](x):=-\frac{1}{2\pi}\int_{D}u(y)\mathrm{d}y\quad\text{ and
}\quad K^{(n)}_{D}[u](x):=\int_{D}\frac{\partial^{n}}{\partial
k^{n}}G(x-y,k)\Big{|}_{k=0}u(y)\mathrm{d}y.$

Then, from the asymptotic expansion of the Hankel function, we have the
following result.

######  Lemma 2.6.

Suppose that $d=2$. Then, for fixed $k_{0}\in\mathbb{C}$, the operator
$K^{\delta k_{0}}_{D}$ satisfies

$\displaystyle K^{\delta k_{0}}_{D}=\log(\delta
k_{0}\hat{\gamma})K^{(-1)}_{D}+K^{(0)}_{D}+(\delta
k_{0})^{2}\log(\hat{\gamma}\delta k_{0})K^{(1)}_{D}+O(\delta^{4}\log\delta),$
(2.26)

as $\delta\to 0$, with convergence in the $L^{2}(D)\to L^{2}(D)$ operator norm
and the constant $\hat{\gamma}$ being given by
$\hat{\gamma}:=\frac{1}{2}k_{0}\exp(\gamma-\frac{i\pi}{2})$, where $\gamma$ is
the Euler constant.

####  2.5.1 Eigenvalue Calculation

We use the same notation for $u_{\delta}$ as in section 2.4.1. Let us consider
the eigenvalue problem

$K^{\delta k_{0}}_{D}u_{\delta}=\lambda_{\delta}u_{\delta},$ (2.27)

where $\lambda_{\delta}$ denotes an eigenvalue of the operator $K^{\delta
k_{0}}_{D}$, and $u_{\delta}$ denotes an associated eigenvector.

######  Proposition 2.7.

Let $\lambda_{\delta}$ denote an eigenvalue of the operator $K^{\delta
k_{0}}_{D}$ in dimension 2. Then, for small $\delta$, it is approximately
given by:

$\displaystyle\lambda_{\delta}\approx\log(\delta
k_{0}\hat{\gamma})\lambda_{-1}+\langle K^{(0)}_{D}u_{-1},u_{-1}\rangle+(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})\langle
K^{(1)}_{D}u_{-1},u_{-1}\rangle,$ (2.28)

where $\lambda_{-1}$ and $u_{-1}$ are an eigenvalue and the associated
eigenvector of the potential $K^{(-1)}_{D}$.

###### Proof.

We start by dropping the $O(\delta^{4}\log(\delta))$ term on the expansion
(2.26). Then, we have that

$\displaystyle K^{\delta k_{0}}_{D}u_{\delta}=\lambda_{\delta}u_{\delta}\ $
$\displaystyle\Leftrightarrow\ (\log(\delta
k_{0}\hat{\gamma})K^{(-1)}_{D}+K^{(0)}_{D}+(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})K^{(1)}_{D})u_{\delta}=\lambda_{\delta}u_{\delta}$
$\displaystyle\Leftrightarrow\ \log(\delta
k_{0}\hat{\gamma})\lambda_{-1}\langle u_{\delta},u_{0}\rangle+\langle
K^{(0)}_{D}u_{\delta},u_{-1}\rangle+(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})\langle
K^{(1)}_{D}u_{\delta},u_{-1}\rangle=\lambda_{\delta}\langle
u_{\delta},u_{-1}\rangle.$

Therefore, assuming that $u_{\delta}\approx u_{-1}$, we can see that

$\lambda_{\delta}\approx\log(\delta k_{0}\hat{\gamma})\lambda_{-1}+\langle
K^{(0)}_{D}u_{-1},u_{-1}\rangle+(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})\langle K^{(1)}_{D}u_{-1},u_{-1}\rangle,$

which is the desired result. ∎

The following corollary is a direct consequence of the above proposition.

######  Corollary 2.7.1.

Let $\lambda_{\delta}$ denote an eigenvalue of the operator $K^{\delta
k_{0}}_{D}$ in dimension 2. Then, for small $\delta$, it is approximately
given by

$\displaystyle\lambda_{\delta}\approx\log(\delta
k_{0}\hat{\gamma})\lambda_{-1}-\frac{\mathbb{P}}{2\pi}-\frac{i(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})\mathbb{G}}{4\pi},$ (2.29)

where $\mathbb{P}$ and $\mathbb{G}$ are constants that depend on $u_{-1}$.

###### Proof.

We observe that

$\displaystyle\langle K^{(0)}_{D}u_{-1},u_{-1}\rangle$
$\displaystyle=\int_{D}\left(-\frac{1}{2\pi}\int_{D}\log|x-y|u_{-1}(y)\mathrm{d}y\right)\overline{u_{-1}}(x)\mathrm{d}x$
$\displaystyle=-\frac{1}{2\pi}\int_{D}\int_{D}\log|x-y|u_{-1}(y)\overline{u_{-1}}(x)\mathrm{d}y\mathrm{d}x=:-\frac{1}{2\pi}\mathbb{P}.$

Then, for $u\in L^{2}(D)$,

$\displaystyle K^{(1)}_{D}[u](x)$
$\displaystyle=\int_{D}\frac{\partial}{\partial
k}G(x-y,k)\Big{|}_{k=0}u(y)\mathrm{d}y=\int_{D}\frac{\partial}{\partial
k}\left(-\frac{i}{4}H_{0}^{(1)}(k|x-y|)\right)\Big{|}_{k=0}u(y)\mathrm{d}y=-\frac{i}{4\pi}\int_{D}\frac{u(y)}{|x-y|}\mathrm{d}y,$

and so, we have

$\displaystyle\langle K^{(1)}_{D}u_{-1},u_{-1}\rangle$
$\displaystyle=\int_{D}\left(-\frac{i}{4\pi}\int_{D}\frac{u_{-1}(y)}{|x-y|}\mathrm{d}y\right)\overline{u_{-1}}(x)\mathrm{d}x=-\frac{i}{4\pi}\int_{D}\int_{D}\frac{u_{-1}(y)\overline{u_{-1}}(x)}{|x-y|}\mathrm{d}y\mathrm{d}x=:-\frac{i}{4\pi}\mathbb{G}.$

Hence, from (2.28), we obtain the desired result. ∎

####  2.5.2 Frequency and Wavenumber

Let us now find the frequency $\omega_{\delta}$ and the wavenumber
$k_{\delta}$ associated to this eigenvalue, which will also constitute the
basis of our analysis of the operator $K^{\delta k_{0}}_{D}$. We see that
(2.8) gives us that

$\displaystyle 1=\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}\ $
$\displaystyle\Leftrightarrow\
1=\delta^{2}\omega^{2}\mu_{0}(\varepsilon(\omega,k)-\varepsilon_{0})\lambda_{\delta}.$

Using the expression (2.28) for $\lambda_{\delta}$, we see that

$\displaystyle\varepsilon(\omega,k)=\frac{1}{\mu_{0}\delta^{2}\omega^{2}\left(\log(\delta
k_{0}\hat{\gamma})\lambda_{-1}-\frac{\mathbb{P}}{2\pi}-\frac{i(\delta
k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})\mathbb{G}}{4\pi}\right)}+\varepsilon_{0}.$

Arguing in the same way as in section 2.4.4 and comparing the two permittivity
expressions, we obtain the following system:

$\displaystyle\begin{cases}4\pi\alpha\delta^{2}\omega^{2}\mu_{0}\log(\delta
k_{0}\hat{\gamma})\lambda_{-1}-2\mathbb{P}\alpha\delta^{2}\omega^{2}\mu_{0}-4\pi\beta+4\pi\omega^{2}-4\pi\eta
k^{2}=0,\\\ -\alpha\delta^{4}\omega^{2}\mu_{0}k_{0}^{2}\log(\delta
k_{0}\hat{\gamma})\mathbb{G}+4\pi\gamma\omega=0.\end{cases}$ (2.30)

From the second equation in (2.30), we see that

$\displaystyle\omega(-\alpha\delta^{4}\omega\mu_{0}k_{0}^{2}\log(\delta
k_{0}\hat{\gamma})\mathbb{G}+4\pi\gamma)=0,$

which shows us that

$\displaystyle\omega=0\ \ \ \text{ or }\ \ \
\omega=\frac{4\pi\gamma}{\alpha\delta^{4}\mu_{0}k_{0}^{2}\log(\delta
k_{0}\hat{\gamma})\mathbb{G}}.$

