Quantum graphs and spectra of flower graphs

Daniel Blixt

September 25, 2015

FysikumDegree 15 HE creditsPhysicsBachelor program in physics (180credits)Spring term 2015Examinators Andreas Rydh andFawad HassanSupervisor Pavel KurasovAssistant supervisor IngemarBengtsson

 

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Abstract

This thesis treats quantum graphs, more precisely, spectra of flower graphs.An equation on the spectrum of such a graph is derived. This equationcontain the number of edges and their lengths as parameters. General so-lutions to the equation cannot be expressed in elementary functions andgets more and more complicated form when the number of edges increases,or when their lengths are different.

Eigenvalues of the flower graph with two edges can be considered asa function of the lengths of the edges and the eight first eigenvalues areordered and plotted. The first three nonzero eigenvalues for a flower graphwith three edges are plotted as well. Provided that the total length of theflower graph’s three edges is fixed, it is proven that the first and the thirdnonzero eigenvalues are maximal when a flower graph of fixed number ofedges and total length is equilateral.

Comparison with a more abstract theorem is made, which proves thatthe quantum graph with a given total length and number of edges, max-imizes the first nonzero eigenvalue when the quantum graph is an equi-lateral flower graph. However, this theorem gives no information abouthigher eigenvalues. Another advantage of our proof is its concreteness,which make it easier to analyze concrete examples. Spectra of some othergraphs are analyzed and compared with the spectra of flower graphs.

KeywordsQuantum graph, Flower graph, Spectrum, Equilateral graph

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Contents

1 Theoretical background 41.1 Metric graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Quantum graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Vertex conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Spectra of quantum graphs . . . . . . . . . . . . . . . . . . . . . 71.5 Sobolev space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Concrete examples 82.1 8-graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Flower graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Equilateral flower graph . . . . . . . . . . . . . . . . . . . . . . . 172.4 Comparison to the star graph . . . . . . . . . . . . . . . . . . . . 182.5 Three petal flower graph . . . . . . . . . . . . . . . . . . . . . . . 212.6 Proposition regarding spectral gap of three petal flower graph . . 282.7 General theorem regarding first nonzero eigenvalue . . . . . . . . 30

3 Conclusion 31

4 References 32

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1 Theoretical background

Many problems in mathematical physics are described by partial differentialequations in R3; we have the Schrodinger equation in quantum physics, waveequation in acoustics, etc.

When the dynamics is localized to a neighborhood of a one-dimensional ob-ject one may, to describe the corresponding system, approximate the waves topropagate in one spatial variable. The corresponding differential equations areusually simple to solve. The mathematical model, called quantum graph, de-scribes these types of phenomena.

Consider a set of one-dimensional intervals joined together at certain end-points. Such object is called metric graph in mathematics. On each interval weconsider a stationary Schrodinger equation. In addition we need vertex condi-tions describing how different waves go from one interval to another.

There are several applications to this model. An example is carbon nan-otubes which are structures formed by rolling up graphene sheet into a cylinder.[7].Electrons travel along the graphene sheet and when the length is much greaterthan the diameter one may approximate the propagation to one spatial variableand hence use the quantum graph description. Nanotubes have been constructedwith length-to-diameter ratio of up to 132,000,000:1[1]. An illustration of a nan-otube is presented in Figure 1.

The goal of my thesis is two-fold: on one side I wanted to learn more aboutquantum graphs - an area of modern research on the border between mathemat-ics and physics. It is fascinating to see how spectral properties of such objectsdepend on their geometry. I was interested in understanding how some estimatesfor eigenvalues can be proven. I realized that in simple cases these estimatescan be proven using methods from mathematical analysis we have learned inone of the basic courses. In this way alternative proof that equilateral (i.e. withedges having the same length) flower graphs possess extremal spectral proper-ties was found. This was done for graphs formed by two and three edges, butthe corresponding proof is not only explicit, but also illustrative. Connectionsbetween spectra of flower and star graphs are also investigated.

In section 1 all concepts needed to understand the content of this thesisare rigorously defined. In section 2 concrete examples are presented and theo-rems related to these examples are stated and proven. Section 3 presents myconclusion and references are given in section 4.

1.1 Metric graphs

In this thesis I will only consider compact finite graphs. Therefore I skip theparts when we have edges of infinite length and graphs with infinitely manyvertices. A metric graph is defined as a set of edges denoted E and a set ofvertices denoted V. Each edge e ∈ E is defined as an interval on the real axisand has a certain length l(e). The edges are parametrized with x taking itssmallest value at the left endpoint of the edge, respectively its largest valueat the right endpoint of the edge. Between these endpoints x simply takesintermediate values.

The vertices are defined using the set of endpoints of the edges. The set ofendpoints is denoted by V = {xj}, j = 1, 2, ...., 2n, where n denote the numberof edges. We divide the set V into m equivalence classes such that

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Figure 1: Illustraion of a nanotube[7]

V = V1 ∪ V2 ∪ ... ∪ VM ,Vi ∩ Vj = ∅, provided i 6= j.

Whenever I refer to a vertex, I refer to one of the equivalence classes definedabove.

1.2 Quantum graphs

A quantum graph is a metric graph Γ with a differential expression on theedges and matching conditions at the vertices. It is defined as a self-adjoint(symmetric) operator in the Hilbert space L2(Γ) (with functions u(x) ∈ L2(Γ))consisting of all square integrable functions. The most common differentialexpressions are the following

− d2

dx2 , Laplace operator,

− d2

dx2 + q(x), Schodinger operator

and (−i d

dx + a(x))2

+ q(x), magnetic Schrodinger operator,

where q(x) ∈ R,∀x is the electrical potential and a(x) ∈ R and continuous ∀xis the magnetic potential.

