L.9

# Surgery transformations and eigenvalue estimates for quantum graphs with
$\delta^{\prime}$ vertex interactions

Aftab Ali and Muhammad Usman  Aftab Ali, Department of Mathematics, Lahore
University of Management Sciences (LUMS), Lahore<EMAIL_ADDRESS>Muhammad Usman, Department of Mathematics, Lahore University of Management
Sciences (LUMS), Lahore<EMAIL_ADDRESS>
###### Abstract.

We extend the surgical tool box for quantum graphs to anti-standard and
$\delta^{\prime}$ vertex conditions. Monotonicity properties of eigenvalues of
graph Laplacian with $\delta^{\prime}$ interactions at vertices depend on the
sign of vertex parameter. Using several interlacing inequalities between
eigenvalues of graph Laplacian with diffeent vertex conditions and surgery
principles we obtain new upper and lower bounds on the eigenvalues of $\delta$
and $\delta^{\prime}$ Laplacians.

###### Key words and phrases:

Metric graphs; Surgery Principles; Bounds on eigenvalues.

###### 2010 Mathematics Subject Classification:

Primary: 34B45, 35P15; Secondary: 05C50

##  1\. Introduction

The quantum graphs - differential operators on metric graphs - is a rapidly
growing branch of mathematical physics lying on the border between
differential equations, spectral geometry and operator theory. During the last
decade, particular attention was paid to study the impact on the spectrum of
several differential operators such as Laplacian and Schrödinger operators on
the metric graphs under the perturbations of the topology and geometry, i.e.,
graph surgery. These surgery principles proved to be a useful tool for the
spectral analysis of quantum graphs, in particular, for eigenvalue estimates.
The usual goal is to derive spectral estimates that depend on the simple
geometric properties of the graph, such as total length, the Betti number, the
diameter or the number of vertices or edges of the underlying metric graph,
see [20, 21, 14, 2, 1, 6, 11, 10, 18, 13] and references therein.

The focus of the current work is to

  * •

extend the existing surgical tool box to the graph Laplacian with
$\delta^{\prime}$ and anti-standard vertex interactions and

  * •

exploit this broader set of surgery principles to obtain new spectral bounds
for graph Laplacians equipped with $\delta$ and $\delta^{\prime}$ vertex
interactions including standard, anti-standard, Neumann and Dirichlet vertex
conditions.

Theorem 4.3 of [3] provides a very useful tool to derive interlacing
inequalities between the eigenvalues of Laplacians with different sets of
vertex conditions for a fixed underlying metric graph. Our derivation of some
new eigenvalue bounds make intensive use of this result. Wherever possible we
derive estimates on the general eigenvalues and not just the lowest non-zero
eigenvalue.

Some of the surgical transformations for Laplacian with $\delta^{\prime}$
vertex conditions appeared in [19]. We will consider three additional (or a
more general form of perturbations that appeared in [19]) types of
perturbations of the graphs for $\delta^{\prime}$-type vertex conditions.

  1. (i)

Increasing or decreasing vertex coupling parameter,

  2. (ii)

a pendant graph is added to a graph, or

  3. (iii)

a graph is inserted at some vertex of other graph.

It is well known that under these transformations, eigenvalues of Laplacian
with different set of vertex interactions behave differently. For the Kirchoff
(Standard) or $\delta$ vertex conditions, increasing the $\delta$-coupling
parameter at a certain vertex of the graph increases the eigenvalues. That is,
a general monotonicity principle is valid. Moreover, under the perturbation
(i), the Theorem (3.4) of [3] and Theorem (3.1.8) of [4] establishes the
interlacing of the eigenvalues of the two graphs. Furthermore, the Dirichelet
eigenvalues of the graph are sandwiched between two consecutive eigenvalues of
same underlying $\delta$-graph. For the anti-standard as well as
$\delta^{\prime}$ type conditions, we show that the effect on the eigenvalues
under the transformation (i), in contrast to standard or $\delta$-conditions,
depends on the sign of the coupling parameters. It turns out that the
perturbation (i) divides the permutation invariant conditions into two
classes: those for which increasing the coupling parameter from a negative
(positive) number to a negative (positive) number leads to the non-increasing
eigenvalues, and those for which increasing the coupling parameter from a
negative number to a positive number lead to the non-decreasing eigenvalues.
Moreover, the eigenvalues of graph equipped with anti-standard conditions are
squeezed between two consecutive eigenvalues of same underlying
$\delta^{\prime}$-graph, see, Theorem (3.1).

The later part of section $3$ is devoted to the surgery transformation which
increases the total length of the metric graph. In Theorem (3.5) it is shown
that for permutation invariant condition such as anti-standard and
$\delta^{\prime}$ conditions the eigenvalues behave non-increasingly if the
length of an edge is scaled by a factor greater than one. Moreover, if graph
and the coupling parameters at each vertex are scaled up by some factor $t>1$
then then eigenvalues are pushed down by the factor $\frac{1}{t^{2}}$, see
Theorem (3.6). In [20, 5, 16, 18] eigenvalue estimates were derived for
certain basic surgical operations of quantum graphs, namely, gluing vertices
and attaching edges. The paper [19] deals with these operations for more
general (permutation invarient vertex conditions) self-adjoint vertex
conditions. More sophisticated surgery operations where the set of edges is
changed were investigated in[6, 13, 2], using the symmetrisation technique
first applied to quantum graphs by L. Friedlander [10] (see also [11]). For
the $\delta$-type vertex conditions it is known that the eigenvalues behave
non-increasing under the surgery transformations (i) and (ii), see [3]. In
Theorem (3.8) of the this paper, we show that for anti-standard and
$\delta^{\prime}$-type vertex conditions, the effect on the eigenvalues under
the transformation (ii) also depend on the sign of the coupling parameters.
This perturbation divides the vertex conditions into two parts: transformation
(ii) that decreases the eigenvalues and those for which the perturbation (ii)
increases the eigenvalues. Moreover, in Proposition (3.12) it is shown that
for $\delta^{\prime}$-type conditions the perturbation (iii) decreases the
eigenvalues. In Section $4$, we provide some lower and upper bounds on the
eigenvalues of the graphs equipped with (among others) $\delta$ and
$\delta^{\prime}$-type conditions. First, we provide some upper bounds on the
lowest eigenvalue of $\delta^{\prime}$-graph using some trial functions from
the domain of the quadratic form. Later we use the surgical transformation to
obtain a few bounds on the eigenvalues. Theorem (4.1) provides an upper bound
in terms of the betti number and the number of edges for the Neumann-Kirchoff
eigenvalues and Theorem (4.2) give bounds on the eigenvalues of
$\delta$-graph. A similar but a better estimate can be found in [1, Lemma
1.5]. However, for any flower graph the estimate in Theorem (4.1) coincides
with the estimate presented in [1, Lemma 1.5]. In Theorem (4.3) we provide the
upper bounds on the eigenvalues of a metric graph with standard conditions at
internal vertices and Dirichlet conditions at the vertices of degree one. For
the lower and upper estimates on the eigenvalues of graph with anti-standard
conditions, see Theorem (4.6) and (4.7), similar estimates for standard
conditions can be found in [7]. Theorem (2.3) and (4.9) provide lower bounds
on the eigenvalues of $\delta^{\prime}$-graph with negative and positive
coupling parameters, respectively. The later part of this section includes a
few more bounds on the eigenvalues of $\delta$ and $\delta^{\prime}$-graph,
these bounds are obtained using the well-known inequalities and especially the
estimates on the eigenvalues of metric tree presented in [20] and [21] for
standard and anti-standard vertex conditions, respectively. We also include
the equivalent form of these estimates expressed in terms of other topological
parameters of the underlying metric graph. In the last part, we have used a
simple relation between the eigenvalues of some bipartite graphs equipped with
standard and anti-standard conditions to first obtain a relation between the
eigenvalues of a bipartite graph and then for general graphs equipped with
$\delta$ and $\delta^{\prime}$ conditions.

We would like to mention that all results depend only on the vertex conditions
at those vertices which are changed by the graph transformation. At all other
vertices, general self-adjoint conditions are allowed. The proofs of our
results are all variational comparing Rayleigh quotients, which is a standard
method in obtaining eigenvalue estimates. In fact, estimates on the quadratic
form associated to the Laplace operator on suitable finite-dimensional
subspaces together with an application of the min-max principle yields the
desired estimates for the eigenvalues.

##  2\. Preliminaries

This section provides some basics of the quantum graph. It also provides the
set of interlacing inequalities between the eigenvalues of the same underlying
graph equipped with different conditions such as Dirichlet, Neumann, standard,
anti-standard, $\delta$ and $\delta^{\prime}$. These inequalities are obtained
using Theorem [3, Theorem 4.3] and help to prove some bounds on the
eigenvalues. For more detail on this subject, we refer to [4].
Let $\Gamma$ be a finite and connected graph having a finite set of edges $E$
and a finite set of vertices $V$. In addition, we identify each edge $e\in E$
with an interval $[0,\ell_{e}]$, where $\ell_{e}>0$ denote the length of the
edge and let $x_{e}$ be coordinate along this edge which induces a metric on
$\Gamma$. The sum of the lengths of each edge denotes the total length of a
metric graph $\Gamma$. That is,

$\mathcal{L}(\Gamma)=\sum\limits_{e\in E}\ell_{e}.$

Let $|E|$ and $|V|$ denote the number of edges and the number of vertices in a
metric graph $\Gamma$. Each edge $e$ connects a pair of vertices $v_{1}$ and
$v_{2}$, where a vertex is viewed as a subset of endpoints of edges. The
vertices are disjoint such that their union coincides with the set containing
each end point of all edges. An edge of $\Gamma$ can be a loop with both
endpoints connected to a single vertex, and we also allow multiple edges to
run between a pair of vertices. We assume that the graph $\Gamma$ is compact,
which means there are a finite number of edges, each having a finite length.
An edge $e$ is said to be an incident to a vertex $v$ if at least one of the
endpoints belongs to that vertex and denote by $e\sim v$, and let $E_{v}$
denote the set of all edges incident to a vertex $v$. The number of edges
incidents to a vertex $v$, denoted by $d_{v}$, is called the degree of a
vertex. A vertex of degree one is the boundary vertex, and an edge incident to
such a vertex is a boundary edge. The first Betti number, the number of
independent cycles, of a metric graph $\Gamma$ is given by,

$\beta=|E|-|V|+1.$

Our main objective is to study the spectrum of the Laplacian
$\frac{-d^{2}}{dx^{2}}$ acting on the functions living on the edges of
$\Gamma$ with vertices equipped with one of the below-mentioned vertex
conditions. The operator is self-adjoint in the Hilbert space

$L^{2}(\Gamma)=\bigoplus\limits_{e\in E}L^{2}(0,\ell_{e}).$

Which consists of functions $\varphi:\Gamma\mapsto\mathbb{C}$, that are
square-integrable along each edge. Let $\varphi_{e}$ be the restriction of
$\varphi$ to some edge $e\in E$. We define this operator on a subspace of the
Sobolev space $W^{2}_{2}(\Gamma\backslash V)$, consisting of functions
$\varphi\in L^{2}(\Gamma)$, whose weak derivatives up to order two exist on
each edge and are also square-integrable. We denote these Sobolev spaces by

$H^{2}(\Gamma)=\bigoplus\limits_{e\in E}H^{2}(0,\ell_{e}).$

equipped with the standard Sobolev norm and inner product. To describe the
vertex conditions of our interest, let us denote the limiting value of a
function $\varphi$ and its derivative at any endpoint of an edge by

$\varphi(x_{i})=\lim\limits_{x\mapsto x_{i}}\varphi(x)$

and

$\partial\varphi(x_{i})=\begin{cases}\varphi^{\prime}(x_{i}),\quad\quad\text{if
$x_{i}$ is the left endpoint},\\\ -\varphi^{\prime}(x_{i}),\quad\ \text{if
$x_{i}$ is the right endpoint.}\end{cases}$

The vertex conditions define the linear relation between values of function
and their derivatives at a vertex. The operator $L=-\frac{d^{2}}{dx^{2}}$ is a
self-adjoint defined on the functions from the Sobolev space
$W^{2}_{2}(\Gamma\backslash V)$ satisfying the following $\delta$-type vertex
conditions.

$\begin{cases}\varphi(x_{i})=\varphi(x_{j})\equiv\varphi(v_{m}),\quad
x_{i},x_{j}\in v_{m},\quad(\text{continuity condition})\\\
\sum\limits_{x_{i}\in
v_{m}}\partial\varphi(x_{i})=\alpha_{m}\varphi(v_{m}).\end{cases}$ (2.1)

Here $\alpha_{m}$ are the fixed real numbers associated with each vertex
$v_{m}$ and are commonly known as interaction strength of $\delta$-conditions.
The corresponding quadratic form and its domain is given by,

$h[\varphi]:=\int_{\Gamma}|\varphi^{\prime}(x)|^{2}\,dx+\sum_{m=1}^{|V|}\alpha_{m}|\varphi(v_{m})|^{2}$
(2.2)

with domain,

$D(h)=\\{\varphi\in H^{1}(\Gamma):\varphi(x_{i})=\varphi(x_{j}),\quad
x_{i},x_{j}\in v_{m}\\}.$

The standard conditions are defined by setting $\alpha_{m}=0$ in (2.1) at all
vertices of $\Gamma$, and the quadratic form is given by
$h[\varphi]:=\displaystyle{\int_{\Gamma}|\varphi^{\prime}(x)|^{2}dx}$ ,
defined on continuous functions in $H^{1}(\Gamma)$. We remark that the
vertices of degree two equipped with standard vertex conditions do not affect
the spectrum of the graph. Thus, any interior point can be regarded as a
vertex of degree two equipped with standard vertex conditions and similarly,
any vertex of degree two can be suppressed.
The Dirichlet vertex conditions are those in which the value of a function
$\varphi$ is zero at each end point of an interval, that is,

$\varphi(x_{i})=0,\quad x_{i}\in v_{m}.$

Equivalently, this means the function vanishes at a vertex $v_{m}$ of a metric
graph $\Gamma$. The Dirichlet conditions corresponds to $\delta$-type
conditions if we let $\alpha_{m}=\infty$ in (2.1). When Dirichlet conditions
are imposed at some vertex, the functions along the incident edges do not
interact (no communication between edges). Suppose all vertices are equipped
with Dirichlet condition. In that case, the metric graph is the collection of
independent intervals. The differential operator decouples into a direct sum
of the operators on separate intervals with Dirichlet condition imposed at
endpoints. The spectrum of the metric graph coincides with the disjoint union
of the spectrum of each interval. That is the operator is not changed if two
Dirichlet vertices are identified to form a single Dirichlet vertex. The
quadratic form is given by the expression
$h[\varphi]:=\int_{\Gamma}|\varphi^{\prime}(x)|^{2}dx$, with domain
$D(h)=\\{\varphi\in H^{1}(\Gamma):\varphi(v)=0\\}.$

In the $\delta^{\prime}$-type condition, the role of the functions and their
derivative are switched. In this case, the Laplacian is still self-adjoint on
the functions from the Sobolev space $W^{2}_{2}(\Gamma\backslash V)$
satisfying the following $\delta^{\prime}$-type vertex conditions.

$\begin{cases}\partial\varphi(x_{i})=\partial\varphi(x_{j}),\quad
x_{i},x_{j}\in v_{m},\quad(\text{continuity condition})\\\
\sum\limits_{x_{i}\in
v_{m}}\varphi(x_{i})=\alpha^{\prime}_{m}\dfrac{d\varphi(v_{m})}{dx},\quad\alpha^{\prime}_{m}\neq
0.\end{cases}$ (2.3)

The corresponding quadratic form is defined on the functions in
$H^{1}(\Gamma)$, and is given by,

$h[\varphi]:=\int_{\Gamma}|\varphi^{\prime}(x)|^{2}\,dx+\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\varphi(x_{i})\right|^{2}.$ (2.4)

Consider the case when $\alpha^{\prime}_{m}=0$ in (2.3) at some vertex $v_{m}$
of a metric graph $\Gamma$. The quadratic form in (2.4) only makes sense if we
force that the sum of the vertex values of functions living on the incidents
edges of $v_{m}$ is zero. The following vertex conditions are known as the
anti-standard vertex conditions,

$\begin{cases}\partial\varphi(x_{i})=\partial\varphi(x_{j}),\quad
x_{i},x_{j}\in v_{m}\quad(\text{continuity condition}),\\\
\sum\limits_{x_{i}\in v_{m}}\varphi(x_{i})=0\hskip
72.26999pt\quad(\text{balance condition}).\end{cases}$ (2.5)

And the quadratic form of a metric graph equipped with anti-standard vertex
conditions is defined by the expression
$h[\varphi]:=\int_{\Gamma}|\varphi^{\prime}(x)|^{2}\,dx$ defined on the
domain.

$D(h)=\left\\{\varphi\in H^{1}(\Gamma):\sum\limits_{x_{i}\in
v_{m}}\varphi(x_{i})=0\right\\}.$

Another type of vertex condition that decouples the graph into disjoints
interval is the Neumann conditions. For these conditions, the derivatives of
functions living on incident edges are zero at each vertex, and no conditions
are assumed on the values of functions. That is,

$\varphi_{e}^{\prime}(v)=0$

Equivalently, this means that the derivatives of functions vanish at each
endpoint of intervals in a metric graph $\Gamma$. The Neumann conditions
corresponds to $\delta^{\prime}$-type conditions if we let
$\alpha^{\prime}_{m}=\infty$ in (2.3). The quadratic form of a metric graph
equipped with such conditions is defined on the function from $H^{1}(\Gamma)$,
given by the following expression.

