Higher Eigenvalues and Topological Perturbations

Abstract

Some fundamental estimates for higher eigenvalues of standard Laplacians have already been derived in Sect. 4.6. The goal of this chapter is twofold: on the one hand considering the standard Laplacian we derive explicit fundamental estimates for higher eigenvalues and describe the behaviour of such eigenvalues under topological perturbations. Here techniques developed in the previous chapter are used. On the other hand, considering Schrödinger operators with most general vertex conditions we analyse the behaviour of the spectrum under topological perturbations and show that intuition gained during our studies of standard Laplacians cannot always be applied: the eigenvalues may depend on topological perturbations in a completely opposite way.

Some fundamental estimates for higher eigenvalues of standard Laplacians have already been derived in Sect. 4.6. The goal of this chapter is twofold: on the one hand considering the standard Laplacian we derive explicit fundamental estimates for higher eigenvalues and describe the behaviour of such eigenvalues under topological perturbations. Here techniques developed in the previous chapter are used. On the other hand, considering Schrödinger operators with most general vertex conditions we analyse the behaviour of the spectrum under topological perturbations and show that intuition gained during our studies of standard Laplacians cannot always be applied: the eigenvalues may depend on topological perturbations in a completely opposite way.

13.1 Fundamental Estimates for Higher Eigenvalues

13.1.1 Lower Estimates

Our aim here will be to derive explicit estimates for all higher eigenvalues of the standard Laplacian. Such estimates were first obtained by L. Friedlander [225] and we use his main ideas here.

Let us try to guess, which metric graph minimises the eigenvalue $\lambda _j$? It is clear that $\lambda _j$ is always greater or equal to $\lambda _{j-1}$, which in turn does not exceed $\lambda _{j-2}$ and so on. Pressing down the eigenvalue $\lambda _j$ we make it degenerate so that $\lambda _j = \lambda _{j-1}.$ Pressing it further we shall make it triple degenerate until we reach $\lambda _j = \lambda _{j-1} = \dots = \lambda _2$, which is strictly larger than $\lambda _1 = 0$ as the ground state is non-degenarate (see Theorem 4.12). Hence our guess is that the j-th eigenvalue is minimised by the graph where $\lambda _j$ has multiplicity $j-1$ and is the first non-trivial eigenvalue.

Consider equilateral star graph on $j$ edges each of length $\mathcal L/j.$ The second eigenvalue is degenerate with multiplicity $j-1$ and coincides with the ground state of the Dirichlet-Neumann interval of length $\mathcal L/j$

$$\displaystyle \begin{aligned} \lambda_2 = \dots = \lambda_j = \Big( \frac{j \pi}{2\mathcal L} \Big)^2. \end{aligned} $$

(13.1)

We therefore suspect that $\Big (\frac {j \pi }{2\mathcal L} \Big )^2$ provides the best lower estimate for $\lambda _j$ on a graph of total length $\mathcal L.$ Our guess can of course not be considered as a rigorous proof, but surprisingly it provides the correct answer to the current problem.

Theorem 13.1

Let $\Gamma$ be a connected metric graph of total length $\mathcal L.$ Then the $j$ -th eigenvalue of the standard Laplacian can be estimated as

$$\displaystyle \begin{aligned} {} \lambda_j (L^{\mathrm{st}} (\Gamma)) \geq \Big( \frac{j \pi}{2 \mathcal L} \Big)^2. \end{aligned} $$

(13.2)

Equality occurs if and only if the graph is an equilateral star with j edges (a segment in the case $j=2$ ).

Proof

To prove the estimate it is enough to consider just trees, since any graph $\Gamma$ can be turned into a tree by chopping few of its vertices. The edges are preserved during this operation, but the domain of the quadratic form is enlarged, since the functions may attain different values at different pieces of chopped vertices. The quadratic forms on the original graph and on the tree are given by exactly the same expression, hence the eigenvalues of the tree do not exceed the corresponding eigenvalues for the original graph.

We first prove that given$j$any tree can be divided by at most$j-1$points into subgraphs of length at most$\mathcal L/j.$ Every point $x_0$ inside an edge naturally divides the tree into two parts each containing points that can be joined by paths not passing through $x_0.$ In a similar way if $x_0$ is a vertex, then the tree is divided into $d$ components, where $d$ is the degree of the vertex.