1.3 Vertex conditions

In order to be able to apply the differential operator to a function u(x) weneed to require that it is from the Sobolev space (see section 1.5). Consider adifferential operator O. In order to get 〈Ou, v〉− 〈u,Ov〉 = 0 (here 〈·, ·〉 denotesthe inner product) for function u, v ∈ D(O), i.e O is symmetric, we need to takevertex conditions into consideration. The vertex conditions connect different

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edges together, whereas the original differential equation is independent of howthe edges are attached to each other. In this thesis we will consider standardvertex conditions. Consider a function u(x) on the graph Γ. At a vertex Vmthe function can be evaluated as lim

x 7→xju(x) := u(Vm), where xj ∈ Vm. This

condition implies continuity which is one of the standard conditions.Normal derivatives are defined as

∂nu(xj) =

limx 7→xj

ddxu(x), xj is the left endpoint,

− limx 7→xj

ddxu(x), xj is the right endpoint.

The second vertex condition is that the sum of all normal derivatives is zero ateach vertex.So the standard conditions are the followingu(x) is continuous at each vertex Vm, m = 1, 2, ...,M∑

xj∈Vm∂u(xj) = 0. (1)

The second condition in (1) is called Neumann condition.Consider the Schrodinger operator and note that the Laplace operator is a

special case of the Schrodinger operator S = − d2

dx2 +q(x), namely when q(x) ≡ 0.An operator is symmetric if and only if 〈Su, v〉 − 〈u,Sv〉 = 0. We now expand

〈Su, v〉 − 〈u,Sv〉 on the graph Γ with edges En, whereN⋃n=1

En = Γ and vertices

as defined above:

〈Su, v〉 − 〈u,Sv〉 =N∑n=1

∫En

([(− d2

dx2 + q(x))u(x)

]v(x)−

N∑n=1

∫En

u(x)[(− d2

dx2 + q(x))v(x)

])dx

=N∑n=1

∫En

(−u′′(x)v(x) + q(x)u(x)v(x) + u(x)v′′(x)− u(x)q(x)v(x)

)dx.

Since q(x) ∈ R, q(x) = q(x) and commutes with u(x) and v(x), so the expressionreduces to

N∑n=1

∫En

(−u′′(x)v(x) + u(x)v′′(x)

)dx.

Now we use partial integration. Note that the limits of En are the left and rightendpoint, which are the vertices in the graph Γ. We use normal derivatives sincewhen taking the upper limit subtracted with the lower limit will result in a sumof equal signs.

M∑m=1

∑xj∈Vm

[∂nu(xj)v(xj)

]−

M∑m=1

∑xj∈Vm

[u(xj)∂nv(xj)

]+

+N∑n=1

∫En

(u′(x)v′(x)− u′(x)v′(x)

)dx

=M∑m=1

( ∑xj∈Vm

[∂nu(xj)v(xj)

])−

M∑m=1

( ∑xj∈Vm

[u(xj)∂nv(xj)

]).

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Requiring continuity condition at the vertices Vm makes it possible to rewritethe expression as

M∑m=1

v(Vm)∑

xj∈Vm

(∂nu(xj)

)−∑Mm=1 u(Vm)

∑xj∈Vm

(∂nv(xj)).

By requiring Neumann condition we see that 〈Su, v〉 − 〈u,Sv〉 = 0, i.e theSchrodinger operator symmetric, which implies that also the Laplace operatoris symmetric under these conditions. One may also show that the operators arenot only symmetric but also self-adjoint.

Observe that it is also possible to imply v(Vm) = u(Vm) = 0, which is calledthe Dirichlet condition, though not treated in this thesis.

1.4 Spectra of quantum graphs

Consider operators, associated to the quantum graphs, acting in the Hilbertspace L2(Γ). Let O be a linear operator in L2(Γ) with domain D(O).

Definition 1 The spectrum σ(O) ⊂ C consists of all λ ∈ C such that theoperator O − λIL2(Γ) does not have a bounded inverse.

In this thesis we will analyze the spectrum of the Laplace operator L for flower

graphs (see section 2.2). The operator − d2

dx2 −λIL2(Γ) does not have a boundedinverse if λ is an eigenvalue, i.e. if ∃ u ∈ D(L) such that:

d2

dx2u(x) + λu(x) = 0. (2)

Note that equation (2) is equivalent to the time independent Schrodninger equa-tion in one dimension for a free particle

− d2

dx2ψ(x) = Eψ(x).

Though in our examples the particle is not free. The particle will be confinedto a certain geometry described by the quantum graph. Still we will not ingeneral get the same solution since there will be different boundary conditionsin consideration when dealing with quantum graphs (see section 1.3).Solutions to equation (2) looks better if one introduce k2 = λ, then equation(2) becomes

d2

dx2u(x) + k2u(x) = 0. (3)

Solutions to equation (3) when λ > 0 are

u(x) = A sin(kx) +B cos(kx), (4)

when λ = 0 we getu(x) = Ax+B (5)

and when λ < 0 we getu(x) = Aeκx +Be−κx, (6)

where κ2 = −λ.Hence we know that the eigenfunctions takes the form of equation (4), (5)

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and (6). Taking into account vertex conditions makes it possible to derive howthe eigenfunctions looks like at the quantum graph in consideration. Note thatwhen applying standard vertex conditions to any metric graph, one always findsu0(x) = B := C, which is a constant function on the entire graph.

Since u0(x) takes the same value (C) on every point on the metric graphit satisfies the continuity condition. Neumann condition is also satisfied since∂nu0(x) = 0,∀x ∈ Γ. Hence

∑xj∈Vm ∂nu(xj) = 0 and Neumann condition is

indeed satisfied. Note that u0(x) = C is a special solution to equation (5), henceλ0 = 0 is always an eigenvalue to any metric graph with the Laplace operatorand standard conditions as vertex conditions.

1.5 Sobolev space

Recall that in our setting the Hilbert space is the space of square integrablefunctions,

L2[a, b] =

{f(x) :

b∫a

|f(x)|2dx <∞

}.

In order to prove Lemma 1 and Theorem 2 one need to restrict the domainfurther and consider the Sobolev space. In order to define the Sobolev spaceit is necessary to first define weak or generalized derivative.

