# Exterior diffraction problems for a triangular lattice

D. Kapanadze<EMAIL_ADDRESS>E. Pesetskaya
<EMAIL_ADDRESS>A. Razmadze Mathematical Institute, TSU, Merab
Aleksidze II Lane 2, Tbilisi 0193, Georgia  Free University of Tbilisi,
Tbilisi 0159, Georgia

###### Abstract

Exterior Dirichlet problems for two-dimensional lattice waves on the semi-
infinite triangular lattice are considered. Namely, we study Dirichlet
problems for the two-dimensional discrete Helmholtz equation in a plane with a
hole. New results are obtained for the existence and uniqueness of the
solution in the case of the real wave number $k\in(0,2\sqrt{2})$ without
passing to a complex wave number. Besides, Green’s representation formula for
the solution is derived with the help of difference potentials. To demonstrate
the results, we propose a method for numerical calculation.

###### keywords:

discrete Helmholtz equation, exterior Dirichlet problem, metamaterials,
triangular lattice model

††journal: arXiv.org

##  1 Introduction

Nowadays there is an increased industrial and scientific interest in the study
of nano- and microstructures of modern materials and composites. Consideration
of discrete structures of the materials is one of the ways to investigate
microstructural processes in them, cf., e.g., [1, 2, 3, 4]. Therefore, we
devote our paper to the study of the exterior Dirichlet problem for the
discrete Helmholtz equation.

Continuum models with sufficiently smooth boundaries are well studied (cf. [5]
and references wherein). However, the derivation of a discrete analogue of the
Rayleigh-Sommerfeld scattering theory for different types of lattices is still
under development and has many applications. There are five two-dimensional
Bravais lattice types which naturally appear in application, and a triangular
lattice is one of them. Some structures of left-handed 2D metamaterials [2]
(which are a host microstrip line network periodically loaded with series
capacitors and shunt inductors for signal processing and filtering), close-
packed planes in some kinds of crystals [6, 7] can be represented by the
triangular lattices. Therefore, in this paper we analyze the exterior
Dirichlet problem for the discrete Helmholtz equation in the triangular
lattice mathematically formulated in Section 2. Although a similar problem for
a square lattice has been studied in [8], its extension to the triangular
lattice model is not direct.

In this paper, we obtain new results on the existence and uniqueness of a
solution in the case of a real wave number $k\in(0,2\sqrt{2})$. It is well
known that for the negative discrete Laplacian in the triangular lattice the
spectrum is (absolutely continuous) $[0,9]$, but there is an exceptional set
$\\{0,8,9\\}$ in $[0,9]$ where the limiting absorption principle fails [9,
10]. Therefore, for “admissible” wave numbers, one can study the problem as
the limit $k+\iota 0$ of the complex wave number. This method is applied by
Sharma in [11], where diffraction on triangular and hexagonal lattices by a
finite crack and rigid constraint is investigated. In the present paper, we
carry out our investigation without passing to the complex wave number. For
this purpose, we use the radiation conditions and asymptotic estimates from
[10] described in Section 3 and Rellich-Vekua type theorem from Isozaki el al
[9]. In Section 4, we prove the unique solvability results and obtain a
representation formula for the solution to the problem under consideration.
Notice that our results will help us to analyze the diffraction of lattice
waves by various types of defects including straight and zigzag rigid
constraints and cracks. To demonstrate purposes, we take a small defect and
present some numerical results in Section 5. For the numerical calculation, we
apply the method developed in [12] which allows us to calculate the lattice
Green’s functions without the need to perform integrals and appears to be much
more effective. Finally, it is worth mentioning that due to the more complex
form of the radiation condition for $k\in(2\sqrt{2},3)$, cf. [10], we have
some difficulties to prove the uniqueness result for this case using the
proposed method and, therefore, we consider only the case $k\in(0,2\sqrt{2})$.

##  2 Formulation of the problem

Let us consider $\\{\mathcal{V},\mathcal{E}\\}$ a periodic simple graph
defining a two-dimensional infinite triangular lattice $\mathfrak{T}$, where

$\mathcal{V}=\\{T(x_{1},x_{2})\subset\mathbb{R}^{2}:(x_{1},x_{2})\in\mathbb{Z}^{2}=\mathbb{Z}\times\mathbb{Z}\\}$
(1)

is a vertex set, $\mathcal{E}$ is an edge set, whose endpoints
$(v,w)\in\mathcal{V}\times\mathcal{V}$ are adjacent points, i.e., $|v-w|=1$,
and $T$ denotes a 2-dimensional coordinate transformation defined as

$T(x_{1},x_{2})=(x_{1}+x_{2}/2,\sqrt{3}x_{2}/2).$

We define for any point $v\in\mathcal{V}$ the 6-neighbourhood $F^{0}_{v}$ as
the set of points $w\in\mathcal{V}$ such that $|v-w|=1$ and the neighbourhood
$F_{v}$ as $F^{0}_{v}\bigcup\\{v\\}$. Recall that $R\subset\mathcal{V}$ is a
region if there exist disjoint nonempty subsets $\mathring{R}$ and $\partial
R$ of $R$ such that

  * (a)

$R=\mathring{R}\cup\partial R$,

  * (b)

if $v\in\mathring{R}$ then $F_{v}\subset R$,

  * (c)

if $v\in\partial R$ then there is at least one point $w\in F^{0}_{v}$ such
that $w\in\mathring{R}$.

As since the subsets $\mathring{R}$ and $\partial R$ are not defined uniquely
by $R$, henceforth, it is assumed for a given region $R$ in $\mathcal{V}$ that
$\mathring{R}$ and $\partial R$ are given and fixed. We say that $v$ is an
interior (boundary) point of $R$ if $v\in\mathring{R}$ ($v\in\partial R$).
Further, a region $R\subset\mathcal{V}$ is said to be connected if for any
$w,\tilde{w}\in R$ there exists a sequence $v^{(1)},\dots,v^{(n)}\in R$ with
$v^{(1)}=w$ and $v^{(n)}=\tilde{w}$ such that for all $0\leq i\leq n-1$,
$|v^{(i)}-v^{(i+1)}|=1$. By definition, a region $R$ with one interior point
$v$ is connected and coincides with $F_{v}$.

Figure 1: Exterior problem in $\mathcal{V}$. Connection between
$x=(x_{1},x_{2})\in\mathbb{Z}^{2}$ and the Euclidean coordinates of the
vertexes is established via $(x_{1},x_{2})\to(x_{1}+x_{2}/2,\sqrt{3}x_{2}/2)$.
For the boundary point $w$, the subset of the “cone” $C_{0}(w)$ is represented
by the red dots.

