Spectral Gap and Dirichlet Ground State

Abstract

This chapter is entirely devoted to the studies of the lowest non-trivial eigenvalue of operators on graphs. For standard Laplacians on connected graphs the lowest eigenvalue is $\lambda _1 = 0$ and we shall be interested in $\lambda _2$, which coincides with the spectral gap $\lambda _2 - \lambda _1$. For Laplacians with Dirichlet vertices it is already non-trivial to calculate the ground state $\lambda _1 > 0$. To study these quantities similar methods can be used: Eulerian path and symmetrisation techniques, Cheeger’s approach, surgery principles. Most of these methods work for Schrödinger operators but in order to illuminate connections between spectrum and topology/geometry we shall focus on standard and Dirichlet Laplacians. The methods developed will be extended to higher eigenvalues in the following chapter.

This chapter is entirely devoted to the studies of the lowest non-trivial eigenvalue of operators on graphs. For standard Laplacians on connected graphs the lowest eigenvalue is $\lambda _1 = 0$ and we shall be interested in $\lambda _2$, which coincides with the spectral gap $\lambda _2 - \lambda _1$. For Laplacians with Dirichlet vertices it is already non-trivial to calculate the ground state $\lambda _1 > 0$. To study these quantities similar methods can be used: Eulerian path and symmetrisation techniques, Cheeger’s approach, surgery principles. Most of these methods work for Schrödinger operators but in order to illuminate connections between spectrum and topology/geometry we shall focus on standard and Dirichlet Laplacians. The methods developed will be extended to higher eigenvalues in the following chapter.

12.1 Fundamental Estimates

Our first step is to obtain a fundamental estimate for the spectral gap in terms of the total graph length $\mathcal L(\Gamma ).$ For standard Laplacians the eigenvalues are decreasing as $\alpha ^{-2}$ as the lengths of the edges are scaled by factor $\alpha > 1,$ hence it is clear that the estimate should contain the factor $\mathcal L^{-2}.$ Alternatively one could study graphs having fixed total length, say $\mathcal L = \pi .$ There are two alternative methods to prove the estimate: via symmetrisation technique or using Eulerian cycles. Both approaches will be presented.

Theorem 12.1

Let $\Gamma$ be a connected finite metric graph with the total length $\mathcal L (\Gamma )$ . The spectral gap for the standard Laplacian on $\Gamma$ can be estimated as follows

$$\displaystyle \begin{aligned} {} \lambda_2 (\Gamma) \geq \left( \frac{\pi}{\mathcal L (\Gamma) } \right)^2. \end{aligned} $$

(12.1)

Remark 12.2

It will be clear from the proof that the obtained estimate is sharp, since the equality is attained if the graph $\Gamma$ is given just by one interval.

This theorem has been proven independently by different authors: S. Nicaise [399, 400], L. Friedlander [225], P. Kurasov and S. Naboko [343]. We are going to present two proofs of the theorem based on two different techniques: Eulerian paths and symmetrisation. The first technique treats a metric graph as a geometric object and is rather illustrative. The second technique is based on the coarea formula and therefore establishes a bridge between quantum graphs and partial differential equations.

12.1.1 Eulerian Path Technique

In this section we follow closely the article [343], where the Eulerian path technique was first presented. Essentially the same method was described by S. Nicaise without exploiting Eulerian paths.

Proof of Theorem 12.1 Using Eulerian Path Technique

The first nontrivial eigenvalue of $L^{\mathrm {st}} (\Gamma )$ can be calculated by minimising the Rayleigh quotient (see Proposition 4.19)

$$\displaystyle \begin{aligned} {} \lambda_2 (\Gamma) = \operatorname*{\mathrm{min}}_{u \perp 1} \frac{\int_{\Gamma} \vert u' (x) \vert^2 dx}{\int_\Gamma \vert u(x) \vert^2 dx}, \end{aligned} $$

(12.2)

where the minimum is taken over all admissible functions $u$: belonging to the Sobolev space $W_2^1$ on every edge and continuous on the whole $\Gamma .$ Note that one may extend the set of admissible functions by allowing continuous piecewise $W_2^1$-functions—this will not change the minimiser.

The first nontrivial eigenfunction $\psi _2$ is a minimiser of (12.2) and therefore satisfies

$$\displaystyle \begin{aligned} {} \lambda_2 (\Gamma) = \frac{\int_\Gamma \vert \psi_2^{\prime} (x) \vert^2 dx}{\int_\Gamma \vert \psi_2 (x) \vert^2 dx} . \end{aligned} $$

(12.3)

Consider the graph $\Gamma ^2$—a certain “double cover” of $\Gamma$—obtained from $\Gamma$ by doubling every edge (see Fig. 12.1). The new edges have the same lengths and connect the same vertices, so that the set of vertices is preserved. The corresponding vertex degrees are doubled—this will become important soon.

Fig. 12.1

Doubling the edges

Let us lift up the function $\psi _2$ from $L_2 (\Gamma )$ to the function $\hat {\psi }_2 \in L_2 (\Gamma ^2)$ in a symmetric way by assigning it the same values on any new pair of edges as on the original edge in $\Gamma .$ More precisely, consider any edge $E_n \in \Gamma$ and let us denote by $E_n^{\prime }$ and $E_n^{\prime \prime }$ the corresponding edge pair in $\Gamma ^2.$ It is natural to use the same parametrisation of the intervals $E_n$, $E_n^{\prime }$, and $E_n^{\prime \prime }.$ Then we have

$$\displaystyle \begin{aligned} \hat{\psi}_2 \vert_{E_n^{\prime}} = \hat{\psi}_2 \vert_{E_n^{\prime\prime}} = \psi_2 \vert_{E_n}.\end{aligned}$$

The function $\hat {\psi }_2$ obtained from $\psi _2$ in this way obviously satisfies

$$\displaystyle \begin{aligned} \lambda_2 (\Gamma) = \frac{\int_{\Gamma^2} \vert {\hat{\psi}_2}^{\prime} (x) \vert^2 dx}{\int_{\Gamma^2} \vert \hat{\psi}_2 (x) \vert^2 dx} ,\end{aligned}$$

where both the numerator and denominator gain factor $2$ compared to (12.3).

Every vertex in $\Gamma ^2$ has even degree and therefore there exists a closed (Eulerian) path $\mathcal P$ on $\Gamma ^2$ coming along every edge in $\Gamma ^2$ precisely one time [181, 267].¹ The path may go through certain vertices several times. The path can be identified with the loop $S_{2 \mathcal L (\Gamma ) }$ of length $2 \mathcal L.$ The loop itself is a metric graph and we consider the corresponding standard Laplacian $L ^{\mathrm {st}} (S_{2 \mathcal L (\Gamma ) })$. As before, the ground state for $L ^{\mathrm {st}}(S_{2 \mathcal L(\Gamma )})$ is $\lambda _1 = 0$ and to calculate the spectral gap the Rayleigh quotient can be used. The set of admissible trial functions consists of $W_2^1 (S_{2 \mathcal L (\Gamma )})$ functions having mean value zero.

The function $\hat {\psi }_2$ defined originally on the graph $\Gamma ^2$ can be considered as a function on the loop $S_{2 \mathcal L(\Gamma ) }.$ It is a continuous and piece-wise $W_2^1$ function with zero mean value and therefore gives an upper estimate for the Laplacian eigenvalue on the loop

$$\displaystyle \begin{aligned} \lambda_2 (S_{2 \mathcal L}) \leq \frac{\int_{S_{2 \mathcal L}} \vert {\hat{\psi}_2}^{\prime} (x) \vert^2 dx}{\int_{S_{2 \mathcal L}} \vert\hat{\psi}_2 (x) \vert^2 dx} = \lambda_2 (\Gamma). \end{aligned}$$

We obtain the result by noticing that $\lambda _2 (S_{2 \mathcal L}) = \left ( \frac {\pi }{\mathcal L(\Gamma )} \right )^2.$ □

In fact we have proven that the minimum of the spectral gap (among all graphs of the same total length) is realised by the single interval of length $\mathcal L$, $I_{\mathcal L(\Gamma )}$. This is due to the fact that

$$\displaystyle \begin{aligned} \lambda_2 (S_{2 \mathcal L(\Gamma)}) = \lambda_2 (I_{ \mathcal L(\Gamma)}) . \end{aligned} $$

(12.4)

Moreover, the graph $I_{\mathcal L}$ is the unique minimiser—this will be proven in Sect. 14.1.2. Remember that the first eigenvalue for the loop is doubly degenerate, while it is simple for the interval.

The cycle $S_{2 \mathcal L}$ can be obtained from $\Gamma ^2$ by chopping its vertices into degree two vertices—this way to obtain spectral inequalities is known under the name surgery of graphs and will be described in details in Sect. 12.5.

12.1.2 Symmetrisation Technique

In this section we follow closely the article [225], where symmetrisation technique [286] was applied to obtain estimates for the spectral gap for the standard Laplacian.

Proof of Theorem 12.1 Using Symmetrisation

The main idea of the method is to introduce a special transformation mapping functions from $L_2 (\Gamma )$ to functions from $L_2 [0, \mathcal L]$ and use it to compare the eigenvalues of the standard Laplacians $L^{\mathrm {st}} (\Gamma )$ and $L^{\mathrm {st}} ([0, \mathcal L]).$

The spectral gap for both operators coincides with the energy $\lambda _2$ of the first excited state. Let $\psi _2$ be such excited state for the operator $L^{\mathrm {st}} (\Gamma ) .$ This is a continuous function on a compact graph $\Gamma$, let us denote the points of minimum and maximum for $\psi _2 (x)$ by $x_{\mathrm {min}}$ and $x_{\mathrm {max}} .$

The symmetrized function $\psi ^*$ on the interval $[0, \mathcal L ]$ is the unique nondecreasing continuous function such that

$$\displaystyle \begin{aligned} \psi^* (0) = \psi_2 (x_{\mathrm{min}}), \; \; \psi^* (\mathcal L) = \psi_2 (x_{\mathrm{max}})\end{aligned}$$

and

$$\displaystyle \begin{aligned} m (t):= \mathrm{measure} \left\{ x \in \Gamma: \psi_2 (x) < t \right\} = \mathrm{measure} \left\{ s \in [0, \mathcal L]: \psi^*(s) < t \right\}.\end{aligned}$$

The function $\psi ^*$ constructed in this way satisfies

$$\displaystyle \begin{aligned} \int_\Gamma \vert \psi_2 (x) \vert^2 dx = \int_0^{\mathcal L} \vert \psi^* (x) \vert^2 dx \end{aligned} $$

(12.5)

and

$$\displaystyle \begin{aligned} {} 0 = \int_\Gamma \psi_2 (x) dx = \int_0^{\mathcal L} \psi^* (x) dx, \end{aligned} $$

(12.6)

where the left equality comes from the fact $\psi _2$ is orthogonal to the constant function, which is the ground state. The measure satisfies

$$\displaystyle \begin{aligned} {} m'(t) = \sum_{x: \psi_2 (x) = t} \frac{1}{\vert \psi_2^{\prime} (x) \vert}. \end{aligned} $$

(12.7)

This formula holds for all t which do not coincide with the local minima and maxima of $\psi _2$ and the values of $\psi _2$ at the vertices. The formula is obtained by summing up contributions from different preimages of $t$ under $\psi _2 (x)$. The number of preimages is finite, since $\psi _2$ satisfies the eigenfunction equation on each interval. Let us denote the number of preimages by $n(t).$ Obviously

$$\displaystyle \begin{aligned} {} n(t) \geq 1, \quad \psi_2 (x_{\mathrm{min}}) < t < \psi_2 (x_{\mathrm{max}}), \end{aligned} $$

(12.8)

since the function $\psi _2$ is continuous.

