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# Spectral analysis of second order operators with sign-changing coefficients on general compact graphs

Serge Nicaise

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Serge Nicaise. Spectral analysis of second order operators with sign-changing coefficients on general compact graphs. 2024. hal-04727986v2

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# Spectral analysis of second order operators with sign-changing coefficients on general compact graphs

Serge Nicaise*

October 16, 2024

## Abstract

In this paper, we first show that second order operators with sign-changing coefficients on general compact graphs are selfadjoint with a discrete spectrum. We secondly give a necessary and sufficient condition for the invertibility of these operators. Furthermore, we give two characterizations of their spectrum (in the case of constant coefficients) and prove Weyl's type formulas for large eigenvalues, adapting approaches developed in the positive definite case.

**AMS (MOS) subject classification:** 35P05, 35P20, 34L05.

**Key Words:** indefinite problems, graphs, spectral theory, asymptotic distributions.

**Acknowledgements:** This work has been carried out in the framework of the Project NQG (ANR-23-CE40-0005-01), funded by the French National Research Agency (ANR). We thank the ANR for its support.

## 1 Introduction

The study of diffusion problems with sign changing coefficients increased recently because these problems appear in several areas of physics, for example in electromagnetism [17, 25, 26, 32, 33, 18]. One typical example corresponds to the coupling between a dielectric material and a metamaterial, where existence and computation of electromagnetic fields are challenging questions. Thus some mathematical investigations have been performed and concern existence results using either potential theory or the notion of T-coercivity [12, 33, 8] and numerical approximations by the finite element methods [33, 10, 11, 9, 31, 14, 7].

One important problem is to look at a nontrivial solution $u \in H_0^1(\Omega)$ of

$$ (1.1) \qquad \operatorname{div}(a\nabla u) + \omega^2 u = 0 \quad \text{in } \Omega, $$

where $\Omega$ is a bounded domain of $\mathbb{R}^d$, $d \ge 1$, $a$ is a function that changes of sign in $\Omega$ and $\omega$ is a positive real number. The existence of such a solution is a difficult question and is related to the Fredholm property of the operator $T$ defined formally by

$$ Tu = -\operatorname{div}(a\nabla u). $$

<footnote>* Université Polytechnique Hauts-de-France, CÉRAMATHS/DMATHS and FR CNRS 2037, F-59313 - Valenciennes Cedex 9 France, Serge.Nicaise@uphf.fr</footnote>
In dimension 1, for $\Omega = (0, L)$, $a = 1$ in $(0, \ell_1)$ and $a = -\kappa < 0$ in $(\ell_1, L)$, with $0 < \ell_1 < L$, it was stated in [13] that this operator is always Fredholm, and therefore the spectrum of $T$ is discrete, yielding non trivial solutions to (1.1) for a discrete set of frequencies $\omega$. In dimension $d \ge 2$ and if the function $a$ is piecewise constant, this Fredholm property only holds if the contrast between positive and negative coefficients is ouside an interval that always contains $-1$, we refer to the papers cited above for the details.

Similar questions were recently considered in [2] for some particular graphs and constant coefficients. In these particular cases, some sufficient conditions on the coefficients are given to guarantee the invertibility of the operator $T$, and the characterization of its spectrum is done. Hence the goal of this paper is to continue and extend this analysis by considering arbitrary compact graphs and, as much as possible, variable coefficients. More precisely, in its full generality, we show that the operator $T$ is always selfadjoint with a discrete spectrum and using a direct approach, find a necessary and sufficient (and explicit) condition for its invertibility. Furthermore, for constant coefficients, we give two characterizations of the spectrum, adapting approaches developed in the positive definite case by J.-P. Roth [35] (see also S. Nicaise [29, §1.3]) and J. von Below [36] (see also S. Nicaise [29, §1.4]). In both cases, these characterizations are based on the use of specific fundamental solutions, consequently the eigenvalues can be obtained with the help of the zeroes of analytic functions. Further, again inspired from the positive definite case [34, 30, 6, 23], assuming that a positive coefficient and a negative one exist, we prove Weyl's type formulas by showing that large positive (resp. negative) eigenvalues of $T$ behaves like $\mathcal{L}_+ n^2$ (resp. $-\mathcal{L}_- n^2$), with $n \in \mathbb{N}$, where $\mathcal{L}_+$ (resp. $\mathcal{L}_-$) is a positive constant that explicitely depends on the coefficients of the edges of the graph where these coefficients are positive (resp. negative). As a result, large positive (resp. negative) eigenvalues of $T$ behaves like the positive (resp. negative) eigenvalues of the “restriction” of $T$ to the subgraph made of edges where the coefficients are positive (resp. negative) with Dirichlet (or Neumann) condition at the exterior nodes. Finally we demonstrate the usefulness of our general results by some illustrative examples.

Let us finally mention indefinite boundary value problems of the form

$$Lu = \lambda Bu,$$

where $L$ is a positive definite and selfadjoint operator and $B$ is indefinite, for which positive and negative eigenvalues $\lambda$ may appear. We refer to [15] where the authors consider the spectral structure of indefinite boundary value problems on graphs.

The paper is organized as follows: in section 2 we briefly recall the notion of compact metric graphs, define a general second order operator $T$ with sign-changing coefficients and prove some of its useful properties. In Section 3, we find a necessary and sufficient condition for its invertibility. Section 4 is devoted to the characterizations of the spectrum, while Weyl's type formulas are obtained in section 5. Finally in section 6 different examples are presented.

## 2 Second order operators with sign-changing coefficients on general compact graphs

Let us first introduce some notations needed to formulate our setting, we refer to [24, 36, 27, 30, 16, 4, 3, 28] for more details.

A metric graph $G$ is a geometric object consisting of intervals (called edges) which are connected via their extremities in a certain way. More precisely, we here assume that we have a finite set $E$
of edges. For each $j \in E$, the edge $e_j$ is identified with a bounded interval $I_j = (0, \ell_j)$ with $\ell_j > 0$. A metric graph $G$ is then obtained by connecting the endpoints of the edges, by introducing an equivalence relation on the set of endpoints and identifying the equivalence classes as vertices, the set of vertices will be denoted by $V$. Without loss of generality we may assume that $G$ is connected. We will not go into more technical details here and refer the reader to [24, 30, 16, 4, 28] for more details.

The degree of a vertex is the number of incident edges at the vertex. A vertex with degree 1 is called an exterior vertex. On the other hand, a vertex with degree greater than 1 is called an interior vertex. The set of exterior (resp. interior) vertices will be denoted by $V_{\text{ext}}$ (resp. $V_{\text{int}}$). Let us also fix a decomposition of $V_{\text{ext}}$ into two distinct subsets $V_{\text{Dir}}$ and $V_{\text{Neu}}$, namely

$$V_{\text{ext}} = V_{\text{Dir}} \cup V_{\text{Neu}},$$

where homogeneous Dirichlet and Neumann conditions will be imposed (note that one of the two sets can be empty, corresponding to full Neumann (resp. Dirichlet) boundary conditions at the exterior vertices).

We now introduce the Hilbert space of square integrable (complex valued) functions on $G$:

$$L^2(G) = \bigoplus_{j \in E} L^2(I_j).$$

Equipped with the usual inner product induced by the inner product in $L^2(I_j)$, it is clearly a Hilbert space. For $\psi \in L^2(G)$, $\psi_j$ will be its restriction to $I_j$.

Similarly for each positive real number $s$, we introduce the Hilbert space

$$PH^s(G) = \bigoplus_{j \in E} H^s(I_j),$$

with the usual inner product induced by the inner product in $H^s(I_j)$, the usual Sobolev space of order $s$ on $I_j$. For future uses, we also denote by $H^1(G)$ as the intersection of $PH^1(G)$ with the set of continuous functions on $G$, while $H_0^1(G)$ is the set of functions in $H^1(G)$ that are zero at the exterior vertices.

For a vertex $v$, we denote by $E_v$ the set of indices of edges having $v$ as endpoint. For two vertices $v$ and $v_1$, we denote by $E(v, v_1)$, the set of indices $j$ such that $v$ and $v_1$ are vertices of the edge $e_j$ (obviously if such an edge does not exist $E(v, v_1) = \emptyset$). We then say that a vertex $v_1$ is a neighbour of the vertex $v$ if $E(v, v_1)$ is not empty. Finally, we denote by $V_v$ the set of vertices that are neighbours of $v$. As we do not exclude edges with only one vertex (a circle), $v$ can be its own neighbour. For further use, we denote by $E_o$, the set of edges having only one vertex.

For all $j \in E$, we denote by $I(j)$ (resp. $F(j)$) the vertex of $e_j$ corresponding to the extremity $x=0$ (resp. $x = \ell_j$).

For a vertex $v$, and $j \in E_v$, we denote the outward normal derivative of a function $u_j \in H^2(I_j)$ at $v$ by

$$\partial_n u_j(v) = \begin{cases} u'_j(\ell_j) & \text{if } F(j) = v, \\ -u'_j(0) & \text{if } I(j) = v, \end{cases}$$

where $u'_j$ means the space derivative of $u_j$. More simply for a vertex $v$, and $j \in E_v$, we sometimes write

$$u_j(v) = \begin{cases} u_j(\ell_j) & \text{if } F(j) = v, \\ u_j(0) & \text{if } I(j) = v. \end{cases}$$

For all $j \in E$, we now assume $k_j \in W^{1,r}(I_j)$ for some $r > 2$, is given a real-valued function that does not change of sign in $I_j$, namely satisfying

$$(2.1) \qquad |k_j(x)| \ge \alpha, \forall x \in I_j, \forall j \in E,$$
for a positive constant $\alpha$.

From now on we denote by $k$, the function defined from $G$ into $\mathbb{R}$ such that its restriction to $I_j$ is equal to $k_j$.

By our assumptions on $k$, we have the following identity

**Lemma 2.1** Under the above assumptions on $k$, it holds

$$ (2.2) \qquad PH^2(G) = \{u \in L^2(G) : ku' \in PH^1(G)\}. $$

**Proof.** Let us denote by $PH_k^2(G)$, the right-hand side of (2.2), namely

$$ PH_k^2(G) := \{u \in L^2(G) : ku' \in PH^1(G)\}. $$

1. Let us show that

$$ (2.3) \qquad PH^2(G) \subset PH_k^2(G). $$

Indeed if $u \in PH^2(G)$, then for all $j \in E$, as $u'_j$ belongs $H^1(I_j)$ and as $k_j \in C(\bar{I}_j)$, we directly obtain that $k_j u'_j \in L^2(I_j)$. Now by the continuous embedding $H^1(I_j) \hookrightarrow L^s(I_j)$ for all $s > 1$, we also deduce that

$$ u'_j \in L^s(I_j), \forall s > 1. $$

Hence by Leibniz's rule

$$ (2.4) \qquad (k_j u'_j)' = k_j u''_j + k'_j u'_j. $$

By the boundedness of $k_j$, the first term belongs to $L^2(I_j)$, while for the second term, we can invoke Hölder's inequality because $k'_j \in L^r(I_j)$, and $u'_j \in L^{r^*}(I_j)$, with $r^* > 2$ such that $1/r + 1/r^* = 1/2$. This proves the inclusion (2.3).

2. For the converse inclusion, we take $u \in PH_k^2(G)$. Then for all $j \in E$, as $k_j u'_j \in H^1(I_j)$, we get that

$$ k_j u'_j \in L^s(I_j), \forall s > 1, $$

and by (2.1) that

$$ (2.5) \qquad u'_j \in L^s(I_j), \forall s > 1, $$

Then by (2.4), we can write

$$ k_j u''_j = (k_j u'_j)' - k'_j u'_j. $$

The first term of this right-hand side is in $L^2(I_j)$ by definition of $PH^2(G)$, while the second one is also in $L^2(I_j)$ using Hölder's inequality (using (2.5) and the assumption on $k_j$). This proves that $u''_j \in L^2(I_j)$ due to (2.1). ■

We are ready to define our operator $T$ on the graph $G$.

**Definition 2.2** With the previous assumptions, we define the unbounded operator $T$ as

$$ (2.6) \qquad Tu = -(k_j u'_j)'_{j \in E}, \forall u \in D(T), $$

with

$$ D(T) = \{u \in PH^2(G) \text{ satisfying (2.7) - (2.9) below}\}, $$

$$ (2.7) \qquad u_j(v) = u_k(v), \forall v \in V_{\text{int}}, $$

$$ (2.8) \qquad u_j(v) = 0, \forall v \in V_{\text{Dir}}, $$

$$ (2.9) \qquad \sum_{j \in E_v} k_j(v) \partial_n u_j(v) = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}}. $$
By Lemma 2.1, the right-hand side of (2.6) belongs to $L^2(G)$, hence $T$ is well-defined on $D(T)$.