For $\omega=\frac{4\pi\gamma}{\alpha\delta^{4}\mu_{0}k_{0}^{2}\log(\delta
k_{0}\hat{\gamma})\mathbb{G}}$, we obtain the equation

$\displaystyle 4\pi\alpha\delta^{2}\mu_{0}\lambda_{-1}\log(\delta
k_{0}\hat{\gamma})\frac{16\pi^{2}\gamma^{2}}{\alpha^{2}\delta^{8}\mu_{0}^{2}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})^{2}\mathbb{G}^{2}}$
$\displaystyle-2\mathbb{P}\alpha\delta^{2}\mu_{0}\frac{16\pi^{2}\gamma^{2}}{\alpha^{2}\delta^{8}\mu_{0}^{2}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})^{2}\mathbb{G}^{2}}$
$\displaystyle\quad-4\pi\beta+4\pi\frac{16\pi^{2}\gamma^{2}}{\alpha^{2}\delta^{8}\mu_{0}^{2}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})^{2}\mathbb{G}^{2}}-4\pi\eta k^{2}=0,$

which has solutions

$\displaystyle
k=\pm\sqrt{-\frac{\beta}{\eta}+\Big{(}2\pi\alpha\delta^{2}\mu_{0}\lambda_{-1}\log(\delta
k_{0}\hat{\gamma})-\alpha\delta^{2}\mu_{0}\mathbb{P}+2\pi\Big{)}\frac{8\pi\gamma^{2}}{\eta\alpha^{2}\delta^{8}\mu_{0}^{2}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})^{2}\mathbb{G}^{2}}}.$

Yet again, we discard the case of $\omega=0$, as there is no physical
interest. Denoting the frequency by $\omega_{\delta}$ and the wavenumber by
$\lambda_{\delta}$, we will work with

$\displaystyle\begin{split}&\omega_{\delta}=\frac{4\pi\gamma}{\alpha\delta^{4}\mu_{0}k_{0}^{2}\log(\delta
k_{0}\hat{\gamma})\mathbb{G}},\\\
&k_{\delta}=\sqrt{-\frac{\beta}{\eta}+\Big{(}2\pi\alpha\delta^{2}\mu_{0}\lambda_{-1}\log(\delta
k_{0}\hat{\gamma})-\alpha\delta^{2}\mu_{0}\mathbb{P}+2\pi\Big{)}\frac{8\pi\gamma^{2}}{\eta\alpha^{2}\delta^{8}\mu_{0}^{2}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})^{2}\mathbb{G}^{2}}},\end{split}$ (2.31)

where we have chosen the wavenumber to be positive.

####  2.5.3 Asymptotic Analysis

Let us consider $\omega$ near $\omega_{\delta}$, and define the coefficients

$\displaystyle c_{n}=\begin{cases}\log(\delta
k_{0}\hat{\gamma}),\quad&n=-1,\\\ 1,&n=0,\\\ (\delta k_{0})^{2n}\log(\delta
k_{0}\hat{\gamma}),&n\geq 1.\end{cases}$

Then, we can write

$K^{\delta k_{0}}_{D}=\sum_{n=-1}^{+\infty}c_{n}K^{(n)}_{D}.$

We are interested in studying the singularities of the operator
$(I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta k_{0}}_{D})^{-1}$. Setting
$B_{n}:=\delta^{2}\omega^{2}\xi(\omega,k)K^{(n)}_{D}$, we find that
$(I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta k_{0}}_{D})^{-1}=0$ can be
written as

$\displaystyle\left(I-\sum_{n=-1}^{+\infty}c_{n}B_{n}\right)^{-1}=0,$

which can be expanded to give

$\displaystyle\left(I-\log(\delta k_{0}\hat{\gamma})B_{-1}-B_{0}-(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})B_{1}-\sum_{n\geq
2}^{+\infty}c_{n}B_{n}\right)^{-1}=0,$

which gives

$\displaystyle\mathfrak{L}+\mathfrak{L}(\delta k_{0})^{4}\log(\delta
k_{0}\hat{\gamma})B_{2}\mathfrak{L}+O(\delta^{6})=0,$

where we have defined,

$\mathfrak{L}:=\Big{(}I-\log(\delta k_{0}\hat{\gamma})B_{-1}-B_{0}-(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})B_{1}\Big{)}^{-1}.$ (2.32)

Using this expression, we have the following proposition.

######  Proposition 2.8.

Let $d=2$ and let $\omega_{\delta}$ be defined by (2.31). Then, as $\delta\to
0$, the $O(\delta^{4})$ approximations of the subwavelength resonant
frequencies $\omega_{s}$ and the associated wavenumbers $k_{s}$ of the single
halide perovskite resonator $\Omega=\delta D+z$ satisfy

$\displaystyle
1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}=-\delta^{6}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})\omega_{s}^{2}\xi(\omega_{s},k_{s})\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle.$ (2.33)

###### Proof.

Applying a pole-pencil decomposition, we obtain

$\Big{(}I-\log(\delta k_{0}\hat{\gamma})B_{-1}-B_{0}-(\delta
k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})B_{1}\Big{)}^{-1}[\cdot]=\frac{\langle\cdot,u_{\delta}\rangle
u_{\delta}}{1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}}+R(\omega_{s})[\cdot],$

where the remainder $R(\omega_{s})$ is analytic in a neighborhood of
$\omega_{\delta}$ and can be dropped. Thus, dropping the $O(\delta^{6})$ term,
we find for $\psi\in L^{2}(D)$ that

$\displaystyle\Big{[}\mathfrak{L}+\mathfrak{L}(\delta k_{0})^{4}\log(\delta$
$\displaystyle k_{0}\hat{\gamma})B_{2}\mathfrak{L}\Big{]}(\psi)=0.$

Applying a pole-pencil decomposition on (2.32), we get

$\displaystyle\frac{\langle\psi,u_{\delta}\rangle
u_{\delta}}{1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}}+\frac{\langle\psi,u_{\delta}\rangle
u_{\delta}}{1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}}$
$\displaystyle(\delta k_{0})^{4}\log(\delta
k_{0}\hat{\gamma})B_{2}\frac{\langle\psi,u_{\delta}\rangle
u_{\delta}}{1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}}=0.$

This implies that

$\displaystyle
1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}=-\delta^{6}k_{0}^{4}$
$\displaystyle\log(\delta
k_{0}\hat{\gamma})\omega_{s}^{2}\xi(\omega_{s},k_{s})\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle,$

which is the desired result. ∎

In order to obtain the associated consequences of this proposition, we observe
that, since $d=2$, we have that

$\displaystyle K^{(2)}_{D}[u_{\delta}](x)$
$\displaystyle=\int_{D}\frac{\partial^{2}}{\partial
k^{2}}G(x-y,k)\Big{|}_{k=0}u_{\delta}(y)\mathrm{d}y=\int_{D}\frac{\partial^{2}}{\partial
k^{2}}\left(-\frac{i}{4}H^{(1)}_{0}(k|x|)\right)\Big{|}_{k=0}$
$\displaystyle=-\frac{i}{4}\int_{D}-\frac{1}{\pi|x-y|^{2}}u_{\delta}(y)\mathrm{d}y=\frac{i}{4\pi}\int_{D}\frac{u_{\delta}(y)}{|x-y|^{2}}\mathrm{d}y.$

Hence,

$\displaystyle\langle K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle$
$\displaystyle=\int_{D}\left(\frac{i}{4\pi}\int_{D}\frac{u_{\delta}(y)}{|x-y|^{2}}\mathrm{d}y\right)\overline{u_{\delta}}(x)\mathrm{d}x=\frac{i}{4\pi}\int_{D}\int_{D}\frac{u_{\delta}(y)\overline{u_{\delta}}(x)}{|x-y|^{2}}\mathrm{d}y\mathrm{d}x=:\frac{i}{4\pi}\mathbb{S}.$
(2.34)

So, we get that (2.33) is equivalent to

$\displaystyle
1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}=\frac{-i\delta^{6}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})\omega_{s}^{2}\xi(\omega_{s},k_{s})\mathbb{S}}{4\pi}.$
(2.35)

######  Remark.

In the Appendix A.2 of this paper, we recover a formula for $\mathbb{S}$.

Then, the next two corollaries follow as immediate results.

