Zeta function of self-adjoint operators onsurfaces of revolution

Tianshi Lu1,3, Thalia Jeffres1 and Klaus Kirsten2

1Department of Mathematics and Statistics, Wichita State University, Wichita, KS67260-0033, USA2GCAP-CASPER, Mathematics Department, Baylor University, Waco, TX 76798-7328, USA

E-mail: lu@math.wichita.edu

Received 29 January 2015Accepted for publication 16 February 2015Published 20 March 2015

AbstractIn this article we analyze the zeta function for the Laplace operator on asurface of revolution. A variety of boundary conditions, separated and unse-parated, are considered. Formulas for several residues and values of the zetafunction as well as for the determinant of the Laplacian are obtained. Theanalysis is based upon contour integration techniques in combination with aWKB analysis of solutions of related initial value problems.

Keywords: zeta function, self-adjoint, contour integral, WKB

1. Introduction

Spectral zeta functions of typically Laplace-type operators are directly related to topics suchas analytic torsion [21], the heat kernel [13, 22], Casimir energies [4, 8, 20] and effectiveactions [5, 6, 9]. It is therefore very desirable to have effective analytical tools available tounderstand specific properties of zeta functions. Whereas in one-dimension closed answersare relatively easily obtained for quantities like the functional determinant, see, e.g.,[7, 10, 18], in higher dimensions the situation is much more involved. However, a contourintegral approach established in [2, 16] has been shown to be very useful as long as theLaplace-type operator separates in a suitable fashion. This approach has been used in a varietyof configurations like the generalized cone [3], the spherical suspension [11], warped productmanifolds [12] and surfaces of revolution [15].

In some detail, in [15] the Laplacian on a surface of revolution was considered withDirichlet boundary conditions imposed. Properties of the zeta function like residues, valuesand its derivative at zero were analyzed. Given that the strictly positive function

Journal of Physics A: Mathematical and Theoretical

J. Phys. A: Math. Theor. 48 (2015) 145204 (22pp) doi:10.1088/1751-8113/48/14/145204

3 Author to whom any correspondence should be addressed.

1751-8113/15/145204+22$33.00 © 2015 IOP Publishing Ltd Printed in the UK 1

∈f C x x( , )20 1 used to generate the surface of revolution is kept general, the analysis is not

based upon known eigenfunctions or known solutions of ordinary differential equations, butinstead on the asymptotic analysis of solutions of an initial value problem related to theboundary conditions imposed. As the boundary conditions change the relevant initial valueproblem changes and so does the pertinent asymptotic analysis. These changes capture howspectral zeta functions depend on the boundary conditions. This is the main subject of thecurrent article, furthermore we consider the influence of kinks on the surface of revolution onspectral properties.

The article is organized as follows. In section 2 we introduce the Laplacian on a surfaceof revolution and find implicit eigenvalue equations when separated or unseparated boundaryconditions are imposed. Furthermore, using the WKB method [1, 19], asymptotic propertiesof solutions of relevant initial value problems are determined. In section 3 we use theseproperties to analyze the spectral zeta function for a variety of separated boundary conditions,whereas in section 4 unseparated boundary conditions are considered. A particular case areperiodic boundary conditions, where as long as the function f and its derivative agree at theendpoints the surface of revolution can be thought of as a smooth torus. However, if thederivative does not agree this introduces a kink point on the torus. This leads to the discussionabout non-smooth surfaces in section 5. The conclusions point to the most important results ofthe article. In the appendix we give an independent proof that the implicit eigenvalueequations do not only capture the value of eigenvalues correctly but also their degeneracies.

2. Spectrum of the self-adjoint Sturm–Liouville equation

Let ∈f C x x( , )20 1 be a strictly positive function from x x[ , ]0 1 to . We consider the

Laplacian on the surface of revolution that is generated by revolving the graph of f around thex-axis. Using separation of variables, the resulting eigenvalue equation for the Laplacian onthis surface of revolution is [15]

λ− ′ ′ + = =+ ′

= + ′puk

pu ru p x

f

fr x f f( ) , ( )

1, ( ) 1 , (1)

2

2

2

where ∈ k is the separation constant entering from the cross-section S1. In rewritingequation (1) as a system of first order differential equations, the quantity = ′v pu isconvenient. The equivalent form of equation (1) then is

λ=

−x

u

v

p

k

pr

u

v

d

d

01

0

. (2)2

⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎛⎝

⎞⎠

We denote the fundamental solution of (2) as

λλ λλ λ

=E xu x u x

v x v x( ; )

( ; ) ( ; )

( ; ) ( ; ), (3)k

k k

k k

N D

N D

⎛⎝⎜⎜

⎞⎠⎟⎟

where the superscripts N and D stand for solutions of the initial value problem λ =u x( ; ) 0,kD

0

λ =v x( ; ) 1kD

0 and λ =u x( ; ) 1,kN

0 λ =v x( ; ) 0.kN

0 In this way λ =E x I( ; )k 0 , furthermoreλ λ= =E x E xdet ( ; ) det ( ; ) 1.k k 0

To guarantee the operator is self-adjoint, the boundary condition must be in one of twocategories [23]. The first category is the separated boundary condition

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

2

+ = + =au x bv x cu x v x( ) ( ) 0, ( ) d ( ) 0, (4)0 0 1 1

where ≠a b( , ) (0, 0) and ≠c d( , ) (0, 0). Following [17], the corresponding eigenvalues arethe zeros of the following function of λ,

λ λ λ= + = −( ) ( ) ( )( ) ( ) ( )F a bc d

E x c d E x ba

( ) det0 0

0 0 ; ; . (5)k k k1 1⎜ ⎟⎛⎝

⎞⎠

The second category is the unseparated boundary condition

= = ( )u x

v xM

u x

v xa bc d

u x

v x

( )( )

( )( )

( )( )

, (6)1

1

0

0

0

0

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

with =Mdet 1. Rewriting

λ= ( )u x

v xE x

u x

v x

( )( )

;( )( )

,k1

11

0

0

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

the corresponding eigenvalues are seen to be zeroes of

λ λ λ

λ λ λ

= − = −

− + +( )( ) ( )

( ) ( ) ( )F E x M du x

av x cu x bv x

( ) det ; 2 ;

; ; ; . (7)

k k k

k k k

1N

1

D1

D1

N1

We prove in the appendix that each zero of λF ( )k has the same multiplicity as that of thecorresponding eigenvalue. In general, the fundamental solution (3) will not be given in termsof known special functions. In order to find certain properties of the zeta function associatedwith eigenvalue problems on surfaces of revolution it will turn out to be sufficient to have aknowledge of the large-λ uniform asymptotic expansion of (3). To this end we need toanalyze u v,k k

N N and u v,k kD D. In order to prepare for the application of the WKB

approximation [1, 19], substituting the ansatz

∫λλ

= ±±±

u xT y

p yy( ; ) exp

( ; )