Definition 2 Consider any test function which can be differentiated infinitelymany times ψ ∈ C∞0 and has compact support inside the interval ]a, b[. Let fbe an integrable function, then we define

f [ψ] :=

b∫a

f(x)ψ(x)dx.

The weak or generalized derivative Df of the function f is then defined as

Df [ψ] := −f [ψ′].

In this thesis we restrict to the Sobolev space denoted W k2 [a, b], k = 1, 2. The

lower limit denotes a space of square integrable functions and the upper limitdenotes that the space also requires that the weak derivative of first order (k = 1)and second order (k = 2) of such functions need to be square integrable.

Definition 3 The Sobolev space W k2 [a, b] mentioned above is defined as

W k2 [a, b] :=

f :

b∫a

|f(x)|2dx <∞,b∫a

|Dkf(x)|2dx <∞

.

2 Concrete examples

In this section concrete examples are treated. The spectrum for the 8-graphconsisting of one vertex, two loops and total length L = 1 is calculated anddisplayed in Figure 3. Equation (13) which all graphs with one vertex and alledges being loops (flower graphs) must satisfy is derived. This equation is solved

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for the case when all edges are at equal length (an equilateral graph). Thesesolutions are compared with the spectrum of an equilateral star graph (all outervertices have degree one and all edges are connected to the unique inner vertex)with the same number of edges.

The rest of the section is devoted to the graph consisting of one vertex andthree loops and a fixed total length L = 1. Plots of the first three nonzeroeigenvalues as a function of two of the edges lengths are presented. Then apreposition stating that the first and third nonzero eigenvalues are maximalwhen the graph described above is equilateral is proven. Finally a theoremregarding the first eigenvalue of any quantum graph is stated and proven.

2.1 8-graph

The 8-graph is the graph consisting of one vertex with two loops of lengths l1respectively l2 and is displayed in Figure 2. The 8-graph is a special case ofthe flower graph, namely the one with two petals (see section 2.2). Consider an8-graph with constant total length which we set to one (L = l1 + l2 = 1). Thiswill make calculations easier and one can simply choose a convenient unit sothat L = 1 for the application in consideration. When L = 1, then l2 = 1 − l1is uniquely determined by l1.

Figure 2: 8-graph

The edge with length l1 is parametrized with the left endpoint at x1 = 0 inFigure 2 and right endpoint at x2 = l1. The edge with length l2 is similarlyparametrized with x3 as left endpoint and x4 as right endpoint.The eigenfunction will have one component for each edge which I denote byu1(x) at the edge with length l1 respectively u2(x) at the edge with length l2.In order to calculate the spectrum for the 8-graph we apply the standard vertexconditions to the eigenfunctions. When λ = 0 the eigenvalues are at the form

u1(x) = A1x+B1

u2(x) = A2x+B2

and the vertex conditions imply{B1 = A1l1 +B1 = B2 = A2l2 +B2

A1 +A2 −A1 −A2 = 0.

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The Neumann condition is automatically fulfilled whilst the continuity conditionimply that {

B1 = B2 := C

A1 = A2 = 0.

The eigenfunction to λ = 0 hence corresponds to

u1(x) = u2(x) = C.

This means that λ = 0 is an eigenvalue of the 8-graph and has multiplicity one,with eigenfunction being the constant function.When λ > 0 the solution to the differential equation look like:

u1 = A1sin(kx) +B1 cos(kx),u2 = A2 sin(kx) +B2 cos(kx).

After taking the standard vertex conditions into account one sees that the eigen-functions are continuous and that the normal derivatives in the vertex add upto zero one arrives at the following equation system

B1 = B2 = B

A1 +B1 (cos(kl1)− 1) = 0

A2 sin(kl2) +B2(cos(kl2)− 1) = 0

A1(1− cos(kl1)) +A2(1− cos(kl2)) +B1 sin(kl1) +B2 sin kl2 = 0.

This linear equation system has a nonzero solution when the following equationis satisfied ∥∥∥∥∥∥

sin(kl1) 0 cos(kl1)− 10 sin(kl2) cos(kl2)− 1

1− cos(kl1) 1− cos(kl2) sin kl1 + sin kl2

∥∥∥∥∥∥ = 0. (7)

After elementary manipulations (7) becomes

sin

(kl12

)sin

(kl22

)sin

(kL

2

)= 0. (8)

Remember that it has been assumed that L = 1. We thus obtain the followingpossible values for k

k = n · 2πl1,

k = m · 2πl2,

k = p · 2πL ,

where n,m and p ∈ 1, 2, 3, ... Which implies that we can obtain the possibleeigenvalues

λ = n2 · 4π2

l21,

λ = m2 · 4π2

l22,

λ = p2 · 4π2

L2 .

The eigenvalues as a function of l1 is shown in Figure 3.When λ < 0 the eigenfunctions are at the form v1(x) = a1e

κx+ b1e−κx, v2(x) =

a2eκx + b2e

−κx (Here κ ∈ R is used instead of k and v instead of u in order

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Figure 3: Eigenvalues for 8-graph divided with 4π2, l2 = 1− l1

to avoid confusion comparing with the case λ > 0). In the case with λ > 0we could have chosen to write the eigenfunction in exponential form instead oftrigonometric form. Equation (8) in exponential form is(

eikl12 − e−i

kl12

2i

)·

(eikl22 − e−i

kl22

2i

)·

(eikL2 − e−i kL2

2i

)= 0. (9)

In the derivation of equation (8) we used u1(x) = A1 sin(kx) + B1 cos(kx),u2(x) = A2 sin(kx) +B2 cos(kx). We could have instead chosen A1, A2, B1 andB2 differently (since they are arbitrary constants) so that we used u′1(x) =A′1e

ikx + B′1e−ikx, u′2(x) = A′2e

ikx + B′2e−ikx. Substituting A′1 = a1, A

′2 =

a2, B′1 = b1, B

′2 = b2 and k = −iκ yields u′1(x) = v1(x) and u′2(x) = v2(x).