The co-area formula (see for example [76]) implies

$$\displaystyle \begin{aligned} \int_\Gamma \vert \psi_2^{\prime}(x) \vert^2 dx = \int_{\psi_2 (x_{\mathrm{min}})}^{\psi_2 (x_{\mathrm{max}})} \sum_{x: \psi_2(x) = t} \vert \psi_2^{\prime} (x) \vert dt .\end{aligned}$$

Using the Cauchy-Schwarz inequality

$$\displaystyle \begin{aligned} \Big(\sum_{j=1}^n \frac{1}{a_j} \Big) \cdot \sum_{j=1}^n a_j \geq n^2\end{aligned}$$

we get the estimate

$$\displaystyle \begin{aligned} {} \sum_{x: \psi_2(x) = t} \vert \psi_2^{\prime}(x) \vert \geq n(t)^2 \displaystyle\left( \sum_{x: \psi_2(x) = t} \frac{1}{\vert \psi_2^{\prime}(x) \vert} \right)^{-1} &\geq \displaystyle \left( \sum_{x: \psi_2(x) = t} \frac{1}{\vert \psi_2^{\prime}(x) \vert} \right)^{-1} \\ &= \frac{1}{l} \frac{1}{m' (t) } , \end{aligned} $$

(12.9)

where we first used (12.8) and then (12.7). Therefore we have

$$\displaystyle \begin{aligned} {} \int_\Gamma \vert \psi_2^{\prime}(x) \vert^2 dx \geq \int_{\psi_2 (x_{\mathrm{min}})}^{\psi_2 (x_{\mathrm{max}})} \frac{dt}{m' (t)} . \end{aligned} $$

(12.10)

Precisely the same argument can be applied to the function $\psi ^*$ with the only difference that all inequalities turn into equalities and there is no need to use (12.8). Finally we get:

$$\displaystyle \begin{aligned} {} \int_\Gamma \vert \psi_2^{\prime}(x) \vert^2 dx \geq \int_{0}^{\mathcal L} \vert {\psi^*}' (s) \vert^2 ds. \end{aligned} $$

(12.11)

Since the norms of the functions coincide, the Rayleigh quotients satisfy the estimates

(12.12)

The left quotient gives us precisely $\lambda _2 (L^{\mathrm {st}}(\Gamma ))$, while the right quotient is an upper estimate for $\lambda _2 (L^{\mathrm {st}}([0, \mathcal L]))$, since $\psi ^*$ is an admissible trial function for the quadratic form of $L^{\mathrm {st}}([0, \mathcal L])$ and is orthogonal to the ground state—the constant function (see (12.6). The precise value of $\lambda _2 (L^{\mathrm {st}}([0, \mathcal L]))$ in accordance with Proposition 4.19 is given by $\lambda _2 (L^{\mathrm {st}}([0, \mathcal L])) = \operatorname *{\mathrm {min}}_{u \perp 1} \frac {\int _{0}^{\mathcal L} \vert u' (x) \vert ^2 dx}{\int _0^{\mathcal L} \vert u(x) \vert ^2 dx} = \Big ( \frac {\pi }{\mathcal L} \Big )^2.$ We get estimate (12.1). □

The proven estimate is fundamental in the sense that it is valid for arbitrary compact finite metric graphs independently of their topological and geometrical properties—everything is determined by the total length $\mathcal L.$ Both methods show that the estimate is sharp and equality is attained for the graph formed by just one interval of length $\mathcal L.$

12.2 Balanced and Doubly Connected Graphs

The obtained estimate can be improved if the original graph $\Gamma$ possesses special properties. For example if we assume that the graph is balanced, i.e. all vertices have even degrees, then there is no need to consider the “double covering” and Euler theorem can be applied to the graph $\Gamma$ directly. ² We say that a metric graph is doubly (edge) connected if to make it disconnected one has to remove at least two edges. Such graphs are also called bridgeless, since a bridge is an edge removing which makes the graph disconnected. Remember that only connected $\Gamma$ are considered now. Every balanced graph is bridgeless, but the opposite implication is not always true.

Problem 50

Construct an explicit example of bridgeless non balanced graph.

Theorem 12.3

Let all assumptions of Theorem 12.1 be satisfied. Assume in addition that the graph $\Gamma$ is bridgeless. Then the spectral gap for the standard Laplacian on $\Gamma$ can be estimated as follows

$$\displaystyle \begin{aligned} {} \lambda_2 (\Gamma) \geq 4 \left( \frac{ \pi}{\mathcal L (\Gamma) } \right)^2. \end{aligned} $$

(12.13)

Proof

Eulerian Path Technique We prove a slightly weaker statement first: every balanced graph possesses estimate (12.13).

Consider the proof Theorem 12.1 via Eulerian path technique. If the original graph $\Gamma$ is balanced then there is no need to create a double cover graph $\Gamma ^2$—the original graph contains an Eulerian cycle and the length of this cycle is of course $\mathcal L$ (instead of $2 \mathcal L$ as for the double cover graph). Repeating the argument we obtain:

$$\displaystyle \begin{aligned} \lambda_2 (S_{ \mathcal L}) \leq \frac{\int_{S_{\mathcal L}} \vert {\psi_2}' (x) \vert^2 dx}{\int_{S_{\mathcal L}} \vert \psi_2 (x) \vert^2 dx} = \lambda_2 (\Gamma). \end{aligned} $$

(12.14)

Taking into account that $\lambda _2 (S_{\mathcal L}) = \left ( \frac {2 \pi }{\mathcal L (\Gamma )} \right )^2$ we get the estimate (12.13).

Symmetrisation Technique Consider the proof of Theorem 12.1 using symmetrisation technique. The proof was given in [282] for balanced graphs, but the same proof holds for bridgeless graphs [52].

If the graph is bridgeless, then any continuous function $\psi$ attains almost any value at least twice. The exceptional values include the minimal and the maximal values $\psi (x_{\mathrm {min}})$, $\psi (x_{\mathrm {max}} )$ as well as the values of $\psi$ at some vertices. Hence the function $n(t)$ satisfies the following inequality almost everywhere

$$\displaystyle \begin{aligned} n(t) \geq 2.\end{aligned}$$

The inequality (12.9) can be written as

$$\displaystyle \begin{aligned} \sum_{x: \psi_2(x) = t} \vert \psi_2^{\prime}(x) \vert \geq n(t)^2 \frac{1}{m' (t) }.\end{aligned}$$

Integration gives us

$$\displaystyle \begin{aligned} \int_\Gamma \vert \psi_2^{\prime}(x) \vert^2 dx \geq 4 \int_{\psi_2 (x_{\mathrm{min}})}^{\psi_2 (x_{\mathrm{max}})} \frac{dt}{m' (t)} \end{aligned}$$

instead of (12.10). We got an extra factor $4$, but no such factor appears for the function $\psi ^*$, hence

$$\displaystyle \begin{aligned} \int_\Gamma \vert \psi_2^{\prime}(x) \vert^2 dx \geq 4 \int_0^{\mathcal L} \vert {\psi^*}' (s) \vert^2 ds\end{aligned}$$

holds instead of (12.11). Considering the Rayleigh quotient and repeating the argument we obtain (12.13) for bridgeless graphs. □

The obtained estimate is again sharp, since we have equality in (12.13) if the graph is a loop. Among all balanced graphs of the same total length the loop has the smallest spectral gap.

The proof via Eulerian path provides a clear pure topological explanation why the lower estimate is multiplied by the factor 4, while such explanation is more hidden if symmetrisation technique is used. On the other hand, symmetrisation technique allows one to easily prove the statement for bridgeless not necessarily balanced graphs.

Problem 51

Calculate the spectrum of the standard Laplacian on the graph $\Gamma _{(3.4)}$ presented in Fig. 12.2 with the lengths of the loop and the outgrowths being equal to $2\mathcal L/3$ and $\mathcal L/6$ respectively.

Fig. 12.2

Graph $\Gamma _{(3.4)}$: loop with two intervals attached

Problem 52

Give a non-trivial example of a metric graph such that the n-th eigenvalue coincides with the n-th eigenvalue for the interval of the same length.

12.3 Graphs with Dirichlet Vertices

Another application of the obtained estimate are graphs with Dirichlet vertices. Consider a metric graph $\Gamma$ with Laplacian $L^{\mathrm {st}, \mathrm {D}}$ defined on functions satisfying Dirichlet conditions at one or more degree one vertices and standard vertex conditions at all other vertices. Considering Dirichlet conditions at just degree one vertices is not a restrictive assumption, since introducing Dirichlet conditions at a vertex of higher degree $d > 1$ decomposes the vertex into $d$ degree one vertices (remember that we agreed to consider only properly connecting vertex conditions). Introducing Dirichlet conditions one has to be careful so that the graph remains connected (introducing Dirichlet conditions at a degree two vertex on a bridge disconnects the graph). The point $\lambda = 0$ is not an eigenvalue anymore, since any constant function satisfying a Dirichlet condition somewhere is identically zero. Therefore instead of the spectral gap $\lambda _2- \lambda _1$ we are going to discuss possible estimates for the lowest eigenvalue $\lambda _1 (L^{\mathrm {st}, \mathrm {D}}(\Gamma )).$

Lemma 12.4

Let $\Gamma$ be a connected finite compact metric graph and let $L^{\mathrm {st}, \mathrm {D}}$ be the Laplace operator defined on the functions satisfying Dirichlet conditions at least one degree one vertices and standard vertex conditions at all other vertices. Then the ground state $\lambda _1 (L^{\mathrm {st}, \mathrm {D}}(\Gamma ))$ satisfies the estimate

$$\displaystyle \begin{aligned} {} \lambda_1 (L^{\mathrm{st}, \mathrm{D}}(\Gamma)) \geq \Big( \frac{\pi}{2 \mathcal L} \Big)^2, \end{aligned} $$

(12.15)

where $\mathcal L$ is the total length of $\Gamma .$

Proof

It is enough to prove Lemma in the case where there is just one Dirichlet vertex, since adding Dirichlet vertices increases the eigenvalues.

Let us denote the Dirichlet vertex by $V^0$. Consider two copies of the graph $\Gamma$ and glue them together into $\Gamma \sqcup _{V^0} \Gamma$ by identifying two distinct vertices $V^0$ belonging to different copies of $\Gamma$ and introducing standard vertex conditions at the new joined vertex (Fig. 12.3).

Fig. 12.3

Gluing two copies of a graph

Let $\psi _1$ be the eigenfunction of the original Laplacian on $\Gamma$ corresponding to $\lambda _1 (L^{\mathrm {st}, \mathrm {D}}(\Gamma )).$ This function is zero at the vertex $V^0$ due to Dirichlet conditions there. Let us extend the function as $- \psi _1 (x)$ to the second copy of $\Gamma .$ The new extended function on $\Gamma \sqcup _{V^0} \Gamma$ is an eigenfunction of the standard Laplacian $L^{\mathrm {st}}( \Gamma \sqcup _{V^0} \Gamma )$ with the same eigenvalue $\lambda _1 (L^{\mathrm {st}, \mathrm {D}}(\Gamma ))$:

• the eigenfunction equation on the edges is satisfied;

• standard conditions at all vertices different from $V^0$ are satisfied on both copies of $\Gamma$;

• standard conditions at the new joined vertex are satisfied:

  • the function is equal to zero at the vertex and therefore is continuous,

  • the sum of normal derivatives is zero, since contributions from the two copies of $\Gamma$ compensate each other.

Hence $\lambda _1 (L^{\mathrm {st}, \mathrm {D}}(\Gamma ))$ can be estimated from below by $\lambda _2 (L^{\mathrm {st}} ( \Gamma \sqcup _{V^0} \Gamma ))$:

$$\displaystyle \begin{aligned} \lambda_1 (L^{\mathrm{st}, \mathrm{D}}(\Gamma)) \geq \Big( \frac{\pi}{2 \mathcal L} \Big)^2,\end{aligned}$$

where we used that the total length of $\Gamma \sqcup _{V^0} \Gamma$ is $2 \mathcal L.$ The constructed eigenfunction is not a ground state for the following reasons:

• the ground state for the standard Laplacian is a non-zero constant function, but the constructed function is zero at the joined vertex;

• moreover, the constructed function is orthogonal to the constant function by construction.

