Direct and Inverse Finite-Dimensional Spectral Problems on Graphs [1st ed.] 9783030604837, 9783030604844

Table of contents :
Front Matter ....Pages i-xvi
Stieltjes strings (Manfred Möller, Vyacheslav Pivovarchik)....Pages 1-60
Vibrations of star graphs (Manfred Möller, Vyacheslav Pivovarchik)....Pages 61-107
Damped vibrations of Stieltjes strings and star graphs (Manfred Möller, Vyacheslav Pivovarchik)....Pages 109-148
Trees (Manfred Möller, Vyacheslav Pivovarchik)....Pages 149-171
Spectral problems on graphs (Manfred Möller, Vyacheslav Pivovarchik)....Pages 173-220
Sturm-Liouville problems on graphs (Manfred Möller, Vyacheslav Pivovarchik)....Pages 221-265
Back Matter ....Pages 267-349

Citation preview

Operator Theory Advances and Applications 283

Manfred Möller Vyacheslav Pivovarchik

Direct and Inverse Finite-Dimensional Spectral Problems on Graphs

Operator Theory: Advances and Applications Volume 283 Founded in 1979 by Israel Gohberg

Series Editors: Joseph A. Ball (Blacksburg, VA, USA) Albrecht Böttcher (Chemnitz, Germany) Harry Dym (Rehovot, Israel) Heinz Langer (Wien, Austria) Christiane Tretter (Bern, Switzerland) Associate Editors: Vadim Adamyan (Odessa, Ukraine) Wolfgang Arendt (Ulm, Germany) B. Malcolm Brown (Cardiff, UK) Raul Curto (Iowa, IA, USA) Kenneth R. Davidson (Waterloo, ON, Canada) Fritz Gesztesy (Waco, TX, USA) Pavel Kurasov (Stockholm, Sweden) Vern Paulsen (Houston, TX, USA) Mihai Putinar (Santa Barbara, CA, USA) Ilya Spitkovsky (Abu Dhabi, UAE)

Honorary and Advisory Editorial Board: Lewis A. Coburn (Buffalo, NY, USA) Ciprian Foias (College Station, TX, USA) J.William Helton (San Diego, CA, USA) Marinus A. Kaashoek (Amsterdam, NL) Thomas Kailath (Stanford, CA, USA) Peter Lancaster (Calgary, Canada) Peter D. Lax (New York, NY, USA) Bernd Silbermann (Chemnitz, Germany) Harold Widom (Santa Cruz, CA, USA)

Subseries Linear Operators and Linear Systems Subseries editors: Daniel Alpay (Orange, CA, USA) Birgit Jacob (Wuppertal, Germany) André C.M. Ran (Amsterdam, The Netherlands) Subseries Advances in Partial Differential Equations Subseries editors: Bert-Wolfgang Schulze (Potsdam, Germany) Michael Demuth (Clausthal, Germany) Jerome A. Goldstein (Memphis, TN, USA) Nobuyuki Tose (Yokohama, Japan) Ingo Witt (Göttingen, Germany)

More information about this series at http://www.springer.com/series/4850

Manfred Möller • Vyacheslav Pivovarchik

Direct and Inverse Finite-Dimensional Spectral Problems on Graphs

Manfred Möller School of Mathematics University of the Witwatersrand Johannesburg, South Africa

Vyacheslav Pivovarchik Department of Higher Mathematics and Statistics South‐Ukrainian National Pedagogical University Odessa, Ukraine

ISSN 0255-0156 ISSN 2296-4878 (electronic) Operator Theory: Advances and Applications ISBN 978-3-030-60483-7 ISBN 978-3-030-60484-4 (eBook) https://doi.org/10.1007/978-3-030-60484-4 Mathematics Subject Classification (2010): 15A18, 15A22, 15A29, 34B45, 39A70, 47B30, 74H45, 74J25, 74K05 © Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This book is published under the imprint Birkhäuser, www.birkhauser-science.com by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Contents Preface

ix

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1 1 3 3 6 10 13 15 22 22 30 36 36 41 57 60

2

Stieltjes strings 1.1 Stieltjes functions in electrical circuits . . . . . . . . . . . . . 1.2 Boundary value problems for Stieltjes strings . . . . . . . . . 1.2.1 Boundary value problems . . . . . . . . . . . . . . . . 1.2.2 Cauer-Fry polynomials . . . . . . . . . . . . . . . . . . 1.2.3 Dirichlet and Neumann boundary conditions . . . . . 1.2.4 Robin-Dirichlet and Robin-Neumann problems . . . . 1.2.5 Lagrange identity and Sturm oscillation theorem . . . 1.3 Inverse problems for Stieltjes strings . . . . . . . . . . . . . . 1.3.1 Two-spectra inverse problems . . . . . . . . . . . . . . 1.3.2 Four-spectra inverse problem . . . . . . . . . . . . . . 1.4 Stieltjes strings with one interior condition . . . . . . . . . . . 1.4.1 The problem and its spectra . . . . . . . . . . . . . . . 1.4.2 The inverse problem . . . . . . . . . . . . . . . . . . . 1.4.3 The finite-dimensional Hochstadt-Lieberman problem 1.5 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Vibrations of star graphs 61 2.1 Star graphs with root at the centre . . . . . . . . . . . . . . . . . . 61 2.1.1 The direct spectral problem . . . . . . . . . . . . . . . . . . 63 2.1.2 The inverse spectral problem . . . . . . . . . . . . . . . . . 70 2.1.3 Restrictions on the multiplicities . . . . . . . . . . . . . . . 74 2.1.4 The inverse problem with given numbers of beads on the edges 82 2.1.5 Comparison with results for tree-patterned matrices . . . . 86 2.2 Symmetric Stieltjes strings . . . . . . . . . . . . . . . . . . . . . . . 88 2.3 Star graphs with root at a pendant vertex . . . . . . . . . . . . . . 90 2.3.1 The direct spectral problem . . . . . . . . . . . . . . . . . 92 2.3.2 The inverse spectral problem . . . . . . . . . . . . . . . . . 96 2.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 2.5 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 v

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Damped vibrations of Stieltjes strings and star graphs 3.1 The direct spectral problem . . . . . . . . . . . . . . . . . . . . . . 3.2 Inverse problems for a string with point-wise damping . . . . . . . 3.3 Comparison of characteristic functions . . . . . . . . . . . . . . . . 3.4 Stieltjes strings with a damping condition at the right end . . . . . 3.5 The inverse problem by parts of spectra of Robin-Regge problems . 3.6 Stieltjes strings with damping at an interior point . . . . . . . . . . 3.6.1 The direct problem . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 The inverse problem . . . . . . . . . . . . . . . . . . . . . . 3.7 Star graphs with damped central vertex . . . . . . . . . . . . . . . 3.7.1 The direct problem . . . . . . . . . . . . . . . . . . . . . . . 3.7.2 The inverse problem . . . . . . . . . . . . . . . . . . . . . . 3.8 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

109 109 115 125 128 130 134 134 136 140 140 144 147

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Trees 4.1 Characteristic polynomials . . . . . . . 4.2 Complementary subtrees . . . . . . . . 4.3 The direct spectral problem on a tree 4.4 Inverse problem: strict interlacing . . . 4.5 Notes . . . . . . . . . . . . . . . . . .

149 149 154 164 168 171

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Spectral problems on graphs 5.1 The problem and its spectrum . . . . . . . . . 5.2 Characteristic polynomials . . . . . . . . . . . . 5.3 Characteristic polynomials for related graphs . 5.4 Series connection of graphs . . . . . . . . . . . 5.5 Parallel connection of graphs . . . . . . . . . . 5.6 Multiplicities of characteristic values of spectral 5.6.1 Cyclically connected graphs . . . . . . . 5.6.2 Trees and generalized quasi-trees . . . . 5.6.3 Connected graphs . . . . . . . . . . . . 5.7 Graphs of symmetric Stieltjes strings . . . . . . 5.7.1 The direct problem . . . . . . . . . . . . 5.7.2 The inverse problem . . . . . . . . . . . 5.8 Notes . . . . . . . . . . . . . . . . . . . . . . .

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Sturm-Liouville problems on graphs 6.1 Selfadjoint differential operators . . . . . . . . . . . . . 6.2 Selfadjoint differential operators on graphs . . . . . . . 6.3 Graphs with Neumann conditions at pendant vertices 6.4 Connected graphs with Dirichlet conditions at pendant 6.5 Spectral problems on perfect binary trees . . . . . . . 6.5.1 The spectral problem . . . . . . . . . . . . . . 6.5.2 Characteristic functions of perfect binary trees

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173 173 183 192 197 200 203 209 210 210 213 214 218 219 221 221 223 225 241 242 242 243

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6.5.3 The eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . 6.5.4 The inverse problem . . . . . . . . . . . . . . . . . . . . . . Sturm-Liouville problems on star graphs . . . . . . . . . . . . . . . 6.6.1 Multiplicities and location of eigenvalues below a fixed value 6.6.2 Sturm-Liouville problems on star graphs with finite edges . 6.6.3 Sturm-Liouville problems on star graphs with infinite edges Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

247 249 256 256 257 260 265

A Rational Nevanlinna and Stieltjes Functions A.1 Rational functions . . . . . . . . . . . A.2 Rational Nevanlinna functions . . . . . A.3 Rational Stieltjes functions . . . . . . A.4 The classes S r . . . . . . . . . . . . . A.5 Hermite-Biehler polynomials . . . . . . A.6 Miscellaneous results . . . . . . . . . .

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267 267 270 275 283 285 290

B Linear and quadratic matrix pencils B.1 Matrix pencils . . . . . . . . . . . . . B.2 The linear pencil L . . . . . . . . . . . B.3 The quadratic pencil T . . . . . . . . . B.4 Schur complements . . . . . . . . . . . B.5 Eigenvalues of real symmetric matrices

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293 293 294 294 298 299

C Graphs C.1 Definitions . . . . . . . . . . . . . . . . . . . . . C.2 Basic properties of graphs . . . . . . . . . . . . C.3 Cyclically connected graphs and subgraphs . . C.4 Maximally connected subgraphs and quasi-trees C.5 Adjacency matrices and incidence matrices . . C.6 T -acyclic matrices . . . . . . . . . . . . . . . .

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305 305 309 314 316 328 332

Bibliography

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Index

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Index of notation

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Preface Vibrating strings and the sound they generate have fascinated humankind ever since homo sapiens inhabited the earth, and almost every culture has constructed musical string instruments. Hence one of the earliest subjects of Mathematical Physics was the investigation of strings, initiated in the 18th century by Euler, d’Alembert and Lagrange. Two of the earliest comprehensive accounts of the theory of sound are the monographs [69] by Helmholtz, published in 1863, and [121] by Rayleigh, published in 1877, see also [70] and [122] for later editions. In particular, Helmholtz considered a string loaded at the midpoint, and Rayleigh calculated the eigenfunctions and the eigenfrequencies as a function of the mass position along the string. Of the many results published in this field, we just mention three more recent ones. In [36] the eigenfrequencies of a loaded string with a concentrated mass were found. In [103] the authors have studied by means of numerical calculations and experimentally the eigenfunctions and the corresponding eigenfrequencies of a string-mass chain. In [41] the theoretically calculated frequencies of strings are compared with experimental data. Mathematicians prefer to deal first with idealized simplified models, and hence M.G. Krein introduced the notion of Stieltjes string, which is a massless thread bearing point (concentrated) masses, which he called beads. A Stieltjes string is also called string-mass chain (see [103]). In his works M.G. Krein considered small transverse vibrations of Stieltjes strings admitting infinitely many beads accumulating at one of the ends of the string. However, in this monograph we restrict our attention to the case of a finite number of beads. The same model is used for vibrations of point masses connected by springs (see, e.g., [120], [62], [63], [93], [123], [37]). This model is also known in the theory of synthesis of electrical circuits (see, e.g., [66], [130], [140], [146], and in particular the Cauer method [33]). Indeed, such strings are widely used as simple models in physics (see, e.g., [84], [54], [55] and [64] where beads are connected with threads of small density). The problem of vibrations of a tree of Stieltjes strings was first considered in [60]. For the case of a finite number of beads the following spectral problems were considered in [59]: the Dirichlet-Dirichlet problem, i.e., the problem with the Dirichlet boundary conditions at both ends, and the Neumann-Dirichlet problem, i.e., the problem with the Neumann condition at the left end and the Dirichlet condition at the right end of the string. Physically, the Dirichlet condition means that the end is fixed and the Neumann condition describes a free end. It was shown that the eigenvalues of each of these problems are simple, positive and interlace. The corresponding inverse problem of recovering the masses of the beads and the lengths of the intervals into which the beads divide the string by the spectra of Dirichlet-Dirichlet and Neumann-Dirichlet problems and the total length of the string was also completely solved: it was shown that for two sequences of positive numbers to be the spectra of a Dirichlet-Dirichlet and a Neumann-Dirichlet problem, respectively, it is necessary and sufficient that the sequences interlace. The algorithm of recovering the string data consists of expanding the quotient of the ix

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characteristic polynomials of the Dirichlet-Dirichlet and the Neumann-Dirichlet problem into a continued fraction. This continued fraction algorithm originates in papers of Cauer [33] and Fry [58]. It was also shown in [59] that the solution of the inverse problem is unique. The investigation of spectra for (over)damped dynamical systems was started in [47], represented as matrix polynomials of degree 2 in the eigenvalue parameter. This investigation was motivated by electrical circuits, but also applies to damped Stieltjes strings. The inverse problem, i.e., the problem of recovering the parameters of the string from the spectrum of its vibrations and from the total length of the string, for a Stieltjes string with a finite number of beads with point-wise (i.e., one-dimensional) damping at the right end and the left end free was solved in [4]. Also this problem can be reduced to the problem of damped oscillators considered in [133], [134]. Another approach to inverse problem for damped finite-dimensional system was developed in [85] where the given data included not only eigenvalues but also the so-called Jordan pairs. An inverse problem for a Stieltjes string with damping at the midpoint was solved in [18]. Direct and inverse spectral problems on graphs is a rapidly developing branch of analysis. Usually, the Sturm-Liouville equation, the Dirac equation or the Stieltjes string equation is considered on the edges of a graph subject to matching and boundary conditions at the vertices. This can be described as a spectral problem for an operator (see, e.g., [50], [119], [42], [117]) which is selfadjoint under certain conditions (see, e.g., [29], [77]). The justification of such models can be found in [83]. There are different approaches to inverse problems on graphs: firstly, recovering the form of a metric graph using spectral or scattering data (see, e.g., [10], [67]), and secondly, the form of the graph is known a priori and the aim is to recover the potentials of the Sturm-Liouville equation on the edges (see, e.g., [107], [108], [143]) or the mass distribution (see, e.g., [17]), using spectral or scattering data. The tradition of using Dirichlet and Neumann spectra to solve an inverse problem on a graph came from the classical results of Borg [13], Levitan and Gasymov [89], and Marchenko [92] for inverse spectral problems on a finite interval. It is possible to consider two boundary value problems, one with Neumann condition at a certain pendant vertex, and one with Dirichlet condition at this vertex (see [27], [143]), and to use their spectra as the given data for solving the inverse problem. However, it is possible (see, e.g., [86]) to introduce analogues of Neumann and Dirichlet conditions at an interior vertex chosen as the root. Such problems were considered in [107], [108], and earlier for the star graph of two edges, i.e., for the three spectra problem, in [106], [61], [72]. Analogous problems generated by the equations of Stieltjes strings were considered in [17] and [18]. In this monograph we present a comprehensive study of direct and inverse spectral problems on Stieltjes strings and on graphs of Stieltjes strings. The boundary conditions at the pendant vertices of the graph will be Dirichlet or Neumann conditions while at interior vertices we apply continuity conditions and the condition of balance of forces (which is the Kirchhoff condition in electrical circuits

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synthesis terms). Sometimes we may allow mixed conditions, also called Robin conditions, at pedant vertices. The direct problem is to describe the eigenvalues, whereas the inverse problem is to determine the data of the Stieltjes string or graph of Stieltjes strings from the eigenvalues. These eigenvalue problems can be described by matrices, that is, roughly said, by a problem (λ2 M − iλK − A)Y = 0, where A, K, M are square matrices, λ is the (complex) eigenvalue parameter, and λ is an eigenvalue if the above equation has a nonzero solution Y . M will normally be a diagonal matrix with real positive entries in the diagonal, and A will be selfadjoint. For undamped problems, K = 0, we will replace λ2 with z, and we will usually call the numbers z characteristic values to distinguish them from the eigenvalues λ. In the undamped case, the spectral problem is then represented by (zM − A)Y = 0, and the characteristic values are therefore real. We also consider some spectral problems with point-wise damping (viscous friction), in which case K 6= 0. The eigenvalues of such problems lie in the upper half-plane. Of course, a number λ or z is an eigenvalue or characteristic value if and only if the corresponding determinant of the matrix problem has the value zero. Hence the main tasks in the study of eigenvalues and characteristic values are to investigate when matrices are singular and to investigate the zeros of characteristic polynomials. We will also study Sturm-Liouville problems on graphs for which the eigenvalues share properties with the characteristic values of Stieltjes string problems on these graphs and where the techniques developed for graphs of Stieltjes strings apply. Now we briefly outline the contents of this monograph. In Chapter 1 we consider boundary value problems which describe small transverse vibrations of Stieltjes strings, the corresponding spectral problems, their characteristic values, and the inverse problems of recovering the string data from the characteristic values. In Section 1.1 we give a brief historical overview on the continued fraction approach in the synthesis of electrical circuits. In Section 1.2 we consider the class of Stieltjes functions which occur in mechanics. We introduce the notion of a Stieltjes string with n beads and describe equations of its motion. Cauer-Fry polynomials are introduced to express the characteristic values of boundary value problems on Stieltjes strings as the zeros of Cauer-Fry polynomials. The Stieltjes functions are realized as quotients of two Cauer-Fry polynomials. At each end of the string, Dirichlet, Neumann or Robin conditions may be imposed. We present each of these problems as a spectral problem for an n×n linear matrix pencil z 7→ zM −A, where M is the diagonal matrix describing the masses of the beads and A is a tridiagonal Jacobi matrix. The characteristic polynomials for each of the Dirichlet-Dirichlet, Dirichlet-Neumann, Neumann-Dirichlet, Neumann-Neumann, Robin-Dirichlet, Robin-Neumann spectral problems are obtained as particular Cauer-Fry polynomials. These are poly-

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nomials of degree n, and their zeros, which are the characteristic values of the corresponding problems, are simple and positive, except for the smallest zero of the Neumann-Neumann problem, which is 0. Also, interlacing properties of the characteristic values of these problems are shown. Furthermore, the Lagrange identity and the Sturm oscillation theorem for Stieltjes strings are proved. In Section 1.3 we consider various inverse problems for Stieltjes strings. First we deal with the two spectra inverse problem. The direct problem has 2n+1 degrees of freedom, namely, the masses of the n beads and the lengths of the subintervals into which the string is divided by the beads. On the other hand, the two spectra consist of 2n arbitrary positive numbers (degrees of freedom) if none of the boundary problems is the Neumann-Neumann problem, and of 2n − 1 arbitrary positive numbers if one of the boundary problems is the Neumann-Neumann problem. It is shown that given any two interlacing sequences of positive numbers which have in total 2n or 2n − 1 terms, respectively, uniquely determine the string data for the Dirichlet-Dirichlet and Dirichlet-Neumann problem, or Dirichlet-Neumann and Neumann-Neumann problems with the given sequences as positive characteristic values, respectively, if one or two of the string data is prescribed. The prescribed data can be, for example, the total length of the string, the distances of an outermost bead from the corresponding end of the string and the total mass of the beads, or the distances of the outermost beads from the corresponding end of the string. It is also shown that not necessarily unique solutions of the inverse problem for the Dirichlet-Dirichlet and Neumann-Neumann or Dirichlet-Neumann and Neumann-Dirichlet conditions, respectively, exist. In the four spectra problem four finite sequences of positive real numbers are given, and the questions is if these sequences can be identified as certain characteristic values of the Dirichlet-Dirichlet, Dirichlet-Neumann, Neumann-Dirichlet and Neumann-Neumann problems for a Stieltjes string. We prove that the four spectra problem has a unique solution if the sequences have certain interlacing properties and if an arbitrary length of the string is prescribed. In Section 1.4 we compare the characteristic values of a spectral problem of a Stieltjes string with the characteristic values of problems on two substrings into which an interior point divides the string. We also solve the so-called three spectra problem: given characteristic values of spectral problems on the substrings and on the whole string together with the lengths of the substrings, find the data of the string. We consider a Stieltjes string analogue of the Hochstadt-Lieberman theorem. Here we prove, roughly speaking, that the knowledge of masses of the beads lying in the left part of the string and the intervals between them together with the spectrum of a problem on the whole interval uniquely determines the masses of the other beads and the intervals between them. Chapter 2 deals with spectral problems which arise in the description of small transverse vibrations of a star graph composed by Stieltjes strings. We consider two different boundary value problems for a star graph of Stieltjes strings with continuity and Kirchhoff conditions at the interior vertex. In each of our two main results we impose conditions on two sequences of real numbers necessary and suffi-

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cient to be the spectra of the Dirichlet and the Neumann problem on a star graph of g Stieltjes strings; in the first case the root lies at the central vertex, in the second case the root is at a pendant vertex. In both cases we establish a constructive method to recover the values of the masses, including the central one, and lengths of the subintervals between them. This method uses the representation of rational functions with interlacing zeros and poles by (possibly branching) continued fractions. If the root is a pendant vertex, then the spectra of the Dirichlet and the Neumann problems together with the total length of the main edge uniquely determine the values of the masses and lengths of the subintervals between them of the main edge. The remaining inverse problem on the subgraph of g − 1 edges may be viewed as an inverse problem with root at the central vertex to which our first result applies. In Section 2.1 we consider the direct and the inverse spectral problem for the case when the root is the central vertex of the star graph with g edges. That is, we impose Dirichlet boundary conditions at all pendant vertices, while at the central vertex we consider Kirchhoff and continuity conditions for the Neumann problem and Dirichlet conditions for the Dirichlet problem (in which case the whole problem decouples into g separate Dirichlet-Dirichlet problems). We allow a mass M to be placed at the central vertex, so that the Dirichlet problem may be viewed as the limit M → ∞ of the Neumann problem. We investigate the spectra of the corresponding Neumann and Dirichlet problems and their relation to each other, including monotonicity in terms of the central mass M . We prove that the two spectra interlace non-strictly, and if they have a characteristic value z in common, then its multiplicity nD,g (z) as a Dirichlet eigenvalue and its multiplicity nN,g (z) as a Neumann eigenvalue satisfy nD,g (z) = nN,g (z) + 1. We then show that the necessary conditions are also sufficient for the solution of the inverse problem: given two sequences satisfying these conditions and the total lengths l1 , . . . , lg of all strings, we construct a mass distribution so that the corresponding star graph of Stieltjes strings with root at the central vertex has these two sequences as Neumann and Dirichlet characteristic values. Since we do not assume strict interlacing of the sequences, this solution need not be unique. The recovering procedure, based on the decomposition of Stieltjes functions into continued fractions, is constructive. The solution of the inverse problem provides sufficient conditions for the existence of a star-patterned matrix with two given sequences being the spectra of the matrix and its first principal submatrix. In Section 2.2 symmetric Stieltjes strings are identified as particular star graphs of Stieltjes strings with two edges. In Section 2.3 we consider the direct and the inverse spectral problem for the case when the root is one of the pendant vertices of the star graph with g edges. That is, we impose Dirichlet boundary conditions at all other pendant vertices, Kirchhoff and continuity conditions at the central vertex, where again a mass M may be placed, and at the pendant vertex chosen as root the Neumann condition for the Neumann problem and the Dirichlet condition for the Dirichlet problem. We prove that the two spectra interlace non-strictly and if they have

xiv

Preface

a characteristic value z in common, then its multiplicity nD,g (z) as a Dirichlet characteristic value and its multiplicity nN,g (z) as a Neumann characteristic value satisfy the inequalities nD,g (z) ≤ g − 1, nN,g (z) ≤ g − 1, and nD,g (z) + nN,g (z) ≤ 2g−3. We also establish a relation of the spectral polynomials of the above Dirichlet and Neumann problems with the boundary value problems for the star subgraph with g − 1 edges obtained from the original graph by deleting the main edge (i.e., the edge incident with the root). We then show that the necessary conditions are also sufficient for the solution of the inverse problem: given two sequences satisfying these conditions together with the lengths of the g strings, we construct a mass distribution so that the corresponding star graph of Stieltjes strings with root at a pendant vertex has these two sequences as Neumann and Dirichlet characteristic values. Moreover, we show that the two spectra and the total length of the main edge uniquely determine the masses of the beads and the lengths of the intervals between them on this main edge; the mass distribution on the other edges cannot be uniquely determined. The recovering procedure is based on the decomposition of the quotient of the characteristic functions of the Dirichlet and the Neumann boundary value problem, which is a Stieltjes function, into branching continued fractions. In fact, the coefficients at the non-branching part of this expansion are the uniquely determined masses and the subintervals between them on the main string, while the mass distribution on the g − 1 other edges may be recovered by our first inverse theorem and the corresponding algorithm. An example in Section 2.4 illustrates that our method, in fact, allows to construct all solutions of the inverse problem. In Chapter 3 we consider spectral problems related to damped vibrations of Stieltjes strings and star graphs of Stieltjes strings. More precisely, we assume that the movement of at least one bead is subject to damping. In Section 3.1 we describe the spectra of small transverse vibrations of a Stieltjes string with n beads, with both ends fixed (Dirichlet-Dirichlet boundary value problem) or with the left end fixed and the right end free (Dirichlet-Neumann boundary value problem). We prove that the problem has 2n eigenvalues, counted with multiplicity, which lie in the closed upper half-plane and are symmetric with respect to the imaginary axis. All eigenvalues lie in the open upper half-plane if at least one of the outermost beads is subject to damping. In Section 3.2 we consider the inverse problem for the case where only the bead closest to the right end of the string is damped. Given a sequence of 2n complex numbers in the open upper half-plane whose terms are symmetric with respect to the imaginary axis, the total length of the string, and the distance of the rightmost bead to the right end of the string, the inverse DirichletDirichlet problem and Dirichlet-Neumann problem, respectively, have a unique solution, i.e., there is a unique Stieltjes string whose corresponding sequence of the eigenvalues is the given sequence. Relations between the characteristic functions of the Dirichlet-Dirichlet problem and the Dirichlet-Neumann problem are derived in Section 3.3. In Sections 3.4 and 3.5 inverse problems are considered for Stieltjes strings with a damped bead at one endpoint and Robin conditions at the other endpoint.

Preface

xv

Direct and inverse problems on Stieltjes strings with damping at an interior point are considered in Section 3.6 and, more generally, on star graphs of Stieltjes strings with damping at the central vertex are considered in Section 3.7. In Chapter 4 we generalize the results from Chapter 2 to trees. An arbitrary vertex v0 of the tree is chosen as the root. At interior vertices, we will impose continuity and Kirchhoff conditions, which we call generalized Neumann conditions, or we will impose generalized Dirichlet conditions. Although we are mainly interested in the case that, except for the root, the boundary conditions at pendant vertices are Dirichlet conditions and the conditions at interior vertices are generalized Neumann conditions, it is convenient for auxiliary results to allow (generalized) Dirichlet conditions or (generalized) Neumann conditions at each of the vertices of the tree. The characteristic values of the boundary value problem are the zeros of the corresponding characteristic polynomial. In Section 4.1, characteristic polynomials for boundary eigenvalue problems on trees are defined and basic properties are established. Cutting trees into subtrees is useful to study these characteristic polynomials, and in Section 4.2, relations between characteristic polynomials of trees and complementary subtrees are obtained. Letting n be the total number of beads on the tree, it is shown that the characteristic polynomial has degree n, its zeros are all real and nonnegative, and 0 is a zero if and only if Neumann conditions are imposed at all pendant vertices. In Sections 4.3 and 4.4 we consider the Dirichlet and Neumann problems which are defined by Dirichlet or (generalized) Neumann conditions at the root, respectively, Dirichlet conditions at all other vertices which are pendant, and generalized Neumann conditions at all other vertices which are interior. In Section 4.3 it is shown that the characteristic values of the Neumann problem and the Dirichlet problem interlace, but in general not strictly. Upper bounds for the multiplicities of the characteristic values are obtained in terms of the structure of the tree. In Section 4.4 the inverse problem is considered for any given rooted tree in case of strict interlacing of the characteristic values. It is shown that for any 2n positive numbers and any given lengths of the edges of the tree, with finitely many exceptions, there are n beads on the tree such that the interlacing characteristic values of the Dirichlet and Neumann problems on the tree are the given 2n positive numbers. The positions of the beads on the tree and their masses are not uniquely determined by the characteristic values. In Chapter 5 we consider spectral problems for general graphs of Stieltjes strings. In Section 5.1 we give a linear matrix polynomial representation z 7→ zM − A of this problem with a diagonal matrix M with nonnegative entries and a symmetric matrix A with real entries. It is shown that if each edge carries at least one bead, then values z for which zM − A is a singular matrix satisfy z ≥ 0. In Section 5.2 an often more convenient matrix polynomial z 7→ Φ(z) of smaller size, but which is not necessarily linear in z, is introduced. It is shown that the characteristic polynomial det Φ(z) is a constant multiple of det(zM − A). In Section 5.3 this characteristic polynomial is expressed in terms of characteristic polynomials for problems on subgraphs. As particular cases, series connection of graphs is considered in Section 5.4, and parallel connection of graphs is considered in Section

xvi

Preface

5.5. In Section 5.6 (sharp) upper bounds for the multiplicities of characteristic values are obtained. These bounds depend on the shape of the graph and possibly on the numbers of (generalized) Dirichlet conditions at pendant vertices and at interior vertices. In particular, cyclically connected graphs, trees and generalized quasi-trees, and problems with only generalized Neumann conditions at interior vertices are considered. In Section 5.7, graphs of symmetric Stieltjes strings are considered. Properties of the characteristic values are obtained, and it is shown that the characteristic values uniquely determine the distribution of the beads. In Chapter 6 we consider self-adjoint differential operators on graphs. In Section 6.1 we briefly recall the definition of selfadjoint systems of differential operators on a bounded interval. We then consider in Section 6.2 selfadjoint differential operators on metric graphs, where all edges of any such graph are supposed to have the same length. Here Neumann or Dirichlet conditions are imposed at the pendant vertices and continuity and Kirchhoff conditions at the interior vertices. In subsequent sections of Chapter 6 it will be assumed that the potential is the same on each edge and that the potential is symmetric. In Sections 6.3 and 6.4 the characteristic values of the boundary value problem on the graph are described as the zeros of entire functions of the form S1s (P ◦S2 ), where s is an integer and P is a polynomial which are determined by the structure of the graph and the boundary conditions at pendant vertices, and S1 and S2 are the characteristic functions of the identical Sturm-Liouville problems on the edges with normalized Dirichlet and Neumann conditions. For the tetrahedral graph, the octahedral graph, the cube graph and for star graphs, the characteristic polynomials are calculated explicitly. In Section 6.5 the particular case of perfect binary trees is considered. In this case, explicit formulas for the polynomial P mentioned earlier in this paragraph are obtained. Finally, in Section 6.6 properties of the eigenvalues are obtained for star graphs with finite edges as well as with infinite edges. In three appendices we provide some background material to the extent it is needed in the main body of this monograph. In Appendix A we present definitions and results with proofs on some polynomials and rational functions such as rational Nevanlinna functions, rational Stieltjes functions and Hermite-Biehler polynomials. Appendix B contains a brief summary on results for linear and quadratic matrix pencils as well as Schur complements. In Appendix C basic definitions and properties in graph theory are given. Then more particular advanced results which are essential in the main body of this monograph are investigated for cyclically connected graphs, quasi-trees and adjacency and incidence matrices.

Chapter 1

Stieltjes strings 1.1

Stieltjes functions in electrical circuits

An important role in the synthesis of electrical circuits is played by the ‘scalar impedance function’ Z(λ) from which the impedance may be derived by writing λ = iω (see [28], [58], [33]), where ω is the cyclic frequency of a sine current. It is known, see [28, Abstract, p. 1], that Z(λ) is a ‘positive real’ function, that is, (i) Z(λ) is a rational function which is real for real values of λ; (ii) the real part of Z(λ) is positive when the real part of λ is positive. O. Brune also proved that for each positive real function Z(λ) a circuit can be found containing resistances Rj , capacities Cj and inductivities Lj such that the given Z(λ) is the impedance function of this circuit. Necessary and sufficient conditions for a function to be positive real are given in [28, p. 33]: (i) no poles lie in the open right half-plane; (ii) pure imaginary poles have positive real finite residues; (iii) Re Z(iω) ≥ 0 for real ω. It should be mentioned that a particular case corresponding to lossless circuits consisting only of Cj ’s and Lj ’s was considered earlier by R. Foster [56].

Figure 1.1

© Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4_1

1

2

Chapter 1. Stieltjes strings

As far as we know, the first who used expansions of rational functions into continued fractions to describe physical phenomena was W. Cauer [33]. In [58] the point of departure was a very simple observation regarding the ladder-type network shown in Figure 1.1. The ordinary laws of circuit theory lead immediately to a continued fraction expansion for the scalar impedance function Z(λ) (see [58, (1)]). For example, if all the series elements in Figure 1 are resistances Rj , j = 1, . . . , n, and all the shunt elements are capacities Cj , j = 1, . . . , n, indexed from right to left, then (see [33, (20)] and [58, (2)]) 1

Z(λ) = Rn +

.

1

Cn λ + Rn−1 +

(1.1.1)

1 Cn−1 λ + . .

1

.+

R1 +

1 C1 λ

If again all the series elements are resistances Rj , j = 1, . . . , n but all the shunt elements are inductances Lj , j = 1, . . . , n, then (see [33, (26)]) 1

Z(λ) = Rn +

,

1

Mn z + Rn−1 +

(1.1.2)

1 Mn−1 z + . .

.+

1 R1 +

1 M1 z

where z = λ−1 and Mj = L−1 j , j = 1, . . . , n. Infinite continued fractions of the form 1 (1.1.3) 1 a1 z + 1 a2 + a3 z + . . 1 .+ 1 a2n + + .. a2n−1 z . with positive coefficients aj , j = 1, 2, . . . , were introduced by T.-J. Stieltjes in [128] and are hence called Stieltjes continued fractions. We quote the following assessment of the culmination of classical network synthesis from [32, p. 7]. “Darlington’s Theorem: Any positive real function Z(λ) admits a realization as the input impedance of a lossless (frequency-dependent) two-port terminated in a positive (frequency-independent) resistor R.

1.2. Boundary value problems for Stieltjes strings

3

In other words, the set of positive real rational functions Z(λ) admits a linear fractional parametrization   A(λ)R + B(λ) A(λ) B(λ) Z(λ) = , T(λ) = C(λ) D(λ) C(λ)R + D(λ) where T(λ) sweeps through the set of rational chain matrices of lossless two-ports. The numerous applications of this lossless embedding of passive impedances stems from its canonical separation of thermodynamical aspects (dissipation of energy) and Hamiltonian concepts (storage of energy, dynamics, holonomic constraints or frequency dependence) within the class of linear passive systems. The theorem was apparently devised by W. Cauer [34], G. Cocci [39], and S. Darlington [45], all independently of the other. Cauer’s proof seems to be not only the shortest (it requires only half a printed page, [34] (p. 232]) but also the most elegant and forward-looking. Instead of dealing with positive real impedances Z(λ), Cauer applies a bilinear transform T (λ) =

g(λ) Zλ) + 1 := Z(λ) − 1 h(λ)

and discusses all rational solutions of a (spectral) factorization of the rational (and for λ2 = −ω 2 positive) function T (λ)T (−λ) − 1 :=

f (λ)f (−λ) . h(λ)h(−λ)

The polynomials f , g, h, in which g(λ) is Hurwitz, are called Betriebskenngr¨oßen by Cauer because they are the key quantities in the design of reactance filters whose preassigned characteristics are specified in terms of insertion loss (Betriebsd¨ ampfung). Cauer’s polynomials f , g, h still survive (in the very same notation!) in Belevitch’s celebrated canonic form of the scattering matrix of a real lossless twoport [9] p. 428 (note 8).” Generalizations of Darlington’s method to the infinite-dimensional case can be found in [4].

1.2

Boundary value problems for Stieltjes strings

1.2.1 Boundary value problems Following M.G. Krein, a Stieltjes string of length l is a massless string bearing n beads of masses m1 , . . . , mn in its interior, where n ∈ N. Selecting one of the endpoints and calling it the left endpoint, the beads are indexed in such a way that the distance from the left endpoint increases with the index. Denoting the distance from the left endpoint to the first bead by l0 , the distance between the j-th bead and the j + 1-th bead by lj , j = 1, . . . , n − 1, and the distance from the

4

Chapter 1. Stieltjes strings

last bead to the right endpoint by ln , it is clear that

n P

lk = l. In other words,

k=0

the beads with masses m1 , . . . , mn divide the string into n + 1 threads (massless strings) of length l0 , . . . , ln . When a Stieltjes string with the location of the beads on the string and the masses of the beads is given, we will briefly call this a distribution of beads (on the string). In particular, the positive numbers m1 , . . . , mn and l0 , . . . , ln determine a unique distribution of beads. In later chapters, we will consider (particular) graphs of Stieltjes strings, and the notion distribution of beads would apply there accordingly. Sometimes it will be convenient to allows strings without beads, i.e., n = 0. The string is assumed to be stretched by a stretching force equal to 1. Taking into account that on the threads the string has zero density, the general solution of the differential equation of the string is a linear function of the position variable for each of the n + 1 threads. Hence small transverse vibrations of such a Stieltjes string are described by the three term recurrence relations vk (t) − vk−1 (t) vk+1 (t) − vk (t) − + mk vk00 (t) = 0 lk−1 lk

(k = 1, . . . , n),

(1.2.1)

where vk (t) is the transverse displacement of the k-th bead for k = 1, . . . , n at time t and v0 (t) and vn+1 (t) are the transverse displacements of the left and right endpoints of the string, respectively. The Dirichlet-Dirichlet problem is generated by the equations (1.2.1) and the boundary conditions v0 (t) = 0, vn+1 (t) = 0,

(1.2.2)

and the Neumann-Dirichlet problem is generated by the equations (1.2.1) and boundary conditions v0 (t) = v1 (t), vn+1 (t) = 0. (1.2.3) Substituting the separation of variables ansatz vk (t) = uk eiλt into (1.2.1), we obtain uk − uk−1 uk+1 − uk − − mk λ2 uk = 0 lk−1 lk

(k = 1, . . . , n, λ ∈ C),

(1.2.4)

for the amplitudes uk of the vibrations of the beads and the endpoints of the string, k = 0, . . . , n + 1, see [59, Supplement II, (13)]. Figure 1.2 illustrates the problem: l0

l1 m1 m2 Figure 1.2

ln mn

5

1.2. Boundary value problems for Stieltjes strings

Remark 1.2.1. The aim is to find nontrivial solution (u0 , u1 , . . . , un+1 ) of (1.2.4) which satisfy certain boundary conditions, one at the left endpoint and one at the right endpoint. The values of λ ∈ C for which such nontrivial solutions exist are called eigenvalues of the corresponding boundary value problem, and the set of all eigenvalues is called the spectrum of the boundary value problem. Since the eigenvalue parameter λ occurs quadratic in (1.2.4), it is convenient to replace λ2 with z. Then we call a number z ∈ C a characteristic value of the corresponding boundary value problem if this problem has a nontrivial solution. Clearly, a number λ ∈ C is an eigenvalue if and only if z = λ2 is a characteristic value, and in that case also −λ is an eigenvalue. Therefore the eigenvalues are uniquely determined by the characteristic values, and in this chapter we will therefore mainly consider characteristic values. That is, when referring to the recursion relation (1.2.4) we will often replace λ2 with z. The boundary conditions (1.2.2) or (1.2.3) would give u0 = 0 or u0 = u1 , respectively, but it is convenient to admit more general boundary condition at the left endpoint of the form u0 = cu1 (c ∈ R). (1.2.5) Similarly, at the right endpoint we admit the boundary condition un+1 = dun (d ∈ R). For c, d ∈ R consider the tridiagonal n × n matrix  1−c 1 − l11 0 l0 + l1 1 1 1 1  − + − 0  l1 l1 l2 l2   0 − l12 l12 + l13 − l13    Adc =       1 1 1  − ln−2 ln−2 + ln−1  1 0 0 − ln−1

(1.2.6)



0 0

0 1 − ln−1 1 ln−1

+

1−d ln

              

(1.2.7) in case n > 1 and

Adc

=



1−c l0

+

1−d l1



in case n = 1.

The following result is easily verified. Proposition 1.2.2. Let c, d ∈ R, z ∈ C, and Y = (u1 , . . . , un )T ∈ Cn \ {0}. Define u0 and un+1 by (1.2.5) and (1.2.6), respectively. Then (1.2.4) is satisfied if and only if (Adc − zM )Y = 0, where M = diag(m1 , . . . , mn ).

6

Chapter 1. Stieltjes strings

By Proposition 1.2.2, problem (1.2.4), (1.2.5), (1.2.6) has a nontrivial solution if and only if det(Adc − zM ) = 0. The polynomials defined by z 7→ det(Adc − zM ) and λ 7→ det(Adc − λ2 M ) and any of their nonzero multiples will be called characteristic polynomial of (1.2.4), (1.2.5), (1.2.6). It will be clear from the context which of those polynomials we mean. Hence the eigenvalues and characteristic values of (1.2.4), (1.2.5), (1.2.6) are the zeros of the characteristic polynomial. In this chapter, the eigenvalues will always occur in pairs (λ, −λ) and it is therefore sufficient and more convenient to consider the characteristic values z = λ2 instead. Proposition 1.2.3. Let c, d ∈ (−∞, 1]. Then the matrix Acd is symmetric and positive semidefinite, and all eigenvalues of the spectral problem for Adc − zM with spectral parameter z are nonnegative and simple. The number 0 is an eigenvalue if and only if c = 1 and d = 1. Proof. The symmetry of Acd is obvious. For the vectors Y = (u1 , . . . , un )T 6= 0 and Y ∗ = (u1 , . . . , un ) we calculate Y ∗ Adc Y =

n−1 X j=1

−

n−1 X 1 1 1−c 1−d |uj |2 + |uj+1 |2 + |u1 |2 + |un |2 lj l l l j 0 n j=1 n−1 X j=1

=

1 (uj uj+1 + uj uj+1 ) lj

n−1 X 1 1−c 1−d |u1 |2 + |un |2 + |uj+1 − uj |2 , l0 ln l j j=1

which means that Adc is positive semidefinite and that Adc is positive definite when c 6= 1 or d 6= 1. Since A11 (1, . . . , 1)T = 0, it follows that 0 is an eigenvalue of the spectral problem when c = 1 and d = 1. Furthermore, the lower left (n−1)×(n−1) submatrix of Adc − zM is upper triangular with nonzero diagonal elements, which implies that for all z ∈ C the rank of Adc − zM is at least n − 1. Therefore all of its eigenvalues are simple. An application of Proposition B.2.1 completes the proof. 

1.2.2 Cauer-Fry polynomials For properties of characteristic values beyond those found in Proposition 1.2.3 we are going to use a different representation of the characteristic polynomials. When u1 = 0, then (1.2.5) would give u0 = 0, and a recursive substitution into (1.2.4) would lead to the trivial solution 0 = u0 = u1 = · · · = un+1 . Therefore, when searching for nontrivial solutions we may assume u1 6= 0. Then, for each z ∈ C and each u1 6= 0, (1.2.4) together with (1.2.5) has a unique solution

7

1.2. Boundary value problems for Stieltjes strings which can be written in the form uk = R2k−2 (z)u1

(k = 0, . . . , n + 1).

(1.2.8)

Substituting (1.2.8) into (1.2.4), we arrive at R2k−2 (z) − R2k−4 (z) R2k (z) − R2k−2 (z) − = mk zR2k−2 (z) lk−1 lk

(1.2.9)

for k = 1, . . . , n. Introducing R2k−1 (z) =

R2k (z) − R2k−2 (z) lk

(k = 0, . . . , n),

(1.2.10)

(1.2.9) is equivalent to R2k−1 (z) = R2k−3 (z) − mk zR2k−2 (z)

(k = 1, . . . , n).

(1.2.11)

The polynomials Rk (k = −2, . . . , 2n) will be called the Cauer-Fry polynomials of (1.2.4), (1.2.5). In order to solve the three term recurrence relations (1.2.10) and (1.2.11) we observe that (1.2.8) for k = 1 gives the natural initial value R0 (z) = 1. Then condition (1.2.5) gives R−2 (z) = c. To account for the dependence on c we will also write Rk (·, c) for the Cauer-Fry polynomials satisfying the initial conditions R0 (z, c) = 1,

(1.2.12)

R−2 (z, c) = c.

(1.2.13)

Then the recurrence relations (1.2.10) and (1.2.11) with the initial values (1.2.12) and (1.2.13) have a unique solution Rk (z, c) for Rk (z), k = −1, 1, 2, . . . , 2n. Indeed, R−1 (z, c) =

1−c , l0

(1.2.14)

and for R1 (z, c), . . . , R2n (z, c), the equation (1.2.11) is used for the odd indices whereas (1.2.10) is used for the even indices. It is easy to see by induction that R2k−1 (·, c) and R2k (·, c) are polynomials of degree k for k = 0, . . . , n. Equation (1.2.11) for z = 0 and (1.2.14) imply that R2n−1 (0, c) = R2n−3 (0, c) = · · · = R−1 (0, c) =

1−c . l0

(1.2.15)

Then (1.2.10) and (1.2.15) lead to R2k (0, c) =

k X

lj R2j−1 (0, c) + R0 (0, c)

j=1

(1.2.16)

k

1−cX =1+ lj l0 j=1

(k = 0, . . . , n).

8

Chapter 1. Stieltjes strings

In particular, l . (1.2.17) l0 Lemma 1.2.4. The recursion for the Cauer-Fry polynomials can be written as      R2k−1 (z, c) 1 −mk z R2k−3 (z, c) = (k = 1, . . . , n). (1.2.18) R2k (z, c) lk 1 − lk mk z R2k−2 (z, c) R2n (0, c) = c + (1 − c)

Proof. From (1.2.10) it follows that     R2k−1 (z, c) R2k−1 (z, c) = R2k (z, c) lk R2k−1 (z, c) + R2k−2 (z, c)    1 0 R2k−1 (z, c) = , lk 1 R2k−2 (z, c) and (1.2.11) gives     R2k−1 (z, c) R2k−3 (z, c) − mk zR2k−2 (z, c) = R2k−2 (z, c) R2k−2 (z, c)    1 −mk z R2k−3 (z, c) = . 0 1 R2k−2 (z, c) Multiplying the two coefficient matrices completes the proof.



Lemma 1.2.4 immediately gives R2k (z, c) lk R2k−3 (z, c) + (1 − lk mk z)R2k−2 (z, c) = R2k−1 (z, c) R2k−3 (z, c) − mk zR2k−2 (z, c) 1 = lk + 1 −mk z + R2k−2 (z, c) R2k−3 (z, c) for k = 2, . . . , n. A recursive substitution and the initial conditions (1.2.12), (1.2.14) lead to the continued fraction expansion R2n (z, c) = ln + R2n−1 (z, c)

1

.

1

−mn z + ln−1 +

1 −mn−1 z + . .

1

.+ l1 +

1 −m1 z +

1−c l0 (1.2.19)

1.2. Boundary value problems for Stieltjes strings

9

Proposition 1.2.5. Let c ∈ (−∞, 1]. Then R2n (·, c) and R2n−1 (·, c) have n simple zeros, which will be denoted by (νj (c))nj=1 and (µj (c))nj=1 , respectively, where νj (c) < νj+1 (c) and µj (c) < µj+1 (c) (j = 1, . . . , n − 1). The zeros of R2n (·, c) and R2n−1 (·, c) interlace as follows: 0 ≤ µ1 (c) < ν1 (c) < µ2 (c) < · · · < µn (c) < νn (c),

(1.2.20)

with µ1 (c) = 0 if and only if c = 1. Proof. Some of the stated properties have already been shown in Proposition 1.2.3. However, since R2n (·, c) and R2n−1 (·, c) are polynomials of degree n and since 2n (·,c) (1.2.19) and Lemma A.3.6 show that RR2n−1 (·,c) is a rational S-function with n 2n (·,c) poles, there is no cancellation of zeros and poles. Hence the poles of RR2n−1 (·,c) are the zeros of R2n−1 (·, c), whereas its zeros are the zeros of R2n (·, c). Therefore all statements of this proposition follows from Lemma A.3.5. 

Proposition 1.2.6. Let c, d ∈ R. Then a characteristic polynomial of (1.2.4), (1.2.5), (1.2.6) is (1 − d)R2n (·, c) + dln R2n−1 (·, c). (1.2.21) Proof. By construction of the Cauer-Fry polynomials, a nontrivial solution (uk )n+1 k=0 of (1.2.4), (1.2.5) is (R2k (z, c))nk=−1 . Then un+1 − dun = (1 − d)un+1 + d(un+1 − un ) = (1 − d)R2n (z, c)u1 + dln R2n−1 (z, c)u1 . Therefore, the (simple) zeros of the polynomials z 7→ det(Adc − zM ) and (1.2.21) of degree n coincide (note that R2n (·, c) and ln R2n−1 (·, c) have the same leading coefficient by (1.2.10)). Hence these two polynomials are constant multiples of each other, which completes the proof in view of the definition of a characteristic polynomial.  Remark 1.2.7. 1. It is clear that the zeros of R2n−1 (·, c), that is, the poles of the continued fraction expansion on the right-hand side of (1.2.19), do not depend on ln . 2. If c = 1, then the zeros of R2n (·, 1) as well as the zeros of R2n−1 (·, 1) do not depend on l0 since the right-hand side of (1.2.19) is independent of l0 . 3. Let the string data (mk )nk=1 and (lk )nk=0 be given, let η > 0, and define m ek = η −1 mk (k = 1, . . . , n) and e lk = ηlk (k = 0, . . . , n) with the corresponding e2n (·, c) and R e2n−1 (·, c). Then it is clear that Cauer-Fry polynomials R e2n (·, c) R R2n (·, c) =η . e R 2n−1 (·, c) R2n−1 (·, c) e2n (·, c) and of R2n (·, c) coincide, and the zeros of In particular, the zeros of R e R2n−1 (·, c) and of R2n−1 (·, c) coincide.

10

Chapter 1. Stieltjes strings

4. For all c ∈ R, the identity Rk (·, c) = (1 − c)Rk (·, 0) + cRk (·, 1) holds when k = −1, 2 in view of (1.2.12) and (1.2.14), and hence the identity is also true for k = 1, . . . , 2n in view of the recursions (1.2.11) and (1.2.10). 5. The coefficients of the Cauer-Fry polynomials Rj (·, c) depend continuously on m1 , . . . , mn , l0 , . . . , ln and c, and for j ≥ 1, their degree is positive and independent of these parameters. Due to Lemma A.6.5 it follows for j = 1, . . . , 2n that the zeros of Rj (·, c) depend continuously on m1 , . . . , mn , l0 , . . . , ln and c.

1.2.3

Dirichlet and Neumann boundary conditions

Dirichlet boundary condition at the left endpoint Here we consider the case that the left end of the string is clamped, that is, v0 (t) = 0.

(1.2.22)

u0 = 0,

(1.2.23)

This gives the Dirichlet condition

which is (1.2.5) with c = 0. Dirichlet-Dirichlet problem. Dirichlet condition

If the right end of the string is clamped, then the un+1 = 0

(1.2.24)

is obtained, which is (1.2.6) with d = 0. Proposition 1.2.8. 1. R2n (·, 0) is the characteristic polynomial of problem (1.2.4) with Dirichlet-Dirichlet conditions (1.2.23), (1.2.24). 2. The zeros of R2n (·, 0) are real, positive and simple. Denoting these zeros by νk (k = 1, . . . , n) with νk < νk+1 for k = 1, . . . , n − 1, it follows that  n  l Y z R2n (z, 0) = 1− . l0 νk

(1.2.25)

k=1

Proof. Statement 1 is Proposition 1.2.6 for c = 0 and d = 0. Then statement 2 follows from Proposition 1.2.5 if we observe that the left-hand side and the right-hand side of (1.2.25) coincide for z = 0 in view of (1.2.17).  Dirichlet-Neumann problem. Here the right end of the string can move freely in the direction orthogonal to the equilibrium position of the string, that is, vn+1 (t) = vn (t).

(1.2.26)

11

1.2. Boundary value problems for Stieltjes strings This gives the Neumann condition un+1 = un ,

(1.2.27)

which is (1.2.6) with d = 1. Proposition 1.2.9. 1. R2n−1 (·, 0) is the characteristic polynomial of problem (1.2.4) with Dirichlet-Neumann conditions (1.2.23), (1.2.27). 2. The zeros of R2n−1 (·, 0) are real, positive and simple. Denoting these zeros by µk (k = 1, . . . , n) with µk < µk+1 for k = 1, . . . , n − 1, it follows that R2n−1 (z, 0) =

 n  1 Y z 1− . l0 µk

(1.2.28)

k=1

Proof. Statement 1 is Proposition 1.2.6 for c = 0 and d = 1 since we can omit the factor ln . Then statement 2 follows from Proposition 1.2.5 if we observe that the left-hand side and the right-hand side of (1.2.28) coincide for z = 0 in view of (1.2.15).  Neumann boundary condition at the left endpoint Now assume that the left end of the string can move freely in the direction orthogonal to the equilibrium position of the string. This gives the Neumann condition u0 = u1 ,

(1.2.29)

that is, (1.2.5) with c = 1. Neumann-Dirichlet problem. If the right end of the string is clamped, then the Dirichlet condition (1.2.24) is obtained. Since the Neumann-Dirichlet problem is identical to the Dirichlet-Neumann problem with reversed indexing of the beads and endpoints, Proposition 1.2.9 and (1.2.17) lead to Proposition 1.2.10. 1. R2n (·, 1) is the characteristic polynomial of problem (1.2.4) with Neumann-Dirichlet conditions (1.2.24), (1.2.29). 2. The zeros of R2n (·, 1) are real, positive and simple. Denoting these zeros by ξk (k = 1, . . . , n) with ξk < ξk+1 for k = 1, . . . , n − 1, it follows that R2n (z, 1) =

n  Y k=1

z 1− ξk

 .

(1.2.30)

12

Chapter 1. Stieltjes strings

Neumann-Neumann problem. If the right end of the string can move freely in the direction orthogonal to the equilibrium position of the string, then the Neumann condition (1.2.27) is obtained. Proposition 1.2.11. 1. R2n−1 (·, 1) is the characteristic polynomial of problem (1.2.4) with Neumann-Neumann conditions (1.2.24), (1.2.29). 2. The zeros of R2n−1 (·, 1) are real, nonnegative and simple, and R2n−1 (0, 1) = 0. Denoting the zeros by ζk , k = 1, . . . , n with ζk < ζk+1 for k = 1, . . . , n − 1, it follows that ζ1 = 0 and ! n   n Y X z mk z R2n−1 (z, 1) = − 1− . (1.2.31) ζk k=2

k=1

Proof. Statement 1 is Proposition 1.2.6 for c = 1 and d = 1 since we can omit the factor ln . Then statement 2 follows from Proposition 1.2.5 and the fact that the derivatives on the left-hand side and the right-hand side of (1.2.31) coincide for z = 0. Indeed, differentiating (1.2.11) and observing (1.2.16) gives 0 0 R2k−1 (0, 1) = R2k−3 (0, 1) − mk

(k = 1, . . . , n).

0 A recursive substitution and R−1 (0, 1) = 0, see (1.2.14), show that

0 R2n−1 (0, 1) = −

n X

mk .



k=1

Proposition 1.2.12. Let Rk− (·, c) be the Cauer-Fry polynomials for the same string, but with opposite orientation. Then − ln R2n (·, 0) = l0 R2n (·, 0), − ln R2n−1 (·, 0) − R2n (·, 1) − R2n−1 (·, 1)

(1.2.32)

= R2n (·, 1),

(1.2.33)

= l0 R2n−1 (·, 0),

(1.2.34)

= R2n−1 (·, 1).

(1.2.35)

− Proof. The zeros of the polynomials R2n (·, 0) and R2n (·, 0) are the characteristic values of the Dirichlet-Dirichlet problem for the original string as well as the string with opposite orientation. Hence these two polynomials are multiples of each other. Then (1.2.32) follows from (1.2.17). − The zeros of the polynomials R2n (·, 1) and R2n−1 (·, 0) are the characteristic values of the Neumann-Dirichlet problem for the original string and the DirichletNeumann problem for the string with opposite orientation, respectively. But these two problems are the same and hence these two polynomials are multiples of each other. Then (1.2.33) follows from (1.2.15) and (1.2.17). The identity (1.2.34) is (1.2.33) for the string with opposite orientation.

13

1.2. Boundary value problems for Stieltjes strings

− The zeros of the polynomials R2n−1 (·, 1) and R2n−1 (·, 1) are the characteristic values of the Neumann-Neumann problem for the original string as well as the string with opposite orientation. Hence these two polynomials are multiples of each other. Then (1.2.35) follows from (1.2.31). 

Comparison of the characteristic values Lemma 1.2.13. The following interlacing holds: 0 < µ1 < ν1 < · · · < µn < νn ;

(1.2.36)

0 < ξ1 < ν1 < · · · < ξn < νn ;

(1.2.37)

0 = ζ1 < µ1 < · · · < ζn < µn ;

(1.2.38)

0 = ζ1 < ξ1 < · · · < ζn < ξn .

(1.2.39)

Proof. The interlacings (1.2.36) and (1.2.39) are (1.2.20) for c = 0 and c = 1, oles of µk respectively. Interchanging the ends of the string will interchange the rˆ and ξk , and hence also (1.2.37) and (1.2.38) hold.  For two sets X and Y of real numbers, the notation X < Y will mean that x < y for all x ∈ X and all y ∈ Y . It will be convenient to put ν0 := 0 and ζn+1 := νn . Then (1.2.36)–(1.2.39) is equivalent to ζ1 = 0 and {νk−1 , ζk } < {µk , ξk } < {νk , ζk+1 }

1.2.4

(k = 1, . . . , n).

(1.2.40)

Robin-Dirichlet and Robin-Neumann problems

In this subsection, a Robin boundary condition v1 (t) − v0 (t) − γv0 (t) = 0 l0

(1.2.41)

with some parameter γ > 0 is imposed at the left end of the string. Physically, this means that the left end of the string is prolonged by a thread that is fixed at distance γ −1 to the left of the string; another interpretation is that Robin conditions simulate an elastic support at the endpoint of the string (see [35, p. 15]). Substituting the separation of variables ansatz vk (t) = uk eiλt (1.2.41), we obtain u1 − u0 − γ u0 = 0. l0

(1.2.42)

The boundary condition (1.2.42) can be written in the form u0 =

u1 =: γ0 u1 . 1 + γl0

(1.2.43)

14

Chapter 1. Stieltjes strings

Clearly, (1.2.43) is of the form (1.2.5) with c = γ0 ∈ (0, 1). In particular, (1.2.15) and (1.2.17) give R2n−1 (0, γ0 ) = R2n−3 (0, γ0 ) = · · · = R−1 (0, γ0 ) = R2n (0, γ0 ) = γ0 + (1 − γ0 )

1 − γ0 γ = , l0 1 + γl0

l 1 + γl = . l0 1 + γl0

(1.2.44) (1.2.45)

Robin-Dirichlet problem. If the right end of the string is clamped, then the Dirichlet condition (1.2.24) is obtained. Robin-Neumann problem. If the right end of the string can move freely in the direction orthogonal to the equilibrium position of the string, then the Neumann condition (1.2.27) is obtained. Theorem 1.2.14. 1. R2n (·, γ0 ) is the characteristic polynomial of problem (1.2.4) with Robin-Dirichlet conditions (1.2.42), (1.2.24). 2. R2n−1 (·, γ0 ) is the characteristic polynomial of problem (1.2.4) with RobinNeumann conditions (1.2.42), (1.2.27). 3. The zeros of R2n (·, γ0 ) as well as the zeros of R2n−1 (·, γ0 ) are real and sim(γ) ple, and the sequence (νk )nk=1 of the zeros of R2n (·, γ0 ) and the sequence (γ) n (µk )k=1 of the zeros of R2n−1 (·, γ0 ) can be indexed in such a way that (γ)

0 < µ1

(γ)

< ν1

(γ) < · · · < µ(γ) n < νn .

(1.2.46)

Proof. Statement 1 and 2 are Proposition 1.2.6 for c = γ0 and d = 0 or d = 1, respectively. Statement 3 follows from Proposition 1.2.5.  Remark 1.2.15. We have mentioned at the beginning of this subsection that the Robin condition can be considered as a Dirichlet condition for a prolonged string. Indeed, defining 1 u1 − u0 b l0 = l0 + , u b0 = u0 − , γ l0 γ a straightforward calculation gives u1 − u b0 u1 − u0 = . b l0 l0 It is easy to see that c ∈ (0, 1) can be written as c = γ0 for γ = l10 ( 1c − 1) > 0. Together with the corresponding procedure at the right end it therefore follows that the problem (1.2.4), (1.2.5), (1.2.6) with c, d ∈ (0, 1) is equivalent to the c Dirichlet-Dirichlet problem for a string which is prolonged by l0 1−c on the left d and by ln 1−d on the right.

15

1.2. Boundary value problems for Stieltjes strings

1.2.5

Lagrange identity and Sturm oscillation theorem

The following theorem is the discrete analogue of of the Lagrange identity in the theory of differential equations. Theorem 1.2.16 (Lagrange identity). For all c1 , c2 ∈ R and k = 1, . . . , n, R2k−1 (z, c1 )R2k (z, c2 ) − R2k (z, c1 )R2k−1 (z, c2 ) =

c2 − c1 . l0

(1.2.47)

Proof. The recurrence relations (1.2.10) and (1.2.11) give for k = 2, . . . , n that R2k−1 (z, c1 )R2k (z, c2 ) − R2k (z, c1 )R2k−1 (z, c2 ) = R2k−1 (z, c1 )R2k−2 (z, c2 ) − R2k−2 (z, c1 )R2k−1 (z, c2 ) = R2k−3 (z, c1 )R2k−2 (z, c2 ) − R2k−2 (z, c1 )R2k−3 (z, c2 ). Proceeding this way and using (1.2.12) and (1.2.14) we arrive at R2k−1 (z, c1 )R2k (z, c2 ) − R2k (z, c1 )R2k−1 (z, c2 ) = R−1 (z, c1 )R0 (z, c2 ) − R0 (z, c1 )R−1 (z, c2 ) =

c2 − c1 . l0



Given three of the values on the left-hand side of the Lagrange identity allows to find the fourth one, provided the other factor in the corresponding product is not zero. The following two examples show that one can find string problems where both factors have the value 0 at the right endpoint for some positive z. Example 1.2.17. Let n ≥ 3. Then there is a string with n beads such that R2n (2, 0) = 0,

R2n−1 (2, 0) = 1,

R2n (2, 1) = 1,

R2n−1 (2, 1) = 0.

(1.2.48)

One can choose the string with lk = 1 (k = 0, . . . , n) and mk = 1 (k = 1, . . . , n) except for the following string data: (i) if n = 4b n − 1 (b n ∈ N), then m2bn = 2; (ii) if n = 4b n (b n ∈ N), then l2bn = l2bn+2 = 12 ; (iii) if n = 4b n + 1 (b n ∈ N), then m2bn−1 = m2bn = m2bn+2 = m2bn+3 = 12 ; (iv) if n = 4b n + 2 (b n ∈ N), then l2bn−1 = l2bn = l2bn+2 = l2bn+3 = 32 . Proof. Defining  Tb(a, b) :=

 1 −2b , a 1 − 2ab

Tn := Tb(ln , mn ) · · · Tb(l1 , m1 ),

it follows from Lemma 1.2.4 that       R2n−1 (2, c) R−1 (2, c) 1−c = Tn = Tn , R2n (2, c) R0 (2, c) 1

16

Chapter 1. Stieltjes strings

where we have used that l0 = 1. Then (1.2.48) will hold if and only if   1 0 Tn = . −1 1

(1.2.49)

Observing Tb(1, 1)2 =



1 1

−2 −1

 1 1

−2 −1



 −1 = 0

 0 −1

n − 1) + n0 with n0 ∈ {3, 4, 5, 6} it follows from lk = 1 and and writing n = 4(b mk = 1 for k = 1, . . . , 2b n − 2 and k = 2b n + n0 − 1, . . . , n that Tn = (Tb(1, 1))2(bn−1) Ten0 (Tb(1, 1))2(bn−1) = (−I)nb−1 Ten0 (−I)nb−1 = Ten0 , where Ten0 = Tb(l2bn+n0 −2 , m2bn+n0 −2 ) . . . Tb(l2bn−1 , m2bn−1 ). Hence (1.2.49) is equivalent to  1 e Tn0 = −1

 0 1

(n0 = 3, 4, 5, 6).

(1.2.50)

A straightforward calculation shows that (1.2.50) is true for the data as given by  (i)–(iv) and the remaining parameters being 1. Example 1.2.18. Let n ∈ N \ {2}. Then there is a string with n beads such that R2n (2, 0) = 1,

R2n−1 (2, 0) = 0,

R2n (2, 1) = 0,

R2n−1 (2, 1) = −1. (1.2.51)

One can choose the string with lk = 1 (k = 0, . . . , n) and mk = 1 (k = 1, . . . , n) except for the following string data: n ∈ N), then l2bn+1 = 3, m2bn−1 = n − 1 (b (i) if n = 4b (ii) if n = 4b n (b n ∈ N), then l2bn+2 = 2 and m2bn+1 =

3 2 and 3 2;

m2bn = 2;

(iii) if n = 4b n + 1 (b n ∈ N ∪ {0}), then m2bn+1 = 12 ; (iv) if n = 4b n + 2 (b n ∈ N), then m2bn−1 = m2bn = m2bn+3 = m2bn+4 = 12 . Proof. With the notation and reasoning from Example 1.2.17 it it clear that (1.2.51) holds if and only if   1 −1 Tn = , (1.2.52) 1 0 and that this holds if and only if  1 Ten0 = 1

−1 0

 (n0 = 1, 3, 4, 6).

(1.2.53)

A straightforward calculation shows that (1.2.53) is true for the data as given by (i)–(iv) and the remaining parameters being 1. 

1.2. Boundary value problems for Stieltjes strings

17

Define a partition X = {0 = x0 < x1 < · · · < xn < xn+1 = l} of [0, l] by xk = xk−1 + lk−1

(k = 1, . . . , n).

(1.2.54)

Then x n + ln =

n−1 X

lj + ln = l = xn+1 ,

j=0

and hence (1.2.54) also holds for k = n + 1. Therefore lk = xk+1 − xk

(k = 0, . . . , n),

and (1.2.4) can be written in the form uk − uk−1 uk+1 − uk − − mk zuk = 0 (k = 1, . . . , n). xk − xk−1 xk+1 − xk

(1.2.55)

Let z ∈ C and let u = (u0 , . . . , un+1 ) be a displacement vector given by a solution of the recurrence relation (1.2.55). It is easy to see that any two consecutive entries uk and uk+1 (k = 0, . . . , n) uniquely determine u. In particular, if u 6= 0, then uk 6= 0 or uk+1 6= 0 for all k = 0, . . . , n. Clearly, the set of solutions u is a two-dimensional vector space. The corresponding movement of the Stieltjes string under the separation ansatz is described by the real or imaginary part of φ(x)eiλt , when λ is a real parameter with z = λ2 and x is the distance of a point on the string in equilibrium position from the left endpoint. The amplitude function φ is piecewise linear and continuous, and its graph is given by the line segments connecting (xk , uk ) and (xk+1 , uk+1 ) for k = 0, . . . , n. Observe that the function φ is uniquely determined by the partition X and the displacement vector u, and we therefore also denote φ by φX,u . The following figures show two examples of the amplitude function with n = 6, l0 = 13, l1 = 18, l2 = 11, l3 = 20, l4 = 22, l5 = 15, l6 = 19, m1 = 1, m2 = 2, m3 = 1, m4 = 2, m5 = 1, m6 = 2, λ = 0.25, where the dotted line is the equilibrium position of the string:

Figure 1.3: u0 = 1 and u1 = 2

Figure 1.4: u0 = 0 and u1 = 1

18

Chapter 1. Stieltjes strings

A point x ∈ [0, l] is called a node of u if φ(x) = 0. A node x ∈ (0, l) is called an internal node. For example, in each of the above two examples there are three internal nodes. It is convenient to consider nodes as partition points of the Stieltjes string. To this end we generalize Stieltjes strings to strings which may have massless e = {0 = x e1 < · · · < x e0 < x em+1 } partition points. Hence let m ≥ n and let X em < x e be a partition of [0, l] with X ⊂ X. This means x em+1 = xn+1 , and e0 = x0 , x eτk that there is a strictly increasing sequence (τk )n+1 k=0 such that such that xk = x e k = 0 if e τk = mk for k = 1, . . . , n and m for k = 0, . . . , n + 1. Then we define m j ∈ {1, . . . , m} but j 6= τk for all k = 1, . . . , n, and consider the recurrence relation u ek−1 ek − u u ek+1 − u ek − −m e k ze uk = 0 (k = 1, . . . , m). x ek − x ek−1 x ek+1 − x ek

(1.2.56)

The recurrence relations (1.2.55) and (1.2.56) have unique solutions u = (u0 , . . . , un+1 ) and u e = (e u0 , . . . , u en+1 ) subject to initial conditions for u0 , u1 and u e0 , u e1 , respectively. Recall that we denote the corresponding piecewise linear continuous functions on [0, l] by φX,u , and φX,e e u , respectively. e be as above and let u Lemma 1.2.19. Let X, u and X e = (e u0 , . . . , u em+1 ) be the solution of (1.2.56) with the initial values u e0 = u0 and u e1 = u e0 + xxe11 (u1 − u0 ). Then φX,e e u = φX,u . Proof. By induction on k = 1, . . . , m + 1 we will show that the two functions φX,e e u and φX,u coincide on the interval [0, xk ] for k = 1, . . . , n. Since u e1 − u e0 u1 − u0 = , x e1 x1 both φX,e e1 ] with the same initial point φX,e e u and φX,u are linear on [0, x e u (0) = u0 = φX,u (0) and the same slope. This proves the induction base if x e1 = x1 . Otherwise, τ1 > 1, and (1.2.56) for j = 1, . . . , τ1 − 1 gives u ej − u ej−1 u ej+1 − u ej = , x ej+1 − x ej x ej − x ej−1 that is, all line segments of the graph of φX,e e u between 0 and x1 have the same slope, and we already know that this slope equals the slope of φX,u on [0, x1 ]. This shows that φX,e e1 6= x1 . e u coincides with φX,u on [0, x1 ] also when x Now let k ∈ {1, . . . , n} and assume that φX,e e u coincides with φX,u on [0, xk ]. Then u eτk +1 − u eτk u eτ − u eτk −1 = k −m e τk ze uτk x eτk +1 − x eτk x eτk − x eτk −1 uk − uk−1 = − mk zuk xk − xk−1 uk+1 − uk = , xk+1 − xk

19

1.2. Boundary value problems for Stieltjes strings

where we have used that by induction hypothesis the slopes of φX,e e u and φX,u coincide on [xk−1 , xk ] and thus on [e xτk −1 , xk ]. Hence φX,e e u and φX,u have the same slope on [xk , x eτk +1 ] and the same initial value uk = u eτk at xk = x eτk . Proceeding as in the first part of the proof it follows that φX,e e u and φX,u coincide on [xk , xk+1 ].  Theorem 1.2.20. Let z1 and z2 be nonnegative real numbers satisfying z2 ≥ z1 and let u = (u0 , . . . , un+1 ) and w = (w0 , . . . , wn+1 ) be solutions of the recurrence relation (1.2.4) with λ2 = z1 and λ2 = z2 , respectively. If z1 = z2 let these solutions be linearly independent. Then between any two nodes of u there exists at least one node of w. Proof. In view of Lemma 1.2.19 we may assume without loss of generality that the nodes of both u and w are included as partition points, that is, we allow mk = 0 for some k. By (1.2.4) we have (uk − uk−1 )wk (uk+1 − uk )wk − − mk z1 uk wk = 0 lk−1 lk and

uk (wk − wk−1 ) uk (wk+1 − wk ) − − mk z2 uk wk = 0 lk−1 lk for k = 1, . . . , n, and therefore (uk − uk−1 )wk (uk+1 − uk )wk uk (wk − wk−1 ) uk (wk+1 − wk ) − − + lk−1 lk lk−1 lk − mk (z1 − z2 )uk wk = 0, which simplifies to uk−1 wk uk+1 wk uk wk−1 uk wk+1 − − + + − mk (z1 − z2 )uk wk = 0. lk−1 lk lk−1 lk

(1.2.57)

Now assume that xk1 and xk2 , 0 ≤ k1 < k2 ≤ n + 1, are adjacent nodes of u. Since u 6= 0 we must have k2 6= k1 + 1, so that k1 + 1 ≤ k2 − 1. Again by Lemma 1.2.19 and its proof we may remove all massless partition points between xk1 and xk2 , and the same argument shows that there is an integer k0 with k1 < k0 < k2 and mk0 > 0. Adding equations (1.2.57) with k = k1 + 1, . . . , k2 − 1 we obtain −

kX 2 −1 uk −1 wk2 uk +1 wk1 uk wk +1 uk2 wk2 −1 + 2 + 1 − 1 1 = (z1 − z2 ) m k uk w k . lk2 −1 lk2 −1 lk1 lk1 k=k1 +1

Assume that w has no node between xk1 and xk2 . Without loss of generality we may assume that uk > 0 and wk > 0 for k = k1 + 1, . . . k2 − 1, replacing u and w with their negatives, if necessary. Then uk1 = 0 = uk2 leads to kX 2 −1 uk2 −1 wk2 uk1 +1 wk1 + = (z1 −z2 ) mk uk wk ≤ (z1 −z2 )mk0 uk0 wk0 . (1.2.58) lk2 −1 lk1 k=k1 +1

20

Chapter 1. Stieltjes strings

By construction and assumption, the left-hand side of (1.2.58) is nonnegative. But the right-hand side is negative if z1 < z2 , so that we have arrived at a contradiction in this case. Finally, let z1 = z2 . Assume that wk1 = 0. Since u and w are uniquely determined by uk1 , uk1 +1 and wk1 , wk1 +1 via the linear recurrence relation (1.2.4), it would follow that w = wk1 +1 uk−1 u, which would contradict the 1 +1 linear independence of u and w. Hence wk1 > 0, which shows that the left-hand side of (1.2.58) is positive, whereas the right-hand side is zero. Therefore also in this case we have a contradiction, and the proof of the theorem is complete.  Corollary 1.2.21. Let u = (u0 , . . . , un+1 ) and w = (w0 , . . . , wn+1 ) be linearly independent solutions of the recurrence relation (1.2.4). Then between any two adjacent nodes of u there exists exactly one node of w. Proof. By Theorem 1.2.20 we know that between two adjacent nodes of u, say at y1 and y2 , there is at least one node of w. If there were at least two such nodes, then interchanging the rˆoles of u and w would give a node y3 of u between these two nodes of w, and y3 would lie between y1 and y2 , contradicting the fact that  y1 and y2 are adjacent nodes of u. Theorem 1.2.22. Let n ∈ N and c, d ∈ [0, 1]. Then an eigenvector of the k-th eigenvalue (k = 1, . . . , n) of the boundary value problem (1.2.4), (1.2.5), (1.2.6) possesses k − 1 internal nodes. Proof. Consider a nontrivial solution vector u with amplitude function φ of the recurrence relation (1.2.4). Recall that the graph of φ consists of n+1 line segments, none of which is in equilibrium position. Hence each line segment can contribute at most one node, which means that the number of nodes of u is at most n + 1. Next consider the initial conditions u0 = 0, u1 = 1 and for z ∈ [0, ∞) denote the amplitude function of the corresponding solution vector u = u(z) by φD (·, z). Let κD (z) be the number of internal nodes of u(z). We already know that the total number of nodes is at most n + 1, and since 0 is a node in this case, it follows that κD (z) ≤ n for all z ∈ [0, ∞). Recall from Proposition 1.2.8 that the characteristic values of the Dirichlet problem are denoted by νk (k = 1, . . . , n) and satisfy 0 < νk < νk+1 (k = 1, . . . , n − 1). Since 0 and l are nodes of u(νn ), it follows that κD (νn ) ≤ n − 1. Now let 0 ≤ z < y. Since 0 is a node of u(z), u(z) has at least κD (z) + 1 nodes and therefore u(y) has at least κD (z) internal nodes by Theorem 1.2.20. This shows that κD is a nondecreasing function on [0, ∞). Furthermore, for each k = 1, . . . , n, both endpoints are nodes of u(νk ), and therefore u(νk ) has κD (νk )+2 nodes, and another application of Theorem 1.2.20 gives that κD (νk ) + 1 ≤ κD (z) for all z > νk . In particular 0 ≤ κD (ν1 ) < κD (ν2 ) < · · · < κD (νn ) ≤ n − 1. But this immediately gives κD (νk ) = k − 1 for k = 1, . . . , n, that is, the statement of the Corollary for the Dirichlet-Dirichlet problem has been proved.

1.2. Boundary value problems for Stieltjes strings

21

Next we observe that the above reasoning completely determines the function κD on [0, ∞), namely, κD (z) = 0 for z ∈ [0, ν1 ], κD (z) = k for z ∈ (νk , νk+1 ] (k = 1, . . . , n), and κD (z) = n for z ∈ (νn , ∞). Then (1.2.36) proves the statement of this corollary for the Dirichlet-Neumann eigenvalues µk (k = 1, . . . , n) and by symmetry also for the Neumann-Dirichlet eigenvalues ξk (k = 1, . . . , n). For the Neumann-Neumann eigenvalues ζk (k = 1, . . . , n) we first observe that ζ1 = 0, and in that case the amplitude function is constant and has therefore no zeros, i.e., κN (ζ1 ) = 0, where κN (z) (z ∈ [0, ∞)) is the number of internal nodes of the solution u = u(z) of (1.2.4) with u0 = u1 = 1. Next we observe that un+1 = R2n (·, 1) has simple zeros by Proposition 1.2.10, so that un+1 changes sign at the Neumann-Dirichlet eigenvalues ξk (k = 1, . . . , n). Since u is constant at z = 0, we have that R2n (0, 1) = 1 > 0. Hence R2n (z, 1) > 0 for z < ξ1 and (−1)k−1 R2n (z, 1) > 0 for ξk−1 < z < ξk (k = 2, . . . , n). Using the notation with ek be such a node. Then (1.2.56) gives possibly massless internal nodes, let u u ek−1 u ek+1 + = 0, x ek − x ek−1 x ek+1 − x ek which shows that u ek−1 and u ek+1 have opposite signs. In other words, u changes sign at each internal node. Therefore the parity of the number of internal nodes equals the sign of un+1 . The above reasoning and (1.2.39) lead to (−1)κN (ζk ) = (−1)k−1

(k = 2, . . . , n).

(1.2.59)

On the other hand, Corollary 1.2.21 shows that κD (z) ≤ κN (z) ≤ κD (z) + 1 for all z ∈ [0, ∞). Recalling that κD is nondecreasing, it follows in view of (1.2.38) that κD (µk−1 ) ≤ κD (ζk ) ≤ κN (ζk ) ≤ κD (ζk ) + 1 ≤ κD (µk ) + 1. Inserting κD (µj ) = j − 1 leads to k − 2 ≤ κN (ζk ) ≤ k. Together with (1.2.59) the required identity κN (ζk ) = k − 1 follows. So far we have proved the theorem for c, d ∈ {0, 1}. If one or both of c, d belong to (0, 1), say d ∈ (0, 1) and c ∈ {0, 1}, then according to Remark 1.2.15 we prolong the string with partition X = (x0 , . . . , xn , xn+1 ) to a string with partition b = (x0 , . . . , xn , x X bn+1 ) with x bn+1 > xn+1 so that the boundary condition (1.2.6) becomes the Dirichlet condition at the right end of the prolonged string. We have already shown that the amplitude function φX,b b corresponding b u for an eigenvector u to the k-th eigenvalue has k−1 internal nodes. Then φX,b b u |[x0 ,xn+1 ] is an eigenvector φX,u of the boundary value problem on the original string corresponding to the kth eigenvalue. Since φX,b bn+1 ] with φX,b xn+1 ) = 0, b u is linear on the interval [xn , x b u (b the function φX,b bn+1 ). Therefore, φX,u has the b u has no internal nodes in [xn , x same number of internal nodes as φX,b b u , that is, k − 1 nodes. The same reasoning can be applied to the remaining cases. 

22

Chapter 1. Stieltjes strings

1.3 Inverse problems for Stieltjes strings Two-spectra inverse problems

1.3.1

Theorem 1.3.1. Let n ∈ N and l > 0 be given together with two sequences of real n and (µk )nk=1 which have the interlacing property (1.2.36). numbers (νk )k=1 Then there exists a unique pair of sequences of positive numbers (mk )nk=1 n P lk = l such that the characteristic values of the Dirichletand (lk )nk=0 with k=0

Dirichlet problem (1.2.4), (1.2.23), (1.2.24) are νk (k = 1, . . . , n) and such that the characteristic values of the Dirichlet-Neumann problem (1.2.4), (1.2.23), (1.2.27) are µk (k = 1, . . . , n). Proof. The function f defined by   z 1− νk   f (z) = l k=1 n Q z 1− µk k=1 n Q

(1.3.1)

is a rational S-function by Corollary A.3.3. In view of Lemma A.3.5 there are positive numbers l0 , . . . , ln and positive numbers m1 , . . . , mn such that 1

f (z) = ln +

.

1

−mn z + ln−1 +

(1.3.2)

1 −mn−1 z + . .

1

.+ l1 +

1 −m1 z +

1 l0

Clearly, l = f (0) =

n X

lk .

k=0

Defining the functions Rk (·, 0) by (1.2.10), (1.2.11), (1.2.12), (1.2.13) with c = 0 and with the above values of lk and mk , it follows from (1.2.19) and (1.3.2) that f (z) =

R2n (z, 0) . R2n−1 (z, 0)

(1.3.3)

But the n zeros of R2n (·, 0) are the characteristic values of the problem (1.2.4), (1.2.23), (1.2.24) by Proposition 1.2.8, and the n zeros of R2n−1 (·, 0) are the characteristic values of the problem (1.2.4), (1.2.23), (1.2.27) by Proposition 1.2.9. This shows that the two constructed problems have the required characteristic values.

23

1.3. Inverse problems for Stieltjes strings

Since the rational S-function f is uniquely determined by its zeros, its poles and f (0), which are given data, the uniqueness part of Lemma A.3.5 and the fact 2n (·,0) n that RR2n−1 (·,0) has the representation (1.2.19) show that the sequences (mk )k=1 n P and (lk )nk=0 with lk = l are uniquely determined by the given data.  k=0

Theorem 1.3.2. Let n ∈ N and l > 0 be given together with two sequences of real numbers (ξk )nk=1 and (νk )nk=1 which have the interlacing property (1.2.37). Then there exists a unique pair of sequences of positive numbers (mk )nk=1 n P and (lk )nk=0 with lk = l such that the characteristic values of the Dirichletk=0

Dirichlet problem (1.2.4), (1.2.23), (1.2.24) are νk (k = 1, . . . , n) and such that the characteristic values of the Neumann-Dirichlet problem (1.2.4), (1.2.29), (1.2.24) are ξk (k = 1, . . . , n). Proof. By Theorem 1.3.1 there is a unique pair of sequences of positive numbers n P b lk = l such that the problem (m b k )nk=1 and (b lk )nk=0 with k=0

u bk+1 − u bk u bk − u bk−1 − −m b k zb uk = 0 b b lk−1 lk

(k = 1, . . . , n),

(1.3.4)

u b0 = 0, u bn+1 = 0 has the characteristic values νk (k = 1, . . . , n) and such that the problem (1.3.4), u b0 = 0, u bn+1 = u bn has the characteristic values ξk (k = 1, . . . , n). Putting mk = m b n+1−k (k = 1, . . . , n), lk = b ln−k (k = 0, . . . , n), uk = u bn+1−k (k = 0, . . . , n + 1), (1.3.4) becomes (1.2.4), and therefore the numbers νk (k = 1, . . . , n) are the characteristic values of (1.2.4), (1.2.23), (1.2.24), whereas the numbers ξk (k = 1, . . . , n) are the characteristic values of (1.2.4), (1.2.29), (1.2.24). The uniqueness is clear from the uniqueness in Theorem 1.3.1 and the fact that the relationship between (mk )nk=1 , (lk )nk=0 , (uk )n+1 b k )nk=1 , (b lk )nk=0 , k=0 and (m n+1 (b uk )k=0 is bijective.  Theorem 1.3.3. Let n ∈ N, m > 0 and l0 > 0 be given together with two sequences of real numbers (ξk )nk=1 and (ζk )nk=1 which have the interlacing property (1.2.39). Then there exists a unique pair of sequences of positive numbers (mk )nk=1 n P and (lk )nk=1 with mk = m such that the characteristic values of the Neumannk=1

Dirichlet problem (1.2.4), (1.2.29), (1.2.24) are ξk (k = 1, . . . , n) and such that the characteristic values of the Neumann-Neumann problem (1.2.4), (1.2.29), (1.2.27) are ζk (k = 1, . . . , n).

24

Chapter 1. Stieltjes strings

Proof. The function f defined by  z 1− ξk 1 k=1   f (z) = − n z m Q z 1− ζk k=2 n Q



(1.3.5)

is a rational S-function by Corollary A.3.3. In view of Lemma A.3.5 there are positive numbers l1 , . . . , ln and positive numbers m1 , . . . , mn such that 1

f (z) = ln +

.

1

−mn z + ln−1 +

(1.3.6)

1 −mn−1 z + . .

1

.+ l1 +

1 −m1 z

Defining the functions Rk (·, 1) by (1.2.10), (1.2.11), (1.2.12), (1.2.13) with c = 1 and with the above values of lk and mk , it follows from (1.2.19) and (1.3.6) that f (z) =

R2n (z, 1) . R2n−1 (z, 1)

Propositions 1.2.10 and 1.2.11 show that

n P

(1.3.7)

mk = m. But the n zeros of R2n (·, 1)

k=1

are the characteristic values of problem (1.2.4), (1.2.29), (1.2.24) by Proposition (1.2.8), whereas the zeros of R2n−1 (·, 1) are the characteristic values of problem (1.2.4), (1.2.29), (1.2.27) by Proposition (1.2.10). This shows that the two constructed problems have the required spectrum. Since the rational S-function f is uniquely determined by its zeros, its poles and limz→0 zf (z), which are given data, the uniqueness part of Lemma A.3.5 and 2n (·,1) n the fact that RR2n−1 (·,1) is an S-function show that the sequences (mk )k=1 and n P (lk )nk=1 with mk = m are uniquely determined by the given data.  k=1

The following theorem is a variant of Theorem 1.3.3 with symmetric assumptions with respect to the prescribed lengths of the intervals, whereas the total mass of the beads is then determined by the given data. Theorem 1.3.4. Let n ∈ N, l0 > 0 and ln > 0 be given together with two sequences of real numbers (ξk )nk=1 and (ζk )nk=1 which have the interlacing property (1.2.39). Then there exists a unique pair of sequences of positive numbers (mk )nk=1 and (lk )n−1 k=1 such that the characteristic values of the Neumann-Dirichlet problem (1.2.4), (1.2.29), (1.2.24) are ξk (k = 1, . . . , n) and such that the characteristic values of the Neumann-Neumann problem (1.2.4), (1.2.29), (1.2.27) are ζk (k = 1, . . . , n).

25

1.3. Inverse problems for Stieltjes strings Proof. The function f defined by n Q

f (z) =

(z − ξk )

ln k=1 n Q z

(1.3.8) (z − ζk )

k=2

is a rational S-function by Corollary A.3.3 and satisfies limz→−∞ f (z) = ln . In view of Lemma A.3.5 there are positive numbers l1 , . . . , ln−1 and positive numbers m1 , . . . , mn such that 1

f (z) = ln +

.

1

−mn z + ln−1 +

(1.3.9)

1 −mn−1 z + . .

1

.+ l1 +

1 −m1 z

We can now continue as in the proof of Theorem 1.3.3 to complete the proof of this theorem.  Theorem 1.3.5. Let n ∈ N, m > 0 and ln > 0 be given together with two sequences of real numbers (µk )nk=1 and (ζk )nk=1 which have the interlacing property (1.2.38). Then there exists a unique pair of sequences of positive numbers (mk )nk=1 n P and (lk )nk=1 with mk = m such that the characteristic values of the Dirichletk=1

Neumann problem (1.2.4), (1.2.23), (1.2.27) are µk (k = 1, . . . , n) and the characteristic values of the Neumann-Neumann problem (1.2.4), (1.2.29), (1.2.27) are ζk (k = 1, . . . , n). Proof. By Theorem 1.3.3 there is a unique pair of sequences of positive numbers n P (m b k )nk=1 and (b lk )nk=1 with m b k = m such that problem k=1

u bk − u bk−1 u bk+1 − u bk − −m b k zb uk = 0 b b lk−1 lk

(k = 1, . . . , n),

(1.3.10)

u ˆ0 = u b1 , u bn+1 = 0 has the characteristic values µk (k = 1, . . . , n) and such that problem (1.3.10), u b0 = u b1 , u bn+1 = u bn has the characteristic values ζk (k = b 1, . . . , n). Note that a value for l0 is not needed here since u b0 = u b1 . Putting mk = m b n+1−k (k = 1, . . . , n), lk = b ln−k (k = 0, . . . , n − 1), uk = u bn+1−k (k = 0, . . . , n + 1), (1.3.4) becomes (1.2.4), and therefore the numbers µk (k = 1, . . . , n) are the characteristic values of (1.2.4), (1.2.23), (1.2.27), whereas the numbers ζk (k = 1, . . . , n) are the characteristic values of (1.2.4), (1.2.29), (1.2.27).

26

Chapter 1. Stieltjes strings

The uniqueness is clear from the uniqueness in Theorem 1.3.3 and the fact n+1 that the relationship between (mk )nk=1 , (lk )n−1 lk )nk=1 , b k )nk=1 , (b k=0 , (uk )k=0 and (m n+1 uk )k=0 is bijective. (b  Applying Theorem 1.3.4 while interchanging the values of l0 and ln gives Theorem 1.3.6. Let n ∈ N, l0 > 0 and ln > 0 be given together with two sequences n and (ζk )nk=1 which interlace as in (1.2.38). of real numbers (µk )k=1 n Then there exists a unique pair of sequences of positive numbers (mk )k=1 n−1 and (lk )k=1 such that the characteristic values of the Dirichlet-Neumann problem (1.2.4), (1.2.23), (1.2.27) are µk (k = 1, . . . , n) and such that the characteristic values of the Neumann-Neumann problem (1.2.4), (1.2.29), (1.2.27) are ζk (k = 1, . . . , n). Theorem 1.3.7. Let n ∈ N. For two nondecreasing sequences (νk )nk=1 and (ζk )nk=1 of real numbers to be sequences of the characteristic values of the Dirichlet-Dirichn let and the Neumann-Neumann problems generated by the same sequences (mk )k=1 n and (lk )k=0 it is necessary and sufficient that the following conditions are satisfied: 1. ζ1 = 0, νk−1 < νk , ζk−1 < ζk (k = 2, . . . , n). 2. The sequences are interlaced as follows: i) ζk < νk (k = 1, . . . , n), ii) νk < ζk+2 (k = 1, . . . , n − 2). Proof. The necessity of these conditions follows from (1.2.37) and (1.2.39). To prove the sufficiency of these conditions we define the polynomials S1 and S2 by  n  Y z 1− S1 (z) = , (1.3.11) νk k=1  n  Y z S2 (z) = −z 1− . (1.3.12) ζk k=2

Let a1 = ζ1 = 0, ak = max{νk−1 , ζk } (k = 2, . . . , n), bk = min{νk , ζk+1 } (k = 1, . . . , n − 1), bn = νn . The conditions 1 and 2 show that a1 < b1 ≤ a2 < b2 ≤ · · · ≤ an < bn , which gives mutually disjoint intervals (ak , bk ) (k = 1, . . . , n). The function S1 is positive on (−∞, ν1 ) and changes sign at each νk (k = 1, . . . , n), and the function S2 is positive on (−∞, ζ1 ) and changes sign at each ζk (k = 1, . . . , n). Therefore, observing in particular that (ak , bk ) = (νk−1 , νk ) ∩ (ζk , ζk+1 )

for k = 2, . . . , n − 1,

it follows that S1 S2 is negative on each of the intervals (ak , bk ) (k = 1, . . . , n). For each k ∈ {1, . . . , n} let ck ∈ (ak , bk ) (e.g., the midpoint of the interval) and choose

27

1.3. Inverse problems for Stieltjes strings a positive number C such that C < −S1 (ck )S2 (ck ) for k = 1, . . . , n. We put T (z) = C + S1 (z)S2 (z).

(1.3.13)

Then T (ak ) = C = T (bk ) and T (ck ) < 0. Therefore T has at least two zeros in each of the n intervals (ak , bk ) (k = 1, . . . , n). On the other hand, T is a polynomial of degree 2n, so that the number of zeros in each of the intervals (ak , bk ) (k = 1, . . . , n) is exactly two. For k = 1, . . . , n let µk be the zero of T in (ak , ck ) and let ξk be the zero of T in (ck , bk ). Therefore the sequences (νk )nk=1 and (µk )nk=1 satisfy the assumptions of Theorem 1.3.1. Choosing now l > 0 arbitrary, it follows from Theorem 1.3.1 that there n P n are sequences of positive numbers (mk )k=1 and (lk )nk=0 with lk = l such that k=0

the characteristic values of problem (1.2.4), (1.2.23), (1.2.24) are νk (k = 1, . . . , n) and such that the characteristic values of problem (1.2.4), (1.2.23), (1.2.27) are µk (k = 1, . . . , n). Since S1 and R2n (·, 0) have the same zeros by Proposition 1.2.8 and since S1 (0) = 1 = ll0 R2n (0, 0) by (1.3.11) and (1.2.25), it follows that S1 = ll0 R2n (·, 0). In view of Proposition 1.2.11 it remains to show that the zeros of S2 and R2n−1 (·, 1) coincide. We recall the Lagrange identity, Theorem 1.2.16, R2n−1 (z, 0)R2n (z, 1) − R2n (z, 0)R2n−1 (z, 1) =

1 . l0

(1.3.14)

Since T and R2n−1 (·, 0) have zeros at µk (k = 1, . . . , n), it follows from (1.3.13) and (1.3.14) that C+

l0 R2n (µk , 0)S2 (µk ) = 0 and l

1 + R2n (µk , 0)R2n−1 (µk , 1) = 0 l0

for k = 1, . . . n. This leads to S2 (µk ) = lCR2n−1 (µk , 1) (k = 1, . . . , n). Furthermore, S2 (0) = 0 and R2n−1 (0, 1) = 0, and hence the polynomial S2 − lCR2n (·, 1) has at least n+1 zeros. Therefore it is the zero polynomial, that is, S2 = lCR2n (·, 1) and it follows that problem (1.2.4), (1.2.29), (1.2.27) has the characteristic values ζk (k = 1 . . . n).  Theorem 1.3.8. Let n ∈ N. For two nondecreasing sequences (µk )nk=1 and (ξk )nk=1 of positive real numbers to be sequences of the characteristic values of the DirichletNeumann and the Neumann-Dirichlet problems generated by the same sequences (mk )nk=1 and (lk )nk=0 it is necessary and sufficient that the following conditions are satisfied: 1. µk−1 < µk , ξk−1 < ξk (k = 2, . . . , n). 2. The sequences are interlaced as follows: µk−1 < ξk , ξk−1 < µk (k = 2, . . . , n). 3. All zeros of the polynomial S are real, where  n   n  Y z z Y 1− − 1. S(z) = 1− µk ξk k=1

k=1

28

Chapter 1. Stieltjes strings

Proof. To prove the necessity of these conditions we observe that properties 1 and 2 follow from (1.2.36) and (1.2.37). By (1.2.28), (1.2.30) and Theorem 1.2.16, S(z) = l0 R2n−1 (z, 0)R2n (z, 1) − 1 = l0 R2n (z, 0)R2n−1 (z, 1). But the zeros of R2n (·, 0) and R2n−1 (·, 1) are real by Propositions 1.2.8 and 1.2.11. Therefore also property 3 holds. To prove the sufficiency of these conditions, let ak = min{µk , ξk } and bk = max{µk , ξk } (k = 1, . . . , n). Then ak ≤ bk (k = 1, . . . , n), and, in view of properties 1 and 2, bk−1 < ak (k = 2, . . . , n). Clearly, T (z) := S(z) + 1 =

n  Y

1−

k=1

z µk

Y  n  z 1− ξk k=1

has the following properties: T (z) > 0 when z < a1 , bk−1 < z < ak (k = 2, . . . , n), bn < z, and T (z) < 0 when ak < z < bk (k = 1, . . . , n). Hence there are numbers dk (k = 1, . . . , n) with ak ≤ dk ≤ bk such that T 0 (dk ) = 0 and numbers ck ∈ (bk−1 , ak ) for k = 2, . . . , n such that T 0 (ck ) = 0. Since T 0 is a polynomial of degree 2n − 1, T 0 has no other zeros. In particular, T and therefore also S are strictly increasing on (bk−1 , ck ) and strictly decreasing on (ck , ak ) for k = 2, . . . , n, strictly decreasing on (−∞, a1 ) and strictly increasing on (bn , ∞). Observing that S(ak ) = S(bk ) = −1 (k = 1, . . . , n), it follows that ζ1 = 0 is the only zero of S on (−∞, a1 ), that S has at most two zeros, counted with multiplicity, on each interval (bk−1 , ak ) (k = 2, . . . , n), that S has exactly one simple zero νn on (bn , ∞) since lim S(z) = ∞, z→∞

and that S(z) ≤ −1 for ak ≤ z ≤ bk (k = 1, . . . , n). By property 3, S has exactly 2n real zeros, and therefore S has exactly two zeros, which we will denote by νk−1 and ζk , in each of the intervals (bk−1 , ak ) (k = 2, . . . , n). For definiteness, we choose νk−1 ≤ ζk . It follows that µk ≤ bk < νk (k = 1, . . . , n) and νk−1 < ak ≤ µk (k = 2, . . . , n). Therefore the sequences (νk )nk=1 and (µk )nk=1 satisfy the assumptions of Theorem 1.3.1. Choosing now l > 0, it follows from Theorem 1.3.1 that there n P are sequences of positive numbers (mk )nk=1 and (lk )nk=0 with lk = l such that k=0

the characteristic values of problem (1.2.4), (1.2.23), (1.2.24) are νk (k = 1, . . . , n) and such that the characteristic values of problem (1.2.4), (1.2.23), (1.2.27) are µk (k = 1, . . . , n). Defining  n  X z S3 (z) = 1− ξk k=1

it remains to show that the zeros of S3 and R2n (·, 1) coincide. For k = 1, . . . , n we have in view of (1.2.28) that R2n−1 (νk , 0)S3 (νk ) =

1 1 T (νk ) = , l0 l0

29

1.3. Inverse problems for Stieltjes strings and by Theorem 1.2.16, R2n−1 (νk , 0)R2n (νk , 1) =

1 . l0

This shows that S3 (νk ) = R2n (νk , 1) (k = 1, . . . , n). Finally, S3 (0) = 1 = R2n (0, 1) by (1.2.30). Therefore S3 = R2n (·, 1) since both polynomials have degree n and since they coincide at n + 1 points. Hence problem (1.2.4), (1.2.29), (1.2.24) has the characteristic values ξk (k = 1, . . . , n).  Theorem 1.3.9. Let n ∈ N and let two sequences of real numbers (b νk )nk=1 and n (b µk )k=1 be given which have the interlacing property (1.2.46). Then there is a number η > 0 such that the following statement is true: For l > 0 and γ > 0 there exists a pair of sequences of positive numbers (mk )nk=1 n P and (lk )nk=0 with lk = l such that the characteristic values of the Robink=0

Dirichlet problem (1.2.4), (1.2.42), (1.2.24) are νbk (k = 1, . . . , n) and such that the characteristic values of Robin-Neumann problem (1.2.4), (1.2.42), (1.2.27) are µ bk (k = 1, . . . , n) if and only if lγ > η. In case such a pair of sequences exists, this pair is unique. Proof. The function f defined by  z 1− νbk   f (z) = k=1 n Q z 1− µ bk k=1 n Q



(1.3.15)

is a rational S0 -function by Corollary A.3.3. In view of Lemma A.3.5 there are positive numbers a0 , . . . , an and positive numbers b1 , . . . , bn such that 1

f (z) = an +

.

1

−bn z + an−1 +

(1.3.16)

1 −bn−1 z + . .

1

.+ a1 +

1 −b1 z +

1 a0

Clearly, 1 = f (0) =

n X

ak ,

k=0

and in particular ak < 1 for k = 0, . . . , n. In view of (1.2.19), (1.2.44) and (1.2.45) the Cauer-Fry polynomials for a solution would have to satisfy   R2n (·, γ0 ) 1 = l+ f. R2n−1 (·, γ0 ) γ

30

Chapter 1. Stieltjes strings

Since the continued fraction representations are unique, it follows from Remark 1.2.7, part 3, that     1 l0 1 lk = l + ak (k = 1, . . . , n), = l+ a0 , γ 1 − γ0 γ  −1 1 mk = l + bk (k = 1, . . . , n). γ Hence there is a solution, which would then be unique, if and only if all lk and all mk satisfying the above identities were positive, i.e., if and only if l0 > 0. But since l0 l0 1 + γl0 1 = = = + l0 , 1 1 − γ0 γ γ 1− 1 + γl0 we arrive at

  1 1 l0 = l + a0 − . γ γ

Hence l0 > 0 if and only if γl + 1 >

1 a0 ,

η=

and the proof is complete by setting 1 − 1. a0

We finally observe that a0 < 1 implies that η > 0.



1.3.2 Four-spectra inverse problem In this subsection we use parts of the four sequences of characteristic values (νk )nk=1 , (µk )nk=1 , (ξk )nk=1 and (ζk )nk=1 together with some other parameters to recover the sequences (mk )nk=1 and (lk )nk=0 . The main theorem of this subsection is Theorem 1.3.10. Let n ≥ 2 be a natural number and let four increasing sequences 1 2 3 4 of natural numbers (kj )nj=1 , (pj )nj=1 , (rj )nj=1 , (sj )nj=1 be given such that kn1 = n, n1 + n2 = n, n3 + n4 = n,

{kj : j = 1, . . . , n1 } ∪ {pj − 1 : j = 1, . . . , n2 } = {1, . . . , n}, {rj : j = 1, . . . , n3 } ∪ {sj : j = 1, . . . , n4 } = {1, . . . , n}.

Let a constant l > 0 be given together with four sequences of positive real numbers 1 2 3 4 (νkj )nj=1 , (ζpj )nj=1 , (µrj )nj=1 , (ξsj )nj=1 such that the sequences (ηk )nk=1 and (γk )nk=1 defined by 1 1 2 2 (ηkj )nj=1 = (νkj )nj=1 , (ηpj −1 )nj=1 = (ζpj )nj=1 , (1.3.17) and 3 3 (γrj )nj=1 = (µrj )nj=1 ,

4 4 (γsj )nj=1 = (ξsj )nj=1 ,

(1.3.18)

31

1.3. Inverse problems for Stieltjes strings interlace: γ1 < η1 < γ2 < η2 < · · · < γn < ηn .

(1.3.19)

Then there exist two unique sequences of positive numbers (mk )nk=1 , (lk )nk=0 P n with k=0 lk = l such that 1. νkj (j = 1, . . . , n1 ) are characteristic values of the Dirichlet-Dirichlet problem (1.2.4), (1.2.23), (1.2.24), 2. µrj (j = 1, . . . , n3 ) are characteristic values of the Dirichlet-Neumann problem (1.2.4), (1.2.23), (1.2.27), 3. ξsj (j = 1, . . . , n4 ) are characteristic values of the Neumann-Dirichlet problem (1.2.4), (1.2.29), (1.2.24), 4. ζpj (j = 1, . . . , n2 ) are characteristic values of the Neumann-Neumann problem (1.2.4), (1.2.29), (1.2.27). 5. In each case, if the characteristic values are indexed in increasing order, then the indices of the given sequences are the corresponding indices in the full sequence of characteristic values; e.g., νkj is the kj -th characteristic value of the Dirichlet-Dirichlet problem. Proof. Due to (1.3.19) and Theorem 1.3.1 we conclude that there exist unique n lk )nk=0 which together with the given and (e e k )k=1 sequences of positive numbers (m l generate problem (1.2.4), (1.2.23), (1.2.24) with the characteristic values (ηk )nk=1 and problem (1.2.4), (1.2.23), (1.2.27) with the characteristic values (γk )nk=1 . Let n (ξek )k=1 be the increasing sequence of the characteristic values of the NeumannDirichlet problem (1.2.4), (1.2.29), (1.2.24) generated by (m lk )nk=0 and e k )nk=1 , (e n let (ζek )k=1 be the increasing sequence of the characteristic values of Neumannn Neumann problem (1.2.4), (1.2.29), (1.2.27) generated by (m e k )nk=1 , (e . From lk )k=0 (1.2.40) it follows that {ηk−1 , ζek } < {γk , ξek } < {ηk , ζek+1 }

(k = 1, . . . , n),

(1.3.20)

1 where η0 = 0 and ζen+1 = ηn . Augment the given sequence (νkj )nj=1 to the sequence n (νk )k=1 by setting 2 2 (νpj −1 )nj=1 = (ζepj )nj=1 (1.3.21) 3 to the sequence (µk )nk=1 by setting and augment the given sequence (µrj )nj=1

n4 4 . = (ξesj )nj=1 (µsj )j=1

(1.3.22)

Since µk = γk or µk = ξek (k = 1, . . . , n) by (1.3.18) and (1.3.22), we have min{γk , ξek } ≤ µk ≤ max{γk , ξek }

(k = 1, . . . , n).

(1.3.23)

32

Chapter 1. Stieltjes strings

From νk = ηk or νk = ζek+1 (k = 1, . . . , n) by (1.3.17) and (1.3.21), it follows that min{ηk , ζek+1 } ≤ νk ≤ max{ηk , ζek+1 }

(k = 1, . . . , n).