( )d ,k

x

xk

0

⎡⎣⎢

⎤⎦⎥

into equation (1), we get

λ λ λ± ′ = −± ±( ) ( )T x p x T x k f x( ; ) ( ) ( ; ) ( ). (8)k k2 2 2

A suitable quantity is

λλ λ

=++ −

T xT x T x

( ; )( ; ) ( ; )

2,k

k k

and we have the identity

∫

∫

λ

λ λ

+ ′ = − ′ ⇒

= −

+ + − −+

( ) ( ) ( ) ( )T p T T p TT x

p xx

T x

p xx

T x

( ; )

( )d

( ; )

( )d

ln ( ; )

2. (9)

k k k kx

xk

x

x k k

x

x

2 2

0

1

0

1

0

1⎡⎣⎢

⎤⎦⎥

Imposing the initial conditions as indicated below equation (3), we can write the elements ofλE x( ; )k 1 as

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

3

λλ λ

λ=

−+ −

( )( ) ( )

( )u x

u x u x

T x;

; ;

2 ;, (10)k

k k

k

D1

1 1

0

λλ λ λ λ

λ=

+− + + −

( )( ) ( ) ( ) ( )

( )u x

T x u x T x u x

T x;

; ; ; ;

2 ;, (11)k

k k k k

k

N1

0 1 0 1

0

λλ λ λ λ

λ=

++ + − −

( )( ) ( ) ( ) ( )

( )v x

T x u x T x u x

T x;

; ; ; ;

2 ;, (12)k

k k k k

k

D1

1 1 1 1

0

λλ λ λ λ λ λ

λ=

−− + + + − −

( )( ) ( ) ( ) ( ) ( ) ( )

( )v x

T x T x u x T x T x u x

T x;

; ; ; ; ; ;

2 ;. (13)k

k k k k k k

k

N1

0 1 1 0 1 1

0

The left-hand sides of equations (10)–(13) will be the needed input for the contour integralformulation of the zeta function. The right-hand sides will be the starting point for thecomputation of the relevant uniform asymptotic expansion.

Next we make some general statements about the zeta function associated withequation (1) supplemented by any choice of boundary conditions. Following [15], we firstcompute

∏λ λλ

= −=

∞

D ( ) 1 ,k

n k n1 ,

⎛⎝⎜

⎞⎠⎟

where λk n, is the nth positive eigenvalue of equation (1) under a certain boundary condition.For ≠F (0) 0k ,

λ λ=D F F( ) ( ) (0). (14)k k k

If λF ( )k has a first order zero instead

λ λ λ= ′( )D F F( ) ( ) (0) . (15)k k k

The zeta function can be represented as

∑ ∑∑ζ ζ ζ λ λ= + = +=

∞−

=

∞

=

∞−s s s( ) ( ) ( ) 2 ,

nns

k nk n

s1 2

10,

1 1,

where, using the contour integral representation

∫ζ ππ

= −∞

− ( )ss

zz

D z z( )sin d

dln d ,s

10

20

2

∫∑ζ ππ

= −=

∞−

∞− ( )s

sk z

zD k z z( ) 2

sin d

dln d .

k

s sk2

1

2

0

2

The functions ζ s( )1 and ζ s( )2 will be analyzed further by subtracting and adding back leadingterms in a suitable asymptotic expansion. Following [15], splitting the asymptotic expansioninto relevant and irrelevant pieces for the computation of the values and residues of ζ s( )1 , wewrite −D z( )0

2 as

− = − + −( )( ) ( ) ( )D z L z R z C , (16)02

02

02

0

where −L z( )02 and −R z( )0

2 are the leading term and the remainder respectively, such that−R z( )0

2 is exponentially small for large z; it therefore will not contribute to ζ − nRes (1/2 )1

or ζ −n( )1 for ⩾n 0. For ζ′(0)1 , in addition to −L zln ( )02 , the contribution from other terms is

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

4

− Cln 0. Similarly, we write −D k z( )k2 as

− = − + − +( )( ) ( ) ( ) ( )D k z L k z R k z L R(0) (0) , (17)k k k k k2 2 2

where −L k z( )k2 and −R k z( )k

2 are the leading term and the remainder respectively, and=L (0) 1k . As before, −R k z( )k

2 is exponentially small for large z and it does not contribute toζRes (1)2 , ζ − nRes (1/2 )2 or ζ −n( )2 for ⩾n 0. In order to compute ζ′(0)2 , the relevant

splitting will be

− = − + + − − − +( )( ) ( ) ( ) ( ) ( )D k z L k z R k z L k z Rln ln ln 1 ln 1 (0) ,k k k k k2 2 2 2

and following [15], in addition to −L k zln ( )k2 , the contribution from other terms to ζ′(0)2 is

− ∑ +=∞ R2 ln (1 (0))k k1 .After this outline of the computation for the general case, let us next consider specific

separated and unseparated boundary conditions.

3. Separated boundary conditions

For separated boundary conditions, we consider the following four cases:Dirichlet–Dirichlet(DD), Neumann–Dirichlet (ND), Dirichlet–Neumann (DN), and Neumann–Neumann (NN).The relevant choices for a b c d, , , in equation (4) are = =a c 1, = =b d 0 for DD,

= − =c b 1, = =a d 0 for ND, = =a d 1, = =b c 0 for DN, and finally = − =d b 1,= =a c 0 for NN. By equation (5), we then find

λ λ λ λ

λ λ λ λ

= =

= =( ) ( )( ) ( )

F u x F u x

F v x F v x

( ) ; , ( ) ; ,

( ) ; , ( ) ; . (18)

k k k k

k k k k

DD D1

ND N1

DN D1

NN N1

The relevant aysmptotic terms in equation (18) are found from equations (10)–(13), once the±Tk have been expanded. A WKB expansion starting with equation (8) shows (following [15])

− = ∓ ′

+ ′+ − ′

+ ′+ ′′

+ ′+± −( ) ( ) ( )

( )T z x zff

f zf

f

f

f f

fO z;

2 1

1

4 2 1 1, (19)0

2

2

2

2 2 22

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

which gives

− = + − ′+ ′

+ ′′

+ ′+ −( ) ( ) ( )

( )T z x zfzf

f

f

f f

fO z;

1

4 2 1 1. (20)0

22

2 2 23

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

Furthermore

− = + ∓ ′

+ ′ +

++

− ′+ ′

−+

+ ′′

+ ′+

±

−

( )

( ) ( )( )

T k z x k tf

f

t

t

t

k t

f

f

t

t

f f

fO k

; 12 1 1

4 ( 1) 2 1

4

1 1, (21)

k2

2

3 2

2

2 2 22

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

5

where =t zf x( )2 . This implies

− = + ++

− ′+ ′

−+

+ ′′

+ ′

+ −

( ) ( ) ( )( )

T k z x k tt

k t

f

f

t

t

f f

f

O k

; 14 ( 1) 2 1

4

1 1

. (22)

k2

3 2

2

2 2 2

3

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

We next apply these expansions to (10)–(13) to deal with the various boundary conditions.