□

The obtained estimate is sharp, since for the interval of length $\ell$ with the Dirichlet and Neumann conditions at the endpoints we have $\lambda _1 = \Big ( \frac {\pi }{2 \ell } \Big )^2.$ Moreover, equality in (12.15) holds only if the graph is a Dirichlet-Neumann segment of length $\mathcal L.$ The reason is rather simple: we have equality in (12.15) only if we have equalities in all estimates used in the proof. In particular, we need that the graph $\Gamma \sqcup _{V^0} \Gamma$ has the smallest spectral gap, hence it coincides with the interval of length $2 \mathcal L.$ Moreover the constructed eigenfunction should coincide with the second eigenfunction on the interval of length $2 \mathcal L$, hence the graph $\Gamma$ is the Dirichlet-Neumann segment of length $\mathcal L$.

12.4 Cheeger’s Approach

An effective method to estimate the spectral gap for a Laplacian on a Riemannian manifold was suggested by J. Cheeger [135]. This method has already become a standard tool in differential geometry. Our goal in this section is to apply Cheeger’s ideas to quantum graphs.

The basic idea of Cheeger is that the first nontrivial eigenfunction $\psi _2$ should attain both positive and negative values, since it is orthogonal to $\psi _1 \equiv 1.$ In other words, $\psi _2$ has at least two nodal domains. The problem is that we do not know much about these domains, but one may obtain estimates without knowing where these domains are situated, or how large they are. The border $S$ between the domains divides the manifold into two or more separate manifolds with boundary. It follows that the border $S$ can be considered as a cut of the original manifold $M$ into two submanifolds $M_1$ and $M_2$ : $M = M_1 \cup M_2.$ One may introduce Cheeger’s quotient $\displaystyle h_S = \frac { L(S)}{\mathrm {min}\,\{ A(M_1), A(M_2) \}},$ where $L(S)$ is the length of the cut $S$ and $A(M_j )$ are the areas of the submanifolds $M_j$. Of course in the case of an $n$-dimensional manifold $M$ one should speak about the $n-1$-dimensional area $L(S) .$ Since we do not know which particular cut $S$ corresponds to the eigenfunction $\psi _2$, an estimate can be obtained by the cut, which minimises the quotient. The corresponding infimum is called Cheeger’s constant

$$\displaystyle \begin{aligned} h (M) := \inf_{S} \frac{ L(S)}{\mathrm{min}\,\{ A(M_1), A(M_2) \}}, \end{aligned} $$

(12.16)

where the infimum is taken over all possible cuts $S$ dividing $M$ into two parts $M_1$ and $M_2.$ The infimum is realised on short cuts dividing $M$ into two parts of approximately equal areas.

Using Cheeger’s constant the following lower estimate for the first nontrivial Laplace eigenvalue $\lambda _2 (M)$ may be proven [135]

$$\displaystyle \begin{aligned} {} \frac{1}{4} (h(M))^2 \leq \lambda_2 (M). \end{aligned} $$

(12.17)

Upper estimates in general are easier to obtain, since any admissible function provides such an estimate via the Rayleigh quotient. The lower estimates are in general much harder to prove. The advantage of Cheeger’s approach is that such estimates are obtained in geometrical terms only. Let us see how this approach can be generalised to quantum graphs. We consider standard Laplacians only and start with the lower estimate.

Lower Estimate

We use Cheeger’s original argument [135] developed for quantum graphs in [266, 400, 436]. It was also the subject of our first common project with R. Suhr, which we follow here. For Cheeger’s approach it is essential that the eigenfunctions are continuous and the ground state eigenfunction can be chosen strictly positive. Therefore we develop our analysis for standard Laplacians. Let $\Gamma$ be a metric graph, consider a set of points $P$ on the edges dividing $\Gamma$ into two subgraphs $M_1$ and $M_2.$ Then we define Cheeger’s constant for the metric graph $\Gamma$ as

$$\displaystyle \begin{aligned} {} h(\Gamma) = \inf_{P} \frac{ \vert P \vert}{\min \left\{ \mathcal L(M_1), \mathcal L(M_2)\right\} }, \end{aligned} $$

(12.18)

where $\vert P \vert$ denotes the number of dividing points and the infimum is taken over the set of possible cuts of $\Gamma$. We formally excluded the possibility to divide the graph along its vertices, but such divisions appear in the limit as the dividing points approach ends of the edges. Taking the infimum incorporates such divisions into our approach.

Let $\lambda _2$ be the spectral gap of the standard Laplacian on $\Gamma$, then consider any corresponding eigenfunction that we denote by $\psi \; (= \psi _2)$. The eigenfunction can be chosen real-valued. Consider the domains $\Gamma ^+$ and $\Gamma ^-$ where the function is positive and negative respectively. Without loss of generality we assume that $\Gamma ^+$ has the lowest volume: if this is not the case the eigenfunction should be multiplied by $-1$.

The restriction of $\psi$ to $\Gamma ^+$ is the ground state eigenfunction for the Laplacian on $\Gamma ^+$ determined by standard vertex conditions at all vertices except the dividing points from $P$ where Dirichlet conditions are assumed, since $\psi$ is an eigenfunction satisfying prescribed vertex conditions and is sign definite.

Using the fact that $\psi$ is an eigenfunction we obtain:

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \lambda_2 (L^{\mathrm{st}} (\Gamma)) & = & \displaystyle \frac{\int_{\Gamma^+} - \psi'' (x) \psi (x) dx }{\int_{\Gamma^+} \psi^2 (x) dx} = \frac{\int_{\Gamma^+} (\psi ' (x) )^2 dx}{\int_{\Gamma^+} \psi^2 (x) dx } \\[5mm] & = & \displaystyle \frac{\int_{\Gamma^+} (\psi ' (x) )^2 dx \; \; \int_{\Gamma^+} (\psi (x))^2 dx }{\Big(\int_{\Gamma^+} \psi^2 (x) dx \Big)^2} \\[5mm] & \geq & \displaystyle \frac{\Big( \int_{\Gamma^+} \vert \psi ' (x) \vert \; \vert \psi (x) \vert dx\Big)^2 }{\Big(\int_{\Gamma^+} \psi^2 (x) dx \Big)^2} \\[5mm] & = & \displaystyle \frac{1}{4} \left( \frac{\int_{\Gamma^+} \vert \frac{d}{dx} (\psi^2 (x)) \vert dx}{\int_{\Gamma^+} \psi^2 (x) dx} \right)^2, \end{array} \end{aligned} $$

(12.19)

where we used the Cauchy-Schwarz inequality. Let us introduce the following notations in order to exploit the co-area formula

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle V(y) & = &\displaystyle \mathrm{measure} \left(\left\{ x \in \Gamma^+: \psi^2 (x) \geq y \right\} \right), \\[3mm] \displaystyle N(y) & = & \displaystyle \vert (\psi^2)^{-1}(y)\cap \Gamma^+ \vert . \end{array} \end{aligned}$$

Then by definition of $h$ we always have

$$\displaystyle \begin{aligned} \frac{N(y)}{V(y)} \geq h (\Gamma),\end{aligned}$$

for any $y> 0$, as any $y$ yields a division of $\Gamma ^+$ and as a result a division of $\Gamma$.

Let us note the following relation

$$\displaystyle \begin{aligned} V(y) = \mathcal L(\Gamma^+) - \int_0^y N(t) dt \end{aligned}$$

implying

$$\displaystyle \begin{aligned} {} \frac{d V(y)}{dy} = - N(y) . \end{aligned} $$

(12.20)

The last equality holds almost everywhere on $\Gamma ^+$, more precisely for all $y$ not equal to the value of $\psi ^2$ at the vertices and local minima and maxima on the edges. Hence $V(y)$ is a piecewise continuously differentiable function.

Subdividing $\Gamma ^+$ into intervals $\varDelta _j$ where $\psi$ is monotone we get

$$\displaystyle \begin{aligned} {} \begin{array}{cclcl} \displaystyle \int_{\Gamma^+} \Big\vert \frac{d}{dx} \psi^2 (x) \Big\vert dx & = & \displaystyle \sum_{\varDelta_j} \int_{\varDelta_j} \Big\vert \frac{d}{dx} \psi^2 (x) \Big\vert dx & = & \displaystyle \sum_{\varDelta_j} \int_{\displaystyle \min_{\varDelta_j} \psi^2}^{\displaystyle \max_{\varDelta_j} \psi^2 } 1 dy \\[3mm] & = & \displaystyle \int_0^{\displaystyle \max_{\Gamma^+} \psi^2} N(y) dy & = & \displaystyle \int_0^{\displaystyle \max_{\Gamma^+} \psi^2} \frac{N(y)}{V(y)} V(y) dy \\[3mm] & \geq & \displaystyle h \; \int_0^{\displaystyle \max_{\Gamma^+} \psi^2} V(y) dy & = & \displaystyle - h \; \int_0^{\displaystyle \max_{\Gamma^+} \psi^2} y \; \frac{d V(y)}{dy} dy \\[3mm] & = & \displaystyle h \int_0^{\displaystyle \max_{\Gamma^+} \psi^2} y \; N(y) dy & = & \displaystyle h \int_{\Gamma^+} \psi^2 (x) dx, \end{array} \end{aligned} $$

(12.21)

where integrating by parts on the third line we used that $V$ is piecewise $C^1$-function and relation (12.20). Finally combining the estimates (12.19) and (12.21) we get the desired lower estimate for standard Laplacians on metric graphs.

Theorem 12.5

Let$\Gamma$be a finite compact connected metric graph and let$h$be the corresponding Cheeger constant defined by (12.18). Then the spectral gap for the standard Laplacian possesses the lower estimate

$$\displaystyle \begin{aligned} {} \lambda_2 \big(L^{\mathrm{st}} (\Gamma) \big) \geq \frac{1}{4} h^2. \end{aligned} $$

(12.22)

One may prove that inequality in (12.22) is in fact strict. Equality occurs only if all inequalities in the proof turn into equalities, in particular if one has equality in (12.19) which holds only if $\psi '$ is proportional to $\psi .$ It follows that $\psi$ is an exponential function and therefore cannot be a real valued solution to the eigenfunction equation. This result is also described in [436, Theorem 6.1].

Problem 53

Check to which extent Cheeger’s approach may be generalised for the case of Laplacians with delta vertex conditions.

Upper Estimates

For a metric graph $\Gamma$ let us delete some of its edges $S = \cup _{j=1}^{s} E_{n_j}.$ If the resulting graph $\Gamma \setminus S$ is not connected, then we say that $S$ is an edge cut of $\Gamma .$ The set $\Gamma \setminus S$ may consist of several connected components. Let us denote by $\Gamma _1$ and $\Gamma _2$ any separation of $\Gamma \setminus S$ into two nonintersecting sets

$$\displaystyle \begin{aligned} \Gamma_1 \cup \Gamma_2 = \Gamma \setminus S, \; \; \Gamma_1 \cap \Gamma_2 = \emptyset.\end{aligned}$$

We assume in this section that $\Gamma$ contains no loops, i.e. edges adjusted to one vertex. This is not an important restriction. Really, consider any graph $\Gamma$ with a loop, mark any point on the loop and put a new vertex at this point. The new metric graph obtained in this way contains no loops but the corresponding Laplace operator is unitarily equivalent to the Laplace operator on the original graph.

With any set $S$ as described above let us associate the following Cheeger-type quotient

$$\displaystyle \begin{aligned} {} \operatorname*{\mathrm{min}}_{\begin{array}{ll}{\Gamma_1, \Gamma_2:}& \Gamma_1 \cup \Gamma_2 = \Gamma \setminus S; \\ & \Gamma_1 \cap \Gamma_2 = \emptyset \end{array}} \frac{ \mathcal L(\Gamma) \sum_{E_n \in S} \ell_n^{-1} }{\mathcal L (\Gamma_1) \mathcal L (\Gamma_2)}, \end{aligned} $$

(12.23)

We are going to prove that this quotient provides an upper estimate for the spectral gap. It resembles Cheeger’s constant (12.17) described above, but is different.