(1.3.24)

Hence (1.2.40), (1.3.23) and (1.3.24) show that µk < νk (k = 1, . . . , n) and νk < µk+1 (k = 1, . . . , n − 1). Therefore the sequences (νk )nk=1 and (µk )nk=1 satisfy n n (1.2.36). By Theorem 1.3.1 there Pn exist two sequences (mk )k=1 and (lk )k=0 of positive numbers which satisfy k=0 lk = l and generate problem (1.2.4), (1.2.23), (1.2.24) with the sequence of characteristic values (νk )nk=1 and problem (1.2.4), (1.2.23), (1.2.27) with the sequence of characteristic values (µk )nk=1 . n Statements 1 and 2 are clear from the construction of (νk )k=1 and (µk )nk=1 . To prove statements 3 and 4 consider the Lagrange identity (1.2.47) for the n lk )nk=0 as well as (mk )nk=1 , (lk )nk=0 : e k )k=1 problem generated by (m , (e e e2n (z, 0)R e2n−1 (z, 0)R e2n (z, 1) − e e2n−1 (z, 1) = 1, l0 R l0 R l0 R2n−1 (z, 0)R2n (z, 1) − l0 R2n (z, 0)R2n−1 (z, 1) = 1, which leads to e2n−1 (z, 0)R e2n (z, 1) − l0 R2n−1 (z, 0)R2n (z, 1) l0 R P (z) := e e2n (z, 0)R e2n−1 (z, 1) − l0 R2n (z, 0)R2n−1 (z, 1). =e l0 R

(1.3.25)

e2n−1 (γr , 0) = 0, For j = 1, . . . , n3 we have µrj = γrj , R2n−1 (µrj , 0) = 0 and R j e2n (ξes , 1) = 0. and for j = 1, . . . , n4 we have µsj = ξesj , R2n−1 (µsj , 0) = 0 and R j This shows that P (µk ) = 0 for k = 1, . . . , n. For j = 1, . . . , n1 we have νkj = ηkj , e2n (ηk , 0) = 0, and for j = 1, . . . , n2 we have νp −1 = R2n (νkj , 0) = 0 and R j j e2n−1 (ζep , 1) = 0. This shows that P (νk ) = 0 for ζepj , R2n (νpj −1 , 0) = 0 and R j e2n−1 (0, 1) = 0 gives P (0) = 0. k = 1, . . . , n. Furthermore, R2n−1 (0, 1) = 0 and R Thus P is a polynomial of degree at most 2n which has 2n + 1 distinct zeros νk (k = 1, . . . , n), µk (k = 1, . . . , n) and 0. Hence P = 0. e2n−1 (·, 0)R e2n (·, 1) and R2n−1 (·, 0)R2n (·, 1) coincide. Therefore the zeros of R Above we have already identified n of the common zeros, which are the zeros µk (k = 1, . . . , n) of R2n−1 (·, 0). The remaining zeros can be identified as the increasing sequence (ξbk )nk=1 of the zeros of R2n (·, 1) on the one hand or the numbers ξerj (j = 1, . . . , n3 ) and ξsj = γsj (j = 1, . . . , n4 ). Hence the numbers ξsj (j = 1, . . . , n4 ) are characteristic values of the Neumann-Dirichlet problem for the n sequences (mk )nk=1 and (lk )k=0 . This completes the proof of statement 3. e2n−1 (·, 1) and e2n (·, 0)R For the proof of statement 4 we use that the zeros of R R2n (·, 0)R2n−1 (·, 1) coincide because P = 0 in (1.3.25). We have already identified n of the common zeros, which are the zeros νk (k = 1, . . . , n) of R2n (·, 0). The remaining zeros can be identified as the increasing sequence (ζbk )nk=1 of the zeros of R2n−1 (·, 1) on the one hand or the numbers ζpj = ηpj −1 (j = 1, . . . , n2 ), 0, and

33

1.3. Inverse problems for Stieltjes strings

ζekj +1 (j = 1, . . . , n1 − 1). Hence the numbers ζpj (j = 1, . . . , n2 ) are charactern istic numbers of the Neumann-Neumann problem for the sequences (mk )k=1 and (lk )nk=0 . This completes the proof of statement 4. (q)

(q)

To prove uniqueness, let two pairs of sequences (mk )nk=1 and (lk )nk=0 (q = 1, 2) be given for which the corresponding characteristic polynomials and characteristic values are also denoted by superscripts (1) and (2) and where 2n of the characteristic values are given according to (1.3.17)–(1.3.19). Below we will also refer to these characteristic values as characteristic values of problems (1) n n and (2), respectively. Let the sequences (m be defined as at the e k )k=1 lk )k=0 and (e beginning of this proof. As in (1.3.25), the Lagrange identity leads to e2n−1 (z, 0)R e2n (z, 1) − l(q) R(q) (z, 0)R(q) (z, 1) P (q) (z) := e l0 R 0 2n−1 2n e2n−1 (z, 1) − l(q) R(q) (z, 0)R(q) (z, 1) e2n (z, 0)R =e l0 R 2n 2n−1 0

(1.3.26)

for q = 1, 2. For j = 1, . . . , n3 , the numbers γrj = µrj are characteristic values of the (q) Dirichlet-Neumann problem for (1) and (2), and therefore R2n−1 (µrj , 0) = 0 e2n−1 (γr , 0) = 0. For j = 1, . . . , n4 , the numbers γs = ξs are charand R j j j acteristic values of the Neumann–Dirichlet problem for (1) and (2), and there(q) e2n−1 (γr , 0) = 0. This shows that P (q) (γk ) = 0 for fore R2n (ξsj , 1) = 0 and R j k = 1, . . . , n. For j = 1, . . . , n1 , the numbers ηkj = νkj are characteristic values (q) of the Dirichlet-Dirichlet problem for (1) and (2), and therefore R2n (νkj , 0) = 0 e2n (ηk , 0) = 0. For j = 1, . . . , n2 , the numbers ηp −1 = ζp are characand R j j j teristic values of the Neumann-Neumann problem for (1) and (2), and therefore (q) e2n (ηp −1 , 0) = 0. This shows that P (q) (ηk ) = 0 for k = R2n−1 (ζpj , 1) = 0 and R j (q) e2n−1 (0, 1) = 0, so that P (q) (0) = 0. 1, . . . , n. Furthermore, R2n−1 (0, 1) = 0 and R Thus P (q) is a polynomial of degree at most 2n which has 2n + 1 zeros γk (k = 1, . . . , n), ηk (k = 1, . . . , n), and 0. Hence P (q) = 0 for q = 1, 2. We obtain that (1) (2) (2) (1) (1) (1) (1) (1) (2) l0 R2n−1 (·, 0)R2n (·, 1) = l0 R2n−1 (·, 0)R2n (·, 1) and l0 R2n (·, 0)R2n−1 (·, 1) = (2) (2) (2) l0 R2n (·, 0)R2n−1 (·, 1). But this implies that (1)

(1)

(2)

(2)

(1)

(1)

(2)

(2)

n n n ∪ (ξk )k=1 ∪ (ξk )nk=1 , (µk )k=1 = (µk )k=1 n n n ∪ (ζk )nk=1 , = (νk )k=1 ∪ (ζk )k=1 (νk )k=1

with, say, nondecreasing order in the concatenations (unions) of the sequences. From (1.2.40) we know that (q)

(q)

(q)

(q)

(q)

(q)

(q)

{0} = {ζ1 } < {µ1 , ξ1 } < {ν1 , ζ2 } < {µ2 , ξ2 } (q)

(q) (q) < · · · < {νn−1 , ζn(q) } < {µ(q) n , ξn } < {νn }

34

Chapter 1. Stieltjes strings

for q = 1, 2. This implies that (1)

(1)

(2)

(2)

{µk , ξk } = {µk , ξk } (1) (1) {νk , ζk+1 }

=

(2) (2) {νk , ζk+1 }

(k = 1, . . . , n),

(k = 1, . . . , n − 1),

νn(1) = νn(2) .

For k = 1, . . . , n, either k = rj or k = sj for some index j, and it follows that (2) (1) (1) (2) µk = µrj = µk or ξk = ξsj = ξk is given, and therefore both identities hold for k = 1, . . . , n. This shows that the Dirichlet-Neumann as well as the NeumannDirichlet characteristic values coincide for both sets of string data. Similarly, for k = 1, . . . , n − 1, either k = kj or k = pj − 1 for some index j, and it follows (2) (2) (1) (1) that νk = νkj = νk or ζk+1 = ζpj = ζk+1 is given, and therefore both identities hold for k = 1, . . . , n − 1. This shows that the Dirichlet-Dirichlet as well as the Neumann-Neumann characteristic values coincide for both sets of string data. The (2) (1) uniqueness part of Theorem 1.3.1 then shows that mk = mk (k = 1, . . . , n) and (1) (2) lk = lk (k = 0, . . . , n). Next we are going to prove statement 5 for the Dirichlet-Dirichlet and the Dirichlet-Neumann problems. We already know that the sequences (νk )nk=1 and (µk )k=1 are increasing sequences of the characteristic values of the DirichletDirichlet and the Dirichlet-Neumann problem, respectively. By definition of these two sequences it follows that νkj (j = 1, . . . , n1 ) is the kj -th eigenvalue of the Dirichlet-Dirichlet problem and that µrj (j = 1, . . . , n3 ) is the rj -th eigenvalue of the Dirichlet-Neumann problem. To prove statement 5 for the Neumann-Dirichlet problem and the NeumannNeumann problem, we are going to use the notation from the existence part of this proof. First observe that the inequality (1.2.40) for the string data (lk )nk=0 and (mk )nk=1 is {νk−1 , ζbk } < {µk , ξbk } < {νk , ζbk+1 }

(k = 1, . . . , n).

(1.3.27)

From (1.3.20), (1.3.24) and (1.3.27) we conclude that

ξbk−1

ξek < min{ηk , ζek+1 } ≤ νk < ξbk+1 (k = 1, . . . , n − 1), < νk−1 ≤ max{ηk−1 , ζek } < min{γk , ξek } ≤ ξek (k = 2, . . . , n).

Hence ξe1 < ξb2 ,

ξbn−1 < ξen ,

ξbk−1 < ξek < ξbk+1

(k = 2, . . . , n − 1).

Since ξerj ∈ {ξb1 , . . . , ξbn } (j = 1, . . . , n3 ), it follows from the above inequalities that ξerj = ξbrj (j = 1, . . . , n3 ). Therefore {ξsj : j = 1, . . . , n4 } = {ξbsj : j = 1, . . . , n4 }, 4 and (ξbsj )4j=1 are increasing sequences, we obtain that and since both (ξsj )j=1 ξsj = ξbsj (j = 1, . . . , n4 ). Hence ξsj (j = 1, . . . , n4 ) is the sj -th characteristic value of the Neumann-Dirichlet problem.

35

1.3. Inverse problems for Stieltjes strings

Observing that µrj = γrj (j = 1, . . . , n3 ) by (1.3.18) and that µsj = ξesj (j = 1, . . . , n4 ) by (1.3.22), the inequalities (1.3.20) show that µk ≤ max{γk , ξek } < ηk < min{γk+1 , ξek+1 } ≤ µk+1

(k = 1, . . . , n − 1).

Together with (1.3.27) and (1.3.17) it follows for j = 1, . . . .n2 that ζbpj −1 < µpj −1 < ηpj −1 = ζpj < µpj < ζbpj +1 , which gives ζbpj −1 < ζpj < ζbpj +1

(j = 1, . . . , n2 ).

Hence ζpj (j = 1, . . . , n2 ) is the pj -th characteristic value of the Neumann-Neumann  problem. Remark 1.3.11. 1. In the proof of Theorem 1.3.10 we have applied Theorem 1.3.1 twice to first find the data of an auxiliary string and then to find the data of the string which solves the four spectra problem. The first application can be replaced by using the Lagrange identity and Lagrange interpolation to e2n−1 (·, 1). This approach e2n (·, 1) and R find the characteristic polynomials R may be easier for explicit calculations, see Example 1.3.12 below. 2. We have already mentioned above that we have used Theorem 1.3.1 to prove Theorem 1.3.10. However, Theorem 1.3.10 can be seen as a generalization of Theorems 1.3.1 and 1.3.2. To this end we observe that in the formulation and the proof of Theorem 1.3.10 we only require explicitly kn1 = n, so that n1 > 0. In order to have really parts of four spectra occurring, n ≥ 2 is needed. However, the statements and proofs still hold if one only assumes that n ≥ 1 and allows n2 , n3 or n4 to have the value 0. In particular, Theorem 1.3.1 is the particular case n2 = n4 = 0, and Theorem 1.3.2 is the particular case n2 = n3 = 0. 3. For Theorems 1.3.1, 1.3.2, 1.3.3, 1.3.4, 1.3.5, 1.3.6, 1.3.9, 1.3.10 it easily follows from Lemma A.6.5 and Corollary A.6.7 that the distribution of the beads depends continuously on the given sequences and the other given data, such as l, m, l0 as they occur in the particular theorem. Furthermore, the number η constructed in the proof of Theorem 1.3.9 depends continuously on the given data of that theorem. Example 1.3.12. Given n = 3, ξ1 = 1, ζ2 = 2, µ2 = 3, ν2 = 4, µ3 = 5, ν3 = 6 and l = 1, find the distribution of the beads. According to Theorem 1.3.10 we set γ1 = ξ1 = 1, η1 = ζ2 = 2, γ2 = µ2 = 3, η2 = ν2 = 4, γ3 = µ3 = 5, η3 = ν3 = 6 and observe that (1.3.19) is satisfied. Then (1.2.25) and (1.2.28) give   z z e6 (z, 0) = 1 1 − z 1− 1− , R e 2 4 6 l0   z e5 (z, 0) = 1 (1 − z) 1 − z R 1− . e 3 5 l0

36

Chapter 1. Stieltjes strings

Now using the Lagrange identity and Lagrange interpolation we arrive at e6 (z, 1) = − 1 (2z 3 − 18z 2 + 37z − 3), R 3 e5 (z, 1) = − 16 z(2z 2 − 12z + 13). R 15 e6 (·, 1) are ξe1 = In the notation of the proof of Theorem 1.3.10, the zeros of R √ √ e5 (·, 1) are ζe1 = 0, 3 − 8.5, ξe2 = 3 and ξe3 = 3 + 8.5, while the zeros of R √ √ √ e e ζ2 = 3 − 2.5 √ and ζ3 = 3 + 2.5. Then it follows that ν1 = ζe2 = 3 − 2.5 and µ1 = ξe1 = 3 − 8.5. This gives   1 z z z √ R6 (z, 0) = 1− 1− 1− , l0 4 6 3 − 2.5   1 z z z √ R5 (z, 0) = 1− 1− 1− . l0 3 5 3 − 8.5 2n (z,0) The continued fraction expansion (1.2.19) of the quotient RR2n−1 (z,0) gives, up to 5 decimal places l0 ≈ 0.52115, l1 ≈ 0.34848, l2 ≈ 0.09314, l3 ≈ 0.03723, m1 ≈ 1.27635, m2 ≈ 4.53523, m3 = 8.05508.

1.4 1.4.1

Stieltjes strings with one interior condition The problem and its spectra

In this section we consider a Stieltjes string S with n interior beads as in Section 1.2 and with an additional partition point P between two beads. The movement of the string at P will be subject to certain restrictions. We may now consider this string l0

ln1

l1 m1 m2

mn1

P

ln mn1 +1

mn

Figure 1.5 as being composed of two strings, S1 and S2 , where S1 starts at the left endpoint and terminates at P and where S2 starts at the right endpoint and also terminates at P . Let nj (j = 1, 2) be the number of beads on Sj . Then n = n1 + n2 . Denoting the masses, lengths between the beads and amplitudes by the superscript (1) for the substring S1 and by the superscript (2) for the substring S2 , we arrive at the (1) (1) (2) notation lk = lk (k = 0, . . . , n1 − 1), mk = mk (k = 1, . . . , n1 ), lk = ln−k (2) (k = 0, . . . , n2 − 1), mk = mn+1−k (k = 1, . . . , n2 ). The interval of length ln1 between the beads at position n1 and n1 + 1 is divided by P into two intervals (1) (2) (1) (2) of lengths ln1 and ln2 and thus ln1 + ln2 = ln1 . Figures 1.5 and 1.6 illustrate

1.4. Stieltjes strings with one interior condition (1)

l0

(1)

(1)

l1 (1)

(1)

l0

ln2 P

(1)

mn1

(2)

(2)

l n1

m1 m2

37

(2)

(2)

m n2

m1

Figure 1.6

this situation. The amplitudes of the string S at the beads and at the endpoints of S are denoted by uk (k = 0, . . . , n + 1), whereas the corresponding amplitudes (1) (2) for the substrings S1 and S2 are denoted by uk (k = 0, . . . , n1 + 1) and uk (k = 0, . . . , n2 + 1), respectively. The connectedness of the string at P gives the (1) (2) natural interface condition un1 +1 = un2 +1 at the partition point P . By Lemma 1.2.19 the recurrence relation (1.2.4) is equivalent to (j)

(j)

uk − uk+1 (j) lk

(j)

+

(j)

uk − uk−1 (j) lk−1

(j)

(j)

− mk λ2 uk = 0, (1)

(k = 1, . . . , nj ; j = 1, 2), (1.4.1)

(2)

un1 +1 = un2 +1 , (1)

(1)

un1 +1 − un1

(2)

+

(1)

(1.4.2) (2)

un2 +1 − un2

= 0.

(2)

ln1

(1.4.3)

ln2

We impose Dirichlet boundary conditions at the ends of the string: (j)

u0 = 0 (j = 1, 2).

(1.4.4)

By Proposition 1.2.8, the characteristic function of this Dirichlet-Dirichlet problem for the whole string is R2n (·, 0). On the other hand, for the substrings with the Dirichlet conditions (1.4.4) the definition (1.2.8) gives, setting again z = λ2 as in Section 1.2, (1)

(1)

(1)

un1 +1 = R2n1 (z, 0)u1 ,

(2)

(2)

(2)

un2 +1 = R2n2 (z, 0)u1 .

Due to (1.4.2), (1.2.10) and (1.4.3) we have (1)

(1)

(2)

(2)

R2n1 (z, 0)u1 = R2n2 (z, 0)u1 , (1)

(1)

(2)

(2)

R2n1 −1 (z, 0)u1 + R2n2 −1 (z, 0)u1 = 0. (1)

(2)

This linear system has a nontrivial solution (u1 , u1 ) if and only if the coefficient matrix is singular. Therefore, the characteristic values of problem (1.4.1)–(1.4.4) coincide with the zeros of the polynomial (1)

(2)

(1)

(2)

φ = R2n1 (·, 0)R2n2 −1 (·, 0) + R2n1 −1 (·, 0)R2n2 (·, 0).

(1.4.5)

38

Chapter 1. Stieltjes strings

The polynomials R2n (·, 0) and φ have the same zeros, which means that they are multiples of each other. Putting l(j) =

nj X

(j)

lk

(j = 1, 2)

(1.4.6)

k=0

it follows from (1.2.15) and (1.2.17) that φ(0) =

l(1) (1) l0

·

1 (2) l0

and therefore φ=

+

1 (1) l0

·

l(2) (2) l0

=

l , l0 ln

1 R2n (·, 0) . ln

(1.4.7)

Hence (1.4.5) can be written in the form h ih i h ih i (1) (1) (1) (2) (2) (1) (1) (2) (2) l0 R2n (·, 0) = l0 R2n1 (·, 0) l0 R2n2 −1 (·, 0) + l0 R2n1 −1 (·, 0) l0 R2n2 (·, 0) . (1.4.8) Lemma 1.4.1. The functions (1)

φ (1)

and

(2)

R2n1 −1 (·, 0)R2n2 −1 (·, 0)

(2)

R2n1 (·, 0)R2n2 (·, 0) φ

are S0 -functions. Proof. In view of (1.2.19) and by Lemmas A.3.6 and A.3.5, both terms on the right-hand side of (1)

φ (1)

(2)

=

R2n1 −1 (·, 0)R2n2 −1 (·, 0)

R2n1 (·, 0) (1)

(2)

+

R2n1 −1 (·, 0)

R2n2 (·, 0) (2)

R2n2 −1 (·, 0)

are S0 -functions. Therefore, the left-hand side is also an S0 -function. The same reasoning and Lemma A.2.4 show that −

(2)

φ (1)

(2)

R2n1 (·, 0)R2n2 (·, 0)

=−

R2n2 −1 (·, 0) (2)

R2n2 (·, 0)

(1)

−

R2n1 −1 (·, 0) (1)

R2n1 (·, 0)

is a rational Nevanlinna function. Furthermore, in view of (1.4.7) and Proposi(1) (2) tion 1.2.8, the real polynomials R2n1 (·, 0), R2n2 (·, 0) and φ are positive at 0 and have no negative zeros. Hence all three functions and therefore also the quotient (1)

(2)

R2n (·,0)R2n (·,0) 1 2 φ

are positive on (−∞, 0], which together with another application of Lemma A.2.4 proves that this quotient is an S0 -function. 

39

1.4. Stieltjes strings with one interior condition

Proposition 1.4.2. 1. If z ∈ R is a zero of at least two of the three functions (1) (2) R2n1 (·, 0), R2n2 (·, 0), φ, then z is a zero of all three functions. (1)

2. If z ∈ R is a zero of at least two of the three functions R2n1 −1 (·, 0), (2) R2n2 −1 (·, 0),

φ, then z is a zero of all three functions. (2)

(1)

R

(1)

(·,0)R

(2)

(·,0)

Proof. 1. Assume that R2n1 (z, 0) = 0 = R2n2 (z, 0). Since 2n1 φ 2n2 is an S0 -function by Lemma 1.4.1, it has only simple zeros, and therefore φ(z) = 0 (1) (1) (2) follows. Now assume that R2n1 (z, 0) = 0 = φ(z). Then R2n1 −1 (z, 0)R2n2 (z, 0) = 0 by (1.4.5). But from either (1.2.36) or the Lagrange identity (1.2.47) we conclude (1) (1) (2) from R2n1 (z, 0) = 0 that R2n1 −1 (z, 0) 6= 0. Hence R2n2 (z, 0) = 0. The remaining (2)

case that R2n2 (z, 0) = 0 = φ(z) follows by interchanging the indices 1 and 2. 2. This is similar to part 1. We only have to observe that also the poles of S0 -functions are simple.  Together with the Dirichlet-Dirichlet problem (1.4.1)–(1.4.4), i.e., (1.2.4), (1.2.23), (1.2.24), for the string S, we will consider boundary value problems associated with the strings S1 and S2 . If we clamp the string at the point P , we obtain Dirichlet-Dirichlet problems for j = 1 and j = 2, that is, (j)

(j)

uk − uk+1 (j) lk

(j)

+

(j)

uk − uk−1 (j) lk−1

(j)

(j)

− mk λ2 uk = 0, (j)

unj +1 = 0, (j)

u0 = 0.

(k = 1, . . . , nj ; j = 1, 2), (1.4.9) (1.4.10) (1.4.11)

Furthermore, we consider the Dirichlet-Neumann problems for the strings S1 and S2 , that is with (j) unj +1 = u(j) (1.4.12) nj instead of (1.4.10). Remark 1.4.3. In view of Propositions 1.2.8 and 1.2.9, the characteristic values νk (k = 1, . . . , n) of the problem (1.4.1)–(1.4.4), and, for j = 1, 2, the characteristic (j) values νk (k = 1, . . . , nj ) of problem (1.4.9)–(1.4.11) as well as the characteristic (j) values µk (k = 1, . . . , nj ) of problem (1.4.9), (1.4.11), (1.4.12) are real, positive and simple. Therefore, the characteristic values will be indexed in increasing order, (j) (j) that is, for admissible indices k < k 0 and j = 1, 2 we have νk < νk0 , νk < νk0 and (j) (j) µk < µk0 , respectively. In view of (1.4.5) and Corollary A.6.7, these characteristic values depend continuously on the distribution of the beads. Theorem 1.4.4. The characteristic values defined in Remark 1.4.3 have the following properties: (1)

1. ν1 < ν1

(2)

and ν1 < ν1 .

40

Chapter 1. Stieltjes strings

2. For each k = 2, . . . , n the following alternative holds: either the interval (1) 1 (ν1 , νk ) contains exactly k − 1 terms of the concatenated sequence (νj )nj=1 ∪ (2)

(1)

(2)

n2 1 2 and νk ∈ ∪ {νj }nj=1 , or the interval (ν1 , νk ) contains (νj )j=1 / {νj }nj=1 (1)

(2)

n2 n1 exactly k − 2 terms of the concatenated sequence (νj )j=1 and ∪ (νj )k=1 (2)

(1)

1 2 νk ∈ {νj }nj=1 ∩ {νj }nj=1 .

3. For each k = 1, . . . , n the following alternative holds: either the interval (1) 1 (0, νk ) contains exactly k terms of the concatenated sequence (µj )nj=1 ∪ (2)

(1)

(2)

n2 1 2 (µj )nj=1 / {µj }nj=1 , or the interval (0, νk ) contains ∪ {µj }j=1 and νk ∈ (1)

(2)

n2 1 and exactly k − 1 terms of the concatenated sequence (µj )nj=1 ∪ (µj )j=1 (2)

(1)

n1 n2 νk ∈ {µj }j=1 ∩ {µj }j=1 . (1)

1 ∪ 4. For each k = 1, . . . , n the following alternative holds: either νk ∈ / {νj }nj=1

(2)

2 {µj }nj=1 , and then the interval (0, νk ) contains k − 1 or k terms of the

(1)

(2)

n1 n2 , or concatenated sequence (νj )j=1 ∪ (µj )j=1

    (2) 2 (1) 1 (1) 1 (2) 2 \ νk ∈ {νj }nj=1 , ∪ {µj }nj=1 νk ∈ {νj }nj=1 ∩ {µj }nj=1 and then the interval (0, νk ) contains k − 2 or k − 1 terms of the concatenated (2) n2 (1) 1 . sequence (νj )nj=1 ∪ (µj )j=1 R

(1)

(·,0)R

(2)

(·,0)

Proof. 1. Let ψ = 2n1 φ 2n2 and recall that φ is a positive multiple of R2n (·, 0), see (1.4.7). Each pole of ψ is a zero of φ, and each zero of φ is either a pole or a zero of ψ by Proposition 1.4.2, part 1. Since ψ is an S0 -function by Lemma 1.4.1, its smallest pole must be smaller than its smallest zero. Therefore the smallest zero ν1 of φ must be a pole of ψ, and all zeros of ψ must be larger (1) (2) than ν1 . This shows that ν1 < ν1 and ν1 < ν1 . 2. By statement 1, ν1 is a pole of ψ, and by Proposition 1.4.2, νk is either a pole or a zero of ψ. Let 0 ≤ k 0 ≤ k − 2 be the number of zero-pole cancellations in the interval (ν1 , νk ) of the quotient representation of ψ. Hence ψ has k − 2 − k 0 poles in (ν1 , νk ). Since the poles and the zeros of ψ interlace, it follows that ψ has k − 1 − k 0 zeros in (ν1 , νk ) if νk is a pole of ψ and k − 2 − k 0 zeros in (ν1 , νk ) if νk is a zero of ψ. In view of Proposition 1.4.2, this results in the following alternative: (1) (2) (1) (2) either R2n1 (νk , 0) 6= 0, R2n2 (νk , 0) 6= 0 and R2n1 (·, 0)R2n2 (·, 0) has k − 1 zeros in (1)

(2)

(1)

(2)

(ν1 , νk ), or R2n1 (νk , 0) = 0 = R2n2 (νk , 0) and R2n1 (·, 0)R2n2 (·, 0) has k − 2 zeros in (ν1 , νk ). In view of the notation introduced in Remark 1.4.3, this completes the proof of statement 2. φ 3. Let ω = (1) . By Remark 1.4.3, all zeros of R2n (·, 0), (2) (1) R2n1 −1 (·, 0)

R2n −1 (·,0)R2n −1 (·,0) 1 2 (2) R2n2 −1 (·, 0) are positive

real numbers. By Proposition 1.4.2, and part 2, νk is either a pole or a zero of ω. Let 0 ≤ k 0 ≤ k − 1 be the number

41

1.4. Stieltjes strings with one interior condition

of zero-pole cancellations within the interval (0, νk ) in the quotient representation of ω. Hence ω has k − 1 − k 0 zeros in (0, νk ). Since the poles and the zeros of ω interlace and since the smallest pole of ω is smaller than any zeros of ω, it follows that ω has k−k 0 poles in (0, νk ) if νk is a zero of ω and k−1−k 0 poles in (0, νk ) if νk is a pole of ω. In view of Proposition 1.4.2, this results in the following alternative: (2) (2) (1) (1) either R2n1 −1 (νk , 0) 6= 0, R2n2 −1 (νk , 0) 6= 0 and R2n1 −1 (·, 0)R2n2 −1 (·, 0) has k (2)

(1)

(1)

(2)

zeros in (0, νk ), or R2n1 −1 (νk , 0) = 0 = R2n2 −1 (νk , 0) and R2n1 −1 (·, 0)R2n2 −1 (·, 0) has k − 1 zeros in (0, νk ). In view of the notation introduced in Remark 1.4.3, this completes the proof of statement 3. (1)

(2)

n1 n2 / {νj }j=1 . We know 4. First let us consider the case that νk ∈ ∪ {µj }j=1 (1)

(2)

1 2 from statement 3 that νk ∈ ∪ {µj }nj=1 / {µj }nj=1 in this case, and therefore the

(1)

(1)

(2)

(1)

(1)

n1 2 ∪ (µj )nj=1 interval (0, νk ) contains k terms of (µj )j=1 . From µj < νj < µj+1 for j = 1, . . . , n1 − 1, see (1.2.36), it follows that the interval (0, νk ) contains k − 1 (2) 2 (1) n1 or k terms of (νj )j=1 ∪ (µj )nj=1 . (1)

(2)

n2 1 , then it follows from statements 2 and 3 that If νk ∈ {νj }nj=1 ∪ {µj }j=1 (1) n1 (2) n2 ∩{νj }j=1 νk ∈ {νj }j=1

(1)

(2)

(1)

(2)

1 2 1 2 ∩{νj }nj=1 or νk ∈ {µj }nj=1 , . If νk ∈ {νj }nj=1 ∩{µj }nj=1 then, in view of statements 1 and 2, the interval (0, νk ) contains k − 2 terms (1) 1 (2) n2 . Then the interlacing (1.2.36) shows that ∪ (νj )j=1 of the sequence (νj )nj=1 the interval (0, νk ) contains k − 2 or k − 1 terms of the concatenated sequence (1) 1 (2) 2 (1) n1 (2) 2 ∪(µj )nj=1 (νj )j=1 , then the interval (0, νk ) contains . If νk ∈ {µj }nj=1 ∩{µj }nj=1

(1)

(2)

n2 1 ∪ (µj )j=1 by statement 3. Again we see that k − 1 terms of the sequence (µj )nj=1 (1)

(2)

n1 2 the interval (0, νk ) contains k−2 or k−1 terms of the sequence (νj )j=1 . ∪(µj )nj=1 (1)

(2)

n2 1 / Finally, if νk ∈ {νj }nj=1 , then νk ∈ {νj }j=1 by statement 2, and νk ∈ (2)

(2)

(1)

n1 2 2 {µj }nj=1 , then νk ∈ {µj }j=1 follows. Similarly, if νk ∈ {µj }nj=1 by statement (1)

1 / {νj }nj=1 3, and νk ∈ follows.



1.4.2 The inverse problem (1)

1 , First we consider the problem of finding the sequences of string data (mk )nk=1 (1) n1 (2) n2 (2) n2 (mk )k=1 , (lk )k=0 , and (lk )k=0 from the sequences of the characteristic values (2) 2 (1) 1 1 +n2 , (νk )nk=1 (νk )nk=1 , (νk )nk=1 , the length of the string S and the length of the string S1 .

l ∈ (0, l). Let Theorem 1.4.5. Let n1 and n2 be positive integers and let l > 0 and b (1) 1 1 +n2 , (νk )nk=1 three strictly increasing sequences of positive real numbers (νk )nk=1 and (2) n2 (2) n2 (1) n1 n1 +n2 (νk )k=1 be given. Let the concatenated sequence (θk )k=1 = (νk )k=1 ∪(νk )k=1 be indexed such that θk ≤ θk0 if k < k 0 . We require that these sequences satisfy the following conditions:

42

Chapter 1. Stieltjes strings

n1 +n2 1 +n2 (i) The sequences (νk )k=1 and (θk )nk=1 interlace as follows:

ν1 < θ1 ≤ · · · ≤ νn1 +n2 ≤ θn1 +n2 .

(1.4.13)

(ii) If θk = νk+1 or νk+1 = θk+1 for some k ∈ {1, . . . , n1 + n2 − 1}, then θk = νk+1 = θk+1 . (1)

(1)

(2)

n1 1 2 , Then there exist sequences of positive real numbers (mk )k=1 , (lk )nk=0 , (mk )nk=1 n2 n1 P P (1) (2) (2) n2 b b and (lk )k=0 such that lk = l − l, which generate the Dirichletlk = l, k=0

k=0

Dirichlet problem (1.4.1)–(1.4.4) on the Stieltjes string S, and, for j = 1, 2, the Dirichlet-Dirichlet problems (1.4.9)–(1.4.11) on the strings Sj , with the sequences (1) 1 (2) 2 1 +n2 and (νk )nk=1 of characteristic values (νk )nk=1 , respectively. The se, (νk )nk=1 (2) n2 (1) n1 (2) n2 (1) n1 quences (mk )k=1 , (mk )k=1 , (lk )k=0 , and (lk )k=0 are uniquely determined by the given data if and only if all inequalities in (1.4.13) are strict. Proof. We introduce the polynomials P0 (z) = l

n1Y +n2

Q1 (z) = b l

k=1 n 1 Y

 1−

1−

z νk



z

!

(1)

,

(1.4.14)

,

(1.4.15)

νk

k=1

and Q2 (z) = (l − b l)

n2 Y

!

z

1−

(2)

.

(1.4.16)

νk

k=1

For all k ∈ {1, . . . , n1 + n2 − 1} with θk = νk+1 = θk+1 choose a number γ(νk+1 ) ∈ (0, 1). Define  (1)  (1)  P0 (νk )  if Q2 (νk ) 6= 0   Q (ν (1) ) 2 k (1) βk = (k = 1, . . . , n1 ), (1.4.17) 0 (1)   (1) P0 (νk ) (1)   if Q2 (νk ) = 0 γ(νk ) 0 (1) Q2 (νk ) and let P1 be the unique polynomial of degree at most n1 satisfying P1 (0) = 1,

(1)

(1)

P1 (νk ) = βk

(k = 1, . . . , n1 ),

(1.4.18)

namely, the Lagrange interpolation polynomial P1 (z) =

n1 (1) X β z

n1 Y k (1) k=1 νk p=1, p6=k

(1)

(z − νp ) (1)

(1)

(νk − νp )

+

n1 (1) Y νk − z (1)

k=1

νk

.

43

1.4. Stieltjes strings with one interior condition (1)

(1)

Let k ∈ {1, . . . , n1 }. If Q2 (νk ) 6= 0, then P0 (νk ) 6= 0 by assumption (ii). There (1) is a number j ≥ k such that νk = θj . In view of (1.4.13), P0 has j simple zeros (1) (2) in the interval (0, θj ). Therefore (−1)j P0 (νk ) > 0. By definition, νm < θj holds (1) (1) exactly for m = 1, . . . , j − k. Hence (−1)j−k Q2 (νk ) > 0. If Q2 (νk ) = 0, then (1) (1) (2) P0 (νk ) = 0 by (1.4.13). There is a number j > k such that νk = νj = νj−k . (1)

(1)

Hence it follows that (−1)j P00 (νk ) > 0 and (−1)j−k Q20 (νk ) > 0. From the above inequalities and (1.4.17) we can now conclude that (1)

(−1)k P1 (νk ) > 0

P1 (0) > 0, (1)

(k = 1, . . . , n1 ) of P1 are all positive and interlace

Consequently, the zeros µk (1) with the zeros νk of Q1 :

(1)

(1)

. < · · · < µn(1) < νn(1) 1 1

0 < µ1 < ν1 Thus, due to

Q1 (0) P1 (0)

=b l > 0 we conclude that

By Lemma A.3.5 the S0 -function Q1 (z) = ln(1) + 1 P1 (z)

(k = 1, . . . , n1 ).

Q1 P1

Q1 P1

is an S0 -function.

can be expanded into continued fractions 1

(1) −mn1 z

.

1

+

1

(1)

ln1 −1 +

(1) −mn1 −1 z

+ ..

1

.+

1

(1)

l1 + (1)

−m1 z +

1 (1) l0

(1.4.19) where and are sequences of positive numbers which we identify with the lengths of the threads and the masses of the beads we are looking for. From (1.4.15), (1.4.18) and (1.4.19) we obtain (1) 1 (lk )nk=0

(1) 1 (mk )nk=1

Q1 (0) (1) (1) (1) b = ln(1) + ln1 −1 + · · · + l1 + l0 . l= 1 P1 (0) Comparing (1.4.19) with (1.2.19) we conclude that (1)

R2n1 (z, 0) = T1 Q1 (z),

(1)

R2n1 −1 (z, 0) = T1 P1 (z),

(1)

(1.4.20)

(1)

where T1 6= 0 is a constant and R2n1 (·, 0), R2n1 −1 (·, 0) are the Cauer-Fry polynomials associated with these two sequences. In the same way as above we construct the interpolation polynomial P2 of degree at most n2 which satisfies P2 (0) = 1,

(2)

(2)

P2 (νk ) = βk

(k = 1, . . . , n2 ),

(1.4.21)

44

Chapter 1. Stieltjes strings

where

(2)

βk

 (2) P0 (νk )     (2)  Q1 (νk ) = i P 0 (ν (2) )  h (2) 0 k   1 − γ(ν )  k (2) Q01 (νk )

(2)

if Q1 (νk ) 6= 0 (k = 1, . . . , n2 ). if

(2) Q1 (νk )

(1.4.22)

=0

As for P1 we conclude that P2 (0) > 0,

(2)

(−1)k P2 (νk ) > 0 (k = 1, . . . , n2 ).

(2)

(2)

Consequently, the zeros µk of P2 are all positive and interlace with the zeros νk of Q2 : (2) (2) (2) 0 < µ1 < ν1 < · · · < µ(2) n2 < νn2 . Thus, due to

Q2 (0) P2 (0)

= l−b l > 0 we conclude that

Q2 P2

is an S0 -function. (2)

2 In the same way as above we obtain sequences of positive numbers (lk )nk=0 (2) n2 and (mk )k=1 and T2 6= 0 such that

(2)

(2)

R2n2 (z, 0) = T2 Q2 (z), R2n2 −1 (z, 0) = T2 P2 (z) and such that

n2 P k=0

(2)

lk

(1.4.23)

= l−b l. Substituting (1.4.20) and (1.4.23) into (1.4.5) we

obtain φ = T1 T2 (Q1 P2 + Q2 P1 ). Let us compare the polynomial P0 with the polynomial Pb0 := Q1 P2 + Q2 P1 . It follows from (1.4.14)–(1.4.17) and (1.4.22) that P0 (0) = l = Pb0 (0), (1) (1) (1) (1) P0 (νk ) = Q2 (νk )P1 (νk ) = Pb0 (νk )

(k = 1, . . . , n1 ),

(2) P0 (νk )

(k = 1, . . . , n2 ).

=

(2) (2) Q1 (νk )P2 (νk )

(2) = Pb0 (νk ) (1)

For j ∈ {1, . . . , n1 } and k ∈ {1, . . . , n2 } with νj (2)

(1)

(1)

(2)

we calculate

(2)

(2)

= νk

P00 (νk ) = γ(νj )P00 (νj ) + [1 − γ(νk )]P00 (νk ) (1)

(1)

(2)

(2)

= βj Q02 (νj ) + βk Q01 (νk ) (1)

(1)

(2)

(2)

= P1 (νj )Q02 (νj ) + P2 (νk )Q01 (νk ) (2) = Pb00 (νk ), (1) (2) where we have used that Q2 (νj ) = 0 = Q1 (νk ). Consequently, P0 = Pb0 . Hence, 1 +n2 φ = T1 T2 P0 , and the sequence of zeros of φ is exactly (νk )nk=1 .

45

1.4. Stieltjes strings with one interior condition

If not all inequalities in (1.4.13) are strict, then there are j ∈ {1, . . . , n1 } and (2) (1) k ∈ {1, . . . , n2 } such that νj = νk . In that case, the polynomial P1 depends on (1)

(1)

γ(νj ), and varying γ(νj ) in the interval (0, 1) we arrive at a family of distinct 1 polynomials P1 . But then also Q P1 has a family of distinct continued fraction ex(1)

1 pansions (1.4.19), so that we arrive at distinct pairs of sequences (mk )nk=1 and (1) n1 (lk )k=1 . Now assume that all inequalities in (1.4.13) are strict. We observe that, by (2) (2) (1) (2) (1) (1) (1.4.7) and (1.2.25), the functions l0 l0 φ, l0 R2n1 (·, 0) and l0 R2n2 (·, 0) are (1) 1 (2) 2 1 +n2 uniquely determined by (νk )nk=1 , (νk )nk=1 , (νk )nk=1 , l and b l. From (1.4.7) and (1.4.8) it follows that the values

(1) (2)

(1)

(1)

(1)

l0 R2n1 −1 (νk , 0) = (1)

(1)

(1)

(1)

(1)

l0 l0 φ(νk ) (2)

(2)

(1)

l0 R2n2 (νk , 0)

(k = 1, . . . , n1 )

and l0 R2n1 −1 (0, 0) = 1 are uniquely determined by the given data, which means that l0 R2n1 −1 (·, 0) and therefore also the function

(1) 1 (1) R2n −1 (·,0) 1

R2n (·,0)

are uniquely deter(1)

1 −1 mined by the given data. Hence Lemma A.3.5 and (1.2.19) show that (lk )nk=1 (1) n1 −1 (2) n2 −1 and (mk )k=0 are unique. The same reasoning also show that (lk )k=1 and (2) 2 (mk )nk=0 are unique. 

(1) 1 Now we consider the problem of finding the sequences (mk )nk=1 , (1) n1 (2) n2 (lk )k=0 , and (lk )k=0 from the sequences of the characteristic values (1) 1 (2) 2 (µk )nk=1 , (µk )nk=1 and the length of the string S.

(2)

2 (mk )nk=1 , n1 +n2 (νk )k=1 ,

Theorem 1.4.6. Let n1 and n2 be positive integers and let l > 0. Let three strictly (1) 1 (2) 2 1 +n2 increasing sequences of positive real numbers (νk )nk=1 , (µk )nk=1 and (µk )nk=1 be (1) n1 (2) n2 n1 +n2 given. Let the concatenated sequence (τk )k=1 = (µk )k=1 ∪ (µk )k=1 be indexed such that τk ≤ τk0 if k < k 0 . We require that these sequences satisfy the conditions: 1 +n2 1 +n2 (i) The sequences (νk )nk=1 and (τk )nk=1 interlace as follows:

τ1 ≤ ν1 ≤ · · · ≤ τn1 +n2 < νn1 +n2 .

(1.4.24)

(ii) If τk = νk or νk = τk+1 for some k ∈ {1, . . . , n1 + n2 − 1}, then τk = νk = τk+1 . (1)

(2)

2 1 Then there exist sequences of positive real numbers (mk )nk=1 , (mk )nk=1 , (1) n1 −1 (2) n2 −1 (lk )k=0 , and (lk )k=0 with

nX 1 −1 k=0

(1)

lk +

nX 2 −1 k=0

(2)

lk < l

(1.4.25)

46

Chapter 1. Stieltjes strings (2)

(1)

such that for each pair of positive numbers ln1 , ln2 with =l− ln(1) + ln(2) 2 1

nX 1 −1

(1)

lk +

k=0

nX 2 −1

(2)

(1.4.26)

lk ,

k=0

the Dirichlet-Dirichlet problem (1.4.1)–(1.4.4) on the Stieltjes string S, and, for j = 1, 2, the Dirichlet-Neumann problems (1.4.9), (1.4.11), (1.4.12) on the Stieltn1 +n2 jes string Sj , generate the three sequences of characteristic values (νk )k=1 , (1) n1 (1) n1 (2) n2 (2) n2 (µk )k=1 , (µk )k=1 . The sequences of Stieltjes string data (mk )k=1 , (mk )k=1 , (1) 1 −1 (2) 2 −1 (lk )nk=0 are uniquely determined by the given data if and only if , and (lk )nk=0 all inequalities in (1.4.24) are strict. Proof. We introduce the polynomials +n2 n1Y



 z P0 (z) = l 1− , νk k=1 ! n1 Y z Q1 (z) = 1 − (1) , µk k=1 ! n 2 Y z Q2 (z) = 1 − (2) . µk k=1

(1.4.27)

(1.4.28)

(1.4.29)

For all k ∈ {1, . . . , n1 +n2 −1} with τk = νk = τk+1 choose a number γ(νk ) ∈ (0, 1). Define  (1) P0 (µk )  (1)   if Q2 (µk ) 6= 0   Q (µ(1) ) 2 k (1) βk = (k = 1, . . . , n1 ), (1.4.30) (1) 0   γ(µ(1) ) P0 (µk ) if Q (µ(1) ) = 0  2 k  k (1) Q02 (µk ) let e l ∈ (0, l) and let P1 be the unique polynomial of degree at most n1 satisfying P1 (0) = e l,

(1)

(1)

P1 (µk ) = βk

(k = 1, . . . , n1 ),

(1.4.31)

namely, the Lagrange interpolation polynomial P1 (z) =

n1 (1) X β z

n1 Y k (1) p=1, p6=k k=1 µk (1)

(1)

(z − µp ) (1)

(1)

(µk − µp )

+e l

(1)

n1 (1) Y µk − z (1)

k=1

.

(1.4.32)

µk

Let k ∈ {1, . . . , n1 }. If Q2 (µk ) 6= 0, then P0 (µk ) 6= 0 by assumption (ii). There is (1) a number j ≥ k such that µk = τj . In view of (1.4.24), P0 has j −1 simple zeros in (1) (2) the interval (0, τj ). Therefore (−1)j−1 P0 (µk ) > 0. By definition, µm < τj holds

47

1.4. Stieltjes strings with one interior condition (1)

(1)

exactly for m = 1, . . . , j − k. Hence (−1)j−k Q2 (µk ) > 0. If Q2 (µk ) = 0, then (1) (1) (2) P0 (µk ) = 0 by (1.4.24). There is a number j ≥ k such that µk = νj = µj+1−k . (1)

(1)

Hence it follows that (−1)j P00 (µk ) > 0 and (−1)j+1−k Q02 (µk ) > 0. From the above inequalities and (1.4.30) we can now conclude that (1)

(−1)k−1 βk > 0

(k = 1, . . . , n1 ),

(1.4.33)

and therefore P1 (0) > 0,

(1)

(−1)k−1 P1 (µk ) > 0

(k = 1, . . . , n1 ).

(1.4.34)

In the same way as above we construct the interpolation polynomial P2 of degree at most n2 which satisfies (2)

(2)

l, P2 (0) = l − e

(k = 1, . . . , n2 ),

P2 (µk ) = βk

(1.4.35)

where

(2)

βk =

 (2)   P0 (µk )    Q (µ(2) ) 1

(2)

if Q1 (µk ) 6= 0

k

h i P 0 (µ(2) )   (2) 0 k    1 − γ(µk ) (2) Q01 (µk )

(k = 1, . . . , n2 ). if

(2) Q1 (µk )

(1.4.36)

=0

As for P1 we conclude that (2)

(−1)k−1 βk > 0 (k = 1, . . . , n2 ), P2 (0) > 0,

(2) (−1)k−1 P2 (µk )

>0

(1.4.37)

(k = 1, . . . , n2 ).

Let us compare the polynomial P0 with the polynomial Pb0 := Q1 P2 + Q2 P1 . It follows from (1.4.27)–(1.4.30) and (1.4.35) that P0 (0) = l = Pb0 (0), (1) (1) (1) (1) P0 (µk ) = Q2 (µk )P1 (µk ) = Pb0 (µk )

(k = 1, . . . , n1 ),

(2) P0 (µk )

(k = 1, . . . , n2 ).

=

(2) (2) Q1 (µk )P2 (µk )

(2) = Pb0 (µk ) (1)

For j ∈ {1, . . . , n1 } and k ∈ {1, . . . , n2 } with µj (2)

(1)

(1)

(2)

= µk we calculate (2)

(2)

P00 (µk ) = γ(µj )P00 (µj ) + [1 − γ(µk )]P00 (µk ) (1)

(1)

(2)

(2)

= βj Q02 (µj ) + βk Q01 (µk ) (1)

(1)

(2)

(2)

= P1 (µj )Q02 (µj ) + P2 (µk )Q01 (µk ) (2) = Pb00 (µk ),

48

Chapter 1. Stieltjes strings

(1) (2) where we have used that Q2 (µj ) = 0 = Q1 (µk ). Consequently, P0 = Pb0 . It is clear from the definition of P0 , Q1 and Q2 that the coefficients of (−z)n1 +n2 of P0 , of (−z)n1 of Q1 and of (−z)n2 of Q2 are positive numbers α0 , α1 and α2 , respectively. From the representation (1.4.32) we see that the coefficient of (−z)n1 of P1 is

−

n1 (1) X β

n1 Y k (1) k=1 µk p=1, p6=k

1 (1)

(1)

(µp − µk )

+e l

n1 Y 1 (1)

k=1 µk

= δ3 + α3e l,

where α3 > 0 and δ3 < 0 in view of (1.4.33). In the same way, we obtain that the coefficient of (−z)n2 of P2 is δ4 + α4 (l − e l ) with α4 > 0 and δ4 < 0. Comparing the coefficients of (−z)n in the identity P0 = Pb0 gives α0 = α1 (δ4 + α4 (l − e l )) + α2 (δ3 + α3e l ). For f (e l ) = α2 (δ3 + α3e l ) we have f (0) = α2 δ3 < 0 and f (l) = α0 − α1 δ4 > α0 . e l ) = α20 Hence there is l ∈ (0, l) such that f (e l ) = α20 . This means that α2 (δ3 + α3e α n and α1 (δ4 + α4 (l − e l )) = 20 , so that the coefficients δ3 + α3e l of (−z) 1 of P1 and n2 e the coefficient δ4 + α4 (l − l ) of (−z) of P2 are positive. We will now continue with this particular value of e l. Then we conclude that P1 (z) → ∞ and P2 (z) → ∞ as z → −∞, that the degree of P1 is n1 , and that the degree of P2 is n2 . We now determine the location of the zeros of P1 and P2 . From above and (1.4.34) it follows that P1 has an even number of zeros in the intervals (−∞, 0) (1) (1) (1) and (0, µ1 ) and an odd number of zeros in the intervals (µk , µk+1 ) for k = (1)

1, . . . , n1 −1. Since the degree of P1 is n1 , P1 has one simple zero νk (k = 1, . . . , n1 ) (1) (1) (1) (1) in the interval (µk , µk+1 ) and one simple zero νn1 in (µn1 , ∞). It follows that the zeros of P1 and Q1 interlace as follows: (1)

(1)

0 < µ1 < ν1 Thus, due to

P1 (0) Q1 (0)

=e l > 0 we conclude that

By Lemma A.3.5 the S0 -function P1 (z) = ln(1) + 1 Q1 (z)

(1) < · · · < µ(1) n1 < νn1 .

P1 Q1

P1 Q1

is an S0 -function.

can be expanded into continued fraction 1

.

1

(1)

−mn1 z +

1

(1)

ln1 −1 +

(1) −mn1 −1 z

+ ..

1

.+

1

(1)

l1 + (1)

−m1 z +

1 (1) l0

(1.4.38)

1.4. Stieltjes strings with one interior condition 1)

49

(1)

1 1 where (lk )nk=0 and (mk )nk=1 are sequences of positive numbers which we identify with the lengths of the threads and the masses of the beads we are looking for. From (1.4.28), (1.4.31) and (1.4.38) we obtain

P1 (0) (1) (1) (1) e l= = ln(1) + ln1 −1 + · · · + l1 + l0 . 1 Q1 (0) Comparing (1.4.38) with (1.2.19) we conclude that (1)

R2n1 (z, 0) = T1 P1 (z),

(1)

R2n1 −1 (z, 0) = T1 Q1 (z),

(1)

(1.4.39)

(1)

where T1 6= 0 is a constant and R2n1 (·, 0), R2n1 −1 (·, 0) are the Cauer-Fry polynomials associated with these two sequences. (2) As for P1 , the zeros νk (k = 1, . . . , n2 ) of P2 interlace with the zeros of Q2 : (2)

(2)

0 < µ1 < ν1 Thus, due to

P2 (0) Q2 (0)

(2) < · · · < µ(2) n2 < νn2 .

= l−e l > 0 we conclude that

P2 Q2

is an S0 -function. (2)

2 In the same way as above we obtain sequences of positive numbers (lk )nk=0 (2) n2 and (mk )k=1 and T2 6= 0 such that

(2)

(2)

R2n2 (z, 0) = T2 P2 (z), R2n2 −1 (z, 0) = T2 Q2 (z) and such that

n2 P k=0

(2)

lk

(1.4.40)

= l−e l. Substituting (1.4.39) and (1.4.40) into (1.4.5) we

obtain φ = T1 T2 (Q1 P2 + Q2 P1 ). 1 +n2 Hence, φ = T1 T2 P0 , and the sequence of zeros of φ is exactly (νk )nk=1 . (1) 1 In view of Remark 1.2.7, part 1, the characteristic values (µk )nk=1 and (2) n2 (1) (2) (1) (2) (µk )k=1 are independent of ln1 and ln2 . Furthermore, ln1 = ln1 + ln2 , see Fig(1) (2) ures 1.5 and 1.6, is uniquely determined by l, lk (k = 0, . . . , n1 − 1) and lk (1) (2) (k = 0, . . . , n2 − 1), and therefore any pair of positive numbers ln1 , ln2 satisfying (1.4.26) gives the same characteristic values. This completes the existence proof. If not all inequalities in (1.4.24) are strict, then there are j ∈ {1, . . . , n1 } and (1) (2) k ∈ {1, . . . , n2 } such that µj = µk . In that case, the polynomial P1 depends on

1)

1)

γ(µj ), and varying γ(µj ) in the interval (0, 1) we arrive at a family of distinct 1 polynomials P1 . But then also Q P1 has a family of distinct continued fraction ex(1)

1 pansions (1.2.19), so that we arrive at distinct pairs of sequences (mk )nk=1 and (1) n1 −1 (lk )k=1 . Now assume that all inequalities in (1.4.24) are strict. By (1.4.7), (1.2.25) and (1) (2) (1) (1) (2) (2) (1.2.28), the functions l0 l0 φ, l0 R2n1 −1 (·, 0) and l0 R2n2 −1 (·, 0) are uniquely

50

Chapter 1. Stieltjes strings (1)

(2)

1 +n2 1 2 determined by (νk )nk=1 , (µk )nk=1 , (µk )nk=1 and l. From (1.4.7) and (1.4.8) it follows that the values

(1)

(1)

(1)

l0 R2n1 (µk , 0) =

(1) (2)

(1)

(2)

(1)

l0 l0 φ(µk ) (2)

(k = 1, . . . , n1 )

l0 R2n2 −1 (µk , 0)

are uniquely determined by the given data. Then the difference of two such func(1) (1) (1) e (1) (1) tions l0 R2n1 (·, 0) and e l0 R 2n1 (·, 0) is zero at µk (k = 1, . . . , n1 ), which means that there is c ∈ R such that (1) (1) (1) e (1) (1) (1) l0 R2n1 (·, 0) − e l0 R 2n1 (·, 0) = cl0 R2n1 −1 (·, 0). (1) (1) (1) e (1) Therefore, in view of l0 R2n1 −1 (·, 0) = e l0 R 2n1 −1 (·, 0), (1) (1) (1) e (1) e e(1) (·, 0) R l0 R2n1 (·, 0) l0 R 2n1 2n1 (·, 0) − (1) = (1) (1) − (1) (1) (1) e e e R2n1 −1 (·, 0) R2n1 −1 (·, 0) l0 R2n1 −1 (·, 0) l0 R2n1 −1 (·, 0) (1)

R2n1 (·, 0)

=

(1) (1) (1) e (1) l0 R2n1 (·, 0) − e l0 R 2n1 (·, 0) (1)

(1)

l0 R2n1 −1 (·, 0) = c. Since the representation (1.2.19) is unique by Lemma A.3.5, it follows that the (1) 1 (1) 1 −1 sequences (mk )nk=1 and (lk )nk=0 are uniquely determined by the given data. (2) (n2 ) (2) 2 −1 In the same way, it also follows that the sequences (mk )k=1 and (lk )nk=0 are uniquely determined by the given data.  Remark 1.4.7. Theorems 1.4.4 and 1.4.6 deal with the direct and inverse problems, respectively, for the Dirichlet-Dirichlet problem of a string and the DirichletNeumann problem of two substrings. It is easy to see that the properties 1 and 3 in Theorem 1.4.4 are equivalent to the conditions (i) and (ii) in Theorem 1.4.6. (1) 1 −1 (2) 1 −1 Furthermore, with (lk )nk=0 and (lk )nk=0 also ln1 is uniquely determined by the given data in Theorem 1.4.6. Only the position of the partition point P between the bead with index n1 and the bead with index n1 +1, see Figures 1.5 and 1.6, can vary according to the range for e l given in Theorem 1.4.6. Hence the assumptions in Theorem 1.4.6 are necessary and sufficient to solve the inverse problem. (1)

(2)

1 2 Now we consider the problem of finding the sequences (mk )nk=1 , (mk )nk=1 , (1) n1 (2) n2 n1 +n2 (lk )k=0 , and (lk )k=0 from the sequences of the characteristic values (νk )k=1 , (1) 1 (2) 2 (νk )nk=1 , (µk )nk=1 and some other parameters. For the remainder of this subsection we will consider some subcases with different behavior regarding uniqueness. Let n1 and n2 be positive integers and (1) 1 (2) 2 1 +n2 let l > 0 and b l ∈ (0, l) and let the sequences (νk )nk=1 , (µk )nk=1 and (νk )nk=1 be

51

1.4. Stieltjes strings with one interior condition given. With these given data, we will associate the polynomials P0 (z) = l

n1Y +n2

P1 (z) = b l Q2 (z) =

k=1 n1 Y

k=1 n2 Y k=1



z νk



z

!

1−

1−

(1)

,

(1.4.41)

,

(1.4.42)

νk 1−

z

! (1.4.43)

(2)

µk

and Φ = P0 − P1 Q2 .

(1.4.44)

Φ(0) = l − b l > 0.

(1.4.45)

Clearly, First we consider a case where the solution of the inverse problem is unique. Theorem 1.4.8. Let n1 and n2 be positive integers and let l > 0 and b l ∈ (0, l). Let (1) 1 n1 +n2 three strictly increasing sequences of positive real numbers (νk )k=1 , (νk )nk=1 and (2) 2 (2) 2 (1) 1 1 +n2 (µk )nk=1 be given. Let the concatenated sequence (ϑk )nk=1 = (νk )nk=1 ∪(µk )nk=1 be indexed such that ϑk ≤ ϑk0 if k < k 0 . We require that these sequences satisfy the following conditions: 1 +n2 1 +n2 (i) The sequences (νk )nk=1 and (ϑk )nk=1 interlace as follows:

ν1 < ϑ1 < · · · < νn1 +n2 < ϑn1 +n2 .

(1.4.46)

(1)

(ii) ϑn1 +n2 = νn1 . (1)

(2)

1 2 Then there exist unique sequences of positive real numbers (mk )nk=1 , (mk )nk=1 , n n 2 1 P (1) b P (2) (1) n1 (2) n2 lk = l, lk = l − b l, which generate the (lk )k=0 , and (lk )k=0 such that

k=0

k=0

Dirichlet-Dirichlet problem (1.4.1)–(1.4.4) for the string S, the Dirichlet-Dirichlet problem (1.4.9)–(1.4.11) for the string S1 and the Dirichlet-Neumann problem (1.4.9), (1.4.11), (1.4.12) for the string S2 with the sequences of characteristic (1) 1 (2) 2 1 +n2 values (νk )nk=1 , (νk )nk=1 and (µk )nk=1 , respectively. Proof. Since Φ(νk ) = −P1 (νk )Q2 (νk ), (1.4.46) implies (−1)k Φ(νk ) > 0 (k = 1, . . . , n1 + n2 ). Since Φ(ϑk ) = P0 (ϑk ), (−1)k Φ(ϑk ) > 0 (k = 1, . . . , n1 + n2 ).

52

Chapter 1. Stieltjes strings

Hence each of the intervals (0, ν1 ), (ϑk , νk+1 ), (k = 1, . . . , n1 + n2 − 1) contains exactly one zero of Φ, and Φ has no other zeros. We denote the zero of Φ lying in (1) (0, ν1 ) by µ1 . For k = 1, . . . , n1 + n2 − 1 denote the zero of Φ lying in (ϑk , νk+1 ) (2) (2) (1) (1) by νr if ϑk = µr and by µr+1 if ϑk = νr . Therefore, for both j = 1 and j = 2, (j) nj (j) nj the sequences (νk )k=1 and (µk )k=1 interlace as follows: (j)

(j)

(j)

(j)

(j) 0 < µ1 < ν1 < µ2 < ν2 < · · · < µ(j) nj < νnj . (j) n

(1.4.47)

(j) n

j j Thus the sequences (νk )k=1 and (µk )k=1 satisfy the assumptions of Theorem (j) nj (j) nj ) 1.3.1, and hence there exists a unique tuple of sequences ((mk )k=1 , (lk )k=0 n n 2 P P1 (1) (2) lk = l − b lk = b l and l, respectively, which generates Dirichletwith

k=0

k=0

(1)

2 Dirichlet and Dirichlet-Neumann problems with the characteristic values (νk )nk=1 (2) n2 and (µk )k=1 , respectively. ih i h

(1)

(1)

(2)

(2)

By the above construction, the zeros of Φ and l0 R2n1 −1 (·, 0) l0 R2n2 (·, 0) i h ih (1) (1) (2) (2) coincide. Since also Φ(0) = l − b l = l0 R2n1 −1 (0, 0) l0 R2n2 (0, 0) , we conclude h ih i (1) (1) (2) (2) that Φ = l0 R2n1 −1 (·, 0) l0 R2n2 (·, 0) . By (1.2.25) and (1.2.28), the polynomials P1 and Q2 defined by (1.4.42) and (1.4.43) have the representations (1)

(1)

P1 = l0 R2n1 (·, 0),

(2)

(2)

Q2 = l0 R2n2 −1 (·, 0). (1)

In view of (1.4.8) it follows that P0 = l0 R2n (·, 0). But this shows that the given n1 +n2 is the sequence of the characteristic values of the Dirichletsequence (νk )k=1 Dirichlet problem (1.4.1)–(1.4.4) for this string. determined by the To prove the uniqueness, we observe h ih that Φ is uniquely i (2) (2) (1) (1) given data and that Φ = l0 R2n1 −1 (·, 0) l0 R2n2 (·, 0) by (1.2.25), (1.2.28) and (1.4.8). In particular, the zeros of Φ are unique. Recall that (1.4.46) is assumed and that (1.4.47) holds because of (1.2.36). We will now show that the zeros (2) of R2n2 (·, 0) are uniquely determined. Indeed, let k ∈ {1, . . . , n1 + n2 − 1} with (2)

(2)

R2n2 −1 (ϑk , 0) = 0. If also R2n2 −1 (ϑk+1 , 0) = 0, then the interlacing property shows (2)

that there is z0 ∈ (ϑk , ϑk+1 ) such that R2n2 (z0 , 0) = 0. Since Φ has no zeros in (2)

[νk+1 , ϑk+1 ), it follows that the zero of Φ in (ϑk , νk+1 ) is a zero of R2n2 (·, 0) = 0. If, (2)

(1)

however, R2n2 −1 (ϑk+1 , 0) 6= 0, then R2n1 (ϑk+1 , 0) = 0. Hence there is a number (2)

(1)

r ∈ {1, . . . , k} such that ϑk = µr , and, by definition of ϑk+1 , ϑk+1 = νk−r+1 (2) R2n2 (·, 0)

(1) R2n1 −1 (·, 0)

follows. Therefore has r − 1 zeros in (0, ϑk ) and has k − r + 1 zeros in (0, ϑk+1 ) by (1.4.47). Since Φ has k zeros in (0, ϑk ), it follows that all zeros (1) of R2n1 −1 (·, 0) in (0, ϑk+1 ) must lie in (0, ϑk ). Hence the zero of Φ in (ϑk , νk+1 ) is (2)

(2)

a zero of R2n2 (·, 0). In this way we have uniquely identified n2 zeros of R2n2 (·, 0),

53

1.4. Stieltjes strings with one interior condition

which are all zeros of this polynomial. The remaining zeros of Φ now determine (1) 1 (2) 2 (1) , (mk )nk=1 uniquely the zeros of R2n1 −1 (·, 0). Finally, by Theorem 1.3.1, (mk )nk=1 , (2)

(1)

n1 2 are uniquely determined by the given data. , and (lk )nk=0 (lk )k=0



It is clear that condition (i) in Theorem 1.4.8 is equivalent to requiring for each k = 1, . . . , n1 + n2 that νk 6= ϑk and that the interval (0, νk ) contains exactly (2) 2 (1) 1 ∪ (µk )nk=1 k − 1 terms of the sequence (νk )nk=1 . This is a particular case of the necessary conditions for the characteristic values derived in Theorem 1.4.4, A solution are not sufficient. statement 4. However, these conditions ih h i of the inverse (2) (2) (1) (1) problem would require that Φ = l0 R2n1 −1 (·, 0) l0 R2n2 (·, 0) . In particular, the coefficient of (−z)n1 +n2 of Φ must be positive. It is easy to see that this condition is n1Y +n2 n1 n2 Y 1 1 Y 1 l − ˆl > 0. (1.4.48) (1) (2) νk ν µ k=1

k=1

k

k=1

k

Under this and another explicit assumption, we can now give a result for the case (1) 1 (2) 2 that each interval (0, νk ) contains k terms of the sequence (νk )nk=1 ∪ (µk )nk=1 . Theorem 1.4.9. Let n1 and n2 be positive integers and let l > 0 and b l ∈ (0, l). Let (1) 1 n1 +n2 three strictly increasing sequences of positive real numbers (νk )k=1 , (νk )nk=1 and (2) 2 (1) 1 (2) 2 1 +n2 (µk )nk=1 be given. Let the concatenated sequence (ϑk )nk=1 = (νk )nk=1 ∪(µk )nk=1 be indexed such that ϑk ≤ ϑk0 if k < k 0 . We require that these sequences satisfy the following conditions: 1 +n2 1 +n2 (i) The sequences (νk )nk=1 and (ϑk )nk=1 interlace as follows:

ϑ1 < ν1 < · · · < ϑn1 +n2 < νn1 +n2 .

(1.4.49)

(2)

(ii) ϑ1 = µ1 . (iii) The inequality (1.4.48) is satisfied. (1)

(2)

1 2 Then there exist unique sequences of positive real numbers (mk )nk=1 , (mk )nk=1 , n1 n2 P P (1) n1 (2) n2 (1) (2) b b (lk )k=0 , and (lk )k=0 such that lk = l, lk = l − l, which generate the

k=0

k=0

Dirichlet Dirichlet problem (1.4.1)–(1.4.4) for the string S, the Dirichlet-Dirichlet problem (1.4.9)–(1.4.11) for the string S1 and the Dirichlet-Neumann problem (1.4.9), (1.4.11), (1.4.12) for the string S2 with the sequences of characteristic (1) 1 (2) 2 1 +n2 , respectively. values (νk )nk=1 , (νk )nk=1 and (µk )nk=1 Proof. We can prove this theorem in the way we have proved Theorem 1.4.8. However, we will transform the problem to a problem with given data as in Theorem 1.4.8 and then apply conclusions in the proof of Theorem 1.4.8. Thus define

54

Chapter 1. Stieltjes strings

n e 2 = n1 , e 1 = n2 , n 1 (k = 1, . . . , n1 + n2 ), νn1 +n2 −k+1 1 (1) νek = (2) (k = 1, . . . , n e1 ), µn2 −k+1 1 (2) µ ek = (1) (k = 1, . . . , n e2 ), νn1 −k+1 1 (k = 1, . . . , n1 + n2 ), ϑek = ζn1 +n2 −k+1 νek =

and α=

n2 Y

(2) µk ,

n1 Y

β=

k=1

(1) νk ,

γ=

n1Y +n2

k=1

l e l= , γ

νk ,

k=1

b l eb l= . αβ

e (1) e1 (2) e2 1 +n2 From (1.4.48) it follows that e l >b l. Then (ζek )nk=1 = (e νk )nk=1 ∪ (e µk )nk=1 and 1 +n2 1 +n2 the sequences (e νk )nk=1 and (ϑek )nk=1 interlace as in (1.4.46). Defining Pe0 , Pe1 , e 2 and Φ e according to (1.4.41)–(1.4.44), the proof of Theorem 1.4.8 gives unique Q (2) e2 (1) e1 sequences of positive numbers (e νk )nk=1 and (e µk )nk=1 which satisfy the interlacing e =Q e 1 Pe2 , where (1.4.47) for the numbers with tilde and such that Φ n e2

e Y Pe2 (z) = (e l−b l)

!

z

1−

,

(2)

νek

k=1

e 1 (z) = Q

n e1 Y

1−

z

! .

(1)

µ ek

k=1

Putting (2)

νk =

1 (1) µ en2 −k+1

1

(1)

(k = 1, . . . , n2 ),

µk =

(k = 1, . . . , n1 ), (1.4.50)

(2) νen1 −k+1

and δ=

n1 Y

(1) µk ,

ε=

k=1

n2 Y

(2)

νk ,

k=1

we arrive at the polynomials

P2 (z) = (l − b l)

n2 Y k=1

1−

z (2)

νk

! ,

Q1 (z) =

n1 Y k=1

1−

z (1)

µk

! .

55

1.4. Stieltjes strings with one interior condition Then    n1Y +n2 n1  Y e 1 z l e e (−z) =l (−z + νk ) = lγ 1− = γ P0 (z), z νk l k=1 k=1   n2 1 eY e (2) (−z)n2 Pe1 =b l (−z + µk ) = αb lQ2 (z), z k=1   Y n1 β (1) e2 1 = (−z)n1 Q (−z + νk ) = P1 (z), b z l k=1 !   Y n2 1 ε e1 1 = −z + (1) = (−z)n2 Q P2 (z), z l−b l µ ek k=1 !   n1 Y 1 1 e e n1 e = (e l−b l) −z + (2) = δ(e l−b l)Q1 (z). (−z) P2 z νek k=1 n1 +n2

Pe0

e 1 (z)Pe2 (z) + Pe1 (z)Q e 2 (z) and then Replacing z by z1 in the identity Pe0 (z) = Q n1 +n2 multiplying the resulting identity with (−z) , the above representations lead to e eb e e l l−b l l γ P0 (z) = εδ Q1 (z)P2 (z) + αβ P1 (z)Q2 (z). b b l l−l l e 1 (z)Pe2 (z)+ Pe1 (z)Q e 2 (z) leads Comparing the coefficients of (−z)n1 +n2 in Pe0 (z) = Q e e to e lγ = (e l−b l )εδ + b l αβ, which shows that εδ

e e eb e l−b l lγ − l αβ = = 1. b l−l l−b l (j) n

j Hence we arrive at P0 = Q1 P2 + P1 Q2 . For j = 1, 2, the zeros (νk )j=1 of Qj and

(j) n

j (µk )j=1 of Pj satisfy the interlacing property (1.4.47). In view of Theorem 1.3.1,

(j)

(j)

(j)

(j)

there are string data such that Pj = l0 R2nj (·, 0) and Qj = l0 R2nj −1 (·, 0), and (j)

(1.4.5) gives P0 = l0 R2n1 +n2 (·, 0). Furthermore, n1 X k=1

(1)

lk =

P1 (0) b = l, Q1 (0)

n2 X k=1

(2)

lk =

P2 (0) = l−b l. Q2 (0)

Hence we have found string data which generate the given characteristic values. To prove uniqueness, we observe that the proof of Theorem 1.4.8 shows that (1) 1 (2) 2 the sequences (e µk )nk=1 and (e νk )nk=1 are uniquely determined by the given data. But by (1.4.50) this means that the characteristic values of the Dirichlet-Neumann problems on the strings S1 and S2 are uniquely determined by the given data, and therefore also the string data are unique by Theorem 1.3.1. 

56

Chapter 1. Stieltjes strings

Remark 1.4.10. Theorems 1.4.4 and 1.4.5 deal with the direct and inverse problems, respectively, for the Dirichlet-Dirichlet problems of a string and two substrings. It is easy to see that the properties 1 and 2 in Theorem 1.4.4 are equivalent to the conditions (i) and (ii) in Theorem 1.4.5. Therefore the assumptions in Theorem 1.4.5 are necessary and sufficient to solve the inverse problem. Remark 1.4.11. In Theorems 1.4.8 and 1.4.9, and in the cases when the distributions of the beads are unique also in Theorems 1.4.5 and 1.4.6, the distribution of the beads depends continuously on the given data in view of Lemma A.6.5 and Corollary A.6.7. Remark 1.4.12. In Theorem 1.4.4, statement 4, we have derived properties for the characteristic values when considering the Dirichlet-Dirichlet problems on S and S1 and the Dirichlet-Neumann problem on S2 . Theorems 1.4.8 and 1.4.9 give sufficient conditions for the solvability of the corresponding inverse problem. We have already seen that apart from the properties stated in Theorem 1.4.4, statement 4, we also need condition (1.4.48) for the existence of a solution of the inverse problem. Furthermore, all zeros of Φ must be real and positive, but this property is not guaranteed. Indeed, consider the particular example n1 = 1, n2 = 1, ν1 = 2, ν2 = 4, ϑ1 = 1, ϑ2 = 8, b l = 1, l > 1. Then (1.4.48) is clearly satisfied, and   z z z Φ(z) = l 1 − 1− − (1 − z) 1 − 2 4  8 1 3 = (l − 1) z 2 − 6 − z+8 . 8 l−1 Then Φ has no real zeros if √ √ 11 11 −3 2 0 there is b δ > 0 such that ll ∈ (0, δ) implies that all zeros νˆk , (k = 1, . . . n1 +n2 ) of Φ are real, simple and satisfy |ˆ νk − νk | < ε. In particular, for sufficiently small ε > 0, all zeros

1.4. Stieltjes strings with one interior condition

57

1 +n2 1 +n2 of Φ are positive. If we additionally assume that {νk }nk=1 ∩ {ϑk }nk=1 = ∅, then 1 +n2 it is clear that for each k ∈ {1, . . . , n1 + n2 } the number of terms of (ϑk )nk=1 in n1 +n2 the interval (0, νk ) equals the number of terms of (ϑk )k=1 in the interval (0, νˆk ). Let us finally consider an example with n1 ≥ 2 such that

(1)

ν1 < ν1 Then choosing l and b l such that of the previous paragraph,

b l l

(1)

< ν2

< ν2 < . . . .

is sufficiently small, we obtain, with the notation

(1)

νb1 < ν1

(1)

< ν2

< νb2 < . . . .

To solve the inverse problem, we need for all k = 1, . . . , n + 1 that there is j ∈ (1) (2) {1, . . . , n1 } such that νbk = µj or that there is j ∈ {1, . . . , n2 } such that νbk = νj . (1)

(2)

When k = 1, we thus have the allocation νb1 = µ1 or νb1 = ν1 . But from (1) (2) (2) (2) (1) νb1 < ν1 < νb2 < µ1 < ν1 we see that νb1 6= ν1 and therefore νb1 = µ1 . Then (1) (2) (2) (2) (1) νb2 = µ2 or νb2 = ν1 . But νb2 < ϑ3 ≤ µ1 shows that νb2 6= ν1 , whereas νb2 = µ2 (1) (1) would lead to the contradiction ν2 < νb2 = µ2 . The above examples and reasonings show that conditions which are necessary and sufficient to solve the inverse problem would be quite cumbersome to derive and verify. Hence we are not going to pursue this problem any further.

1.4.3

The finite-dimensional Hochstadt-Lieberman problem

In this subsection it will be shown that the distribution of the beads on a string is uniquely determined by the length of the string, by the distribution of the beads on half of the string, and by the characteristic values of the Dirichlet-Dirichlet problem. Figure 1.7 illustrates the notation used in the following theorem. l0

l1 m1 m2

l0

l1 m1 m2

(1)

(1)

mp

(1)

(2)

mp

Figure 1.7

mp−1

(1)

m1

(2)

(2)

lp−1

(2)

mp

l0 (1)

mp

l p + lp

(1)

lp−1

l p + lp

l0 (2)

mp−1

(2)

m1

58

Chapter 1. Stieltjes strings

Theorem 1.4.13. Let p ∈ N and put n = 2p. Let sequences of positive numbers (1) (1) (2) (2) (mk )pk=1 , (lk )pk=0 , (mk )pk=1 , (lk )pk=0 , (mk )pk=1 , (lk )pk=0 be given such that p p P P (1) (2) lk = lk . For j = 1, 2, let Sj be the string generated by (mk )pk=1 , (lk )pk=0 , k=0 k=0 (j) (j) (mk )pk=1 , (lk )pk=0

as indicated in Figure 1.7. For j = 1, 2, let σj be the sequence of the characteristic values of the Dirichlet-Dirichlet problem (1.2.4), (1.2.23), (1.2.24) for the string Sj . (1) (2) (1) (2) Then σ1 = σ2 implies mk = mk for k = 1, . . . , p and lk = lk for k = 0, . . . , p. Proof. The strings Sj (j = 1, 2) are composed of two substrings Sj,1 and Sj,2 , where Sj,1 is the string with the given data (mk )pk=1 , (lk )pk=0 , and Sj,2 is the (j) (j) string with the given data (mk )pk=1 , (lk )pk=0 . In particular, S1,1 = S2,1 . Denoting (j) (j,1) the characteristic polynomials of Sj , Sj,1 and Sj,2 by Rk (·, 0), Rk (·, 0) and (j,2) Rk (·, 0), respectively, we have from (1.4.8) that h i h i (j) (j,1) (j) (j,2) (j,1) (j) (j,2) R2n (·, 0) = R2p (·, 0) l0 R2p−1 (·, 0) + R2p−1 (·, 0) l0 R2p (·, 0) . Since the total length of S1 equals the total length of S2 and since σ1 = σ2 , it (1) (2) follows from (1.2.25) that R2n (·, 0) = R2n (·, 0). Also, since S1,1 = S2,1 , it follows (1,1) (2,1) that Rk (·, 0) = Rk (·, 0). Hence taking the differences of the above identity for j = 1 and j = 2 it follows that h i (1,1) (1) (1,2) (2) (2,2) 0 = R2p (·, 0) l0 R2p−1 (·, 0) − l0 R2p−1 (·, 0) h i (1,1) (1) (1,2) (2) (2,2) + R2p−1 (·, 0) l0 R2p (·, 0) − l0 R2p (·, 0) . (1,1)

(1,1)

Since the polynomials R2p (·, 0) and R2p−1 (·, 0) do not have common zeros, see (1)

(1,2)

(2)

(2,2)

(1.2.36), it follows that l0 R2p (·, 0) − l0 R2p (·, 0) is zero at the p zeros of (1,1)

R2p (·, 0). Hence there is a constant C such that (1)

(1,2)

(2)

(2,2)

(1,1)

l0 R2p (·, 0) − l0 R2p (·, 0) = CR2p (·, 0). Then (1)

(1,2)

(2)

(2,2)

(1,1)

l0 R2p−1 (·, 0) − l0 R2p−1 (·, 0) = −CR2p−1 (·, 0). Evaluating the latter identity at 0 gives in view of (1.2.28) that C = 0. Since the lengths of S1 and S2 are equal and since the lengths of S1,1 and S1,2 are equal, also the lengths of S1,2 and S2,2 are equal. It follows from (1.2.25) and (1.2.28) that the characteristic values of the Dirichlet-Dirichlet and Dirichlet-Neumann problems for S1,2 and S2,2 coincide. An application of the uniqueness part of Theorem 1.3.1 completes the proof. 

1.4. Stieltjes strings with one interior condition

59

The uniqueness result in Theorem 1.4.13 raises the following question. Let p ∈ N and put n = 2p. Given a sequence (νk )nk=1 of distinct positive real numbers, two sequences of positive real numbers (mk )pk=1 and (lk )pk=0 and a positive real p P number l > lk , does there exist a Stieltjes string S of total length l with n k=0

beads such that the sequence of the characteristic values of the Dirichlet-Dirichlet problem for S is (νk )nk=1 and such that the lengths the threads and the masses of the beads of the left part of S are (mk )pk=1 and (lk )pk=0 , respectively? The following example for the simplest case n = 2 shows that restrictions are needed and that these restrictions are most likely difficult to describe for n > 2. We will use the (1) (1) notation from Figure 1.7 and set l2 := l0 and m2 = m1 . (1)

Example 1.4.14. Let S be a string of length 4 with 2 beads and l0 = 1, l1 = 1, m1 = 1 and 0 < ν1 < ν2 . Then there are l2 ∈ (0, 2) and m2 > 0 such that ν1 and ν2 are the characteristic values of the Dirichlet-Dirichlet problem on S if and only if 16 < 12(ν1 + ν2 ) − 9ν1 ν2 < 24. Proof. A straightforward calculation with the given data of the string shows that R4 (z, 0) = m2 (2 + l2 − l22 )z 2 − (3 + 4m2 − m2 l22 )z + 4. On the other hand, by (1.2.25),    z z 4 2 4(ν1 + ν2 ) R4 (z, 0) = 4 1 − 1− = z − z + 4. ν1 ν2 ν1 ν2 ν1 ν2 Comparing the coefficients gives the following two equations which describe the relation between the string data and the spectral data: 4 , ν1 ν2 4(ν1 + ν2 ) 3 + m2 (4 − l22 ) = . ν1 ν2

m2 (2 + l2 − l22 ) =

The first equation gives m2 =

4 , ν1 ν2 (2 + l2 − l22 )

(1.4.51)

and substituting (1.4.51) into the second equation results in 3+

4(2 + l2 ) 4(ν1 + ν2 ) = . ν1 ν2 (1 + l2 ) ν1 ν2

(1.4.52)

For any admissible values of l2 , ν1 and ν2 , (1.4.51) gives a (unique) positive number m2 , and therefore the restrictions on ν1 and ν2 are completely described by (1.4.52), which can be rewritten as [4(ν1 + ν2 ) − 3ν1 ν2 − 4](1 + l2 ) = 4.

60

Chapter 1. Stieltjes strings

Since l2 ∈ (0, 2), this leads to the admissible range 4 < 12(ν1 + ν2 ) − 9ν1 ν2 − 12 < 12.



1.5 Notes M.G. Krein [78], [79] and L. de Branges [19], [20], [21], [22], [23], [24], [25], [26] considered equations occurring in the theory of small transverse vibrations of a string. They investigated a very wide class of strings (Krein strings) including those which do not possess density in its usual meaning, i.e., the mass distribution function M is not differentiable. The spectra of two boundary value problems describing small transverse vibrations of such a string with different (NeumannDirichlet and Dirichlet-Dirichlet) boundary conditions together with the length of the string uniquely determine the mass distribution function M of the string. This result was stated by M.G. Krein and proved by L. de Branges (see also [48], p. 252). Moreover, these authors found necessary and sufficient conditions on two sequences of real numbers to be the spectra of two boundary value problems generated by a string of this class. These conditions include strict interlacing of the two sequences (see also [75]). However, a constructive algorithm of recovering the mass distribution function from the two spectra and the length of the string in the general case is unknown. It is known for two particular cases: firstly for the case of ‘smooth’ strings where the density ρ = M 0 exists and is twice continuously differentiable and is strictly positive and secondly for Stieltjes strings. In the first case, using the Liouville transform (see [40], p.292, [96], p.40) one can reduce the string equation to the Sturm-Liouville equation and use results of B.M. Levitan and G.M. Gasymov [89] or V.A. Marchenko [92]. In the second case, the solution of two spectra problems for Stieltjes strings is read off from the coefficients of the continued fraction expansion of a rational function whose zeros and poles are the two given spectra. This method was used by Gantmacher and Krein (see [59, p. 287]). In particular, Theorems 1.3.1 and 1.3.3 were proved in [59]. It was stated in Section 1.1 that Cauer and Fry expressed the impedance function as a Stieltjes function. For this reason the numerator and denominator of such Stieltjes functions are called Cauer-Fry polynomials. The Stieltjes string equations derived in Section 1.2 also occur in the theory of longitudinal vibrations of point masses joined by springs, see [93] and [62]. The Sturm oscillation theorem for Stieltjes strings can be found in [131]. The problem considered in Subsection 1.4.2 (based on [16]) is a finite-dimensional analogue of the three spectra problem for Sturm-Liouville equation solved in [105], [67], see also [72] and for Jacobi matrices in [95]. The result in Subsection 1.4.3 is a finite-dimensional analogue of HochstadtLieberman theorem (see [71] for the original infinite-dimensional version).

Chapter 2

Vibrations of star graphs 2.1

Star graphs with root at the centre

In this section, we consider a plane star graph of g (≥ 2) Stieltjes strings, that is, g Stieltjes strings which are joined at a common vertex, called the central vertex of the star graph. In this section, the central vertex will be called the root of the star graph. At the root, a bead of mass M > 0 may be placed. Since the cases with and without a bead at the central vertex can often be considered simultaneously, we put M = 0 when there is no bead at the central vertex. We assume that this graph is stretched and study its small transverse vibrations in two different cases: (D1) the central vertex is fixed (Dirichlet problem), (N1) the central vertex is free to move in the direction orthogonal to the equilibrium position of the strings (Neumann problem). We investigate the relation of the characteristic values of the Neumann problem (N1) to those of the problem (D1) which decouples completely into q Dirichlet problems on the edges of the star graph. In the sequel, we label the edges of the star graph by j = 1, . . . , g (g ≥ 2) and we assume that each edge is a Stieltjes string. We suppose that the j-th edge (j) consists of nj + 1 (nj ≥ 0) threads of length lk (k = 0, . . . , nj ) with beads of mass (j) mk (k = 1, . . . , nj ) separating them, both counted from the exterior towards the Pnj (j) centre. The length of the j-th edge is denoted by l(j) := k=0 lk . (j)

By vk (t) (k = 1, . . . , nj , j = 1, . . . , g) we denote the transverse displacement (j) (j) of the k-th bead on the j-th string at time t, and by v0 (t), vnj +1 (t) those of the ends of the j-th string. If we assume the threads to be stretched by forces each equal to 1, the Lagrange equations for the small transverse vibrations of the graph © Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4_2

61

62

Chapter 2. Vibrations of star graphs (3)

l0 (3)

m1

(3)

l1 (2)

(2)

(2)

l0

l1

(j)

(2)

m1 m2

mnj

M

root

(2)

(j)

l2

lnj

(1)

(1) m1

(j)

(j)

m2 m1

... ...

(j)

(j)

l1

l0

l(j)

l1

(1)

l0

Figure 2.1: Star graph with root at the central vertex are given by (see (1.2.1)) (j)

(j)

vk (t) − vk−1 (t) (j) lk−1

(j)

−

(j)

vk+1 (t) − vk (t) (j) lk

(j) (j)00

+ mk vk

(t) = 0

(k = 1, . . . , nj , j = 1, . . . , g). At the central vertex joining the edges, the continuity of the graph requires that (1)

(2)

(g)

vn1 +1 (t) = vn2 +1 (t) = · · · = vng +1 (t), and the balance of forces implies that (j) (j) g X vnj +1 (t) − vnj (t) (j) l nj

j=1

(1) 00

+ M vn1 +1 (t) = 0.

In case where a bead is absent at the central vertex, the balance of forces condition may be called Kirchhoff condition because it is called so in the theory of electrical circuits. For the pendant vertices we assume the general boundary conditions (j)

(j)

v0 (t) = cj v1 (t) (j)

(cj ∈ R, j = 1, . . . , g).

(j)

Separation of variables vk (t) = uk eiλt leads to the following difference equations (j) for the displacement amplitudes uk (k = 0, . . . , nj , j = 1, . . . , g) for the Neumann and Dirichlet problem: Dirichlet problem (D1). If we clamp the central vertex, then the problem decouples and consists of the g separate problems on the edges with Dirichlet boundary

63

2.1. Star graphs with root at the centre condition at the central vertex end, (j)

(j)

uk − uk−1 (j) lk−1

(j)

−

(j)

uk+1 − uk (j) lk

(j)

(j)

− mk λ2 uk = 0

(k = 1, . . . , nj ),

(j)

(2.1.1)

unj +1 = 0,

(2.1.2)

(j)

(2.1.3)

(j)

u0 = cj u1 for j = 1, . . . , g.

Neumann problem (N1). If the central vertex is allowed to move freely in the direction orthogonal to the equilibrium position of the strings, we obtain (j)

(j)

uk − uk−1 (j) lk−1

(j)

−

(j)

uk+1 − uk (j) lk

(j)

(j)

(k = 1, . . . , nj , j = 1, . . . , g),

= ... =

(g) ung +1 ,

(2.1.5)

(1)

(2.1.6)

− mk λ2 uk = 0

(2.1.4) (1) un1 +1

=

(2) un2 +1

(j) (j) g X unj +1 − unj (j) lnj

j=1 (j)

(j)

u0 = cj u1

− M λ2 un1 +1 = 0, (j = 1, . . . , g).

(2.1.7)

Note that the Dirichlet problem (D1) and the Neumann problem (N1) share the equations (2.1.1) and (2.1.3). If M tends to ∞, then the Neumann problem (N1) becomes the Dirichlet problem (D1). Indeed, in this case the condition (2.1.6) (1) becomes un1 +1 = 0 and, together with (2.1.5), it follows that (2.1.2) holds for j = 1, . . . , g. We will denote by n :=

g X

nj

j=1

the number of beads on the star graph without the possible bead at the centre. As in Chapter 1 we will replace λ2 with z and we will call a complex number z a characteristic value of (D1) or (N1) when the corresponding problem has a nontrivial solution for that value of z. The number of linearly independent solutions, that is, the dimension of the vector space of solutions, will be called the geometric multiplicity of the characteristic value. The dimension of the corresponding solution space will be denoted by nD,g (z) and nN,g (z), respectively.

2.1.1 The direct spectral problem In this subsection we investigate the interlacing properties and multiplicities of the characteristic values of the Dirichlet problem (D1) and the Neumann problem (N1).

64

Chapter 2. Vibrations of star graphs

For simplicity we will assume for the remainder of this section that each pendant vertex obeys either the Dirichlet condition or the Neumann condition. This assumption is in most of what follows in this section not necessary, cj ∈ [0, 1] may suffice. In other parts of this monograph we will make use of notation and results which include the case g = 1. Therefore, in this subsection, we will consider the problems (D1) and (N1) for g ≥ 1. Problem (D1) in case g = 1 has been considered in Sections 1.2 and 1.3. For (N1) in case g = 1, the condition (2.1.5) is void, and for M > 0, the boundary condition (2.1.6) has not been considered in Chapter 1. According to Subsection 1.2.1, for each j = 1, . . . , g, one may obtain the (j) (j) (j) solutions uk (k = 1, . . . , nj + 1) of (2.1.1) with initial condition u0 = cj u1 as in (2.1.3) successively in the form (j)

(j)

(j)

uk = R2k−2 (z, cj )u1

(2.1.8)

(k = 1, . . . , nj + 1).

From (1.2.10) we recall the notation (j)

(j)

R2k−1 (·, cj ) =

(j)

R2k (·, cj ) − R2k−2 (·, cj )

(k = 1, . . . , nj ).

(j)

lk

We may therefore use all the properties of the Cauer-Fry polynomials Rk which were derived in Chapter 1. Some of the expressions involving these polynomials are quite lengthy, and since the cj will be fixed, we will often omit these parameters in (j) (j) the presentations of the Cauer-Fry polynomials and write Rk instead of Rk (·, cj ). The discussion of the Stieltjes string problems on page 10 and Propositions 1.2.8–1.2.11 give the following Remark 2.1.1. For j = 1, . . . , g, the characteristic values of problem (2.1.1)–(2.1.3) on the j-th edge are given by the zeros of the polynomial (j)

(j)

1. R2nj for the Dirichlet condition unj +1 = 0, (j)

(j)

(j)

2. R2nj −1 for the Neumann condition unj +1 = unj , and all characteristic values are real, positive (except for the Neumann-Neumann problem, where the smallest characteristic value is 0) and simple and interlace according to (1.2.36)–(1.2.39). We introduce the polynomials φD,g and φN,g defined by φD,g (z) =

g Y

(j)

R2nj (z),

(2.1.9)

j=1

   g X M (j) (j)  R2n φN,g (z) = (z) − zR (z) 2nj j −1 g j=1

g Y k=1, k6=j

 (k)

R2nk (z) ,

(2.1.10)

65

2.1. Star graphs with root at the centre and we introduce the rational functions φg =

φD,g φN,g

(2.1.11)

and (j)

φ(j) =

R2nj

(j = 1, . . . , g).

(j)

(2.1.12)

R2nj −1 Note that the polynomial φN,g may also be written as   (j) g X R2nj −1 (z) φN,g (z) =  − M z  φD,g (z). (j) R (z) j=1 2nj

(2.1.13)

Theorem 2.1.2. The function φg has the representation φg (z) =

1 1 − Mz (j) j=1 φ (z)

(2.1.14)

g P

and is an S-function, and an S0 -function if at least one pendant vertex is clamped. Proof. The representation (2.1.14) of φg follows from (2.1.11), (2.1.13), (2.1.12). The representation (1.2.19) and Lemmas A.3.6 and A.3.5 show that φ(j) is a rational S-function for all j = 1, . . . , g and therefore a rational Nevanlinna function. Furthermore, φ(j) is a rational S0 -function for those j for which the j-th edge is clamped at the pendant vertex. In particular, φ(j) (z) > 0 for z ∈ (−∞, 0) and φ(j) (0) > 0 for those j for which the j-th edge is clamped at the pendant vertex. Using Lemma A.2.4 and the fact that a finite sum of rational Nevanlinna functions is a rational Nevanlinna function, it follows that z 7→ −

g X j=1

1 + Mz φ(j) (z)

is a rational Nevanlinna function. Another application of Lemma A.2.4 shows that φg is a Nevanlinna function. Finally, φg (z) > 0 for z ∈ (−∞, 0) and φg (0) > 0 if at least one pendant vertex is clamped, which shows that φg is a rational S-function or S0 -function.  Theorem 2.1.3. The characteristic values of the Dirichlet problem (D1), counted with multiplicity, are the zeros of φD,g .  g (j) nj Proof. The conditions (2.1.2) imply that (uk )k=1 is a solution of (D1) if j=1

(j) n

j and only if for each j = 1, . . . , g, (uk )k=1 is a solution of the Dirichlet-Dirichlet

66

Chapter 2. Vibrations of star graphs

or Neumann-Dirichlet problem on the j-th edge. In particular, the dimension of the solution space of problem (D1) equals the sum of the dimensions of the solution spaces on the edges, if we observe that any of these dimensions is 0 if the corresponding problem has only the trivial solution, for which all displacements are 0. Taking Remark 2.1.1 into account completes the proof.  Theorem 2.1.4. The characteristic values of the Neumann problem (N1), counted with multiplicity, are the zeros of the polynomial φN,g . Let z ∈ C. If nD,g (z) = 0, then nN,g (z) = 0 or nN,g (z) = 1. If nD,g (z) > 0, then nN,g (z) = nD,g (z) − 1. Proof. First let g = 1 and z ∈ C. Then (N1) has a nontrivial solution if and only if (1) (1) (1) (1) (1) φN,1 (z)u1 = R2n1 −1 (z)u1 − M zR2n1 (z)u1 = 0 (1)

for u1 6= 0, which means that the characteristic values of (N1) are the zeros of (1) (1) (1) (1) φN,1 . Because (u0 , u1 , . . . , un+1 ) satisfying (N1) is uniquely determined by u1 , (1) the multiplicity of each characteristic value is 1. By (1.2.36) and (1.2.39), R2n1 −1 (1)

has only simple zeros which are distinct from the zeros of φD,1 = R2n1 , which are also simple. In particular, the zeros of φN,1 are distinct from the zeros of φD,1 . Since φ1 is a Nevanlinna function by Theorem 2.1.2, its poles are simple, which implies that the zeros of φN,1 are simple. This completes the proof of Theorem 2.1.4 for g = 1. Now let g ≥ 2. Then the conditions (2.1.5) and (2.1.6) at the central vertex (j) yield the following system of linear equations for u1 (j = 1, . . . , g): (1)

(1)

(2)

(g)

(2)

(g)

R2n1 (z)u1 = R2n2 (z)u1 = · · · = R2ng (z)u1 , g X

(j)

(j)

(1)

(1)

R2nj −1 (z)u1 − M zR2n1 (z)u1 = 0.

(2.1.15) (2.1.16)

j=1

In view of (2.1.15), the equation (2.1.16) can be written in the form  g  X M (j) (j) (1) R2nj −1 (z) − zR2nj (z) uj = 0. g j=1 Then the system of equations has a nontrivial solution if and only if its coefficient matrix  (1)  (2) R2n1 (z) −R2n2 (z) 0 ··· 0 0   (2) (3) R2n2 (z) −R2n3 (z) · · · 0 0  0    . . . . . .   .. .. .. .. .. .. A(z) =     (g−1) (g)  0 0 0 · · · R2ng−1 (z) −R2ng (z) ag,1 (z) ag,2 (z) ag,3 (z) · · · ag,g−1 (z) ag,g (z)

67

2.1. Star graphs with root at the centre is singular, where (j)

ag,j (z) = R2nj −1 (z) −

M (j) zR2nj (z) (j = 1, . . . , g). g

Expanding the determinant of A(z) with respect to the last row, it follows that det A(z) = φN,g (z). We still have to prove the statements about the multiplicities. Let z ∈ C. Clearly, nN,g (z) equals the defect of the matrix A(z). From Theorem 2.1.3 it (j) follows that the number of indices j ∈ {1, . . . , g} for which R2nj (z) = 0 is nD,g (z). b of A(z) is a lower triangular matrix The upper right (g−1)×(g−1) submatrix A(z) (2)

(g)

with diagonal entries −R2n2 (z), . . . , −R2ng (z). If nD,g (z) = 0, then it follows from b Theorem 2.1.3 that these diagonal entries are nonzero. Therefore A(z) has full rank g − 1, which implies nN,g (z) = 0 or nN,g (z) = 1. If nD,g (z) > 0, then we (1) will assume without loss of generality that R2n1 (z) = 0, see Theorem 2.1.3. Then the first g − 1 entries of the first column of A(z) are 0, whereas the last entry (1) (1) is ag,1 (z) = R2n1 −1 (z) 6= 0 because the zeros of R2n1 −1 and R2n1 interlace (see (1.2.36) and (1.2.39)). Hence the last row of A(z) is linearly independent of the (2) (g) first g − 1 rows of A(z). Since nD,g (z) − 1 of the values of R2n2 (z), . . . , R2ng (z) are ˆ 0 (see Theorem 2.1.3), the rank of A(z) is (g − 1) − (nD,g (z) − 1). It follows that A(z) has rank g − nD,g (z) + 1, which means that nN,g (z) = nD,g (z) − 1. Finally, it remains to verify the multiplicities of φN,g . Therefore, let z ∈ C such that nN,g (z) > 0. By Theorem 2.1.2 we know that φg is a rational S-function. In particular, its zeros and poles are simple. If nD,g (z) = 0, it follows that z is a simple zero of φN,g , and the multiplicity of the zero z of φN,g therefore equals the multiplicity nN,g (z) of the characteristic value z of (N1). Now let p = nD,g (z) > 0. (j) Without loss of generality we may assume that Rnj (z) = 0 for j = 1, . . . , p and (j) Rnj (z) 6= 0 for j = p + 1, . . . , g. Since each product in the definition (2.1.10) of (k) φN,g contains at least p − 1 factors R2nk which are zero at z, it is clear that φN,g has a zero of multiplicity at least p − 1 at z. To prove that this order is exactly p − 1, we are going to show that (φN,g )(p−1) (z) 6= 0. Firstly it is clear that

(φN,g )

(p−1)

(z) = (p − 1)!

p X j=1

 (j) R2n (z) j −1



p Y

(k)0 R2nk (z)

k=1, k6=j (k)

g Y

(k)

R2nk (z),

k=p+1

since all other summand have at least one factor R2nk (z) with k ≤ p. We rewrite this as   (j) p p g X Y Y R (z) 2nj −1 (k)0 (k)  (φN,g )(p−1) (z) = (p − 1)!  R (z) R2nk (z). (2.1.17) 2nk (j)0 j=1 R2nj (z) k=1 k=p+1

68

Chapter 2. Vibrations of star graphs

For j = 1, . . . , p we find that (j)0

R2nj (z) (j)

R2nj −1 (z)

(j)0

=

(j)

(j)

(j)0

R2nj (z)R2nj −1 (z) − R2nj (z)R2nj −1 (z) (j)

(R2nj −1 (z))2

(j)

=

R2nj (z) (j)

!0 >0

R2nj −1 (z)

by Lemma A.2.5. Hence the sum in (2.1.17) is positive, whereas all other factors are nonzero. Therefore the multiplicity of the zero z of φN,g equals nN,g (z).  We can now prove the following relations between the characteristic values of the Dirichlet problem (D1) and the characteristic values of the Neumann problem (N1). Theorem 2.1.5. Let n0 = n + 1 if M > 0 and n0 = n if M = 0. Then the sequence 0 of characteristic values (µk )nk=1 of the Neumann problem (N1) and the sequence of the characteristic values (νk )nk=1 of the Dirichlet problem (D1) are real and have the following properties, when indexed in nondecreasing order: 1. 0 ≤ µ1 < ν1 ≤ · · · ≤ µn ≤ νn and νn < µn+1 if M > 0, where µ1 = 0 if and only if all pendant vertices are freely moving; 2. νk−1 = µk if and only if µk = νk (k = 2, . . . , n); 3. the multiplicity of νk does not exceed g (k = 1, . . . , n). Proof. From its definition it is clear that φD,g is a polynomial of degree n, and all zeros of φD,g are real and positive by Propositions 1.2.8 and 1.2.10. Hence the characteristic values of the Dirichlet problem (D1) can be indexed as indicated. From the definition of φN,g in (2.1.10) it is clear that φN,g is a polynomial of degree at most n if M = 0 and a polynomial of degree exactly n + 1 if M > 0. But since φg is an S-function by Theorem 2.1.2, the degree of φN,g is at least as large as the degree of φD,g , which is n. Hence all characteristic values of the Neumann problem (N1) are real and nonnegative and can be indexed as indicated, with µ1 = 0 as stated in part 1. By Lemmas A.3.5 and A.4.4, the interlacing given in statement 1 follows, however only in the weaker form µ1 ≤ ν1 and νn ≤ µn+1 in case M > 0. Here we note that Lemma A.4.4 becomes applicable in the case M > 0 if we introduce an auxiliary number νn+1 > µn+1 . (k) Since the zeros of R2nk (k = 1, . . . , g) are simple, it follows that nD,g (z) ≤ g for all z ∈ C. This proves statement 3. If νk−1 = µk or if µk = νk for indices k for which these characteristic values are defined, then µk is a common zero of φN,g and φD,g . But then nD,g (µk ) = nN,g (µk ) + 1 by Theorem 2.1.4, so that there are indices j1 and j2 with 1 ≤ j1 < j2 ≤ n such that µj1 < νj1 = µj1 +1 = · · · = µj2 = νj2 (2.1.18) with νj2 < µj2 +1 in case M > 0 or j2 < n. This proves statement 2 and the strict inequalities in statement 1. 

69

2.1. Star graphs with root at the centre

Corollary 2.1.6. At least one of two neighbouring characteristic values µk < µk+1 (k = 2, . . . , n − 1) is simple. Proof. If both µk and µk+1 were not simple, it would follow from (2.1.18) that  µk = νk and νk = µk+1 , which would give the contradiction µk = µk+1 . 0

n of the Neumann problem (N1) Proposition 2.1.7. The characteristic values (µk )k=1 have the following monotonicity properties:

1. µk is a nonincreasing function of M ∈ [0, ∞) for k = 1, . . . , n + 1, where we have set µn+1 := ∞ if M = 0; 2. µk → νk−1 (k = 2, . . . , n + 1) and µ1 → 0 as M → ∞. Proof. For the purpose of this proof, we write µk (M ) and φN,g (·, M ) to indicate the dependence on M ∈ [0, ∞). When νk−1 = νk for some k ∈ {2, . . . , n}, then νk−1 = µk (M ) = νk for all M ∈ [0, ∞), and the statements of this proposition are trivial. Now let k ∈ {1, . . . , n+1} with νk−1 6= νk , where ν0 := 0 and νn+1 := ∞. Then Theorem 2.1.5, statement 2, gives that νk−1 < µk (M ) < νk for all M ∈ [0, ∞), except for the case when M = 0 and k = n + 1 where µn+1 (0) = νn+1 = ∞. From (2.1.13) we conclude for M, M 0 ∈ [0, ∞), M 0 < M , that −

φN,g (z, M ) φN,g (z, M 0 ) =− + (M − M 0 )z. φD,g (z) φD,g (z)

Then −

(2.1.19)

φN,g (µk (M 0 ), M ) = (M − M 0 )µk (M 0 ) > 0 φD,g (µk (M 0 )) φ

(·,M )

and the fact that the rational function − N,g is a Nevanlinna function (see φD,g Theorem 2.1.2 and Lemma A.2.4) and hence increasing between its poles (see Lemma A.2.5) shows that νk−1 < µk (M ) < µk (M 0 ). This completes the proof of part 1. Finally, to prove part 2 in case νk−1 6= νk , let w ∈ (νk−1 , νk ). Then (2.1.19) gives φN,g (w, M ) φN,g (w, 0) − =− + M w, φD,g (w) φD,g (w) which is positive for sufficiently large M , say M ≥ M0 . We already know that φ (·,M ) is increasing on (νk−1 , νk ), and therefore − N,g φD,g −

φN,g (z, M ) >0 φD,g (z)

when z ∈ [w, νk ) and M ≥ M0 . Since φN,g (µk (M ), M ) = 0 and µk (M ) ∈ (νk−1 , νk ) for M > 0, it follows that µk (M ) ∈ (νk−1 , w) for M ≥ M0 . This completes the proof of part 2. 