$h[\varphi]:=\int_{\Gamma}|\varphi^{\prime}(x)|^{2}\,dx.$ (2.6)

We remark that at the vertex of degree one, the anti-standard vertex
conditions coincide with Dirichlet conditions, and the standard condition
simplifies to Neumann conditions. Provided the length of an edge is preserved,
the eigenvalues of operators corresponding to the above mentioned vertex
conditions are independent of the choice of parametrization of edges in
$\Gamma$. We further remark that these operators are self-adjoint and semi-
bounded, and the underlying metric graph is compact, which guarantees that the
spectrum of Laplacians consists of a sequence of real eigenvalues of finite
multiplicity and can be listed as follows.

$\lambda_{1}\leq\lambda_{2}\leq\lambda_{3}\leq\cdots,$

and each eigenvalue is repeated according to its multiplicity, and the
corresponding eigenfunctions form an orthonormal basis of the Hilbert space
$L^{2}(\Gamma)$. The eigenvalues and the corresponding eigenfunctions depend
on the length, vertex conditions and the topological structure of the
underlying metric graph [3, 19, 4, 18]. For this reason, we will write
$\lambda_{k}=\lambda_{k}(\Gamma)$ to reflect the spectrum’s dependence on the
quantum graph- a metric graph, differential operator and the vertex
conditions.
Generally, the explicit computation of spectrum is challenging to calculate,
except in the case of a few simple graphs, impossible to calculate explicitly.
Because the relevant secular equation is transcendental to a very high degree.
The zeros of these secular equations are the eigenvalues of a quantum graph,
and it is not straightforward to find reliable analytical and numerical
solutions to such equations. Therefore, the interesting question is whether we
can find universal bounds on the eigenvalues depending on a few simple
parameters of the underlying metric graph. To obtain these bounds and the
corresponding extremal graphs, a better understanding of how the geometry of
the graph influences the eigenvalues is required. Therefore, we attempt to
understand how the eigenvalues change subject to specific changes in the graph
structure. For example, one may be interested to know if a particular graph
minimizes or maximizes a given eigenvalue among all graphs with specific fixed
geometric quantities such as length, number of edges, diameter, Betti number.
To answer such questions, one needs to be able to make comparisons between two
or more graphs. For this, we need to make some fundamental changes in the
geometry of the underlying graph, such as lengthening (or shortening) an edge,
gluing together vertices to obtain a new graph etc. These changes are made so
that the impact on the eigenvalues are predictable; such alterations in the
graph were referred to as surgery in [3].
A standard method in obtaining eigenvalue estimates is based on variational
arguments. The variational argument is based on comparing the Rayleigh
quotients on suitable finite-dimensional subspaces. The Rayleigh quotient
$R(\varphi)$ associated with Laplacian $L=-\dfrac{d^{2}}{dx^{2}}$ is defined
as

$R(\varphi)=\frac{h[\varphi]}{||\varphi||^{2}}$ (2.7)

on the space $L^{2}(\Gamma)$, the inner product is given by

$<\varphi,\psi>=\int_{\Gamma}\varphi(x)\overline{\psi(x)}dx.$ (2.8)

Here, $h[\varphi]=<L\varphi,\varphi>$ is the quadratic form of the operator
$L$. It is evidenced that computing the Rayleigh quotient for certain trial
functions from the domain of the quadratic form leads to estimates for
eigenvalues, and by minimizing the Rayleigh quotient, the eigenvalues and
eigenfunctions of an operator can be obtained. Thus, the Rayleigh quotient is
related to eigenvalues via the following min-max principle,

$\displaystyle\lambda_{k}(\Gamma)=\min_{\underset{\dim(X)=k}{X\subset
D\left(h\right)}}{\max_{0\neq\varphi\in
X}{\frac{h[\varphi]}{||\varphi||^{2}}}}.$ (2.9)

If a $k$-dimensional subspace realises the minimum $\lambda_{k}(\Gamma)$ in
(2.9), we call it a minimising subspace for $\lambda_{k}(\Gamma)$. Let $H$ and
$\tilde{H}$ be two self-adjoint operators with discrete spectrum and $h$ and
$\tilde{h}$ be their corresponding semi-bounded from below quadratic forms. We
say that $\tilde{h}$ is a positive rank-n perturbation of $h$ if $\tilde{h}=h$
on some $Y\subset_{n}\mathrm{dom}(h)$ and either
$Y=\mathrm{dom}(\tilde{h})\subset_{n}\mathrm{dom(h)}$ or $\tilde{h}\geq h$
with $\mathrm{dom}(\tilde{h})=\mathrm{dom}(h)$. Here, the symbol $\subset_{n}$
denotes the subspace such that the quotient space $\mathrm{dom}(h)/Y$ is
$n$-dimensional.

The following theorem provides the interlacing with equality characterisation
between the eigenvalues of the two forms where one form is the rank
$1$-perturbation of the other.

######  Theorem 2.1.

[3, Theorem 4.3]. Let $H$ and $\tilde{H}$ be two self-adjoint operators with
discrete spectrum and $h$ and $\tilde{h}$ be their corresponding semi-bounded
from below quadratic forms. If form $\tilde{h}$ is a positive rank-1
perturbation of the form $h$ then, eigenvalues of $H$ and $\tilde{H}$ satisfy

$\lambda_{k}(\Gamma)\leq\lambda_{k}(\tilde{\Gamma})\leq\lambda_{k+1}(\Gamma)\leq\lambda_{k+1}(\tilde{\Gamma}),\quad
k\geq 1.$

Moreover, if $\lambda$ is a common eigenvalue of $H$ and $\tilde{H}$, then
their respective multiplicities $m$ and $\tilde{m}$ differ by at most $1$ and
the dimension of the intersection of their $\lambda$-eigenspaces is
$\mathrm{min}(m,\tilde{m})$.

We will use the following notation to represent the metric graph equipped with
respective vertex conditions.

Metric Graph | Vertex Conditions
---|---
$\Gamma^{s}$ | Standard
$\Gamma^{a}$ | Anti-Standard
$\Gamma^{d}$ | Dirichlet
$\Gamma^{n}$ | Neumann
$\Gamma^{\delta}$ | $\delta$
$\Gamma^{\delta^{\prime}}$ | $\delta^{\prime}$

Furthermore, if arbitrary vertex conditions are assumed at all other vertices
of $\Gamma$ except $v_{1}$ where Dirichlet conditions are imposed, we will use
the notation $\lambda_{k}(\Gamma;v_{1})$ to represent the eigenvalues for this
graph. Moreover, if more than one vertex is equipped with Dirichlet condition,
we will denote the eigenvalues by $\lambda_{k}(\Gamma;D)$, where $D$ is the
subset of vertex set $V$ containing all Dirichlet vertices. We will only
consider the case when the degree one vertex are endowed with Dirichlet
conditions.

Let $\Gamma^{n}$ be the metric graph obtained from $\Gamma^{a}$ by imposing
Neumann conditions at a vertex $v_{0}$. Then the corresponding quadratic form
for $\Gamma^{a}$ and $\Gamma^{n}$ are defined by the same expression (2.6).
However, the domain $D(h^{a})$ is a co-dimension one subspace of $D(h^{n})$.
Thus by Theorem (2.1),

$\lambda_{k}(\Gamma^{n})\leq\lambda_{k}(\Gamma^{a})\leq\lambda_{k+1}(\Gamma^{n})\leq\lambda_{k+1}(\Gamma^{a}).$

Moverover, for any Neumann graph, graph equipped with Neumann conditions,
obtained from $\Gamma^{a}$ the domain $D(h^{a})$ is a co-dimension $|V|$
subspace of $D(h^{n})$, then applying the Theorem (2.1) we obatined the
following inequality.

$\lambda_{k}(\Gamma^{n})\leq\lambda_{k}(\Gamma^{a})\leq\lambda_{k+|V|}(\Gamma^{n}).$

Let $\Gamma^{\delta^{\prime}}$ be the graph obtained from the Neumann graph
$\Gamma^{n}$ by imposing the $\delta^{\prime}$-type conditions at vertex
$v_{0}$ in $\Gamma^{n}$ with strength $\alpha^{\prime}_{0}>0$, then the
domains of the quadratic forms are same, $D(h^{n})=D(h^{\delta^{\prime}})$,
and

$h^{n}[\varphi]=\int_{\Gamma}|\varphi^{\prime}(x)|^{2}\,dx\leq\int_{\Gamma}|\varphi^{\prime}(x)|^{2}\,dx+\frac{1}{\alpha^{\prime}_{0}}\left|\sum\limits_{x_{i}\in
v_{0}}\varphi(x_{i})\right|^{2}=h^{\delta^{\prime}}[\varphi].$

Since $D(h^{a})\subset_{1}D(h^{n})$ and on this domain
$h^{n}=h^{\delta^{\prime}}$. Thus, $h^{\delta^{\prime}}$ is a positive
rank-$1$ perturbation of the form $h^{n}$. Therefore,

$\lambda_{k}(\Gamma^{n})\leq\lambda_{k}(\Gamma^{\delta^{\prime}})\leq\lambda_{k+1}(\Gamma^{n}).$

Similarly if $\alpha^{\prime}_{0}<0$, then

$h^{\delta^{\prime}}[\varphi]=\int_{\Gamma}|\varphi^{\prime}(x)|^{2}\,dx+\frac{1}{\alpha^{\prime}_{0}}\left|\sum\limits_{x_{i}\in
v_{0}}\varphi(x_{i})\right|^{2}\leq\int_{\Gamma}|\varphi^{\prime}(x)|^{2}\,dx=h^{n}[\varphi],$

and $D(h^{a})\subset_{1}D(h^{\delta^{\prime}})$ and on this domain
$h^{n}=h^{\delta^{\prime}}$. Thus, $h^{n}$ is a positive rank-$1$ perturbation
of the form $h^{\delta^{\prime}}$. Therefore,

$\lambda_{k}(\Gamma^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{n})\leq\lambda_{k+1}(\Gamma^{\delta^{\prime}}).$

An equality between a Neumann and a $\delta^{\prime}$\- eigenvalue is possible
only if the eigenspace of the Neumann eigenvalue contains a function which
satisfy the balance condition at $v_{0}$, or, equivalently, the eigenspace of
the $\delta^{\prime}$-eigenvalue contains a function whose derivative vanishes
at $v_{0}$.

Let $\Gamma^{n}$ be the Neumann graph and obtain the graph $\Gamma^{d}$ by
prescribing the Dirichlet conditions at a vertex $v_{0}$ in $\Gamma^{n}$.
Since the quadratic forms of $\Gamma^{n}$ and $\Gamma^{d}$ are defined by the
same expression and agree on the domain $D(h^{d})\subset_{1}D(h^{n})$. Thus,

$\lambda_{k}(\Gamma^{n})\leq\lambda_{k}(\Gamma^{d};v_{0})\leq\lambda_{k+1}(\Gamma^{n}).$

For the Dirichlet graph $\Gamma^{d}$ and the graph $\Gamma^{a}$ obtained from
$\Gamma^{d}$ by imposing the anti-standard vertex condition at $v_{0}$ in
$\Gamma^{d}$, the respective forms $h^{d}$ and $h^{a}$ coincides on
$D(h^{d})\subset_{1}D(h^{a})$. Hence,

$\lambda_{k}(\Gamma^{a})\leq\lambda_{k}(\Gamma^{d})\leq\lambda_{k+1}(\Gamma^{a}).$

The equality is attained when the eigenspace of the Dirichlet eigenvalue
contains a function such that its derivatives are continuous at $v_{0}$, or,
equivalently, the eigenspace of the eigenvalue of $\Gamma^{a}$ contains a
function that vanishes at $v_{0}$.
Moreover, for the Dirichlet graph $\Gamma^{d}$ and the graph
$\Gamma^{\delta^{\prime}}$ obtained from $\Gamma^{d}$ by imposing the
$\delta^{\prime}$-type condition at $v_{0}$ in $\Gamma^{d}$, the respective
forms $h^{d}$ and $h^{\delta^{\prime}}$ coincides on
$D(h^{d})\subset_{1}D(h^{\delta^{\prime}})$. Hence,

$\lambda_{k}(\Gamma^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{d})\leq\lambda_{k+1}(\Gamma^{\delta^{\prime}}).$

Let $\Gamma^{\delta^{\prime}}$ be finite compact metric graph with arbitrary
self-adjoint vertex conditions at each vertex, except $v_{0}$ where
$\delta^{\prime}$-condition are imposed with strength $\alpha^{\prime}_{0}$,
and obtain a graph $\Gamma^{\delta}$ from $\Gamma^{\delta^{\prime}}$ by
imposing $\delta$-condition at $v_{0}$ with strength $\alpha_{0}$. Let
$h^{\delta^{\prime}}$ and $h^{\delta}$ denote the quadratic forms with domain
$D(h^{\delta^{\prime}})$ and $D(h^{\delta})$, respectively. Then the domain
$D(h^{\delta})$ is a subspace of $D(h^{\delta^{\prime}})$, and on
$D(h^{\delta})$, we have

$\displaystyle h^{\delta}-h^{\delta^{\prime}}$
$\displaystyle=\alpha_{0}|\varphi(v_{0})|^{2}-\frac{1}{\alpha^{\prime}_{0}}\left|\sum\limits_{x_{i}\in
v_{0}}\varphi(x_{i})\right|^{2}$
$\displaystyle=\alpha_{0}|\varphi(v_{0})|^{2}-\frac{1}{\alpha^{\prime}_{0}}\left|d_{v_{0}}\varphi(v_{0})\right|^{2}$
$\displaystyle=|\varphi(v_{0})|^{2}\left(\alpha_{0}-\frac{d_{v_{0}}^{2}}{\alpha^{\prime}_{0}}\right)$

Now, if $\alpha_{0}-\frac{d_{v_{0}}^{2}}{\alpha^{\prime}_{0}}\geq 0$, then
$h^{\delta}\geq h^{\delta^{\prime}}$.
Moreover, if $\alpha_{0}=\frac{d_{v_{0}}^{2}}{\alpha^{\prime}_{0}}$, then
$h^{\delta}=h^{\delta^{\prime}}$ on
$D(h^{\delta})\subset_{1}D(h^{\delta^{\prime}})$, and by rank one perturbation
the following interlacing inequalities holds

$\lambda_{k}(\Gamma^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k+1}(\Gamma^{\delta^{\prime}}).$

For our later purpose, we provide the explicit eigenvalues of an interval $I$
and the path graph $P$.

######  Example 2.2.

Let $I$ be a metric graph consist of single edge of length $\ell$, the
eigenvalues for the operator $-\frac{d}{dx^{2}}$ on an interval $[0,\ell]$ are
given by,

$\displaystyle\lambda_{k}(I,v_{1},v_{2})$
$\displaystyle=\frac{k^{2}\pi^{2}}{\ell^{2}},\quad k\geq
1,\quad\text{Dirichlet Conditions}$ $\displaystyle\lambda_{k+1}(I^{s})$
$\displaystyle=\frac{k^{2}\pi^{2}}{\ell^{2}},\quad k\geq 0,\quad\text{Standard
Conditions}$ $\displaystyle\lambda_{k+1}(I^{s})$
$\displaystyle=\frac{k^{2}\pi^{2}}{\ell^{2}},\quad k\geq 0,\quad\text{Neumann
Conditions}$ $\displaystyle\lambda_{k}(I^{a})$
$\displaystyle=\frac{k^{2}\pi^{2}}{\ell^{2}},\quad k\geq 1,\quad\text{Anti-
standard Conditions}$

######  Example 2.3.

Let $P$ be the path graph with standard vertex condition prescribed at each
vertex. Due to standard vertex condition the eigenvalue problem for Laplacian
$-\frac{d^{2}}{dx_{e}^{2}}$ on the path $P$ can be identified with eigenvalue
problem on an interval of length $L(P)$ for the operator
$-\frac{d^{2}}{dx^{2}}$ subject to Neumann boundary condition prescribed at
both endpoints. Thus the eigenvalues of path graph $P$ are given by

$\lambda_{k+1}(P^{s})=\frac{k^{2}\pi^{2}}{L(P)^{2}},\quad k\geq 0.$

######  Example 2.4.

The eigenvalues for $-\frac{d^{2}}{dx_{e}^{2}}$ on the path graph $P$ of
length $L(P)$ with anti standard vertex conditions imposed at each vertex
coincides with the eigenvalues of $-\frac{d^{2}}{dx^{2}}$ on an interval of
length $L(P)$ with Dirichlet conditions at both endpoints. Thus the
eigenvalues of $P$ are given by

$\lambda_{k}(P^{a})=\frac{k^{2}\pi^{2}}{L(P)^{2}},\quad k\geq 1.$

##  3\. Surgical Transformations

In this section, we present the tools that make some fundamental changes in
the geometry of the underlying metric graph, such as lengthening (or
shortening) an edge, gluing together vertices; these tools will alter the
given metric graph into some other graph for which the spectrum is known. We
discuss the effects of surgery principles on the spectrum of a Laplacian on a
finite compact quantum graph with arbitrary self-adjoint vertex conditions
($\delta$-type, $\delta^{\prime}$-type). For the metric graph equipped with
$\delta$-conditions, these surgical transformations and their effect on the
spectrum were studied in [3]. Here, we would like to investigate how some
surgical operations affect the spectrum of a given metric graph $\Gamma$ and
whether we can use a set of surgical operations to transform the graph into a
simpler one by precisely predicting the corresponding change in the spectrum.