Let $T$ be a tree. Consider all pendant edges in $T$ - the edges connected to vertices of degree one and pick up any star subgraph of degree $d$ containing at least $d-1$ pendant edges. We have three possibilities:

• 1. If the length of at least one of the pendant edges in the star subgraph is greater than $\mathcal L/j$, then put the point $x_0$ at the distance $\mathcal L/j$ from the degree one vertex. We have divided the tree into a segment of length $\mathcal L/j$ and a tree of length $\mathcal L (j-1)/j .$

• 2. If the sum of the lengths of the $d-1$ pendant edges in the star graph is greater than or equal to $\mathcal L/j$ but the length of each pendant edge is less than or equal to $\mathcal L/j$, then put $x_0$ at the central vertex in the star subgraph. As a result we get a division of $T$ into $d-1$ segments of lengths less than or equal to $\mathcal L/j$ and a tree of length less than $\mathcal L (j-1)/j .$

• 3. If the sum of the lengths of the $d-1$ pendant edges in the star graph is less than $\mathcal L/j$, then substitute the $d-1$ pendant edges in the star subgraph by a single edge of the same total length and repeat the procedure. Do not forget to restore the pendant edges in the star graphs after the division is accomplished.

We repeat the described procedure cutting at most $j-1$ times until the tree is divided into at least $j$ components of lengths at most $\mathcal L/j.$ Let us denote the points dividing $T$ by $x_1, x_2, \dots , x_m$, $m \leq j-1.$ We restore all star graphs that were substituted by edges. This does not affect the sizes of the components which we denote by $T_i.$

Consider now the first $j$ eigenfunctions $\psi _i, i = 1,2, \dots , j$, which are linearly independent. In the linear span of $\psi _i, i=1, \dots , j$ there exists a non-zero function $\phi (x)$ vanishing at all points $x_i, \; i = 1,2, \dots , m \leq j-1.$ On every component $T_i$ of the tree, where $\phi$ is not identically zero, it satisfies a Dirichlet condition at at least one edge, hence the corresponding Rayleigh quotient is greater than or equal to the first eigenvalue of the Dirichlet-Neumann interval of the same length as the component

$$\displaystyle \begin{aligned} \frac{\int_{T_i} \vert \phi'(x) \vert^2 dx }{\int_{T_i} \vert \phi (x) \vert^2 dx} \geq \Big( \frac{j \pi}{2 \mathcal L } \Big)^2, \end{aligned} $$

(13.3)

where we have taken into account (12.15) and the fact that the length of $T_i$ does not exceed $\mathcal L/j.$ Summing up the contributions from all components where $\phi$ is not identically zero we get the same estimate for the Rayleigh quotient on the whole tree:

$$\displaystyle \begin{aligned} \frac{\int_{T} \vert \phi'(x) \vert^2 dx }{\int_{T} \vert \phi (x) \vert^2 dx} \geq \Big( \frac{j \pi}{2 \mathcal L } \Big)^2. \end{aligned} $$

(13.4)

The function $\phi$ belongs to the linear span of the first $j$ eigenfunctions and therefore satisfies

$$\displaystyle \begin{aligned} \frac{\int_{T} \vert \phi'(x) \vert^2 dx }{\int_{T} \vert \phi (x) \vert^2 dx} \leq \lambda_j (L^{\mathrm{st}} (T)). \end{aligned} $$

(13.5)

Comparing the last two inequalities we get

$$\displaystyle \begin{aligned} \lambda_j (L^{\mathrm{st}} (\Gamma)) \geq \lambda_j (L^{\mathrm{st}} (T)) \geq \Big( \frac{j \pi}{2 \mathcal L } \Big)^2.\end{aligned}$$

To show uniqueness one may use mathematical induction with respect to $j$. Details can be found in [225]. □

We have already shown that the estimate is sharp and turns into equality for equilateral star on $j$ edges. An interesting observation is that to cut this star into pieces of length at most $\mathcal L/j$ one needs just one cutting point - the central vertex.

Equality in the estimate (13.2) is realised precisely if the graph is an equilateral star graph with j edges. Such a graph has $j$ Neumann degree one vertices and its first Betti number is zero $\beta _1 = 0$ ($\chi = 1$). Hence it is natural to expect that, given a graph, estimate (13.2) is not sharp for sufficiently large eigenvalues. For example, let us assume that the graph has $N$ pendant vertices and $\beta _1 > 0$ cycles, then for large enough $j$ the estimate (13.2) can be improved as follows:

Theorem 13.2 (Following Theorem 4.7 from [87])

Let $\Gamma$ be a metric graph with $|N|\geq 0$ vertices of degree one with the Neumann condition. Assume that $\Gamma$ is not a cycle. Then for all $j \geq 2$

$$\displaystyle \begin{aligned} {} \lambda_{j} (\Gamma) \geq \begin{cases} \left(j - \frac{|N|+\beta_1}2 \right)^2 \dfrac{\pi^2}{\mathcal L^2} & \mathit{\mbox{if }} j \geq |N| + \beta_1\\[10pt] \dfrac{j^2 \pi^2}{4 \mathcal L^2} & \mathit{\mbox{otherwise}}, \end{cases} \end{aligned} $$

(13.6)

where $\beta _1$ is the first Betti number of the graph.