Let $\mathcal{W}$ be a connected region in $\mathcal{V}$ such that its
complement $\mathcal{W}^{c}=\mathcal{V}\backslash\mathcal{W}$ is an empty set
or a set with a finite number of lattice points. If
$\mathcal{W}^{c}=\varnothing$ then we have the case
$\mathring{\mathcal{W}}\cup\partial\mathcal{W}=\mathcal{V}$. When
$\mathcal{W}^{c}\neq\varnothing$ then we additionally require that
$\mathcal{W}^{c}$ is a finite region such that
$\partial\mathcal{W}^{c}=\partial\mathcal{W}$. Consequently, we have a
disjoint decomposition

$\mathring{\mathcal{W}}\cup\partial\mathcal{W}\cup\mathring{\mathcal{W}}^{c}=\mathcal{V}.$

To guarantee the uniqueness of the solution, we suppose that $\mathcal{W}$
satisfies the cone condition. Namely, for any $w=(w_{1},w_{2})\in\mathcal{W}$,
there is a “cone” $C_{i}(w)$, $i=0,1,\dots,5,$ such that
$C_{i}(w)\subset\mathcal{W}$. Here, $C_{i}(w)$ is defined as follows

$\displaystyle C_{i}(w)=\big{\\{}v=(v_{1},v_{2})\in\mathcal{V}:$
$\displaystyle|-\sin(\pi i/3)(v_{1}-w_{1})+\cos(\pi i/3)(v_{2}-w_{2})|$
$\displaystyle\leq\sqrt{3}(\cos(\pi i/3)(v_{1}-w_{1})+\sin(\pi
i/3)(v_{2}-w_{2}))\big{\\}}.$

In particular, when $i=0$, we have

$C_{0}(w)=\big{\\{}v=(v_{1},v_{2})\in\mathcal{V}:|(v_{2}-w_{2})|\leq\sqrt{3}(v_{1}-w_{1})\big{\\}}.$

The time-harmonic discrete waves in
$\mathcal{W}=\mathring{\mathcal{W}}\cup\partial\mathcal{W}$ can be described
by solutions of the following discrete Helmholtz equation

$(\Delta_{d}+k^{2})U(w)=0,\quad w=(w_{1},w_{2})\in\mathcal{W},$ (2)

where $\Delta_{d}$ denotes the discrete (a 7-point) Laplacian

$\displaystyle\Delta_{d}U(w)=\sum_{v\in F^{0}_{w}}U(v)-6U(w).$ (3)

We state the problem to find a unique solution $U$ to the discrete Helmholtz
equation in $\Omega$ satisfying the non-homogeneous Dirichlet problem:

$\displaystyle(\Delta_{d}+k^{2})U(w)$ $\displaystyle=0,\quad\quad\ \
\textup{in}\ \mathring{\mathcal{W}},$ (4a) $\displaystyle U(w)$
$\displaystyle=f(w),\quad\textup{on}\ \partial\mathcal{W}.$ (4b)

Here, $f:\partial\mathcal{W}\rightarrow\mathbb{C}$ is a given function, and
the wave number $k\in(0,2\sqrt{2})$ is real, cf. Figure 1.

For convenience, we prefer to work in a simplified coordinate system
$\mathbb{Z}^{2}$ used in (1) and, therefore, for describing the content in
terms of integer coordinates we tacitly use all the definitions introduced
above. Notice that our problem can be written as follows:

$\displaystyle(\Delta_{d}+k^{2})u(x)$ $\displaystyle=0,\quad\quad\ \
\textup{in}\ \mathring{\Omega},$ (5a) $\displaystyle u(x)$
$\displaystyle=f(x),\quad\textup{on}\ \partial\Omega.$ (5b)

Here, $u(x)=U(w)$ with $x=(x_{1},x_{2})\in\mathbb{Z}^{2}$, and
$w=T(x_{1},x_{2})\in\mathcal{W}$. The discrete Laplacian is given by the
following expression

$\displaystyle\Delta_{d}u(x)=$ $\displaystyle
u(x+e_{1})+u(x-e_{1})+u(x+e_{2})+u(x-e_{2})$ (6)
$\displaystyle+u(x+e_{1}-e_{2})+u(x-e_{1}+e_{2})-6u(x),$

where $e_{1}=(1,0)$, $e_{2}=(0,1)$ stand for the standard base of
$\mathbb{Z}^{2}$, cf. Figure 2.

Figure 2: Exterior problem in $\mathbb{Z}^{2}$. The subset of the “cone”
$C_{0}(w)$ for the boundary point $w=T(x)$ is represented by the red dots.

Thus, we are interested in studying the problem of the existence and
uniqueness of the function $u:\Omega\to\mathbb{C}$ such that $u(x)$ satisfies
the discrete Helmoltz equation (5a) with $k\in(0,2\sqrt{2})$ and the boundary
condition (5b). From now on we will refer to this problem as Problem
$\mathcal{P}_{\mathrm{ext}}$.

##  3 Green’s representation formula

Denote by $\mathcal{G}(x;y)$ the Green’s function for the discrete Helmholtz
equation (5a) centered at $y$ and evaluated at $x$. Then, the function
$\mathcal{G}(x;y)$ satisfies the equation

$(\Delta_{d}+k^{2})\mathcal{G}(x;y)=\delta_{x,y},$ (7)

where $\delta_{x,y}$ is the Kronecker delta. For brevity, we use the notation
$\mathcal{G}(x)$ for $\mathcal{G}(x;0)$. Notice that
$\mathcal{G}(x;y)=\mathcal{G}(x-y)$.

Using the discrete Fourier transform and the inverse Fourier transform we get

$\mathcal{G}(x)=\frac{1}{4\pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{e^{\iota(x\cdot\xi)}}{\sigma(\xi;k)}d\xi,\quad\xi=(\xi_{1},\xi_{2}),$
(8)

where

$\displaystyle\sigma(\xi;k^{2})$
$\displaystyle=e^{\iota\xi_{1}}+e^{-\iota\xi_{1}}+e^{\iota\xi_{2}}+e^{-\iota\xi_{2}}+e^{\iota\xi_{1}}e^{-\iota\xi_{2}}+e^{-\iota\xi_{1}}e^{\iota\xi_{2}}-6+k^{2}$
(9) $\displaystyle=k^{2}-6+2\cos\xi_{1}+2\cos\xi_{2}+2\cos(\xi_{1}-\xi_{2}).$

The lattice Green’s function $\mathcal{G}$ is quite well known when
$k^{2}\in\mathbb{C}\backslash[0,9]$ (cf., e.g., [13]). Notice that if
$k^{2}\in\mathbb{C}\backslash[0,9]$ then $\sigma\neq 0$ and, consequently,
$\mathcal{G}$ in (8) is well defined. In this case $\mathcal{G}(x)$ decays
exponentially when $|x|\to\infty$.