###  3.1. Operation changing vertex conditions

In this part of a section, we study the eigenvalue’s dependence on the vertex
conditions. The operation changing the vertex conditions does not affect the
total length of a metric graph. However, it changes its connectivity in case
of gluing of vertices. First, we study the dependence of eigenvalues of
Laplacian on the $\delta^{\prime}$-coupling parameter $\alpha^{\prime}_{m}$.
This operation of changing the $\delta$-potential at a vertex $v$ is well
understood. Let $\Gamma^{\delta}_{\tilde{\alpha}}$ be obtained from
$\Gamma^{\delta}_{\alpha}$ by changing the interaction strength at a vertex
$v$ from $\alpha$ to $\tilde{\alpha}$. If
$\infty<\alpha\leq\tilde{\alpha}\leq\infty$, then the following interlacing
inequalities were established in [3,  Theorem 3.4] and [4, Theorem 3.1.8].

$\lambda_{k}(\Gamma^{\delta}_{\alpha})\leq\lambda_{k}(\Gamma^{\delta}_{\tilde{\alpha}})\leq\lambda_{k}(\Gamma^{\delta};v)\leq\lambda_{k+1}(\Gamma^{\delta}_{\alpha})$
(3.1)

The repeated application of (3.1) leads to the following inequalities.

$\lambda_{k}(\Gamma^{\delta}_{\alpha})\leq\lambda_{k}(\Gamma^{d})\leq\lambda_{k+|V|}(\Gamma^{\delta}_{\alpha})$
(3.2)

Let $\Gamma$ and $\tilde{\Gamma}$ be two quantum graphs with the same
underlying metric graph but with different strengths $\alpha^{\prime}_{m}$ in
the vertex conditions. Then the domains of the quadratic forms for $\Gamma$
and $\tilde{\Gamma}$ are the same. That is, $D(h)=D(\tilde{h})$. Thus, any
$\varphi\in D(h)$ which minimizes the Rayleigh quotient for
$\lambda_{k}(\Gamma)$ will also be in $D(\tilde{h})$ and will also minimize
the Rayleigh quotient for $\lambda_{k}(\tilde{\Gamma})$. The following theorem
shows the spectrum’s dependence on the vertex parameter $\alpha^{\prime}_{m}$.

######  Theorem 3.1.

Let $\Gamma$ be finite, compact and connected metric graph with local, self-
adjoint vertex condition at all vertices of $\Gamma$ except $v_{0}$, where
$\delta^{\prime}$-condition is imposed with strength $\alpha_{0}^{\prime}$.
Let $\tilde{\Gamma}$ be a graph obtained from $\Gamma$ by changing the
strength from $\alpha_{0}^{\prime}$ to $\tilde{\alpha}^{\prime}$, and let
$\Gamma_{0}$ be obtained from $\Gamma$ by imposing anti-standard vertex
conditions at $v_{0}$.

  1. (1)

If $0<\alpha_{0}^{\prime}<\tilde{\alpha}_{0}^{\prime}$, or
$\alpha^{\prime}_{0}<\tilde{\alpha}_{0}^{\prime}<0$, then

$\lambda_{k}(\tilde{\Gamma})\leq\lambda_{k}(\Gamma)\leq\lambda_{k}(\Gamma_{0})\leq\lambda_{k+1}(\tilde{\Gamma}).$
(3.3)

  2. (2)

If $\alpha_{0}^{\prime}<0$ and $\tilde{\alpha}_{0}^{\prime}>0$, then

$\lambda_{k}(\Gamma)\leq\lambda_{k}(\tilde{\Gamma})\leq\lambda_{k}(\Gamma_{0})\leq\lambda_{k+1}(\Gamma).$
(3.4)

###### Proof.

Let $h,\tilde{h}\ \text{and}\ h_{0}$ be the quadratic forms of corresponding
Laplacian associated to graphs $\Gamma,\tilde{\Gamma}\ \text{and}\ \Gamma_{0}$
respectively, then we have

$h[\varphi]=h_{0}[\varphi]+\frac{1}{\alpha_{0}^{\prime}}\left|\sum\limits_{x_{i}\in
v_{0}}\varphi(x_{i})\right|^{2}\text{and}\ \
\tilde{h}[\varphi]=h_{0}[\varphi]+\frac{1}{\tilde{\alpha}_{0}^{\prime}}\left|\sum\limits_{x_{i}\in
v_{0}}\varphi(x_{i})\right|^{2},$

on the appropriate sub-spaces of $H^{1}(\Gamma)=\bigoplus\limits_{e\in
E}H^{1}(e)$. Since the underlying metric graph is the same and the domain of
the quadratic form does not depend on strength, therefore

$D(h)=D(\tilde{h})\ \text{and}\ D(h_{0})=\left\\{\varphi\in
D(h):\sum\limits_{x_{i}\in v_{0}}\varphi(x_{i})=0\right\\},$

and

$h[\varphi]-\tilde{h}[\varphi]=\left|\sum\limits_{e_{i}\in
E_{v}}\varphi_{i}(v)\right|^{2}\left(\frac{1}{\alpha_{0}^{\prime}}-\frac{1}{\tilde{\alpha}_{0}^{\prime}}\right).$

Since $0<\alpha_{0}^{\prime}<\tilde{\alpha}_{0}^{\prime}$, or
$\alpha^{\prime}_{0}<\tilde{\alpha}_{0}^{\prime}<0\implies\frac{1}{\alpha_{0}^{\prime}}-\frac{1}{\tilde{\alpha}_{0}^{\prime}}>0$.
Thus $h\geq\tilde{h}$ for all $\varphi$, and the inequality follows from the
min-max description of eigenvalues. Furthermore, we have

$h[\varphi]=h_{0}[\varphi]\ \text{and}\
\tilde{h}[\varphi]=h_{0}[\varphi]\quad\text{for all}\quad\varphi\in D(h_{0}).$

The domain $D(h)$ is larger than the domain $D(h_{0})$ and the corresponding
quadratic forms $h$ and $h_{0}$ agree on $D(h_{0})$. Therefore, minimizing the
Rayleigh quotient over smaller space results in larger eigenvalues. The last
inequality follows from the fact that $D(h_{0})$ is a subspace of $D(h)$ of
co-dimension one.
If $\alpha_{0}^{\prime}<0$ and $\tilde{\alpha}_{0}^{\prime}>0$, then
$\frac{1}{\alpha_{0}^{\prime}}<0$ and
$\frac{1}{\tilde{\alpha}_{0}^{\prime}}>0\implies\left(\frac{1}{\alpha_{0}^{\prime}}-\frac{1}{\tilde{\alpha}_{0}^{\prime}}\right)<0$.
Thus, $h\leq\tilde{h}$ and hence by min-max principle,
$\lambda_{k}(\Gamma)\leq\lambda_{k}(\tilde{\Gamma}).$ ∎

Equality between an eigenvalue of $\Gamma$ and $\Gamma_{0}$ is only possible
if the eigenspace of $\lambda_{k}(\Gamma)$ contains an eigenfunction $\varphi$
such that at vertex $v_{0}$ the sum of components of $\varphi$ living on the
incident edges to $v_{0}$ is zero.

######  Definition 3.2.

(Gluing). The gluing of vertices is a surgical transformation in which a graph
$\tilde{\Gamma}$ is obtained by identifying some vertices
$v_{1},v_{2},\cdots,v_{n}$ of a graph $\Gamma$. If the strengths $\alpha_{m}$
for $m=1,2,\cdots,n$ were associated to the vertices
$v_{1},v_{2},\cdots,v_{n}$, then the glued vertex $v_{0}$ will be specified
with strength $\alpha_{0}=\sum\limits_{m=1}^{n}\alpha_{m}$.

The reverse operation to gluing of vertices is named as splitting, in which a
vertex $v_{0}$ of $\tilde{\Gamma}$ is cut into $n$ descendent vertices
$v_{1},v_{2},\cdots,v_{n}$ to obtain a new graph $\Gamma$. The obtained graph
$\Gamma$, using this surgery transformation, is not unique in general, as the
incident edges to $v_{0}$ may be assigned to descendent vertices in several
different ways. Furthermore, if $\varphi$ is a function on $\tilde{\Gamma}$
satisfying $\delta$-type conditions at vertex $v_{0}$, one can lift this
function on the graph $\Gamma$ by assigning the strengths

$\alpha_{m}=-\dfrac{\sum\limits_{e\sim
v_{m}}\partial\varphi_{e}(v_{0})}{\varphi(v_{0})},$

to the descendent vertices $v_{m}$ for $m=1,2,\cdots,n$ in $\Gamma$. If the
function $\varphi$ vanishes at $v_{0}$, then the descendent vertices will be
endowed with Dirichlet conditions [3]. Similarly, if the graph
$\tilde{\Gamma}$ is equipped with $\delta^{\prime}$-type conditions then any
function on $\tilde{\Gamma}$ can be lifted to $\Gamma$ by specifying the
strengths $\alpha^{\prime}_{m}$

$\alpha^{\prime}_{m}=-\dfrac{\sum\limits_{e\sim
v_{m}}\varphi_{e}(v_{0})}{\partial\varphi(v_{0})},$

to the descendent vertices $v_{m}$ for $m=1,2,\cdots,n$ in $\Gamma$.
Let $\tilde{\Gamma}^{\delta}$ be the graph obtained by gluing two vertices of
$\Gamma^{\delta}$, then the following interlacing inequalities can be easily
proved using variational principles. (c.f., e.g., [3, Theorem 3.4], [4,
Theorem 3.1.10]).

$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\tilde{\Gamma}^{\delta})\leq\lambda_{k+1}(\Gamma^{\delta})\leq\lambda_{k+1}(\tilde{\Gamma}^{\delta}).$
(3.5)

For each metric graph there exist a unique graph obtained by identifying all
the vertices to a single vertex, such a graph is known as flower graph. Note
that, there is a one-to-one correspondence between a unique flower graph and
the class of graphs with fix total number of edges, total length of metric
graph and length of each edge, $(|E|,\mathcal{L},\ell_{e})$. Let
$\Gamma_{f}^{\delta}$ be the flower graph corresponding to metric graph
$\Gamma^{\delta}$, then the repeated applications of (3.5) leads to the
following inequalities.

$\lambda_{k-|V|+1}(\Gamma_{f}^{\delta})\leq\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma_{f}^{\delta})\leq\lambda_{k+|V|-1}(\Gamma^{\delta}).$
(3.6)

The gluing operation does not depend on the sign of the strengths $\alpha_{m}$
for the $\delta$-graph. However, for the $\delta^{\prime}$-graph we have the
following theorem, which shows that the effect of gluing on the spectrum
depends on the sign of the strengths $\alpha^{\prime}_{m}$.

######  Theorem 3.3.

[19, Theorem 4.2] . Let $\Gamma^{\delta^{\prime}}$ be the metric graph
equipped with local, arbitrary self-adjoint vertex conditions at each vertex,
except $v_{1}$ and $v_{2}$, where $\delta^{\prime}$-type conditions are
imposed with strengths $\alpha^{\prime}_{1}$ and $\alpha^{\prime}_{2}$. Let
$\tilde{\Gamma}^{\delta^{\prime}}$ be graph obtained by gluing vertices
$v_{1}$ and $v_{2}$ of $\Gamma^{\delta^{\prime}}$, producing new vertex
$v_{0}$ with strength
$\alpha^{\prime}_{0}=\alpha^{\prime}_{1}+\alpha^{\prime}_{2}$. Then for all
$k\geq 1$ the following assertions hold.

  1. (1)

If $\alpha^{\prime}_{1},\alpha^{\prime}_{2}>0$ then
$\lambda_{k}(\tilde{\Gamma}^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{\delta^{\prime}}).$

  2. (2)

If $\alpha^{\prime}_{1},\alpha^{\prime}_{2}<0$ then
$\lambda_{k}(\Gamma^{\delta^{\prime}})\leq\lambda_{k}(\tilde{\Gamma}^{\delta^{\prime}}).$

  3. (3)

If $\alpha^{\prime}_{1}\cdot\alpha^{\prime}_{2}<0$ and $\alpha_{0}>0$ then
$\lambda_{k}(\Gamma^{\delta^{\prime}})\leq\lambda_{k}(\tilde{\Gamma}^{\delta^{\prime}}).$

  4. (4)

If $\alpha^{\prime}_{1}\cdot\alpha^{\prime}_{2}<0$ and $\alpha^{\prime}_{0}<0$
then
$\lambda_{k}(\tilde{\Gamma}^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{\delta^{\prime}}).$

  5. (5)

If $\alpha^{\prime}_{1}\cdot\alpha^{\prime}_{2}<0$ and $\alpha^{\prime}_{0}=0$
then
$\lambda_{k}(\Gamma^{\delta^{\prime}})\leq\lambda_{k}(\tilde{\Gamma}^{\delta^{\prime}}).$

  6. (6)

If $\alpha^{\prime}_{1}\cdot\alpha^{\prime}_{2}=0$ then
$\lambda_{k}(\tilde{\Gamma}^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{\delta^{\prime}}).$

######  Remark 3.4.

Let $\Gamma^{\delta^{\prime}}$ be the metric graph and
$\Gamma^{\delta^{\prime}}_{f}$ be its corresponding flower graph. Then the
following statements hold.

  1. (1)

If $\alpha^{\prime}_{m}>0$ for all $m=1,2,\cdots,|V|$ in
$\Gamma^{\delta^{\prime}}$, then

$\lambda_{k}(\Gamma^{\delta^{\prime}}_{f})\leq\lambda_{k}(\Gamma^{\delta^{\prime}}).$

  2. (2)

If $\alpha^{\prime}_{m}<0$ for all $m=1,2,\cdots,|V|$ in
$\Gamma^{\delta^{\prime}}$, then

$\lambda_{k}(\Gamma^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{\delta^{\prime}}_{f}).$

###  3.2. Operations Increasing the total length

One of the most important quantities of a metric graph is its total length,
which contains information about the eigenvalues. Naturally, any change in the
length of a metric graph will shift the spectrum of the respective metric
graph in some direction. A more profound study on how the change in the length
of the metric graph effect the spectrum can be found in [8, 19, 3]. For the
standard vertex condition, the scaling of a metric graph by a factor of $t>1$
reduce the eigenvalues by $t^{2}$. If
$\mathcal{L}(\tilde{\Gamma}^{s})=t\mathcal{L}(\Gamma^{s})$, then

$\lambda_{k}(\tilde{\Gamma}^{s})=\frac{\lambda_{k}(\Gamma^{s})}{t^{2}}.$

Moreover, for general $\delta$-conditions, if the metric graph
$\tilde{\Gamma}^{\delta}$ is obtained by scaling each edge of the graph
$\Gamma^{\delta}$ by a factor of $t>0$ and coupling parameters $\alpha_{m}$ at
each vertex are scaled with the factor $\frac{1}{t}$, then

$\lambda_{k}(\tilde{\Gamma}^{\delta})=\frac{\lambda_{k}(\Gamma^{\delta})}{t^{2}}.$

Since the quadratic form and its domain depend on the graph’s length, the
following proposition show that increasing any edge’s length of
$\Gamma^{\delta^{\prime}}$ lowers all eigenvalues.