3.1. Dirichlet boundary condition

This case was studied in [15], but for convenience we include the results of [15] here. ForDirichlet boundary condition, it is easy to show that

= = ≠( ) ( )u x A u x kA k k0; , 0; sinh ( ) , 0,k0D

1D

1

where

∫=+ ′

Af

fx

1d . (23)

x

x 2

0

1

Substituted into equation (14) for λF ( )kDD , this shows

− =−

− =−

( ) ( ) ( ) ( )D z

u z x

AD k z

u k z x

kA k

;,

;

sinh ( ).k

k0DD 2 0

D 21 DD 2

D 21

Noting that upper indices + and − correspond to exponentially growing and decaying terms,substituting the WKB expansion of −u z x( ; )0

D 21 obtained from equation (10) into −D z( )0

DD 2 ,and following equation (16), we obtain

− =−

−− = −

−

−=

+ −

( ) ( )( ) ( ) ( )

( )L z

u z x

T z xR z

u z x

T z xC A

;

2 ;,

;

2 ;, .0

DD 2 02

1

02

00DD 2 0

21

02

00DD

Taking the logarithm, and using equation (9), shows

∫− =−

−− + −

−( ) ( ) ( ) ( )L z

T z x

p xx

T z x T z xln

;

( )d

ln ; ln ;

2ln 2.

x

x

0DD 2

02

02

0 02

1

0

1

Making the asymptotic terms explicit, by equation (20)

∫

∫

−

= + ′ + ′

+ ′+ ′

+ ′+ −

( )

( )

T z x

p xx

z ff

zf fx

f

zf fO z

;

( )d

18 1

d4 1

, (24)

x

x

x

x

x

x

02

22

2 2 2

3

0

1

0

1

0

1⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

− = + + − ′+ ′

+ ′′

+ ′+ −( ) ( ) ( )

( )T z x z f xz f

f

f

f f

fO zln ; ln ln ( )

1

4 2 1 1. (25)0

22 2

2

2 2 24

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

The information gathered so far is sufficient to obtain the following properties of the functionζ s( )1 associated with k = 0 (see [15])

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

6

∫ζπ

= + ′f x xRes1

2

1

21 ( ) d ,

x

x

1DD 2

0

1⎜ ⎟⎛⎝

⎞⎠

ζ = −(0)1

2,1

DD

∫ζπ

− = ′

+ ′+

′

+ ′−

′

+ ′

f x

f f

f x

f x f x

f x

f x f xRes

1

2

1

2

1

8

d

1

( )

4 ( ) 1 ( )

( )

4 ( ) 1 ( ),

x

x

1DD

2

2 2

1

12

1

0

02

00

1⎜ ⎟⎛⎝

⎞⎠

⎡

⎣⎢⎢

⎤

⎦⎥⎥

ζ ′ = −+

−f x f x

A(0)ln ( ) ln ( )

2ln (2 ).1

DD 0 1

For ≠k 0, substituting −u k z x( ; )kD 2

1 from equation (10) into −D k z( )kDD 2 , and following

equation (17), the splitting reads

− =−

−

− = −−

−= −

+ −

− −−

( ) ( )( )

( ) ( )( )

L k zu k z x

T k z x k

R k zu k z x

T k z x kR

; e

;,

; e

;, (0) e .

kk

kA

k

kk

kA

kk

kA

DD 22

1

20

DD 22

1

20

DD 2

Taking the logarithm, and using equation (9)

∫− =− −

−− + −( ) ( )

( ) ( )

( ) ( )

L k zT k z x k

p xx

T k z x k T k z x k

ln;

( )d

ln ; ln ;

2,

kx

x k

k k

DD 22

20

21

0

1

asymptotic terms are found from equation (22)

∫ ∫− −

= + − ++

′

+ ′

++

′

+ ′+ −

( )

( )

( )T k z x k

p xx

k

pt

t

k t

f

f fx

t

k t

f

fO k

;

( )d 1 1

8 ( 1) 1d

4 ( 1) 1, (26)

x

x k

x

x

x

x

2 2

5 2

2

2

3 2 2

3

0

1

0

1

0

1

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−= + +

+

× −+

′+ ′

+ ′′

+ ′+ −

( )

( )( )

T k z x

kt

t

k t

t

t

f

f

f f

fO k

ln;

ln 14 ( 1)

4

2( 1) 1 1. (27)

k2

2 2

2

2 2 24

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

From here, one can show

∫ζ = + ′f x f x xRes (1)1

2( ) 1 ( ) d ,

x

x

2DD 2

0

1

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

7

∫

∫

∫ ∫

ζπ

ζ

ζπ

π π

π π

ζ ϕ π

= − + ′ − +

=

− = − ′

+ ′

−′

+ ′+

′

+ ′

−′

+ ′−

′′

+ ′

−′

+ ′+

′

+ ′

−′

+ ′−

′′

+ ′

′ = − + + + +

+ ′

+ ′+ ′′

+ ′

−

( ) ( )

( ) ( )( )

( )

[ ]

( )

f x x f x f x

f x

f x f xx

f x

f x f x

f x

f x f x

f x

f x f x

f x

f x

f x

f x f x

f x

f x f x

f x

f x f x

f x

f x

Af x f x

f x

f x f xx

f x

f xx

Res1

2

1

21 ( ) d

1

4( ) ( ) ,

Res (0)1

2,

Res1

2

1

16

( )

( ) 1 ( )d

1

8

( )

( ) 1 ( )

1

8

( )

( ) 1 ( )

1

256

( )

( ) 1 ( )

1

32

( )

1 ( )

1

8

( )

( ) 1 ( )

1

8

( )

( ) 1 ( )

1

256

( )

( ) 1 ( )

1

32

( )

1 ( ),

(0) 2 ln e6

ln (2 )1

2ln ( ) ln ( )

1

6

( )

( ) 1 ( )d

1

2

( )

1 ( )d ,

x

x

x

x

A

x

x

x

x

2DD 2

0 1

2DD

2DD

2

2 2

1

1 12

0

0 02

02

0 02

0

02 2

0

0 02

1

1 12

12

1 12

1

12 2

2DD 2

0 1

2

2 2 3 2

0

1

0

1

0

1

0

1

⎜ ⎟

⎜ ⎟

⎛⎝

⎞⎠

⎛⎝

⎞⎠

where ϕ is the Euler function, ϕ = ∏ −=∞q q( ) (1 )k

k1 . Adding up, ζ ζ ζ= +s s s( ) ( ) ( )1 2 , we

confirm the result in [15]

∫ζ = + ′f x f x xRes (1)1

2( ) 1 ( ) d , (28)

x

xDD 2

0

1

ζ = − −f x f x

Res1

2

( )

4

( )

4, (29)DD 0 1⎜ ⎟⎛

⎝⎞⎠

ζ =(0) 0, (30)DD

ζ − = −′

+ ′−

′′

+ ′

−′

+ ′−

′′

+ ′

( ) ( )

( ) ( )

f x

f x f x

f x

f x

f x

f x f x

f x

f x

Res1

2

( )

256 ( ) 1 ( )

( )

32 1 ( )

( )

256 ( ) 1 ( )

( )

32 1 ( ), (31)

DD2

0

02

0

0

20

2

21

12

1

1

21

2

⎜ ⎟⎛⎝

⎞⎠

∫ζ ϕ

π

′ = − + + ′

+ ′

−′

+ ′+

′

+ ′+ −

−( ) A f x

f x f xx

f x

f x

f x

f xA

(0) 2 ln e6

1

6

( )

( ) 1 ( )d

( )

2 1 ( )

( )

2 1 ( )ln ln .