For $k\in(0,2\sqrt{2})$ we define the lattice Green’s function as a pointwise
limit of

$(R_{\lambda+\iota\varepsilon}\delta_{x,0})(x):=\frac{1}{4\pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{e^{\iota
x\cdot\xi}d\xi_{1}d\xi_{2}}{\sigma(\xi;k^{2}+\iota\varepsilon)}$ (10)

as $k^{2}+\iota\varepsilon\to k^{2}+\iota 0$ and denote it again by
$\mathcal{G}(x)$, i.e., $\mathcal{G}(x)=(R_{\lambda+\iota 0}\delta_{x,0})(x)$,
cf. [10]. Notice that $\mathcal{G}(x)$ is a solution to equation (7) and
satisfies equalities

$\mathcal{G}(x_{1},x_{2})=\mathcal{G}(x_{2},x_{1})=\mathcal{G}(-x_{1},-x_{2})=\mathcal{G}(x_{1}+x_{2},-x_{2})$
(11)

for all $x=(x_{1},x_{2})\in\mathbb{Z}^{2}$.

In order to simplify further arguments, let us introduce the following
vectors:

$\displaystyle e_{3}=e_{1}-e_{2},\quad e_{4}=-e_{1},\quad e_{5}=-e_{2},\quad
e_{6}=-e_{3}.$

Consider a region $R$ in $\mathbb{Z}^{2}$. Denote by $(\partial R)_{j}$,
$j=1,...,6$, a set of all boundary points $y\in\partial R$ such that
$y-e_{j}\in\mathring{R}$ and call it the sides of the boundary $R$. Clearly,
$\partial R$ is the union of its six sides: $\partial
R=\cup_{j=1}^{6}(\partial R)_{j}$. Notice that a boundary point $y$ can
simultaneously belong to all six sides of $R$. However, in our arguments
presented below it will be always clear which side is needed to be considered.
Under this condition, we define the discrete derivative in the outward normal
direction $e_{j}$, $j=1,\dots,6$,

$\mathcal{T}u(y)=u(y)-u(y-e_{j}),\quad y\in(\partial R)_{j}.$ (12)

Let us introduce the following set $H_{0}=\\{(0,0)\\}$ and then define
$H_{N}$, $N\in\mathbb{N}$, with the help of recurrence formula

$H_{N}:=\bigcup_{x\in H_{N-1}}F_{x}$ (13)

with $\mathring{H}_{N}:=H_{N-1}$, and $(\partial
H)_{N}:=H_{N}\backslash\mathring{H}_{N}$.

Recall a Green’s representation formula for a finite region, cf. [10].

###### Theorem 3.1

Let $R$ be a finite region. Then, for a given function $u:R\to\mathbb{C}$ and
any point $x\in\mathring{R}$, we have a discrete Green’s representation
formula

$u(x)=\sum_{y\in\partial
R}\big{(}u(y)\mathcal{T}\mathcal{G}(x-y)-\mathcal{G}(x-y)\mathcal{T}u(y)\big{)}+\sum_{y\in\mathring{R}}\mathcal{G}(x-y)(\Delta_{d}+k^{2})u(y).$

In particular, if $u$ is a solution to the discrete Helmholtz equation

$(\Delta_{d}+k^{2})u(x)=0\quad\textup{in}\ \mathring{R},$

then

$u(x)=\sum_{y\in\partial
R}\big{(}u(y)\mathcal{T}\mathcal{G}(x-y)-\mathcal{G}(x-y)\mathcal{T}u(y)\big{)}.$
(14)

Finally, for a finite region $R$, recall a discrete analogue of Green’s second
identity

$\sum_{x\in\mathring{R}}(u(x)\Delta_{d}v(x)-v(x)\Delta_{d}u(x))=\sum_{y\in\partial
R}(u(y)\mathcal{T}v(y)-v(y)\mathcal{T}u(y)).$ (15)

Now let us give a definition of a radiating solution on $\Omega$ when
$k^{2}\in(0,8)$. We say that $u:\Omega\to\mathbb{C}$ satisfies the radiation
condition at infinity if

$\left\\{\begin{aligned} u(x)&=O(|x|^{-\frac{1}{2}}),\\\
u(x+e_{j})&=e^{\iota\xi^{*}_{j}(\alpha,k)}u(x)+O(|x|^{-\frac{3}{2}}),\quad
j=1,2,\end{aligned}\right.$ (16)

with the remaining term decaying uniformly in all directions $x/|x|$, where
$x$ is characterized as $x_{1}=|x|\cos\alpha$, $x_{2}=|x|\sin\alpha$,
$0\leq\alpha<2\pi$. Here, $\xi^{*}_{j}(\alpha,k)$ is the $j$th coordinate of
the point $\xi^{*}(\alpha,k)$. Recall that the point
$\xi^{*}(\alpha,k)=\xi^{*}=(\xi^{*}_{1},\xi^{*}_{2})$ is a unique solution to
the following system of equations

$\displaystyle 2\zeta(\sin\xi_{1}+\sin(\xi_{1}-\xi_{2}))$ $\displaystyle=$
$\displaystyle\cos\alpha,$ $\displaystyle
2\zeta(\sin\xi_{2}-\sin(\xi_{1}-\xi_{2}))$ $\displaystyle=$
$\displaystyle\sin\alpha,$ $\displaystyle
k^{2}-6+2\cos\xi_{1}+2\cos\xi_{2}+2\cos(\xi_{1}-\xi_{2})$ $\displaystyle=$
$\displaystyle 0,$

where $\zeta$ is a positive constant, cf. [10].

###### Lemma 3.2

Let $k^{2}\in(0,8)$, and the function $u$ satisfies the radiation condition at
infinity (16). Then, for any boundary point $y\in(\partial H_{N})$, we have

$\mathcal{T}u(y)=\zeta(y,k)u(y)+O(|y|^{-\frac{3}{2}})\quad\textup{as}\
N\to\infty$ (17)

such that $\mathrm{\Im m}\,\zeta(y,k)>0$.