######  Corollary 2.8.1.

Let $d=2$. Then, as $\delta\to 0$, the $O(\delta^{10}\log(\delta)^{3})$
approximation of the subwavelength resonant frequencies of the halide
perovskite resonator $\Omega=\delta D+z$ can be computed as

$\displaystyle
1-\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}\cdot\frac{16\pi^{2}\gamma^{2}\lambda_{\delta}\Big{(}\varepsilon(\omega_{s},k_{s})-\varepsilon_{0}\Big{)}}{\alpha^{2}\delta^{6}\mu_{0}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})^{2}\mathbb{G}^{2}}=-\frac{i\delta^{6}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})\omega_{s}^{2}\xi(\omega_{s},k_{s})\mathbb{S}}{4\pi}.$
(2.36)

###### Proof.

By a direct computation, we observe that

$\displaystyle
1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}$
$\displaystyle\ =\
1-\delta^{2}\omega_{s}^{2}\mu_{0}\Big{(}\varepsilon(\omega_{s},k_{s})-\varepsilon_{0}\Big{)}\lambda_{\delta}\frac{\omega_{\delta}^{2}}{\omega_{\delta}^{2}}$
$\displaystyle\overset{(\ref{w_e k_e for d=2})}{=}\
1-\delta^{2}\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}\mu_{0}\Big{(}\varepsilon(\omega_{s},k_{s})-\varepsilon_{0}\Big{)}\lambda_{\delta}\frac{16\pi^{2}\gamma^{2}}{\alpha^{2}\delta^{8}\mu_{0}^{2}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})^{2}\mathbb{G}^{2}}$ $\displaystyle\ =\
1-\frac{\omega_{s}^{2}}{\omega_{\delta}^{2}}\cdot\frac{16\pi^{2}\gamma^{2}\lambda_{\delta}\Big{(}\varepsilon(\omega_{s},k_{s})-\varepsilon_{0}\Big{)}}{\alpha^{2}\delta^{6}\mu_{0}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})^{2}\mathbb{G}^{2}},$

and thus, (2.35) gives the desired result. ∎

######  Corollary 2.8.2.

Let $d=2$. Then, as $\delta\to 0$, the $O(\delta^{4}\log(\delta))$
approximation of the subwavelength resonant frequencies of the halide
perovskite resonator $\Omega=\delta D+z$ can be computed as

$\displaystyle
1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\lambda_{\delta}=-\frac{\omega_{s}^{2}}{\omega^{2}_{\varepsilon}}\cdot\frac{4\pi
i\mathbb{S}\gamma^{2}\Big{(}\varepsilon(\omega_{s},k_{s})-\varepsilon_{0}\Big{)}}{\alpha^{2}\delta^{2}\log(\delta
k_{0}\hat{\gamma})\mu_{0}\mathbb{G}^{2}}.$ (2.37)

###### Proof.

Again, to show this, we need to make a straightforward calculation:

$\displaystyle\frac{i\delta^{6}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})\omega_{s}^{2}\xi(\omega_{s},k_{s})\mathbb{S}}{4\pi}$
$\displaystyle\ =\ \frac{i}{4\pi}\ \delta^{6}\ k_{0}^{4}\ \log(\delta
k_{0}\hat{\gamma})\ \frac{\omega^{2}_{s}}{\omega_{\delta}^{2}}\ \mathbb{S}\
\mu_{0}\ \Big{(}\varepsilon(\omega_{s},k_{s})-\varepsilon_{0}\Big{)}\
\omega_{\delta}^{2}$ $\displaystyle\overset{(\ref{w_e k_e for d=2})}{=}\
\frac{i}{4\pi}\ \delta^{6}\ k_{0}^{4}\ \log(\delta k_{0}\hat{\gamma})\
\frac{\omega^{2}_{s}}{\omega_{\delta}^{2}}\ \mathbb{S}\ \mu_{0}\
\Big{(}\varepsilon(\omega_{s},k_{s})-\varepsilon_{0}\Big{)}\
\frac{16\pi^{2}\gamma^{2}}{\alpha^{2}\delta^{8}\mu_{0}^{2}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})^{2}\mathbb{G}^{2}}$ $\displaystyle\ =\
\frac{\omega_{s}^{2}}{\omega^{2}_{\varepsilon}}\cdot\frac{4\pi
i\mathbb{S}\gamma^{2}\Big{(}\varepsilon(\omega_{s},k_{s})-\varepsilon_{0}\Big{)}}{\alpha^{2}\delta^{2}\log(\delta
k_{0}\hat{\gamma})\mu_{0}\mathbb{G}^{2}}.$

Hence, (2.35) gives the desired result. ∎

We continue our analysis in the same way as in the previous case.

######  Proposition 2.9.

Let $d=2$. For $\omega$ real close to the resonant frequency $\omega_{s}$, the
following approximation for the field scattered by the halide perovskite nano-
particle $\Omega=\delta D+z$ holds:

$\displaystyle
u(x)-u_{in}(x)\approx\frac{4\pi\Big{(}1+\delta^{2}\omega^{2}\xi(\omega,k)G(x-y,\delta
k_{0})\Big{)}}{i\delta^{6}k_{0}^{4}\omega^{2}\log(\delta
k_{0}\hat{\gamma})\omega^{2}\xi(\omega,k)\mathbb{S}}\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}.$ (2.38)

###### Proof.

As we mentioned at the beginning or our analysis, our goal is to find $u\in
L^{2}(D)$, $u\neq 0$, such that (2.8) is satisfied. Using the pole-pencil
decomposition on this Lippmann-Schwinger formulation of the problem, we can
rewrite, as in Corollary 2.3 in [2],

$\displaystyle u(x)-u_{in}(x)\ \approx-\delta^{2}\omega^{2}\xi(\omega,k)G(x-$
$\displaystyle y,\delta k_{0})\frac{\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}$
$\displaystyle+\delta^{6}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})\omega^{2}\xi(\omega,k)\frac{\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{\Big{(}1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}\Big{)}^{2}},$

which, using (2.35), becomes:

$\displaystyle u(x)-u_{in}(x)$ $\displaystyle\ \approx\ \frac{G(x-y,\delta
k_{0})\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{\delta^{4}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle}+\frac{\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta}}{\delta^{6}k_{0}^{4}\omega^{2}\log(\delta
k_{0}\hat{\gamma})\omega^{2}\xi(\omega,k)\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle}.$

From (2.34), this gives

$\displaystyle u(x)-u_{in}(x)$ $\displaystyle\ \approx\
\frac{4\pi\Big{(}1+\delta^{2}\omega^{2}\xi(\omega,k)G(x-y,\delta
k_{0})\Big{)}}{i\delta^{6}k_{0}^{4}\omega^{2}\log(\delta
k_{0}\hat{\gamma})\omega^{2}\xi(\omega,k)\mathbb{S}}\langle
u_{in},u_{\delta}\rangle\int_{D}u_{\delta},$

which is the desired result. ∎

We finish our analysis with the following result.

######  Proposition 2.10.

Let $d=2$ and $\delta$ be small enough. Then, the $o(\delta^{4})$
approximation of the subwavelength resonant frequencies $\omega_{s}$ of the
halide perovskite nano-particle $\Omega=\delta D+z$ satisfies

$\displaystyle
1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\left(-\frac{|D|}{2\pi}\log(\delta
k_{0}\hat{\gamma})+\langle
K^{(0)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle+\delta^{2}k_{0}^{2}\log(\delta)\langle
K^{(1)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle\Big{)}\right)=O\big{(}\delta^{4}\big{)},$
(2.39)

where $|D|$ is the volume of $D$ and
$\hat{\mathbb{I}}_{D}=\mathbb{I}_{D}/\sqrt{|D|}$.