If $k_j$ is positive, for all $j \in E$, this operator is non negative (positive if $V_{\text{Dir}}$ is non empty) and self-adjoint in $L^2(G)$, see [29, 30, 4, 28] for instance. When $|k_j| = 1$, for all $j \in E$, this self-adjoint property still holds as proved in [21, Theorem 3.1], see also [2, Proposition 5.1] for a star graph and $k_j$ constant function, for all $j \in E$. Let us show that this property remains valid in the variable coefficients case and arbitrary finite graphs.

**Theorem 2.3** The operator $T$ is a self-adjoint operator in $L^2(G)$.

**Proof.** Since the operator $T$ does not change if we cut an edge $e_j$ into a finite union of subintervals, keeping the same functions $k_j$ on each new edges and adding the new vertices in $V_{\text{int}}$, in this proof we assume that the two extremities of any edge do not correspond to the same vertex.

First of all, since for all $u \in D(T)$, $k_j u'_j$ belongs to $H^1(I_j)$, and $u_j$ belongs to $H^1(I_j)$, by Green's formula, we directly see that $T$ is symmetric. Further as $\oplus_{j \in E} \mathcal{D}^\infty(I_j)$ is included into $D(T)$, $T$ is densely defined.

Since the symmetry of $T$ implies that $D(T) \subset D(T^*)$, it then remains to show that

$$D(T^*) \subset D(T).$$

For that purpose, let $g \in D(T^*)$, which means that $g \in L^2(G)$ and there exists $f \in L^2(G)$ such that

$$(2.10) \qquad (g, Tu)_{L^2(G)} = (f, u)_{L^2(G)}, \forall u \in D(T).$$

1. For an arbitrary $j \in E$, take an arbitrary $\varphi \in \mathcal{D}(I_j)$. Then $u$ defined on $G$ by

$$u_j = \varphi \text{ on } I_j,$$

$$u_\ell = 0 \text{ on } I_\ell, \forall \ell \in E \setminus \{j\},$$

clearly belongs to $D(T)$. This choice of $u$ in (2.10) implies that

$$-\int_{I_j} g_j(k_j \varphi)' dx = \int_{I_j} f_j \varphi dx, \forall \varphi \in \mathcal{D}(I_j).$$

This shows that

$$(2.11) \qquad -(k_j g_j')' = f_j \text{ in } \mathcal{D}'(I_j).$$

As $f_j$ belongs to $L^2(I_j)$, we deduce that

$$(2.12) \qquad k_j g_j' = p_j + c_j,$$

for a constant $c_j$ and the function $p_j$ defined by

$$p_j(x) = - \int_0^x f_j(y) dy, \forall x \in I_j.$$

Since $p_j$ also belongs to $L^2(I_j)$, by (2.11) and (2.12), we deduce that $k_j g_j' \in H^1(I_j)$. Therefore by Lemma 2.1, $g$ belongs to $PH^2(G)$.
2. Fix a vertex $v \in V_{\text{Dir}}$ and let $e_{j_v} \in E$ be the unique edge having $v$ as extremity. Without loss of generality, we can assume that $v$ corresponds to the extremity $x = 0$. Then we define $u$ on $G$ by

$$u_{j_v} = x\varphi(x) \text{ on } I_{j_v},$$

$$u_{\ell} = 0 \text{ on } I_{\ell}, \forall \ell \in E \setminus \{j_v\},$$

where $\varphi \in \mathcal{D}(-\ell_{j_v}, \ell_{j_v})$ such that $\varphi(0) = 1$. We readily check that $u \in D(T)$, that it is zero in a neighbourhood of all vertices except $v$ and satisfies

$$u'_{j_v}(v) = 1.$$

Hence using this test-function in (2.10), applying Green's formula and using the mentioned property of $u$ and (2.11), we obtain

$$g(v)k(v) = 0,$$

and since $k(v) \neq 0$, we deduce that

$$g(v) = 0.$$

3. Let us fix $v \in V_{\text{int}}$ and two arbitrary and different edges $e_{j_1}$ and $e_{j_2}$ that have $v$ as vertex. Without loss of generality, we can assume that $v$ corresponds to the extremity $x = 0$ for each edge. Then as before we can define $u$ on $G$ by

$$u_{j_1} = \frac{1}{k_m(v)}x\varphi_{j_1}(x) \text{ on } I_{j_1}$$

$$u_{j_2} = -\frac{1}{k_{j_1}(v)}x\varphi_{j_2}(x) \text{ on } I_{j_2},$$

$$u_{\ell} = 0 \text{ on } I_{\ell}, \forall \ell \in E \setminus \{j_1, j_2\},$$

where for $i = 1$ and $2$, $\varphi_{j_i} \in \mathcal{D}(-\ell_{j_i}, \ell_{j_i})$ such that $\varphi_{j_i}(0) = 1$. Again we readily check that $u \in D(T)$, that it is zero in a neighbourhood of all vertices except $v$ and

$$u'_{j_i}(v) \neq 0, i = 1, 2.$$

Hence using this test-function in (2.10), applying Green's formula and using the mentioned property of $u$ and (2.11), we obtain

$$\sum_{i=1,2} g_{j_i}(v)k_{j_i}(v)u'_{j_i}(v) = 0.$$

Since by construction, $\sum_{i=1,2} k_{j_i}(v)u'_{j_i}(v) = 0$, we deduce

$$k_{j_1}(v)u'_{j_1}(v)(g_{j_1}(v) - g_{j_2}(v)) = 0.$$

The factor $k_{j_1}(v)u'_{j_1}(v)$ being different from 0, we conclude that

$$g_{j_1}(v) = g_{j_2}(v).$$

Since $j_1$ and $j_2$ were arbitrary edges adjacent to $v$, and $v$ was arbitrary in $V_{\text{int}}$, $g$ is continuous on $G$.

4. Fix $v \in V_{\text{int}} \cup V_{\text{Neu}}$. Without loss of generality, we can assume that $v$ corresponds to the
extremity $x = 0$ for all edges having $v$ as extremity. Then for all $j \in E_v$, fix $\varphi_j \in D(-\ell_j, \ell_j)$ such that $\varphi_j = 1$ in a neighbourhood of 0 and define $u$ on $G$ by

$$u_j = \varphi_j \text{ on } I_j, \forall j \in E_v,$$

$$u_{\ell} = 0 \text{ on } I_{\ell}, \forall \ell \in E \setminus E_v.$$

Again $u$ belongs to $D(T)$ and with this test-function in (2.10), applying Green's formula and the previous properties, we find

$$\sum_{j \in E_v} k_j(v) \partial_n g_j(v) = 0.$$

In other words, $g$ satisfies (2.9).

In conclusion, we have shown that $g$ belongs to $D(T)$, and the conclusion follows. ■

**Corollary 2.4** The spectrum $\sigma(T)$ of $T$ is discrete. Further the set of normalized eigenvectors forms an orthonormal basis of $L^2(G)$.

**Proof.** As $T$ is selfadjoint in $L^2(G)$, $T + i\mathbb{I}$ is an isomorphism from $D(T)$ into $L^2(G)$, see [20, Theorem 8.3] for instance. Since $D(T)$ is compactly embedded into $L^2(G)$, we deduce that $(T + i\mathbb{I})^{-1}$ is a compact operator. Hence the spectrum $(T + i\mathbb{I})^{-1}$ is discrete, and so it is the case for $T$. The second statement directly follows from the fact that for a fixed $\lambda \in \mathbb{R}$ that is not an eigenvalue of $T$, the operator $(\lambda\mathbb{I} - T)^{-1}$ is a selfadjoint and compact operator. ■

**Remark 2.5** Let us denote by $T_-$, the operator defined by (2.6)-(2.9) replacing $k$ by $-k$. Since the Kirchhoff condition (2.9) is invariant by a multiplication by the factor $-1$, we directly deduce that $T_- = -T$. Hence $\lambda \in \sigma(T)$ if and only if $-\lambda \in \sigma(T_-)$.

Let us end up this section by studying the unbounded character of the spectrum. For that purpose, let us denote by

$$\sigma_+(T) = \sigma(T) \cap (0, \infty), \text{ and } \sigma_-(T) = \sigma(T) \cap (-\infty, 0).$$

**Theorem 2.6** 1. If there exists $j \in E$ such that

$$(2.13) \qquad k_j(x) \ge \alpha, \forall x \in I_j,$$

then $\sigma_+(T)$ is unbounded.

2. Similarly, if there exists $\ell \in E$ such that

$$k_{\ell}(x) \le -\alpha, \forall x \in I_{\ell},$$

then $\sigma_-(T)$ is unbounded.

**Proof.** 1. Fix one $j \in E$ such that (2.13) holds and assume that $\sigma_+(T)$ is bounded, or equivalently

$$(2.14) \qquad \lambda \le C, \forall \lambda \in \sigma(T).$$

for some positive constant $C$. Hence by the spectral representation theorem and (2.14), we find

$$(2.15) \qquad (Au, u)_{L^2(G)} \le C \|u\|_{L^2(G)}^2, \forall u \in D(T).$$
Now fix a non zero cut-off function $\varphi \in D(\mathbb{R})$ with a support included in $[-1, 1]$ and for all $m \in \mathbb{N}$, define the function $u_{(m)}$ in $G$ as follows:

$$ \begin{cases} u_{(m),j}(x) = \varphi(m(x - \frac{\ell_j}{2})), & \forall x \in I_j, \\ u_{(m),j'} = 0, & \forall j' \in E \setminus \{j\}. \end{cases} $$

For $m > \frac{2}{\ell_j}$, we see that $u_{(m)}$ is zero in a neighborhood of the extremities of $I_j$ and therefore $u_{(m)}$ belongs to $D(T)$. Now Green's formula yields

$$ (Au_{(m)}, u_{(m)})_{L^2(G)} = \int_0^{\ell_j} k_j |u'_{(m)}|^2 dx = m^2 \int_0^{\ell_j} k_j(x) |\varphi'(m(x - \frac{\ell_j}{2}))|^2 dx. $$

Hence by a change of variable, we find

$$ (Au_{(m)}, u_{(m)})_{L^2(G)} = m \int_{-1}^{1} k_j \left(\frac{\ell_j}{2} + \frac{y}{m}\right) |\varphi'(y)|^2 dy. $$

By the assumption (2.13), we deduce that

$$ (Au_{(m)}, u_{(m)})_{L^2(G)} \geq \alpha m \int_{-1}^{1} |\varphi'(y)|^2 dy. $$

Similarly for $m > \frac{2}{\ell_j}$, by a change of variable we find that

$$ \|u_{(m)}\|_{L^2(G)}^2 = m^{-1} \int_{-1}^{1} |\varphi(y)|^2 dy, $$

and therefore

$$ \frac{(Au_{(m)}, u_{(m)})_{L^2(G)}}{\|u_{(m)}\|_{L^2(G)}^2} \geq \alpha m^2 \frac{\int_{-1}^{1} |\varphi'(y)|^2 dy}{\int_{-1}^{1} |\varphi(y)|^2 dy}, $$

which contradicts (2.15) by letting $m$ goes to infinity.

2. The second statement follows from the first one and Remark 2.5. ■

### 3 Invertibility of $T$

In this section, we want to find a necessary and sufficient condition on the coefficients $k_j$ that guarantee that $T$ is invertible, or equivalently that $0$ is not an eigenvalue of $T$.