######  Proposition 3.5.

Let $\tilde{\Gamma}^{\delta^{\prime}}$ be the metric graph obtained from
$\Gamma^{\delta^{\prime}}$ by scaling one of the edge say $e_{0}$ of
$\Gamma^{\delta^{\prime}}$ by a factor $t>1$, then

$\lambda_{k}(\tilde{\Gamma}^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{\delta^{\prime}}).$
(3.7)

###### Proof.

We parametrize each edge $e_{j}$ by $[0,\ell_{j}]$, let the function $\psi(x)$
minimizes the expression (2.4) for the $\Gamma^{\delta^{\prime}}$. Let
$\varphi(x)\in D(\tilde{h}^{\delta^{\prime}})$ be defined by

$\varphi(x)=\begin{cases}\psi\left(\frac{x}{t}\right)\quad\text{for}\ x\in
e^{\prime}_{0}=[0,t\ell_{0}]\\\ \psi(x)\quad\text{ for }\ x\in\\{e\in E:e\neq
e_{0}\\}\end{cases}$ (3.8)
$\displaystyle\int_{e^{\prime}_{0}}|\varphi(x)|^{2}dx$
$\displaystyle=\int^{tl_{0}}_{0}|\varphi(x)|^{2}dx=\int^{tl_{0}}_{0}\left|\psi\left(\frac{x}{t}\right)\right|^{2}dx=t\int_{e_{0}}|\psi(x)|^{2}dx,$
$\displaystyle\implies\int_{e^{\prime}_{0}}|\varphi(x)|^{2}dx>\int_{e_{0}}|\psi(x)|^{2}dx,$
$\displaystyle\int_{e^{\prime}_{0}}|\varphi^{\prime}(x)|^{2}dx$
$\displaystyle=\int^{tl_{0}}_{0}|\varphi^{\prime}(x)|^{2}dx=\frac{1}{t^{2}}\int^{tl_{0}}_{0}\left|\psi^{\prime}\left(\frac{x}{t}\right)\right|^{2}dx=\frac{1}{t}\int_{e_{0}}|\psi^{\prime}(x)|^{2}dx,$
$\displaystyle\implies\int_{e^{\prime}_{0}}|\varphi^{\prime}(x)|^{2}dx<\int_{e_{0}}|\psi^{\prime}(x)|^{2}dx.$

We have constructed $\varphi(x)$ in such a way that the values at the vertices
are unchanged $\psi(x_{i})=\varphi(x_{i})$, and hence

$\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\psi(x_{i})\right|^{2}=\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\varphi(x_{i})\right|^{2}$ $\displaystyle R(\psi)$
$\displaystyle=\frac{\sum\limits_{e\in
E}\int_{e}|\psi^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\psi(x_{i})\right|^{2}}{\sum\limits_{e\in E}\int_{e}|\psi(x)|^{2}dx}$
$\displaystyle=\frac{\sum\limits_{e\in E:e\neq
e_{0}}\int_{e}|\psi^{\prime}(x)|^{2}dx+\int_{e_{0}}|\psi^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\psi(x_{i})\right|^{2}}{\sum\limits_{e\in E:e\neq
e_{0}}\int_{e}|\psi(x)|^{2}dx+\int_{e_{0}}|\psi(x)|^{2}dx}$
$\displaystyle\geq\frac{\sum\limits_{e\in E:e\neq
e_{0}}\int_{e}|\psi^{\prime}(x)|^{2}dx+\frac{1}{t}\int_{e_{0}}|\psi^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\psi(x_{i})\right|^{2}}{\sum\limits_{e\in E:e\neq
e_{0}}\int_{e}|\psi(x)|^{2}dx+t\int_{e_{0}}|\psi(x)|^{2}dx}$
$\displaystyle=\frac{\sum\limits_{e\in E:e\neq
e_{0}}\int_{e}|\varphi^{\prime}(x)|^{2}dx+\int_{e^{\prime}_{0}}|\varphi^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\varphi(x_{i})\right|^{2}}{\sum\limits_{e\in E:e\neq
e_{0}}\int_{e}|\varphi(x)|^{2}dx+\int_{e^{\prime}_{0}}|\varphi(x)|^{2}dx}=R(\varphi)$

Since $R(\psi)\geq R(\varphi)$, so by max-min description of eigenvalues we
obtain the required result. Note that $\varphi(x)\not\in
D(L^{\delta^{\prime}}),$ but $\varphi(x)\in D(h^{\delta^{\prime}})$. ∎

The result of the above theorem still holds when each edge $e$ of the graph
$\Gamma^{\delta^{\prime}}$ is scaled by a factor $t_{e}>1$. This show that the
spectrum of a quantum graph depends on its total length. Increasing the
graph’s total length also decreases the eigenvalues, so one can shift the
eigenvalues in some direction by scaling the graph’s total length. When
$0<t<1$, the above result is reversed. The following theorem describes the
relationship between the Rayleigh quotients of the graphs with total lengths
$\mathcal{L}$ and $t\mathcal{L}$.

######  Theorem 3.6.

Let $\Gamma^{\delta^{\prime}}$ be the metric graph with total length
$\mathcal{L}(\Gamma^{\delta^{\prime}})$. Obtain a graph
$\tilde{\Gamma}^{\delta^{\prime}}$ with length
$t\mathcal{L}(\Gamma^{\delta^{\prime}})$ by scaling each edge of
$\Gamma^{\delta^{\prime}}$ by $t>0$ and scaling each strength
$\alpha^{\prime}_{m}$ by same factor, then

$\lambda_{k}(\tilde{\Gamma}^{\delta^{\prime}})=\frac{1}{t^{2}}\lambda_{k}(\Gamma^{\delta^{\prime}}).$
(3.9)

###### Proof.

Let $\psi(x)$ with $||\psi||_{L^{2}(\Gamma^{\delta^{\prime}})}=1$ minimizes
the Rayleigh quotient for graph $\Gamma^{\delta^{\prime}}$. Then the function
defined by $\varphi(x)=\frac{1}{\sqrt{t}}\psi(\frac{x}{t})$ with
$||\varphi||_{L^{2}(\tilde{\Gamma}^{\delta^{\prime}})}=1$ minimizes the
Rayleigh quotient for the graph $\tilde{\Gamma}^{\delta^{\prime}}$. Let
$e_{j}=[0,\ell_{j}]$ and $\tilde{e_{j}}=[0,t\ell_{j}]$ be the parametrization
of edges in $\Gamma^{\delta^{\prime}}$ and $\tilde{\Gamma}^{\delta^{\prime}}$,
respectively.

$\displaystyle\tilde{R}(\varphi(\tilde{x}))$
$\displaystyle=\int_{\tilde{\Gamma}^{\delta^{\prime}}}|\varphi^{\prime}(\tilde{x})|^{2}d\tilde{x}+\sum\limits_{m=1}^{|V|}\frac{1}{\tilde{\alpha}^{\prime}_{m}}\left|\sum\limits_{\tilde{x}_{i}\in\tilde{v}_{m}}\varphi(\tilde{x}_{i})\right|^{2}$
$\displaystyle=\sum\limits_{j=1}^{|\tilde{E}|}\int_{\tilde{e_{j}}}|\varphi^{\prime}(\tilde{x})|^{2}d\tilde{x}+\sum\limits_{m=1}^{|V|}\frac{1}{\tilde{\alpha}^{\prime}_{m}}\left|\sum\limits_{\tilde{x}_{i}\in\tilde{v}_{m}}\varphi(\tilde{x}_{i})\right|^{2}$
$\displaystyle=\sum\limits_{j=1}^{|\tilde{E}|}\int^{t\ell_{j}}_{0}\left|\frac{1}{\sqrt{t}}\psi^{\prime}\left(\frac{\tilde{x}}{t}\right)\frac{1}{t}\right|^{2}d\tilde{x}+\sum\limits_{m=1}^{|V|}\frac{1}{\tilde{\alpha}^{\prime}_{m}}\left|\sum\limits_{\tilde{x}_{i}\in\tilde{v}_{m}}\frac{1}{\sqrt{t}}\psi\left(\frac{\tilde{x}_{i}}{t}\right)\right|^{2}$

Let $x=\frac{\tilde{x}}{t}\implies tdx=d\tilde{x}$, and the limits of
integration become $x=0,x=\ell_{e}.$

$\displaystyle=\frac{1}{t^{2}}\left[\sum\limits_{j=1}^{|E|}\int_{e_{j}}|\psi^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{{\alpha}^{\prime}_{m}}\left|\sum\limits_{x_{i}\in{v}_{m}}\psi(x_{i})\right|^{2}\right]$

The expression inside the bracket is the Rayleigh quotient for the graph
$\Gamma^{\delta^{\prime}}$. Therefore, by min-max description of eigenvalues.

$\displaystyle\lambda_{k}(\tilde{\Gamma}^{\delta^{\prime}})$
$\displaystyle=\min_{\underset{\dim(X)=k}{X\subset
D\left(\tilde{h}\right)}}{\max_{0\neq\varphi\in X}}\tilde{R}(\varphi)$
$\displaystyle=\min_{\underset{\dim(X)=k}{X\subset
D\left(\tilde{h}\right)}}{\max_{\varphi\in
X:||\varphi||=1}}\frac{1}{t^{2}}\left[\sum\limits_{j=1}^{|E|}\int_{e_{j}}|\psi^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{{\alpha}^{\prime}_{m}}\left|\sum\limits_{x_{i}\in{v}_{m}}\psi(x_{i})\right|^{2}\right]$
$\displaystyle=\frac{1}{t^{2}}\left(\min_{\underset{\dim(X)=k}{X\subset
D\left(h\right)}}{\max_{\psi\in
X:||\psi||=1}}\left[\sum\limits_{j=1}^{|E|}\int_{e_{j}}|\psi^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{{\alpha}^{\prime}_{m}}\left|\sum\limits_{x_{i}\in{v}_{m}}\psi(x_{i})\right|^{2}\right]\right)$
$\displaystyle=\frac{1}{t^{2}}\lambda_{k}(\Gamma^{\delta^{\prime}})$

If we let $t$ approaches infinity, the values of a Rayleigh quotient, and thus
the eigenvalues are also pushed towards zero. And letting $t$ go to zero will
force the Rayleigh quotient approaches to $\pm\infty$. Furthermore, the result
is still valid, if both the graphs are equipped anti-standard vertex
conditions.

∎

We now consider the surgical transformation that will increase the volume of a
given metric graph by attaching a new subgraph to it.

######  Definition 3.7.

(Attaching a pendant graph). Given two metric graphs $\Gamma$ and
$\hat{\Gamma}$ and let $v_{1}\in\Gamma$ and $w_{1}\in\hat{\Gamma}$. If the
metric graph $\tilde{\Gamma}$ is obtained by gluing vertices $v_{1}$ and
$w_{1}$, then we speak of attaching a pendant graph $\hat{\Gamma}$ to
$\Gamma$.

Let $\tilde{\Gamma}^{\delta}$ be obtained by attaching a pendant graph
$\hat{\Gamma}^{\delta}$ to $\Gamma^{\delta}$, then it was proved in [3,
Theorem 3.10] that if for some $r$ and $k$,
$\lambda_{r}(\hat{\Gamma}^{\delta})\leq\lambda_{k}(\Gamma^{\delta})$ then

$\lambda_{k+r-1}(\tilde{\Gamma}^{\delta})\leq\lambda_{k}(\Gamma^{\delta}).$

For the graphs equipped with $\delta^{\prime}$-type condition it was
established in [19, Theorem 3.5] that attaching a pendant edge to some vertex
of $\Gamma^{\delta^{\prime}}$ to obtained a graph
$\tilde{\Gamma}^{\delta^{\prime}}$ leads to the following inequality,

$\lambda_{k}(\tilde{\Gamma}^{\delta})\leq\lambda_{k}(\Gamma^{\delta}).$

This inequality also hold when the graphs are equipped with anti-standard
vertex conditions. Moreover, the repeated application of this result also
shows that the same result is true when the graph
$\tilde{\Gamma}^{\delta^{\prime}}$ is formed by attaching a graph
$\hat{\Gamma}^{\delta}$ to $\Gamma^{\delta^{\prime}}$.

######  Theorem 3.8.

Let $\Gamma$ be a finite, compact and connected metric graph, let $v_{1}$ be a
vertex of $\Gamma$ and let $L=-\frac{d^{2}}{dx^{2}}$ be the Laplacian in
$L^{2}(\Gamma)$ subject to arbitrary local, self-adjoint vertex conditions at
each vertex $v_{m}\in V\setminus\\{v_{1}\\}$, with $\delta^{\prime}$ vertex
condition at $v_{1}$ , with strength $\alpha^{\prime}_{1}$. Let $\hat{\Gamma}$
be a finite, compact and connected metric graph, with arbitrary local, self-
adjoint vertex conditions at each vertex
$w_{m}\in\hat{V}\setminus\\{w_{1}\\}$, with $\delta^{\prime}$ vertex condition
at $w_{1}$, with strength $\hat{\alpha}^{\prime}_{2}$. Let $\tilde{\Gamma}$ be
graph formed by attaching the pendant graph $\hat{\Gamma}$ to $\Gamma$, if for
some $r,k_{0}$, we have,

$\lambda_{r}(\hat{\Gamma})\leq\lambda_{k_{0}}(\Gamma),$

then the following assertions hold.

  1. (1)

If $\alpha^{\prime}_{1},\hat{\alpha}^{\prime}_{2}>0$, then

${\lambda_{k+r}(\tilde{\Gamma})\leq\lambda_{k}(\Gamma),\quad k\geq k_{0}}.$

  2. (2)

If $\alpha^{\prime}_{1}\cdot\hat{\alpha}^{\prime}_{2}=0$ then,

${\lambda_{k+r}(\tilde{\Gamma})\leq\lambda_{k}(\Gamma),\quad k\geq k_{0}}.$

  3. (3)

If $\alpha^{\prime}_{1}\cdot\hat{\alpha}^{\prime}_{2}<0\ \text{and}\
\alpha^{\prime}_{1}+\hat{\alpha}^{\prime}_{2}<0$ then,

${\lambda_{k+r}(\tilde{\Gamma})\leq\lambda_{k}(\Gamma),\quad k\geq k_{0}}.$

  4. (4)

If $\alpha^{\prime}_{1},\hat{\alpha}^{\prime}_{2}<0$, then

${\lambda_{k+r}(\tilde{\Gamma})\geq\lambda_{k}(\Gamma),\quad k\geq k_{0}}.$

  5. (5)

If $\alpha^{\prime}_{1}\cdot\hat{\alpha}^{\prime}_{2}<0\ \text{and}\
\alpha^{\prime}_{1}+\hat{\alpha}^{\prime}_{2}>0$ then

${\lambda_{k+r}(\tilde{\Gamma})\geq\lambda_{k}(\Gamma),\quad k\geq k_{0}}.$

  6. (6)

If $\alpha^{\prime}_{1}\cdot\hat{\alpha}^{\prime}_{2}<0\ \text{and}\
\alpha_{1}+\hat{\alpha}^{\prime}_{2}=0$ then

${\lambda_{k+r}(\tilde{\Gamma})\geq\lambda_{k}(\Gamma),\quad k\geq k_{0}}.$

###### Proof.

  1. (1)

Let $\Gamma^{\prime}$ be the disconnected graph whose components are $\Gamma$
and $\hat{\Gamma}$, that is, $\Gamma^{\prime}=\Gamma\cup\hat{\Gamma}$. The
assumption $\lambda_{r}(\hat{\Gamma})\leq\lambda_{k_{0}}(\Gamma)$, and the
fact that the spectrum of $\Gamma^{\prime}$ is equal to the union of the
spectra of $\Gamma$ and $\hat{\Gamma}$ implies that every $k$-th eigenvalue of
$\Gamma$ is some $m$-th eigenvalue of $\Gamma^{\prime}$,

$\lambda_{k_{0}}(\Gamma)=\lambda_{m}(\Gamma^{\prime})$

for some $m=m(r,k_{0})\geq r+k_{0}$. Now attach the vertex $w_{1}$ of
$\hat{\Gamma}$ with a vertex $v_{1}$ of $\Gamma$ to form $\tilde{\Gamma}$,
then by Theorem (3.3), when $\alpha^{\prime}_{1},\hat{\alpha}^{\prime}_{2}>0$,
we obtain

$\lambda_{m}(\tilde{\Gamma})\leq\lambda_{m}(\Gamma^{\prime}),$

combining this inequality with the estimate $m\geq r+k_{0}$, we obtain

$\lambda_{r+k}(\tilde{\Gamma})\leq\lambda_{k}(\Gamma),\quad k\geq k_{0}.$

(4) Let $\Gamma^{\prime}$ be the disconnected graph whose components are
$\Gamma$ and $\hat{\Gamma}$, that is,
$\Gamma^{\prime}=\Gamma\cup\hat{\Gamma}$. The assumption
$\lambda_{r}(\hat{\Gamma})\leq\lambda_{k_{0}}(\Gamma)$, and the fact that the
spectrum of $\Gamma^{\prime}$ is equal to the union of the spectra of $\Gamma$
and $\hat{\Gamma}$ implies that

$\lambda_{k_{0}}(\Gamma)=\lambda_{m}(\Gamma^{\prime})$

for some $m=m(r,k_{0})\geq r+k_{0}$. Now attach the vertex $w_{1}$ of
$\hat{\Gamma}$ with a vertex $v_{1}$ of $\Gamma$ to form $\tilde{\Gamma}$,
then by Theorem (3.3), when $\alpha^{\prime}_{1},\hat{\alpha}^{\prime}_{2}<0$,
we obtain

$\lambda_{m}(\tilde{\Gamma})\geq\lambda_{m}(\Gamma^{\prime}),$

combining this inequality with the estimate $m\geq r+k_{0}$, we obtain

$\lambda_{r+k}(\tilde{\Gamma})\geq\lambda_{k}(\Gamma),\quad k\geq k_{0}.$

The other parts of the theorem can be proved correspondingly. ∎

######  Definition 3.9.

The inserting a graph at vertex is a surgical operation in which a metric
graph $\tilde{\Gamma}$ is obtained by removing a vertex $v_{0}$ of $\Gamma$
and attaching its incident edges to some vertices of $\hat{\Gamma}$. Let
$e_{1},e_{2},\cdots,e_{n}$ be the incident edges to a vertex $v_{0}$ of a
metric graph $\Gamma$ and $w_{1},w_{2},\cdots,w_{m},m\leq n$ be the vertices
in $\hat{\Gamma}$ to which an edge has been so attached. If $\alpha_{0}$ and
$\hat{\alpha}_{1},\hat{\alpha}_{2},\cdots,\hat{\alpha}_{m}$ are the coupling
parameters associated to the vertices $v_{0}$ and $w_{1},w_{2},\cdots,w_{m}$,
respectively. Then the coupling parameters should be placed at the vertices of
$\tilde{w}_{1},\tilde{w}_{2},\cdots,\tilde{w}_{m},$ of $\tilde{\Gamma}$ in
such a way that their sum is equal to
$\alpha_{0}+\hat{\alpha}_{1}+\hat{\alpha}_{2}+\cdots+\hat{\alpha}_{m}$.

A similar result to the following theorem was proved in [3, Theorem $3.10$], ,
in which it was assumed that if the vertices of the graphs $\Gamma$ and
$\hat{\Gamma}$ are equipped with $\delta$-type and standard vertex conditions,
respectively. Then, for all $k$ such that $\lambda_{k}(\Gamma)\geq 0$,

$\lambda_{k}(\tilde{\Gamma})\leq\lambda_{k}(\Gamma).$

The reason for imposing standard conditions on $\hat{\Gamma}$ is that we want
that the sum of the strengths assigned to the new vertices of $\tilde{\Gamma}$
must equal to the strength assigned at $v_{0}$ of $\Gamma$. The following
result omits this restriction. Therefore, the eigenvalues of $\tilde{\Gamma}$
are bounded above by the eigenvalues of the graphs with different strengths at
$v_{0}$ each strength equal to the sum of the strengths of vertices in
$\tilde{\Gamma}$ to which the incident edges of $v_{0}$ are attached.
Moreover, we believe that the following theorem gives an improved result when
$r\geq m$.

