On a Functional Equation of Ruijsenaars

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On a Functional Equation of Ruijsenaars J.G.B.Byatt-Smith ?and H.W.Braden ?Department of Mathematics and Statistics,The University of Edinburgh,Edinburgh,UK Submitted December,2001Abstract We obtain the general solution of the functional equation I ?{1,2,...,n }|I |=k i ∈I j ∈I h (x j ?x i )h (x i ?x j ?iβ)? i ∈I j ∈I h (x i ?x j )h (x j ?x i ?iβ) =0.This equation,introduced by Ruijsenaars,guarantees the commutativity of n operators associated with the quantum Ruijsenaars-Schneider models.2000AMS Subject Classi?cation :Primary 39B3230D0533E05Key Words :Integrability,Functional Equations

1Introduction

The purpose of this paper is to investigate the functional equation

I?{1,2,...,n} |I|=k

i∈I

j∈I

h(x j?x i)h(x i?x j?iβ)? i∈I j∈I h(x i?x j)h(x j?x i?iβ) =0.(1)

Hereβis an arbitrary positive number and the sum is over all subsets with k elements.As we will describe shortly,this equation underlies the quantum integrability of the Ruijsenaars-Schneider models.We will establish

Theorem1The general solution of the functional equation(1)analytic in a neighbourhood of the real axis with either a simple pole at the origin or an array of such poles at np on the real axis(n∈Z)is given by

h(x)=b

σ(x+ν)

2 j=k cosh(p j?p k) l=j f(x j?x l) m=k f(x k?x m) ?j ln f(x k?x j)+?k ln f(x j?x k)

and for the independent terms proportional to cosh(p j?p k)to vanish we require that f′(x)/f(x) be odd.This entails that f(x)is either even or odd1and in either case F(x)=f2(x)is even. Supposing that f(x)is so constrained,then the?nal Poisson bracket is equivalent to the functional equation

{H,P}=0??

n

j=1?j k=j f2(x j?x k)=0.(4)

Observe that upon dividing (1)by βand letting β→0this yields (4)with F (x )=h (x )h (?x )when k =1.For n =3equation (4)may be written in the form 111F (x )F (y )F (z )F ′(x )F ′(y )F ′(z ) =0,x +y +z =0,(5)where F (x )=f 2(x ).Ruijsenaars and Schneider [23]showed that F (x )=?(x )+c satis?es

(5)and further satis?es (4)for all n .This same functional equation (without assumptions on the parity of the function F (x ))has arisen in several settings related to integrable systems.It arises when characterising quantum mechanical potentials whose ground state wavefunction (of a given form)is factorisable [13,24,19].2More recently it has been shown [1]to characterise the Calogero-Moser system [25],which is a scaling limit of the Ruijsenaars-Schneider system.There appear deep connections between functional equations and integrable systems [12,8,21,20,6,14,4,5,15,16,17,18].The analytic solutions to (5)were characterised by Buchstaber and Perelomov [9]while more recently a somewhat stronger result with considerably simpler proof was obtained by the authors [2].One has

Theorem 2([2])Let F be a three-times di?erentiable function satisfying the functional equa-tion (5).Then,up to the manifest invariance F (z )→αF (δz )+β,the solutions of (5)are one of F (z )=?(z +d ),F (z )=e z or F (z )=z .Here ?is the Weierstrass ?-function and 3d is a lattice point of the ?-function.