First introduce,

$$ \kappa_j := \frac{1}{\int_0^{\ell_j} \frac{1}{k_j(y)} dy}, $$

and the square (and symmetric) matrix $M_k := (m_{v,v_1})_{v,v_1 \in V_{\text{int}} \cup V_{\text{Neu}}}$, where for all $v, v_1 \in V_{\text{int}} \cup V_{\text{Neu}}$,

$$ m_{v,v} = \sum_{j \in E_v \setminus E_o} \kappa_j, $$

$$ m_{v,v_1} = - \sum_{j \in E_{v,v_1}} \kappa_j, \text{ if } v_1 \in V_v \setminus \{v\}, $$

$$ m_{v,v_1} = 0, \text{ else.} $$

With these notations, we have the next result.
**Theorem 3.1** It holds

$$
\begin{array}{ll}
(3.1) & \dim \ker T = \dim \ker M, \\
\textit{and consequently} & \ker T = \{0\} \Leftrightarrow \det M \neq 0.
\end{array}
$$

**Proof.** Let $u \in \ker T$, or equivalently $u \in D(T)$ be such that

$$
(k_j u'_j)' = 0 \text{ on } I_j, \forall j \in E.
$$

Then for all $j \in E$, there exists a constant $c_j$ such that

$$
k_j(x)u'_j(x) = c_j, \forall x \in I_j.
$$

By the assumptions on $k_j$, we can divide this identity by $k_j(x)$ and therefore there exists another constant $d_j$ such that

$$
(3.2) \qquad u_j(x) = c_j \int_0^x \frac{1}{k_j(y)} dy + d_j, \forall x \in I_j.
$$

This expression is actually equivalent to

$$
(3.3) \qquad u_j(x) = \kappa_j \left( a_j \int_0^x \frac{1}{k_j(y)} dy + b_j \int_x^{\ell_j} \frac{1}{k_j(y)} dy \right), \forall x \in I_j,
$$

for some constants $a_j$ and $b_j$. Indeed (3.2) is equivalent to (3.3) with (recalling that $\kappa_j$ is different from 0)

$$
a_j = c_j \int_0^{\ell_j} \frac{1}{k_j(y)} dy + d_j \text{ and } b_j = d_j.
$$

The interest of (3.3) is that

$$
(3.4) \qquad u_j(0) = b_j, \text{ and } u_j(\ell_j) = a_j.
$$

Now by the continuity of $u$, we can set $x_v := u(v)$ for all $v \in V$ and by the Dirichlet boundary condition (2.8), we will have

$$
(3.5) \qquad x_v = 0, \forall v \in V_{\text{Dir}},
$$

and the remaining unknowns are $x_v$ for all $v \in V_{\text{int}} \cup V_{\text{Neu}}$. Now we may notice that (3.4) can be written

$$
(3.6) \qquad b_j = x_{I(j)}, \text{ and } a_j = x_{F(j)}, \forall j \in E.
$$

It then remains to check the condition (2.9). For that purpose, by (3.3), we notice that

$$
u'_{j}(x) = \kappa_{j} \left( \frac{a_{j} - b_{j}}{k_{j}(x)} \right), \forall x \in I_{j}.
$$

Therefore, we have

$$
\partial_n u_j(I(j)) = -u'_j(0) = \kappa_j \frac{b_j - a_j}{k_j(0)} = \frac{\kappa_j}{k_j(I(j))} (x_{I(j)} - x_{F(j)}),
$$
as well as

$$ \partial_n u_j(F(j)) = u'_j(\ell_j) = \kappa_j \frac{a_j - b_j}{k_j(\ell_j)} = \frac{\kappa_j}{k_j(F(j))} (x_{F(j)} - x_{I(j)}), $$

Note that these two last expressions are symmetric in $I(j)$ and $F(j)$, which means that (2.9) can be written as

$$ \sum_{v_1 \in V_v} \sum_{j \in E_{v,v_1}} \kappa_j (x_v - x_{v_1}) = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}}, $$

or equivalently

$$ \sum_{v_1 \in V_v \setminus \{v\}} \sum_{j \in E_{v,v_1}} \kappa_j (x_v - x_{v_1}) = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}}. $$

Splitting the difference, we get equivalently,

$$ \left( \sum_{v_1 \in V_v \setminus \{v\}} \sum_{j \in E_{v,v_1}} \kappa_j \right) x_v - \sum_{v_1 \in V_v \setminus \{v\}} \left( \sum_{j \in E_{v,v_1}} \kappa_j \right) x_{v_1} = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}}. $$

Finally using again (3.5), we obtain

$$ \left( \sum_{j \in E_v \setminus E_o} \kappa_j \right) x_v - \sum_{v_1 \in (V_v \setminus \{v\}) \cap (V_{\text{int}} \cup V_{\text{Neu}})} \left( \sum_{j \in E_{v,v_1}} \kappa_j \right) x_{v_1} = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}}. $$

By the definition of the matrix $M_k$, this system is equivalent to

$$ M_k X = 0, $$

where $X = (x_v)_{v \in V_{\text{int}} \cup V_{\text{Neu}}}$ is the unknown vector. This proves (3.1). ■

Some illustrative examples of this Theorem are presented in section 6.

# 4 Spectral characterization

In this section, we mainly want to characterize the spectrum of $T$ as the zeroes of a transcendental equation in the constant coefficients case. In the positive definite case, two approaches can be used: a first one due to J.-P. Roth [35] (see also S. Nicaise [29, §1.3]), where a system of size $2\#E \times 2\#E$ is involved, and a second one due to J. von Below [36] (see also S. Nicaise [29, §1.4]), where, at the exception of canonical eigenvalues, a system of size $(\#V_{\text{int}} + \#V_{\text{Neu}}) \times (\#V_{\text{int}} + \#V_{\text{Neu}})$ is involved. Since $\#E \ge \#V - 1$ [19, Corollary 4.5 (b)], $\#V_{\text{int}} + \#V_{\text{Neu}}$ is quite smaller than $2\#E$, consequently the second approach seems to be more attractive than the first one.

Our goal is to adapt each approach to the indefinite case in the constant coefficients case. Let us nevertheless shortly discuss the general case.

## 4.1 General coefficients

In the one-dimensional situation, the characterisation of the spectrum is obtained as soon as two linearly independent solutions of the differential equation

$$ (4.1) \qquad -(k_j u')' = \lambda u, \text{ in } (0, \ell_j), $$

exist, for all $\lambda \in \mathbb{R}$ and $j \in E$. Under our assumptions on $k_j$, this is always possible as the next lemma shows.
**Lemma 4.1** For any $j \in E$ and $\lambda \in \mathbb{R}$, there exist two linearly independent solutions $u_{j,1}(\lambda), u_{j,2}(\lambda) \in H^2(0, \ell_j)$ of (4.1).

**Proof.** We combine the standard reduction of order method with the Cauchy-Lipschitz theorem. Namely fix $j \in E$ and $\lambda \in \mathbb{R}$, and assuming that a solution $u$ of (4.1) exists, then $V_j = (u, k_j u')^\top$ satisfies

$$V'_j(x) = F_j(x, V_j(x)), \forall x \in (0, \ell_j),$$

where

$$F_j(x, V) = \begin{pmatrix} 0 & \frac{1}{k_j(x)} \\ -\lambda & 0 \end{pmatrix} V.$$

Since $k_j$ can be extended into the function, still denoted by $k_j$, that belongs to $\in W^{1,r}(-\ell_j, 2\ell_j)$, $r > 2$, and is defined by

$$k_j(x) = \begin{cases} k_j(x) & \text{if } x \in (0, \ell_j), \\ k_j(\ell_j) & \text{if } x \in [\ell_j, 2\ell_j), \\ k_j(0) & \text{if } x \in (-\ell_j, 0], \end{cases}$$

we can look at a solution $V_j$ of

$$(4.2) \qquad V'_j(x) = F_j(x, V_j(x)), \forall x \in (-\ell_j, 2\ell_j).$$

By the regularity of $k_j$, we can check that $F_j$ is continuous on $(-\ell_j, 2\ell_j) \times \mathbb{R}^2$ and is globally Lipschitz with respect to the second variable, hence by the Cauchy-Lipschitz theorem, for any $(x_0, V_0) \in (-\ell_j, 2\ell_j) \times \mathbb{R}^2$, there exists a unique global solution $V_j \in C^1(-\ell_j, 2\ell_j)$ of (4.2) satisfying $V_j(x_0) = V_0$. Taking $x_0 = 0$ and $V_0 = (1, 0)^\top$ or $V_0 = (0, 1)^\top$, we find two linearly independent solutions $V_{j,1} \in C^1(-\ell_j, 2\ell_j)$ and $V_{j,2} \in C^1(-\ell_j, 2\ell_j)$ of (4.2) satisfying

$$V_{j,1}(0) = (1, 0)^\top, \quad V_{j,2}(0) = (0, 1)^\top.$$

Now for $m = 1$ and $2$, writting $V_{j,m} = (u_{j,m}, v_{j,m})^\top$, we directly see that $v_{j,m} = k_j u'_{j,m}$ and that $u_{j,m}$ is solution of (4.1). The linear independence of $u_{j,1}$ and $u_{j,2}$ follows from the one of $V_{j,1}$ and $V_{j,2}$. Finally their regularity $H^2(0, \ell_j)$ follows from their regularity $C^1([0, \ell_j])$, (4.1) and Lemma 2.1. $\blacksquare$

Once linearly independent solutions $u_{j,1}(\lambda)$ and $u_{j,2}(\lambda)$ of (4.1) are available, we can characterize the spectrum of $T$ as follows. The case $\lambda = 0$ being treated in Theorem 3.1, we can restrict ourselves to the case $\lambda \neq 0$.

**Theorem 4.2** $\lambda \in \mathbb{R} \setminus \{0\}$ is an eigenvalue of $T$ if and only if $\det N_\lambda = 0$, where $N_\lambda$ is a $2\#E \times 2\#E$ matrix $N_\lambda$ described below, furthermore its multiplicity is always $\leq 2\#E$.

**Proof.** Fix $\lambda \in \mathbb{R} \setminus \{0\}$, and let $u \in D(T)$ be an eigenvector of eigenvalue $\lambda$. Then by Lemma 4.1, for all $j \in E$, there exist two constants $a_{j,1}$ and $a_{j,2}$ such that

$$u_j(x) = a_{j,1}u_{j,1}(\lambda, x) + a_{j,2}u_{j,2}(\lambda, x), \forall x \in I_j.$$

Let us now check that the boundary conditions defining $D(T)$ are equivalent to a $2\#E \times 2\#E$ homogeneous linear system whose unknowns are the $a_{j,m}$'s.

1. The Dirichlet boundary conditions can be written:

$$(4.3) \qquad a_{j,1}u_{j,1}(\lambda, v) + a_{j,2}u_{j,2}(\lambda, v) = 0, \forall j \in I_v, v \in V_{\text{Dir}}.$$
2. Similarly the continuity conditions can be written

$$ (4.4) \qquad a_{j,1}u_{j,1}(\lambda, v) + a_{j,2}u_{j,2}(\lambda, v) = a_{j',1}u_{j',1}(\lambda, v) + a_{j',2}u_{j',2}(\lambda, v), \forall j, j' \in I_v, v \in V_{\text{int}}. $$

3. Finally, the Kirchhoff condition (2.9) is equivalent to

$$ (4.5) \qquad \sum_{j \in I_v} k_j (a_{j,1} \partial_n u_{j,1}(\lambda, v) + a_{j,2} \partial_n u_{j,2}(\lambda, v)) = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}}. $$

Now we notice that (4.3) yields $\#V_{\text{Dir}}$ homogeneous linear equations in the $a_{j,m}$'s, (4.4) yields $\sum_{v \in V_{\text{int}}} (\#I_v - 1)$ homogeneous linear equations in the $a_{j,m}$'s, while (4.5) yields $\#V_{\text{int}} + \#V_{\text{Neu}}$ homogeneous linear equations in the $a_{j,m}$'s. Since

$$ \#V_{\text{Dir}} + \sum_{v \in V_{\text{int}}} (\#I_v - 1) + \#V_{\text{int}} + \#V_{\text{Neu}} = \#V_{\text{ext}} + \sum_{v \in V_{\text{int}}} \#I_v, $$

and since in this right-hand side each edge is counted twice, (4.3)-(4.5) correspond to 2\#E homogeneous linear equations in the $a_{j,m}$'s, that can be written as

$$ N_{\lambda}((a_{j,1}, a_{j,2})^{\top})_{j \in E} = 0. $$

Hence a non trivial solution exists if and only if $\det N_{\lambda} = 0$, and in such a case, the multiplicity of $\lambda$ cannot exceed 2\#E. ■

The second conclusion of this Theorem, consequence of the one-dimensional nature of graphs, means that the multiplicity of all eigenvalues is bounded. This will be used to prove Weyl's laws. The first conclusion is useful as soon as two explicit linearly independent solutions of the differential equation (4.1) are available. This is the case for constant coefficients $k_j$, but let us mention that the solutions $u_{j,1}(\lambda)$ and $u_{j,2}(\lambda)$ built in Lemma 4.1 are not optimal, because they do not take into account the boundary conditions from $D(T)$ and the symmetry of $k_j$ with respect to the midpoint. As a consequence, the matrix $N_{\lambda}$ is not appropriate for the characterization of the spectrum. In the next subsections we will show the advantage of the use of better choices.

In the remainder of this section, we suppose that $k_j$ is a constant function, for all $j \in E$.

## 4.2 Roth's approach

We first introduce the following notations. To each edge $e_j$, $j \in E$, we associate an edge $e_{-j}$ of opposite orientation than $e_j$ (hence $\ell_{-j} = \ell_j$) and denote by $\bar{E} = \{j, -j | j \in E\}$. Obviously, we then have $F(j) = I(-j)$, for all $j \in \bar{E}$. We further set $k_{-j} = k_j$, for all $j \in E$.