Proof

If $\Gamma$ is not a tree, we find an edge whose removal would not disconnect the graph. Let $V^0$ be a vertex to which this edge is incident; since $\Gamma$ is not a cycle, without loss of generality we can assume its degree is 3 or larger (otherwise this vertex can be absorbed into the edge). We disconnect the edge from this vertex, reducing $\beta _1$ by one and creating an extra vertex of degree one where we impose the Neumann condition, see Fig. 13.1. We keep standard conditions at $V^0$. Then the new graph is not a cycle, as a new vertex of degree 1 was created. We may therefore repeat the process inductively until we obtain a tree $\mathbf T$ with $|N'| = |N|+\beta _1$ Neumann vertices.

Fig. 13.1

Disconnecting the edge $E_0$ from the vertex $V^0$ in the proof of Theorem 13.2. This operation reduces $\beta _1$ by 1 at the expense of increasing the number of Neumann vertices by 1

Since the eigenvalues are reduced at every step, $\lambda _k(\Gamma ) \geq \lambda _k(\mathbf T)$. It is therefore enough to verify the inequality for trees.

Given a tree $\mathbf T$ we can find an arbitrarily small perturbation under which the k-th eigenvalue is simple and its eigenfunction is nonzero on vertices [82]. In these circumstances the k-th eigenfunction has exactly k nodal domains [50, 433, 465] (see also [80, Thm. 6.4] for a short proof). Each nodal domain is a subtree $\mathbf T_j$, and with vertex conditions inherited from $\Gamma$ (plus Dirichlet conditions on the nodal domain boundaries), $\lambda _k(\Gamma )$ is the first eigenvalue of the subtree.

There are at most $|N|$ subtrees with some Neumann conditions on their pendant vertices. Since these are nodal subtrees ($k>1$), there are also some pendant vertices with Dirichlet conditions and we can use estimate (12.15) in the form $\mathcal L_j k \geq \pi /2$. The same conclusion is true if $k=1$ and $\Gamma$ has at least one Dirichlet vertex.

If $k \geq |N|$, we also have at least $k-|N|$ subtrees with only Dirichlet conditions at the pendant vertices. For such trees the ground state energy satisfies the estimate: $\mathcal L_j \sqrt {\lambda } \geq \pi .$ To see this it is enough to realise that the corresponding eigenfunction has a maximum and the point where it is attained divides the tree $\mathbf T$ into at least two pieces. Each of the pieces is a tree with at least one Dirichlet vertex and estimate (12.15) can be used.

Summing up we have

$$\displaystyle \begin{aligned} \mathcal L \sqrt{\lambda_k(\Gamma)} = \sum_{j=1}^k \mathcal L_j \sqrt{\lambda_1(\mathbf T_j)} \geq |N| \frac{\pi}{2} + (k-|N|) \pi = \left(k - \tfrac{|N|}{2}\right) \pi.\ \end{aligned}$$

When $k < |N|$, we use estimate (12.15) for each of the k nodal subtrees, obtaining Friedlander’s bound. □

13.1.2 Upper Bounds

Let us look at the upper estimates for the eigenvalues.

Theorem 13.3 (Following Theorem 4.9 from [87] Inspired by Ariturk [32])

Let $\Gamma$ be a connected metric graph with Dirichlet or Neumann conditions at the vertices of degree one and standard condition elsewhere. If $\Gamma$ is not a cycle, then for all $k\in \mathbb N$

$$\displaystyle \begin{aligned} {} \lambda_k (\Gamma) \leq \left(k - 2 + \beta_1 + |D| + \tfrac{|N|+\beta_1}2\right)^2 \frac{\pi^2}{\mathcal L^2}, \end{aligned} $$

(13.7)

where the set of Dirichlet vertices is denoted by D and the set of Neumann vertices of degree one is denoted by N.

Proof

If $\Gamma$ is not a tree (i.e. if $\beta _1 > 0$) and not a cycle, we repeat the process described at the beginning of the proof of Theorem 13.2, disconnecting $\beta _1$ edges at vertices and creating a tree $\mathbf T$ with $\beta _1$ additional Neumann vertices of degree one. At every step, the eigenvalue goes down, but not further than the next eigenvalue. The reason is very simple: from any two eigenfunctions of the Laplacian on the graph with the chopped vertex, one may always glue together a trial function for the original graph with the Rayleigh quotient not exceeding the maximum of the Rayleigh quotient for each of the functions in the pair, therefore we have $\lambda _k(\Gamma ) \leq \lambda _{k+\beta _1}(\mathbf T)$ and the bound for general graphs follows from the bound for trees, $\beta _1=0$ (see also [80, Thm 3.1.10]).

It is enough to prove the theorem for trees $\mathbf T$ and we shall use induction on the number of edges. The inequality turns into equality for a single edge with either Dirichlet or Neumann or mixed conditions.