Thus the even solutions of (5)are precisely those obtained by Ruijsenaars and Schneider.Until this year the general solution to (4)remained unknown when the authors established Theorem 3([11])The general even solution of (4)amongst the class of meromorphic func-tions whose only singularities on the Real axis are either a double pole at the origin,or double poles at np (p real,n ∈Z )is:

a)for all odd n given by the solution of Ruijsenaars and Schneider while

b)for even n ≥4there are in addition to the Ruijsenaars-Schneider solutions the following:

F 1(z )=

σ2(z )=θ3(v )θ4(v )

4ω2θ3(0)θ4(0)=b

dn(u )(?(z )?e 1)(?(z )?e 3)=

σ1(z )σ3(z )θ21(v )θ′21(0)sn 2(u )

F 3(z )= σ2(z )

=

θ2(v )θ3(v )4ω2θ2(0)θ3(0)=b cn(u )dn(u )2As an aside we remark that the

delta function potential aδ(x )of many-body quantum mechanics on the line,which has a factorisable

ground-state wavefunction,can be viewed as the α→0limit of ?b/αsinh 2(?x/α+πi/3)with πaα=

6b .Thus all of the known quantum mechanical problems with fac-torisable ground-state wavefunction are included in (5).

Here

σα(z)=σ(z+ωα)

e1?e3z,v=z

2exp ?√2

pairwise commute if and only if(1)held for all k and n≥1.Further he was able to show(2) led to a solution of(1),the solution being related to the earlier Ruijsenaars-Schneider solution via

σ(x+ν)σ(x?ν)

The Fourier transform of(1)will be in terms of the Fourier transform g(k,β)of g(z,β).We will show how,after a judicious choice of parametrization,the Fourier transform of(1)when k=1leads to precisely the same equation encountered when studying the Fourier transform of(4).Theorem3then gives the general solutions for g(z,β).We?nd that we can write g(z,β)≡h(?iβ) F1(z?iβ)?F1(z)?F1(ν?iβ)+F1(ν) where F1(z)= F(z)dz,F(ν)=0and F(x)is any solution given in theorem3.Now given g(z,β)we wish to factorise this in the form(7).The new solutions of theorem3do not have the factorisation property(7),whereby we establish that for k=1the only solutions to(1)are given by(2).It is known however that these solutions satisfy(1)for all k,and so the theorem will be proved.Our strategy will be to take the Fourier transform for functions of increasing complexity,?rst considering those functions with only a pole at the origin and vanishing at in?nity;next we consider similar functions decaying to a constant at in?nity;?nally we consider those functions with a periodic array of poles along the real axis including the origin.

Before we derive the equation for g we look at the properties of g and the conditions that these properties demand of g.The original solutions of Ruijsenaars and Schneider for F(z) can be expressed as

F(z)=A?(z,g2,g3)+B,(8) where A,B,g2and g3are constants.These are even functions of z with a double pole at the origin.The new solutions of[11]have a similar structure but do not possess the addition of the arbitrary constant B.However,in both cases we can use the constant A and the scaling properties of?(z)to restrict our choice of F(z),without loss of generality,so that z2F(z)→?1as z→0and that when the solution has?nite real period we take this to be 2π.Letνbe a zero of F(z).Thus for the solutions(8)we may express the constant B in terms ofνas

F(z)=?(ν)??(z).(9) The condition F(ν)=0then requires h(ν)h(?ν)=0.We?x h(z)by demanding that h(z)=0at z=?ν.This entails g(ν,β)=g(?ν+iβ,β)=0.Also since F(z)has a double pole at z=0,h(z)must have a simple pole at z=0and we choose h(z)so that zh(z)→+1 as z→0.Thus g(z,β)must have simple poles at z=0and z=iβwith zg(z)→?h(?iβ) as z→0and(z?iβ)g(z)→h(?iβ)as z→iβ.

We will now obtain an equation for g by taking the Fourier transform of(1).Set z j=x j+iy j and denote by E(k,z n)the k=1equation(1),where z is the vector(z1,z2,...z n?1).We de?ne the(n?1)-dimensional Fourier transform by

E(k,z n,β)= R n?1E(z,z n,β)e?i k.z d z.(10)

However,since g(z,β)has a pole at the origin,we replace z j by z j+i?j and assume that ?1>?2>...>?n>0and that?1is small.We then assume in the de?nition of E in(10) that we integrate along the Real axis in the complex x j+iy j plane.