  * Proof.  First, let us consider the case $y\in(\partial H_{N})_{1}$. Then from (16), we get

$\mathcal{T}(y)=u(y)-u(y-e_{1})=\zeta(y,k)u(y)+O(|y|^{-\frac{3}{2}}),$

where $\zeta(y,k)=1-e^{-\iota\xi^{*}_{1}(\alpha,k)}$, and
$\alpha\in(-\pi/4,\pi/2)$ for $y\in(\partial H_{N})_{1}$. Thus,
$\sin(\xi^{*}_{1})>0$ and, consequently, $\mathrm{\Im m}\,\zeta(y,k)>0$.

For the case $y\in(\partial H_{N})_{2}$ the discrete derivative in the outward
normal direction takes the form $\mathcal{T}u(y)=u(y)-u(y-e_{2})$. From (16)
we easily derive

$\mathcal{T}(y)=\zeta(y,k)u(y)+O(|y|^{-\frac{3}{2}}),$

where $\zeta(y,k)=1-e^{-\iota\xi^{*}_{2}(\alpha,k)}$. In this case
$\alpha\in(0,3\pi/4)$ which implies that $\mathrm{\Im m}\,\zeta(y,k)>0$.

For $y\in(\partial H_{N})_{3}$ we need to consider
$\mathcal{T}u(y)=u(y)-u(y-e_{3})$. From (16) we have

$u(y^{\prime}+e_{1}-e_{2})=e^{\iota\xi^{*}_{1}}u(y^{\prime}-e_{2})+O(|y-e_{2}|^{-\frac{3}{2}}),$

and

$u(y^{\prime}-e_{2})=e^{-\iota\xi^{*}_{2}}u(y^{\prime})+O(|y^{\prime}|^{-\frac{3}{2}}),$

where $y^{\prime}=y-e_{1}+e_{2}$. Further, we get

$u(y^{\prime}+e_{1}-e_{2})=e^{\iota(\xi^{*}_{1}-\xi^{*}_{2})}u(y^{\prime})+O(|y^{\prime}|^{-\frac{3}{2}})$

which can be written as

$u(y)=e^{\iota(\xi^{*}_{1}-\xi^{*}_{2})}u(y-e_{3})+O(|y^{\prime}|^{-\frac{3}{2}}).$

Consequently, we derive

$\mathcal{T}(y)=\zeta(y,k)u(y)+O(|y|^{-\frac{3}{2}}),$

where $\zeta(y,k)=1-e^{-\iota(\xi^{*}_{1}-\xi^{*}_{2})}$. For $y\in(\partial
H_{N})_{3}$ we have $\alpha\in(-\pi/2,0)$ and, in this case,
$0<\xi^{*}_{1}-\xi^{*}_{2}<\pi$ which again gives us $\mathrm{\Im
m}\,\zeta(y,k)>0$.

Similar arguments applied to the remaining sides of the boundary $\partial
H_{N}$ complete the proof. $\Box$

For a fixed point $x\in\mathbb{Z}^{2}$ and any point $y\in\partial H_{N}$ the
radiation conditions (16) implies

$\sum_{y\in\partial
H_{N}}(u(y)\mathcal{T}\mathcal{G}(x-y)-\mathcal{G}(x-y)\mathcal{T}u(y))\to
0,\quad N\to\infty.$

Indeed, for instance, since $\alpha(y-x)$ tends to $\alpha=\alpha(y)$, and,
consequently, $\zeta(y-x,k)$ tends to $\zeta(y,k)$ as $|y|\to\infty$ then, for
sufficiently large $N$, we have

$\displaystyle u(y)\mathcal{T}\mathcal{G}(x-y)$
$\displaystyle-\mathcal{G}(x-y)\mathcal{T}u(y)$ $\displaystyle=u(y)\cdot
O(N^{-\frac{3}{2}})+\mathcal{G}(x-y)\cdot O(N^{-\frac{3}{2}})=O(N^{-2}).$

Further, Theorem 3.1 applied for $\Omega\cap H_{N}$, where $N\in\mathbb{N}$ is
sufficiently large, and then passing to the limit $N\to\infty$, yields the
following Green’s formula for a radiating solution $u$ to the discrete
Helmholtz equation (5a)

$u(x)=\sum_{y\in\partial\Omega}(u(y)\mathcal{T}\mathcal{G}(x-y)-\mathcal{G}(x-y)\mathcal{T}u(y)).$
(18)

From (18), using results obtained in [10], we can conclude that every
radiating solution $u$ to the discrete Helmholtz equation (4a) has the
following asymptotic expansion

$u(x)=-\frac{e^{\iota\mu(\alpha,k)|x|}}{|x|^{\frac{1}{2}}}\left\\{u_{\infty}(\hat{x})+O\left(\frac{1}{|x|}\right)\right\\},\quad|x|\to\infty,$
(19)

where $\mu(\alpha,k):=\xi^{*}(\alpha,k)\cdot\hat{x}$, $\hat{x}:=x/|x|$, and
$\xi^{*}(\alpha,k)=(\xi^{*}_{1}(\alpha,k),\xi^{*}_{2}(\alpha,k))$. Here, the
function $u_{\infty}(\hat{x})$, known as the far field pattern of $u$, can be
expressed with the help of formula (12) from [10].

Now we can formulate the following statement.

###### Theorem 3.3

The Problem $\mathcal{P}_{\mathrm{ext}}$ has at most one radiating solution.

  * Proof.  It is sufficient to show that corresponding homogeneous problem has only the trivial solution.

A discrete analogue of the Green’s first identity applied in $\Omega\cap
H_{N}$ has a form (see [10])

$\sum_{x\in\Omega\cap
H_{N}}(\nabla^{+}_{d}u(x)\cdot\nabla^{+}_{d}v(x)+\nabla^{-}_{d}u(x)\cdot\nabla^{-}_{d}v(x)+u(x)\Delta_{d}v(x))=\sum_{y\in\partial
H_{N}}u(y)\mathcal{T}v(y).$ (20)

Taking $v:=\overline{u}$, we get

$\sum_{x\in\Omega\cap
H_{N}}\left(|\nabla^{+}_{d}u(x)|^{2}+|\nabla^{-}_{d}u(x)|^{2}-k^{2}|u(x)|^{2}\right)=\sum_{y\in\partial
H_{N}}\overline{u}(y)\mathcal{T}u(y).$ (21)

Using Lemma 3.2, we rewrite equality (21) as follows

$\sum_{x\in\Omega\cap
H_{N}}\left(|\nabla^{+}_{d}u(x)|^{2}+|\nabla^{-}_{d}u(x)|^{2}-k^{2}|u(x)|^{2}\right)=\sum_{y\in\partial
H_{N}}\zeta(y)|u(y)|^{2}+O(N^{-1}).$