We also denote by

$$ E_k^+ := \{j \in E : k_j > 0\}, \quad \bar{E}_k^+ := \{j \in \bar{E} : k_j > 0\} \quad \text{and} \quad E_k^- := \{j \in E : k_j < 0\}, \quad \bar{E}_k^- := \{j \in \bar{E} : k_j < 0\}. $$

Now we set

$$ (4.6) \qquad \nu_j(k) = \frac{\ell}{\sqrt{k_j}}, \forall j \in \bar{E}_k^+, \quad \nu_j(k) = \frac{1}{\sqrt{-k_j}}, \forall j \in \bar{E}_k^- $$

For all $v \in V_{\text{int}} \cup V_{\text{Neu}}$, we also set

$$ (4.7) \qquad q_k(v) = \sum_{j \in \bar{E}: I(j)=v} k_j \nu_j(k). $$
Let us remark that $q_k(v)$ is always different from zero, since

$$q_k(v) = \sum_{j \in \bar{E}^+: I(j) = v} \sqrt{k_j} - \sum_{j \in \bar{E}^-: I(j) = v} \sqrt{-k_j},$$

hence

$$|q_k(v)|^2 = \left( \sum_{j \in \bar{E}^+: I(j) = v} \sqrt{k_j} \right)^2 + \left( \sum_{j \in \bar{E}^-: I(j) = v} \sqrt{-k_j} \right)^2,$$

which is always positive.

Let us finally introduce the square matrix $M_k$ as well as the diagonal matrices $D$ and $N_k$ of size $2\#E \times 2\#E$ as follows:

$$(M_k)_{j',j} = \mathbb{1}_{V_{\text{int}} \cup V_{\text{Neu}}} (I(j')) \left\{ -\delta_{j',-j} + \frac{2k_j \nu_j(k)}{q_k(I(j'))} \delta_{I(j'),F(j)} \right\} - \mathbb{1}_{V_{\text{Dir}}} (I(j')) \delta_{j',-j}, \forall i, j \in \bar{E}, \\ D = \text{diag}(\ell_j)_{j \in \bar{E}}, \text{ and } N_k = \text{diag}(\nu_j(k))_{j \in \bar{E}}.$$

**Theorem 4.3** Let $\lambda$ be a positive real number. Then $\lambda$ is an eigenvalue of $T$ if and only if

$$(4.8) \qquad \det(\mathbb{I} - M_k e^{\sqrt{\lambda}DN_k}) = 0.$$

**Proof.** Let $u$ be an eigenvector of $T$ of eigenvalue $\lambda$. Then on each edge $e_j$, $u_j$ is in the form

$$(4.9) \qquad u_j(x) = a_j e^{\sqrt{\lambda}\nu_j d(I(j),x)} + a_{-j} e^{\sqrt{\lambda}\nu_j d(F(j),x)}, \forall x \in I_j, j \in E,$$

where for simplicity we drop the dependency of $\nu_j(k)$ with respect to $k$ in (4.6) (because $k$ is fixed). Note that we find the same expression by replacing $j$ by $-j$ in the above expression, so (4.9) remains valid for $j \in \bar{E}$.

It then remains to impose the boundary conditions.

1. The Dirichlet boundary conditions can be written:

$$(4.10) \qquad a_j + a_{-j} e^{\sqrt{\lambda}\nu_j \ell_j} = 0, \forall j \in \bar{E} : I(j) \in V_{\text{Dir}}.$$

2. Similarly the continuity conditions can be written:

$$(4.11) \qquad a_j + a_{-j} e^{\sqrt{\lambda}\nu_j \ell_j} = a_{j'} + a_{-j'} e^{\sqrt{\lambda}\nu_{j'} \ell_{j'}}, \forall j, j' \in \bar{E} : I(j) = I(j').$$

3. Finally, we easily see that the Kirchhoff condition (2.9) may be written:

$$(4.12) \qquad \sum_{j \in \bar{E}: I(j)=v} k_j \nu_j (a_j - a_{-j} e^{\sqrt{\lambda} \nu_j \ell_j}) = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}}.$$

Let us now show that (4.11) and (4.12) are equivalent to:

$$(4.13) \qquad a_{j'} + a_{-j'} e^{\sqrt{\lambda}\nu_{j'}\ell_{j'}} = \frac{2}{q_k(v)} \sum_{j \in \bar{E}: I(j)=v} k_j \nu_j a_{-j} e^{\sqrt{\lambda}\nu_j \ell_j}, \forall j' \in E : I(j') = v, v \in V_{\text{int}} \cup V_{\text{Neu}}.$$

a. Assume that (4.11) and (4.12) hold. Then for $v \in V_{\text{int}} \cup V_{\text{Neu}}$, we fix $j' \in \bar{E}$ such that $I(j') = v$, then for any $j \in \bar{E}$ such that $I(j) = v$, we multiply (4.11) by $k_j \nu_j$ and sum on $j$ to find:

$$q_k(v)(a_{j'} + a_{-j'} e^{\sqrt{\lambda}\nu_{j'}\ell_{j'}}) = \sum_{j \in \bar{E}: I(j)=v} k_j \nu_j (a_j + a_{-j} e^{\sqrt{\lambda}\nu_j \ell_j}),$$
and by using (4.12), we obtain (4.13).

b. Conversely, assume that (4.13) holds. Then (4.11) trivially holds. To obtain (4.12), we multiply (4.13) by $k_{j'}\nu_{j'}$ and sum on $j'$ such that $I(j') = v \in V_{\text{int}} \cup V_{\text{ext}}$ and find (since the right-hand side is independent of $j'$)

$$ \sum_{j' \in \bar{E}: I(j')=v} k_{j'} \nu_{j'} (a_{j'} + a_{-j'} e^{\sqrt{\lambda} \nu_{j'} \ell_{j'}}) = 2 \sum_{j \in \bar{E}: I(j)=v} k_j \nu_j a_{-j} e^{\sqrt{\lambda} \nu_j \ell_j}, $$

which is nothing else than (4.12).

Consequently a non trivial eigenvector $u$ exists if and only if a non trivial solution $a_j, j \in \bar{E}$ of (4.10) and (4.13) exists. But by the definition of $M_{k,\lambda}$, (4.10) and (4.13) are equivalent to

$$ (\mathbb{I} - M_k e^{\sqrt{\lambda}DN_k})A = 0, $$

where $A = (a_j)_{j \in \bar{E}}$ is the unknown vector. The proof is complete. ■

This theorem and Remark 2.5 directly allow to characterize $\sigma_-(T)$.

**Corollary 4.4** Let $\lambda$ be a negative real number. Then $\lambda$ is an eigenvalue of $T$ if and only if

$$ \det(\mathbb{I} - M_{-k} e^{\sqrt{-\lambda}DN_{-k}}) = 0. $$

## 4.3 von Below's approach

For any positive real number $\lambda$, we define

$$ (4.14) \qquad \begin{cases} c_j(k, \lambda) := \cos(\sqrt{\frac{\lambda}{k_j}} \ell_j), & s_j(k, \lambda) := \sin(\sqrt{\frac{\lambda}{k_j}} \ell_j), \forall j \in E_k^+, \\ c_j(k, \lambda) := \cosh(\sqrt{\frac{\lambda}{|k_j|}} \ell_j), & s_j(k, \lambda) := \sinh(\sqrt{\frac{\lambda}{|k_j|}} \ell_j), \forall j \in E_k^-, \end{cases} $$

Let $\lambda$ be a positive real number such that $\sin(\sqrt{\frac{\lambda}{k_j}} \ell_j) \neq 0$, for all $j \in E_k^+$, then we can introduce the square matrix $M_{k,\lambda}$ of size $\#V_{\text{int}} + \#V_{\text{Neu}}$ defined by

$$ (M_{k,\lambda})_{v,v} = \sum_{j \in E_v \setminus E_o} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)} + 2 \sum_{j \in E_v \cap E_o} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda) - 1}{s_j(k, \lambda)}, $$

$$ (M_{k,\lambda})_{v,v_1} = - \sum_{j \in E_{v,v_1}} \frac{\operatorname{sgn}(k_j) \sqrt{|k_j|}}{s_j(k, \lambda)}, \quad \text{if } v_1 \in V_v \setminus \{v\}, $$

$$ (M_{k,\lambda})_{v,v_1} = 0, \text{ if } v_1 \notin \{v\} \cup V_v. $$

Note that using standard trigonometric formulas, one can show that

$$ 2 \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda) - 1}{s_j(k, \lambda)} = -2 \sqrt{|k_j|} \frac{c_j(4k, \lambda)}{s_j(4k, \lambda)}, $$

hence $(M_{k,\lambda})_{v,v}$ can be expressed as

$$ (M_{k,\lambda})_{v,v} = \sum_{j \in E_v \setminus E_o} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)} - 2 \sum_{j \in E_v \cap E_o} \sqrt{|k_j|} \frac{c_j(4k, \lambda)}{s_j(4k, \lambda)}. $$

Now we can state the following result.
**Theorem 4.5** Let $\lambda$ be a positive real number such that

$$ (4.15) \qquad \lambda \neq k_j \left(\frac{n\pi}{\ell_j}\right)^2, \forall n \in \mathbb{N}^*, j \in E_k^+. $$

Then $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is a root of the characteristic equation

$$ (4.16) \qquad \det M_{k,\lambda} = 0. $$

**Proof.** Clearly our assumption (4.15) is equivalent to

$$ \sin\left(\sqrt{\frac{\lambda}{k_j}}\ell_j\right) \neq 0, \forall j \in E_k^+. $$

Hence setting for shortness $\iota_j = \frac{1}{\sqrt{|k_j|}}$, for all $j \in E$, an eigenvector $u$ of eigenvalue $\lambda$ may be expressed in the form

$$ (4.17) \qquad u_j(x) = \frac{a_j \sin(\iota_j \sqrt{\lambda} x) + b_j \sin(\iota_j \sqrt{\lambda} (\ell_j - x))}{\sin(\iota_j \sqrt{\lambda} \ell_j)}, \forall x \in I_j, j \in E_k^+, $$

$$ (4.18) \qquad u_j(x) = \frac{a_j \sinh(\iota_j \sqrt{\lambda} x) + b_j \sinh(\iota_j \sqrt{\lambda} (\ell_j - x))}{\sinh(\iota_j \sqrt{\lambda} \ell_j)}, \forall x \in I_j, j \in E_k^-, $$

for some constants $a_j$ and $b_j$, $j \in E$. As in the proof of Theorem 3.1, we notice that (3.4) holds.

Again, by the continuity of $u$, we can set $x_v := u(v)$ for all $v \in V$ and the Dirichlet boundary condition (2.8) is again equivalent to (3.5). The remaining unknowns are then $x_v$, for all $v \in V_{\text{int}} \cup V_{\text{Neu}}$, and as before (3.6) holds.