A

x

xDD 2

2

2

0

02

1

12

0

1

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

8

The residues and the value at s = 0 can be verified from known heat kernel asymptotics[13, 16]. Let us denote the surface of revolution by M and its boundary by ∂M . The flat space3 induces a metric tensor on the surface of revolution, which is given by

=+ ′

gf x

f x

1 ( ) 0

0 ( ).

2

2

⎛⎝⎜⎜

⎞⎠⎟⎟

We then have

ζπ

= MRes (1)1

4vol( ),DD

which agrees with (28) as the Riemannian volume element of M is ∣ ∣ = + ′g f x f x( ) 1 ( )1 2 2 .Also,

ζπ

= − ∂MRes1

2

1

8vol( ),DD ⎜ ⎟⎛

⎝⎞⎠

which is seen to agree with (29). To compare ζ (0)DD and ζ −Res ( 1/2)DD with the resultsknown from the heat kernel coefficients we need some curvature tensors of the surface ofrevolution. In particular, the Riemann scalar reads

= − ′′

+ ′( )R

f x

f x f x

2 ( )

( ) 1 ( ), (32)

2 2

and the second fundamental form for the boundary at x1, respectively x0, is

=′

+ ′= −

′

+ ′K

f x

f x f xK

f x

f x f x

( )

( ) 1 ( ),

( )

( ) 1 ( ). (33)x x

1

1 12

0

0 02

1 0

Use of these in the local formula for ζ (0)DD shows (30)

∫ ∫ζπ

θ θ= + =∂{ }R g x K h(0)

1

4·

1

6d d 2 d 0,

M M

DD 1 2 1 2

once the induced Riemannian volume element on the boundary is realized as ∣ ∣ =h f1 2 .Finally, to verify ζ −Res ( 1/2)DD , note that the normal component of the Ricci tensor is

=R R1/2mm , and so

∫ζπ

θ− = −∂

( )R K hRes1

2

1

153612 3 d ,

M

DD 2 1 2⎜ ⎟⎛⎝

⎞⎠

in complete agreement with (31), once the expressions (32) and (33) have been substituted.Note, that also for other boundary conditions we will find ζ ζ=Res (1) Res (1)DD , as this

residue is proportional to the volume of the surface, so it is independent of boundaryconditions.

The structure of the computation for the other boundary conditions is as just presented.The numerical coefficients in front of most terms will be different, but the strategy outlinedworks equally well.

3.2. Dirichlet–Neumann boundary conditions

Next we consider the Neumann condition at x0 and Dirichlet condition at x1 (ND). In this case

= = ≠( ) ( )u x u x kA k0; 1, 0; cosh ( ), 0.k0N

1N

1

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

9

Substituted into equation (14) for λF ( )kND ,

− = − − =−

( ) ( ) ( ) ( )D z u z x D k z

u k z x

kA; ,

;

cosh ( ).k

k0ND 2

0N 2

1ND 2

N 21

Substituting −u z x( ; )0N 2

1 from equation (11) into −D z( )0ND 2 , and following equation (16),

shows

− = − + − =−( )( ) ( ) ( )L z L z T z x Cln ln ln ; , 1.0ND 2

0DD 2

02

0 0ND

For ≠k 0, substituting −u k z x( ; )kN 2

1 from equation (11) into −D k z( )kND 2 , and following

equation (17), gives

− = − + − =− −( )( ) ( ) ( )L k z L k z T k z x k Rln ln ln ; , (0) e .k k k kkAND 2 DD 2 2

0ND 2

This reduces the calculation to the DD case, except for two terms, which follow fromequations (19) and (21), namely

− = + ∓ ′

+ ′

+ − ′+ ′

+ ″

+ ′+

±

−

( )

( )( )

T z x z f xf

zf f

z f

f

f

f f

fO z

ln ; ln ln ( )2 1

1

4 1 1,

02

2

2 2

2

2 2 23

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

−= + ∓ ′

+ ′ +

++

−+

′+ ′

+ ″

+ ′+

±

−

( )

( )( )

T k z x

kt

f

k f

t

t

t

k t

t

t

f

f

f f

fO k

ln;

ln 12 1 ( 1)

4 ( 1)

2

1 1 1.

k2

2 3 2

2 2

2

2 2 23

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

Using the above expressions, we then find

ζ = −f x f x

Res1

2

( )

4

( )

4,ND 0 1⎜ ⎟⎛

⎝⎞⎠

ζ =(0) 0,ND

ζ − = −′

+ ′+

″

+ ′

−′

+ ′−

″

+ ′

( ) ( )

( ) ( )

f x

f x f x

f x

f x

f x

f x f x

f x

f x

Res1

2

5 ( )

256 ( ) 1 ( )

( )

32 1 ( )

( )

256 ( ) 1 ( )

( )

32 1 ( ),

ND2

0

02

0

0

20

2

21

12

1

1

21

2

⎜ ⎟⎛⎝

⎞⎠

∫ζ ϕ ϕ′ = − − + + ′

+ ′− −( )( ) ( ) A f x

f x f xx(0) 2 ln e ln e

6

1

6

( )

( ) 1 ( )dA A

x

xND 4 2

2

20

1

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

10

+′

+ ′+

′

+ ′−

f x

f x

f x

f x

( )

2 1 ( )

( )

2 1 ( )ln 2,0

02

1

12

where the residues at =s 1 2, = −s 1 2, and the value at s = 0 compare favorably with theknown results as they follow from known heat kernel coefficients [13, 16].