Choose an arbitrary vertex $V^0$ of degree $d_0$ and divide the corresponding equivalence class into two, denoted by $V'$ and $V''$ and having $d_0-1$ and $1$ elements respectively. We introduce standard conditions at $V'$ and Dirichlet conditions at $V''$. This process corresponds to chopping off from $\mathbf T$ the branch growing from $V''$ and introducing Dirichlet conditions at the root $V''$. We denote the resulting trees by $\mathbf T'$ and $\mathbf T''$ respectively. It is clear that

$$\displaystyle \begin{aligned} \lambda_k (\mathbf T) \leq \lambda_{k+1} (\mathbf T' \cup \mathbf T''),\end{aligned}$$

where Dirichlet condition is assumed at the root of $\mathbf T''$. Here $\mathbf T' \cup \mathbf T''$ denotes the union of two metric trees. This inequality follows from the fact that from any $k+1$ eigenfunctions on $\mathbf T' \cup \mathbf T''$ one may always build k continuous functions on the original tree $\mathbf T$, and the Rayleigh quotient for these trial functions does not exceed $\lambda _{k+1} (\mathbf T' \cup \mathbf T'')$.

We denote by $\mathcal L'$ and $\mathcal L''$ the total lengths of the subtrees $\mathbf T'$ and $\mathbf T''$ respectively. The numbers of Dirichlet and Neumann vertices satisfy

$$\displaystyle \begin{aligned} |D| = |D'| + |D''| -1, \quad |N| = |N'| + |N''|,\end{aligned}$$

as a new Dirichlet condition is introduced at $V''$.

The $k+1$-st eigenvalue of $\mathbf T' \cup \mathbf T''$ coincides with the eigenvalue either on $\mathbf T'$ or on $\mathbf T''$. Assume without loss of generality that $\lambda _{k+1} (\mathbf T' \cup \mathbf T'') = \lambda _j (\mathbf T'')$ for a certain j, then we get

$$\displaystyle \begin{aligned} \mathcal L \sqrt{\lambda_k(\mathbf T)} & \leq \mathcal L \sqrt{\lambda_{k+1} (\mathbf T' \cup \mathbf T'')} \\ &\leq \mathcal L' \sqrt{\lambda_{k-j+1} (\mathbf T')} + \mathcal L'' \sqrt{\lambda_{j}(\mathbf T'')} \\ &\leq \pi \left(k-j+1 - 2 + |D'| + \tfrac{|N'|}2\right) + \pi \left(j - 2 + |D''| + \tfrac{|N''|}2\right) \\ &= \pi \left(k - 2 + |D| + \tfrac{|N|}2\right). \end{aligned} $$

This completes the proof. □

13.1.3 Graphs Realising Extremal Eigenvalues

For large indices the lower (13.6) and upper bounds (13.7) give the following two-sided estimate:

$$\displaystyle \begin{aligned} {} \frac{\pi^2}{\mathcal L^2} \left( j - \frac{|N|}{2} - \frac{\beta_1}{2} \right) \leq \lambda_j (\Gamma) \leq \frac{\pi^2}{\mathcal L^2} \left(j-2 + |D| + \frac{|N|}{2} + \frac{3}{2} \beta_1 \right). \end{aligned} $$

(13.8)

We already know that the estimates are sharp since there exist graphs realising both the lower and upper bounds [353]. What is more remarkable is that there exist graphs where both estimates are realised simultaneously. Since the upper bound is never equal to the lower one, we need highly degenerate eigenvalues to realise the estimates. This is possible if one considers looptrees—tree graphs with loops attached to some of the degree one vertices. One assumes that Dirichlet or Neumann conditions are introduced at the other degree one vertices. Carefully adjusting the lengths of the Dirichlet and Neumann pendant edges and of the loops allows one to create graphs with degenerate eigenvalues $\lambda _{j_{\mathrm {min}}} = \lambda _{j_{\mathrm {min}} +1} = \dots = \lambda _{j_{\mathrm {max}}}$, where

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \lambda_{j_{\mathrm{max}}} & = & \displaystyle \frac{\pi^2}{\mathcal L^2} \left( j_{\mathrm{min}} - \frac{|N|}{2} - \frac{\beta_1}{2} \right) \\[3mm] \displaystyle \lambda_{j_{\mathrm{min}}} & = & \displaystyle \frac{\pi^2}{\mathcal L^2} \left(j_{\mathrm{max}} -2 + |D| + \frac{|N|}{2} + \frac{3}{2} \beta_1 \right). \end{array} \end{aligned}$$

The construction of such graphs is described in [469]. One of the simplest looptrees is presented in Fig. 13.2.