It then remains to check the condition (2.9). For that purpose, by (4.17) and (4.18), we notice that

$$ \begin{aligned} u'_j(x) &= \iota_j \sqrt{\lambda} \frac{a_j \cos(\iota_j \sqrt{\lambda} x) - b_j \cos(\iota_j \sqrt{\lambda} (\ell_j - x))}{\sin(\iota_j \sqrt{\lambda} \ell_j)}, \forall x \in I_j, j \in E_k^+, \\ u'_j(x) &= \iota_j \sqrt{\lambda} \frac{a_j \cosh(\iota_j \sqrt{\lambda} x) - b_j \cosh(\iota_j \sqrt{\lambda} (\ell_j - x))}{\sinh(\iota_j \sqrt{\lambda} \ell_j)}, \forall x \in I_j, j \in E_k^-. \end{aligned} $$

Therefore, for $j \in E_k^+$, we have

$$ \begin{aligned} \partial_n u_j(I(j)) &= -u'_j(0) = -\iota_j \sqrt{\lambda} \frac{a_j - b_j \cos(\iota_j \sqrt{\lambda} (\ell_j))}{\sin(\iota_j \sqrt{\lambda} \ell_j)} \\ &= -\iota_j \sqrt{\lambda} \frac{x_{F(j)} - x_{I(j)} \cos(\iota_j \sqrt{\lambda} (\ell_j))}{\sin(\iota_j \sqrt{\lambda} \ell_j)}, \end{aligned} $$

as well as

$$ \partial_n u_j(F(j)) = u'_j(\ell_j) = -\iota_j \sqrt{\lambda} \frac{x_{I(j)} - x_{F(j)} \cos(\iota_j \sqrt{\lambda} \ell_j)}{\sin(\iota_j \sqrt{\lambda} \ell_j)}. $$

Since these two last expressions are symmetric in $I(j)$ and $F(j)$, we then have

$$ \partial_n u_j(v) = -\iota_j \sqrt{\lambda} \frac{x_{v_1} - x_v \cos(\iota_j \sqrt{\lambda} \ell_j)}{\sin(\iota_j \sqrt{\lambda} \ell_j)}, \forall j \in E_{v,v_1} \cap E_k^+, v_1 \in V_v, v \in V_{\text{int}} \cup V_{\text{Neu}}. $$
We can prove similarly that

$$ \partial_n u_j(v) = -\iota_j \sqrt{\lambda} \frac{x_{v_1} - x_v \cosh(\iota_j \sqrt{\lambda} \ell_j)}{\sinh(\iota_j \sqrt{\lambda} \ell_j)}, \forall j \in E_{v,v_1} \cap E_k^-, v_1 \in V_v, v \in V_{\text{int}} \cup V_{\text{Neu}}. $$

With the notation (4.14), the two above properties can be written

$$ \partial_n u_j(v) = \iota_j \sqrt{\lambda} \frac{x_v c_j(k, \lambda) - x_{v_1}}{s_j(k, \lambda)}, \forall j \in E_{v,v_1}, v_1 \in V_v, v \in V_{\text{int}} \cup V_{\text{Neu}}. $$

This means that (2.9) can be written as

$$ \sum_{v_1 \in V_v} \sum_{j \in E_{v,v_1}} k_j \iota_j \frac{x_v c_j(k, \lambda) - x_{v_1}}{s_j(k, \lambda)} = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}}. $$

Splitting the difference and using that $x_v = 0$ if $v \in V_{\text{Dir}}$, we get equivalently

$$ \left( \sum_{j \in E_v} k_j \iota_j \frac{c_j(k, \lambda)}{s_j(k, \lambda)} \right) x_v - \sum_{v_1 \in V_v \cap (V_{\text{int}} \cup V_{\text{Neu}})} \left( \sum_{j \in E_{v,v_1}} \frac{k_j \iota_j}{s_j(k, \lambda)} \right) x_{v_1} = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}}, $$

By the definition of the matrix $M_{k,\lambda}$, this system is equivalent to

$$ M_{k,\lambda} X = 0, $$

where $X = (x_v)_{v \in V_{\text{int}} \cup V_{\text{Neu}}}$ is the unknown vector. This proves (4.16). ■

As in the previous subsection, this result and Remark 2.5 directly allow to treat the case of negative eigenvalues.

**Corollary 4.6** Let $\lambda$ be a negative real number such that

$$ (4.19) \qquad \lambda \neq k_j \left(\frac{n\pi}{\ell_j}\right)^2, \forall n \in \mathbb{N}, j \in E_k^-. $$

Then $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is a root of the characteristic equation

$$ \det M_{-k,-\lambda} = 0. $$

Let us now treat the case when (4.15) or (4.19) does not hold.

**Theorem 4.7** Let $\lambda$ be a positive real number such that

$$ (4.20) \qquad \lambda = k_{j_0} \left( \frac{n_{j_0} \pi}{\ell_{j_0}} \right)^2, $$

for some $n_{j_0} \in \mathbb{N}^*$, and some $j_0 \in E_k^+$. Then $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is a root of the characteristic equation

$$ (4.21) \qquad \det \tilde{M}_{k,\lambda} = 0, $$

where $\tilde{M}_{k,\lambda}$ is defined below.
**Proof.** Due to our assumption (4.20), the set

$$E_{k,e}^{+}(\lambda) := \{j \in E_{k}^{+} | \exists! n_{j}(k, \lambda) \in \mathbb{N}^{*}: \lambda = k_{j} \left( \frac{n_{j}(k, \lambda)\pi}{\ell_{j}} \right)^{2}\}$$

is not empty. Hence, for $j \in E_{k,e}^{+}$ (for the sake of simplicity we skip its dependency in $\lambda$, similarly we skip the dependency in $k$ and $\lambda$ of $n_j(k, \lambda)$), we cannot use the expression (4.17) for the restriction to $e_j$ of an eigenvector $u$ associated with $\lambda$. Then setting $E_{k,g}^{+} = E_{k}^{+} \setminus E_{k,e}^{+}$, an eigenvector $u$ of eigenvalue $\lambda$ is in the form

$$(4.22) \qquad u_j(x) = a_j \sin(\iota_j \sqrt{\lambda} x) + b_j \cos(\iota_j \sqrt{\lambda} x), \forall x \in I_j, j \in E_{k,e}^{+},$$

$$(4.23) \qquad u_j(x) = \frac{a_j \sin(\iota_j \sqrt{\lambda} x) + b_j \sin(\iota_j \sqrt{\lambda} (\ell_j - x))}{\sin(\iota_j \sqrt{\lambda} \ell_j)}, \forall x \in I_j, j \in E_{k,g}^{+},$$

while $u_j$ is in the form (4.18) for all $j \in E_k^-$, for some constants $a_j$ and $b_j$, $j \in E$. Now for $j \in E_{k,e}^+$, since

$$\iota_j \sqrt{\lambda} \ell_j = n_j \pi,$$

we directly check that

$$(4.24) \qquad u_j(0) = b_j, \text{ and } u_j(\ell_j) = b_j(-1)^{n_j}.$$

Similarly one has

$$(4.25) \qquad u'_{j}(0) = \iota_{j} \sqrt{\lambda} a_{j}, \text{ and } u'_{j}(\ell_{j}) = \iota_{j} \sqrt{\lambda} a_{j} (-1)^{n_{j}}.$$

Hence setting $x_v = u(v)$ for all $v \in V$, we see that (2.7) holds if and only if

$$(4.26) \qquad b_j = x_{I(j)}, \text{ and } a_j = x_{F(j)}, \forall j \in E \setminus E_{k,e}^{+},$$

and

$$(4.27) \qquad b_j = x_{I(j)}, \text{ and } b_j = x_{F(j)}(-1)^{n_j}, \forall j \in E_{k,e}^{+},$$

while (2.8) holds if and only if (3.5) holds.

Hence using as unknowns $x_v = u(v)$ for all $v \in V_{\text{int}} \cup V_{\text{Neu}}$ and $a_j, j \in E_{k,e}^+$, we see that (2.9) holds if and only if

$$\left( \sum_{j \in E_v \setminus E_{k,e}^+} k_j \ell_j \frac{c_j(k, \lambda)}{s_j(k, \lambda)} \right) x_v - \sum_{v_1 \in V_v \cap (V_{\text{int}} \cup V_{\text{Neu}})} \left( \sum_{j \in E_{v,v_1} \setminus E_{k,e}^+} \frac{k_j \ell_j}{s_j(k, \lambda)} \right) x_{v_1} + \sum_{j \in E_v \cap E_{k,e}^+} k_j \ell_j (-1)^{\sigma_j(k,v)} a_j = 0, \forall v \in V_{\text{int}} \cup V_{\text{Neu}},$$

where for all $j \in E_{k,e}^+$,

$$\sigma_j(k, v) = \begin{cases} 1 & \text{if } I(j) = v, \\ n_j & \text{if } F(j) = v. \end{cases}$$

Now we notice that (4.27) holds if and only if

$$(4.28) \qquad x_{I(j)} = (-1)^{n_j} x_{F(j)}, \forall j \in E_{k,e}^{+},$$
while this property and (3.5) also leads to

$$ (4.29) \quad (x_{I(j)} = 0 \text{ if } j \in E_{k,e}^{+} \text{ and } F(j) \in V_{\text{Dir}}), \text{ or } (x_{F(j)} = 0 \text{ if } j \in E_{k,e}^{+} \text{ and } I(j) \in V_{\text{Dir}}), $$

as well as

$$ (4.30) \qquad x_{I(j)} = x_{F(j)} = 0 \text{ if } j \in E_{k,e}^{+} \cap E_{\circ} \text{ and } n_j \text{ odd.} $$

In summary, defining the matrix $\tilde{M}_{k,\lambda}$ of size $\#V_{\text{int}} + \#V_{\text{Neu}} + \#E_{k,e}^{+}$ by

$$ (M_{k,\lambda})_{v,v} = \sum_{j \in E_v \setminus (E_{k,e}^{+} \cup E_{\circ})} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)} - 2 \sum_{j \in (E_v \cap E_{\circ}) \setminus E_{k,e}^{+}} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(4k, \lambda)}{s_j(4k, \lambda)}, $$

$$ (M_{k,\lambda})_{v,v_1} = - \sum_{j \in E_{v,v_1} \setminus E_{k,e}^{+}} \frac{\operatorname{sgn}(k_j) \sqrt{|k_j|}}{s_j(k, \lambda)}, \text{ if } v_1 \in V_v \setminus \{v\}, $$

$$ (M_{k,\lambda})_{v,v_1} = 0, \text{ if } v_1 \notin \{v\} \cup V_v, $$

$$ (M_{k,\lambda})_{v,j} = k_j \ell_j (-1)^{\sigma_j(k,v)}, \text{ if } j \in E_v \setminus E_{\circ}, $$

$$ (M_{k,\lambda})_{v,j} = -2, \text{ if } j \in E_v \cap E_{\circ}: n_j \text{ odd,} $$

$$ (M_{k,\lambda})_{v,j} = 0, \text{ else,} $$

$$ (M_{k,\lambda})_{j,v} = 1, \text{ if } I(j) = v, j \notin E_{\circ} \text{ or } j \in E_{\circ}: n_j \text{ odd,} $$

$$ (M_{k,\lambda})_{j,v} = (-1)^{1+n_j}, \text{ if } F(j) = v, j \notin E_{\circ}, $$

$$ (M_{k,\lambda})_{j,v} = 0, \text{ else,} $$

$$ (M_{k,\lambda})_{j,j_1} = 0, $$

for all $v, v_1 \in \#V_{\text{int}} + \#V_{\text{Neu}}$, $j, j_1 \in E_{k,e}^{+}$, then (4.28), (4.28), (4.29), and (4.30) hold if and only if

$$ (4.31) \qquad \tilde{M}_{k,\lambda} \left( (x_v)_{v \in V_{\text{int}} \cup V_{\text{Neu}}}, (a_j)_{j \in E_{k,e}^{+}} \right)^{\top} = 0. $$

Hence a non trivial eigenvector associated with $\lambda$ exists if and only if (4.21) holds. ■

This theorem and Remark 2.5 directly lead to

**Corollary 4.8** Let $\lambda$ be a negative real number such that

$$ (4.32) \qquad \lambda = k_{j_0} \left( \frac{n_{j_0} \pi}{\ell_{j_0}} \right)^2, $$

for some $n_{j_0} \in \mathbb{N}^*$, and some $j_0 \in E_k^+$. Then $\lambda$ is an eigenvalue of $T$ if and only if

$$ \det \tilde{M}_{-k,-\lambda} = 0. $$

**Remark 4.9** Note that the previous Theorem and Corollary show that if $G$ has an edge $e_j$ that is a circle, then for all $n \in \mathbb{N}^*$, $k_j (\frac{2n\pi}{\ell_j})^2$ is always an eigenvalue of $T$, with at least the associated eigenvector $u$ given by

$$ u_j(x) = \sin(2nx), \forall x \in I_j, $$

$$ u_{\ell} = 0, \forall \ell \in E \setminus \{j\}. $$

**Remark 4.10** In the positive definite case, namely if $E_k^- = \emptyset$, Theorem 4.7 covers any value of $\ell_j$ and $k_j$, while in [36] (resp. [29]) only the case $\ell_j^2 = k_j$, for all $j \in E$ (resp. $\ell_j = k_j = 1$, for all $j \in E$) was treated.
# 5 Weyl's formulas

Our goal is to give the asymptotic behavior of large eigenvalues of $T$ in the spirit of the one obtained in the positive definite case in [34, 30, 6]. Here we use a method developed in [23, §4.3] based on an abstract operator theory result, namely [5, Theorem 3, p. 215]. Since the positive definite case seems to be treated only in the constant coefficients case, we assume in this section that $k$ could be positive in $G$, in other words, that $E_k^+$ is nonempty, while $E_k^-$ may be empty.

Due to Theorem 2.6, we can write

$$ \sigma_+(T) = \{\lambda_{+,n}\}_{n \in \mathbb{N}^*}, $$

where the eigenvalues $\lambda_{+,n}$ are repeated according to their multiplicity and are enumerated in a non-decreasing order in $n$. Of course, if $E_k^-$ is nonempty, then we can write

$$ \sigma_-(T) = \{\lambda_{-,n}\}_{n \in \mathbb{N}^*}, $$

where the eigenvalues $\lambda_{-,n}$ are repeated according to their multiplicity and are enumerated in a non-increasing order in $n$.