Similarly, for Dirichlet condition at x0 and Neumann condition at x1 (DN), one has

= = ≠( ) ( )v x v x kA k0; 1, 0; cosh ( ), 0.k0D

1D

1

Substituted into equation (14) for λF ( )kDN , gives

− = − − =−

( ) ( ) ( ) ( )D z v z x D k z

v k z x

kA; ,

;

cosh ( ).k

k0DN 2

0D 2

1DN 2

D 21

Substituting in v y x( ; )kD

1 from equation (12), and following equations (16) and (17), oneverifies

− = − + − =+( ) ( ) ( )L z L z T z x Cln ln ln ; , 1,0DN 2

0DD 2

02

1 0DN

− = − + − =+ −( )( ) ( ) ( )L k z L k z T k z x k Rln ln ln ; , (0) e .k k k kkADN 2 DD 2 2

1DN 2

Again skipping to write out answers for ζ1 and ζ2, the results are

ζ = − +f x f x

Res1

2

( )

4

( )

4,DN 0 1⎜ ⎟⎛

⎝⎞⎠

ζ =(0) 0,DN

ζ − = −′

+ ′−

″

+ ′

−′

+ ′+

″

+ ′

( ) ( )

( ) ( )

f x

f x f x

f x

f x

f x

f x f x

f x

f x

Res1

2

( )

256 ( ) 1 ( )

( )

32 1 ( )

5 ( )

256 ( ) 1 ( )

( )

32 1 ( ),

DN2

0

02

0

0

20

2

21

12

1

1

21

2

⎜ ⎟⎛⎝

⎞⎠

∫ζ ϕ ϕ′ = − − + + ′

+ ′

−′

+ ′−

′

+ ′−

− −( )( ) ( ) A f x

f x f xx

f x

f x

f x

f x

(0) 2 ln e ln e6

1

6

( )

( ) 1 ( )d

( )

2 1 ( )

( )

2 1 ( )ln 2.

A A

x

xDN 4 2

2

2

0

02

1

12

0

1

The value and residue are again in agreement with expectations from the heat kernelcoefficients. Also, ζ s( )ND and ζ s( )DN are symmetric as expected.

3.3. Neumann boundary conditions

Unlike in previous cases, zero is an eigenvalue for the Neumann boundary condition, whichmakes certain modifications necessary. First we note that

= = −( ) ( )v xy

v x B0; 0,d

d0; ,0

N1 0

N1

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

11

where

∫= + ′B f x f x x( ) 1 ( ) d . (34)x

x2

0

1

Substituted into equation (15) for F y( )0NN , this yields

− =−

( ) ( )D z

v z x

z B

;.0

NN 2 0N 2

1

2

For ≠k 0, instead

=( )v x k kA0; sinh ( ),kN

1

and substituted into equation (14) for F y( )kNN shows

− =−

( ) ( )D k z

v k z x

k kA

;

sinh ( ).k

kNN 2

N 21

Using the WKB expansion of v y x( ; )kN

1 in equation (13), and following equations (16) and(17), the relevant pieces are

− = − + − + − =− +( ) ( ) ( ) ( )L z L z T z x T z x C z Bln ln ln ; ln ; , ,0NN 2

0DD 2

02

0 02

1 0NN 2

− = − + −

+ − = −

−

+ −

( )( )

( ) ( ) ( )( )

L k z L k z T k z x k

T k z x k R

ln ln ln ;

ln ; , (0) e .

k k k

k kkA

NN 2 DD 2 20

21

NN 2

From here, one verifies the final answers are

ζ = +f x f x

Res1

2

( )

4

( )

4,NN 0 1⎜ ⎟⎛

⎝⎞⎠

ζ = −(0) 1,NN

ζ − = −′

+ ′+

″

+ ′

−′

+ ′+

″

+ ′

( ) ( )

( ) ( )

f x

f x f x

f x

f x

f x

f x f x

f x

f x

Res1

2

5 ( )

256 ( ) 1 ( )

( )

32 1 ( )

5 ( )

256 ( ) 1 ( )

( )

32 1 ( ),

NN2

0

02

0

0

20

2

21

12

1

1

21

2

⎜ ⎟⎛⎝

⎞⎠

∫ζ ϕ

π

′ = − + + ′

+ ′

+′

+ ′−

′

+ ′− −

−( ) A f x

f x f xx

f x

f x

f x

f xB

(0) 2 ln e6

1

6

( )

( ) 1 ( )d

( )

2 1 ( )

( )

2 1 ( )ln (4 ) ln ,

A

x

x

2NN 2

2

2

0

02

1

12

0

1

and the same remarks as for the other boundary conditions hold.

4. Unseparated boundary conditions

For unseparated boundary conditions, we first consider two special cases, namely, periodicboundary conditions (P)

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

12

= =u x u x v x v x( ) ( ), ( ) ( ),1 0 1 0

and anti-periodic boundary conditions (AP)

= − = −u x u x v x v x( ) ( ), ( ) ( ).1 0 1 0

We assume =f x f x( ) ( )0 1 for both conditions. The periodic boundary condition represents atorus. The associated functions are from (7)

= − − = + +F y u y x v y x F y u y x v y x( ) 2 ( ; ) ( ; ), ( ) 2 ( ; ) ( ; ). (35)k k k k k kP N

1D

1AP N

1D

1

4.1. Periodic and antiperiodic conditions

For periodic boundary conditions, zero is an eigenvalue.The relevant information needed is

= =Fy

F AB(0) 0,d

d(0) ,0

P0P

where A and B are defined in (23) and (34). Substituted into equation (15) for F y( )0P ,

− =− + − −

( ) ( ) ( )D z

u z x v z x

z AB

; ; 2.0

P 2 0N 2

1 0D 2

1

2

For ≠k 0,

= −F kA(0) 2(1 cosh ( )).kP

Substituted into equation (14) for F y( )kP ,

− =− + − −

−( ) ( ) ( )D k z

u k z x v k z x

kA

; ; 2

2(cosh ( ) 1).k

k kP 2

N 21

D 21

Using the WKB expansions of u y x( ; )kN

1 and v y x( ; )kD

1 in D y( )kP , and following

equations (16) and (17), one obtains

− = − + − + − =− +( ) ( ) ( ) ( )L z L z T z x T z x C z ABln ln ln ; ; , ,0P 2

0DD 2

02

0 02

1 0P 2⎡⎣ ⎤⎦

− = − +− + −

+ = −

− +

−

( ) ( ) ( ) ( )

( )

L k z L k zT k z x T k z x

k

L R

ln ln ln; ;

2,

(0) (0) 1 e .

k kk k

k kkA

P 2 DD 22

02

1

P P 2

Using the assumption that =f x f x( ) ( )0 1 , and writing t x( )0 as t0, we have

∫ δ− = + ′ + ′

+ ′− +( )L z z f

f

zf fx

z f xO

zln 1

8 1d

32 ( )

1,

x

x

0P 2 2

2

2 2

02

2 20

30

1⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎛⎝⎜

⎞⎠⎟

∫

δ

− = + − ++

′

+ ′

−+

+

( ) ( )

( )

L k zk

pt

t

k t

f

f fx

t

k tO

k

ln 1 18 ( 1) 1

d

32 1

1,

kx

xP 2

2

5 2

2

2

02

02

20

3 3

0

1⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎛⎝⎜

⎞⎠⎟

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

13

where

δ =′

+ ′−

′

+ ′

f x

f x

f x

f x

( )

1 ( )

( )