Taking the imaginary part of the last identity and passing to the limit as
$N\to\infty$, we get

$\sum_{y\in\partial H_{N}}|u(y)|^{2}\to 0\quad\textup{as}\ N\to\infty.$

Further, due to Rellich type theorem [9, 14, 8], we get $u\equiv 0$ outside of
$H_{N}$ for sufficiently large $N$. Since $\Omega$ satisfies the cone
condition then it has the unique continuation property [9, Theorem 5.7] which
implies $u\equiv 0$ in $\Omega$. $\Box$

##  4 Difference potentials and existence of solution

For any function $\varphi:\partial R\to\mathbb{C}$ we define difference single
layer and double layer potentials as follows

$V\varphi(x)=\sum_{y\in\partial R}\mathcal{G}(x-y)\varphi(y),\quad\textup{for
all}\ x\in\mathbb{Z}^{2},$ (22)

and

$W\varphi(x)=\sum_{y\in\partial
R}\big{(}\mathcal{T}\mathcal{G}(x-y)+\delta_{x,y}\big{)}\varphi(y),\quad\textup{for
all}\ x\in\mathbb{Z}^{2},$ (23)

respectively. Since $\delta_{x,y}=0$ for every $x\in\mathring{R}$ and
$y\in\partial R$ then (18) can be written as

$u(x)=Wu(x)-V(\mathcal{T}u)(x),\quad x\in\mathring{\Omega}.$

The role of the summand $\delta_{x,y}$ is clarified by the following result.

###### Lemma 4.1

For every $x\in\mathring{R}$ we have

$(\Delta_{d}+k^{2})V\varphi(x)=0,\quad\textup{and}\quad(\Delta_{d}+k^{2})W\varphi(x)=0.$

  * Proof.  For the difference single layer potential we have

$(\Delta_{d}+k^{2})V\varphi(x)=\sum_{y\in\partial
R}[(\Delta_{d}+k^{2})\mathcal{G}(x-y)]\varphi(y)=\sum_{y\in\partial
R}\delta_{x,y}\varphi(y)=0$

for all $x\notin\partial R$. Similarly we can show the result for the discrete
double layer potential when $x\in\mathring{R}$ and $F_{x}\cap\partial
R=\varnothing$. Indeed, we have
$(\Delta_{d}+k^{2})\mathcal{T}\mathcal{G}(x-y)=0$, and
$(\Delta_{d}+k^{2})\delta_{x,y}=0$. Thus, it remains to consider the case
$x\in\mathring{R}$ and $F_{x}\cap\partial R\neq\varnothing$. Let $y\in
F_{x}\cap(\partial R)_{1}$. Then, $y_{1}+1=x_{1}$, $y_{2}=x_{2}$, and we have

$\displaystyle(\Delta_{d}+k^{2})(\mathcal{T}\mathcal{G}(x_{1}-y_{1}$
$\displaystyle,x_{2}-y_{2})+\delta_{x_{1},y_{1}}\delta_{x_{2},y_{2}})$
$\displaystyle=(\Delta_{d}+k^{2})(\mathcal{G}(x_{1},x_{2};y_{1},y_{2})-\mathcal{G}(x_{1},x_{2};y_{1}+1,y_{2}))$
$\displaystyle\ +(\Delta_{d}+k^{2})\delta_{x_{1},y_{1}}\delta_{x_{2},y_{2}}$
$\displaystyle=\delta_{x_{1},y_{1}}\delta_{x_{2},y_{2}}-\delta_{x_{1},y_{1}+1}\delta_{x_{2},y_{2}}+\delta_{x_{1}-1,y_{1}}\delta_{x_{2},y_{2}}$
$\displaystyle=0-1+1=0.$

Arguing analogously for the other sides of the boundary $\partial R$, we
finally obtain

$(\Delta_{d}+k^{2})W\varphi(x)=0,\quad x\in\mathring{R}.$

$\Box$

As a consequence of Lemma 4.1, we have

$V\varphi(x)=\sum_{y\in\partial\Omega}\mathcal{G}(x-y)\varphi(y),\quad
x\in\mathring{\Omega},$

and

$W\varphi(x)=\sum_{y\in\partial\Omega}\left(\mathcal{T}\mathcal{G}(x-y)+\delta_{x,y}\right)\varphi(y),\quad
x\in\mathring{\Omega},$

are radiating solutions to the equation (4a) for any function
$\varphi:\partial\Omega\to\mathbb{C}$. From the proof of Lemma 4.1, it also
follows that if $\Omega^{c}\neq\varnothing$ and
$\partial\Omega=\partial\Omega^{c}$ then

$W^{\prime}\varphi(x)=\sum_{y\in\partial\Omega^{c}}\mathcal{T}\mathcal{G}(x-y)\varphi(y)$

is a radiating solution to the equation (5a).

Case I: $\Omega=\mathbb{Z}^{2}$. If $y_{i}$ is a point of intersection of
several sides, we choose and fix only one side of the boundary in order to
reduce number of numerical computations. Let $m$ be a number of points of
$\partial\Omega$. Then, $\partial\Omega$ can be represented as a sequence of
points $y_{1},y_{2},...,y_{m}$ such that $y_{i}=y_{j}$ if and only if $i=j$
for all $1\leq i,j\leq m$. Thus, in (22) we will have only one summand
connected with the boundary point $y_{i}$ and denote by $\widetilde{V}$ the
corresponding difference potential. Further, from the given function $f$ on
$\partial\Omega$, we form a vector $F=(f_{1},\dots,f_{m})^{\top}$,
$f_{i}:=f(y_{i})$. Similarly, for an unknown function $\varphi$, we write
$\Phi=(\varphi_{1},\dots,\varphi_{m})^{\top}$, $\varphi_{i}:=\varphi(y_{i})$.