Theorem 12.6

Let$\Gamma$be a connected metric graph without loops, then the spectral gap for the standard Laplacian is estimated from above by the Cheeger-type quotient (12.23) as follows

$$\displaystyle \begin{aligned} {} \lambda_2 (\Gamma) \leq C (\Gamma) := \operatorname*{\mathrm{inf}}_{S} \operatorname*{\mathrm{min}}_{\begin{array}{ll}{\Gamma_1, \Gamma_2:}& { \Gamma_1 \cup \Gamma_2 = \Gamma \setminus S}\\ & \Gamma_1 \cap \Gamma_2 = \emptyset \end{array}} \frac{ \mathcal L (\Gamma) \sum_{E_n \in S} \ell_n^{-1} }{\mathcal L (\Gamma_1) \mathcal L (\Gamma_2)}, \end{aligned} $$

(12.24)

where the infimum is taken over all edge cuts $S$ of $\Gamma .$

Proof

Consider the function $g$ defined as follows

$$\displaystyle \begin{aligned} {} g(x) = \left\{ \begin{array}{ll} 1, & x \in \Gamma_1; \\ -1, & x \in \Gamma_2; \\ \ell_n^{-1} \Big( - \mathrm{dist}\,(x, \Gamma_1) + \mathrm{dist}\, (x, \Gamma_2) \Big), & x \in E_n \subset S, \end{array} \right. \end{aligned} $$

(12.25)

where the distances $\mathrm {dist}\,(x, \Gamma _j )$, $j=1,2,$ are calculated along the corresponding interval $E_n.$ The continuous function $g$ is constructed in such a way, that it is equal to $\pm 1$ on $\Gamma _1$ and $\Gamma _2$ and is linear on the edges connecting $\Gamma _1$ and $\Gamma _2.$ The mean value of the function might be different from zero. In that case the function $g$ has to be modified so that it will be orthogonal to the ground state. Consider then the function $f$ which is not only continuous, but also orthogonal to the ground state:

$$\displaystyle \begin{aligned} f (x) = g(x) - \mathcal L (\Gamma)^{-1} \langle g, 1 \rangle_{L_2 (\Gamma)} = g(x) - \frac{\mathcal L(\Gamma_1) - \mathcal L (\Gamma_2)}{\mathcal L (\Gamma)}.\end{aligned}$$

The Rayleigh quotient for the function $f$ gives an upper estimate for the spectral gap.

To determine the Rayleigh quotient we calculate the Dirichlet integral and the norm of $f$:

$$\displaystyle \begin{aligned} \begin{array}{ccl} \parallel f' \parallel^2_{L_2(\Gamma)} & = & \parallel g' \parallel^2_{L_2(\Gamma)} = \sum_{E_n \in S} \int_{E_n} (-2 \ell_n^{-1})^2 dx = 4 \sum_{E_n} \ell_n^{-1}; \\[3mm] \parallel f \parallel^2_{L_2(\Gamma)} & = & \parallel g \parallel_{L_2(\Gamma)}^2 - \mathcal L(\Gamma)^{-1} \langle g, 1 \rangle^2 \\ & \geq & \mathcal L (\Gamma_1) + \mathcal L (\Gamma_2) - \mathcal L(\Gamma)^{-1} \left( \mathcal L (\Gamma_1) - \mathcal L(\Gamma_2) \right)^2 \\ & \geq & 4 \frac{\mathcal L (\Gamma_1) \mathcal L (\Gamma_2)}{\mathcal L(\Gamma)}. \end{array} \end{aligned} $$

(12.26)

This gives the following upper estimate for $\lambda _2 (\Gamma )$

$$\displaystyle \begin{aligned} {} \lambda_2 (\Gamma) \leq C (\Gamma), \end{aligned} $$

(12.27)

where we use (12.24) and the fact that the set $S$ dividing $\Gamma$ into disconnected components is arbitrary. □

The derived estimate shows that the spectral gap is small if the metric graph can be cut into two approximately equal parts by deleting few long edges. Of course choosing long edges to delete makes the rest of the graph smaller, but to get the best estimate one has to find the best balance between these two tendencies.

The estimate we have just proven is rather explicit, but not exact in the sense that we do not know any graph for which the equality holds. We shall now present a modified estimate, which has a more complicated form, but with the advantage that there are graphs for which the estimate is exact.

The graphs we consider are finite and compact, hence the number of possible cuts is always finite. Therefore it is tempting to substitute the infimum in (12.24) by the minimum. On the other hand, every point on an edge can be seen as a dummy degree two vertex and the number of possible cuts becomes infinite making it unavoidable to use the infimum.

An Improved Upper Estimate

The function $g$ used in the proof of Theorem 12.6 can be chosen equal to

$$\displaystyle \begin{aligned} g(x) = \left\{ \begin{array}{ll} 1, & x \in \Gamma_1; \\ \cos \frac{\mathrm{dist}\, (x, \Gamma_1)}{\ell_n} \pi = - \cos \frac{\mathrm{dist}\, (x, \Gamma_2)}{\ell_n} \pi, & x \in E_n \subset S;\\ -1, & x \in \Gamma_2. \end{array} \right. \end{aligned} $$

(12.28)

We again shift the function by a constant to satisfy the orthogonality condition

$$\displaystyle \begin{aligned} f(x) = g(x) - \frac{\mathcal L(\Gamma_1) - \mathcal L(\Gamma_2)}{\mathcal L(\Gamma)} . \end{aligned} $$

(12.29)

Calculating the Dirichlet integral and the norm

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \parallel f' \parallel^2_{L_2(\Gamma)} & = & \displaystyle \sum_{E_n \subset S} \left( \frac{\pi}{\ell_n} \right)^2 \int_{E_n} \sin^2 \frac{\mathrm{dist}\, (x, \Gamma_1)}{\ell_n} \pi dx = \displaystyle \frac{\pi^2}{2} \sum_{E_n \subset S} \ell_n^{-1}; \\[5mm] \displaystyle \parallel f \parallel^2_{L_2 (\Gamma)} & = & \displaystyle \parallel g \parallel^2_{L_2(\Gamma)} - \frac{(\mathcal L(\Gamma_1) - \mathcal L(\Gamma_2))^2}{\mathcal L(\Gamma) }; \\[5mm] \displaystyle \parallel g \parallel^2_{L_2(\Gamma)} & = & \displaystyle \mathcal L (\Gamma_1) + \mathcal L(\Gamma_2) + \frac{1}{2} \mathcal L(S) \end{array}\end{aligned}$$

and substituting into the Rayleigh quotient, we get the following estimate for the first nontrivial eigenvalue

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \lambda_2 (\Gamma)& \leq & \displaystyle \frac{\parallel f' \parallel^2_{L_2(\Gamma)} }{\parallel f \parallel^2_{L_2(\Gamma)}} \\ & = & \displaystyle \frac{\frac{\pi^2}{2} \sum_{E_n \subset S} \ell_n^{-1}}{\mathcal L(\Gamma_1) + \mathcal L(\Gamma_2) + \frac{1}{2} \mathcal L(S) - \frac{(\mathcal L(\Gamma_1) - \mathcal L(\Gamma_2))^2}{\mathcal L(\Gamma)}}\\ & = & \displaystyle \frac{\pi^2 \mathcal L (\Gamma) \sum_{E_n \subset S} \ell_n^{-1} }{8 \mathcal L(\Gamma_1) \mathcal L(\Gamma_2) + 3 (\mathcal L (\Gamma_1) + \mathcal L(\Gamma_2)) \mathcal L (S) + \mathcal L^2(S)}. \end{array} \end{aligned} $$

(12.30)

Here $\mathcal L(S)$ denotes the total length of all deleted edges forming the set $S.$ The obtained estimate can be used even in the case where the graphs $\Gamma _1$ and $\Gamma _2$ have zero lengths—the denominator is still different from zero in that case. This will be used in the proof of the following theorem.

Theorem 12.7

The spectral gap for the Laplace operator on a metric graph $\Gamma$ satisfies the following lower and upper estimates

$$\displaystyle \begin{aligned} \frac{\pi^2}{\mathcal L^2(\Gamma)} \leq \lambda_2 (\Gamma) \leq \frac{\pi^2}{\mathcal L^2(\Gamma)} \; \; 4 \mathcal L(\Gamma) \sum_{E_n \in \Gamma} \ell_n^{-1}. \end{aligned} $$

(12.31)

If the metric graph$G$is bipartite,³then the upper estimate can be improved by a factor of 4 as follows

$$\displaystyle \begin{aligned} \quad \quad \quad \quad \lambda_2 (\Gamma) \leq \frac{\pi^2}{\mathcal L^2(\Gamma)} \; \; \mathcal L(\Gamma) \sum_{E_n \in \Gamma} \ell_n^{-1}. \end{aligned} $$

(12.32)

Proof

The lower estimate has been already proven in Theorem 12.1, it remains to show the upper one. We start from the second formula. Assume that the graph $\Gamma$ is bipartite. Then the graphs $\Gamma _1$ and $\Gamma _2$ appearing in Cheeger’s estimate (12.30) can be chosen equal to the two disjoint sets of vertices $\mathbb V^1$ and $\mathbb V^2$ appearing in the definition of a bipartite graph. These sets consist of vertices only, hence $\mathcal L(\Gamma _1) = 0 = \mathcal L(\Gamma _2).$ Every edge in $\Gamma$ connects vertices from the two sets. Therefore the proper cut $S$ contains all edges. In other words we cut all edges in the graph $\Gamma$ and we have $\mathcal L(S) = \mathcal L(\Gamma )$ leading to the following estimate

$$\displaystyle \begin{aligned} \lambda_2 (\Gamma) \leq \frac{\pi^2}{\mathcal L(\Gamma)} \sum_{E_n \subset \Gamma} \ell_n^{-1}.\end{aligned}$$

The upper estimate for arbitrary graphs can be proven by the following trick: any metric graph $\Gamma$ can be turned into a bipartite graph by introducing new vertices in the middle of every edge. Then the sets $\mathbb V^1$ and $\mathbb V^2$ can be chosen equal to the unions of old and new vertices respectively. Then the previous estimate gives

$$\displaystyle \begin{aligned} \lambda_2 (\Gamma) \leq \frac{\pi^2}{\mathcal L(\Gamma)} \; \; 2 \sum_{E_n \subset \Gamma} (\ell_n/2)^{-1}.\end{aligned}$$

The factor $2$ in front of the sum appears due to the fact that every edge in $\Gamma$ is divided into two smaller edges of lengths $\ell /2.$ □

Example 12.8

Consider the equilateral complete graph $K_M$ of total length $\mathcal L.$ To get upper estimates for the spectral gap we are going to cut in two different ways:

Cut A:

Assume that $M$ is even $M= 2m$. Let $\Gamma _1$ and $\Gamma _2$ be two complete graphs on $m$ vertices.

Cut B:

Let $\Gamma _1$ and $\Gamma _2$ contain one and $M-1$ vertices of $K_M.$ From each edge connecting $\Gamma _1$ and $\Gamma _2$ we cut away an interval of a length $a$ attached to one of the vertices in $\Gamma _2.$

Intuitively the first cut seems to be better, since the second cut appears to be very asymmetric.