2.1Functions with in?nite real period vanishing at in?nity

In the?rst instance we consider the class of solutions g(z,β)which have in?nite real period and tend to zero at in?nity with no other singularities on the real axis other than the pole

at the origin.Hence,when z j has been replaced by z j+i?j,the integrand will have no other singularities in the domain of integration provided?1<βwhich we take to be real and positive. The reduction of E=0as?1→0,to an equation involving the generalised Fourier transform g(k,β)follows the lines of[11].The de?nition of g is given by

g(k,β)=1

Thus for example(relevant to the n=3case)

g(z1?z2)g(z1?z3)= R2g(z1?z2+i(?1??2))g(z1?z3+i(?1??3))e?ik1z1?ik2z2dz1dz2

=e?i(k1+k2)z3 ∞?∞dvg(v+i?′)e?i(k1+k2)v ∞?∞dug(?u+i?′′)e?ik2u

=e?i(k1+k2)z3 g M(k1+k2)+iπh(?iβ)?iπh(?iβ)eβ(k1+k2)

× g M(?k2)+iπh(?iβ)?iπh(?iβ)e?βk2 .

Here we have set u=z2?z1,v=z1?z3,?′=?1??3and?′′=?1??2.The common factor of h(?iβ)in these expressions suggests the rewriting g M(k,β)=?iI(k,β)βh(?iβ).Then in the limitβ→0, g M(k,0)=I(k,0).If we use this we?nd that the n=3equation(10)can

3 1J j=0,where for example

be written as

J1=? I(k1+k2,β)βπ?1+e?βk2

+ I(?k1?k2,β)βπ+1?eβk2 .(15)

g(z2?z1)g(z3?z1)in the sum,with similar de?-g(z1?z2)g(z1?z3)?

This corresponds to

nitions for J2and J3:

J2= I(?k1?k2,β)βπ?1+eβk1

? I(k1+k2,β)βπ+1?e?βk1 ,

J3= I(k1,β)βπ?1+eβk2

? I(?k1,β)βπ+1?e?βk2 .

At this point we introduce what,with hindsight,will prove a judicious change of variable: set

I(k,β)= G(k,β) eβk?1 /(kβ).(16)

This change of variable is suggested by a careful examination of the series for I(k,β)= ∞0I j(k)βj that results from3 1J j=0with the expressions above.Whatever,whenβ→0,

I→ G(k,0)and we know(as F(x)is even)that G(k,0)is even.An examination of the equation for G(k,β)(say by a series expansion)shows further that G(k,β)itself is even.This fact leads to simpli?cations.When G(k,β)is even J1simpli?es to

J1=? eβk1?1 eβk?1 1?e?β(k1+k2)

and similarly for J2,J3.We then?nd the exponential factors involvingβare common to all J i and so(10)can?nally be expressed,in the case n=3as

k2 G(k1,β)+k1 G(k2,β) G(k1+k2,β)?(k1+k2) G(k1,β) G(k2,β)=π2k1k2(k1+k2).(17) Now this is precisely the equation for the Fourier transform of F(z)obtained in[11][eqn.

5.5]when studying(4)for n=3.The only solutions of the required from are G(k,β)≡ G(k)= F(k)=πk coth(πk/a)and its limit as a→0namelyπ|k|.Hence

g M(k,β)=?iβh(?iβ) eβk?1

4

a2/sinh2 1

2

h(?iβ) coth12az

=

a2iaβ

2

az sinh1

2

a/sinh 1

z?iβ?1

(?z)(z?iβ)≡h(?z)h(z?iβ),(22)

with h(z)=1/z and F(z)=h(?z)h(z).Both these factorisations are only de?ned up to the shift by the exponential h(z)→h(z)eαz given in(2).

For the cases n≥4we also?nd that the same set of transformations of g M reduce equation (10)to the original equation for the transform F encountered in the study of(4).For odd n we only have the solutions(19)and(22)while for even n we have in addition to these solutions the solution

F(k)=πk tanh πk sinh2az.(23) Now however we have

g(z,β)=ah(?iβ) 1sinh az ,(24)

which cannot be written in the form g(z,β)=h(?z)h(z?iβ).Thus these do not yield solutions to(1).