1 ( ). (36)0

0

02

1

12

From the expansion we obtain for the k = 0 mode

∫ζ = + ′f x xRes1

2

1

21 ( ) d , (37)

x

x

1P 2

0

1⎜ ⎟⎛⎝

⎞⎠

ζ = −(0) 1, (38)1P

∫ζπ

− = ′

+ ′

f x

f x f xxRes

1

2

1

16

( )

( ) 1 ( )d , (39)

x

x

1P

2

2 20

1⎜ ⎟⎛⎝

⎞⎠

ζ ′ = − −A B(0) ln ln , (40)1P

and for the ≠k 0 modes

∫ζ = − + ′f x xRes1

2

1

21 ( ) d , (41)

x

x

2P 2

0

1⎜ ⎟⎛⎝

⎞⎠

ζ =(0) 0, (42)2P

∫ζπ

δ− = − ′

+ ′−f x

f x f xx

f xRes

1

2

1

16

( )

( ) 1 ( )d

3

512 ( ), (43)

x

x

2P

2

2 2 002

0

1⎜ ⎟⎛⎝

⎞⎠

∫ζ ϕ′ = − + + ′

+ ′−( ) A f x

f x f xx(0) 4 ln e

6

1

6

( )

( ) 1 ( )d . (44)A

x

x

2P

2

20

1

Adding up, ζ ζ ζ= +s s s( ) ( ) ( )1 2 , we get

ζ =Res1

20,P ⎜ ⎟⎛

⎝⎞⎠

ζ = −(0) 1,P

ζ δ− = −f x

Res1

2

3

512 ( ),P

002⎜ ⎟⎛

⎝⎞⎠

and

∫ζ ϕ′ = − + + ′

+ ′− −−( ) A f x

f x f xx A B(0) 4 ln e

6

1

6

( )

( ) 1 ( )d ln ln .A

x

xP

2

20

1

The first of the above equations reflects that the torus does not have a boundary. The secondequation says that its Euler characteristic is zero, taking into account that the one zero mode isnot included in the zeta function.

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

14

For antiperiodic boundary condition, the corresponding equations are

− =− + − +

− =− + − +

+

( ) ( ) ( )

( ) ( ) ( )D z

u z x v z x

D k zu k z x v k z x

kA

; ; 2

4,

; ; 2

2(cosh ( ) 1),k

k k

0AP 2 0

N 21 0

D 21

AP 2

N 21

D 21

which imply that

− = − =( ) ( )L z L z Cln ln , 4,0AP 2

0P 2

0AP

− = − + − = + −( ) ( ) ( ) ( )L k z L k z L R k zln ln , (0) 1 e .k k k kkAAP 2 P 2 AP AP 2 2

The resulting residues of ζ s( )1 and ζ s( )2 are the same as those for periodic boundaryconditions. Their values and derivatives at s = 0 are

ζ =(0) 0, (45)1AP

ζ ′ = −(0) ln 4, (46)1AP

ζ =(0) 0, (47)2AP

∫ζ ϕ ϕ′ = − − + + ′

+ ′− −( )( ) ( ) A f x

f x f xx(0) 4 ln e ln e

6

1

6

( )

( ) 1 ( )d . (48)A A

x

x

2AP 2

2

20

1

They lead to

ζ =Res1

20,AP ⎜ ⎟⎛

⎝⎞⎠

ζ =(0) 0,AP

ζ δ− = −f x

Res1

2

3

512 ( ),AP

002⎜ ⎟⎛

⎝⎞⎠

and

∫ζ ϕ ϕ′ = − − + + ′

+ ′−− −( )( ) ( ) A f x

f x f xx(0) 4 ln e ln e

6

1

6

( )

( ) 1 ( )d ln 4.A A

x

xAP 2

2

20

1

Similar remarks as those made above for periodic boundary conditions apply.

4.2. Klein bottle

We consider a special unseparated boundary condition that involves the azimuthal angle θ,

θ θ θ θ= − = −( ) ( ) ( ) ( )u x u x v x v x, , , , , .1 0 1 0

We assume =f x f x( ) ( )0 1 . The boundary condition represents a Klein bottle (K). For k = 0, itis the same as periodic boundary condition. Therefore

ζ ζ=s s( ) ( ).K1 1

P

For ≠k 0, the eigenfunction are no longer in the form of ϕ θx k( ) exp (i ). Instead, theeigenfunction is either

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

15

θ ϕ θ=u x x k( , ) ( ) cos ( ),

with periodic ϕ x( ), or

θ ϕ θ=u x x k( , ) ( ) sin ( ),

with antiperiodic ϕ x( ). As a result

ζζ ζ

=+

ss s

( )( ) ( )

2,K

22P

2AP

and

ζ =Res1

20,K ⎜ ⎟⎛

⎝⎞⎠

ζ = −(0) 1,K

ζ δ− = −f x

Res1

2

3

512 ( ),K

002⎜ ⎟⎛

⎝⎞⎠

∫ζ ϕ′ = − + + ′

+ ′− −−( ) A f x

f x f xx A B(0) 2 ln e

6

1

6

( )

( ) 1 ( )d ln ln .K A

x

x2

2

20

1

Once again, observations made earlier regarding the first two equations are valid also for theKlein bottle.

5. Nonsmooth surfaces

For the examples with unseparated boundary conditions, a smooth surface requires that′ = ′f x f x( ) ( )0 1 , which implies that ζ − =Res ( 1/2) 0P . On the other hand, if ′ ≠ ′f x f x( ) ( )0 1 ,

the kink points on the torus would generate a nonzero residue of the zeta function at −1/2. Inthis section we will study the effect of kink points in f(x) inside the interval x x[ , ]0 1 on the zetafunction for various boundary conditions. For clarity we assume f(x) has only one kink pointat xK. Let

= ⩽ ⩽ = < ⩽f x f x x x x f x f x x x x( ) ( ), , and ( ) ( ), ,K K1 0 2 1

where f x( )1 and f x( )2 are smooth, and

= ′ ≠ ′f x f x f x f x( ) ( ), ( ) ( ).K K K K1 2 1 2

The WKB method cannot be applied to nonsmooth f(x) directly, but it can be applied to f1 andf2 respectively. Introducing the fundamental solution for the intervals x x[ , ]K0 and x x[ , ]K 1 as

λλ λ

λ λλ

λ λ

λ λ= =E x

u x u x

v x v xE x

u x u x

v x v x( ; )

( ; ) ( ; )

( ; ) ( ; )( ; )

( ; ) ( ; )

( ; ) ( ; ),k

k k

k kk

k k

k k,1

,1N

,1D

,1N

,1D ,2

,2N

,2D

,2N

,2D

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

with the conditions λ =E x I( ; )k,1 0 and λ =E x I( ; )k K,2 , the fundamental solution from x0 tox1 is

=E y x E y x E y x( ; ) ( ; ) ( ; ). (49)k k k K1 ,2 1 ,1

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

16

First consider Dirichlet boundary condition. By equation (49)

= +u x u x u x u x v x( ) ( ) ( ) ( ) ( ).K KD

1 2N

1 1D

2D

1 1D

Following equation (16)

= ++ +

+ −Lu x u x

T x T xT x T x

( ) ( )

2 ( )2 ( )( ) ( ) .K

KK K

DD 1 2 1

1 0 21 2

⎡⎣ ⎤⎦For k = 0, taking the logarithm

∫ ∫− =−

+−

−− + −

− +

( ) ( ) ( )

( ) ( )

L zT z x

p xx

T z x

p xx

T z x T z xK z

ln;

( )d

;

( )d

ln ; ln ;

2ln 2 ( ),

x

x

x

x

0DD 2

02

02

02

0 02

1

K

K0

1

where

=− + −

−− + −− +( ) ( ) ( ) ( )

K zT z x T z x T z x T z x

( ) ln; ;

2

ln ; ln ;

2.