Cut A The lengths of $\Gamma _1$ and $\Gamma _2$ and the size of the cut are given by:

• the length of each edge in $K_M$ is: $\ell = \mathcal L / N(K_{2m}) = \frac {2 \mathcal L }{2m (2m-1)}= \frac {\mathcal L }{m (2m-1)};$

• the number of edges that are deleted is: $m^2;$

• the volumes of the two parts $\mathcal L (\Gamma _1) = \mathcal L (\Gamma _2) = \ell \frac {m(m-1)}{2} = \frac {m-1}{2 (2m-1)} \mathcal L;$

Applying (12.24) we get the following upper estimate:

$$\displaystyle \begin{aligned} \lambda_2 \leq \frac{\mathcal L \frac{M^2}{4} \frac{M (M-1)}{2 \mathcal L}}{\left( \frac{M(M-2)}{8} \frac{2 \mathcal L}{M (M-1)} \right)^2} = \frac{ 2 M^4}{\mathcal L} \frac{(1-1/M)^3}{(1-2/M)^2} \sim_{M \rightarrow \infty} 2 \frac{M^4}{\mathcal L^2}. \end{aligned} $$

(12.33)

Cut B The lengths needed for the estimate are:

• the length of each edge in $K_M$ is: $\ell = \mathcal L / N(K_{M}) = \frac {2 \mathcal L }{M (M-1)};$

• the number of edges that are cut: $M-1;$

• the volumes of the two parts $\mathcal L (\Gamma _1) = (\ell - a) (M-1), \quad \mathcal L (\Gamma _2) = \frac {M-2}{M} \mathcal L;$

The estimate will be

$$\displaystyle \begin{aligned} \lambda_2 \leq \frac{\mathcal L (M-1) a^{-1}}{(\ell-a) (M-1) \frac{M-2}{M} \mathcal L} = \frac{1}{a (\ell -a)} \frac{M}{M-2}. \end{aligned} $$

(12.34)

Here $a$ is arbitrary and we get the best estimate if we choose $a = \ell /2$. This gives us:

$$\displaystyle \begin{aligned} \lambda_2 \leq \frac{M^3 (M-1)^2}{M-2} \frac{1}{\mathcal L^2} \sim_{M \rightarrow \infty} \frac{M^4}{\mathcal L^2}. \end{aligned} $$

(12.35)

The precise value of the spectral gap for complete graphs has already been calculated:

$$\displaystyle \begin{aligned} \lambda_2 (K_M) = \frac{M^2 (M-1)^2}{4 \mathcal L^2} \left( \arccos (- \frac{1}{M-1}) \right)^2 \sim_{M \rightarrow \infty} \frac{\pi^2}{16} \frac{M^4}{\mathcal L^2}.\end{aligned}$$

We see that Cut B provides an estimate with an almost correct asymptotic behaviour (factor $1$ instead of $\pi ^2/16$). To get a reasonable estimate we were forced to introduce dummy degree two vertices and use new edges in the cut.

Another interesting upper estimate [292] may be obtained using the fact that flower graphs maximise the second eigenvalue [294, Theorem 4.2]

$$\displaystyle \begin{aligned} \lambda_2 \leq \frac{\pi^2 N^2}{\mathcal L^2} \leq \pi^2 N^2 \frac{h^2(\Gamma)}{4}, \end{aligned} $$

(12.36)

where $N$ is the number of edges and one uses the trivial estimate $h(\Gamma ) \geq \frac {2}{\mathcal L} .$

12.5 Topological Perturbations in the Case of Standard Conditions

Quantum graphs for us are primarily geometric objects, therefore it is important to understand how their spectral properties depend on topological perturbations of the underlying metric graph. We are going to discuss here just two such perturbations when two vertices are glued together or when an edge or an interval is cut. We call these perturbations gluing and cutting respectively. Starting from standard Laplacians we proceed to most general Schrödinger operators on graphs. The method of topological perturbations will help us to understand how spectral properties depend on the topology and obtain new explicit spectral estimates. The method was first introduced in [343] and soon became a standard tool in spectral analysis of quantum graphs. For a comprehensive survey of this method one may consult [88]. In fact we have already used this approach when Eulerian path technique was applied. Our goal here will be to provide a more systematic study presenting explicit examples.

As already mentioned, the spectral gap has been extensively investigated for discrete graphs where it is referred to as algebraic connectivity [221]. Therefore our goal here will be to understand behaviour of the spectral gap under topological and geometrical perturbations of the underlying metric graph. Such perturbations include:

• joining two vertices together;

• adding a pendant edge (or a pendant graph) introducing a new vertex at the same time;

• adding an edge (or a graph) between two or more existing vertices.

In the last two cases the total length of the graph increases, therefore it is not surprising that the spectral gap has a tendency to decrease (Theorems 12.11 and 12.12). Hence let us start our studies with the first case.

In this section we are going to compare our results with the corresponding statements for discrete graphs described in detail in Chap. 24. To understand most of the comments it is not necessary to read that chapter in advance, but one may consult it if necessary. The main message we want to deliver is that metric and discrete graphs are similar as far as properties of their lowest eigenvalues are concerned. The difference is lying in the fact that the role of vertices and edges is interchanged. Thus the total length of a metric graph plays a role similar to the number of vertices for discrete graphs; adding an edge to a discrete graph is similar to joining two vertices in metric graphs, and so on.

12.5.1 Gluing Vertices Together

Our first step is to formalise the observations on what happens to the spectrum when two vertices with standard conditions are joined together into one common vertex. We have already used this observation in Sect. 12.1.1, where Eulerian path technique was introduced.

Theorem 12.9

Let $\Gamma$ be a connected 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 holds:

• (1) The spectral gap for the standard Laplacian satisfies the inequality

  $$\displaystyle \begin{aligned} {} \lambda_2 (\Gamma) \leq \lambda_2 (\Gamma'). \end{aligned} $$

  (12.37)

• (2) The equality $\lambda _2 (\Gamma ) = \lambda _2 (\Gamma ')$ holds if and only if the eigenfunction $\psi _2$ corresponding to the first excited state can be chosen such that it attains the same values in the vertices to be joined:

  $$\displaystyle \begin{aligned} {} \psi_2 (V^1) = \psi_2 (V^2). \end{aligned} $$

  (12.38)

Proof

Consider the standard Laplacians $L^{\mathrm {st}}(\Gamma )$ and $L^{\mathrm {st}} (\Gamma ')$. Their quadratic forms are given by the same expression—the Dirichlet integral $\int \vert u'(x) \vert ^2 dx$, where the integration is over the corresponding metric graph. The integration is over the edges and therefore it is irrelevant whether the vertices $V^1$ and $V^2$ are joined together or not. The domain of the quadratic form is given by all functions from $W_2^1$ on the edges satisfying continuity conditions at the vertices. The functions from $L_2(\Gamma )$ and $L_2 (\Gamma ')$ can be identified, therefore one may say that functions from the domain of the quadratic form on $\Gamma '$ satisfy the additional continuity condition

$$\displaystyle \begin{aligned} u (V^1 ) = u(V^2),\end{aligned}$$

(compared to functions from the domain of the quadratic form on $\Gamma$).

The first excited state is calculated by minimising the Rayleigh quotient $\displaystyle \frac { \int _\Gamma \vert u' (x) \vert ^2 dx }{ \int _\Gamma \vert u (x) \vert ^2 dx}$ over the set of functions from the domain of the quadratic form which in addition are orthogonal to the ground state eigenfunction $\psi _1 (x) \equiv 1.$ The set of admissible functions for $\lambda _2 (\Gamma )$ is larger than that for $\lambda _2 (\Gamma ')$, hence inequality (12.37) for the corresponding minima follows.

To prove the second statement we first note that if the minimising function $\psi _2$ for $\Gamma$ satisfies in addition (12.38), then the same function is a minimiser for $\Gamma '$ and the corresponding eigenvalues coincide. Conversely if $\lambda _2 (\Gamma ) = \lambda _2 (\Gamma ')$, then the eigenfunction for $L^{\mathrm {st}}( \Gamma ')$ is also a minimiser for the Rayleigh quotient for $\Gamma$ and therefore is an eigenfunction for $L^{\mathrm {st}} (\Gamma )$ (satisfying in addition (12.38)). □

It is interesting to compare spectral behaviour of quantum graphs and discrete graphs; as we already mentioned, the role of vertices and edges is exchanged, hence it is natural to compare Theorem 12.9 with Proposition 24.10, which describes what happens to the spectral gap as an edge is added to a discrete graph. These two statements may appear to be rather similar at first glance. But the reasons for the spectral gap to increase are different. In the case of discrete graphs the difference between the Laplace operators is a nonnegative matrix, hence we have an explicit inequality for the quadratic forms having identical domains. For quantum graphs the quadratic forms are given by identical expressions, but inequality (12.37) is valid due to the fact that the opposite inclusion holds for the domains of the quadratic forms.

Corollary 12.10

Theorem12.9implies in particular that the flower graph formed by loops, all attached to one vertex, has the largest spectral gap among all graphs formed by a given set of edges (Fig.12.4).

Fig. 12.4

A flower graph

12.5.2 Adding an Edge

Our goal in this section is to study behaviour of the spectral gap as an extra edge is added to the metric graph. There are two possibilities:

• adding an edge between two existing vertices,

• adding a pendant edge introducing a new vertex at the same time.

In both cases the total length of the graph increases, therefore it is not surprising that the spectral gap has a tendency to decrease (Theorem 12.11). Therefore it is particularly interesting to find cases when the spectral gap is growing. Nevertheless we start by investigating what happens when pendant edges are added—the spectral gap cannot increase in this case. The main reason is to compare our findings with the corresponding result for discrete graphs (Proposition 24.11).

Theorem 12.11

Let$\Gamma$be a connected metric graph with a vertex$V^1$and let$\Gamma '$be another graph obtained from$\Gamma$by adding a pendant edge, i.e. one vertex and one edge between the new vertex with the vertex$V^1$.

• (1) The spectral gap for the standard Laplacians satisfies the following inequality:

  $$\displaystyle \begin{aligned} \lambda_2(\Gamma)\geq \lambda_2(\Gamma').\end{aligned}$$

• (2) The equality$\lambda _2 (\Gamma ) = \lambda _2 (\Gamma ')$holds only if every eigenfunction$\psi _2$corresponding to$\lambda _2 (\Gamma )$is equal to zero at$V^1$:

  $$\displaystyle \begin{aligned} \psi_2 (V^1) = 0.\end{aligned}$$

Proof

The graph $\Gamma$ is naturally considered as a subset of $\Gamma '.$ Let us introduce the following function on $\Gamma '$:

$$\displaystyle \begin{aligned} f(x)=\left\{ \begin{array}{ll} \psi_2(x), & x \in \Gamma,\\ \psi_2(V^1)& x \in \Gamma'\backslash \Gamma. \end{array}\right.\end{aligned}$$

This function coincides with $\psi _2$ on the original graph $\Gamma$ and is extended to the new edge by a constant preserving continuity at $V^1.$ The function is not necessarily orthogonal to the ground state on $\Gamma '.$ Therefore consider the nonzero function g differed from $f$ by a constant

$$\displaystyle \begin{aligned} g(x):=f(x)+c,\end{aligned}$$

where c is chosen so that the orthogonality condition in $L_2(\Gamma ')$ holds

where $\ell$ and $\mathcal L'$ are the length of the added edge and the total length of $\Gamma '$ respectively. This implies $c = -\frac {\psi _2(V^1) \ell }{\mathcal L'}.$ Using the new function the following estimate for the spectral gap is obtained:

Here $\mathcal L$ denotes the total length of the metric graph $\Gamma .$ The last inequality follows from the fact that

$$\displaystyle \begin{aligned} \| \psi^{\prime}_2 \|{}^2_{L_2(\Gamma)} = \lambda_2(\Gamma) \|\psi_2\|{}^2.\end{aligned}$$

Note that in the last expression the equality holds if and only if $c=0$ and $|\psi _2(V^1) + c|{ }^2=0$ implying $\psi _2(V^1)=0.$ This proves the second assertion. □

The proven statement has a direct analogy in the theory of discrete Laplacians—Proposition 24.11 below. The analogy is complete, since the transformation we analyse consists of adding one new edge and one new vertex simultaneously. It has similar effect on discrete and metric graphs.