70

2.1.2

Chapter 2. Vibrations of star graphs

The inverse spectral problem

In this subsection we investigate the inverse problem, that is, the question if two sequences of nonnegative real numbers satisfying properties 1–3 in Theorem 2.1.5 can be realized as sequences of the characteristic values of the Neumann problem (N1) and the Dirichlet problem (D1) on a star graph when all pendant vertices carry the Dirichlet condition. More precisely, suppose that g ∈ N, and that a sequence of lengths l(j) > 0 n n0 , (νk )k=1 with n0 = n or n0 = n + 1 hav(j = 1, . . . , g) as well as sequences (µk )k=1 ing the properties 1–3 in Theorem 2.1.5 are given. Can we determine nonnegative (j) nj , a mass M and sequences of lengths integers nj , sequences of masses (mk )k=1 (j) nj (lk )k=0 for j = 1, . . . , g so that the corresponding star graph has the sequences 0 (µk )nk=1 , (νk )nk=1 as Neumann and Dirichlet characteristic values, respectively? Theorem 2.1.8. Let n, g ∈ N and let n0 = n or n0 = n + 1. Let (l(j) )gj=1 be n0 a sequence of positive real numbers and let sequences of real numbers (µk )k=1 , n (νk )k=1 be given such that (i) 0 < µ1 < ν1 ≤ · · · ≤ µn ≤ νn , and additionally νn < µn+1 if n0 = n + 1; (ii) νk−1 = µk if and only if µk = νk (k = 2, . . . , n); (iii) the multiplicity of each νk in (νk )nk=1 does not exceed g. Then there exists a star graph of g Stieltjes strings, i.e., a sequence (nj )gj=1 of nonnegative integers, M ≥ 0 with M = 0 if and only if n0 = n, and sequences Pg (j) nj (j) nj (mk )k=1 (j = 1, . . . , g) of positive numbers with n = , (lk )k=0 j=1 nj and Pnj (j) (j) Neumann problem that the l such (2.1.4)–(2.1.7), has the = (N1), l k=0 k n0 characteristic values (µk )k=1 and the Dirichlet problem (D1), (2.1.1)–(2.1.3), has the characteristic values (νk )nk=1 , with cj = 0 in (2.1.3) and (2.1.7) for all j. If n ≥ g, all nj can be chosen to be positive. If n < g, n of the numbers nj can be chosen to be 1, and g − n of the nj must be 0. Proof. Let g 0 = min{n, g}. Trivially, the multiplicity of νk in (νk )nk=1 does not exceed n, so that assumption (iii) holds with g replaced by g 0 , and we can choose g0 0 P a sequence of positive integers (nj )gj=1 such that n = nj and such that the j=1

n can be written as the concatenation sequence (νk )k=1 0

(νk )nk=1

=

g [

n

j (νκ(j) )κ=1

(2.1.20)

j=1 (j) n

(j)

(j)

j with sequences (νκ )κ=1 so that νκ > νκ0 for κ > κ0 and j = 1, . . . , g 0 . For (j) nj example, we may define the sequences (νκ )κ=1 as follows. Writing n in the form 0 e + 1 for e ≥ 1 and 0 ≤ n b with integers n n = n eg + n b ≤ g 0 − 1, we put nj = n

71

2.1. Star graphs with root at the centre (j)

j = 1, . . . , n e for j = n b , nj = n b + 1, . . . , g 0 , and νκ = ν(κ−1)g0 +j for j = 1, . . . , g 0 and κ = 1, . . . , nj . We consider the rational function ψg defined by   n0 z   Q 1− g X µk 1  k=1  . ψg (z) =  n (j) Q z l j=1 1− νk k=1

(2.1.21)

The function ψ1g is a rational S0 -function by Lemma A.3.5 since ψg (0) > 0 and since its poles and zeros, after cancellation of common factors, are all simple in view of condition (ii), interlace strictly and are ordered as in (i). In particular, each νk (k = 1, . . . , n) is a simple pole of ψg , although any pole of ψg may be represented by νk with up to g different indices k. Since ψ1g is a Nevanlinna function, so is −ψg (see Lemma A.2.4). In view of Lemma A.2.7 there are constants A0 ≥ 0, A1 , . . . , An > 0, and B ∈ R so that ψg has the partial fractions expansion ψg (z) = −A0 z +

n X k=1

Ak + B. z − νk

(2.1.22)

We note that the coefficients are not unique when νi has the same value for more than one index. But we can still choose them to be positive, say by requiring Aj = Ak when νj = νk . Putting z = 0 in (2.1.22) we find B=

g n X X 1 Ak + . (j) νk l j=1 k=1

We now define (j = 1, . . . g 0 , κ = 1, . . . nj , k = 1, . . . , n),

(j) A(j) κ := Ak if νκ = νk

Bj := ψ (j) (z) :=

1 l(j)

+

nj (j) X Aκ (j)

κ=1

nj X

Aκ

κ=1

z − νκ

(j = 1, . . . g),

(2.1.23)

νκ

(j) (j)

+ Bj

(j = 1, . . . g).

Therefore we have ψg (z) = −A0 z +

g X

ψ (j) (z).

(2.1.24)

(2.1.25)

j=1 (j)

Now consider j ∈ {1, . . . , g 0 }. Since Aκ = Ak > 0, the derivative of the rational 1 function ψ (j) is negative on the real axis. Observing further that ψ (j) (0) = l(j) >0

72

Chapter 2. Vibrations of star graphs (j)

(j)

(j)

it follows that ψ (j) has one simple zero µ1 > 0 in (0, ν1 ), one simple zero µκ (j) (j) in each of the intervals (νκ−1 , νκ ) for κ = 2, . . . , nj and therefore no other zeros. Hence the zeros and poles of ψ (j) interlace as follows: (j)

(j)

(j)

(j)

(j) 0 < µ1 < ν1 < µ2 < ν2 < · · · < µ(j) nj < νnj .

Therefore,

1 ψ (j)

is an S0 -function and by Lemma A.3.5 there exist unique sequences (j) n

(j) n

j j of positive numbers (mκ )κ=1 and (lκ )κ=0 such that

1 ψ (j) (z)

1

= ln(j) + j (j) −mnj z

.

1

+

1

(j)

lnj −1 +

1

(j)

−mnj −1 z + · · · +

1

(j)

l1 +

1

(j)

−m1 z +

(j) l0

(2.1.26) From (2.1.24) and (2.1.26) with z = 0 we see that l(j) =

1 (j) (j) (j) = ln(j) + lnj −1 + . . . + l1 + l0 , j ψ (j) (0) (j)

and this is clearly also true when n < g and j ∈ {g 0 +1, . . . , g} if we put l0 := l(j) . Hence the above sequences of masses and lengths yield a star graph of Stieltjes strings with g edges of length l(j) (j = 1, . . . , g). For this star graph we find in view of (2.1.26) and (1.2.19) and with the notation (2.1.12) that 1 ψ (j) (z)

= φ(j) (z)

(j = 1, . . . , g).

We set M = A0 and observing (2.1.25) and (2.1.14) we arrive at ψg (z) = −M z +

g X j=1

1 φ(j) (z)

=

1 . φg (z)

(2.1.27)

From (2.1.9) and (1.2.25) we find φD,g (z) =

=

g Y

(j) R2nj (z)

j=1 g (j) Y

l

=

n  Y

(j) j=1 l0 k=1

nj g Y l(j) Y (j) j=1 l0 k=1

1−

z νk

1−

z

!

(j)

νk

 .

(2.1.28)

73

2.1. Star graphs with root at the centre

Hence Theorem 2.1.3 shows that the sequence of the characteristic values of the n Dirichlet problem (D1) is (νk )k=1 . Finally, in view of (2.1.11), (2.1.27) and (2.1.21), φD,g (z) = φD,g (z)ψg (z) φg (z)    g g n0  (j) Y X Y 1 l z  = 1 − , µk l(j) j=1 l0(j) k=1 j=1

φN,g (z) =

(2.1.29)

which shows in view of Theorem 2.1.4 that the sequence of the characteristic values 0 of the Neumann problem (N1) is (µk )nk=1 , where n0 = n if and only if M = 0.  Theorem 2.1.9. Let the assumptions and notation be as in Theorem 2.1.8. 1. The number M in Theorem 2.1.8 is uniquely determined by the given data. 0

2. In addition we assume that the sequences (µk )nk=1 and (νk )nk=1 interlace strictly and that the distribution (2.1.20) of the latter characteristic values (j) nj (j) nj onto the g strings is prescribed. Then the sequences (mk )k=1 and (lk )k=0 (j = 1, . . . , g) are unique. Proof. Consider a solution of the inverse problem and the function φg defined in (2.1.11) associated with this solution. Then the zeros and poles of φ1g coincide with the zeros and poles of ψg defined in (2.1.21). Furthermore, in view of (2.1.13), (1.2.25) and (1.2.28), g

1 φN,g (0) X 1 = = = ψg (0). φg (0) φD,g (0) j=1 l(j) Hence φg is uniquely determined by the given data. But (2.1.14) gives g

ψg (z) =

X 1 1 = − M z, φg (z) j=1 φ(j) (z)

(2.1.30)

and therefore ψ(z) z is uniquely determined by the given data. This completes the proof of statement 1. For the proof of statement 2 we observe that the functions φ(j) (j = 1, . . . , g) 1 defined in (2.1.12) are Nevanlinna functions, and hence also the functions − φ(j) are Nevanlinna functions. Therefore Lemma A.2.7 gives the partial fractions representation nj (j) X 1 aκ = + bj , φ(j) (z) κ=1 z − νκ(j) M = − lim

z→∞

74

Chapter 2. Vibrations of star graphs (j)

where aκ > 0 (j = 1, . . . , q, κ = 1, . . . , nj ) and bj ∈ R (j = 1, . . . , q). By assump(j) tion, the numbers νκ are uniquely allocated as terms of the sequence (νk )nk=1 , and since the νk are pairwise distinct, the coefficients in the partial fractions expansion (2.1.22) are uniquely determined by ψg . Comparing the partial fractions (j) expansions on the left and right hand sides of (2.1.30) shows that aκ = Ak when (j) 1 νκ = νk . Furthermore, ψ(j)1(0) = l(j) in view of (2.1.12), (1.2.24) and (1.2.26). Therefore φ(j) (j = 1, . . . , g) is uniquely determined by the given data, and an application of Lemma A.3.5 completes the proof. 

2.1.3 Restrictions on the multiplicities In this subsection, let g ≥ 2. It will also be assumed without loss of generality that the edges are labelled in such a way that n1 ≥ n2 ≥ · · · ≥ ng . We augment the notation which was used earlier in this section. Notation 2.1.10. We denote by 0

1. ΛN = (µk )nk=1 , µk ≥ µk0 > 0 for k > k 0 , the nondecreasing sequence of the characteristic values of the Neumann problem (N1), (2.1.4)–(2.1.7), on the star graph, 2. ΛD = (νk )nk=1 =

g S j=1

(j) n

j (νκ )κ=1 , νk ≥ νk0 > 0 for k > k 0 , the nondecreasing

sequence of the characteristic values of the Dirichlet problem (D1) on the star graph, where (j) n

(j)

(j)

j 3. (νκ )κ=1 , νκ > νκ0 for κ > κ0 , are the characteristic values of the DirichletDirichlet or Neumann-Dirichlet problem (2.1.1)–(2.1.3) on the j-th edge for j = 1, . . . , g, N ˜ N = (e 4. Λ µk )rk=1 , µ ek > µ ek0 for k > k 0 the sequence of distinct characteristic values of the Neumann problem (N1), D e D = (e 5. Λ νk )rk=1 , νek > νek0 for k > k 0 the set of distinct characteristic values of the Dirichlet problem (D1), N D 6. (pk (N ))rk=1 and (pk (D))rk=1 the vectors of multiplicities of the characteristic values µ ek (k = 1, . . . rN ) of problem (N1) and νek (k = 1, . . . , rN ) of problem (D1).

Remark 2.1.11. that

1. By definition of pk (N ) and pk (D) as multiplicities, it is clear rN X k=1

pk (N ) = n0 ,

rD X

pk (D) = n.

(2.1.31)

k=1

2. The number rD of different positive Dirichlet eigenvalues satisfies rD ≥ n1 (1) (1) (1) since ν1 < ν2 < · · · < νn1 .

75

2.1. Star graphs with root at the centre

To derive necessary conditions on the multiplicities of the characteristic values of the Neumann and Dirichlet problems, we need the notion of majorization, which goes back to Muirhead [99] for the case of integers and was generalized to nonnegative numbers by Hardy, Littlewood, and Poly´ a (see [68, 2.18]). s t be two vectors with nonnegative and y = (yi )i=1 Definition 2.1.12. Let x = (xi )i=1 entries ordered nonincreasingly, that is, x1 ≥ x2 ≥ · · · ≥ xs ≥ 0 and y1 ≥ y2 ≥ · · · ≥ yt ≥ 0. Then x is said to majorize y,

xy

:⇐⇒

s X

xi =

i=1

t X

yi ,

i=1

τ X

xi ≥

i=1

τ X

yi

(τ = 1, . . . , min{s − 1, t}).

i=1

(2.1.32) t Remark 2.1.13. If a vector x = (xi )si=1 majorizes a vector (yi )i=1 , then the number of nonzero entries of x is less or equal to the number of nonzero entries of y, i. e.,

x  y =⇒ #{i ∈ {1, . . . , s} : xi > 0} ≤ #{i ∈ {1, . . . , t} : yi > 0}. In fact, denote the numbers of nonzero entries of x and y by s0 and t0 , respectively, and assume s0 > t0 . By (2.1.32) this implies t0 X i=1

xi ≤

s X i=1

xi =

t X i=1

yi =

t0 X i=1

yi ≤

t0 X

xi ,

i=1

where the last inequality is true in view of t0 ≤ min{s − 1, t}. Hence we have equality everywhere. Since all xi are nonnegative, this shows that xi = 0 for i = t0 + 1, . . . , s, a contradiction to the assumption. The following elementary lemma on the inversion of the nonincreasing function {1, . . . , g} → N, j 7→ nj , will be useful. Lemma 2.1.14. Let g ∈ N, g ≥ 2, (nj )gj=1 ∈ Ng with n1 ≥ n2 ≥ · · · ≥ ng , and set Pg n = j=1 nj , ng+1 := 0. For i = 1, . . . , n1 define Ni := #{j ∈ {1, . . . , g} : nj ≥ i} = max{j ∈ {1, . . . , g} : nj ≥ i}. Then N1 ≥ N2 ≥ · · · ≥ Nn1 and Pn1 1. i=1 Ni = n; 2. Ni > 1 (i = 1, . . . , n2 ), Ni = 1 (i = n2 + 1, . . . , n1 ); 3. #{i ∈ {1, . . . , n1 } : Ni = j} = nj − nj+1 (j = 1, . . . , g); 4. nj = #{i ∈ {1, . . . , n1 } : Ni ≥ j} = max{i ∈ {1, . . . , n1 } : Ni ≥ j} (j = 1, . . . , g). Proof. Statements 1, 2 and 3 follow from the definition of Ni , and statement 4  follows from statement 3.

76

Chapter 2. Vibrations of star graphs • • • • • N1

• • • •

• • • •

• • • •

• • •

•

•

N2

N3

N4

N5

N6

N7

n1 n2 n3 n4 n5

Figure 2.2

Figure 2.2 illustrates Lemma 2.1.14, where the j-th row shows the beads of the j-th string, and where nj and Nk is the number of beads in the j-th row and k-th column, respectively. Lemma 2.1.15. Let x = (xi )si=1 and y = (yi )ti=1 be two vectors whose entries are nonnegative integers ordered nonincreasingly, x1 ≥ x2 ≥ · · · ≥ xs ≥ 0 and y1 ≥ y2 ≥ · · · ≥ yt ≥ 0. Assume that xs > 0, that x1 > 1 and that x  y. b Let sb = max{i ∈ {1, . . . , s} : xi > 1} and define x b = (b xi )si=1 by x bi = xi − 1 (i = 1, . . . , sb). Let t1 = min{i ∈ {1, . . . , t} : yi = ys },

t2 = max{i ∈ {1, . . . , t} : yi = ys },

and define yb = (b yi )ti=1 by ybi ybi ybi ybi

= yi − 1 for i = 1, . . . , t1 − 1, = yi for i = t1 , . . . , t1 + t2 − s − 1, = yi − 1 for i = t1 + t2 − s, . . . , t2 , = yi for i = t2 + 1, . . . , t.

Then x b and yb are nonincreasingly ordered vectors of nonnegative integers, x bsb > 0, and x b  yb. Proof. Since x1 > 1, sb is well-defined, and x bsb = xsb − 1 > 0. Clearly x bi = xi − 1 ≥ xi+1 − 1 = x bi+1 for i = 1, . . . , sb − 1. Hence x b is nonincreasingly ordered. By Remark 2.1.13 there is t0 ≥ s such that yt0 > 0. Then ys ≥ yt0 > 0. Therefore, for i = 1, . . . , t2 , yi ≥ yt2 = ys > 0. This shows that ybi ≥ 0 for i = 1, . . . , t. Next we find for i = 1, . . . , t1 − 2 and i = t1 + t2 − s, . . . , t2 − 1 that ybi = yi − 1 ≥ yi+1 − 1 = ybi+1 . For i = t1 , . . . , t1 + ts − s − 2 and i = t2 + 1, . . . , t − 1 we have ybi = yi ≥ yi+1 = ybi+1 . In case t1 > 1 and t2 > s we have yt1 −1 6= ys = yt1 and yt1 −1 ≥ yt1 , so that yt1 −1 > yt1 . It follows that ybt1 −1 = yt1 −1 − 1 ≥ yt1 = ybt1 . In case t1 > 1 and t2 = s we have ybt1 −1 = yt1 −1 − 1 ≥ yt1 − 1 = ybt1 . If t2 6= t, then yt2 +1 6= ys = yt2 , and therefore yt2 +1 ≤ yt2 gives yt2 +1 < yt2 . Hence ybt2 +1 = yt2 +1 ≤ yt2 −1 = ybt2 . Altogether we have shown that yb is a nonincreasingly ordered sequence of nonnegative integers.

77

2.1. Star graphs with root at the centre

We have ybi = yi − 1 for (t1 − 1) + (t2 − (t1 + t2 − s − 1)) = s terms and ybi = yi for the remaining terms. Therefore, observing that xi = 1 for i = sb + 1, . . . , s, we get t t b s s X X X X ybi , yi − s = x xi − s = bi = i=1

i=1

i=1

i=1

s − 1, t1 − 1}, and, similarly for τ = 1, . . . , min{b τ X

x bi =

τ X

xi − τ ≥

i=1

i=1

τ X

yi − τ =

τ X

ybi .

i=1

i=1

This completes the proof if sb ≤ t1 . If sb > t1 , assume that there is τ ∈ {t1 , . . . , sb − 1} such that τ X

bi < x

τ X

ybi .

i=1

i=1

Choosing the smallest such τ we have −1 τ X

x bi ≥

−1 τ X

ybi ,

i=1

i=1

where in case τ = t1 > 1 this holds by what we have already shown and where the case τ = 1 is trivial. Taking the difference of these two inequalities shows that x bτ < ybτ , and therefore, in view of t1 ≤ τ ≤ sb − 1 < s ≤ t2 , xτ ≤ yτ = ys . bi ≤ ybi . Hence For i = τ + 1, . . . , sb it follows that xi ≤ xτ ≤ ys = yi and therefore x s b X

x bi =

−1 τ X i=1

i=1

x bi +

s b X i=τ

x bi
1 and τ = 1, . . . n1 − 1. For the purpose of this proof we may consider g as a parameter, and we will prove the inequalities in (2.1.32) by induction on g, where g = 1 is admitted. For convenience, we will write N1 (g) and p↓i (D, g) to indicate the dependence on g. For g = 1 we have Ni (1) = 1 = p↓i (D, 1), (i = 1, . . . , n1 ), and the inequalities in (2.1.32) are indeed equalities for all τ = 1, . . . , n1 − 1. Now let g ≥ 2 and assume that the inequalities in (2.1.32) hold for g − 1. Clearly, ( Ni (g − 1) + 1 for i = 1, . . . , ng , Ni (g) = Ni (g − 1) for i = ng + 1, . . . , n1 , whereas for i = 1, . . . , n1 −1 the number p↓i (D, g) is p↓i (D, g −1) or pi↓ (D, g −1)+1, (g) where the latter case occurs for exactly ng of these indices since the numbers νκ (κ = 1, . . . , ng ) are distinct. Therefore τ X

Ni (g) =

i=1

and

τ X

p↓i (D, g) ≤

i=1

τ X

Ni (g − 1) + min{τ, ng }

i=1 τ X

p↓i (D, g − 1) + min{τ, ng },

i=1

and the inequalities in (2.1.32) for g follow from the induction hypothesis. Proposition 2.1.18.



1. Let

κN (j) := #{k ∈ {1, . . . , rN } : pk (N ) = j}

(j = 1, . . . , g − 1),

κD (j) := #{k ∈ {1, . . . , rD } : pk (D) = j}

(j = 1, . . . , g).

Then ( κN (1) =

κD (2) + rD + 1 κD (2) + rD

κN (j) = κD (j + 1)

if M > 0. if M = 0,

(j = 2, . . . , g − 1).

(2.1.34)

79

2.1. Star graphs with root at the centre

2. Let reD be the number of distinct multiple characteristic values of the Dirichlet problem (D1). Then ( rN =

rD + reD + 1 rD + reD

if M > 0, if M = 0,

(2.1.35)

and      ↓ p↓1 (N ) p↓1 (N ) + 1 p1 (N )    ..      ..         .        ↓ . ↓      p (N ) + 1  preD (N )         reD  (2.1.36)  =   = ↓     ↓    p preD +1 (N )      reD +1 (N )           .. .. .        .     .     ↓ prD (D)   p↓rD (N )   p↓rD (N )          1           .. r eD     .  ↓ 1 prD +erD (N )



p↓1 (D)







3. N1 , N2 , . . . , Nn1  p↓1 (N ), p↓2 (N ), . . . , p↓rD +erD (N ) . 4. pj (D) ≤ g (j = 1, . . . , rD ),

pj (N ) ≤ g − 1 (j = 1, . . . , rN ).

Proof. 1. By Theorem 2.1.4 we know for each z ∈ C and j ≥ 2 that nN,g (z) = j if and only if nD,g (z) = j + 1, which proves κN (j) = κD (j + 1) for j = 2, . . . , g − 1. Since (N1) has n0 characteristic values and (D1) has n characteristic values, it follows that g−1 X

jκN (j) = n

0

and

j=1

g X

jκD (j) = n.

(2.1.37)

j=1

Therefore 0

κN (1) = n −

g−1 X

0

jκN (j) = n −

j=2

g−1 X

0

jκD (j + 1) = n −

j=2

= n0 − n + κD (1) + 2κD (2) +

j=3 g X j=3

= n0 − n + κD (2) + rD .

g X

κD (j)

(j − 1)κD (j)

80

Chapter 2. Vibrations of star graphs

2. Using the above relations for κN (j) we find that rN =

g−1 X

0

κN (j) = κD (2) + rD + n − n +

g−1 X

κD (j + 1)

j=2

j=1

= rD + n0 − n +

g X

κD (j) = rD + reD + n0 − n.

j=2

To prove (2.1.36), we first conclude from the definition of reD that pj↓ (D) > 1 if and only if j ≤ reD . Furthermore, by Theorem 2.1.4, p↓j (N ) = 1 for j = reD + 1, . . . , rN . But then it is clear from Theorem 2.1.4 that p↓j (D) > 1,

p↓j (N ) = pj↓ (D) − 1

p↓j (D) p↓j (N )

p↓j (N )

= 1, =1

=

pj↓ (D)

(j = 1, . . . , reD ),

(2.1.38)

(j = reD + 1, . . . , rD ),

(2.1.39)

(j = rD + 1, . . . , rN ).

(2.1.40)

This implies the first and the last equality in (2.1.36) if we note (2.1.35). The majorization property in (2.1.36) is obvious since the number of components where 1 is added on the left hand side is equal to reD and coincides with the number of new components 1 added on the right hand side. Statement 3 immediately follows from Theorem 2.1.17 and from (2.1.36),  whereas statement 4 follows from Theorem 2.1.5. Theorem 2.1.19. Let rN be the number of distinct characteristic values of the N Neumann problem (N1), let (pi (N ))ri=1 the vector of their multiplicities, and let ↓ r ↓ N p (N ) = (pi (N ))i=1 be the corresponding vector of multiplicities in nonincreasing order, see Notation 2.1.10. Then: 1. p1 (N ) = 1; if pi (N ) > 1, then pi−1 (N ) = pi+1 (N ) = 1 (i = 2, . . . , rN − 1); if M > 0, then prN (N ) = 1; if M = 0 and prN (N ) > 1, then prN −1 (N ) = 1; 2. rN ≥ n1 + n2 + 1 if M > 0 and rN ≥ n1 + n2 if M = 0; 0 3. (N1 − 1, N2 − 1, . . . , Nn1 − 1)  (p↓1 (N ), p↓2 (N ), . . . , p↓r0 −n1 (N )) where rN = N 0 rN − 1 if M > 0 and rN = rN if M = 0.

Proof. The statements in part 1 follow from Theorem 2.1.5, part 1, and from Corollary 2.1.6. For statement 2 we observe that, starting counting at the first edge, each characteristic value of the Dirichlet problem on the first edge increases rD by 1, whereas each characteristic value of the Dirichlet problem on the second edge then increases either rD or reD by 1. This shows that rD + reD = n1 + n2 if g = 2, and therefore rD + reD ≥ n1 + n2 for any g ≥ 2. Taking (2.1.35) into account completes the proof of statement 2.

81

2.1. Star graphs with root at the centre

To prove statement 3 we first recall that rD ≥ n1 since the characteristic 0 values for (D1) are distinct on each edge. We further note that rN = rD + reD by (2.1.35). Therefore, in view of Proposition 2.1.18, part 3, (2.1.39) and (2.1.40), rD +e rD rDX +e rD −n1 n1 n1 X X X (Ni − 1) = Ni − n 1 = pi↓ (N ) − n1 = p↓i (N ), i=1

(2.1.41)

i=1

i=1

i=1

rD , n1 − 1}, and, in view of Theorem 2.1.17 and (2.1.38), for τ = 1, . . . , min{e τ τ τ τ X X X X Ni − τ ≥ pi↓ (D) − τ = p↓i (N ). (Ni − 1) = i=1

i=1

(2.1.42)

i=1

i=1

This completes the proof in case reD ≥ n1 − 1. If reD < n1 − 1, assume there is rD + 1, . . . , min{n1 − 1, rD + reD − n1 }} such that τ ∈ {e τ τ X X p↓i (N ). (Ni − 1) < i=1

i=1

Letting τ be the smallest number with this property, we have τ −1 X

τ −1 X

i=1

i=1

(Ni − 1) ≥

pi↓ (N ),

which is trivially true if τ = 1. Taking the difference gives Nτ − 1 < pτ↓ (N ) = 1 in view of (2.1.39) and (2.1.40). That is, Nτ < 2 and therefore Nτ = 1. It follows that Ni = 1 for i = τ, . . . , n1 . But this would lead to rD −n1 rD +e n1 τ τ X X X X (Ni − 1) < pi↓ (N ) ≤ (Ni − 1) = p↓i (N ), i=1

i=1

which contradicts (2.1.41).

i=1

i=1



Remark 2.1.20. 1. If M > 0, then the number of simple characteristic values of the Neumann problem (N1) is at least max{n1 , [ r2N ]} + 1. 2. If M = 0, then the number of simple characteristic values of the Neumann problem (N1) is at least max{n1 , [ rN2+1 ]}. Proof. Recall from Proposition 2.1.18 that the number of simple characteristic values of (N1) is denoted by κN (1). Since rD ≥ n1 , Proposition 2.1.18 gives κN (1) ≥ n1 + 1 if M > 0 and κN (1) ≥ n1 if M = 0. Recall from Theorem 2.1.5 that µ1 and, in case M > 0, µn+1 are simple. 0 Hence it follows that #{µk : k = 2, . . . , n} = rN − 1. By Corollary 2.1.6, multiple characteristic values µk (k = 2, . . . , n) are separated by simple characteristic r 0 −1 values. Therefore at least [ N2 ] of these characteristic values are simple, which gives κN (1) ≥ r2N + 1 if M > 0 and κN (1) ≥ rN2+1 if M > 0. 

82

Chapter 2. Vibrations of star graphs

2.1.4

The inverse problem with given numbers of beads on the edges

In this subsection we consider the inverse problem under the additional assumption that the number of beads on each edge is prescribed. Furthermore, the problems (N1) and (D1) will always to be considered with cj = 0 for all j, i. e., all pendant vertices are assumed to be clamped. Theorem 2.1.21. Let g ∈ N, g ≥ 2, (l(j) )gj=1 ∈ (0, ∞)g , and n ∈ N. Let n0 = n + 1 0 n of positive real or n0 = n. Suppose that sequences ΛN = (µk )nk=1 , ΛD = (νk )k=1 g g numbers are given and let (nj )j=1 ∈ N , n1 ≥ n2 ≥ · · · ≥ ng , be given with Pg j=1 nj = n. Define Ni := #{j ∈ {1, . . . , g} : nj ≥ i} for i = 1, . . . , n1 , let rD be D be the vector of noninthe number of distinct entries in ΛD and let (pk↓ (D))rk=1 creasingly ordered multiplicities of the distinct entries in ΛD . Then the conditions

(i) 0 < µ1 < ν1 ≤ · · · ≤ µn ≤ νn and νn < µn+1 if n0 = n + 1, (ii) νk−1 = µk if and only if µk = νk (k = 2, . . . , n),

(iii) N1 , N2 , . . . , Nn1  p↓1 (D), p↓2 (D), . . . , p↓rD (D) , are necessary and sufficient for the existence of a sequence of (positive) masses (j) nj (j) nj (mk )k=1 (j = 1, . . . , q) with , a mass M ≥ 0, and sequences of lengths (lk )k=0 Pnj (j) (j) = l such that the sequences of characteristic values of the spectral k=0 lk problems (N1) and (D1) on the corresponding star graph are ΛN and ΛD . Then M is uniquely determined by the given data, and M > 0 if and only if n0 = n + 1. Proof. The necessity of the conditions (i) and (ii) follows from Theorem 2.1.5, and the necessity of condition (iii) follows from Theorem 2.1.17. For the sufficiency we observe that (iii) implies g = N1 ≥ p1↓ (D). Therefore condition (iii) of Theorem 2.1.8 is satisfied. The sufficiency of the conditions (i)–(iii) then follows from Theorem 2.1.8 and its proof if we show that (2.1.20) can be satisfied, i. e., that the (j) nj n sequence (νk )k=1 can be partitioned into g subsequences (νk )k=1 (j = 1, . . . , g) such that g [ (j) nj n (2.1.43) ΛD = (νk )k=1 = (νk )k=1 j=1 (j)

(j)

and νk 6= νk0 for k 6= k 0 .

The vectors x = N1 , N2 , . . . , Nn1 and y = p1↓ (D), p↓2 (D), . . . , pr↓D (D) satisfy the assumption of Lemma 2.1.15 in view of condition (iii). Then it follows from Lemma 2.1.15 that numbers pb↓j (D) (j = 1, . . . , r(D)) can be chosen such that pb↓j (D) = pj↓ (D) − 1 ≥ 0 for exactly n1 indices j and that pb↓j (D) = pj↓ (D) for the remaining indices and such that

N1 − 1, N2 − 1, . . . , Nn2 − 1  pb1↓ (D), pb2↓ (D), . . . , pb↓rD (D) .

83

2.1. Star graphs with root at the centre

e D for which pb↓ (D) = p↓ (D) − 1 and arranging Choosing the n1 numbers νej ∈ Λ j j (1)

n1 . them in increasing order, we have constructed the sequence (νk )k=1 The remaining numbers νk ∈ ΛD with νk = νej have the multiplicities pbj↓ (D), where pbj↓ (D) = 0 if a number with multiplicity 1 has been taken. The numbers g , and successive appliNj − 1 (j = 1, . . . , n2 ) correspond to the sequence (nj )j=2 (j) n

j cation of the above algorithm gives the sequences (νk )k=1 for j = 2, . . . , g − 1 of e elements from ΛD . In the last step we have Ni = 1 for i = 1, . . . , ng and a vector
pe1↓ (D), pe2↓ (D), . . . , per↓D (D) of the multiplicities of the remaining elements from ΛD . Because the recursion preserves majorization, we have
(1, . . . , 1)  pe↓1 (D), pe↓2 (D), . . . , per↓D (D) ,

where the vector on the left has ng components. Then pe↓ng (D) > 0 in view of Remark 2.1.13. Since the entries of the vector on the right-hand side are sorted in nonincreasing order and sum up to ng , it follows that pe↓k (D) = 1 for k = 1, . . . , ng and pek↓ (D) = 0 for k = ng +1, . . . , rD . That means, exactly the numbers νe1 , . . . , νeng ng are left from ΛD , and sorting them into an increasing sequence (νkg )k=1 gives (j) (j) (2.1.43) with νk 6= νk0 for k 6= k 0 and j = 1, . . . , g.  In the next theorem we derive sufficient conditions for a sequence of numbers to be the characteristic values of the Neumann problem on a star graph. Theorem 2.1.22. Let g ∈ N, g ≥ 2, (l(j) )gj=1 ∈ (0, ∞)g , n ∈ N, and n0 = n + 1 n0 or n0 = n. Suppose that a sequence ΛN := (µk )k=1 of positive real Pgnumbers is g given, and let (nj )j=1 ⊂ N, n1 ≥ n2 ≥ · · · ≥ ng , be given with j=1 nj = n. Define Ni := #{j ∈ {1, . . . , g} : nj ≥ i} for i = 1, . . . , n1 . Denote by rN the 0 N ek in the sequence (µk )nk=1 , by (pi (N ))ri=1 number of different elements µ the vector ↓ rN ↓ of their multiplicities, and by p (N ) = (pi (N ))i=1 the corresponding vector of multiplicities in nonincreasing order. Then the conditions (i) 0 < µ1 < µ2 ≤ µ3 ≤ . . . ≤ µn−2 ≤ µn−1 ≤ µn ; µn < µn+1 if n0 = n + 1; (ii) if pi (N ) > 1 then pi−1 (N ) = pi+1 (N ) = 1 for i = 2, . . . , rN − 1, and if n0 = n and prN (N ) > 1, then prN −1 (N ) = 1; 0 = (iii) (N1 − 1, N2 − 1, . . . , Nn1 − 1)  (p1↓ (N ), p↓2 (N ), . . . , p↓r0 −n1 (N )) where rN N 0 0 0 rN − 1 if n = n + 1 and rN = rN if n = n;

are necessary and sufficient such that there exist sequences of (positive) masses Pnj (j) (j) nj (j) nj (j = 1, . . . , q) with k=0 lk = l(j) such that the (mk )k=1 and lengths (lk )k=0 spectral problem (N1) on the corresponding star graph with central mass M ≥ 0 has the sequence ΛN as characteristic values. Then M = 0 if and only if n0 = n. Proof. The necessity of the conditions has been shown in Theorems 2.1.5 and 2.1.19.

84

Chapter 2. Vibrations of star graphs

To prove sufficiency of the conditions (i)–(iii) we will first show that the n conditions (i)–(iii) ensure the existence of a sequence ΛD := (νk )k=1 such that the two sequences ΛN and ΛD satisfy the assumptions of Theorem 2.1.21. To this end we set µn+1 := ∞ in the case n0 = n for convenience. We will call a term of a sequence a simple value of that sequence if it occurs exactly once in that sequence and we will call it a multiple value of that sequence if it occurs more than once in that sequence. Let k ∈ {1, . . . , n} and consider µk and µk+1 . If µk < µk+1 and both µk and µk+1 are simple values of ΛN , then we choose νk ∈ (µk , µk+1 ). If µk < µk+1 and at least one of µk and µk+1 is not simple, then exactly one is a multiple value of ΛN in view of condition (ii). In this case, if µk is a multiple value, we set νk = µk , and if µk+1 is a multiple value, we set νk = µk+1 . Finally, if µk = µk+1 , we set νk = µk . Now we are going to verify that the sequences ΛD and ΛN satisfy the assumptions of Theorem 2.1.21. Assumption (i) holds by construction and since µ1 and µn+1 are simple values of ΛN . To verify condition (ii) in Theorem 2.1.21 let k ∈ {2, . . . , n}. If νk−1 = µk , then µk is a multiple value of ΛN , and therefore νk = µk by definition of νk . And if µk = νk , then µk is a multiple value of ΛN , and therefore νk−1 = µ(k−1)+1 = µk by definition of νk−1 . D . To verify condition (iii) in Theorem 2.1.21 we first find the vector (p↓k (D))rk=1 We note that by construction of νk (k = 1, . . . , n) we have νk ∈ ΛN if and only if νk is a multiple value of ΛN . In this case there are unique numbers j1 , j2 ∈ {2, . . . , n} such that j1 6= j2 and µj1 −1 < µj1 = νk = µj2 < µj2 +1 . By construction of νj it follows that νj = νk = µj1 if and only if j = j1 − 1, . . . , j2 . Thus we have shown that the number reD of multiple values of ΛD is greater or equal than the number of multiple values of ΛN and that (p↓1 (N ) + 1, . . . , pr↓eD (N ) + 1) is a vector of the multiplicities of the distinct multiple values of ΛD . Since this vector is in nonincreasing order, it follows that (p1↓ (D), . . . , p↓reD (D)) = (p1↓ (N ) + 1, . . . , p↓reD (N ) + 1). 0 Note that rN denotes the number of distinct terms in the sequence (µk )nk=1 . Hence r eD   X 0 rN =n− p↓i (N ) − 1 . i=1

Similarly, the number of distinct terms in ΛD is rD = n −

eD  r X

 pi↓ (D) − 1 .

i=1

It therefore follows that 0 rN = rD +

r eD  X i=1

 pi↓ (D) − p↓i (N ) = rD + reD .

(2.1.44)

85

2.1. Star graphs with root at the centre We have that the number of terms in the sequence (µk )nk=1 is 0

rN X

p↓i (N ) = n,

i=1

and condition (iii) and Lemma 2.1.14 give 0 −n1 rN

X

pi↓ (N ) =

i=1

n1 n1 X X Ni − n1 = n − n1 . (Ni − 1) = i=1

i=1

Taking the difference shows that 0

rN X

pi↓ (N ) = n1 .

0 −n +1 i=rN 1

0 0 0 But since p↓i (N ) ≥ 1 for i = 1, . . . , rN − n1 + 1, . . . , rN , it follows for i = rN that ↓ 0 pi (N ) = 1, which gives reD ≤ rN − n1 . With the aid of (2.1.44) we conclude that 0 rD = rN − reD ≥ n1 .

Now we are able to prove condition (iii) in Theorem 2.1.21. The equality in the majorization condition holds since n1 X

Ni = n =

rD X

p↓i (D).

(2.1.45)

i=1

i=1

For τ = 1, . . . , n2 we calculate τ X i=1

Ni =

τ τ τ τ   X X X X (Ni − 1) + τ ≥ p↓i (N ) + τ = pi↓ (D), pi↓ (N ) + 1 ≥ i=1

i=1

i=1

i=1

and for τ = n2 + 1, . . . , n1 − 1 it is clear in view of (2.1.45), rD ≥ n1 and Ni = 1 for i = n2 + 1, . . . , n1 that τ X

Ni =

i=1

=

rD X i=1 τ X i=1

p↓i (D) − p↓i (D) +

n1 X

Ni ≥

i=τ +1 n1 X

n1 X

p↓i (D) −

i=τ +1

Ni

i=τ +1

i=1

(p↓i (D) − 1) ≥

n1 X

τ X

pi↓ (D).

i=1

The statement of this theorem follows now from Theorem 2.1.21.



86

Chapter 2. Vibrations of star graphs

2.1.5

Comparison with results for tree-patterned matrices

Interlacing conditions of finite sequences of real numbers also play a rˆ ole in the theory of symmetric matrices. In Section C.6, interlacing results of eigenvalues of tree-patterned matrices have been proved. In this subsection we show how the above results on star graphs of Stieltjes strings can be used to prove the existence of a star-patterned symmetric matrix and submatrix with prescribed interlacing spectra, thus giving an alternative proof of one direction of Theorem C.6.2 for star-patterned matrices. To relate the results about characteristic values of star graphs of Stieltjes strings to star-patterned matrices, we reformulate the direct Neumann and Dirichlet problems with cj = 0 for all j considered in Section 2.1.1 in the case M > 0 as matrix eigenvalue problems. To this end, let L be the (n + 1) × (n + 1) matrix defined by 

g X 1

  j=1 ln(j) j   0   .  ..   0   1   − (1)  l n1   0  .  .  .   0   1 L :=   − l(2) n2              0   ..  .   0  1  − (g) l ng

0 ··· 0 −

1 (1)

0 ··· 0 −

ln1

1

0 ··· 0 −

(2)

ln2

L1

0

0

0

0

L2

0

0

0

0

0

0

0

0

Lq

1



(g) lng                       ,                      

where the matrices Lj occurring in the block diagonal are nj × nj matrices defined

87

2.1. Star graphs with root at the centre by 

1 (j) l0

+

1 (j) l1

1   − l(j) 1    0                 0

− 1 (j) l1

1 (j) l1

+

−

0

1 (j) l2

1 (j) l2

− 1 (j) l2

1 (j) l2

+

0

1 (j) l3

−

0



0

                    

1 (j) l3

0 −

1 (j) j −2

ln

1 (j) j −2

ln

−

0

+

1

−

(j) j −1

ln

1

1 (j) j −1

ln

1

(j) j −1

ln

(j) j −1

ln

+

1 (j) lnj

for j = 1, . . . g. Furthermore, let M be the (n + 1) × (n + 1) diagonal matrix with diagonal elements (1)

(2)

(g)

(2) (g) M, m1 , . . . , m(1) n1 , m1 , . . . , mn2 , . . . , m1 , . . . , mng .

Then the Neumann problem (N1) in (2.1.4)–(2.1.7) with z = λ2 is the eigenvalue problem for the matrix 1 1 Le := M− 2 LM− 2 ,

while the Dirichlet problem (D1) given by (2.1.1)–(2.1.3) is the eigenvalue problem for the principal submatrix Le1,1 . The matrix Le is T -acyclic, where the tree T is derived from the star graph under consideration with the beads and vertices of the star graph being identified as the vertices of T and with the threads of the star graph being identified as the edges of T . Theorem 2.1.8 means that, under the assumptions therein with M > 0, there exists a real Hermitian star-patterned (n + 1) × (n + 1) matrix such that its sequence of eigenvalues coincides with the sequence (µk )n+1 k=1 and the sequence of the eigenvalues of the submatrix obtained by deleting the first row and the first column coincides with the sequence (νk )nk=1 . Thus Theorem 2.1.8 with M > 0 provides sufficient conditions for two n sequences (µk )n+1 k=1 and (νk )k=1 to be the eigenvalues of a real Hermitian starpatterned matrix and its first principal submatrix, respectively. Observe that stronger assumptions are made for star-shaped string problems, where the total lengths of the strings are given, which has no meaning in graph theory since no length is associated to the edges. Furthermore, the eigenvalues of the Dirichlet problems on the strings are simple, whereas in the notation of Theorem C.6.2, the eigenvalues of the matrices AΓk (j) do not need to be simple.

88

Chapter 2. Vibrations of star graphs

2.2 Symmetric Stieltjes strings In this section we consider Stieltjes strings with clamped ends and with at least one bead which are symmetric with respect to their midpoints. With the notation from Section 1.2 this means that mk = mn−k+1 lk = ln−k

(k = 1, . . . , n0 ),

(k = 0, . . . , n0 ),

  where n0 = n2 is the integer part of n2 . For symmetric Stieltjes strings, the Dirichlet-Neumann problem is identical to the Neumann-Dirichlet problem, and therefore the corresponding characteristic functions are multiples of each other. More precisely, (1.2.34) gives Lemma 2.2.1. If the Stieltjes string is symmetric, then R2n−1 (·, 0) =

1 R2n (·, 1). l0

(2.2.1)

Using (2.2.1) we see that the Lagrange identity (1.2.47) for a symmetric Stieltjes string attains the form 1 2 1 R2n (z, 1) − = R2n (z, 0)R2n−1 (z, 1). l0 l0

(2.2.2)

Splitting the symmetric Stieltjes string S at its midpoint P into two substrings S1 and S2 , we arrive at a special case of a star graph with two edges and root at the central vertex. Each of the two strings has n0 beads in its interior, and if n is odd, there is a bead of mass M = mn0 +1 at the midpoint P . We put M = 0 when n is even. The Dirichlet-Dirichlet problem for the string S can therefore be considered as a special case of the Neumann problem (N1) considered in Section 2.1. Since the strings S1 and S2 are identical, they have the same (1) Cauer-Fry polynomials, which will be denoted by Rk (·, 0) (k = 0, . . . , 2n0 ). We (1) note that Rk (·, 0) = Rk (·, 0) for k = 0, . . . , 2n0 − 2, but that they may differ for k = 2n0 − 1 and k = 2n0 . However, it is important to note that the lengths of the first subinterval is l0 for S, S1 and S2 . Lemma 2.2.2. For a symmetric string, 1 (1) (1) (1) R2n (z, 0) = [2R2n0 −1 (z, 0) − M zR2n0 (z, 0)]R2n0 (z, 0). l0

(2.2.3)

Proof. By Theorem 2.1.4, the function on the right hand side of (2.2.3) is a characteristic function for the Dirichlet-Dirichlet problem on S, and therefore the functions on the left hand side and on the right hand side have the same (simple) zeros. Comparing the coefficients at z = 0 completes the proof. 

89

2.2. Symmetric Stieltjes strings

Theorem 2.2.3. Let n ∈ N. Then each increasing sequence of positive numbers n (νk )k=1 is the sequence of the characteristic values of the Dirichlet-Dirichlet problem (1.2.4), (1.2.23), (1.2.24) for a symmetric Stieltjes string with n beads and of n a prescribed length l > 0, and the sequences of the masses (mk )k=1 and the lengths (lk )nk=0 are uniquely determined by (νk )nk=1 and l. Proof. In the case of even n, Theorem 1.3.1 shows that there is a unique string S1 (1) 0 (1) 0 of length 2l with sequences of masses (mk )nk=0 and lengths (lk )nk=0 such that the (1) sequences of the zeros of the corresponding Cauer-Fry polynomial R2n0 −1 (·, 0) and (1)

(1)

0 0 R2n0 (·, 0) are (ν2k−1 )nk=1 and (ν2k )nk=1 , respectively. Putting mk = mn+1−k = mk

(1)

(1)

for k = 1, . . . , n0 , lk = ln−k = lk for k = 0, . . . , n0 − 1 and ln0 = 2ln0 we arrive at a symmetric string S, with S1 being one of the two identical copies obtained by splitting S at the midpoint. Since 1 (1) (1) R2n (·, 0) = 2R2n0 −1 (·, 0)R2n0 (·, 0) l0

(2.2.4)

by Lemma 2.2.2, the sequence (νk )nk=1 is the sequence of characteristic values of the string S. Due to (2.2.4) and the interlacing property (1.2.36), the zeros of (1) (1) R2n0 −1 (·, 0) and R2n0 (·, 0) are uniquely determined by the given sequence and since l is given, also the length 2l of S1 is given. Hence the uniqueness statement in Theorem 1.3.1 shows that S1 , and therefore also the symmetric string S, is uniquely determined by the given data. In the case of odd n, we define f1 (z) :=

n0 Y

(ν2k − z),

f2 (z) :=

k=1

nY 0 +1

(ν2k−1 − z),

h(z) := f2 (z) + zf1 (z).

k=1

Since the leading term of f1 is (−z)n0 and the leading term of f2 is (−z)n0 +1 , it follows that the degree of the polynomial h is at most n0 . Furthermore, the fact that (νk )nk=1 is an increasing sequence of positive numbers shows for k = 1, . . . , n0 that (−1)k h(ν2k−1 ) = (−1)k ν2k−1 f1 (ν2k−1 ) < 0 and (−1)k h(ν2k ) = (−1)k f2 (ν2k ) > 0, the function h has a zero τk ∈ (ν2k−1 , ν2k ) (k = 1, . . . , n0 ). Hence h has degree n0 , and its simple zeros τk (k = 1, . . . , n0 ) satisfy 0 < τ1 < ν2 < τ2 < ν4 < · · · < τn0 < ν2n0 . By Theorem 1.3.1 there is a unique string S1 of length 2l with sequences of (1) 0 (1) 0 masses (mk )nk=0 and lengths (lk )nk=0 such that the sequences of the zeros of (1) (1) 0 the corresponding Cauer-Fry polynomials R2n0 −1 (·, 0) and R2n0 (·, 0) are (τk )nk=1 (1)

0 and (ν2k )nk=1 , respectively. Putting mk = mn+1−k = mk (1) lk = ln−k = lk for k = 0, . . . , n0 and

mn0 +1 =

n0 nY 0 +1 1 2Y ν2k , l ν2k−1 k=1

k=1

for k = 1, . . . , n0 ,

(2.2.5)

90

Chapter 2. Vibrations of star graphs

we arrive at a symmetric string S, with S1 being one of the two identical copies obtained by splitting S at the midpoint. It is clear that (1)

R2n0 (z, 0) =

l f1 (z) , 2l0 f1 (0)

(1)

R2n0 −1 (z, 0) =

1 h(z) . l0 h(0)

It follows that (1)

(1)

1 h(z) l f1 (z) − mn0 +1 z l0 h(0) 2l0 f1 (0)   1 h(0) h(z) − mn0 +1 l = zf1 (z) l0 h(0) 2f1 (0) f2 (z) = . l0 h(0)

2R2n0 −1 (z, 0) − mn0 +1 zR2n0 (z, 0) =

In view of Lemma 2.2.2 it follows that the generated string problem has the given sequence (νk )nk=1 as the sequence of its characteristic values, and the string has length l. (1) For the uniqueness, we observe that the zeros of R2n0 (·, 0) are necessarily (1)

the numbers ν2k (k = 1, . . . , n) since the function

(R2n )2 (·,0) 0 R2n (·,0)

is an S0 -function by

Lemma 2.2.2 and by Theorem 2.1.2. Furthermore, R2n (0, 0) = l 2l0 show that R2n (z, 0) (1)

l0 (R2n0 )2 (z, 0)

l l0

(1)

and R2n0 (0, 0) =

(1)

=2

R2n0 −1 (z, 0) (1)

R2n0 (z, 0)

− mn0 +1 z (1)

is uniquely determined by the given data. Thus mn0 +1 and the zeros of R2n0 −1 (·, 0) are unique. Together with the uniqueness of the length of S1 another application of Theorem 1.3.1 completes the proof.  Remark 2.2.4. In view of (2.2.5), the mass of the central bead for odd n is explicitly given by the characteristic values and the length of the string. Conversely, given the length of the string and the mass of the central bead, (2.2.5) gives an upper bound for ν1 and a lower bound for νn : ν1 ≤ lmn2 +1 ≤ νn , where equality occurs 0 if and only if n = 1.

2.3

Star graphs with root at a pendant vertex

In this section we consider a plane star graph of g (≥ 2) Stieltjes strings joined at the central vertex where a bead of mass M > 0 may be placed. One pendant vertex is chosen to be the root of the star graph. If there is no bead at the central vertex, then we put M = 0. We suppose that the star graph is stretched and study its small transverse vibrations with continuity conditions and balance of forces at the central vertex in two different cases:

2.3. Star graphs with root at a pendant vertex

91

(D2) the root is fixed (Dirichlet problem), (N2) the root is free to move in the direction orthogonal to the equilibrium position of the strings (Neumann problem). We investigate the characteristic values of both problems and their relations to each other in order to be able to establish necessary and sufficient conditions for the solution of the corresponding inverse problem. Observe that both (D2) and (N2) are special cases of the Neumann problem (N1). However, to compare the characteristic values of (D2) and (N2), the approach differs from that taken in Section 2.1. We are going to use the notation for the star graph which was introduced in Section 2.1. In the sequel, the string incident to the root is called the main edge of the string. The main edge will be labelled with j = g, and therefore the other edges of the star graph will be labelled with j = 1, . . . , g − 1. (2)

l0 (2) m1

(2)

l1 (1)

(j)

(1)

m1 m2 (1)

(1)

l0

Neumann condition

l1

mnj

M (1)

l2

(j)

lnj

(q)

(q) m1

... ...

(j)

(j)

m2 m1 (j)

l1

(j)

l0

l(j)

l1

main edge

(q)

l0

root

Figure 2.3: Star graph with root at a pendant vertex Under the assumption that the threads are stretched by forces each equal to 1, we have derived in Section 2.1 the following difference equations for the (j) amplitudes uk : (j)

(j)

uk − uk−1 (j) lk−1

(j)

−

(j)

uk+1 − uk (j) lk

(j)

(j)

− mk λ2 uk = 0

(k = 1, . . . , nj , j = 1, . . . , g), (2.3.1)

and the following conditions at the vertices which are not the root: (1)

(2)

(g)

un1 +1 = un2 +1 = . . . = ung +1 ,

(2.3.2)

92

Chapter 2. Vibrations of star graphs (j) (j) g X unj +1 − unj j=1

(j) lnj

(j)

(j)

u0 = cj u1

(1)

− M λ2 un1 +1 = 0,

(2.3.3)

(j = 1, . . . , g − 1).

(2.3.4)

For the main edge, the boundary condition at the root is: Dirichlet problem (D2). If we clamp the root, we have to consider (2.3.1)–(2.3.4) together with (g)

u0 = 0.

(2.3.5)

Neumann problem (N2). If the root is allowed to move freely, we have to consider (2.3.1)–(2.3.4) together with (g)

(g)

u0 = u1 .

(2.3.6)

As in Section 2.1 we will denote by n :=

g X

nj

j=1

the number of beads on the star graph without the possible bead at the central vertex. As in Chapter 1 and Section 2.1 we will replace λ2 with z and we will call a complex number z a characteristic value of (D2) or (N2) when the corresponding problem has a nontrivial solution for that value of z. The number of linearly independent solutions, that is, the dimension of the vector space of solutions, will be called the geometric multiplicity of the characteristic value. The dimension of the corresponding solution space will be denoted by n eD,g (z) and n eN,g (z), respectively.

2.3.1 The direct spectral problem In this subsection we investigate the relations of the characteristic values of the Dirichlet problem (D2) with those of the Neumann problem (N2). Defining (g)

φg (z, c) := R2ng (z, c)

g−1 h X

(j)

R2nj −1 (z)

j=1 (g)

+ R2ng −1 (z, c)

g−1 Y

i (k) R2nk (z)

k=1, k6=j g−1 Y

(k)

(g)

R2nk (z) − M zR2ng (z, c)

k=1

g−1 Y

(k)

R2nk (z)

(2.3.7)

k=1 (g)

(g)

it is clear that Theorem 2.1.4 and (2.1.10) with Rk (z) replaced by Rk (z, c) for k = 2ng , k = 2ng − 1, c = 0, c = 1 lead to the following result.

93

2.3. Star graphs with root at a pendant vertex

Theorem 2.3.1. 1. The characteristic values of the Dirichlet problem (D2), counted with multiplicity, are the zeros of the polynomial φg (·, 0). 2. The characteristic values of the Neumann problem (N2), counted with multiplicity, are the zeros of the polynomial φg (·, 1). Proposition 2.3.2. Let φD,g−1 , φN,g−1 be defined as in (2.1.9), (2.1.10) for the subgraph of the g − 1 edges that are not the main edge. Then (g)

(g)

φg (·, c) = R2ng (·, c)φN,g−1 + R2ng −1 (·, c)φD,g−1 (g)

φD,g−1 = l0 φN,g−1 =

(g) l0

(c = 0, 1),
(g) (g) R2ng (·, 1)φg (·, 0) − R2ng (·, 0)φg (·, 1) ,
(g) (g) −R2ng −1 (·, 1)φg (·, 0) + R2ng −1 (·, 0)φg (·, 1) .

(2.3.8) (2.3.9) (2.3.10)

Proof. The identities (2.3.8) for c = 0 and c = 1 follow immediately from the definitions of φg (·, c), φD,g−1 and φN,g−1 . Writing (2.3.8) in matrix form, we have !    (g) (g) R2ng −1 (·, 0) R2ng (·, 0) φg (·, 0) φD,g−1 (2.3.11) . = (g) (g) φg (·, 1) φN,g−1 R (·, 1) R (·, 1) 2ng −1

2ng

(g)

(g)

Then 

φD,g−1 φN,g−1



R2ng −1 (·, 0) R2ng (·, 0)

!−1 

 φg (·, 0) (g) (g) φg (·, 1) R2ng −1 (·, 1) R2ng (·, 1) !  (g) (g) R2ng (·, 1) −R2ng (·, 0) φg (·, 0) (g) = l0 (g) (g) φg (·, 1) (·, 1) R −R (·, 0)

=

2ng −1

2ng −1

due to the Lagrange identity (1.2.47). This proves (2.3.9) and (2.3.10).



Theorem 2.3.3. The function φg (·, 0) φg (·, 1) is an S-function and an S0 -function if at least one of the pendant vertices which is distinct from the root is clamped, and can be represented as a continued fraction (g) φg (z, 0)

l0

φg (z, 1)

1

(g)

= l0 + (g) −m1 z

,

1

+

1

(g)

l1 +

(g) −m2 z

1

+ ··· + (g) −mng z

+

1 (g) lng

+ φg−1 (z) (2.3.12)

where φg−1 is defined as in (2.1.11) for the subgraph of the g − 1 edges that are not the main edge.

94

Chapter 2. Vibrations of star graphs

Proof. Let (g)

! (g) R2k (·, 0)

(g)

R2k (·, 1)

R2k−1 (·, 0)

Ψk =

(g)

R2k−1 (·, 1)

(k = 0, . . . , ng ).

In view of Lemma 1.2.4 we have (k = 1, . . . , ng ),

Ψk = Ψk−1 Θk where Θk (z) =

(g)

−mk z

!

(g)

1

lk

(g)

(k = 1, . . . , ng , z ∈ C).

(g)

1 − lk m k z

Hence it follows in view of (2.3.11) that 

φg (·, 0) φg (·, 1)



 = Ψng

φD,g−1 φN,g−1

 = Ψ0 Θ1 · · · Θng

  φD,g−1 . φN,g−1

Let     f1 φD,g−1 . := Θ1 · · · Θng φN,g−1 f2 From (2.1.13) and Theorem 2.1.2 and recalling Definition A.4.1 it follows that φ ∈ S. A recursive application of Lemma (φD,g−1 , φN,g−1 ) ∈ S 0 as well as φD,g−1 N,g−1 A.4.6 shows that (f1 , f2 ) ∈ S 0 with 

   φg (·, 0) f = Ψ0 1 = φg (·, 1) f2

f1 f2

∈ S. Finally, 1 (g) l0

0

!  f1 = f2 1 1

1 (q) f1 l0

+ f2

!

f2

shows that φg (·, 0) 1 f1 = (g) +1 φg (·, 1) l0 f2 belongs to the class S. The representation (2.3.12) now follows by a recursive application of (A.4.1) and from (2.1.11). In the above reasoning we may replace S with S0 if at least one of the pendant vertices which is distinct from the root is clamped.  Proposition 2.3.4. For all z ∈ R, min{e nD,g (z), n eN,g (z)} = min{nD,g−1 (z), nN,g−1 (z)}

(2.3.13)

and n eD,g (z)+e nN,g (z) ≤ 2g−3. Furthermore, if n eD,g (z) ≥ 1 and n eN,g (z)} ≥ 1, then nN,g−1 (z) = min{e nD,g (z), n eN,g (z)} and nD,g−1 (z) = min{e nD,g (z), n eN,g (z)} + 1.

95

2.3. Star graphs with root at a pendant vertex Proof. By Theorem 2.3.1, (2.3.9) and (2.3.10), it follows that eN,g (z)} ≤ min{nD,g−1 (z), nN,g−1 (z)}, nD,g (z), n min{e while (2.3.8) yields nD,g (z), n min{e eN,g (z)} ≥ min{nD,g−1 (z), nN,g−1 (z)}.

eD,g (z) ≥ 1 and n This proves (2.3.13). If n eN,g (z)} ≥ 1, then nD,g−1 (z) ≥ 1, and Theorem 2.1.4 gives nD,g−1 (z) = nN,g−1 (z) + 1. Hence (2.3.13) implies that nD,g (z), n nD,g (z), n nN,g−1 (z) = min{e eN,g (z)} and nD,g−1 (z) = min{e eN,g (z)} + 1. φg (·,0) Since φg (·,1) is a Nevanlinna function by Theorem 2.3.3, its zeros and poles are simple and interlace. Therefore |e nD,g (z) − n eN,g (z)| ≤ 1, which gives that n eD,g (z)+e nN,g (z) ≤ 2 min{e nD,g (z), n eN,g (z)}+1 = 2 min{nD,g−1 (z), nN,g−1 (z)}+1. If nD,g−1 (z) = 0, then it follows that n eD,g (z) + n eN,g (z) ≤ 1 ≤ 2g − 3, whereas if nD,g−1 (z) > 0, then n eD,g (z) + n eN,g (z) ≤ 2nD,g−1 (z) − 1 ≤ 2(g − 1) − 1 = 2g − 3 in view of Theorem 2.1.4. 0

 0

Theorem 2.3.5. Let n = n if M = 0 and let n = n + 1 if M > 0. Then both the Neumann problem (N2) and the Dirichlet problem (D2) have n0 characteristic values, counted with multiplicity. The nondecreasing sequence of the characteristic values (λk )nk=1 of the Neumann problem (N2) and the nondecreasing sequence of the characteristic values (µk )nk=1 of the Dirichlet problem (D2) have the following properties: 1. 0 < λ1 < µ1 ≤ λ2 ≤ · · · ≤ λn0 ≤ µn0 , where λ1 = 0 if and only if all pendant vertices distinct from the root are freely moving; 2. the multiplicities of λk and µk do not exceed g − 1; 3. if µk = λk or µk = λk+1 , then the sum of multiplicities of µk and λk or µk and λk+1 does not exceed 2g − 3; 4. if µk = λk or µk = λk+1 , then µk is a zero of φg−1 =

φD,g−1 φN,g−1 .

Proof. From Theorem 2.1.5 we know that (N1) has n0 characteristic values, and therefore also (D2) and (N2) have n0 characteristic values, counted with multiφ (·,0) plicity. Since φgg (·,1) belongs to the class S0 by Theorem 2.3.3, Lemmas A.3.5 and A.4.4 prove statement 1, except for the strict inequality λ1 < µ1 therein. 2. If the multiplicity of a characteristic value λk or µk is 1, then it does not exceed

96

Chapter 2. Vibrations of star graphs

g − 1 since g ≥ 2. Now assume that z = µk = µk+1 or z = λk = λk+1 . Then eN,g (z) ≥ 1. We have z = λk+1 or z = µk , which implies that n eD,g (z) ≥ 1 and n nD,g (z) − n eN,g (z)| ≤ 1, which already noted in the proof of Proposition 2.3.4 that |e gives ˜ N,g (z)} + 1 = nD,g−1 (z) ≤ g − 1 max{˜ nD,g (z), n nD,g (z), n ˜ N,g (z)} ≤ min{˜ in view of Proposition 2.3.4 and Theorem 2.1.5, part3. Statement 3 is part of Proposition 2.3.4, and statement 4 is immediately clear since nD,g−1 (λk ) > nN,g−1 (λk ) by Proposition 2.3.4. To complete the proof of statement 1, it remains to be shown that λ1 < µ1 . By proof of contradiction, assume that λ1 = µ1 . In view of (2.3.12), the ng -th tail (g) (g) 0 of l0 φ(·,0) φ(·,1) is fn −ng := lng + φg−1 . By statement 4 we have φg−1 (µ1 ) = 0, and (g)

we can conclude that fn0 −ng (µ1 ) = lng > 0. Since φg−1 ∈ S by Theorem 2.1.2, we know by Proposition A.3.2 that φg−1 has at least one pole which is smaller than µ1 . Let z0 be the largest pole of φg−1 which is smaller than µ1 . Clearly, z0 is also a pole of fn0 −ng . By Lemma A.2.5, the function φg−1 is increasing on (z0 , µ1 ], and therefore also fn0 −ng is increasing on (z0 , µ1 ]. Hence there is z1 ∈ (z0 , µ1 ) such that fn0 −ng (z1 ) = 0. But this contradicts Lemma A.3.5, part 5, which states that (g) any zero of the n0 − ng -th tail of l0 φ(·,0) φ(·,1) must be larger than the smallest zero of (g) φ(·,0) φ(·,1) .

l0

This contradiction proves that λ1 < µ1 .



One more property of the direct problem is needed for the inverse problem. Proposition 2.3.6. If the number of poles of

φg (·,0) φg (·,1)

is ng , then M = 0 and ng = n.

Proof. The assumption, the uniqueness of the representation (A.3.8) of S-functions and (2.3.12) show that φg−1 is constant. The representation (2.1.14) of φg−1 imPg−1 1 plies that M = 0 and that j=1 φ(j) is constant. But the functions φ(j) belong 1 to S0 , and it follows from Lemma A.2.5 that each function φ(j) is nonincreasing on (−∞, 0). Hence each function φ(j) must be constant, which means nj = 0 (j = 1, . . . , g − 1). This shows that n = ng . 

2.3.2 The inverse spectral problem In this subsection we investigate the inverse problem, that is, the question if two sequences of positive real numbers satisfying properties 1–4 in Theorem 2.3.5 can be realized as sequences of the characteristic values of the Neumann problem (N2) and the Dirichlet problem (D2) on a star graph in the case that all pendant vertices distinct from the root are clamped. More precisely, suppose that g ∈ N (g ≥ 2) is fixed and a set of lengths l(j) > 0 (j = 1, . . . , g) as well as sequences (µk )nk=1 , (λk )nk=1 of positive real numbers are given. Under which conditions can we determine numbers nj ∈ N0 (j = 1, . . . , g), the mass M of a possible bead at the central vertex, sequences of

97

2.3. Star graphs with root at a pendant vertex (j) n

(j) n

j j so that the corresponding star masses (mk )k=1 , and sequences of lengths (lk )k=0 n and (λk )nk=1 as characteristic graph of Stieltjes strings has the sequences (µk )k=1 values of the Dirichlet problem (D2) and the Neumann problem (N2), respectively?

Lemma 2.3.7. Let g ∈ N (g ≥ 2), l(j) > 0, (j = 1, . . . , g), n ∈ N. Suppose that n n (λk )nk=1 , (µk )k=1 are sequences of real numbers such that (λk )k=1 6= (µk )nk=1 and 0 < λ1 ≤ µ1 ≤ λ2 ≤ · · · ≤ λn ≤ µn ,

(2.3.14)

and let 

 z 1− µk  , Φ(z) := γ k=1 n Q z 1− λk k=1 n Q

γ := l

(g)

+

g−1 X 1 (k) l k=1

!−1 .

(2.3.15)

Then there exist unique p ∈ N, ng ∈ N0 with ng ≤ p and ak > 0 (k = 0, . . . , p), bk > 0 (k = 1, . . . , p), such that ng −1

X

ng X

ak < l(g) ,

k=0

ak ≥ l(g)

(2.3.16)

k=0

with a1ng :=

ng X

ak − l(g) ≥ 0,

k=0

and Φ has the representation 1

Φ(z) = a0 +

1

−b1 z +

1

a1 +

1

−b2 z + · · · +

1

ang −1 + −bng z +

1 ang −

a1ng

+ fbng (z) (2.3.17)

with 1

fbng (z) := a1ng +

.

1

−bng +1 z +

1

ang +1 +

1

−bng +2 z + · · · + ap−1 +

1 −bp z +

1 ap (2.3.18)

98

Chapter 2. Vibrations of star graphs

n n Proof. In view of (λk )k=1 6= (µk )k=1 and (2.3.14), Φ has at least one pole, its smallest pole is positive and less than its smallest zero, the poles and zeros of Φ are simple and interlace, and the number p of poles of Φ equals the number of zeros of Φ. Hence Φ is an S0 -function with limz→−∞ Φ(z) > 0. By Lemma A.3.5 there are unique ak > 0 (k = 0, . . . , p), bk > 0 (k = 1, . . . , p) such that

1

Φ(z) = a0 +

.

1

−b1 z +

1

a1 +

1

−b2 z + · · · +

1

ap−1 +

−bp z +

Since we have

p P

1 ap

ak = Φ(0) = γ > l(g) , there exists a unique ng ∈ N0 such that

k=0

(2.3.16) holds and all claims follow.



The next theorem shows that the properties obtained in Subsection 2.3.1 are also sufficient to find a star graph of Stieltjes strings which have these properties. Theorem 2.3.8. Let g ∈ N, g ≥ 2, l(j) > 0 (j = 1, . . . , g), n ∈ N, and let λk (k = 1, . . . , n), µk (k = 1, . . . , n) be real numbers such that (i) 0 < λ1 < µ1 ≤ λ2 ≤ · · · ≤ λn ≤ µn ; (ii) the multiplicities of λk in the sequence (λk )nk=1 and of µk in the sequence (µk )nk=1 do not exceed g − 1; (iii) if µk = λk or µk = λk+1 , then the sum of multiplicities of µk and λk or µk and λk+1 does not exceed 2g − 3. Let ng and fbng be defined as in Lemma 2.3.7. Further assume: (iv) if µk = λk (or µk = λk+1 ), then fbng (µk ) = 0; (v) if fbng is constant, then n = ng . Then there exists a star graph of g Stieltjes strings, i. e., numbers nj ∈ N0 (j = (j) 1, . . . , g − 1), masses mk > 0 (j = 1, . . . , g, k = 1, . . . , nj ), M ∈ [0, ∞), and Pnj (j) (j) lengths lk > 0 (j = 1, . . . , g, k = 0, . . . , nj ) with k=0 lk = l(j) (j = 1, . . . , g), and  g P   nj , if a1ng > 0, M = 0, n = j=1

  M > 0,

n=1+

g P j=1

nj

if a1ng = 0,

99

2.3. Star graphs with root at a pendant vertex

with a1ng as defined in Lemma 2.3.7, so that Dirichlet problem (D2) has the characteristic values µk (k = 1, . . . , n) and the Neumann problem (N2) has the characteristic values λk (k = 1, . . . , n). Proof. Due to assumptions 1 and 2, the given data yield integers p ∈ N, ng ∈ N0 and the functions Φ and fbng as in Lemma 2.3.7. The star graph we search for will be constructed as follows. For the main edge we have already defined the number ng of beads on this edge and with the coefficients from the continued fraction (g) expansion of Φ given by (2.3.17) we let mk := bk (k = 1, . . . , ng ) be their masses, (g) lk := ak (k = 0, . . . , ng − 1) and ln(g) g

:= ang −

an1 g

= ang −

ng X

ng −1

ak + l

(g)

=l

(g)

−

k=0

X

ak > 0

(2.3.19)

k=0

be the lengths of the threads between them, while the function fbng from (2.3.18) will be used to construct the subgraph of the other g − 1 edges using the inverse (g) Theorem 2.1.8. Observe that the definition of the lk and the identity in (2.3.19) Png (g) shows that k=0 lk = l(g) . For any such subgraph with total lengths l(j) (j = 1, . . . , g − 1) of its g − 1 strings we know from (1.2.25) and (1.2.28) that (j)

R2nj (0)

φ(j) (0) =

(j)

= l(j) ,

R2nj −1 (0) and then (2.1.14) and (2.3.15) show that  g−1 X φg−1 (0) =  j=1

−1 −1  g−1 X 1  1  = l φ(j) (0) j=1 j

= γ − l(g) . On the other hand, (2.3.18) and (2.3.19) give p X

fbng (0) = a1ng +

k=ng +1

=γ−l

(g)

ak =

p X

ak − l(g) = Φ(0) − l(g)

k=0

.

Altogether, we obtain φg−1 (0) = fˆng (0) > 0.

(2.3.20)

First consider the case ng = p, which is the case that fbng is constant. But then n = ng by assumption (v). We put M = 0 and nj = 0 (j = 1, . . . , g − 1).

100

Chapter 2. Vibrations of star graphs

Then φg−1 is constant, and (2.3.20) gives φg−1 = fbng . Together with (2.3.19) it follows that (2.3.17) is identical to (2.3.12). Since Φ has ng = n poles and zeros, it therefore follows that the characteristic values of (D2) and (N2) are given by the sequences (µk )nk−1 and (λk )nk=1 . Now consider the case that ng < p. Then fbng is not constant, and from the continued fraction expansion (2.3.18) of fbng it follows that fbng is an S0 -function; moreover, in view of Lemmas A.3.6 and A.3.5, fbng is the quotient of two polynomials h1,ng and h2,ng , ( h p − ng if a1ng > 0, 1,n g fbng = , deg h1,ng = deg h2,ng = p − ng . h2,ng p − ng − 1 if a1ng = 0, By Lemma A.3.5, the zeros and poles of fbng , i. e., the zeros of h1,ng and h2,ng , interlace strictly. bk )p , (b Since Φ has p poles and zeros, there are sequences (λ µk )pk=1 and k=1 n−p n n b p µk )pk=1 ∪ (γk )n−p (γk )n−p k=1 , k=1 such that (λk )k=1 = (λk )k=1 ∪ (γk )k=1 , (µk )k=1 = (b p p bk ) (λ is the sequence of the poles of Φ and (b µ ) is the sequence of the zeros k k=1 k=1 of Φ. Now set e h1,ng (z) := h1,ng (z)

n−p Y

(z − γk ),

e h2,ng (z) := h2,ng (z)

k=1

n−p Y

(z − γk ).

k=1

It follows that ( deg e h1,ng =

n − ng n − ng − 1

if a1ng > 0, if a1ng = 0,

deg e h2,ng = n − ng .

For convenience we introduce the notation n eh1 = deg e h1,ng and n eh2 = deg e h2,ng . n e h 1 e Denote the nondecreasing sequence of the zeros of h1,n by (τk ) , and denote the g

k=1

n eh2 nondecreasing sequence of the zeros of e h2,ng by (θk )k=1 . Then these two sequences satisfy the interlacing property

0 < θ1 ≤ τ1 ≤ θ2 ≤ · · · ≤ θn˜ h1 ≤ τn˜ h1 and τneh1 ≤ θneh2 if a1ng = 0, see Lemma A.4.4. We are going to show that these sequences satisfy the assumptions of Theorem 2.1.8 with g − 1 instead of g; more precisely, θk will take the role of µk in Theorem 2.1.8 and τk the role of νk in Theorem 2.1.8, and we will have M = 0 if a1ng > 0 and M > 0 if a1ng = 0. Statement A. If τk = θk or τk = θk+1 , then the maximal interlacing sequence of h2,ng which equal τk is of the form τr = θr+1 = · · · = θs = those zeros of e h1,ng and e τs for some r < s. Indeed, from the assumption it follows that τk = γj for some j ∈ {1, . . . , n0 − p} since h1,ng and h2,ng do not have common zeros. But γj = λt with λt = µt or

2.3. Star graphs with root at a pendant vertex

101

λt = µt+1 for some index t, and it follows from assumption (iv) that fbng (τk ) = 0. h2,ng which h1,ng and e Hence the maximal interlacing sequence of those zeros of e e equal τk must start and end with zeros τr and τs of h1,ng . This completes the verification of statement A. Observing the above interlacing properties and that statement A implies θ1 < τ1 and τneh1 < θneh2 if an1 g > 0, it follows that the sequences satisfy the interlacing conditions in Theorem 2.1.8 (i). Condition (ii) in Theorem 2.1.8 holds because of statement A. To verify condition (iii) of Theorem 2.1.8 consider some τk and let κ be its multiplicity. If κ = 1, then κ ≤ g − 1 since g ≥ 2. Now let κ > 1. By statement A, eh is κ − 1. With τk = γj it therefore the multiplicity of θs in the sequence (θk )nk=1 n−p is κ−1. This implies that follows that the multiplicity of γj in the sequence (γk )k=1 the sum of the multiplicity of γj in the sequence (λk )nk=1 and of the multiplicity n of γj in the sequence (µk )k=1 is 2κ − 2 or 2κ − 1. By assumption (iii) it follows that 2κ − 2 ≤ 2g − 3, which proves κ ≤ g − 1. By Theorem 2.1.8 there is a star graph of g − 1 Stieltjes strings, i. e., a 1 sequence (nj )g−1 j=1 of positive integers, M ≥ 0 with M = 0 if and only if ang > 0, (j) n

(j) n

j , (l ) j (j = 1, . . . , g − 1) of positive numbers with and sequences (mk )k=1 Pnj (j)k k=0 Pg−1 n eh1 = j=1 nj and k=0 lk = l(j) such that the Neumann problem (N1) for this

n e

h1 star graph has the characteristic values (θk )k=1 and the Dirichlet problem (D1) n e h2 for this star graph has the characteristic values (τk )k=1 . Hence fbng and φg−1 for this star graph have the same poles and zeros, and therefore fbng = φg−1 in view of (2.3.20).

With the notation of this proof it follows with (2.1.28) and (2.1.29) that the identities

φD,g−1 (z) =

h1 g−1 Y (j) neY

l

(j) j=1 l0 k=1

 1−

z τk

 ,

   h2  g g−1 X Y lj neY 1 z   φN,g−1 (z) = 1− . θk l(j) j=1 l0(j) k=1 j=1 hold. The functions h1,ng and h2,ng are only unique up to a constant factor. But the polynomials φD,g−1 and e h1,ng have the same zeros, counted with multiplicity, and the polynomials φN,g−1 and e h2,ng have the same zeros, counted with multiplicity. We may therefore assume that φD,g−1 = e h1,ng , which implies in view of (2.1.11), ˜ 2,n that φN,g−1 = e (2.3.20) and the definitions of h1,ng , h2,ng , e h1,ng and h h2,ng . g

102

Chapter 2. Vibrations of star graphs

Let φg (·, c) (c = 0, 1) be the functions defined in (2.3.7) with respect to these string data. Then (2.3.8) gives for c = 0, 1 that (g) (g) h1,ng h2,ng + R2ng −1 (·, c)e φg (z, c) = R2ng (·, c)e

 n−p  Y (g) (g) = R2ng (·, c)h2,ng + R2ng −1 (·, c)h1,ng (z − γk ) . k=1

From Theorem 2.3.5 we know that both φg (·, 0) and φg (·, 1) are polynomials of (g) (g) degree n. Therefore, for c = 0, 1, R2ng (·, c)h2,ng + R2ng −1 (·, c)h1,ng is a polynomial of degree p. But from (2.3.12) and (2.3.17) we know that (g) φg (·, 0)

l0

φg (·, 1)

=Φ

(2.3.21)

has p (simple) zeros, which we denoted by µ bk (k = 1, . . . , p) and p (simple) bk (k = 1, . . . , p). Therefore the sequence (b poles, which we denoted by λ µk )pk=1 (g) (g) is the sequence of the zeros of R2ng (·, 0)h2,ng + R2ng −1 (·, 0)h1,ng and the sebk )p is the sequence of the zeros of R(g) (·, 1)h2,n + R(g) (·, 1)h1,n . quence (λ 2ng

k=1

g

2ng −1

g

Thus (µk )nk=1 = (b µk )pk=1 ∪ (γk )n−p k=1 is the sequence of the zeros of φg (·, 0) and bk )p ∪ (γk )n−p is the sequence of the zeros of φg (·, 1), that is, for (λk )nk=1 = (λ k=1 k=1 the constructed data of the star graph, (µk )nk=1 is the sequence of the characteristic values of the Dirichlet problem (D2), whereas (λk )nk=1 is the sequence of the characteristic values of the Neumann problem (N2).  Corollary 2.3.9. The characteristic values of the Dirichlet problem (D2) and of the (g) Neumann problem (N2) together with the total length lg and the length l0 of the main edge uniquely determine the Stieltjes string data on the main edge, i. e., the (g) (g) number ng of beads, their masses mk and the lengths lk (k = 1, . . . , ng ). φ (·,0)

Proof. The poles and zeros of φgg (·,1) are uniquely determined by the given characteristic values, and hence this function is uniquely determined up to a constant (g) multiple. Prescribing l0 therefore shows that the continued fraction expansion (2.3.12) is uniquely determined. Then ng given by (2.3.16) is unique, and all the other data of the main string are now uniquely given by the continued fraction expansion (2.3.12).  Remark 2.3.10. 1. If all inequalities in assumption (i) in Theorem 2.3.8 are strict, then assumption (ii) is satisfied, and the assumptions (iii) and (iv) are void. Therefore the assumptions (ii)–(iv) can be omitted in this case. 2. In the notation of Lemma 2.3.7, let b ak = γ −1 ak (k = 0, . . . , p). That is, the numbers b ak are the numbers ak in case γ = 1 and are therefore uniquely

2.4. Examples

103

n and (λk )nk=1 . For convenience we put determined by (µk )k=1 −1  g−1 r X X (j) b   l= l , e b ar = ak (r = 0, . . . , p). j=1

k=0

Then 0 < e ap = 1, a0 < · · · < e l(g) −

r X

ar l )e ar = l(g) (1 − e ak = l(g) − (l(g) + b ar ) − b le

(r = 0, . . . , p),

k=0

and therefore, for r = 0, . . . , p − 1, r X

ak < l(g) ⇔ l(g) (1 − e le ar ) > b ar

k=0

e ar l ⇔ l(g) > b =: cr 1−e ar as well as r X

ak ≥ l(g) ⇔ l(g) ≤ cr .

k=0

We introduce the intervals I0 = (0, c0 ], Ir = (cr−1 , cr ] (r = 1, . . . , p − 1), and Ip = (cp−1 , ∞). Then ng = r if and only if l(g) ∈ Ir . Hence, for any two sequences (µk )nk=1 and (λk )nk=1 satisfying the assumptions of Lemma 2.3.7 and any numbers r ∈ {0, . . . , p} and l(j) > 0 (j = 1, . . . , g − 1) one can always find a number l(g) > 0 such that ng = r. 3. In order to find a star graph realization with at least one bead on each of its g strings it is necessary and sufficient that ng > 0 and, in view of Theorem 2.1.8 and the proof of Theorem 2.3.8, that n eh1 ≥ g − 1. By part 2 above the largest possible value of n eh1 with nh1 > 0 is n − 1, which is obtained if (and only if) lg is in the interior of I1 . Therefore n ≥ g is necessary and sufficient to find a realization with at least one bead on each of the g edges of the star graph.

2.4

Examples

In this section we illustrate the inverse Theorems 2.1.8 and 2.3.8 and their constructive proofs by means of a simple example. Example 2.4.1. Construct all star graphs with root at the central vertex with 2 edges and edge lengths l(1) = 2, l(2) = 1 such that the characteristic values (µk )2k=1 of the Neumann problem (N1) and the characteristic values (νk )2k=1 of the Dirichlet problem (D1) are given by µ1 = 1,

ν1 = µ2 = ν2 = 2.

(2.4.1)

104

Chapter 2. Vibrations of star graphs

Solution. In the notation of Theorem 2.1.8 we have n = g = 2. It is easy to see that the numbers µ1 , µ2 , ν1 , ν2 satisfy the assumptions of Theorem 2.1.8, and hence there exist such star graphs. In order to find all of them we first observe that M = 0 by Theorem 2.1.5. From Theorem 2.1.3 we know that the characteristic values of (D1) are the zeros of φD,2 , and (2.1.9) and (1.2.25) therefore shows n1 = n2 = 1 and (j)

R2 (z) =

l(j)  (j)

1−

l0

z 2

(j = 1, 2)

for any solution of the problem. In view of (1.2.28) and Lemma 1.2.13, ! 1 z (j) R1 (z) = (j) 1 − (j) (j = 1, 2) l0 µ1 (j)

with 0 < µ1 < 2 is necessary. Then 1−

(j)

φ(j) (z) =

R2 (z) (j)

= (3 − j)

R1 (z)

1−

z 2 z

(j = 1, 2)

(j) µ1

due to l(1) = 2 and l(2) = 1. In view of (2.1.11) and Theorems 2.1.2, 2.1.3 and 2.1.4, the function 1 1 1 + (2) = φ2 φ(1) φ has a simple zero at 1 and a simple pole at 2. But " ! !# 1 1 1 1 z z + = z 2 1 − (1) + 1 − (2) φ(1) (z) φ(2) (z) µ1 µ1 1− 2 h   i 1 1 (1) (2) (1) (2) = 3µ µ − 2µ + µ z , 1 1 1 1 z (1) (2) 1 − 2µ1 µ1 2

(2.4.2)

which shows that (1) (2)

(1)

(2)

3µ1 µ1 = 2µ1 + µ1 is necessary. Putting a = (1)

2 (1) µ1

µ1 =

(2.4.3)

− 1, the necessary conditions can be written as

2 , a+1

(2)

µ1 =

4 , 5−a

a ∈ (0, 3).

Taking Theorem 1.3.1 into account, it follows that under these conditions there is a unique pair of strings for each a ∈ (0, 3). The string data are found from

2.4. Examples

105

the continued fraction expansions (1.3.2). For j = 1, 2 we have (j)

φ(j) (z) =

(3 − j)µ1 2

z−2 (j)

z − µ1

(j)

=

(3 − j)µ1 + 2

1 −

2z 

(j)

(3 − j)µ1

(j)

2 − µ1

+

2   (j) (3 − j) 2 − µ1

,

which gives (j)

l0 =

  (j) (3 − j) 2 − µ1 2

(j)

,

(j)

l1 =

(3 − j)µ1 , 2

(j)

m1 =

(3 −

2 

(j) j)µ1

(j)

2 − µ1



for j = 1, 2. Hence the locations and masses of the beads for all star graphs with root at the central vertex and 2 edges having the Neumann and Dirichlet characteristic values (2.4.1) are given by M = 0 and (1)

2a , a+1 3−a = , 5−a

(1)

2 (a + 1)2 (1) , m1 = , a+1 4a 2 (5 − a)2 (2) = , m1 = , 5−a 4(3 − a)

l0 =

l1 =

(2) l0

(2) l1

(2.4.4)

where a ∈ (0, 3) is a free parameter. Example 2.4.2. Find all star graphs with root at a pendant vertex with 3 edges and (3) edge lengths l(1) = 2, l(2) = 1, l(3) = 2, l0 = 1 so that the characteristic values (λk )3k=1 of the Neumann problem (N2) and the characteristic values (µk )3k=1 of the Dirichlet problem are given by λ1 = 0.5,

λ2 = 1.5,

λ3 = 2,

µ1 = 1,

µ2 = µ3 = 2?

(2.4.5)

Solution. First we check if the assumptions of Theorem 2.3.8 are satisfied. To this end we note that n = g = 3. Since 0 < λ1 < µ1 < λ2 < µ2 = λ3 = µ3 , (i) is satisfied. The multiplicity of all λk is 1, whereas the multiplicity of the µk is 1 or 2. Therefore (ii) is satisfied. Also (iii) holds since the maximal such multiplicity is 3 = 2g − 3. In order to verify assumptions (iv) and (v) we first have to find n3 and fbn3 . Note that γ and Φ defined by (2.3.15) with the representation (2.3.17)

106

Chapter 2. Vibrations of star graphs

are −1 1 1 8 γ = l3 + + = , l1 l2 3  z 1 8 (1 − z) 1 − 2 z 2 − 3z + 2 =  Φ(z) = =1+ . 3 2z 3 1 z 2 − 2z + (1 − 2z) 1 − −z + 4 3 1 4 + 3 1 −3z + 1



3 4 3

4 3

(3)

Hence a0 = 1, a1 = and so a0 = 1 < l = 2 < 1 + = a0 + a1 . Thus, by (2.3.16), it follows that n3 = 1. Moreover, we have a1n3 = a0 + a1 − l(3) = 13 , an3 − a1n3 = 43 − 13 = 1 and hence, by (2.3.17), (2.3.18), Φ(z) = 1 +

1 1 −z + 1 + fb1 (z)

,

1 1 fb1 (z) = + 3 −3z +

1

=

1 3

1 (2 − z) . 3 (1 − z)

(2.4.6)

Since fb1 (µ3 ) = fb1 (2) = 0, assumption (iv) holds, and (v) is trivially true since fb1 is not constant. Therefore a graph as required does exist by Theorem 2.3.8. In order to construct such graphs, we observe that (2.3.21), (2.3.12) and (3) (3) (3) (2.4.6) give that l0 = l1 = 1, m1 = 1 and φ2 = fb1 . Observing that a11 6= 0, it also follows that M = 0. In view of (2.1.14) we thus have 1 φ(1) (z)

+

1 φ(2) (z)

=

1 fb1 (z)

=3

z−1 . z−2

But this is exactly (2.4.2) satisfying (2.4.3), and hence the star graphs with the data (3) (3) (3) M = 0, l0 = l1 = 1, m1 = 1, (2.4.4) (2.4.7) are exactly those graphs which have the desired characteristic values.

2.5

Notes

In this chapter we have dealt with star graphs of Stieltjes strings, which are special cases of trees and graphs of Stieltjes strings, which will be considered in Chapters 4 and 5, respectively. However, we have decided to first devote a whole chapter to this special type of trees for two reasons. Firstly, no graph theory is needed for the results of Chapter 2. Secondly, most of the results on inverse problems in this chapter are not special cases of results in Chapters 4 or 5. Inverse problem on star graphs were first considered for the Sturm-Liouville equations in [107] for g = 3 and in [108] for arbitrary g ≥ 3. The inverse spectral

2.5. Notes

107

problem for Stieltjes string equations on a star graph without a bead at the central vertex and with strict interlacing of the given spectra was solved in [18]. In all these papers the central vertex was considered as the root, i. e., the spectra of boundary value problems with generalized Dirichlet and Neumann conditions at the central vertex were used as the given data to solve the inverse problem of reconstructing the Stieltjes string data on the edges. In this monograph we have extended those results by allowing pendant vertices to be clamped or freely moving. The full solution of the inverse spectral problem generated by the Stieltjes string equations on a star graph rooted at the central vertex was given in [114], where also an inverse spectral problem generated by the Stieltjes string equations on a star graph rooted at a pendant vertex was considered. In these papers the numbers of beads on the edges were not prescribed in the given data. The inverse spectral problem generated by the Stieltjes string equations on a star graph rooted at the central vertex with prescribed numbers of beads on the edges was solved in [116]. The inverse problem where not just the sequence of all Neumann eigenvalues, but the sequences of the Dirichlet eigenvalues for all edges are given was considered in [87] for trees with M = 0 at the root. In the case when the tree is a star graph, Theorem 2.1.8 is an immediate consequence of [87, Theorem 3.3]. In [46] a three spectra inverse problem was considered for the following three spectra: the first spectrum is the spectrum of a problem on a star graph of three edges with the Dirichlet boundary conditions at the pendant vertices and the generalized Neumann conditions at the centre; the second is the spectrum of the Dirichlet-Dirichlet problem on the union of two of the graphs edges; the third is the spectrum of the Dirichlet-Neumann problem on the third edge. The presentation in Subsection 2.1.5 generalizes results for star-patterned matrices, in particular [73, Theorem 9] (see also [47], [101]). Many properties of the characteristic functions and characteristic values of string problems on star graphs are determined by the star graph itself and shared by various string problems, with finite or infinite spectrum, see [117], [49].

Chapter 3

Damped vibrations of Stieltjes strings and star graphs 3.1

The direct spectral problem

As in Subsection 1.2.1 we consider a Stieltjes string bearing n beads in its interior. But in this section these beads and the ends of the string will be subject to possible viscous friction. From Subsection 1.2.1 we recall the notation mk > 0 (k = 1, . . . , n) for the P masses of the beads, lk > 0 (k = 0, . . . , n) for the lengths of the threads and n l = k=0 lk for the total length of the string (see Figure 1.2). We denote by αk ≥ 0 the coefficient of damping (viscous friction) of the k-th bead (k = 1, . . . , n). As in Chapter 1, the string is assumed to be stretched by a stretching force equal to 1. Then small transverse vibrations of such a Stieltjes string are described by the three term recurrence relations vk (t) − vk+1 (t) vk (t) − vk−1 (t) + +mk vk00 (t)+αk vk0 (t) = 0 lk lk−1

(k = 1, . . . , n), (3.1.1)

where vk (t) is the transverse displacement of the k-th bead for k = 1, . . . , n at time t and v0 (t) and vn+1 (t) are the transverse displacements of the left and right endpoints of the string, respectively. The Dirichlet-Dirichlet problem is generated by equations (3.1.1) and boundary conditions v0 (t) = 0, vn+1 (t) = 0, (3.1.2) and the Dirichlet-Neumann problem is generated by equations (3.1.1) and boundary conditions v0 (t) = 0, vn+1 (t) = vn (t). (3.1.3) © Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4_3

109

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

110

Substituting vk (t) = uk eiλt we obtain uk − uk+1 uk − uk−1 + − λ2 mk uk + iλαk uk = 0 (k = 1, . . . , n) lk lk−1

(3.1.4)

with Dirichlet-Dirichlet boundary conditions u0 = 0, un+1 = 0

(3.1.5)

or Dirichlet-Neumann boundary conditions u0 = 0, un+1 = un .

(3.1.6)

Similar to (1.2.5) we consider the initial condition (c ∈ R)

(3.1.7)

(k = 0, . . . , n + 1).

(3.1.8)

u0 = cu1 and write uk = S2k−2 (λ, c)u1 Defining S2k−1 (λ, c) implicitly by

S2k (λ, c) = lk S2k−1 (λ, c) + S2k−2 (λ, c)

(k = 0, . . . , n),

(3.1.9)

and observing that then (3.1.4) can be written as S2k−1 (λ, c) = S2k−3 (λ, c) + (−mk λ2 + iαk λ)S2k−2 (λ, c)

(k = 1, . . . , n), (3.1.10)

we see that (3.1.10), (3.1.9) is a three term recurrence relation with the natural initial condition S0 (λ, c) = 1 and the initial condition (3.1.7), which gives S−2 (λ, c) = c. Because S−1 can be found from S0 and S−2 , we have the threeterm recurrence relation (3.1.10), (3.1.9) for Sj (λ, c) (j = 1, . . . , 2n) with the initial terms 1−c , S0 (λ, c) = 1. (3.1.11) S−1 (λ, c) = l0 It is immediately clear from the recurrence relation and the initial conditions that S2k (·, c) and S2k−1 (·, c) are polynomials of degree 2k for k = 0, . . . , n, with the exception of S−1 (·, 1) = 0. We complement the condition (3.1.7) for c ≥ 0 with un+1 = dun

(3.1.12)

for d ≥ 0, we recall the matrix Ac,d from (1.2.7) and we put M = diag(m1 , . . . , mn ),

K = diag(α1 , . . . , αn ).

(3.1.13)

The quadratic matrix pencil Tcd is defined by Tcd (λ) = λ2 M − iλK − Adc . The following result is easily verified.

(3.1.14)

3.1. The direct spectral problem

111

Proposition 3.1.1. Let c, d ≥ 0, λ ∈ C, and Y = (u1 , . . . , un )T ∈ Cn \ {0}. Define u0 and un+1 by (3.1.7) and (3.1.12), respectively. Then (3.1.4) is satisfied if and only if Tcd (λ)Y = 0. A complex number λ is called an eigenvalue of the problem (3.1.4), (3.1.7), (3.1.12), if this system of equations has a nontrivial solution, which happens if and only if there is a solution with u1 6= 0. Recalling the definitions from Section B.1, this means that the eigenvalues of the problem (3.1.4), (3.1.7), (3.1.12) are the eigenvalues of the matrix pencil Tcd . Hence we make the following Definition 3.1.2. The zeros of the function det Tcd , counted with multiplicity, are called the eigenvalues of the problem (3.1.4), (3.1.7), (3.1.12). It is evident from (3.1.14) that the problem (3.1.4), (3.1.7), (3.1.12) has 2n eigenvalues, counted with multiplicity. Proposition 3.1.3. Let c, d ≥ 0. Then   n Y 1 det Tcd (λ) = (−1)n  [S2n (λ, c) − dS2n−2 (λ, c)] l j=1 j   n Y 1 = (−1)n  [(1 − d)S2n (λ, c) + dlk S2n−1 (λ, c)] l j=1 j

(3.1.15)

(λ ∈ C).

Proof. We will prove the first identity by induction. The second identity is then a direct consequence of (3.1.9). For integers 1 ≤ k ≤ n let k Tcd (λ) be the upper left k × k submatrix of Tcd (λ). Also let d0 = d if k = n and d0 = 0 if k < n. We are going to prove by induction on k that   k Y 1  [S2k (λ, c) − d0 S2k−2 (λ, c)] (λ ∈ C). (3.1.16) det k Tcd (λ) = (−1)k  l j j=1 For k = 1 we calculate 1 − c 1 − d0 det 1 Tcd (λ) = m1 λ2 − iα1 λ − − l0 l   1 0 1 − d = m1 λ2 − iα1 λ − S0 (λ, c) − S−1 (λ, c) l1 1 d0 = −S1 (λ, c) − S0 (λ, c) + S0 (λ, c) l1 l1 1 0 = − [S2 (λ, c) − d S0 (λ, c)] , l1 which proves (3.1.16) for k = 1.

112

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

Now let k ≥ 2 and assume (3.1.16) holds for all smaller indices. When k ≥ 3, then Proposition B.3.7 and the induction hypothesis give   1 1 − d0 1 det k Tcd (λ) = mk λ2 − iαk λ − − det k−1 Tcd (λ) − 2 det k−2 Tcd (λ) lk−1 lk lk−1     k−1 0 Y 1 1−d 1 = mk λ2 − iαk λ − − (−1)k−1  S2k−2 (λ, c) lk−1 lk l j=1 j   k−2 Y 1 1  S2k−4 (λ, c). − 2 (−1)k  lk−1 l j=1 j In view of (3.1.16) for k = 1 and S0 (λ, c) = 1, this is also true when k = 2. Using (3.1.9) and (3.1.10), we find   k Y lk (−1)k  lj  det k Tcd = lk (−mk λ2 + iαk λ)S2k−2 (λ, c) + S2k−2 (λ, c) l k−1 j=1 lk + (1 − d0 )S2k−2 (λ, c) − S2k−4 (λ, c) lk−1 = lk (S2k−1 (λ, c) − S2k−3 (λ, c)) + lk S2k−3 (λ, c) + S2k−2 (λ, c) − d0 S2k−2 (λ, c) = S2k (λ, c) − d0 S2k−2 (λ, c). Hence (3.1.16) has been shown, and (3.1.15) is the case k = n.



The eigenvalues of the Dirichlet-Dirichlet problem (3.1.4), (3.1.5), i.e., the eigenvalues of the matrix pencil T00 , counted with algebraic multiplicity, will be denoted by (νk )nk=−n,k6=0 . The eigenvalues of the Dirichlet-Neumann problem (3.1.4), (3.1.6), i.e., the eigenvalues of the matrix pencil T01 , counted with algebraic multiplicity, will be denoted by (µk )nk=−n,k6=0 . Remark 3.1.4. Since the coefficients of Sj (·, c) (j = 1, . . . , 2n) depend continuously on the distribution of the beads and on c, Corollary A.6.7 shows that the zeros of det Tcd depend continuously on the distribution of the beads and on c and d. Theorem 3.1.5. 1. The sequences (νk )nk=−n,k6=0 and (µk )nk=−n,k6=0 are the zeros of S2n (·, 0) and S2n−1 (·, 0), counted with multiplicity, respectively, and {νk }nk=−n,k6=0 ∩ {µk }nk=−n,k6=0 = ∅.

(3.1.17)

2. The sequences (νk )nk=−n,k6=0 and (µk )nk=−n,k6=0 can be indexed in such a way that for all k ∈ {±1, . . . , ±n}, ν−k = −νk , µ−k = −µk if µk or νk is not pure imaginary, respectively.

3.1. The direct spectral problem

113

3. For all k ∈ {±1, . . . , ±n}, νk 6= 0, µk 6= 0, Im µk ≥ 0 and Im νk ≥ 0. If α1 > 0, or αn > 0, or αj > 0 and αj+1 > 0 for some j ∈ {2, . . . , n − 2}, then Im µk > 0 and Im νk > 0 for all k ∈ {±1, . . . , ±n}. Proof. 1. The first statement is clear from Proposition 3.1.3. Suppose that S2n (·, 0) and S2n−1 (·, 0) have a common zero, say λ0 . Substitution into (3.1.9) with k = n gives S2n−2 (λ0 , 0) = 0. Then using (3.1.10) we arrive at S2n−3 (λ0 , 0) = 0 and so on. Finally, we obtain S0 (λ0 , 0) = 0, a contradiction. Statement 2 follows from Lemma B.3.2. 3. By Proposition 1.2.3, the matrices A00 and A10 are positive definite. Hence all eigenvalues of T00 and T01 lie in the closed upper half-plane by Lemma B.3.5, part 1. They are also nonzero since A00 and A10 are invertible. Now assume additionally that α1 > 0, or αn > 0, or αj > 0 and αj+1 > 0 for some j ∈ {2, . . . , n − 2}. Let Y = (u1 , . . . , un ) ∈ Cn with KY = 0. Then u1 = 0 or un = 0 or uk = 0 and uk+1 = 0. For each (real) λ and d = 0 or d = 1, the matrix λ2 M −Ad0 is a tridiagonal matrix whose elements in the diagonals below and above the main diagonal are nonzero. Then it is easy to see that (λ2 M −Ad0 )Y = 0 would lead to Y = 0. Therefore Y 6= 0 and KY = 0 give (λ2 M − A0d )Y 6= 0, and an  application of Lemma B.3.5, part 3, completes the proof. Corollary 3.1.6. If α1 > 0, or αn > 0, or αj > 0 and αj+1 > 0 for some j ∈ {2, . . . , n − 2}, then the polynomials S2n (·, 0) and S2n−1 (·, 0) are symmetric Hermite-Biehler polynomials. Proof. Since T00 and T01 do not have real eigenvalues by Theorem 3.1.5, part 3, it follows from Proposition B.3.6 that det T00 and det T01 belong to the SHB class.  An application of Proposition 3.1.3 completes the proof. Proposition 3.1.7. The eigenvalues of the Dirichlet-Dirichlet and Dirichlet-Neumann problems satisfy the following identities: n Y

l

n Y

µk = ln

k=−n, k6=0 n X

µk =

k=−n, k6=0

νk ,

(3.1.18)

k=−n, k6=0 n X

νk ,

(3.1.19)

k=−n, k6=0

and both sides of (3.1.19) are pure imaginary. Proof. It is evident that, with the notation from Subsection 1.2.1, Sj (0, 0) = Rj (0, 0), so that S2n (0, 0) = ll0 in view of (1.2.17) and S2n−1 (0, 0) = l10 in view of (1.2.15). Hence S2n (λ, 0) =

l l0

n Y k=−n, k6=0

 1−

λ νk

 ,

S2n−1 (λ, 0) =

1 l0

n Y k=−n, k6=0

 1−

λ µk

 .

(3.1.20)

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

114

Observing that S2n (·, 0) − ln S2n−1 (·, 0) = S2n−2 (·, 0) is a polynomial of degree 2n − 2, a comparison of the coefficients for λ2n and λ2n−1 in the representations of S2n (·, 0) and S2n−1 (·, 0) in (3.1.20) leads to l l0 and l l0

n Y k=−n, k6=0

n X

n Y

j=−n, j6=0 k=−n, k6=0,j

1 ln = νk l0

1 ln = νk l0

n Y k=−n, k6=0

1 µk

n X

n Y

j=−n, j6=0 k=−n, k6=0,j

1 . µk

Rearranging the first equation gives (3.1.18). Substituting the first equation into the second equation we arrive at (3.1.19). It follows from Theorem 3.1.5, part 1, that ν−k + νk and µ−k + µk are pure imaginary for k = 1, . . . , n, and hence both sides of (3.1.19) are pure imaginary.  Theorem 3.1.8. Let (νk )nk=−n, k6=0 and (µk )nk=−n, k6=0 be two sequences of complex numbers and let l be a positive number. For (νk )nk=−n, k6=0 and (µk )nk=−n, k6=0 to be the eigenvalues of problems (3.1.4), (3.1.5) and (3.1.4), (3.1.6),P respectively, with n mk > 0, αk ≥ 0 (k = 1, . . . , n), lk > 0 (k = 0, . . . , n) and k=0 lk = l it is    n −1 Q necessary and sufficient that the product l 1 − νλk 1 − µλk can be k=−n,k6=0

represented as a continued fraction of the form: l

n Y k=−n,k6=0

  −1 λ λ 1− 1− νk µk 1

= an +

1

bn λ2 + icn λ +

1

an−1 +

1

bn−1 λ2 + icn−1 λ + · · · +

1

a1 +

1 a0 (3.1.21)

b1 λ2 + ic1 λ +

with ak > 0 (k = 0, . . . , n), bk < 0, ck ≥ 0 (k = 1, . . . , n). Proof. Let (νk )nk=−n, k6=0 and (µk )nk=−n, k6=0 be the spectra of problems (3.1.4), (3.1.5) and (3.1.4), (3.1.6), respectively. Then l

n Y k=−n,k6=0



λ 1− νk

 −1 λ S2n (λ, 0) 1− = , µk S2n−1 (λ, 0)

3.2. Inverse problems for a string with point-wise damping

115

see (3.1.20). Using (3.1.10) and (3.1.9) we obtain (3.1.21), if we formally replace mk z in (1.2.19) with mk λ2 − iαk λ (k = 1, . . . , n). In this representation, ak = lk (k = 0, . . . , n), bk = −mk , ck = αk (k = 1, . . . , n). Conversely, if (3.1.21) holds, then the two spectral problems with lk = ak Pn (k = 0, . . . , n), mk = −bk , αk = ck (k = 1, . . . , n) satisfy k=0 lk = l, which is (3.1.21) at λ = 0, and their spectra are the zeros and poles of the left-hand side of (3.1.21).  Corollary 3.1.9. The eigenvalues µk (k = ±1, . . . , ±n) and νk (k = ±1, . . . , ±n) together with the given total length l uniquely determine mk , αk (k = 1, . . . , n) and lk (k = 0, . . . , n). Proof. The left-hand side of (3.1.21) is uniquely determined by the given data. It is easy to see that the right-hand-side of (3.1.21) is of the form an + rn (λ), where rn (λ) → 0 as n → ∞. Hence an is uniquely determined. Subtracting an from the right-hand side of (3.1.21) and then taking the inverse, one arrives at an 1 expression of the form bn λ2 + icn λ + ren (λ), where ren (λ) → an−1 as λ → ∞. This shows that bn and cn are uniquely determined. Continuing in this way, one sees that all coefficients in the continued fractions expansion are uniquely determined. But these are the numbers mk , αk and lk , which proves the stated uniqueness. 

3.2

Dirichlet-Dirichlet and Dirichlet-Neumann inverse problems for a string with point-wise damping

In this section we will find sufficient conditions for a sequence of 2n complex numbers to be the sequence of eigenvalues of a string with n beads where exactly the n-th bead is subject to viscous friction. Theorem 3.2.1. Let n ∈ N, l > 0 and ln ∈ (0, l) be given together with the sequence (νk )nk=−n, k6=0 of complex numbers which satisfy the conditions: (i) Im νk > 0 for k = ±1, . . . , ±n; (ii) ν−k = −νk for not pure imaginary ν−k . Let DP be either the Dirichlet-Dirichlet problem (3.1.4), (3.1.5) or the DirichletNeumann problem (3.1.4), (3.1.6). Then there exists a unique Stieltjes string, i.e., a unique sequence of posn−1 P itive numbers (lk )n−1 lk = l − ln , a unique sequence of positive numk=0 with k=0

bers (mk )nk=1 and a unique positive number α which generate DP with αk = 0 for k ≤ n − 1 and αn = α such that its eigenvalues coincide with the sequence (νk )nk=−n, k6=0 .

116

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

It is convenient to have the initial condition at the left endpoint generalized to u0 = cu1 with c ∈ [0, 1). Therefore, consider u0 = cu1 , un+1 = 0

(c ∈ [0, 1))

(3.2.1)

(c ∈ [0, 1)).