###### Proof.

We want to find $\omega_{s}\in\mathbb{C}$ such that

$\displaystyle\Big{(}I-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})K^{\delta
k_{0}}_{D}\Big{)}[u](x)=0,$

which, for small $\delta$, can be written as

$\displaystyle\Big{(}I-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\Big{(}$
$\displaystyle\log(\delta k_{0}\hat{\gamma})K^{(-1)}_{D}+K^{(0)}_{D}+(\delta
k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})K^{(1)}_{D}\Big{)}\Big{)}[u](x)=O\Big{(}\delta^{6}\log(\delta)\Big{)}.$

Let us denote

$M^{\delta k_{0}}_{D}:=\log(\delta
k_{0}\hat{\gamma})K^{(-1)}_{D}+K^{(0)}_{D}+(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})K^{(1)}_{D}.$

We take $\nu(\delta)\in\sigma(M^{\delta k_{0}}_{D})$ and consider the
eigenvalue problem for $M^{\delta k_{0}}_{D}$:

$\displaystyle M^{\delta k_{0}}_{D}[\Psi]=\nu(\delta)\Psi.$ (2.40)

We employ the ansatz

$\displaystyle\Psi(\delta)$
$\displaystyle=\Psi_{0}+O\left(\frac{1}{\log(\delta)}\right),$
$\displaystyle\nu(\delta)$
$\displaystyle=\log(\delta)\nu_{0}+\nu_{1}+\delta^{2}\log(\delta)\nu_{2}+O\Big{(}\delta^{2}\Big{)}.$

From (2.40) and the fact that $\delta^{2}\log(\delta
k_{0}\hat{\gamma})=\delta^{2}\log(\delta)+O(\delta^{2})$ we have that

$\displaystyle\Big{(}\log(\delta)K^{(-1)}_{D}+\log(k_{0}\hat{\gamma})K^{(-1)}_{D}+K^{(0)}_{D}$
$\displaystyle+(\delta k_{0})^{2}\log(\delta)K^{(1)}_{D}\Big{)}[\Psi_{0}]=$
$\displaystyle\Big{(}\log(\delta)\nu_{0}+\nu_{1}+\delta^{2}\log(\delta)\nu_{2}\Big{)}[\Psi_{0}]+O\Big{(}\delta^{2}\Big{)}.$

Equating the $O(\log\delta)$ terms gives

$\displaystyle K^{(-1)}_{D}[\Psi_{0}]=\nu_{0}\Psi_{0}$
$\displaystyle\Rightarrow\nu_{0}\hat{\mathbb{I}}_{D}=K^{(-1)}_{D}[\hat{\mathbb{I}}_{D}]\Rightarrow\nu_{0}\hat{\mathbb{I}}_{D}=-\frac{|D|}{2\pi}\hat{\mathbb{I}}_{D}\Rightarrow\nu_{0}=-\frac{|D|}{2\pi}.$

Then, equating the $O(1)$ terms gives

$\displaystyle\log(k_{0}\hat{\gamma})K^{(-1)}_{D}[\hat{\mathbb{I}}_{D}]$
$\displaystyle+K^{(0)}_{D}[\hat{\mathbb{I}}_{D}]=\nu_{1}\hat{\mathbb{I}}_{D}\Rightarrow\nu_{1}\hat{\mathbb{I}}_{D}=-\frac{|D|}{2\pi}\log(k_{0}\hat{\gamma})\hat{\mathbb{I}}_{D}+K^{(0)}_{D}[\hat{\mathbb{I}}_{D}]$
$\displaystyle\Rightarrow\nu_{1}=-\frac{|D|}{2\pi}\log(k_{0}\hat{\gamma})+\langle
K^{(0)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle.$

Using the same reasoning for the $O\Big{(}\delta^{2}\log(\delta)\Big{)}$
terms, we get

$\displaystyle\nu_{2}\hat{\mathbb{I}}_{D}=k_{0}^{2}K^{(1)}_{D}[\hat{\mathbb{I}}_{D}]\Rightarrow\nu_{2}=k_{0}^{2}\langle
K^{(1)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle.$

Thus,

$\displaystyle\nu(\delta)$
$\displaystyle=\log(\delta)\nu_{0}+\nu_{1}+\delta^{2}\log(\delta)\nu_{2}+O\Big{(}\delta^{2}\Big{)}$
$\displaystyle=-\frac{|D|}{2\pi}\log(\delta)-\frac{|D|}{2\pi}\log(k_{0}\hat{\gamma})+\langle
K^{(0)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle+\delta^{2}k_{0}^{2}\log(\delta)\langle
K^{(1)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle+O\Big{(}\delta^{2})\Big{)}$
$\displaystyle=-\frac{|D|}{2\pi}\log(\delta k_{0}\hat{\gamma})+\langle
K^{(0)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle+\delta^{2}k_{0}^{2}\log(\delta)\langle
K^{(1)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle+O\Big{(}\delta^{2}\Big{)}.$

Using these expressions,
$1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})M^{\delta
k_{0}}_{D}=O(\delta^{6}\log(\delta))$ can be rewritten as

$\displaystyle
1-\delta^{2}\omega_{s}^{2}\xi(\omega_{s},k_{s})\Big{(}-\frac{|D|}{2\pi}\log(\delta
k_{0}\hat{\gamma})+\langle
K^{(0)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle+\delta^{2}k_{0}^{2}\log(\delta)\langle
K^{(1)}_{D}[\hat{\mathbb{I}}_{D}],\hat{\mathbb{I}}_{D}\rangle$
$\displaystyle\hskip
312.9803pt+O\Big{(}\delta^{2}\Big{)}\Big{)}=O\Big{(}\delta^{6}\log(\delta)\Big{)},$

from which the result follows. ∎

##  3 Hybridization of Two Resonators

###  3.1 Three Dimensions

Let us consider two identical halide perovskite resontators $D_{1}$ and
$D_{2}$ (e.g. the speres in Figure 1), made from the same material. From now
on, we will denote the permittivity by $\xi(\omega,k)$, where $\omega$ is the
frequency and $k$ the associated wavenumber. In order to generalize our
results, we will define it by

$\displaystyle\xi(\omega,k):=\frac{\mu_{0}\alpha}{\beta-\omega^{2}+\eta
k^{2}-i\gamma\omega},$ (3.1)

where the positive constants $\alpha,\beta,\gamma$ and $\eta$ characterise the
material.

Figure 1: Two identical spherical halide perovskite resontators $D_{1}$ and
$D_{2}$ of radius $\delta$, made from the same material, placed at a distance
$d$ from each other.

Then, since there is an interaction between the two resonators, the field
$u-u_{in}$ scattered by the two particles will be given by the following
representation formula:

$\displaystyle(u-u_{in})(x)=-\delta^{2}\omega^{2}\xi(\omega,k)\left[\int_{D_{1}}G(x-y,\delta
k_{0})u(y)\mathrm{d}y+\int_{D_{2}}G(x-y,\delta k_{0})u(y)\mathrm{d}y\right]\ \
\ \text{ for }x\in\mathbb{R}^{d}.$ (3.2)

######  Definition 3.1.

We define the integral operators $K^{\delta k_{0}}_{D_{i}}$ and $R^{\delta
k_{0}}_{D_{i}D_{j}}$, for $i,j=1,2$, by

$\displaystyle K^{\delta k_{0}}_{D_{i}}:u\big{|}_{D_{i}}\in
L^{2}(D_{i})\longmapsto-\int_{D_{i}}G(x-y,\delta
k_{0})u(y)\mathrm{d}y\Big{|}_{D_{i}}\in L^{2}(D_{i})$

and

$\displaystyle R^{\delta k_{0}}_{D_{i}D_{j}}:u\big{|}_{D_{i}}\in
L^{2}(D_{i})\longmapsto-\int_{D_{i}}G(x-y,\delta
k_{0})u(y)\mathrm{d}y\Big{|}_{D_{j}}\in L^{2}(D_{j}).$

Then, the following lemma is a direct consequence of these definitions.

######  Lemma 3.2.

The scattering problem (3.2) can be restated, using the Definition 3.1, as

$\displaystyle\begin{pmatrix}1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{1}}&-\delta^{2}\omega^{2}\xi(\omega,k)R^{\delta
k_{0}}_{D_{2}D_{1}}\\\ -\delta^{2}\omega^{2}\xi(\omega,k)R^{\delta
k_{0}}_{D_{1}D_{2}}&1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{2}}\end{pmatrix}\begin{pmatrix}u|_{D_{1}}\\\
u|_{D_{2}}\end{pmatrix}=\begin{pmatrix}u_{in}|_{D_{1}}\\\
u_{in}|_{D_{2}}\end{pmatrix}.$ (3.3)

Thus, the scattering resonance problem is to find $\omega$ such that the
operator in (3.3) is singular, or equivalently, such that there exists
$(u_{1},u_{2})\in L^{2}(D_{1})\times L^{2}(D_{2})$, $(u_{1},u_{2})\neq 0$,
such that

$\displaystyle\begin{pmatrix}1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{1}}&-\delta^{2}\omega^{2}\xi(\omega,k)R^{\delta
k_{0}}_{D_{2}D_{1}}\\\ -\delta^{2}\omega^{2}\xi(\omega,k)R^{\delta
k_{0}}_{D_{1}D_{2}}&1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{2}}\end{pmatrix}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}=\begin{pmatrix}0\\\ 0\end{pmatrix}.$ (3.4)

######  Theorem 3.3.

Let $d=3$. Then, the hybridized subwavelength resonant frequencies $\omega$
satisfy

$\Big{(}1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}\Big{)}^{2}-\delta^{4}\omega^{4}\xi(\omega,k)^{2}\langle
R^{\delta
k_{0}}_{D_{1}D_{2}}\phi_{1}^{(\delta)},\phi_{2}^{(\delta)}\rangle\langle
R^{\delta
k_{0}}_{D_{2}D_{1}}\phi_{2}^{(\delta)},\phi_{1}^{(\delta)}\rangle=0,$ (3.5)

where $\phi_{i}^{(\delta)}$, for $i=1,2$, is the eigenfunction associated to
the eigenvalue $\lambda_{\delta}$ of the potential $K^{\delta k_{0}}_{D_{i}}$.