In the proof of the theorem we did not use that $\Gamma ' \setminus \Gamma$ is an edge. It is straightforward to generalise the theorem for the case where $\Gamma ' \setminus \Gamma$ is an arbitrary finite connected graph joined to $\Gamma$ at a single vertex $V^1.$ One may even show that joining together any two graphs $\Gamma _1$ and $\Gamma _2$ at one vertex leads to a new graph $\Gamma '$ with the spectral gap satisfying the estimate:

$$\displaystyle \begin{aligned} \mathrm{min} \left\{ \lambda_2 (\Gamma_1), \lambda_2 (\Gamma_2) \right\} \geq \lambda_2 (\Gamma'). \end{aligned} $$

(12.39)

One may also prove this statement using general perturbation theory, taking into account that the difference between the resolvents of the standard Laplacians on $\Gamma _1 \cup \Gamma _2$ and $\Gamma '$ has rank one. The operator $L^{\mathrm {st}} (\Gamma _1 \cup \Gamma _2)$ has at least three eigenvalues in the interval $[0, \mathrm {min} \left \{ \lambda _2 (\Gamma _1), \lambda _2 (\Gamma _2) \right \}],$ hence the operator $L^{\mathrm {st}} (\Gamma ')$ has at least two eigenvalues in the same interval.

The approach using general perturbation theory illuminates one important point: the statement holds only if two graphs are joined at one vertex: in that case the difference between the resolvents has rank one.

We return now to our original goal and investigate the behaviour of the spectral gap when an edge between two vertices is added to a metric graph.

Theorem 12.12

Let$\Gamma$be a connected metric graph and$L^{st}$—the corresponding standard Laplace operator. Let$\Gamma '$be a metric graph obtained from$\Gamma$by adding an edge between the vertices$V^1$and$V^2$. Assume that the eigenfunction$\psi _2$corresponding to the first excited eigenvalue can be chosen such that

$$\displaystyle \begin{aligned} {} \psi_2(V^1) = \psi_2(V^2). \end{aligned} $$

(12.40)

Then the following inequality for the spectral gap holds:

$$\displaystyle \begin{aligned} {} \lambda_2(\Gamma) \geq \lambda_2(\Gamma'). \end{aligned} $$

(12.41)

Proof

Consider the eigenfunction $\psi _2(\Gamma )$ for $L^{st}(\Gamma )$ and extend it to the new edge by a constant, which is possible due to (12.40)

$$\displaystyle \begin{aligned} f(x)=\left\{ \begin{array}{ll} \psi_2(x), & x \in \Gamma,\\ \psi_2(V^1) \; \; (=\psi_2(V^2)), & x \in \Gamma'\backslash \Gamma. \end{array}\right.\end{aligned}$$

This function is not orthogonal to the constant function. Let us, as before, adjust the constant c so that the function $g(x) = f(x) + c$ is orthogonal to $1$ in $L_2(\Gamma ')$:⁴

where we keep notations from the proof of the previous Theorem. We have used that the eigenfunction $\psi _2$ has mean value zero, i.e. is orthogonal to the ground state on $\Gamma .$ This implies $c= -\frac {\psi _2(V^1) \ell }{\mathcal L'}.$ Now we are ready to get an estimate for $\lambda _2(\Gamma ')$ using the Rayleigh quotient

$$\displaystyle \begin{aligned} \lambda_2(\Gamma') \leq \frac{\| g' \|{}^2_{L_2(\Gamma')}}{\|g\|{}^2_{L_2(\Gamma')}}.\end{aligned}$$

The numerator and denominator can be evaluated as follows

$$\displaystyle \begin{aligned} \| g' \|{}^2_{L_2(\Gamma')} &= \| \psi^{\prime}_2 \|{}^2_{L_2(\Gamma)} = \lambda_2(\Gamma)\|\psi_2\|{}^2_{L_2(\Gamma)},\\[3mm] \|g\|{}^2_{L_2(\Gamma')} &= \|\psi_2 +c \|{}^2_{L_2(\Gamma)} + | \psi_2(V^1) + c|{}^2 \ell = \\ & = \|\psi_2\|{}^2_{L_2(\Gamma)} + c^2 \mathcal L + |\psi_2(V^1) + c|{}^2 \ell \\ & \geq \|\psi_2\|{}^2_{L_2(\Gamma)} \end{aligned} $$

leading to (12.41). □

One may think that the above theorem is rather artificial due to the presence of condition (12.40). To see that this condition is natural, let us consider a couple of examples:

Example 12.13

Let $\Gamma$ be the graph formed by one edge of length $a$. The spectrum of $L^{\mathrm {st}} (\Gamma )$ is $\Sigma (L^{\mathrm {st}} (\Gamma )) = \left \{\left (\frac {\pi }{a}\right )^2 n^2\right \}_{n=0}^\infty .$ All eigenvalues have multiplicity one.

Consider the graph $\Gamma '$ obtained from $\Gamma$ by adding edge of length $b,$ so that $\Gamma '$ is formed by two intervals of lengths a and b connected in parallel (see Fig. 12.5). The graph $\Gamma '$ is equivalent to the circle of length $a+b$. The spectrum is: $\Sigma (L^{\mathrm {st}}(\Gamma ')) = \left \{\left (\frac {2\pi }{a+b}\right )^2 n^2\right \}_{n=0}^\infty ,$ where all the eigenvalues except for the ground state have double multiplicity.

Fig. 12.5

Graphs $\Gamma = \Gamma _{(1.1)}$, $\Gamma '= \Gamma _{(2.3)}$, and $\Gamma ''=\Gamma _{(3.9)}$

Let us study the relation between the spectral gaps:

$$\displaystyle \begin{aligned} \lambda_2 (\Gamma) = \frac{\pi^2}{a^2}, \; \, \lambda_2 (\Gamma') = \frac{4 \pi^2}{(a+b)^2}.\end{aligned}$$

Any relation between these values is possible:

$$\displaystyle \begin{aligned} b>a & \: \Rightarrow \: \lambda_2(\Gamma) > \lambda_2(\Gamma') ,\\ b<a & \: \Rightarrow \: \lambda_2(\Gamma) < \lambda_2(\Gamma') . \end{aligned} $$

Therefore the first eigenvalue is not in general a monotonously decreasing function of the set of edges. The spectral gap decreases only if certain additional conditions are satisfied.

Example 12.14

Consider, in addition to the graph $\Gamma '$ discussed in Example 12.13, the graph $\Gamma ''$ obtained from $\Gamma '$ by adding another one edge of length $c$ between the same two vertices (see Fig. 12.5). Hence $\Gamma ''$ is formed by three parallel edges of lengths $a,b$ and c. The first eigenfunction for $L^{\mathrm {st}} (\Gamma ')$ can always be chosen so that its values at the vertices are equal. Then, in accordance with Theorem 12.12, the first eigenvalue for $\Gamma ''$ is less or equal to the first eigenvalue for $\Gamma '$:

$$\displaystyle \begin{aligned} \lambda_2(\Gamma'')\leq \lambda_2(\Gamma').\end{aligned}$$

This fact can easily be supported by explicit calculations.

The above examples and theorems show that the spectral gap has a tendency to decrease, when a new sufficiently long edge is added. It is not surprising, since addition of an edge increases the total length of the graph, but the eigenvalues satisfy Weyl’s law and therefore are asymptotically close to $(\pi n)^2/\mathcal L^2.$ This is in contrast to discrete graphs, for which addition of an edge never decreases the spectral gap.

Condition (12.40) in Theorem 12.12 is not easy to check for non-trivial graphs and therefore it might be interesting to obtain other explicit sufficient conditions. In what follows we would like to discuss one such geometric condition ensuring that the spectral gap drops as a new edge is added to a graph. The main idea is to compare the length $\ell$ of the new edge with the total length of the original graph $\mathcal L (\Gamma ).$ It turns out that if $\ell > \mathcal L(\Gamma )$, then the spectral gap always decreases. We have already observed this phenomenon when discussing Example 12.13, where behaviour of $\lambda _2$ depended on the ratio between the lengths $a$ and $b.$ If $b = \ell > a = \mathcal L (\Gamma )$, then the gap decreases. It is surprising that the same explicit condition holds for arbitrary connected graphs $\Gamma .$

Theorem 12.15

Let $\Gamma$ be a connected finite compact metric graph of length $\mathcal L(\Gamma )$ and let $\Gamma '$ be a graph constructed from $\Gamma$ by adding an edge of length $\ell$ between certain two vertices. If

$$\displaystyle \begin{aligned} {} \ell > \mathcal L(\Gamma) , \end{aligned} $$

(12.42)

then the spectral gaps of the corresponding standard Laplacians satisfy the estimate

$$\displaystyle \begin{aligned} {} \lambda_2 (\Gamma) \geq \lambda_2 (\Gamma'). \end{aligned} $$

(12.43)

Proof

Let $\psi _2$ be any eigenfunction corresponding to the first excited eigenvalue $\lambda _2 (\Gamma )$ of $L^{\text{st}}(\Gamma )$. It follows that the minimum of the Rayleigh quotient is attained at $\psi _2$:

$$\displaystyle \begin{aligned} \lambda_2(\Gamma) =\min_{u \in \stackrel{c}{W^1_2}(\Gamma): u \perp 1} \frac{\|u'\|{}^2_{L_2(\Gamma)}}{\|u\|{}^2_{L_2(\Gamma)}} = \frac{\|\psi_2^{\prime}\|{}^2_{L_2(\Gamma)}}{\|\psi_2\|{}^2_{L_2(\Gamma)}},\end{aligned}$$

where $\stackrel {c}{W^1_2}(\Gamma )$ denotes the set of continuous $W_2^1$-functions on the graph $\Gamma$:

$$\displaystyle \begin{aligned} \stackrel{c}{W^1_2}(\Gamma) = W^1_2(\Gamma \setminus \mathbf V) \cap C(\Gamma).\end{aligned}$$

Let us denote by $V^1$ and $V^2$ the vertices in $\Gamma$ where the new edge $E$ of length $\ell$ is attached.

The eigenvalue $\lambda _2(\Gamma ')$ can again be estimated using the Rayleigh quotient

$$\displaystyle \begin{aligned} {} \lambda_2(\Gamma') =\min_{u \in \stackrel{c}{W^1_2}(\Gamma'): u \perp 1} \frac{\|u'\|{}^2_{L_2(\Gamma')}}{\|u\|{}^2_{L_2(\Gamma')}} \leq \frac{\|g'\|{}^2_{L_2(\Gamma')}}{\|g\|{}^2_{L_2(\Gamma')}}, \end{aligned} $$

(12.44)

where $g(x)$ is any function in $\stackrel {c}{W^1_2}(\Gamma ')$ orthogonal to the constant functions in $L_2 (\Gamma ').$ Let us choose a trial function $g$ of the form $g(x) = f(x) + c$ where

$$\displaystyle \begin{aligned} f(x):=\left\{ \begin{array}{ll} \psi_2(x), & x\in \Gamma,\\ \gamma_1+\gamma_2\sin{\left(\frac{\pi x}{\ell}\right)} & x\in \Gamma'\backslash \Gamma = E = [-\ell/2, \ell/2], \end{array}\right. \end{aligned} $$

(12.45)

with $\gamma _1 = (\psi _2 (V^1) + \psi _2(V^2))/2$ and $\gamma _2 = (\psi _2(V^2)- \psi _2 (V^1))/2$. Here we assumed that the left endpoint of the interval is connected to $V^1$ and the right endpoint to $V^2.$ The function $f$ obviously belongs to $\stackrel {c}{W^1_2}(\Gamma ')$, since it is continuous at $V^1$ and $V^2$ , but it is not necessarily orthogonal to the ground state eigenfunction 1. The constant c is adjusted in order to ensure the orthogonality condition is satisfied

$$\displaystyle \begin{aligned} {} \Rightarrow c = - \frac{\gamma_1 \ell}{\mathcal L'}. \end{aligned} $$

(12.46)

The function $g$ can be used as a trial function in (12.44) to estimate the spectral gap. Let us begin by computing the denominator using the fact that $g$ is orthogonal to $1$ and Pythagoras theorem can be used

$$\displaystyle \begin{aligned} {} \| g \|{}^2_{L_2(\Gamma')} &= \|f + c\|{}_{L_2(\Gamma')}^2 = \| f \|{}^2_{L_2(\Gamma')} - \| c \|{}^2_{L_2 (\Gamma')} \\ &= \|\psi_2\|{}_{L_2(\Gamma)}^2 + \int_{-\ell/2}^{\ell/2}\left(\gamma_1+\gamma_2\sin{\left(\frac{\pi x}{\ell}\right)}\right)^2\: dx - c^2 \mathcal L' \\ &= \|\psi_2\|{}_{L_2(\Gamma)}^2 + \ell \gamma_1^2 + \frac{\ell}{2}\gamma_2^2- c^2 \mathcal L' \end{aligned} $$