K K K K0,22

0,12

0,12

0,22

Note, that K(z) would vanish if ′ = ′f x f x( ) ( )K K1 2 . For ′ ≠ ′f x f x( ) ( )K K1 2 ,

δ δ= − + −( )K z

zf x z f xO z( )

4 ( ) 32 ( ),K

K

K

K

2

2 23

where

δ =′

+ ′−

′

+ ′

f x

f x

f x

f x

( )

1 ( )

( )

1 ( ). (50)K

K

K

K

K

2

22

1

12

Using equation (24)

∫

δ

− = + ′ + ′

+ ′+ ′

+ ′

−− + −

− − + −

( )

( ) ( ) ( )

L z z ff

zf fx

f

zf f

T z x T z x

z f xO z

ln 18 1

d4 1

ln ; ln ;

2ln 2

32 ( ).

x

x

x

x

K

K

0DD 2 2

2

2 2 2

02

0 02

1 2

2 23

0

1

0

1⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

The formula for ζRes (1/2)1DD , ζ (0)1

DD , ζ −Res ( 1/2)1DD and ζ ′(0)1

DD are not affected, thoughζ −( 1)1

DD changes. Similarly

∫ ∫− =− −

+− −

−− + −

+( ) ( )

( ) ( ) ( )

( ) ( )

L k zT k z x k

p xx

T k z x k

p xx

T k z x k T k z x kK k z

ln;

( )d

;

( )d

ln ; ln ;

2( , ),

kx

x k

x

x k

k k

DD 22 2

20

21

K

K0

1

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

17

where

δ δ=

+−

++ −

( )( )

( )K k z

t

k t

t

k tO k( , )

4 1 32 1,K K

K

K K

K2 3 2

2 2

2 33

in which =t zf x( )K K2 . With equation (26), this shows

∫

δ

− = + − ++

′

+ ′

++

′

+ ′

−− + −

−+

+ −

( ) ( )

( )

( ) ( )

( )

( )

( )

L k zk

pt

t

k t

f

f fx

t

k t

f

f

T k z x k T k z x k

t

k tO k

ln 1 18 ( 1) 1

d

4 ( 1) 1

ln ; ln ;

2

32 1.

kx

x

x

x

k k

K K

K

DD 22

5 2

2

2

3 2 2

20

21

2 2

2 33

0

1

0

1

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

The formula for ζRes (1/2)2DD , ζ (0)2

DD and ζ ′(0)2DD are not affected. However,

ζ −Res ( 1/2)2DD changes, and we find

ζ

δ

− = −′

+ ′−

′′

+ ′

−′

+ ′−

′′

+ ′

−

( ) ( )

( ) ( )

f x

f x f x

f x

f x

f x

f x f x

f x

f x

f x

Res1

2

( )

256 ( ) 1 ( )

( )

32 1 ( )

( )

256 ( ) 1 ( )

( )

32 1 ( )

3

512 ( ). (51)K

K

DD2

0

02

0

0

20

2

22

22

2

2

22

2

2

⎜ ⎟⎛⎝

⎞⎠

If there is more than one kink point, ζ −Res ( 1/2)DD will have an extra termδ− f x3 (512 ( ))K K

2 for each kink point. We will prove that the effect on the zeta function isthe same for other boundary conditions, by showing that the ratio between L y( )DD and theL(y) for a given boundary condition is unaffected by kink points. For clarity we drop thedependence on k and y in the derivation and denote the leading term of u x( )D by u xˆ ( )D , etc.Then for a kink point in f(x) at xK,

= +−( )E x

T xT x u xˆ ( )

1( )

( ) 1 ˆ ( ),KK

K11

1 0 1D

⎛⎝⎜

⎞⎠⎟

= +−( )E x

T xT x u xˆ ( )

1( )

( ) 1 ˆ ( ),K2 12 1

2 2D

1⎛⎝⎜

⎞⎠⎟

which gives

= = +−( )E x E x E x

T xT x u xˆ ( ) ˆ ( ) ˆ ( )

1( )

( ) 1 ˆ ( ),K1 2 1 12 1

1 0D

1⎛⎝⎜

⎞⎠⎟

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

18

where

= +− +u x u x u x T x T xˆ ( ) ˆ ( ) ˆ ( ) ( ) ( ) .K K KD

1 2D

1 1D

2 1⎡⎣ ⎤⎦

It indicates that the ratios between u xˆ ( )N1 , v xˆ ( )D

1 , v xˆ ( )N1 and u xˆ ( )D

1 are unaffected by thekink point, therefore the corresponding zeta functions will change as much as ζ s( )DD .

6. Conclusions

This paper provides the analysis of the spectral zeta function for the Laplacian on a surface ofrevolution with a variety of boundary conditions imposed. Explicit results for several residuesand values of the zeta function are given; all are in agreement with results known for moregeneral geometries [13]. Furthermore, surprisingly simple results for the determinant arefound. Our analysis allowed for the introduction of kink points such that the effect of non-smoothness could be studied. Additional contributions to some properties due to the kinkpoint were found as was expected from a general perspective [14]. In some detail, denoting byΣ the circle of the surface located at xK, in the notation of [14] our continuity assumptions onthe eigenfunctions along Σ imply U = 0. Then, ζ (0)DD and ζ −Res ( 1/2)DD obtain additionalcontributions due to the fact that the surface is not smooth, namely (see theorem 2.3 in [14]restricted to the surface of revolution)

∫ ∫ ∫

∫ ∫

ζπ

θ θ θ

ζπ

θ θ

= + + +

− = − + +

ΣΣ

ΣΣ

∂+ −

∂+ −

{ }{ ( )

( )

( )

R g x K h K K h

R K h K K h

(0)1

4·

1

6d d 2 d 2 d ,

Res1

2

1

153612 3 d

9

2d ,

M M

M

DD 1 2 1 2 1 2

DD 2 1 2 2 1 2⎜ ⎟⎛⎝

⎞⎠

⎫⎬⎭where +K , respectively −K , are the second fundamental forms as induced from the surface tothe left, respectively to the right, of xK, and ∣ ∣Σh 1 2 is the Riemannian volume element of thecircle at xK, namely ∣ ∣ =Σh f x( )K

1 2 . The additional contribution along Σ in ζ (0)DD

guarantees that still ζ =(0) 0DD , as this piece is needed to compensate a contribution comingfrom the integral along M because of the non-smoothness at xK.