###### Proof.

We observe that (3.4) is equivalent to

$\displaystyle\begin{pmatrix}1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{1}}&0\\\ 0&1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{2}}\end{pmatrix}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}-\delta^{2}\omega^{2}\xi(\omega,k)\begin{pmatrix}0&R^{\delta
k_{0}}_{D_{2}D_{1}}\\\ R^{\delta
k_{0}}_{D_{1}D_{2}}&0\end{pmatrix}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}=0,$

which gives

$\displaystyle\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}-\delta^{2}\omega^{2}\xi(\omega,k)\begin{pmatrix}\Big{(}1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{1}}\Big{)}^{-1}&0\\\
0&\Big{(}1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{2}}\Big{)}^{-1}\end{pmatrix}\begin{pmatrix}R^{\delta
k_{0}}_{D_{2}D_{1}}u_{2}\\\ R^{\delta
k_{0}}_{D_{1}D_{2}}u_{1}\end{pmatrix}=0.$ (3.6)

Let us now apply a pole-pencil decomposition on the operators
$\Big{(}1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{i}}\Big{)}^{-1}$, for $i=1,2$. We see that

$\displaystyle\Big{(}1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{1}}\Big{)}^{-1}(\cdot)=\frac{\langle\cdot,\phi_{1}^{(\delta)}\rangle\phi_{1}^{(\delta)}}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}+R_{1}[\omega](\cdot)$

and

$\displaystyle\Big{(}1-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{2}}\Big{)}^{-1}(\cdot)=\frac{\langle\cdot,\phi_{2}^{(\delta)}\rangle\phi_{2}^{(\delta)}}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}+R_{2}[\omega](\cdot),$

where the remainder terms $R_{1}[\omega](\cdot)$ and $R_{2}[\omega](\cdot)$
are holomorphic for $\omega$ in a neighborhood of $\omega_{\delta}$, so can be
neglected. Then, (3.6), is equivalent to

$\displaystyle\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}-\delta^{2}\omega^{2}\xi(\omega,k)\begin{pmatrix}\frac{\langle\cdot,\phi_{1}^{(\delta)}\rangle\phi_{1}^{(\delta)}}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}&0\\\
0&\frac{\langle\cdot,\phi_{2}^{(\delta)}\rangle\phi_{2}^{(\delta)}}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}\end{pmatrix}\begin{pmatrix}R^{\delta
k_{0}}_{D_{2}D_{1}}u_{2}\\\ R^{\delta
k_{0}}_{D_{1}D_{2}}u_{1}\end{pmatrix}=0.$

This gives us the system

$\displaystyle\begin{cases}u_{1}-\frac{\delta^{2}\omega^{2}\xi(\omega,k)}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}\langle
R^{\delta
k_{0}}_{D_{2}D_{1}}u_{2},\phi_{1}^{(\delta)}\rangle\phi_{1}^{(\delta)}=0,\\\
u_{2}-\frac{\delta^{2}\omega^{2}\xi(\omega,k)}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}\langle
R^{\delta
k_{0}}_{D_{1}D_{2}}u_{1},\phi_{2}^{(\delta)}\rangle\phi_{2}^{(\delta)}=0.\end{cases}$

Applying the operator $R^{\delta k_{0}}_{D_{1}D_{2}}$ (resp. $R^{\delta
k_{0}}_{D_{2}D_{1}}$) to the first (resp. second) equation, and then applying
$\langle\cdot,\phi_{2}^{(\varepsilon)}\rangle$ (resp.
$\langle\cdot,\phi_{1}^{(\varepsilon)}\rangle$), we find that

$\displaystyle\begin{cases}\langle R^{\delta
k_{0}}_{D_{1}D_{2}}u_{1},\phi_{2}^{(\delta)}\rangle-\frac{\delta^{2}\omega^{2}\xi(\omega,k)}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}\langle
R^{\delta
k_{0}}_{D_{1}D_{2}}\phi_{1}^{(\delta)},\phi_{2}^{(\delta)}\rangle\langle
R^{\delta k_{0}}_{D_{2}D_{1}}u_{2},\phi_{1}^{(\delta)}\rangle=0,\\\ \langle
R^{\delta
k_{0}}_{D_{2}D_{1}}u_{2},\phi_{1}^{(\delta)}\rangle-\frac{\delta^{2}\omega^{2}\xi(\omega,k)}{1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}}\langle
R^{\delta
k_{0}}_{D_{2}D_{1}}\phi_{2}^{(\delta)},\phi_{1}^{(\delta)}\rangle\langle
R^{\delta k_{0}}_{D_{1}D_{2}}u_{1},\phi_{2}^{(\delta)}\rangle=0.\end{cases}$

This system has a solution only if its determinant is zero. That is, if

$\displaystyle
1-\frac{\delta^{4}\omega^{4}\xi(\omega,k)^{2}}{\Big{(}1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}\Big{)}^{2}}\langle
R^{\delta
k_{0}}_{D_{1}D_{2}}\phi_{1}^{(\delta)},\phi_{2}^{(\delta)}\rangle\langle
R^{\delta
k_{0}}_{D_{2}D_{1}}\phi_{2}^{(\delta)},\phi_{1}^{(\delta)}\rangle=0,$

which gives the desired result. ∎

The following corollary is a direct result of Theorem 3.3.

######  Corollary 3.3.1.

Let $d=3$. Then, the hybridized subwavelength resonant frequencies are given
by

$\displaystyle\omega=\frac{i\gamma\pm\sqrt{-\gamma^{2}-4\Gamma(\beta+\eta
k^{2})}}{2\Gamma},$ (3.7)
$\displaystyle\text{where}\qquad\Gamma=-1-\delta^{2}\alpha\lambda_{\delta}\pm\alpha\delta^{2}\sqrt{\langle
R^{\delta
k_{0}}_{D_{1}D_{2}}\phi_{1}^{(\delta)},\phi_{2}^{(\delta)}\rangle\langle
R^{\delta k_{0}}_{D_{2}D_{1}}\phi_{2}^{(\delta)},\phi_{1}^{(\delta)}\rangle}.$

where $\phi_{i}^{\delta}$, for $i=1,2$, is the eigenfunction associated to the
eigenvalue $\lambda_{\delta}$ of the potential $K^{\delta k_{0}}_{D_{i}}$ and
the $\pm$ in the two expressions do not have to agree.

###### Proof.

We introduce the notation $\mathbb{K}:=\langle R^{\delta
k_{0}}_{D_{1}D_{2}}\phi_{1}^{(\delta)},\phi_{2}^{(\delta)}\rangle$ and
$\mathbb{M}:=\langle R^{\delta
k_{0}}_{D_{2}D_{1}}\phi_{2}^{(\delta)},\phi_{1}^{(\delta)}\rangle$. Then,
(3.5) becomes

$\displaystyle\Big{(}1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}$
$\displaystyle\Big{)}^{2}-\delta^{4}\omega^{4}\xi(\omega,k)^{2}\mathbb{K}\mathbb{M}=0\
\Leftrightarrow\
1-\delta^{2}\omega^{2}\xi(\omega,k)\lambda_{\delta}\pm\delta^{2}\omega^{2}\xi(\omega,k)\sqrt{\mathbb{K}\mathbb{M}}=0\
\Leftrightarrow$
$\displaystyle\left(-1-\delta^{2}\alpha\lambda_{\delta}\pm\alpha\delta^{2}\sqrt{\mathbb{K}\mathbb{M}}\right)\omega^{2}-i\gamma\omega+\beta+\eta
k^{2}=0,$

and the roots to this second degree polynomial are given by

$\displaystyle\omega=\frac{i\gamma\pm\sqrt{-\gamma^{2}-4\Gamma(\beta+\eta
k^{2})}}{2\Gamma}\quad\text{where}\quad\Gamma=-1-\delta^{2}\alpha\lambda_{\delta}\pm\alpha\delta^{2}\sqrt{\mathbb{K}\mathbb{M}},$

with the two $\pm$ not necessarily agreeing. Finally, substituting the
expressions for $\mathbb{K}$ and $\mathbb{M}$, we obtain the result. ∎

###  3.2 Two Dimensions

Let us move on to the case of dimension $d=2$. For simplicity, we again
consider two identical halide perovskite resontators $D_{1}$ and $D_{2}$, made
from the same material with permittivity given by the formula (3.1). We define
the operators $K^{\delta k_{0}}_{D_{i}}$ and $R^{\delta k_{0}}_{D_{i}D_{j}}$,
for $i,j=1,2$, as in Definition 3.1 and we continue by defining the following
integral operators.