Fig. 13.2

Looptree with 4 Dirichlet and 2 Neumann vertices and 2 loops

Problem 56

Determine the lengths of the edges and loops, so that the eigenvalue estimates (13.8) are sharp for the graph depicted in Fig. 13.2.

13.2 Gluing and Cutting Vertices with Standard Conditions

The results presented here are completely analogous to those proven in Sect. 12.5. Therefore we skip the proofs leaving them for interested readers to work on.

In Theorem 12.9 we assumed that the graph is connected, since the estimate was trivial for non-connected graphs: for graphs with at least two connected components we have formally $\lambda _1 (\Gamma ) = \lambda _2 (\Gamma ) = 0$. It is natural to drop the connectivity requirement if one is interested in higher eigenvalues:

Theorem 13.4

Let $\Gamma$ be a metric graph and let $\Gamma '$ be another metric graph obtained from $\Gamma$ by joining together two of its vertices, say $V^1$ and $V^2.$ Then the following inequality for the eigenvalues of the standard Laplacian holds:

$$\displaystyle \begin{aligned} {} \lambda_n (\Gamma) \leq \lambda_n (\Gamma'). \end{aligned} $$

(13.9)

Problem 57

Use the minmax principle (Proposition 4.19) for general $n$ to prove Theorem 13.4.

The above theorem can be reformulated speaking about cutting vertices instead of gluing. In particular Theorem 12.16 can be generalised as follows:

Theorem 13.5

Let $\Gamma$ be a connected metric graph and let $\Gamma ^*$ be another graph obtained from $\Gamma$ by cutting one of the edges at an internal point $x^*$ producing two new vertices $V^{1*}$ and $V^{2*}.$ Then the eigenvalues of the standard Laplacian satisfy the following inequality

$$\displaystyle \begin{aligned} \lambda_n (L^{\mathrm{st}} (\Gamma)) \geq \lambda_n (L^{\mathrm{st}} (\Gamma^*)). \end{aligned} $$

(13.10)

Problem 58

Prove Theorem 13.5. How to modify the second statement in Theorem 12.16 in order to cover higher eigenvalues?

One may conclude that our intuition acquired investigating the spectral gap can be applied to higher eigenvalues of the standard Laplacian. It is straightforward to include Schrödinger operators with standard conditions, but one has to be careful when vertex conditions are not standard. To understand what should be modified we look first at scaling-invariant vertex conditions, leaving the most general conditions for the last section.

13.3 Gluing Vertices with Scaling-Invariant Conditions

13.3.1 Scaling-Invariant Conditions Revisited

The ideas previously developed can be applied to vertices with arbitrary vertex conditions. Let us start our analysis by discussing gluing of scaling-invariant conditions. Let us recall that scaling-invariant vertex conditions at a vertex $V$ of degree $d$ correspond to a parameter $S$ in (3.21), which is not only unitary but also Hermitian. Every such $d \times d$ matrix $S$ has just eigenvalues $-1$ and $1$ so that the corresponding eigensubspaces span the space $\mathbb C^{d}$:

$$\displaystyle \begin{aligned} P_{-1} + P_1 = \mathbb I_{\mathbb C^d}.\end{aligned}$$

The vertex condition can be written as two projectors (3.33). One may say that scaling-invariant conditions are a combination of Dirichlet and Neumann conditions on two mutually orthogonal subspaces. Denoting the eigensubspace of $S$ associated with $1$ by $\mathcal D$, the same vertex conditions (3.33) can be written as

$$\displaystyle \begin{aligned} \left\{ \begin{array}{ccc} P_{\mathcal D}^\perp \vec{u} & = & 0, \\[3mm] P_{\mathcal D} \partial \vec{u} & = & 0. \end{array} \right. \end{aligned}$$

The Dirichlet data at the vertex span the subspace $\mathcal D$, while the Neumann data span the orthogonal complement $\mathcal D^\perp$:

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{l} \vec{u} \in \mathcal D, \\[3mm] \partial \vec{u} \in \mathcal D^\perp. \end{array} \right. \end{aligned} $$

(13.11)

Note that the vertex conditions are properly connecting if and only if both the Neumann subspace $\mathcal D$ and the Dirichlet subspace $\mathcal D^\perp$ do not contain basis vectors from the standard basis in $\mathbb C^d$.

Assume for example that $\mathcal D$ contains the vector $\vec {e}_j$ from the standard basis in $\mathbb C^d.$ Then every vector from the orthogonal complement $\mathcal D^\perp$ has zero j-th component and therefore condition (13.11) implies that

$$\displaystyle \begin{aligned} \partial u (x_j ) = 0.\end{aligned}$$

At the same time no restriction on $u(x_j)$ is imposed by the requirement $\vec {u} \in \mathcal D.$ In other words, the limit values $u(x_j)$ and $\partial u(x_j)$ are not related to other limit values. The vertex can be split into two vertices implying that such vertex conditions are not properly connecting.