(3.2.2)

and u0 = cu1 , un+1 = un

Then Theorem 3.2.1 is the special case c = 0 of the following theorem. l > 0, ln ∈ (0, e l ) and c ∈ [0, 1) be given together with Theorem 3.2.2. Let n ∈ N, e n the sequence (νk )k=−n, k6=0 of complex numbers which satisfy the conditions: (i) Im νk > 0 for k = ±1, . . . , ±n; (ii) ν−k = −νk for not pure imaginary ν−k . Let BP be either the problem (3.1.4), (3.2.1) or the problem (3.1.4), (3.2.2). Then there exists a unique Stieltjes string, i.e., a unique sequence of positive n P l0 lk = e l, a unique sequence of positive numbers numbers (lk )n−1 + k=0 with 1 − c k=1 (mk )nk=1 and a unique positive number α which generate BP with αk = 0 for k ≤ n−1 and αn = α such that its eigenvalues coincide with the sequence (νk )nk=−n, k6=0 . l0 Amongst these numbers, only l0 depends on c, and 1−c is independent of c. Proof. In view of assumptions (i) and (ii), Proposition A.5.4 shows that the polynomial Φ defined by   n Y λ Φ(λ) = 1− (3.2.3) νk k=−n, k6=0

e such that is of SHB class. Hence there are real polynomials Pe and Q e 2 ), Φ(λ) = Pe(λ2 ) + iλQ(λ

(3.2.4)

see (A.5.1). Proposition A.5.4 gives Pe(0) = Φ(0) = 1 and

e Q(0) =

n X k=−n,k6=0

i > 0. νk

(3.2.5)

For convenience we introduce a parameter τ with τ = ln−1 for the Dirichlet condition at the right endpoint and τ = 0 for the Neumann condition at the right endpoint. Setting   1 e α = Q(0) +τ (3.2.6) e l − ln it follows from (3.2.5) that   X n 1 i α= +τ > 0. (3.2.7) e ν k l − ln k=−n,k6=0

3.2. Inverse problems for a string with point-wise damping

117

From Φ(λ)

=

|λ|→∞

λ2n

n Y j=−n,j6=0

1 − λ2n−1 νj

n X

n Y

k=−n,k6=0 j=−n,j6=0,k

1 + O(λ2n−2 ) νj

it follows that Pe(z) e Q(z)

=

|z|→∞

=

|z|→∞

zn

iz

n Y

1 + O(z n−1 ), νj

j=−n,j6=0 n X n−1

n Y

k=−n,k6=0 j=−n,j6=0,k

Therefore  Pe(z) =  e |z|→∞ z Q(z)

n X

1 + O(z n−2 ). νj

−1 iνk 

+ O(z −1 ).

(3.2.8)

k=−n,k6=0

Then Pe(z) e |z|→∞ z Q(z)

mn := −α lim

(3.2.9)

is real and positive since Re(iνk ) = − Im νk < 0 for all k = ±1, . . . , ±n. For real numbers η, ζ we define the polynomial φ(·, η, ζ) by e 2 ) + iλα−1 Q(λ e 2 ). φ(λ, η, ζ) = Pe(λ2 ) + α−1 (ηλ2 − ζ)Q(λ Since Φ is of SHB class, it follows from Theorem A.5.2 that the polynomial φ(·, 0, 0) is also of SHB class. Since φ is a polynomial in all variables, its zeros in the λplane are piecewise analytic and continuous functions of η and of ζ, respectively, see [96, Theorem 9.1.1]. The zeros do not cross the real axis when η moves from 0 to mn and ζ moves from 0 to τ if τ = ln−1 and when η moves from 0 to mn and ζ = 0 is fixed if τ = 0. Otherwise, there would be λ ∈ R, η ∈ [0, mn ] and ζ ∈ [0, τ ] such that e 2) = 0 Pe(λ2 ) + α−1 (ηλ2 − ζ)Q(λ

and

e 2 ) = 0. λα−1 Q(λ

e 2 ) = 0 and, consequently, Pe(λ2 ) = 0 and Φ(λ) = 0, which If λ 6= 0, then Q(λ e e contradicts condition (i). If λ = 0, then 0 = Pe(0) − α−1 ζ Q(0) = 1 − α−1 ζ Q(0), which is impossible for ζ ∈ [0, τ ] due to (3.2.6). In view of the representation " # 2e 2 λ Q(λ ) φ(λ, η, ζ) = Pe(λ2 ) 1 + α−1 η + O(λ2n−1 ) |λ|→∞ Pe(λ2 ) it follows from (3.2.8) and (3.2.9) that the coefficient of λ2n of φ(·, η, ζ) is a nonzero
−1 −1 multiple of 1 − α η αmn = 1 − ηm−1 n . Therefore the degree of the polynomial

118

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

e has φ(·, η, ζ) is 2n for each η ∈ [0, mn ) and each ζ ∈ [0, τ ]. Since the polynomial Q degree n − 1, it is clear that the degree of φ(·, mn , ζ) is 2n − 1 for each ζ ∈ [0, τ ]. This means that the zeros of φ(·, η, ζ) do not come from infinity when η moves from 0 to mn and ζ moves from 0 to τ . Therefore, φ(·, η, ζ) is of SHB class for each ζ ∈ [0, τ ] and each η ∈ [0, mn ]. In view of Theorem A.5.5 this implies that e τ ) defined by the function R(·, e τ) = R(z,

e α−1 Q(z)

(3.2.10)

e Pe(z) + (mn z − τ )α−1 Q(z)

is a rational S0 -function, where the polynomial in the numerator has degree n − 1. Since the degree of the polynomial in the denominator is at most n − 1 and since a rational S-function has at least as may poles as zeros, see Corollary A.3.3, also the polynomial in the denominator has degree n − 1. Then, according to Lemma e τ ) has the representation A.3.5, R(·, 1

e τ ) = an−1 + R(z,

(3.2.11)

1

−bn−1 z +

1

an−2 +

1

−bn−2 z + · · · +

1

a1 +

−b1 z +

1 a0

with ak > 0 and bk > 0 for each k. The coefficients on the right-hand side depend on τ . Recall that τ = ln−1 corresponds to the Dirichlet condition at the right endpoint and τ = 0 corresponds to the Neumann condition at the right endpoint. We put l0 = (1 − c)a0 , and for k = 1, . . . , n − 1, we identify ak with the length e τ) lk of k-th interval and bk with the k-th mass mk of a Stieltjes string. Since R(·, is independent of c, it is clear that amongst these numbers only l0 depends on c l0 and that 1−c = a0 is independent of c. Furthermore, (1.2.19) shows that e α−1 Q(z) e Pe(z) + (mn z − τ )α−1 Q(z)

=

R2n−2 (z, c) , R2n−3 (z, c)

(3.2.12)

where R2n−2 (·, c) and R2n−3 (·, c) are the Cauer-Fry polynomials for this Stieltjes string as defined in Subsection 1.2.1. Consequently, e α−1 Q(z) = CR2n−2 (z, c), e e P (z) + (mn z − τ )α−1 Q(z) = CR2n−3 (z, c), where e α−1 Q(0) C= = R2n−2 (0, c)



1 e l − ln

−1 +τ

n−1 1−c X 1+ lk l0 k=1

!−1

3.2. Inverse problems for a string with point-wise damping

119

in view of (3.2.6) and (1.2.16). Therefore, e 2) Φ(λ) = Pe(λ2 ) + iλQ(λ = C[R2n−3 (λ2 , c) + (−mn λ2 + τ + iλα)R2n−2 (λ2 , c)].

(3.2.13)

Since we require αk = 0 for k ≤ n − 1, it follows from (3.1.9)–(3.1.11) by induction that Sk (λ, c) = Rk (λ2 , c) for k = −1, . . . , 2n − 2. Hence, letting αn := α, ln Φ(λ) = ln S2n−3 (λ, c) + ln (−mn λ2 + τ + iλαn )S2n−2 (λ, c) C = ln S2n−1 (λ, c) + τ ln S2n−2 (λ, c).

(3.2.14)

When τ = ln−1 , then (3.2.14) shows that ln Φ(λ) = S2n (λ, c). C

(3.2.15)

In view of (1.2.16), (3.1.9), (3.1.10), (3.1.11) and Φ(0) = 1 we find l0 + (1 − c)

n X

lk = l0 R2n−2 (0, c) + (1 − c)ln = l0 S2n−2 (0, c) + ln l0 S2n−1 (0, c)

k=1

!   n−1 l0 ln 1 1−c X 1 = l0 ln 1+ lk = l0 S2n (0, 0) = + e C ln l0 l − ln k=1 !   n X e lln (1 − c) ln = +1 l0 + (1 − c) lk − . e e l − ln l − ln k=1 Pn l0 l. Since the eigenvalues of the problem (3.1.4), It follows that 1−c + k=1 lk = e (3.2.1) are the zeros of S2n (·, c) by (3.2.15) and (3.2.3), the existence proof is complete when τ = ln−1 . When τ = 0, then (3.2.13) shows that ln Φ(λ) = ln S2n−1 (λ, c). C

(3.2.16)

In view of (3.2.11), (3.2.10) and (3.2.6) we find n−1

X l0 e 0) = α−1 Q(0) e + lk = R2n−2 (0, c) = R(0, =e l − ln . 1−c k=1

Since the eigenvalues of the problem (3.1.4), (3.2.2) are the zeros of S2n−1 (·, c) by (3.2.16) and (3.2.3), the existence proof is also complete when τ = 0. To prove uniqueness consider any two strings, indexed with 1 and 2, which solve this inverse problem. We denote the corresponding data and functions with

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

120

upper indices (j) for j = 1 and j = 2, respectively. That is, these Stieltjes strings (j) n−1 P (j) l satisfy 0 + l + ln = e l, the length of the interval between the right end 1 − c k=1 k (j) and the first bead from the right is ln , the remaining intervals have lengths lk (j) (k = 0, . . . , n − 1), the masses of the beads are mk (k = 1, . . . , n), the damping coefficient of the damped bead is α(j) > 0, and the sequence of the eigenvalues of (j) the two strings is (νk )nk=−n,k6=0 . Let τ = ln−1 and Sb(j) = S2n (·, c) for the problem (j) (3.1.4), (3.2.1) and let τ = 0 and Sb(j) = ln S2n−1 (·, c) for the problem (3.1.4), (3.2.2). Then the polynomials Sb(j) have the same sequences of zeros (νk )nk=−n,k6=0 for j = 1 and j = 2. We know from (3.2.15), (3.2.16) and (3.2.13) that (j) (j) 2 (j) Sb(j) (λ) = ln R2n−3 (λ2 , c) + ln (−m(j) )R2n−2 (λ2 , c). n λ + τ + iλα

(3.2.17)

In particular, (j) (j) Sb(j) (0) = ln R2n−3 (0, c) + τ ln R2n−2 (0, c)

= (1 − c)

ln (j)

l0

+ τ ln

1+

n−1 1−c X (j)

l0

! (j) lk

= (1 − c)

b l (j)

l0

k=1

by (1.2.15) and (1.2.16), where b l = ln for the problem (3.1.4), (3.2.2) and n−1

(j)

X (j) l b l = ln + 0 + lk = e l 1−c k=1

for the problem (3.1.4), (3.2.1) is independent of j. It follows that (2) (1) l0 Sb(1) = l0 Sb(2) .

(3.2.18)

Taking the real parts and the imaginary parts of the polynomials in (3.2.18) we arrive at (1)

(1)

(1)

l0 [R2n−3 (z, c) + (−m(1) n z + τ )R2n−2 (z, c)] (2)

(2)

(2)

= l0 [R2n−3 (z, c) + (−m(2) n z + τ )R2n−2 (z, c)], (1)

(1)

(2)

(2)

l0 α(1) R2n−2 (z, c) = l0 α(2) R2n−2 (z, c).

(3.2.19) (3.2.20)

Then (3.2.19) and (3.2.20) imply (1)

(2)

(1)

R2n−3 (z, c)

(2)

R2n−3 (z, c) mn z τ mn z τ − (1) + (1) = − (2) + (2) . (1) (1) (1) (2) α α α α α R2n−2 (z, c) α R2n−2 (z, c) (j)

(j)

(3.2.21)

The polynomials R2n−2 (·, c) and R2n−3 (·, c) are of equal degree n−1, and therefore (3.2.21) implies (1) (2) mn mn = (3.2.22) α(1) α(2)

3.2. Inverse problems for a string with point-wise damping and

(1)

121

(2)

R2n−3 (z, c) (1)

α(1) R2n−2 (z, c)

+

R2n−3 (z, c) τ τ = + (2) (1) (2) α α(2) R2n−2 (z, c) α

(3.2.23)

and, consequently, 1 α(1)

!

(1)

R2n−3 (0, c) (1)

+τ

R2n−2 (0, c)

1 = (2) α

!

(2)

R2n−3 (0, c) (2)

+τ

.

(3.2.24)

R2n−2 (0, c)

From (1.2.15) and (1.2.16) it follows that (j)

R2n−2 (0, c)

n−1

(j)

X (j) l0 = + lk = e l − ln > 0 (j) 1−c R2n−3 (0, c) k=1 is independent of j, and therefore (3.2.24) and (3.2.22) imply that α(1) = α(2) and (1) (2) mn = mn . Finally, (3.2.23) gives (1)

R2n−2 (·, c) (1)

R2n−3 (·, c)

(2)

=

R2n−2 (·, c) (2)

,

R2n−3 (·, c)

and since the continued fraction expansion (1.2.19), with n replaced by n − 1, (1) (2) is unique by Lemma A.3.5, it follows that lk = lk (k = 0, . . . , n − 1) and (1) (2) mk = mk (k = 1, . . . , n − 1).  Corollary 3.2.3. Let n ∈ N, e l > 0, α > 0 and c ∈ [0, 1) be given together with the sequence (νk )nk=−n, k6=0 of complex numbers which satisfy the conditions: (i) Im νk > 0 for k = ±1, . . . , ±n; (ii) ν−k = −νk for not pure imaginary ν−k . Let DP be either the problem (3.1.4), (3.2.1) or the problem (3.1.4), (3.2.2). (iii) If DP is the problem (3.1.4), (3.2.1), then it is additionally assumed that Pn αe l ≥ 4 k=−n,k6=0 νik . (iv) If DP is the problem (3.1.4), (3.2.2), then it is additionally assumed that Pn αe l > k=−n,k6=0 νik . Then there exists a Stieltjes string, i.e., a sequence of positive numbers (lk )nk=0 with n P l0 + lk = e l and a sequence of positive numbers (mk )nk=1 which generate DP 1 − c k=1 with αk = 0 for k ≤ n − 1 and αn = α such that its eigenvalues coincide with the sequence (νk )nk=−n, k6=0 . The string data are uniquely determined by the given Pn i data in the case of problem (3.1.4), (3.2.1) when αe l=4 and in the k=−n,k6=0 νk

case of problem (3.1.4), (3.2.2). However, in the case of problem (3.1.4), (3.2.1)

122

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

Pn for αe l > 4 k=−n,k6=0 νik , there are two solutions, one with ln < 12 e l and one with l0 1e ln > 2 l. In each case, l0 is the only string data which depends on c, and 1−c is independent of c. Pn i Proof. We introduce the notation β = k=−n,k6=0 νk and recall that we have shown in the proof of Theorem 3.2.2 that the assumptions (i) and (ii) imply that β > 0. In view of Theorem 3.2.2 and its proof it suffices to prove that the equation (3.2.7) has exactly one or two solutions ln ∈ (0, e l) in the respective cases. Then the existence and uniqueness statements of Theorem 3.2.3 would complete the proof of this corollary. First consider the condition (3.2.1), i.e., τ = ln−1 . Then (3.2.7) becomes αβ −1 =

1 1 + , e l n l − ln

which can be rewritten as 

2 e 1 l l = (e l − 4α−1 β). ln − e 2 4

This equation has a solution if and only if αe l ≥ 4β, in which case each solution e e e ln is in (0, l ). When αl = 4β, then the solution ln = 2l is unique, whereas for e e αe l > 4β, there is exactly one solution ln in each of the intervals (0, 2l ) and ( 2l , e l ). For the condition (3.2.2), i.e., τ = 0, the equation (3.2.7) becomes α(e l − ln ) = β, which has a solution ln ∈ (0, e l ) if and only if αe l > β, and then this solution is unique.  Corollary 3.2.4. Let n ∈ N, ln > 0, α > 0 and c ∈ [0, 1) be given together with the sequence (νk )nk=−n, k6=0 of complex numbers which satisfy the conditions: (i) Im νk > 0 for k = ±1, . . . , ±n; (ii) ν−k = −νk for not pure imaginary ν−k . Let BP be either the problem (3.1.4), (3.2.1) or the problem (3.1.4), (3.2.2). (iii) If PnBP is thei problem (3.1.4), (3.2.1), we additionally assume that αln > k=−n,k6=0 νk . Then there exists a unique Stieltjes string, i.e., a unique sequence of positive numn bers (lk )n−1 k=0 and a sequence of positive numbers (mk )k=1 which generate BP with αk = 0 for k ≤ n − 1 and αn = α such that its eigenvalues coincide with the sequence (νk )nk=−n, k6=0 . All generated string data are independent of ln , l0 is the l0 only string data which depends on c, and 1−c is independent of c. Proof. Using a reasoning similar to that in the proof of Corollary 3.2.3 it follows that (3.2.7) gives a unique 1 e l = ln + , −1 αβ − τ

3.2. Inverse problems for a string with point-wise damping

123

and the existence and uniqueness statement of Theorem 3.2.3 immediately implies e given by the statement of this corollary if we further observe that the function R (3.2.10), which generates the data, does not depend on ln .  Remark 3.2.5. Observing Corollary A.6.7 it is easy to see that the distribution of the beads depends continuously on the given data in Theorem 3.2.2 and in Corollaries 3.2.3 and 3.2.4. Example 3.2.6. For the sequence (−2 + i, −1 + i, i, 2i, 1 + i, 2 + i) find strings of total length l = 4.2 with three beads and l3 = 1.3 which are damped at the third bead with mass m3 such that the given sequence is the sequence of the eigenvalues of the Dirichlet-Dirichlet and Dirichlet-Neumann problem, respectively. e defined by (3.2.4) are Solution. The functions Pe and Q 1 5 37 Pe(z) = − z 3 + z 2 − z + 1, 20 4 10

7 2 11 29 e Q(z) = z − z+ , 20 4 10

and therefore Pe(z) 1 =− . e 7 |z|→∞ z Q(z) lim

We first consider the Dirichlet-Dirichlet problem. We find     1 1 29 10 10 42 6 e + = + = , m3 = . α = Q(0) l − ln ln 10 29 13 13 13 Then   e 10 α−1 Q(z) 7z 2 − 105z + 58 e R z, = = e 13 50z 2 − 170z + 20 Pe(z) + (m3 z − 10)α−1 Q(z) 1 7 = 50 + 1 − 125 78 z + 1 507 1325 + 1 − 2809 1638 z + 126 53

shows that the string data for the Dirichlet-Dirichlet problem with the given data are l0 =

126 507 7 2809 125 6 42 , l1 = , l2 = , m1 = , m2 = , m3 = , α= . 53 1325 50 1638 78 13 13

For the Dirichlet-Neumann problem we have e α = Q(0)

1 = 1, l − ln

m3 =

1 . 7

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

124 Then

e α−1 Q(z)

e 0) = R(z,

=

e Pe(z) + m3 zα−1 Q(z) 1 49 = 120 + 720 − 1183 z+ 2197 2760 +

49z 2 − 385z + 406 120z 2 − 460z + 140 1 1 − 529 507 z +

1 39 23

shows that the string data for the Dirichlet-Neumann problem with the given data are l0 =

2197 49 529 720 1 39 , l1 = , l2 = , m1 = , m2 = , m3 = , α = 1. 23 2760 120 507 1183 7



Example 3.2.7. For the sequence (i, i, i, i) find strings of total length l = 3 with two beads and l2 = 1 which are damped at the second bead with mass m2 such that the given sequence is the sequence of the eigenvalues of the Dirichlet-Dirichlet and Dirichlet-Neumann problem, respectively. e defined by (3.2.4) are Solution. The functions Pe and Q Pe(z) = z 2 − 6z + 1,

e Q(z) = −4z + 4,

and therefore Pe(z) 1 =− . e 4 |z|→∞ z Q(z) lim

We first consider the Dirichlet-Dirichlet problem. We find     1 1 1 3 e α = Q(0) + =4 + 1 = 6, m2 = . l − ln ln 2 2 Then e 1) = R(z,

e α−1 Q(z)

e Pe(z) + (m2 z − 1)α−1 Q(z) 1 2 = 13 + 1 − 169 24 z + 24

=

2z − 2 13z − 1

13

shows that the string data for the Dirichlet-Dirichlet problem with the given data are 24 2 169 3 l0 = , l1 = , m1 = , m2 = , α = 6. 13 13 24 2

3.3. Comparison of characteristic functions

125

For the Dirichlet-Neumann problem we have e α = Q(0)

1 1 = 4 · = 2, l − ln 2

m2 =

1 . 2

Then e 0) = R(z,

e α−1 Q(z)

e Pe(z) + m2 zα−1 Q(z) 1 = 25 + 1 − 25 8 z+ 8

=

2z − 2 5z − 1

5

shows that the string data for the Dirichlet-Neumann problem with the given data are 2 25 1 8 , m2 = , α = 2.  l0 = , l1 = , m1 = 5 5 8 2

3.3

Comparison of characteristic functions

In this section we will give necessary and sufficient conditions for two polynomials of even degree 2n to be the characteristic functions S2n (·, 0) and S2n−1 (·, 0) of the problem (3.1.4) with Dirichlet-Dirichlet boundary conditions (3.1.5) and DirichletNeumann boundary conditions (3.1.6), respectively, for the same string which is damped only at the n-th bead. Theorem 3.3.1. Let n ∈ N and let φ1 and φ2 be two polynomials of degree 2n. Then there is a string as considered in Section 3.1 with n beads, where only the n-th bead is damped, such that φ1 = υS2n−1 (·, 0) and φ2 = υS2n (·, 0) for some υ > 0 if and only if the following properties are satisfied: (i) φ1 and φ2 are real on the imaginary axis, i.e., there are real polynomials Pe1 , e 1 , Pe2 , Q e 2 such that Q e j (λ2 ), φj (λ) = Pej (λ2 ) + iλQ

j = 1, 2, λ ∈ C;

(ii) The polynomials φ1 and φ2 do not have common zeros; (iii) φ1 (0) > 0 and φ2 (0) > 0; e1 = (iv) Q 6 0 and there is a positive number β such that e 1 − Pe1 Q e2 = β Q e 21 Pe2 Q

(v)

e2 P e1 P

is a rational S0 -function.

and

e e 2 = lim P2 (z) Q e ; Q e1 (z) 1 |z|→∞ P

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

126

Proof. To prove necessity we assume that there is a string with n beads such that only the n-th bead is damped and that φ1 = υS2n−1 (·, 0) and φ2 = υS2n (·, 0) for some υ > 0. Since the properties (i)–(iv) are independent of the choice of υ, we may assume that υ = 1. From (3.1.9)–(3.1.11) it follows that S2n−1 (·, 0) and S2n (·, 0) are real on the imaginary axis, so that property (i) is satisfied. Property (ii) is satisfied because if S2n (λ, 0) = 0 = S2n−1 (λ, 0), then (3.1.9)– (3.1.10) would lead to S0 (λ, 0) = 0, which is impossible since S0 (λ, 0) = 1 by (3.1.11). Property (iii) is clear since S2n−1 (0, 0) = l0−1 and S2n (0, 0) = ll0−1 . The first n − 1 beads are undamped and therefore the recurrence relations (1.2.10)–(1.2.14) and (3.1.9)–(3.1.11) show that S2n−1 (λ, 0) = (−λ2 mn + iαλ)R2n−2 (λ2 , 0) + R2n−3 (λ2 , 0), S2n (λ, 0) = ln S2n−1 (λ, 0) + R2n−2 (λ2 , 0), whence e 1 (z) = αR2n−2 (z, 0), Pe1 (z) = −mn zR2n−2 (z, 0) + R2n−3 (z, 0), Q e 2 (z) = ln Q e 1 (z). Pe2 (z) = ln Pe1 (z) + R2n−2 (z, 0), Q It follows that e 1 − Pe1 Q e1 e 2 = (ln Pe1 + R2n−2 (·, 0))Q e 1 − Pe1 ln Q Pe2 Q e2 , e 1 = α−1 Q = R2n−2 (·, 0)Q 1

which proves the first identity in (iv) with β = α−1 . The second identity holds e2 (z) P e because lim P e (z) = ln since P1 has degree n whereas R2n−2 (·, 0) has degree |z|→∞

1

n − 1. From R2n−1 (z, 0) = −mn zR2n−2 (z, 0) + R2n−3 (z, 0) = Pe1 (z) and R2n (z, 0) = ln R2n−1 (z, 0) + R2n−2 (z, 0) = ln Pe1 (z) + R2n−2 (z, 0) = Pe2 (z) it follows that R2n (·, 0) Pe2 = . R2n−1 (·, 0) Pe1 2 The representation (1.2.19) and Lemma A.3.6 show that P e1 is a rational SP function. In view of Pe1 (0) 6= 0 we conclude that also property (v) holds. For the proof of the sufficiency let φ1 and φ2 be given which satisfy properties (i)–(v). Since φ1 and φ2 are polynomials of degree 2n, it follows from the

e

3.3. Comparison of characteristic functions

127

representation in (i) that the polynomials Pe1 and Pe2 have degree n. Assume that Pe1 and Pe2 have a common zero. In view of property (iv) this would also be a zero e 1 and Q e 2 . Therefore φ1 and φ2 would have common zeros by property (i); but of Q e2 P this is impossible by property (ii). Therefore P e is a rational S0 -function with n 1

2 poles and n zeros by assumption (iv). By Lemma A.3.5 the function P e1 has the P representation (A.3.8) with c0 > 0 and an > 0. With the notation from (A.3.8) we define a string with n beads whose masses are mk = bk (k = 1, . . . , n) and whose threads have lengths lk = ak (k = 1, . . . , n) and l0 = c10 . By (1.2.19) the Cauer-Fry polynomials for this string satisfy

e

R2n (·, 0) Pe2 = . R2n−1 (·, 0) Pe1 Therefore there is a nonzero complex number υ such that Pe2 = υR2n (·, 0),

Pe1 = υR2n−1 (·, 0),

and property (iii) implies υ > 0. Furthermore, the representation (1.2.19) gives Pe2 (z) R2n (z, 0) = ln , = lim e1 (z) |z|→∞ R2n−1 (z, 0) |z|→∞ P lim

and therefore e 2 = ln Q e1 Q by property (iv). Putting α = β

−1

, property (iv) implies

e e 1 = Pe2 − Pe1 Q2 = Pe2 − ln Pe1 α−1 Q e1 Q = υ [R2n (·, 0) − ln R2n−1 (·, 0)] = υR2n−2 (·, 0). Therefore e 1 (λ2 ) φ1 (λ) = Pe1 (λ2 ) + iλQ   = υ R2n−1 (λ2 , 0) + iλαR2n−2 (λ2 , 0)   = υ (−mn λ2 + iλα)R2n−2 (λ2 , 0) + R2n−3 (λ2 , 0) = υS2n−1 (λ, 0) and e 2 (λ2 ) φ2 (λ) = Pe2 (λ2 ) + iλQ   = υ R2n (λ2 , 0) + iλαln R2n−2 (λ2 , 0)    = υ ln R2n−1 (λ2 , 0) + iλαR2n−2 (λ2 , 0) + R2n−2 (λ2 , 0) = υ {ln S2n−1 (λ, 0) + S2n−2 (λ, 0)} = υS2n (λ, 0).



128

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

Remark 3.3.2. 1. The number υ in Theorem 3.3.1 can be determined once the string data have been found, e.g., υ=

φ1 (0) = l0 φ1 (0). S2n−1 (0, 0)

2. Theorem 3.3.1 is most useful when prescribing the eigenvalues of the Dirichlet-Dirichlet and the Dirichlet-Neumann problems, i.e., the zeros of φ1 and φ2 . Then one can assume without loss of generality that φ1 (0) > 0 and φ2 (0) > 0, and if for one such pair (φ1 , φ2 ) properties (i)–(v) are satisfied, then these conditions hold for all pairs (υ1 φ1 , υ2 φ2 ) (υ1 > 0, υ2 > 0), and we obtain a one-parameter family of strings, where the parameter is the total −1 length of the string υ2 φ2 (0) (υ1 φ1 (0)) . 3. Given two sequences of complex numbers which are the eigenvalues of the Dirichlet-Dirichlet and Dirichlet-Neumann problem, respectively, one can uniquely determine which of the sequences belongs to which problem. Indeed, if we have that the zeros of φ2 are the eigenvalues of the Dirichlet-Dirichlet problem and the zeros of φ1 are the eigenvalues of the Dirichlet-Neumann e1 e2 P P problem, then P e1 is a rational S0 -function by property (v). But then P e2 is a not a rational S0 -function, so that it cannot be true that the zeros of φ1 are the eigenvalues of the Dirichlet-Dirichlet problem and the zeros of φ2 are the eigenvalues of the Dirichlet-Neumann problem.

3.4

Stieltjes strings with a damping condition at the right end

In this section we consider small transverse vibrations of a Stieltjes string with Robin boundary condition (1.2.42) at the left end and with an additional bead of mass mn+1 > 0 at the right end that can move in the direction orthogonal to the equilibrium position of the string subject to damping with coefficient α > 0. In this case the condition at the right end is given by vn+1 (t) − vn (t) 00 0 + mn+1 vn+1 (t) + αvn+1 (t) = 0. ln Substituting vk (t) = uk eiλt we obtain un+1 − un − mn+1 λ2 un+1 + iαλun+1 = 0. ln

(3.4.1)

A mathematically equivalent formulation of (3.4.1) is obtained by introducing un+2 and any ln+1 > 0 and writing (3.4.1) as un+1 − un+2 un+1 − un + − mn+1 λ2 un+1 + iαλun+1 = 0, (3.4.2) ln+1 ln un+2 = un+1 . (3.4.3)

3.4. Stieltjes strings with a damping condition at the right end

129

With the notation γ0 from (1.2.43) it follows that problem (1.2.4), (3.4.2), (1.2.42), (3.4.3) is a special case of problem (1.2.4), (1.2.42), (3.1.12) with n replaced by n + 1, with c = γ0 , and with d = 1. Here α1 = · · · = αn = 0 and αn+1 = α > 0. In particular, S2n+1 (λ, γ0 ) is defined by (3.1.10) for k = n + 1. Recalling the notation Tcd from (3.1.14), we have therefore that the sequence of the 2n + 2 eigenvalues (λk )n+1 k=−(n+1),k6=0 , counted with algebraic multiplicity, is the sequence of the eigenvalues of the problem (1.2.4), (1.2.42), (3.4.1). Theorem 3.4.1. 1. The sequence (λk )n+1 k=−(n+1),k6=0 is the sequence of the zeros of S2n+1 (·, γ0 ), counted with multiplicity. 2. The sequence (λk )n+1 k=−(n+1),k6=0 can be indexed in such a way that for all k ∈ {±1, . . . , ±(n + 1)}, λ−k = −λk if λk is not pure imaginary. 3. For all k ∈ {±1, . . . , ±(n + 1)}, Im λk > 0. Proof. In the proof of Theorem 3.1.5 we may replace T01 with Tγ10 .



Theorem 3.4.2. Let n ∈ N, l > 0 and α > 0 be given together with a string of length l and the eigenvalue sequence (λk )n+1 k=−(n+1), k6=0 of the corresponding problem (1.2.4), (1.2.42), (3.4.1). Then the sequence (lk )nk=0 , the sequence (mk )n+1 k=1 and the number γ are uniquely determined by (λk )n+1 k=−(n+1),k6=0 , l and α. Theorem 3.4.3. Let n ∈ N and α > 0 be given together with the sequence of complex numbers (λk )n+1 k=−(n+1), k6=0 which satisfy the conditions: (i) Im λk > 0 for k = ±1, . . . , ±(n + 1); (ii) λ−k = −λk for not pure imaginary λ−k . Pn+1 Let β := k=−n+1,k6=0 λik . Then there exists a number b l ∈ (0, α−1 β) such that for all l ∈ (b l, α−1 β) there exist a unique number γ > 0, a unique sequence of positive n P numbers (lk )nk=0 with lk = l and a unique sequence of positive numbers (mk )n+1 k=1 k=0

such that (λk )n+1 k=−(n+1),k6=0 is the sequence of the eigenvalues of the corresponding problem (1.2.4), (1.2.42), (3.4.1). Proof of Theorems 3.4.2 and 3.4.3. We are going to apply Corollary 3.2.4 with n replaced by n + 1. To this end we first consider an arbitrary c ∈ (0, 1) and an auxiliary ln+1 > 0. By Corollary 3.2.4 there are a sequence of positive numbers (lk )nk=0 and a sequence of positive numbers (mk )n+1 k=1 which generate (3.1.4), (3.2.2) with αk = 0 for k ≤ n and αn+1 = α such that its eigenvalues coincide with the sequence (λk )n+1 k=−(n+1), k6=0 . Furthermore, l1 , . . . , ln , m1 , . . . , mn+1 are independent l0 is independent of c and ln+1 . These of c and ln+1 , and also the number a0 = 1−c numbers are therefore uniquely determined by the given data. When l is given,

130

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

also l0 is uniquely determined, and l# := a0 +

n X k=1

lk =

n

n

k=1

k=0

X X l0 + lk = lk + ca0 = l + ca0 1−c

(3.4.4)

is independent of c and ln+1 . But since we have to satisfy γ0 = c we obtain 1 + γl0 = 1c by (1.2.43), and therefore γ=

1−c 1 1 = − # cl0 ca0 l −l

is uniquely determined by the given data. This completes the proof of Theorem 3.4.2 and the uniqueness part of Theorem 3.4.3. For the existence part in Theorem 3.4.3 we require c ∈ (0, 1), and (3.4.4) leads to the condition l ∈ (l# − a0 , l# ). In the notation of Theorem 3.2.2 we have e l = ln+1 + l# , and it follows from (3.2.7) that l# = α−1 β. Putting b l = l # − a0 =

n X

lk

k=1

completes the proof.

3.5



The inverse problem by parts of spectra of RobinRegge problems

In this section we consider two problems for the Stieltjes string equations (1.2.4) with the same damping condition (3.4.1) at the right end, but with possibly different Robin conditions at the left end, u1 − u0 − γ1 u0 = 0, l0 u1 − u0 − γ2 u0 = 0, l0

(3.5.1) (3.5.2)

where γ1 > 0 and γ2 > 0. According to Theorem 3.4.1, the characteristic functions of these problems are S2n+1 (·, γ1,0 ) and S2n+1 (·, γ2,0 ), where γj,0 = (1 + γj l0 )−1 (j = 1, 2) by (1.2.43). Furthermore, the sequence of the zeros of S2n+1 (·, γj,0 ), i.e., the eigenvalues of problem (1.2.4), (3.4.1) with (3.5.1) for j = 1 and with (3.5.2) for j = 2, can be indexed as (λk (γj ))n+1 k=−(n+1),k6=0 so that the properties 2 and 3 of Theorem 3.4.1 are satisfied. Recalling from Section 3.4 that (3.1.10) also holds for k = n + 1 here, it follows from α1 = · · · = αn = 0 that S2n+1 (λ, γj,0 ) = (−mn+1 λ2 + iαλ)R2n (λ2 , γj,0 ) + R2n−1 (λ2 , γj,0 ).

(3.5.3)

3.5. The inverse problem by parts of spectra of Robin-Regge problems

131

Proposition 3.5.1. For all α > 0, γ1 > 0, γ2 > 0 and λ ∈ C it is true that S2n+1 (λ, γ2,0 )S2n+1 (−λ, γ1,0 ) − S2n+1 (−λ, γ2,0 )S2n+1 (λ, γ1,0 )   1 1 = 2iαλ − S2n+1 (0, γ1,0 )S2n+1 (0, γ2,0 ). (3.5.4) γ2 γ1 Proof. Proposition A.6.1 and the Lagrange identity (1.2.47) show that S2n+1 (λ, γ2,0 )S2n+1 (−λ, γ1,0 ) − S2n+1 (−λ, γ2,0 )S2n+1 (λ, γ1,0 ) = 2iαλ[R2n−1 (λ2 , γ1,0 )R2n (λ2 , γ2,0 ) − R2n−1 (λ2 , γ2,0 )R2n (λ2 , γ1,0 )] γ2,0 − γ1,0 . = 2iαλ l0 But in view of (3.5.3) and (1.2.44), γ2,0 − γ1,0 1 − γ1,0 1 − γ2,0 = − l0 l l  0  0   1 − γ1,0 1 − γ2,0 l0 l0 = − l0 l0 1 − γ2,0 1 − γ1,0   1 + γ 2 l0 1 + γ 1 l0 = S2n+1 (0, γ1,0 )S2n+1 (0, γ2,0 ) − γ2 γ1   1 1 = S2n+1 (0, γ1,0 )S2n+1 (0, γ2,0 ) − . γ2 γ1



If γ1 6= γ2 , then (3.5.4) implies that {λk (γ1 ) : 0 < |k| ≤ n + 1} ∩ {λk (γ2 ) : 0 < |k| ≤ n + 1} = ∅.

(3.5.5)

The information given in Theorem 3.4.1, Proposition 3.5.1 and (3.5.5) suffices to recover string data from parts of the eigenvalues of two problems, which are given together with the total length of the string, the damping constant and the difference of the inverses of the Robin parameters. Theorem 3.5.2. Let n ∈ N, l > 0 and α > 0 be given together with a string of length l. Let p ∈ N0 with 0 ≤ p ≤ n + 1, let (θk )0 0), and let (τk )p 0). If additionally {θk : 0 < |k| ≤ p} ∩ {τk : p < |k| ≤ n + 1} = ∅ and γ2−1 − γ1−1 is n known, then the string data (mk )n+1 k=1 , (lk )k=0 , γ1 , γ2 are uniquely determined by l, α, (θk )0 0 (k = 0, . . . , nj , j = 1, 2) and > 0 (k = 1, . . . , nj , j = 1, 2). (j) Denoting the Cauer-Fry polynomials for these string data by Rk (·, 0), we know (j) lk

(j) mk

from (1.2.19) that the right-hand side of (3.6.16) is equal to

(j) 1 (j) R2n −1 (z,0) 1

R2n (z,0)

, (j = 1, 2).

Therefore it follows from (3.6.12), (3.6.15) and (3.6.16) that (1) (2) R2n1 −1 (·, 0) R2n2 −1 (·, 0) αRPe αPe αP2 αP1 = + = = + (1) (2) e e Q1 Q2 RQ Q R2n1 (·, 0) R2n2 (·, 0) (1)

=

(2)

(1)

(2)

R2n1 −1 (·, 0)R2n2 (·, 0) + R2n1 (·, 0)R2n2 −1 (·, 0) (1)

(2)

.

R2n1 (·, 0)R2n2 (·, 0) (j)

Since Qj and R2nj (·, 0) have the same degree nj and the same zeros, there are (j)

numbers Tj ∈ C\{0} such that Qj = Tj R2nj (·, 0). With T = α−1 T1 T2 we conclude (1)

(2)

e = Q1 Q2 = T αR (·, 0)R (·, 0) RQ 2n1 2n2 and then (1)

(2)

(1)

(2)

RPe = T [R2n1 −1 (·, 0)R2n2 (·, 0) + R2n1 (·, 0)R2n2 −1 (·, 0)]. We conclude that e 2) Φ(λ) = R(λ2 )Pe(λ2 ) + iλR(λ2 )Q(λ (1)

(2)

(1)

(2)

= T [R2n1 (λ2 , 0)R2n2 −1 (λ2 , 0) + R2n1 −1 (λ2 , 0)R2n2 (λ2 , 0) (1)

(2)

+ iαλR2n1 (λ2 , 0)R2n2 (λ2 , 0)]. In Subsection 3.6.1 we have seen that the zeros of the function on the right-hand side are the eigenvalues of problem (3.6.1)–(3.6.4) associated with the generated 1 +n2 data. Hence these eigenvalues coincide with the sequence (λk )nk=−(n . 1 +n2 ) Finally, (3.6.13)–(3.6.15) gives αP1 (0) 1 = , b Q1 (0) l

αP2 (0) 1 = . Q2 (0) l−b l

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

140

According to (3.6.16) and Propositions 1.2.8 and 1.2.9 this implies that and

k=0

n2 P

(2)

k=0

3.7 3.7.1

n1 P

lk = l − b l.

(1)

l lk = b 

Star graphs with damped central vertex The direct problem

We consider a plane star graph of g ≥ 2 Stieltjes strings. We assume all pendant vertices to be fixed. A bead of mass M > 0 may be placed at the central vertex; in case there is no bead at the central vertex we put M = 0. In Section 2.1 we considered two problems of small transverse vibrations of the graph: one with the the central vertex fixed, called Dirichlet problem (D1), and one with the central vertex free to move in the direction orthogonal to the equilibrium position of the strings, called Neumann problem (N1). In this section it is assumed that the central vertex can move subject to damping with constant coefficient of damping α > 0, and we call this the damped Neumann problem. It is clear that the damping only affects the balance of forces condition, and therefore the damped Neumann problem is described by (2.1.4), (2.1.5), (2.1.7) and (j) (j) g X unj +1 − unj

(1)

− (M λ2 − iαλ)un1 +1 = 0.

(j)

lnj

j=1

(3.7.1)

If we clamp the central vertex, then the damping has no effect, so that we obtain the Dirichlet problem (D1). According to (1.2.8) we look for solutions of (2.1.4) and (2.1.7) in the form (j)

(j)

(j)

uk = R2k−2 (λ2 )u1 (j)

(k = 1, . . . , nj ),

(j)

where R2k−2 := R2k−2 (·, 0) are the Cauer-Fry polynomials of degree k − 1 corresponding to the j-th string. Since (2.1.5) and (3.7.1) have to be satisfied simultaneously, they are equivalent to (2.1.5) and (j) (j) g X unj +1 − unj (j)

j=1

=

l nj

g X 1 j=1

g

(j)

(M λ2 − iαλ)unj +1 .

(3.7.2)

Then (2.1.5) and (3.7.2) can be written as (1)

(1)

(g)

(g)

R2n1 (λ2 )u1 = · · · = R2ng (λ2 )u1 , g X j=1

(j)

(j)

R2nj −1 (λ2 )u1 =

g X j=1

1 (j) (j) (M λ2 − iαλ)R2nj (λ2 )u1 . g

(3.7.3) (3.7.4)

3.7. Star graphs with damped central vertex

141

(j) g,n +1

j is a nontrivial solution of the As in Subsection 2.1.1 we find that (uk )j=1,k=0 damped problem (2.1.4), (2.1.5), (2.1.7), (3.7.1) if and only if the matrix   (1) (2) R2n1 (λ2 ) −R2n2 (λ2 ) ··· 0 0 0   (2) (3) 0 0 0 R2n2 (λ2 ) −R2n3 (λ2 ) · · ·     . .. .. .. ..  ... Aα (λ) =  . . .   .. .   (g−1) (g)  0 0 0 · · · R2ng−1 (λ2 ) −R2ng (λ2 ) aα,g (λ) aα,1 (λ) aα,2 (λ) aα,3 (z) · · · aα,g−1 (λ)

is singular, where 1 (j) (j) aα,j (λ) = R2nj −1 (λ2 ) − (M λ2 − iαλ)R2nj (λ2 ) (j = 1, . . . , g). g Expanding the determinant of Aα (λ) with respect to the last row and recalling the definitions of φD,g and φN,g in (2.1.9) and (2.1.10), it follows that φ(λ) := det Aα (λ) = φN,g (λ2 ) + iαλφD,g (λ2 ).

(3.7.5)

We have shown Proposition 3.7.1. The eigenvalues of (2.1.4), (2.1.5), (2.1.7), (3.7.1) are the zeros of the polynomial φ given by (3.7.5). Remark 3.7.2. From Theorem 2.1.5 and its proof know that the degree of the Pg Pwe g polynomial φN,g is 1 + j=1 nj if M > 0 and j=1 nj if M = 0, and that the Pg Pg degree of φD,g is j=1 nj in any case. Therefore the degree of φ is 2 + 2 j=1 nj Pg if M > 0 and 1 + 2 j=1 nj if M = 0. Next we consider the damped problem in the case that none of the threads has an interior bead. Example 3.7.3. Assume that nj = 0 for all j = 1, . . . , g. 1. If M = 0, then the damped problem has exactly one simple eigenvalue g P i 1 . α l(j) j=1

2. If M > 0, then the damped problem has exactly two eigenvalues, counted with multiplicity, namely   12 g 2 X iα 1 1 α  ± − . 2M M j=1 l(j) 4M 2 (j)

(j)

(j)

Proof. For nj = 0 we have R2nj = R0 = 1 and R2nj −1 = R−1 = and (1.2.14). Therefore φD,g (λ2 ) = 1 and  X g  g X 1 M 2 1 2 − λ = − M λ2 , φN,g (λ ) = (j) (j) g l l j=1 j=1

1 , l(j)

see (1.2.12)

142

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

which gives φ(λ) = −M λ2 +

g X 1 + iαλ. (j) l j=1

Solving φ(λ) = 0 completes the proof.



Theorem 3.7.4. The polynomial φ belongs to the class SHB. Proof. Let φ0 be the monic polynomial whose zeros (if any) are the common zeros of φN,g and φD,g , counted with multiplicity, and let φN,g , φeN = φ0

φD,g φeD = . φ0

Then φeN and φeD have no common zeros, and φeD φD = φN φeN is an S0 -function by Theorem 2.1.2. Setting P (λ) = φeN (λ2 ) and Q(λ) = αλφeD (λ2 ) it follows that the zeros of P and Q are real and interlace and that Q0 (0)P (0) − Q(0)P 0 (0) = αφeD (0)φeN (0) > 0. Hence Theorem A.5.2 gives that e φ(λ) := φeN (λ2 ) + iαλφeD (λ2 ) e By Theorems 2.1.3, 2.1.4 and 2.1.5, the describes a Hermite-Biehler polynomial φ. zeros of φN,g and φD,g are real and positive, and therefore all zeros of λ 7→ φ0 (λ2 ) e are real. Since φ(λ) = φ0 (λ2 )φ(λ), it follows that φ is a generalized Hermite-Biehler polynomial. It is clear that the polynomial φ is symmetric.  The following factorization of φ immediately follows from Theorem 3.7.4, (2.1.9), (2.1.10), (1.2.15) and (1.2.17). Corollary 3.7.5. There are p1 , p2 , p3 ∈ N0 such that the zeros of φ, counted with multiplicity, can be represented as the union of (possibly zero length) sequences 1 2 3 (βk )pk=−p , (γk )pk=1 , and (δk )pk=−p with the following properties: 1 ,k6=0 3 ,k6=0 (i) βk ∈ R \ {0} and β−k = βk for k = 1, . . . , p1 ; (ii) γk lies on the positive imaginary semiaxis for k = 1, . . . , p2 ; (iii) Re δk 6= 0, Im δk > 0 and δ−k = −δk for k = 1, . . . , p3 .

3.7. Star graphs with damped central vertex

143

Then φ(λ) = φ(0)

p1 Y

 1−

k=−p1 ,k6=0

where

λ βk

p3 Y



k=−p3 ,k6=0

  p2   λ Y λ 1− 1− , δk γk k=1



  g g (j) Y X l 1  . φ(0) =  (j) l(j) j=1 l0 j=1 0

n We recall from Theorem 2.1.5 that the zeros (µP k )k=1 of φN,g and the zeros g of φD,g are positive real numbers, where n = j=1 nj and n0 = n if M = 0 or n = n + 1 if M > 0. For any of those zeros λ, m(λ) denotes their multiplicity as a zero of the corresponding function.

(νk )nk=1 0

Theorem 3.7.6.

0

1 = (µk )nk=1 ∩ (νk )nk=1 . 1. (βk2 )pk=0

2. If βk2 = µp = νp , then m(µp ) = m(βk ), m(νp ) = m(βk )+1 and m(βk ) ≤ g−1. 3. Let pe1 be the number of distinct positive real zeros of φ. Then 2e p1 ≤ p2 + 2p3 − 1. 4. The number p2 is odd if and only if M = 0. 5. Im φ(s) (βk ) = 0 for s = 0, . . . , m(βk ). Proof. 1, 2. Since φN,g and φD,g are real polynomials, it follows for real λ that φ(λ) = 0 if and only if φN,g (λ2 ) = 0 and φD,g (λ2 ) = 0. In this case, the multiplicity of λ as a zero of φ is clearly the minimum of the multiplicities of λ2 as a zero of φN,g and φD,g . This proves statement 1. If βk2 = µp is such a common zero, we may assume without loss of generality that p is the smallest index such that βk2 = µp = νp , and we let r be the largest index such that βk2 = µr = νr . In view of Theorem 2.1.5, parts 1 and 2, we have p ≥ 2 and µp−1 < νp−1 = µp = νp = · · · = µr = νr . Hence m(νp ) = m(µp ) + 1, which proves the first part of statement 2. Since m(νk ) ≤ g by Theorem 2.1.5, part 3, also m(βk ) ≤ g − 1 has been shown. 3. By part 2, every real zero of φ is the square P root of a zero of multiplicity g at least 2 of φD,g . In view of Remark 3.7.2 and n = j=1 nj it follows that 2e p1 ≤ n < Therefore

1 (2p1 + p2 + 2p3 ). 2

1 1 pe1 < p1 − pe1 + p2 + p3 ≤ p2 + p3 , 2 2 which implies 2e p1 ≤ p2 + 2p3 − 1.

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

144

4. Since the degree 2p1 + p2 + 2p3 of φ is odd if and only if p2 is odd, it follows from Remark 3.7.2 that p2 is odd if and only if M = 0. 5. By (3.7.5), Im φ(λ) = αλφD,g (λ2 ) for real λ. By part 2, every βk is a zero of multiplicity m(βk ) + 1 of φD,g , and therefore Im φ(s) (βk ) = 0 for s = 0, . . . , m(βk ). The representation of φ is clear from φ(0) = φN,g (0). 

3.7.2 The inverse problem In this subsection it will be shown that the necessary properties of the eigenvalues which were derived in Subsection 3.7.1 are also sufficient to find a star graph which has those eigenvalues. Furthermore, apart from the eigenvalues, also the number of edges and their total lengths can be prescribed. Theorem 3.7.7. Let g ≥ 2 be an integer and let positive numbers l(j) , j = 1, . . . , g, p3 1 2 , (γk )pk=1 and (δk )k=−p be given. Let the sequences of numbers (βk )pk=−p 3 ,k6=0 1 ,k6=0 be given, where p1 , p2 , p3 ∈ N0 . Assume that the following conditions are satisfied: p1 ≤ p2 + 2p3 − 1, where pe1 is (i) For all indices k, βk ∈ R \ {0}, β−k = −βk , 2e p1 the number of distinct elements of the sequence (βk )k=0 . (ii) For each k, m(βk ) does not exceed g − 1, where m(βk ) is the number of times p1 . βk occurs in the sequence (βj )j=−p 1 ,k6=0 (iii) γk = i|γk | and γk 6= 0 for all k. (iv) Im δk > 0, Re δk 6= 0 and δ−k = −δk for all k. (v) Im Φ(s) (βk ) = 0 for all k and for s = 0, . . . , m(βk ), where p1 Y

Φ(λ) =

k=−p1 ,k6=0

  λ 1− βk

p3 Y

 1−

k=−p3 ,k6=0

λ δk

Y p2 

1−

k=1

λ γk

 .

(3.7.6)

(j) n

j Then there exists a collection of sequences of positive real numbers (mk )k=1 and P n n (j) j (j) j (j) (lk )k=0 with = l for j = 1, . . . , g, numbers M ≥ 0 and α > 0 k=0 lk such that problem (2.1.4), (2.1.5), (2.1.7), (3.7.1) with these data has eigenvalues 1 2 3 (βk )pk=−p ∪ (γk )pk=1 ∪ (δk )pk=−p , counted with multiplicity. Here M = 0 if 1 ,k6=0 3 ,k6=0 and only if p2 is odd.

Proof. For λ ∈ C define R(λ) =

p1  Y k=1

1−

λ βk2

 ,

ω(λ) =

p3 Y k=−p3 ,k6=0

  p2   λ Y λ 1− 1− . δk γk k=1

By definition, Φ(λ) = R(λ2 )ω(λ) for λ ∈ C, and it is clear that ω is a symmetric e such that Hermite-Biehler polynomial. Hence there are real polynomials Pe and Q

3.7. Star graphs with damped central vertex

145

e 2 ) for λ ∈ C. By Theorems A.5.2 and A.5.5 and Propoω(λ) = Pe(λ2 ) + iλQ(λ e do not have common zeros and Qe is an S0 -function with sition A.5.4, Pe and Q e P e Pe(z) > 0 and Q(z) > 0 for z ∈ (−∞, 0]. Case I: p2 is even. The numbers n = p1 + 12 p2 + p3 − 1 and n − p1 are nonnegative integers because 21 p2 + p3 > 0 by condition (i). First we consider the case that n − p1 = 0. Then p2 + 2p3 = 2(n − p1 ) + 2 = 2 and 2e p1 ≤ p2 + 2p3 − 1 = 1 shows that pe1 = 0 and therefore p1 = 0. Hence      λ λ 1 1 λ2 Φ(λ) = 1 − 1− =1−λ + + , λ1 λ2 λ1 λ2 λ1 λ2 where λ1 and λ2 are nonzero complex numbers satisfying λ1 = i|λ1 | and λ2 = i|λ2 | or λ2 = −λ1 and Im λ1 > 0. Then problem (2.1.4), (2.1.5), (2.1.7), (3.7.1) with M =−

g 1 X 1 > 0, λ1 λ2 j=1 l(j)

 α=i

1 1 + λ1 λ2

X g j=1

1 l(j)

> 0,

and nj = 0 for j = 0, . . . , g has the two eigenvalues (λ1 , λ2 ) by Example 3.7.3. Now let n − p1 > 0. Since ω is a polynomial of even degree p2 + 2p3 , the e is smaller degree of the polynomial Pe is 12 p2 + p3 = n + 1 − p1 and the degree of Q e and Pe do not have common zeros and that than the degree of Pe. Recalling that Q e Q e is an S0 -function, we conclude from Proposition A.3.2 and Theorem A.2.6 that P there are positive numbers µ ek (k = 1, . . . , n + 1 − p1 ) and νek (k = 1, . . . , n − p1 ) n+1−p1 1 such that (e µk )k=1 is the sequence of the zeros of Pe, such that (e νk )n−p k=1 is the e and such that sequence of the zeros of Q, 0 0. Due to condition (v) we see that each In particular, the degree of Q 2 e whereas β 2 is a zero of R of βk is a zero of multiplicity at least m(βk ) + 1 of RQ, k 2 e multiplicity m(βk ). Therefore Q(βk ) = 0, and each βk2 coincides with some νer . Let 1 1 (µk )n+1 µk )n+1−p ∪ (βk2 )pk=1 , k=1 := (e k=1

2 p1 1 (νk )nk=1 := (e νk )n−p k=1 ∪ (βk )k=1 ,

with µk ≤ µk+1 and νk ≤ νk+1 . n We claim that these two sequences (µk )n+1 k=1 and (νk )k=1 satisfy the assump0 tions of Theorem 2.1.8 with n = n + 1. For k ∈ {1, . . . , p1 } we have νe1 ≤ βk2 ≤ νen−p1 and therefore µ1 = µ e1 < νe1 = ν1 and νn = νen−p1 < µ en+1−p1 = µn+1 . Hence condition (i) of Theorem 2.1.8 holds (see Lemma A.4.4). Condition (ii) then also holds if we observe that µr = νr holds if and only if there is k such that µr = νr = βk2 , and in this case m(µr ) = m(βk ) = m(νr ) − 1. Finally, condition (iii) of Theorem 2.1.8 is satisfied because of assumption (ii).

146

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

Thus, in view of Theorem 2.1.8, there exists a star graph of g Stieltjes strings with all quantities as indicated in the statement of that theorem, and in particular n+1 and (νk )nk=1 are the sequences of the M > 0, except for α, such that (µk )k=1 eigenvalues of the Neumann and Dirichlet problems, respectively. Let φN,g and φD,g be the corresponding characteristic functions, see (2.1.9) and (2.1.10). The polynomials RPe and φN,g have the same zeros, counted with multiplicity, and e and φD,g have the same zeros, counted with multiplicity. the polynomials RQ e > 0 by (A.5.2), φN,g (0) > 0 and Observing that R(0) = 1, Pe(0) = 1, Q(0) φD,g (0) > 0, we obtain positive real numbers αN and αD such that RPe = αN φN,g e = αD φD,g . Putting α = αD α−1 , it follows that and RQ N e 2 ) = αN [φN,g (λ2 ) + iαλφD,g (λ2 )], Φ(λ) = R(λ2 )Pe(λ2 ) + iλR(λ2 )Q(λ

(3.7.7)

and hence Proposition 3.7.1 shows that the eigenvalues of the damped problem for this star graph and this α coincide with the zeros of Φ, that is, the given sequences. Case II: p2 is odd. The numbers n = p1 + 21 p2 + p3 − 12 and n − p1 are nonnegative integers because 12 p2 + p3 − 12 ≥ 0 by condition (i). First we consider the case that n − p1 = 0. Then p2 + 2p3 = 2(n − p1 ) + 1 = 1 and 2e p1 ≤ p2 + 2p3 − 1 = 0 shows that p2 = 1, p3 = 0 and pe1 = 0, and therefore p1 = 0. Hence λ , Φ(λ) = 1 − γ1 where δ1 = i|δ1 | and δ1 6= 0. Then problem (2.1.4), (2.1.5), (2.1.7), (3.7.1) with M = 0,

α=

q i X 1 > 0, γ1 j=1 l(j)

and nj = 0 for j = 0, . . . , g has the single eigenvalue γ1 by Example 3.7.3. Now let n − p1 > 0. Since ω is a polynomial of odd degree p2 + 2p3 , the e is 1 p2 + p3 − 1 = n − p1 and the degree of Pe does not exceed the degree of Q 2 2 e Recalling that Q e and Pe do not have common zeros and that Qe is an degree of Q. e P S0 -function, we conclude from Proposition A.3.2 that there are positive numbers 1 µ ek (k = 1, . . . , n−p1 ) and νek (k = 1, . . . , n−p1 ) such that (e µk )n−p k=1 is the sequence n−p 1 e and such of the zeros of Pe, such that (e νk )k=1 is the sequence of the zeros of Q, that 0 0, such that the corresponding damped problem has the zeros of Φ, that is, the given sequences of numbers as eigenvalues.  Remark 3.7.8. The star graphs and the damping coefficient in Theorem 3.7.7 can be constructed explicitly. A construction of star graphs is given in the proof of [114, Theorem 2.9] for n − p1 > 0 and in the proof of Theorem 3.7.7 for n − p1 = 0.

3.8. Notes

147

To find α, we deduce from the proof of Theorem 3.6.3 and from (A.5.2), (2.1.13), (1.2.15), (1.2.17) that e R(0)Q(0) φN,g (0) φD,g (0) R(0)Pe(0)  p3 p2 q X X 1 1 X 1 . + δk γ k j=1 l(j)

−1 = α = αD αN

 = i

k=−p3 ,k6=0

k=1

Remark 3.7.9. We have seen in Remark 3.7.2 that the number of eigenvalues uniquely determines the total number of beads in the interiors of the strings and whether or not there is a bead at the joint vertex. We have also seen in Remark 3.7.8 that the damping coefficient α is uniquely determined by the spectrum. Finally, we know from (3.7.7) that φN,g (λ2 ) Φ(λ) + Φ(−λ) = iαλ . φD,g (λ2 ) Φ(λ) − Φ(−λ) Hence we conclude from (2.1.13) that φN,g (λ2 ) λ→∞ λ2 φD,g (λ2 ) Φ(λ) + Φ(−λ) = −iα lim , λ→∞ λ(Φ(λ) − Φ(−λ))

M = − lim

which means that M is uniquely determined by the eigenvalues of the damped problem.

3.8

Notes

The investigation of spectral problems for damped systems was started in [47]. As far as we know first results on inverse problems for damped strings were obtain in [4] and [80], [81]. In these papers the left end of the string was supposed to be free and the right end damped or in other words the right end of the string could move with viscous friction in the direction orthogonal to the equilibrium position of the string. Very wide classes of strings were considered. In [4] the class of the so-called regular strings was used, i.e., strings of finite mass and length, while in [80], [81] the class of so-called S-strings, i.e., strings of finite lengths and finite first momentum of mass distribution. Necessary and sufficient conditions for a sequence of complex numbers to be the spectrum of a damped string were given in [4] in implicit form and in [80] and [81] explicitly. If the string is so smooth that its density ρ satisfies ρ ∈ W22 (0, l) and ρ(s) ≥  > 0 where l is the length of the string, then one can apply the Liouville transformation (see [40, p. 202]) to transform the equation of the string into the Sturm-Liouville equation. The corresponding

148

Chapter 3. Damped vibrations of Stieltjes strings and star graphs

boundary value problem with the spectral parameter in the boundary conditions was considered in many publications (see [35], [125], [104], [105] and references therein). In [65] and [112] necessary and sufficient conditions are given such that a sequence of complex numbers is the spectrum of a smooth inhomogeneous string damped at one end for any value of the damping parameter except of one crucial value. The inverse problem, i.e., the problem of recovering the parameters of the string from the spectrum of its vibrations and from the total length of the string, for a Stieltjes strings with a finite number of beads with point-wise (i.e., onedimensional) damping at the right end and the left end free was solved in [4]. Also, this problem can be reduced to the problem of damped oscillators considered in [133], [134]. Another approach to inverse problem for damped finite-dimensional system was developed in [85] where the given data included not only eigenvalues but also the so-called Jordan pairs. Inverse problems for Stieltjes strings damped at an interior point were solved in [16] and [142]. The inverse problem for star graphs of Stieltjes strings with damping at the central vertex was solved in [97]. In Sections 3.4 and 3.5 we have considered inverse problems with damping at one endpoint and one or two Robin conditions at the other endpoint, where in the case of two Robin conditions parts of the spectra for each of the two problems are known. In [141] the inverse spectral problem is solved in the case of one Robin condition, where part of the string data and part of the spectrum are known. In Section 3.1 we have considered the matrix operator pencil Tcd (λ) = λ2 M − iλK − Adc where a symmetric nonnegative matrix K describes damping. The quadratic matrix operator pencil λ2 M − λ(B − iK) − A with symmetric B also occurs in physics where the term λB with symmetric B describes gyroscopic forces (see [144], [7]).

Chapter 4

Trees 4.1

Characteristic polynomials

Let T be a plane tree with g straight edges. Then this tree has g + 1 vertices, and an arbitrary of those vertices will be chosen as the root. For convenience, the root will be denoted by v0 and the tree will be considered as being directed towards the root (see Lemma C.2.6). The other vertices will be denoted by vj (j = 1, . . . , g). For each j = 1, . . . , g there is a unique simple path from vj to v0 , and the first edge in this path, which is the unique outgoing edge from vj , will be denoted by ej . In this way we have uniquely labelled the g edges of T from the given labelling of the vertices. We recall some notation from Appendix C. For j = 0, . . . , g, Ij+ := I + (vj ) is the set of those indices k for which ek is an incoming edge into the vertex vj . The set of the indices of the outgoing edges from the vertex vj is I − (vj ) = {j} for j = 1, . . . , n and I − (v0 ) = ∅. Obviously, vj (j = 1, . . . , g) is a pendant vertex if and only if Ij+ = ∅. For j = 0, . . . , g, Ij := Ij+ ∪ Ij− is the set of indices of those edges of T which are incident with vj , and d(vj ) = #Ij is the degree of vj . Clearly, d(v0 ) = #I0+ and d(vj ) = 1 + #Ij+ (j = 1, . . . , g). Also let I00 := {1, . . . , g} \ I0+ be the index set of all edges which are not incident with the root v0 . For each j ∈ {1, . . . , g}, the vertex vj is the entrance v+ (ej ) of ej , whereas the exit v− (ej ) of ej can be written as vσ(j) for some σ(j) ∈ {0, . . . , g} \ {j}. Note that ej is an incoming edge at vσ(j) , and therefore a pair (r, j) satisfies r ∈ {0, . . . , g} and j ∈ Ir+ if and only if j ∈ {1, . . . , g} and r = σ(j). Each edge ej (j = 1, . . . , g) is divided into nj + 1 segments of the lengths (j) (j) (j) (j) (j) (j) l0 , . . . , lnj by beads with masses m1 , . . . , mnj (lk > 0, mk > 0). Here n j P (j) nj ≥ 0 and, clearly, l(j) := lk is the length of ej . The indexing of the beads k=0

and the segments are such that the lower indices increase in the direction of the edge. This means that for j = 1, . . . , g, each vertex vj is located at the beginning © Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4_4

149

Chapter 4. Trees

150 (j)

of the segment of length l0 . For j = 0, . . . , g with Ij+ 6= ∅, each incoming edge (r)

er (r ∈ Ij+ ) ends at vj with a segment of length lnr . It is assumed that the tree is stretched. The edges of the tree are strings which can vibrate in the direction orthogonal to the equilibrium position of the strings. We denote the transverse (j) (j) displacement of the j-th bead with mass mk by vk (t). For j = 1, . . . , g, the (j) displacement of the entrance vj of the edge ej is denoted by v0 (t), whereas the (j) displacement of the exit of the edge ej is denoted by vnj +1 (t). With this notation, vibrations of the graph can be described by the system of equations (j)

(j)

vk (t) − vk−1 (t) (j)

(j)

−

(j)

vk+1 (t) − vk (t) (j)

lk−1

lk

(j) ∂

+ mk

2 (j) vk (t) ∂t2

=0

(4.1.1)

(j = 1, . . . , g, nj ≥ 1, k = 1, . . . , nj ). The vibrating tree is still assumed to be connected. For the interior vertices we therefore impose the continuity conditions (r)

(j)

v0 (t) = vnj +1 (t) (j ∈ Ir+ ) for r = 1, . . . , g with Ir+ 6= ∅, (r)

(s)

vnr +1 (t) = vns +1 (t) (r, s ∈ I0+ ) if d(v0 ) > 1.

(4.1.2) (4.1.3)

Balance of forces at an interior vertex which is not the root implies (j)

(j)

v1 (t) − v0 (t) (j)

=

l0

X vn(r)+1 (t) − vn(r) r (t) r (r)

lnr

r∈Ij+

for j = 1, . . . , g with Ij+ 6= ∅. (4.1.4)

For a pendant vertex (except for the root if it is pendant) we impose a mixed boundary condition (j)

(j)

v0 (t) = cj v1 (t) for j = 1, . . . , g with Ij+ = ∅,

(4.1.5)

with cj ∈ R. We need to impose one more condition at the root. We consider two cases: the Dirichlet case with (r)

vnr +1 (t) = 0 for some r ∈ I0+

(4.1.6)

and the Neumann case X vn(r)+1 (t) − vn(r) r (t) r (r)

r∈I0+

= 0.

(4.1.7)

lnr

In (4.1.2) we observe that due to the discussion at the beginning of this section we may write (r, j) with r ∈ {1, . . . , g} and j ∈ Ir+ as j ∈ I00 and r = σ(j). Furthermore, we observe that there is ambiguity in (4.1.3) since there are exactly

4.1. Characteristic polynomials

151

d(v0 ) − 1 independent conditions. Also, we want to make the formulation of the conditions independent of any reference to the details of the indexing of the vertices and edges. Clearly, in the Dirichlet case, (4.1.3) and (4.1.6) are equivalent to (j)

vnj +1 (t) = 0 (j ∈ I0 ).

(4.1.8)

In the Neumann case, (4.1.3) and (4.1.7) imply (r)

(j) d(v0 )vnj +1 (t)

(r) vnr +1 (t)

X

=

−

(r)

vnr +1 (t) − vnr (t)

!

(r)

lnr

r∈I0

(j ∈ I0 ).

(4.1.9)

Conversely, since the right-hand sides in (4.1.9) are independent of j, (4.1.3) immediately follows from (4.1.9), whereas (4.1.7) follows from adding up the d(v0 ) equations in (4.1.9). Later in this section we will discuss the reason for this choice and possible simplifications. (j) (j) Substituting vk (t) = eiλt uk into (4.1.1)–(4.1.7) we obtain the Dirichlet problem (j)

(j)

(j)

uk − uk−1 (j) lk−1

(j)

uk+1 − uk

−

(j) lk

(j)

(j)

− mk λ2 uk = 0

(4.1.10)

(j = 1, . . . , g, nj ≥ 1, k = 1, . . . , nj ), (σ(j)) u0 (j) u1

=

(j) unj +1

(j) − u0 (j) l0

(j)

=

(j ∈ I00 ),

X r∈Ij+

(r) unr +1 − (r) l nr

(4.1.11) (r) unr

for j = 1, . . . , g with Ij+ 6= ∅,

(j)

u0 = cj u1 for j = 1, . . . , g with Ij+ = ∅,

(4.1.12) (4.1.13)

(j)

unj +1 = 0 (j ∈ I0 ),

(4.1.14)

and the Neumann problem (4.1.10)–(4.1.13), (r)

(j) d(v0 )unj +1

=

X

(r) unr +1

r∈I0

−

(r)

unr +1 − unr (r)

lnr

! (j ∈ I0 ).

(4.1.15)

The conditions (4.1.14) are called Dirichlet conditions at the root, whereas the conditions (4.1.15) are called Neumann conditions at the root. If the root is a pendant vertex, i.e., d(v0 ) = 1, then (4.1.15) is equivalent to the usual Neumann condition. In what follows, problem (4.1.10)–(4.1.14) is called the Dirichlet problem for the tree T and problem (4.1.10)–(4.1.13), (4.1.15) is called the Neumann problem for the tree T . We observe that both the Dirichlet and the Neumann problem are wellg P (j) posed. Indeed, letting n := nj , there are n + 2g unknown uk (j = 1, . . . , g, j=1

Chapter 4. Trees

152

k = 0, . . . , nj + 1), (4.1.10) consists of n equations, (4.1.11) consists of g − d(v0 ) equations, (4.1.12) and (4.1.13) together consist of g equations, and both (4.1.14) and (4.1.15) consist of d(v0 ) equations. Therefore the Dirichlet problem as well as the Neumann problem consists of a system of n + 2g linear equations for n + 2g unknown variables. (j) However, we can immediately eliminate the n unknowns uk for k ≥ 2 by using the n equations in (4.1.10). Indeed, for fixed j ∈ {1, . . . , g}, the nj equation (j) in (4.1.10) are of the form (1.2.4). With the functions Rk (·, c) for k = −2, . . . , 2nj defined according to the recurrence relations (1.2.10) and (1.2.11) with the initial conditions (1.2.12) and (1.2.13), also the linear combinations e(j) := R(j) (·, 0)(u(j) − u(j) ) + R(j) (·, 1)u(j) R 1 0 0 k k k satisfy these recurrence relations (1.2.10) and (1.2.11), whereas the initial cone(j) for e(j) (z) = u(j) and R e(j) (z) = u(j) . It follows that u(j) = R ditions are R −2 0 1 0 k 2k−2 k = 0, . . . , nj + 1, that is, (j)

(j)

(j)

(j)

(j)

(j)

uk = R2k−2 (·, 0)(u1 − u0 ) + R2k−2 (·, 1)u0

(j = 1, . . . , g; k = 0, . . . , nj + 1). (4.1.16) Therefore, we are left with 2g unknown variables (j)

(j)

(j)

u2j−1 := u0 and u2j := u1 − u0

(j = 1, . . . , g),

(4.1.17)

and in view of (4.1.16) the conditions (4.1.11)–(4.1.13) can be expressed in terms of linear functionals operating on U := (u1 , . . . , u2g ) which are defined as follows: (j)

(j)

L2j (z)U := R2nj (z, 1)u2j−1 + R2nj (z, 0)u2j − u2σ(j)−1 (j ∈ I00 ), X (r) u2j (r) L2j−1 (z)U := − (j) + (R2nr −1 (z, 1)u2r−1 + R2nr −1 (z, 0)u2r ) l0 r∈Ij+ for j = 1, . . . , g with Ij+ 6= ∅,

(4.1.19)

L2j−1 (z)U := (1 − cj )u2j−1 − cj u2j for j = 1, . . . , g with Ij+ = ∅.

(4.1.20)

(4.1.18)

For the Dirichlet problem, the conditions at the root lead to the functionals (j)

(j)

LD 2j (z)U := R2nj (z, 1)u2j−1 + R2nj (z, 0)u2j (j ∈ I0 ),

(4.1.21)

and for the Neumann problem, the conditions at the root lead to the functionals D LN 2j (z)U := L2j (z)U −

1 X D L2r (z)U d(v0 ) r∈I0

+

X r∈I0

(r) R2nr −1 (z, 1)u2r−1

(r)

+ R2nr −1 (z, 0)u2r



(j ∈ I0 ). (4.1.22)

4.1. Characteristic polynomials

153

D For convenience, we set LN j := Lj := Lj for j = 1, . . . , 2g with j 6∈ 2I0 . Defining the (2g) × (2g) matrices ΦD (z) and ΦN (z) by  D    N L1 (z)U L1 (z)U     .. . (4.1.23) ΦD (z)U =  ,  , ΦN (z)U =  . ..

LD 2g (z)U

N L2g (z)U

it follows that the Dirichlet problem (4.1.10)–(4.1.14) is equivalent to ΦD (z)U = 0, and the Neumann problem (4.1.10)–(4.1.13), (4.1.15) is equivalent to ΦN (z)U = 0. We call the function φD defined by (4.1.24)

φD (z) = det(ΦD (z))

the characteristic polynomial of the Dirichlet problem (4.1.10)–(4.1.14). Similarly, the function φN defined by φN (z) = det(ΦN (z))

(4.1.25)

is called the characteristic polynomial of the Neumann problem (4.1.10)–(4.1.13), (4.1.15). The zeros of the characteristic polynomials will be called the characteristic values of the Dirichlet and the Neumann problem, respectively. Proposition 4.1.1. φD and φN are real polynomials which are independent of the indexing of the vertices and edges of the tree T . Proof. Since all entries of the matrices ΦD and ΦN are real polynomials, it is clear that also their determinants φD and φN are real polynomials. Reindexing the edges and vertices would result in simultaneous permutations of rows and columns of  the matrices ΦD (z) and ΦN (z), which leaves their determinants unchanged. The linear functionals (4.1.22) for the Neumann conditions at the root have complicated right-hand sides, but with this definition we obtain symmetry, that is, the conditions are invariant under permutation of indices in I0 and therefore lead to an unambiguously defined characteristic polynomial. However, it may be more convenient to consider functionals which are directly derived from (4.1.3) and (4.1.7). But for d(v0 ) > 1 such choice is ambiguous, and the characteristic polynomial obtained in this way may be the negative of the one we have defined in (4.1.25). In the proposition below we list a class of such choices which always give the correct characteristic polynomial. Proposition 4.1.2. Let τ : {1, . . . , d(v0 )} → I0 be bijective and define D N D (j = 1, . . . , d(v0 ) − 1), Le2τ (j) (z)U := L2τ (j) (z)U − L2τ (j+1) (z)U  X  (r) (r) N Le2τ (d(v0 )) (z)U := R2nr −1 (z, 1)u2r−1 + R2nr −1 (z, 0)u2r .

(4.1.26) (4.1.27)

r∈I0 τ Let ΦN be derived from ΦN by replacing the linear functionals LN 2τ (j) (z) in (4.1.23) N e with the linear functionals L (z) (j = 1, . . . , d(v0 )). Then det(Φτ (z)) = φN (z). 2τ (j)

N

154

Chapter 4. Trees

Proof. We will show that ΦτN is derived from ΦN by elementary row operations which do not change the determinant. For j = 1, . . . , d(v0 ) − 1 it immediately follows from (4.1.22) that N N LeN 2τ (j) = L2τ (j) − L2τ (j+1) .

Together with d(v0 )−1

X

d(v0 )−1

j LeN 2τ (j) =

j=1

X

d(v0 )

jLD 2τ (j) −

j=1

X

(j − 1)LD 2τ (j)

j=1 d(v0 )

=

−d(v0 )LD 2τ (d(v0 ))

+

X

LD 2τ (j)

j=1

and (4.1.22) we obtain LeN 2τ (d(v0 )) (z) =

X

(r)

(r)

R2nr −1 (z, 1)u2r−1 + R2nr −1 (z, 0)u2r



r∈I0

= LN 2τ (d(v0 )) (z) +

4.2

d(v0 )−1 X 1 j LeN 2τ (j) (z). d(v0 ) j=1



Relations between characteristic polynomials for a tree and for complementary subtrees

Assume that the root v0 of the tree T is an interior vertex. We cut the tree T at v0 into two subtrees T1 and T2 (see Figure 4.1). We say that T1 and T2 are

v0

T1

v0 v0 T T2 Figure 4.1: Complementary subtrees for an interior root complementary subtrees of T . In the notation of Section 4.1 let I (ι) be the set of

4.2. Complementary subtrees

155

all indices j for which ej is an edge of the subtree Tι (ι = 1, 2). With g1 := #I (1) we have 1 ≤ g1 < g and #I (2) = g − g1 . For convenience, the indexing of the edges (ι) of T will be chosen in such a way that I (1) = {1, . . . , g1 }. We also let I0 = I0 ∩I (ι) (ι = 1, 2). Next we are going to introduce notation for the Dirichlet and Neumann problems on the subtrees T1 and T2 . Let LjD,1 (z), LN,1 j (z) (j = 1, . . . , 2g1 ) and N,2 D,2 Lj (z), Lj (z) (j = 2g1 +1, . . . , 2g) be the linear functionals as defined in Section 4.1 associated with the trees T1 and T2 , respectively. With uj (j = 1, . . . , 2g) 2g1 defined in Section 4.1 we let U (1) = (uj )j=1 and U (2) = (uj )2g j=2g1 +1 . It is clear that LjD (z)U = LjD,1 (z)U (1) LD j (z)U

=

(2) LD,2 j (z)U

(j = 1, . . . , 2g1 ),

(4.2.1)

(j = 2g1 + 1, . . . , 2g).

(4.2.2)

Hence the Dirichlet problem for T splits into two Dirichlet problems for T1 and (2) (1) T2 , and the corresponding coefficient matrices ΦD (z) and ΦD (z) are given by

(1)

ΦD (z)U (1)

  D,1 L1 (z)U (1)   ... = ,

(2)

ΦD (z)U (2)

D,1 (z)U (1) L2g 1

  D,2 L2g1 +1 (z)U (2)   .. = . .

(4.2.3)

(2) LD,2 2g (z)U

(ι)

Then the characteristic polynomials φD of the Dirichlet problems on the subtrees Tι (ι = 1, 2) are (ι)

(ι)

φD (z) = det(ΦD (z)) (1)

(ι = 1, 2)

(4.2.4)

(2)

The coefficient matrices ΦN (z) and ΦN (z) for the Neumann problems on T1 and T2 are given by

(1)

ΦN (z)U (1)

 N,1  L1 (z)U (1)   .. = , .

(2)

ΦN (z)U (2)

N,1 L2g (z)U (1) 1

  N,2 L2g1 +1 (z)U (2)   ... = , N,2 (2) L2g (z)U

(4.2.5)

D,ι where LN,ι j (z) = Lj (z) is as in (4.2.1), (4.2.2) when j 6∈ 2I0 since the conditions (4.1.18) and (4.1.19) are the same for the Dirichlet and the Neumann case. For (ι) j ∈ 2I0 , the functionals LN,ι 2j (z) (j ∈ I0 , ι = 1, 2) are obtained from (4.1.22) (ι)

(ι)

by replacing I0 in the sums by I0 and d(v0 ) by d(ι) (v0 ) := #I0 . Then the (ι) the characteristic polynomials φN of the Neumann problems on the subtrees Tι (ι = 1, 2) are (ι)

(ι)

φN (z) = det(ΦN (z)) (ι = 1, 2).

(4.2.6)

Chapter 4. Trees

156

Theorem 4.2.1. Let the root v0 of a tree T be an interior vertex. Let T1 and T2 be complementary subtrees of T having v0 as the only common vertex and being directed towards the root. Then (1)

(2)

(4.2.7)

(2) (1) φD (z)φN (z).

(4.2.8)

φD (z) = φD (z)φD (z), φN (z) =

(2) (1) φN (z)φD (z)

+

Proof. It immediately follows from (4.2.3) and (4.1.23) that ! (1) 0 ΦD (z) , ΦD (z) = (2) 0 ΦD (z) which proves (4.2.7). For the proof of (4.2.8) we observe that according to Proposition 4.1.1 we may choose any arbitrary indexing of the edges. It is convenient to choose the indices in such a way that j ∈ I0 if and only if g1 − d(1) (v0 ) < j ≤ g1 + d(2) (v0 ). The function τ defined by ( g1 − d(1) (v0 ) + j for j = 1, . . . , d(1) (v0 ), τ (j) = g1 + d(v0 ) + 1 − j for j = d(1) (v0 ) + 1, . . . , d(v0 ), is a bijection from {1, . . . , d(v0 )} to I0 , and therefore Proposition 4.1.2 shows that φN (z) = det(ΦτN (z)).

(4.2.9)

τ Writing ΦN (z) as a block matrix

τ ΦN (z)

=

! e 1,2 (z) e 1,1 (z) Φ Φ e 2,1 (z) Φ e 2,2 (z) Φ

with respect to the decomposition C2g = C2g1 ⊕ C2(g−g1 ) , it is clear that the only e 1,2 and Φ e 2,1 are in the last row of Φ e 1,2 and in the second row nonzero elements of Φ τ 2,1 e . Observing (4.2.9), an expansion of det(Φ (z)) with respect to the 2g1 -th of Φ N row gives e 1 (z)) + det(Φ e 2 (z)), φN (z) = det(Φ where e 1 (z) = Φ

e 1,1 (z) Φ 0 e 2,2 (z) e 2,1 (z) Φ Φ

!

e 2 (z) is obtained from Φτ (z) by setting the first 2g1 eleand where the matrix Φ N e 1,1 (z) by the ments in its 2g1 -th row equal 0, that is, by replacing the last row of Φ zero row.

4.2. Complementary subtrees

157

e 2,2 (z)). In view of Proposition 4.1.2, e 1 (z)) = det(Φ e 1,1 (z)) det(Φ Clearly, det(Φ 1,1 e the last row of Φ is given by  (1)   (1)  U U D,1 N D e L2g1 (z) = L2g (z)U (1) , = L2g1 (z) 1 0 0 e 1,1 (z) to and, in case d(1) (v0 ) > 1, recursively adding the new 2(j + 1)-th row in Φ (1) its (2j)-th row for j = 2(g1 − 1), . . . , 2(g1 − d(1) (v0 ) + 1) one arrives at ΦD (z). e 1,1 (z)) = det(Φ(1) (z)). This shows that det(Φ D With τ2 (j) = g1 + d(2) (v0 ) + 1 − j for j = 1, . . . , d(2) (v0 ) it is easy to verify e 2,2 (z) = Φ2,τ2 (z), and det(Φ e 2,2 (z)) = φ(2) (z) follows in view of Proposition that Φ N N 4.1.2. We have shown that (2)

(1)

e 1 (z)) = φ (z)φ (z). det(Φ D N e 2 (z) is given by The (2g1 )-th row of Φ   0 D N (z) (z)U. Le2g = −L2(g 1 1 +1) U (2) e 2 (z) with its negative and then interchanging the Replacing the (2g1 )-th row of Φ 2g1 -th row with the 2(g1 + 1)-th row leads to a matrix ! b 1,2 (z) b 1,1 (z) Φ Φ b Φ(z) = b 2,2 (z) Φ 0 e 2 (z)). With τ1 (j) = g1 − d(1) (v0 ) + j for j = 1, . . . , d(1) (v0 ) b with det(Φ(z)) = det(Φ b 1,1 (z)) = b 1,1 (z) = Φ1,τ1 (z), and hence det(Φ it follows from Proposition 4.1.2 that Φ N (1) (1) b 2,2 (z) is given by the functional det(ΦN (z)) = φN (z). The second row of Φ D,2 (2) (2) L2(g1 +1) (z)U , and if d (v0 ) > 1, its 2j-th row for j = 2, . . . , d(2) (v0 ) is given by D,2 (2) the functional (LD,2 . Recursively adding the 2(j − 1)-th row 2j (z) − L2(j−1) (z))U (2)

of this matrix to its 2j-th for j = 2, . . . , d(2) (v0 ), the resulting matrix is ΦD (z). b 2,2 (z)) = φ(2) (z). This completes the proof of (4.2.8).  Hence det(Φ D Corollary 4.2.2. Let the root v0 of a tree T be an interior vertex, let 2 ≤ p ≤ d(v0 ) and let Tj (j = 1, . . . , p) be complementary subtrees of T having v0 as the (j) only common vertex and being directed towards the root. For j = 1, . . . , p let φD (j) and φN be the characteristic polynomials of the Dirichlet and Neumann problem, respectively, on the subtree Tj . Then φD (z) =

p Y

(j)

φD (z),

(4.2.10)

j=1

φN (z) =

p X j=1

(j) φN (z)

p Y k=1, k6=j

(k)

φD (z).

(4.2.11)

158

Chapter 4. Trees

Remark 4.2.3. 1. If one takes p = d(v0 ) in Corollary 4.2.2, then the root v0 is a pendant vertex for all subtrees Tj (j = 1, . . . , d(v0 )). 2. Let ej be an edge of the tree T which is incident with v0 such that vj is a (j) pendant vertex of T . We consider ej as a subtree Tj of T . Then ΦD (z) and (j) ΦN (z) are 2 × 2 matrices, and the first functional is given by (4.1.20) for (j) (j) both ΦD (z) and ΦN (z), whereas the second functional is given by (4.1.21) (j) (j) for ΦD (z) and by (4.1.22) for ΦN (z). Therefore, for j = 1, . . . , g, ! ! 1 − cj −cj 1 − cj −cj (j) (j) ΦD (z) = , ΦN (z) = , (j) (j) (j) (j) R2nj (z, 1) R2nj (z, 0) R2nj −1 (z, 1) R2nj −1 (z, 0) and taking their determinants gives (j)

(j)

φD (z) = R2nj (z, cj ),

(j)

(j)

φN (z) = R2nj −1 (z, cj )

in view of part 4 of Remark 1.2.7. 3. Let the tree T be a star graph with the centre being the root and consider Corollary 4.2.2 with p = d(v0 ) = g. Then for each subtree, the only edge (j) (j) is incident with the root, and φD and φN are as found in part 2 of this remark. Therefore the characteristic polynomials (4.2.10) and (4.2.11) with cj = 0 (j = 1, . . . , g) are identical to the characteristic polynomials (2.1.9) and (2.1.10) for the star graph with central mass M = 0. That is, for a star graph without central mass and with Dirichlet conditions at the pendant vertices, the characteristic polynomials obtained in this section agree with those obtained in Section 2.1. Next we consider trees with the root being a pendant vertex. Removing the edge which is incident with v0 , a new tree T 0 is obtained (see Figure 4.2).

ep v0

vp

vp

T Figure 4.2: Subtree for a pendant root

T0

4.2. Complementary subtrees

159

Proposition 4.2.4. Let T be a tree with g ≥ 2 edges and assume that v0 is a pendant vertex. Removing the edge ep which is incident with v0 , a subtree T 0 is obtained, (0) (0) where vp becomes the root of T 0 . Let φD and φN be the characteristic functions of the Dirichlet and the Neumann problem for T 0 . Then φD = φN =

1 (p) l0

1 (p) l0

(p)

(0)

(p)

(0)

R2np (·, 1)φD + R2np (·, 0)φN , (p)

(0)

(p)

(4.2.12) (0)

R2np −1 (·, 1)φD + R2np −1 (·, 0)φN .

(4.2.13)

Proof. Without loss of generality we may assume that p = 1. We first introduce a tree Tb which is obtained from the tree T by choosing v1 as the root of Tb. To fix notation, we put ebj = ej for j = 1, . . . , g but with eb1 directed from v0 to v1 . Then, in the notation of Section 4.1, vb0 = v1 , vb1 = v0 , vbj = vj for j = 1, . . . , g. (1) (1) (1) (1) Furthermore, ˆl0 = ln0 and ˆln0 = l0 . The amplitudes of the displacements (j) (j) (1) (1) are u bk = uk (j = 2, . . . , g; k = 0, . . . , nj + 1), whereas u bk = un1 +1−k (k = 0, . . . , n1 + 1) due to the reversal of the orientation of e1 . Clearly, Ibj = Ij for j = 2, . . . , g, Ib1 = I0 = {1}, Ib0 = I1 , Ib00 = I00 \ I1 , d(b v0 ) = d(v1 ). Observe that the Dirichlet and Neumann boundary conditions for T are given (1) (1) by un1 +1 = c0 un1 with c0 = 0 for the Dirichlet problem and c0 = 1 for the Neumann problem. Hence, most of the following reasoning will be the same for the Dirichlet and the Neumann case, and we briefly write Φ for the matrix functions ΦD , ΦN given by (4.1.23) and φ for the characteristic functions φD and φN given by (4.1.24), (4.1.25) with a generic value for c0 . In particular, we will write Lj (z) N for the functionals LD j (z) and Lj (z) (j = 1, . . . , 2g). Since this notation is only used in this proof, no confusion with the definition of Lj (z) in Section 4.1 should arise. (j) (j) (j) Introducing the notation u b2j−1 := u b0 , u b2j := u b1 − u b0 (j = 1, . . . , g) b = (b (see (4.1.17)), and U uj )2g j=1 , we are going to express the linear functionals associated with the tree T in terms of the linear functionals associated with the tree Tb. Since the indexing of the edges does not influence the characteristic functions (see Proposition 4.1.1), we will assume now that I1 = {1, . . . , d(v1 )}. Clearly, (4.1.11) for j ∈ Ib00 as well as (4.1.12), (4.1.13) for j = 2, . . . , g are identical for both trees, so that b (j = d(b L2j (x)U = Lb2j (z)U v0 ) + 1, . . . , g), b b L2j−1 (x)U = L2j−1 (z)U (j = 2, . . . , g).

(4.2.14) (4.2.15) (1)

(j)

The conditions (4.1.11) for T and j ∈ I1 can be written in the form u bn1 +1 =

u bnj +1 for j = 2, . . . , d(b v0 ), whereas (4.1.12) for j = 1 gives (r) (r) X u bnr +1 − u bnr = 0, (r) b lnr

r∈Ib0

Chapter 4. Trees

160

which means that the condition at the root of Tb is a Neumann condition. We define τ : {1, . . . , d(b v0 )} → Ib0 = {1, . . . , d(b v0 )} by τ (j) = d(b v0 ) + 1 − j. Therefore it follows by Proposition 4.1.2, (4.1.21), (4.1.16) and (4.1.17) that (j−1) (j) N b =u Lb2j bnj−1 +1 (j = 2, . . . , d(b (z)U v0 )), bnj +1 − u r r X u bnr +1 − u bnr b= Lb2N (z)U . (r) b lnr r∈Ib 0

On the other hand, by (4.1.18), (4.1.16) and (4.1.17), (j)

(1)

L2j (z)U = unj +1 − u0

(j = 2, . . . , d(b v0 )),

which gives (j) (j−1) bN b (L2j (z) − L2(j−1) )U = unj +1 − unj−1 +1 = L v0 )), (4.2.16) 2j (z)U (j = 3, . . . , d(b

whereas (2) (1) (2) (1) b 4 (z)U b. L4 (z)U = un2 +1 − u0 = u bn2 +1 − u bn1 +1 = L

(4.2.17)

Next we find in view of (4.1.19), (4.1.16) and (4.1.17) that L1 (z)U = −

=

u2 (1)

(r) X u(r) nr +1 − unr

+

l0

(r)

lnr

r∈I1+

(r) (r) X u bnr +1 − u bnr (r) b lnr

r∈Ib0

b = LbN 2 (z)U .

(4.2.18)

The functional (4.1.20) for Tb and j = 1 gives b = (1 − c0 )b Lb1 (z)U u1 − c0 u b2 . Then, observing (4.1.16) and (4.1.17), the Dirichlet condition (4.1.21) or the Neumann condition (4.1.22) at the root of T give (1)

(1)

L2 (z)U = (1 − c0 )un1 +1 + c0 = (1 − c0 )b u1 − =

1 − c0 +

c0 (1)

ln1 !

c0

(1)

ln1

(1)

un1 +1 − un1 (1)

ln1

u b2 b. Lb1 (z)U

(4.2.19)

4.2. Complementary subtrees

161

The linear combinations of the functionals on the left-hand side of (4.2.16) are described by a 2g × 2g matrix Ψ with det Ψ = 1, whereas the transformations b with (4.2.18) and (4.2.19) are described by a 2g × 2g matrix Ψ b = 1 − c0 + c0 . − det Ψ (1) ln1 From (4.2.14)–(4.2.19) it follows that bΦ b τN (z)U b. ΨΦ(z)U = Ψ

(4.2.20)

Furthermore,  b= U

 Ψ1 (z) 0 U, 0 I2g−2

(4.2.21)

where I2g−2 is the (2g − 2) × (2g − 2) identity matrix. Therefore, b b τ (z)))(det(Ψ1 (z))). det(Φ(z)) = (det Ψ)(det( Φ N From

    u b1 u = Ψ1 (z) 1 , u b2 u2 (1)

(1)

(1)

u b1 = un1 +1 , u b2 = un1 − un1 +1 , (4.1.16) and (4.1.17) it follows that ! (1) (1) R2n1 (z, 1) R2n1 (z, 0) Ψ1 (z) = . (1) (1) (1) (1) −ln1 R2n1 −1 (z, 1) −ln1 R2n1 −1 (z, 0) Hence

(1)

det Ψ1 (z) = −

ln1

(1)

l0

by the Lagrange identity (1.2.47). Therefore, (1)

(1 − c0 )ln1 + c0

(1)

(1 − c0 )ln1 + c0 b φN (z). (1) (1) l0 l0 (4.2.22) (1) (1) Denoting the characteristic functions for the string eb1 by φbD (·, c0 ) and φbN (·, c0 ) and observing that eb1 and T (0) are complementary subtrees of Tb with common vertex vb0 , it follows from Theorem 4.2.1 and Remark 4.2.3 that φ(z) = det(Φ(z)) =

b τN (z)) = det(Φ

(1) (0) (1) (0) φbN = φbN (·, c0 )φD + φbD (·, c0 )φN (1)

(0)

(1)

(0)

b b =R 2n1 −1 (·, c0 )φD + R2n1 (·, c0 )φN ,

(4.2.23)

b(1) are the Cauer-Fry polynomials for the string eb1 . The relations where the R k between the Cauer-Fry polynomials of a string for opposite orientations stated in Proposition 1.2.12 immediately leads to (4.2.12) and (4.2.13) for p = 1. 

Chapter 4. Trees

162

Remark 4.2.5. The star graph considered in Section 2.3 is a particular case of the tree Tb introduced in the proof of Proposition 4.2.4. With the obvious identification of notation in this section and in Section 2.3 it is clear that (2.3.8) is a particular case of (4.2.23). Lemma 4.2.6. Let the assumptions and notation be as in Proposition 4.2.4. Then φD (z) = ln(p) + p φN (z)

1 (p) −mnp z

.

1

+

1

(p)

lnp −1 +

(p) −mnp −1 z

1

+ ··· + (p) −m1 z

1

+

(0)

(p)

l0 +

φD (z) (0)

φN (z) (4.2.24)

Proof. This proof is similar to the proof of (2.3.12). Defining Ψk =

b(p) (·, 0) R 2k−1 b(p) (·, 1) R 2k−1

! b(p) (·, 0) R 2k b(p) (·, 1) R

(k = 0, . . . , np ),

2k

it follows in view of Lemma 1.2.4 that Ψk = Ψk−1 Θk

(k = 1, . . . , np ),

where Θk (z) =

(p) b lk

1 (p)

!

(p) (p) 1−b lk m bk z

−m bk z

(k = 1, . . . , np , z ∈ C).

Hence (p) l0



φD φN



 =

(p)

lnp 0

 0 Ψnp 1

(0)

φD

(0)

φN

!

 =

(p)

lnp 0

 0 Ψ 0 Θ1 · · · Θ ng 1

(0)

φD

(0)

! (4.2.25)

φN

by (4.2.22) and (4.2.23) with edge index p. Observing that 

(p)

lnp 0

  0 1 Ψ0 = 1 0

(p)

l np 1

 ,

the representation (4.2.24) now follows by a recursive application of (A.4.1), observing that m b k = mnp +1−k and b lk = lnp −k . 

4.2. Complementary subtrees

163

Theorem 4.2.7. Let T be a tree with g ≥ 2 edges such that T is directed towards its root. Let n be the number of beads on T and assume that the numbers cj in the boundary conditions (4.1.13) at the pendant vertices vj (j = 1, . . . , g with Ij+ = ∅) satisfy 0 ≤ cj ≤ 1. Then the characteristic functions φD and φN of the tree defined in (4.1.24) and (4.1.25) are polynomials of degree n and satisfy φD (z) > 0 for z ∈ (−∞, 0] and φN (z) > 0 for z ∈ (−∞, 0). Furthermore, φN (0) = 0 if and only D if cj = 1 for all pendant vertices vj . The rational function φφN is an S-function. Proof. We are going to prove the statement by induction on the height h of T , that is, the maximal number of edges traversed from any vertex towards the root. If h = 1, then T is a star graph, and Corollary 4.2.2 and Remark 4.2.3 show that φD =

g Y

(j)

R2nj (·, cj ),

j=1

φN =

g X j=1

(j)

R2nj −1 (·, cj )

g Y

(k)

R2nk (·, ck ).

k=1, k6=j

The statements of this theorem, except for the last one, now immediately follow in this case h = 1 from Remark 1.2.7, part 4, and Propositions 1.2.8–1.2.11. Observing that ! (j) g X R2nj −1 (·, cj ) φN − = − (j) , φD R2nj (·, cj ) j=1 it follows from (1.2.19) and Lemma A.2.4 that each summand on the right-hand side is a rational Nevanlinna function. Hence the right-hand side and therefore also the left-hand side is a rational Nevanlinna function, and by Lemma A.2.4 and what we already have shown the proof in the case h = 1 is complete. Now assume that h > 1 and that the statement of this theorem is true for trees of height less than h. First let v0 be a pendant vertex. In the notation of Proposition 4.2.4 with p = 1, T 0 is a tree of height h − 1 since each path from vj (j = 2, . . . , g) to v0 has to pass through v1 = vb0 and is therefore shorter by one edge than the path from vj to v0 . In view of (4.2.12) and (4.2.13) the statement of this theorem follows with a reasoning similar to that for the case h = 1 if we take note of the following. The pendant vertices which are not the root are the same for (1) (0) T and T 0 . Hence φN (0) = R2n1 −1 (0, 0)φN (0) is 0 by induction hypothesis if and only if all cj = 1 for the pendant vertices vj of T 0 , and hence of T , which are not (0)

the root. By induction hypothesis, the rational function

φD

(0)

φN

is an S-function and

has therefore a representation of the form (A.3.8). Then, by Lemma 4.2.6, also

φD φN

D is a rational S-function. has the form (A.3.8), and Lemma A.3.6 shows that φφN Finally, if the root of T is not a pendant vertex, then the statements of this theorem hold for T in view of Corollary 4.2.2, Remark 4.2.3, part 1, and the fact

Chapter 4. Trees

164

that the theorem has already been proved for trees of height h whose roots are pendant vertices. In particular, ! g (j) X φN (·, cj ) φN − = − (j) φD φ (·, cj ) j=1 D

is a rational S-function.

4.3



The direct spectral problem on a tree

In this section, let T be a rooted tree with g ≥ 2 edges as considered in Section 4.1 where each of its pendant vertices which is not the root is considered with Dirichlet conditions. We will use the notation introduced in Sections 4.1 and 4.2. Recall that n is the number of beads on the tree. Theorem 4.3.1. All zeros of φD and φN are real, and the nondecreasing sequence (νk )nk=1 of zeros of φD interlaces with the nondecreasing sequence (µk )nk=1 of zeros of φN : 0 < µ1 ≤ ν1 ≤ µ2 ≤ ν2 ≤ · · · ≤ νn . Proof. By Theorem 4.2.7, φD and φN are polynomials of degree n, φN (0) 6= 0 and ΦD ΦD ΦN is an S-function. Hence the poles and zeros of ΦN are positive real numbers and interlace strictly by Corollary A.3.3. Observing that φD and φN may have common zeros, an application of Lemma A.4.4 completes the proof.  Theorem 4.3.2. Let z ∈ C and denote by nD (z) the multiplicity of z as a zero of φD and by nN (z) the multiplicity of z as a zero of φN . Let κ be the number of interior vertices of the tree T which are not the root. Then 1. |nN (z) − nD (z)| ≤ 1; 2. nD (z) ≤ g − d(v0 ) − κ if nD (z) ≤ nN (z); 3. nD (z) ≤ g − κ if nD (z) > nN (z). Proof. The estimate in part 1 immediately follows from Theorem 4.3.1. The statements 2 and 3 will be proved by induction on g. Since we might have to apply the induction hypothesis to the case g = 1 and since the case g = 1 is easy, we take the induction base to be g = 1. In this case, κ = 0, d(v0 ) = 1, and Remark 4.2.3 gives φD = R2n (·, 0) and φN = R2n−1 (·, 0). But the zeros of these two polynomials are simple and distinct (see (1.2.36)), so that nN (z) ≥ nD (z) is only possible if nD (z) = 0, and in this case nD (z) = 0 = g − d(v0 ) − κ, whereas nD (z) = 1 = g − κ if nD (z) > nN (z). Now let g ≥ 2. Our induction hypothesis is that statements 2 and 3 are true for any tree with at most g − 1 edges. First assume that the root is an interior vertex. Then T can be cut into two complementary subtrees T1 and T2 , as considered in Theorem 4.2.1. Denoting by gj the number of edges of Tj , by d(j) (v0 )

4.3. The direct spectral problem on a tree

165

the number of edges of Tj incident with v0 and by κj the number of interior edges of Tj distinct from the root v0 (j = 1, 2), it is clear that g = g1 + g2 , κ = κ1 + κ2 , (j) and d(v0 ) = d1 (v0 ) + d2 (v0 ). Also, let nD (z) be the multiplicity of z as a zero (j) (j) (j) of φD and let nN (z) be the multiplicity of z as a zero of φN . Without loss (2) (1) (1) (2) of generality we may assume that nN (z) − nD (z) ≥ nN (z) − nD (z). Clearly, (1) (2) (4.2.7) shows that nD (z) = nD (z) + nD (z). Furthermore, the right-hand side of (4.2.8) is a sum of two polynomials for which the multiplicity of the zero z is (1) (1) (1) (2) (1) (1) (2) (1) nN (z) + nD (z) = nD (z) + nD (z) + nN (z) − nD (z) = nD (z) + nN (z) − nD (z) (2) (2) and nD (z) + nN (z) − nD (z), respectively. If these two numbers are distinct, then nN (z) equals the minimum of these two numbers, whereas nN (z) may be larger otherwise. Taking statement 1 into account, we have to consider the following 9 cases: Case (1) (1) nN (z) − nD (z) (2) (2) nN (z) − nD (z) nN (z) − nD (z)

I 1 1 1

II 1 0 0

III 1 −1 −1

IV 0 0 1

V 0 0 0

VI 0 −1 −1

VII −1 −1 1

VIII −1 −1 0

IX −1 −1 −1

In cases I, II, IV and V we have (2)

(1)

nD (z) = nD (z) + nD (z) ≤ (g1 − d(1) (v0 ) − κ1 ) + (g2 − d(2) (v0 ) − κ2 ) = g − d(v0 ) − κ, which proves statement 2 for these cases. In cases III, VI and IX we have (1)

(2)

nD (z) = nD (z) + nD (z) ≤ (g1 − κ1 ) + (g2 − κ2 ) = g − κ, which proves statement 3. This part of the proof will be complete if we show that cases VII and VIII cannot occur. Indeed, by (4.2.7) and (4.2.8), (1)

−

(2)

φ φ φN − N , =− N (1) (2) φD φD φD

where z is a pole of both summands on the right-hand side, and these summands are rational Nevanlinna functions by Theorem 4.2.7 and Lemma A.2.4. Since rational Nevanlinna functions are locally increasing on the real axis by Lemma A.2.5, these poles cannot cancel, and hence z is a pole of the left-hand side, which means that Cases VII and VIII are impossible. Finally, let v0 be a pendant vertex of T . With the notation from Proposition 4.2.4 and from (4.2.25) we know that !    (p)  (0) φD (p) φD lnp 0 l0 = Ψnp (0) . φN 0 1 φN

Chapter 4. Trees

166

Therefore, if z is a zero of multiplicity at least ν (we may include ν = 0) of both (0) (0) φD and φN , then z is also a zero of multiplicity at least ν of both φD and φN , and vice versa because w 7→ (Ψnp (w))−1 is a matrix polynomial as the determinant of Ψnp is constant by the Lagrange identity. It follows that (0)

(0)

min{nD (z), nN (z)} = min{nD (z), nN (z)}. Taking statement 1 into account, we have to consider the following 9 cases, with the obvious notation for the numbers associated with the tree T 0 : Case (0) (0) nN (z) − nD (z) nN (z) − nD (z)

I 1 1

II 1 0

III 1 −1

IV 0 1

V 0 0

VI 0 −1

VII −1 1

VIII −1 0

IX −1 −1

We observe that g (0) = g − 1 since one edge was removed, κ (0) = κ − 1 since one (0) interior vertex has become the new root, and d(v0 ) ≥ 1 = d(v0 ). In cases I, II, IV and V, (0)

(0)

(0)

nD (z) = min{nD (z), nN (z)} = min{nD (z), nN (z)} = nD (z), and therefore (0)

(0)

nD (z) = nD (z) ≤ g (0) − d(v0 ) − κ (0) ≤ g − d(v0 ) − κ, which proves statement 2 for these cases. In cases III, VI and IX we have (0)

(0)

nD (z) = min{nD (z), nN (z)} + 1 = min{nD (z), nN (z)} + 1. For cases III and VI this leads to (0)

(0)

nD (z) = nD (z) + 1 ≤ g (0) − d(v0 ) − κ (0) + 1 ≤ g − κ, whereas in case IX we get (0)

nD (z) = nD (z) ≤ g (0) − κ (0) = g − κ. Thus statement 2 in cases III and VI and statement 3 in case IX has been proved. Finally, statement 2 in cases VII and VIII follows from (0)

(0)

(0)

nD (z) = min{nD (z), nN (z)} = min{nD (z), nN (z)} = nD (z) − 1 ≤ g (0) − κ (0) − 1 = g − κ − 1 = g − d(v0 ) − 1. Proposition 4.3.3. Under the assumptions of this section, the number depends on the shape of the rooted tree and the lengths of its edges.