######  Theorem 3.10.

Let $\tilde{\Gamma}^{\delta}$ be obtained by inserting vertices
$w_{1},w_{2},\cdots,w_{m}$ of $\hat{\Gamma}^{\delta}$ at vertex $v_{0}$ of
$\Gamma^{\delta}$. If for some $r\geq m$ and $k$, we have,
$\lambda_{r}(\hat{\Gamma}^{\delta})\leq\lambda_{k}(\Gamma^{\delta}).$ Then,

$\lambda_{k+r-m}(\tilde{\Gamma}^{\delta})\leq\lambda_{k}(\Gamma^{\delta}).$
(3.10)

###### Proof.

Let $\check{\Gamma}^{\delta}$ be a graph obtained by gluing the vertices
$w_{1},w_{2},\cdots,w_{m}$ of $\hat{\Gamma}^{\delta}$ to form a single vertex
$w^{*}$, then

$\lambda_{r-m+1}(\check{\Gamma}^{\delta})\leq\lambda_{r}(\hat{\Gamma}^{\delta}).$

The assumption
$\lambda_{r}(\hat{\Gamma}^{\delta})\leq\lambda_{k}(\Gamma^{\delta}),$ implies
that
$\lambda_{r-m+1}(\check{\Gamma}^{\delta})\leq\lambda_{k}(\Gamma^{\delta})$.
Obtain the graph $\check{\check{\Gamma}}^{\delta}$ by attaching the vertex
$w^{*}$ of pendant graph $\check{\Gamma}^{\delta}$ at vertex $v_{0}$ of
$\Gamma^{\delta}$, this implies,

$\lambda_{k+r-m}(\check{\check{\Gamma}}^{\delta})\leq\lambda_{k}(\Gamma^{\delta}).$

Now, cut through the vertex $v_{0}$ of $\check{\check{\Gamma}}^{\delta}$ to
restore the vertices $\tilde{w}_{1},\tilde{w}_{2},\cdots,\tilde{w}_{m}$ to
form a graph $\tilde{\Gamma}^{\delta}$, thus

$\lambda_{k+r-m}(\tilde{\Gamma}^{\delta})\leq\lambda_{k+r-m}(\check{\check{\Gamma}}^{\delta}).$

∎

######  Corollary 3.11.

Let $\Gamma$ be a finite, compact and connected metric graph with arbitrary
self adjoint vertex conditions at each vertex. Let $\tilde{\Gamma}$ be a
metric graph obtained by lengthening an edge of $\Gamma$. If for some $k_{0}$,
$\lambda_{k_{0}}(\Gamma)\geq 0$, then

$\lambda_{k}(\tilde{\Gamma})\leq\lambda_{k}(\Gamma),\quad k\geq k_{0}.$ (3.11)

###### Proof.

Let $\Gamma^{s}$ be an edge of finite length with endpoints equipped with
standard vertex conditions, and $\mathcal{C}^{s}$ be a loop obtained by
joining the boundary points of $\Gamma^{s}$. Consider any interior point
$x^{*}$ on the edge of $\Gamma$ as a vertex of degree two equipped with
standard condition. Since

$0=\lambda_{1}(\mathcal{C}^{s})\leq\lambda_{k_{0}}(\Gamma)$

Obtain a graph $\hat{\Gamma}$ by attaching $\mathcal{C}^{s}$ as a pendant
graph to $\Gamma$ at an interior point $x^{*}$ then.

$\lambda_{k}(\hat{\Gamma})\leq\lambda_{k}(\Gamma),\quad k\geq k_{0}.$

Now restore the endpoints of $\Gamma^{s}$ in accordance with the definition of
splitting of the vertex to obtained a graph $\tilde{\Gamma}$, having two
interior points as a vertex of degree two each equipped with standard
conditions. Which can easily be removed and considered as interior points of
an edge of $\Gamma$. Thus,

$\lambda_{k}(\tilde{\Gamma})\leq\lambda_{k}(\hat{\Gamma}),\quad k\geq k_{0}.$

∎

######  Proposition 3.12.

Suppose $\tilde{\Gamma}$ is formed by inserting a graph $\Gamma^{a}$ at a
vertex $v_{0}$ of $\Gamma$ in a way that incident edges of $v_{0}$ are
attached to only two vertices , say $w_{1}$ and $w_{2}$, of $\Gamma^{a}$.
Assume that arbitrary local self-adjoint vertex conditions are imposed on each
vertex $v_{m}$ of $\Gamma$ except $v_{0}$, where $\delta^{\prime}$-condition
is imposed with strength $\alpha^{\prime}_{0}$, and anti-standard conditions
are prescribed at each vertex of $\Gamma^{a}$ prior to insertion. If for some
$k_{0}$, $\lambda_{k_{0}}(\Gamma)\geq 0$, then

$\lambda_{k+1}(\tilde{\Gamma})\leq\lambda_{k}(\Gamma),\quad k\geq k_{0}.$

###### Proof.

Let $\hat{\Gamma}^{a}$ be a graph obtained by gluing the vertices $w_{1}$ and
$w_{2}$ of $\Gamma^{a}$ to form a single vertex $w^{*}$ with total strength
zero. The assumption $\lambda_{k}(\Gamma)\geq 0$, for all $k\geq k_{0}$, along
with the fact that the spectrum of $\hat{\Gamma}^{a}$ is non-negative, implies
that $\lambda_{1}(\hat{\Gamma}^{a})=0\leq\lambda_{k}(\Gamma)$, for all $k\geq
k_{0}$. Let $\check{\Gamma}$ be obtained by attaching the vertex $w^{*}$ of
pendant graph $\hat{\Gamma}^{a}$ at vertex $v_{0}$ of $\Gamma$, then by
Theorem (3.8)(2),

$\lambda_{k+1}(\check{\Gamma})\leq\lambda_{k}(\Gamma),\quad k\geq k_{0}.$

Now, cut through the vertex $v_{0}$ of $\check{\Gamma}$ to restore the
vertices $w_{1}$ and $w_{2}$ to obtain the graph $\tilde{\Gamma}$, the
restoration of vertices is performed in such a way that the sum of strengths
$\tilde{\alpha}^{\prime}_{1}$, $\tilde{\alpha}^{\prime}_{2}$, assigned to
$\tilde{w}_{1}$ and $\tilde{w}_{2}$, is $\alpha^{\prime}_{0}$. Moreover, the
strengths are assigned such that the the operation of splitting the vertex
$v_{0}$ lowers all eigenvalues of $\check{\Gamma}$. Thus,

$\lambda_{k+1}(\tilde{\Gamma})\leq\lambda_{k+1}(\check{\Gamma}).$

∎

##  4\. Eigenvalue Estimates

This section provides upper and lower estimates for Laplacian on a metric
graph equipped with self-adjoint vertex conditions ($\delta$ and
$\delta^{\prime}$-type conditions). This section is divided into two
subsections.

###  4.1. Estimates on the eigenvalues of $\delta$-graph

In this part of the section we provide some upper and lower estimates on the
eigenvalues of the $\delta$-graph.

######  Theorem 4.1.

Let $\Gamma^{s}$ be finite, compact connected metric graph with standard
conditions, then

$\lambda_{k}(\Gamma^{s})\leq\left(\frac{(k-1+\beta+|E|)\pi}{\mathcal{L}(\Gamma^{s})}\right)^{2}.$
(4.1)

###### Proof.

Consider an interval $I^{s}$ of length $\mathcal{L}(\Gamma^{s})$ equipped with
standard vertex conditions. Obtain a path graph $P^{s}$ by creating vertices
of degree two at interior points of $I^{s}$ such that the number of edges and
their lengths are same as in $\Gamma^{s}$. Cut through the degree two vertices
of $P^{s}$ to form a new graph $\hat{\Gamma}^{s}$ which consists of $|E|$
disconnected edges. Create a metric tree $T^{s}$ by gluing the vertices of
$\hat{\Gamma}^{s}$ in such a way that the graph $\Gamma^{s}$ can be obtained
from $\Gamma^{s}$ by pairwise gluing of $\beta$ pair of vertices of $T^{s}$.
Thus,

$\left(\frac{k\pi}{\mathcal{L}(\Gamma^{s})}\right)^{2}=\lambda_{k}(I^{s})=\lambda_{k}(P^{s})\geq\lambda_{k}(\hat{\Gamma}^{s})\geq\lambda_{k-|E|+1}(T^{s})\geq\lambda_{k-|E|+1-\beta}(\Gamma^{s}).$
(4.2)

Equivalently, this estimates can be written as,

$\lambda_{k}(\Gamma^{s})\leq\left(\frac{(k-2+2\beta+|V|)\pi}{\mathcal{L}(\Gamma^{s})}\right)^{2}.$

∎

We can also obtain a lower bound on $\lambda_{k}(\Gamma^{s})$ as follows.
However, this bound is not a better estimate as it only makes sense when
$k\geq|E|-1$.

$\left(\frac{(k-|E|+1)\pi}{\mathcal{L}(\Gamma^{s})}\right)^{2}=\lambda_{k-|E|+1}(I)=\lambda_{k-|E|+1}(P)\leq\lambda_{k}(\hat{\Gamma}^{s})\leq\lambda_{k}(T^{s})\leq\lambda_{k}(\Gamma^{s}).$

We use this estimate to obtain a few bounds on the eigenvalues of
$\Gamma^{\delta}$. Let $\alpha=\sum\limits_{m=1}^{|V|}\alpha_{m}$ denote the
total strength of $\Gamma^{\delta}$ and $\Gamma^{\delta}_{f}$ be the
corresponding flower graph. Let $\Gamma^{s}_{f}$ be the same underlying flower
graph equipped with standard conditions. If $\alpha\leq 0$, then the above
result implies that

$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{\delta}_{f})\leq\lambda_{k}(\Gamma^{s}_{f})\leq\left(\frac{(k-1+2|E|)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}.$

Similarly, if $\alpha>0$, then

$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{\delta}_{f})\leq\lambda_{k+1}(\Gamma^{s}_{f})\leq\left(\frac{(k+2|E|)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}.$

######  Theorem 4.2.

Let $\Gamma^{\delta}$ be a metric graph with $|E|$ number of edges and $|V|$
number of vertices. Then the following assertions hold.

  1. (1)

If $\alpha=0$, then

$\left(\frac{(k-|V|+1)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}\leq\lambda_{k}(\Gamma^{\delta})\leq\left(\frac{(k+|E|)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}.$

  2. (2)

If $\alpha>0$, then

$\left(\frac{(k-|V|+1)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}\leq\lambda_{k}(\Gamma^{\delta})\leq\left(\frac{(k+|E|+1)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}.$
(4.3)

  3. (3)

If $\alpha<0$, then

$\left(\frac{(k-|V|)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}\leq\lambda_{k}(\Gamma^{\delta})\leq\left(\frac{(k+|E|)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}.$
(4.4)

###### Proof.

Let $I^{s}$ be an interval of length $\mathcal{L}(\Gamma^{\delta})$ equipped
with standard vertex conditions and let $P^{s}$ be a path graph obtained by
creating vertices of degree two at interior points of $I^{s}$ such that the
number of edges and their lengths are same as in $\Gamma^{\delta}$. Then,

$\lambda_{k}(I^{s})=\lambda_{k}(P^{s}).$

Obtain a flower graph $P^{s}_{f}$ from $P^{s}$, then (3.5) implies that

$\lambda_{k-|E|}(P^{s}_{f})\leq\lambda_{k}(P^{s})\leq\lambda_{k}(P^{s}_{f})\leq\lambda_{k+|E|}(P^{s}).$

Thus,

$\left(\frac{k\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}=\lambda_{k}(I^{s})=\lambda_{k}(P^{s})\leq\lambda_{k}(P^{s}_{f})\leq\lambda_{k+|E|}(P^{s})=\lambda_{k+|E|}(I^{s})=\left(\frac{(k+|E|)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}.$

  1. (1)

If $\alpha=0$, then

$\left(\frac{(k-|V|+1)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}\leq\lambda_{k-|V|+1}(P^{s}_{f})\leq\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(P^{s}_{f})\leq\left(\frac{(k+|E|)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}.$

  2. (2)

If $\alpha>0$, then

$\left(\frac{(k-|V|+1)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}\leq\lambda_{k-|V|+1}(P^{s}_{f})\leq\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k+1}(P^{s}_{f})\leq\left(\frac{(k+|E|+1)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}.$

  3. (3)

If $\alpha<0$, then

$\left(\frac{(k-|V|)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}\leq\lambda_{k-|V|}(P^{s}_{f})\leq\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(P^{s}_{f})\leq\left(\frac{(k+|E|)\pi}{\mathcal{L}(\Gamma^{\delta})}\right)^{2}.$

∎

######  Theorem 4.3.

Let $\Gamma$ be finite compact connected metric graph with pendant vertices
equipped with Dirichlet conditions and all other internal vertices the
standard conditions are imposed. Let $|D|$ and $|S|$ denote the number of
Dirichlet and number of standard vertices in $\Gamma$. Then

$\lambda_{k}(\Gamma)\leq\left(\frac{(k-2+2\beta+2|D|+|S|)\pi}{\mathcal{L}(\Gamma)}\right)^{2}.$
(4.5)

###### Proof.

Let $\Gamma^{s}$ be same underlying metric graph as $\Gamma$, but equipped
with standard vertex conditions. Then by (4.1)

$\lambda_{k}(\Gamma^{s})\leq\left(\frac{(k-1+\beta+|E|)\pi}{\mathcal{L}(\Gamma^{s})}\right)^{2}.$

Obtain the graph $\Gamma$ by imposing Dirichlet conditions at required
vertices of $\Gamma^{s}$. Then by (3.1)

$\lambda_{k-|D|}(\Gamma)\leq\lambda_{k}(\Gamma^{s})\leq\lambda_{k}(\Gamma)\leq\lambda_{k+|D|}(\Gamma^{s}).$
$\lambda_{k}(\Gamma)\leq\left(\frac{(k-2+2\beta+|D|+|V|)\pi}{\mathcal{L}(\Gamma^{s})}\right)^{2}.$

Equivalently, this estimates can be written as,

$\lambda_{k}(\Gamma)\leq\left(\frac{(k-2+2\beta+2|D|+|S|)\pi}{\mathcal{L}(\Gamma)}\right)^{2}.$

∎

For the metric graphs in Theorem (4.1) and Theorem (4.3), similar but improved
estimates on the eigenvalues can be found in [1, Lemma 1.5]. However, for any
flower graph equipped with standard conditions the estimate (4.1) coincides
with the estimate presented in [1, Lemma 1.5].

###  4.2. Estimates on the eigenvalues of $\delta^{\prime}$-graph

We establish a few estimates on the lowest eigenvalue of the
$\delta^{\prime}$-graph using some trials functions from the domain of the
quadratic form and also on general eigenvalues using some established
inequalities.
Since the domain of the quadratic form of the $\delta^{\prime}$-graph
$\Gamma^{\delta^{\prime}}$ contain a constant function, in particular it
contains
$\hat{\varphi}(x)=\frac{1}{\sqrt{\mathcal{L}(\Gamma^{\delta^{\prime}})}}$.
Therefore,

$\displaystyle\lambda_{1}(\Gamma^{\delta^{\prime}})$
$\displaystyle=\underset{\varphi\in
D(h^{\delta^{\prime}})}{\min}h^{\delta^{\prime}}[\varphi]$
$\displaystyle=\underset{\varphi\in
D(h^{\delta^{\prime}})}{\min}\left(\int_{\Gamma^{\delta^{\prime}}}|\varphi^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\varphi(x_{i})\right|^{2}\right)$
$\displaystyle\leq\int_{\Gamma^{\delta^{\prime}}}|\hat{\varphi}^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\hat{\varphi}(x_{i})\right|^{2}$
$\displaystyle=\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\frac{1}{\sqrt{\mathcal{L}(\Gamma^{\delta^{\prime}})}}\right|^{2}$
$\displaystyle=\frac{1}{\mathcal{L}(\Gamma^{\delta^{\prime}})}\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}1\right|^{2}$
$\displaystyle=\frac{1}{\mathcal{L}(\Gamma^{\delta^{\prime}})}\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}|d_{v_{m}}|^{2}$
$\displaystyle=\frac{1}{\mathcal{L}(\Gamma^{\delta^{\prime}})}\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}|d_{v_{m}}|^{2}$

######  Remark 4.4.

Even if there are some vertices where $\alpha^{\prime}_{m}=0$, we can still
use this upper bound and in this case the summation is taken over the vertices
where strengths $\alpha^{\prime}_{m}$ are non zero.