The case where $\mathcal D^\perp$ contains a vector from the standard basis is completely similar - the roles of function values and normal derivatives are interchanged.

Two examples of properly connecting scaling-invariant conditions are:

• one-dimensional conditions corresponding to $\mathcal D$ spanned by a certain fixed vector $\vec a \in \mathbb C^{d}$ with all non-zero coordinates;

• hyperplanar conditions given by $\mathcal D = \{ \vec {u} \in \mathbb C^{d}: \vec {u} \perp \vec {b} \}$, where $\vec b \in \mathbb C^{d}$ is a certain fixed vector with all non-zero coordinates.

If at least one of the coordinates in $\vec {a}$ or $\vec {b}$ is zero, then the orthogonal subspace contains one of the vectors from the standard basis in $\mathbb C^d$ and therefore the corresponding vertex condition is not properly connecting.

13.3.2 Gluing Vertices

Consider two vertices $V^1$ and $V^2$ with scaling-invariant conditions. Let us denote by $\mathcal D_1 \subset \mathbb C^{d_1}$ and $\mathcal D_2 \subset \mathbb C^{d_2}$ the corresponding Neumann subspaces and assume that they are canonically embedded into $\mathbb C^d = \mathbb C^{d_1 +d_2}$ by assigning zero to all coordinates not related to the corresponding subvertex.

We wish to connect these vertices into one common vertex $V$ by assigning scaling-invariant conditions. In other words, we need to select a new Neumann subspace $\mathcal D' \in \mathbb C^{d}= \mathbb C^{d_1 +d_2}.$ The subsbace $\mathcal D_1 + \mathcal D_2$ cannot be used directly without any modification - the corresponding vertex conditions are obviously not properly connecting by construction. In principle, the subspace $\mathcal D'$ can be chosen arbitrarily, but it is natural to look for a procedure giving $\mathcal D'$ as a certain modification of $\mathcal D_1 + \mathcal D_2.$

There are two obvious modifications

• 1. Gluing by restriction

  Select a subspace $\mathcal E \subset \mathcal D_1 + \mathcal D_2$ and define the new Neumann subspace by requiring that all its elements are orthogonal to $\mathcal E$:

  $$\displaystyle \begin{aligned} {} \mathcal D' = \Big\{ \vec{u} \in \mathcal D_1 + \mathcal D_2 : \vec{u} \perp \mathcal E \Big\}. \end{aligned} $$

  (13.12)

• 2. Gluing by extension

  Select a subspace $\mathcal F \subset \big (\mathcal D_1 + \mathcal D_2 \big )^\perp$ and define the new Neumann subspace by adding $\mathcal F$ to $\mathcal D_1 + \mathcal D_2$:

  $$\displaystyle \begin{aligned} {} \mathcal D' = \mathcal D_1 + \mathcal D_2+ \mathcal F . \end{aligned} $$

  (13.13)

One has to be careful selecting the subspaces $\mathcal E$ and $\mathcal F$ and check that the new vertex conditions are properly connecting.

It is straightforward to generalise the developed methods for the case where more than two vertices are joined together.

Problem 59

How to describe gluing vertices with scaling-invariant conditions for the case of several vertices.

Problem 60

Is it possible to glue vertices by combining extension and restriction procedures? Provide explicit examples to support your answer.

Let us discuss how these modifications work when applied to one-dimensional and hyperplanar conditions.

13.3.2.1 Gluing Vertices with One-Dimensional Vertex Conditions

We plan to preserve the character of vertex conditions, so that new vertex conditions are also of one-dimensional type.

Assume that $\mathcal D_j$ are spanned by $\vec {a}_j.$ Then we have $\mathcal D_1 + \mathcal D_2$ is spanned by $\vec {a}_1$ and $\vec {a}_2$. Obviously $\dim (\mathcal D_1 + \mathcal D_2) = 2$ and the first gluing method should be used. We get vertex conditions of one-dimensional type if we select a vector $\vec {a} \in \mathcal D_1 + \mathcal D_2 .$ Every such vector is a linear combination of $\vec {a}_j$ but should be different from the basis vectors

$$\displaystyle \begin{aligned} \vec{a} = h_1 \vec{a}_1 + h_2 \vec{a}_2, \quad h_1 h_2 \neq 0.\end{aligned}$$

We get vertex conditions of one-dimensional type with

$$\displaystyle \begin{aligned} \mathcal D' \; \mbox{spanned by} \; \vec{a} = h_1 \vec{a}_1 + h_2 \vec{a}_2.\end{aligned}$$

Standard conditions is a special case of one-dimensional scaling-invariant conditions with $\vec {a} = (1,1, \dots , 1).$ Therefore when gluing standard vertices it is natural to choose $h_1 = h_2 = 1$ so that