 φN (0) φD (0)

only

4.3. The direct spectral problem on a tree

167

Proof. We are going to prove this result by induction on g, the number of edges of T . Let l(j) be the length of the edge ej (j = 1, . . . , g). For g = 1 it follows from Remark 4.2.3, part 2, (1.2.25) and (1.2.28) that φD (0) = l(1) . φN (0)

(4.3.1)

Now let g > 1 and assume that the statement is true for trees with fewer than g edges. First consider the case that v0 is an interior vertex. Then Corollary 4.2.2 gives d(v0 ) (j) X φ (0) φN (0) N = , (4.3.2) (j) φD (0) j=1 φD (0) where each of the summands on the right-hand side satisfies the statement by induction hypothesis. Hence the statement of this proposition holds in this case. Finally, if v0 is a pendant vertex, then Proposition 4.2.4 and (1.2.25), (1.2.28), (1.2.30), (1.2.31) show that (0)

φ (0) φD (0) = l(p) + D , (0) φN (0) φN (0)

(4.3.3)

which completes the proof since the tree T 0 satisfies the induction hypothesis.  (0) is completely determined by the Remark 4.3.4. Proposition 4.3.3 shows that φφN D (0) shape of the tree, the lengths of its edges and the location of the root v. We call this number the shape characteristic of the rooted tree T and denote it by χT,v . The recursions (4.3.1), (4.3.2) and (4.3.3) give a constructive way to find the shape characteristic of a rooted tree T . Cutting T at the root into d(v0 ) subtrees Tj (j ∈ I0 ) such that ej is an edge of Tj , v0 becomes a pendant vertex of Tj . In case Ij+ 6= ∅, let Tj0 be the tree with root vj obtained by removing the vertex v0 and the edge ej from Tj . Then it follows from (4.3.2), (4.3.3) and (4.3.1) that X X X 1 1 χT,v0 = χTj ,v0 = + . (4.3.4) (r) 1 l r∈I0 r∈I0 r∈I0 l(r) + χTr0 ,vr I + (vr )6=∅ I + (vr )=∅

As an example, consider a metric star graph with central vertex v0 . If v0 is the root, then g X 1 χT,v0 = . (r) l r=1 If, however, the pendant vertex v1 is the root, then χT,v1 =

1 l(1) + χ

1 0 ,v T1 0

=

1 l(1) +

1 g X r=2

. 1

l(r)

Chapter 4. Trees

168

4.4 The inverse problem for trees of Stieltjes strings: strict interlacing of the characteristic values Here we give a positive answer to the following question: given two sequences of interlacing real positive numbers, a rooted metric tree of a prescribed shape with prescribed lengths of edges, does there exist a distribution of beads on the tree such that the two sequences are Neumann and Dirichlet characteristic values for this tree? n Theorem 4.4.1. Let n be a nonnegative integer, let (µk )nk=1 and (νk )k=1 be sequences of real numbers which interlace:

0 < µ1 < ν1 < · · · < µn < νn . Let T be a tree with g ≥ 2 edges rooted at a vertex v0 and let positive numbers l(j) (j = 1, . . . g) be given. Then there exist nonnegative integers nj (j = 1, . . . , g) with g P (j) nj (j) nj and (lk )k=0 (j = 1, . . . , g) of positive numbers nj = n, sequences (mk )k=1

j=1

with

nj P k=0

(j)

lk

= l(j) , and for each j = 1, . . . , g with Ij+ 6= ∅ there is a set Λj of

positive numbers with #Λj ≤ n such that the following holds. If l(j) ∈ / Λj for j = 1, . . . , g with Ij+ 6= ∅, then the sequence of the characteristic values of the Dirichlet problem (4.1.10)–(4.1.14) for T with Dirichlet conditions, i.e., cj = 0, at the pendant vertices vj which are not the root coincides with (νk )nk=1 , and the sequence of the characteristic values of the Neumann problem (4.1.10)–(4.1.13), (4.1.15) for T with Dirichlet conditions, i.e., cj = 0, at the pendant vertices vj which are not the root coincides with (µk )nk=1 . Proof. We are going to prove the statement by induction on g. It is convenient to include the case g = 1 since trees with one edge may occur in the induction step of the proof. But for g = 1 the statement of this theorem is the existence statement of Theorem 1.3.1. We note that here we do not need any restrictions on l(1) , which implies that in the induction step no restrictions on l(j) are needed when Ij+ = ∅. Now assume that g ≥ 2 and that the statement of this theorem is true for every tree with fewer than g edges. Consider the rational function φ defined by z n 1− Y µk φ(z) = χT,v0 z , k=1 1 − νk

(4.4.1)

where the shape characteristic χT,v0 > 0 of the tree is known because the shape of T as well as the lengths of its edges are given (see Proposition 4.3.3 and Remark 4.3.4). We will construct a distribution of beads on the interior of the edges of the tree T such that the function φ defined in (4.4.1) equals φφN for our given D graph and the constructed distribution of beads. By Corollary A.3.3, the function

4.4. Inverse problem: strict interlacing

169

1 φ

belongs to S0 and therefore −φ is a Nevanlinna function by Lemma A.2.4. In view of Lemma A.2.7 it follows that n X

φ(z) =

k=1

where Ak > 0 and B = χT,v0 +

n P k=1

Ak + B, z − νk

(4.4.2)

Ak νk .

First assume that d(v0 ) > 1. Then we cut the tree into d(v0 ) complementary subtrees Tι which are rooted at v0 (ι = 1, . . . , d(v0 )). Denote by I (ι) the set of those indices j ∈ {1, . . . , g} for which ej is an edge of Tι (ι = 1, . . . , d(v0 )) and d(v P0 ) choose nonnegative integers Nι (ι = 1, . . . , d(v0 )) such that Nι = n. We ι=1

arrange the set {1, . . . , n} in an arbitrary manner as the union of disjoint sets (ι) {kj : j = 1, . . . , Nι } (ι = 1, . . . , d(v0 )). Put φι (z) :=

Nι X

Ak(ι)

j=1

z − νk(ι)

j

+ Bι

(ι = 1, . . . , d(v0 )),

(4.4.3)

j

where Bι := χTι ,v0 +

Nι A (ι) X kj j=1

Recall from (4.3.2) that χT,v0 =

d(v P0 )

νk(ι)

(ι = 1, . . . , d(v0 )).

j

χTι ,v0 . Each summand on the right-hand side

ι=1

of (4.4.3) is the negative of a Nevanlinna function (see Theorem A.2.6), so that for each ι = 1, . . . , d(v0 ), −φι is a Nevanlinna function with Nι simple positive real poles. Furthermore, Bι > 0, and therefore φι has also Nι zeros, which are real and interlace with its poles. It follows that φ1ι ∈ S0 . In view of φι (0) = χTι ,v0 > 0 and lim φι (z) = Bι > 0, all zeros of φι are positive. Hence, by induction hypothesis z→−∞

(j) n

(j) n

j j there exist sequences of positive numbers (mk )k=1 (j = 1, . . . , g) and (lk )k=0 n j P (j) such that lk = l(j) , the sequence of the characteristic values of the Dirichlet

k=0

problem (4.1.10)–(4.1.14) for Tι coincides with the poles of φι , and the sequence of the characteristic values of the Neumann problem (4.1.10)–(4.1.13), (4.1.15) (ι)

for Tι coincides with the zeros of φι . The functions φι and

φN

(simple) zeros and poles, and together with φι (0) = χTι ,v0 = that φι =

(ι) φN (ι) φD

(ι = 1, . . . , d(v0 )). Hence φ =

φN φD

(ι)

φD

have the same

(ι) φN (0) (ι) φD (0)

this means

by Corollary 4.2.2, and therefore

the generated problem for T has the requested characteristic values.

Chapter 4. Trees

170

1 φ

Now consider the case that v0 is a pendant vertex of T . Since function, Lemma A.3.5 gives a continued fraction expansion 1 = a0 + φ(z)

1

.

1

−b1 z +

is an S0 -

(4.4.4)

1

a1 +

1

−b2 z + · · · +

−bn z +

1 an

Without loss of generality we may assume that e1 is incident with v0 . With the notation from Proposition 4.2.4 and in view of (4.3.3) we have n

1

0 < l(1) < l(1) +

= χT,v0 =

χT 0 ,v1

Letting Λ1 be the set of the numbers

X 1 = ak . φ(0) k=0

j P

ak (j = 0, . . . , n − 1), it follows that there

k=0

is a number n1 ∈ {0, . . . , n} such that nX 1 −1

ak < l(1)
0 (j = 1, . . . , g). In our constructions below we may encounter degenerate cases, and therefore we only require that G has at least one vertex. In particular, it will be allowed that the graph has no edges. Each edge ej is a Stieltjes string of the total length l(j) bearing nj beads. (j) The beads on ej have the masses mk > 0 (k = 1, . . . , nj ) and divide the edge (j) into threads of lengths lk > 0 (k = 0, . . . , nj ), where the lower index is chosen in such a way that it increases in the direction of the edge. The amplitudes of the vibrations of the beads and the ends of the strings are subject to the equations (1.2.4), and with the indexing as in (4.1.10) these (j) (j) amplitudes will be denoted by u0 , . . . , unj +1 on the edges ej (j = 1, . . . , g). We recall from Appendix C that Vb denotes the set of all vertices of G which are not isolated. Each vertex v ∈ Vb may be clamped or freely moving. The set of clamped vertices is denoted by V c and the set of freely moving vertices is denoted by V f . For each interior vertex v we require the continuity condition that all values (j) (j) u0 with j ∈ I − (v) and unj +1 with j ∈ I + (v) are equal. Defining (j)

(j)

U (v) := {u0 : j ∈ I − (v)} ∪ {unj +1 : j ∈ I + (v)}, this condition can be written as #U (v) = 1

(v ∈ V int ).

(5.1.1)

Hence each vertex v ∈ Vb has a unique amplitude, which will be denoted by u(v). © Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4_5

173

Chapter 5. Spectral problems on graphs

174

If v ∈ V c , then u(v) = 0, and we arrive at (j)

(j)

u0 = 0 (j ∈ I − (v)),

unj +1 = 0 (j ∈ I + (v)).

(5.1.2)

For an interior vertex v, the condition (5.1.2) is called the generalized Dirichlet condition at v. If, however, v ∈ V f , then (j)

(j)

(j)

X

u1 − u0

j∈I − (v)

l0

(j)

=

(j)

X

unj +1 − unj

j∈I + (v)

lnj

(5.1.3)

(j)

is imposed, which is the balance of forces condition when the vertex v is an interior vertex. We will call the pairs of conditions (5.1.1), (5.1.3) generalized Neumann conditions at the interior vertex v. For pendant vertices the conditions (5.1.2) and (5.1.3) are the usual Dirichlet and Neumann conditions, respectively. In this chapter we will consider the problem (4.1.10), (5.1.1), (5.1.2), (5.1.3), i.e., (4.1.10) with (generalized) Dirichlet conditions at each v ∈ V c and with (generalized) Neumann conditions at each v ∈ V f . We are now going to write the problem (4.1.10), (5.1.1) (5.1.2), (5.1.3) as a matrix eigenvalue problem (A − zM )Y = 0.

(5.1.4)

For convenience, vk (k = 1, . . . , ng+1 ) is an arbitrary indexing of the freely moving (g+1) vertices in V with ng+1 := #V f and uk := u(vk ) (k = 1, . . . , ng+1 ) is the (j) amplitude of the vibration at vk . Then Y will be written as Y = (uk )j=1,...,g+1 . k=1,...,nj

Although the indices (j, k) may be arranged in an arbitrary manner, for definiteness we will use lexicographic order, that is, (j, k) < (r, s) if and only if j < r or j = r and k < s. Denoting by n=

g X

nj

j=1

the total number of beads on the graph G, we formally allow that n = 0. Of course, also ng+1 = 0 is possible. In order for Y to have at least one component we will require that n + ng+1 > 0. (j)

We observe that (5.1.1) and (5.1.2) mean that the values of u0 (g+1)

(j)

and unj +1

are either 0 or some uk . Therefore the solutions of (4.1.10), (5.1.1), (5.1.2), (5.1.3) are uniquely determined by (4.1.10) and (5.1.3) for Y . Hence A and M will be (n + ng+1 ) × (n + ng+1 ) matrices. The component of the vector (A − zM )Y

5.1. The problem and its spectrum

175

with index (j, k) (j = 1, . . . , g + 1, k = 1, . . . , nj ) will be (j)

(j)

uk − uk−1 (j)

(j)

−

(j)

uk+1 − uk (j)

lk−1

lk

(j)

(j)

− mk zuk

(j = 1, . . . , g, nj ≥ 1, k = 1, . . . , nj ), (5.1.5)

X u(g+1) − k (ι) l nι ι∈I + (vk )

(ι) unι

(ι) u1

X

−

ι∈I − (vk )

(g+1) − uk (ι) l0

(j = g + 1, k = 1, . . . , ng+1 ). (5.1.6)

(j)

(j)

(g+1)

Of course, for u0 and unj +1 the appropriate value of us for freely moving ends and the value 0 for clamped ends has to be substituted in (5.1.5) and also in (5.1.6) if nr = 0 for some r = 1, . . . , g. Definition 5.1.1. The problem (4.1.10), (5.1.1) (5.1.2), (5.1.3) on the graph G is called degenerate if G has a component G1 with at least one edge such that no edge of G1 carries a bead and such that all vertices of G1 are freely moving. Theorem 5.1.2. Let n + ng+1 > 0. Then the matrix A is symmetric and positive semidefinite and the matrix M is a positive semidefinite diagonal matrix. The matrix A − zM is singular for all z ∈ C if and only if the problem is degenerate. If the problem is not degenerate, then the eigenvalue problem z 7→ (A − zM ) has n eigenvalues, counted with multiplicity, all eigenvalues are real and nonnegative, and the geometric multiplicity of each eigenvalue coincides with its algebraic multiplicity. The number 0 is an eigenvalue if and only if there is a component of G with at least one edge such that each vertex of this component is freely moving. The multiplicity of the eigenvalue 0 is the number of components with at least one edge in which each vertex is freely moving. Proof. We will first show that the matrix A is symmetric. We will denote it by A =: (aj,k,r,s ), where (j, k) is the row index and (r, s) is the column index. To show the symmetry of A we have to verify that aj,k,r,s = ar,s,j,k for all (j, k) < (r, s). We will consider several cases: Case 1: j < r ≤ g. In this case it is clear that (5.1.5) with index (j, k) does (r) (j) not contain us and (5.1.5) with index (r, s) does not contain uk . Therefore aj,k,r,s = 0 = ar,s,j,k . Case 2: j = r ≤ g and s ≥ k + 2. Here it is obvious from (5.1.5) that aj,k,j,s = 0 = aj,s,j,k . Case 3: j = r ≤ g and s = k + 1. It is clear by (5.1.5) that aj,k,j,k+1 = −

1 (j) lk

= aj,k+1,j,k .

Chapter 5. Spectral problems on graphs

176

(j)

(g+1)

Case 4: j ≤ g, r = g + 1, k = 1, j ∈ I − (vs ). Then us follows in view of (5.1.5) and (5.1.6) that aj,k,g+1,s = −

1 (j) l0

= u0

= ag+1,s,j,k . (g+1)

Case 5: j ≤ g, r = g + 1, k = nj , j ∈ I + (vs ). Then us it follows in view of (5.1.5) and (5.1.6) that aj,k,g+1,s = −

1 (j) lnj

and therefore it

(j)

= unj +1 and therefore

= ag+1,s,j,k .

Case 6: j ≤ g, r = g + 1, and k and s are different from the cases 4 and 5. Then it follows in view of (5.1.5) and (5.1.6) that aj,k,g+1,s = 0 = ag+1,s,j,k . Case 7: j = r = g + 1. For l, m ∈ {1, . . . , ng+1 } let b m) = {ι ∈ {1, . . . , g} : nι = 0, ι ∈ I + (vl ) ∩ I − (vm )}. I(l, Then ag+1,k,g+1,s = −

1

X

(ι) l0 b b ι∈I(k,s)∪ I(s,k)

= ag+1,s,g+1,k .

Thus we have shown that A is a symmetric matrix, and we have also explicitly calculated the off-diagonal entries of A. The diagonal elements of A are easily obtained from (5.1.5) and (5.1.6), namely, aj,k,j,k = ag+1,k,g+1,k =

1

+

(j) lk−1

1

(j = 1, . . . , g, k = 1, . . . , nj ),

(j) lk

1

X

(ι) lnι b ι∈I + (vk )\I(k,k)

+

1

X

(ι) l0 b ι∈I − (vk )\I(k,k)

(k = 1, . . . , ng+1 ).

To show that A is positive semidefinite we observe that the definition of AY in (5.1.5) and (5.1.6) gives Y ∗ AY = S1 + S2 , where S1 =

nj g X X

(j)

(j) uk

j=1 k=1

S2 =

k=1

(g+1)

uk

−

uk+1 − uk (j)

lk−1



ng+1

X

(j)



uk

,

(ι)

− unι

(ι)

ι∈I + (vk )

!

lk (g+1)

X

(j)

(j)

(j)

uk − uk−1

lnι

(ι)

−

X

(g+1)

u1 − uk (ι)

ι∈I − (vk )

l0

 .

5.1. The problem and its spectrum

177

We split S1 =: S11 + S12 + S13 with S11 : =

=

g X 1  (j) j=1 l0 nj >0

X

(j)

(j) (j)

|u1 |2 − u1 u0

1

X

ng+1 (j)

|u |2 + (j) 1

v∈V c j∈I − (v) l0 nj >0

S12 : =

=

g X 1  (j) j=1 lnj nj >0

X



X

(j) k=1 j∈I − (vk ) l0 nj >0

(j) (j)

2 |u(j) nj | − unj unj +1

|u(j) |2 + (j) nj

v∈V c j∈I + (v) lnj nj >0

(j)



,



,

(j) (g+1)

|u1 |2 − u1 uk



ng+1

1

X

1 

X

X

1 

X

(j) k=1 j∈I + (vk ) lnj nj >0

(j) (g+1)

2 |u(j) nj | − unj uk

and S13 :=

g nX j −1 X 1

(j)

(j) j=1 k=1 lk

(j)

|uk+1 − uk |2 .

We can now write S11 + S12 + S2 = S21 + S22 , where S21 =

X

X

ng+1

S22 =

1

(j) v∈V c j∈I − (v) l0 nj >0

αj

(j)

|u1 |2 +

(j) |u (j) 1 k=1 j∈I − (vk ) l0

X

X

X

X

1

(j) v∈V c j∈I + (v) lnj nj >0

2 |u(j) nj | ,

ng+1

−

(g+1) 2 uk |

+

X

X

αj

(j) k=1 j∈I + (vk ) lnj

(g+1) 2

|u(j) nj − uk

| ,

where αj = 12 if nj = 0, v− (ej ) ∈ V f and v+ (ej ) ∈ V f , whereas αj = 1 otherwise. This identity is clear for all terms with nj > 0. Now let nj = 0. We have to compare terms in S2 and S22 . If both v− (ej ) and v+ (ej ) are clamped vertices, then this edge ej does not contribute to S2 and S22 . If v+ (ej ) belongs to V f , say v+ (ej ) = vk , then j ∈ I + (vk ) gives the term (g+1)

(g+1)

uk

uk

− uj0

(j)

l0

in S2 . Similarly, if v− (ej ) belongs to V f , say v− (ej ) = vl , then j ∈ I − (vl ) gives the term (g+1) j (g+1) u1 − ul −ul (j) l0

Chapter 5. Spectral problems on graphs

178

in S2 . We are going to consider three cases. (j) If v+ (ej ) ∈ V f and v− (ej ) ∈ V c , then u0 = 0, and the contribution from (g+1) 2 (j) (g+1) 2 1 1 j in S2 is (j) |uk | , whereas the contribution in S22 is (j) |u0 − uk | = 1

l0 (g+1) 2

(j) |uk

l0

| . Therefore, the terms in S2 and S22 for such j are the same. Similarly,

l0

(j)

if v+ (ej ) ∈ V c and v− (ej ) ∈ V f , then u1 (g+1) 2 1 | in S2 as well as S22 . (j) |ul

= 0, and j gives the contribution

l0

Finally, if v+ (ej ) ∈ V f and v− (ej ) ∈ V f , then the contribution of j in S2 is 1 h (j)

l0

(g+1)

uk

(g+1)

(uk

(g+1)

− ul

(g+1)

) − ul

(g+1)

(uk

(g+1)

− ul

2 i 1 (g+1) (g+1) ) = (j) uk − ul , l0

which coincides with the contribution in S22 . From S13 ≥ 0, S21 ≥ 0, S22 ≥ 0 and Y ∗ AY = S13 + S21 + S22 it follows that A is positive semidefinite. (j) Clearly, M is a diagonal matrix with positive diagonal entries mk in the row and column with index (j, k) (j = 1, . . . , g, k = 1, . . . , nj ), whereas all other diagonal entries are 0. Now assume that the problem is degenerate and let G1 be as in Definition (g+1) 5.1.1. Define the vector Y by setting uk = 1 for all k = 1, . . . , ng+1 with (j) G1 vk ∈ V and by setting uk = 0 for all other components. Then clearly S13 = 0 (g+1) and S21 = 0. For the terms in S22 we observe the following. If uk = 0 in the (j) (g+1) 2 (j) (g+1) term |u1 − uk | and if nj > 0, then also u1 = 0. If uk = 0 in the term (j) (g+1) 2 (j) (j) |u1 − uk | and if nj = 0, then u1 = unj +1 is the amplitude at v+ (ej ). But since vk is not a vertex of the component G1 , also v+ (ej ) is not a vertex of G1 since (j) vk and v+ (ej ) are ends of the edge ej . Hence u1 = 0. A corresponding reasoning (j) (g+1) 2 (g+1) (j) (g+1) 2 holds for |unj − uk | . If uk = 1 then the term |u1 − uk | and the term (j) (g+1) 2 (g+1) (g+1) 2 |unj − uk | are of the form |ul − uk | with vl ∈ V G1 since vk ∈ V G1 implies that ej is an edge in G1 . Altogether, it follows that all summands in S22 are 0. Therefore we have shown that Y ∗ AY = 0, which implies AY = 0 since A is positive semidefinite. Obviously, M Y = 0, and hence (A − zM )Y = 0. Thus A − zM is a singular matrix for all z ∈ C. For the remainder of this proof let the problem be nondegenerate. If n > 0 and ng+1 = 0, then M is a positive definite diagonal matrix, and by Proposition B.2.1, all eigenvalues are real and nonnegative, and the geometric multiplicity of each eigenvalue coincides with its algebraic multiplicity. Now let ng+1 > 0 and let A22 be the lower right ng+1 × ng+1 submatrix of A. (g+1) ng+1 We are going to show that A22 is  invertible. To this end let Y2 = (uk )k=1 such  0 that Y2∗ A22 Y2 = 0. Putting Y = , it follows that Y ∗ AY = Y2∗ A22 Y2 = 0 and Y2 in particular S22 = 0. Hence each summand in S22 is zero.

5.1. The problem and its spectrum

179

Let k ∈ {1, . . . , ng+1 } and let ej be an edge of G which is incident with vk . (j) (g+1) 2 αj (j) (g+1) 2 α Then at least one of (j)j |u1 − uk | , (j) |unj − uk | is a summand in A22 . l0

lnj

Let G0 be the component of G with vk ∈ V G0 and let V0 be the set of those vertices of G0 which are either clamped vertices or freely moving vertices having at least one incident edge ej with nj > 0. Since G is nondegenerate, V0 6= ∅. If vk ∈ V0 , then there is an incident edge ej with nj > 0, and in view of the definition (j) (j) (g+1) of Y , u1 = 0 and unj = 0, so that also uk = 0. If, however, vk 6∈ V0 , then we choose a path (vk = w0 , ej1 , w1 , . . . , ejl , wl ) in G0 with wl ∈ V0 . Without loss of generality we may assume that wi 6∈ V0 for i < l. By definition of V0 , nji = 0 for i = 1, . . . , l and wi is a freely moving vertex for i = 1, . . . , l − 1. Therefore (j1i ) |u(wi−1 ) − u(wi )|2 occurs in S22 for i = 1, . . . , l, l0

(g+1)

which shows that uk = u(w0 ) = · · · = u(wl ). If wl is a clamped vertex, then u(wl ) = 0 by definition of clamped vertices, and if wl is a freely moving vertex, then u(wl ) = 0 has already been shown for freely moving vertices in V0 . Therefore (g+1) uk = 0 also follows for vk 6∈ V0 . We have shown that Y2 = 0, which means that A22 is invertible. In case n = 0, the matrix A−zM becomes A22 . Therefore the problem has no eigenvalues, formally in accordance with the statement of this theorem. Now let n > 0 and ng+1 > 0. Writing A and M as (n + ng+1 ) × (n + ng+1 ) block matrices we get   A11 − zM1 A12 A − zM = . A21 A22 The Schur factorization (see (B.4.2))   I A12 A−1 A11 − A12 A−1 22 22 A21 − zM1 A − zM = 0 I 0

0 A22



I A−1 22 A21

0 I



shows that all spectral data are given by A11 − A12 A−1 22 A21 − zM1 . But since A is symmetric and positive semidefinite, also A11 and A22 are symmetric and positive semidefinite, and A21 = A∗12 . Therefore A11 − A12 A−1 22 A21 is symmetric and positive semidefinite in view of the Schur factorization, and an application of Proposition B.2.1 shows that all eigenvalues are real and that the geometric multiplicity of the eigenvalues coincides with their algebraic multiplicity also in case n > 0 and ng > 0. Since M has exactly n nonzero entries it is clear in all cases that there are exactly n eigenvalues, counted with multiplicity. If the graph has a component without clamped vertices, let Ie be the index set (j) of the edges in that component. For j = 1, . . . , g and k = 1, . . . , nj put uk = 1 if (j) e We put u(g+1) = 1 if there is j ∈ Ie such that j ∈ I(vk ) j ∈ I˜ and uk = 0 if j 6∈ I. k (g+1) and uk = 0 otherwise. Clearly, S21 = 0. On each edge of the graph, either all

Chapter 5. Spectral problems on graphs

180

amplitudes are 0 or all amplitudes are 1. Hence S13 = 0 and S22 = 0. Therefore 0 is an eigenvalue in this case. Let Y be such that Y ∗ AY = 0. Note that this implies S13 = 0, S21 = 0 and (j) (j) S22 = 0. Let j ∈ {1, . . . , g}. Then S13 = 0 gives uk+1 = uk (j = 1, . . . , g, k = (j)

(j)

1, . . . , nj − 1). If v− (ej ) ∈ V c , then u0 = 0, and u1 = 0 if nj > 0 in view of (j) (j) S21 = 0. If v− (ej ) ∈ V f , then u0 = u1 due to S22 = 0. Similarly, if v+ (ej ) ∈ V c , (j) (j) then unj+1 = 0, and unj = 0 if nj > 0 in view of S21 = 0. If v+ (ej ) ∈ V f , then (j) (j) (j) (j) (j) unj +1 = unj due to S22 = 0. It follows that u0 = u1 = . . . . = unj +1 for all j = 1, . . . , g. For any two edges (including loops) in a component of G there is a walk G which contains these two edges. Hence the identities from the previous paragraph (j) and the continuity condition (5.1.1) show that Y ∗ AY = 0 implies that all uk have the same value for all j for which ej is in the same component and all k = 0, . . . , nj +1. If a component has a clamped vertex, then this value must be 0 in view of (5.1.2). This proves the statement on the multiplicity of the eigenvalue 0.  Remark 5.1.3. Reversing the direction of an edge, say ej , results in reversing the order of the rows and columns with index (j, k) (k = 1, . . . , nj ). Hence the eigenvalue problem is invariant under changing the direction of the edges, and in particular det(A−zM ) does not depend on the direction of the edges. When n > 0 and ng+1 > 0, i.e., when the graph carries at least one bead and has freely moving vertices, then we set e1 , A1 = A11 − A

e1 = A12 A−1 A21 , A 22

recalling the notation from the proof of Theorem 5.1.2. Then it is clear that also det(A1 − zM1 ) does not depend on the direction of the edges. For convenience we may also use this notation when ng+1 = 0, in which case A1 = A and M1 = M . Proposition 5.1.4. Assume that the graph G carries at least one bead. Let v be a freely moving vertex of G with d(v) = 2 and two incident edges. Let G0 be the graph obtained by concatenating the two edges incident with v to one edge. Then, with appropriate indexing, A1 − zM1 is the same for G as for G0 . Proof. Observe that the assumption implies that there are exactly two edges which are incident with v; that is, v is not the vertex of a loop. We choose an indexing and orientation of the edges and freely moving vertices in such a way that v = vng+1 , I + (v) = {g − 1} and I − (v) = {g}. Therefore the edges of G0 are e1 , . . . , eg−2 and 0 obtained by concatenating eg−1 and eg . Then the amplitudes of the edge eg−1 (g−1)

(g−1)

(g)

(g)

0 , . . . , ung−1 , u1 , . . . , ung +1 . Writing A as an are u0 the ends and beads of eg−1 ([n + ng+1 − 1] + 1) × ([n + ng+1 − 1] + 1) block matrix, ! b11 A b12 A A =: b b22 , A21 A

5.1. The problem and its spectrum

181

we define b12 A b b11 − A b−1 A A† = A 22 21 and claim that A0 = A† . b12 , that is, the elements aj,k,g+1,n To prove this claim, we first evaluate A g+1 of A with (j, k) < (g + 1, ng+1 ). According to the cases 4 to 7 in the proof of b1,2 are Theorem 5.1.2, the nonzero elements of A ag,1,g+1,ng+1 = −

1 (g) l0

ag−1,ng−1 ,g+1,ng+1 = − ag+1,k,g+1,ng+1 = − ag+1,k,g+1,ng+1 = −

1 (g) l0

1

if ng−1 > 0,

(g−1) lng−1

if ng = 0, v+ (eg ) = vk and (ng−1 > 0 or v− (eg−1 ) 6= vk ),

1 (g−1) lng−1

ag+1,k,g+1,ng+1 = −

if ng > 0,

1 (g−1) lng−1

if ng−1 = 0, v− (eg−1 ) = vk and (ng > 0 or v+ (eg ) 6= vk ), −

1 (g) l0

if ng−1 = ng = 0 and v− (eg−1 ) = v+ (eg ) = vk .

Furthermore, ag+1,ng+1 ,g+1,ng+1 =

1 (g−1) lng−1

+

1 (g) l0

.

The variables in the linear functionals defining A0 are the same as the varib11 , and the corresponding coefficients in the linear functionals are the ables for A b11 and A0 if they do not contain l(g) or lg−1 . We may assume without same for A ng−1 0 loss of generality that ng−1 ≥ ng . Due to the symmetry of A0 and A† we only have to consider indices (j, k, r, s) with (j, k) ≤ (r, s). The first row to consider is b11 it is the row with index (g − 1, ng−1 ), which is present when ng−1 > 0. In A represented by the linear functional (g−1)

(g−1)

ung−1 − ung−1 −1 (g−1)

(g−1)

+

lng−1 −1

ung−1

(g−1)

,

lng−1

whereas its representation in A0 is (g−1)

(g−1)

ung−1 − ung−1 −1 (g−1)

(g)

−

lng−1 −1 Then a0g−1,ng−1 ,g−1,ng−1 =

(g−1)

u1 − ung−1 (g−1)

.

(g)

lng−1 + l0 1 (g−1) lng−1 −1

+

1 (g−1) lng −1

(g)

+ l0

,

Chapter 5. Spectral problems on graphs

182 whereas

ag−1,ng−1 ,g+1,ng+1 ag+1,ng+1 ,g−1,ng−1 ag+1,ng+1 ,g+1,ng+1

† = ag−1,ng−1 ,g−1,ng−1 − ag−1,n g−1 ,g−1,ng−1

=

1 (g−1)

lng−1 −1 =

1

+

1 (g−1) lng−1 −1

−

(g−1)

lng−1 +

(g−1)

(g−1)

(g)

(lng−1 )2 lng −1 + l0

1 (g−1) lng −1

(g−1) (g)

lng−1 l0

1

(g)

,

+ l0

which proves a0g−1,ng−1 ,g−1,ng−1 = a†g−1,ng−1 ,g−1,ng−1 . These are all relevant entries (g)

in this row if ng = 0 and v+ (eg ) ∈ V c . Otherwise, the coefficient of u1 be considered. If ng > 0 then we find a0g−1,ng−1 ,g,1 = −

needs to

1 (g−1) lng −1

(g)

+ l0

and a†g−1,ng−1 ,g,1 = − −

ag−1,ng−1 ,g+1,ng+1 ag+1,ng+1 ,g,1 ag+1,ng+1 ,g+1,ng+1 (g−1) (g)

1

lng−1 l0

(g−1) (g)

lng −1 + l0

lng−1 l0

(g−1)

(g)

,

which proves a0g−1,ng−1 ,g,1 = a†g−1,ng−1 ,g,1 . When ng = 0 and v+ (eg ) ∈ V f , say v+ (eg ) = vs , then 1 a0g−1,ng−1 ,g+1,s = − (g−1) (g) lng −1 + l0 and a†g−1,ng−1 ,g+1,s = − =−

ag−1,ng−1 ,g+1,ng+1 ag+1,ng+1 ,g+1,s ag+1,ng+1 ,g+1,ng+1 (g−1) (g)

1

lng−1 l0

(g−1) (g)

lng −1 + l0

lng−1 l0

(g−1)

(g)

,

which proves a0g−1,ng−1 ,g+1,s = a†g−1,ng−1 ,g+1,s . The next row to consider is the row with index (g, 1), which is present when ng > 0. This can be formally obtained from the previous case by reorienting and reindexing. Hence also this row is the same for A0 and A† . We still have to consider rows with index (g + 1, k) for which ng−1 = 0 and v− (eg−1 ) = vk or ng = 0 and v+ (eg ) = vk . First let ng = 0 and v+ (eg ) = vk but

5.2. Characteristic polynomials

183

ng−1 > 0 or ng−1 = 0 and g − 1 6∈ I − (vk ). In view of (5.1.6), the difference of the linear functionals for A11 and A0 is (g+1)

(g+1)

uk

uk

−

(g)

(g−1)

− ung−1

(g−1)

l0

.

(g)

lng−1 + l0

The only relevant component with index (g + 1, k, r, s) and (g + 1, k) ≤ (r, s) is when (r, s) = (g + 1, k). From (g−1) (g) ag+1,k,g+1,ng+1 ag+1,ng+1 ,g+1,k lng−1 l0 1 = (g) (g−1) (g) ag+1,ng+1 ,g+1,ng+1 (l0 )2 lng −1 + l0

it follows that a†g+1,k,g+1,k − a0g+1,k,g+1,k =

1 (g)

−

l0

1 (g−1)

(g)

(g−1) (g)

lng−1 l0

1

−

(g)

(g−1)

(g)

(l0 )2 lng −1 + l0

lng−1 + l0

= 0.

The case that ng−1 = ng = 0, v− (eg−1 ) = vk and g 6∈ I + (vk ) is similar; in the two (g−1) (g) formulas above, the numbers lng−1 and l0 just have to be interchanged. Finally, let ng−1 = ng = 0, v− (eg−1 ) = v+ (eg ) = vk . In view of (5.1.6), the difference of the linear functionals for A11 and A0 is (g+1)

(g+1)

uk

(g)

l0

+

uk

(g−1)

(g+1)

−2

lng−1

uk

(g+1)

− uk

(g−1)

.

(g)

lng−1 + l0

From ag+1,k,g+1,ng+1 ag+1,ng+1 ,g+1,k = ag+1,ng+1 ,g+1,ng+1

1 (g−1)

+

lng−1

!2

1 (g)

l0

(g−1) (g)

lng−1 l0 (g−1)

(g)

lng −1 + l0

it follows that a†g+1,k,g+1,k − a0g+1,k,g+1,k =

1 (g)

l0

1

+

(g−1)

lng−1

(g−1)

−

(g)

lng −1 + l0

(g−1) (g)

= 0.

lng−1 l0

Thus we have shown that A0 = A† . Since A†1 = A1 in view of Lemma B.4.1, it follows that A01 = A1 . 

5.2

Characteristic polynomials

Unless otherwise stated we are going to assume in this section that g > 0. (j) The vector of all amplitudes of problem (4.1.10) has the components uk (j = 1, . . . , g, k = 0, . . . , nj +1). In Section 5.1 we have eliminated some redundant

Chapter 5. Spectral problems on graphs

184

amplitudes at the vertices to arrive at a matrix representation A − zM of the problem. This procedure has been appropriate to prove that the characteristic values of the problem are real and nonnegative. For further results, we will use the Cauer-Fry polynomials for (4.1.10), which are described in (4.1.16) with (j)

(j)

(j)

u2j−1 := u0 and u2j := u1 − u0

(j = 1, . . . , g),

U = (uj )2g j=1 .

(5.2.1)

For an interior vertex, there is in general no canonical way to write the boundary conditions, and we therefore observe that similar to (4.1.9) it follows that for the freely moving interior vertices v of G the identities (5.1.1) and (5.1.3) imply that ! ! (r) (r) (r) (r) (r) (r) X X unr +1 unr +1 − unr u0 u1 − u0 (j) u0 = − + + (r) (r) d(v) d(v) lnr l + − r∈I (v)

0

r∈I (v)

−

(j ∈ I (v)), (j)

unj +1 =

X

(r) unr +1

d(v)

r∈I + (v)

−

(r) unr +1 − (r) lnr

(r) u nr

! +

X r∈I − (v)

(r) u0

d(v)

+

(r) u1

(5.2.2) !

(r) − u0 (r) l0

(j ∈ I + (v)).

(5.2.3)

Conversely, since the right-hand sides in (5.2.2) and (5.2.3) are independent of j, (5.1.1) immediately follows from (5.2.2) and (5.2.3), whereas (5.1.3) follows from adding up the d(v) equations in (5.2.2) and (5.2.3). The left-hand sides in (5.2.2) and (5.2.3) are the same as in (5.1.2), whereas the right-hand sides only depend on v. Hence it is convenient to introduce the following linear functionals: Lc2j−1 (z)U := u2j−1 Lc2j (z)U

:=

(j = 1, . . . , g),

(j) R2nj (z, 1)u2j−1

+

(5.2.4)

(j) R2nj (z, 0)u2j

(j = 1, . . . , g),  (r) (r) R2nr (z, 1)u2r−1 + R2nr (z, 0)u2r

1 d(v) r∈I + (v)  X  (r) (r) + R2nr −1 (z, 1)u2r−1 + R2nr −1 (z, 0)u2r X 

Lfv (z)U := −

(5.2.5) (5.2.6)

r∈I + j(v)

−

X r∈I − (v)

u2r−1 u2r + (r) d(v) l0

! (v ∈ V f ).

It is clear that being imposed at a pendant clamped vertex, (5.1.2) just becomes the usual Dirichlet condition (condition (1.2.23) or (1.2.24)), whereas imposed at a freely moving pendant vertex, (5.2.2), (5.2.3) reduces to exactly one condition, which is the usual Neumann condition (condition (1.2.27) or (1.2.29)).

5.2. Characteristic polynomials

185

As in Section 4.1, we have 2g unknown variables, and the conditions (5.1.2), (5.2.2), (5.2.3) can be expressed in terms of the linear functionals c L2j−1 (z)U := L2j−1 (z)U

(j ∈ I − (v), v ∈ V c ),

c (z)U (j ∈ I + (v), v ∈ V c ), L2j Lc2j−1 (z)U + Lfv (z)U (j ∈ I − (v), Lc2j (z)U + Lvf (z)U (j ∈ I + (v), v

L2j (z)U := L2j−1 (z)U := L2j (z)U :=

(5.2.7) (5.2.8) f

v ∈ V ), f

∈ V ).

(5.2.9) (5.2.10)

Defining the (2g) × (2g) matrix Φ(z) by 

 L1 (z)U   ... Φ(z)U =  , L2g (z)U

(5.2.11)

it follows that the system of equations (Lj (z)U )2g j=1 = 0 can be written in the form Φ(z)U = 0. The matrix function Φ is called the characteristic matrix function of (4.1.10), (5.1.2), (5.2.2), (5.2.3), and the function φ defined by φ(z) := det(Φ(z))

(5.2.12)

is called the characteristic polynomial of problem (4.1.10), (5.1.2), (5.2.2), (5.2.3). For convenience we let φ = 1 when g = 0. It is clear that the characteristic polynomials are independent of the indexing of the edges since a permutation if the edges would result in simultaneous permutations of columns and rows of Φ, which would leave the corresponding determinants unchanged. Although it appears by the above reasoning that the spectral data of A − zM and Φ(z) are the same, we will now formally state and prove it. Proposition 5.2.1. There is a constant C 6= 0 such that det(A − zM ) = CΦ(z) for all z ∈ C. The multiplicity of each zero z of φ equals the defect of the matrix Φ(z). (j)

Proof. Let Y0 = (uk )

j=1,...,g k=0,...,nj +1

and let A0 − zM0 be the (n + 2g) × (n + 2g)

matrix for which the components of (A0 − zM0 )Y0 = 0 are given by (4.1.10), (5.1.1), (5.1.2), (5.1.3). Here the order of the components will be arbitrary. Indeed, we will not compare the spectral data of A − zM and Φ(z) directly, but instead we will compare them with conveniently chosen matrices A0 − zM0 in the class of matrices obtained from one particular representation by permutation of rows and columns, and in different parts of this proof we may use different matrices from this class. In order to simplify the notation, we will denote any such matrix with A0 − zM0 . Since the determinants of the matrices in this class are unique up to a multiplicative factor −1 and since solutions Y0 of (A0 − zM0 )Y0 = 0 would only differ by a permutation of components, it is clear that we can take any A0 − zM0 for this comparison.

Chapter 5. Spectral problems on graphs

186

For each vertex v ∈ V choose an enumeration I(v) = {j1 (v), . . . , jd(v) (v)} and put sr (v) = 0 if jr (v) ∈ I − (v) and sr (v) = njr (v) + 1 if jr (v) ∈ I + (v). Note that for each v ∈ V int , (5.1.1) consists of d(v) − 1 conditions. Then we can write (5.1.1) as (j (v)) (j (v)) usrr(v) − us11(v) = 0 (v ∈ V int , r = 2, . . . , d(v)), (5.2.13) whereas (5.1.2) becomes (j (v))

usrr(v) = 0

(v ∈ V c , r = 1, . . . , d(v)).

(5.2.14)

First we will compare the spectral data of A − zM and A0 − zM0 . For each v ∈ V c consecutive components of (A0 − zM0 )Y0 = 0 will be defined by (5.2.14), and for each v ∈ V int ∩ V f consecutive components of (A0 − zM0 )Y0 = 0 will be defined by (5.2.13). The remaining components of (A0 − zM0 )Y0 are defined by (4.1.10) and (5.1.3) in the same order as in (A − zM )Y = 0. We now arrange (j (v)) the columns of A0 − zM0 such that the column index of usrr(v) equal the row (j (v))

index of usrr(v) = 0 for v ∈ V c and r = 1, . . . , d(v), and that the column index of (j (v))

usrr(v)

for v ∈ V int ∩ V f and r = 2, . . . , d(v) coincides with the row index given (j (v))

the equation from (5.2.13) which contains usrr(v) . For k = 1, . . . , ng+1 we may (g+1)

(j (v ))

identify us11(vkk) with uk to a block representation

T . That is, we can write Y0T = (Y00 , Y T ), which leads

 (A0 − zM0 )Y0 =

I B2

B1 e A0 − zM

  Y00 , Y

where the matrix I is an identity matrix. The matrix B1 has entries −1 when the column corresponds to some vk (k = 1, . . . , g) and the row index corresponds to an equation (5.2.13) for vk , and all other entries of B1 are 0. By the Schur factorization,     I 0 I B1 I 0 A0 − zM0 = . e0 − B2 B1 − zM 0 I B2 I 0 A e0 and A their columns corresponding to amplitudes By definition of the matrices A at the beads are equal, whereas those columns of B1 and therefore also of B2 B1 (g+1) . But in A are zero. It remains to compare the columns corresponding to uk (j) these columns are obtained by identifying all ul which are amplitudes at vk with (g+1) e0 are the columns corresponding uk , whereas the corresponding columns of A (j1 (vk )) to us1 (vk ) , and the corresponding columns of −B2 B1 are the sums of the columns (v)) (j (j (v)) e0 −B2 B1 = A has been shown. corresponding to u 2 , . . . , u d(v) . Therefore A s2 (v)

sd(v) (v)

The above Schur factorization implies det(A0 − zM0 ) = det(A − zM ), and (A0 − zM0 )Y0 = 0 is equivalent to (A − zM )Y = 0 with Y0T = (−(B1 Y )T , Y T ).

5.2. Characteristic polynomials

187

We are going to use (4.1.16) to relate A0 − zM0 to Φ(z). To this end, let (j) (j) (j) j ∈ {1, . . . , g} with nj > 0, let u(j) = (u0 , . . . , unj +1 )T , define Lbk (z)u(j) by the 2 left-hand side of (4.1.10), with λ = z, that is, !

(j)

(j) (j) lk Lbk (z)u(j)

=

(j) −uk+1

+

lk

1+

−

(j)

lk−1

(j) (j) lk mk z

(j)

lk

(j)

uk −

(j)

lk−1

(j)

uk−1

and let (j)

(j)

(j)

(j)

(j)

(j)

(j)

Lk (z)u(j) := −uk+1 + R2k (z, 0)(u1 − u0 ) + R2k (z, 1)u0

(k = 1, . . . , nj )

be the linear functionals corresponding to (4.1.16). We are going to prove that for j = 1, . . . , g with nj > 0 and for k = 1, . . . , nj (j) (j) 1 and r = 1, . . . , k there are polynomials αk,r with αk,k (z) = (j) such that the lk

identities (j) Lbk (z)u(j) =

k X

(j)

(j) αk,r (z)L(j) r (z)u

(k = 1, . . . , nj )

(5.2.15)

r=1

hold. To prove (5.2.15) we observe that the recursion (1.2.9) for R2k gives !

(j)

(j) R2k (z, c)

=

1+

lk

(j) lk−1

−

(j) (j) lk m k z

(j)

lk

(j)

R2k−2 (z, c) −

(j)

(j) lk−1

R2k−4 (z, c).

In view of (1.2.12) and (1.2.13) it follows that (j)

(j)

R2 (z, c) = 1 +

l1

(j)

l0

(j)

(j)

(j)

− l1 m 1 z −

l1

(j)

c.

l0

Hence !

(j)

(j) (j) l1 Lb1 (z)u(j)

=

(j) −u2

+

1+

l1

(j) l0

−

(j) (j) l1 m 1 z

(j)

(j)

(u1 − u0 )

  (j) (j) (j) + 1 − l1 m1 z u0 (j)

(j)

(j)

(j)

(j)

(j)

= −u2 + R2 (z, 0)(u1 − u0 ) + R2 (z, 1)u0 (j)

= L1 (z)u(j) ,

Chapter 5. Spectral problems on graphs

188

which shows that (5.2.15) holds for k = 1. For k ≥ 2 we calculate ! (j) lk (j) b(j) (j) (j) (j) (j) lk Lk (z)u = −uk+1 + 1 + (j) − lk mk z lk−1   (j) (j) (j) (j) (j) (j) × −Lk−1 (z)u(j) + R2k−2 (z, 0)(u1 − u0 ) + R2k−2 (z, 1)u0 (j)

−

lk

(j)

(j)

lk−1

uk−1 !

(j)

=

(j) −uk+1

−

lk

1+

−

(j)

lk−1

(j) (j) lk mk z

!

(j)

+

(j) R2k (z, 0)

+

R2k (z, 1) +

(j)

+

(j)

Lk−1 (z)u(j)

lk

(j) (j) R2k−4 (z, 0) (u1 (j) lk−1 ! (j) lk (j) (j) R2k−4 (z, 1) u0 (j) lk−1

(j)

− u0 )

(j)

−

lk

(j)

(j) lk−1

uk−1 !

(j)

=

(j) Lk (z)u(j) (j)

+

lk

(j)

lk−1

−

1+

lk

(j) lk−1

−

(j) (j) lk mk z

(j)

Lk−1 (z)u(j)

  (j) (j) (j) (j) (j) (j) −uk−1 + R2k−4 (z, 0)(u1 − u0 ) + R2k−4 (z, 1)u0 .

Hence !

(j)

(j) (j) l2 Lb2 (z)u(j)

=

(j) L2 (z)u(j)

−

1+

l2

(j) l1

−

(j) (j) l2 m2 z

−

(j) (j) lk mk z

(j)

L1 (z)u(j)

and !

(j)

(j) (j) lk Lbk (z)u(j)

=

(j) Lk (z)u(j)

−

1+

lk

(j)

lk−1

(j)

Lk−1 (z)u(j)

(j)

+

lk

(j) lk−1

(j)

Lk−2 (z)u(j)

when k > 2. Thus we have shown (5.2.15). Now the rows of A0 − zM0 will be arranged in such a way that we first take the conditions (4.1.10), that is, the rows corresponding to the linear func(j) tionals Lbk (z) (j = 1, . . . , g, k = 1, . . . , nj ) with lexicographic order in (j, k)

5.2. Characteristic polynomials

189

and then the rows corresponding to (5.1.1), (5.1.2) and (5.1.3). The columns of T T T A0 − zM0 are arranged according to the permutation (Y01 , Y02 ) of Y0 , where (1) (1) (g) (g) (1) (g) (g) (1) T T T Y01 = (u2 , . . . , un1 +1 , . . . , u2 , . . . , ung +1 ) and Y02 = (u0 , u1 , . . . , u0 , u1 )T . (j)

In accordance with the indexing of the rows, the components uk (j = 1, . . . , g, k = 2, . . . , nj+1 ) of Y01 will be given the index (j, k − 1). Then this representation of A0 − zM0 can be written as     0 Ψ11 (z) I 0 A0 − zM0 = Φ0 (z) 0 Ψ12 0 Ψ2 in the following  way. The matrix Ψ2 is a block diagonal matrix with g diagonal 1 0 blocks , and therefore Ψ2 Y02 = U with U as defined in (5.2.1). The −1 1 matrix Ψ11 (z) consists of the coefficients of (5.2.15). In particular, Ψ11 (z) is a 1 triangular matrix with diagonal elements (j) > 0. At the beginning of this section lk

we have seen that the conditions (5.1.1) and (5.1.3) can be written as (5.2.2) and (5.2.3) via an invertible linear transformation. Hence we arrive at an invertible matrix Ψ12 , where we may assume that the rows corresponding to (5.1.2), (5.2.2) and (5.2.3) in Φ0 (z) are ordered as in Φ(z). This shows that there is a constant C0 such that det(A0 − zM0 ) = C0 det Φ0 (z) for all z ∈ C and that the defect of A0 − zM0 equals the defect of Φ0 (z) for all z ∈ C. (j) The first n rows of Φ0 (z) are given by the linear functionals Lk , and therefore   −I Φ01 (z) Φ0 (z) = , Φ02 (z) Φ03 (z) where the entries of Φ01 (z) with row index (j, k) and column index 2j −r (r = 0, 1) (j) are R2k (z, r), and all other entries of Φ01 (z) are 0. With the above indexing of the first n rows of Φ0 (z) and the components of Y01 we let ej,k be the corresponding unit vectors in Cn . Hence we can write the above description of Φ01 (z) as (j)

(j)

eT j,k Φ01 (z)U = R2k (z, 1)u2j−1 + R2k (z, 0)u2j

(j = 1, . . . , g, k = 1, . . . , nj ).

We will have to express the conditions (5.1.2), (5.2.2) and (5.2.3) in terms (j) of the components of Y01 and U . Let j ∈ {1, . . . , g}. Clearly, u0 = u2j−1 and (j) (j) u1 − u0 = u2j by definition of U . When nj = 0, then (j)

(j)

(j)

(j)

unj +1 = u1 = u2j + u2j−1 = R2nj (z, 1)u2j−1 + R2nj (z, 0)u2j and (j)

(j)

(j)

unj +1 − u(j) nj = u1 − u0 = u2j . When nj = 1, then (j)

(j)

(j)

(j)

(j)

(j)

unj +1 − u(j) nj = u2 − u1 = unj +1 − R2nj (z, 1)u2j−1 − R2nj (z, 0)u2j .

Chapter 5. Spectral problems on graphs

190 (j)

(j)

(j)

In all other cases, unj +1 and unj +1 − unj are written in terms of components of Y01 . T With Y1 := (Y01 , U T )T define (j) c (z)Y1 := u0 Le2j−1 (j) Lec2j (z)Y1 := unj +1

Lefv (z)Y1 := −

X r∈I + (v)

(r) unr +1

d(v)

−

(r) unr +1 − (r) lnr

(j = 1, . . . , g), (j = 1, . . . , g), ! X −

(r) unr

r∈I − (v)

(r)

(r)

(r)

u0 u −u + 1 (r) 0 d(v) l0

!

(v ∈ V f ), (j)

where the uk have to be expressed in terms of the components of Y1 as discussed above. Putting Le2j−1 (z)Y1 := Lec2j−1 (z)Y1 Le2j (z)Y1 := Lec2j (z)Y1

(j ∈ I − (v), v ∈ V c ), (j ∈ I + (v), v ∈ V c ), (j ∈ I − (v), v ∈ V f ),

Le2j−1 (z)Y1 := Lec2j−1 (z)Y1 + Lefv (z)Y1 Le2j (z)Y1 := Lec2j (z)Y1 + Lefv (z)Y1

(j ∈ I + (v), v ∈ V f )

it follows that eT l Φ02 (z)


Φ03 (z) Y1 = Lel (z)Y1

(l = 1, . . . , 2g),

where el is the l-th unit vector in C2g . Now we are going to compare the matrix Φ(z) with the Schur complement Φ04 (z) := Φ02 (z)Φ01 (z) + Φ03 (z) of Φ0 (z). From above we conclude that   Φ01 (z)U T e el Φ04 (z)U = Ll (z) (l = 1, . . . , 2g). U Let j ∈ {1, . . . , g}. Then   Φ01 (z)U c e L2j−1 (z) = u2j−1 = Lc2j−1 (z)U. U When nj = 0, then   Φ01 (z)U (j) (j) Lec2j (z) = R2nj (z, 1)u2j−1 + R2nj (z, 0)u2j = Lc2j (z)U, U and when nj > 0, then   Φ01 (z)U c c e L2j (z) = eT j,nj Φ01 (z)U = L2j (z)U. U

5.2. Characteristic polynomials For v ∈ V f we have   X Φ01 (z)U =− Levf (z) U +

r∈I (v) nr =0

X

+

191

X u2r + u2r−1 − d(v) +

r∈I (v) nr >0

X eT r,nr Φ01 (z)U + d(v) +

u2r

(r) l r∈I (v) nr nr =0

eT r,nr Φ01 (z)U − u2r − u2r−1 (r)

lnr

r∈I + (v) nr =1

X

+

T eT r,nr Φ01 (z)U − er,nr −1 Φ01 (z)U (r)

lnr

r∈I + (v) nr >1

X

−

r∈I − (v)

u2r−1 u2r + (r) d(v) l0

!

(r)

(r)

R2nr (z, 1)u2r−1 + R2nr (z, 0)u2r d(v) r∈I + (v)  X  (r) (r) + R2nr −1 (z, 1)u2r−1 + R2nr −1 (z, 0)u2r

=−

X

r∈I + (v)

X

−

r∈I − (v)

u2r−1 u2r + (r) d(v) l0

!

= Lfv (z)U. Hence e eT l Φ04 (z)U = Ll (z)



Φ01 (z)U U



= Lc2j−1 (z)U = eT l Φ(z)

(l = 1, . . . , 2g),

which shows that Φ04 (z) = Φ(z) for all z ∈ C. Since det Φ0 (z) = (−1)n det Φ04 (z) and since the defects of Φ0 (z) and Φ04 (z) are equal, the proof is complete.  In Chapter 1 we have found characteristic polynomials for string problems in terms of Cauer-Fry polynomials. Vibrations on strings are special cases of the vibrations considered in this chapter. However, since characteristic polynomials are only unique up to multiplication by a constant factor and since in this chapter some constructions require to take sums of (products of) characteristic polynomials, these multiplicative constants are not redundant, in general. Considering the (j) metric edge ej with the distribution of beads as defined in Section (5.1), let ϕAB with A, B ∈ {D, N } be the characteristic polynomial of the string problem on ej as defined by (5.2.12), where A is the condition at the entrance, B is the condition at the exit, D stands for the Dirichlet condition, and N stands for the Neumann condition.

Chapter 5. Spectral problems on graphs

192

(j)

The following proposition gives the explicit representation of ϕAB in terms of Cauer-Fry polynomials. Proposition 5.2.2. The characteristic polynomials for the string problems on ej (j = 1, . . . , g) have the representations (j)

(j)

ϕN D (z) =

1 (j) l0

(j)

(j)

ϕDN (z) = R2nj −1 (z, 0),

(j)

ϕN N (z) =

ϕDD (z) = R2nj (z, 0), R2nj (z, 1),

(j)

1

(j)

(j) l0

(j)

R2nj −1 (z, 1).

Proof. Without loss of generality we let j = 1. Then the characteristic matrix function is a 2 × 2 matrix function whose entries are determined by L1 and L2 as given by (5.2.7)–(5.2.10). Denoting these functionals by LA k for k = 1, 2 and (1) A ∈ {D, N }, it follows that ϕAB is the determinant of the matrix whose rows are B given by LA 1 and L2 . We calculate T LD 1 (z)(u1 , u2 ) = u1 , T LN 1 (z)(u1 , u2 ) = − T LD 2 (z)(u1 , u2 ) = T LN 2 (z)(u1 , u2 ) =

1

u , (1) 2 l0 (1) R2n1 (z, 1)u1 + (1) R2n1 −1 (z, 1)u1

(1)

R2n1 (z, 0)u2 , (1)

+ R2n1 −1 (z, 0)u2 .

Hence 1 0 = R(1) (z, 0), = (1) (1) 2n1 R2n1 (z, 1) R2n1 (z, 1) 1 0 (1) = R(1) (z, 0), ϕDN (z) = (1) (1) 2n1 −1 R2n1 −1 (z, 1) R2n1 −1 (z, 1) 1 0 − (1) 1 (1) (1) l0 ϕN D (z) = (1) R (z, 1), = R (z, 1) R(1) (z, 1) l(1) 2n1 2n1 2n1 0 1 0 − (1) 1 (1) (1) l0 ϕN N (z) = (1) = (1) R2n1 −1 (z, 1). (1) R l 2n1 −1 (z, 1) R2n1 −1 (z, 1) 0 (1) ϕDD (z)

5.3



Characteristic polynomials for related graphs

In this section we will represent characteristic polynomials of graphs via characteristic polynomials of subgraphs. Lemma 5.3.1. Let G = G1 ∪˙ G2 , and let φ, φ1 and φ2 be the corresponding characteristic polynomials. Then φ = φ1 φ2 .

5.3. Characteristic polynomials for related graphs

193

Proof. When G1 or G2 has no edges, then φ1 = 1 or φ2 = 1, and φ = φ1 φ2 is trivially true. Now let both G1 and G2 have at least one edge. With a suitable indexing of the edges, the corresponding characteristic matrices can be written in the block diagonal form   Φ1 (z) 0 . Φ(z) = 0 Φ2 (z) Taking determinants completes the proof.



With some abuse of notation, an arbitrarily chosen vertex v0 ∈ V will be called the root of G. At the root we will consider both the (generalized) Dirichlet and the (generalized) Neumann problem. Hence it is convenient for (generalized) Neumann conditions at the root to use the notation VNc := V c \ {v0 } as well as VNf := V f ∪ {v0 } for the clamped and freely moving vertices, respectively. For (generalized) Dirichlet conditions at the root the notation VDc := V c ∪ {v0 } and VDf := V f \ {v0 } for the clamped and freely moving edges, respectively, will be used. For A ∈ {D, N }, the corresponding characteristic matrix ΦA (z) and the characteristic polynomial φA are obtained by replacing V c and V f in (5.2.7)– (5.2.10) with VAc and VAf . For A = D, the function φD is called the characteristic polynomial for the Dirichlet problem at the root, and for A = N , the function φN is called the characteristic polynomial for the Neumann problem at the root. If v0 is an interior vertex, we can cut the graph G in the following way. The (1) (2) root v0 is replaced by two vertices v0 and v0 , and each edge incident with v0 (2) (1) is attached to exactly one of v0 or v0 , preserving the orientation of the edges, (1) (1) (2) (2) i.e., I − (v0 ) = I − (v0 ) ∪˙ I − (v0 ) and I + (v0 ) = I + (v0 ) ∪˙ I + (v0 ). It is assumed that at least one edge is attached to each of the new vertices. Denote this new graph by G0 , where the orientation of the edges is the same as for the graph G. For A, B ∈ {D, N } we denote by ΦAB (z) and φAB the characteristic matrix and the characteristic polynomial, respectively, where A corresponds to the vertex (1) (2) v0 and B corresponds to the vertex v0 . For example, φND is the characteristic polynomial of the graph G0 with (generalized) Neumann condition at the vertex (2) (1) v0 and with (generalized) Dirichlet condition at the vertex v0 . Lemma 5.3.2. Let v0 be an interior vertex and let G0 be as defined above. Then φD = φDD .

(5.3.1)

Proof. The edges and linear functionals Lk (k = 1, . . . , 2g) are identical for the problems on G and G0 . Therefore ΦD = ΦDD , and taking the determinants completes the proof.  e For v ∈ G (or v ∈ G0 ) let I(v) = {2j − 1 : j ∈ I − (v)} ∪ {2j : j ∈ I + (v)}. (s) e (s) ). We deLet s = 1, 2 and choose bijective maps τs : {1, . . . , d(v0 )} → I(v 0 (s) fine the matrix function Φv0 by replacing the rows of the characteristic matrix

Chapter 5. Spectral problems on graphs

194

(s)

ΦN (z) with index τs (j) for j = 1, . . . , d(v0 ) with the corresponding rows for the (s) (generalized) Neumann conditions at v0 for G0 , whereas the rows with index (3−s) ) are replaced with the rows given by the linear τ3−s (j) for j = 1, . . . , d(v0 for j = 1 and s = 1, by Lτc2 (1) for j = 1 and s = 2, and functionals Lc (1) τ1 (d(v0 ))

(3−s)

by Lcτ3−s (j) − Lcτ3−s (j−1) for j = 2, . . . , d(v0 proof of the next lemma for more details.

(s)

(s)

). We put φv0 = det Φv0 . See the (s)

Remark 5.3.3. The construction of the matrix functions Φv0 (s = 1, 2) is quite (1) (2) technical, but becomes easier when d(v0 ) = 2, i.e., d(v0 ) = d(v0 ) = 1. In this e (s) ) consists of a single index, say ks . Then Φ(s) case, I(v v0 is obtained from ΦN 0 by replacing the two rows with indices k1 and k2 , which describe the generalized Neumann conditions at v0 for G, as follows. (i) The row with index ks becomes a row which describes the Neumann condition (s) at v0 in G0 . (ii) The row with index k3−s becomes a row which describes the Dirichlet con(s) dition at v0 in G0 . Lemma 5.3.4. Let v0 be an interior vertex of G and let G0 be as defined above. Then φN = φND + φDN − φv(1) − φ(2) v0 . 0

(5.3.2) (1)

e 0 ) by τ (j) = τ1 (j) for j = 1, . . . , d(v ) and Proof. Define τ : {1, . . . , d(v0 )} → I(v 0 (1) (2) by τ (d(v0 ) + j) = τ2 (j) for j = 1, . . . , d(v0 ). Then let Lbτ (1) =

d(v0 ) 1 X Lτ (r) , d(v0 ) r=1

Lbτ (j) = Lτ (j) − Lτ (j−1) (j = 2, . . . , d(v0 )), e 0 )). Lbj = Lj (j ∈ {1, . . . , 2g} \ I(v b Let Φ(z) be the matrix whose rows are represented by Lbj (z) (j = 1, . . . , 2g). b Clearly, Φ(z) = TbΦ(z) with a 2g × 2g matrix Tb. Recursively expanding the dee 0 ) and simultaneous terminant of Tb with respect to the rows with indices j 6∈ I(v permutation of the remaining rows and columns show that det Tb = det T , where T is the d(v0 ) × d(v0 ) matrix  1 1 1     T =  

d(v0 )

d(v0 )

−1 0 .. .

1 −1 .. .

0 1 .. .

··· ···

d(v0 )

0 0 .. .

0

···

0

−1

1

   .  

5.3. Characteristic polynomials for related graphs

195

Expanding the determinant of T with respect to its first row it follows that det T =

d(v0 ) 1 X (−1)j−1 det Tj , d(v0 ) j=1

where Tj is a 2×2 block diagonal matrix with the first block of size (j −1)×(j −1) being upper triangular with diagonal elements −1 and the second block of size (d(v0 ) − j) × (d(v0 ) − j) being lower triangular with diagonal elements 1. Hence b det T = 1, so that det T = 1 and therefore det(Φ(z)) = det(ΦN (z)). It is clear from (5.2.9), (5.2.10) and (5.2.6) that Lbτ (j) = Lcτ (j) − Lcτ (j−1)

(j = 2, . . . , d(v0 ))

and (1) (2) Lbτ (1) = Lbτ (1) + Lbτ (1) ,

where (s) Lbτ (1) (z)U =

X



(r)

(r)

R2nr −1 (z, 1)u2r−1 + R2nr −1 (z, 0)u2r



(s)

r∈I + (v0 )

−

u2r

X

(r)

(s) r∈I − (v0 )

(s = 1, 2).

l0

By expansion rules for determinants, it follows that b 11 + det Φ b 12 + det Φ b 21 + det Φ b 22 , φN = det Φ

(5.3.3)

b ik (i, k = 1, 2) are obtained from Φ b by substituting Lbτ (1) where the matrices Φ (i) with Lbτ (1) and Lbτ (d(v(1) )+1) with (−1)k Lc . (1) 0

τ (d(v0 )+k−1)

b 1k (k = 1, 2) we may perform the inverse of the above row operations, In Φ (1) now corresponding to τ1 , and therefore the rows with indices τ (1), . . . , τ (d(v0 )) (1) represent (generalized) Neumann conditions at v0 . In particular, replacing the row with index τk (1) with its negative for k = 1, we arrive at Φv(1) , and therefore 0

b 11 = det Φ

−φ(1) v0 .

(5.3.4)

(2) For k = 2 and j = d(v0 ) + 2, . . . , d(v0 ) we replace Lbτ (j) with Lcτ (j) , which b 12 by a triangular matrix whose diagonal is similar to a matrix multiplication of Φ elements are 1. Thus we have arrived at ΦND and therefore

b 12 = φND . det Φ

(5.3.5)

b 2k (k = 1, 2) we take a cyclic permutation of the rows with indices In Φ (1) τ (j) for j = 1, . . . , d(v0 ) + 1 such that the row with index τ (1) becomes the

Chapter 5. Spectral problems on graphs

196

(1)

(1)

row with index τ (d(v0 ) + 1). Note that this cyclic permutation of d(v0 ) + 1 (1) rows is realized by a permutation matrix with determinant (−1)d(v0 ) . With the b 1k the rows with indices τ2 (j) (j = 1, . . . , d(v (2) )) same reasoning as above for Φ 0 (2) (1) correspond to (generalized) Neumann conditions at v0 . For j = 1, . . . , d(v0 ), the (1) new rows with index τ1 (j) are given by Lτc1 (j+1) −Lτc1 (j) (j = 1, . . . , d(v0 )−1) and (1) (j = d(v0 )). For k (1) τ (d(v0 )+k−1) (1) Lcτ (j) for j = 1, . . . , d(v0 ), which

(−1)k Lc

= 1 we replace these linear functionals

with

is achieved by a transformation which (1)

is similar to an upper triangular matrix with d(v0 ) diagonal entries being −1, whereas the remaining diagonal entries are 1. Therefore this transformation matrix (1) has determinant (−1)d(v0 ) . We have arrived at ΦDN , and hence b 21 = det ΦDN . det Φ

(5.3.6)

For k = 2 we take the cyclic permutation of the rows with indices τ1 (j) for (1) j = 1, . . . , d(v0 ) such that the row with index τ1 (j) is given by Lcτ2 (1) , whereas the (1)

rows with indices τ1 (j) for j = 2, . . . , d(v0 ) are given by Lcτ1 (j) − Lcτ1 (j−1) . Hence (2)

we have arrived at Φv0 . Since this permutation is achieved by a transformation (1) matrix with determinant (−1)d(v0 )−1 , it follows that b 22 = −φ(2) . det Φ v0

(5.3.7)

Substituting (5.3.4)–(5.3.7) into (5.3.3) completes the proof. (s)

Definition 5.3.5. Let s ∈ {1, 2}. Then v0 set Vs of vertices of G0 such that (3−s)

(i) v0



is called locally clamped if there is a

6∈ Vs ; (s)

(ii) if e is an edge of G0 which is incident with v0 , then both ends of e belong to Vs ; (iii) if v ∈ Vs has an incident edge whose other end does not belong to Vs , then the vertex v carries the (generalized) Dirichlet condition. Remark 5.3.6. 1. Let s ∈ {0, 1} and let V (s) be the set of vertices v of G0 such (s) that v 6= v0 and such that there is an edge e in G0 with the ends of e being (3−s) (s) 6∈ V (s) and each v ∈ V (s) is a pendant vertex or carries v and v0 . If v0 (s) generalized Dirichlet conditions, then v0 is locally clamped. (1)

(2)

(1)

2. If v0 and v0 belong to different components of G0 , then v0 locally clamped. (s)

(2)

and v0

are

Proof. In statement 1 put Vs = V (s) ∪ {v0 }, and in statement 2 let Vs be the (s) component of G0 which contains v0 (s = 1, 2). Then it is clear that Vs satisfies  the conditions of Definition 5.3.5.

5.4.

Series connection of graphs

197 (s)

Proposition 5.3.7. Let s ∈ {1, 2}. If v0

(s)

is locally clamped, then φv0 = 0.

Proof. Let Is be the set of all j ∈ {1, . . . , g} such that both ends of ej belong to (s) Vs . In view of condition (ii) in Definition 5.3.5 we have I(v0 ) ⊂ Is . The rows (s) of Φv0 with indices 2j − 1 and 2j for j ∈ Is only depend on u2k−1 and u2k for k ∈ Is . Indeed, this is clear when v+ (ej ) or v− (ej ) carries the (generalized) Dirichlet condition, and it follows from condition (iii) in Definition 5.3.5 when v+ (ej ) or v− (ej ) carries the (generalized) Neumann condition. By condition (i) in Definition 5.3.5, the row with index τ3−s (1) is not one of those rows listed (s) above, and by definition of Φv0 this row only depends on u2k−1 and u2k for some (s) (s) k ∈ I(v0 ). Therefore Φv0 has a (2#Is + 1) × (2g) submatrix of rank at most (s) (s)  2#Is , which proves that φv0 = det(Φv0 ) = 0.

5.4

Series connection of graphs

Let us consider two connected graphs Gj (j = 1, 2) with at least two vertices in j j and vout in each of them as the each of them. We choose two distinct vertices vin entrance and exit vertices. In this section we investigate the series connection of j these graphs. For A, B ∈ {D, N } and j = 1, 2 we denote by φAB the characteristic j polynomial of the boundary value problem on the graph G , where A refers to the j j boundary condition at vin , B refers to the boundary condition at vout , D stands for (generalized) Dirichlet boundary conditions and N stands for (generalized) Neumann boundary conditions. 2 1 , then we obtain a new graph G = G1 ∪ G2 If we connect vout with vin which is assumed to have the generalized Neumann condition at the cut vertex 1 2 vout with all other conditions inherited from G1 and G2 . For A, B ∈ {D, N } = vin the functions φNN , φND , φDN and φDD denote the characteristic polynomials on 2 1 , B corresponds to vout and D and N denote the G, where A corresponds to vin (generalized) Dirichlet and Neumann conditions, respectively. Theorem 5.4.1. For A, B ∈ {D, N } the following identities are true: φAB = φ1AN φ2DB + φ1AD φ2NB .

(5.4.1)

(1) 1 Proof. In the notation of Section 5.3 we have G0 = G1 ∪˙ G2 with v0 = vout (2) 2 . Observing part 2 of Remark 5.3.6 and Proposition 5.3.7, (5.4.1) and v0 = vin immediately follows from Lemmas 5.3.4 and 5.3.2. 

Using these identities we immediately obtain an analogue of the Lagrange identity. Corollary 5.4.2. It is true that φND φDN − φNN φDD = (φ1ND φ1DN − φ1NN φ1DD )(φ2ND φ2DN − φ2NN φ2DD ).

(5.4.2)

Chapter 5. Spectral problems on graphs

198 Proof. Defining  Θ=

φND φDD

 φNN , φDN

Θj =



φjND φjDD

φjNN j φDN

 (j = 1, 2),

the identities (5.4.1) are equivalent to Θ = Θ1 Θ2 . Taking determinants gives  (5.4.2). Above we have considered the series connection of two graphs, which is obtained by joining the graphs at two vertices. An alternative method would be to add an edge which joins these two vertices. The reverse of this operation is to remove an edge of the graph so that the two ends of this edge belong to different components of the resulting graph. Below we are going to consider the case of a tree with repeated removal of edges, which gives a particularly easy Lagrange type formula. Let T be a directed tree, choose any two distinct vertices of T and denote them by vin and vout such that the unique path P in T connecting the two vertices is directed from vin to vout . Let p + 1 (p ≥ 1) be the number of vertices of the path and denote these vertices by vin = v0 , v1 , . . . , vp = vout in the order they are traversed. Denote the edge from vj−1 to vj by ej . After removing the edges ej (j = 1, . . . , p) from T the remaining graph has p+1 components (see Lemma C.2.3), which can be uniquely indexed as Tj by the requirement that vj ∈ Tj (j = 0, . . . , p). Each Tj is either an isolated vertex or a tree, including the degenerate case of a tree with only one edge. At each vertex v of T which is distinct from v0 , . . . , vp the tree T is equipped with (generalized) Dirichlet boundary conditions or (generalized) Neumann boundary conditions, whereas the tree T carries generalized Neumann boundary conditions at v1 , . . . , vp−1 for p > 1. For A, B ∈ {D, N } we denote by φAB the characteristic polynomial of the boundary value problem on the tree T , where A refers to the boundary condition at vin , B refers to the boundary condition at vout , D stands for the (generalized) Dirichlet boundary conditions (j) and N stands for the (generalized) Neumann boundary conditions. Let l0 have (j) the usual meaning for the string ej and let φD be the characteristic polynomial for Tj with root vj (j = 0, . . . , p) and (generalized) Dirichlet boundary conditions at vj whereas the boundary conditions at all other vertices of Tj are inherited from the boundary conditions for T . Proposition 5.4.3. For the directed tree T as considered above, the following Lagrange type formula holds: 2  p p Y Y 1 (j) φD  . (5.4.3) φND φDN − φNN φDD =  (j) j=1 l0 j=0 (j)

Proof. Recall the definition of the characteristic polynomials ϕAB from Section (j) 5.2 (see Proposition 5.2.2) and recall that φD = 1 if Tj is an isolated vertex. We

5.4.

Series connection of graphs

199

will prove the statement by induction on p. Hence first consider the case p = 1. Applying Lemmas 5.3.2 and 5.3.1 twice if v0 and v1 are interior vertices it follows that (0) (1) (1) φDD = φD ϕDD φD . Clearly, the above reasoning also applies, mutatis mutandis, when v0 or v1 are pendant vertices. Therefore the above representation of φDD holds in all cases. For j = 0, 1 let Tej consist of Tj augmented by the vertex v1−j and the edge e1 , with the orientation inherited from the orientation of T . For A, B ∈ {D, N } (j) we denote by φeAB the characteristic polynomial of the boundary value problem on Tej , where A refers to the boundary condition at vj , B refers to the boundary condition at v1−j , D stands for the (generalized) Dirichlet boundary conditions and N stands for the (generalized) Neumann boundary conditions. Another application of Lemma 5.3.2, part 2 of Remark 5.3.6 and Proposition 5.3.7 shows that (0) (1) φDN = φD φeND ,

(1) (0) φND = φD φeND .

If v0 is an interior vertex of T , then T can be considered as a series connection of 1 1 2 6= v0 G1 = T0 and G2 = Te1 by letting vout = v0 , vin = v0 and choosing vertices vin 2 and vout of T0 and T1 , respectively. Then Theorem 5.4.1 gives (0) (1) (0) (1) φNN = φD φeNN + φN φeND . (j)

Putting φN := 0 if Tj is an isolated vertex, this representation also holds when v0 is a pendant vertex of T . Similarly, if T0 or T1 is not an isolated vertex, then Theorem 5.4.1 shows that (1) (0) (1) (0) (0) φeND = ϕDD φN + ϕND φD , (1) (1) (1) (1) (1) φeND = ϕDD φN + ϕDN φD , (1) (1) (1) (1) (1) φeNN = ϕND φN + ϕNN φD ,

which clearly extends to the case when T0 or T1 is an isolated vertex. Therefore the left-hand side of (5.4.3) can be written as (0) (1)

φND φDN − φNN φDD = φD φD ψ, where i h (0) (1) (0) (1) (0) (1) (1) ψ = φeND φeND − ϕDD φD φeNN + φN φeND i h (1) (1) (0) (1) (1) (0) (0) = φeND − ϕDD φN φeND − ϕDD φD φeNN (1) (0) (1) (1) (0) (1) = ϕND φD φeND − ϕDD φD φeNN i h (1) (1) (1) (1) (1) (1) (1) (1) (1) (0) (1) (1) (1) = φD ϕND ϕDD φN + ϕND ϕDN φD − ϕDD ϕND φN − ϕDD ϕNN φD i h (1) (1) (1) (1) (0) (1) = φD φD ϕND ϕDN − ϕDD ϕNN .

200

Chapter 5. Spectral problems on graphs

In view of Proposition 5.2.2 and the Lagrange identity, Theorem 1.2.16, it follows that i 1 (0) (1) h (1) (1) (1) (1) ψ = (1) φD φD R2n1 (·, 1)R2n1 −1 (·, 0) − R2n1 (·, 0)R2n1 −1 (·, 1) l0 !2 1 (0) (1) = φD φD . (1) l0 This complete the proof in case p = 1. For p > 1, let T 1 be the tree obtained from T by removing all vertices of Tp and all edges which are incident with a vertex of of Tp . The tree T 1 with path from v0 to vp−1 satisfies the induction hypothesis with p − 1. Therefore, in the notation of Corollary 5.4.2, 

p−1 Y

φ1ND φ1DN − φ1NN φ1DD = 

p−1 1 Y (j)

j=1

l0

2 (j) φD  .

j=0

Let T 2 be the tree Tp , augmented by the vertex vp−1 and the edge ep . Applying the case p = 1 to the tree T 2 and the path from vp−1 to vp it follows that φ2ND φ2DN

−

φ2NN φ2DD

=

1

(p) φ (p) D l0

!2 ,

where we have used that removing the edge ep from T 2 produces an isolated vertex 1 1 2 2 and the tree Tp . Setting vin = v0 , vout = vp−1 , vin = vp−1 , vout = vp and observing that T has generalized Neumann conditions at vp−1 , T can be considered as the series connection of T 1 and T 2 . An application of Corollary 5.4.2 completes the proof. 

5.5

Parallel connection of graphs

Now we consider parallel connection of two graphs G1 and G2 , i.e., with the 1 notation from the previous section, a graph G obtained by connecting vin with j 2 1 2 vin and vout with vout . This construction is illustrated in Figure 5.1, where vin j and vout are pendant vertices of Gj for j = 1, 2, respectively, where all edges are j j outgoing from vin and incoming into vout . However, this is not required in general. For A, B ∈ {D, N } let φAB be the characteristic polynomial of the problem 1 2 on G, where A corresponds to the vertex vin := vin = vin , B corresponds to the 1 2 vertex vout := vout = vout , D stands for the generalized Dirichlet conditions and N stands for the generalized Neumann conditions. Corresponding notation is used for G1 and G2 .

5.5. Parallel connection of graphs

201

e1g1

e2g2

e11

1 2 vout = vout

e21 1 2 vin = vin

Figure 5.1

Theorem 5.5.1. For parallel connected graphs with the notation introduced above, φDD = φ1DD φ2DD , φDN = φND =

φ1DN φ2DD φ1ND φ2DD

+ +

φ1DD φ2DN , φ1DD φ2ND .

(5.5.1) (5.5.2) (5.5.3)

Proof. The identity (5.5.1) follows by applying Lemma 5.3.2 twice and by then 1 applying Lemma 5.3.1. Let G0 be the graph obtained by only joining vout and 2 1 2 vout . In (5.5.2) we have the generalized Dirichlet condition at vin = vin , and therefore Lemma 5.3.2 shows that φDN is the characteristic polynomial of G0 with 1 2 1 generalized Neumann condition at vout = vout and Dirichlet conditions at vin 2 and vin . An application of Lemma 5.3.4, Proposition 5.3.7 and Lemma 5.3.1 gives (5.5.2). The same reasoning with incoming and outgoing vertices interchanged proves (5.5.3).  The formula for φNN is much more involved, and therefore we are going to state and prove it separately. But first we will introduce some notation. Let ΦNN , Φ1NN and Φ2NN be characteristic matrix functions of G, G1 and G2 with (generalized) Neumann conditions at the entrance and exit vertices, respectively. Adapting the construction in the paragraph before Lemma 5.3.4 to parallel connected graphs, let s ∈ {1, 2} and κ ∈ {in, out}, let gs be the number of edges (s) e κ(s) ). For the matrices in Gs , and choose bijective maps τκ,s : {1, . . . , d(vκ )} → I(v

Chapter 5. Spectral problems on graphs

202

s ΦAB it is convenient to keep the row and column indices inherited from ΦAB . Since these matrices are only unique up to a simultaneous permutation of rows and columns, we may take, for convenience, 1, . . . , g1 as the row and column indices 1 ˜ = out when for ΦAB and g1 +1, . . . , g as the row and column indices for Φ2AB . Let κ κ = in and κ ˜ = in when κ = out. We define the (2gs ) × (2gs ) matrix function Φsκ (s) obtained from ΦsNN by replacing the rows with index τκ˜ ,s (j) for j = 1, . . . , d(vκ˜ ) c for j = 1 and s = 1, with the rows given by the linear functionals L (s)

Lτcκ,s (1)

for j = 1 and s = 2, and by by Finally put φκs = det Φsκ .

Lcτ˜ (j) k,s

τκ,s (d(vκ )) − Lcτ˜ (j−1) k,s

(s)

for j = 2, . . . , d(vκ˜ ).

Theorem 5.5.2. The characteristic polynomial of the Neumann-Neumann problem on G is 2 1 2 φNN = φ1NN φ2DD + φND + φ1DN φ2ND + φ1DD φ2NN − φ1in φ2out − φ1out φin φDN . (5.5.4) (j)

j . Due to Proof. The assumptions of Lemma 5.3.4 are satisfied with v0 = vin 0 . some clash of notation, we will denote the functions φAB in Lemma 5.3.4 by φAB 2 1 Applying Lemma 5.3.4 first to G and v0 being vin = vin and then to G0 and v0 1 2 being vout and observing part 2 of Remark 5.3.6 and Proposition 5.3.7 = vout shows that (2) 0 0 − φ(1) + φDN φNN = φND vin − φvin 1 1 − φv(2) φ2DN + φDN = φ1NN φ2DD + φND . φ2ND + φ1DD φ2NN − φv(1) in in (1)

(2)

(1)

We still have to find the representations of φvin and φvin . First consider φvin . (1) b 11 from the proof of Lemma 5.3.4 with Recalling the construction of φv0 = − det Φ 0 b v0 = vin , we let Φ be the matrix function Φ11 as defined by (5.3.3) with respect to v0 = vin . Applying now (5.3.3) with respect to v0 = vout results in a representation 0 det Φ0 = det Φ11 + det Φ012 + det Φ021 + det Φ022 .

e in ) ∩ I(v e out ) = ∅, it is clear that the transformations Since vin 6= vout implies I(v b 11 and then to the above representation of Φ0 are independent of from ΦNN to Φ e in ) ∪ I(v e out ), the rows with each other. For indices k = 1, . . . , 2g1 with k 6∈ I(v 0 0 , and Φ022 and only depend on , Φ21 index k are the same for ΦNN , Φ0 , Φ011 , Φ12 the variables ur with r = 1, . . . , 2g1 . By construction this extend to k = 1, . . . , 2g1 e 1 ) e 1 ) for the matrix function Φ0 and then additionally to #I(v with k ∈ I(v out in e out ) for the matrix functions Φ0 , Φ0 and Φ0 . Since rows with indices in I(v 11 12 21 b 11 , and therefore also of Φ0 , Φ0 and Φ0 , corresponding to also one row of Φ 21 11 12 2 the boundary conditions at vin only depends on the variables corresponding to 1 0 0 , Φ012 and Φ21 , each of the three matrices Φ11 the edges incident with vin has a (2g1 + 1) × (2g) submatrix of rank at most 2g1 . Hence their determinants are 0. From (5.3.7) we therefore see that  (2)  (1) (1) . φvin = det Φin out

5.6. Multiplicities of characteristic values of spectral problems on graphs

203

(2)  (1) is obtained from ΦNN by replacing the rows with indices By definition, Φin out

(1)

(2)

e e in I(v in ) and I(vout ) with rows corresponding to (generalized) Neumann condi(2) (1) tions at vin for G1 and vout for G2 , respectively, whereas the remaining changes (1) (2) 1 2 , result in rows corresponding to the conditions at vout and vin as in Φin and Φout respectively. The unchanged rows correspond to boundary conditions at vertices of G which are distinct from vin and vout and can therefore be identified with rows of 2 Φ1in and Φout , respectively. The new row with index τin,2 (1) is given by the linear (1) c functional Lτin,1 (d(vin )), whereas the new row with index τout,1 (1) is given by the linear functional Lcτin,1 (1). Interchanging these two rows, the resulting matrix may be written as a block diagonal matrix, with one block being Φ1in and the other (1) 1 2 2 φout . Interchanging the indices 1 and 2 . This shows that φv0 = −φin block is Φout (2) gives the representation for φv0 .  Remark 5.5.3. 1. The construction of the matrix functions Φκs (κ ∈ {in, out}, (1) s ∈ {1, 2}) is quite technical, but becomes easier when d(vκ ) = 2, i.e., d(vκ ) = (2) (s) e κ ) consists of a single index, say kκ,s . Then Φs is d(vκ ) = 1. In this case, I(v κ s obtained from ΦNN by replacing the row which describes the Neumann condition (s) (s) at vκ˜ by a row which describes the Dirichlet condition at vκ in Gs . i 2. The parallel connection G of m ≥ 2 connected graphs G (i = 1, . . . , m) with at i i and vout of each graph least two vertices is obtained by choosing distinct vertices vin i i i G and identifying all vertices vin (i = 1, . . . , m) and all vertices vout (i = 1, . . . , m). A recursive application of Theorems 5.5.1 and 5.5.2 to the case m ≥ 3 gives formulas which express the characteristic functions of G in terms of characteristic i functions of Gi and functions of the form φiin and φout . Theorems 5.5.1 and 5.5.2 immediately lead to Corollary 5.5.4. The following Lagrange type formula holds:
2
2 1 1 1 1 − φNN φDD φDD φND φDN − φNN φDD = φND φDN

2 2 φ1DD + φ2ND φ2DN − φ2NN φDD 2 2 1 2 + φ1DD φ1in φ2DD φout + φDD φ1out φDD φin .

5.6

(5.5.5)

Multiplicities of characteristic values of spectral problems on graphs

In this section we consider connected graphs with at least one edge. Whenever needed the graphs will be assumed to be digraphs. We recall from Remark 5.1.3 that the spectral data of the problem (4.1.10), (5.1.1), (5.1.2), (5.1.3) do not depend on the orientation, and therefore a convenient orientation of the edges may be chosen when considering spectral data of this problem. We start this section with a lemma, for which we need the following notation.

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204

Notation 5.6.1. Let G be a digraph with a given indexing of the edges. 1. The set V ? is the set of all v ∈ V int ∩ V f satisfying I − (v) 6= ∅, I + (v) 6= ∅, min I − (v) < min I + (v) and max I − (v) < max I + (v). Set p? = #V ? . 2. The set V † is the set of all v ∈ V int ∩ V f satisfying I − (v) 6= ∅, I + (v) 6= ∅ and max I − (v) ≥ max I + (v). Set p† = #V † . Lemma 5.6.2. Let G be a connected digraph. Then the maximal multiplicity of a characteristic value of the problem (4.1.10), (5.1.1), (5.1.2), (5.1.3) on G does not exceed g + p† − p? . Proof. First we observe that for each vertex v ∈ V int ∩V f we can choose a bijective e map τv : {1, . . . , d(v)} → I(v) with τv ({1, . . . , d− (v)}) = {2j − 1 : j ∈ I − (v)}. We may further assume that τv is increasing on {1, . . . , d− (v)} when d− (v) > 1 and decreasing on {d− (v) + 1, . . . , d(v)} when d+ (v) > 1. Let z ∈ C. For each v ∈ V int ∩ V f define the linear functionals Lbτv (j) (z) := Lτv (j) (z) − Lτv (j+1) (z) d(v) P 1 (j ∈ {1, . . . , d(v) − 1}) and Lbτv (d(v)) (z) := d(v) Lτv (r) (z). For all remaining r=1

indices 1 ≤ k ≤ 2g let Lbk (z) := Lk (z). b Then define the (g + p? − p† ) × (g + p? − p† ) matrix Φ(z) as follows. For convenience, we take a subset Λ of the integers from 1 to 2g as the index set, defined as follows. For v ∈ V c take all indices 2j − 1 with j ∈ I − (v). For v ∈ V ? take all indices 2j − 1 with j ∈ I − (v) and the index τv (1) + 1. For v ∈ V † take all indices τv (1), . . . , τv (d− (v) − 1). For all v ∈ V f \ (V ? ∪ V † ) take all indices τv (1), . . . , τv (min{d− (v), d(v) − 1}), and the index τv (1) + 1 if I + (v) = ∅. We are now going to verify that all chosen indices are mutually distinct. Indeed, indices of the form 2j − 1 are odd and are mutually distinct indices from e I(v). On the other hand, indices of the form τv (1)+1 are all even and the numbers τv (1) + 1 are mutually distinct. We conclude that the number of chosen indices is d− (v) + all v ∈ V ? , d− (v) − 1 for all v ∈ V † , and d− (v) for all other v ∈ V . P1 for − Since d (v) = g, it follows that the set Λ of these indices has g + p? − p† v∈V

b elements, which will be the ordered set of row and column indices of Φ(z). b The rows of Φ(z) will now be determined by the coefficients of the uk (k ∈ Λ) of the following linear functionals. For an odd index k, the row with index k is determined by Lbk (z). For an even index k, which is of the form k = τv (1) + 1 for b some v ∈ V f , the row with index k is determined by Lbτv (d(v)) (z). In this way, Φ(z) has been obtained as a submatrix of a linear transformation of Φ(z). Therefore b the rank of Φ(z) does not exceed the rank of Φ(z). b Now we are going to prove that Φ(z) has rank g +p? −p† by showing that it is an upper triangular matrix with nonzero diagonal entries. This will complete the proof in view of Proposition 5.2.1. For v ∈ V c and j ∈ I − (v) we have Lb2j−1 (z)U = b Lc2j−1 (z)U = u2j−1 , that is, this row with index 2j − 1 of Φ(z) has entry 1 in the f diagonal and all other entries are 0. For v ∈ V and j = 1, . . . , min{d− (v), d(v)−1}

5.6. Multiplicities of characteristic values of spectral problems on graphs

205

we have Lbτv (j) (z)U = Lτcv (j) (z)U − Lτcv (j+1) (z)U = uτv (j) − Lcτv (j+1) (z)U. When j < d− (v), then Lcτv (j+1) (z)U = uτv (j+1) with τv (j + 1) > τv (j), and therefore the row with index τv (j) has entry 1 in the diagonal, an entry −1 in the column with index τv (j + 1) if τv (j + 1) ∈ Λ, and all other entries 0. However, (r) (r) when j = d− (v), then Lτcv (j+1) (z)U = R2nr (z, 1)u2r−1 + R2nr (z, 0)u2r with 2r = τv (d− (v) + 1)). By definition of Λ we have v 6∈ V † , I − (v) 6= ∅ and I + (v) 6= ∅. Hence τv (d− (v)) = max I − (v) < max I + (v) = τv (d− (v) + 1). Therefore the row with index τv (d− (v)) has entry 1 in the diagonal, possibly two nonzero entries in two columns to the right of the diagonal, and all other entries 0. The remaining rows, which have row index τv (1) + 1, are given by d(v)

1 X Lτ (r) (z)U d(v) r=1 v  X  (r) X (r) = R2nr −1 (z, 1)u2r−1 + R2nr −1 (z, 0)u2r −

Lbτv (d(v)) (z)U =

r∈I + (v)

u2r

(r) r∈I − (v) l0

for v ∈ V ? and v ∈ V f with I + (v) = ∅. This gives in either case that this row has 1 the entry − (r) in the columns with indices 2r and r ∈ I − (v). In particular, for l0

2r − τv (1) + 1 this is a nonzero entry in the diagonal, whereas all other of these nonzero entries are to the right of the diagonal. Further nonzero entries can only occur for the case v ∈ V ? , namely possibly in columns with indices 2r − 1 and 2r for r ∈ I + (v). But these indices are larger than τv (1) + 1 by definition of V ∗ . b Hence we have indeed shown that Φ(z) is an upper triangular matrix with nonzero diagonal elements. b Since the rank of Φ(z) does not exceed the rank of Φ(z), it follows that the rank of Φ(z) is at least g + p∗ − p† . Therefore the multiplicity of a characteristic value z of the problem (4.1.10), (5.1.1), (5.1.2), (5.1.3) on G, which equals the defect of Φ(z), does not exceed 2g − (g + p∗ − p† ) = g + p† − p∗ .  The estimate for the maximal multiplicity of characteristic values in Lemma 5.6.2 depends on the choice of the directions of the edges and will in general be larger than the maximal possible multiplicity. However, below we will apply Lemma 5.6.2 to some graphs where a suitably chosen orientation gives indeed this maximal value for the multiplicity of some characteristic values. For a vibration problem on a graph G with an indexing ej (j = 1, . . . , g) pen of the edges of G let IN := ∅ if G has at most one pendant vertex or if G is a string, and if all vertices of G carry (generalized) Neumann conditions. Otherwise, pen IN denotes the set of indices j ∈ {1, . . . , g} for which one end of ej is a pendant vertex of G with a Neumann condition. Note that if G has an edge where both ends pen are pendant vertices, then the connected graph G is a string. Therefore j ∈ IN implies that exactly one end of ej is a pendant vertex with a Neumann condition.

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206

Definition 5.6.3. Let G be a connected digraph with edges ej (j = 1, . . . , g). Then a problem (4.1.10), (5.1.1), (5.1.2), (5.1.3) on G is called rigged if (j)

(j)

(j)

(j)

pen ), R2nj (2, 0) = 0, R2nj −1 (2, 0) = 1, R2nj (2, 1) = 1, R2nj −1 (2, 1) = 0 (j 6∈ IN (5.6.1) (j)

(j)

(j)

(j)

pen ), R2nj (2, 0) = 1, R2nj −1 (2, 0) = 0, R2nj (2, 1) = 0, R2nj −1 (2, 1) = −1 (j ∈ IN (5.6.2) (j)

pen and l0 = 1 holds for j = 1, . . . , g. For j ∈ IN it will always be assumed that the + edge ej is directed in such a way that v (ej ) is the pendant vertex with Neumann condition.

We recall from Examples 1.2.17 and 1.2.18 that every connected digraph with at least 3 beads on each edge has a rigged problem. The identities (5.6.1) and (5.6.2) immediately lead to Lemma 5.6.4. Let G be a rigged digraph and U ∈ C2g . Then Lc2j−1 (2)U = u2j−1 Lc2j (2)U Lc2j (2)U

= u2j−1 = u2j

(j = 1, . . . , g), pen (j ∈ {1, . . . , g} \ IN ), pen (j ∈ IN ),

and 

Lfv (2)U = −

1  d(v)

 X r∈I + (v)

u2r−1 +

X

u2r−1  +

r∈I − (v)

X r∈I + (v)

u2r −

X

u2r

r∈I − (v)

pen for v ∈ V f when v 6∈ {v + (ej ) : j ∈ IN }, whereas

Lfv+ (ej ) (2)U = −u2j−1 − u2j

pen (j ∈ IN ).

In Appendix C we have introduced the matrix of cycles, which will now be augmented to the matrix of paths. c = (M cP,j ), where the row Definition 5.6.5. Let G be a digraph. The matrix M index P runs through all oriented paths in G and j = 1, . . . , g, called the matrix of paths of G, is defined as follows. cP,j = 0. 1. For an edge ej which does not belong to the path P , set M 2. For an edge ej which belongs to the path P and whose direction coincides cP,j = 1. with the orientation of the path, set M 3. For an edge ej which belongs to the path P and whose direction is opposite cP,j = −1. to the orientation of the path, set M

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207

Notation 5.6.6. For a rigged problem on G and a directed path P in G, the vector P P c U P = (urP )2g r=1 is defined by u2j := MP,j and u2j−1 := 0 (j = 1, . . . , g). Also, let (0)

(0)

(0)

U (0) = (ur )2g r=1 be defined by u2j−1 = 1 and u2j = 0 (j = 1, . . . , g). Lemma 5.6.7. Let G be a rigged digraph and let P be a path in the graph G. Then Lfv (2)U P = 0 for all vertices v ∈ V f which are not ends of P . Proof. When the vertex v of G is not a vertex of P , then its incident edges are not edges of P , and therefore Lvf (2)U P = 0. However, for a vertex v of P which is not an endpoint of P , there are, in the direction of P , one incoming edge ej1 and one outgoing edge ej2 of P at v. We have to note that j1 = j2 when P is a loop; but what follows also holds in this case. Only these two edges may give a contribution to Lfv (2)U P . By Lemma 5.6.4, for s ∈ {1, 2} the contribution of the edge ejs is − P + P it follows uP 2js if js ∈ I (v) and −u2js if js ∈ I (v). Due to the definition of U that the edge ej1 gives the contribution 1 and the edge ej2 gives the contribution −1 to Lvf j (2)U P , and therefore Lvf (2)U P = 0.  Corollary 5.6.8. Let G be a rigged digraph. 1. When P is a cycle in G, then Φ(2)U P = 0. 2. Let P be a chain in G with each of its ends being a pendant vertex of G. If P is not a string with Neumann conditions at both ends, then Φ(2)U P = 0. 3. If all vertices of G carry (generalized) Neumann conditions, then Φ(2)U (0) = 0. Proof. Recalling the definition of Φ, statement 1 is an immediate consequence of Lemmas 5.6.4 and 5.6.7 if we observe that Lfv occurs in Lr only if v ∈ V f . Statement 2 follows in a similar manner, but we still have to investigate the ends of P . Therefore, let v be an end of P . There is a unique index j such that v is pen , then v = v + (ej ) carries the Neumann condition and incident with ej . If j ∈ IN − v (ej ) is an interior vertex or a pendant vertex with Dirichlet boundary condition. Then Lemmas 5.6.4 and 5.6.7 show that L2j−1 (2)U P = 0 and L2j (2)U P = u2j + (−u2j−1 − u2j ) = −u2j−1 = 0. pen If j 6∈ IN , then L2j−1 (2)U P = 0 and L2j (2)U P = 0 immediately follow from Lemmas 5.6.4 and 5.6.7. pen For the proof of statement 3 observe that IN = ∅. Then Lrc (2)U (0) = 1 for r = 1, . . . , 2g and Lemma 5.6.4 shows that Lvf (2)U (0) = −1 for all vertices v of G. Hence Φ(2)U (0) = 0 by definition of Φ. 

Let ppen and pint denote the number of pendant vertices and interior vertices of a graph G, respectively. Corollary 5.6.9. Let G be a rigged digraph such that all interior vertices carry generalized Neumann conditions. Then 2 is a characteristic value whose multiplicity is at least µ + 1 if all vertices carry Neumann conditions and at least µ + ppen − 1 otherwise.

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208

Proof. If µ > 0, then choose a fundamental set C of µ cycles. If ppen ≥ 2, fix one pendant vertex and choose a set P of ppen − 1 paths in G which connect this vertex to exactly one of the remaining pendant vertices. Each of these pendant vertices has exactly one incident edge, say eji (i = 1, . . . , ppen − 1). Hence the rows c with row indices P ∈ C ∪ P are linearly independent of the matrix of paths M since exactly one row of this submatrix has a nonzero entry in the column with index ji (i = 1, . . . , ppen − 1). It follows that the vectors U P (P ∈ C ∪ P) and U (0) are linearly independent. If all vertices carry Neumann conditions, then Corollary 5.6.8, parts 1 and 3, shows that the dimension of the null space of Φ(2) is at least µ + 1. Otherwise, Corollary 5.6.8, parts 1 and 2, shows that the dimension of the null space of Φ(2) is at least µ + ppen − 1. An application of Proposition 5.2.1  completes the proof.

Observe that the multiplicity obtained in Corollary 5.6.9 is at least 1. Indeed, if not all vertices carry (generalized) Neumann conditions, then ppen ≥ 1. If µ = 0, then G is a tree, and property (iii) of Proposition C.2.5 shows that ppen ≥ 2. Finally, we consider the case that the vibration problem has generalized Dirichlet conditions at interior vertices. Lemma 5.6.10. Let G be a rigged digraph such that the number δ of interior vertices of G with generalized Dirichlet conditions is positive. Then 2 is a characteristic value whose multiplicity is at least µ + ppen + δ − 1. Proof. Cutting the graph G at those interior vertices v which carry generalized Dirichlet conditions generates a graph GN where each such v is replaced by d(v) pendant vertices with Dirichlet conditions. Then all interior vertices of GN have generalized Neumann conditions, and GN has the same characteristic function as G, see Lemma 5.3.2. Let VD be the set of interior vertices of G with generalized Dirichlet conditions. Then the number of vertices of GN is pGN = pG +

X

(d(v) − 1)

v∈VD

since each of the d(v) edges in G which is incident with v ∈ VD gets its own vertex in GN . The number of edges does not change, i.e., g GN = g G . However, GN may no longer be connected, so that κGN may be larger than 1. Furthermore, the number of pendant vertices of GN is clearly pN pen = ppen +

X v∈VD

d(v).

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209

From Lemma C.2.3 one concludes that µGN = g GN + κGN − pGN = g G + κGN − pG −

X

(d(v) − 1)

v∈VD

= µG − 1 + κGN −

X

(d(v) − 1).

v∈VD

Since G and each component of GN have at least one vertex with (generalized) Dirichlet condition, each of the components Gj (j = 1, . . . , κGN ) of GN is a rigged digraph and has at least one pendant vertex with Dirichlet conditions. By Corollary 5.6.9 the multiplicity of 2 as a characteristic value of the problem on Gj is (j) (j) µGj + ppen − 1, where ppen is the number of pendant vertices of Gj . Since the union of fundamental sets of cycles in Gj (j = 1, . . . , κGN ) is clearly a fundamental set of cycles in GN and since the union of the sets of pendant vertices of Gj (j = 1, . . . , κGN ) is clearly the set of the pendant vertices of GN , it follows that GN κX

GN GN (µGj + p(j) + pN pen − 1) = µ pen − κ

j=1 N = µG − 1 + ppen −

X

(d(v) − 1)

v∈VD

= µG − 1 + ppen + δ. From Lemmas 5.3.2 and 5.3.1 we can now conclude that 2 is a characteristic value  of the problem on G whose multiplicity is at least µ + ppen + δ − 1.

5.6.1

Cyclically connected graphs

The maximum of the multiplicities of characteristic values for cyclically connected graphs is given in the following theorem. Theorem 5.6.11. Let G be a cyclically connected graph with generalized Dirichlet conditions at δ ≤ p vertices and generalized Neumann condition at the remaining vertices. Then the maximal multiplicity of a characteristic value is µ + |δ − 1|. The given maximal value is achieved for problems with at least 3 beads on each edge and for a suitable choice of the lengths of the edges and the distribution of the beads. Proof. Let m be the maximal multiplicity of a characteristic value. We choose a digraph structure on G according to Proposition C.4.9. First let δ = 0. Then V ∗ = V0× and hence p? ≥ p − 1 and p† ≤ 1 by Proposition C.4.9. Observing Lemmas 5.6.2 and C.2.3, it follows that m ≤ g + p† − p? ≤ g + 2 − p = µ − κ + 2 = µ + 1.

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210

If δ > 0, then we may choose the vertex w in Proposition C.4.9 to be a vertex with generalized Dirichlet conditions. Then p? = p − δ and p† = 0, and therefore m ≤ g + p† − p? ≤ g − p + δ = µ − κ + δ = µ + δ − 1. In view of Example 1.2.17, the graph G can be made into a rigged digraph with the given number of beads on each edge. Then Corollary 5.6.8 for δ = 0 and Lemma 5.6.10 for δ > 0 show that the upper bounds are indeed the maximal  possible values.

5.6.2

Trees and generalized quasi-trees

The maximum of the multiplicities of characteristic values for trees and generalized quasi-trees is given in the following theorem. Theorem 5.6.12. Let G be a tree or a generalized quasi-tree with generalized Dirichlet conditions at δ ≤ pint interior vertices and generalized Neumann condition at the remaining interior vertices. Then the maximal multiplicity of a characteristic value is µ + ppen + δ − 1 = g − pint + δ. The maximum is achieved for problems with at least 3 beads on each edge and for a suitable choice of the lengths of the edges and the distribution of the beads. Proof. Observing that ppen + pint = p, the identity µ + ppen = g − pint + 1 follows immediately from Lemma C.2.3. Let m be the maximal multiplicity of a characteristic value. Since each maximal cyclically connected subgraph of G has at least two vertices in common with its complement, it is clear that the vertices in the tree GC representing maximal cyclically connected subgraphs in G are interior vertices of GC . Therefore any pendant vertex of GC is also a pendant vertex of G. Since GC is a tree, it follows that ppen > 0. Any pendant vertex of G will be chosen as the root. With an indexing and orientation of the edges according to Lemma C.4.8 it follows that V0× = V int , which implies p? = pint − δ and p† = 0. Here we have to observe that the proof of Lemma C.4.6 provides the required indexing if G is a tree. Therefore Lemma 5.6.2 gives m ≤ g + p† − p? ≤ g − pint + δ. It follows from Corollary 5.6.8 for δ = 0 and Lemma 5.6.10 for δ > 0 that the given upper bound can be achieved for a suitable digraph structure on G. 

5.6.3

Connected graphs

In this subsection we consider connected graphs G which are not cyclically connected. We exclude the case g = 0, i.e., the graphs have at least one edge. Fixing now such a graph G and denoting by C the set of all maximal cyclically connected subgraphs of G, we know from Proposition C.4.4 and Corollary C.4.5 that T := GC is a tree.