Note also that equation (9) can be derived from both the exponential form andthe trigonometric form. Substituting k = −iκ in equation (9) and doing someelementary operations yields(

eκl12 − e−

κl12

)·(eκl22 − e

κl22

)·(eκL2 − eκL2

)= 0, (10)

which has no solution since κ 6= 0 and κ ∈ R. Hence there are no solutionsλ < 0 and those found by equation (8) together with λ0 = 0 are all eigenvaluesto the 8-graph.

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2.2 Flower graphs

A flower graph is defined as a graph with only one vertex and all edges beingloops starting and ending at this vertex. An arbitrary flower graph will haven loops with lengths l1, l2, ..., ln. The 8-graph displayed in Figure 2 is a flowergraph and another example of a flower graph is displayed in Figure 4.

Figure 4: Flower graph with four petals

A general quantum graph can be transformed into a flower by gluing togetherall vertices. By fusing all vertices while preserving the length of all edges turnsany compact quantum graph into a flower graph. This makes the flower graphsextra interesting. More detailed studies about how flower graphs relates to thestar graphs, which describes all quantum graphs locally is described in section2.4. The following theorem helps us to find the eigenvalues faster.

Theorem 1 The eigenvalues λ to a quantum graph Γ with standard conditions

and the Laplace operator L = − d2

dx2 are always non-negative λ ≥ 0.

Proof: Consider the quadratic form

〈Lu, u〉 =N∑n=1

∫En

(−u′′(x) · u(x)

)dx.

Then using partial integration and that the sum of boundary terms vanishesdue to the standard conditions (see section 1.3) one gets∑N

n=1

∫En

|u′(x)|2 ≥ 0.

So the quadratic form of the Laplace operator on a graph Γ with standardconditions is positive semidefinite and hence its eigenvalues are always non-negative, λ ≥ 0.

�

We have already seen that the 8-graph has no negative solutions, which is pre-dicted by this theorem. Due to this theorem we no longer need to consider thecase λ < 0 since there are no such solutions.

When λ = 0 the function

u1(x) = u2(x) = ... = un(x) = C

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is always an eigenfunction. Hence we can note that λ = 0 always belong to thespectrum of any quantum graph with standard vertex condition. Although themultiplicity might be more than one. Eigenfunctions are at the form

ui(x) = Aix+Bi,

where i = 1, 2, 3, ..., n. The continuity condition implies

Bi = Aili +Bi.

Hence Ai = 0,∀i ∈ 1, 2, 3, ..., n. Continuity condition also implies

B1 = B2 = ... = Bn := C.

Hence the multiplicity for λ0 = 0 is one.The eigenfunctions for a general flower graph with n petals and λ > 0 are

u1 = A1 sin kx+B1 cos(kx)u2 = A2 sin(kx) +B2 cos(kx)

···

un = An sin(kx) +Bn cos(kx)

By applying the standard vertex condition on u(x) one gets the following equa-tion system

B1 = B2 = ... = Bn := B

A1 sin kl1 +B(cos(kl1)− 1) = 0

A2 sin(kl2) +B(cos(kl2)− 1) = 0...

An sin(kln) +B(cos(kln)− 1) = 0

kn∑i=1

Ai(1− cos(kli)) + k ·Bn∑i=1

sin(kli) = 0.

This system has a unique solution if and only if equation (11) is satisfied.

0 =∥∥∥∥∥∥∥∥∥∥∥∥∥∥∥∥

sin(kl1) 0 0 0 0 0 cos(kl1)− 10 sin(kl2) 0 0 0 0 cos(kl2)− 10 0 · 0 0 0 ·0 0 0 · 0 0 ·0 0 0 0 · 0 ·0 0 0 0 0 sin(kln) cos(kln)− 1

k(1− cos(kl1)) k(1− cos(kl2)) · · · k(1− cos(kln)) kn∑i=1

sin(kli)

∥∥∥∥∥∥∥∥∥∥∥∥∥∥∥∥.

(11)

13

If we calculate the determinant by column operation to the very right columnand factor out k from the last row we get

0 =

k · (−1)n(cos(kl1)− 1)·

·

∥∥∥∥∥∥∥∥∥∥0 sin(kl2) 0 0 0 00 0 . 0 0 00 0 0 . 0 00 0 0 0 . 0

cos(kl1)− 1 cos(k2)− 1 . . . cos(kln)− 1

∥∥∥∥∥∥∥∥∥∥+ k · (−1)n+1(cos(kl2 − 1))·

·

∥∥∥∥∥∥∥∥∥∥∥∥

sin(kl1) 0 0 0 0 00 0 sin(kl3) 0 0 00 0 0 . 0 00 0 0 0 . 00 0 0 0 0 .

cos(kl1)− 1 k(cos(k2)− 1) . . . cos(kln)− 1

∥∥∥∥∥∥∥∥∥∥∥∥+ ...+

+ k · (−1)2n−1(cos(kln − 1))·

·

∥∥∥∥∥∥∥∥∥∥∥∥∥∥

sin(kl1) 0 0 0 0 00 sin(kl2) 0 0 0 00 0 . 0 0 00 0 0 . 0 00 0 0 0 . 00 0 0 0 0 sin(kln−1)

cos(kl1)− 1 cos(k2)− 1 . . . cos(kln)− 1

∥∥∥∥∥∥∥∥∥∥∥∥∥∥+ k · (−1)2n

n∑i=1

sin(kli)

·

∥∥∥∥∥∥∥∥∥∥∥∥

sin(kl1) 0 0 0 0 0 00 sin(kl2) 0 0 0 0 00 0 0 . 0 0 00 0 0 0 . 0 00 0 0 0 0 . 00 0 0 0 0 0 sin(kln)

∥∥∥∥∥∥∥∥∥∥∥∥.

The last term is easy to calculate, but for the other n terms we wish to makethe determinants on diagonal form. Hence, we use column operation to the i-th

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term where i = 1, 2, 3, ..., n.