13.3.2.2 Gluing Vertices with Hyperplanar Vertex Conditions

Gluing hyperplanar vertices we want to preserve their hyperplanarity. Assume $\mathcal D_j = \{ \vec {u} \in \mathbb C^{d_j}: \vec {u} \perp \vec {b}_j \},$$\dim \mathcal D_j = d_j-1.$ Then $\dim \left ( \mathcal D_1 + \mathcal D_2 \right ) = d-2:$

$$\displaystyle \begin{aligned} \mathcal D_1 + \mathcal D_2 = \Big\{ \vec{u} \in \mathbb C^{d}: \langle \vec{u}, \vec{b}_1 \rangle = 0 = \langle \vec{u}, \vec{b}_2 \rangle \Big\}. \end{aligned} $$

(13.14)

We need to use the second gluing method to increase the dimension of the supspace. One has to select a single vector $\vec {b} \in \mathcal L \{ \vec {b}_1, \vec {b}_2 \}$ without zero coordinates. Every such vector is of the form

$$\displaystyle \begin{aligned} \vec{b} = h_1 \vec{b}_1 + h_2 \vec{b}_2, \quad h_1 h_2 \neq 0.\end{aligned}$$

The corresponding conditions are given by

$$\displaystyle \begin{aligned} \mathcal D' = \left\{ \vec{u} \in \mathbb C^{d}: \vec{u} \perp \vec{b}= h_1 \vec{b}_1 + h_2 \vec{b}_2 \right\}. \end{aligned} $$

(13.15)

An important class of hyperplanar conditions is given by vectors $\vec {b}_j = (1,1, \dots , 1) \in \mathbb C^{d_j}.$ The situation is similar to standard conditions with the only one natural choice for $\vec {b} = (1,1, \dots , 1) \in \mathbb C^d$ (see [450]).

13.3.3 Spectral Gap and Gluing Vertices with Scaling-Invariant Conditions

Let us generalise Theorem 13.4 by allowing scaling-invariant conditions and non-zero potentials (considering Schrödinger operators instead of Laplacians):

Theorem 13.6

Let$\Gamma$be a metric graph with selected vertices$V^1$and$V^2$, and let$\Gamma '$be the metric graph obtained from$\Gamma$by joining the vertices$V^1$and$V^2$into one vertex$V.$Let the vertex conditions (determined by the matrices$\mathbf S$, $\mathbf S'$) on$\Gamma$and$\Gamma '$be scaling-invariant and obtained from each other either by restriction (13.16) or by extension (13.17). Assume that the potential$q$is absolutely integrable$q \in L_1 (\Gamma )$, then the eigenvalues of the Schrödinger operators on$\Gamma$and$\Gamma '$satisfy the following inequalities:

• 1. If the gluing is given by restriction as described in (13.12), then

  $$\displaystyle \begin{aligned} {} \lambda_j (L_q^{\mathbf S} (\Gamma)) \leq \lambda_j (L_q^{\mathbf S'} (\Gamma')). \end{aligned} $$

  (13.16)

• 2. If the gluing is given by extension as described in (13.13), then

  $$\displaystyle \begin{aligned} {} \lambda_j (L_q^{\mathbf S} (\Gamma)) \geq \lambda_j (L_q^{\mathbf S'} (\Gamma')). \end{aligned} $$

  (13.17)

Proof

Presence of the potential does not affect the proof so much, since the quadratic forms for the Schrödinger operators on $\Gamma$ and $\Gamma '$ are given by the same expression (see (11.9)):

$$\displaystyle \begin{aligned} \int_\Gamma \vert u' (x) \vert^2 dx + \int_\Gamma q(x) \vert u(x) \vert^2 dx ,\end{aligned}$$

provided the functions on the graphs $\Gamma$ and $\Gamma '$ are identified. For both forms the function $u$ should belong to the Sobolev space $W_2^1 (E_n)$ on every edge. The difference between the form domains lies in the conditions the function $u$ satisfies at the vertices affected by the gluing. Let $\vec {u}$ denote the vector of function values at the vertices $V^1$ and $V^2$ or at the vertex $V$. Then the vertex conditions for $\Gamma$ and $\Gamma '$ are

$$\displaystyle \begin{aligned} \vec{u} \in \mathcal D_1 + \mathcal D_2 , \quad \vec{u} \in \mathcal D',\end{aligned}$$

respectively.

If the vertices are glued by restriction then obviously

$$\displaystyle \begin{aligned} \vec{u} \in \mathcal D' \Rightarrow \vec{u} \in \mathcal D_1 + \mathcal D_2 ,\end{aligned}$$

i.e. the domain corresponding to $\Gamma '$ is smaller, hence all eigenvalues are larger due to the minmax principle (Proposition 4.19).