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211

Theorem 5.6.13. Let G be a connected but not cyclically connected graph with at least one edge. It is supposed that each interior vertex carries a generalized Neumann condition. Then the maximal possible multiplicity of a characteristic T − 1. value is µ + ppen Proof. Again we are going to apply Lemma 5.6.2. To this end we have to find a lower bound for p? and an upper bound for p† . Let C1 be the set of all Γ ∈ C which have exactly one common vertex vΓ with their complement in G and such that vΓ has exactly one incident edge in the complement of Γ. Let C2 be the set of all Γ ∈ C which have exactly one common vertex vΓ with their complement in G and such that vΓ has at least two incident edges in the complement of Γ. Let C3 be the set of all Γ ∈ C which have at least two common vertices with their complement in G. Clearly, each Γ ∈ C belongs to exactly one of C1 , C2 or C3 . By Proposition C.4.4, the graph G3 := GC1 ∪C2 (see (C.4.1)) has C3 as the set of its maximal cyclically connected subgraphs. Since each Γ ∈ C1 ∪ C2 has exactly one vertex in common with its complement, it is clear that GC1 ∪C2 is the subgraph of G which consists of all edges in G which are not edges of any Γ ∈ C1 ∪ C2 and all vertices which are incident with these edges. In particular, any vertex of Γ ∈ C3 which is a vertex of the complement of Γ in G is also a vertex of the complement of Γ in G3 . But by definition of C3 , each Γ ∈ C3 has at least two vertices in common with its complement in G and therefore also in G3 . By definition C.4.7 this means that G3 is a generalized quasi-tree if C3 6= ∅ or a tree if C3 = ∅. Furthermore, the pendant vertices of G3 are exactly the pendant vertices of T . Let any pendant vertex of G3 be its root. From the proof of Lemma C.4.6 and from Lemma C.4.8 we know that there is an orientation and indexing of the edges of G3 satisfying properties 1, 2 and 3 of Lemma C.4.8. For convenience, we allow the indices of the edges of G3 and of G to be arbitrary real numbers. We now reverse the orientation and the indexing of all edges of G3 ; for example, if an edge is originally indexed by er , we could now write e−r . Then it is clear that property 3 of Lemma C.4.8 still holds, whereas I3+ (v) = ∅ for the root v of G3 and I3− (v) = ∅ for all other pendant vertices of G3 . Here and for other notation a lower index 3 indicates that this quantity is taken with respect to the graph G3 , whereas a lower index Γ will indicate that it is taken with respect to a corresponding Γ ∈ C. The maximal cyclically connected subgraphs Γ ∈ C2 are attached at interior vertices vΓ of G3 , and therefore vΓ ∈ V3? , which implies that jΓ := min I3− (vΓ ) < min(I3 \ {jΓ }) =: jΓ0 . Taking vΓ as the vertex w from Lemma C.4.9 with respect to Γ, the edges of the graph Γ will be oriented and indexed according to Lemma C.4.9, where we may assume that all indices of the edges of Γ lie between jΓ and jΓ0 . Then it follows from Lemma C.4.9 that v ∈ V ? for v ∈ Γ \ {vΓ }. Furthermore, it is not hard to see that vΓ ∈ V3? gives vΓ ∈ V ? . On the other hand, the maximal cyclically connected subgraphs Γ ∈ C1 are attached at pendant vertices vΓ of G3 . Then there is a unique index jΓ such that

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212

I3 (vΓ ) = {jΓ }. Taking vΓ as the vertex w from Lemma C.4.9 with respect to Γ, the edges of the graph Γ will be oriented and indexed according to Lemma C.4.9, where we may assume that all indices of the edges of Γ lie below jΓ . Then it follows from Lemma C.4.9 that v ∈ V ? for v ∈ Γ \ {vΓ }. Furthermore, it is not hard to see that vΓ 6∈ V † when vΓ is not the root of G. We first assume that V † = ∅. In the last two paragraphs it was shown that all vertices which are not pendant vertices of T belong to V ? . Therefore, p† = 0 T , and hence Lemma 5.6.2 shows that the multiplicity of a and p? ≥ pG − ppen T T characteristic value of the problem does not exceed g G − pG + ppen . = µ − 1 + ppen The second case is that V † 6= ∅. By the above considerations this can only happen when the root of G3 is some vΓ for Γ ∈ C1 . We now attach an edge eg+1 to G at vΓ , oriented in such a way that vΓ is the exit of eg+1 . The entrance of eg+1 will be denoted by v+ . Denoting this augmented graph by G+ it is clear that the cyclomatic numbers of G and G+ are the same. The corresponding tree T+ is obtained by attaching eg+1 at vΓ , so that vΓ becomes an interior vertex of T+ , whereas v+ becomes a pendant vertex of T+ , which we will choose as the root. Therefore V+† = ∅, and by what has already been shown it follows that the multiplicity of each characteristic value for the problem on G+ does not exceed µ + pTpen − 1. Now let z be a characteristic value of the problem on G with multiplicity m. By Theorem 1.3.1 there is a distribution of beads on eg+1 such that z is a characteristic value of the Dirichlet-Neumann problem on eg+1 . If now U is a solution (g+1) corresponding to the problem on G, then there is u2g such that R2ng+1 (z, 0)u2g is the unique value given by the continuity condition (5.1.1) at vΓ for the prob(g+1) lem on G. Here we recall that the Neumann condition R2ng+1 −1 (z, 0) = 0 implies (g+1)

that R2ng+1 (z, 0)u2g 6= 0. Hence augmenting U by (0, u2g ) leads to a solution of the problem on G+ . Therefore the multiplicity of the characteristic value z for the problem on G+ is at least m. Then the discussion in the previous paragraph shows that m ≤ µ + pTpen − 1. To show that the bound can be reached we start with a rigged problem on G3 with Neumann conditions at all vΓ for Γ ∈ C1 and consider data according to Example 1.2.17 on all the remaining edges. Note that the indexing and orientation may be different from that chosen in the first part of this proof. In particular, each Γ ∈ C is a rigged digraph, and Corollary 5.6.8 shows that Φ(2)U P = 0 for every cycle P in such Γ with U P considered in Γ. But then a reasoning as in the proof of Lemma 5.6.7 implies that also Φ(2)U P = 0 when U P is considered in G. Hence a fundamental set of cycles in G contributes µ linearly independent vectors in the null space of Φ(2). If G3 is a string, then G 6= G3 since G is not a tree. Hence at least one end of G3 carries a Neumann condition. If both ends carry Neumann conditions, then all vertices of G carry Neumann conditions, and we already know from Corollary 5.6.8 that Φ(2)U (0) = 0. Therefore the null space of Φ(2) has dimension at least

5.7. Graphs of symmetric Stieltjes strings

213

µ + pTpen − 1. Finally we consider the case that G3 is either a string with one end carrying a Dirichlet condition and the other end carrying a Neumann condition, or that G3 has at least 3 vertices. Let P be a chain in G3 connecting a pendant vertex in G3 to the root of G3 . Then we know that the vector corresponding to this chain in G3 is in the null space of the characteristic matrix function at 2. For j such that P P are as defined for G3 . We are going ej is an edge of G3 , the values u2j−1 and u2j to define values corresponding to all other edges of G. If an end of P is vΓ with P := uP Γ ∈ C1 and with incident edge ejΓ in G3 and hence in P , then let u2r−1 2jΓ for all r for which er is an edge of Γ. For all remaining indices s = 1, . . . , 2g we put uP s := 0. With a reasoning as in the proofs of Lemma 5.6.7 and Corollary 5.6.8 we conclude that to show Φ(z)U P = 0, only Ls (2)U P = 0 has still to be shown for indices s = 2k − 1 or s = 2k when ek is incident with an end vΓ of P with Γ ∈ C1 and k ∈ I − (vΓ ) or k ∈ I + (vΓ ), respectively. For such an end let v = vΓ and j = jΓ . The vibration problem satisfies (5.6.2) on the edge ej and (5.6.1) on all other edges ek which are incident with v. Since the edge ej is an incoming edge into v and the only edge incident with v which is not an edge of Γ, it follows that   Lfv (2)U = −

X X  uP 1  2j   uP uP = −uP 2r−1 + 2r−1  − 2j .  d(v) d(v) + − r∈I (v) r6=j

r∈I (v)

P Hence L2j (2)U P = Lc2j (2)U P + Lfv (2)U P = uP 2j − u2j = 0, whereas for the other edges ek which are incident with v we have Ls (2)U P = Lcs (2)U P + Lfv (2)U P = P uP 2k−1 − u2j = 0. Summarizing the above construction, we have found for q = T ppen − 1 pendant vertices v1 , . . . , vq vectors U (1) , . . . U (q) in the null space of Φ(2) (k)

such that u2ji 6= 0 if and only if k = i (i, k = 1, . . . , q), where ji is the unique index of the edge in G3 which is incident with the vertex vi (i = 1, . . . , q). Therefore the µ vectors U P corresponding to a fundamental set of cycles and the vectors U (1) , . . . , U (q) are linearly independent. We have shown in each of the above cases that the matrix Φ(2) has a null space of dimension at least µ − 1 + pTpen . In view of Proposition 5.2.1 this means that 2 is a characteristic value of multiplicity at least µ − 1 + pTpen . 

5.7

Graphs of symmetric Stieltjes strings

In this section let G be a simple connected metric graph with at least two edges. It is assumed that all edges have the same length and the same distribution of beads (j) and that this distribution is symmetric. This means that if mk (k = 1, . . . , nj ) (j) denotes the masses of the beads and lk (k = 0, . . . , nj ) the lengths of the threads for the j-th edge (j = 1, . . . , g), then these numbers, as well as nj , are independent

Chapter 5. Spectral problems on graphs

214 (j)

(j)

(j)

(j)

of j, mk = mnj −k+1 (k = 1, . . . , nj ) and lk = lnj −k (k = 0, . . . , nj ). We may therefore remove the upper index for the masses and lengths of the threads, as well as the upper index for the corresponding Cauer-Fry polynomials. However, we will use n1 for the number of beads on any of the edges. Then the total number of beads on the graph is n = gn1 . Throughout this section, we will consider the Neumann problem, i.e., all vertices of G are freely moving.

5.7.1

The direct problem

Because of the special structure of the Stieltjes strings on the edges, the characteristic polynomial can be described in terms of Cauer-Fry polynomials for the common Stieltjes string and the (weighted) adjacency matrix of the graph. We are going to consider first generic z ∈ C with R2n1 (z, 0) 6= 0. Then in (5.2.11) we e whose component are defined by u replace U with the vector U e2j−1 = u2j−1 and c e u e2j = L2j (z)U for j = 1, . . . , g. In this notation we suppress the dependence of U e on z since it is not important for what follows. We further define Φ(z) by e = Φ(z)U. e U Φ(z) c For j = 1, . . . , g it follows from the definition of L2j in (5.2.5) that      e2j−1 u 1 0 u2j−1 = , u2j u e2j R2n1 (z, 1) R2n1 (z, 0)

e which shows that Φ(z) is well defined and that g e (z, 0) det Φ(z). φ(z) = det Φ(z) = R2n 1

(5.7.1)

e Next we will associate with Φ(z) a matrix of smaller size but with the same determinant. To this end we introduce for each vertex v of the graph G the unique e For any interior vertex v increasing surjective function τv : {1, . . . , d(v)} → I(v). of G and j = 2, . . . , d(v) we subtract the row with index τv (1) from the row with index τv (j). In view of (5.2.9) and (5.2.10) the new row is represented by eτv (j) − u eτv (1) =: u bτv (j) . Lτv (j) U − Lτv (1) (z)U = Lcτv (j) (z)U − Lτcv (1) (z)U = u eτv (1) for each v ∈ V . The row operations do not change bτv (1) := u We also define u 2g e to U b := (b is uj )j=1 the determinant of the matrix, and the transformation from U given by a triangular matrix whose diagonal elements are 1. Hence the corresponde b has the same determinant as Φ(z) and its ing coefficient matrix with respect to U rows with index τv (2), . . . , τv (d(v)) have 1 in the diagonal position and zeros everywhere else. Deleting all rows and columns with indices τv (j) with j = 2, . . . , d(v) e for interior vertices v leads to a matrix Φ† (z) with det Φ† (z) = det Φ(z). † † To describe the entries of the matrix Φ , we note that Φ (z) can be indexed by the vertices, with indices τv (1) inherited from the original matrix. It is convenient

5.7. Graphs of symmetric Stieltjes strings

215

to write uv† := u bτv (1) (v ∈ V ) and U † = (uv† )v∈V . Since we are interested in the determinant, any arrangement of the rows and columns gives the same result, and therefore the order is irrelevant. For vertices v, w ∈ V let Φ† (v, w) denote the element of Φ† in row v and column w. Since the above row operations did not change the rows of Φ(z) with indices τv (1), we have to express Lτv (1) (z)U in terms † b ; then Φ† (v, w) is its coefficient of u of U . bτw (1) = uw We need some preparation. From u2j =

1 (e u2j − R2n1 (z, 1)e u2j−1 ) R2n1 (z, 0)

(j = 1, . . . , g)

we conclude that R2n1 −1 (z, 1)u2j−1 + R2n1 −1 (z, 0)u2j 1 = (R2n1 −1 (z, 1)R2n1 (z, 0) − R2n1 −1 (z, 0)R2n1 (z, 1)) u e2j−1 R2n1 (z, 0) R2n1 −1 (z, 0) + u e2j R2n1 (z, 0) 1 R2n1 −1 (z, 0) =− u e2j−1 + u e2j l0 R2n1 (z, 0) R2n1 (z, 0) in view of the Lagrange identity (1.2.47). For each v ∈ V it follows from (5.2.9) and (5.2.10) that Lτv (1) (z)U = u eτv (1) − + −

1 d(v)

X

u e2r −

r∈I + (v)

R2n1 −1 (z, 0) R2n1 (z, 0) 1

X

u e2r − u e2r +

r∈I − (v)

X

l0 R2n1 (z, 0)

r∈I + (v)

X

l0 R2n1 (z, 0)

1 1 d(v)

u e2r−1

r∈I + (v)

X

u e2r−1

r∈I − (v)

R2n1 (z, 1) l0 R2n1 (z, 0)

X

u e2r−1 .

r∈I − (v)

In view of u eτv (1) = u bτv (1) and u eτv (j) = u bτv (j) + u bτv (1) (j = 2, . . . , d(v)) we find d(v)

X r∈I + (v)

u e2r +

X r∈I − (v)

u e2r−1 =

X j=1

d(v)

u eτv (j) = d(v)u†v +

X

u bτv (j) .

j=2

e − (er )), and therefore the contribution of u For r ∈ I + (v) we have 2r − 1 ∈ I(v e2r−1 e + (er )), and therefore the to U † is u†v− (er ) , whereas for r ∈ I − (v) we have 2r ∈ I(v contribution of u e2r to U † is u†v+ (er ) . But the vertices v− (er ) for r ∈ I + (v) and v+ (er ) for r ∈ I − (v) are exactly the vertices which are adjacent with v, and each

Chapter 5. Spectral problems on graphs

216

adjacent vertex occurs exactly once since the graph is simple. In view of Lemma 2.2.1 it follows for v, w ∈ V that  d(v) R2n1 (z, 1)  if w = v,   l0 R2n1 (z, 0)  1 Φ† (v, w) = if w 6= v and w is adjacent to v, −  l R (z, 0)  0 2n 1   0 if w 6= v and w is not adjacent to v. Putting ζ := R2n1 (z, 1), T := diag{d(v) : v ∈ V } and denoting by A the adjacency matrix of G, it follows that Φ† (z) =

1 (ζT − A). l0 R2n1 (z, 0)

(5.7.2)

e = T − 12 AT − 12 the following result is With the weighted adjacency matrix A an immediate consequence of (5.7.1) and (5.7.2). Proposition 5.7.1. The characteristic polynomial of the Neumann problem on the graph G of identical symmetric Stieltjes strings satisfies Y g−p e φ(z) = l0−p d(v)R2n (z, 0) det(R2n1 (z, 1)Ip − A). 1 v∈V

The following two theorems now immediately follow from Lemmas C.5.4 and C.5.5, where we observe that A is irreducible since G is connected. Theorem 5.7.2. The characteristic polynomial of the Neumann problem on the graph G of identical symmetric Stieltjes strings is of the form g−p φ(z) = R2n (z, 0)(R2n1 (z, 0) − 1)Pep−1 (R2n1 (z, 1)), 1

where Pep−1 is a polynomial of degree p − 1 with P˜p−1 (1) 6= 0. Theorem 5.7.3. The characteristic polynomial of the Neumann problem on the graph G of identical symmetric Stieltjes strings is of the form g−p 2 m 2 φ(z) = R2n (z, 0)(R2n (z, 1) − 1)R2n (z, 1)Pbp−m −1 (R2n (z, 1)), 1 1 1 1 2

where m ∈ N0 , p − m ∈ N is an even number, Pbp−m −1 is a polynomial of degree 2 p−m bp−m (1) 6= 0, if and only if the graph is bipartite. − 1 with P −1 2 2

For the following results we recall the notation νk and ξk (k = 1, . . . , n1 ) and the properties of the zeros of R2n1 (·, 0) and R2n1 (·, 1) from Section 1.2. Proposition 5.7.4. 1. Under the assumptions of this section, let a ∈ [−1, 1]. Then the zeros of the polynomial R2n1 (·, 1) − a are real and nonnegative and will be denoted by ξ1,a ≤ ξ2,a ≤ · · · ≤ ξn1 ,a .

5.7. Graphs of symmetric Stieltjes strings

217

2. The zeros are simple for a ∈ (−1, 1), ξ1,a = 0 if and only if a = 1, and νn1 = ξn1 ,(−1)n1 . 3. Let −1 ≤ a < b ≤ 1 and j ∈ {1, . . . , n1 }. Then ξj,a < ξj,b when j is even and ξj,a > ξj,b when j is odd. 4. One has ν2j−1 ∈ {ξ2j−1,−1 ξ2j,−1 } for j = 1, . . . , [ n12+1 ] and ν2j ∈ {ξ2j,1 , ξ2j+1,1 } for j = 1, . . . , [ n21 ]. Proof. For convenience, we put ν0 := 0. Then R2n1 (ν0 , 1) = 1 by (1.2.17) and 2 R2n (νj , 1) = 1 (j = 1, . . . , n1 ) by (2.2.2). Since the zeros ξj = ξj,0 (j = 1, . . . , nj ) 1 are simple, we have that (−1)j R2n1 (z, 1) > 0 for z ∈ (ξj , ξj+1 ) (j = 1, . . . , n1 − 1). The interlacing property (1.2.37) gives that R2n1 (νj , 1) = (−1)j (j = 1, . . . , n1 ) and that R2n1 (·, 1) − 1 as well as R2n1 (·, 1) + 1 have an even number of zeros in (ξj , ξj + 1) (j = 1, . . . , n1 − 1). For j = 1, . . . , [ n21 ] it follows that R2n1 (·, 1) + 1 has at least 2 zeros, counted with multiplicity, in (ξ2j−1 , ξ2j ). But this accounts for at least n1 − 1 zeros of the polynomial R2n1 (·, 1) + 1, which means that R2n1 (·, 1) + 1 has exactly two zeros in (ξ2j−1 , ξ2j ) (j = 1, . . . , [ n21 ]) and one more zero νn1 ∈ (ξn1 , ∞) if n1 is odd. This means that all zeros of R2n1 (·, 1) + 1 are real, and with the indicated indexing ξ2j−1 < ξ2j−1,−1 ≤ ν2j−1 ≤ ξ2j,−1 < ξ2j

(j = 1, . . . , [

nj 2 ]),

which proves part 1 for a = −1 and the first statement of part 4. Similarly, it follows for j = 1, . . . , [ n12−1 ] that R2n1 (·, 1) − 1 has at least 2 zeros, counted with multiplicity, in (ξ2j , ξ2j+1 ), one zero at 0, and one zero at νn1 ∈ (ξn1 , ∞) if n1 is even. But this already accounts for all zeros of the polynomial R2n1 (·, 1) − 1, so that ξ1,1 = 0, ξn1 ,1 = νn1 if n1 , and ξ2j < ξ2j,1 ≤ ν2j ≤ ξ2j+1,1 < ξ2j+1

(j = 1, . . . , [

nj −1 2 ]).

This proves part 1 for a = 1 and completes the proof of part 4. Now let a ∈ (−1, 1). Then ξj,(−1)j−1 < ξj < ξj,(−1)j (j = 1, . . . , nj ) shows that R2n1 (·, 1) − a has exactly one zero in each of these n1 intervals. In particular, R2n1 (·, 1) is monotonic on these intervals, which proves part 3 and completes the proof of parts 1 and 2.  1 Theorem 5.7.5. Let (λk )gn k=1 be the nondecreasing sequence of the characteristic values of the Neumann problem on a simple connected metric graph G with cyclomatic number µ ≥ 2 and with the same symmetric distribution of beads on each edge. Then λ1 = 0 and λgj = νj for all j = 1, . . . , n1 .

Proof. Let p0 = p − 2 if G is bipartite and p0 = p − 1 otherwise. In view of Lemmas e has exactly p0 eigenvalues in C.5.4 and C.5.5 the weighted adjacency matrix A (−1, 1), counted with multiplicity, and therefore Proposition 5.7.4 shows that φ has p0 zeros, counted with multiplicity, in each of the intervals (νj−1 , νj ) (j = 1, . . . , n1 ); recall that ν0 = 0. By Theorems 5.7.2 and 5.7.3, Proposition 5.7.4

Chapter 5. Spectral problems on graphs

218

and Lemmas C.5.4, C.5.5 and C.2.3 the remaining zeros are νj with multiplicity µ − 1 > 0, ξj,1 , and ξj,−1 in case of a bipartite graph, (j = 1, . . . , n1 ). It is also clear that λ1 ≥ 0 and that ξ1,1 = 0 gives λ1 = 0. In view of Proposition 5.7.4 we can therefore allocate ξj,1 to the interval [νj−1 , νj ] for j = 1, . . . , nj , and, in case of a bipartite graph, we also allocate ξj,−1 to the interval [νj−1 , νj ] for j = 1, . . . , nj . In this way, each of the intervals [νj−1 , νj ] (j = 1, . . . , n1 ) has been allocated p zeros of φ. The g − p-fold zeros νj are allocated to [νj−1 , νj ] (j = 1, . . . , n1 ). Thus we have allocated g zeros of φ to each of the intervals, and νj is the biggest of  them for j = 1, . . . , n1 ; this means that λgj = νj (j = 1, . . . , n1 ).

5.7.2

The inverse problem

Theorem 5.7.6. Let G be a simple connected metric graph with the same edges and with cyclomatic number µ ≥ 2. Assume that each edge has the same distribution of beads. Then the characteristic values of the Neumann problem on G and the length of an edge uniquely determine the distribution of beads. Proof. By Theorem 5.7.5, the characteristic values of the Dirichlet-Dirichlet problem on an edge of G are uniquely determined by the characteristic values of the Neumann problem on G. An application of Theorem 2.2.3 completes the proof.  Remark 5.7.7. 1. In addition to the string data on the edges, also some information on the graph can be recovered from the spectrum of the Neumann problem on the graph G. If it is known that µ ≥ 1 but the actual value is not known, and if it is known that the graph is bipartite, then λgn1 = νn1 is a characteristic value of multiplicity g − p + 1 = µ, and therefore µ can be recovered from the multiplicity of the largest characteristic value of the Neumann problem on G. The same is true when G is not bipartite but n1 is even. If G is not bipartite, n1 is odd and µ ≥ 2, then the multiplicity of λgn1 is g − p = µ − 1, so that also in this case µ can be recovered from the multiplicity of the largest eigenvalue of the Neumann problem on G. 2. As a particular case consider simple connected graphs G with µ ≥ 2 and with a minimal number of vertices. Hence G has at least two independent cycles. Since G is simple, any cycle has at least 3 vertices, and any two independent cycles must have at least 4 vertices. Hence the minimal number of vertices is p =
4. On the other hand, since G is simple, the maximal number of edges is 42 = 6. Therefore the cases g = 5 and g = 6 are possible, which gives the diamond graph (g = 5, µ = 2) or the tetrahedral graph (g = 6, µ = 3). Observe that both graphs are nonbipartite since they contain cycles with three edges (see Lemma C.5.3). Hence the characteristic values, p = 4 and µ ≥ 2 uniquely determine the graph by part 1 of this remark, and then Theorem 5.7.6 shows that the distribution of the beads is also unique. 3. If we know two of the numbers g, p, µ or two of the numbers p, µ, n1 and if the characteristic values (λk )nk=1 of the Neumann problem on G are known,

5.8. Notes

219

then all four of g, p, µ, n1 can be found from g − p = µ − 1 and gn1 = n. In the case µ ≥ 2 the distribution of the beads is uniquely determined by the characteristic values, and therefore R2n1 (·, 1) is known, and so are R2n1 (αk , 1)

(k = 1, . . . , p).

From the results in Subsection 5.7.1 we know that these p values are the e Now it remains to investieigenvalues of the weighted adjacency matrix A. e has those gate for which graphs with g edges and p vertices the matrix A eigenvalues. 4. We could even consider as little information as possible. That is, we may even consider an arbitrary simple connected graph (including g = 1) and the characteristic values (λk )nk=1 with n ∈ N. For n = 1 this is the string problem with just one bead, which has as unique solution. For n > 1, we consider all pairs (g, n1 ) with gn√ 1 = n and then all simple connected graphs with p vertices, where 21 (1 + 8g + 1) ≤ p ≤ g + 1. Then the results from subsection 5.7.1 may give information about the shape of the graph. As a trivial example we can state that if n is a prime number and not all characteristic values are simple, then the graph has n edges, and each edge carries exactly one bead, positioned at the midpoint of the edge.

5.8

Notes

Inverse spectral and scattering problems on graphs is a rapidly developing branch of analysis. Usually the Sturm-Liouville or the Dirac equation or the string equation is considered on the edges of a graph subject to matching and boundary conditions at the vertices. The tradition of using Dirichlet and Neumann spectra to solve an inverse problem on a graph (see, e.g., [27], [143]) came from the classical results of Borg [15], Levitan and Gasymov [89] and Marchenko [92] for inverse spectral problems on a finite interval. However, in case of graphs it is possible (see, e.g., [86]) to introduce analogues of Neumann and Dirichlet conditions at an interior vertex chosen as the root. Analogues of Theorem 5.4.1 for Sturm-Liouville problem on a graph were obtained in [129] and [86]. To describe the spectrum of a boundary value problem on a metric graph we use characteristic polynomials, i.e., determinants of characteristic matrix functions defined by (5.2.11). This is similar to characteristic functions of Sturm-Liouville problems on graphs (see, e. g. [118]). A result similar to Theorem 5.5.2 for Sturm-Liouville problems on graphs was obtained in [110], [111]. The condition that the number of beads on each edge be at least 3 can be explained as follows. If it is less than 3 then in case of a loop there are one or two eigenvalues. Since the first eigenvalue is simple (see [53] for the case of SturmLiouville problem), the second one is also simple and maximal multiplicity is 1.

220

Chapter 5. Spectral problems on graphs

Analogues of Theorems 5.6.12 and 5.6.13 for Sturm-Liouville problems were obtained in [91] and [76]. Absence of beads at the interior vertices of a graph makes the spectral problem a bit more complicated than in [62], [88], [101] because instead of the equation (λI − L)Y = 0, where L is the discrete Laplacian, we have the equation (λM − L)Y = 0, where M ≥ 0 is a diagonal matrix which has some entries zero when not all interior vertices carry beads. Theorem 5.6.13 and a particular case of Theorem 5.6.11 were obtained in [14].

Chapter 6

Sturm-Liouville problems on graphs 6.1

Selfadjoint differential operators

Let I = (a, b) or I = R. First we recall the definition of the Lebesgue spaces Lp (I) for 1 ≤ p ≤ ∞. For Lebesgue measurable functions f, h : I → C we write f = h (f ≤ h, etc.) if f (x) = h(x) (f (x) ≤ h(x), etc.) for almost all x ∈ I, i.e., if there is a Lebesgue measurable set X ⊂ I such that I \ X has Lebesgue measure zero and such that f (x) = h(x) (f (x) ≤ h(x), etc.) for all x ∈ X. The function f is called essentially bounded if there is such a Lebesgue measurable set X such that f is bounded on X. The relation f = h in the above sense defines equivalence classes of Lebesgue measurable functions, and for 1 ≤ p < ∞, Lp (I) is defined as the set of equivalence classes of measurable functions f on I for which Z |f (x)|p dx < ∞, I

whereas L∞ (I) is the set of equivalence classes of essentially bounded functions on I. As is customary, we will identify equivalence classes with any of the functions representing it. In particular, when such an equivalence class contains a continuous functions, then this continuous function is unique, and we will identify the equivalence class with its continuous representative. The sesquilinear form Z (f, g) := f (x)g(x) dx, f, g ∈ L2 (I), I

defines an inner product on L2 (I) which makes L2 (I) a Hilbert space. We recall some basic properties of Sturm-Liouville operators. We consider the Sturm-Liouville expression `y = −y 00 + qy © Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4_6

(6.1.1) 221

222

Chapter 6. Sturm-Liouville problems on graphs

on an interval [a, b], where q ∈ L1 (a, b) is a real valued function. The domain D(`) of this differential expression is the set of all y ∈ C 1 [a, b] such that y 0 is absolutely continuous. Then y 00 ∈ L1 (a, b) and therefore `y ∈ L1 (a, b) is well-defined. Hence ` with domain D(`) can be considered as an operator from L2 (a, b) to L1 (a, b). Let Lmax be the maximal differential operator in L2 (a, b) associated with `. That is, D(Lmax ) = {y ∈ D(`) : `y ∈ L2 (a, b)},

Lmax y = `y,

(y ∈ D(Lmax )). (6.1.2)

Let l > 0, g ∈ N and for j = 1, . . . , g let `j be formally selfadjoint differential expressions on the interval [0, l] as defined in (6.1.1), i.e., `j yj = −yj00 + qj yj ,

(6.1.3)

where qj ∈ L1 (0, l) (j = 1, . . . , g) are real valued functions. We are going to use the notation   y1   y =  ...  yg for elements in the Hilbert space H=

g M

L2 (0, l).

j=1

Furthermore, when yj ∈ D(`j ) (j = 1, . . . , g) we use the notation     y Y (0) = . , Y Y = R Y (l) y0 It is clear that YR ∈ C4g . Finally, let U be a (2g) × (4g) matrix of rank 2g. Then the operator LU in H is defined by     `1 y1 g   M   LU y =  ...  , D(LU ) = y ∈ (6.1.4) D(Lj,max ) : U YR = 0 .   j=1 `g yg From the results in [98, Section 2] it follows that Proposition 6.1.1. The operator LU is selfadjoint if and only if U J1 U ∗ = 0, where     0 −Ig 0 J0 J1 = . , J0 = Ig 0 −J0 0 Remark 6.1.2. Writing U = (U1 U2 U3 U4 ) with (2g)×g matrices Uj (j = 1, 2, 3, 4), it is clear that the operator LU is self-adjoint if and only if U1 U2∗ − U2 U1∗ − U3 U4∗ + U4 U3∗ = 0.

6.2. Selfadjoint differential operators on graphs

223

6.2 Selfadjoint differential operators on graphs Let G be a metric digraph, where it is assumed that all g edges of G have the same length l > 0. Thus each edge ej (j = 1, . . . , g) of G will be identified with the interval [0, l], and on each edge, the local coordinate x increases in the direction of the edge. On each edge of the graph we consider a Sturm-Liouville problem (6.1.3). Now we are ready to define the operator L. With the notation from the g previous section, its domain D(L) is the set of vector functions y = (yj )j=1 with yj ∈ D(Lj,max ) subject to the following conditions at the vertices of G. At pendant vertices, Dirichlet or Neumann conditions are imposed: yj (0) = 0 or yj0 (0) = 0 if I − (v) = {j} and v ∈ V pen , yj (l) = 0 or

yj0 (l)

+

= 0 if I (v) = {j} and v ∈ V

pen

(6.2.1) (6.2.2)

.

At each interior vertex, a continuity condition #({yj (0) : j ∈ I − (v)} ∪ {yj (l) : j ∈ I + (v)}) = 1 as well as the Kirchhoff condition X yj0 (l) = j∈I + (v)

X

yj0 (0)

(v ∈ V int ),

(v ∈ V int )

(6.2.3)

(6.2.4)

j∈I − (v)

are imposed. The conditions (6.2.1)–(6.2.4) can be written in the form U YR = 0, where U is defined as follows (the indexing of the rows is arbitrary). We use the 4g and observe that notation YR = (yR,j )j=1 yR,j = yj (0), yR,g+j = yj0 (0), yR,2g+j = yj (l), yR,3g+j = yj0 (l),

(j = 1, . . . , g).

Each of the conditions (6.2.1) and (6.2.2) gives rise to a component of U YR of the form YR,αv g+j with αv ∈ {0, 1, 2, 3}. Condition (6.2.3) for a particular v ∈ V int involves the components YR,j with j ∈ I0 (v) := I − (v) ∪ {2g + k : k ∈ I + (v)}. Choosing any j1 (v) ∈ I0 (v) it is clear that the continuity condition at v gives rises to d(v) − 1 components of U YR of the form YR,j − YR,j1 (v) (j ∈ I0 (v), j 6= j1 (v)). Finally, each condition (6.2.4) gives rise to a component of U YR of the form X X YR,g+j (v ∈ V int ). YR,3g+j − (6.2.5) j∈I + (v)

j∈I − (v)

Theorem 6.2.1. The operator L is selfadjoint. Proof. Since the sum of the degrees d(v) (v ∈ V G ) is 2g, the matrix U defined above is a (2g) × (4g) matrix. Since each of the terms YR,k related to any of the conditions (6.2.1) and (6.2.2) as well as YR,j (j ∈ I0 (v), j 6= j1 (v), v ∈ V int ) and any YR,k occurring in (6.2.5) is present in exactly one component of U YR it is clear that the rank of U is 2g.

224

Chapter 6. Sturm-Liouville problems on graphs According to Remark 6.1.2 the proof will be complete if we have shown that W := U1 U2∗ − U2 U1∗ − U3 U4∗ + U4 U3∗ = 0.

2g ∗ Using the notation W = (wr,s )2g r,s=1 , Uj Uk = (wj,k,r,s )r,s=1 , we have

wr,s = w1,2,r,s − w1,2,s,r − w3,4,r,s + w3,4,s,r

(r, s = 1, . . . , 2g)

and, for (j, k) = (1, 2) or (j, k) = (3, 4), wj,k,r,s =

g X

ur,(j−1)g+t us,(k−1)g+t

(r, s = 1, . . . , 2g).

t=1

Hence wj,k,r,s 6= 0 is only possible for pairs of boundary conditions which contain yj (0) and yj0 (0) or yj (l) and yj0 (l). So let r be a row index which corresponds to a component YR,j − YR,j1 (v) of U YR , and let s be a row index which corresponds to a component of the form (6.2.5), with the same v ∈ V int . Without loss of generality we may assume that j1 (v) ∈ I − (v); the case j1 (v) − 2g ∈ I + (v) is similar. If also j ∈ I − (v), then all nonzero entries in the r-th row of U are entries of the matrix U1 , and it is therefore clear that wr,s = w1,2,r,s in this case. But then ur,j = 1, ur,j1 (v) = −1, ur,t = 0 for t = 1, . . . , g with t 6= j and t 6= j1 (v), us,g+j = −1, us,g+j1 (v) = −1 shows that wr,s = w1,2,r,s = −1 + 1 = 0. Similarly, if j − 2g ∈ I + (v), then ur,j = 1, ur,j1 (v) = −1, ur,t = 0 for t = 1, . . . , 4g with t 6= j and t 6= j1 (v), us,g+j = 1, us,g+j1 (v) = −1 shows that w1,2,r,s = 1 and w3,4,r,s = 1.  This shows that wr,s = w1,2,r,s − w3,4,r,s = 1 − 1 = 0. Let v1 , . . . , vp be an enumeration of the vertices of G and put T = diag{d(v1 ), . . . , d(vp )}. Recall that a graph is called k-regular if d(v) = k for all v ∈ V G . Thus T = kIp for a k-regular graph G. For a simple graph, the matrix ( 1 if vi and vj are adjacent, p A = (Ai,j )i,j=1 , Ai,j = 0 otherwise, is called the adjacency matrix, and the matrix e = T −1/2 AT −1/2 A is called the weighted adjacency matrix, which can be written as e = 1A A k if G is k-regular.

(6.2.6)

6.3. Graphs with Neumann conditions at pendant vertices

225

6.3 Graphs with Neumann conditions at pendant vertices In this section we assume the potentials on the g edges of the graph G are the same, i.e., q1 = · · · = qg = q. Furthermore, it will be assumed that q is symmetric with respect to the midpoint of the interval (0, l), i.e., (x ∈ (0, l)).

q(l − x) = q(x)

Denote by s(λ, ·) and c(λ, ·) the solutions of −y 00 + qy = λ2 y

(λ ∈ C)

(6.3.1)

on [0, l] which satisfy the initial conditions s(λ, 0) = 0, s0 (λ, 0) = 1, c(λ, 0) = 1, c0 (λ, 0) = 0. Note that for q = 0, the functions s and c are simply (scaled) sine and cosine functions, respectively. Proposition 6.3.1. For all λ ∈ C, the functions s and c satisfy s(λ, l − x) = s(λ, l)c(λ, x) − s0 (λ, l)s(λ, x) 0

c(λ, l − x) = c(λ, l)c(λ, x) − c (λ, l)s(λ, x) 0

s (λ, l) = c(λ, l), 0

2

c (λ, l)s(λ, l) = c (λ, l) − 1.

(x ∈ [0, l]),

(6.3.2)

(x ∈ [0, l]),

(6.3.3) (6.3.4) (6.3.5)

Proof. It is easy to see that due to the symmetry of the potential q also the functions x 7→ s(λ, l − x) and x 7→ c(λ, l − x) are solutions of (6.3.1). Hence these functions can be expressed as linear combinations of c(λ, ·) and s(λ, ·). Since in (6.3.2) as well as in (6.3.3) the functions and their derivatives on the left-hand side and on the right-hand side have the same values at x = 0, the identities (6.3.2) and (6.3.3) follow. When x = l, (6.3.2) gives s(λ, l)(s0 (λ, l) − c(λ, l)) = 0. Since the spectrum of the Dirichlet problem (6.3.1), y(0) = 0, y(l) = 0 has discrete (real) spectrum, the analytic function s(·, l) is not the zero function, and it follows that the analytic function s0 (·, l) − c(·, l) is the zero function. Hence (6.3.4) is true. Taking x = l in (6.3.3) gives (6.3.5).  Theorem 6.3.2. Let G be a connected simple graph with at least one edge and L with Neumann conditions at all pendant vertices. √ consider the operator Then √ λ ∈ C with s( λ, l) 6= 0 is an eigenvalue of the operator L if and only if c( λ, l) e and the multiplicity of λ as is an eigenvalue of the weighted adjacency matrix A, √ e eigenvalue of L equals the multiplicity of c( λ, l) as eigenvalue of the matrix A. e are selfadjoint, geometric multiplicities equal algebraic Proof. Since both L and A √ e λ, l)I multiplicities, and we have to show that the null spaces of L−λI and A−c( have the same dimension. Given f1,0 , . . . , fg,0 ∈ C and f1,l , . . . , fg,l ∈ C it is well

226

Chapter 6. Sturm-Liouville problems on graphs

known and easy to verify that, for j = 1, . . . , g, the boundary value problem `j yj = λyj , yj (λ, 0) = fj,0 , yj (λ, l) = fj,l (λ ∈ C) (6.3.6) √ has a unique solution if s( λ, l) 6= 0 and that this solution is given by √ √ fj,l − fj,0 c( λ, l) √ √ yj (λ, x) = s( λ, x) + fj,0 c( λ, x). (6.3.7) s( λ, l) Hence y is a solution of Ly = λy if and only if y = (yj (λ, ·))gj=1 is given by (6.3.7) and satisfies the conditions (6.2.1)–(6.2.4). These solutions are uniquely given by f1,0 , . . . , fg,0 , f1,l , . . . , fg,l . Taking (6.2.3) and (6.3.6) into account, these values can be uniquely expressed in terms of vectors F = (fi )pi=1 , where fj,0 = fi when vi = v− (ej ) and fj,l = fi when vi = v+ (ej ). We calculate √ fj,l − fj,0 c( λ, l) 0 √ yj (λ, 0) = s( λ, l) and, observing (6.3.4) and (6.3.5), √ √ fj,l − fj,0 c( λ, l) 0 √ 0 √ s ( λ, l) + fj,0 c0 ( λ, l) yj (λ, l) = s( λ, l) √ √ √ √ √ 0 fj,l s ( λ, l) − fj,0 (c( λ, l)s0 ( λ, l) − c0 ( λ, l)s( λ, l)) √ = s( λ, l) √ fj,l c( λ, l) − fj,0 √ = . s( λ, l) Hence the conditions (6.2.1), (6.2.2) and (6.2.4) can be written as √ fj,l − fi c( λ, l) = 0 (vi ∈ V pen , I − (vi ) = {j}), √ fj,0 − fi c( λ, l) = 0 (vi ∈ V pen , I + (vi ) = {j}), X X √ fj,0 + fj,l − d(v)c( λ, l)fi = 0 (v ∈ V int ). j∈I + (vi )

(6.3.8) (6.3.9) (6.3.10)

j∈I − (vi )

But these are exactly the p components of the vector equation √ AF − c( λ, l)T F = 0, which is equivalent to

√ e − c( λ, l))T 12 F = 0. (A

Hence the theorem is proved.



Recall that for convenience we call a complex number an eigenvalue of multiplicity 0 if it is no eigenvalue.

6.3. Graphs with Neumann conditions at pendant vertices

227

Theorem 6.3.3. Let G be a connected graph with at least two edges and consider Neumann conditions at all vertices. Let λ ∈ C the operator L with (generalized) √ √ with s( λ, l) = 0. Then (c( λ, l))2 = 1, and the following holds: √ √ 1. If c0 ( λ, l) = 0 and G is bipartite or c( λ, l) = 1, then λ is an eigenvalue of the operator L of multiplicity g − p + 2. √ √ 2. If c0 ( λ, l) 6= 0 and G is bipartite or c( λ, l) = 1, then λ is an eigenvalue of the operator L of multiplicity g − p + 1. √ 3. If G is not bipartite and c( λ, l) = −1, then λ is an eigenvalue of the operator L of multiplicity g − p. 0 0 Proof. Given f1,0 , . . . , fg,0 ∈ C and f1,0 , . . . , fg,0 ∈ C it is well known that, for j = 1, . . . , g, the initial value problem `j yej = λe yj ,

yej (λ, 0) = fj,0 ,

0 yej0 (λ, 0) = fj,0

(λ ∈ C),

(6.3.11)

has the unique solution √ √ 0 yej (λ, x) = fj,0 c( λ, x) + fj,0 s( λ, x).

(6.3.12)

Hence y is a solution of Ly = λy if and only if y = (e yj (λ, ·))gj=1 is given by (6.3.12) and satisfies the conditions (6.2.1)–(6.2.4). These solutions are uniquely 0 0 determined by f1,0 , . . . , fg,0 , f1,0 , . . . , fg,0 . We calculate √ yej (λ, l) = fj,0 c( λ, l) (6.3.13) and, taking (6.3.4) into account, √ √ √ √ 0 0 yej0 (λ, l) = fj,0 c0 ( λ, l) + fj,0 s0 ( λ, l) = fj,0 c0 ( λ, l) + fj,0 c( λ, l).

(6.3.14)

The continuity condition (6.2.3) implies that (fj,0 )gj=1 is uniquely given by √ F = (fi )pi=1 , where fj,0 = fi with vi = v−√ (ej ). The property (c( λ, l))2 = 1 is obvious in view of (6.3.5). Putting ε := −c( λ, l) it follows that ε = 1 or ε = −1. Since G is connected and has at least one interior vertex, it follows in case ε = −1 from (6.3.11) and (6.3.13) that the continuity conditions (6.2.3) are satisfied if and only if (fi )pi=1 is a multiple of 1p , the vector in Cp with all components being 1. If ε = 1 and G is bipartite with decomposition G1 and G2 of the vertices, let F b = (fib )pi=1 be defined by fib = −1 if vi ∈ G1 and fib = 1 if vi ∈ G2 . In view of (6.3.11) and (6.3.13) the continuity conditions (6.2.3) are satisfied if and only if (fi )gi=1 is a multiple of F b . Finally, if ε = 1 and G is not bipartite, then the continuity conditions (6.2.3) are satisfied if and only if (fi )gi=1 = 0. 0 Let F 0 = (fj,0 )gj=1 . The conditions (6.2.1), (6.2.2) and (6.2.4) can be written as X X X √ 0 0 fj,0 + εfj,0 = εc0 ( λ, l) fj,0 (i = 1, . . . , p). (6.3.15) j∈I + (vi )

j∈I − (vi )

j∈I + (vi )

228

Chapter 6. Sturm-Liouville problems on graphs

In case ε = −1 we have seen that all fj,0 (j = 1, . . . , g) are equal. In case ε = 1 and G is bipartite, we have seen that fj,0 = −fi for j ∈ I(vi ). Finally, when ε = 1 and G is not bipartite, then fj,0 = 0 for all j = 1, . . . , p. Therefore (6.3.15) can be written as X X √ 0 0 fj,0 + εfj,0 = −c0 ( λ, l)d+ (v)fi (i = 1, . . . , p). (6.3.16) j∈I + (vi )

j∈I − (vi )

Recalling the definition of the (signed) incidence matrix Uε , the system of equations (6.3.16) has the matrix form √ Uε F 0 = −c0 ( λ, l)T + F, (6.3.17) where T + := diag(d+ (v1 ), . . . , d+ (vp )). Hence the multiplicity of the eigenvalue λ of L is the dimension of all vectors (F, F 0 ) which satisfy (6.3.17) with F a multiple of 1p when ε = −1, with F a multiple of F b when ε = 1 and G is bipartite, and with F = 0 when ε = 1 and G is not √ bipartite. First consider the case that c0 ( λ, l) = 0 or that ε = 1 and G is not bipartite. In this case, (6.3.17) is a homogeneous equation, and therefore ( g − rank Uε + 1 if ε = −1 or G is bipartite, dim N (L − λI) = g − rank Uε if ε = 1 and G is not bipartite. By Proposition C.5.1 and Lemma C.5.5, the rank of Uε is p − 1 when ε = −1 or G is bipartite, and the rank of Uε is p when ε = 1 and G is not bipartite. This proves parts 1 and 3. √ Now consider the case that c0 ( λ, l) 6= 0. When ε = −1, then each column of Uε has two nonzero elements, one is −1 and the other one is 1. Therefore 1T p Uε = 0, + whereas 1T T 1 = g > 0. This shows that p p dim N (L − λI) = g − rank Uε = g − p + 1, which proves part 2 in this case. Similarly, when ε = 1 and G is not bipartite, then each column of Uε has two nonzero elements, which are both 1. If the corresponding row indices of these nonzero elements are r and s, then vr ∈ G1 and vs ∈ G2 , or vice versa. Therefore (F b )T Uε = 0, whereas (F b )T T + F b = g > 0. This shows that dim N (L − λI) = g − p + 1 also in this case.  √ Remark 6.3.4. √ In Theorem 6.3.3√we have considered s( λ, l) = 0 and distinguished the cases c0 ( λ, l) = 0√and c0 ( λ, l) 6= √ 0. We will show that each of these cases can occur, both for c( λ, l) = 1 and c( λ, l) = −1. For simplicity, we will write √ λ instead of λ. First consider (6.3.1) on [0, 1] with q = 0. For λ 6= 0 it is obvious that s(λ, x) = λ−1 sin(λx), s0 (λ, x) = cos(λx), c(λ, x) = cos(λx), c0 (λ, x) = −λ sin(λx). Clearly, s(λ, 1) = 0 implies c0 (λ, 1) = 0.

6.3. Graphs with Neumann conditions at pendant vertices

229

Now consider the case that l = 4 and ( 0, x ∈ (0, 1) ∪ (3, 4), q(x) = 1, x ∈ (1, 3). The potential is clearly symmetric, and a straightforward calculation gives for x ∈ [1, 3] that √ √ √ p  λ2 − 1 sin λ sin λ2 − 1 + λ cos λ cos λ2 − 1 √ s(λ, x) = sin λ2 − 1x λ λ2 − 1 √ √ √ p  λ2 − 1 sin λ cos λ2 − 1 − λ cos λ sin λ2 − 1 √ + λ2 − 1x , cos λ λ2 − 1 √ √ √ p  λ2 − 1 cos λ cos λ2 − 1 + λ sin λ sin λ2 − 1 √ c(λ, x) = cos λ2 − 1x λ2 − 1 √ √ √ p  2 λ − 1 cos λ sin λ2 − 1 − λ sin λ cos λ2 − 1 √ + λ2 − 1x . sin λ2 − 1 This gives 1 √ 2  λ −1 = √ − λ2 − 1

! √ 1  cos λ sin λ2 − 1 λ √ sin λ cos λ2 − 1 −λ

(6.3.18)

 √ 2 λ −1  0  − s (λ, 2) λ =  c(λ, 2) λ −√ 2 λ −1

 ! 1 sin λ sin √λ2 − 1   cos λ cos √λ2 − 1 . 1

(6.3.19)



s(λ, 2) c0 (λ, 2)

and





It is well known that s(·, 2) has infinitely many real zeros which tend to ∞, see, e.g., [96, Lemma 12.5.1]. Therefore choose any λ ∈ (1, ∞) such that s(λ, 2) = 0. With x = 2 in (6.3.2) it follows that 0 = s(λ, 2) = s(λ, 4)c(λ, 2) − s0 (λ, 4)s(λ, 2) = s(λ, 4)c(λ, 2). But c(λ, 2) 6= 0 since s(λ, ·) and c(λ, ·) are linearly independent solutions of (6.3.1). Hence s(λ, 4) = 0. With x = 2 in (6.3.3) it follows that c(λ, 2) = c(λ, 4)c(λ, 2) and therefore√c(λ, 4) = 1. The matrix in (6.3.18) is invertible and hence in√view of 0 max (2C1 )−2 (1 − a2 )l4 , (4C1 C2 )2 (1 − a2 )−2 l−4 let p 1 rk := (2C1 )−1 1 − a2 l|k|− 2 . Then rk < 1, and for |ζ| = rk it follows that p | cos((ζ + µ2k )l) − a| ≥ 1 − a2 lrk − C1 rk2   p 1 −1 1p 2 2 = 1 − a lrk 1 − |k| 1 − a2 lrk ≥ 2 2 1 1 = C1−1 (1 − a2 )l2 |k|− 2 4 C2 > ≥ |(c(ζ + µ2k , l) − a) − (cos((ζ + µ2k )l) − a)|. |k| Then Rouch´e’s theorem, applied to the circle with centre µ2k and radius rk , shows that c(·, l) − a has exactly one simple zero inside this circle, which we are going to denote by µ e2k . Let f (λ) := c(λ, l) − a. Then Z µ2k f (µ2k ) = f (µ2k ) − f (e µ2k ) = f 0 (λ) dλ µ e2k 0

= f (µ2k )(µ2k − µ e2k ) + = f 0 (µ2k )(µ2k − µ e2k ) +

Z

µ2k

µ e2k Z µ2k µ e2k

(f 0 (λ) − f 0 (µ2k )) dλ Z

λ

µ2k

f 00 (µ) dµ dλ.

232

Chapter 6. Sturm-Liouville problems on graphs

From (6.3.25) we find that cos λl sin λl −Q + 2λ 2λ2

f 0 (λ) = −l sin λl + Ql Therefore

f 0 (µ2k ) = −l

p

1 − a2 +

Z

l

tB(t) 0

cos λt dt − λ

Z

l

B(t) 0

sin λt dt. λ2

(1) (2) βb βb Qla + 2k + 2k , 2µ2k µ2k µ2k

where (1) βb2k =

Z

l

tB(t) cos µ2k t dt, 0

1 (2) βb2k = µ2k

! Z l 1 p 2 − Q 1−a − B(t) sin µ2k t dt . 2 0

(2) (1) Clearly, the sequence (βb2k ) is an l2 sequence. But also (βb2k ) is an l2 sequence, see, e.g., [96, Lemma 12.2.1]. Hence we have ! (3) βb2k 1 1 Qa √ =− √ 1+ + f 0 (µ2k ) µ2k l 1 − a2 2µ2k 1 − a2 (3)

with an l2 sequence (βb2k ). Furthermore, (6.3.25) gives √ Q 1 − a2 1 b(4) f (µ2k ) = + β 2µ2k µ2k 2k (4)

with an l2 sequence (βb2k ). Differentiating f 0 again shows that the entire function f 00 is bounded on the strip {λ ∈ C : | Im λ| ≤ l−1 }. Hence Z µ2k Z λ f 00 (µ) dµ dλ = O((e µ2k − µ2k )2 ) = O(rk2 ) = O(|k|−1 ). µ e2k

µ2k

Altogether, it follows that µ e2k − µ2k = −

f (µ2k ) + O((e µ2k − µ2k )2 ) . 0 f (µ2k )

The above representations of f (µ2k ) and f 0 (µ2k ) show that |e µ2k −µ2k | = O(|k|−1 ), and therefore f (µ2k ) + O(|k|−2 ) f 0 (µk )   1 Q b(5) = µ2k + + β2k 2µ2k l Q 1 (6) = µ2k + + βb 4kπ k 2k

µ e2k = µ2k −

(6.3.26)

6.3. Graphs with Neumann conditions at pendant vertices

233

(5) (6) with l2 sequences (βb2k ) and (βb2k ). Next we are going to compare the zeros of λ 7→ cos λl−a and c(·, l)−a globally. For k ∈ N let Γk be the boundary of the square with vertices ±l−1 kπ ± il−1 kπ. A standard estimate shows that there is Na ∈ N such that

max{| sin λx|, | cos λx| : x ∈ [0, l], λ ∈ Γk } ≤

2| cos λl − a| 1 − |a|

(λ ∈ Γk , k ≥ Na ). (6.3.27)

Indeed, writing λ = u + iv (u, v ∈ R) for λ ∈ C, the formula cos λx = cos ux cosh vx − i sin ux sinh vx shows that | cos λx| ≤ cosh(Im λl)

(λ ∈ C, x ∈ [0, l]).

Similarly, sin λx = sin ux cosh vx + i cos ux sinh vx shows that | sin λx| ≤ cosh(Im λl)

(λ ∈ C, x ∈ [0, l]).

Putting u = ±l−1 kπ it follows that cos λl = (−1)k cosh vl and therefore | cos λl − a| ≥ (1 − |a|) cosh(Im λl)

(| Re λ| = l−1 kπ).

Observing | cos λl − a|2 ≥ sinh2 vl − 2|a| cosh vl gives for v = ±l

−1

(λ ∈ C)

kπ that

| cos λl − a|2 ≥ cosh2 kπ



 sinh2 kπ 2|a| , − cosh2 kπ cosh kπ

where the term in square brackets is larger than there is Na ∈ N such that | cos λl − a| ≥

1 2

1 (1 − |a|) cosh(Im λl) 2

for sufficiently large k. Therefore

(λ ∈ Γk , k ≥ Na ).

Combining the above estimates proves (6.3.27). From (6.3.25) and (6.3.27) we conclude that | cos λl − a| > |c(λ, l) − cos λl| for all λ ∈ Γk and sufficiently large k. By Rouch´e’s theorem, the number of zeros of λ 7→ cos λl − a and λ 7→ c(λ, l) − a inside those Γk coincide, counted with multiplicity. The functions λ 7→ cos λl − a and c(·, l) − a are even functions, and µ2k−1 = −µ−2k (k ∈ Z). Due to the asymptotic representation (6.3.25) we can index the zeros of c(·, l) − a, counted with multiplicity, as a sequence (e µj )j∈Z , where µ e2k for large |k| is given by (6.3.26) and µ e√ µ−2k (k ∈ Z). 2k−1 = −e It is now clear that the zeros of λ 7→ c( λ, l) − a are given by the √ sequence (e µ2k )∞ . On the other hand, it is well known that the function s( λk , ·) k=0

234

Chapter 6. Sturm-Liouville problems on graphs

has√k − 1 simple zeros in (0, √l), see, e.g., [139, Theorem 13.2]. It follows from s0 (√ λk , 0) = 1 that (−1)k s0 ( λk , l) > 0. Therefore √ (6.3.4) and (6.3.5) show that c( λk , l) = (−1)k . Hence the function λ 7→ c( λ, l) − a has a zero ζa,k in the interval (λk−1 , λk ) for k = 2, √ 3, . . . . For λ < 0 and λ → −∞, √ the asymptotic √ √ 1 1 |λ|t representations cos( λt) = 2 e + o(1) and | sin( λt)| = 2 e |λ|t + o(1) show √ √ √ that c( λ, l) → ∞. From c( λ1 , l) = −1 it follows that λ 7→ c( λ, l) − a has a zero ζa,1 in the interval (−∞, λ1 ). The asymptotic representation (6.3.26) shows that λ2k < µ e22k < λ2k+1 and λ2k−1 < µ e2−2k = µ e22k−1 < λ2k for sufficiently large k. Thus the numbers ζa,k and µ e2k−1 can be identified for k ≥ 1, √ ∞ and hence the sequence (ζa,k )k=1 is the sequence of all zeros of λ 7→ c( λ, l) − a. Finally, (6.3.26) proves (6.3.24), applying − arccos a = arccos a − π for odd k.  Furthermore, we will need a representation of the zeros of the two functions √ λ 7→ c( λ, l) − ε with ε ∈ {−1, 1}. Due to √ the identity (6.3.5) it is more convenient to state properties of the zeros of λ 7→ c0 ( λ, l). √ Proposition 6.3.7. Let q ∈ L2 (0, l). Then the zeros of λ 7→ c0 ( λ, l) are real, simple and can be arranged as an increasing sequence (ξk )∞ k=0 which obeys ξ0 < ζ0,1 < ξ1 < ζ0,2 < . . .

(6.3.28)

and the asymptotic formula ξk =

π2 k2 Q + + γ1,k , l2 l

(6.3.29)

√ k where (γ1,k )∞ k=0√∈ l2 and Q is given by (6.3.22). Furthermore, c( λk , l) = (−1) k (k ∈ N) and c( ξk , l) = (−1) (k ∈ N0 ). Proof. The representations s(λ, l) =

cos λl sin λl −Q + λ 2λ2

Z

l

B1 (t) 0

1 c0 (λ, l) = −λ sin λl + Q cos λl + 2

Z

cos λt dt, λ2

(6.3.30)

l

B2 (t) cos λt dt,

(6.3.31)

0

where B1 , B2 ∈ L2 (0, l) are real-valued functions,√see, e.g., [96, (12.2.22) and √ (12.2.25)], show that λ 7→ λs( λ, l) and λ 7→ −c0 ( λ, l) have the same leading terms. Therefore the asymptotic representation (6.3.21) implies that the zeros of the latter function can be written as a sequence (ξk )∞ k=0 of complex numbers satifying√the asymptotic representation (6.3.29). For each k ∈ N, the function λ 7→ c2 ( λ, l)−1 has the value −1 at the endpoints of the interval [ζk , ζk+1 ]. Hence this function has an even number of zeros in (ζk , ζk+1 ), counted with multiplicity.

6.3. Graphs with Neumann conditions at pendant vertices

235

√ By (6.3.20), λk is the only zero of λ 7→ s( λ, l) in (ζk , ζk+1 ). Hence it follows √ from (6.3.5) that λ 7→ c0 ( λ, √ l) has an odd number of zeros in (ζk , ζk+1 ) (k ∈ N). Also, the function λ 7→ c2 ( λ, l) − 1 has √ an odd number of zeros, counted with multiplicity, in (−∞, ζ1 ) because of c( λ, l) → ∞ as λ → −∞, which has been shown √ in the proof of Proposition 6.3.6. Since λ1 > ζ1 (see (6.3.20)) it is clear√that s( λ, l) 6= 0 for all z ∈ (−∞, ζ1 ). Again from (6.3.5) it follows that λ 7→√c0 ( λ, l) has an odd number of zeros in the interval (−∞, ζ1 ). Hence λ 7→ c0 ( λ, l) has at least one zero ξ0 in (−∞, ζ1 ) and at least one zero ξk in (ζk , ζk+1 ) (k ∈ N). In view of ζk = ζ0,k and the asymptotic representations (6.3.24) and (6.3.29) √ 0 it follows that the zeros of λ 7→√c ( λ, l) are real, simple and can be indexed such that √ (6.3.28) holds. Since c( λ, l) →√∞ as λ → −∞ and since the zeros of λ 7→ c( √λ, l) are simple, it is clear that c( λ, l) > 0 when λ ∈ (−∞, √ ζ1 ) and that (−1)k c( λ, l)√> 0 for k ∈ N and λ ∈ (ζk , ζk+1 ). In view of c2 ( ξk , l) = 1 this shows that c( ξk , l) = (−1)k (k ∈ N0 ).  In view of Lemma C.5.5 the spectrum of the weighted adjacency matrix A˜ consists of p eigenvalues −1 ≤ α1 ≤ α2 ≤ · · · ≤ αp−1 < αp = 1, counted with multiplicity, where α2 > −1 and α1 = −1 if and only if G is bipartite. We are now in a position to describe the asymptotic behaviour of the eigenvalues of L. It is convenient to state two theorems, one for bipartite graphs and one for nonbipartite graphs. But both theorems will be proved together. We will use the notation for the sequences considered in Propositions 6.3.5, (j) 6.3.6 and 6.3.7 with λk := λk (k, j ∈ N). Theorem 6.3.8. Let G be a bipartite connected simple graph with at least two edges. Then the spectrum of the operator L on G given by the differential equation (6.3.1), the Neumann conditions (6.2.1), (6.2.2) and the conditions (6.2.3), (6.2.4), consists of g sequences of eigenvalues, counted with multiplicity, as follows: (i) the sequence (ξk )∞ k=0 with the asymptotic representation (6.3.29); (j)

(ii) if G is not a tree, the sequences (λk )∞ k=1 (j = 1, . . . , g − p + 1) with the asymptotic representation (6.3.21); (iii) the sequences (ζαj ,k )∞ k=1 (j = 2, . . . , p − 1) with the asymptotic representation (6.3.24). Theorem 6.3.9. Let G be a nonbipartite connected simple graph with at least two edges. Then the spectrum of the operator L on G given by the differential equation (6.3.1), the Neumann conditions (6.2.1), (6.2.2) and the conditions (6.2.3), (6.2.4), consists of g + 1 sequences of eigenvalues, counted with multiplicity, as follows: (i) the sequence (ξ2k )∞ k=0 with the asymptotic representation (6.3.29) and the sequence (λ2k )∞ k=1 with the asymptotic representation (6.3.21); (j)

(ii) if g > p, the sequences (λk )∞ k=1 (j = 1, . . . , g − p) with the asymptotic representation (6.3.21);

236

Chapter 6. Sturm-Liouville problems on graphs

∞ (j = 1, . . . , p − 1) with the asymptotic representation (iii) the sequences (ζαj ,k )k=1 (6.3.24).

Proof. Theorems 6.3.2 and 6.3.3 describe all eigenvalues of L. Taking Proposition 6.3.6 into account, √ it is clear that part (iii) is a representation √ of the eigenvalues λ for which |c( λ, l)| < 1. All other eigenvalues λ satisfy√|c( λ, l)| = 1. Therefore, let now λ be an eigenvalue of L with √ c2 ( λ, l) = 1. √ 0 First consider the case that G is bipartite. If c ( λ, √ l) = 0 and s( √λ, l) 6= 0, then λ is a simple eigenvalue by Theorem 6.3.2, and if c0 ( λ, l) = 0 and s( λ, l) = 0, then λ is an eigenvalue by Theorem 6.3.3, part 1. This proves that the sequence in part (i) of Theorem 6.3.8 contributes to the spectrum. It is therefore clear from Theorem 6.3.3 parts 1 and 2 that all remaining eigenvalues are described in part (ii) of Theorem 6.3.8. √ 0 √ Finally, consider the case that G is not bipartite. If c ( λ, l)√= 0 and s( λ,√l) 6= 0, then λ is a√simple eigenvalue by Theorem 6.3.2 when c( λ, l) = 1. If c0 ( λ, l) = 0√and s( λ, l) = 0, then λ is an eigenvalue by Theorem 6.3.3, part 1, when c( λ, l) = 1. Furthermore, apart from this sequence, also (λ2k )∞ k=1 contributes to the eigenvalues by Theorem 6.3.3, parts 1 and 2, and Proposition 6.3.7. This proves that the two sequences in part (i) of Theorem 6.3.9 contribute to the spectrum. It is therefore clear from Theorem 6.3.3 that all remaining eigenvalues are described in part (ii) of Theorem 6.3.9.  For a connected simple graph G with adjacency matrix A and matrix of degree T let PG (z) := det(zT − A) (z ∈ C). (6.3.32) e counted The zeros of PG are the eigenvalues of the weighted adjacency matrix A, e − z) = with multiplicity. It might appear to be more natural to consider det(A det(−T −1 ) det(zT − A), but the function PG has integer coefficients and a positive leading coefficients, which looks nicer when PG is calculated explicitly. The characteristic function φ of the operator L is defined by √ √ φL (z) := sg−p ( λ, l)PG (c( λ, l)) (z ∈ C). (6.3.33) Corollary 6.3.10. Let G be a simple connected graph. Then the eigenvalues of the operator L on G given by the differential equation (6.3.1), the Neumann conditions (6.2.1), (6.2.2) and the conditions (6.2.3), (6.2.4), are the zero of the entire function φL , counted with multiplicity. Proof. If G is bipartite, we use Theorem 6.3.8 for the proof. The eigenvalues √ 0 in part (i) are the zeros of λ → 7 c ( λ, l), the eigenvalues in part (ii) are the √ zeros of λ √7→ sg−p+1√ ( λ, l), and the eigenvalues in part (iii) are the zeros of e by λ 7→ PG (c( λ, l))(c2 ( λ, l) − 1))−1 since −1 and 1 are simple eigenvalues of A Lemma C.5.5. In view of (6.3.5), √ √ √ √ √ √ c0 ( λ, l)sg−p+1 ( λ, l)PG (c( λ, l))(c2 ( λ, l) − 1))−1 = sg−p ( λ, l)PG (c( λ, l)),

6.3. Graphs with Neumann conditions at pendant vertices

237

which proves the statement for bipartite graphs. If G is not bipartite, we use Theorem 6.3.9 for the proof. In view of (6.3.5) and Propositions 6.3.5 and√6.3.7, the eigenvalues in part (i) are the √ √ zeros of λ 7→ c2 ( λ, l) − 1 with c( λ, l) = 1, that is, the zeros of λ → 7 c( λ, l) − 1. √ g−p The eigenvalues in part (ii) are the zeros of λ → 7 s ( λ, l), and the eigenvalues √ √ in part (iii) are the zeros of λ 7→ PG (c( λ, l))(c( λ, l) − 1))−1 since −1 is not an e by Lemma C.5.5. Then the identity eigenvalue and 1 a simple eigenvalue of A √ √ √ √ √ √ (c( λ, l) − 1)sg−p ( λ, l)PG (c( λ, l))(c( λ, l) − 1))−1 = sg−p ( λ, l)PG (c( λ, l)) completes the proof.



Remark 6.3.11. 1. It was convenient to consider g sequences in Theorem 6.3.8 and g + 1 sequences in Theorem 6.3.9. However, if we combine the two sequences in Theorem 6.3.9, part (i), into one sequence, then one arrives in both cases at g sequences where the k-th term, say ak , satisfies the asymptotic estimate √ ak − πk ≤ π + o(1) as k → ∞. l l Denoting the nondecreasing sequence of the eigenvalues of L, counted with multiplicity, by (λL,k )∞ k=1 , Weyl’s law g=

π k lim p l k→∞ λL,k

follows (see [3], [100] or [10, Corollary 1]).

√ 2. The smallest eigenvalue is ξ0 , which is the smallest zero of λ 7→ c( λ, l), and this eigenvalue is simple. Example 6.3.12. In this example we will find the characteristic functions as defined in (6.3.33) for the graphs associated with the platonic solids tetrahedron, cube and octahedron. All three graphs are planar graphs. 1. The tetrahedral graph is not a bipartite graph since it has cycles with three edges.

It has 4 vertices and 6 edges, that is, g − p = 2, and   B4 (z) −A4 zT − A = , −A4 B4 (z) where

 B4 (z) =

3z −1

 −1 , 3z

 1 A4 = 1

 1 . 1

238

Chapter 6. Sturm-Liouville problems on graphs Denoting the polynomial PG in (6.3.32) by P4 , it follows for det B4 (z) 6= 0 from the Schur factorization (B.4.2) that P4 (z) = det(zT − A) = det B4 (z) det(B4 (z) − A4 B4−1 (z)A4 ). A straightforward calculation shows that A4 B4−1 (z)A4 =

6z + 2 A4 , 9z 2 − 1

and hence 2  2 ! 6z + 2 6z + 2 P4 (z) = (9z − 1) 3z − 2 − 1+ 2 9z − 1 9z − 1   4(3z + 1) 2 = (9z − 1)(3z + 1) 3z − 1 − 9z 2 − 1  3 1 = 34 (z − 1) z + . 3 

2

This shows that the characteristic function as defined in (6.3.33) for the tetrahedral graph is  3 √ √ √ 1 φ4 (λ) = 34 s2 ( λ, l)(c( λ, l) − 1) c( λ, l) + . 3 2. The octahedral graph is not a bipartite graph since it has cycles with three edges. 2 1

5

6

4

3 It has 6 vertices and 12 edges, that is, g −p = 6. With respect to the indexing of the vertices as indicated in the figure above we have   B6 (z) −A6 zT − A = , −A6 B6 (z) where



4z B6 (z) = −1 −1

−1 4z −1

 −1 −1 , 4z

 1 A6 = 1 0

1 0 1

 0 1 . 1

Denoting the polynomial PG in (6.3.32) by P6 , it follows for det B6 (z) 6= 0 from the Schur factorization (B.4.2) that P6 (z) = det(zT − A) = det B6 (z) det(B6 (z) − A6 B6−1 (z)A6 ).

6.3. Graphs with Neumann conditions at pendant vertices

239

A straightforward calculation shows that det B6 (z) = 2(2z − 1)(4z + 1)2 ,   4z − 1 1 1 1  1 4z − 1 1  B6−1 (z) = 2(4z + 1)(2z − 1) 1 1 4z − 1 and then 

4z 1 2z + 1 A6 B6−1 (z)A6 = (4z + 1)(2z − 1) 2z + 1

2z + 1 4z 2z + 1

 2z + 1 2z + 1 . 4z

The diagonal elements of the matrix (4z + 1)(2z − 1)(B6 (z) − A6 B6−1 (z)A6 ) are 4z(4z +1)(2z −1)−4z = 8z(4z 2 −z −1), whereas its off-diagonal elements are −(4z + 1)(2z − 1) − (2z + 1) = −8z 2 . Then elementary row operations give det(B6 (z) − A6 B6−1 (z)A6 )  2  4z − z − 1 −z 0 2 z (4z − 1) −2z 4z 2 − 2z − 1 0 det  (4z + 1)3 (2z − 1)3 −z −z 1 9 3 2  2 z (4z − 1)  2 (4z − z − 1)(4z 2 − 2z − 1) − 2z 2 3 3 (4z + 1) (2z − 1) 29 z 3 (2z + 1) (4z 2 − 1)(4z 2 − 3z − 1) (4z + 1)3 (2z − 1)2 29 z 3 (2z + 1)2 (z − 1), (4z + 1)2 (2z − 1) 9 3

=

= = =

which shows that

2

 2 1 P6 (z) = 46 (z − 1)z 3 z + . 2

Therefore the characteristic function as defined in (6.3.33) for the octahedral graph is  2 √ √ √ √ 1 6 6 3 φ6 (λ) = 4 s ( λ, l)(c( λ, l) − 1)c ( λ, l) c( λ, l) + . 2 3. The cube graph is a bipartite graph since every edge has one black end and one white end. 3 7 5 1 2 8

6 4

240

Chapter 6. Sturm-Liouville problems on graphs It has 8 vertices and 12 edges, that is, g −p = 4. With respect to the indexing of the vertices as indicated in the figure above we have   3zI4 −A8 zT − A = , −A8 3zI4 where

  1 1 1 0 1 1 0 1  . A8 =  1 0 1 1 0 1 1 1

The Schur factorization (B.4.2) shows that P8 (z) := det(zT − A) = det(9z 2 I4 − A82 ). But

  3 2 2 2 2 3 2 2  , A28 =  2 2 3 2 2 2 2 3

and letting ζ = 21 (3z 2 − 1) it follows that 9z 2 I4 − A28 equals 2(ζT − A) with A and T from the tetrahedral graph. Therefore part 1 leads to  3 2 P8 (z) = 24 P4 (ζ) = 34 (2ζ − 2) 2ζ + 3  3 1 = 38 (z 2 − 1) z 2 − , 9 which gives the characteristic function as defined in (6.3.31) for the cube graph as  3 √ √ √ 1 8 4 2 2 φ8 (λ) = 3 s ( λ, l)(c ( λ, l) − 1) c ( λ, l) − . 9 Example 6.3.13. For a star graph with g ≥ 2 edges we have   gz 1T g zT − A = , 1g zIg and the Schur factorization gives
g−1 2 det(zT − A) = z g gz − z −1 1T (z − 1). g 1g = gz It follows in view of (6.3.5) that the characteristic function as defined in (6.3.31) is √ √ √ √ √ φ(λ) = gs−1 ( λ, l)(c2 ( λ, l) − 1)cg−1 ( λ, l) = gc0 ( λ, l)cg−1 ( λ, l).

6.4. Connected graphs with Dirichlet conditions at pendant vertices

6.4

241

Connected graphs with Dirichlet conditions at pendant vertices

In this section it is assumed that Dirichlet conditions are imposed at r pendant vertices of the connected simple graph G with at least two edges. The graph b is defined by removing the pendant vertices with Dirichlet conditions and G b will be denoted by their incident edges in G. For convenience, the vertices of G v1 , . . . , vp−r , whereas the remaining vertices of G will be denoted by vp−r+1 , . . . , vp . b p−r )) and let b 1 ), . . . , d(v b let Tb = diag(d(v b be the adjacency matrix of G, Let A b i ) is the degree of the vertex vi in G, b TbG = diag(d(v1 ), . . . , d(vp−r )), where d(v whereas d(vi ) is the degree of the vertex vi in G. Proposition 6.4.1. Let r > 0. Then the zeros of the polynomial PG,Gb defined by b PG,Gb (z) := det(z TbG − A)

(z ∈ C)

are real and lie in the interval (−1, 1). If G is bipartite, then PG,Gb is an even or an odd polynomial. b is bipartite b is a connected simple graph and that G Proof. It is clear that G b if and only if G is bipartite. Furthermore, when G is an isolated vertex v, then PG,Gb (z) = d(v)z, which proves the statement in this case. From now on we consider b is not an isolated vertex. Since A b is a symmetric matrix, also the case that G − 12 b b− 12 b TG ATG is a symmetric matrix, and all its eigenvalues are therefore real. From b we conclude for y ∈ Cp−r that Lemma C.5.4 applied to G b y) ≤ (Tby, y) ≤ (TbG y, y). −(TbG y, y) ≤ −(Tby, y) ≤ (Ay, b is not This shows that the zeros of PG,Gb lie in [−1, 1]. Furthermore, when G b y), which shows that −1 is not a zero. bipartite and y 6= 0, then −(Tby, y) < (Ay, b However, when G is bipartite, then it follows as in the proof of Lemma C.5.4 that the zeros are symmetric, and it therefore remains to show that 1 is not a zero. b − Tb)1p−r = 0. Since 1 is a simple eigenvalue of the weighted Observe that (A b y) < (Tby, y) whenever y is not a multiple of adjacency matrix it follows that (Ay, b i ) ≤ d(vi ) for all i = 1, . . . , p − r and d(v b i ) < d(vi ) 1p−r . On the other hand, d(v b b for at least one i, and therefore it is clear that (T 1p−r , 1p−r ) < (TG 1p−r , 1p−r ). Altogether, it follow that 1 is not a zero of PG,Gb .  Theorem 6.4.2. Let G be a connected simple graph with at least two edges. Assume that all edges have the same length and the same symmetric potential. Also assume that L is the operator corresponding to Dirichlet conditions at r pendant vertices. Then the eigenvalues of L, counted with multiplicity, are the zeros of the characteristic function φL defined by √ √ φL (λ) = sg−p+r ( λ, l)PG,Gb (c( λ, l)). (6.4.1)

242

Chapter 6. Sturm-Liouville problems on graphs

Proof. When r = 0, this √ theorem is exactly Corollary 6.3.10. Now let r > 0 and let λ ∈ C such that s( λ, l) 6= 0. We are going to adapt the proof of Theorem 6.3.2 to the case r > 0. Then the Dirichlet conditions give fp−r+1 = · · · = fp = 0, whereas that remaining conditions from (6.2.1), (6.2.2) and (6.2.4) can be written as (6.3.8) and (6.3.9) restricted to 1 ≤ i ≤ p − r and X X √ fj,l − d(v)c( λ, l)fi = 0 (v ∈ V int ), fj,0 + j∈Ib+ (vi )

j∈Ib− (vi )

where Ib+ (vi ) and Ib− (vi ) are the indegrees and outdegrees with respect to the b But these are exactly the p − r components of the vector equation graph G. √ bFb − c( λ, l)TbG Fb = 0, A √ where Fb = (f1 , . . . , fp−r )T . Hence λ ∈ C such that √ s( λ, l) 6= 0 is an eigenvalue of L if and only if λ is a zero of λ 7→ PG,Gb (c( λ, l)), and the multiplicities are √ √ equal. Here we have to observe that PG,Gb (c( λ, l)) = 0 implies c2 ( λ, l) < 1 √ by Proposition 6.4.1 and therefore s( λ, l) 6= 0 by (6.3.5) and that the zeros of √ λ 7→ c( λ, l) − a (|a| < 1) are simple by Proposition √ 6.3.6. Finally let r > 0 and let λ ∈ C such that s( λ, l) = 0. Adapting the proof of Theorem 6.3.3 to this situation we have fp−r+1 = · · · = fp = 0 and in particular F = 0. The remaining conditions (6.2.1), (6.2.2) and (6.2.4) can then be written as bε F 0 = 0, U bε consists of the first q − r rows of Uε . From the proof of Proposition C.5.1 where U bε U bε∗ = TbG + εA. b By Proposition 6.4.1, the matrix TbG + εA b is it is clear that U invertible both for ε = 1 and ε = −1, so that bε = rank U bε U bε∗ = rank(TbG + εA) b = p − r. rank U √ Since the zeros of λ√7→ s( λ, l) are simple (see Proposition 6.3.5), it follows that each λ ∈ C with s( λ, l) = 0 is an eigenvalue of L of multiplicity g − (p − r). 

6.5

Spectral problems on perfect binary trees

6.5.1 The spectral problem In this section we consider equilateral perfect binary trees, which will be denoted by Tn (n ∈ N). The trees Tn will be rooted and directed towards the root and can be constructed recursively as follows. The tree T1 consists of two edges joined at the root. Tn will be obtained from Tn−1 by attaching two edges to each of the pendant vertices of Tn−1 . Thus, the degree of the root is 2 and the degree of each other interior vertex is 3. It is easy to see that the number of pendant vertices

6.5. Spectral problems on perfect binary trees

243

of Tn is ppen = 2n , that the number of all vertices of Tn is pn = 2n+1 − 1, and n that the number of edges of Tn is gn = 2n+1 − 2. We enumerate the vertices by v0 , . . . , vg in such a way that the distance to the root is a nondecreasing function of the edge index. The enumeration is arbitrary in other respects. In particular, v0 is the root. Each vertex vj with j = 1, . . . , gn has a unique outgoing edge, which will be indexed as ej . In the following, n will be a generic natural number unless specified otherwise. Recall that the local coordinate on Tn identifies each directed edge ej of Tn with the interval [0, l] and that the coordinate increases in the direction of the edge. On each edge of Tn consider the Sturm-Liouville equation −yj00 + qyj = λyj ,

(j = 1, . . . , gn ),

(6.5.1)

where q ∈ L2 (0, l) is a symmetric real-valued function. We impose the Dirichlet boundary conditions yj (0) = 0

(j = gn−1 + 1, . . . , gn )

(6.5.2)

at each pendant vertex. At each interior vertex vj which is not the root we impose the continuity conditions yj1 (l) = yj2 (l) = yj (0)

(j = 1, . . . , gn−1 )

(6.5.3)

and Kirchhoff’s condition yj0 1 (l) + yj0 2 (l) − yj0 (0) = 0

(j = 1, . . . , gn−1 ).

(6.5.4)

Here j is the index of the edge outgoing from the vertex vj and j1 and j2 are the indices of the edges incoming into vj . Note that j1 and j2 depend on j and that j1 > j and j2 > j. For definiteness we may take j1 < j2 . At the root, the continuity and Kirchhoff conditions are y1 (l) = y2 (l), 0 y1 (l) + y20 (l) = 0.

(6.5.5) (6.5.6)

With the notation from the first paragraph of Section 6.4 we have Tbn = Tn−1 (n ∈ N), where T0 denotes the subgraph of T1 consisting of the isolated vertex v0 .

6.5.2

Characteristic functions of perfect binary trees (n)

For the tree Tn we denote by φN the corresponding Neumann characteristic func(n) tion and by φD (λ) the corresponding Dirichlet characteristic function with respect (n) to the root. This means, that φN is given by the problem (6.5.1)–(6.5.6), whereas (n) φD is given by (6.5.1)–(6.5.4) and y1 (l) = 0, y2 (l) = 0.

244

Chapter 6. Sturm-Liouville problems on graphs (n)

Recall that the characteristic function φN has been defined in Theorem (n) 6.4.2. The characteristic function φD is not directly covered by previous work. However, one can split the tree Tn at the root into two subtrees, say Tn,1 and Tn,2 , where ej becomes an edge of Tn,j for j = 1, 2. Clearly, the problem with Dirichlet condition at the root splits into two separate problems for the trees Tn,1 and Tn,2 . Note that the trees Tn,1 and Tn,2 are identical. Hence we can define (n)

2 , φD := φD,n

(6.5.7)

where φD,n is the characteristic function for the problem on Tn,1 with Dirichlet conditions at its pendant vertices and continuity and Kirchhoff conditions at its interior vertices. For n > 1 let Tn,0 be the graph obtained by removing all pendant vertices and their adjacent edges from Tn,1 . Clearly, Tn,0 is an isolated vertex when n = 2 and Tn,0 is a copy of Tn−2 when n > 2. In the notation from the first paragraph of Section 6.4 we can also write Tn,0 = Tbn,1 . Proposition 6.5.1. For each n ∈ N,

√ √ ( λ, l)PTn ,Tn−1 (c( λ, l)) √ n √ (n) φD (λ) = s2 ( λ, l)PT2n,1 ,Tn,0 (c( λ, l)), (n)

n

φN (λ) = s2

−1

(6.5.8) (6.5.9)

where PT1,1 ,T1,0 := 1. Proof. The identities (6.5.8) and (6.5.9) for n > 1 follow immediately from the above discussion and Theorem 6.4.2. When n = 1, then √ Tn,1 is an edge with  Dirichlet conditions at both ends, so that φD,1 (λ) = s( λ, l). For k ∈ N let Ak be the adjacency matrix of Tk and let A0 = 0 and p0 = 1. For α = 2, 3 and k ∈ N define the (1, pk−1 , pk−1 ) block matrix   αz −eT −eT 1 1  (z ∈ C), 0 Φk,α (z) := −e1 3zI − Ak−1 (6.5.10) −e1 0 3zI − Ak−1 where e1 is the first unit vector in Cpk−1 . Rearranging the indexing of the vertices of Tn in such a way that v0 and v1 are still the first two vertices, but then followed by all vertices connected to v1 by a path which does not pass through v0 , then the vertex v2 and then all remaining vertices, it is clear that Φk,α (0) = −Ak for n ≥ 2 and that the first row and column in Ak−1 in (6.5.10) is taken with respect to the root of Tk−1 . Hence we have PTn ,Tn−1 = det Φn−1,2

(n ≥ 2),

(6.5.11)

PTn,1 ,Tn,0 = det Φn−2,3

(n ≥ 3).

(6.5.12)

We note that PT1 ,T0 and PT2,1 ,T2,0 correspond to isolated vertices obtained by removing two or three pendant vertices from T1 and T2,1 , respectively. Hence it follows that PT1 ,T0 (z) = 2z,

PT2,1 ,T2,0 (z) = 3z

(z ∈ C).

(6.5.13)

6.5. Spectral problems on perfect binary trees

245

Proposition 6.5.2. The functions det Φk,α satisfy the following recurrence relations: det Φ1,α (z) = 9αz 3 − 6z, 3

(6.5.14) 2

3

det Φ2,α (z) = (27z − 6z)(αz(27z − 6z) − 18z ),

(6.5.15) 2

det Φk,α (z) = det Φk−1,3 (z)[αz det Φk−1,3 (z) − 2(det Φk−2,3 (z)) ] (k ≥ 3). (6.5.16) Proof. For k = 1 we find 

αz det Φ1,α (z) = det −1 −1

−1 3z 0

 −1 0 3z

= 9αz 3 − 6z, which proves (6.5.14). Now let k ≥ 2. From the Schur factorization (B.4.2) it follows for z ∈ C with det(3zI − Ak−1 ) 6= 0 that det Φk,α (z) = (det(3zI − Ak−1 ))2 × "  −1  #
3zI − Ak−1 e1 0 T T × αz − e1 e1 e1 0 3zI − Ak−1   −1 e1 . = (det(3zI − Ak−1 ))2 αz − 2eT 1 (3zI − Ak−1 ) We observe that 3zI − Ak−1 = Φk−1,3 (z) and that −1 det(3zI − Ak−1 )eT e1 1 (3zI − Ak−1 )

is the minor of the matrix 3zI −Ak−1 with respect to its top left entry. We therefore have ( 9z 2 when k = 2, T −1 det(3z − Ak−1 )e1 (3zI − Ak−1 ) e1 = 2 (det Φk−2,3 (z)) when k > 2. This proves (6.5.15) and (6.5.16).



We introduce the numbers 1 bk = (2k − (−1)k ), 3

( δk =

1 if k is odd, 0 if k is even,

(k ∈ N0 )

(6.5.17)

and for α = 2, 3 the polynomials P−1,3 (z) = −1, δk

P0,α (z) = α,

Pk,α (z) = Pk−1,3 (z)[αz Pk−1,3 (z) −

2 2Pk−2,3 (z)]

(6.5.18) (k ∈ N).

(6.5.19)

246

Chapter 6. Sturm-Liouville problems on graphs

Proposition 6.5.3. For k ∈ N and α = 2, 3, the functions det Φk,α are odd polynomials and have the representation det Φk,α (z) = z bk+1 Pk,α (z 2 ).

(6.5.20)

The polynomial Pk,α has degree bk+1 + δk − 1, (−1)k Pk,α (0) > 0, and the zeros of Pk,α are located in the interval (0, 1). Proof. We are going to prove the statement by induction on k. For k = 1 and k = 2, the result is obvious from (6.5.14) and (6.5.15). Now let k ≥ 3 and assume that the statement is true for indices less than k. Then (6.5.16) gives 2 det Φk,α (z) = z bk Pk−1,3 (z 2 )[αz bk +1 Pk−1,3 (z 2 ) − 2z 2bk−1 Pk−2,3 (z 2 )].

We calculate 1 k 2 (2 − (−1)k ) + 1 − (2k−1 − (−1)k−1 ) = 1 − (−1)k = 2δk , 3 3 1 k 2 bk + 2bk−1 = (2 − (−1)k ) + (2k−1 − (−1)k−1 ) = bk+1 . 3 3

bk + 1 − 2bk−1 =

Hence the representation (6.5.20) holds in view of (6.5.19) with an odd number bk+1 . Furthermore, 2 (−1)k Pk,α (0) = (−1)k−1 Pk−1,3 (0)[α(1 − δk )(−1)k−1 Pk−1,3 (0) + 2Pk−2,3 (0)] > 0.

Since all zeros of det Φk,α lie in (−1, 1), it follows that the zeros of Pk,α lie in (0, 1). Finally, by definition of Φk,a in (6.5.10) it is clear that the degree of the polynomial det Φk,α is 2k+1 − 1, and bk+1 + 2(bk+1 + δk − 1) = 2k+1 + (−1)k + 1 − (−1)k − 2 = 2k+1 − 1 shows that the degree of Pk,α is bk+1 + δk − 1.



Theorem 6.5.4. With the notation from Proposition 6.5.3, the characteristic functions on Tn (n ∈ N) have the representations √ √ √ n (n) (6.5.21) φN (λ) = s2 −1 ( λ, l)cbn ( λ, l)Pn−1,2 (c2 ( λ, l)), √ √ n √ (n) 2 φD (λ) = s2 ( λ, l)c2bn−1 ( λ, l)Pn−2,3 (c2 ( λ, l)), (6.5.22) where the functions Pk,α are defined in (6.5.18) and (6.5.19). The zeros of the functions Pk,α are located in the interval (0, 1). Proof. For n = 1, (6.5.21) follows from (6.5.8), (6.5.13) and (6.5.18). For n ≥ 2, (6.5.21) follows from (6.5.8), (6.5.11) and (6.5.20). For n = 1, (6.5.22) follows from (6.5.9) and (6.5.18). For n = 2, (6.5.22) follows from (6.5.9), (6.5.11) and (6.5.18). For n ≥ 3, (6.5.22) follows from (6.5.9), (6.5.12) and (6.5.20). The location of the zeros of the polynomials Pk,α has already been stated in Proposition 6.5.3. 

6.5. Spectral problems on perfect binary trees

6.5.3

247

The eigenvalues

The results of this chapter so far allow us to give a representation of the eigenvalues of the problem for perfect binary trees. We will make use of the notation (6.5.17)– (6.5.19). √ (1) Remark 6.5.5. We observe that for n = 1 we have φN (λ) = 4s( λ, l), and therefore all eigenvalues of (6.5.1)–(6.5.6) are simple and are given by λk (k ∈ N) from Proposition 6.3.5. The next theorem is concerned with the general case n ≥ 2. Theorem 6.5.6. Let n ∈ N, n ≥ 2. Then there is a natural number p0 (n) such that the spectrum of problem (6.5.1)–(6.5.6) on Tn with a real-valued symmetric function q ∈ L2 (0, l) can be represented as the set of the terms of increasing sequences (1) (2) ∞ (j,+) ∞ (j,−) (λk )∞ )k=1 , (λk )∞ k=1 , (λk )k=1 , (λk k=1 (j = 1, . . . , p0 (n)) of pairwise distinct real numbers with the following properties. (1)

1. The eigenvalues λk (k ∈ N) have multiplicity 2n −1 and have the asymptotic representation Q π2 k2 (1) (1) (6.5.23) λ k = 2 + + βk l l Rl (1) with Q = 21 0 q(x)dx and (βk )∞ k=1 ∈ l2 . (2)

2. The eigenvalues λk resentation

(k ∈ N) have multiplicity bn and have asymptotic rep(2)



λk =

π(k − 1/2) l

2 +

Q (2) + βk , l

(6.5.24)

(2)

with (βk )∞ k=0 ∈ l2 . (j,+)

3. The eigenvalues λk pj (n) > 0 with

(j,−)

and λk

(k ∈ N, j = 1, . . . , p0 (n)) have multiplicity

p0 (n)

X

pj (n) = bn + δn−1 − 1

j=1

and have the asymptotic representation (j,ε) λk

=

√ !2 kπ − arccos((−1)k ε zj ) Q (j,ε) + + βk , l l

(6.5.25)

where ε ∈ {+, −}, z1 , . . . , zp0 (n) are the distinct zeros of the polynomial (j,ε) ∞ )k=1

Pn−1,2 , and (βk

∈ l2 .

248

Chapter 6. Sturm-Liouville problems on graphs (1)

(2)

4. For j = 1, . . . , p0 (n) and ε ∈ {+, −}, the eigenvalues λk , λk (k ∈ N) are interlaced as follows: (j,+)

λ1

(2)

(j,−)

< λ1 < λ1

(1)

(j,−)

< λ1 < λ2

(2)

(j,+)

< λ2 < λ2

(1)

(j,+)

< λ2 < λ3

(j,ε)

and λk

< .... (6.5.26)

Proof. The eigenvalues of problem (6.5.1)–(6.5.6), counted with multiplicity, are (n) the zeros of the characteristic function φN . By (6.5.21) the eigenvalues are as √ (1) stated, with (λk )∞ of λ 7→ s( λ, l), k=1 being the increasing sequence of zeros √ (2) (j,ε) ∞ (λk )∞ k=1 being the increasing sequence of zeros of √ λ 7→ c( √λ, l), and (λk )k=1 being the increasing sequence of zeros of λ 7→ c( λ, l) − ε zp , where 0 < z1 < · · · < zp0 (n) < 1 are the distinct zeros of Pn−1,2 . The asymptotics in part 3 and the interlacing properties have been shown in Propositions 6.3.5 and 6.3.6. To q  (1) prove part 4 we recall from Proposition 6.3.7 that c λk , l = (−1)k , using √ the notation of this theorem. Also, for a ∈ (−1, 1) the function λ 7→ c( λ, l) − a has a unique zero in each of the intervals (−∞, λ1 ) and (λk , λk+1 ) (k ∈ N) by Proposition 6.3.6. The interlacing (6.5.26) immediately follows from the above observations.  For solving the inverse problem, we observe that there is a huge degree of redundancy in the spectrum of problem (6.5.1)–(6.5.6). We will consider both cases, that only the spectrum is known and that the eigenvalues, with multiplicity, are known. We will start with the latter case. Theorem 6.5.7. Let (λk )∞ k=1 be the nondecreasing sequence of eigenvalues of problem (6.5.1)–(6.5.6) on Tn (n ∈ N, n ≥ 2). Then the largest segment of equal (j) ∞ n numbers in (λk )∞ k=1 is of length 2 − 1. The sequences (λk )k=1 (j = 1, 2) in The(1) (2) orem 6.5.6 are given by λk = λk(2n+1 −2) and λk = λ(k−1)(2n+1 −2)+2n−1 (k ∈ N). Furthermore, ! ∞ Y 0 λ ψ0 (λ) := λm 1 − (2) (λ ∈ C), λk k=1 Q0 defines an entire function, where means that the term with k0 is omitted in (2) case λk0 = 0 for some k0 , and in this case m = 1, whereas m = 0 otherwise. With Cr := ψ0 (2πrl−1 ) (r ∈ N) the limit C := lim Cr 6= 0 r→∞

exists, and ψ := C −1 ψ0 (1) ψ(λk )

satisfies of the sequence

= (−1) (k ∈ N). The sequence (λk )∞ k=1 consists (1) ∞ (1) (λk )k=1 with λk(2n+1 −2)−2n +2 = λk (k ∈ N), k

(6.5.27) n

of 2 − 1 copies bn copies of the

6.5. Spectral problems on perfect binary trees

249

(2)

∞ and all zeros of the function Pn−1,2 ◦ ψ 2 , counted with multisequence (λk )k=1 plicity.

Proof. By (6.5.26), a segment of equal numbers in the sequence of eigenvalues (j,ε) (1) (2) consists of copies of one of λk , λk , λk (k ∈ N, j ∈ {1, . . . , p0 (n)}, ε ∈ {+, −}. n Because of 2 − 1 > bn ≥ bn + δn−1 − 1 the largest segment of equal numbers in n (λk )∞ k=1 is of length 2 − 1, and exactly each segment consisting of all copies of (1) λk has this maximal length. The eigenvalues are grouped into segments λ(k−1)(2n+1 −2)+1 , . . . , λk(2n+1 −2) (k ∈ N). By (6.5.26), the first bn + δn−1 + 1 eigenvalues in such a segment are of (2) (j,±) the form λk , followed by bn copies of λk , then again bn + δn−1 + 1 eigenvalues (j,±) (1) of the form λk , followed by 2n − 1 copies of λk . In particular, the first and last (1) term of the latter are λk(2n+1 −2)−2n +2 = λk(2n+1 −2) = λk . It is now clear that there are 2n+1 −2−(2n −1) = 2n −1 consecutive terms starting at λ(k−1)(2n+1 −2)+1 (2)

(2)

with λk in the middle, which shows that λk = λ(k−1)(2n+1 −2)+2n−1 . The function c(·, l) is a sine type function, see [96, Corollary 12.2.11]. Then [96, Lemma 11.2.29] shows that the function λ 7→ ψ0 (λ2 ) is a constant multiple of c(·, l). Hence (6.3.25) shows that c(λ, l) = ψ(λ2 ) (λ ∈ C). This completes the  proof in view of Theorem 6.5.4.

6.5.4

The inverse problem

Here we will show that the properties of the eigenvalues of (6.5.1)–(6.5.6) obtained in Theorem 6.5.7, together with the asymptotic properties (6.3.21) and (6.3.24) are also sufficient to realize such a sequence of real numbers as the eigenvalues, counted with multiplicity, of a problem (6.5.1)–(6.5.6) on an equilateral perfect binary tree with symmetric potential. The size of the tree, the lengths of the edges and the potential on the edges are then uniquely determined by the eigenvalues. ∞ Theorem 6.5.8. Let (λk )k=1 be a nondecreasing sequence of real numbers such that there is a number n ∈ N, n ≥ 2, with

λk(2n+1 −2)−2n +1 < λk(2n+1 −2)−2n +2 = λk(2n+1 −2) < λk(2n+1 −2)+1

(k ∈ N). (6.5.28)

Define (1)

(2)

λk = λk(2n+1 −2) , λk = λ(k−1)(2n+1 −2)+2n−1 Assume that

πk l := lim q k→∞ (1) λk

(k ∈ N).

(6.5.29)

(6.5.30)

exists and is positive. Furthermore, assume that there is a real number Q such that (1) (2) ∞ (λk )∞ k=1 satisfies the asymptotic representation (6.3.21) and that (λk )k=1 satisfies the asymptotic representation (6.3.24). Let the function ψ defined by (6.5.27)

250

Chapter 6. Sturm-Liouville problems on graphs (1)

(2)

satisfy ψ(λk ) = (−1)k (k ∈ N). Assume that the terms of the sequence (λk ) occur bn times in the sequence (λk )∞ k=1 and assume that the terms of the sequence (j) ∞ (λk )k=1 which are different from λs (j = 1, 2, s ∈ N) are exactly the zeros of the the function λ 7→ Pn−1,2 (ψ 2 (λ)), counted with multiplicity. Then there is a real-valued symmetric q ∈ L2 (0, l) such that (λk )∞ k=1 is the sequence of the eigenvalues of (6.5.1)–(6.5.6) on the equilateral perfect binary tree Tn with edge length l. The numbers n and l and the function q are uniquely determined by ∞ (λk )k=1 . (1)

Proof. Formula (6.5.28) implies that each of the numbers λk (k ∈ N) occurs ex∞ . The remaining eigenvalues are therefore actly 2n −1 times in the sequence (λk )k=1 located in nonadjacent segments of length 2n+1 − 2 − (2n − 1) = 2n − 1. Each of (2) these segments contains a value λk of multiplicity bn < 2n − 1. Hence also the remaining terms have multiplicity less than 2n − 1. By [96, Theorem 12.6.2] there is a unique real-valued function q ∈ L2 (0, l) (1) (2) and (λk )∞ such that sequences √ (λk )∞ k=1 are the zeros of the functions √ the

k=1 λ 7→ s λ, l and λ 7→ c λ, l associated with (6.3.1), respectively. Next we are going to show that q is symmetric. In the proof of Theorem 6.5.7 we have shown that c(λ, l) = ψ(λ2 ) (λ ∈ C) (the fact that q is symmetric in Theorem 6.5.7 is irrelevant for this identity). Using the Lagrange identity c(λ, l)s0 (λ, l) − c0 (λ, l)s(λ, l) = 1 q  q  q  (1) (1) (1) we arrive at c λk , l s0 λk , l = 1. Then the assumption c λk , l =  q  q (1) (1) (1) λk , l = s0 λk , l . Hence all singularities of ψ(λk ) = (−1)k shows that c the function c(·, l) − s0 (·, l) h := s(·, l) are removable singularities, which means that h can be identified with an entire function. By [96, Corollary 12.2.11], the functions λ 7→ λs(λ, l), c(·, l) and s0 (·, l) are sine type functions of type l, which implies by [96, Lemma 11.2.6, part 2] that there is a number M > 0 such that |c(λ, l) − s0 (λ, l)| < M el| Im λ| (λ ∈ C). Furthermore, by [96, Remark 11.2.21], for each δ > 0 there is a constant cδ > 0 such that the estimate |λs(λ, l)| ≥ cδ el| Im λ| holds for all λ ∈ C \ Λδ , where Λδ = {λ ∈ C : ∃ ζ ∈ C, ζs(ζ, l) = 0 and |λ − ζ| < δ}. It follows that |h(λ)| ≤ M c−1 δ |λ| for λ ∈ C \ Λδ . When δ > 0 is sufficiently small, then all components of Λδ are bounded, and the maximum principle shows that there are positive numbers c and d such that |h(λ)| ≤ c|λ| + d for λ ∈ C. Therefore the entire function h is a polynomial of degree at most 1. Observing the representations (6.3.25) and Z l sin λt sin λl + B3 (t) dt s0 (λ, l) = cos λl + Q 2λ λ 0

6.5. Spectral problems on perfect binary trees

251

with B3 ∈ L2 (0, l) (see [96, 12.2.23)], it follows that Z l 0 (B(t) − B3 (t)) sin λt dt = o(1) for λ ∈ R with λ → ∞ λ (c(·, l) − s (·, l)) = 0

by the Riemann-Lebesgue lemma. On the other hand, the representation (6.3.30) shows that  1  −1  1  −1  2r + πl s 2r + πl = 1 + O(r−1 ) for r ∈ N with r → ∞. 2 2

These last two estimates show that h 2r + 12 πl−1 = o(1). Since h is a polynomial, it follows that h = 0 and therefore c(·, l) = s0 (·, l). In particular, c(·, l) and s0 (·, l) have the same (simple) zeros. This means that the spectrum of the eigenvalue problem for −y 00 + qy = λy

(6.5.31)

0

with Neumann-Dirichlet boundary conditions y (0) = y(l) = 0 coincides with the spectrum of the eigenvalue problem for (6.5.31) with Dirichlet-Neumann boundary conditions y(0) = y 0 (l) = 0. Putting qb(x) = q(l − x), the latter problem can be identified with the spectral problem −y 00 + qby = λy, y 0 (0) = y(l) = 0. Clearly, the Dirichlet-Dirichlet problem y(0) = y(l) = 0 is the same for −y 00 +qy = λy and for −y 00 + qby = λy. We have already used at the beginning of this proof that the potential q is uniquely determined by the spectra of the Dirichlet-Dirichlet problem and the Dirichlet-Neumann problem. This shows that q = qb, which means that the function q is symmetric. The statement about the remaining terms of the sequence (λk )∞ k=1 follows from Theorem 6.5.7. By Theorem 6.5.6, the length of the largest segments of equal numbers in (1) (2) n (λk )∞ k=1 is 2 − 1 and therefore uniquely determines n. Hence also λk and λk are uniquely determined. This shows that l and q are unique as well.  We do not need to know the multiplicity of the eigenvalues. All we need to know is the spectrum. For this it is advantageous to have a more explicit representation of the polynomials Pk,α . We note that (6.5.19) is essentially a three term recurrence relation for the polynomials Pk,3 ; Pk,2 is then obtained from Pk−1,3 and Pk−2,3 . We will now rewrite this recurrence relation as a formally more complex two term recurrence relation for the pairs of polynomials (Pk,2 , Pk−1,3 ). Proposition 6.5.9. For k ∈ N, the recurrence relations    2 2 Pk,2 (z) = 2 Pk−1,2 (z) + z δk−1 Pk−2,3 (z) z δk Pk−1,2 (z) + (z − 1)Pk−2,3 (z) , (6.5.32) 2 Pk−1,3 (z) = Pk−1,2 (z) + z δk−1 Pk−2,3 (z)

hold, and the initial terms are P0,2 (z) = 2 and P−1,3 (z) = −1.

(6.5.33)

252

Chapter 6. Sturm-Liouville problems on graphs

Proof. The initial terms have been stated in (6.5.18), and then (6.5.33) follows for k = 1 because δ0 = 0. The recurrence relation (6.5.19) shows for k ≥ 2 that Pk−1,3 (z) − Pk−1,2 (z) = Pk−2,3 (z)z δk−1 Pk−2,3 (z), which proves (6.5.33) for k ≥ 2. Similarly, (6.5.19) and (6.5.33) give for k ∈ N that 2 (z)] Pk,2 (z) = Pk−1,3 (z)[2z δk Pk−1,3 (z) − 2Pk−2,3 2 2 2 = [Pk−1,2 (z) + z δk−1 Pk−2,3 (z) − 2Pk−2,3 (z)] (z)][2z δk Pk−1,2 (z) + 2zPk−2,3 2 2 = 2[Pk−1,2 (z) + z δk−1 Pk−2,3 (z)], (z)][z δk Pk−1,2 (z) + (z − 1)Pk−2,3

which is (6.5.32).



Theorem 6.5.10. For n ∈ N0 the polynomials Pn,2 and Pn−1,3 can be written in the form Pn,2 = 2An2 Bn Cn ,

(6.5.34)

Pn−1,3 = An Bn ,

(6.5.35)

where An , Bn , and Cn are polynomials which satisfy the recurrence relations An+1 (z) = A2n (z)Bn (z), δn

Bn+1 (z) = 2Cn (z) + z Bn (z), Cn+1 (z) = 2z

δn+1

Cn (z) + (z − 1)Bn (z),

(6.5.36) (6.5.37) (6.5.38)

with the initial terms being A0 (z) = −1,

B0 (z) = 1,

C0 (z) = 1.

(6.5.39)

Proof. It is clear from Proposition 6.5.9 and (6.5.39) that (6.5.34) and (6.5.35) hold for n = 0. Now let n ∈ N0 and assume that (6.5.34) and (6.5.35) hold for this n. Then, in view of (6.5.33), An+1 (z)Bn+1 (z) = An2 (z)Bn (z)(2Cn (z) + z δn Bn (z)) = 2A2n (z)Bn (z)Cn (z) + z δn A2n (z)Bn2 (z) 2 (z) = Pn,2 (z) + z δn Pn−1,3

= Pn,3 (z), which proves (6.5.35) for n + 1, and together with An+1 (z)Cn+1 (z) = A2n (z)Bn (z)(2z δn+1 Cn (z) + (z − 1)Bn (z)) = (2z δn+1 An2 (z)Bn2 (z)Cn (z) + (z − 1)A2n (z)Bn2 (z)) 2 (z) = z δn+1 Pn,2 (z) + (z − 1)Pn−1,3

6.5. Spectral problems on perfect binary trees

253

it follows in view of (6.5.32) that 2A2n+1 (z)Bn+1 (z)Cn+1 (z) 2 2 = 2[Pn,2 (z) + z δn Pn−1,3 (z)][z δn+1 Pn,2 (z) + (z − 1)Pn−1,3 (z)]

= Pn+1,2 (z), which proves (6.5.34) for n + 1.