0 =

k · (−1)n(cos(kl1)− 1) · (−1)n−1(1− cos(kl1))·

·

∥∥∥∥∥∥∥∥∥∥sin(kl2) 0 0 0 0

0 . 0 0 00 0 . 0 00 0 0 . 00 0 0 0 sin(kln)

∥∥∥∥∥∥∥∥∥∥+ k · (−1)n+1(cos(kl2)− 1) · (−1)n(1− cos(kl2))·∥∥∥∥∥∥∥∥∥∥∥∥

sin(kl1) 0 0 0 0 00 sin(kl3) 0 0 0 00 0 . 0 0 00 0 0 . 0 00 0 0 0 . 00 0 0 0 0 sin(kln)

∥∥∥∥∥∥∥∥∥∥∥∥+ ...+

+ k · (−1)2n−1(cos(kln)) · (−1)2n−2(1− cos(kln))·

·

∥∥∥∥∥∥∥∥∥∥∥∥

sin(kl1) 0 0 0 0 00 sin(kl2) 0 0 0 00 0 . 0 0 00 0 0 . 0 00 0 0 0 . 00 0 0 0 0 sin(kln−1)

∥∥∥∥∥∥∥∥∥∥∥∥+ k

(n∑i=1

sin(kli)

n∏i=1

sin(kli)

).

Now all determinants are diagonal and we can simplify the equation to

0 = k

n∑i=1

(1− cos(kli))2∏j 6=i

sin(klj)

+ k

(n∑i=1

sin(kli)

n∏i=1

sin(kli)

).

This expression can be rewritten as

0 = k

n∑i=1

(1− cos(kli))2∏j 6=i

sin(klj)

+ k

n∑i=1

sin2(kli)∏j 6=i

sin(klj)

,

which is further simplified to

0 = k

n∑i=1

(sin2(kli) + 1− 2 cos(kli) + cos2(kli))∏j 6=i

sin(klj)

.

By using the trigonometric identity, sin (klj) = 2 sin(klj2

)cos(klj2

)and cos(kli) =

cos2(kli2

)− sin2

(kli2

)we get

0 = k

n∑i=1

2 sin2

(kli2

)∏j 6=i

2 sin

(klj2

)cos

(klj2

) ,

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which can be written as

0 = k · 2n(

n∏i=1

sin

(kli2

))·

n∑i=1

sin

(kli2

)∏j 6=i

cos

(klj2

) (12)

and can be further simplified to

0 =

(n∏i=1

sin

(kl12

))·

n∑i=1

sin

(kli2

)∏j 6=i

cos

(klj2

) , (13)

since it is assumed that k 6= 0.One sees directly that k can take the following values

k = m12πl1

k = m22πl2

...

k = mn2πln,

where m1,m2, ...,mn ∈ 1, 2, 3, ... This implies that we get the eigenvalues

λ =(m1

2πl1

)2

λ =(m2

2πl2

)2

...

λ =(mn

2πln

)2

.

However other eigenvalues except λ0 = 0 cannot be found in general.The first nonzero eigenvalue λ1 can be obtained by minimizing the Rayleigh

quotient subject to the constraint of L2-orthogonality to the eigensubspace cor-responding to λ0 = 0:[6],(see also [4])

λ1(Γ) = inf

∫

Γ|u′(x)|2dx∫

Γ|u(x)|2dx

: u ∈W 12 (Γ),

∫Γ

u(x)dx = 0

. (14)

The following Lemma from [4] with sketch of proof illustrates why flowergraphs are important to study and will be used when proving Theorem 2.

Lemma 1 If Γ′ is formed from Γ by gluing together two vertices of Γ (say,x1, x2 are replaced by a new vertex x0 and each edge having x1 or x2 as anendpoint is replaced with loops around x0), then λ1(Γ) ≤ λ1(Γ′).

Proof. (This is only a sketch of the proof). This follows immediately fromthe fact that W 1

2 (Γ′) may be identified with a subspace of W 12 (Γ), since the con-

tinuity conditions imposed on functions in the former space are more restrictive;but the Rayleigh quotient is given by the same formula.[4]

�

This Lemma is important since it shows that the first excited eigenvalue ofany quantum graph under standard vertex conditions is always less or equal tothe corresponding flower graph, formed by gluing together vertices. Hence anupper bound for the first excited eigenvalue for any quantum graph is providedby calculating the corresponding flower graph.

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2.3 Equilateral flower graph

Remember that for a flower graph λ0 = 0 is always part of the spectrum withmultiplicity one. When all edges are of equal length the graph is called equilat-eral. When a flower graph is equilateral equation (13) reduces to

0 =

[sinn

(kL

2n

)]·[sinn−1

(kL

2n

)cos

(kL

2n

)]. (15)

This equation is satisfied if and only if either

Case 1

sin

(kL

2n

)= 0 (16)

or

Case 2

cos

(kL

2n

)= 0. (17)

This yields two different sets of eigenvalues corresponding to solutions to equa-tions (16) and (17) respectively. We write the solution of the differential equa-tion as:

ui(x) = Ai sin(kx) +Bi cos(kx),

where i = 1, 2, 3, ..., n. In order to calculate the degeneracy it is useful toparametrize each petal from − l

2 to l2 , where l = L

n . Then the continuity condi-tion gives

A1 sin

(kl

2

)+B1 cos

(kl

2

)= A1 sin

(−kl

2

)+B1 cos

(−kl

2

)...

= An sin

(kl

2

)+Bn cos

(kl

2

)= An sin

(−kl

2

)+Bn cos

(−kl

2

).

(18)

The Neumann condition yields

0 =A1k cos

(kl

2

)−B1k sin

(kl

2

)−A1k cos

(kl

2

)−B1k sin

(kl

2

)+ ...

+Ank cos

(kl

2

)−Bnk sin

(kl

2

)−Ank cos

(kl

2

)−Bnk sin

(kl

2

).

(19)

Since a lot of terms cancel and k 6= 0 (19) can be rewritten as(n∑i=1

Bi

)sin

(kl

2

)= 0. (20)

17

Consider now Case 1. Equation (20) will be automatically fulfilled due to(16). Equation (18) reduces to

B1 cos

(kl

2

)= B1 cos

(−kl

2

)...