(12.47)

The numerator yields

$$\displaystyle \begin{aligned} {} \|g'\|{}^2_{L_2(\Gamma')} &= \|f'\|{}^2_{L_2(\Gamma')} = \|\psi_2^{\prime}\|{}^2_{L_2(\Gamma)} + \int_{-\ell/2}^{\ell/2} \left(\gamma_2^2 \frac{\pi^2}{\ell^2}\cos^2\left(\frac{\pi x}{\ell}\right)\right)\: dx \\ &= \lambda_2(\Gamma)\|\psi_2\|{}^2_{L_2(\Gamma)} + \gamma_2^2 \frac{\pi^2}{2\ell}. \end{aligned} $$

(12.48)

After plugging (12.47) and (12.48) into the Rayleigh quotient (12.44) we obtain

$$\displaystyle \begin{aligned} \lambda_2(\Gamma')\leq \frac{\lambda_2(\Gamma)\|\psi_2\|{}^2_{L_2(\Gamma)} + \gamma_2^2 \frac{\pi^2}{2\ell}}{\|\psi_2\|{}_{L_2(\Gamma)}^2 + \ell \gamma_1^2 + \frac{\ell}{2}\gamma_2^2- c^2 \mathcal L'}. \end{aligned}$$

Using (12.46) the last estimate implies

$$\displaystyle \begin{aligned} {} \lambda_2(\Gamma')\leq \frac{\lambda_2(\Gamma)\|\psi_2\|{}^2_{L_2(\Gamma)} + \gamma_2^2 \frac{\pi^2}{2\ell}}{\|\psi_2\|{}_{L_2(\Gamma)}^2 + \ell \gamma_1^2 \left( 1- \frac{\ell}{\mathcal L'} \right) + \frac{\ell}{2}\gamma_2^2} \leq \frac{\lambda_2(\Gamma)\|\psi_2\|{}^2_{L_2(\Gamma)} + \gamma_2^2 \frac{\pi^2}{2\ell}}{\|\psi_2\|{}_{L_2(\Gamma)}^2 + \frac{\ell}{2}\gamma_2^2} , \end{aligned} $$

(12.49)

where we used that $\ell < \mathcal L' = \mathcal L + \ell .$ It remains to take into account the fundamental estimate (12.1)

$$\displaystyle \begin{aligned} \lambda_2 (\Gamma) \geq \left( \frac{\pi}{\mathcal L} \right)^2. \end{aligned}$$

Then taking into account that $\ell > \mathcal L(\Gamma )$ the estimate (12.49) can be written as

$$\displaystyle \begin{aligned} {} \lambda_2(\Gamma') \leq \frac{\lambda_2(\Gamma)\|\psi_2\|{}^2_{L_2(\Gamma)} + \lambda_2 (\Gamma) \gamma_2^2 \ell /2 }{\|\psi_2\|{}_{L_2(\Gamma)}^2 + \gamma_2^2 \ell/2 } = \lambda_2 (\Gamma). \end{aligned} $$

(12.50)

The theorem is proven. □

The fundamental estimate (12.1) was crucial for the proof. It relates the spectral gap and the total length of the metric graph, i.e. geometric and spectral properties of metric graphs. It might be interesting to prove an analogue of the theorem for discrete graphs. Proposition 24.10 states that the spectral gap increases if one edge is added to a discrete graph. Adding a long edge should correspond to adding a chain to a discrete graph.

The above theorem can again be proven using perturbation theory methods. The standard Laplacian $L^{\mathrm {st}} (\Gamma \cup [0,\ell ] )$ has at least four eigenvalues in the interval $[0, \lambda _2 (\Gamma )]$: the double eigenvalue $0 = \lambda _1 (\Gamma ) = \lambda _1 ([0,\ell ])$, $\lambda _2 (\Gamma )$ and $\lambda _2 ([0,\ell ]) = \frac {\pi ^2}{\ell ^2} < \frac {\pi ^2}{\mathcal L^2} \leq \lambda _2 (\Gamma ).$ The difference between the resolvents of $L^{\mathrm {st}} (\Gamma \cup [0,\ell ])$ and $L^{\mathrm {st}} (\Gamma ')$ has rank two, hence (12.50) holds.

The previous theorem gives us a sufficient geometric condition for the spectral gap to decrease. Let us study now the case where the spectral gap is increasing. Similarly, as we proved that adding one edge that is long enough always makes the spectral gap smaller (Theorem 12.15), we claim that an edge that is short enough makes it not to decrease. We have already seen in Theorem 12.9 that adding an edge of zero length (joining two vertices into one) may lead to an increase of the spectral gap. It turns out that the criterion for a gap to decrease can be formulated explicitly in terms of the eigenfunction on the larger graph. Therefore let us change our point of view and study the behaviour of the spectral gap as an edge is deleted.

12.6 Bonus Section: Further Topological Perturbations

12.6.1 Cutting Edges

In the following subsection we are going to study the behaviour of the spectral gap when one of the edges is deleted. The result of such a procedure is not obvious, since deleting an edge decreases the total length of the metric graph and one expects that the first excited eigenvalue increases. On the other hand deleting an edge decreases the graph’s edge connectivity and therefore the spectral gap is expected to decrease. It is easy to construct examples when one of these two tendencies prevails: Example 12.13 shows that the spectral gap may both decrease and increase when an edge is deleted.

Let us discuss first what happens when one of the edges is cut at a certain internal point. Let $\Gamma ^*$ be a connected metric graph obtained from a metric graph $\Gamma$ by cutting one of the edges, say $E_1 = [x_1, x_2]$ at a point $x^* \in (x_1, x_2).$ It will be convenient to denote by $x_1^*$ and $x_2^*$ the points on the two sides of the cut. In other words, the graph $\Gamma ^*$ has precisely the same set of edges and vertices as $\Gamma$ except that the edge $[x_1, x_2 ]$ is substituted by two edges $[x_1, x_1^*]$ and $[ x_2^*, x_2 ]$ and two new vertices $V^{1*} = \{ x_1^*\}$ and $V^{2*} = \{ x_2^* \}$ are added to the set of vertices. It is irrelevant whether the new graph is still connected or not.

One may change point of view and consider the point $x^*$ as a degree two vertex, then Theorem 12.9 can be reformulated as:

Theorem 12.16

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 $V^{2*}.$

• 1. Then the first excited eigenvalues satisfy the following inequality

  $$\displaystyle \begin{aligned} \lambda_2 (\Gamma) \geq \lambda_2 (\Gamma^*). \end{aligned} $$

  (12.51)

• 2. If$\lambda _2 (\Gamma ^*) = \lambda _2 (\Gamma )$then every eigenfunction of$L^{\mathrm {st}}(\Gamma )$corresponding to$\lambda _2 (\Gamma )$satisfies Neumann condition at the cut point$x^*$: $\psi ^{\prime }_1 (x^*) = 0.$If at least one of the eigenfunctions on$\Gamma ^*$satisfies$\psi _2^* (V^{1*}) = \psi _2^* (V^{2*}),$then$\lambda _2 (\Gamma ^*) = \lambda _2 (\Gamma ).$

This theorem implies that the spectral gap has a tendency to decrease as an edge is cut in an internal point. Note that the total length of the graph is preserved.

12.6.2 Deleting Edges

Let us study now what happens if an edge is deleted, or if an interval of non-zero length is cut away from an edge (without gluing the remaining sides together). Every point inside an edge can be seen as a degree two vertex, hence it is enough to study what happens if an edge is deleted.

The following theorem proves a sufficient condition that guarantees that the spectral gap is decreasing as one of the edges is deleted.

Theorem 12.17

Let $\Gamma$ be a connected finite compact metric graph of the total length $\mathcal L$ and let $\Gamma ^*$ be another connected metric graph obtained from $\Gamma$ by deleting one edge of length $\ell$ between certain vertices $V^1$ and $V^2$ . Assume in addition that

$$\displaystyle \begin{aligned} {} \left( \max_{\psi_2: L^{\mathrm{st}} (\Gamma) \psi_2 = \lambda_2 \psi_2 } \frac{(\psi_2 (V^1) - \psi_2 (V^2))^2}{(\psi_2 (V^1) + \psi_2 (V^2))^2} \cot^2 \frac{k_2 \ell}{2} - 1 \right) \frac{k_2}{2} \cot \frac{k_2 \ell}{2} \geq (\mathcal L - \ell)^{-1}, \end{aligned} $$

(12.52)

where $\lambda _2 (\Gamma ) = k_2^2, \; k_2 > 0 ,$ is the first excited eigenvalue of $L^{\mathrm {st}} (\Gamma )$ , then

$$\displaystyle \begin{aligned} \lambda_2 (\Gamma) \geq \lambda_2 (\Gamma^*). \end{aligned} $$

(12.53)

The inequality holds even in the special case where there exists a function $\psi _2$ with $\psi _2 (V^1 ) = - \psi _2 (V^2) ,$ provided $\ell < \pi /k_2.$

Proof

It will be convenient to denote the edge to be deleted by $E = \Gamma \setminus \Gamma ^*$ as well as to introduce notation $\mathcal L^* = \mathcal L-\ell$ for the total length of $\Gamma ^*.$

Let us consider any eigenfunction $\psi _2$ on $\Gamma$ corresponding to the eigenvalue $\lambda _2(\Gamma )$. We then define the function $g \in \stackrel {c}{W^1_2}(\Gamma ^*)$ by

$$\displaystyle \begin{aligned} g = \psi_2|{}_{\Gamma^*} + c, \end{aligned}$$

where the constant c is to be adjusted so that g has mean value zero on $\Gamma ^*$:

$$\displaystyle \begin{aligned} {} \langle g,1\rangle_{L_2(\Gamma^*)} = 0. \end{aligned} $$

(12.54)

Straightforward calculations lead to

$$\displaystyle \begin{aligned} {} 0 = \langle \psi_2, 1 \rangle_{L_2(\Gamma^*)} + c \mathcal L^* = - \langle \psi_2, 1 \rangle_{L_2(E)} + c \mathcal L^* \Rightarrow c = \frac{\int_E \psi_2 (x) dx}{\mathcal L^* }. \end{aligned} $$

(12.55)

The function $g$ can be used to estimate the first excited eigenvalue $\lambda _2(\Gamma ^*)$:

$$\displaystyle \begin{aligned} {} \lambda_2(\Gamma^*) = \min_{u \in \stackrel{c}{W^1_2}(\Gamma^*): u\perp 1} \frac{\|u'\|{}^2_{L_2(\Gamma^*)}}{\|u\|{}^2_{L_2(\Gamma^*)}} \leq \frac{\|g'\|{}^2_{L_2(\Gamma^*)}}{\|g\|{}^2_{L_2(\Gamma^*)}}. \end{aligned} $$

(12.56)

Bearing in mind that $\langle \psi _2,1 \rangle _{L_2(\Gamma )} = 0$ and using (12.55) we evaluate the denominator in (12.56) first:

$$\displaystyle \begin{aligned} {} \|g\|{}^2_{L_2(\Gamma^*)} &= \|\psi_2 + c \|{}^2_{L_2(\Gamma^*)} = \int_{\Gamma}{(\psi_2 + c)^2}\: dx - \int_{E}{(\psi_2 + c)^2} \: dx \\ & = \|\psi_2\|{}^2_{L_2(\Gamma)} - \int_E {\psi_2}^2 \: dx - \frac{1}{\mathcal L^*} \left(\int_E \psi_2 \: dx \right)^2. \end{aligned} $$

(12.57)

The numerator similarly yields

$$\displaystyle \begin{aligned} {} \|g'\|{}^2_{L_2(\Gamma^*)} = \int_{\Gamma}{(\psi_2^{\prime})^2} \: dx - \int_{E} (\psi_2^{\prime})^2 \: dx = \lambda_2(\Gamma)\|\psi_2\|{}^2_{L_2(\Gamma)} - \int_{E} (\psi_2^{\prime})^2 \: dx. \end{aligned} $$