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{1}{\mathcal{L}(\Gamma^{\delta^{\prime}})}\left(\sum\limits_{\\{v\in
V|\alpha^{\prime}_{v}\neq 0\\}}\frac{d^{2}_{v}}{\alpha^{\prime}_{v}}\right)$

  1. (1)

If the degree of each vertex is $n$ then,

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{n^{2}}{\mathcal{L}(\Gamma^{\delta^{\prime}})}\sum\limits^{|V|}_{m=1}\frac{1}{\alpha^{\prime}_{m}}.$

  2. (2)

If the strength at each vertex is same, say $\alpha^{\prime}$, then

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{1}{\alpha^{\prime}\mathcal{L}(\Gamma^{\delta^{\prime}})}\sum\limits^{|V|}_{m=1}|d_{v_{m}}|^{2}.$

  3. (3)

In addition, if the total number of vertices are $|V|$ and degree of each
vertex is $n$ then,

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{n^{2}\
|V|}{\alpha^{\prime}\mathcal{L}(\Gamma^{\delta^{\prime}})}.$

  4. (4)

If the strengths are evenly distributed, then

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{n^{2}\
|V|^{2}}{\alpha^{\prime}\mathcal{L}(\Gamma^{\delta^{\prime}})}.$

The question now is how close one can push the lowest eigenvalue towards this
upper bound. The quadratic form does not converge very close to lowest
eigenvalue when the strengths are evenly distributed. So, evenly distribution
does not increase the eigenvalue. How can we get better estimation?
Total degree of a graph is defined as the sum of degree of each vertex.
Handshaking theorem states that the degree of a graph $\Gamma$ equals twice
the number of edges in $\Gamma$ and the total degree of a graph is even.

$degree\ of\ a\ graph=\sum_{v\in V}deg(v)=2\ (number\ of\ edges\ in\ \Gamma)$

######  Remark 4.5.

We express the upper bound on the lowest eigenvalue in terms of total degree
of a graph and total number of edges. Although, the following is not a good
approximation as compare to earlier one. Let $|V|$ and $|E|$ denote the total
number of vertices and total number of edges in a graph $\Gamma$,
respectively. Since

$\displaystyle|d_{v_{1}}|^{2}+|d_{v_{2}}|^{2}+\cdots+|d_{v_{|V|}}|^{2}$
$\displaystyle\leq(|d_{v_{1}}|+|d_{v_{2}}|+\cdots+|d_{v_{|V|}}|)^{2}$
$\displaystyle=[deg(\Gamma)]^{2}$ $\displaystyle=4\ |E|^{2}.$

If the strengths are evenly distributed, then

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{|V|}{\beta\mathcal{L}(\Gamma^{\delta^{\prime}})}\
[deg(\Gamma^{\delta^{\prime}})]^{2},$

and in terms of number of edges,

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{4\ |V|\
|E|^{2}}{\beta\mathcal{L}(\Gamma^{\delta^{\prime}})}.$

If we let $\alpha^{\prime}_{m}=\infty$ at each vertex then,
$\lambda_{1}(\Gamma)=0$. Maybe we can increase the eigenvalue and push it
towards the upper bound by adding the maximum number of edges with a large
length and minimum vertices.

Since there is no restriction on the domain of the quadratic form, so we could
have used other functions that belongs to $H^{1}(\Gamma^{\delta^{\prime}})$ as
a test functions i.e. $\cos(x),\sin(x)$, but the bounds that we get using
these functions are not so explicitly nice. Let us use the functions
$\varphi(x)=\sin(\frac{\pi x}{\ell_{j}}),\quad x\in e_{j}=[0,\ell_{j}]$ and
$\varphi(x)=\cos(\frac{\pi x}{\ell_{j}}),\quad x\in
e_{j}=[\frac{-\ell_{j}}{2},\frac{\ell_{j}}{2}]$. Then,

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{\pi^{2}}{L(\Gamma^{\delta^{\prime}})}\sum\limits_{j=1}^{|E|}\frac{1}{\ell_{j}}.$

If the length of each edge is equal, say, $\ell$, then

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\left(\frac{\pi}{\ell}\right)^{2}.$

We can also use the function $\varphi(x)=\sin(\frac{2\pi x}{\ell_{j}}),\quad
x\in e_{i}=[0,\ell_{j}]$, as a test function and we get an upper bound

$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{4\pi^{2}}{L(\Gamma^{\delta^{\prime}})}\sum\limits_{j=1}^{|E|}\frac{1}{\ell_{j}}.$

Let us consider the function $\varphi(x)=\cos(\frac{2\pi x}{\ell_{j}}),\quad
x\in e_{j}=[0,\ell_{j}]$. Clearly, this function is in the domain of the
quadratic form and can be used as a test function to get an upper bound on
first eigenvalue and $\varphi(0)=\varphi(\ell_{j})=1$ implies that
$\sum\limits_{x_{i}\in v_{m}}\varphi(x_{i})=d_{v_{m}}$, for each
$m=1,2,\cdots,|V|.$

$\displaystyle h^{\delta^{\prime}}[\varphi]$
$\displaystyle=\int_{\Gamma^{\delta^{\prime}}}|\varphi^{\prime}(x)|^{2}dx+\sum\limits_{m=1}^{|V|}\frac{1}{\alpha^{\prime}_{m}}\left|\sum\limits_{x_{i}\in
v_{m}}\varphi(x_{i})\right|^{2},$
$\displaystyle=2\pi^{2}\sum\limits_{j=1}^{|E|}\frac{1}{\ell_{j}}+\sum\limits_{m=1}^{|V|}\frac{d^{2}_{v_{m}}}{\alpha^{\prime}_{m}}.$
$<\varphi,\varphi>=\int_{\Gamma^{\delta^{\prime}}}|\varphi(x)|^{2}dx=\sum\limits_{j=1}^{|E|}\frac{\ell_{j}}{2}=\frac{L(\Gamma^{\delta^{\prime}})}{2}.$
$\lambda_{1}(\Gamma^{\delta^{\prime}})\leq\frac{2}{L(\Gamma^{\delta^{\prime}})}\left(2\pi^{2}\sum\limits_{j=1}^{|E|}\frac{1}{\ell_{j}}+\sum\limits_{m=1}^{|V|}\frac{d^{2}_{v_{m}}}{\alpha^{\prime}_{m}}\right).$

For any fix real number $\lambda$, the eigenvalue counting function
$N_{\Gamma}(\lambda)$ is defined as the number of eigenvalues of a graph
$\Gamma$ smaller than $\lambda$. Since the quantum graph $\Gamma$ is finite
compact, and the operator is self-adjoint with a discrete spectrum bounded
from below. Therefore the value of function $N_{\Gamma}(\lambda)$ is finite.

$N_{\Gamma}(\lambda)=\\#\\{\lambda_{i}\in\sigma(\Gamma):\lambda_{i}\leq\lambda\\}.$

######  Theorem 4.6.

Let $\Gamma^{a}$ be finite compact graph of total length
$\mathcal{L}(\Gamma^{a})$ with $|V|$ the number of vertices and $|E|$ the
number of edges. Let each vertex of $\Gamma^{a}$ be equipped with anti-
standard vertex conditions. Then

$\left(\frac{\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}(k-|V|)^{2}\leq\lambda_{k}(\Gamma^{a})\leq\left(\frac{\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}(k+|E|-1)^{2}.$
(4.6)

###### Proof.

Since the interaction strengths are zero at each vertex, therefore the
quadratic form is non-negative, and thus the eigenvalues $\lambda\geq 0$.
Hence the lower estimate in (4.6) is interesting only if $k>|V|$.
Let $\Gamma^{d}$ be the graph obtained from $\Gamma^{a}$ by imposing the
Dirichlet Conditions at each vertex of $\Gamma^{a}$, and let $h^{a}$ and
$h^{d}$ denotes the quadratic form of the graphs $\Gamma^{a}$ and $\Gamma^{d}$
with domains $D(h^{a})$ and $D(h^{d})$, respectively. Since the expression of
the quadratic forms $h^{a}$ and $h^{d}$ are same; moreover, if a function
$\varphi$ satisfy Dirichlet conditions at some vertex $v$, then it also
satisfies the anti-standard vertex condition that vertex. Therefore, the
domain $D(h^{d})$ is a subspace of $D(h^{a})$ and the quadratic forms $h^{a}$
and $h^{d}$ agree on $D(h^{d})$, thus minimizing over a smaller domain results
in large eigenvalues. Furthermore, the domain $D(h^{d})$ is a co-dimension one
subspace of $D(h^{a})$. Thus, by the rank one nature perturbation, the
following interlacing inequalities holds.

$\lambda_{k}(\Gamma^{a})\leq\lambda_{k}(\Gamma^{d})\leq\lambda_{k+|V|}(\Gamma^{a})$

Let for some $\lambda\in\mathbb{R}$, $N_{\Gamma^{a}}(\lambda)$ and
$N_{\Gamma^{d}}(\lambda)$ denotes the eigenvalue counting functions for the
graphs $\Gamma^{a}$ and $\Gamma^{d}$, respectively. Then the above interlacing
inequalities implies the following relation between eigenvalue counting
functions.

$N_{\Gamma^{d}}(\lambda)\leq N_{\Gamma^{a}}(\lambda)\leq
N_{\Gamma^{d}}(\lambda)+|V|.$ (4.7)

Consider the Laplace operator acting on a interval of length $\ell$ , and
assume that the end points are equipped with Dirichlet Conditions. The
eigenvalues are $\lambda_{k}=\left(\frac{k\pi}{\ell}\right)^{2}$, for
$\lambda\geq 0$ the value of eigenvalue counting function is
$N_{[0,\ell]}(\lambda)=\left[\frac{\sqrt{\lambda}}{\pi}\ell\right].$ Where
square brackets mean to take the integer part of the argument. Since Dirichlet
conditions imposed at a vertex of degree two or more does not connect the
individual function living on the incident edge in any way. The Dirichlet
conditions have the effect of disconnecting the vertex of degree $d_{v}$ into
$d_{v}$ vertices of degree one, so the graph $\Gamma^{d}$ is now decoupled
into a set of intervals, and the set of eigenvalues is just the union of
eigenvalues of each interval (counting multiplicities). Moreover,

The counting function is given by

$N_{\Gamma^{d}}(\lambda)=\sum\limits_{j=1}^{|E|}N_{[0,\ell_{j}]}(\lambda)=\left[\frac{\sqrt{\lambda}}{\pi}\ell_{1}\right]+\left[\frac{\sqrt{\lambda}}{\pi}\ell_{2}\right]+\cdots+\left[\frac{\sqrt{\lambda}}{\pi}\ell_{|E|}\right]\leq\left[\frac{\sqrt{\lambda}}{\pi}\mathcal{L}(\Gamma^{a})\right].$
(4.8)

Since for any $a$ and $b$, $[a]+[b]\leq[a+b]$, therefore taking integer part
of the sum of the terms is increased at most by the number of terms minus one
as compared to adding integer parts only. As the number of terms in (4.8) are
equal to number of edges $|E|$, thus

$\left[\frac{\sqrt{\lambda}}{\pi}\mathcal{L}(\Gamma^{a})\right]=\left[\frac{\sqrt{\lambda}}{\pi}(\ell_{1}+\ell_{2}+\cdots+\ell_{|E|})\right]\leq\left[\frac{\sqrt{\lambda}}{\pi}\ell_{1}\right]+\left[\frac{\sqrt{\lambda}}{\pi}\ell_{2}\right]+\cdots+\left[\frac{\sqrt{\lambda}}{\pi}\ell_{|E|}\right]+|E|-1,$

and we have

$\left[\frac{\sqrt{\lambda}}{\pi}\mathcal{L}(\Gamma^{a})\right]-|E|+1\leq
N_{\Gamma^{d}}(\lambda).$ (4.9)

∎

Thus, from (4.7),(4.8) and (4.9), we obtain the following inequalities.

$\left[\frac{\sqrt{\lambda}}{\pi}\mathcal{L}(\Gamma^{a})\right]-|E|+1\leq
N_{\Gamma^{d}}(\lambda)\leq N_{\Gamma^{a}}(\lambda)\leq
N_{\Gamma^{d}}(\lambda)+|V|\leq\left[\frac{\sqrt{\lambda}}{\pi}\mathcal{L}(\Gamma^{a})\right]+|V|.$

Setting $\lambda=\left(\dfrac{k\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}$ we
get

$k-|E|+1\leq
N_{\Gamma^{a}}\left(\frac{k\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}\leq
k+|V|,$

so

$\lambda_{k-|E|+1}\leq\left(\frac{k\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}\leq\lambda_{k+|V|}.$

This estimate implies that the multiplicity of the eigenvalues is uniformly
bounded by $|E|+|V|$. setting $\tilde{k}=k-|E|+1$ we get
$\lambda_{\tilde{k}}\leq\left(\dfrac{\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}(\tilde{k}+|E|-1)^{2}$
and similarly setting $\tilde{k}=k+|V|$ we get
$\left(\dfrac{\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}(\tilde{k}-|V|)^{2}\leq\lambda_{\tilde{k}}$.
Since $\tilde{k}$ is a dummy subscript so we can replace it with $k$ and
finally, we obtain,

$\left(\frac{\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}(k-|V|)^{2}\leq\lambda_{k}(\Gamma^{a})\leq\left(\frac{\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}(k+|E|-1)^{2}.$

Another type of vertex condition that decouples the graph into disjoints
interval is the Neumann conditions. For these conditions, the derivatives of
functions living on incident edges are zero at each vertex, and no conditions
are assumed on the values of functions. That is,

$\varphi_{e}^{\prime}(v)=0$

Let $\Gamma^{n}$ be the finite compact metric graph with Neumann vertex
conditions, then the quadratic form is defined by,

$h^{n}(\varphi)=\int_{\Gamma^{n}}|\varphi^{\prime}|^{2}dx$

with domain $D(h^{n})=H^{1}(\Gamma^{n}\backslash V)$. The domain $D(h^{a})$ is
a co-dimension one subspace of $D(h^{n})$, and the quadratic forms $h^{a}$ and
$h^{n}$ agree on $D(h^{a})$. Thus, by rank one perturbation, the following
interlacing inequalities holds.

$\lambda_{k}(\Gamma^{n})\leq\lambda_{k}(\Gamma^{a})\leq\lambda_{k+|V|}(\Gamma^{n})$

Let for some $\lambda\in\mathbb{R}$, $N_{\Gamma^{a}}(\lambda)$ and
$N_{\Gamma^{n}}(\lambda)$ denotes the eigenvalue counting functions for the
graphs $\Gamma^{a}$ and $\Gamma^{n}$, respectively. Then the above interlacing
inequalities implies the following relation between eigenvalue counting
functions.

$N_{\Gamma^{a}}(\lambda)\leq N_{\Gamma^{n}}(\lambda)\leq
N_{\Gamma^{a}}(\lambda)+|V|.$ (4.10)

In the previous theorem, if we impose Neumann conditions instead of Dirichlet
conditions to decouple the graph $\Gamma^{a}$, then one can prove the
following theorem correspondingly.

######  Theorem 4.7.

Let $\Gamma^{a}$ be finite compact graph of total length
$\mathcal{L}(\Gamma^{a})$ with $|V|$ the number of vertices and $|E|$ the
number of edges. Let each vertex of $\Gamma^{a}$ be equipped with anti-
standard vertex conditions. Then

$\left(\frac{k\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}\leq\lambda_{k}(\Gamma^{a})\leq\left(\frac{\pi}{\mathcal{L}(\Gamma^{a})}\right)^{2}(k+|E|+|V|-1)^{2}.$
(4.11)

######  Theorem 4.8.

Let $\Gamma^{\delta^{\prime}}$ be metric graph with negative strengths
$\alpha^{\prime}_{m}<0$ at each vertex. Then,

$\left(\frac{(k-2|V|)\pi}{2\mathcal{L}(\Gamma^{\delta^{\prime}})}\right)^{2}\leq\lambda_{k}(\Gamma^{\delta^{\prime}}),\quad
k\geq 2|V|.$ (4.12)

###### Proof.

Let $\hat{\Gamma}^{\delta^{\prime}}$ be a double cover obtained by doubling
each edge in $\Gamma^{\delta^{\prime}}$ and assigning strengths
$2\alpha^{\prime}_{m}$ at each vertex. Then each eigenvalue of
$\Gamma^{\delta^{\prime}}$ is also an eigenvalue of
$\hat{\Gamma}^{\delta^{\prime}}$. However, the converse is not true. Let
$\lambda(\Gamma^{\delta^{\prime}})$ be an eigenvalue corresponding to the
eigenfunction $\varphi$. The eigenfunction $\varphi$ can be lifted on
$\hat{\Gamma}^{\delta^{\prime}}$ by letting it assume the same values on the
new edges as on edges of $\Gamma^{\delta^{\prime}}$. Then newly constructed
function $\hat{\varphi}$ is an eigenfunction for
$\hat{\Gamma}^{\delta^{\prime}}$ corresponding to eigenvalue
$\lambda(\Gamma^{\delta^{\prime}})$, because for
$\lambda(\Gamma^{\delta^{\prime}})$ this function satisfies the eigenvalue
equation and vertex conditions on $\hat{\Gamma}^{\delta^{\prime}}$. Thus,

$\lambda_{k}(\hat{\Gamma}^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{\delta^{\prime}}).$

Let $\mathcal{C}^{\delta^{\prime}}$ be the graph obtained from
$\hat{\Gamma}^{\delta^{\prime}}$ by splitting each of the vertices of
$\hat{\Gamma}^{\delta^{\prime}}$ to form a cycle of length
$2\mathcal{L}(\mathcal{C}^{\delta^{\prime}})$, since each vertex has strength
$2\alpha^{\prime}_{m}<0$, therefore [Theorem (3.3)] implies,

$\lambda_{k}(\mathcal{C}^{\delta^{\prime}})\leq\lambda_{k}(\hat{\Gamma}^{\delta^{\prime}}).$

Obtain a graph $\mathcal{C}^{a}$ by assigning the zero at all vertices of
$\mathcal{C}^{\delta^{\prime}}$. Then by Theorem (3.1),

$\lambda_{k-2|V|}(\mathcal{C}^{a})\leq\lambda_{k}(\mathcal{C}^{\delta^{\prime}}).$

The graph $\mathcal{C}^{a}$ is a bipartite metric graph, therefore [15,
Theorem 3.6] implies that

$\lambda_{k-2|V|}(\mathcal{C}^{s})=\lambda_{k-2|V|}(\mathcal{C}^{a})$

The graph $\mathcal{C}$ is a loop of length
$2\mathcal{L}(\Gamma^{\delta^{\prime}})$, and now we cut the graph
$\mathcal{C}^{s}$ to form an edge $I$ of length
$2\mathcal{L}(\Gamma^{\delta^{\prime}}).$ Thus,

$\left(\frac{(k-2|V|)\pi}{2\mathcal{L}(\Gamma^{\delta^{\prime}})}\right)^{2}=\lambda_{k-2|V|}(I)\leq\lambda_{K-2|V|}(\mathcal{C}^{s}).$

∎

A similar lower bound on the eigenvalues of the Laplacian acting on the edges
of a star graph with anti-standard vertex condition was proved in [21, Theorem
1 ]. We will use it to get a lower estimate on the metric graph $\Gamma$ with
$\delta^{\prime}$-conditions.