= Bn cos

(kl

2

)= Bn cos

(−kl

2

).

(21)

and since cos(kl2

)6= 0 we need to require

B1 = B2 = ... = Bn := B. (22)

Note that neither equation (18) nor (20) gives us any restrictions on Ai, i =1, 2, 3, ..., n. So both Ai and B can be chosen arbitrarily. Hence each solutionto (16) corresponds to an eigenvalue with multiplicity n+ 1.

Consider now Case 2. Since sin(kl2

)6= 0, equation (18) reduces to

A1 = −A1 = A2 = −A2 = ... = An = −An = 0 (23)

and equation (20) reduces ton∑i=1

Bi = 0, (24)

which has n − 1 linearly independent solutions. Hence each solution to (17)corresponds to an eigenvalue with multiplicity n− 1.

So the eigenvalues of an equilateral flower graph with n petals are the fol-lowing:

λ0 = 0, (25)

λi+2nm =

(2πmn

L+πn

L

)2

, m = 0, 1, 2, ... and i = 1, 2, ..., n− 1 (26)

and

λj+n(2p−1) =

(2πpn

L

)2

, p = 1, 2, ... and j = 0, 1, ..., n. (27)

2.4 Comparison to the star graph

A general compact star graph has n edges and n+1 vertices v0, v1, ..., vn. Vertexv0 is connected to all other vertices, though v1, v2, ..., vn are not connected toeach other and thus have degree one. These types of graphs are important sincethey describe a complicated quantum graph locally around every vertex. In thequantum graph one choose a vertex which will be v0 in the star graph and theneighbors to this vertex will be the other vertices in the star graph. An exampleof a star graph is displayed in Figure 5. A star graph with equal edge lengths

18

Figure 5: Star graph with five edges

is called equilateral.For an equilateral (l1, l2, ..., ln = l) n-star graph we parametrize the edges

with zero at the outer vertices (those of degree one) and l at the inner vertexv0. Let us show that the multiplicity for λ0 = 0 is one. When λ = 0, solutionto equation (2) becomes

uj(x) = Ajx+Bj , j = 1, 2, ..., n.

Taking Neumann condition into account on the outer edges yields Aj = 0, ∀ j.Though we get no information from continuity on the outer edges. Hence

uj(x) = Bj

and since u(x) must be continuous at v0, so

B1 = B2 = ... = Bn := C.

Note that Neumann condition is satisfied at v0. Hence we get the constantfunction on the compact star graph

u(x) = C,

with eigenvalue λ0 = 0 and degeneracy 1.Consider now λ > 0. The solution to (2) is then

uj(x) = Aj sin(kx) +Bj cos(kx).

Taking Neumann condition into account on the outer vertices yields

kAj cos(k · 0)− kBj sin(k · 0) = 0 =⇒ Aj = 0, ∀j

since k 6= 0. Souj(x) = Bjcos(kx).

Again continuity on the outer edges does not give us any further restriction.Applying continuity at the inner vertex gives

B1 cos(kl) = B2 cos(kl) = ... = Bn cos(kl).

Case 1: Consider now the case when cos(kl) 6= 0. Then

B1 = B2 = ... = Bn := B

19

andu1(x) = u2(x) = ... = un(x) = B cos(kx).

Applying Neumann condition yields

n∑j=1

∂nuj(l) = 0,

which can be rewritten as−nk sin(kl) = 0.

So the solutions for k arek =

πm

l

and the eigenfunction becomes

u(x) = B cos(πm

lx),

with eigenvalues

λ =π2m2

l2,

with degeneracy 1.Case 2: Consider now the case when cos(kl) = 0. Then continuity at theinner vertex is automatically satisfied. The eigenvalues λ can easily be calculatedfrom k and are

λ =( π

2l+πm

l

)2

,

The Neumann condition gives

−B1 sin(kl)−B2 sin(kl)− ...−Bn sin(kl) = 0.

Since cos(kl) = 0 =⇒ sin(kl) = ±1 we can cancel − sin(kl) and get

B1 +B2 + ...+Bn = 0,

which has n− 1 independent solutions. Hence the degeneracy of the eigenvalueconsidered in this case is n− 1.

If the total length of the star graph is L2 (half of the flower graph with the

same number of edges), we get l = L2n and the eigenvalues will be

λ0 = 0, (28)

λi+mn =

(2πmn

L− πn

L

)2

, m = 0, 1, 2, ... and i = 1, 2, ..., n− 1 (29)

and

λnp =

(2πpn

L

)2

, p = 1, 2, ... (30)

They are the same eigenvalues as those for the flower graph with the samenumber of edges and double total length. However, the degeneracy of equation(30) is less. The reason is that eigenfunctions on the star graph come from eveneigenfunctions on the flower graph, hence there are no odd eigenfunctions on

20

the star graph.Consider the equilateral compact star graph with n edges and total length L

2 .

Then from each outer vertex one can add an edge with length l = L2n to the inner

vertex v0. We call this new graph Γ. When standard conditions are used thosevertices that were the outer vertices in the star graph do not contribute withnew information since continuity and Neumann condition are always satisfiedat these vertices. Hence one can remove these vertices which produces the newgraph Γ which is an equilateral flower graph with n petals.

2.5 Three petal flower graph

In order to be able to study the eigenvalues for a flower graph of three petals ithelps to rewrite (13) as

0 = sin

(kl12

)sin

(kl22

)sin

(kl32

)·

·[sin

(kL

2

)+ sin

(kl12

)sin

(kl22

)sin

(kl32

)].