Corollary 6.5.11. The explicit forms of the polynomials An , Bn and Cn for n ∈ N0 are: 1 B2n (z) = F (z)(α2n (z) − α1n (z)) + (α1n (z) + α2n (z)), 2 3 B2n+1 (z) = (F (z) + 2G(z))(α2n (z) − α1n (z)) + (α1n (z) + α2n (z)), 2 1 C2n (z) = G(z)(α2n (z) − α1n (z)) + (α1n (z) + α2n (z)), 2 3z −1 n C2n+1 (z) = H(z)(α2n (z) − α1n (z)) + (α1 (z) + α2n (z)), 2 n−1 Y n−1−k An (z) = Bk2 (z) (n > 0),

(6.5.40) (6.5.41) (6.5.42) (6.5.43) (6.5.44)

k=1

where F (z) =

9z − 6 27z 2 − 21z G(z) = , H(z) = , 2(α2 (z) − α1 (z)) 2(α2 (z) − α2 (z)) √ √ 9z − 4 + 3 9z 2 − 8z 9z − 4 − 3 9z 2 − 8z α1 (z) = , α2 (z) = . 2 2

9z , 2(α2 (z) − α1 (z))

Proof. The recurrence relations (6.5.37), (6.5.38) can be written in matrix form    δ   Bn+1 (z) z n 2 Bn (z) = . (6.5.45) Cn+1 (z) Cn (z) z − 1 2z δn+1 With

 D(z) :=

it follows that

z z−1 

   2 1 2 3z − 2 = 2 z − 1 2z 3(z − 1)

   B2n+2 (z) B2n (z) = D(z) C2n+2 (z) C2n (z)

for n ∈ N0 , and this together with (6.5.39) gives     B2n (z) 1 n = D (z) . C2n (z) 1

 6z 6z − 2

254

Chapter 6. Sturm-Liouville problems on graphs

A straightforward calculation shows that the eigenvalues of D(z) are α1 (z) and α2 (z). For the generic case when the eigenvalues are distinct, a diagonalization of D(z) has the form D(z) = E(z)∆(z)E −1 (z), with ∆(z) = diag (α1 (z), α2 (z)) and where a matrix of eigenvectors E(z) and its inverse E −1 (z) are given by   √ 4z √ 4z E(z) = , z − 9z 2 − 8z z + 9z 2 − 8z √   1 z + √9z 2 − 8z −4z −1 E (z) = √ . 8z 9z 2 − 8z −z + 9z 2 − 8z 4z Then 

   B2n (z) 1 = E(z)∆n (z)E −1 (z) C2n (z) 1 √   n  1 α1 (z) 0 −3z +√ 9z 2 − 8z = √ , E(z) 0 α2n (z) 3z + 9z 2 − 8z 8z 9z 2 − 8z

and a straightforward calculation gives (6.5.40) and (6.5.42). Then (6.5.41) and (6.5.43) easily follow from (6.5.45). Observing that A1 (z) = A20 (z)B0 (z) = 1, the identity (6.5.44) easily follows by induction.  For a polynomial P with at least one real zero let wP be its largest real zero. Corollary 6.5.12. The polynomials Bn and Cn defined in Theorem 6.5.10 for n = 1 and n = 2 are B1 (z) = 3, C1 (z) = 3z − 1, B2 (z) = 9z − 2, C2 (z) = 9z − 5. For n ≥ 2, the zeros of Bn and Cn are real, lie in the interval (0, 1) and satisfy wCn > wBn and wCn > wCn−1 . Proof. The representations for n = 1 and n = 2 immediately follow from the recurrence relations (6.5.37) and (6.5.38). From wC2 = 95 > 29 = wB2 and wC2 = 5 1 9 > 3 = wC1 we see that the inequalities are satisfied for n = 2. Since the zeros of Bn and Cn are also zeros of Pn,2 by (6.5.34), the statement on the location of the zeros of Bn and Cn follows from Proposition 6.5.3. We are going to show by induction that wCn > wBn and that wCn > wBn implies wCn+1 > wCn for n ≥ 2. The recurrence relations (6.5.37) and (6.5.38) show that the leading coefficients of Bn and Cn are positive. Now let n ≥ 2 and assume that wCn > wBn . Then Bn+1 is positive on [wCn , ∞) by (6.5.37), which means that wBn+1 < wCn . On the other hand, (z − 1)Bn (z) < 0 for z ∈ [wCn , 1), and then (6.5.38) shows that Cn+1 (wCn ) < 0. Therefore Cn+1 has a least one zero in (wCn , ∞), which proves wCn+1 > wCn . In particular, wCn+1 > wCn > wBn+1 . 

6.5. Spectral problems on perfect binary trees

255

Corollary 6.5.13. For n ∈ N, wPn,2 < wPn+1,2 . Proof. We know from (6.5.34), (6.5.44) and Corollary 6.5.12 that wPn,2 = max{wB2 , . . . , wBn , wCn } = wCn , and another application of Corollary 6.5.12 completes the proof.



In the last theorem of this section we will show that the data of the string problem on a perfect tree are uniquely determined by its spectrum. Theorem 6.5.14. For n ≥ 2, the numbers n and l and the symmetric function q are uniquely determined by the spectrum of problem (6.5.1)–(6.5.6). Proof. We enumerate the spectrum as an increasing sequence of real numbers ∞ (σk )k=1 . Since p0 (n) is the number of distinct zeros of Pn−1,2 , it is clear from (1) (2) (6.5.27) that λk = σ2kN and λk = σ(2k−1)N where N = p0 (n) + 1. Hence it follows as in Theorem 6.5.6 that l and q are uniquely determined by the sequence (σk )∞ k=1 and the number n. We still have to show the uniqueness of n. Therefore let n ∈ N, n ≥ 2, be ∞ an integer for which (σk )k=1 enumerates the spectrum on the perfect tree Tn . Let ∞ be the nondecreasing sequence of the eigenvalues of the problem under (λk )k=1 consideration, counted with multiplicity. Then Theorem 6.5.6, parts 3 and 4, shows that k2 π2 2kπαn σ2kN −1 = 2 − + O(1) as k → ∞, l l √ where αn = arccos ze and ze is the largest zero of Pn−1,2 . If n1 and n2 are two numbers for which (σk )∞ k=1 enumerates the spectrum on Tn1 and Tn2 , and denoting the numbers N and l for n1 and n2 by l1 , l2 and N1 , N2 , respectively, it follows that (kN2 )2 π 2 2kN2 παn1 − + O(1) l12 l1 2kN1 παn2 (kN1 )2 π 2 = − + O(1) 2 l2 l2

σ2kN1 N2 −1 =

as k → ∞,

σ2kN1 N2 −1

as k → ∞.

This shows that N2 l1−1 = N1 l2−1 and then αn1 = αn2 . But by Corollary 6.5.13, αn is strictly decreasing when n increases, and therefore n1 = n2 .  Remark 6.5.15. A combination of Theorems 6.5.6, 6.5.7, 6.5.8 and 6.5.14 gives necessary and sufficient conditions for a set of real numbers to be the spectrum of problem (6.5.1)–(6.5.6) on a perfect equilateral tree with more than two edges and equal symmetric potentials.

256

Chapter 6. Sturm-Liouville problems on graphs

6.6 Sturm-Liouville problems on star graphs In this section we consider a plane star graph of g strings (g ≥ 2), joined at the central vertex called the root and with all g pendant vertices fixed. We recall that spectral problems for a star graph of Stieltjes strings have been considered in Section 2.1. In what follows we need a slightly modified notion of majorization. s and y = (yi )ti=1 be two vectors Definition 6.6.1 (see [94, p. 12]). Let x = (xi )i=1 with nonnegative entries ordered nonincreasingly, x1 ≥ x2 ≥ · · · ≥ xs ≥ 0 and y1 ≥ y2 ≥ · · · ≥ yt ≥ 0. If s = t, then x is said to weakly majorize y if

x w y :⇔

τ X

xi ≥

τ X

yi

(τ = 1, . . . , t).

(6.6.1)

i=1

i=1

max{s,t}

and If s 6= t, we fill up the shorter vector with zeros, i.e., we set x e = (xi )i=1 max{s,t} ye = (yi )i=1 with xi = 0 for i = s + 1, . . . , max{s, t}, yi = 0 for i = t + 1, e majorizes ye, . . . , max{s, t}. Then x is said to weakly majorize y, x w y, if x x e w ye. We will otherwise use the same notation as in Subsection 2.1.3.

6.6.1

Multiplicities and location of eigenvalues below a fixed value

In this subsection we revisit the string problem on a star graph whose edges are labelled by j = 1, . . . , g such that the j-th edge carries nj > 0 beads with mass (j) mk in its interior (k = 1, . . . , nj ) and n1 ≥ n2 ≥ · · · ≥ ng . Recall that (µk )nk=1 denotes the nondecreasing sequence of the characteristic values of the Neumann problem (N1) on the star graph. The results of Subsection 2.1.3 can be interpreted in the following way. We notice that by Theorem 2.1.5, µk is a characteristic value of problem (N1) of multiplicity pk ≥ 2 if and only if it is a characteristic value of the problem (D1) of multiplicity pk + 1. This means that µk is a characteristic value of the Dirichlet-Dirichlet problem on exactly pk + 1 amongst the g edges. (j) nj Let (νk )k=1 be the increasing sequence of the (positive) characteristic values of the Dirichlet-Dirichlet problem on the edges (j = 1, . . . , g). For E ∈ (0, ∞] let (j)

nj (E) := #{k ∈ {1, . . . , nj } : νk ≤ E}. (j)

Clearly, nj (E) = 0 if and only if ν1 > E, otherwise, nj (E) is the largest integer (j) (j) k ∈ {1, . . . , nj } for which νk ≤ E, i.e., nj (E) = max{k ∈ {1, . . . , nj } : νk ≤ E}

6.6. Sturm-Liouville problems on star graphs

257

(j)

if ν1 ≤ E, and nj (∞) = nj . We define n(E) :=

g X

nj (E),

(6.6.2)

j=1

Ni (E) := #{j ∈ {1, . . . , g} : nj (E) ≥ i},

(6.6.3)

nN (E) := #{k ∈ {1, . . . , n} : µk ≤ E}.

(6.6.4)

Then N1 (E) ≥ N2 (E) ≥ · · · ≥ Nn1 (E). With the above notation, the following theorem immediately follows from Theorems 2.1.5 and 2.1.19 and the proof of Theorem 2.1.19. Theorem 6.6.2. Let (µk )nk=1 be the nondecreasing sequence of the characteristic values of the Neumann problem (N1), counted with multiplicity, let E > 0, let r(E) be the number of distinct characteristic values of (N1) which are less or r(E) r(E) equal E, let (pi )i=1 be the vector of their multiplicities, and let p↓ (E) = (p↓i )i=1 be the corresponding vector of multiplicities in nonincreasing order. Then: 1. nN (E) =

r(E) P

pi ; nN (E) = n(E) or nN (E) = n(E) + 1;

i=1

2. 0 < µ1 < µ2 ≤ µ3 ≤ . . . ≤ µnN (E)−1 ≤ µnN (E) ≤ E; 3. if pi > 1, then pi−1 = pi+1 = 1 (i = 2, . . . , r(E) − 1), and if pr(E) > 1, then pr(E)−1 = 1; ↓ ) in case 4. (N1 (E) − 1, N2 (E) − 1, . . . , Nn1 (E) (E) − 1)  (p↓1 , p2↓ , . . . , pr(E)−n 1 (E) nN (E) = n(E);

40 . (N1 (E) − 1, N2 (E) − 1, . . . , Nn1 (E) (E) − 1)  (p↓1 , p↓2 , . . . , p↓r(E)−n1 (E)−1 ) in case nN (E) = n(E) + 1. Although Theorem 6.6.2 is formulated for the characteristic values of a star graph of Stieltjes strings, in its essence it is purely algebraic, only depending on the properties of two sequences as stated in Theorem 2.1.5. Therefore being independent of the initial physical meaning, it can be extended to problems which have infinitely many characteristic values.

6.6.2

Sturm-Liouville problems on star graphs with finite edges

The following problem appears in quantum mechanics as well as in the theory of small transverse vibrations of a star graph of smooth strings with g edges of

258

Chapter 6. Sturm-Liouville problems on graphs

lengths aj > 0 (j = 1, . . . , g) (see, e.g., [96, p. 238]): yj00 + λyj − qj (x)yj = 0, yj (0) = 0

(j = 1, . . . , g, x ∈ [0, aj ]), (j = 1, . . . , g),

(6.6.6)

y1 (a1 ) = · · · = yg (ag ), g X

(6.6.5) (6.6.7)

yj0 (aj ) = 0.

(6.6.8)

j=1

We assume that qj ∈ L2 (0, aj ) (j = 1, . . . , n) are real-valued functions and that qj ≥ Qj for some Qj ∈ R. The problem (6.6.5)–(6.6.8) is called the Neumann problem on the star graph. We also consider the corresponding Dirichlet problem (6.6.5), (6.6.6), y1 (a1 ) = · · · = yg (ag ) = 0. (6.6.9) √ With the notation from Section 6.3 and sj ( λ, ·) being the solution of yj00 + λyj − qj (x)yj = 0,

(x ∈ [0, aj ]),

yj (0) = yj (aj ) = 0, (j)

the increasing sequence (νk )∞ k=1 of the√eigenvalues of this problem is the increas√ ing sequence of the zeros of λ 7→ sj ( λ, aj ) (j = 1, . . . , g). Also, let sbj ( λ, ·) (j) be the solution of this problem with qj replaced by Qj and let (b νk )∞ k=1 be the increasing sequence of the corresponding eigenvalues. The well-know Sturm Com(j) (j) parison Theorem shows the inequalities νbk ≤ νk (j = 1, . . . , g, k ∈ N); these inequalities are, e.g., a special case of [145, Theorem 4.9.1(1)], or easily follow from [139, Theorem 13.1]. Let E ∈ R and define (j)

nj (E) := #{k ∈ N : νk ≤ E},

(j)

n bj (E) := #{k ∈ N : νbk ≤ E}

(j = 1, . . . , g).

It is clear that ( p aj nj (E) ≤ n bj (E) :=

π

E − Qj

0



if E > Qj , if E ≤ Qj ,

(6.6.10)

where bbc means the integer part of a nonnegative number b. The nondecreasing (j) ∞ union of the sequences (νk )∞ j=1 (j = 1, . . . , g) will be denoted by (νk )k=1 . In view of [96, (6.4.5)], the characteristic function of the Neumann problem (6.6.5)–(6.6.8) is g g X Y √ √ φN (λ) = s0j ( λ, aj ) sr ( λ, ar ). (6.6.11) j=1

r=1 r6=j

6.6. Sturm-Liouville problems on star graphs

259

Proposition 6.6.3. The eigenvalues of problem (6.6.5)–(6.6.8) are real and can be arranged in a nondecreasing sequence (µk )∞ k=1 . This sequence satisfies µ1 < ν1 ≤ µ2 ≤ ν2 ≤ · · · ≤ µk−1 ≤ νk−1 ≤ µk ≤ . . . .

(6.6.12)

Proof. With √ g Y √ s0j ( λ, aj ) φN (λ) √ ψ(λ) := , φD (λ) := sr ( λ, ar ), ψj (λ) := (j = 1, . . . , g), φD (λ) sj ( λ, aj ) r=1 it is clear that ψ(λ) =

g X

ψj (λ).

j=1

√ The zeros of λ 7→ sj ( λ, aj ) are the eigenvalues of the √ Dirichlet-Dirichlet problem on the j-th string, whereas the zeros of λ 7→ s0j ( λ, aj ) are the eigenvalues of the Dirichlet-Neumann problem on the j-th string. Reversing the direction of the strings, it follows from Proposition 6.3.5 that (j)

(j)

(j)

(j)

ξ1 < ν1 < ξ2 < ν2 < . . .

(j = 1, . . . , g),

√ (j) 0 where (ξk )∞ k=1 is the increasing sequence of the zeros of λ 7→ sj ( λ, aj ). By [96, Corollary 12.2.11], the functions λ 7→ λsj (λ, aj ) and sj (·, aj ) are sine type functions. From √ the representation [96, √ Lemma 11.2.29], from [96, Remark 11.1.6] and from sj ( λ, aj ) → ∞ and s0j ( λ, aj ) → ∞ as λ → −∞ it is clear that −ψj (j = 1, . . . , g) are Nevanlinna functions since their smallest zero is smaller than their smallest pole. Hence also their sum −ψ is a Nevanlinna function. Let (0) (j) (0) (j) ν1 := min{ν1 : j = 1, . . . , g} and ξ1 := min{ξ1 : j = 1, . . . , g}. Then −ψ < 0 (0) (0) on (−∞, ξ1 ) and ν1 is the smallest pole of −ψ. Since Nevanlinna functions are (0) (0) locally increasing, ξ1 ≤ µ1 < ν1 = ν1 holds. The remaining inequalities in (6.6.12) follow as the zeros and poles of the Nevanlinna function −ψ interlace.  For E ∈ R let r(E) be the number of distinct eigenvalues of the Neumann problem which are less or equal E. Let (p1 , . . . , pr(E) ) be the vector and (p↓1 , . . . , p↓r(E) ) the ordered vector of multiplicities of the distinct eigenvalues of problem (6.6.7) which are less or equal E. Theorem 6.6.4. Let E ∈ R. Then the eigenvalues of (6.6.7) have the following properties: 1. µ1 < µ2 ≤ µ3 ≤ . . . ≤ µn(E,N )−2 Pq ≤ µn(E,N )−1 ≤ µn(E,N ) ≤ E < µn(E,N )+1 , where n(E, N ) ≤ n b(E) + 1 := j=1 n bj (E) + 1; 2. if pi > 1, then pi−1 = pi+1 = 1 (i = 2, . . . , r(E) − 1), and if pr(E) > 1, then pr(E)−1 = 1;

260

Chapter 6. Sturm-Liouville problems on graphs

↓ ). 3. (N1 (E) − 1, N2 (E) − 1, . . . , Nn1 (E) (E) − 1) w (p↓1 , p2↓ , . . . , pr(E)−n 1 (E)−1

Proof. 1. Part 1 immediately follows from (6.6.10) and (6.6.12). 2. Let k ∈ N be such that µk is the i-th distinct eigenvalue of the Neumann problem and assume that pi > 1. Then µk = νk−1 or µk = νk so that µk is a pole of at least one ψj . As we have seen for ν1 in the proof of Proposition 6.6.3, a pole of some ψj is also a pole of ψ, which means that m(φN , µk ) < m(φD , µk ) and the interlacing in part 1 implies that νk−1 = µk = νk . It is clear that in view of the interlacing from part 1, no two adjacent distinct eigenvalue µk can have this property. This proves part 2. 3. If µk ≤ E < νk , then µk−1 ≤ νk−1 < µk by the proof of part 2. In the notation of Subsection 2.1.3 we identify nj with nj (E). Then the results of Subsection 2.1.3 are applicable, and part 3 of Theorem 2.1.19 completes the proof. Here we note that the weak majorization w is the majorization  if µk ≤ E < νk for some k. Otherwise, majorization is obtained by part 3 of Theorem 2.1.19 if p↓r(E)−n1 (E) is added as the last term in the vector on the right-hand side of the  statement of part 3.

6.6.3

Sturm-Liouville problems on star graphs with infinite edges

In this subsection we consider a star graph with g infinite edges, with SturmLiouville problems on the edges and with continuity and Kirchhoff conditions at the central vertex. Thus, we have yj00 + λyj − qj (x)yj = 0,

(j = 1, . . . , g, x ∈ [0, ∞]),

y1 (0) = · · · = yg (0), g X

yj0 (0) = 0,

(6.6.13) (6.6.14) (6.6.15)

j=1

where R ∞ qj (j = 1, . . . , g) are real valued measurable functions on (0, ∞) satisfying (1 + x)|qj (x)| dx < ∞. Problem (6.6.13)–(6.6.15) is called Neumann problem 0 on the star graph. Let Q = diag(q1 , . . . , qg ). The operator TN associated with the Neumann problem on the star graph is defined by TN Y = −Y 00 + QY, where its domain D(TN ) is the set of all Y = (y1 , . . . , yg ) ∈ (L2 (0, ∞))g which satisfy −Y 00 + QY ∈ (L2 (0, ∞))g , (6.6.14) and (6.6.15). We also consider the auxiliary Dirichlet problem (6.6.13), y1 (0) = · · · = yg (0) = 0,

(6.6.16)

where the differential operator TD associated with the Dirichlet problem on the

6.6. Sturm-Liouville problems on star graphs

261

star graph is defined by TD Y = −Y 00 + QY, D(TD ) = {Y ∈ (L2 (0, ∞))g : −Y 00 + QY ∈ (L2 (0, ∞))g , Y (0) = 0)}, The Dirichlet problem on the star graph is the direct product of the Dirichlet problems on the edges, where the differential operator TD,j in L2 (0, ∞) on the j-th edge is defined by TD,j y = −y 00 + qj y, D(TD,j ) = {y ∈ L2 (0, ∞) : −y 00 + qj y ∈ L2 (0, ∞), y(0) = 0)}. Furthermore, the Neumann problem on the j-th edge is realized by the operator operator TN,j in L2 (0, ∞) defined by TN,j y = −y 00 + qj y, D(TN,j ) = {y ∈ L2 (0, ∞) : −y 00 + qj y ∈ L2 (0, ∞), y 0 (0) = 0)}. Here we note that y as well as y 0 are continuous on [0, ∞) for each y ∈ L2 (0, ∞) which satisfies −y 00 + qj y ∈ L2 (0, ∞). Next we will state three propositions and one theorem, which will be proved together after the statement of the theorem. Proposition 6.6.5. For j = 1, . . . , g, the operator TD,j is selfadjoint, its essential spectrum is [0, ∞), and it has nj eigenvalues, all simple and located in (−∞, 0), where Z ∞

xqj− (x) dx < ∞

0 ≤ nj ≤

(6.6.17)

0

and qj− (x) := max{0, −q(x)}. Proposition 6.6.6. For j = 1, . . . , g, the operator TN,j is selfadjoint, its essential spectrum is [0, ∞), and it has at most nj + 1 eigenvalues, all simple and located in (−∞, 0). Proposition 6.6.7. The operator TD is selfadjoint, its essential spectrum is [0, ∞), and it may have finitely many eigenvalues, all located in (−∞, 0). The total multiplicity n of the eigenvalues satisfies 0≤n=

g X

nj .

j=1

Theorem 6.6.8. The operator TN is selfadjoint, its essential spectrum is [0, ∞), and it may have finitely many eigenvalues, all located in (−∞, 0).

262

Chapter 6. Sturm-Liouville problems on graphs

Proof of Propositions 6.6.5, 6.6.6, 6.6.7 and Theorem 6.6.8. All statements follow from the summary given in the introduction of [137], which also summarizes results from [135] and [1], taking [136] and [138] into account. In particular, [135, Theorem 6.3] states that the essential spectrum of any such selfadjoint operator is [0, ∞), that the operator is bounded below and that its eigenvalues, if any, are located in (−∞, 0). For Propositions 6.6.5 and 6.6.6 we may also use, e.g., [8, Corollaries 6.3.16 and 6.17], except for the finiteness of the eigenvalues and that 0 is not an eigenvalue. Clearly, Proposition 6.6.7 follows immediately from Proposition 6.6.5 since TD is the direct product of the operators TD,j (j = 1, . . . , g). To complete the proof we still have to show that the operator TN is selfadjoint, which will be accomplished by verifying the conditions (1.2)–(1.4) in [137]. In the notation [137, (1.2)], the boundary conditions (6.6.14) and (6.6.15) can be g are written as −B ∗ Y (0) + A∗ Y 0 (0) = 0, where A = (aj,k )gj,k=1 and B = (bj,k )j,k=1 g × g matrices whose nonzero entries are given by aj,j = 1 and aj+1,j = −1 for j = 1, . . . , g − 1 and bj,g = 1 for j = 1, . . . , g. A straightforward calculation gives B ∗ A = 0 and therefore the condition −B ∗ A+A∗ B is satisfied, which is [137, (1.3)]. Furthermore, the matrix A∗ A+B ∗ B is positive semidefinite, and a straightforward calculation shows that it is a Jacobi matrix with diagonal elements 2, . . . , 2, g and the elements above and below the diagonal are −1, . . . , −1, 0. By Proposition B.3.7 it is now easy to see that its determinant is g 2 . Therefore, A∗ A + B ∗ B > 0, which is the condition [137, (1.4)].  In [137, Theorem 1.1] an upper bound for the total multiplicity of the eigenvalues of TN in (−∞, 0) is given, but we will obtain a better estimate in our main theorem below. √ √ For λ ∈ C \ {0} let λ be the unique square root with 0 ≤ arg λ < π. For each j = 1, . . . , g there √ is a unique solution of (6.6.13), called the Jost solution and denoted by ej ( λ, ·), which satisfies √ √ (6.6.18) ej ( λ, x) = ei λx (1 + o(1)) as x → ∞, √ √ √ e0j ( λ, x) = i λei λx (1 + o(1)) as x → ∞, (6.6.19) see, e.g., [8, Lemma 6.13.1]. For each λ ∈ C \ [0, ∞) the multiples of the Jost solution are those solutions of (6.6.13) which belong to L2 (0, ∞).√ √ The construction of ej shows that the functions λ 7→ ej ( λ, 0) and λ 7→ e0j ( λ, 0) are analytic in C \ [0, ∞). Proposition 6.6.9. For j√= 1, . . . , g, the eigenvalues of TD,j and TN,j in (−∞, 0) √ are the zeros of λ 7→ ej ( λ, 0) and λ 7→ e0j ( λ, 0) in (−∞, 0), respectively, these zeros are simple and interlace, and the functions φj defined by √ e0j ( λ, 0) √ φj (λ) := ej ( λ, 0)

6.6. Sturm-Liouville problems on star graphs

263

are Nevanlinna functions. If φj has a pole in (−∞, 0), then φj has at least one zero there, and the smallest zero of φj is smaller than the smallest pole of φj . Proof. Let u1 (λ, ·) and u2 (λ, ·) be the solutions of (6.6.13) satisfying the initial conditions u1 (λ, 0) = 1, u01 (λ, 0) = 0, u2 (λ, 0) = 0, u02 (λ, 0) = 1. By [8, (6.13.9)], the Weyl function M is the unique function such that u1 (λ, ·) + M (λ)u2 (λ, ·) ∈ L2 (0, ∞) (λ ∈ C \ R). Hence there is a function α such that √ u1 (λ, ·) + M (λ)u2 (λ, ·) = α(λ)ej ( λ, ·) (λ ∈ C \ R). (6.6.20) √ Substituting the initial value 0 in (6.6.20) gives that 1 = α(λ)ej ( λ, 0). Differenti√ ating (6.6.20) and then substituting the initial value 0 gives M (λ) = α(λ)e0j ( λ, 0). This shows that M = φj , and therefore φj is a Nevanlinna function by [8, Corollary 2.3.7]. √ It is clear that the eigenvalues of TD,j are the√zeros of λ 7→ ej ( λ, 0) and that the eigenvalues of TN,j are the zeros of λ 7→ e0j ( λ, 0). Although [96, Lemma 11.1.3] is stated for meromorphic functions on C, the proof shows that it is true for functions which are for example meromorphic on C \ [0, ∞). Hence the zeros and poles of φj in (−∞, 0) are simple and interlace. Finally, since solutions √ y of 0 (6.6.13) are uniquely determined by y(0) and y (0), it is clear that λ → 7 e ( λ, 0) j √ 0 and λ 7→ ej ( λ, 0) cannot have common zeros. The simplicity of the poles and √ √ zeros of φj therefore shows that the zeros of λ 7→ ej ( λ, 0) and λ 7→ e0j ( λ, 0) are simple and interlace. By [8, (6.13.18)], the representation √ R∞ √ i λ − 0 ei λt qj (t)u1 (λ, t) dt φj (λ) = R∞ √ 1 + 0 ei λt qj (t)u2 (λ, t) dt holds, and by [8, (6.13.18)], the integral in the denominator tends to 0 as λ → −∞ and the integral in the numerator remains bounded when λ → −∞. Hence φj (λ) → −∞ as λ → −∞, and therefore the smallest zero of φj , if any, is smaller than its smallest pole, if any, because Nevanlinna functions are increasing on each interval in their domain (see [96, Lemma 11.1.3]).  Let φN (λ) :=

g X

g Y √ √ e0j ( λ, 0) er ( λ, 0),

j=1

r=1 r6=j

φD (λ) :=

g Y

√ er ( λ, 0).

(6.6.21)

(6.6.22)

r=1

For λ ∈ (−∞, 0) let nD (λ) and nN (λ) be the multiplicity of λ as an eigenvalue of TD and TN , respectively.

264

Chapter 6. Sturm-Liouville problems on graphs

Theorem 6.6.10. The eigenvalues, counted with multiplicity, of LD and LN lie in (−∞, 0) and are the zeros of φD and φN , counted with multiplicity. They can be n written as finite (possibly void) nondecreasing sequences (ζk )k=1 for the Dirichlet Pg n0 problem and (λk )k=1 for the Neumann problem with n := j=1 nj and n0 = n or n0 = n + 1 and satisfy λ1 < ζ1 ≤ λ2 ≤ · · · ≤ ζn < 0

(6.6.23)

λ1 < ζ1 ≤ λ2 ≤ · · · ≤ ζn < λn+1 < 0

(6.6.24)

when n0 = n > 0 and

when n0 = n + 1. The multiplicities satisfy nN (ζj ) = nD (ζj ) − 1. Proof. The function n

X φN φ := = φj φD j=1 is a sum of Nevanlinna functions and therefore a Nevanlinna function. A reasoning as in the proof of Proposition 6.6.3 shows that λ1 < ζ1 . Observing that we already know the zeros of ej (·, 0) and e0j (·, 0) are simple and distinct, all the statements √ (j) follow from the proof of Theorem 2.1.4 with R2nj (z) replaced with ej ( λ, 0) and √ (j) R2nj −1 (z) replaced with e0j ( λ, 0).  Let (p1 , p2 , . . . , pr ) be the vector and (p↓1 , p↓2 , . . . , p↓r ) the nonincreasingly ordered vector of distinct eigenvalues of problem (6.6.13)–(6.6.15). Theorem 6.6.11. Let Ni := #{j ∈ {1, . . . , g} : nj (D) ≥ i} = max{j ∈ {1, . . . , g} : nj (D) ≥ i}. Then (N1 − 1, N2 − 1, . . . , Nn1 − 1) w (p↓1 , p↓2 , . . . , p↓r−n1 −1 ). If pi > 1, then pi−1 = pi+1 = 1 (i = 2, . . . , r − 1). Proof. The last statement immediately follows from Theorem 6.6.10. The weak majorization can be proved as for part 2 of Theorem 6.6.4.  Example qj (x) = −γj e−αj x with αj > 0, γj > 0 for j = 1, . . . , g. Then R ∞ − 6.6.12. Let−2 xqj (x) dx = γj αj and therefore 0 nj ≤ bγj αj−2 c.

(6.6.25)

It follows from Theorem 6.6.10 that the number of eigenvalues of TN , counted with g g P P multiplicity, is less or equal 1 + nj ≤ 1 + bγj αj−2 c. j=1

j=1

6.7. Notes

265

6.7 Notes Usually the term ‘quantum graphs’ means metric graphs considered as quasi-onedimensional domains with differential equations defined on these domains (see, e.g., [67], [11], [82], [83], [5], [43]). In quantum mechanics the Sturm-Liouville and the Dirac equation and in vibration theory the string equation are considered on the edges of a graph subject to matching and boundary conditions at the vertices. These are Dirichlet or Neumann or Robin conditions at pendant vertices and continuity conditions together with Kirchhoff conditions at interior vertices. Such models are often used in problems of free-electron theory of conjugate molecules in chemistry and in the theory of quantum wires and thin wave-guides. The differential equations together with the matching and boundary conditions define an operator which is usually called continuous Laplacian (see, e.g., [33]). There are different definitions of the so-called discrete (combinatorial) Laplacian (see [38], [12]); we have used the one in [38] (see (6.2.6)). The notion of discrete Laplacian is closely related to the notion of adjacency matrix of the classical spectral graph theory (see, e.g., [44]), namely, for a simple connected graph with no loops the discrete Laplacian is L = I − T −1/2 AT −1/2 . The question of connection between continuous and discrete Laplacian was pointed out in [2] and developed in a rigorous manner in [51], [30], [10], [57]. This connection exists in the case of the free continuous Laplacian or under restrictive conditions on the potentials of the Sturm-Liouville equations on the edges of the graph; the edges must be of the same length and the potential must be the same. The correspondence between the spectra for the discrete (combinatorial) and continuous Laplacians for a large class of the equilateral graphs was given in [6], [33]. It was proved there that if the potentials on all edges are zero, then the spectrum of the continuous Laplacian on the graph consists of the Dirichlet π 2 n2 spectrum √ (n = 1, 2, . . . , and l is length of an edge) and those numbers z for l2 which cos z belongs to the spectrum of the corresponding discrete Laplacian. Some of the results of Section 6.4 were obtained in [115]. Amongst the variety of graphs an important role is played by the perfect trees (see, e.g., [127]). The results of Section 6.5 are based on [113], but equations (6.5.37), (6.5.38) are given in [113] with errors. The publication [102] considered the case of the same L2 potential on each edge and δ-type boundary conditions at the vertices. The spectrum of the quantum graph is the union of the Dirichlet spectrum on an edge and of the set η −1 (σ(2∆)), where σ(2∆) is the spectrum of 2∆, ∆ is the corresponding discrete Laplacian, and η is the discriminant (Lyapunov function) of the Hill equation on an edge. The estimate (6.6.17) of the number of eigenvalues of the Sturm-Liouville problem on the semiaxis is called the Bargmann-Birman-Schwinger inequality. Therefore the estimate of the number of eigenvalues of the Sturm-Liouville problem on a star graph is called a Bargmann-Birman-Schwinger type inequality.

Appendix A

Rational Nevanlinna and Stieltjes Functions In this appendix we will state and prove results for the class of rational Nevanlinna functions and its Stieltjes subclass. The results on the Stieltjes class are a slight generalization of the results in [59, Supplement II]. The interested reader can find corresponding results for analytic or meromorphic Nevanlinna functions for example in [96].

A.1

Rational functions

Definition A.1.1. A rational function f is the quotient of two polynomials q and p with p 6= 0: q(z) f (z) = (z ∈ C, p(z) 6= 0). p(z) Remark A.1.2. By the fundamental theorem of algebra, a nonconstant polynomial q assumes each value, i.e., for each b ∈ C there is at least one complex number a ∈ C such that q(a) = b. It follows that any polynomial q of degree m ∈ N has a representation m Y q(z) = C (z − βj ) (z ∈ C) (A.1.1) j=1

with βj ∈ C (j = 1, . . . , m) and C ∈ C \ {0}. This representation extends to nonzero constant polynomials, which have degree 0, if one observes the convention that the product over an empty index set is defined to be 1. A zero of q is a complex number z0 such that q(z0 ) = 0, i.e., z0 = βj0 for some j0 ∈ {1, . . . , m}. If z = βj0 and βj 6= βj0 for all j 6= j0 , then the zero z0 of q is called a simple zero of q. © Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4

267

Appendix A. Rational Nevanlinna and Stieltjes Functions

268

Remark A.1.3. Let f be a nonzero rational function with the representation f = pq . If q and p would have common zeros, then we can cancel corresponding factors from q and p in the product representations (A.1.1) of q and p, so that we may assume that q and p do not have common zeros, that is, that there are no z ∈ C such that q(z) = p(z) = 0. Then z0 ∈ C is called a (simple) zero of f if z0 is a (simple) zero of q, whereas z0 ∈ C is called a (simple) pole of f if z0 is a (simple) zero of p. There are m, n ∈ N ∪ {0}, complex numbers α1 , . . . , αn , β1 , . . . , βm with αj 6= βk for j = 1, . . . , n and βk = 1, . . . , m, and C 6= 0 such that f (z) = C

m Y

(z − βj )

j=1

n Y

(z − αj )−1

(z ∈ C, p(z) 6= 0).

(A.1.2)

j=1

The polynomials q and p are unique up to a constant factor. When z0 ∈ C is a zero of f , the number of indices j for which βj = z0 is called the multiplicity of the zero z0 of f and denoted by m(f, z0 ). Similarly, when z0 is a pole of f , the number of indices j for which αj = z0 is called the pole order of the pole z0 of f . In any case, for each z0 ∈ C there is a unique integer k such that f (z) = (z − z0 )k h(z) where h is a rational function for which z0 is neither a zero nor a pole. The complex number z0 is a zero of f if and only if k > 0, and then k is the multiplicity of the zero z0 of f . The complex number z0 is a pole of f if and only if k < 0, and then |k| is the order of the pole z0 of f . Defining the function f by f (z) = f (z) it follows from (A.1.2) that f (z) = C

m Y

(z − βj )

j=1

n Y

(z − αj )−1 .

j=1

Hence also f is a rational function. Definition A.1.4. A rational function f is called a real rational function if it maps real numbers to real numbers, i.e., if f (z) ∈ R for z ∈ R which are not poles of f . Proposition A.1.5. If f is a nonzero real rational function, then its zeros and poles are symmetric with respect to the real axis, f = f , and f can be written as f = pq , where q and p are real polynomials without common zeros. Proof. From the definition of f one concludes that f (z) = f (z) = f (z) for real z which are not poles of f or f . Hence the rational function f − f has infinitely many (real) zeros, and f = f follows. In particular, if βj is a nonreal zero of f , also βj is a zero of f , and if αj is a nonreal pole of f , also αj is a pole of f . Then the factors (z − βj )(z − βj ) and (z − αj )(z − αj ) are real for real z. If, however, βj or αj are real, then the factors z − βj and z − αj are real for real z. When all these factors are exhausted, we see that also C is real. Defining q(z) = C

m Y

(z − βj ),

j=1

p(z) =

n Y

(z − αj ),

j=1

it follows that q and p are real polynomials such that f = pq .



A.1. Rational functions

269

It is sometimes convenient to consider a rational function f on C as a function from C into C = C ∪ {∞} by putting f (z) = ∞ for poles z of f . Lemma A.1.6. Let f be a nonconstant rational function. Then C \ f (C) has at most one element. Proof. If ∞ 6∈ f (C), then f has no poles, so that f is a nonconstant polynomial, and f (C) = C follows from Remark A.1.2. Now let ∞ ∈ f (C) but b 6∈ f (C) for some b ∈ C. Writing f = pq with polynomials q and p without common zeros it follows that q(z) 6= bp(z) for all z ∈ C. Hence the polynomial q − bp has no zeros and must therefore be constant by the fundamental theorem of algebra, i.e., there is c ∈ C \ {0} such that q(z) − bp(z) = c for all z ∈ C. Then it follows for all w ∈ C with w 6= b that q(z) − wp(z) = (b − w)p(z) + c. Since f has poles, p is nonconstant. Hence there is z0 ∈ C such that p(z0 ) = and thus q(z0 ) − wp(z0 ) = 0. Therefore p(z0 ) 6= 0 and f (z0 ) =

c w−b

q(z0 ) q(z0 ) − wp(z0 ) = +w =w p(z0 ) p(z0 )

shows that all w ∈ C \ {b} belong to f (C).



Proposition A.1.7. Let f be a nonconstant rational function, let a ∈ C be a zero of f or a pole of f , and let ϕ1 , ϕ2 ∈ R with ϕ1 < ϕ2 . For the representation f (z) = (z − a)j h(z) with h(a) 6= 0 as in Remark A.1.3 let ψ0 = arg h(a). Let ψ1 ∈ R be such that jψ1 ∈ (ϕ1 − ψ0 , ϕ2 − ψ0 ). Then there is r0 > 0 such that for r ∈ (0, r0 ) the complex number w = a + reiψ1 is neither a zero nor a pole of f and satisfies f (w) = |f (w)|eiϕ0 with ϕ0 ∈ (ϕ1 , ϕ2 ). Proof. Recall that j 6= 0 in the representation of f where j > 0 if a is a zero of f and j < 0 if a is a pole of f . Let η0 = min{jψ1 − ϕ1 + ψ0 , ϕ2 − ψ0 − jψ1 , π6 } and put ρ = |h(a)| sin η0 . Im ρ h(a) η0 ψ0

Re

Due to the continuity of h at a there is r0 > 0 such that |h(z) − h(a)| < ρ and h(z) = (|h(a)| + δ(z))ei(ψ0 +η(z)) whenever 0 < |z − a| < r0 , where δ(z) ∈ (−ρ, ρ)

Appendix A. Rational Nevanlinna and Stieltjes Functions

270

and η(z) ∈ (−η0 , η0 ). For r ∈ (0, r0 ) and w = a + reiψ1 one has f (w) = rj eijψ1 (h(a) + δ(w))ei(ψ0 +η(w)) = |f (w)|eiϕ0 , where ϕ0 = jψ1 + ψ0 + η(w) ∈ (jψ1 + ψ0 − η0 , jψ1 + ψ0 + η0 ) ⊂ (ϕ1 , ϕ2 ).

A.2



Rational Nevanlinna functions

Definition A.2.1. A rational function f is called a Nevanlinna function if (i) f has no poles in C+ ∪ C− , (ii) f (z) = f (z) for Im z 6= 0, (iii) Im z · Im f (z) ≥ 0 for Im z 6= 0. Note that property (ii) is equivalent to f being a real rational function. In this monograph only rational functions are considered. We recall the general definition of a Nevanlinna function: f is called a Nevanlinna function if f is analytic on C+ and C− and if f has properties (ii) and (iii) of definition A.2.1. Nevanlinna functions are called R-functions in [74, §1]. Lemma A.2.2. Let p and q be real polynomials without common nonreal zeros. Consider the functions ω = p + iq,

f=

q , p

g=

ω . ω

1. The following statements are equivalent: (i) |g(z)| < 1 for all z ∈ C+ such that ω(z) 6= 0; (ii) ω has no zeros in the open lower half-plane, and |g(z)| < 1 for all z ∈ C+ ; (iii) p 6= 0 and Im f (z) > 0 for all z ∈ C+ with p(z) 6= 0; (iv) All zeros of p and q are real, f (C+ ) ⊂ C+ and f (C− ) ⊂ C− . 2. Assume that p 6= 0. Then the following statements are equivalent: (i) |g(z)| ≤ 1 for all z ∈ C+ such that ω(z) 6= 0; (ii) ω has no zeros in the open lower half-plane, and |g(z)| ≤ 1 for all z ∈ C+ ; (iii) Im f (z) ≥ 0 for all z ∈ C+ with p(z) 6= 0; (iv) All zeros of p are real, q = 0 or all zeros of q are real, f (C+ ) ⊂ C+ and f (C− ) ⊂ C− .

A.2. Rational Nevanlinna functions

271

Proof. 1. (i)⇒(ii): From    ω 1 = ω 1

i −i

  p q

and the assumption that p and q do not have common nonreal zeros it follows that also ω and ω have no common nonreal zeros. In particular, ω 6= 0. Since g is bounded in the open upper half-plane, g does not have a pole there, and hence ω has no zeros in the open upper half-plane, which means that ω has no zeros in the open lower half-plane. (ii)⇒(iii): Observing that p and q are real polynomials it follows for z ∈ C that ω(z) + ω(z) = p(z) + iq(z) + p(z) − iq(z) = 2p(z). Since |g(z)| < 1 for z ∈ C+ gives ω(z) 6= −ω(z), it follows that p(z) 6= 0 for z ∈ C+ . In particular, p 6= 0. Similarly, ω − ω = 2iq so that f=

1−g q ω−ω =i . =i ω+ω 1+g p

(A.2.1)

Since the right-hand side is a M¨obius transformation in g which maps the interior of the unit disk onto the upper half-plane, it is clear that |g(z)| < 1 implies Im f (z) > 0 for z ∈ C+ with p(z) 6= 0. (iii)⇒(iv): Since f (z) 6= 0 for all z ∈ C+ with p(z) 6= 0, q has no zeros in the open upper half-plane. Assume that p has a zero a in C+ . Then f has a pole (of order k) at a. By Proposition A.1.7 with ϕ1 = −π, ϕ2 = 0 and ψ0 as given there we can find 0 < r < Im a and ψ1 ∈ ((ψ0 − ϕ2 )k −1 , (ψ0 − ϕ1 )k −1 ) such that f (w) = |f (w)|eiϕ0 6= 0 with w = a + reiψ1 and ϕ0 ∈ (−π, 0). But then w ∈ C+ and f (ω) ∈ C− , which contradicts the assumption that Im f (w) > 0. Therefore p has no zeros in C+ . By assumption (iii) it follows that f (C+ ) ⊂ C+ . Since f is a real rational function we have f = f , which gives f (C− ) ⊂ C− . In particular, f has only real zeros or poles, if any, and therefore all zeros of p and q are real. (iv)⇒(i) is an immediate consequence of (A.2.1), see the proof of (ii)⇒(iii). 2. The difference to part 1 is that |g(z)| = 1 and Im f (z) = 0 is possible for some z ∈ C+ . In this case, f and, equivalently, g are constant. Indeed, assume that there is a ∈ C+ such that f (a) ∈ R but that f is not constant. Then f˜(z) =

1 f (a) − f (z)

defines a rational function f˜ which has a pole at a and which satisfies Im f˜(z) ≥ 0 for all z ∈ C+ which are not poles of f˜. As in the proof of part 1, (iii)⇒(iv), this leads to a contradiction. But if f is a constant function, then also p and q are constant functions whose values are real. In this case, (i)–(iv) are satisfied.  Remark A.2.3. Let f be a rational Nevanlinna function. Due to Definition A.2.1 (i), f = f on C \ R and therefore everywhere. Hence every rational Nevanlinna

Appendix A. Rational Nevanlinna and Stieltjes Functions

272

function is a real rational function and is thus characterized by property (iv) in part 2 of Lemma A.2.2. Indeed, in the proof of Lemma A.2.2 we have seen that a rational function as in part 2 is as in part 1 if and only if it is not constant. The equivalence of (iii) and (iv) in part 1 shows that a nonconstant rational function is a rational Nevanlinna function if and only if it is a real rational function and maps C+ into itself. Clearly, a constant function is a rational Nevanlinna function if and only if its value is real. Lemma A.2.4. If f = 6 0 is a rational Nevanlinna function, then so are the functions − f1 and ( f1 + c)−1 for each real constant c unless f = − 1c . Proof. For each w ∈ C \ {0} the identity   1 Im f (z) = Im − f (z) |f (z)|2

1 w

= 

and

Im

w |w|2

shows that

1 +c f (z)

−1

1 Im(− f (z) ) = 2 . 1 f (z) + c

Thus the statement immediately follows from Remark A.2.3.



Lemma A.2.5. Let f be a real rational function which satisfies f (C+ ) ⊂ C+ . Then f has at least one zero or pole, f (C+ ) = C+ , and all zeros and poles of f lie on the real axis, are simple and interlace, i.e., between two poles of f lies a zero of f and between two zeros of f lies a pole of f . For each z ∈ R which is not a pole of f , the inequality f 0 (z) > 0 holds. Proof. Since f (C+ ) ⊂ C+ , f has neither poles nor zeros in the open upper halfplane, and hence also not in the open lower half-plane since f is real. Therefore all zeros and poles of f are real. If f would have no zeros or poles, then f would be constant, see Remark A.1.3. Since f is real on the real axis, f (z) would be real for all z ∈ C, which contradicts f (C+ ) ⊂ C+ . Assume that f has a pole or zero a which is not simple. Then there are an integer j with |j| > 1 and a rational function h with h(a) 6= 0 such that f (z) = (z − a)j h(z). Since f is real on the real axis, also h is real on the real axis, so that h(a) = |h(a)|eiψ0 with ψ0 = 0 or ψ0 = π. If j > 1, put ψ1 = 1j ( 3π 2 − ψ0 ). Then π 3 3 0< ≤ ψ1 ≤ π ≤ π < π 2j 2j 4 and jψ1 = If j < −1, put ψ1 =

1 π |j| ( 2

3 π − ψ0 ∈ (π − ψ0 , 2π − ψ0 ). 2

+ ψ0 ). Then 0 < ψ1 ≤

3 3 π≤ π 0 such that w = a + reiψ1 gives a representation f (w) = |f (w)|eiϕ0 with ϕ0 ∈ (π, 2π) if j > 0 and ϕ0 ∈ (−π, 0) if j < 0. Thus we have arrived at the contradiction w ∈ C+ and f (w) ∈ C− . Therefore all zeros and poles of f must be simple. Now assume that the zeros and poles of f do not interlace. If there are two adjacent poles z1 < z2 of f without a zero of f between them, then f is either positive or negative on the interval (z1 , z2 ). Hence either f (z) tends to ∞ as z tends to z1 from the right and as z tends to z2 from the left, or f (z) tends to −∞ as z tends to z1 from the right and as z tends to z2 from the left. Therefore f must have a local extremum at some a ∈ R between these two poles. If there are two adjacent zeros of f without a pole of f between them, then f is continuous on the interval between these two zeros. Therefore f must have a local extremum at some a ∈ R between these two zeros. In either case we have a point a ∈ R with f 0 (a) = 0. Since f is real on the real axis, f (a) ∈ R, and therefore the function f1 defined by f1 (z) = f (z) − f (a) is a real rational function with f1 (a) = f10 (a) = 0 and with f1 (C+ ) ⊂ C+ . This would mean that the zero a of f1 were not simple. But by what we have already shown in the previous paragraph, zeros of f1 have to be simple. This contradiction shows that zeros and poles of f interlace. Assume there is w ∈ C+ \ f (C). From Lemma A.1.6 it would follow that {w} = C \ f (C); in particular, w = f (z) for some z ∈ C. Then w ∈ C− , f real and f (C+ ) ⊂ C+ would imply that z ∈ C− . Therefore z ∈ C+ and w = f (z) = f (z) would lead to the contradiction w ∈ f (C+ ). Hence f (C+ ) = C+ . Finally, when a ∈ R is not a pole of f , then also f − f (a) is a real rational function which maps C+ into itself. Hence, by what we have already proved, a is a simple zero of f − f (a). Therefore f (z) = f (a) + (z − a)h(z), where h is a real rational function and h(a) 6= 0. Since Im f (a + iε) > 0

for ε > 0,

it follows that Re h(a + iε) =

1 Im f (a + iε) > 0. ε

By continuity, f 0 (a) = h(a) = lim Re h(a + iε) ≥ 0, ε&0

and f 0 (a) > 0 follows since h(a) 6= 0.



Appendix A. Rational Nevanlinna and Stieltjes Functions

274

Theorem A.2.6. A real rational function f satisfies f (C+ ) ⊂ C+ if and only if f is represented in the form f (z) = C

m Y

(z − βk )

n Y k=1

k=1

1 , z − αk

(A.2.2)

where m, n ∈ N0 , m + n > 0, the sequences (α1 , . . . , αn ) and (β1 , . . . , βm ) are interlacing sequences of real numbers, C > 0 if f has at least one zero and the largest zero of f is larger than any pole of f , and C < 0 if f has at least one pole and the largest pole of f is larger than any zero of f . Proof. If f (C+ ) ⊂ C+ , then the representation (A.2.2) with C ∈ R \ {0} and the interlacing properties of its zeros and poles follows from (A.1.2) and Lemma A.2.5. Let a be the largest of the numbers α1 , . . . , αn , β1 , . . . βm . By Lemma A.2.5 we know that f 0 (x) > 0 for x ∈ (a, ∞), i.e., f is increasing on the interval (a, ∞). It follows that f is positive on (a, ∞) if a is a zero of f and that f is negative on (a, ∞) if a is a pole of f . Since all factors z − αk and z − βk in the product (A.2.2) are positive there, C must be positive if a is a zero of f and C must be negative if a is a pole of f . Assume that f is of the form (A.2.2) with m+n > 0 and interlacing sequences of zeros and poles. Clearly, f is a real rational function. Without loss of generality we may also assume that α1 < α2 < · · · < αn and β1 < β2 < · · · < βm . We will first consider the case that the largest zero of f , if any, is larger than the largest pole of f , if any. In case the number of poles equals the number of zeros, i.e., m = n, the interlacing property gives α1 < β1 < α2 < β2 < · · · < αn < βn . Denoting α = arg(z − αk ) and β = arg(z − βk ) for generic k, the sketch z β−α

π−β β βk

α αk−1

βk−1

αk

αk+1

βk+1

shows that for Im z > 0 the difference β − α is the angle subtended by the segment [αk , βk ] of the real axis from the point z. Since C > 0 in this case and since arg f (z) =

n X

[arg(z − βk ) − arg(z − αk )],

k=1

it follows that 0 < arg f (z) < π if Im z > 0, i.e., f (C+ ) ⊂ C+ .

A.3. Rational Stieltjes functions

275

If now the number of poles is different from the number of zeros, still in the case that βm is larger than any pole of f , the interlacing property gives m = n + 1. Observing that 0 < arg(z − x1 ) < arg(z − x2 ) < π for all z ∈ C+ and all real numbers x1 , x2 with x1 < x2 , it follows for z ∈ C+ that 0 < arg(z − β1 ) ≤ arg(z − β1 ) +

n X

[arg(z − βk+1 ) − arg(z − αk )]

k=1

= arg f (z) =

n X

[arg(z − βk ) − arg(z − αk )] + arg(z − βm )

k=1

≤ arg(z − βm ) < π. Therefore f (C+ ) ⊂ C+ also in this case. It remains to consider the case that the largest pole of f is larger than the largest zero of f . Then − f1 has zeros αk and poles βk interlacing as above, and the constant factor is − C1 > 0. Therefore − f1 maps the upper half-plane into itself by what we have already shown. In view of Remark A.2.3 and Lemma A.2.4, also f maps the upper half-plane into itself.  Lemma A.2.7. Let n ∈ N and let f be a rational Nevanlinna function with n poles. Then there are real numbers A ≥ 0, B, α1 < · · · < αn , Cj > 0 (j = 1, . . . , n) such that n X Cj f (z) = Az + B − . z − αj j=1 Proof. Referring to Remark A.2.3 and Lemma A.2.5 we know that f has n−1, n, or n + 1 real zeros and n real simple poles. Hence the partial fraction decomposition is clear with A, B ∈ R and Cj ∈ R \ {0}, j = 1, . . . , n. Since f (z) increases as z → ∞ on the real axis, A ≥ 0 follows, and similarly, since f increases on the real axis to the left and the right of αj , it follows that Cj > 0 for j = 1, . . . , n. 

A.3

Rational Stieltjes functions

In this section we are going to introduce two classes of Nevanlinna functions and we will show that these functions can be characterized by particular continued fraction expansions. Definition A.3.1. A rational function f is called a rational Stieltjes function or S-function if it is a rational Nevanlinna function and satisfies in addition that (i) f has no poles in (−∞, 0), (ii) f (z) > 0 for z ∈ (−∞, 0).

276

Appendix A. Rational Nevanlinna and Stieltjes Functions

A rational S-function f is called a rational S0 -function if (iii) 0 is not a pole of f . We also say that an S-function (S0 -function) f belongs to the class S (S0 ) and we write f ∈ S (f ∈ S0 ). Proposition A.3.2. A nonconstant rational Nevanlinna function is a rational Sfunction if and only if it has at least one pole, and the smallest pole is nonnegative and smaller that all zeros, if any. Proof. Let f be a nonconstant rational S-function. Then f has at least one zero or pole, and let a be the smallest of the zeros and poles. By assumption and Lemma A.2.5, f (z) > 0 for z ∈ (−∞, 0) and f 0 (z) > 0 for z ∈ (−∞, a), which shows that a ≥ 0 and that a is a pole. On the other hand, if f is a rational Nevanlinna function with at least one pole and such that the smallest pole is nonnegative and smaller than all zeros, then f 0 (z) > 0 and f (z) 6= 0 for all z ∈ (−∞, 0) implies f (z) > 0 for all z ∈ (−∞, 0).  Corollary A.3.3. A nonconstant real rational function f is a rational S-function if and only if f is represented in the form f (z) = C0

m Y

(βk − z)

k=1

n Y k=1

1 , αk − z

(A.3.1)

where n ∈ N, m ∈ N0 , the increasing sequences (α1 , . . . , αn ) and (β1 , . . . , βm ) are interlacing sequences of real numbers, α1 ≥ 0, α1 < β1 if m > 0, and C0 > 0. Proof. This follows immediately from Proposition A.3.2 and Theorem A.2.6 with C0 = (−1)m+n C if we observe that m = n or m = n − 1 and that m = n if and only if C > 0.  Lemma A.3.4. Let r ∈ N and let f be a rational S-function with r poles. Then: 1. There are unique numbers a, b and a unique rational function g which is either identically zero or has at most r − 1 zeros such that f (z) = a +

1 ; −bz + g(z)

(A.3.2)

2. a = limz→−∞ f (z) ≥ 0, b > 0, and the number of zeros of f is r if a > 0 and r − 1 if a = 0. 3. The poles αk and zeros βk of f are all simple and interlace, 0 ≤ α1 < β1 < α2 < · · · < βr−1 < αr < βr 0 ≤ α1 < β1 < α2 < · · · < βr−1 < αr

if a > 0, if a = 0.

(A.3.3) (A.3.4)

A.3. Rational Stieltjes functions

277

4. The rational function f is strictly increasing between its poles, i.e., in the intervals (−∞, α1 ), (αk , αk+1 ), k = 1, . . . , r − 1, and (αr , ∞). 5. If r = 1, then g is a nonnegative constant function, and g = 0 if and only if 0 is a pole of f . 6. If r > 1, then g 6= 0 and h = g1 is a rational S-function with r − 1 poles and r − 1 zeros βh,k , k = 1, . . . , r − 1 which are simple and interlace, 0 is a pole of h if and only if 0 is a pole of f , and for k = 1, . . . , r − 1 if a > 0,

βk < βh,k < βk+1 βk = βh,k

for k = 1, . . . , r − 1 if a = 0.

Proof. By Lemma A.2.5 and Remark A.2.3 we know that the zeros and the poles of f are simple and interlace, and together with Proposition A.3.2 this shows that f has r or r − 1 zeros. Hence f can be written as f (z) =

q(z) p(z)

with p(z) = z r +

r−1 X

cj z j ,

j=0

q(z) =

r X

dj z j ,

j=0

where p and q have no common zeros, their coefficients are real, and dr 6= 0 or dr−1 6= 0. Below we will show that a = dr , which would complete the proof of part 3, again taking Proposition A.3.2 into account. Since the function f is not constant, (A.3.2) is equivalent to 1 f (z) − a bz(q(z) − ap(z)) + p(z) = q(z) − ap(z) pg (z) = qg (z)

g(z) = bz +

with qg (z) = q(z) − ap(z),

pg (z) = bzqg (z) + p(z).

Clearly, qg is not identically zero since f is not constant. We observe that qg and pg do not have common zeros because qg (z) = pg (z) = 0 would lead to the contradiction p(z) = 0 and q(z) = ap(z) = 0.

Appendix A. Rational Nevanlinna and Stieltjes Functions

278

In order to prove statement 1 it is therefore necessary and sufficient to find unique a and b such that the degree of pg does not exceed r − 1. This means that the coefficients of z r+1 and z r of the Taylor expansion of pg about 0 must be zero, that is, we must satisfy b(dr − a) = 0

and

b(dr−1 − acr−1 ) + 1 = 0.

(A.3.5)

The second equation in (A.3.5) gives b 6= 0, and then the first equation has the unique solution a = dr . It is clear that a = dr = limz→−∞ f (z) ≥ 0. To solve for b in the second equation we first consider the case dr = 0. Then dr−1 6= 0 and f (z) > 0 for z ∈ (−∞, 0) gives dr−1 < 0. Hence we have a unique b = −d−1 r−1 , and b > 0 is satisfied. Now let dr 6= 0. Then cr−1 = −

r X

αj ,

j=0

r X dr−1 =− βj , dr j=0

and (A.3.3) gives dr−1 < cr−1 . dr Hence the second equation in (A.3.5) has a unique solution b, and dr = a > 0 shows that 1  > 0. b=  dr−1 a cr−1 − dr This completes the proof of the statements 1 to 4. Observe that the above proof shows that both qg and pg are polynomials of degree not exceeding r − 1 and that qg is not identically zero since f is not constant. Furthermore, pg (0) = 0 if and only if p(0) = 0. If 0 is not a pole of f , then the fact that f increases on (−∞, 0] gives f (0) > limx&−∞ f (x) = a and therefore 1 g(0) = > 0. (A.3.6) f (0) − a In case r = 1 both qg and pg are constant. Hence g is constant. Above we have seen that this constant is zero if 0 is a pole of f and positive otherwise, see (A.3.6). This proves statement 5. Now we turn our attention to statement 6 and we henceforth assume that r > 1. From qg (αk ) = q(αk ) and the interlacing property of the zeros and poles of f it follows that qg has opposite signs at consecutive αk . Hence qg has an odd number of zeros, counted with multiplicity, in each of the intervals (αk , αk+1 ), k = 1, . . . , r − 1. We already know that the degree of qg does not exceed r − 1, and therefore qg has degree r − 1 and all its zeros are real and simple, so that they can be indexed as βh,k , k = 1, . . . , r − 1 and satisfy αk < βh,k < αk+1 ,

k = 1, . . . , r − 1.

(A.3.7)

A.3. Rational Stieltjes functions

279

If a = 0 then qg = q, and hence βh,k = βk for j = 1, . . . , r − 1. If a > 0, then qg (βk ) = −ap(βk ), so that qg has opposite signs at consecutive βk , which means that qg has an odd number of zeros, counted with multiplicity, in each of the intervals (βk , βk+1 ), k = 1, . . . , r − 1. Hence it follows that βk < βh,k < βk+1 for k = 1, . . . , r − 1. From pg (αk ) = bαk qg (αk ) it follows that the sign of pg (αk ) alternates with k (except for αk = 0). In any case, it follows that pg has r − 1 simple zeros. For k = 1, . . . , r − 1, pg (βh,k ) = p(βh,k ). Since the zeros of p are simple, p has opposite signs on adjacent intervals (αk−1 , αk ), (αk , αk+1 ). In view of (A.3.7), this means that pg (βh,k ) has opposite signs for consecutive k. Hence between two zeros of qg there is an odd number of zeros of pg . Since qg and pg have the same number of zeros, counted with multiplicity, it follows that the zeros of qg and pg interlace. If α1 = 0, then pg (0) = p(0) = 0 and 0 = α1 < βh,1 shows that the smallest zero of pg is smaller than the smallest zero of qg . If α1 > 0, then 0 < α1 < βh,1 < α2 shows that pg (0) = p(0) and pg (βh,1 ) = p(βh,1 ) have opposite signs, that is, pg has a positive zero which is smaller than the smallest zero βh,1 of qg . Hence it follows in any case that the smallest zero of pg is nonnegative and is smaller than the smallest zero of qg . Altogether, we have shown so far that h = g1 has a representation (A.3.1) with all properties as stated in Corollary A.3.3 satisfied except possibly C0 > 0. That is, h or −h is a rational S-function. Therefore consider g 0 (z) = b −

f 0 (z) . (f (z) − a)2

The pole βh,1 of g is a zero of z 7→ f (z)−a, and therefore limx→βh,1 (f (x)−a)2 = 0, whereas f 0 (βh,1 ) > 0 in view of Lemma A.2.5. Hence limx→βh,1 g 0 (x) = −∞, which gives that h0 (x) is positive for real x near βh,1 . In view of Lemma A.2.5, −h cannot be a rational Nevanlinna function, and therefore h must be a rational S-function.  Lemma A.3.5. Let n ∈ N and let f be a rational S-function with n poles. Then: 1. f admits a unique continued fraction expansion 1

f (z) = an +

(A.3.8)

1

−bn z + an−1 +

1 −bn−1 z + . .

1

.+ a1 +

1 −b1 z + c0

with an = limz→−∞ f (z) ≥ 0, ak > 0 for k = 1, . . . , n − 1, bk > 0 for k = 1, . . . , n, c0 ≥ 0 with c0 = 0 if and only if 0 is a pole of f . 2. The number of zeros of f is n if an > 0 and n − 1 if an = 0.

Appendix A. Rational Nevanlinna and Stieltjes Functions

280

3. The poles αk and zeros βk of f are all simple and interlace, 0 ≤ α1 < β1 < α2 < · · · < βn−1 < αn < βn 0 ≤ α1 < β1 < α2 < · · · < βn−1 < αn

if an > 0,

if an = 0.

(A.3.9) (A.3.10)

4. The rational function f is strictly increasing between its poles, i.e., in the intervals (−∞, α1 ), (αk , αk+1 ), k = 1, . . . , n − 1, and (αn , ∞). 5. If fj (z) denotes the n − j-th tail of the continued fraction expansion (A.3.8) for j = 1, . . . , n, i.e., 1

fj (z) = aj +

(A.3.11)

1

−bj z + aj−1 +

1 −bj−1 z + . .

1

.+ a1 +

1 −b1 z + c0

(j)

(j)

then all zeros of fj are real and simple, written β1 < · · · < βmj with mj = j if j < n or j = n and an > 0, and mn = n − 1 if an = 0, and the following holds: (j)

(j−1)

βk < βk (n)

βk

(n−1)

= βk

(j)

< βk+1

for j = 2, . . . , mn , k = 1, . . . , j − 1,

for k = 1, . . . , n − 1 if an = 0; (n)

here we observe that βk = βk . Proof. Parts 2, 3, and 4 follow from Lemma A.3.4 with r = n and an = a. We will prove parts 1 and 5 by induction on n. For n = 1, Lemma A.3.4 with r = 1, a1 = a, b1 = b and c0 = g shows that indeed properties 1 and 5 are satisfied with f1 = f . Now let n > 1. From Lemma A.3.4 with r = n we know that there are an = a ≥ 0, bn = b > 0 and a rational S-function h with n − 1 zeros and n − 1 poles such that 1 f (z) = an + . 1 −bn z + h(z) By induction hypothesis, h has the representation (A.3.8) with n replaced by n − 1 and an−1 > 0. Therefore h = fn−1 as in (A.3.11). This proves part 1 for n. The remaining statements of part 5 follow by induction hypothesis for j < n and from Lemma A.3.4 for j = n.

A.3. Rational Stieltjes functions

281

We still have to prove the uniqueness of the representation (A.3.8). To this end observe that for any such representation with (complex) numbers aj , bj and c0 and corresponding representations (A.3.11) we have 1

fj (z) = aj +

−bj z +

(A.3.12)

1 fj−1 (z)

q

for j = 2, . . . , n. We can write the rational functions fj in the form pjj with polynomials pj and qj without common zeros. Then, recursively, for j = 2, . . . , n, fj (z) = aj +

1 qj−1 (z) − aj bj zqj−1 (z) + aj pj−1 (z) = . −bj zqj−1 (z) + pj−1 (z) pj−1 (z) −bj z + qj−1 (z)

Observing that f1 has one pole and at most one zero, it follows by inductions that fj has at most j poles and zeros. In particular, for n > 1, fn = f has n poles and fn−1 has at most n − 1 poles. In view of Lemma A.3.4 the numbers an and bn as well as the rational function fn−1 are unique, and fn−1 is a rational S-function. By induction it follows that also the numbers an−1 , . . . , a2 , bn−1 , . . . , b2 as well as the rational S-function f1 are unique. Another application of Lemma A.3.4 finally shows that also a1 , b1 and c0 are unique.  Lemma A.3.6. Let n ∈ N and let f be given by (A.3.8) with an ≥ 0, ak > 0 for k = 1, . . . , n − 1, bk > 0 for k = 1, . . . , n and c0 ≥ 0. Then f is a rational S-function with n poles. Proof. For j = 1, . . . , n define fj by (A.3.11). We are going to show by induction that fj is a rational S-function. We have that f1 (z) = a1 + c0 b1

has a pole at z0 =

1 −b1 z + c0

≥ 0. Then  1 1   b z − z 1 0 f1 (z) = z + 1 −z   a1 0 a1 b1 z0 − z

if a1 = 0, if a1 > 0,

shows by Corollary A.3.3 that f1 is a rational S-function. Now let n > 1, 2 ≤ j ≤ n and assume that fj−1 is a rational S-function. Then fj has the representation (A.3.12). Lemma A.2.4 and the fact that a sum of Nevanlinna functions is a Nevanlinna function show that, in turn, with fj−1 also −

1 fj−1

,

z 7→ bj z −

1 fj−1 (z)

,

z 7→

1 −bj z +

1 fj−1 (z)

,

fj

282

Appendix A. Rational Nevanlinna and Stieltjes Functions

are rational Nevanlinna functions, and fj−1 (z) > 0 for z < 0 implies that also fj (z) > 0 for z < 0. If k is the number of poles of f , then n = k by Lemma A.3.5.  Definition A.3.7. A rational function f is called a positive real function if (i) f (z) is real for real z, (ii) Re f (z) > 0 for Re z > 0. Lemma A.3.8. If all zeros and poles of a nonconstant positive real function f lie on the imaginary axis, then the function h defined by h(z) = if (−iz) is a rational Nevanlinna function and there is a rational S-function g such that f (z) = zg(−z 2 ). In this case, g belongs to S0 if and only if 0 is not a pole of f . Proof. By condition (i) of Definition A.3.7, f is a real rational function and hence f = f by Proposition A.1.5. In particular, zeros and poles of f are symmetric with respect to the real axis. Since zeros and poles of f lie on the imaginary axis, there are nonnegative integers m and n, an integer j, positive real numbers α1 , . . . , αn and β1 , . . . , βm , and a nonzero complex number C such that f (z) = Cz j

m Y

z 2 + βk

k=1

n
Y

z 2 + αk


−1

.

(A.3.13)

k=1

Without loss of generality we assume that αk ≤ αk+1 and βk ≤ βk+1 . Observe that at least one of m, n, j is nonzero since f is nonconstant and that C > 0 since f (z) > 0 for z > 0. Assume that |j| > 1. Then f has the representation f (z) = z j f1 (z) where π π f1 is a rational function with f1 (0) > 0. Let ψ1 = 3π 4j ∈ (− 2 , 2 ) and observe that
3π π π jψ1 = 4 ∈ 2 , π . With ψ0 = 0, ϕ1 = 2 and ϕ2 = π it follows from Proposition iψ1 A.1.7 that there we have f (w) = |f (w)|eiϕ0 6= 0
is r > 0 such that for w = re π and ϕ0 ∈ 2 , π . This would lead to the contradiction Re w > 0 and Re f (w) < 0. If j = 0, then the representation (A.3.13) would give f (z) = f (−z) for all z which are not poles of f . When Re z < 0 then Re(−z) = − Re z = − Re z > 0, and hence condition (ii) of Definition A.3.7 would give Re f (z) = Re f (−z) > 0. Hence Re f (z) ≥ 0 for all z ∈ C which are not poles of f . But this is impossible by Lemma A.1.6. It follows from the previous two paragraphs that j = 1 or j = −1. In both cases, f maps the imaginary axis to itself, and h is therefore a rational function which maps the upper half-plane into itself and is real on the real axis. By Remark A.2.3 and Lemma A.2.5, h is a Nevanlinna function. In particular, all the numbers αk , k = 1, . . . , n, and βk , k = 1, . . . , m, are different and interlace. If j = −1, then 0 is a pole of f , and n ≤ m as well as β1 < α1 if n > 0 follows. If j = 1, then 0 is a zero of f , and m ≤ n as well as α1 < β1 if m > 0 follows. Define m n Y j−1 Y 1 g(z) = C(−z) 2 (βk − z) . αk − z k=1

k=1

A.4. The classes S r

283

Then poles and zeros of g interlace and the smallest pole is smaller than the smallest zero if g has any zeros. Hence g is a rational S-function by Corollary A.3.3. The identity f (z) = zg(−z 2 ) is clear by definition of g, and 0 is a pole of g if and only if j = −1. 

A.4

The classes S r

Definition A.4.1. Let r ∈ N. A pair (q, p) of real polynomials is said to belong to S r if q(z) 6= 0 and p(z) 6= 0 for all z ∈ C \ [0, ∞) and if there exist nonzero real polynomials p1 , . . . , pr such that q p1 pr−1 pr ∈ S, ∈ S, . . . , ∈ S, ∈ S. p1 p2 pr p A pair (q, p) of real polynomials is said to belong to S 0 if q(z) 6= 0 and p(z) 6= 0 for all z ∈ C \ [0, ∞) and if pq ∈ S. Remark A.4.2. 1. Let r, s ∈ N with r < s and let (q, p) ∈ S r with p, q, q1 , . . . , qr as in Definition A.4.1. Then putting ps = ps−1 = · · · = pr+1 = pr and observing that the constant function 1 belongs to S it is clear that (q, p) ∈ S s . Thus we have seen that S r ⊂ S s for s > r. 2. Starting with either p or q and observing that the zeros and poles of rational S-functions are nonnegative real numbers, it follows by induction that the zeros of the polynomials p1 , . . . , pr in Definition A.4.1 are also nonnegative real numbers. Lemma A.4.3. Let X = (xj )nj=1 and Y = (yj )nj=1 be two finite sequences of real numbers with xj ≤ xj+1 and yj ≤ yj+1 (j = 1, . . . , n − 1). For d ∈ R let N (X, d) = #{j ∈ {1, . . . , n} : xj ≤ d}. Then yj ≤ xj for j = 1, . . . , n if and only if N (X, d) ≤ N (Y, d) for all d ∈ R. Proof. If yj ≤ xj for j = 1, . . . , n, and if d ∈ R, then xj ≤ d implies yj ≤ d, and therefore N (X, d) ≤ N (Y, d). Conversely, if there is j0 ∈ {1, . . . , n} such that xj0 < yj0 , then let d = xj0 . In this case, xj ≤ xj0 = d for j = 1, . . . , j0 , so that N (X, d) ≥ j0 . But d = xj0 < yj0 ≤ yj for j = j0 , . . . , n shows that N (Y, d) < j0 , and therefore N (Y, d) < N (X, d).  (0)

(0)

(0)

(0)

Lemma A.4.4. Let m, k ∈ N, let X (0) = (x1 , . . . , xm ) and Y (0) = (y1 , . . . , ym ) be sequences of real numbers such that (0)

(0)

(0)

(0)

(0) y1 ≤ x1 ≤ y2 ≤ x2 ≤ · · · ≤ ym ≤ x(0) m .

Let C = (c1 , . . . , ck ) be a sequence of real numbers and let X = (x1 , . . . , xm+k ) and (0) (0) Y = (y1 , . . . , ym+k ) be the sequences consisting of the terms x1 , . . . , xm , c1 , . . . , ck

Appendix A. Rational Nevanlinna and Stieltjes Functions

284 (0)

(0)

and y1 , . . . , ym , c1 , . . . , ck , respectively, ordered in such a way that xj ≤ xj+1 and yj ≤ yj+1 for j = 1, . . . , k + m − 1, respectively. Then y1 ≤ x1 ≤ y2 ≤ x2 ≤ · · · ≤ ym+k ≤ xm+k . Proof. Note that the statement of the lemma is equivalent to yj ≤ xj for j = 1, . . . , m + k and xj ≤ yj+1 for j = 1, . . . , m + k − 1. Then xj ≤ xj+1 and yj ≤ yj+1 for j = 1, . . . , k+m−1 is satisfied as well so that these inequalities do not need to be shown separately. With the notation of Lemma A.4.3 we observe that N (X, d) = N (X (0) , d) + N (C, d) and N (Y, d) = N (Y (0) , d) + N (C, d) for all d ∈ R. By Lemma A.4.3, N (X (0) , d) ≤ N (Y (0) , d), which then implies N (X, d) ≤ N (Y, d) for all d ∈ R. Again by Lemma A.4.3, this shows yj ≤ xj for j = 1, . . . , m + k. (0) (0) (0) (0) Choose numbers x0 < min{y1 , c1 , . . . , ck } and ym+1 > max{xm , c1 , . . . , ck } and put X

(0)

(0)

= (xj )m j=0 and Y (0)

(0)

(0)

(0)

= (yj )m+1 j=1 . Since xj

(0)

≤ yj+1 for j = 0, . . . , m,

(0)

Lemma A.4.3 gives N (Y , d) ≤ N (X , d) for all d ∈ R, and N (Y , d) ≤ N (X, d) follows. Then Lemma A.4.3 shows that xj ≤ yj+1 for j = 1, . . . , k + m − 1.  Theorem A.4.5. Let r ∈ N and let (q, p) ∈ S r where the polynomials p and q have the same degree n. Then the nondecreasing sequence of zeros (ak )nk=1 of q and the nondecreasing sequence of zeros (bk )nk=1 of p have the following properties: 1. For s = 1, . . . , n the interval (−∞, bs ) contains at most s−1 terms of (ak )nk=1 . 2. For s = r + 2, . . . , n the interval (−∞, bs ] contains at least s − 1 − r terms of (ak )nk=1 . Proof. Observe that it is allowed that two adjacent polynomials in the sequence (q, p1 , . . . , pr , p) have common zeros. Since quotients of adjacent polynomials are rational S-functions, the degree of the polynomials in the above sequence is nondecreasing, and it follows that all polynomials have the same degree n. Hence the ze(j) ros of pj (j = 1, . . . , r) can be written as a nondecreasing sequence (αk )nk=1 . Now (1) (r) consider any consecutive pair of the sequences (ak )nk=1 , (αk )nk=1 , . . . , (αk )nk=1 , n (bk )k=1 . After removal of common terms, the remaining two sequences are sequences of the zeros and poles of a rational S-function, which interlace like (A.3.3). Therefore, making use of Lemma A.4.4, statement 1 is obvious due to bs ≤ αs(r) ≤ · · · ≤ αs(1) ≤ as . Now let s ≥ r + 2 and j ∈ {1, . . . , s − 1 − r}. Observing j + r + 1 ≤ s it follows that (1) (r) aj ≤ αj+1 ≤ · · · ≤ αj+r ≤ bj+r+1 ≤ bs , which proves statement 2.



Lemma A.4.6. Let (q, p) ∈ S 0 , let l, m > 0, and define the polynomials f and g by      f (z) 1 l q(z) = . g(z) −mz 1 − lmz p(z)

A.5. Hermite-Biehler polynomials

285

Then (f, g) ∈ S 0 and f (z) = g(z)

If additionally

q p

∈ S0 , then also

f g

1 1 −mz + q(z) l+ p(z)

.

(A.4.1)

∈ S0 .

Proof. Clearly, f and g are real polynomials, and f (z) q(z) + lp(z) = g(z) −mzq(z) + (1 − lmz)p(z) 1 . = 1 −mz + q(z) l+ p(z) If pq is constant, then fg ∈ S by Lemma A.3.6. Otherwise, if pq is not constant, then, by Lemma A.3.5, pq has a representation of the form (A.3.8), and it follows immediately that also fg has such a representation. Hence fg ∈ S by Lemma A.3.6. To complete the proof that (f, g) ∈ S 0 let z ∈ C \ [0, ∞). We have to show that f (z) 6= 0 and g(z) 6= 0. This is obvious if pq is constant. Otherwise, if pq is not constant, assume that f (z) = 0. Then

q(z) p(z)

= −l, which contradicts

q p

∈ S. Hence

f g

f (z) 6= 0. Since ∈ S, this function has no pole at z, and hence also g(z) 6= 0. If pq ∈ S0 , then we may assume that p(0) 6= 0, and hence g(0) = p(0) 6= 0 proves that

A.5

f g

∈ S0 .



Hermite-Biehler polynomials

1 (ω −ω) it is immediately Let ω be a polynomial. Defining P = 21 (ω +ω) and Q = 2i clear that P and Q are real polynomials and that the identity ω = P + iQ holds. We call P the real part of ω and Q the imaginary part of ω.

Definition A.5.1. A nonconstant polynomial is said to be Hermite-Biehler (HB) if it has no zeros in the closed lower half-plane. Theorem A.5.2. Let ω be a polynomial with real and imaginary parts P and Q, i.e., real polynomials P and Q such that ω = P + iQ. Then ω is a Hermite-Biehler polynomial if and only if the following two properties hold: (i) the zeros of the polynomials P and Q are real, simple and interlace, i.e., between any two zeros of one of these polynomials there lies one zero of the other;

Appendix A. Rational Nevanlinna and Stieltjes Functions

286 (ii) for some λ0 ∈ R,

Q0 (λ0 )P (λ0 ) − Q(λ0 )P 0 (λ0 ) > 0. Proof. Let ω be a Hermite-Biehler polynomial. Let n ∈ N be its degree and let aj (j = 1, . . . , n) be its zeros. Since ω is a Hermite-Biehler polynomial, Im aj > 0 (j = 1, . . . , n). There is a nonzero complex number a such that ω(λ) = a

n Y

(λ − aj )

(λ ∈ C).

j=1

For Im λ > 0 we calculate |λ − aj |2 = (Re λ − Re aj )2 + (Im λ − Im aj )2 < (Re λ − Re aj )2 + (Im λ + Im aj )2 = |λ − aj |2 . We therefore conclude that |ω(λ)| = |a|

n Y j=1

|λ − aj | < |a|

n Y

|λ − aj | = |ω(λ)| (Im λ > 0).

j=1

Hence condition (ii) in part 1 of Lemma A.2.2 holds. Therefore, condition (iv) P in part 1 of Lemma A.2.2 holds, which shows that f = Q maps C+ into itself. Furthermore, since ω does not have real zeros, P and Q do not have common real zeros. Then the properties (i) and (ii) immediately follow from Lemma A.2.5. P Conversely, let properties (i) and (ii) be satisfied. By Theorem A.2.6, f = Q + or −f maps C into itself. In view of (ii) and Lemma A.2.5, this is not true for −f . Therefore f (C+ ) ⊂ C+ . Hence f satisfies property (iv) in part 1 of Lemma A.2.2, and property (ii) in part 1 of Lemma A.2.2 shows that ω has no zeros in the open lower half-plane. But since P and Q are real on the real axis, a real zero of ω would be a common real zero of P and Q, which is impossible since property (i) holds.  Definition A.5.3. A polynomial ω is said to be symmetric if ω(−λ) = ω(λ) for all λ ∈ C. The polynomial ω is said to belong to the class SHB if it is symmetric and belongs to the Hermite-Biehler class. For a symmetric polynomial ω, we have ω(−λ) = ω(λ) = ω(λ) (λ ∈ C). Hence, with P and Q the real and imaginary parts of ω and λ ∈ C it follows that P (−λ) = and Q(−λ) =

1 1 (ω(−λ) + ω(−λ)) = (ω(λ) + ω(λ)) = P (λ) 2 2

1 1 (ω(−λ) − ω(−λ)) = (ω(λ) − ω(λ)) = −Q(λ). 2i 2i

A.5. Hermite-Biehler polynomials

287

This means that P is an even real polynomial and Q is an odd real polynomial. e such that P (λ) = Pe(λ2 ) and Therefore there exist real polynomials Pe and Q 2 e Q(λ) = λQ(λ ) (λ ∈ C). Hence we have e 2) ω(λ) = Pe(λ2 ) + iλQ(λ

(λ ∈ C).

(A.5.1)

Proposition A.5.4. Let n ∈ N and let ω be a polynomial of degree n. Let J = {−k, . . . , k} if n = 2k + 1 and J = {−k, . . . , −1, 1, . . . , k} if k = 2n. Then ω is of SHB class if and only if there are C ∈ R \ {0} and λj ∈ C (j ∈ J) such that (i) Im λj > 0 for j ∈ J; (ii) λ−j = −λj for not pure imaginary λ−j ;  Y λ (iii) ω(λ) = C (λ ∈ C). 1− λj j∈J

If ω is of SHB class, then, in the representation (A.5.1), Pe(0) = C and e C −1 Q(0) =

X i > 0. λj

(A.5.2)

j∈J

Proof. Under the conditions (i), (ii) and C ∈ R\{0}, the polynomial defined by (iii) belongs to the SHB class. Conversely, the zeros of a Hermite-Biehler polynomial ω satisfy (i) by definition, and the symmetry of ω implies that its zeros, counted with multiplicity, are symmetric with respect to the imaginary axis. Hence they can be indexed in such a way that (ii) holds. Finally, in the representation (iii) of ω we have C = ω(0) ∈ R \ {0} due to (i) and the symmetry of ω. Now let ω be of SHB class. The representations (A.5.1) and (iii) give Pe(0) = ω(0) = C

e and Q(0) = −iω 0 (0) = C

X i . λj

(A.5.3)

j∈J

e is a real polynomial, it follows that Q(0) e Since Q is real, and in view of (i), Re

i iλj Im λj = Re = > 0, λj |λj |2 |λj |2

e which gives C −1 Q(0) > 0.

(j ∈ J), 

Theorem A.5.5. Let ω be a polynomial of SHB class with the representation e (A.5.1). Then Q e is a rational S0 -function. P Proof. By Theorem A.5.2, the zeros of P and Q are real, simple, and interlace. From e 2) λQ(λ Q(λ) = (A.5.4) P (λ) Pe(λ2 )

288

Appendix A. Rational Nevanlinna and Stieltjes Functions

e are real, positive, simple, and interlace, we conclude that the zeros of Pe and Q e if any. Since and that the smallest zero of Pe is smaller than the smallest zero of Q, P and Q satisfy property (ii) of Theorem A.5.2, it follows that  0 e Q(0) Q = (0) > 0. P Pe(0) P By Theorem A.2.6, either Q e or Q e is a Nevanlinna function. But since Nevanlinna P functions are locally increasing by Lemma A.2.5, this Nevanlinna function cannot e have a zero between 0 and its smallest pole, and therefore indeed Q e is a Nevanlinna P function, which is also positive on (−∞, 0) since it has no zeros there. Altogether, e  it follows that Q e is a rational S0 function. P e

e

Proposition A.5.6. Let ω be a monic symmetric polynomial of degree 2n (n ∈ N) e in the representation (A.5.1) of ω is a positive constant. If with ω(0) 6= 0, where Q ω(0) > 0, then ω has n zeros in the open upper half-plane and n zeros in the open lower half-plane, counted with multiplicity. If ω(0) < 0, then ω has n − 1 zeros in the open upper half-plane and n + 1 zeros in the open lower half-plane, counted with multiplicity. Proof. Because ω is symmetric, it is evident that ω(0) is a nonzero real number. It is also clear from the assumptions and the representation (A.5.1) that ω(λ) 6∈ R whenever λ ∈ R \ {0}. For η ∈ R let ωη (λ) = λ2n + ω(0) + iηλ. We observe that also ωη (λ) 6∈ R whenever η 6= 0 and λ ∈ R \ {0}. First assume that ω(0) < 0. Then ω0 has two distinct simple real zeros. Let λ0 be such a zero. Then ω(0) = −λ2n 0 , and for sufficiently small η > 0, say η ∈ (0, η0 ], ωη has a zero λη ∈ C \ R which depends continuously on η ∈ [0, η0 ]. We may write λη = ρ(η)eiψ(η) where ρ and ψ are real functions depending continuously on η with ρ(0) = λ0 and ψ(0) = 0. We may also assume that η0 is chosen so small that π 0 < |ψ(η)| < 2n for η ∈ (0, η0 ). Then 2n 0 = ωη (λη ) = λ2n η − λ0 + iηλη ,

and taking imaginary parts gives 2n 0 = Im λ2n η + η Re λη = ρ (η) sin(2nψ(η)) + ηρ(η) cos ψ(η).

Therefore Im λη = ρ(η) sin ψ(η) = −ρ2n (η) 0, then ω0 has n zeros in the open upper half-plane and n zeros in the open lower half-plane, and the same reasoning as above shows that ωQe has n zeros in the open upper half-plane and n zeros in the open lower half-plane when ω(0) > 0. Finally, observing that ω − ωQe is real on the real axis and has degree at most 2n − 2, it follows that the zeros of ωQe + t(ω − ωQe ) depend continuously on t ∈ [0, 1] and do not cross the real axis. Therefore the number of zeros in the open upper  (lower) half-plane of ω and ωQe are equal. Definition A.5.7. A nonconstant polynomial is said to belong to the generalized Hermite-Biehler class (HB) if it has no zeros in the open lower half-plane. The class of all symmetric generalized Hermite-Biehler polynomials is denoted by SHB. Proposition A.5.8. Let ω be be nonconstant symmetric polynomial. If the polynoe in the representation (A.5.1) have no common zeros in C \ [0, ∞) mials Pe and Q e Q e have and if Pe is an S-function, then ω is of SHB class. If additionally Pe and Q no common zeros and

e Q e P

is an S0 -function, then ω is of SHB class.

Proof. It is convenient to write the formula (A.3.1) for the S-function form m n Y Y e Q(z) 1 = (−1)m+n C0 (z − b2k ) z − a2k Pe(z) k=1 k=1

e Q e P

in the

with 0 ≤ a1 < b1 < a2 < . . . and C0 > 0. Then the representation (A.5.4) gives m

n

k=1

k=1

Y Y Q(λ) 1 = (−1)m+n C0 λ (λ − bk )(λ + bk ) . P (λ) (λ − ak )(λ + ak ) Clearly, the zeros and poles of Q P are simple and interlace. If the largest pole is larger than the largest zero, then m = n − 1, and in this case (−1)m+n C0 < 0. If the largest zero is larger than the largest pole, then m = n, and in this case (−1)m+n C0 > 0. Hence Remark A.2.3 and Theorem A.2.6 shows that Q P is a e e Nevanlinna function. The assumption that common zeros of P and Q can only be nonnegative real numbers implies that P and Q have no nonreal common zeros. Therefore, the equivalence of (ii) and (iii) in Lemma A.2.2, part 1 or part 2, shows that ω ∈ SHB. Under the additional assumption, the real functions P and Q have no common zeros. Hence ω ∈ SHB. 

Appendix A. Rational Nevanlinna and Stieltjes Functions

290

A.6

Miscellaneous results

A straightforward calculation proves the following result. Proposition A.6.1. Let ψj , Pj , Qj (j = 1, 2) be polynomials such that ψj (λ) = Pj (λ2 ) + iλQj (λ2 ) (λ ∈ C). Then ψ2 (λ)ψ1 (−λ) − ψ2 (−λ)ψ1 (λ) = 2iλ[P1 (λ2 )Q2 (λ2 ) − P2 (λ2 )Q1 (λ2 )]. Proposition A.6.2. Let η ∈ C, let k ∈ N and let ω be a polynomial of degree 2k such that λ ∈ C is not a zero of ω whenever −λ is a zero of ω. Then there is a unique polynomial ψ of degree not exceeding 2k such that ω(0)ψ(0) = 1 and ω(λ)ψ(−λ) − ω(−λ)ψ(λ) = iηλ

(λ ∈ C).

The coefficients of ψ depend linearly on η, and there is at most one value of η ∈ C for which the degree of ψ is less than 2k. Proof. Let ψ be any polynomial of degree 2k. Decomposing each of ω and ψ into a sum of an even and odd polynomial, we find unique polynomials Pe of degree k, e of degree at most k − 1, Ye of degree k and Ze of degree at most k − 1 such that Q e 2 ), ω(λ) = Pe(λ2 ) + iλQ(λ

e 2) ψ(λ) = Ye (λ2 ) + iλZ(λ

(λ ∈ C).

(A.6.1)

By the Euclidean algorithm, there are a unique constant c1 ∈ C, a unique polynob0 , and a unique polynomial mial Yb of degree at most k − 1, a unique polynomial Z b e Z whose degree is less than the degree of Q such that Ye = c1 Pe + Yb ,

e + Z. b b0 Q e=Z Z

(A.6.2)

e do not have common zeros because otherwise there The polynomials Pe and Q e 2 ) = 0, which would would be a complex number λ such that Pe(λ2 ) = 0 and Q(λ e lead to the contradiction ω(λ) = 0 = ω(−λ). Also, P (0) = ω(0) 6= 0. By Proposition A.6.1, e 2 )] e 2 )Ye (λ2 ) − Pe(λ2 )Z(λ ω(λ)ψ(−λ) − ω(−λ)ψ(λ) = 2iλ[Q(λ

(λ ∈ C).

Therefore we have to show that there are unique polynomials Ye of degree at most k with Pe(0)Ye (0) = 1 and Ze of degree at most k − 1 such that e Ye − PeZe = η , Q 2 which is equivalent to e Ye Z η − = . e e e e P Q 2P Q

(A.6.3)

A.6. Miscellaneous results

291

b0 and The representation in (A.6.2) gives that (A.6.3) is equivalent to c1 = Z b Yb Z η − = . e e e e P Q 2P Q

(A.6.4)

In view of the uniqueness of the partial fraction decomposition and the fact that e do not have common zeros it follows that (A.6.4) has a unique pair of Pe and Q b If Y and Z denote the solutions for η = 1, then clearly Yb = ηY solutions Yb , Z. b and Z = ηZ. Therefore ω(0)ψ(0) = Pe(0)Ye (0) = c1 Pe2 (0) + η Pe(0)Y (0), and the requirement ω(0)ψ(0) = 1 gives a unique c1 =

1 − η Pe(0)Y (0) , Pe2 (0)

(A.6.5)

b0 = c1 is unique. It is now obvious that all coefficients of Ye and and hence also Z e depend linearly on η. When Y (0) = 0, then c1 6= 0 for all η ∈ C. However, when Z Y (0) 6= 0, then there is exactly one η ∈ C such that c1 = 0. This completes the proof since the degree of ψ is 2k if and only if c1 6= 0.  Corollary A.6.3. In addition to the assumptions of Proposition A.6.2 assume that η ∈ R and ω is symmetric. Then also ψ is symmetric. For fixed ω let Jω be the set of all η ∈ R for which the coefficient of λ4k in ω(λ)ψ(λ) is positive. Then Jω is an unbounded open interval with 0 ∈ Jω . Proof. Since ω is symmetric, i.e., ω(−λ) = ω(λ), it follows that ω(λ)ψ(λ) − ω(−λ)ψ(−λ) = −[ω(λ)ψ(−λ) − ω(−λ)ψ(λ)] = −[ω(λ)ψ(−λ) − ω(−λ)ψ(λ)] = −iηλ = iηλ. The solution ψ of the functional equation in Proposition A.6.2 is unique, and therefore ψ(−λ) = ψ(λ), i.e., ψ is symmetric. e are real when ω is symmetric (see p. 287), also Since the polynomials Pe and Q the polynomial Y considered in the proof of Proposition A.6.2 is real. In particular, Pe(0) is a nonzero real number and Y (0) is a real number. Then the statement on Jω immediately follows from (A.6.5) since the coefficient of λ4k in ω(λ)ψ(λ) equals the leading coefficient of λ2k in Pe(λ)Ye (λ), which equals the leading coefficient of λ2k in c1 Pe2 (λ) by (A.6.2).  Definition A.6.4. Let U be a metric space, e.g., a subset of Cr (r ∈ N). Then a rational function f (·, u) whose coefficients depend on u ∈ U is said to be strongly

292

Appendix A. Rational Nevanlinna and Stieltjes Functions

continuous in u ∈ U if it has a representation f (·, u) = h(·, u) are polynomials with the following properties:

g(·,u) h(·,u) ,

where g(·, u) and

(i) the degrees of g(·, u) and h(·, u) are independent of u ∈ U ; (ii) the coefficients of the polynomials g(·, u) and h(·, u) depend continuously on u ∈ U; (iii) for all u ∈ U , the polynomials g(·, u) and h(·, u) do not have common zeros. Lemma A.6.5. Let f (·, u) be a rational function which is strongly continuous in u ∈ U . Then the zeros and poles of f (·, u) depend continuously on u ∈ U . Proof. Due to property (iii) in Definition A.6.4 it suffices to consider g and h separately. Hence we only need to consider, say, the polynomial g(·, u). In this case, the statement of this lemma immediately follows from Rouch´e’s theorem for simple zeros, and for multiple (branching) zeros if one clusters them as a group. More effort is needed to show that there are continuous branches, see, e.g., [96, Theorem 9.1.1].  Corollary A.6.6. In addition to the assumptions of Proposition A.6.2 let U be a metric space and assume that the coefficients of ω depend continuously on u ∈ U . For u ∈ U let η0 (u) ∈ C be the exceptional value of η for which c1 = 0 in (A.6.5), with η0 (u) := ∞ if c1 6= 0 for all η ∈ C. Then η0 depends continuously on u ∈ U , and ψ is strongly continuous in u ∈ U and η ∈ C \ {η0 (u)}. Proof. It is well known that the coefficients in the partial fraction decomposition (A.6.4) can be obtained as solutions of a linear system of equations, which is obtained by comparing coefficients of powers of λ (after multiplying through by e The given data of this linear system (for η = 1) are polynomials of the PeQ). e Hence the solution of this linear system and therefore coefficients of Pe and Q. also Y and Z depend continuously on u ∈ U , and so does η0 ∈ C. In particular, the coefficients of the polynomial ψ satisfy property (ii) in Definition A.6.4 with respect to u ∈ U and η ∈ C, whereas property (i) is clear from the definition of η0 (u). When η0 (u0 ) = ∞ for some u0 ∈ U , then Y (0) = 0 for this u0 , and the continuous dependence of Y (0) and Pe(0) on u ∈ U implies that η0 (u0 ) → ∞ as u → u0 .  Corollary A.6.7. Let f (·, u) be a rational S-function which is strongly continuous in u ∈ U . Then the numbers a1 , . . . , an , b1 , . . . , bn and c0 in the continued fraction expansion (A.3.8) depend continuously on u ∈ U . Proof. From the proof of Lemma A.3.4 we see that the numbers a and b there would depend continuously on u ∈ U and that the function g there would be strongly continuous in u ∈ U . Mathematical induction completes this proof. 

Appendix B

Linear and quadratic matrix pencils B.1

Matrix pencils

In this appendix we will state and prove results for quadratic matrix pencils of the form T (λ) = λ2 M − iλK − A (λ ∈ C), (B.1.1) where M , K, A are n × n symmetric matrices with real entries (n ∈ N), K being positive semidefinite and M being positive definite. We write M > 0 and K ≥ 0. The trivial case n = 1 is included for convenience. We shall assume throughout this appendix that M is a diagonal matrix with positive diagonal entries since all pencils considered in this monograph are of this form. A complex number λ is called an eigenvalue of T if the equation T (λ)y = 0 has a nontrivial solution y ∈ Cn \ {0}, and each such y is called an eigenvector of T at λ. The set of all such solutions, including 0, is a linear subspace of Cn and its dimension is called the geometric multiplicity of the eigenvalue λ. It is well known that the equation T (λ)y = 0 has a nontrivial solution if and only if the determinant det T (λ) = 0. The multiplicity of the zero λ of the function det T will be called the algebraic multiplicity of the eigenvalue λ and denoted by m(λ, T ). We may write m(λ, T ) = 0 when λ is not an eigenvalue of T . Since M is invertible, det T is a polynomial of degree 2n, and therefore the sum of the algebraic multiplicities of all eigenvalues of T equals 2n. When K = 0 one can set λ2 = z, and (B.1.1) becomes the linear matrix pencil L(z) = zM − A (z ∈ C). (B.1.2) The notion of eigenvalue and multiplicity as well as the notation m(z, L) applies accordingly. The sum of the algebraic multiplicities of all eigenvalues of L equals n. © Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4

293

294

Appendix B. Linear and quadratic matrix pencils

B.2 The linear pencil L In this section we assume that K = 0, which leads to the pencil L defined in (B.1.2). Then T (λ) = L(λ2 ) (λ ∈ C), and λ is an eigenvalue of T if and only if λ2 is an eigenvalue of L; in this case also −λ is an eigenvalue of T . Denoting the zeros of det L, counted with multiplicity, with z1 , . . . , zn , it follows that det L(z) = det M

n Y

(z − zk ).

k=1

Clearly, m(λ, T ) = m(−λ, T ) = m(λ2 , L) when λ 6= 0 and m(0, T ) = 2m(0, L). Hence we will only deal with the linear pencil L in this section. Proposition B.2.1. All eigenvalues of L are real and the geometric multiplicity of each eigenvalue equals its algebraic multiplicity. If A is positive semidefinite, then all eigenvalues of L are nonnegative, and if A is positive definite, then all eigenvalues of L are positive. Proof. Since M is a diagonal matrix with positive entries, there are positive numbers mk (k = 1, . . . , n) such that M = diag(m1 , . . . , mn ). With the matrix 1

1

1

M 2 = diag(m12 , . . . , mn2 ) it is clear that   1 1 1 1 L(z) = zM − A = M 2 z − M − 2 AM − 2 M 2 ,   1 1 1 and therefore L(z)y = 0 for y ∈ Cn if and only M − 2 AM − 2 − z M 2 y = 0. The 1

1

matrix M − 2 AM − 2 is symmetric, and it is positive (semi)definite if and only if A is positive (semi)definite. Hence the result follows from well-known properties of the eigenvalues of Hermitian matrices. 

B.3

The quadratic pencil T

For a matrix C let C be its conjugate complex and let C ∗ be its conjugate complex transpose, i.e., for C = (cj,k )j=1,...,n; k=1,...,m define C = (cj,k )j=1,...,n; k=1,...,m and C ∗ = (ck,j )j=1,...,m; k=1,...,n . Since the matrices M , K, A are real symmetric, M ∗ = M , K ∗ = K, A∗ = A. For x, y ∈ Cn , x = (xj )nj=1 , y = (yj )nj=1 , the inner product is defined by (x, y) =

n X

xj yj .

j=1

Observe that (C ∗ x, y) = (x, Cy) for n × n matrices C. Lemma B.3.1. The geometric multiplicity of an eigenvalue of T does not exceed its algebraic multiplicity.

B.3. The quadratic pencil T

295

Proof. Let λ0 be an eigenvalue of T with geometric multiplicity k. Then the matrix T (λ0 ) has defect k and therefore there is an invertible n × n matrix C such that the first k columns of CT (λ0 ) are zero. Since the entries of CT are polynomials (of degree 2 or less), there are polynomial vector functions a1 , . . . , an in Cn such that CT (λ) = ((λ − λ0 )a1 (λ), . . . , (λ − λ0 )ak (λ), ak+1 (λ), . . . , an (λ)) . Hence det T (λ) = (λ − λ0 )k (det C)−1 det(a1 (λ), . . . , an (λ)) shows that k ≤ m(λ0 , T ).



An eigenvalue of T is called semisimple if its algebraic multiplicity equals its geometric multiplicity. An eigenvalue of T is called simple if its algebraic multiplicity is 1. Lemma B.3.2. T (λ)∗ = T (λ) = T (−λ) for all λ ∈ C. In particular, the eigenvalues of T are symmetric with respect to the imaginary axis, for each eigenvalue λ0 of T , the geometric multiplicity of λ0 equals the geometric multiplicity of the eigenvalue −λ0 , and the algebraic multiplicity of λ0 equals the algebraic multiplicity of −λ0 . Proof. For all λ ∈ C, 2

T (λ)∗ = (λ2 M − iλK − A)∗ = λ M + iλK − A = T (−λ), which proves T (λ)∗ = T (λ) = T (−λ). It follows that det T (λ) = det T (−λ). Hence, if det T (λ) = (λ − λ0 )k f (λ) with k = m(λ0 , T ) and some polynomial f , then det T (λ) = det T (−λ) = (−λ − λ0 )k f (−λ). Therefore m(−λ0 , T ) ≥ m(λ0 , T ), and applying this inequality with λ0 replaced by −λ0 , it follows that m(−λ0 , T ) = m(λ0 , T ). From T (−λ0 )y = T (λ0 ) y = T (λ0 )y for y ∈ Cn it is clear that T (λ0 )y = 0 if and only if T (−λ0 ) y = 0. Therefore the geometric multiplicity of an eigenvalue λ0 equals the geometric multiplicity of the eigenvalue −λ0 .  Lemma B.3.3. The eigenvalues of the pencil T are located in the closed upper half-plane and on the imaginary axis. Proof. Let y0 be an eigenvector corresponding to an eigenvalue λ0 of T . Then (T (λ0 )y0 , y0 ) = 0, and consequently, taking real and imaginary parts, ((Re λ0 )2 − (Im λ0 )2 )(M y0 , y0 ) + Im λ0 (Ky0 , y0 ) − (Ay0 , y0 ) = 0

(B.3.1)

296

Appendix B. Linear and quadratic matrix pencils

and Re λ0 [2 Im λ0 (M y0 , y0 ) − (Ky0 , y0 )] = 0.

(B.3.2)

If Re λ0 6= 0, then (B.3.2) reduces to 2 Im λ0 (M y0 , y0 ) − (Ky0 , y0 ) = 0,

(B.3.3)

and Im λ0 ≥ 0 follows since (M y0 , y0 ) > 0 and (Ky0 , y0 ) ≥ 0 in view of M > 0  and K ≥ 0. Lemma B.3.4. If K > 0, then the eigenvalues of the pencil T located in the closed lower half-plane lie on the imaginary axis. Proof. In view of Lemma B.3.3 we have to show that the pencil T has no nonzero real eigenvalues. Hence, assume that the pencil T has a nonzero real eigenvalue λ0 with corresponding eigenvector y0 . Then (B.3.2) leads to (Ky0 , y0 ) = 0, which contradicts K > 0.  Lemma B.3.5. 1. If A ≥ 0, then the spectrum of the pencil T is located in the closed upper half-plane. 2. If A > 0 and K > 0, then the spectrum of the pencil T is located in the open upper half-plane. 3. If A > 0 and λ2 M y − Ay 6= 0 for all real λ and all nonzero y ∈ N (K), then the spectrum of the pencil T is located in the open upper half-plane. Proof. 1. In view of Lemma B.3.3 we have to show that there are no eigenvalues on the negative imaginary semiaxis. Hence, let y0 be an eigenvector corresponding to a pure imaginary eigenvalue λ0 . Then Re λ0 = 0, and in view of M > 0, K ≥ 0, A ≥ 0, equation (B.3.1) would imply that (M y0 , y0 ) = 0, (Ky0 , y0 ) = 0, and (Ay0 , y0 ) = 0 if Im λ0 < 0. But this contradicts M > 0. 2. From A > 0 it follows that 0 is not an eigenvalue of T , and hence the statement of part 2 follows in view of part 1 and Lemma B.3.4. 3. The same proof as in part 2 above shows that 0 is not an eigenvalue of the pencil T . If the pencil T would have a nonzero real eigenvalue λ0 with corresponding eigenvector y0 , then (B.3.2) would imply Ky0 = 0, and therefore λ20 M y0 − Ay0 = T (λ0 )y0 = 0, which contradicts the assumptions of this case.



Proposition B.3.6. Let A ≥ 0. Then there are a unique integer k ≥ 0, a unique monic real polynomial R with R(0) 6= 0 and a unique monic polynomial ω which is either 1 or a Hermite-Biehler polynomial such that det T (λ) = (det M )λk R(λ2 )ω(λ).

(B.3.4)

If R is not constant, then all zeros of R are positive real numbers. If ω is not constant, then ω belongs to the SHB class.

B.3. The quadratic pencil T

297

Proof. Let w1 , . . . , wm be the nonreal zeros of det T , counted with multiplicity, and define m Y (λ − wj ). ω(λ) = j=1

Let z1 , . . . , zl be the positive real zeros of det T , counted with multiplicity, and define l Y R(z) = (z − zj2 ). j=1

Putting k = m(0, T ) it is clear from the symmetry statement in Lemma B.3.2 and from det M being the coefficient of λ2n in T that the representation (B.3.4) holds, that all zeros of R, if any, are positive, that ω = 1 if det T has only real zeros, and that ω is an SHB polynomial on account of Lemma B.3.5, part 1, if det T has nonreal zeros. For the uniqueness we first observe that k = m(0, T ) since R(0) 6= 0 and since a Hermite-Biehler polynomial does not have real zeros. Since det T has no zeros in the open lower half-plane by Lemma B.3.5, part 1, and since λ 7→ R(λ2 ) is a real polynomial, all its zeros must be real because nonreal zeros would occur in conjugate complex pairs. On the other hand, all zeros of ω are nonreal, so that the monic polynomials P and ω are uniquely determined by the real and nonreal zeros of det T , respectively.  ∞ ∞ and (bj )j=1 Proposition B.3.7. Let (aj )j=1 be sequences of complex number. Then the determinants of the Jacobi matrices

 a1  b1   0    Jk =       0 0

b1 a2

0 b2

0 0

b2

a3

b3

0

bk−2 0

ak−1 bk−1

 0 0            0   bk−1  ak

(k ∈ N)

satisfy the recursion relations det J1 = a1 ,

det J2 = a1 a2 − b21 ,

det Jk = ak det Jk−1 − b2k−1 det Jk−2 (k ≥ 3).