######  Theorem 4.9.

Let $\Gamma^{\delta^{\prime}}$ be finite compact connected metric graph with
$|E|\geq 2$, and $\delta^{\prime}$-conditions equipped at each vertex with
total strength $\alpha^{\prime}=\sum\limits^{|V|}_{m=1}\alpha^{\prime}_{m}$.
Let $\ell_{1}\geq\ell_{2}\geq\cdots\geq\ell_{|E|}$ denotes the length of each
edge $e_{j}$, then for $k,|E|\in\mathbb{N}$, and $1\leq j\leq|E|$, we have

$min\left\\{\frac{\left(2\left(k-\left(1+\frac{|V|+1}{|E|}\right)\right)+3\right)^{2}\pi^{2}}{4\ell_{1}^{2}},\frac{\left(2\left(k-\left(1+\frac{|V|+1}{|E|}\right)\right)+1\right)^{2}\pi^{2}}{4\ell_{j}^{2}}\right\\}\leq\lambda_{k|E|+j}(\Gamma^{\delta^{\prime}})$
(4.13)

###### Proof.

  1. Let $S^{a}$ denote the star graph constructed from the edges of $\Gamma^{\delta^{\prime}}$ and let each vertex of $S^{a}$ be equipped with anti-standard vertex condition. Then by [21, Theorem 1 ]

$min\left\\{\frac{(2k+3)^{2}\pi^{2}}{4\ell_{1}^{2}},\frac{(2k+1)^{2}\pi^{2}}{4\ell_{j}^{2}}\right\\}\leq\lambda_{k|E|+j}(S^{a}).$
(4.14)

Construct a new star graph $S^{\delta^{\prime}}$ by assigning the negative
strengths at vertices of $S^{a}$ . Then by Theorem (3.1),

$\lambda_{k|E|+j}(S^{a})\leq\lambda_{k|E|+j+|E|+1}(S^{\delta^{\prime}}).$

Now, first create a flower graph $\Gamma^{\delta^{\prime}}_{f}$ from
$S^{\delta^{\prime}}$ in accordance to Theorem(3.3)(2). Thus,

$\lambda_{k|E|+j+|E|+1}(S^{\delta^{\prime}})\leq\lambda_{k|E|+j+|E|+1}(\Gamma^{\delta^{\prime}}_{f}).$

and then obtain a graph $\Gamma^{a}_{f}$ by prescribing anti-standard
conditions to $\Gamma^{\delta^{\prime}}_{f}$.
We restore the vertices of $\Gamma^{\delta^{\prime}}$ by cutting, step by
step, the vertex of flower graph $\Gamma^{a}_{f}$, then by Theorem(3.3)(6),
the obtained graph is same as $\Gamma^{\delta^{\prime}}$ but equipped with
anti-standard conditions. Call this new graph $\Gamma^{a}$, Thus

$\lambda_{k|E|+j+|E|+1}(\Gamma^{\delta^{\prime}}_{f})\leq\lambda_{k|E|+j+|E|+1}(\Gamma^{a}_{f})\leq\lambda_{k|E|+j+|E|+1}(\Gamma^{a}).$

Now, assign the strengths at all vertices of $\Gamma^{a}$ to obtain the graph
$\Gamma^{\delta^{\prime}}$. Thus

$\lambda_{k|E|+j+|E|+1}(\Gamma^{a})\leq\lambda_{k|E|+j+|E|+|V|+1}(\Gamma^{\delta^{\prime}}).$
$\displaystyle k|E|+j+|E|+|V|+1$ $\displaystyle=k|E|+|E|+|V|+1+j$
$\displaystyle=k|E|+\frac{(|E|+|V|+1)|E|}{|E|}+j$
$\displaystyle=\left(k+\frac{(|E|+|V|+1)}{|E|}\right)|E|+j$

Let $k^{\prime}=\left(k+\frac{(|E|+|V|+1)}{|E|}\right)$ $\implies$
$k=k^{\prime}-\left(1+\frac{|V|+1}{|E|}\right).$

$min\left\\{\frac{(2k+3)^{2}\pi^{2}}{4\ell_{1}^{2}},\frac{(2k+1)^{2}\pi^{2}}{4\ell_{j}^{2}}\right\\}\leq\lambda_{k|E|+j+|E|+|V|+1}(\Gamma^{\delta^{\prime}}).$
(4.15)
$min\left\\{\frac{\left(2\left(k^{\prime}-\left(1+\frac{|V|+1}{|E|}\right)\right)+3\right)^{2}\pi^{2}}{4\ell_{1}^{2}},\frac{\left(2\left(k^{\prime}-\left(1+\frac{|V|+1}{|E|}\right)\right)+1\right)^{2}\pi^{2}}{4\ell_{j}^{2}}\right\\}\leq\lambda_{k^{\prime}|E|+j}(\Gamma^{\delta^{\prime}})$

∎

######  Definition 4.10.

A spanning tree $T$ of a graph $\Gamma$ is an undirected subgraph of $\Gamma$
containing maximal edges of $\Gamma$ with no cycles, or it contains all the
vertices of $\Gamma$ connected by a minimal set of edges of $\Gamma$.

A graph $\Gamma$ can have more than one spanning tree, each having an equal
number of edges, let $|E_{T}|$ denotes the cardinality of the edge set of
spanning tree $T$. Each spanning tree has it’s own length so we can compare
their lengths, we define a new term for our later use.

######  Definition 4.11.

A maximal spanning tree $T_{m}$ is a spanning tree for a graph $\Gamma$ having
the largest length among all the spanning trees of $\Gamma$.

For a finite, compact connected metric tree $T^{s}$ with standard conditions,
let $P^{s}$ be the largest path inside $T^{s}$. Since the spectrum of
Laplacian on the path graph $P^{s}$ with standard conditions consists of
discrete points

$\lambda_{k+1}(P^{s})=\left(\frac{k\pi}{\mathcal{L}(P^{s})}\right)^{2},\quad
k=0,1,2,\cdots$

The tree graph $T^{s}$ can be obtained by attaching pendant edges with
standard conditions to the vertices of $P^{s}$ in accordance to [3, Theorem
3.10]. Thus,

$\lambda_{k+1}(T^{s})\leq\lambda_{k+1}(P^{s})=\left(\frac{k\pi}{\mathcal{L}(P^{s})}\right)^{2},\quad
k=0,1,2,\cdots$ (4.16)

Now consider a finite compact connected metric tree $T^{a}$ with anti-standard
conditions and let $P^{a}$ be the longest path inside $T^{a}$. Since the
spectrum of Laplacian on the path graph $P^{a}$ with anti-standard conditions
consists of discrete points

$\lambda_{k}(P^{a})=\left(\frac{k\pi}{\mathcal{L}(P^{a})}\right)^{2},\quad
k=1,2,\cdots$

The tree graph $T^{a}$ can be obtained by attaching pendant edges with anti-
standard conditions to the vertices of $P^{a}$ in accordance to [19, Theorem
3.5]. Thus

$\lambda_{k}(T^{a})\leq\lambda_{k}(P^{a})=\left(\frac{k\pi}{\mathcal{L}(P^{a})}\right)^{2},\quad
k=1,2,\cdots$

Or using Theorem(3.8)

$\lambda_{k+1}(T^{a})\leq\lambda_{k}(P^{a})=\left(\frac{k\pi}{\mathcal{L}(P^{a})}\right)^{2},\quad
k=1,2,\cdots$

One can obtain these estimates with any path graph inside the metric tree,
even an interval would give an estimate. However, we choose the longest path
to get better estimates. The weak version of these bounds were established in
[20, Theorem 3.2] for standard conditions and in [21, Theorem 3] for anti-
standard conditions, in which a particular path graph $P$ was chosen which
starts with the longest edge and ends at second largest edge. These estimates
were further expressed in terms of average length of metric tree. For standard
conditions,

$\lambda_{k+1}(T^{s})\leq\lambda_{k+1}(P^{s})=\left(\frac{k\pi}{\mathcal{L}(P^{s})}\right)^{2}\leq\left(\frac{k\pi|E|}{2\mathcal{L}(T^{s})}\right)^{2},\quad
k=0,1,2,\cdots$ (4.17)

Similarly, for anti-standard conditions,

$\lambda_{k}(T^{a})\leq\lambda_{k}(P^{a})=\left(\frac{k\pi}{\mathcal{L}(P^{a})}\right)^{2}\leq\left(\frac{k\pi|E|}{2\mathcal{L}(T^{a})}\right)^{2},\quad
k=1,2,\cdots$ (4.18)

The following simple but useful relation was proved in [15, Corollary 3.8 ],
which connects the eigenvalues of metric tree with standard and anti-standard
vertex conditions.

$\lambda_{k}(T^{a})=\lambda_{k+1}(T^{s}),\quad k\ \in\mathbb{N}.$ (4.19)

We use these estimates to obtain some bounds on the spectrum of the Laplacian
acting on a finite compact connected metric graph $\Gamma$ with standard and
anti-standard vertex conditions, and then we will generalize these estimates
for $\delta$ and $\delta^{\prime}$\- type vertex conditions. Most of the
following estimates are equal in magnitude but are expressed in terms of
different geometrical parameters of the metric graph $\Gamma$.
For the anti-standard vertex conditions, we know that either attaching a
pendant graph (particularly an edge), or attaching an edge between existing
vertices lowers all eigenvalues [19, Theorem 3.2 , 3.5 ]. Therefore,

$\lambda_{k}(\Gamma^{a})\leq\lambda_{k}(T^{a})\leq\lambda_{k}(P^{a})=\left(\frac{k\pi\gamma}{\mathcal{L}(\Gamma^{a})}\right)^{2},\quad
k=1,2,\cdots$

Where $\gamma=\frac{\mathcal{L}(\Gamma^{a})}{\mathcal{L}(P^{a})}$. If we let
$\gamma=\mathcal{L}(\Gamma^{a})-\mathcal{L}(P^{a})$, then the same estimate
obtained by the path graph can be rewritten as

$\lambda_{k}(\Gamma^{a})\leq\lambda_{k}(P^{a})=\left(\frac{k\pi}{\mathcal{L}(\Gamma^{a})-\gamma}\right)^{2},\quad
k=1,2,\cdots$

Note that the above bounds are equal to
$\lambda_{k}(P^{a})=\left(\frac{k\pi}{\mathcal{L}(P^{a})}\right)^{2}$ in
magnitude, but are expressed in terms of total length of the graph.
Let $T^{a}$ be the tree graph inside $\Gamma^{a}$ with the longest length
among all the tree inside $\Gamma^{a}$. Then we use the estimate (4.18) on
$T^{a}$ to obtain an estimate on $\Gamma^{a}$.

$\lambda_{k}(\Gamma^{a})\leq\lambda_{k}(T^{a})\leq\left(\frac{k\pi|E_{T}|}{2\mathcal{L}(T^{a})}\right)^{2},\quad
k=1,2,\cdots$

Moreover, if the graph $\Gamma^{a}$ with $|E|$ number of edges is obtained by
from metric tree $T^{a}$ by either attaching edges or loops of lengths
$\ell_{1},\ell_{2},\cdots,\ell_{m}$, then

$\lambda_{k}(\Gamma^{a})\leq\left(\frac{k\pi(|E|-m)}{2\left(\mathcal{L}(\Gamma^{a})-\sum\limits_{i=1}^{m}\ell_{i}\right)}\right)^{2},\quad
k=1,2,\cdots$

If $T^{a}_{m}$ is the maximal spanning tree inside $\Gamma^{a}$, then

$\lambda_{k}(\Gamma^{a})\leq\lambda_{k}(T^{a}_{m})\leq\left(\frac{k\pi|E_{T_{m}}|}{2\mathcal{L}(T^{a}_{m})}\right)^{2},\quad
k=1,2,\cdots$

The number of vertices in $T^{a}_{m}$ are equal to the number of vertices in
$\Gamma^{a}$, and the number of edges in $T^{a}_{m}$ are one less than the
number of vertices in $\Gamma^{a}$. Let $|V|$ denotes the number of vertices
in $\Gamma^{a}$, and the graph $\Gamma^{a}$ can be obtained from $T^{a}_{m}$
by attaching pendant loops or multiple edges between existing vertices. Thus

$\lambda_{k}(\Gamma^{a})\leq\lambda_{k}(T^{a}_{m})\leq\left(\frac{k\pi\gamma(|V|-1)}{2\mathcal{L}(\Gamma^{a})}\right)^{2},\quad
k=1,2,\cdots$ (4.20)

The above bounds can be expressed in terms of the first Betti number of the
graph $\Gamma^{a}$. Since the first Betti number of maximal spanning tree is
zero, and while constructing $\Gamma^{a}$ from $T^{a}_{m}$ each attachment of
either loop or an edge between existing vertices increases the Betti number by
one. Let $\beta$ be number of cycles in graph $\Gamma^{a}$ and let
$\ell_{1},\ell_{2},\cdots,\ell_{\beta}$ be the number of required edges to
obtain $\Gamma^{a}$ from $T^{a}_{m}$. Then,

$\lambda_{k}(\Gamma^{a})\leq\left(\frac{k\pi(|E|-\beta)}{2\left(\mathcal{L}(\Gamma^{a})-\sum\limits_{i=1}^{m}\ell_{i}\right)}\right)^{2},\quad
k=1,2,\cdots$

In addition, if $\ell_{1}=\ell_{2}=\cdots=\ell_{\beta}$, then

$\lambda_{k}(\Gamma^{a})\leq\left(\frac{k\pi(|E|-\beta)}{2\left(\mathcal{L}(\Gamma^{a})-\beta\ell\right)}\right)^{2},\quad
k=1,2,\cdots$

For different types of metric tree inside the graph $\Gamma^{a}$, we obtain
different expression for bounds, some of these were equal in magnitude but
expressed in different topological parameters. Now we chose a particular type
of a metric tree inside $\Gamma^{a}$, and will use the estimate (4.18). Let
$T^{a}$ be the tree inside $\Gamma^{a}$ such that the number of edges and the
length of each edge in $T^{a}$ and $\Gamma^{a}$ are equal. Then $\Gamma^{a}$
can be obtained by gluing some vertices of $T^{a}$ and each gluing produces a
cycle. Thus

$\lambda_{k}(\Gamma^{a})\leq\lambda_{k}(T^{a})\leq\left(\frac{k\pi|E|}{2\mathcal{L}(\Gamma^{a})}\right)^{2},\quad
k=1,2,\cdots$ (4.21)

Similarly, for the graph $\Gamma^{s}$ with standard vertex condition the
following bounds hold. Let $P^{s}$ be the longest path inside $\Gamma^{s}$,
then we attach the pendant edges, step by step, with standard condition to the
vertices of $P^{s}$ unless we obtain a tree $T^{s}$ with number of edges equal
to the number of edges $|E|$ of graph $\Gamma^{s}$. Now $\Gamma^{s}$ can be
obtained by pairwise gluing of $\beta$ pair of vertices.

$\lambda_{k+1-\beta}(\Gamma^{s})\leq\lambda_{k+1}(T^{s})\leq\lambda_{k+1}(P^{s})=\left(\frac{k\pi}{\mathcal{L}(P^{s})}\right)^{2},\quad
k=1,2,\cdots,$
$\lambda_{k+1}(\Gamma^{s})\leq\left(\frac{(k+\beta)\pi\gamma}{\mathcal{L}(\Gamma^{s})}\right)^{2},\quad
k=1,2,\cdots$

where $\gamma=\frac{\mathcal{L}(\Gamma^{s})}{L(P^{s})}$. If we consider the
longest metric tree $T^{s}$, then we can use the estimate (4.17),

$\lambda_{k+1}(\Gamma^{s})\leq\lambda_{k+1}(T^{s})\leq\left(\frac{(k+\beta)\pi|E_{T}|}{2\mathcal{L}(T^{s})}\right)^{2}=\left(\frac{(k+\beta)\pi|E_{T}|\gamma}{2\mathcal{L}(\Gamma^{s})}\right)^{2},\quad
k=1,2,\cdots$

If we choose a tree $T^{s}$ such that each vertex of $T^{s}$ is equipped with
standard vertex conditions and the number of edges and the length of each edge
is same as of $\Gamma^{s}$. However, the only difference between $T^{s}$ and
$\Gamma^{s}$ is that $\Gamma^{s}$ can be obtained from $T^{s}$ by pairwise
gluing of $\beta$ pair of vertices, then

$\lambda_{k+1}(\Gamma^{s})\leq\left(\frac{(k+\beta)\pi|E|}{2\mathcal{L}(\Gamma^{s})}\right)^{2}$
(4.22)

If $T^{s}_{m}$ is the maximal spanning tree inside $\Gamma^{s}$, then

$\lambda_{k+1}(\Gamma^{s})\leq\left(\frac{(k+\beta)(|V|-1)\pi}{2\left(\mathcal{L}(\Gamma^{s})-\sum\limits_{1}^{m}\ell_{i}\right)}\right)^{2}.$

Or

$\lambda_{k+1}(\Gamma^{s})\leq\left(\frac{(k+\beta)(|E|-\beta)\pi}{2\left(\mathcal{L}(\Gamma^{s})-\sum\limits_{1}^{m}\ell_{i}\right)}\right)^{2}.$

Where $|E|$, $|V|$ and $\beta$ denotes the number of edges, number of vertices
and the first Betti number of the graph $\Gamma^{s}$.