(31)

Equation (31) is satisfied if and only if one of the the four terms is equal to zero

sin

(kl12

)= 0, (32)

sin

(kl22

)= 0, (33)

sin

(kl32

)= 0 (34)

or

sin

(kL

2

)+ sin

(kl12

)sin

(kl22

)sin

(kl32

)= 0. (35)

Equation (32), (33) and (34) are elementary and their solutions are

k =2π

lj· n, n ∈ N, j = 1, 2, 3. (36)

Equation (35) is harder to solve in general and can be rewritten as

sin

(kL

2

)= − sin

(kl12

)sin

(kl22

)sin

(kl32

). (37)

Assume again the total length to be equal to one. By analyzing equations(32), (33), (34) and (35) together with the solution λ0 = 0, the spectrum forthe Laplace operator on the three petal flower graph can be derived and furtherstudied. The equations are symmetric with respect to permutation of l1, l2, l3;therefore without loss of generality we demand that

l1 ≤ l2 ≤ l3. (38)

One can make 3D plots of the eigenvalues. by knowing l1 and l2 and since weassumed L = 1, the value of l3 is uniquely determined. All possible values of l1

21

and l2 span a triangle in the (l1, l2)-plane. The length combination (l1, l2, l3) =( 1

3 ,13 ,

13 ) gives the maximal value of l1 and will be a vertex on the triangle.

The length combination (l1, l2, l3) = (0, 12 ,

12 ) gives the maximal value of l2 and

will be the second vertex on the triangle. The third vertex of the triangle is(l1, l2, l3) = (0, 0, 1) and corresponds to l3 being maximal. The sides of thetriangle are given by the segments:

l1 = 0, 0 ≤ l2 ≤1

2,

0 ≤ l2 = l1 ≤1

3

and1

3≤ l2 =

1

2− l1

2≤ 1

2.

The triangle describing the possible values of l1 and l2 is displayed in Figure 6.

Figure 6: Possible values for l1 and l2

Plots of the first three nonzero eigenvalues as a function of l1 and l2 arepresented in Figure 7, Figure 8 and Figure 9, with resolution 0.052L2 and theextra points where one or both of the lengths l1 and l2 are equal to 1

3 , since thecase where all lengths are equal is interesting. Points outside the domain areput to zero.

To get better resolution implicit plots are also made and presented Figure 10and Figure 11. For a given eigenvalue λi(l1, l2) it is possible that λi+1(l′1, l

′2) ≤

λi(l1, l2). Hence the eigenvalues are difficult to separate in a plot. Although λ1

22

is successfully isolated from higher eigenvalues and presented in Figure 10. InFigure 7, Figure 8 and Figure 9 the resolution is bad, still the plots are usefulto identify the eigenvalues in Figure 11.

Figure 7: λ1 for three petal flower graph, l3 = 1− l1 − l2

23

Figure 8: λ2 for three petal flower graph, l3 = 1− l1 − l2

24

l1l2

λ3

λ3

Figure 9: λ3 for three petal flower graph, l3 = 1− l1 − l2

25

l1

l2

λ1

Figure 10: λ1 for three petal flower graph, l3 = 1− l1 − l2

26

λ

l2

l1

Figure 11: λ1,λ2 and λ3 for three petal flower graph, l3 = 1− l1 − l2 (λ4 is alsoin the figure since for some l1 and l2, λ4 ≤ 36π2).

27

In the Figure 7, 9, 10 and 11 it is clear that λ1 and λ3 are maximal atl1 = l2 = l3 = L

3 . This motivates Proposition 1 in section 2.6.

2.6 Proposition regarding spectral gap of three petal flowergraph

Proposition 1 The eigenvalues λ1 and λ3 for the three petal flower graph aremaximal at l1 = l2 = l3 = L

3 .

Proof. Assume again L = 1. Since l1 ≤ l2 ≤ l3, the first nonzero eigenvalue

will either be λ1 =(

2πl3

)2

or the first nonzero solution to equation (37). The

value of the solution λ =(

2πl3

)2

is minimally (2π)2 (l3 = 1) and maximally (6π)2

(l1 = l2 = l3 = 13 ). We need to compare this with the first nonzero solution to

the equation (37). We introduce the function

f(k) = sin

(k

2

),

which correspond to the left hand side of equation (37) and

g(k) = − sin

(kl12

)sin

(kl22

)sin

(kl32

),

which correspond to the right hand side of equation (37). We know that f(ε)is continuous and positive for small enough ε > 0. We know that f(0) = 0 andf(3π) = −1. The function g(k) is also continuous. We know that g(k) ≥ −1and g(ε) < 0 for small enough ε.

Therefore ∃ k∗ ∈ [ε, 3π] : g(k∗) = f(k∗). We introduce the function

h(k) = f(k)− g(k).

When h(k) = 0 we have a solution to (35). Since f(k) and g(k) are continuous,it follows that h(k) is continuous as well. Note that h(ε) > 0 and h(3π) ≤ 0then ∃ k∗ : h(k∗) = 0, ε ≤ k∗ ≤ 3π. Since l1 = l2 = l3 = 1

3 gives k∗ = 3π weknow that λ1 = (k∗)2 is maximal when l1 = l2 = l3 = 1

3 with λ1 = 9π2.

Let us show that there are always at least three solutions to equation (34)and (35) on the interval [ε, 6π]:1. One solution is on the interval [ε, 3π] as shown above.2. The function f satisfies f(3π) = −1 and f(5π) = 1. The function g satis-fies |g(k∗)| ≤ 1 and g(3π) ≥ −1. Also g is continuous. Hence h(3π) ≤ 0 andh(5π) ≥ 0. Therefore ∃ k∗ ∈ [3π, 5π] : h(k∗) = 0. Note that the solution in 1and 2 could be the same if the solution is k∗ = 3π. It is only for an equilateralthree petal flower graph that h(3π) = 0 is satisfied. Although this is a doublepoint as shown in section 2.3.3. There is also a solution to equation (34) on the interval [2π, 6π].It has hence been shown that there exists at least three solutions to equation(32), (33), (34) and (35) on the interval [0, 6π]. Hence k3 ≤ 6π. For the equi-lateral graph l1 = l2 = l3 = 1

3 the solution to equation (34) is k = 3π and thesolutions to equation (35) are the two at k = 3π mentioned above and one at

28

Figure 12: Plot of sin(k2

)with a kπ scale on the x-axis

k = 6π. The functions f and g for the equilateral graph is displayed in Figure13. By ordering these solution one gets k1 = 3π, k2 = 3π, k3 = 6π and k4 = 6π.Hence it has been shown that λ3 is maximal when l1 = l2 = l3 = 1

3 with thevalue λ3 = 36π2.