(12.58)

Plugging (12.57) and (12.58) into (12.56) we arrive at

$$\displaystyle \begin{aligned} {} \lambda_2(\Gamma^*) \leq \frac{\lambda_2(\Gamma)\|\psi_2\|{}^2_{L_2(\Gamma)} - \int_{E} (\psi_2^{\prime})^2 \: dx}{\|\psi_2\|{}^2_{L_2(\Gamma)} - \int_E {\psi_2}^2 \: dx - \frac{1}{\mathcal L^*} \left(\int_E \psi_2 \: dx \right)^2}. \end{aligned} $$

(12.59)

Let us evaluate the integrals appearing in (12.59) taking into account that $\psi _2$ is a solution to Eq. (2.30) on the edge $E$ which can be parameterized as $E = [-\ell /2, \ell /2]$ so that $x = - \ell /2$ belongs to $V^1$ and $x = \ell /2$ to $V^2$

$$\displaystyle \begin{aligned} {} \psi_2|{}_E(x) = \alpha \sin{(k_2 x)} + \beta \cos{(k_2 x)}, \end{aligned} $$

(12.60)

where

$$\displaystyle \begin{aligned} {} \alpha = - \frac{\psi_2(V^1)-\psi_2(V^2)}{2 \sin{}(k_2 \ell /2)}, \quad \beta = \frac{\psi_2(V^1)+\psi_2(V^2)}{2 \cos{(k_2 \ell/2)}}.\end{aligned} $$

(12.61)

Direct calculations imply

$$\displaystyle \begin{aligned} \begin{array}{lcl} \displaystyle \int_E \psi_2(x) dx & = & \displaystyle \frac{2 \beta}{k_2} \sin \left(\frac{k_2 \ell}{2} \right); \\[2mm] \displaystyle \int_E (\psi_2(x))^2 dx & = & \displaystyle \frac{\alpha^2 + \beta^2}{2} \ell - \frac{\alpha^2 - \beta^2}{2} \frac{\sin (k_2 \ell)}{k_2} ; \\[2mm] \displaystyle \int_E (\psi_2^{\prime} (x))^2 dx & = & \displaystyle k_2^2 \left( \frac{\alpha^2 + \beta^2}{2} \ell + \frac{\alpha^2-\beta^2}{2} \frac{\sin ( k_2 \ell)}{k_2} \right). \end{array}\end{aligned}$$

Inserting calculated values into (12.59) we get

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} &&\displaystyle \lambda_2 (\Gamma^*) \leq \displaystyle \lambda_2 (\Gamma)\\ & & \\ && \quad \times \displaystyle \frac{\displaystyle \parallel \psi_2 \parallel^2_{L_2(\Gamma)} - \frac{\alpha^2+\beta^2}{2} \ell - \frac{\alpha^2 - \beta^2}{2} \frac{\sin (k_2 \ell)}{k_2}} {\displaystyle \parallel \psi_2 \parallel^2_{L_2(\Gamma)} - \frac{\alpha^2+\beta^2}{2} \ell + \frac{\alpha^2 - \beta^2}{2} \frac{\sin (k_2 \ell)}{k_2} - \frac{1}{\mathcal L^*} \frac{4 \beta^2}{\lambda_2 (\Gamma)} \sin^2 \left(\frac{k_2 \ell}{2} \right)} . \end{array} \end{aligned} $$

(12.62)

To guarantee that the quotient is not greater than $1$ and therefore $\lambda _2 (\Gamma ^*) \leq \lambda _2 (\Gamma )$ it is enough that

$$\displaystyle \begin{aligned} \frac{\alpha^2 - \beta^2}{2} \frac{\sin (k_2 \ell)}{k_2} \geq - \frac{\alpha^2 - \beta^2}{2} \frac{\sin (k_2 \ell)}{k_2} + \frac{1}{\mathcal L^*} \frac{4 \beta^2}{\lambda_2 (\Gamma) } \sin^2 \left(\frac{k_2 \ell}{2} \right)\end{aligned}$$

$$\displaystyle \begin{aligned} {} \Longleftrightarrow \; \; \frac{k_2}{2} \left( \frac{\alpha^2}{\beta^2} -1 \right) \cot \left(\frac{k_2 \ell}{2}\right) \geq (\mathcal L^*)^{-1}. \end{aligned} $$

(12.63)

Using (12.61) the last inequality can be written as

$$\displaystyle \begin{aligned} \left( \frac{(\psi_2 (V^1) - \psi_2 (V^2))^2}{(\psi_2 (V^1) + \psi_2 (V^2))^2} \cot^2 \left(\frac{k_2 \ell}{2}\right) - 1 \right) \frac{k_2}{2} \cot \left(\frac{k_2 \ell}{2}\right) \geq (\mathcal L^*)^{-1}. \end{aligned}$$

Remembering that the eigenfunction $\psi _2$ could be chosen arbitrary we arrive at (12.52).

It remains to study the special case where $\psi _2 (V^1) = - \psi _2 (V^2).$ It follows that $\beta = 0$ and

$$\displaystyle \begin{aligned} \alpha = - \frac{\psi_2(V^1)}{\sin k_2 \ell/2} .\end{aligned}$$

Instead of (12.62) we arrive at the following inequality

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \displaystyle \lambda_2 (\Gamma^*) & \leq & \displaystyle \lambda_2 (\Gamma) \frac{\displaystyle \parallel \psi_2 \parallel^2_{L_2(\Gamma)} - \frac{\alpha^2}{2} \ell - \frac{\alpha^2}{2} \frac{\sin (k_2 \ell)}{k_2}} {\displaystyle \parallel \psi_2 \parallel^2_{L_2(\Gamma)} - \frac{\alpha^2}{2} \ell + \frac{\alpha^2 }{2} \frac{\sin (k_2 \ell)}{k_2} } . \end{array} \end{aligned} $$

(12.64)

The quotient is always less than $1$ provided $\sin k_2 \ell > 0$, which is true if $\ell < \pi /k_2.$ □

Roughly speaking, condition (12.52) means that the length $\ell$ is sufficiently small, of course provided $\psi _2 (V^1) \neq \psi _2 (V^2).$ Indeed, for small $\ell$ the cotangent term is of order $1/\ell$. Therefore the left-hand side of (12.52) is of order $1/\ell ^3$ and thus growing to infinity as $\ell$ decreases, while the right-hand side remains bounded.

Let us apply the above theorem to obtain an estimate for the length of the piece that can be cut from an edge so that the spectral gap still decreases. Consider any edge in $\Gamma$, say $E_1 = [x_1, x_2]$ and choose an arbitrary internal point $x^* \in (x_1, x_2).$ Assume that we cut away an interval of length $\ell$ centred at $x^*.$ Of course the length $\ell$ should satisfy the obvious geometric condition: $x_1 \leq x^*- \ell /2$ and $x^* + \ell /2 \leq x_2.$ We assume in addition that

$$\displaystyle \begin{aligned} {} \ell < \frac{\pi}{2 k_2} \end{aligned} $$

(12.65)

guaranteeing in particular that the cotangent function in (12.52) is positive.

The function $\psi _2$ on the edge $E_1$ can be written in a form similar to (12.60)

$$\displaystyle \begin{aligned} \psi_2 (x) = \alpha \sin k_2 (x-x^*) + \beta \cos k_2 (x-x^*). \end{aligned}$$

Then formula (12.63) implies that the spectral gap decreases as the interval is cut away if

$$\displaystyle \begin{aligned} {} \vert \alpha \vert > \vert \beta \vert. \end{aligned} $$

(12.66)

and the following estimate is satisfied

$$\displaystyle \begin{aligned} \cot \left( \frac{k_2 \ell}{2} \right) \geq \frac{2}{k_2 \mathcal L^* \left( \frac{\alpha^2}{\beta ^2} -1 \right)}. \end{aligned} $$

(12.67)

Condition (12.66) means that the eigenfunction does not satisfy Neumann condition at $x^*.$ This condition was expected, since if $\psi _2$ is symmetric with respect to $x^*$, then the spectral gap may increase for any $\ell .$ Really, one may imagine that deleting the interval is performed in two steps. One cuts the edge $E_1$ at the point $x^*$ first. Then one deletes the intervals $[x^*-\ell /2, x^*_1]$ and $[x^*_2, x^* + \ell /2]$. If $\alpha = 0$ (symmetric function), then the spectral gap may be preserved in accordance to Theorem 12.16. Deleting the pendant edges (intervals $[x^*-\ell /2, x^*_1]$ and $[x^*_2, x^* + \ell /2]$) always increases the spectral gap due to Theorem 12.12.

Using the fact that under condition (12.65) we have $\cot \left ( \frac {k_2 \ell }{2} \right ) \geq \frac {2}{ k_2 \ell }$ the following explicit estimate on $\ell$ can be obtained

$$\displaystyle \begin{aligned} {} \ell \leq (\mathcal L - \ell) \left( \frac{\alpha^2}{\beta ^2} -1 \right) \Rightarrow \ell \leq \frac{\displaystyle \left( \frac{\alpha^2}{\beta ^2} -1 \right) \mathcal L}{ \displaystyle 1+ \left( \frac{\alpha^2}{\beta ^2} -1 \right)}, \end{aligned} $$

(12.68)

of course under condition (12.66). For the spectral gap not to increase it is enough that estimate (12.68) is satisfied for at least one eigenfunction $\psi _2$:

$$\displaystyle \begin{aligned} \ell \leq \min \left\{ \frac{\pi}{2 k_2}, \; \; \max_{\psi_2: L^{\mathrm{st}}(\Gamma) \psi_2 = \lambda_2 \psi_2} \frac{ \left( \frac{\alpha^2}{\beta ^2} -1 \right) \mathcal L}{ 1+ \left( \frac{\alpha^2}{\beta ^2} -1 \right)} \right\}, \end{aligned} $$

(12.69)

where we have taken into account (12.65).

We see that if the eigenfunction $\psi _2$ is asymmetric with respect to the point $x^*$ (i.e. (12.66) is satisfied), then a certain sufficiently small interval can be cut from the edge ensuring that the spectral gap decreases despite the total length decreases.

We have shown that deleting not so long edges or cutting away short intervals from the edges may lead to a decrease of the spectral gap despite the fact that total length of the graph decreases. One can see an analogy between these results and the phenomenon observed in [193], where the behaviour of the spectral gap under extension of edges was discussed for graphs with delta couplings at the vertices. It was shown that the lowest eigenvalue may increase when the edge lengths also increase, provided the ground state has certain special properties.

Problem 54

Consider the complete graph $K_M$ with $M$ vertices connected by $\frac {M(M-1)}{2}$ edges of equal length. What happens to the spectral gap if

• One of the edges is cut at the middle;

• One of the edges is deleted;

• One of the vertices is chopped into two;

• Two vertices are glued together.

Problem 55

Consider the flower graph depicted in Fig. 12.4. Describe the behaviour of the spectral gap if

• The central vertex is chopped into two;

• One of the edges is cut at the middle;

• One of the edges is deleted.

Consider both the cases where the edges have equal and different lengths.

Four different approaches to get spectral estimates have been described in the current chapter:

• Eulerian path technique,

• symmetrisation technique;

• Cheeger approach;

• topological perturbations.

The main ingredient of all proofs are estimates on the quadratic forms, hence it is straightforward to take care of higher eigenvalues. In many cases the same proofs may be applied. Taking into account not only standard vertex conditions may lead to certain difficulties. For example, symmetrisation technique and the Cheeger approach require that the functions from the quadratic form domain are continuous, which is the case for standard and delta vertex conditions. Only approaches based on topological perturbations (and Eulerian path technique, which uses topological perturbations) can be extended to arbitrary vertex conditions. We discuss this direction of research in the following chapter. Starting from special classes of scaling-invariant conditions we accomplish our studies including most general vertex conditions.