2.2Functions with in?nite real period constant at in?nity

The solutions(8)of Ruijsenaars and Schneider contain the addition of an arbitrary constant that,in the hyperbolic limit,corresponds to the function not vanishing at in?nity.The new solutions of[11],of which(23)is an example,do not have this degree of freedom:the hyperbolic degenerations of these solutions all tend to zero at in?nity.To deal with functions which do not tend to zero at in?nity we must deal with distributional Fourier transforms.The addition of a constant to F(z)requires an addition of a similar constant to g(z,β),since for example if F→a2as z→∞then g→a2as z→∞.For n=3we can easily verify that g(z,β)→g(z,β)+constant leaves(1)invariant.However,this is not automatically the case when n>3.We?nd that for n>3the transformation g(z,β)→g(z,β)+A,requires extra conditions on g to leave the equation invariant.When n=4,there is only one extra condition which is automatically satis?ed for the solution g corresponding to the solution F given by Ruijsenaars and Schneider,but not for the solution F corresponding to the new solutions of [11].We believe that for n>4,the Ruijsenaars and Schneider solution automatically satisfy all the extra conditions but that the new solutions given in theorem3do not.The addition of an arbitrary constant A,to g,requires the addition of Aδ(k)to g(k,β).For the solution g which are otherwise well behaved functions of k we can easily obtain the solution for g(z,β) and the corresponding solution for h(z),when they exist.For example,instead of(20)we add an arbitrary constant to F(z)and,demanding that F(ν)=0,we have

F(z)=1

sinh2 1sinh2 1

2ah(?iβ)

sinh ?1

sinh ?12a(z?iβ)+A.(26)

Since by de?nition g(z,β)=h(?z)h(z?iβ)and h(z)has a zero at z=?ν,we have g(ν,β)=0,as indicated earlier.Hence

A≡A(β,ν)=?

12aiβ

2

aν sinh1

2ah(?iβ)

sinh1

sinh(1

2

az sinh1sinh12aiβ

2

a(ν?iβ)

≡h(?z)h(z?iβ),(28) where

h(z)=

12a(z+ν)

2

az sinh(1

4sinh1

2

a(z+ν)

2

az sinh2 14

1

2

aν ?12az

,(30)

as required.Also observe that asν→∞we have h(z)tending to the solutions of the previous section times the exponential factor e az/2,and our factorisation is only unambiguous up to such terms.

A similar but easier calculation for F(z)=1/ν2?1/z2gives

g(z,β)=h(?iβ)ν(?iβ)

(?z)ν

z+ν?iβ

2π∞

p=?∞a p e ipz for g(z,β).The equation satis?ed by?g is the same as that

for the non-periodic case,but it is solved only at integer values of the{k j}.Thus,if?g(k,β)is the solution of the continuous case we have the solution a p(β)=?g(p,β).The corresponding solution,g2π(z,β),is the periodic extension of the continuous case expressed in the form

g2π(z,β)=

∞

p=?∞g(z?2πp)(32)

where g(z)is the nonperiodic solution determined by(19)with the particular forms(21),(22), (24),(28)and(31).

For the function F(z)=1/sinh2z this corresponds to the?function.Since without loss of generality we have taken F(z)to satisfy z2F(z)=?1as z→0,we can use the scaling property ?(z,g2,g3)=a2? az,g2/a4,g3/a6 to take F(z)=??(z).The addition of a constant to F gives rise to the addition of a constant to g,thus

F→F+B? F→ F+Bz+B1?g→g?iBβh(?iβ).(33) Hence a periodic solution corresponding to(21)is

g(z,β)=?h(?iβ) ζ(z)?ζ(z?iβ)+C(β) (34) where again C(β)is an arbitrary function ofβ,which also depends on the parameterν,and is determined by the condition g(ν,β)=0.Thus

g(z,β)=?h(?iβ){ζ(z)?ζ(z?iβ)?ζ(ν)+ζ(ν?iβ)}

=h(?iβ)σ(ν)σ(?iβ)