For $\lambda > 0$, let us denote by

$$ N_+(\lambda) = \#(\sigma_+(T) \cap (0, \lambda)). $$

We further introduce the operator $T_{\text{Dir}}$ on $L^2(G)$ as follows:

$$ T_{\text{Dir}}u = -(k_j u_j')'_{j \in E}, \forall u \in D(T_{\text{Dir}}) := \bigoplus_{j \in E} (H_0^1(I_j) \cap H^2(I_j)). $$

This operator is clearly selfadjoint in $L^2(G)$ with a compact resolvent and since by the Dirichlet condition at all extremities, the edges are completely separated, its spectrum $\sigma(T_{\text{Dir}})$ is discrete and made of a positive part $\sigma_+(T_{\text{Dir}})$ and a negative part $\sigma_-(T_{\text{Dir}})$ (that is non empty if $E_k^-$ is non empty), that can be characterized as follows:

$$ \sigma_+(T_{\text{Dir}}) = \bigcup_{j \in E_k^+} \sigma(T_{j, \text{Dir}}), \quad \sigma_-(T_{\text{Dir}}) = \bigcup_{j \in E_k^-} \sigma(T_{j, \text{Dir}}), $$

where

$$ T_{j, \text{Dir}}u = -(k_j u_j')', \forall u \in D(T_{j, \text{Dir}}) = H_0^1(I_j) \cap H^2(I_j), \forall j \in E. $$

As before, for $\lambda > 0$, let us denote by

$$ N_{+, \text{Dir}}(\lambda) = \#(\sigma_+(T_{\text{Dir}}) \cap (0, \lambda)). $$

$$ N_{j, \text{Dir}}(\lambda) = \#(\sigma_+(T_{j, \text{Dir}}) \cap (0, \lambda)), \forall j \in E_k^+. $$

Let us now state the following property satisfied by $N_{+, \text{Dir}}(\lambda)$.

**Lemma 5.1** Under the assumptions of this section, for all $\lambda > 0$, we have

$$ (5.1) \qquad N_{+, \text{Dir}}(\lambda) \sim \frac{\mathcal{L}_+}{\pi} \sqrt{\lambda} \quad \text{for } \lambda \to \infty, $$

in the sense that

$$ \lim_{\lambda \to \infty} \frac{N_{+, \text{Dir}}(\lambda)}{\frac{\mathcal{L}_+}{\pi} \sqrt{\lambda}} = 1, $$

$$ \text{where } \mathcal{L}_+ = \sum_{j \in E_k^+} \int_0^{\ell_j} \frac{1}{\sqrt{k_j(x)}} dx. $$
**Proof.** By [1, Theorem 8.2] (see also [22, (3.2)]), we know that

$$N_{j,\mathrm{Dir}}(\lambda) \sim \frac{\int_0^{\ell_j} \frac{1}{\sqrt{k_j(x)}} dx}{\pi} \sqrt{\lambda} \quad \text{for } \lambda \to \infty,$$

for all $j \in E_k^+$. (5.1) directly follows from this equivalence and the identity

$$N_{+,\mathrm{Dir}}(\lambda) = \sum_{j \in E_k^+} N_{j,\mathrm{Dir}}(\lambda).$$

<figure><img src="image_3.png" /></figure>

**Theorem 5.2** Under the assumptions of this section, the following Weyl's formula holds

$$(5.2) \qquad \lim_{n \to \infty} \frac{\lambda_{+,n}}{n^2} = \frac{\pi^2}{\mathcal{L}_+^2}.$$

Further if $E_k^-$ is non empty, then by setting $\mathcal{L}_- = \sum_{j \in E_k^-} \int_0^{\ell_j} \frac{1}{\sqrt{|k_j(x)|}} dx$, one has

$$(5.3) \qquad \lim_{n \to \infty} \frac{\lambda_{-,n}}{n^2} = -\frac{\pi^2}{\mathcal{L}_-^2}.$$

**Proof.** We start by proving the asymptotic behaviour (5.2). According to [5, Theorem 3, p. 215], for a fixed positive real number $\xi \in \rho(T) \cap \rho(T_{\mathrm{Dir}})$, we consider the operator

$$D = (\xi \mathbb{I} - T)^{-1} - (\xi \mathbb{I} - T_{\mathrm{Dir}})^{-1},$$

and need to show that $\dim R(D)$ is finite. For that purpose, we fix $f \in L^2(G)$ and consider

$$d = u - u_{\mathrm{Dir}},$$

where $u = (\xi \mathbb{I} - T)^{-1}f$ and $u_{\mathrm{Dir}} = (\xi \mathbb{I} - T_{\mathrm{Dir}})^{-1}f$. Hence by definition, we have

$$\xi u_j + (k_j u'_j)' = \xi u_{\mathrm{Dir},j} + k_j (u'_{\mathrm{Dir},j})' = f_j, \quad \text{in } I_j, \forall j \in E.$$

Therefore for all $j \in E$, $d_j$ is solution of the homogeneous differential equation

$$\xi d_j + (k_j d'_j)' = 0 \quad \text{in } I_j, \forall j \in E.$$

Hence by Lemma 4.1,

$$d_j(x) = a_{1,j} u_{1,j}(\xi, x) + a_{2,j} u_{2,j}(\xi, x), \forall x \in I_j, j \in E,$$

where $a_{1,j}, a_{2,j}$ are arbitrary constants. Therefore $R(D)$ is included into a finite dimensional space of dimension $2\#E$ and therefore it is closed and of finite dimension $r \le 2\#E$.

For any $\lambda > 0$, we deduce by [5, Theorem 3, p. 215] that

$$N_{+,\mathrm{Dir}}(\lambda) - r \le N_+(\lambda) \le N_{+,\mathrm{Dir}}(\lambda) + r.$$

Using the bounds from (5.1) and the fact that $r \le 2\#E$, we obtain

$$N_+(\lambda) \sim \frac{\mathcal{L}_+}{\pi} \sqrt{\lambda}, \quad \text{for } \lambda \to \infty.$$
In this equivalence, for $n \in \mathbb{N}^*$, if we take $\lambda = \lambda_{+,n}$, we find

$$n - m(\lambda_{+,n}) \sim \frac{\mathcal{L}_+}{\pi} \sqrt{\lambda_{+,n}}, \text{ for } n \to \infty,$$

where $m(\lambda_{+,n})$ is the multiplicity of $\lambda_{+,n}$. By Theorem 4.2, $m(\lambda_{+,n}) \le 2\#E$, and therefore

$$n \sim \frac{\mathcal{L}_+}{\pi} \sqrt{\lambda_{+,n}}, \text{ for } n \to \infty,$$

that is nothing else than (5.2).

The asymptotic behavior (5.3) follows from (5.2) and Remark 2.5. ■

**Remark 5.3** If $E_k^-$ is empty, the previous theorem extends to the variable coefficients case the results from [34, 30, 6, 23].

**Remark 5.4** 1. Let us define by $T_{G_+,Dir}$ (resp. $T_{G_-,Dir}$) on $G_+ = \oplus_{j \in E_k^+} e_j$ (resp. $G_- = \oplus_{j \in E_k^-} e_j$) as the operator $T$ but with Dirichlet boundary conditions at the nodes that belongs to $\bar{G}_+ \cap \bar{G}_-$. Then the previous theorem tells us that the eigenvalues of $T_{G_+,Dir}$ (resp. $T_{G_-,Dir}$) satisfy (5.2) (resp. (5.3)). In other words, large positive (resp. negative) eigenvalues of $T$ and $T_{G_+,Dir}$ (resp. $T_{G_-,Dir}$) have the same asymptotic distribution.

2. Since (5.2) and (5.3) hold with Dirichlet or Neumann boundary conditions at the exterior nodes, the same asymptotic distributions hold if we replace the Dirichlet boundary condition at some nodes of $\bar{G}_+ \cap \bar{G}_-$ by the Neumann boundary condition.

# 6 Examples

In the whole section, we illustrate our results on some examples. On each subsection, we suppose that $k_j \in W^{1,r}(0, \ell_j)$, for some $r > 2$, and $j \in E$ are given real-valued functions satisfying (2.1) in the first subsection, while they will be constant in the following subsections.

## 6.1 A star-shaped graph with mixed boundary conditions

Consider a graph made of $N$ edges $e_j$ of length $\ell_j$, $j = 1, \dots, N \ge 2$. We assume that these edges form a star-shaped graph by attaching them at a vertex $v_0$. Hence $V_{\text{int}} = \{v_0\}$ and $V_{\text{ext}}$ is made of $N$ vertices. We also split up $V_{\text{ext}}$ into a Dirichlet part $V_{\text{Dir}}$ and a Neumann part $V_{\text{Neu}}$ (one of them may be empty). Without loss of generality, we can denote by $v_j$, $j = 1, \dots, N_n = \#V_{\text{Neu}}$ the set of vertices of $V_{\text{Neu}}$, that are here extremities of $e_j$ (if $V_{\text{Neu}}$ is empty, we set $N_n = 0$).

### 6.1.1 Invertibility of the operator

In the above setting, the matrix $M_k$ is of size $1 + N_n$ and is given by

$$m_{v_0,v_0} = \sum_{j=1}^{N} \kappa_j,$$

$$m_{v_0,v_j} = \kappa_j, \forall j = 1, \dots, N_n,$$

$$m_{v_j,v_\ell} = 0, \forall j, \ell = 1, \dots, N_n.$$
By recurrence on $N_n$, one can show that

$$ \det M_k = \prod_{j=1}^{N_n} \kappa_j \left( \sum_{j=N_n+1}^{N} \kappa_j \right), $$

if $N_n \ge 1$, otherwise,

$$ \det M_k = \sum_{j=1}^{N} \kappa_j. $$

Therefore $T$ is invertible if and only if $\sum_{j=N_n+1}^{N} \kappa_j \ne 0$. This result is in accordance with Theorems 2.1 and 3.1 of [2] obtained in the constant coefficients case and improves [2, Theorem 2.3].

### 6.1.2 Spectral characterisation: Dirichlet boundary conditions

In this subsection, we assume that $V_{\text{Dir}} = V_{\text{ext}}$.

In the setting of Theorem 4.5, for $\lambda > 0$ satisfying (4.15), the matrix $M_{k,\lambda} = (m_{k,\lambda})$ is a real number given by

$$ (6.1) \quad m_{k,\lambda} = \sum_{j=1}^{N} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)} = \sum_{\substack{j=1 \\ k_j > 0}}^{N} \sqrt{k_j} \cot \left( \sqrt{\frac{\lambda}{k_j}} \ell_j \right) - \sum_{\substack{j=1 \\ k_j < 0}}^{N} \sqrt{|k_j|} \coth \left( \sqrt{\frac{\lambda}{|k_j|}} \ell_j \right), $$

hence $\lambda$ is an eigenvalue of $T$ if and only if

$$ (6.2) \quad \sum_{\substack{j=1 \\ k_j > 0}}^{N} \sqrt{k_j} \cot \left( \sqrt{\frac{\lambda}{k_j}} \ell_j \right) - \sum_{\substack{j=1 \\ k_j < 0}}^{N} \sqrt{|k_j|} \coth \left( \sqrt{\frac{\lambda}{|k_j|}} \ell_j \right) = 0. $$

Similarly $\lambda < 0$ satisfying (4.19) is an eigenvalue of $T$ if and only if

$$ (6.3) \quad \sum_{\substack{j=1 \\ k_j > 0}}^{N} \sqrt{k_j} \coth \left( \sqrt{-\frac{\lambda}{k_j}} \ell_j \right) - \sum_{\substack{j=1 \\ k_j < 0}}^{N} \sqrt{|k_j|} \cot \left( \sqrt{-\frac{\lambda}{|k_j|}} \ell_j \right) = 0. $$

In the particular case $k_1 = -k_\ell = 1$, for all $\ell \ge 2$, $\ell_j = 1$, for all $j = 1, \dots, N$, (6.2) and (6.3) reduce to

$$ (6.4) \quad \cot \sqrt{\lambda} - (N-1) \coth \sqrt{\lambda} = 0, $$

and

$$ (6.5) \quad \coth \sqrt{-\lambda} - (N-1) \cot \sqrt{-\lambda} = 0. $$

As $\coth \sqrt{\lambda}$ is close to 1 as $\sqrt{\lambda}$ is large, we deduce that the large roots of (6.4) (resp. (6.5)) are close to $(x_+ + \ell\pi)^2$ (resp. $-(x_- + \ell\pi)^2$), for $\ell$ large and $x_+, x_- \in (0, \frac{\pi}{2})$ are such that $\cot x_+ = N-1$ and $\cot x_+ = \frac{1}{N-1}$.