We look for a solution to the Problem $\mathcal{P}_{\mathrm{ext}}$ in the form

$u(x)=\widetilde{V}\varphi(x)=\sum_{i=1}^{m}\mathcal{G}(x-y_{i})\varphi_{i},\quad
x\in\mathring{\Omega}.$ (24)

As in the proof of Lemma 4.1, it can be easily shown that $u$ is a radiating
solution to the equation (5a), and it only need to satisfy the boundary
conditions (5b). Then, (5b) implies the following linear system of boundary
equations

$\mathcal{H}\Phi=F,$ (25)

where

$\mathcal{H}=\begin{pmatrix}\mathcal{G}(y_{1}-y_{1})&\mathcal{G}(y_{1}-y_{2})&\mathcal{G}(y_{1}-y_{3})&\dots&\mathcal{G}(y_{1}-y_{m})\\\
\mathcal{G}(y_{2}-y_{1})&\mathcal{G}(y_{2}-y_{2})&\mathcal{G}(y_{2}-y_{3})&\dots&\mathcal{G}(y_{2}-y_{m})\\\
{5}\\\
\mathcal{G}(y_{m}-y_{1})&\mathcal{G}(y_{m}-y_{2})&\mathcal{G}(y_{m}-y_{3})&\dots&\mathcal{G}(y_{m}-y_{m})\end{pmatrix}.$

###### Lemma 4.2

The linear system of boundary equations (25) is uniquely solvable.

  * Proof.  Due to the Rouché-Capelli theorem it is sufficient to proof that the homogeneous system

$\mathcal{H}\Phi=0$ (26)

has only the trivial solution. Let
$\Phi^{*}=(\varphi^{*}_{1},\dots,\varphi^{*}_{m})^{\top}$ be a solution to
(26). Then,

$u(x)=\widetilde{V}\varphi^{*}(x)$

is a radiating solution to the homogeneous Problem
$\mathcal{P}_{\mathrm{ext}}$. Therefore, due to Theorem 3.3, we have $u\equiv
0$ in $\Omega$. Since $\Omega=\mathbb{Z}^{2}$ then at any boundary point
$y_{i}\in\partial\Omega$ we have

$0=(\Delta_{d}+k^{2})u(y_{i})=(\Delta_{d}+k^{2})\widetilde{V}\varphi^{*}(y_{i})=\sum_{i=1}^{m}\delta_{y_{i},y}\varphi^{*}(y)=\varphi^{*}(y_{i})=\varphi^{*}_{i}$

for all $1\leq i\leq m$. $\Box$

Case II:
$\mathring{\Omega}\cup\partial\Omega\cup\mathring{\Omega}^{c}=\mathbb{Z}^{2}$.
By our assumption on $\partial\Omega^{c}$ (cf., Section 2) we can represent it
as a sequence of points $y_{1},y_{2},...,y_{m}$ such that $y_{i}=y_{j}$ if and
only if $i=j$ for all $1\leq i,j\leq m$. Denote by $n_{i}$ the number of the
sides of the boundary $\partial\Omega^{c}$ that $y_{i}$ belongs to. Then,
using the same notation $F=(f_{1},\dots,f_{m})^{\top}$, $f_{i}:=f(y_{i})$, and
$\Phi=(\varphi_{1},\dots,\varphi_{m})^{\top}$, $\varphi_{i}:=\varphi(y_{i})$,
we look for a solution to the Problem $\mathcal{P}_{\mathrm{ext}}$ in the form

$\displaystyle u(x)$ $\displaystyle=W^{\prime}\varphi(x)+\iota\eta
V\varphi(x)$ (27)
$\displaystyle=\sum_{j=1}^{m}\sum_{l=1}^{n_{j}}\big{(}\mathcal{G}(x-y_{j})-\mathcal{G}(x-y^{-}_{jl})\big{)}\varphi_{j}+\iota\eta\sum_{j=1}^{m}n_{j}\mathcal{G}(x-y_{j})\varphi_{j}$
$\displaystyle=(1+\iota\eta)\sum_{j=1}^{m}n_{j}\mathcal{G}(x-y_{j})\varphi_{j}-\sum_{j=1}^{m}\sum_{l=1}^{n_{j}}\mathcal{G}(x-y^{-}_{jl})\varphi_{j},$

where $\eta\neq 0$ is a real coupling parameter, and $y^{-}_{jl}$ ,
$l=1,...,n_{j}$, are interior points of $\Omega^{c}$ which we encounter in the
expression $\mathcal{T}\mathcal{G}(x-y)$ that is the discrete derivative in
the outward normal direction with respect to $\Omega^{c}$. Due to Lemma 4.1,
$u$ is a radiating solution to the equation (5a), and we only need to satisfy
the boundary conditions (5b). From (5b) we get the following linear system of
boundary equations

$(1+\iota\eta)\mathcal{H}\mathcal{N}\Phi-\mathcal{K}\Phi=F,$ (28)

where $\mathcal{H}$ is defined as above,
$\mathcal{N}=\mathrm{diag}(n_{1},n_{2},\dots,n_{m})$ is diagonal matrix and

$\mathcal{K}=\begin{pmatrix}\sum_{l=1}^{n_{1}}\mathcal{G}(y_{1}-y^{-}_{1l})&\sum_{l=1}^{n_{2}}\mathcal{G}(y_{1}-y^{-}_{2l})&\dots&\sum_{l=1}^{n_{m}}\mathcal{G}(y_{1}-y^{-}_{ml})\\\
\sum_{l=1}^{n_{1}}\mathcal{G}(y_{2}-y^{-}_{1l})&\sum_{l=1}^{n_{2}}\mathcal{G}(y_{2}-y^{-}_{2l})&\dots&\sum_{l=1}^{n_{m}}\mathcal{G}(y_{2}-y^{-}_{ml})\\\
{4}\\\
\sum_{l=1}^{n_{1}}\mathcal{G}(y_{m}-y^{-}_{1l})&\sum_{l=1}^{n_{2}}\mathcal{G}(y_{m}-y^{-}_{2l})&\dots&\sum_{l=1}^{n_{m}}\mathcal{G}(y_{m}-y^{-}_{ml})\end{pmatrix}.$

###### Lemma 4.3

The linear system of boundary equations (28) is uniquely solvable.