σ(?z)σ(ν)

σ(z+ν?iβ)

2.4Another Functional Equation

At this stage we have proven the theorem.We have shown that the Fourier transform of(1) with k=1leads to studying

g(z,β)≡h(?iβ)(F1(z?iβ)?F1(z)+C(β))=h(?z)h(z?iβ).(37) Here F1(z)= F(z)dz and C(β)=?F1(ν?iβ)+F1(ν),since we stipulated g(ν,β)=0. We found that F(z)had to be a solution given by theorem3,and that amongst these the only such solutions allowing the desired factorisation were given by(2).We shall conclude by showing that an analysis of(37)directly yields this result.

We view(37)as a functional equation for C,h and F1,with g being consequently de-termined.We solve this subject to appropriate conditions that we inherit from the original problem:

zh(z)→a,zF1(z)→a,and g(ν,β)=0.(38) The last condition means that C(β)=?F1(ν?iβ)+F1(ν),and so C(0)=0.We observe that(37)is invariant under

F1(z)→F1(z)+Bz+B1,C(β)→C(β)+iβB.

We may?x this freedom by further requiring that

C(0)=C′(0)=0.(39) To solve(37)we expand the equation as a power series inβ.The?rst non-trivial terms give su?cient equations to eliminate h(z)and h(?z)and derive a third order equation for F1(z)in terms of the coe?cients{C0,C1,b0,b1,b2}in the expansions

C(β)=

∞

j=0C jβj+2and h(?iβ)=∞ j=0b j(?iβ)j?1.(40)

The resulting third order equation for F1,

(2F′′′1(x)F′1(x)?3F′′1(x)2)b20+12C20b20+12F′1(x)2(b21?2b2b0)+24iC1b20F′1(x)=0,(41) can be integrated to give

F1(z)=b0ζ(z,g2,g3)+z b0?(ν,g2,g3),(42) with C determined as

C(β)=b0ζ(ν)?b0ζ(ν?iβ)+iβb0?(ν).(43) (The remaining constants in this are de?ned in terms of C0,C1,b1and b2below.) The function h(z)is then determined by the equation

h′

2F′′1

2

b0?′(ν)+

b1

σ(z)σ(ν)

eαz.(45)

The four constants{C0,C1,b1,b2}de?ne the four constants appearing in(42-43)namely ?(ν),ξ(ν),g2and?′(ν),with g3given by{?(ν),?′(ν)and g2}.Of course,to satisfy(38) we also require b0=a.The relations between the two sets of constants are given by

g2 ,

C0=?1

6

b1=b0(ζ(ν)+α),b2=1

[11]J.G.B.Byatt-Smith and H.W.Braden,Functional Equations and Poincar′e Invariant Me-

chanical Systems,Preprint EMPG-01-12.

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ticles of unequal mass,(Russian)Uspekhi Mat.Nauk49207-208(1994);translation in Russian Math.Surveys49233-234(1994).

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一般过去式 时间状语：yesterday just now (刚刚) the day before three days ag0 a week ago in 1880 last month last year 1. I was in the classroom yesterday. I was not in the classroom yesterday. Were you in the classroom yesterday. 2. They went to see the film the day before. Did they go to see the film the day before. They did go to see the film the day before. 3. The man beat his wife yesterday. The man didn’t beat his wife yesterday. 4. I was a high student three years ago. 5. She became a teacher in 2009. 6. They began to study english a week ago 7. My mother brought a book from Canada last year. 8.My parents build a house to me four years ago . 9.He was husband ago. She was a cooker last mouth. My father was in the Xinjiang half a year ago. 10.My grandfather was a famer six years ago. 11.He burned in 1991

伯努利方程推导

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初中英语造句

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