Proof. Clearly, det J1 = a1 and det J2 = a1 a2 − b21 . For k ≥ 3 we use the block

298

Appendix B. Linear and quadratic matrix pencils

decompositions    Jk−2   Jk =    0 · · · 0 bk−2 0 ··· 0 0 and we define

 0 ..  .   0 0   bk−2 0   ak−1 bk−1  bk−1 ak

  Jk−2  Jbk−1 =   0 · · · 0 bk−2

0 .. .

 0 ..  .  . 0  bk−1

Expanding the determinant of Jk with respect to the last row it follows that det Jk = ak det Jk−1 − bk−1 det Jbk−1 = ak det Jk−1 − b2k−1 det Jk−2 .

B.4



Schur complements

Let A be a square matrix, say of size n × n. For 0 < k < n we write A as an ((n − k) + k) × ((n − k) + k) block matrix   Ak,1 Ak,2 A= . Ak,3 Ak,4 If Ak,4 is invertible, then S(A, k) := Ak,1 − Ak,2 A−1 k,4 Ak,3

(B.4.1)

is called the Schur complement of Ak,4 in A. In this case, the Schur factorization     I 0 I Ak,2 A−1 S(A, k) 0 k,4 A= (B.4.2) A−1 0 Ak,4 0 I k,4 Ak,3 I holds. In particular, det A = det S(A, k) det Ak,4 , A is invertible when S(A, k) is invertible, and then    I 0 S(A, k)−1 0 I −1 A = −1 −A−1 A I 0 A 0 k,4 k,3 k,4

(B.4.3)

 −Ak,2 A−1 k,4 . I

(B.4.4)

The following lemma shows that the Schur complement can be iterated.

B.5. Eigenvalues of real symmetric matrices

299

Lemma B.4.1. Let A be an n × n matrix and let 0 < k, l such that k + l < n. If Ak,4 and Ak+l,4 are invertible, then S(A, k + l) = S(S(A, k), l). Proof. Writing A as an ((n − k − l) + l + k) × ((n − k − l) + l + k) block matrix   B11 B12 B13 A = B21 B22 B23  B31 B32 B33 it follows from the assumptions that Ak,4 = B33 and   B22 B23 Ak+l,4 = B32 B33 are invertible. By definition,    
B13 B11 B12 −1 B31 B32 − S(A, k) = B33 B21 B22 B23   −1 −1 B31 B12 − B13 B33 B32 B11 − B13 B33 = −1 −1 B31 B22 − B23 B33 B21 − B23 B33 B32 and S(A, k + l) = B11 − B12

−1 B13 Ak+l,4




Furthermore, the Schur factorization   −1 −1 I B23 B33 B32 B22 − B23 B33 Ak+l,4 = 0 0 I



0 B33

 B21 . B31



I −1 B33 B32

0 I



−1 B32 is invertible and and the invertibility of Ak+l,4 show that B22 − B23 B33     −1 −1 I 0 (B22 − B23 B33 0 B32 )−1 I −B23 B33 . A−1 = −1 −1 k+l,4 B32 I I −B33 0 B33 0

Therefore −1 B31 S(A, k + l) = B11 − B13 B33 −1 −1 −1 B32 )(B22 − B23 B33 B32 )−1 (B21 − B23 B33 B31 ) − (B12 − B13 B33

= S(S(A, k), l).

B.5



Eigenvalues of real symmetric matrices

The necessity of (B.5.1) in the following well-known equivalence result is attributed to Cauchy [31], whereas its sufficiency was proved in [52]. A proof can be found for example in [124, Section 71]. However, since the interlacing (B.5.1) below is

300

Appendix B. Linear and quadratic matrix pencils

related to poles and zeros of rational Nevanlinna functions, which are extensively used in this monograph, we will give a short proof making use of properties of rational Nevanlinna functions. In this section, let H be a real symmetric (n+1)×(n+1) matrix (n ∈ N), and denote by H1,1 the n × n first principal submatrix obtained from H by deleting its first row and first column. Proposition B.5.1. 1. Let n ∈ N, let H be an (n + 1) × (n + 1) real symmetric matrix H such that the eigenvalues of H are µ1 ≤ µ2 ≤ · · · ≤ µn+1 and the eigenvalues of the submatrix H1,1 are ν1 ≤ ν2 ≤ · · · ≤ νn . Then µ1 ≤ ν1 ≤ µ2 ≤ ν2 ≤ · · · ≤ νn ≤ µn+1 .

(B.5.1)

n+1 n 2. Let (µk )j=1 be sequences of real numbers satisfying (B.5.1). and (νk )k=1 Then there exists an (n + 1) × (n + 1) real symmetric matrix H such that the eigenvalues of H are µ1 , . . . , µn+1 and the eigenvalues of the submatrix H1,1 are ν1 , . . . , νn .

Proof. 1. Define the rational function χH by χH (λ) = −

det(H − λ) . det(H1,1 − λ)

Let U be an orthogonal matrix such that U T H1,1 U = diag{ν1 , . . . , νn }. For the orthogonal block matrix   1 0 V = 0 U b := V T HV is a real symmetric matrix of the form it follows that H   b0 b1 · · · bn  b1 ν1 0 0  b = H  .. ..  . . . . . 0 bn 0 · · · νn

(B.5.2)

b − λ) with respect with real numbers b0 , . . . , bn . Expanding the determinant of (H to the first row, we have   n n Y X b2j b − λ) =  det(H − λ) = det(H (νk − λ) (b0 − λ) − ν − λ j j=1 k=1

b 1,1 − λ), and then, in view of det(H1,1 − λ) = det(H χH (λ) = λ − b0 −

n X j=1

b2j . λ − νj

B.5. Eigenvalues of real symmetric matrices

301

Because of Theorem A.2.6 and the fact that the sum of Nevanlinna functions is a Nevanlinna function it follows that χH is a Nevanlinna function. Denote the n0 +1 and the increasing sequence of µk )k=1 increasing sequence of the zeros of χH by (b n−n0 n0 such that (µk )n+1 νk )k=1 . Then there is a sequence (τk )k=1 the poles of χH by (b k=1 = n n0 +1 n−n n−n0 n 0 νk )k=1 ∪ (τk )k=1 0 . If n0 > 0, then (b µk )k=1 ∪ (τk )k=1 and (νk )k=1 = (b µ bn0 +1 b1 < νb1 < · · · < µ by Theorem A.2.6, and (B.5.1) follows in view of Lemma A.4.4. 2. With the notation from the proof of part 1 of this proof we define χ(λ) =

n+1 Y

n Y

k=1

k=1

(λ − µk )

(λ − νk )−1 =

nY 0 +1

n0 Y

k=1

k=1

(λ − µ bk )

(λ − νbk )−1 .

(B.5.3)

Since (B.5.1) holds, χ is a Nevanlinna function by Theorem A.2.6. In view of Lemma A.2.7 there are real numbers b0 , . . . , bn such that χ(λ) = λ − b0 −

n X j=1

b2j , λ − νj

(B.5.4)

where we put bj = 0 when νj corresponds to some τk . Hence, defining H by the right-hand side of (B.5.2) it follows that χH = χ. Clearly det(H1,1 − λ) =

n Y

(νk − λ),

j=1

and therefore det(H − λ) = −χ(λ) det(H1,1 − λ) =

n+1 Y

(µk − λ).

j=1

Hence the eigenvalues of H are µ1 , . . . , µn+1 and the eigenvalues of H1,1 are ν1 , . . . , νn .  Let n ∈ N and let σ be a permutation of {1, . . . , n}, i.e., σ : {1, . . . , n} → {1, . . . , n} is a bijective map. The n × n matrix Pσ := (δi,σ(j) )ni,j=1 is called a permutation matrix. We will also write Pσ =: (pσ,i,j )ni,j=1 . Lemma B.5.2. Let n ∈ N and let σ, τ be permutations. Then Pτ Pσ = Pτ ◦σ and Pσ∗ = PσT = Pσ−1 = Pσ−1 . Proof. For i, j = 1, . . . , n we calculate n X s=1

pτ,i,s pσ,s,j =

n X s=1

δi,τ (s) δs,σ(j) = δi,τ (σ(j)) = pτ ◦σ,i,j ,

302

Appendix B. Linear and quadratic matrix pencils

which proves Pτ Pσ = Pτ ◦σ . Since the entries of Pσ are real, Pσ∗ = PσT is obvious. For i, j = 1, . . . , n the identities pσ,j,i = δj,σ(i) = δi,σ−1 (j) = pσ−1 ,i,j show that PσT = Pσ−1 . Finally, the  first statement gives Pσ−1 Pσ = In , and Pσ−1 = Pσ−1 follows. n and a permutation σ of {1, . . . , n} let For an n × n matrix A = (ai,j )i,j=1

PσT APσ =: Aσ =: (aσ,i,j )ni,j=1 .

(B.5.5)

For i, j = 1, . . . , n, Lemma B.5.2 gives aσ,i,j =

n X n X

δi,σ−1 (s) as,t δt,σ(j) =

n X

δi,σ−1 (s) as,σ(j)

s=1

s=1 t=1

= aσ(i),σ(j) . This shows that Aσ = (aσ(i),σ(j) )ni,j .

(B.5.6)

Definition B.5.3. Let m ∈ N, m ≥ 2. An m × m matrix A = (ai,j )m i,j=1 is called reducible if there are two nonempty subsets ∆1 , ∆2 of {1, . . . , m} satisfying ∆1 ∪˙ ∆2 = {1, . . . , m} such that ai,j = 0 for i ∈ ∆1 and j ∈ ∆2 . The matrix A is called irreducible if it is not reducible. Proposition B.5.4. Let m ≥ 2. A Hermitian m × m matrix A is reducible if and only if there is a permutation σ of {1, . . . , m} with σ(1) = 1 and a number k ∈ N with 1 ≤ k ≤ m − 1 such that Aσ has the representation   0 Aσ,1 Aσ = (B.5.7) 0 Aσ,2 with Aσ,1 ∈ Mk (C) and Aσ,2 ∈ Mm−k (C). Proof. If (B.5.7) holds, put ∆1 = σ({1, . . . , k}) and ∆2 = σ({k + 1, . . . , m}). For i ∈ ∆1 and j ∈ ∆2 we have 1 ≤ σ −1 (i) ≤ k and k + 1 ≤ σ −1 (j) ≤ m, and then (B.5.6) gives ai,j = aσ−1 (i),σ−1 (j) = 0. Therefore A is reducible. Conversely, let A be reducible. Since A is Hermitian, ai,j = 0 for i ∈ ∆1 and j ∈ ∆2 implies aj,i = 0 for i ∈ ∆1 and j ∈ ∆2 , that is, one can interchange ∆1 and ∆2 . Therefore we may assume that 1 ∈ ∆1 . Let k = #∆1 . Then 1 ≤ k ≤ n − 1. Choose a permutation σ of {1, . . . , m} such a that σ({1, . . . , k}) = ∆1 and σ(1) = 1. Then σ({k + 1, . . . , m}) = ∆2 . By (B.5.6) it follows that aσ,i,j = aσ(i),σ(j) = 0

(i = 1, . . . , k, j = k + 1, . . . , m).

Since Aσ is Hermitian, (B.5.7) follows.



Corollary B.5.5. Under the assumptions of Proposition B.5.1 consider the special case that (B.5.8) µ1 < ν1 < µ2 < ν2 < · · · < νn < µn+1 . Then the matrix H is irreducible.

B.5. Eigenvalues of real symmetric matrices

303

Proof. Assume that H is reducible. By Proposition B.5.4 there are an n × n permutation matrix P , integers 1 ≤ m1 , m2 ≤ n with m1 + m2 = n + 1, an m1 × m1 matrix F and an m2 × m2 matrix G such that       1 0 1 0 F 0 = H . 0 G 0 P 0 PT Recalling P T = P −1 from Lemma B.5.2 it follows that det(H − λIn+1 ) = det(F − λIm1 ) det(G − λIm2 ). If m1 = 1, then H1,1 = P GP T , and therefore det(H1,1 − λIn ) = det(G − λIn ). If m1 > 1, then  H1,1 = P

F1,1 0

 0 P T, G

and det(H1,1 − λIn ) = det(F1,1 − λIm1 −1 ) det(G − λIm2 ) follows. In any case, it would follow that the eigenvalues of G are eigenvalues of H as well as H1,1 , which contradicts (B.5.8). 

Appendix C

Graphs C.1

Definitions

In this section we fix our notation and give some elementary definitions and facts about graphs, as far as they are needed for the main subject of this monograph. Nearly 40 years ago Tutte wrote in the first edition of his monograph [132, p. 1] “The terminology of graph theory is not yet standardized”, and the same is still true today and will most likely be true for a long time. Hence we have chosen to use notation which seems to be commonly used in the context of our subject. The notation is closely related to that of Tutte and Lov´asz [90], and in particular the first sentence of the next paragraph is almost verbally taken from [132, p. 1]. A graph G is defined by a set V = V G of elements called vertices, a set E = E G of elements called edges, and a relation of incidence, which associates with each edge either one or two vertices called its ends. A vertex v will be called incident with an edge e and the edge e will be called incident with the vertex v if v is any of the ends of e. We will only consider finite graphs, that is, all graphs in this monograph will satisfy #V < ∞ and #E < ∞. Otherwise, we will allow arbitrary graphs and even the null graph, which has no vertices and no edges. Definition C.1.1. A walk in a graph is a sequence (v0 , e1 , v1 , e2 , v2 , . . . , vk−1 , ek , vk ) of vertices vj and edges ej . Here vj−1 and vj are the ends of ej if ej has two ends, whereas vj−1 = vj is the single end of ej if ej has only one end. A walk is called open when v0 6= vk , and it is called closed when v0 = vk . A walk in which all the edges are distinct is called a trail. A trail in which all vertices v0 , v1 , . . . , vk are distinct (except possibly for vk = v0 ) is called a path; an open path is called a chain; a closed path is called a cycle. Any edge joining a vertex v to itself, i.e., a cycle consisting of exactly one edge, is called a loop. © Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4

305

306

Appendix C. Graphs

A graph is called simple if it has no loops and if for any two vertices there is is at most one edge which is incident with these two vertices. We may briefly write (v0 , . . . , vk ) for a walk, trail, etc., if its edges are not essential in the corresponding context or uniquely determined by the vertices. If there is a walk from a vertex v in G to a vertex w 6= v in G which is not a path, then there are indices j < k in the walk such that vj = vk . Removing ej+1 , vj+1 , . . . , ek , vk from the walk, one arrives at another walk starting at v and ending at w. Repeating this procedure, if necessary, finitely many times gives a chain from v to w. A graph is called connected if for each pair of distinct vertices (v, w) there is 0 a walk from v to w. A graph G0 is called a subgraph of a graph G if V G ⊂ V G , G0 G 0 0 E ⊂ E and each edge of G has the same ends in G as in G. For a set V1 of vertices of a graph G let G|V1 be the subgraph of G whose set of vertices is V1 and whose edges are all edges in G whose ends are in V1 . For vertices v, w in G define v ∼ w ⇔ (v = w or there is a walk from v to w). It is clear that this is an equivalence relation and that each equivalence class of vertices together with the edges in G which connect these vertices defines a (maximal) connected subgraph, called a component of G. Each graph G is the disjoint union of its components, and G is connected if and only if it has exactly one component. For each vertex v in G, the degree d(v) is the number of edges incident with v. If v is the common endpoint of a loop, then this loop is counted twice in d(v). A vertex v with d(v) = 1 is called a pendant vertex of G, and a vertex v with d(v) ≥ 2 is called an interior vertex of G. For convenience we denote the set of pendant vertices of G by V pen and the set of interior vertices of G by V int . We let Vb = V pen ∪˙ V int be the set of all vertices of G which are incident with at least one edge of G. We recall that we may allow isolated vertices v, i.e., d(v) = 0. It is convenient to index the edges of G by ej (j = 1, . . . , g), where g (or g G ) is the number of edges of G. Unless otherwise stated, the indexing of the edges may be chosen arbitrarily. A graph is called a metric graph if a length lj > 0 is assigned to each of its edges ej . A graph with at least one edge is called a digraph if a direction is associated to each of its edges. If v and w are the vertices incident with an edge e and if e is directed from v to w, then v is called the entrance of e, denoted by v− (e), w is called the exit of e, denoted by v+ (e), and e is called outgoing from v and incoming into w. For a vertex v of G let I + (v) be the set of indices j ∈ {1, . . . , g} such that ej is incoming into v, and let I − (v) be the set of indices j ∈ {1, . . . , g} such that ej is outgoing from v. Then I(v) := I − (v) ∪ I + (v) is the set of indices j for which ej is an edge incident with v. It is clear that j ∈ I − (v) ∩ I + (v) if and only if ej is a loop with end v. For each vertex v of G, the numbers d− (v) := #I − (v) and d+ (v) := #I + (v) denote the number of outgoing and incoming edges which are incident with v, respectively. Hence d(v) = d− (v) + d+ (v) (v ∈ V ).

C.1. Definitions

307

Since each edge in a digraph has exactly one vertex for which it is an incoming edge and exactly one vertex for which it is an outgoing edge, it is clear that [ ˙

{I − (v) : v ∈ V } =

[ ˙

{I + (v) : v ∈ V } = {1, . . . , g},

i.e., {1, . . . , g} equals the disjoint union of the sets I − (v) taken over all vertices of G as well as the disjoint union of the sets I + (v) taken over all vertices of G. Definition C.1.2. Two distinct vertices v and w of a graph G are said to be cyclically connected if there is a finite number of cycles C1 , . . . , Ck in G such that v is a vertex in C1 , w is a vertex in Ck , and each pair (Cj , Cj+1 ) (j = 1, . . . , k − 1) possesses at least one common vertex. Definition C.1.3. A graph with at least one edge is said to be cyclically connected if each pair of distinct vertices in it is cyclically connected. It is easy to verify that a cyclically connected graph is connected. Also note that a graph with one vertex and at least one edge is cyclically connected and that all its edges are loops. Definition C.1.4. A cyclically connected subgraph G0 of a graph G is said to be maximal cyclically connected if it is not a nontrivial subgraph of another cyclically connected subgraph of G. Definition C.1.5. A connected graph with at least one edge is said to be a tree if it has no cycles. Let G be a graph and let E 0 ⊂ E G . With V 0 ⊂ V G being the set of all vertices of G which are incident with at least one e ∈ E one obtains a unique subgraph G0 0 0 of G without isolated vertices such that V G = V 0 and E G = E 0 . In particular, 0 0 each subgraph G of G without isolated vertices is uniquely determined by E G . Definition C.1.6. Let G be a graph without isolated vertices and let G1 be a subgraph of G without isolated vertices satisfying ∅ 6= E G1 6= E G . Then the unique subgraph G2 of G without isolated vertices satisfying E G2 = E G \ E G1 is called the complement of G1 , and G1 and G2 are called complementary subgraphs of G. From the above definition of complementary subgraphs, the following remark is obvious. Remark C.1.7. Let G be a graph without isolated vertices and let G1 and G2 be subgraphs of G without isolated vertices. Then G1 and G2 are complementary subgraphs of G if and only if E G1 6= ∅, E G2 6= ∅ and E G = E G1 ∪˙ E G2 . Let G be a digraph without isolated vertices and let G1 and G2 be complementary subgraphs of G. Then a common vertex v of G1 and G2 is called an entrance of Gj (j = 1, 2) if there is at least one edge in Gj for which v is an entrance. Similarly, v is called an exit of Gj if there is at least one edge in Gj for

308

Appendix C. Graphs

which v is an exit. Clearly, such a vertex could be both an entrance and an exit. If there is a loop with vertex v in Gj , then v is surely an entrance and an exit of Gj . We say that the graph G is the disjoint union of the graphs G1 and G2 , written G = G1 ∪˙ G2 , if the edges and vertices of G are the disjoint unions of the edges and vertices of G1 and G2 , respectively. Clearly, if V G = V1 ∪˙ V2 , then G = G|V1 ∪˙ G|V2 if and only if there is no edge in G which has one incident vertex in V1 and the other incident vertex in V2 . We note that G is the disjoint union of two complementary subgraphs G1 and G2 if and only if the two complementary subgraphs have no common vertices. Definition C.1.8. A connected graph with at least one edge is said to be a quasitree if it is neither a tree nor cyclically connected and if each of its cyclically connected subgraphs has at least two vertices in common with the complement of this subgraph. Clearly, a quasi-tree does not have any loops. To every cycle we ascribe any of its two possible orientations. In a digraph the direction of an edge in a cycle may be opposite to the orientation of that cycle. Definition C.1.9. Let G be a graph with g edges and let s be the number of cycles of the graph G. A direction is assigned to each edge of the graph and an orientation is assigned to each cycle of the graph. Indexing the cycles from 1 to s in an arbitrary manner, the s × g matrix M = (Mk,j )k=1,...,s , called the matrix of j=1,...,g

cycles of G, is defined as follows. 1. For an edge ej which does not belong to the k-th cycle, set Mk,j = 0. 2. For an edge ej which belongs to the k-th cycle and whose direction coincides with the orientation of the cycle, set Mk,j = 1. 3. For an edge ej which belongs to the k-th cycle and whose direction is opposite to the orientation of the cycle, set Mk,j = −1. In the above definition it may not be completely clear what the number of cycles is. By Definition C.1.1 of a cycle as a path (v0 , e1 , v1 , e2 , v2 , . . . , vk−1 , ek , vk ) with v0 = vk (k ≥ 2) we may ask if the path (v1 , e2 , v2 , . . . , vk−1 , ek , vk , e1 , v1 ) describes the same cycle or whether the cycle with opposite orientation is the same cycle. We take here the pragmatic view that it does not matter since any such ‘modified’ cycle would contribute the same row or the negative of it to the matrix M and therefore would not change the rank of M . Hence we will consider cycles as uniquely determined by their edges. Therefore no two rows of M are multiples of each other. A square matrix is called a signed permutation matrix if it has exactly one entry in each row and each column whose value is 1 or −1 and if all its other entries have the value 0. If A is an r × s matrix and J an r × r permutation matrix, then JA is obtained from A by a signed permutation of the rows of A, that is, by a

C.2. Basic properties of graphs

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permutation of rows, and rows may be replaced by their negatives. Conversely, each matrix obtained from A by a signed permutation of the rows of A is of the form JA with a signed permutation matrix J. If A is an r ×s matrix and J an s×s signed permutation matrix, then AJ is obtained from A by a signed permutation of columns of A. Conversely, each matrix obtained from A by a signed permutation of the columns of A is of the form AJ with a signed permutation matrix J. It is clear that a signed permutation of a matrix keeps its rank unchanged. The matrix M introduced in Definition C.1.9 is unique up to signed permutations of rows and columns, and all properties of M we are going to use are invariant under these signed permutations. In particular, the rank of M is invariant under signed permutations. Therefore it is justified to call M ‘the matrix of cycles’. Definition C.1.10. If the graph G has at least one cycle, then the rank µ =: µG of the matrix of cycles of a graph G is called the cyclomatic number of the graph G. A set of cycles in a graph G is said to be linearly independent if the corresponding set of rows in the matrix of cycles is linearly independent. Each set of µ linearly independent cycles is said to be fundamental. If the graph G has no cycles, then we set µG := 0. For a concrete representation of a matrix of cycles one needs a digraph and orientations of the cycles. However, each graph G can be written as a digraph with arbitrarily chosen directions of the edges, and as was seen above, the cyclomatic number is independent of the chosen directions and orientations. Hence the cyclomatic number is well-defined for any graph.

C.2

Basic properties of graphs

We generalize the construction of the rows of the cyclomatic matrix to walks in graphs with at least one edge. To this end let G be a graph with at least one edge, which is considered as a digraph by assigning an arbitrary but fixed direction to each edge. The edges will be indexed by e1 , . . . , eg in an arbitrary but fixed manner. Each walk W = (v0 , ee1 , v1 , . . . , vk−1 , eek , vk ) has a natural orientation, with the edge eej directed from the vertex vj−1 towards the vertex vj if vj−1 6= vj . Only if eej is a loop, the direction is not determined by the ends of this edge. g as For an oriented walk W define the row vectors m± (W ) = (mj± (W ))j=1 + follows: mj (W ) is the number of times the edge ej is traversed in W in the same direction as in the digraph, and m− j (W ) is the number of times the edge ej is traversed in W in the direction opposite to that in the digraph. Then we define m(W ) := m+ (W ) − m− (W ) and |m|(W ) = m+ (W ) + m− (W ). In particular, if Ck is the k-th cycle in G, then m(Ck ) is the k-th row of the matrix of cycles of G. For two walks W1 and W2 we write |m|(W1 ) ≥ |m|(W2 ) if this inequality holds for all components of the two row vectors, and |m|(W1 ) > |m|(W2 ) means that

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additionally at least one component of |m|(W1 ) is larger than the corresponding component of |m|(W2 ). Traversing the walk W = (v0 , ee1 , v1 , . . . , vk−1 , eek , vk ) with the opposite orientation gives the walk W − := (vk , eek , vk−1 . . . , v1 , ee1 , v0 ). In particular, when e is a directed edge, then e− will denote the edge with the opposite direction. Let W1 = (v1,0 , e1,1 , . . . , e1,j , v1,j ) and W2 = (v2,0 , e2,1 , . . . , e2,k , v2,k ) be two walks in a graph G. If v1,j = v2,0 then (W1 , W2 ) := (v1,0 , e1,1 , . . . , e1,j , v1,j , e2,1 , . . . , e2,k , v2,k ) is the walk obtained by first traversing the walk W1 and then the walk W2 . Clearly, m(W1 , W2 ) = m(W1 ) + m(W2 ) and |m|(W1 , W2 ) = |m|(W1 ) + |m|(W2 ). Clearly, a walk W is a trail if and only if each component of |m|(W ) has the value 0 or 1. Lemma C.2.1. Let W be a closed walk in a graph G with m(W ) 6= 0 and assume that W is not a cycle. Then there are two closed walks W1 and W2 in G such that W = (W1 , W2 ), m(W ) = m(W1 ) + m(W2 ) and |m|(W ) = |m|(W1 ) + |m|(W2 ). If W is a trail, then also W1 and W2 are trails. Proof. There is a vertex v0 of W which is traversed at least twice. Indeed, if W is a trail or contains a loop, then there is such a vertex because W is not a cycle. And if W is not a trail and does not contain loops, then there is at least one edge which is traversed twice. Since m(W ) 6= 0 it follows that W has more than two edges and therefore at least one of the vertices of those edges which are traversed at least twice is also traversed at least twice. Using this vertex as the starting point of the closed walk we have a representation W = (v0 , ee1 , . . . , eej , vj , eej+1 , . . . , eek , vk ) with 1 ≤ j < k and v0 = vj = vk . Defining W1 = (v0 , ee1 , . . . , eej , vj ) and W2 = (vj , eej+1 , . . . , eek , vk ) it is clear that W1 and W2 are closed walks with W = (W1 , W2 ). If W is a trail, then all components of |m|(W ) are 1 or 0, and hence also all components of |m|(W1 ) and |m|(W2 ) have this property. It follows that W1 and  W2 are trails. Corollary C.2.2. Let W be a closed walk in a graph G satisfying m(W ) 6= 0. k P m(Cj ) and Then there are cycles C1 , . . . , Ck in G (k ≥ 1) such that m(W ) = j=1

|m|(W ) ≥

k P

|m|(Cj ).

j=1

Proof. We are going to prove the statement by induction on the number of edges in the walk W , that is, the sum of the components of the vector |m|(W ). If W has only one edge, then W is a loop and hence the statement follows with k = 1 and C1 = W . Now let n ∈ N with n > 1 and assume that the statement holds for all closed walks with less than n edges. Let W be a closed walk in G with exactly n edges satisfying m(W ) 6= 0. If W is a cycle, then the statement of this corollary

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follows with k = 1 and C1 = W . If W is not a cycle, then by Lemma C.2.1 there are closed walks W1 and W2 in G with m(W ) = m(W1 ) + m(W2 )

and |m|(W ) = |m|(W1 ) + |m|(W2 ).

(C.2.1)

In particular, n is the sum of the number of edges of W1 and the number of edges of W2 , so that both the number of edges of W1 and the number of edges of W2 are less than n. Since m(W ) 6= 0, it follows from (C.2.1) that at least one of m(W1 ) and m(W2 ) is different from the zero vector, say m(W1 ) 6= 0. By induction hypothesis, there are cycles C1 , . . . , Ck1 such that m(W1 ) =

k1 X

m(Cj )

and |m|(W1 ) ≥

j=1

k1 X

|m|(Cj ).

(C.2.2)

j=1

If m(W2 ) = 0, then (C.2.1) implies m(W ) = m(W1 ) and, together with (C.2.2), |m|(W ) = |m|(W1 ) + |m|(W2 ) > |m|(W1 ) ≥

k1 X

|m|(Cj ),

j=1

which proves the statement for W in this case. If m(W2 ) 6= 0, the induction hypothesis gives cycles Ck1 +1 , . . . , Ck such that m(W2 ) =

k X j=k1 +1

m(Cj )

and |m|(W2 ) ≥

k X

|m|(Cj ).

(C.2.3)

j=k1 +1

Then the statement of this corollary for W in this case follows from (C.2.1), (C.2.2)  and (C.2.3). Lemma C.2.3. Let G be a graph, let pG be the number of its vertices, and let κG be the number of its components. Then g G + κG = pG + µG . Proof. We will prove the statement by induction on g = g G . When g = 0, then G has no edges and therefore no cycles, which means that g G = 0 = µG . Also, every vertex is isolated, and therefore κG = pG , which proves the formula for g = 0. Now let g > 0 and assume the formula holds for g − 1. We will consider three different cases, but in each case we will remove one edge e0 from G to obtain a subgraph 0 0 G0 of G, which obviously satisfies g G = g G − 1 and pG = pG . Therefore the graph G0 will satisfy the induction hypothesis. Firstly assume that G contains a loop, and any such loop will be chosen as 0 the edge e0 to be removed. Clearly, κG = κG . The loop cannot be part of any other cycle, and hence the loop corresponds to a row and a column in the matrix of cycles of G whose entries are 0 except for the entry where this row and this column meet. Removing this row and this column from the matrix of cycles of G, its rank decreases by 1 (with the obvious interpretation if the loop is the only cycle in G).

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But this new matrix is the matrix of cycles of G0 , and therefore µG = µG − 1. 0 0 0 0 Altogether, it follows that g G + κG = g G + 1 + κG = pG + 1 + µG = pG + µG . Next assume that G has no loops and choose any edge e0 of G for removal. Let v1 and v2 be the ends of e0 . Obviously, each component of G which does not contain e0 is also a component of G0 . Let G0 be the component of G which contains e0 (and thus v1 and v2 ), and let Gk0 be the component of G0 containing vk (k = 1, 2). Let v 6= v2 be a vertex in G0 which does not belong to G01 , if any. Since G0 is connected, there is a chain from v1 to v in G. But this chain must contain the edge e0 because otherwise it would be a chain in G0 . Therefore this chain must be of the form (v1 , e0 , v2 , . . . , v), giving a chain (v2 , . . . , v) in G20 . This 0 0 shows that V G1 ∪ V G2 = V G0 . We will consider two cases. Here we assume that G10 6= G20 , in which case G10 and G02 are components 0 of G0 . Therefore κG = κG + 1. Furthermore, e0 is not an edge of any cycle in G. Because if there were such a cycle, say (v1 , e0 , v2 , . . . , v1 ), then (v2 , . . . , v1 ) 0 would be a path connecting v2 and v1 in G0 . Therefore µG = µG , and hence G0 G0 G0 G G G G0 G g +κ =g +κ =p +µ =p +µ . 0 Finally, assume that G01 = G20 . Then κG = κG and there is a path (v2 , . . . , v1 ) 0 in G . But then (v2 , . . . , v1 , e0 , v2 ) is a cycle in G. Assume that there are at least two cycles in G with edge e0 . With suitably chosen orientation two such cycles are of the form C (1) = (v1 , e0 , v2 , e1,1 , . . . , e1,j , v1 ) and C (2) = (v1 , e2,1 , . . . , e2,r , v2 , e0 , v1 ) and satisfy m(C (1) ) + m(C (2) ) 6= 0 if they are distinct. Removing the edge e0 from each of the cycles the concatenation W := (v2 , e1,1 , . . . , e1,j , v1 , e2,1 , . . . , e2,r , v2 ) is a walk in G which satisfies m(W ) = m(C (1) ) + m(C (2) ) 6= 0. By Corollary C.2.2 k k P P |m|(Cj ). m(Cj ) and |m|(W ) ≥ there are k cycles C1 , . . . , Ck with m(W ) = j=1

j=1

Since W does not traverse the edge e0 , the last inequality implies that none of the Cj traverses the edge e0 . Therefore a fundamental set of cycles of G can be chosen in such a way that exactly one of the cycles traverses the edge e0 . Removing this 0 cycle then gives a fundamental set of cycles of G0 , that is, µG = µG − 1, and 0 0 0 0 therefore g G + κG = g G + 1 + κG = pG + 1 + µG = pG + µG .  Corollary C.2.4. A connected graph G has at most g G + 1 vertices. Proposition C.2.5. Let G be a graph with g ≥ 1 edges and p vertices. Then the following properties are equivalent. (i) G is a tree. (ii) The graph does not have loops and for every two distinct vertices v and w of G there is a unique path from v to w. (iii) G is connected and g = p − 1. (iv) G is connected and g < p. Proof. Corollary C.2.4 states that p − 1 ≤ g for a connected graph, and therefore (iii) and (iv) are equivalent.

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313

In the notation of Lemma C.2.3, (i) means κG = 1 and µG = 0, whereas (iii) means κG = 1 and g G = pG − 1. In view of Lemma C.2.3 we conclude that (i) and (iii) are equivalent. Now assume that (i) holds. Since trees do not have cycles, they do not have loops. Since a tree is connected, each two distinct vertices can be connected by a path. If there would be distinct vertices v and w in G which are connected by two different paths, then taking one path from v to w followed by the other path from w to v we would obtain a closed walk with m(W ) 6= 0. But in view of Corollary C.2.2 this is impossible since G has no cycles. Hence (ii) has been shown. Now assume that (ii) holds. Then G has no loops, and G is connected since each two distinct vertices can be connected by a path. If G would contain a cycle with at least two edges, then it has also at least two distinct vertices, and the cycle could be split to connect these vertices by two distinct paths, contradicting (ii). Therefore (i) is true.  Lemma C.2.6. Let T be a tree and let v0 be an arbitrarily chosen fixed vertex of T , called the root of T . For each v ∈ V T \ {v0 } let Pv be the oriented path from v to v0 . Then the following statements hold. 1. Each edge of T has a unique direction such that the direction of each edge in Pv (v ∈ V T \ {v0 }) coincides with its direction in T . 2. There is an injective function wT : V T → R such that wT (v− (e)) > wT (v+ (e)) for each edge e of T . 3. I − (v0 ) = ∅ and #I − (v) = 1 for each v ∈ V T \ {v0 }. Proof. By Proposition C.2.5, the paths Pv are well-defined. For v ∈ V T \ {v0 }, the number of edges in Pv will be denoted by δ(v), and for convenience we put δ(v0 ) = 0. Let wT : V T → [0, ∞) be an injective function such that wT (u) < wT (v) whenever u, v ∈ V T satisfy δ(u) < δ(v). For example, we may choose wT such that wT (u) ∈ [δ(u), δ(u) + 1) with wT (u) 6= wT (v) when u 6= v and δ(u) = δ(v). Let e be an edge of T and let u and v be its ends. Since wT (u) 6= wT (v), we may label the ends in such a way that wT (u) < wT (v). Directing e in such a way that v− (e) = v and v+ (e) = u it follows that statement 2 holds. If e is an edge which is incident with v0 , then v0 = v+ (e) by statement 2 since wT (v0 ) is the smallest value of the function wT in view of δ(v0 ) = 0 < δ(v) for all v ∈ V T \ {v0 }. Hence each edge which is incident with the root is an incoming edge, which means that I − (v0 ) = ∅. Now let v ∈ V T \ {v0 }. Then the first edge of Pv is outgoing from v, which shows that #I − (v) ≥ 1. For any edge e which is outgoing from v consider the walk W = ((v = v− (e), e, v+ (e)), Pv+ (e) ). If W were not a path, then v− (e) would be a vertex of Pv+ , and therefore Pv− (e) ⊂ Pv+ (e) . This would imply that δ(v− (e)) < δ(v+ (e)) and therefore the contradiction wT (v− (e)) < wT (v+ (e)) would follow. Hence W is a path and therefore W = Pv , which means that e is uniquely determined as the first edge of Pv . This proves statement 3.

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To prove statement 1 let v ∈ V T \ {v0 } and let e be an edge of Pv . Then there are vertices v1 , v2 ∈ V T such that Pv = (v, . . . , v1 , e, v2 , . . . , v0 ), where v = v1 and v2 = v0 are allowed. The edge e is directed from v1 to v2 in Pv . On the other hand, it is clear from the representation of Pv that δ(v1 ) = 1 + δ(v2 ), and therefore wT (v1 ) > wT (v2 ), which means that e is also directed from v1 to v2 in T . Finally, if e is any edge of T , then it is the unique outgoing edge from v− (e), and in the previous paragraph we have shown that e is an edge of Pv− (e) . Therefore the direction of e is uniquely determined by its direction in Pv− (e) . 

C.3

Cyclically connected graphs and subgraphs

Lemma C.3.1. Let C1 and C2 be two cycles in a graph G with #(V C1 ∩ V C2 ) ≥ 2 and assume there are v ∈ V C1 \ V C2 and w ∈ V C2 \ V C1 . Then there is a cycle C in G with E C ⊂ E C1 ∪ E C2 and v, w ∈ V C . Proof. Choosing an orientation of C1 , we can write C1 = (v, . . . , v1 , . . . , v2 , . . . , v), where v1 is the first vertex of C1 which is also a vertex of C2 and v2 is the last vertex of C1 which is also a vertex of C2 , that is, there are vertices v1 6= v2 in C2 such that C1 = (P1,1 , P1,2 , P1,3 ), with chains P1,1 = (v, . . . , v1 ), P1,2 = (v1 , . . . , v2 ), P1,3 = (v2 , . . . , v) such that the only vertex of P1,1 which is also a vertex of C2 is v1 and such that the only vertex of P1,3 which is also a vertex of C2 is v2 . With a suitable orientation, we can write C2 = (P2,1 , P2,2 , P2,3 ) with P2,1 = (w, . . . , v2 ), P2,2 = (v2 , . . . , v1 ), P2,3 = (v1 , . . . , w). Then C := (P1,3 , P1,1 , P2,3 , P2,1 ) is a closed walk. By construction, P1 := (P1,3 , P1,1 ) and P2 := (P2,3 , P2,1 ) are chains and the only common vertices of P1 and P2 are v1 and v2 since all other vertices of P1 are no vertices of C2 . Therefore each vertex of C is traversed exactly once, which  means that C is a cycle. By construction, v and w are vertices of C. Lemma C.3.2. Let G0 be a cyclically connected subgraph of a graph G. 0

1. Let C be a cycle in G with V G ∩V C 6= ∅. Then G0 ∪C is cyclically connected. 0

2. Let P = (v, . . . , w) be a chain in G such that V G ∩V P = {v, w}. Then G0 ∪P is cyclically connected. 0

Proof. 1. Let v ∈ V G ∩ V C . Then v is cyclically connected to each vertex in G0 \ {v} and to each vertex in C \ {v}. Therefore G0 ∪ C is cyclically connected. 2. Since G0 is connected, there is a chain P 0 = (w, . . . , v) in G0 . By assump0 0 tion, V P ∩ V P ⊂ V G ∩ V P = {v, w}. Therefore (P, P 0 ) is a cycle which satisfies G0 (P,P 0 ) V ∩V 6= ∅. By statement 1, G0 ∪P = G0 ∪(P, P 0 ) is cyclically connected.  Definition C.3.3. Let v and w be two cyclically connected vertices of a graph G. A sequence (C1 , . . . , Ck ) (k ≥ 1) of cycles in G is called a minimal sequence of cycles for (v, w) if (i) v ∈ C1 , w ∈ Ck ;

C.3. Cyclically connected graphs and subgraphs

315

(ii) if k ≥ 2, then V Cj ∩ V Cj+1 6= ∅ (j = 1, . . . , k − 1); (iii) the number k is minimal amongst all sequences of cycles having the properties (i) and (ii). By definition of cyclically connected, the conditions (i) and (ii) in Definition C.3.3 are satisfied for at least one sequence of cycles. Hence there is at least one minimal sequence of cycles for each cyclically connected pair of vertices. Lemma C.3.4. Let (C1 , . . . , Ck ) (k ≥ 2) be a minimal sequence of cycles for a cyclically connected pair (v, w) of vertices in a graph G. Then: 1. v 6∈ C2 and w 6∈ Ck−1 ; 2. V Cr ∩ V Cs = ∅ for r, s = 1, . . . , k with r + 1 ≤ s − 1; 3. #(V Cj ∩ V Cj+1 ) = 1 (j = 1, . . . , k − 1). Proof. 1. Assume that v ∈ C2 . Then the sequence of cycles (C2 , . . . , Ck ) would satisfy properties (i) and (ii) of Definition C.3.3, contradicting the minimality of k. Hence v 6∈ C2 has been shown. Similarly one proves w 6∈ Ck−1 . Assume that statement 2 is false. Then there would be numbers 1 ≤ r < s ≤ k with r + 1 ≤ s − 1 and V Cr ∩ V Cs 6= ∅, and we could remove all cycles Cj with r + 1 ≤ j ≤ s − 1. Hence the sequence of cycles (C1 , . . . , Cr , Cs , . . . , Ck ) would satisfy properties (i) and (ii) of Definition C.3.3, contradicting the minimality of k. Therefore statement 2 is true. Assume that statement 3 is false. Then there is some j ∈ {1, . . . , k − 1} such that Cj and Cj+1 have more that one common vertex. Put v0 = v and vk = w. By property (ii) of Definition C.3.3 we can choose vr ∈ V Cr ∩ V Cr+1 for r = 1, . . . , k − 1. From statements 1 and 2 we know that vj−1 6∈ V Cj+1 and vj+1 6∈ V Cj . An application of Lemma C.3.1 would show that we can replace Cj and Cj+1 with a single cycle still having vertices vj−1 and vj+1 . But this would contradict the minimality of k.  Lemma C.3.5. Let v and w be two cyclically connected vertices of a graph G. Then there is a closed trail W in G which has the following properties. There are a positive integer k, mutually distinct vertices v0 = v, v1 , . . . , vk = w and chains P1,j = (vj−1 , . . . , vj ), P2,j = (vj , . . . , vj−1 ) (j = 1, . . . , k) such that P1 = (P1,1 , . . . , P1,k ) and P2 = (P2,k , . . . , P2,1 ) are chains with V P1 ∩V P2 = {v0 , . . . , vk } and such that W = (P1 , P2 ). Proof. Let (C1 , . . . , Ck ) (k ≥ 1) be a minimal sequence of cycles for (v, w). When k = 1, both v and w are vertices of the cycle C1 , which can be written in the form C1 = (P1 , P2 ) with chains P1 = (v, . . . , w) and P2 = (w, . . . , v) where V P1 ∩ V P2 = {v, w}. Hence W = C1 is a trail of the requested form with k = 1. Now let k > 1. We put v0 := v, vk := w and let vj ∈ V Cj ∩ V Cj+1 for j = 1, . . . , k − 1. It is clear from Lemma C.3.4 that these k + 1 vertices are mutually distinct and uniquely determined. Due to the minimality of k, none of

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Appendix C. Graphs

the cycles is a loop. Therefore each edge of the graph G occurs at most once in C1 , . . . , Ck . Writing Cj = (P1,j , P2,j ) (j = 1, . . . , k), where P1,j is a chain from vj−1 to vj and P2,j is a chain from vj to vj−1 , it is clear that no two distinct chains  have a common edge. In view of Lemma C.3.4 the proof is complete. Proposition C.3.6. Let G be a graph with at least two vertices and let v and w be two distinct vertices of G. Then v and w are cyclically connected if and only if there is a closed trail in G which traverses both v and w. Proof. If v and w are cyclically connected, then Lemma C.3.5 shows that there is a closed trail in G which traverses both v and w. Conversely, let W be a closed trail in G such that v and w are vertices of W . We will prove that v and w are cyclically connected by induction on the number of edges in W . The minimal number of edges in a closed trail with at least two vertices is 2, and in this case the closed trail is a cycle. Now assume that W has more than 2 edges. If W is a cycle, nothing has to be shown. Therefore assume now that W is not a cycle. Then there is a vertex v0 such that W = (W1 , W2 ) and W1 and W2 are closed trails starting and ending at v0 . If v and w are vertices of the same subtrail, say W1 , then v and w are cyclically connected by induction hypothesis since W1 has fewer edges than W . If, on the other hand, v and w do not belong to the same subtrail, then the induction hypothesis shows that v and v0 as well as v0 and w are cyclically connected since both W1 and W2 have fewer edges that W . Therefore also v and w are cyclically connected.  Lemma C.3.7. Each cyclically connected graph with cyclomatic number 1 is a cycle. Proof. For a cyclically connected graph, its matrix of cycles M contains a nonzero entry in each column. But the rank of M is the cyclomatic number of the graph and therefore equal to 1. Hence M has exactly one row (see the discussion after Definition C.1.9), and all entries of this row are different from zero. But this means that the only cycle in the graph traverses all edges, that is, the graph is a cycle. 

C.4 Maximally connected subgraphs and quasi-trees For each vertex v of G let C(v) be the set of all w ∈ V G for which there is a closed trail W such that v and w are vertices of W . Obviously, if C(v) 6= ∅, then v ∈ C(v). For any two distinct vertices v and w of G it is clear by Proposition C.3.6 that C(v) = C(w) if v and w are cyclically connected and that C(v) ∩ C(w) = ∅ if v and w are not cyclically connected. For each vertex v of G let Gv be the subgraph of G consisting of the vertices in C(v) and those edges of G for which both ends are vertices in C(v). Lemma C.4.1. 1. Let v be a vertex of the graph G for which Gv is not the null graph. Then Gv is a maximal cyclically connected subgraph of G.

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317

2. Each cyclically connected subgraph of G is a subgraph of a maximal cyclically connected subgraph of G. Proof. 1. If C(v) = {v}, then all edges in Gv are loops, and therefore Gv is clearly cyclically connected. Now consider the case that #C(v) ≥ 2 and let w ∈ C(v) with w 6= v. Let (C1 , . . . , Ck ) be a minimal sequence of cycles for the pair (v, w). Then it is clear that each vertex in each of these cycles is cyclically connected to v in G. By definition of Gv it follows that with all vertices of these cycles being in Gv also all edges of these cycles are in Gv . Therefore v and w are cyclically connected in Gv . This proves that the graph Gv is cyclically connected. Let G0 be a cyclically connected subgraph of G with Gv ⊂ G0 . Since no vertex in Gv is cyclically connected in G to a vertex in G which does not belong to C(v), the graphs Gv and G0 have the same vertices. But then edges in G0 are edges in G with both ends being vertices in C(v). It follows that edges in G0 are also edges in Gv . Therefore G0 = Gv , which shows that Gv is maximal. 2. Let G1 be a cyclically connected subgraph of G and let v be a vertex in G1 . Then the vertices of G1 belong to C(v), and therefore also the edges of G1  belong to Gv . This proves G1 ⊂ Gv . Lemma C.4.2. Let G be a quasi-tree and let G0 be a maximal cyclically connected subgraph of G. Let v0 , . . . , vn (n ≥ 1) be an arbitrary indexing of the common vertices of G0 and its complement in G. The remaining vertices of G0 (if any) are indexed as vn+1 , . . . , vr in an arbitrary manner. Let G0 be the null graph and put w(v0 ) = 0. Then there is a sequence (Gj )∞ j=1 of cyclically connected subgraphs of G0 and an injective function   ∞ [ V Gj  \ {v0 } → (0, ∞) w: j=1

with the following properties: (i) Gj−1 ⊂ Gj (j ∈ N); (ii) if k ∈ {1, . . . , r} and j ∈ N such that vk ∈ V Gj \ V Gj−1 , then there is a number i ∈ {0, . . . , r} such that vi ∈ V Gj , w(vk ) > w(vi ) and such that there is an edge in Gj which is incident with vk and vi ; (iii) if k ∈ {n + 1, . . . , r} and j ∈ N such that vk ∈ V Gj \ V Gj−1 , then there is i ∈ {0, . . . , r} such that vi ∈ V Gj , w(vk ) < w(vi ) and such that there is an edge in Gj which is incident with vk and vi . Proof. For j ∈ N, the subgraphs Gj and the function w on Gj \ Gj−1 satisfying (i), (ii) and (iii) will be constructed recursively. First let j = 1. In view of Lemma (1) (1) (1) C.3.5 and Proposition C.3.6 there are two chains P1 = (v0 = v1,1 , . . . , v1,r1 = v1 ) (1)

and P2

(1)

(1)

= (v0 = v2,1 , . . . , v2,s1 = v1 ) and two increasing sequences of integers (1)

(1)

k such that v1,t(1,ι) = v2,t(2,ι) (ι = 0, . . . , k) as well as t((1, ι))kι=0 and t((2, ι))ι=0

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(1)

(1)

(1)

(1)

(1)−

V P1 ∩ V P2 = {v1,t(1,ι) : ι = 0, . . . , k}, and such that (P1 , P2 ) is a closed trail. Therefore also G1 := G| P (1) P (1) is cyclically connected. Positive weights V

1

∪V (1)

2

(1)

will be attributed to the vertices v1,ι (ι = 1, . . . , r1 ) of P1 (1)

(1)

such that

(1)

w(v1,1 ) < w(v1,2 ) < · · · < w(v1,r1 ) = w(v1 ). For i = 1, . . . , k with t(2, i − 1) + 1 < t(2, i) and ι ∈ N with t(2, i − 1) < ι < t(2, i), (1) weights for v2,ι will be chosen such that that (1)

(1)

(1)

w(v2,t(2,i−1) ) < w(v2,t(2,i−1)+1 ) < · · · < w(v2,t(2,i) ) and such that these weights are distinct from the already chosen weights. That is, (1) (1) the weights of the vertices of G1 are chosen in such a way that w(v1,ι ) and w(v2,ι ) are increasing with ι and such that w is injective on G1 . Clearly, (i) is satisfied for j = 1. For k ∈ {1, . . . , r} with vk ∈ V G1 there is (1) (1) ι ∈ {1, . . . , r1 } such that vk = v1,ι or there is ι ∈ {1, . . . , s1 } such that vk = v2,ι . (1)

(1)

With i = v1,ι−1 or i = v2,ι−1 , respectively, it is now clear that (ii) holds for j = 1. Similarly, for k ∈ {n + 1, . . . , r} with vk ∈ V G1 there is ι ∈ {1, . . . , r1 − 1} such (1) (1) (1) that vk = v1,ι or there is ι ∈ {1, . . . , s1 − 1} such that vk = v2,ι . With i = v1,ι+1 (1)

or i = v2,ι+1 , respectively, it is now clear that (iii) holds for j = 1. Now assume that G1 , . . . , Gj−1 have been constructed for some j > 1. We consider three cases. The first case is that there exists l ∈ {2, . . . , n} such that vl is not a vertex of Gj−1 . As for j = 1 earlier in this proof we conclude from Lemma C.3.5 and (j) (j) (j) Proposition C.3.6 that there are two chains P1 = (v0 = v1,1 , . . . , v1,rj = vl ) and (j)

P2

(j)

(j)

= (v0 = v2,1 , . . . , v2,sj = vl ) in G0 without common edges and two increasing (j)

(j)

k k and t((2, ι))ι=0 sequences of integers t((1, ι))ι=0 such that v1,t(1,ι) = v2,t(2,ι) (ι = (j)

0, . . . , k) and V P1

(j)

∩ V P2

(j)

= {v1,t(1,ι) : ι = 0, . . . , k}. Since v0 is a vertex of (j)

Gj−1 but vl is not, the numbers rej := max{ι ∈ {0, . . . , rj } : v1,ι ∈ V Gj−1 } as

(j) well as sej := max{ι ∈ {0, . . . , sj } : v2,ι ∈ V Gj−1 } are well-defined and satisfy (j) (j) (j) (j) (j) (j) rej < rj and sej < sj . Then Pe1 = (v1,erj , . . . , v1,rj ) and Pe2 = (v2,esj , . . . , v2,sj ) (j) (j) are subchains of P1 and P2 , respectively. The first vertex of each of these two

chains belongs to Gj−1 , whereas all their other vertices do not belong to Gj−1 . k be the smallest index ι ∈ {1, . . . , k} such Define Gj := G0 | Gj−1 Pe (j) Pe (j) . Let e V

∪V

1

∪V

2

(j) k, . . . , k} we assign weights that t(1, ι) > rej and t(2, ι) > sej . For v1,t(1,ι) with ι ∈ {e (j)

(j)

max{w(v1,erj ), w(v2,esj )} < w(v

(j) ) k) 1,t(1,e

(j)

< · · · < w(v1,t(1,k) ).

With the already assigned weights we now define weights to all the remain(j) (j) (j) ing vertices in Pe1 and Pe2 such that the values w(v1,ι ) (ι = rej , . . . , rj ) and

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319

(j)

w(v2,ι ) (ι = sej , . . . , sj ) are increasing in ι. Furthermore, the values of w on Gj \ Gj−1 can and will be chosen in such a way that w is injective on Gj . If (j) (j) (j)− (j) (j) (j) v1,erj = v2,esj , then define W (j) := (Pe1 , Pe2 ). If v1,erj 6= v2,esj , then, in view of (j) (j) (j) and we define the connectedness of Gj−1 , there is a chain Pe from v to v 3

2,e sj

rj 1,e

(j) (j) (j)− W (j) := (Pe1 , Pe2 , Pe3 ). In both cases, W (j) is a closed trail connecting each vertex in Gj \ Gj−1 to a vertex in Gj−1 . Since Gj−1 is cyclically connected, it follows that also Gj is cyclically connected. With the same reasoning as for G1 one shows that the properties (ii) and (iii) holds. The second case is that {v0 , . . . , vn } ⊂ V Gj−1 and that there is a chain (j) (j) (j) (j) (j) P (j) = (v0 , . . . , vk ) with k ≥ 2 such that V Gj−1 ∩ V P = {v0 , vk }. Since (j) (j) v0 6= vk , we may choose the indexing of the vertices of this chain in such a (j) (j) (j) way that w(v0 ) < w(vk ). Then we choose weights w(vι ) (ι = 1, . . . , k − 1) (j) k such that (w(vι ))ι=0 is a increasing sequence and such that w is injective on Gj := G0 |V Gj−1 ∪V P (j) . It follows as above that Gj is cyclically connected and that properties (ii) and (iii) hold. The final case is that {v0 , . . . , vn } ⊂ V Gj−1 and that each chain in G0 with both ends in Gj−1 must have all vertices in Gj−1 . In this case we set Gj =  Gj−1 .

Lemma C.4.3. Let G be a quasi-tree and let G0 be a maximal cyclically connected subgraph of G. Let v0 , . . . , vn (n ≥ 1) be an arbitrary indexing of the common vertices of G0 and its complement in G. The remaining vertices of G0 (if any) are indexed as vn+1 , . . . , vr in an arbitrary manner. Then the edges of G0 can be directed and indexed such that G0 becomes a digraph with the following properties for k = 0, . . . , r: 1. I − (vk ) = ∅ if and only if k = 0; 2. I + (vk ) 6= ∅ for k = n + 1, . . . , r; 3. if I − (vk ) 6= ∅, I + (vk ) 6= ∅, j ∈ I − (vk ) and i ∈ I + (vk ), then j < i. ∞ Proof. Let a sequence (Gj )j=1 of cyclically connected subgraphs of G0 and an injective function w be chosen according to Lemma C.4.2. Since G0 is a finite graph, there is an index j0 such that G00 := Gj0 = Gj for all j > j0 . Assume that 00 00 0 G00 6= G0 . For u ∈ V G \ V G and v ∈ V G let P (u) be a chain from v to u. Writing k := max{ι ∈ {0, . . . , k} : vei ∈ G00 } is P (u) = (v = ve0 , . . . , vek = u), the number e k < k. Therefore Pe(u) = (e well-defined and satisfies e vek , . . . , vek = u) is a well-defined 00 chain whose first vertex v(P (u)) := veek belongs to V G , whereas all other vertices 0 00 of Pe(u) are in V G \ V G . Let P1 (u) and P2 (u) be two such chains (with possibly distinct v) and assume that v(P1 (u)) 6= v(P2 (u)). Then W = (Pe1 (u), Pe2 (u)− ) is a 0 00 walk from v(P1 (u)) to v(P2 (u)) with its interior vertices belonging to V G \ V G . We know that there is a chain P ⊂ W from v(P1 (u)) to v(P2 (u)). For ι = 1, 2, the 00 00 0 first edge eι,1 of Peι (u) has one end v(P1 (u)) in V G and the other end in V G \V G ,

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which implies that e1,1 6= e2,1 . Hence e1,1 and e2,1 are edges of P , which therefore must have at least one interior vertex. But by construction of the Gj , such an interior vertex would belong to Gj0 +1 , contradicting Gj0 = Gj0 +1 . Therefore the vertex v(P (u)) is uniquely determined by u, and we write v(u) := v(P (u)). Since G0 is cyclically connected, Lemma C.3.5 gives a closed trail W with vertices u and v(u) such that W = (P1 (u), P2 (u)− ), where P1 (u) and P2 (u) are f = (Pe1 (u), Pe2 (u)− ) is a chains from v(u) to u without common edges. Then W closed trail whose only vertex in G00 is v(u) and which passes through u. 0 0 00 00 Now choose any u0 ∈ V G \ V G and let Ve be the set of all u ∈ V G \ V G e := G0 | e with v(u) = v(u0 ). Then each u ∈ Ve is cyclically connected in G V ∪{v(u0 )}

e is cyclically connected. Since G is a quasi-tree, to v(u0 ), which means that G e G must therefore have at least two vertices in common with its complement in e G. Hence there is u1 ∈ Ve such that u1 is a vertex of the complement of G. e Then there is an edge e of G which is not an edge of G and which is incident e From with u1 . Let u2 be the other end of e. Then u2 6∈ Ve by definition of G. G0 G0 G00 e V ⊂ V \V ⊂ V \ {v0 , . . . , vn } it follows that u1 is not a vertex of the complement of G0 in G, which implies that e is not an edge of this complement. 0 Therefore e is an edge of G0 , which gives u2 ∈ V G . Choose a chain Pe(u1 ) from 00 0 v(u1 ) = v(u0 ) to u1 whose vertices other than v(u1 ) belong to V G \ V G and 00 consider the chain P = (Pe(u1 ), (u1 , e, u2 )). Since u2 6∈ Ve , it follows that u2 ∈ V G and P would be a chain from v(u0 ) to u2 with u1 and all other interior vertices (if 00 0 any) belonging to V G \ V G . But since G00 = Gj0 = Gj0 +1 , this contradicts the construction of Gj0 +1 in the penultimate paragraph of the proof of Lemma C.4.2. This contradiction shows that G00 = G0 . We have therefore constructed an injective function w : G0 → [0, ∞) with 0 = w(v0 ) < w(vi ) for i = 1, . . . , r. Recall that a quasi-tree does not have loops. Hence for each edge e of G0 there are unique indices τe,1 and τe,2 in {0, . . . , r} such that the ends of e are vτe,1 and vτe,2 and such that w(vτe,1 ) < w(vτe,2 ). Then we choose an indexing of the edges with respect to the lexicographic order of the 0 pairs (w(vτe,1 ), w(vτe,2 )). For definiteness, letting g 0 := #E G , we can (uniquely) index the edges of G0 as ej (j = 1, . . . , g 0 ) such that w(vτej+1 ,1 ) ≥ w(vτej ,1 ) and w(vτej+1 ,2 ) > w(vτej ,2 ) if w(vτej+1 ,1 ) = w(vτej ,1 ) (j = 1, . . . , g 0 − 1). The edges e of G0 will be directed from vτe,2 to vτe,1 . This means that v− (e) = vτe,2 and v+ (e) = vτe,1 . Finally, we are going to verify that properties 1, 2 and 3 are satisfied for this choice of the indexing of the edges of G0 . Since w(v0 ) < w(vi ) for all i ∈ {1, . . . , r}, it is clear that each edge which is incident with v0 is an incoming edge, which shows that I − (v0 ) = ∅. On the other hand, let k ∈ {1, . . . , r}. Since G0 is the null graph and G0 = Gj0 for some j0 ∈ N, it follows that there is j ∈ N such that vk ∈ V Gj \ V Gj−1 . By property (ii) in Lemma C.4.2 there are i ∈ {0, . . . , r} and an edge e of G0 such that vk and vi are ends of e and w(vk ) > w(vi ). In particular, the edge e is outgoing from vk , and therefore I − (vk ) 6= ∅. Similarly, statement 2

C.4. Maximally connected subgraphs and quasi-trees

321

follows from property (iii) in Lemma C.4.2. To prove statement 3, let j ∈ Ik− and i ∈ Ik+ . Then τej ,2 = k = τei ,1 . Therefore w(vej ,1 ) < w(vej ,2 ) = w(vk ) = w(vei ,1 ),  which gives j < i. Let G be a graph and let CG := {Gv : v ∈ V G , C(v) 6= ∅} be the set of maximal cyclically connected subgraphs of G. If CG 6= ∅, then the graph GC0 is defined for any nonempty subset C0 of CG by contracting each maximal cyclically connected subgraph in C0 into a vertex, i.e., writing C0 =: {Γ1 , . . . , Γr }, we identify each Γj as a new vertex vej (j = 1, . . . , r) and put     r r [ [ C0 C0 E Γj  . (C.4.1) v1 , . . . , ver }, E G = E G \ V Γj  ∪ {e V G = V G \ j=1

j=1 C0

We still have to define the incidence relationin GC0S. Let e ∈E G and let v be a r vertex of G which is incident with e. If v ∈ V G \ j=1 V Γj , then this end of e remains v in GC0 . If however v ∈ V Γj for some j ∈ {1, . . . , r}, then the end v of e becomes vej in GC0 . Since a vertex in G cannot belong to two different maximal cyclically connected subgraphs, the incidence relation is well-defined. Note that CG = {G} and GCG is an isolated vertex when G is cyclically connected. The above construction can formally be extended to the case C0 = ∅, in which case G∅ = G. Proposition C.4.4. Let G be a connected graph with at least one edge which is 6 C0 ⊂ CG . Then GC0 is a connected neither a tree nor cyclically connected. Let ∅ = C0 graph with at least one edge, G is not cyclically connected, and CG \ C0 is the set of maximal cyclically connected subgraphs of GC0 . Proof. First consider the case C0 = {Γ1 }. Assume there is a cycle in GΓ1 := G{Γ1 } v1 , ee1 , v2 , . . . , v3 , ee2 , ve1 ), which passes through ve1 . Such a cycle is represented as (e where (v2 , . . . , v3 ) is a chain in G with all vertices in V G \ V Γ1 . In G, the cycle with the same edges is a path P1 = (v1 , ee1 , v2 , . . . , v3 , ee2 , v4 ) with v1 , v4 ∈ V Γ1 . If v1 = v4 , then P1 is a cycle. If v1 6= v4 , then there is a path P2 = (v4 , . . . , v1 ) in Γ1 since Γ1 is connected. Hence (P1 , P2 ) is a cycle in G. In either case, there would be a cycle in G with and edge ee1 6∈ V Γ1 but with one of its ends v1 ∈ V Γ1 . But this is impossible since Γ1 is a maximal cyclically connected subgraph of G. Therefore ve1 is not cyclically connected to any other vertex in GΓ1 , which implies that GΓ1 is not cyclically connected. Furthermore, the cycles in GΓ1 are exactly the cycles in G which are not cycles in Γ1 , and it follows that C \ {Γ1 } is the set of maximal cyclically connected subgraphs of GΓ1 . In a similar way one Γ1 Γ1 can show that GΓ1 is connected. Alternatively, g G = g G − g Γ1 , µG = µG − µΓ1 , Γ1 Γ1 pG = pG − pΓ1 + 1, κG = 1, κΓ1 = 1 and Lemma C.2.3 give κG = 1.

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Appendix C. Graphs

Finally, since Γ1 6= G, there is v0 ∈ V G \ V Γ1 , and since GΓ1 is connected, there is a path from v0 to ve1 in GΓ1 . This shows that GΓ1 has at least one edge. A recursive application of the case r = 1 gives the result for general r.  Corollary C.4.5. Let G be a connected graph which is not the null graph. Then GCG is not the null graph, is connected, and has no cycles. Proof. We have already noticed above that GCG = G when G has no cycles and that GCG is an isolated vertex if G is cyclically connected. In all other cases, the result follows from Proposition C.4.4 with C0 = CG .  Let CG = {Γ1 , . . . , Γr } be the set of maximal cyclically connected subgraphs of a connected graph G. For j = 1, . . . , r let Vbj be the set of all vertices of Γj which are also vertices of the complements of Γj in G and let   r r [ [ G G Γ j b  ∪ V = V \ (C.4.2) V Vbj . j=1

j=1

If G is a neither the null graph nor cyclically connected, then Vb G 6= ∅. Choosing and fixing any vertex v ∈ Vb G , G will be called a rooted graph with root v. Lemma C.4.6. The set of all edges of any rooted quasi-tree G can be oriented and indexed in such a way that for each vertex v of G the following properties hold: 1. I − (v) = ∅ if and only if v is the root; 2. I + (v) = ∅ if and only if v is a pendant vertex but not the root; 3. if j ∈ I − (v) and i ∈ I + (v), then j < i. Proof. By Corollary C.4.5, T := GCG is a tree. Let CG =: {Γ1 , . . . , Γr }. If the root of G is a vertex of some Γj0 , we set v0∗ := vej0 . Otherwise, let v0∗ be the root of G. Then T is a rooted tree with root v0∗ . The edges of T will be oriented according to Lemma C.2.6. Also let wT be as in Lemma C.2.6. For each edge e in T let (wT (v− (e)), 0) be the index of e. In view of statements 2 and 3 in Lemma C.2.6, indices of distinct edges are distinct. For each j = 1, . . . , r, the edges of Γj are indexed and directed according to Lemma C.4.3, and if the index of such an edge of Γj is i, then we give it the index (wT (e vj ), i) as an edge of G. In this way we have given each edge in G a unique index and orientation, and the edges in G will be ordered with respect to the lexicographic ordering of their indices. Of course, the indices can now be identified in an order preserving manner with the numbers 1, . . . , g G , but for the purpose of this proof, we will use the above notation. The vertices of Γj (j = 1, . . . , r) will be indexed as vj,k (k = 0, . . . , rj ), where the indices k are as in Lemma C.4.3. Here vj,k for k = 0, . . . , nj are the common vertices of Γj and its complement. The vertices vj,0 play a particular role, and we

C.4. Maximally connected subgraphs and quasi-trees

323

are now going to identify them. If the root of G belongs to Γj , we put vj,0 := v0 . If the root of G does not belong to Γj , then vej is not the root of T , and by Lemma C.2.6 there is a unique outgoing edge from vej in T , and its entrance in G is a vertex of Γj and will be the vertex vj,0 of G. We are now in a position to prove the statements of this lemma. To this end, we partition the set of vertices of G into five classes, and we will consider each of these classes in turn. The first class is the set of all vertices vj,k with j = 1, . . . , r, k = nj +1, . . . , rj . In this case, all edges of G which are incident with such a vertex of G are also edges of Γj , and the statements follow from Lemma C.4.3 if we observe that a cyclically connected subgraph has no pendant vertices. The second class only consists of the root. In case the root belongs to some Γj , it is vj,0 . But in any case, an edge incident with the root would be an incoming edge to v0 by Lemma C.2.6 or Lemma C.4.3. The third class consists of all vertices vj,0 which are not the root. All edges which are incident with such vj,0 and which are edges of Γj are incoming edges by vj ), 0), and vj ), i) > (wT (e statement 1 of Lemma C.4.3, their indices satisfy (wT (e there is at least one such edge. Any remaining incoming edges would be edges of vj ), 0). By choice of vj,0 , T and would therefore have indices (wT (v ∗ ), 0) > (wT (e the only outgoing edge from it is the unique outgoing edge from vej in T , which vj ), 0). Hence the statements of the lemma also hold for this class has index (wT (e of vertices. The fourth class consists of all vertices vj,k with j = 1, . . . , r and k = 1, . . . , nj . By statement 1 of Lemma C.4.3 each such vertex has at least one outgoing edge in Γj . Furthermore, it follows from Lemma C.4.3 that there is a positive vj ), ι) of any edge of Γj which is incomreal number ij,k such that the index (wT (e vj ), ij,k ), whereas the index (wT (e ing into vj,k is larger than (wT (e vj ), ι) of any edge of Γj which is outgoing from vj,k is smaller than (wT (e vj ), ij,k ). As we have seen above when choosing vj,0 , there are no edges of T outgoing from vj,k . But since vj,k has at least one incident edge in T , vj,k has at least one incoming edge, and each incoming edge in T has index (wT (v ∗ ), 0) with wT (v ∗ ) > wT (e vj ), which implies vj ), ij,k ). This shows that the statement that its index satisfies (wT (v ∗ ), 0) > wT (e of the lemma also holds for the fourth class. The fifth class consists of all remaining vertices, that is, all vertices of G which are not vertices of any Γj (j = 1, . . . , r) and not the root. It is clear that all edges of G which are incident with such a vertex v are edges of T . Observe that this class includes all pendant vertices of G since a pendant vertex cannot be a vertex of any cycle. Each vertex v of T in this fifth class has exactly one outgoing vertex by Lemma C.2.6 and is therefore a pendant vertex if and only if I + (v) = ∅. Finally, statement 3 for vertices in this fifth class which are not pendant vertices follows from statement 2 of Lemma C.2.6. 

324

Appendix C. Graphs

Definition C.4.7. A connected graph with at least one edge is said to be a generalized quasi-tree if it is neither a tree nor cyclically connected and if each of its cyclically connected subgraphs with a vertex in Vb G has at least two vertices in common with the complement of this subgraph. Lemma C.4.8. The set of all edges of any rooted generalized quasi-tree G can be oriented and indexed in such a way that for each vertex v of G the following properties hold: 1. I − (v) = ∅ if and only if v is the root; 2. I + (v) = ∅ if and only if v is a pendant vertex but not the root; 3. if v is neither a pendant vertex nor the root of G, then min I − (v) < min I + (v) and max I − (v) < max I + (v). Proof. We are going to prove the statement by induction on the cyclomatic number µ. Since µG = 0 implies that G is a tree, there are no generalized quasi-trees with µG = 0, so that the statement is trivially true for µ = 0. Now let µ > 0 and assume 0 that the statement is true for all generalized quasi-trees G0 with µG < µ. Let G be a generalized quasi-tree with µG = µ. If G is a quasi-tree, then the statements of this lemma are true by Lemma C.4.6. It remains to consider the case that G is not e of G which a quasi-tree, and in this case there is a cyclically connected subgraph G e has only one vertex w in common with its complement in G. Then V G ∩ Vb G = ∅. e is a subgraph of a maximally connected subgraph Γ1 of G. By Lemma C.4.1, G e Γ1 b Observing V ∩ V G ⊃ Vb1 6= ∅ it follows that V G $ V Γ1 . We will now cut the graph G at w into two vertices to arrive at a graph G0 as follows. Each edge of G which is not incident with w will have the same ends in G0 . Since w 6∈ Vb G , it is clear that w 6∈ Vb1 and therefore each edge of G incident e is cyclically connected, there is a closed with w must be an edge of Γ1 . Since G e trail (w, ee1 , . . . , ee2 , w) in G, possibly (w, ee1 , w), which will be identified with a trail e Pe = (w1 , ee1 , . . . , ee2 , w2 ) in G0 , thus determining the ends of ee1 and ee2 in G0 . If G 0 has any other edges with end w, this end will become one of w1 or w2 in G (this e choice is arbitrary, for definiteness we may choose w1 ). Now let v1 ∈ V Γ1 \ V G . Since Γ1 is cyclically connected, there is a closed trail P = (w, eb1 , . . . , v1 , . . . eb2 , w) e then there would be a subtrail in in Γ1 . If this trail would traverse an edge of G, e e and G. b After removing G entering and exiting at the only common vertex w of G b any such closed subtrail of P we may therefore assume that P is a trail in G. b Since P is a closed trail with v1 6= w, it follows that eb1 and eb2 are edges of G with eb1 6= eb2 . Then we identify P with the trail Pb = (w1 , eb1 , . . . , v1 , . . . , eb2 , w2 ) in G0 , b has any other edges with end thus determining the ends of eb1 and eb2 in G0 . If G 0 w, this end will become one of w1 or w2 in G . This completes the construction of G0 . To show that G0 is connected choose v ∈ G0 \ {w1 , w2 }. Since G is connected, there is a path from v to w in G0 , which corresponds to a path from v to w1 or from v to w2 in G0 . But we have already seen that there is a trail Pe from w1 to w2

C.4. Maximally connected subgraphs and quasi-trees 0

325 0

in G0 . Therefore G0 is indeed connected. Since g G = g G , pG = pG + 1 and both 0 G and G0 are connected, it follows from Lemma C.2.3 that µG = µG − 1. Next we are going to show that G0 is a generalized quasi-tree. To this end let 0 Γ1 be the subgraph of G0 which is generated by all edges of Γ1 , that is, Γ10 is the graph obtained from Γ1 by cutting the vertex w in the manner described above. Let Γ2 , . . . , Γr be the other maximal cyclically connected subgraphs of G, if any. We are going to show that Γ01 , Γ2 , . . . , Γr are the maximal cyclically connected subgraphs of G0 . Let P be a closed trail in G0 . If P traverses w1 or w2 , we replace all occurrences of w1 and w2 with w to arrive at a closed trail P0 in G. But this trail must lie in one of Γ1 , . . . , Γr . Since for j = 2, . . . r any closed trail in Γj is the same for G and G0 , it is clear that Γ2 , . . . , Γr are maximal cyclically connected in G0 . Furthermore, any other maximal cyclically connected subgraph of G0 must be a subgraph of Γ10 . Finally, we are going to show that each vertex v ∈ Γ01 \ {w1 } is cyclically connected to w1 , which would show that Γ10 is maximal cyclically b respectively, they e and G, connected. Since Pe and Pb are trails from w1 to w2 in G − e b have no common edges. Hence (P , P ) is a closed trail, and therefore w1 and w2 are cyclically connected. It remains to show that each vertex v of Γ10 which is different from w1 and w2 is cyclically connected to one of w1 and w2 . If v is e choose a closed trail through v and w. This becomes a trail P1 a vertex in G, from w3 to w4 in Γ01 with w3 , w4 ∈ {w1 , w2 }. If w3 = w4 , then P1 is a closed trail through v and w1 . If w3 6= w4 , we may assume that w3 = w2 and w4 = w1 . In this case, the trail (P1 , Pb) is a closed trail through v, w1 and w2 . The proof when v is e is similar. Since w 6∈ Vb G it is clear that Vb G0 = Vb G . not a vertex of G By induction hypothesis, we may choose an indexing of the edges of the 0 generalized quasi-tree G0 with µG = µ − 1 such that the statements of this lemma hold for G0 . Hence the statements also hold for all vertices v of G which are different from w. From min I − (wj ) < min I + (wj ) and max I − (wj ) < max I + (wj ) (j = 1, 2) it follows that min I − (w) = min(I − (w1 ) ∪ I − (w2 )) < min(I + (w1 ) ∪ I + (w2 )) = min I + (w), max I − (w) = max(I − (w1 ) ∪ I − (w2 )) < max(I + (w1 ) ∪ I + (w2 )) = max I + (w). Therefore property 3 also holds for w.



Let G be a digraph and denote by V × the set of all vertices v ∈ V int such that I − (v) 6= ∅ and I + (v) 6= ∅ and denote by V0× the set of all v ∈ V × satisfying min I − (v) < min I + (v) and max I − (v) < max I + (v). Lemma C.4.9. Let G be a cyclically connected graph and let w be a vertex of G. Then the edges of G can be oriented and indexed in such a way that v ∈ V0× for all v ∈ V G \ {w}.

326

Appendix C. Graphs e w

v1

v0 w1

w2

w0

Figure C.1

Proof. Let e be any edge of G with is incident with w. We denote the other vertex e by adding of e by w2 , where w2 = w is allowed. We augment G to a graph G two vertices v0 and v1 in the interior of the edge e and then two new vertices w0 and w1 and edges from v0 to w0 and v1 to w1 , see Figure C.1. It is clear that G e augmented by v0 and v1 is the only maximal cyclically connected subgraph of G e G e and that Vb = {v0 , v1 , w0 , w1 }. Clearly, if a cyclically connected subgraph of G e is a generalized quasi-tree traverses one of v0 or v1 , then it traverses both. Hence G e can be indexed and oriented in with root at v0 . By Lemma C.4.8 the edges of G e e × G G e b such a way that v ∈ V0 for all v ∈ V \ V . In G, the edge e will be given the index and orientation of the edge from w2 to v0 (which is directed from w2 to v0 since v0 has no outgoing edges by Lemma C.4.8, and therefore e is directed from w2 to w.) All other edges of G will be given the indices and orientations they have e For each v ∈ V G \ {w} it follows that I − (v) = Ie− (v) and I + (v) = Ie+ (v) and in G. therefore v ∈ V0× .  We have defined several special classes of connected graphs, namely trees, cyclically connected graphs, quasi-trees and generalized quasi-trees. Clearly, there are trees and cyclically connected graphs, and these two classes of graphs are disjoint. In the next proposition we will show that the classes of quasi-trees and generalized quasi-trees cannot be expressed in terms of the other classes. Recall that it is clear by definition that generalized quasi-trees are neither trees nor cyclically connected and that each quasi-tree is a generalized quasi-tree. Proposition C.4.10.

1. There are quasi-trees.

2. There are generalized quasi-trees which are not quasi-trees. 3. There are connected graphs which are neither trees nor cyclically connected nor generalized quasi-trees. Proof. We are going to present a simple example for each of the three statements. In each of these cases we will denote the graph by G. The edges of G will be denoted by numbers, and its vertices by bold numbers. For a subset E 0 of the set E of the edges of G, we let G(E 0 ) denote the subgraph of G whose edges are the edges in E 0 and whose vertices are the ends of the edges in the set E 0 .

C.4. Maximally connected subgraphs and quasi-trees

327

1. The graph G depicted below is a quasi-tree. 3 2 1

1

2

5 5

4 3

7

6

6 4

Indeed, the cyclically connected subgraphs if G are Gj = G(Ej ) (j = 1, . . . , 4) with E1 = {2, 3, 4}, E2 = {4, 5, 6}, E3 = {2, 3, 5, 6}, E4 = {2, 3, 4, 5, 6}. It is easy to see that the common vertices of Gj and its complement in G are {2, 3, 4} for j = 1, {3, 4, 5} for j = 2, {2, 3, 4, 5} for j = 3, and {2, 5} for j = 4. Since each of these sets contains at least two vertices, the graph G is a quasi-tree. 2. The graph G depicted below is a generalized quasi-tree but not a quasi-tree. 1

1

3

6 4 5

3 2

10

8 11 7

5

2 4

12

6 9

8

9 13 7

Indeed, the graph G(E1 ) with E1 = {6, . . . , 13} is a cyclically connected graph, but 5 is the only common vertex of G(E1 ) and its complement in G. Hence G is not a quasi-tree. On the other hand, G(E2 ) with E2 = {3, . . . , 13} is the only maximal cyclically connected subgraph of G, and 3 and 4 are the common vertices of G(E2 ) and its complement in G. Therefore, Vb = {1, 2, 3, 4}. But any cyclically connected subgraph G0 of G which has one of these vertices must contain the edge 3, and hence the vertices 3 and 4 are vertices of G0 . This shows that G is a generalized quasi-tree. 3. The graph G depicted below is not a quasi-tree. 3 1

1 2

6 5 5 6

2 4 3 4

11

7 9

9 12 8

13 10 14

8

7

Indeed, the subgraph G(E1 ) of the graph G with E1 = {2, . . . , 14} is the only maximal cyclically connected subgraph of G. Clearly, Vb = {1, 2} and 2 is the only vertex of G(E1 ) in Vb .  In the above proposition we have chosen the examples in such a way that the graphs are simple plane graphs with interior vertices v satisfying d(v) ≥ 3.

328

Appendix C. Graphs

If these assumptions are relaxed, then even simpler graphs can be found for the three statements of Proposition C.4.10. For example, the graphs

are, in turn, a quasi-tree, a generalized quasi-tree which is not a quasi-tree, and a graph which is neither a tree nor a cyclically connected graph nor a generalized quasi-tree.

C.5

Adjacency matrices and incidence matrices

In this section, G is a simple graph without isolated vertices. Therefore p ≥ 2 and g ≥ 1. Two vertices of G are called adjacent if they are the two ends of an edge of G. Let v1 , . . . , vp be an arbitrary but fixed indexing of the vertices of G. The adjacency matrix A = (aj,k )pj,k=1 is defined by aj,k = 1 if there is an edge in G whose ends are vj and vk and aj,k = 0 otherwise (j, k = 1, . . . , p). We observe that p p P P dj := d(vj ) = aj,k = ak,j (j = 1, . . . , p). Note that A is a symmetric matrix. k=1

k=1

Conversely, it is easy to see that each symmetric p × p matrix A = (aj,k )pj,k=1 with p P aj,j = 0, aj,k ∈ {0, 1} and aj,k > 0 (j, k = 1, . . . , p) is the adjacency matrix k=1

of a simple graph G without isolated vertices. Let T = diag(d1 , . . . , dp ). Then e := T − 12 AT − 12 is called the weighted adjacency matrix. A Let IA := {(j, k) : 1 ≤ j < k ≤ p, ak,j = 1} and let ε ∈ {−1, 1} and εj,k , εk,j ∈ {−1, 1} with εj,k εk,j = ε for (j, k) ∈ IA . Clearly, #IA = g. Define the p × g matrix Uε by   εj,k if j = i, (j, k) ∈ IA ui,(j,k) = εk,j if k = i, (j, k) ∈ IA (C.5.1)   0 otherwise. Up to the order of the enumeration of IA , the matrix Uε is well-defined. For ε = 1, the matrix U1 is called the incidence matrix of G. For ε = −1, the matrix Uε = U−1 can be considered as the signed incidence matrix for a digraph structure on G, where εj,k < 0 means that the corresponding edge is directed from vj to vk . A permutation of the indices of the vertices of G is equivalent to a permutation of the rows of Uε and a simultaneous permutation of the rows and columns e Since all results in this section are invariant under these permutaof A, T and A. tions, we may assume particular enumerations of the vertices and edges of G if it is convenient to do so. Proposition C.5.1. For ε ∈ {−1, 1}, the identities Uε Uε∗ = T + εA hold.

C.5. Adjacency matrices and incidence matrices

329

Proof. Since the matrix Uε Uε∗ is symmetric, we only have to evaluate the entries X (Uε Uε∗ )r,s = ur,(j,k) us,(j,k) (j,k)∈IA

with 1 ≤ r ≤ s ≤ p. For r = s it is clear that (Uε Uε∗ )r,s =

X

(ur,(j,k) )2 =

(j,k)∈IA

p X

ar,t = dr .

t=1

For r < s with (r, s) ∈ IA , the sum contains exactly one nonzero element, which is (Uε Uε∗ )r,s = εr,s εs,r = ε = εar,s . For r < s with (r, s) 6∈ IA , all terms in the sum are 0, which shows that (Uε Uε∗ )r,s = 0 = ar,s in this case.  Definition C.5.2. A graph G is called bipartite if V G is the disjoint union of two sets V1 and V2 such that each edge of G has one end belonging to V1 and the other end belonging to V2 . The length of a walk W in a graph G, denoted by len(W ), is the number of edges traversed in that walk and therefore equals the sum of the components of |m|(W ). Let G be a connected graph and choose any vertex v0 of G as the root of G. For each v ∈ V \ {v0 } let Distv0 (v) be the set of all numbers len(P ), where P runs through all chains from v0 to v. For convenience we put Distv0 (v0 ) = {0}. If G is a simple graph it is clear that # Distv0 (v) = 1 for all v if and only if G does not have cycles. The number distv0 (v) := min Distv0 (v) is called the distance of v from v0 . Lemma C.5.3. A graph is bipartite if and only if it has no cycle with an odd number of edges. Proof. Clearly, it suffices to consider the components of G separately. Therefore we may assume for this proof that G is a connected digraph which has a root v0 . First assume that G has no cycle with an odd number of edges. Let W be a closed walk k P m(Ck ), in G. If m(W ) 6= 0, then there are cycles C1 , . . . , Ck such that m(W ) = j=1

see Corollary C.2.2. Hence |m|(W ) = m(W ) + 2m− (W ) =

k X j=1

|m|(Ck ) − 2

k X

m− (Ck ) + 2m− (W ),

j=1

and therefore len(W ) is an even number, which is also true when m(W ) = 0. Let v 6= v0 be a vertex of G, let P1 and P2 be two chains from v0 to v and put W = (P1 , P2− ). Then W is a closed walk with len(W ) = len(P1 ) + len(P2 ) and therefore len(P1 ) − len(P2 ) = len(W ) + 2 len(P2 ) is even. Hence the set Distv0 (v) consists either of odd numbers or of even numbers. Define G1 to be the set of all

330

Appendix C. Graphs

v ∈ G for which Distv0 (v) consists of odd numbers and G2 to be the set of all v ∈ G for which Distv0 (v) consists of even numbers. Let e be an edge of G with ends v and w. If one of the ends is the root, say v = v0 , then v ∈ G2 , whereas len(v0 , e, w) = 1 shows that w ∈ G1 . Now let both v and w be distinct from v0 and choose chains Pv and Pw from v0 to v and w, respectively. Then W = (Pv , (v, e, w), Pw− ) is a closed walk, and therefore len(Pv ) + len(Pw ) = len(W ) − 1 is an odd number, which implies that one end of e belongs to G1 , whereas the other end belongs to G2 . This shows that G is bipartite. Now let G have a cycle with an odd number of edges and assume that G is bipartite. Traversing a cycle with an odd number of edges starting at a vertex, say, in G1 , the traversed vertices must be alternatively in G1 and G2 , and the terminal vertex must therefore be in G2 . But since the terminal vertex is the starting vertex,  this is impossible. e lie in the interval [−1, 1] and 1 is Lemma C.5.4. All eigenvalues of the matrix A e an eigenvalue of A. If G is bipartite, then the eigenvalues, with multiplicities, are e symmetric with respect to 0. If g < p, then −1 is an eigenvalue of A. e is real and symmetric, its eigenvalues are real. The Proof. Since the matrix A representations of T − A and T + A in Proposition C.5.1 show that T − A ≥ 0 and e are nonnegative. This e and I + A T + A ≥ 0, and therefore the eigenvalues of I − A e lie in the interval [−1, 1]. By definition means that all eigenvalues of the matrix A of T it is clear that (T − A)(1, . . . , 1)T = 0, which proves that 1 is an eigenvalue e of A. If G is bipartite let v1 , . . . , vp1 be the edges of G1 and let vp1 +1 , . . . , vp be the edges of G2 . Then there is a p1 × (p − p1 ) matrix A1 such that   0 A1 A= , A∗1 0 e1 such that and therefore a p1 × (p − p1 ) matrix A ! e1 A 0 e= A e∗ 0 . A 1 For λ 6= 0, the Schur factorization (B.4.2) gives   −1 e e∗ − λI e1 A λ−1 A 1 e − λI = I −λ A1 A 0 I 0

0 −λI



I

e∗ −λ−1 A 1

 0 , I

e if and only if λ2 is an eigenvalue of A e∗ , e1 A which shows that λ is an eigenvalue of A 1 e with equal multiplicities. This proves that λ is an eigenvalue of A if and only if e and their multiplicities are equal. −λ is an eigenvalue of A, If g < p, then the rank of the p × g matrix U1 is at most g. It follows that

C.5. Adjacency matrices and incidence matrices

331

also the rank of T + A = U1 U1∗ is at most g, which means that the p × p matrix e in this case.  T + A is not invertible. Thus −1 is an eigenvalue of A e If G is Lemma C.5.5. Let A be irreducible. Then 1 is a simple eigenvalue of A. e and if G is not bipartite, then −1 bipartite, then −1 is a simple eigenvalue of A, e is not an eigenvalue of A. Proof. It is well-known that the rank of Uε Uε∗ = T +εA equals the rank of Uε . Since A is assumed to be irreducible, the (di)graph which describes Uε is connected. We are going to reduce the sizes of Uε by deleting rows and columns until we arrive at a case where it is easy to prove the claim. Let us agree to still denote the reduced matrix Uε by Uε (it will correspond to subgraph of the (di)graph G) and to call the number p − rank Uε the defect of Uε . First we consider the case that G has at least two edges and at least one pendant vertex. We may assume that v1 is a pendant vertex and that the unique edge ee which is incident with v1 is represented by the first column in Uε . Therefore the first row of Uε has the value 1 or −1 in the first column and the value 0 everywhere else. Since each vertex v 6= v0 is incident with at least one edge which is different from ee, all other rows have a nonzero entry outside the first column. It follows that the first row of Uε is linearly independent from the other rows. Hence removing the vertex v1 and its incident edge leads to a new connected graph G of reduced size and the same cyclomatic number, and the defect of the (signed) incidence matrix is unchanged. If the starting graph G was a tree, then all subsequent graphs are trees, and after a finite number of steps we arrive at a graph with two vertices and an edge eε , is connecting them. But in this case the final Uε , which we will denote by U eε = (e eε is 1, which shows in ε1 , εe2 )T with εe1 , εe2 ∈ {−1, 1}. Clearly, the defect of U U e in this case. this case that 1 and −1 are simple eigenvalues of A If the cyclomatic number µ is positive, then the above procedure stops at a graph with cyclomatic number µ and without pendant vertices. Choosing any cycle in G and removing one edge of the cycle, we arrive at a connected graph with cyclomatic number lowered by 1. Removal of that edge could only increase the defect of Uε . If µ ≥ 1 and the new graph has pendant vertices, we repeat the first procedure and the second procedure and so on until we have arrived at a graph with µ = 0. But then this graph is a tree, and we already know that the corresponding Uε has defect 1. Hence the original U−1 has defect at most 1. By e and it follows that the eigenvalue 1 is simple. Lemma C.5.4, 1 is an eigenvalue of A, e if the graph G is bipartite. By Lemma C.5.4, −1 is a simple eigenvalue of A It remains to consider the case ε = 1 when G is not bipartite. By Lemma C.5.3, the graph G has a cycle C1 with an odd number of edges. We follow the same procedure as above, except that we do not remove any edge of C1 and that b with cyclomatic number µ b = 1 and we stop once we have arrived at a graph G b with no pendant vertices. But then G = C1 . Indexing the vertices and edges in the order they are traversed, it follows that the corresponding incidence matrix is

332

Appendix C. Graphs  1 1  0  b1 =  U .  ..  0 0

0 1 1

0 1

0 1 0

1 1

 1 0  0  .. , .  0 1

which is a square matrix with an odd number of rows and columns. Expanding b1 is the sum of its determinant with respect to the first row shows that det U the determinant of a lower triangular matrix and the determinant of an upper b1 = 2, which triangular matrix, both with all diagonal entries being 1. Thus det U b shows that U1 and hence also U1 has defect 0. This proves that −1 is not an e when G is not bipartite. eigenvalue of A 

C.6 T -acyclic matrices Definition C.6.1. [88] Let n ∈ N, let T be a tree with n + 1 vertices v1 , . . . , vn+1 and let A = (aj,k )n+1 j,k=1 be an (n + 1) × (n + 1) matrix (with entries aj,k from some ring). Then A is called T -acyclic if aj,k = ak,j = 0 whenever j 6= k and the vertices vj , vk are not adjacent. Let T be a tree with n + 1 vertices, let A be T -acyclic matrix and let T 0 be a subgraph of T . Then AT 0 denotes the principal submatrix of A consisting of all rows and columns whose indices are indices of the vertices of T 0 . For j, k ∈ {1, . . . , n+1}, j 6= k, we denote by T (j) the subgraph obtained from T by deleting the vertex vj and all the edges incident with vj , and by Tk (j) the connected component of T (j) that has vk as a vertex. Put Nj (T ) := {k ∈ {1, . . . , n+1} : vk adjacent to vj in T }. It follows for each j ∈ {1, . . . , n + 1} that the set of the vertices of T is the disjoint union of {vj } and the sets of vertices of Tk (j) with k ∈ Nj (T ). Furthermore, each Tk (j) is either a tree or an isolated vertex. The proof of the following theorem is along the lines of the original proof. However, we are making use of properties of rational Nevanlinna functions rather than giving ad hoc proofs. Theorem C.6.2. [88] Let T be a tree with n + 1 vertices, let j ∈ {1, . . . , n + 1} and write Nj (T ) = {k1 , . . . , km }. Let h1 , . . . , hm be monic polynomials with real zeros and deg hr equal to the number of vertices of Tkr (j) (r ∈ {1, . . . , m}). Denote the zeros of the product h := h1 · · · hm , counted with multiplicity, by ν1 ≤ · · · ≤ νn and let µ1 ≤ µ2 ≤ · · · ≤ µn+1 be real numbers. Then there exists a real symmetric T -acyclic matrix A possessing the eigenvalues µ1 , . . . , µn+1 such that for each k ∈ Nj (T ) the eigenvalues of the submatrix ATk (j) are the zeros of hk if and only if µ1 ≤ ν1 ≤ µ2 ≤ ν2 ≤ · · · ≤ νn ≤ µn+1

(C.6.1)

C.6. T -acyclic matrices

333

holds. If all inequalities are strict, i.e., if µ1 < ν1 < µ2 < ν2 < · · · < νn < µn+1 .

(C.6.2)

holds, then A is irreducible. Proof. The statements of this theorem are independent of the indexing of the vertices. We may therefore choose the indices in such a way that the ordered set of vertices of the tree starts with vj followed by the vertices of Tk1 (j) and then up to those of Tkm (j). Furthermore, we assume that kr becomes the smallest index amongst all indices of vertices in Tkr (j). In other words, we assume that j = 1, that the set of the vertices of Tkr (1) is {vk : kr ≤ k < kr+1 } where we put km+1 := n+2 for convenience. To prove that (C.6.1) is necessary we observe that A1,1 =

m M

ATkr (1) .

(C.6.3)

r=1

By assumption, λ1 , . . . , λn+1 are the eigenvalues of A and µ1 , . . . , µn are the eigenvalues of A1,1 . Hence Proposition B.5.1 shows that (C.6.1) holds. Furthermore, Corollary B.5.2 implies that A is irreducible if (C.6.2) is satisfied. The sufficiency of (C.6.1) will be proved by induction on n. For n = 1, the tree T consists of two vertices and the edge connecting them, and trivially the two vertices of T are adjacent to each other. Any real symmetric matrix A chosen according to Proposition B.5.1 is therefore T -acyclic. Now let n ≥ 2 and assume that the sufficiency statement is true for trees with at most n edges. With the notation from the proof of Proposition B.5.1 we hr are (possibly constant) hr e hr and b hr (r = 1, . . . , m), where the e can write hr = b hm are νb1 , . . . , νbn0 and such h := b h1 · · · b monic polynomials such that the zeros of b h1 · · · e that the zeros of e h=e hm are τ1 , . . . , τn−n0 . Without loss of generality we may assume that the above reindexing is done in such a way that there is an integer hr is constant if and only if r > m0 . The function χ m0 , 0 ≤ m0 ≤ m, such that b defined in (B.5.3) has the representation (B.5.4) and can be written in the form χ(λ) = λ − b0 −

m0 X r=1

γr

fbr (λ) , b hr (λ)

where the fbr are (possibly constant) monic polynomials, the γr are nonzero real numbers, and the r-th summand is the sum of those summands in (B.5.4) whose poles are zeros of b hr . Since each summand in (B.5.4), including the minus sign, is a Nevanlinna function, also each summand in the above representation, including the minus sign, is a Nevanlinna function. Clearly, the degree of fbr is smaller than the degree of b hr . In view of Theorem A.2.6 we conclude that the polynomial fbr b has deg hr − 1 zeros which interlace with the zeros of b hr , that the largest zero of

334

Appendix C. Graphs

b hr hr is larger than the largest zero of fbr , if any, and that γr > 0. Putting fr = fbr e for r = 1, . . . , m0 , we consider two cases. If deg hr ≥ 2, then Tkr (1) is a tree, and by induction hypothesis there is a Tkr (1)-acyclic matrix Ar whose eigenvalues are the zeros of hr such that the eigenvalues of Ar,1,1 are the zeros of fr . Here we have to observe that we take a representation as in (C.6.3) for Tkr (1), which is split into the subgraphs (Tkr (1))κs (kr ) for certain κs ∈ {kr + 1, . . . , kr+1 − 1}. If deg hr = 1, then we define Ar to be the zero of hr . For r = m0 + 1, . . . , m let Ar be a diagonal matrix whose diagonal entries are the zeros of hr and let γr = 0. √ hl For r = 1, . . . , m, we let cr be the vector cr = ( γr δ1,l )deg l=1 . We now can define A as a block matrix,   b0 cT . . . cT m 1  c1 A1 0 0    A= . ..  . . .  .. . 0 .  cm 0 · · · Am It is clear by construction that A is T -acyclic and that the eigenvalues of ATkr (1) = Ar are the zeros of hr for r = 1, . . . , m. Finally, using the multilinearity of the determinant, it follows that det(A − λ) = (b0 − λ) det(A1,1 − λ) +

m0 X

det Br,s (λ),

r,s=1

where the matrices Br,s (λ) are obtained from (A − λ) by replacing all entries in the first row and first column by 0, except for cr and cT s . For r = 1, . . . , m0 we find that m Y det Br,r (λ) = det Cr (λ) det(At − λ) t=1 t6=r

with

 Cr (λ) =

0 cr

 cT r . Ar − λ

Hence det Cr (λ) = −γr det(Ar,1,1 − λ). For r, s = 1, . . . , m0 with r 6= s we have that m Y

det Br,s (λ) = ± det Cr,s (λ)

det(At − λ),

t=1 t6=r,s

where

cT s Cr,s (λ) = As − λ 0 

0 0 cr

 0 0 . Ar − λ

C.6. T -acyclic matrices  Since

335

 csT has more rows than columns, det Cr,s (λ) = 0. Therefore As − λ

det(A − λ) = (b0 − λ) det(A1,1 − λ) − (

m0 X

γr det(Ar,1,1 − λ)

m Y

r=1

t=1 t6=r

m0 X

)

det(Ar,1,1 − λ) = det(A1,1 − λ) b0 − λ − γr det(Ar − λ) r=1 ( ) m0 X f (λ) r = (−1)n+1 h(λ) λ − b0 − γr hr (λ) r=1

det(At − λ)

= (−1)n+1 h(λ)χ(λ) =

n+1 Y

(µn − λ).

r=1

Hence the eigenvalues of A are µ1 , . . . , µn+1 , and the proof is complete.



In [101, Theorem 1.3] the above result with the stronger condition (C.6.2) was reproved by another method: Corollary C.6.3. Let the assumptions of Theorem C.6.2 be satisfied with strict inequalities (C.6.2). Then any matrix A satisfying the statement of Theorem C.6.2 has the property that aj,k = ak,j = 0 for j, k = 1, . . . , n + 1 with j 6= k if and only if vj and vk are adjacent. Proof. In view of Theorem C.6.2 we still have to prove that aj,k 6= 0 whenever j 6= k and vj and vk are adjacent. Hence let j 6= k such that vj and vk are adjacent. Since Tk (j) is a component of T (j), it follows that ar,s = 0 for all vertices vr of Tk (j) and all vertices vs of T which do not belong to Tk (j). But A is irreducible by Theorem C.6.2, and therefore aj,k = ak,j 6= 0. 

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Index T -acyclic, 332 adjacency matrix, 224, 328 weighted, 224, 328 amplitude, 4 amplitude function, 17 balance of forces, 62 bead, 3 boundary value problem, 5 Cauer-Fry polynomial, 7 central vertex, 61 chain, 305 characteristic matrix function, 185 characteristic polynomial, 6, 153, 185, 193 characteristic value, 5 geometric multiplicity, 63 characteristic values comparison of, 13 complementary subtrees, 154, 156, 157 continued fraction, 8 continuity condition, 62 cut vertex, 197 cycle, 305 cycles fundamental set of, 309 matrix of, 308 minimal sequence of, 314 cyclomatic number, 309 damping, 109

degenerate problem, 175 digraph, 306 Dirichlet boundary condition, 10 Dirichlet condition, 10 generalized, 174 Dirichlet problem, 151, 168 Dirichlet problem (D1), 62, 65, 68, 70, 74, 77 Dirichlet problem (D2), 91, 92, 93, 95, 96, 99, 102 Dirichlet-Dirichlet problem, 4, 10, 22, 23, 26, 31, 46, 51, 53, 58, 109, 112, 113 Dirichlet-Neumann problem, 10, 22, 25–27, 31, 46, 51, 53, 109, 112, 113 displacement vector, 17 distribution of beads, 4 edge, 305 ends of an, 305 entrance, 306 exit, 306 incoming, 306 main, 91 outgoing, 306 eigenvalue, 5, 293 algebraic multiplicity, 293 geometric multiplicity, 293 normal, 295 semisimple, 295 simple, 295 eigenvector, 293

© Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4

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346 formally selfadjoint differential expression, 222 friction viscous, 109 graph, 305 bipartite, 329 complement, 307 component, 306 connected, 306 cyclically connected, 307 diamond, 218 k-regular, 224 metric, 306 rooted, 322 simple, 306 tetrahedral, 218 Hermite-Biehler polynomial, 285, 296 generalized, 289 symmetric, 286 symmetric generalized, 289 Hochstadt-Lieberman problem, 57 incidence, 305 incidence matrix, 328 incident, 305 inner product, 294 interlacing poles and zeros, 272 zeros, 13, 14 inverse problem four spectra, 30 two spectra, 22 irreducible matrix, 302 Kirchhoff condition, 62 Lagrange equation, 62 Lagrange identity, 15 Lagrange interpolation polynomial, 42, 46 lexicographic order, 174, 320 loop, 305

Index majorization, 75 weak, 256 matrix of paths, 206 matrix pencil, 293 linear, 293 quadratic, 293 Neumann boundary condition, 10 Neumann condition, 11 generalized, 174 Neumann problem, 151, 168 damped, 140 Neumann problem (D2), 93, 95, 102 Neumann problem (N1), 63, 66, 68– 70, 74, 80, 257 Neumann problem (N2), 91, 92, 96, 99 Neumann-Dirichlet problem, 4, 11, 23, 24, 27, 31 Neumann-Neumann problem, 12, 23–26, 31 Nevanlinna function rational, 270 node, 18 internal, 18, 20 null graph, 305 operator maximal, 222 oscillation theorem, 20 path, 305 permutation, 301 permutation matrix, 301 signed, 308 pole, 268 simple, 268 pole order, 268 polynomial imaginary part, 285 real part, 285 positive real function, 1 quasi-tree, 308 generalized, 324

Index rational function, 267 positive real, 282 real, 268, 274, 276 reducible matrix, 302 rigged digraph, 207 Robin boundary condition, 13 Robin-Dirichlet problem, 14, 29 Robin-Neumann problem, 14, 29 root, 149, 193 Schur complement, 298 Schur factorization, 298 spectrum, 5 star graph with root at a pedant vertex, 90 with root at the centre, 61 Stieltjes continued fractions, 2 Stieltjes function rational, 275, 276 Stieltjes string, 3 symmetric, 88 string massless, 3 strongly continuous, 292 subgraph, 306 complementary, 307 disjoint union, 308 entrance, 307

347 exit, 307 maximal cyclically connected, 307 trail, 305 transverse displacement, 4 transverse vibrations, 4 tree, 307 rooted shape characteristic, 167 vertex, 305 degree, 306 interior, 306 isolated, 306 locally clamped, 196 pendant, 306 vertices adjacent, 328 cyclically connected, 307 distance of, 329 walk, 305 closed, 305 open, 305 zero, 267, 268 multiplicity, 268 simple, 267, 268

Index of Notation C C C+ C+ C− C− C C∗ det M ej I Ir (, ) Lp (a, b) #S N N0 N (A) R Z Za

the set of complex numbers C ∪ {∞} the open upper half-plane the closed upper half-plane the open lower half-plane the closed lower half-plane the conjugate complex of a matrix C the conjugate complex transpose of a matrix C the determinant of a square matrix M the j-th unit vector in Cn the identity operator or the identity matrix the r × r identity matrix or a particular index set, depending on the context the inner product the Banach space of all equivalence classes of measurable functions f on (a, b) such that x 7→ |f (x)|p is Lebesgue integrable on (a, b) the magnitude of a set S the set of positive integers N ∪ {0}, the set of nonnegative integers the nullspace (kernel) of an operator, or a matrix, A the set of real numbers the set of integers the set of integer multiplies of the number a

Classes of functions S, 277 S0 , 278 S r , 285

SHB, 288

, 75

CG , 321

(D1), 61, 62

Adc ,

χT,v , 167

(D2), 91, 92

HB, 287 HB, 291 Lp (I), 223

SHB, 291

Special Symbols

5

C, 294

∗

C , 294

d(v), 306

© Springer Nature Switzerland AG 2020 M. Möller, V. Pivovarchik, Direct and Inverse Finite-Dimensional Spectral Problems on Graphs, Operator Theory: Advances and Applications 283, https://doi.org/10.1007/978-3-030-60484-4

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Index distv0 (v), 329 d− (v), 306 d+ (v), 306 E G , 305 f , 268 ΦD , 153 φD , 153 φD,g , 64 φg , 65 φ(j) , 65 ΦN , 153 φN , 153 φN,g , 64 (s) Φv0 , 193 (s) φv0 , 194 G0 , 193 GC , 321 g G , 306 Gv , 316 G|V1 , 306 I(v), 306 Ij+ , 149 I − (v), 306 pen IN , 205 I + (v), 306 e I(v), 193 I00 , 149 κD (j), 78 κG , 311 κN (j), 78 l, 4

349 ΛD , 74 e D , 74 Λ ΛN , 74 e N , 74 Λ len(W ), 329 Lj , 152, 185 `j , 222 Lcj , 184 Lfj , 184 m(f, z), 268 m(λ, T ), 293 µG , 309 m(W ), 309 |m|(W ), 309 m− (W ), 309 m+ (W ), 309 (N1), 61, 63 (N2), 91, 92 nD,g (z), 63 n eD,g (z), 92 ng+1 , 174 Ni , 75 nN,g (z), 63 n eN,g (z), 92 N (x, d), 283 p† , 204 pG , 311 PG,Gb , 241 pint , 207 Pk,α , 245 pk (D), 74

pk (N ), 74 ppen , 207 Pσ , 301 p? , 204 rD , 74 reD , 79 Rk (·, c), 7 Rk− (·, c), 12 rN , 74 0 rN , 80 Sk (·, c), 110 Uε , 328 U (v), 173 u(v), 173 V c , 173 VDc , 193 VNc , 193 v− (e), 306 v+ (e), 306 V G , 305 V int , 306 V pen , 306 V f , 173 VDf , 193 VNf , 193 Vb G , 322 j vin , 197 j vout , 197 V × , 325 V0× , 325 wP , 254