  * Proof.  As above it is sufficient to proof that the homogeneous system

$(1+\iota\eta)\mathcal{H}\mathcal{N}\Phi-\mathcal{K}\Phi=0$ (29)

has only the trivial solution. Let
$\Phi^{*}=(\varphi^{*}_{1},\dots,\varphi^{*}_{m})^{\top}$ be a solution to
(29). Then, a function

$u_{+}(x)=W^{\prime}\varphi^{*}(x)+\iota\eta V\varphi^{*}(x)$

is a radiating solution to the homogeneous Problem
$\mathcal{P}_{\mathrm{ext}}$. Therefore, due to Theorem 3.3, we have
$u^{+}\equiv 0$ in $\Omega$. In particular, $u^{+}(y_{i})=0$ for all
$i=1,\dots,m$. Further, a function

$u_{-}(x)=W\varphi^{*}(x)+\iota\eta V\varphi^{*}(x)$

satisfies the discrete Helmholtz equation in $\Omega^{c}$ (cf., Lemma 4.1).
Notice that for any $y_{i}\in\partial\Omega^{c}=\partial\Omega$, we have

$u_{-}(y_{i})=u_{+}(y_{i})+\sum_{j=1}^{m}\sum_{l=1}^{n_{i}}\delta_{y_{j},y_{i}}\varphi_{j}^{*}=u_{+}(y_{i})+n_{i}\varphi_{i}^{*}=n_{i}\varphi_{i}^{*}.$

Since $(\Delta_{d}+k^{2})u_{+}(y_{i})=(1+\iota\eta)n_{i}\varphi_{i}^{*}$ and
$u_{+}\equiv 0$ in $\Omega$ we get

$\sum_{l=1}^{n_{i}}\mathcal{T}u_{+}(y_{i})=0-(1+\iota\eta)n_{i}\varphi_{i}^{*}=-(1+\iota\eta)n_{i}\varphi_{i}^{*}.$

Noting that $u_{+}(x)=u_{-}(x)$ in $\mathring{\Omega}^{c}$, we obtain

$\sum_{l=1}^{n_{i}}\mathcal{T}u^{-}(y_{i})=n_{i}\varphi_{i}^{*}-(1+\iota\eta)n_{i}\varphi_{i}^{*}=-\iota\eta
n_{i}\varphi_{i}^{*}.$

Hence, using (20) for $u_{-}$ and its complex conjugation $\bar{u}_{-}$, we
get

$\displaystyle\sum_{x\in\mathring{\Omega}^{c}}\big{(}|\nabla^{+}_{d}u_{-}(x)|^{2}$
$\displaystyle+|\nabla^{-}_{d}u_{-}(x)|^{2}-k^{2}|u_{-}(x)|^{2}\big{)}$
$\displaystyle=\sum_{y\in\partial\Omega^{c}}\bar{u}_{-}(y)\mathcal{T}u_{-}(y)=-\iota\eta\sum_{i=1}^{m}n^{2}_{i}|\varphi_{i}^{*}|^{2}.$

Taking the imaginary part of the last equation, we obtain that
$\varphi_{i}^{*}=0$ for all $i=1,\dots,m$, so $\Phi^{*}=0$. $\Box$

Due to a direct combination of the results obtained above now we have the main
conclusions of the present work.

###### Theorem 4.4

The Problem $\mathcal{P}_{\mathrm{ext}}$ has a unique radiating solution which
is represented as (24) if $\Omega=\mathbb{Z}^{2}$ or as (27) if
$\mathring{\Omega}\cup\partial\Omega\cup\mathring{\Omega}^{c}=\mathbb{Z}^{2}$,
where $\Phi$ is a unique solution to the system of linear equations (25) or
(28), respectively.

##  5 Numerical results

The main difficulty for numerical evaluation of solutions to (25) and (28) is
to compute the lattice Green’s function. For this purpose we apply the method
developed in [12]. Using 8-fold symmetry, we need only to compute the lattice
Green’s function $\mathcal{G}(i,j)$ with $i\geq j\geq 0$. Following to [12],
let us introduce the vectors
$\mathcal{V}_{2p}=(\mathcal{G}(2p,0),\mathcal{G}(2p-1,1),\dots,\mathcal{G}(p,p))^{\top}$
and
$\mathcal{V}_{2p+1}=(\mathcal{G}(2p+1,0),\mathcal{G}(2p,1),\dots,\mathcal{G}(p+1,p))^{\top}$
that collect all distinct Green’s functions $\mathcal{G}(i,j)$ with “Manhattan
distances” $\mid i\mid+\mid j\mid$ of $2p$ and $2p+1$, respectively. For any
Manhattan distance larger than 1, equation

$(\Delta_{d}+k^{2})\mathcal{G}(x)=\delta_{x,0}$ (30)

can be written in the matrix form
$\gamma_{n}(k)\mathcal{V}_{n}=\alpha_{n}(k)\mathcal{V}_{n-1}+\beta_{n}(k)\mathcal{V}_{n+1}$
where $\alpha_{n}(k)$, $\beta_{n}(k)$ and $\gamma_{n}(k)$ are sparse matrices
(cf., Appendix A). Notice that only the dimensions of these matrices depend on
$n$. It is shown in [12] that, for any $n\geq 1$, we have

$\mathcal{V}_{n}=A_{n}(k)\mathcal{V}_{n-1},$ (31)

where the matrices $A_{n}(k)$ are defined by the following recurrence formula

$A_{n}(k)=[\gamma_{n}(k)-\beta_{n}(k)A_{n+1}]^{-1}\alpha_{n}(k).$ (32)

They can be computed starting from a sufficiently large $N$ with
$A_{N+1}(k)=0$. Here, it is worth mentioning that for $k=2$ we need to choose
a better “initial guess” than $A_{N+1}(k)=0$, since in this case
$\mathrm{det}\gamma_{n}(k)=0$, and the matrix
$\gamma_{n}(k)-\beta_{n}(k)A_{n+1}$ is not invertible.

Once $A_{n}(k)$ are known, we have $\mathcal{V}_{n}=A_{n}(k)\dots
A_{1}(k)\mathcal{V}_{0}$, where $\mathcal{V}_{0}=\mathcal{G}(0,0)$. In
particular, $\mathcal{V}_{1}=\mathcal{G}(1,0)=A_{1}(k)\mathcal{G}(0,0)$ which,
together with $6\mathcal{G}(1,0)-(6-k^{2})\mathcal{G}(0,0)=1$, gives
$\mathcal{G}(0,0)=1/[6A_{1}(k)-6+k^{2}]$. This completes the calculation of
the Green’s function using elementary operations and no integrals. Notice also
one more important advantage of this method. The $A_{n}(k)$ matrices are
calculated coming down from asymptotically large Manhattan distances. As they
are propagated towards smaller Manhattan distances, it definitely gives us the
physical solution.