For $N = 50$, Figures 1 and 2 show the graph of $f_+(x) = \cot x - (N-1) \coth x$, and of $f_-(x) = \coth x - (N-1) \cot x$, whose zeros yield the solutions of (6.4) and (6.5). These Figures confirm the asymptotic behavior mentioned above. Note even that Newton method gives that the
<figure><img src="image_4.png" /><figcaption>Figure 1: The function $f_+$ for positive eigenvalues for a star-graph with Dirichlet bc for $N = 50$</figcaption></figure>

<figure><img src="image_5.png" /><figcaption>Figure 2: The function $f_-$ for negative eigenvalues for a star-graph with Dirichlet bc for $N = 50$</figcaption></figure>
first three positive roots $x_1, x_2, x_3$ of $f_+$ are $x_1 \sim 3.16192496, x_2 \sim 6.30359050$ and $x_3 \sim 9.44518329$ that are already close to the roots of $\cot x - 49 = 0$ whose values, also computed by Newton method, are 3.16199798, 6.30359064 and 9.44518329.

Now for a fixed $j_0 \in \{1, \dots, N\}$ such that $k_{j_0} > 0$ and any $n_0 \in \mathbb{N}$, let

$$ \lambda = k_{j_0} \left( \frac{n_0 \pi}{\ell_{j_0}} \right)^2. $$

Owing to Theorem 4.7, as the Dirichlet boundary conditions imply that $x_{v_0} = 0$, we see that the system (4.31) reduced to

$$ \sum_{j \in E_{k,e}^{+}(\lambda)} \sqrt{k_j} (-1)^{\sigma_j(k,v_0)} a_j = 0. $$

Hence $\lambda$ is an eigenvalue of $T$ if and only if $\#E_{k,e}^{+}(\lambda) \ge 2$ and its multiplicity is $\#E_{k,e}^{+}(\lambda) - 1$. Note that this situation appears if and only if there exist two different indices $j, j' \in \{1, \dots, N\}$ with $k_j, k_{j'} > 0$ such that

$$ \frac{\sqrt{k_j}}{\frac{\ell_j}{\ell_{j'}}} \in \mathbb{Q}. $$

Hence if

$$ \frac{\sqrt{k_j}}{\frac{\ell_j}{\ell_{j'}}} \notin \mathbb{Q}, \forall j, j' \in \{1, \dots, N\}: j' \neq j, k_j, k_{j'} > 0, $$

the set $E_{k,e}^{+}(\lambda)$ is reduced to a singleton, and the positive eigenvalues of $T$ are therefore reduced to the roots of (6.2).

In the same manner, Corollary 4.8 yields that $k_{j_0} \left( \frac{n_0 \pi}{\ell_{j_0}} \right)^2$ for some $n_0 \in \mathbb{N}$ and some $j_0 \in \{1, \dots, N\}$ such that $k_{j_0} < 0$ if and only if there exists $j \in E_k^- \setminus \{j_0\}$ and $n \in \mathbb{N}^*$ such that

$$ \frac{\sqrt{k_{j_0}}}{\frac{\ell_{j_0}}{\ell_{j'}}} = \frac{n}{n_0}. $$

In such a case, its multiplicity is $\#E_{k,e}^{-}(\lambda) - 1$.

Note that the results concerning the spectrum of $T$ of this subsection are in accordance with the results from [2, §5.1] that consider the case of equilateral star-graphs ($\ell_j = 1$, for all $j \in \{1, \dots, N\}$) and $k_j = 1, j = 1, \dots, D$ for some $D < N$ and $k_j = k^- < 0$ else.

Let us finally illustrate our Weyl's formulas on the particular example considered before, namely $k_1 = -k_\ell = 1$, for all $\ell \ge 2$, $\ell_j = 1$, for all $j = 1, \dots, N$. Then by our previous considerations, we have only one sequence of positive eigenvalues that behave like $(x_+ + n\pi)^2$, for $n$ large, which is in accordance with (5.2), while we have a first sequence of negative eigenvalues that behave like $-(x_- + n\pi)^2$, for $n$ large, and a second sequence of eigenvalues $-n^2\pi^2$, $n \in \mathbb{N}^*$, of multiplicity $N-2$. This allows to recover (5.3) because for $\lambda < 0$, the number of negative eigenvalues larger than $\lambda$ is asymptotically $(N-1)n\pi$.
### 6.1.3 Spectral characterisation: mixed boundary conditions

Here we suppose that Neumann conditions are applied on a non empty part of $V_{\text{ext}}$, i. e., we suppose that $N_n = \#V_{\text{Neu}} \ge 1$.

In the setting of Theorem 4.5, for $\lambda > 0$ satisfying (4.15), the matrix $M_{k,\lambda}$ is now a square matrix of size $1 + N_n$ given by

$$ (M_{k,\lambda})_{v_0,v_0} = m_{k,\lambda}, $$

$$ (M_{k,\lambda})_{v_0,v_j} = (M_{k,\lambda})_{v,v_0} = -\operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{1}{s_j(k, \lambda)}, \forall j = 1, \dots, N_n, $$

$$ (M_{k,\lambda})_{v_j,v_j} = \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)}, \forall j = 1, \dots, N_n, $$

$$ (M_{k,\lambda})_{v_j,v_j'} = 0, \forall j, j' = 1, \dots, N_n, j \neq j', $$

where we use the notation from (6.1). By recurrence on $N_n$, one can show that

$$ \det M_{k,\lambda} = \prod_{j=1}^{N_n} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)} - \sum_{j=1}^{N_n} \frac{|k_j|}{s_j(k, \lambda)^2} \prod_{\substack{l=1 \\ l \neq j}}^{N_n} \operatorname{sgn}(k_l) \sqrt{|k_l|} \frac{c_l(k, \lambda)}{s_l(k, \lambda)}, $$

with the convention that

$$ \prod_{\substack{l=1 \\ l \neq j}}^{N_n} \operatorname{sgn}(k_l) \sqrt{|k_l|} \frac{c_l(k, \lambda)}{s_l(k, \lambda)} = 1, \text{ if } N_n = 1. $$

Hence $\lambda$ is an eigenvalue of $T$ if and only if $\det M_{k,\lambda} = 0$.

Similarly $\lambda < 0$ satisfying (4.19) is an eigenvalue of $T$ if and only if $\det M_{-k,-\lambda} = 0$.

In the particular case, when $N_n = 1$, $k_1 = -k_\ell = 1$, for all $\ell \ge 2$, $\ell_j = 1$, for all $j = 1, \dots, N$, one directly checks that $\det M_{k,\lambda} = 0$ if and only if $f(\sqrt{\lambda}) = 0$, where

$$ f(x) = \sin^2 x + (N-1) \cos x \coth x, \forall x \in \mathbb{R}. $$

In Figure 3, the graph of $f$ on $x > 0$ is presented with $N = 50$. There we can see that the large positive roots of $f$ are close to the roots of the cosine function (which is natural since for $x$ large, $f(x)$ is close to $\sin^2 x + (N-1) \cos x$, hence dividing by $N-1$, for a root $x$ of $f$, $\cos x$ is close to zero if $N$ is large). But Newton method yields that the first three positive roots $x_1, x_2, x_3$ of $f$ are $x_1 \sim 1.5912, x_2 \sim 4.6920$ and $x_3 \sim 7.8744$ that are already close to $\frac{\pi}{2} \sim 1.5708$, $\frac{3\pi}{2} \sim 4.7124$ and $\frac{5\pi}{2} \sim 7.8540$. Similarly, one shows that $\det M_{-k,-\lambda} = 0$ if and only if

$$ \sin \sqrt{-\lambda} + (N-1) \frac{\cosh \sqrt{-\lambda} \cos \sqrt{-\lambda}}{\sinh^2 \sqrt{-\lambda}} = 0. $$

The graph of this left-hand side for $N = 50$ is shown in Figure 4, and confirms the asymptotic behavior of these eigenvalues in $-n^2\pi^2$.

For the illustration of Theorem 4.7 and Corollary 4.8, since many configurations may appear, let us restrict ourselves to the previous particular case, when $N_n = 1$, $k_1 = -k_\ell = 1$, for all $\ell \ge 2$, $\ell_j = 1$, for all $j = 1, \dots, N$. In such a case, one easily deduces that $n^2\pi^2$ is not an eigenvalue of $T$, while $-n^2\pi^2$ is an eigenvalue of $T$, for all $n \in \mathbb{N}^*$.
<figure><img src="image_6.png" /><figcaption>Figure 3: The function $f$ for positive eigenvalues for a star-graph with mixed bc, $N_n = 1$ and $N = 50$</figcaption></figure>

<figure><img src="image_7.png" /><figcaption>Figure 4: The function $f$ for negative eigenvalues for a star-graph with mixed bc, $N_n = 1$ and $N = 50$</figcaption></figure>
<figure><img src="image_8.png" /><figcaption>Figure 5: A tadpole</figcaption></figure>

## 6.2 The tadpole with a Dirichlet boundary condition

Consider the bounded tadpole made of a circle $C_1$ of length $\ell_1 > 0$ and an interval $I_2$ of length $\ell_2$, that will form a graph by attaching one of the extremity of $I_2$ at a point, say 0 of $C_1$, see Figure 5. Then the set of vertices is made of the connecting point $v$ and the other extremity $v_1$ of $I_2$. Hence $V_{\text{int}} = \{v\}$ and $V_{\text{ext}} = \{v_1\}$. We consider the operator $T$ with the choice of $V_{\text{Dir}} = V_{\text{ext}}$. Since $\sigma(\alpha T) = \alpha\sigma(T)$, in the constant coefficients case, we can always assume that $k_1 = 1$. Further since we are mainly interested in the indefinite case, we here assume that $k_2$ is negative.

### 6.2.1 Invertibility of the operator

By Theorem 3.1, the matrix $M_k = (\kappa_2)$ is a real number which is always different from zero. Therefore the operator $T$ is always invertible. This result is in accordance with [2, Theorem 4.1] obtained in the constant coefficients case.

### 6.2.2 Spectral characterisation

Now in the setting of Theorem 4.5, for $\lambda > 0$ satisfying

$$ (6.6) \qquad \lambda \neq \left(\frac{n\pi}{\ell_1}\right)^2, \forall n \in \mathbb{N}^*, $$

the matrix $M_{k,\lambda} = (m_{k,\lambda})$ is a real number with

$$ (6.7) \qquad m_{k,\lambda} = -\sqrt{|k_2|} \coth \left( \sqrt{\frac{\lambda}{|k_2|}} \ell_2 \right) - 2 \cot \left( \frac{\sqrt{\lambda} \ell_1}{2} \right). $$

Hence $\lambda$ is an eigenvalue of $T$ if and only if

$$ (6.8) \qquad 2 \cot \left( \frac{\sqrt{\lambda} \ell_1}{2} \right) + \sqrt{|k_2|} \coth \left( \sqrt{\frac{\lambda}{|k_2|}} \ell_2 \right) = 0. $$

As $\coth x$ is close to 1 for $x$ large, we directly deduce that large positive eigenvalues are close to $\frac{4(x_++\ell\pi)^2}{\ell_1^2}$, with $\ell \in \mathbb{N}^*$ large enough and $x_+ \in (0, \pi)$ is the unique solution of $\cot x_+ = -\frac{\sqrt{|k_2|}}{2}$, see Figure 6 for the graph of $2\cot(\frac{x\ell_1}{2}) + \sqrt{|k_2|}\coth(\frac{x}{\sqrt{|k_2|}}\ell_2)$ with $\ell_1 = 2$, $\ell_2 = 1$, and $k_2 = -100$. For this example, Newton method gives a first root $\sim 3.0819$ yielding the smallest positive eigenvalue around $3.0819^2$.

Similarly $\lambda < 0$ satisfying

$$ (6.9) \qquad \lambda \neq k_2 \left(\frac{n\pi}{\ell_2}\right)^2, \forall n \in \mathbb{N}, $$
<figure><img src="image_9.png" /><figcaption>Figure 6: A characteristic equation for positive eigenvalues for the tadpole with Dirichlet bc</figcaption></figure>

<figure><img src="image_10.png" /><figcaption>Figure 7: A characteristic equation for negative eigenvalues for the tadpole with Dirichlet bc</figcaption></figure>
is an eigenvalue of $T$ if and only if

$$ (6.10) \qquad 2 \coth \left( \frac{\sqrt{-\lambda} \ell_1}{2} \right) + \sqrt{|k_2|} \cot \left( \sqrt{\frac{-\lambda}{|k_2|}} \ell_2 \right) = 0. $$

Hence large negative eigenvalues are close to $-\frac{k_2(x_- + \ell\pi)^2}{\ell_2^2}$, with $\ell \in \mathbb{N}^*$ large enough and $x_- \in (0, \pi)$ is the unique solution of $\cot x_- = -\frac{2}{\sqrt{|k_2|}}$, see Figure 7 for the graph of $\sqrt{|k_2|} \cot(\frac{x}{\sqrt{|k_2|}} \ell_2) + 2 \coth(\frac{x\ell_1}{2})$ with $\ell_1 = 2$, $\ell_2 = 1$, and $k_2 = -100$. For this example, Newton method gives a first root $\sim 17.6819$ yielding the largest negative eigenvalue around $-17.6819^2$.