######  Definition 3.4.

We define the integral operators $M^{\delta k_{0}}_{D_{i}}$ and $N^{\delta
k_{0}}_{D_{i}D_{j}}$ for $i,j=1,2$ as

$\displaystyle M^{\delta k_{0}}_{D_{i}}:=\hat{K}^{\delta
k_{0}}_{D_{i}}+K^{(0)}_{D_{i}}+(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})K^{(1)}_{D_{i}},$

and

$\displaystyle N^{\delta k_{0}}_{D_{i}D_{j}}:=\hat{K}^{\delta
k_{0}}_{D_{i}D_{j}}+R^{(0)}_{D_{i}D_{j}}+(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})R^{(1)}_{D_{i}D_{j}},$

where

$\displaystyle K^{(0)}_{D_{i}}:u\Big{|}_{D_{i}}\in L^{2}(D_{i})$
$\displaystyle\longmapsto\int_{D_{i}}G(x-y,0)u(y)\mathrm{d}y\Big{|}_{D_{i}}\in
L^{2}(D_{i}),$ $\displaystyle\hat{K}^{\delta
k_{0}}_{D_{i}}:u\Big{|}_{D_{i}}\in L^{2}(D_{i})$
$\displaystyle\longmapsto\log(\hat{\gamma}\delta
k_{0})\hat{K}_{D_{i}}[u]\Big{|}_{D_{i}}\in L^{2}(D_{i}),$
$\displaystyle\hat{K}_{D_{i}}:u\Big{|}_{D_{i}}\in L^{2}(D_{i})$
$\displaystyle\longmapsto-\frac{1}{2\pi}\int_{D_{i}}u(y)\mathrm{d}y\Big{|}_{D_{i}}\in
L^{2}(D_{i}),$ $\displaystyle K^{(1)}_{D_{i}}:u\Big{|}_{D_{i}}\in
L^{2}(D_{i})$ $\displaystyle\longmapsto\int_{D_{i}}\frac{\partial}{\partial
k}G(x-y,k)\Big{|}_{k=0}u(y)\mathrm{d}y\Big{|}_{D_{i}}\in L^{2}(D_{i}),$

and

$\displaystyle R^{(0)}_{D_{i}D_{j}}:u\Big{|}_{D_{i}}\in L^{2}(D_{i})$
$\displaystyle\longmapsto\int_{D_{i}}G(x-y,0)u(y)\mathrm{d}y\Big{|}_{D_{j}}\in
L^{2}(D_{j}),$ $\displaystyle\hat{K}^{\delta
k_{0}}_{D_{i}D_{j}}:u\Big{|}_{D_{i}}\in L^{2}(D_{i})$
$\displaystyle\longmapsto\log(\hat{\gamma}\delta
k_{0})\hat{K}_{D_{i}D_{j}}[u]\Big{|}_{D_{j}}\in L^{2}(D_{j}),$
$\displaystyle\hat{K}_{D_{i}D_{j}}:u\Big{|}_{D_{i}}\in L^{2}(D_{i})$
$\displaystyle\longmapsto-\frac{1}{2\pi}\int_{D_{i}}u(y)\mathrm{d}y\Big{|}_{D_{j}}\in
L^{2}(D_{j}),$ $\displaystyle R^{(1)}_{D_{i}D_{j}}:u\Big{|}_{D_{i}}\in
L^{2}(D_{i})$ $\displaystyle\longmapsto\int_{D_{i}}\frac{\partial}{\partial
k}G(x-y,k)\Big{|}_{k=0}u(y)\mathrm{d}y\Big{|}_{D_{j}}\in L^{2}(D_{j}).$

We observe the following result.

######  Proposition 3.5.

For the integral operators $K^{\delta k_{0}}_{D_{i}}$ and $R^{\delta
k_{0}}_{D_{i}D_{j}}$, we can write

$\displaystyle K^{\delta k_{0}}_{D_{i}}=M^{\delta
k_{0}}_{D_{i}}+O\Big{(}\delta^{4}\log(\delta)\Big{)},\quad\text{and}\quad
R^{\delta k_{0}}_{D_{i}D_{j}}=N^{\delta
k_{0}}_{D_{i}D_{j}}+O\Big{(}\delta^{4}\log(\delta)\Big{)},$ (3.8)

as $\delta\to 0$ and with $k_{0}$ fixed.

###### Proof.

The proof is a direct result of the expansion of the Green’s function in
dimension $d=2$. Indeed, for $u|_{D_{i}}\in L^{2}(D_{i})$, we observe that

$\displaystyle K^{\delta k_{0}}_{D_{i}}[u](x)$
$\displaystyle=-\int_{D_{i}}G(x-y,\delta k_{0})u(y)\mathrm{d}y\Big{|}_{D_{i}}$
$\displaystyle=-\int_{D_{i}}\Big{(}\log(\hat{\gamma}\delta
k_{0})\frac{1}{2\pi}+G(x-y,0)+(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})\frac{\partial}{\partial k}G(x-y,k)\Big{|}_{k=0}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \
+O\Big{(}\delta^{4}\log(\delta)\Big{)}\Big{)}u(y)\mathrm{d}y\Big{|}_{D_{i}}$
$\displaystyle=\Big{(}\hat{K}^{\delta k_{0}}_{D_{i}}+K^{(0)}_{D_{i}}+(\delta
k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})K^{(1)}_{D_{i}}\Big{)}[u](x)+O\Big{(}\delta^{4}\log(\delta)\Big{)}$
$\displaystyle=M^{\delta
k_{0}}_{D_{i}}[u](x)+O\Big{(}\delta^{4}\log(\delta)\Big{)}.$

Similarly, for $u|_{D_{i}}\in L^{2}(D_{i})$,

$\displaystyle R^{\delta k_{0}}_{D_{i}D_{j}}[u](x)$
$\displaystyle=-\int_{D_{i}}G(x-y,\delta
k_{0})u(y)\mathrm{d}y\Big{|}_{D_{j}}\in L^{2}(D_{j})$
$\displaystyle=-\int_{D_{i}}\Big{(}\log(\hat{\gamma}\delta
k_{0})\frac{1}{2\pi}+G(x-y,0)+(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})\frac{\partial}{\partial k}G(x-y,k)\Big{|}_{k=0}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \
+O\Big{(}\delta^{4}\log(\delta)\Big{)}\Big{)}u(y)\mathrm{d}y\Big{|}_{D_{j}}$
$\displaystyle=\Big{(}\hat{K}^{\delta
k_{0}}_{D_{i}D_{j}}+R^{(0)}_{D_{i}D_{j}}+(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})R^{(1)}_{D_{i}D_{j}}\Big{)}[u](x)+O\Big{(}\delta^{4}\log(\delta)\Big{)}$
$\displaystyle=N^{\delta k_{0}}_{D_{i}D_{j}}[u](x)+O(\delta^{4}\log(\delta)).$

∎

Therefore, our problem is to determine the frequencies $\omega$ and the
associated wavenumber $k$, for which the following holds:

$\displaystyle\begin{pmatrix}I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{1}}&-\delta^{2}\omega^{2}\xi(\omega,k)R^{\delta
k_{0}}_{D_{2}D_{1}}\\\ -\delta^{2}\omega^{2}\xi(\omega,k)R^{\delta
k_{0}}_{D_{1}D_{2}}&I-\delta^{2}\omega^{2}\xi(\omega,k)K^{\delta
k_{0}}_{D_{2}}\end{pmatrix}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}=\begin{pmatrix}0\\\ 0\end{pmatrix}$ (3.9)

for nontrivial $u:=(u_{1},u_{2})$, such that $u|_{D_{i}}\in L^{2}(D_{i})$, for
$i=1,2$.