Finally, we demonstrate our theoretical and numerical approaches on the
following Problem $\mathcal{P}_{\mathrm{ext}}$: let $\mathring{\Omega}^{c}$
and $\partial\Omega^{c}$ be the sets of the following points $(2,2)$, $(3,2)$,
$(3,3)$ and $(2,1)$, $(3,1)$, $(4,1)$, $(4,2)$, $(4,3)$, $(3,4)$, $(2,4)$,
$(2,3)$, $(1,3)$, $(1,2)$, respectively. Further, we set
$\Omega=\mathbb{Z}^{2}\backslash\mathring{\Omega}^{c}$ with
$\partial\Omega=\partial\Omega^{c}$, cf. Figure 2. Clearly $\Omega$ satisfies
the cone condition, and we have a decomposition
$\mathring{\Omega}\cup\partial\Omega\cup\mathring{\Omega}^{c}=\mathbb{Z}^{2}$.
For this example, we take $f(y)\equiv 1$ on $\partial\Omega$ and $k=\sqrt{2}$.
Due to Theorem 4.4, the problem is uniquely solvable, and the solution can be
found as

$u(x)=W^{\prime}\varphi(x)+\iota V\varphi(x).$

Here, we take $\eta=1$. One can try to minimize the condition number of
corresponding matrices by the proper choice of $\eta$, but it is not our goal
at the moment. The vector $\Phi=(\varphi_{1},\dots,\varphi_{10})^{\top}$ is a
unique solution to equation (28). In order to solve obtained system of linear
equations and then find the solution $u$, we have developed MATLAB code that
uses the efficient method described above to compute Green’s functions. As a
technical aside, these data were obtained in several minutes on a regular
desktop. Some results of numerical evaluations are plotted in Figure 3. Some
key features of numerical solutions can be immediately observed. Namely, due
to the small hole, we notice some kind of symmetry of $\mathrm{\Re e}\,u$ and
$|u|$ in the macro level, however, as the plot on Figure 3 (e) shows, we do
not have an exact symmetry at the micro level. Besides, we also see some
interference effects on the density plots of $\mathrm{\Re e}\,u$ and $|u|$.

(a) The density plot of $\mathrm{Re}\,u$.

(b) The density plot of $|u|$ in $\mathbb{Z}^{2}$.

(c) The density plot of $\mathrm{Re}\,u$.

(d) The density plot of $|u|$.

(e) The graph of $\mathrm{Re}\,u(\cdot,2)$ .

Figure 3: Plots in (a), (b), and (e) are represented in $\mathbb{Z}^{2}$,
while (c) and (d) are represented in original coordinates of the triangular
lattice $\mathfrak{T}$. $k=\sqrt{2}$.

##  6 Discussion

In this paper, we have constructed the discrete scattering theory for the two-
dimensional Helmholtz equation with the real wave number $k\in(0,2\sqrt{2})$
for the triangular lattice. Similarly to the continuum theory, we used the
notion of radiating solution for the continuous Helmholtz equation and solved
the exterior Dirichlet problem without passing to the complex wave number.
Here, it is worth mentioning that due to the more complex form of the
radiation condition for $k\in(2\sqrt{2},3)$ than (16) we have some
difficulties to prove the uniqueness result for this case using the presented
approach. Clearly, we can introduce a small dumping parameter, pass it to the
complex wave number, and then use the limiting absorption principle to get the
desired solution, but this is not our goal and, therefore, we have restricted
ourselves only to the case $k\in(0,2\sqrt{2})$.

Finally, depending on the objectives of the investigation one can consider
different spaces on lattices, but for our purposes it is sufficient to take
the space $\ell^{\infty}_{R}(\Omega)$, which is a Banach space of all bounded
sequences on $\Omega\subset\mathbb{Z}^{2}$ that satisfy the radiation
condition (16).

##  7 Appendix A : Sparse matrices

The sparse matrices $\alpha_{n}(k)$, $\beta_{n}(k)$ and $\gamma_{n}(k)$ are
defined as follows: if $n=2p$ then $\alpha_{2p}(k)$ is a $(p+1)\times p$
matrix such that $\alpha_{2p}(k)\mid_{i,i}=1$, $i=\overline{1,p}$,
$\alpha_{2p}(k)\mid_{i,i-1}=1$, $i=\overline{2,p}$, while
$\alpha_{2p}(k)\mid_{p+1,p}=2$, and all other matrix elements are zero. The
$\beta_{2p}(k)$ is a $(p+1)\times(p+1)$ matrix such that
$\beta_{2p}(k)\mid_{i,i}=1$, $i=\overline{1,p}$,
$\beta_{2p}(k)\mid_{i,i+1}=1$, $i=\overline{2,p}$, while
$\beta_{2p}(k)\mid_{p+1,p+1}=\beta_{2p}(k)\mid_{1,2}=2$, and all other matrix
elements are zero. The $\gamma_{2p}(k)$ is a $(p+1)\times(p+1)$ matrix such
that $\gamma_{2p}(k)\mid_{i,i}=6-k^{2}$, $i=\overline{1,p+1}$,
$\gamma_{2p}(k)\mid_{i,i+1}=\gamma_{2p}(k)\mid_{i,i-1}=-1$,
$i=\overline{2,p}$, and
$\gamma_{2p}(k)\mid_{1,2}=\gamma_{2p}(k)\mid_{p+1,p}=-2$.

If $n=2p+1$ then $\alpha_{2p+1}(k)$ is a $(p+1)\times(p+1)$ matrix such that
$\alpha_{2p+1}(k)\mid_{i,i}=1$, $i=\overline{1,p+1}$,
$\alpha_{2p+1}(k)\mid_{i,i-1}=1$, $i=\overline{2,p+1}$, and all other matrix
elements are zero. The $\beta_{2p+1}(k)$ is a $(p+1)\times(p+2)$ matrix such
that $\beta_{2p+1}(k)\mid_{i,i}=1$, $i=\overline{1,p+1}$,
$\beta_{2p+1}(k)\mid_{i,i+1}=1$, $i=\overline{2,p+1}$, while
$\beta_{2p+1}(k)\mid_{1,2}=2$, and all other matrix elements are zero. The
$\gamma_{2p+1}(k)$ is a $(p+1)\times(p+1)$ matrix such that
$\gamma_{2p+1}(k)\mid_{i,i}=6-k^{2}$, $i=\overline{1,p}$,
$\gamma_{2p+1}(k)\mid_{p+1,p+1}=5-k^{2}$, while
$\gamma_{2p+1}(k)\mid_{i,i+1}=-1$, $i=\overline{2,p}$,
$\gamma_{2p+1}(k)\mid_{i,i-1}=-1$, $i=\overline{2,p+1}$, and
$\gamma_{2p+1}(k)\mid_{1,2}=-2$. Finally, $\gamma_{1}(k)$ is a $1\times 1$
matrixs with an element $4-k^{2}$.

## Acknowledgments

This work was supported by Shota Rustaveli National Science Foundation of
Georgia (SRNSFG) [FR-21-301]

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