Now by Remark 4.9, for all $n \in \mathbb{N}^*$, $(\frac{2n\pi}{\ell_1})^2$ is always an eigenvalue of $T$. Further if $n \in \mathbb{N}^*$ is odd, then for $\lambda = (\frac{n\pi}{\ell_1})^2$, by Theorem 4.7, $\tilde{M}_{k,\lambda}$ is given by

$$ \tilde{M}_{k,\lambda} = \begin{pmatrix} \sqrt{|k_2|} \coth(\sqrt{\frac{\lambda}{|k_2|}} \ell_2) & -2 \\ 0 & 1 \end{pmatrix}, $$

and since its determinant is different from zero, $\lambda$ is not an eigenvalue of $T$. Finally for $\lambda = k_2(\frac{n\pi}{\ell_2})^2$, for some $n \in \mathbb{N}^*$, then

$$ \tilde{M}_{-k,-\lambda} = \begin{pmatrix} \coth(\sqrt{-\lambda} \ell_1) & (-1)^{\sigma_1(-k,v)} \\ 1 & 0 \end{pmatrix}, $$

and since its determinant is different from zero, by Corollary 4.8, $\lambda$ is not an eigenvalue of $T$. Note that our results are in accordance with the ones from [2, §7.1].

## 6.3 The tadpole with a Neumann boundary condition

Here we consider the same configuration than in the previous subsection but with $V_{\text{Neu}} = V_{\text{ext}}$.

### 6.3.1 Invertibility of the operator

Here by Theorem 3.1, the matrix $M_k$ is a $2 \times 2$ matrix with

$$ m_{v,v} = m_{v_1,v_1} = \kappa_2 = -m_{v,v_1}. $$

This implies that $\det M_k = 0$, therefore the operator $T$ is never invertible and its kernel is one-dimensional (because the rank of $M_k$ is one) made of constant functions.

### 6.3.2 Spectral characterisation

Now in the setting of Theorem 4.5, for $\lambda > 0$ satisfying (6.6), the matrix $M_{k,\lambda}$ is given by

$$ M_{k,\lambda} = \begin{pmatrix} m_{k,\lambda} & -\frac{\sqrt{|k_2|}}{\sinh(\sqrt{\frac{\lambda}{|k_2|}} \ell_2)} \\ -\frac{\sqrt{|k_2|}}{\sinh(\sqrt{\frac{\lambda}{|k_2|}} \ell_2)} & -\sqrt{|k_2|} \coth(\sqrt{\frac{\lambda}{|k_2|}} \ell_2) \end{pmatrix}, $$
where we used the definition (6.7). Some easy calculations show that

$$ \det M_{k,\lambda} = \sqrt{|k_2|} \left( 2 \cot \left( \frac{\sqrt{\lambda} \ell_1}{2} \right) \coth \left( \sqrt{\frac{\lambda}{|k_2|}} \ell_2 \right) + \sqrt{|k_2|} \right). $$

Hence $\lambda$ is an eigenvalue of $T$ if and only if

$$ 2 \cot \left( \frac{\sqrt{\lambda} \ell_1}{2} \right) + \sqrt{|k_2|} \tanh \left( \sqrt{\frac{\lambda}{|k_2|}} \ell_2 \right) = 0. $$

Similarly $\lambda < 0$ satisfying (6.9) is an eigenvalue of $T$ if and only if

$$ 2 \coth \left( \frac{\sqrt{-\lambda} \ell_1}{2} \right) + \sqrt{|k_2|} \tan \left( \sqrt{\frac{-\lambda}{|k_2|}} \ell_2 \right) = 0. $$

In comparison with (6.8) and (6.10), only the function coth (resp. cot) is replaced by tanh (resp. tan), hence the asymptotic behavior of large eigenvalues are the same than for the Dirichlet boundary condition.

Now by Remark 4.9, $(\frac{2n\pi}{\ell_1})^2$ is always an eigenvalue of $T$, for all $n \in \mathbb{N}^*$, while if $n \in \mathbb{N}^*$ is odd, then for $\lambda = (\frac{n\pi}{\ell_1})^2$, one can show that $\det \tilde{M}_{k,\lambda} = -\sqrt{|k_2|} \frac{\coth(\sqrt{-\frac{\lambda}{|k_2|}} \ell_2)}{\sinh(\sqrt{\frac{-\lambda}{|k_2|}} \ell_2)}$ that is different from zero, and by Theorem 4.7, $\lambda$ is not an eigenvalue of $T$. On the other hand, for $\lambda = k_2(\frac{n\pi}{\ell_2})^2$, for some $n \in \mathbb{N}^*$, then

$$ \tilde{M}_{-k, -\lambda} = \begin{pmatrix} -2 \coth \left( \frac{\sqrt{-\lambda} \ell_1}{2} \right) & 0 & (-1)^{\sigma_1(-k, v)} \\ 0 & 0 & (-1)^{\sigma_1(-k, v_1)} \\ -1 & (-1)^n & 0 \end{pmatrix}, $$

and since its determinant is different from zero, by Corollary 4.8, $\lambda$ is not an eigenvalue of $T$.

## 6.4 A 5 edges graph with a spine, two arms and two legs with Dirichlet boundary conditions

Consider a graph made of 5 edges $e_j$ of length $\ell_j > 0$, $j = 1, \dots, 5$ such that $e_1, e_2$ and $e_3$ (resp. $e_3, e_4$ and $e_5$ have a common vertex $v_1$ (resp. $v_2$), while the other extremities of $e_1, e_2, e_4$ and $e_5$ belong to the set of exterior vertices, see Figure 8. Hence $V_{\text{int}} = \{v_1, v_2\}$. We consider the operator $T$ with the choice $V_{\text{Dir}} = V_{\text{ext}}$.

### 6.4.1 Invertibility of the operator

Using Theorem 3.1, the matrix $M_k$ is a $2 \times 2$ matrix with

$$ m_{v_1,v_1} = \kappa_1 + \kappa_2 + \kappa_3, m_{v_2,v_2} = \kappa_3 + \kappa_4 + \kappa_5, m_{v_1,v_2} = -\kappa_3. $$

Therefore $T$ is invertible if and only if $\det M_k = \kappa_3(\kappa_1 + \kappa_2 + \kappa_3, +\kappa_4 + \kappa_5) + (\kappa_1 + \kappa_2)(\kappa_4 + \kappa_5)$, is different from 0. Note that if $\det M_k$ is equal to zero, then 0 is of multiplicity one since the rank of $M_k$ is one.
<figure><img src="image_11.png" /><figcaption>Figure 8: A 5 edges graph</figcaption></figure>

### 6.4.2 Spectral characterisation

First for $\lambda > 0$ satisfying (4.15), the matrix $M_{k,\lambda}$ is a $2 \times 2$ matrix given by

$$M_{k,\lambda} = \left( \begin{array}{cc} \sum_{j=1}^{3} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)} & - \frac{\operatorname{sgn}(k_3) \sqrt{|k_3|}}{s_3(k, \lambda)} \\ - \frac{\operatorname{sgn}(k_3) \sqrt{|k_3|}}{s_3(k, \lambda)} & \sum_{j=3}^{5} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)} \end{array} \right).$$

One directly has

$$\det M_{k,\lambda} = \left( \sum_{j=1}^{3} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)} \right) \left( \sum_{j=3}^{5} \operatorname{sgn}(k_j) \sqrt{|k_j|} \frac{c_j(k, \lambda)}{s_j(k, \lambda)} \right) - \frac{|k_3|}{s_3(k, \lambda)^2},$$

and by Theorem 4.5, $\lambda$ is an eigenvalue of $T$ if and only if $\det M_{k,\lambda} = 0$. In the same way, $\lambda < 0$ satisfying (4.19) is an eigenvalue of $T$ if and only if $\det M_{-k,-\lambda} = 0$.

In the particular case, when all lengths are 1, $k_3 < 0$ and $k_j = 1$ for all $j \neq 3$, $\det M_{k,\lambda}$ reduces to

$$\det M_{k,\lambda} = \left( 2 \cot \sqrt{\lambda} - \sqrt{|k_3|} \coth \left( \frac{\sqrt{\lambda}}{\sqrt{|k_3|}} \right) \right)^2 - \frac{|k_3|}{\sinh^2 \left( \frac{-\sqrt{\lambda}}{\sqrt{|k_3|}} \right)} \\ = f_+(\sqrt{\lambda}) f_-(\sqrt{\lambda}),$$

where

$$f_{\pm}(x) = 2 \cot x - \sqrt{|k_3|} \coth \left( \frac{x}{\sqrt{|k_3|}} \right) \pm \frac{\sqrt{|k_3|}}{\sinh \left( \frac{x}{\sqrt{|k_3|}} \right)}.$$

Since for $x$ large, $f_+(x)$ and $f_+(x)$ are close to $2 \cot x - \sqrt{|k_3|}$, we can conclude that $\det M_{k,\lambda}$ has two sequences of roots $\lambda_{+,n}$ and $\lambda_{-,n}$, $n \in \mathbb{N}$ that behaves like $(x_+ + n\pi)^2$ with $x_+ \in (0, \frac{\pi}{2})$ such that $\cot x_+ = \frac{\sqrt{|k_3|}}{2}$. On the other hand, using a trigonometric formula, one has

$$\det M_{-k,-\lambda} = \left( 2 \coth \sqrt{-\lambda} - \sqrt{|k_3|} \coth \left( \frac{\sqrt{-\lambda}}{\sqrt{|k_3|}} \right) \right)^2 - \frac{|k_3|}{\sin^2 \left( \frac{\sqrt{-\lambda}}{\sqrt{|k_3|}} \right)} \\ = 4 \coth^2 \sqrt{-\lambda} - 4 \sqrt{|k_3|} \coth \sqrt{-\lambda} \cot \left( \frac{\sqrt{-\lambda}}{\sqrt{|k_3|}} \right) - |k_3|$$
<figure><img src="image_12.png" /><figcaption>Figure 9: The functions $f_+$ (in green) and $f_-$ (in blue) for positive eigenvalues for the 5-edges graph with Dirichlet bc</figcaption></figure>

and therefore large negative eigenvalues are close to the roots of $4 - |k_3| - 4\sqrt{|k_3|} \cot(\frac{\sqrt{-\lambda}}{\sqrt{|k_3|}})$ that have the form $k_3(x_- + n\pi)^2$, with $n \in \mathbb{N}$ and $x_- \in (0, \pi)$ such that $\cot x_- = \frac{4-|k_3|}{4\sqrt{|k_3|}}$.

An illustration of the graphs of $f_+$ and $f_-$, with $k_3 = -100$ are given in Figure 9 and confirm the asymptotic behavior mentioned above. Similarly in Figure 10, the graph of $4\coth^2 x - 4\sqrt{|k_3|}\coth x \cot(\frac{x}{\sqrt{|k_3|}}) - |k_3|$, with $k_3 = -100$ illustrates the asymptotic behavior of negative eigenvalues.

For the illustration of Theorem 4.7 and Corollary 4.8, again many configurations may appear, hence we restrict ourselves to the previous particular case. In such a case, by the proof of Theorem 4.7, one easily deduces that $n^2\pi^2$ is an eigenvalue of $T$, while for any $n \in \mathbb{N}^*$, $\lambda = k_3n^2\pi^2$ is not an eigenvalue of $T$, because (assuming that $I(e_3) = v$)

$$ \tilde{M}_{-k, -\lambda} = \begin{pmatrix} -2 \coth \sqrt{-\lambda} & 0 & -\sqrt{|k_3|} \\ 0 & -2 \coth \sqrt{-\lambda} & (-1)^n \sqrt{|k_3|} \\ -1 & (-1)^n & 0 \end{pmatrix}, $$

whose determinant is equal to $4\sqrt{|k_3|} \coth \sqrt{-\lambda}$.

Let us finally mention that our considerations allow to recover our Weyl's formulas on the previous particular examples treated after subsubsections 6.1.2, we have skipped the details for shortness.

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