######  Proposition 3.6.

Let $d=2$. Then, the hybridized subwavelength resonant frequencies $\omega$
satisfy

$\displaystyle\begin{split}1-\delta^{2}\omega^{2}\xi(\omega,k)&\Big{(}-\frac{|D_{1}|}{2\pi}\log(\hat{\gamma}\delta
k_{0})(1\pm 1)+\langle
K^{(0)}_{D_{1}}[\hat{\mathbb{I}}_{D_{1}}],\hat{\mathbb{I}}_{D_{1}}\rangle+(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})\langle
K^{(1)}_{D_{1}}[\hat{\mathbb{I}}_{D_{1}}],\hat{\mathbb{I}}_{D_{1}}\rangle\\\
&\pm\langle
R^{(0)}_{D_{2}D_{1}}[\hat{\mathbb{I}}_{D_{2}}],\hat{\mathbb{I}}_{D_{1}}\rangle\pm(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})\langle
R^{(1)}_{D_{2}D_{1}}[\hat{\mathbb{I}}_{D_{2}}],\hat{\mathbb{I}}_{D_{1}}\rangle\Big{)}=0,\end{split}$
(3.10)

where the $\pm$ symbols coincide.

###### Proof.

The first thing that we do is to observe that, by applying the expansion (3.8)
to (3.9), we reach the problem

$\displaystyle\begin{pmatrix}I-\delta^{2}\omega^{2}\xi(\omega,k)M^{\delta
k_{0}}_{D_{1}}&-\delta^{2}\omega^{2}\xi(\omega,k)N^{\delta
k_{0}}_{D_{2}D_{1}}\\\ -\delta^{2}\omega^{2}\xi(\omega,k)N^{\delta
k_{0}}_{D_{1}D_{2}}&I-\delta^{2}\omega^{2}\xi(\omega,k)M^{\delta
k_{0}}_{D_{2}}\end{pmatrix}\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}=\begin{pmatrix}O\Big{(}\delta^{4}\log(\delta)\Big{)}\\\
O\Big{(}\delta^{4}\log(\delta)\Big{)}\end{pmatrix}.$

We note that $|D_{1}|=|D_{2}|$. Then, using the symmetries of the dimer, let
us denote

$\displaystyle\hat{\nu}(\delta):$
$\displaystyle=-\frac{|D_{1}|}{2\pi}\log(\delta k_{0}\hat{\gamma})+\langle
K^{(0)}_{D_{1}}[\hat{\mathbb{I}}_{D_{1}}],\hat{\mathbb{I}}_{D_{1}}\rangle+(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})\langle
K^{(1)}_{D_{1}}[\hat{\mathbb{I}}_{D_{1}}],\hat{\mathbb{I}}_{D_{1}}\rangle$
$\displaystyle=-\frac{|D_{2}|}{2\pi}\log(\delta k_{0}\hat{\gamma})+\langle
K^{(0)}_{D_{2}}[\hat{\mathbb{I}}_{D_{2}}],\hat{\mathbb{I}}_{D_{2}}\rangle+(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})\langle
K^{(1)}_{D_{2}}[\hat{\mathbb{I}}_{D_{2}}],\hat{\mathbb{I}}_{D_{2}}\rangle,$

and

$\displaystyle\hat{\eta}:$ $\displaystyle=\langle N^{\delta
k_{0}}_{D_{1}D_{2}}[\hat{\mathbb{I}}_{D_{1}}],\hat{\mathbb{I}}_{D_{2}}\rangle=\langle
N^{\delta
k_{0}}_{D_{2}D_{1}}[\hat{\mathbb{I}}_{D_{2}}],\hat{\mathbb{I}}_{D_{1}}\rangle.$

In addition, we have that

$\displaystyle\hat{K}^{\delta
k_{0}}_{D_{i}D_{j}}[\hat{\mathbb{I}}_{D_{i}}]=\hat{K}^{\delta
k_{0}}_{D_{j}}[\hat{\mathbb{I}}_{D_{j}}].$

Now, we define the quantity $\nu(\delta)$ to be the eigenvalues of the
operator $M^{\delta k_{0}}_{D_{i}}$, that is,

$\displaystyle\nu(\delta)=\langle M^{\delta
k_{0}}_{D_{1}}[\Psi_{D_{1}}],\Psi_{D_{1}}\rangle=\langle M^{\delta
k_{0}}_{D_{2}}[\Psi_{D_{2}}],\Psi_{D_{2}}\rangle,$

for the eigenfunctions $\Psi_{D_{i}}(\delta)\in L^{2}(D_{i})$,
$\Psi_{D_{i}}(\delta)=\hat{\mathbb{I}}_{D_{i}}+O\left(\frac{1}{\log(d)}\right)$.
Thus, we have that (3.9) is equivalent to

$\displaystyle\begin{pmatrix}u_{1}\\\
u_{2}\end{pmatrix}-\delta^{2}\omega^{2}\xi(\omega,k)\begin{pmatrix}\Big{(}I-\delta^{2}\omega^{2}\xi(\omega,k)M^{\delta
k_{0}}_{D_{1}}\Big{)}^{-1}&0\\\
0&\Big{(}I-\delta^{2}\omega^{2}\xi(\omega,k)M^{\delta
k_{0}}_{D_{2}}\Big{)}^{-1}\end{pmatrix}\begin{pmatrix}N^{\delta
k_{0}}_{D_{2}D_{1}}u_{2}\\\ N^{\delta
k_{0}}_{D_{1}D_{2}}u_{1}\end{pmatrix}=0.$ (3.11)

Applying a pole-pencil decomposition, we observe that

$\displaystyle\Big{(}I-\delta^{2}\omega^{2}\xi(\omega,k)M^{\delta
k_{0}}_{D_{i}}\Big{)}^{-1}[\cdot]=\frac{\langle\cdot,\hat{\mathbb{I}}_{D_{i}}\rangle\hat{\mathbb{I}}_{D_{i}}}{1-\delta^{2}\omega^{2}\xi(\omega,k)\nu(\delta)}+R[\omega](\cdot),$

where the remainder terms $R[\omega](\cdot)$ can be neglected. Hence, (3.11)
is equivalent to

$\displaystyle\begin{cases}u_{1}-\delta^{2}\omega^{2}\xi(\omega,k)\frac{\langle
N^{\delta
k_{0}}_{D_{2}D_{1}}u_{2},\hat{\mathbb{I}}_{D_{1}}\rangle\hat{\mathbb{I}}_{D_{1}}}{1-\delta^{2}\omega^{2}\xi(\omega,k)\nu(\delta)}=0,\\\
u_{2}-\delta^{2}\omega^{2}\xi(\omega,k)\frac{\langle N^{\delta
k_{0}}_{D_{1}D_{2}}u_{1},\hat{\mathbb{I}}_{D_{2}}\rangle\hat{\mathbb{I}}_{D_{2}}}{1-\delta^{2}\omega^{2}\xi(\omega,k)\nu(\delta)}=0,\end{cases}$

which is equivalent to

$\displaystyle\begin{cases}\langle N^{\delta
k_{0}}_{D_{1}D_{2}}u_{1},\hat{\mathbb{I}}_{D_{2}}\rangle-\frac{\delta^{2}\omega^{2}\xi(\omega,k)}{1-\delta^{2}\omega^{2}\xi(\omega,k)\nu(\delta)}\langle
N^{\delta
k_{0}}_{D_{1}D_{2}}\hat{\mathbb{I}}_{D_{1}},\hat{\mathbb{I}}_{D_{2}}\rangle\langle
N^{\delta k_{0}}_{D_{2}D_{1}}u_{2},\hat{\mathbb{I}}_{D_{1}}\rangle=0,\\\
\langle N^{\delta
k_{0}}_{D_{2}D_{1}}u_{2},\hat{\mathbb{I}}_{D_{1}}\rangle-\frac{\delta^{2}\omega^{2}\xi(\omega,k)}{1-\delta^{2}\omega^{2}\xi(\omega,k)\nu(\delta)}\langle
N^{\delta
k_{0}}_{D_{2}D_{1}}\hat{\mathbb{I}}_{D_{2}},\hat{\mathbb{I}}_{D_{1}}\rangle\langle
N^{\delta
k_{0}}_{D_{1}D_{2}}u_{1},\hat{\mathbb{I}}_{D_{2}}\rangle=0.\end{cases}$

For this to have a solution, we need the determinant of the matrix induced by
this system to be zero. This gives