In the case of gluing by extension, the quadratic form associated with $\Gamma '$ has larger domain, hence the eigenvalues are smaller leading to (13.17). □

In particular, it follows that gluing vertices with standard conditions the eigenvalues may only grow, while gluing vertices with hyperplanar conditions the eigenvalues may only decrease [450].

13.4 Gluing Vertices with General Vertex Conditions

Let us briefly discuss the most general case of gluing vertices. Consider a quantum graph with selected vertices $V^j, \; j= 1,2,$ of degree $d_j.$ Assume following Sect. 3.8.2 that the vertex conditions are determined by selecting subspaces $\mathcal D_j \subset \mathbb C^{d_j}$ and Hermitian matrices $A_j$ acting in $\mathcal D_j.$ Assume that gluing the vertices into one vertex $V$ the vertex conditions are given by a subspace $\mathcal D' \subset \mathbb C^{d} = \mathbb C^{d_1+d_2}$ and a Hermitian matrix $A'.$ We are interested in selecting conditions that guarantee that the eigenvalues of the glued graph behave monotonically upon gluing. The answer can be given in terms of the quadratic forms $\langle u, A u \rangle _{\mathcal D}$ associated with the vertex conditions.

We are going to say that a quadratic form $a (u,u)$ with the domain $D_a$subordinates a quadratic form $b (u,u)$ with the domain $D_b$ if and only if:

• $D_a \subset D_b,$

• $a(u,u) \geq b(u,u)$ for any $u \in D_a.$

Under the same conditions we are going to say that $b$ is subordinated by $a.$ Both conditions above are important, since the two quadratic forms can be compared only if one of the forms is defined on the intersection of the form domains.

Theorem 13.7

Let $L_q^{\mathbf S(A)} (\Gamma )$ be a Schrödinger operator on a metric graph $\Gamma$ with vertex conditions at the vertices $V^1$ and $V^2$ determined by the subspaces $\mathcal D_1$ and $\mathcal D_2$ and Hermitian matrices $A_1$ and $A_2$ respectively. Let $\Gamma '$ be the metric graph obtained from $\Gamma$ by joining together the two vertices introducing vertex conditions determined by the subspace $\mathcal D'$ and Hermitian matrix $A'.$ Then the eigenvalues of the Schrödinger operators on $\Gamma$ and $\Gamma '$ satisfy the following inequalities:

• 1. If the quadratic form $\langle \vec {u}, A' \vec {u} \rangle _{\mathcal D'}$ for the joined vertex is subordinated to the sum of the quadratic forms $\langle \vec {u}, A_1 \vec {u} \rangle _{\mathcal D_1} + \langle \vec {u}, A_2 \vec {u} \rangle _{\mathcal D_2}$ corresponding to the vertices to be joined, then

  $$\displaystyle \begin{aligned} {} \lambda_j ( L_q^{\mathbf S(A)} (\Gamma)) \leq \lambda_j ( L_q^{\mathbf S(A')} (\Gamma')). \end{aligned} $$

  (13.18)

• 2. If the sum of quadratic forms $\langle \vec {u}, A_1 \vec {u} \rangle _{\mathcal D_1} + \langle \vec {u}, A_2 \vec {u} \rangle _{\mathcal D_2}$ associated with the vertices $V^1$ and $V^2$ is subordinated by the quadratic form $\langle \vec {u}, A' \vec {u} \rangle _{\mathcal D'}$ for the joined vertex, then

  $$\displaystyle \begin{aligned} {} \lambda_j ( L_q^{\mathbf S(A)} (\Gamma)) \geq \lambda_j ( L_q^{\mathbf S(A')} (\Gamma')). \end{aligned} $$

  (13.19)

Defining sums of quadratic forms above we assume that each of the terms is canonically extended to the common domain $\mathcal D_1 + \mathcal D_2$ as

$$\displaystyle \begin{aligned} \langle \vec{u}, A_j \vec{u} \rangle_{\mathcal D_1+ \mathcal D_2} = \langle P_{\mathcal D_j} \vec{u}, A_j P_{\mathcal D_j} \vec{u} \rangle_{\mathcal D_j}, \quad j = 1,2.\end{aligned}$$

The theorem can be proven by just repeating the arguments used to prove Theorem 13.6. The only difference is that one should not only take into account conditions determining the set of admissible functions, but look at the corresponding quadratic forms that are going to satisfy suitable inequalities. Observe that the theorem gives just sufficient conditions for the eigenvalues not to increase/not to decrease; moreover we do not study under which conditions the eigenvalues are preserved.

Problem 61

Prove Theorem 13.7 using the minmax principle.

Problem 62

Reformulate Theorem 13.7 for the case of delta conditions at the vertices.

Problem 63

Generalise Theorem 13.7 for the case where more than two vertices are glued together. Is it always possible to consider such gluing as a sequence of pair-wise gluings?