The contribution along Σ in ζ −Res ( 1/2)DD generates exactly the last term in (51), so thatour result is in agreement with [14]. This was not completely clear as in [14] continuity of themetric is assumed which here is not given. Also in agreement with our findings, [14] predictsthat these additional contributionas are independent of the boundary conditions imposed at∂M .

Appendix. Spectral function of Sturm–Liouville equation

In this appendix we give an independent proof that equations (5) and (7) not only determinethe eigenvalues but also the degeneracy correctly.

Consider the Sturm–Liouville problem.

λ= − ′ ′ + = ⩽ ⩽ u Pu Qu Ru t( ) ( ) , 0 1,

where >P 0, Q, >R 0, and u are functions of t, and for simplicity we have chosen theinterval [0, 1]. With = ′v Pu , the equation can be written as

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

19

λ′ =

−−( ) ( )u

vP

Q Ruv

00

.1⎛

⎝⎜⎞⎠⎟

P(t), Q(t), and R(t) are not necessarily continuous. We write the fundamental solution λE t( ; )as in equation (3). As described in section 2, to guarantee that is self-adjoint, the boundarycondition can be chosen as separated, equation (4), or unseparated ones, equation (6). Forseparated boundary conditions, the corresponding eigenvalues are the zeros of the followingfunction of λ, see equation (5)

λ λ λ= + = −( ) ( ) ( )( )F a bc d

E c d E ba

( ) det0 0

0 0 ( ; 1) ( ; 1) , (52)⎜ ⎟⎛⎝

⎞⎠

whereas for unseparated conditions the corresponding eigenvalues are the zeros of, seeequation (7),

λ λ λ λ λ λ= − = − − + +F E M u av cu bv( ) det ( ( ; 1) ) 2 d ( ; 1) ( ; 1) ( ; 1) ( ; 1). (53)N N D N

For the separated boundary condition, each eigenvalue is simple [23]. We will prove that thecorresponding λF ( ) also has only simple zeros. For the unseparated boundary condition, theeigenvalues can be simple or double. For example, for P = 1, Q = 0, R = 1, and the periodicboundary condition, all eigenvalues are double except for λ = 0. We will prove that each zeroof λF ( ) has the same multiplicity as that of the corresponding eigenvalue.

Taking the derivative with respect to λ on both sides of the following equation,

λ=( )u Ru ,N N

we have

λλ

λ∂∂

= ∂∂

+ uR

uRu .

N NN

⎛⎝⎜

⎞⎠⎟

The solution is

∫ ∫λλ

λ τ λ τ λ τ τ λ τ λ τ τ∂∂

= − ( )u tu t R u u u t R u

( ; )( ; ) ( ) ( ; ) ( ; )d ( ; ) ( ) ( ; ) d .

t tNN

0

N D D

0

N 2

Similarly

∫ ∫λλ

λ τ λ τ τ λ τ λ τ λ τ τ∂∂

= −( )u tu t R u u t R u u

( ; )( ; ) ( ) ( ; ) d ( ; ) ( ) ( ; ) ( ; )d .

t tDN

0

D 2 D

0

N D

Setting t = 1, we have

∫ ∫∫ ∫

λλλ

λλλ

λ λ λ

λ λ λ

λλ

=

=−

−

×

( )

( )

u

uK

u

u

R t u t u t t R t u t t

R t u t t R t u t u t t

u

u

d

d

( ; 1)

( ; 1)( )

( ; 1)

( ; 1)

( ) ( ; ) ( ; )d ( ) ( ; ) d

( ) ( ; ) d ( ) ( ; ) ( ; )d

( ; 1)

( ; 1). (54)

N

D

N

D

0

1N D

0

1N 2

0

1D 2

0

1N D

N

D

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎛⎝⎜⎜

⎞⎠⎟⎟

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

20

It is easy to see that

λλλ

λλλ

=v

vK

v

v

d

d

( ; 1)

( ; 1)( )

( ; 1)

( ; 1),

N

D

N

D

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

and therefore

λλ λ λ=E E K

d

d( ; 1) ( ; 1) ( ).T

For separated the boundary condition, if λ is a zero of F, by equation (52) there exists ≠k 0such that

λ =( ) ( )c d E k a b( ; 1) ,

≠k 0, because ≠c d( , ) (0, 0) and λE ( ; 1) is nonsingular. Then

∫λ

λ λ λ λ

λ λ

= − = −

= −

( ) ( )( ) ( )F c d E K ba

k a b K ba

k R t au t bu t t

d

d( ) ( ; 1) ( ) ( )

( ) ( ; ) ( ; ) d ,

T T

0

1 D N 2⎡⎣ ⎤⎦which is nonzero since λu t( ; )N and λu t( ; )D are linearly independent. Therefore λ is a simplezero of F.

For the unseparated boundary condition, if λ is a double eigenvalue, λ =E M( ; 1) , andso λ is a zero of each element of λ −E M( ; 1) . By equation (53) λ is a zero of λF ( ) withmultiplicity at least 2. We prove that the multiplicity is indeed 2 by noticing

λλ λ

λλ λ λ= = =F

EE K K

1

2

d

d( ) det

d ( ; 1)

ddet ( ; 1)det ( ) det ( ),T T

2

2

which is positive by the Cauchy–Schwarz inequality and because λu t( ; )N and λu t( ; )D arelinearly independent. Finally, to prove that a single eigenvalue of L must be a single zero of F,we will show that if λ is a zero of F with multiplicity more than 1, λ =E M( ; 1) must hold.Indeed, if λ is a zero of F with multiplicity at least 2

λϵ

λ ϵ λ− = + − =ϵ=

( )( )E M E I K Mdet ( ( ; 1) ) 0,d

ddet ( ; 1) ( ) 0.T

0

Let λ= − −A I Me ( ; 1)1 , we have

ϵϵ λ= + =

ϵ=( )A A Kdet 0,

d

ddet ( ) 0.T

0

Combining =Adet 0 with − = =I A Mdet ( ) det 1, we have =Atr 0. Notice that we alsohave λ =Ktr ( ) 0T . Denoting the elements of A and K by aij and kij, we have

+ = = +a k a k a k a k a k2 . (55)11 22 22 11 11 22 12 12 21 21

On the other hand, =Adet 0 implies

− =a a a ,12 21 112

and >Kdet 0T implies

− >k k k .12 21 222

J. Phys. A: Math. Theor. 48 (2015) 145204 T Lu et al

21

Consequently

+ ⩾ ⩾a k a k a k a k a k2 2 .12 12 21 21 12 12 21 21 11 22

Therefore equation (55) only holds when =a 011 , which implies A = 0, or λ =E M( ; 1) .

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