$\displaystyle
1-\frac{\delta^{4}\omega^{4}\xi(\omega,k)^{2}}{(1-\delta^{2}\omega^{2}\xi(\omega,k)\nu(\delta))^{2}}\langle
N^{\delta
k_{0}}_{D_{1}D_{2}}\hat{\mathbb{I}}_{D_{1}},\hat{\mathbb{I}}_{D_{2}}\rangle\langle
N^{\delta
k_{0}}_{D_{2}D_{1}}\hat{\mathbb{I}}_{D_{2}},\hat{\mathbb{I}}_{D_{1}}\rangle=0.$

Given the symmetry of our setting, we have that

$\langle N^{\delta
k_{0}}_{D_{1}D_{2}}\hat{\mathbb{I}}_{D_{1}},\hat{\mathbb{I}}_{D_{2}}\rangle=\langle
N^{\delta
k_{0}}_{D_{2}D_{1}}\hat{\mathbb{I}}_{D_{2}},\hat{\mathbb{I}}_{D_{1}}\rangle,$

and hence, we get

$\displaystyle
1-\delta^{2}\omega^{2}\xi(\omega,k)\nu(\delta)\pm\delta^{2}\omega^{2}\xi(\omega,k)\langle
N^{\delta
k_{0}}_{D_{1}D_{2}}\hat{\mathbb{I}}_{D_{1}},\hat{\mathbb{I}}_{D_{2}}\rangle=0.$

This is equivalent to

$\displaystyle
1-\delta^{2}\omega^{2}\xi(\omega,k)\Big{(}-\frac{|D_{1}|}{2\pi}\log(\delta
k_{0}\hat{\gamma})(1\pm 1)+$ $\displaystyle\langle
K^{(0)}_{D_{1}}[\hat{\mathbb{I}}_{D_{1}}],\hat{\mathbb{I}}_{D_{1}}\rangle+(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})\langle
K^{(1)}_{D_{1}}[\hat{\mathbb{I}}_{D_{1}}],\hat{\mathbb{I}}_{D_{1}}\rangle$
$\displaystyle\pm\langle
R^{(0)}_{D_{2}D_{1}}[\hat{\mathbb{I}}_{D_{2}}],\hat{\mathbb{I}}_{D_{1}}\rangle\pm(\delta
k_{0})^{2}\log(\delta k_{0}\hat{\gamma})\langle
R^{(1)}_{D_{2}D_{1}}[\hat{\mathbb{I}}_{D_{2}}],\hat{\mathbb{I}}_{D_{1}}\rangle\Big{)}=0,$

which is the desired result. ∎

##  4 Example: Circular Resonators

Figure 2: Behaviour of the subwavelength resonances for small circular nano-
particles of radius $\delta$. The resonant frequency $\omega_{s}$ of a single
circular methylammonium lead chloride nano-particle is shown. For two circular
nano-particles, made from the same material, we see how the hybridization
causes the frequencies $\omega_{d}$ (dipole) and $\omega_{m}$ (monopole) to
shift either side of $\omega_{s}$.

In this section, we illustrate our results for the case of two-dimensional
circular halide perovskite resonators. We can find the resonant frequencies of
a single particle $\omega_{s}$ by solving (2.35). Similarly, the hybridized
resonant frequencies of a pair of circular resonators can be found by solving
(3.10). The two solutions of (3.10) are denoted by $\omega_{m}$ and
$\omega_{d}$, to describe their monopolar and dipolar characteristics. As is
expected from other hybridized systems, it holds that $\omega_{m}<\omega_{d}$.
We plot these three frequencies as a function of the particle size $\delta$ in
Figure 2. We use the parameter values from [14] to model resonators made from
methylammonium lead chloride ($\text{MAPbCl}_{3}$), which is a popular halide
perovskite.

The first thing we observe from Figure 2 is that in the $\delta\to 0$ limit,
the frequencies coincide. This is because the nano-particles behave as
isolated, identical resonators when $\delta$ is very small. Then, as $\delta$
increases the single-particle resonance $\omega_{s}$ always stays between the
monopole and dipole frequencies of the hybridized case. The phenomenon of the
dipole frequency $\omega_{d}$ being shifted above $\omega_{s}$ and the
monopole frequency $\omega_{m}$ being shifted below $\omega_{s}$ is a typical
behaviour of hybridized resonator systems, see e.g. [3].

##  5 Conclusion

We have established a new mathematical model for halide perovskite resonators.
This is a significant development of the existing theory of subwavelength
resonators [2, 3], as it generalizes the techniques to dispersive settings
where the permittivity of the material depends on both the frequency and the
wavenumber. Given the rapidly growing use of halide perovskites in engineering
applications, this theory will have a significant impact on the design of
advanced devices [11, 13]. The integral methods used here are able to describe
a very broad class of resonator shapes, so are an ideal approach for studying
complex geometries, such as the biomimetic eye developed by [8].

## Data availability

There are no associated data, arising from this work. The only data used in
this work are the material parameter values for methylammonium lead chloride,
which are stated in [14].

##  Appendix A Appendix

###  A.1 Calculation of Three-dimensional Constants

We derive a formula for $\mathbb{F}$, which was a crucial quantity in section
2.4.3, in the case of a single three-dimensional halide perovskite resonator.
We have that

$\langle K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle=\frac{1}{8\pi}\mathbb{F}.$

From (2.22), we observe that

$\displaystyle\delta^{2}\omega^{2}\xi(\omega,k)$
$\displaystyle=\frac{8\pi}{8\pi\lambda_{\delta}-\delta^{2}k_{0}^{2}\mathbb{F}}.$

Also, we know that
$\xi(\omega,k)=\mu_{0}(\varepsilon(\omega,k)-\varepsilon_{0})$, and we have
shown that

$\varepsilon(\omega,k)=\varepsilon_{0}+\frac{1}{\mu_{0}\delta^{2}\omega^{2}\left(\lambda_{0}-\frac{i}{4\pi}\delta
k_{0}\mathbb{B}\right)}.$

Substituting this into the above equation, we get

$\displaystyle\mathbb{F}=\frac{8\pi}{\delta^{2}k_{0}^{2}}\Big{(}$
$\displaystyle\lambda_{\delta}-\lambda_{0}+\frac{i}{4\pi}\delta
k_{0}\mathbb{B}\Big{)}.$

Therefore, we obtain that

$\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle=\frac{1}{\delta^{2}k_{0}^{2}}\Big{(}\lambda_{\delta}-\lambda_{0}+\frac{i}{4\pi}\delta
k_{0}\mathbb{B}\Big{)}.$

###  A.2 Calculation of Two-dimensional Constants

We derive a formula for $\mathbb{S}$, which was a crucial quantity in section
2.5.3, in the case of a single two-dimensional halide perovskite resonator. We
have that

$\langle K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle=\frac{i}{4\pi}\mathbb{S}.$

From (2.35), we can obtain an expression for $\mathbb{S}$. Indeed, (2.35) is
equivalent to

$\displaystyle
4\pi=\delta^{2}\omega^{2}\xi(\omega,k)\Big{(}4\pi\lambda_{\delta}-i\delta^{4}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})\mathbb{S}\Big{)}.$

We know that
$\xi(\omega,k)=\mu_{0}\Big{(}\varepsilon(\omega,k)-\varepsilon_{0}\Big{)}$ and
we have shown that

$\varepsilon(\omega,k)=\frac{1}{\mu_{0}\delta^{2}\omega^{2}\left(\log(\delta
k_{0}\hat{\gamma})\lambda_{-1}-\frac{\mathbb{P}}{2\pi}-\frac{i(\delta
k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})\mathbb{G}}{4\pi}\right)}+\varepsilon_{0}.$

Substituting this into the above equality, we get

$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \
\mathbb{S}=\frac{-i}{\delta^{4}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})}\Big{(}4\pi(\lambda_{\delta}-\log(\delta
k_{0}\hat{\gamma})\lambda_{-1})+2\mathbb{P}+i(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})\mathbb{G}\Big{)}.$

Therefore, we obtain that

$\langle
K^{(2)}_{D}[u_{\delta}],u_{\delta}\rangle=\frac{1}{4\pi\delta^{4}k_{0}^{4}\log(\delta
k_{0}\hat{\gamma})}\Big{(}4\pi(\lambda_{\delta}-\log(\delta
k_{0}\hat{\gamma})\lambda_{-1})+2\mathbb{P}+i(\delta k_{0})^{2}\log(\delta
k_{0}\hat{\gamma})\mathbb{G}\Big{)}.$

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