######  Definition 4.12.

Let $\Gamma$ be a finite compact connected metric graph. Then diameter of
$\Gamma$ is defined as,

$d(\Gamma):=\max\\{d(x,y):x,y\in\Gamma\\}.$

Let $T^{s}$ be a finite, connected metric tree and let $d(T^{s})$ denote the
diameter of $T^{s}$, then the following inequality was proved in [20, Theorem
3.4]

$\lambda_{k+1}(T^{s})\leq\left(\frac{k\pi}{d(T^{s})}\right)^{2}.$ (4.23)

A similar inequality for $T^{a}$ was proved in [21, Theorem 3],

$\lambda_{k}(T^{a})\leq\left(\frac{k\pi}{d(T^{a})}\right)^{2}.$ (4.24)

In order to establish better estimates on the eigenvalues of metric trees
$T^{s}$ and $T^{a}$, we require the following lemma which is also contained in
[2, Lemma 4.2]. However, we provide a brief proof for the completeness of the
presentation.

######  Lemma 4.13.

Let $T$ be a metric tree of total length $\mathcal{L}(T)$ with $|E_{p}|\geq 2$
the pendant edges, then

$\frac{2\mathcal{L}(T)}{|E_{p}|}\leq d(T).$ (4.25)

###### Proof.

We will show that the diameter of a metric tree is the distance between two
pendant points. Suppose that $d(\Gamma)=d(x_{1},x_{2})$ for some
$x_{1},x_{2}\in T$. To prove by contradiction we assume that $x_{1}$ is not a
pendant point of $T$, then $x_{1}$ is some interior point and the graph
$T\backslash x_{1}$ has at least two connected components. Let $T_{1}$ denote
of the component such that $x_{2}\notin T_{1}$ and $x_{1}\neq y\in T_{1}$.
Then

$d(y,x_{2})>d(x_{1},x_{2})=d(T).$

This holds because $T$ is a tree and any path inside $T$ that between $x_{2}$
and $y$ contains the point $x_{1}$. This contradicts our assumption of
$d(x_{1},x_{2})=d(T)$. Let $x_{0}$ be a middle point of a shortest path $P$
between $x_{1}$ and $x_{2}$. Thus,

$d(x_{1},x_{0})=\frac{d(T)}{2}=d(x_{0},x_{2}).$

Since the tree $T$ has $E_{p}$ pendant edges, therefore there are exactly
$E_{p}$ path of length at most $d(x_{1},x_{0})$, that connect the point
$x_{0}$ and a pendant vertex of $T$. Moreover, these paths covers the metric
tree $T$, thus

$\mathcal{L}(T)\leq\sum\limits_{v\text{ is a
pendant}}d(x_{0},v)\leq\sum\limits_{v\text{ is a
pendant}}d(x_{0},x_{1})=\frac{E_{p}d(T)}{2}.$

∎

The bounds (4.23) and (4.24) implies that

$\lambda_{k+1}(T^{s})\leq\left(\frac{k\pi|E_{p}|}{2\mathcal{L}(T^{s})}\right)^{2}$
(4.26)

and,

$\lambda_{k}(T^{a})\leq\left(\frac{k\pi|E_{p}|}{2\mathcal{L}(T^{a})}\right)^{2}.$
(4.27)

Thus, we get improved estimates as compared to the estimates (4.17) and (4.18)
on the eigenvalues of a metric trees $T^{s}$ and $T^{a}$. Therefore, the above
listed bounds can be improved when the estimates (4.17) and (4.18) are
replaced by (4.26) and (4.27).
We now use previously established estimates on metric graph with standard
conditions to obtain some estimates on the metric graph $\Gamma^{\delta}$. For
two same underlying metric graph $\Gamma^{\delta}_{\alpha}$ and
$\Gamma^{\delta}_{\alpha^{\prime}}$ with strengths $\alpha$ and
$\alpha^{\prime}$ at some vertex $v_{0}$, respectively, the following
interlacing inequality holds.

$\lambda_{k}(\Gamma^{\delta}_{\alpha})\leq\lambda_{k}(\Gamma^{\delta}_{\alpha^{\prime}})\leq\lambda_{k+1}(\Gamma^{\delta}_{\alpha}).$

Let $\Gamma^{\delta}$ be finite compact connected metric graph, and let
$|V^{p}|$ and $|V^{n}|$ denotes the number of vertices of $\Gamma^{\delta}$
with positive and negative strengths, respectively, and let $\Gamma^{s}$ be
the same underlying metric graph with standard conditions. If $\Gamma^{p}$ is
a graph obtained from $\Gamma^{s}$ by replacing the interaction strengths from
zero to positive at those vertices where positive strengths were specified in
$\Gamma^{\delta}$. Now $\Gamma^{\delta}$ can be obtained from this new graph
$\Gamma^{p}$ by assigning negative strengths at some vertices, where negative
strengths were assigned in $\Gamma^{\delta}$. Then following inequalities
holds.

$\lambda_{k}(\Gamma^{s})\leq\lambda_{k}(\Gamma^{p})\leq\lambda_{k+|V^{p}|}(\Gamma^{s}),$
$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{p})\leq\lambda_{k+|V^{n}|}(\Gamma^{\delta})$
$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{p})\leq\lambda_{k+|V^{p}|}(\Gamma^{s})\leq\left(\frac{(k+\beta+|V^{p}|-1)\pi|E|}{2\mathcal{L}(\Gamma^{\delta})}\right)^{2}$

Where last inequality is due to (4.22). If we replace the zero strengths with
negative strengths first and then with the positive strengths even then we
obtain the same bound.

$\lambda_{k}(\Gamma^{n})\leq\lambda_{k}(\Gamma^{s})\leq\lambda_{k+|V^{n}|}(\Gamma^{n})$
$\lambda_{k}(\Gamma^{n})\leq\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k+|V^{p}|}(\Gamma^{n})$
$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k+|V^{p}|}(\Gamma^{n})\leq\lambda_{k+|V^{p}|}(\Gamma^{s})\leq\left(\frac{(k+\beta+|V^{p}|-1)\pi|E|}{2\mathcal{L}(\Gamma^{\delta})}\right)^{2}$
(4.28)

The other way to establish an upper estimate for eigenvalues of the metric
graph $\Gamma^{\delta}$ is using the application of the corresponding flower
graph. Let $\Gamma^{\delta}$ be the metric graph with total interaction
strength $\alpha=\sum\limits_{m=1}^{|V|}\alpha_{m}$, and $\Gamma^{s}$ be same
graph with standard conditions at each vertex. Let $\Gamma^{s}_{f}$ and
$\Gamma^{\alpha}_{f}$ be the corresponding flower graph with strengths zero
and $\alpha$.
If $\alpha>0$, then

$\lambda_{k}(\Gamma^{s})\leq\lambda_{k}(\Gamma^{s}_{f})\leq\lambda_{k+|V|-1}(\Gamma^{s})$
$\lambda_{k}(\Gamma^{s}_{f})\leq\lambda_{k}(\Gamma^{\alpha}_{f})\leq\lambda_{k+1}(\Gamma^{s}_{f})$

Now restore the vertices of $\Gamma^{\delta}$ from $\Gamma^{\alpha}_{f}$, then

$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{\alpha}_{f})\leq\lambda_{k+|V|-1}(\Gamma^{\delta})$

$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{\alpha}_{f})\leq\lambda_{k+1}(\Gamma^{s}_{f})\leq\lambda_{k+|V|}(\Gamma^{s})\leq\left(\frac{(k-1+\beta+|V|)|E|\pi}{2L(\Gamma^{\delta})}\right)^{2}.$

If $\alpha\leq 0$,

$\lambda_{k}(\Gamma^{s})\leq\lambda_{k}(\Gamma^{s}_{f})\leq\lambda_{k+|V|-1}(\Gamma^{s})$
$\lambda_{k}(\Gamma^{\alpha}_{f})\leq\lambda_{k}(\Gamma^{s}_{f})\leq\lambda_{k+1}(\Gamma^{\alpha}_{f})$

Now restore the vertices of $\Gamma^{\delta}$ from $\Gamma^{\alpha}_{f}$, then

$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{\alpha}_{f})\leq\lambda_{k+|V|-1}(\Gamma^{\delta})$

$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{\alpha}_{f})\leq\lambda_{k}(\Gamma^{s}_{f})\leq\lambda_{k+|V|-1}(\Gamma^{s})\leq\left(\frac{(k-2+\beta+|V|)|E|\pi}{2L(\Gamma^{\delta})}\right)^{2}.$

Let $\Gamma_{a}^{\delta^{\prime}}$ be the finite compact connected metric
graph equipped with $\delta^{\prime}$-type conditions and possibly anti-
standard conditions at some vertices. Let $\Gamma^{a}$ be the same underlying
metric graph as $\Gamma_{a}^{\delta^{\prime}}$ with anti-standard vertex
conditions at each vertex. Then, with the help of Theorem (3.1), and the
estimate (4.21), we obtain

$\lambda_{k}(\Gamma_{a}^{\delta^{\prime}})\leq\lambda_{k}(\Gamma^{a})\leq\left(\frac{k|E|\pi}{2L(\Gamma_{a}^{\delta^{\prime}})}\right)^{2}.$

Let $B$ denote a finite compact connected bipartite metric graph, and let
$B^{a}$ and $B^{s}$ be the bipartite metric graphs equipped with anti-standard
and standard vertex conditions at each vertex of both graphs, respectively.
The following relation was established in [15, Theorem 3.6].

$\lambda_{k+\beta}(B^{a})=\lambda_{k+1}(B^{s}).$ (4.29)

We use this relation and the Theorem (3.1) to obtain a relation between the
eigenvalues of a graph with $\delta^{\prime}$-condition and the same graph
equipped with $\delta$-conditions. Let $B^{\delta^{\prime}}$ be the bipartite
metric graph and $B^{s}$ be same underlying bipartite metric graph equipped
with standard vertex conditions. Then from Theorem (3.1), we know that either
increasing or decreasing the strength at vertex from zero, decreases the
eigenvalues. Thus,

$\lambda_{k+\beta}(B^{\delta^{\prime}})\leq\lambda_{k+\beta}(B^{a})=\lambda_{k+1}(B^{s})\leq\lambda_{k+\beta+|V|}(B^{\delta^{\prime}}).$

Let $B^{\delta}$ be a bipartite graph with non-negative strengths at each
vertex, then

$\lambda_{k+\beta}(B^{\delta^{\prime}})\leq\lambda_{k+1}(B^{\delta}).$

Let $|V^{p}|$ and $|V^{n}|$ denote the number of vertices in $B^{\delta}$ with
positive and negative strengths, respectively, and $B^{s}$ be the same
underlying metric graph. Then the graph $B^{\delta}$ can be obtained from
$B^{s}$ in two steps. First, obtain a graph $B^{p}$ from $B^{s}$ by assigning
positive strengths to those vertices where positive strengths are assigned in
$B^{\delta}$, and then obtain $B^{\delta}$ by assigning negative strengths to
some vertices of $B^{p}$ wherever is required. Thus

$\lambda_{k+1}(B^{s})\leq\lambda_{k+1}(B^{p})\leq\lambda_{k+1+|V^{p}|}(B^{s})$
$\lambda_{k+1}(B^{\delta})\leq\lambda_{k+1}(B^{p})\leq\lambda_{k+1+|V^{n}|}(B^{\delta})$
$\lambda_{k+\beta}(B^{\delta^{\prime}})\leq\lambda_{k+1}(B^{s})\leq\lambda_{k+1}(B^{p})\leq\lambda_{k+1+|V^{n}|}(B^{\delta}).$

Therefore,

$\lambda_{k+\beta}(B^{\delta^{\prime}})\leq\lambda_{k+1+|V^{n}|}(B^{\delta})\quad\text{and}\quad\lambda_{k+1}(B^{\delta})\leq\lambda_{k+\beta+|V^{p}|}(B^{a}).$
(4.30)

Moreover, for any metric graph $\Gamma$, not necessarily bipartite, one can
obtain several bipartite metric graphs $B$ by creating vertices of degree two
equipped with standard conditions at some interior points of $\Gamma$.
Therefore, for any metric graph $\Gamma^{\delta}$ with non-negative strengths
at each vertex, the following relation hold.

$\lambda_{k+\beta}(B^{\delta^{\prime}})\leq\lambda_{k+\beta}(B^{a})=\lambda_{k+1}(B^{s})=\lambda_{k+1}(\Gamma^{s})\leq\lambda_{k+1}(\Gamma^{\delta}).$
(4.31)

Note that in the above relation the total length of the graphs
$B^{\delta^{\prime}}$ and $\Gamma^{\delta}$ are equal. That is,
$\mathcal{L}(B^{\delta^{\prime}})=\mathcal{L}(\Gamma^{\delta}).$ Let
$\Gamma^{\delta}$ and $\Gamma^{\delta^{\prime}}$ be same underlying metric
graphs with strengths $\alpha_{m}$ and $\alpha^{\prime}_{m}$, respectively.
Let $\Gamma^{\delta}_{f}$ be the corresponding flower graph obtained from
$\Gamma^{\delta}$, then $\Gamma^{\delta}_{f}$ can be turned into a
$\mathfrak{B}^{s}$ bipartite graph by introducing the degree two vertices at
its leave. Therefore,

$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{\delta}_{f})\leq\lambda_{k+1}(\Gamma^{s}_{f})=\lambda_{k+1}(\mathfrak{B}^{s})=\lambda_{k+\beta}(\mathfrak{B}^{a})$

Let us impose the Dirichlet conditions at degree two vertices of
$\mathfrak{B}^{a}$ to obtain a graph $\mathfrak{B}^{a}_{d}$. Now again impose
the standard conditions at the Dirichlet vertices in $\mathfrak{B}^{a}_{d}$.
Thus

$\lambda_{k+\beta}(\mathfrak{B}^{a})\leq\lambda_{k+\beta}(\mathfrak{B}^{a}_{d})\leq\lambda_{k+2\beta}(\mathfrak{B}^{a}_{s})=\lambda_{k+2|E|}(\Gamma^{a}_{f})\leq\lambda_{k+2|E|+1}(\Gamma^{\delta^{\prime}}_{f}).$
(4.32)

Where $\Gamma^{\delta^{\prime}}_{f}$ has strength
$\alpha=\sum\limits_{m=1}^{|V|}\alpha^{\prime}_{m}$, and if for all
$m=1,2,\cdots,|V|$, the strengths $\alpha^{\prime}_{m}\geq 0$ in
$\Gamma^{\delta^{\prime}}$, then restoring the vertices of
$\Gamma^{\delta^{\prime}}$ from $\Gamma_{f}^{\delta^{\prime}}$ implies that

$\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k+2|E|+1}(\Gamma^{\delta^{\prime}}_{f})\leq\lambda_{k+2|E|+1}(\Gamma^{\delta^{\prime}}).$

If $|V^{p^{\prime}}|$ and $|V^{n^{\prime}}|$ denote the number vertices in
$\Gamma^{\delta^{\prime}}$ with positive and negative strengths, respectively.
Then,

$\displaystyle\lambda_{k}(\Gamma^{\delta})\leq\lambda_{k}(\Gamma^{p})$
$\displaystyle\leq\lambda_{k+|V^{p}|}(\Gamma^{s})$
$\displaystyle=\lambda_{k+\beta}(B^{s})$
$\displaystyle=\lambda_{k+\beta+|V^{p}|-1}(B^{a})$
$\displaystyle\leq\lambda_{k+\beta+|V^{p}|+|V^{p^{\prime}}|+|V^{n^{\prime}}|-1}(B^{\delta^{\prime}})$

Since for the graph endowed with $\delta^{\prime}$-type conditions, attaching
a pendant edge, attaching an edge between existing vertices, or gluing two
vertices with at least one equipped with anti-standard vertex conditions
lowers all eigenvalues [19, Theorem 3.2, 3.5 and 4.2]. Therefore, for any
metric graph $\Gamma^{\delta^{\prime}}$ with length
$\mathcal{L}(\Gamma^{\delta^{\prime}})\geq\mathcal{L}(\tilde{\Gamma}^{\delta})$,
we obtain,

$\lambda_{k+\beta}(\Gamma^{\delta^{\prime}})\leq\lambda_{k+\beta}(B^{\delta^{\prime}})\leq\lambda_{k+1+|V^{n}|}(\tilde{\Gamma}^{\delta}).$

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