�

29

Figure 13: Functions f(k) and g(k) with a kπ scale on the x-axis

2.7 General theorem regarding first nonzero eigenvalue

The following theorem with proof is found in [4]

Theorem 2 Let Γ be a quantum graph having length L > 0 and number ofedges n ≥ 2. Then

λ1(Γ) ≤ π2n2

L2, (39)

with equality if and only if(1) Γ is equilateral and(2) there exists an eigenfunction corresponding to λ1(Γ) which takes on the valuezero at all vertices on Γ.If n = 1, then

λ1(Γ) =

{4π2/L2 if Γ is a loop,

π2/L2 if Γ is a path.

Proof. Assume Γ is any quantum graph as in the statement of the theorem,with L > 0 and n ≥ 2 given, and denote by Γ the corresponding flower graphhaving the same number of edges with the same lengths as Γ. i.e., Γ is thegraph which may be formally obtained from Γ by gluing together all vertices ofthe latter. Then W 1

2 (Γ) may be canonically identified with the subset of W 12 (Γ)

consisting of all functions u ∈ W 12 (Γ) such that u(v1) = ... = u(vn) for all

vertices x1, ..., xn of Γ. It follows from Lemma 1 that λ1(Γ) ≤ λ1(Γ).

30

We now show that the equilateral flower graph F (L, n) is the (unique) max-imizer of λ1 among all flower graphs of fixed total length L > 0 and num-ber of edges n ≥ 2, which will then complete the proof of Theorem 2, sinceλ1(F (L, n)) = π2n2/L2 by equation (17). To that end, let e1, e2 with lengthsl1, l2 respectively, be the two longest edges of the arbitrary flower graph Γ (orany two longest if these are not uniquely determined); by the pigeonhole prin-ciple, M := l1 + l2 ≥ 2L/n. Denote by Γ12 the flower graph consisting of thesetwo longest edges. By Lemma 1,

λ1(Γ12) ≤ λ1(Γ).

But we see immediately that λ1(Γ12) ≤ 4π2/M2 ≤ π2n2/L2, since we mayuse any eigenfunction belonging to the first eigenvalue of a circle of length Mas a test function on Γ12, provided it is rotated appropriately so as to satisfythe continuity condition at the points corresponding to the vertex of Γ12. Thisestablishes the inequality.

To prove the case of equality, we first note that this is only possible if M =2L/n, that is, if the two longest edges (and all other edges) have length L/neach, and so Γ must already be equilateral in this case.

Moreover, equality requires that the identification of all vertices of Γ notaffect λ1(Γ). This is only possible if there is already an eigenfunction u1 ∈W 1

2 (Γ) corresponding to λ1(Γ) for which u1(x1) = ... = u1(xn), since otherwiseW 1

2 (F (L, n)) - which may be canonically identified with a subspace of W 12 (Γ) -

does not contain any minimizer achieving equation (14), so that the infinum inW 1

2 (F (L, n)) must be larger. But since u1 corresponds to an eigenfunction onthe flower graph F (L, n), and any eigenfunction of λ1(F (L, n)) must take on thevalue zero at the unique vertex of F (L, n) it follows that ux1

= ... = u1(xn) = 0.The case n = 1 is trivial.

�

Note that the theorem with its proof is slightly modified from the reference inorder to match the notation used in this thesis.

The advantages with this theorem is that it can be applied to any graph.Also the proof provides a canonical transformation from a general quantumgraph Γ to a flower graph Γ. This illustrates the importance of studying flowergraphs. Although this theorem tells us nothing about the higher eigenvalues incontrast to Proposition 1. The proof of Proposition 1 is relatively simple andcould be useful to provide proofs about maximizing even higher eigenvalues.

3 Conclusion

Calculating the spectrum of a flower graph is the same as solving the equation

0 =

(n∏i=1

sin

(kl12

))·

n∑i=1

sin

(kli2

)∏j 6=i

cos

(klj2

)and also determine the degeneracies of the eigenvalues. And also includingλ0 = 0 which always have degeneracy one. This is quite easy for the flowerwith two petals and is displayed in Figure 3. The importance of studying flower

31

graphs is motivated by Lemma 1, which shows that an upper bound for the firstexcited eigenvalue is given by the first excited eigenvalue to the correspondingflower graph, formed by gluing together all vertices.

The eigenvalues for the equilateral star graph are the same as the eigen-values for the equilateral flower graph with double total length. However alleigenfunctions that exists on the flower graph do not lead to eigenfunctions onthe star graph, namely those that are odd.

One may create an equilateral flower graph from a star graph by adding oneedge from every outer vertex towards the inner vertex.

Proposition 1 proves that, for a given total length, the first and the thirdnonzero eigenvalues of a flower graph with three petals are maximal when thegraph is equilateral. This can be applied to concrete examples, where the geo-metrical structure is the same as the three petal flower graph.

The technique used to prove Proposition 1 can be useful when analyzinghigher eigenvalues.

4 References

[1] http://en.wikipedia.org/wiki/Carbon nanotube#cite note-Longest-1.

[2] G. Berkolaiko and P. Kuchment, Introduction to Quantum Graphs,AMS,2013.

[3] L. Friedlander, External properties of eigenvalues for a metric graph, An-nales de L’institut Fourier, n. 1 (2005), p. 199-211.

[4] J.B. Kennedy, P. Kurasov, G. Melanova and D. Mugnolo, On the spectralgap of a quantum graph, Research reports in mathematics, n. 5 (2015).

[5] P. Kurasov, Quantum Graph, Spectral Theory and Inverse Problems, inpreperation.

[6] http://web.stanford.edu/class/math220b/handouts/eigenvalues.pdf

[7] A. Iantchenko, E. Korotyaev, Schrodinger Operator on the Zigzag Half-Nanotube in Magnetic Field, Math. Model. Nat. Phenom. Vol. 5, n. 4 (2010),p. 175-197.

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