The parametrization of the Marchenko-Ostrovsky mapping in terms of the Dirichlet eigenvalues

0.25 MB·English

by  Maria Evgenievna Korotyaeva

The parametrization of the Mar
henko-Ostrovsky
mapping in terms of the Diri
hlet eigenvalues
8 ∗
0 Maria Evgenievna Korotyaeva
0
2
February 2, 2008

n
a
J

5

Abstra
t
]
P
S We
onsider the inverse spe
tral problem for periodi
Ja
obi matri
es in terms of the
. verti
al slits on the quasi-momentum domain plus the Diri
hlet eigenvalues, i.e., the
h
t Mar
henko-Ostrovsky mapping. Moreover, we show that the gradients of the Diri
hlet
a
m eigenvalues and of the so-
alled norming
onstants are linear independent.
[

1
v
7 1 Introdu
tion
0 ℓ2 = ℓ2(Z)
8
0 We
onsid(eryt)hn∈eZse=lf-aandyjno+in1t+Nb-npyenri+odain
−J1ay
no−b1i opeyra=to(rynJ)n∈oZn aℓH2ilbert spa
e N
1. given by Ja = exn > 0, x ,b R. for ∈ and for the -periodi
n n n
0 sequen
es ∈ Furthermore, we assume
8
N
v:0 q = (x,b) ≡ (xn,bn)n∈ZN ∈ H 2, H ≡ {b ∈ RN : Xbn = 0}. (1.1)
n=1
i
X
Z = 1,2,...,N , N N,
r N
a For simpli
ity, we use the notation { } ∈ throughout this paper. To
ϕ = ϕ(λ,q) = (ϕn(λ,q))n∈Z
begin, we re
all some well known fa
ts (see, e.g., [vM℄). Let and
ϑ = ϑ(λ,q) = (ϑn(λ,q))n∈Z
denote two fundamental solutions of the equation
a y +b y +a y = λy , (λ,n) C Z,
n−1 n−1 n n n n+1 n
∈ × (1.2)
ϕ ϑ 0, ϕ ϑ 1. ∆(λ,q) =
0 1 1 0
w1(iϕth th(eλi,nqi)ti+alϑ
on(dλi,tqio))ns ≡ ≡ ≡ ≡ The Lyapunov fun
tion
2 N+1 σ(q) =Nλ R :is∆t(hλe,qd)is
ri1minant.of the equation (1.2) and
hara
terizes the
spe
trum N { ∈ σ |= [λ+ |,λ≤−],}nof JZ T,he spe
trum of J is aγbso=lu(tλel−y,
λo+n)t,inuous
aλn±d=
oλn±si(sqt)sof bands n ∆2n(−λ1,q)n= 1 ∈ N separλat+edbyλ+th<egλa−ps nλ+ < n... <n λ−where
λ+n <nλ−.are the roots of γ and satisfy N ≡ 0 1 ≤ 1 σ ,N−σ1 ≤
N−1 N That is, if a gap n degenerates, then the
orresponding segments n n+1
∗
Mathematis
hes Institut, Humboldt-Universität zu Berlin, e-mail: korotiaevayahoo.de
1
λ = λ (q) [λ−,λ+] n Z
merge. Moreover, there is exa
tly one point n n ∈ n n for ea
h ∈ N−1 su
h
that ∆′(λ ,q) = 0, ∆′′(λ ,q) = 0, ( 1)N−n∆(λ ,q) 1.
n n n
6 − ≥ (1.3)
(′) = ∂/∂λ
Here and below, .
Traditionally sin
e [vM℄, the inverse spe
tralµpr=obµlem(q)f,orntheZperio,di
Ja
obi operator
n n N−1
has been solved using the Neumann eigenvalues ∈ given by the zeroes
ϑ (λ,q) = 0.
N+1
of The main goal of this pνape=r iνs (tqo),sonlve tZhe in,verse spe
tral problem
n n N−1
alternatively using the Diri
hlet eigenvalues ∈ given by the zeroes of
ϕ (λ,q) = 0.
N
That is, we de(cid:28)ne theµau,xνiliary[λs−p,eλ
t+r]u,mn byZthe D. iri
hlet spe
trum instead of
the Neumann spe
trum. Note that n n ∈ n n ∈ N−1
To outline the plan of this note, we re
all that the inverse spe
tral problem
onsists of
the following four parts, namely,
i) The uniqueness. Prove that the spe
tral data uniquely determines the potential.
ii) The re
onstru
tion. Give an algorithm to re
over the potential from the spe
tral data.
iii) The
hara
terization. Give the
onditions for some data to be the spe
tral data of some
potential.
iv) The stability. Give the two-sided a priori estimates of the potential in terms of the
spe
tral data. ψ : H 2 R2N−2
oWpeer
aotnosrtirnu
tteramMs oafr
thheenDkoir-iO
hsltertovesigkeynvmaalupepsinνgn by ψ = (ψ→n)n∈ZN−1,foψrnt=he(ψp1e,nri,oψd2i,
n)J∈a
Rob2i,
where ψ1,n = log((−1)N−nϕN+1(νn)), ψ2,n = (|ψn|2 −ψ12,n)21 sign(λn −νn), (1.4)
cosh ψ = ( 1)N−n∆(λ ).
n n
| | − (1.5)
ψ ψ 2 ψ 2 0
1,n n 1,n
Here, is the so-
alled norming
onstant. It is easy to verify | | − | | ≥ sin
e
ϕ ϑ ϕ ϑ = 1
N+1 N N N+1
(1.4), (1.5) and the Wronskian identity − together imply
( 1)N−n∆(ν ,q) = coshψ (q).
n 1,n
− (1.6)
ψ
Note in
identally that the mapping is an analogue of the Mar
henko-Ostrovski mapping
[MO℄ for the
ontinuous
ase and has similar properties (see [MO℄, [Ko℄).
Firstly, we will prove the
hara
terization and the uniqueness showing that the mapping
{potential} → {spe
tral data} is a homeomorphism.
ψ : H 2 R2N−2
Theorem 1.1H. T2he maRp2pNin−g2 → is a real analyti
isomorphism between the
Hilbert spa
es and .
H ,H
0
Remark. Wefre:
aHll someHne
essary de(cid:28)qnitioHns. Let be Hilbert spa
es.HThe derHiva-
0 0
tive of a map d→f at a fpo:inHt ∈ H is a bounded linear map from into H,
q 0
whi
Hh we dfenote by . A map f f→−1 is a real analyti
isomorphism between
0
and if is bije
tive and both and are real analyti
maps of the spa
e.
Se
ondly, we will obtain the re
onstru
tion and the stability. We use for it the geometri
interpretation of the Mar
henko-Ostrovsky mapping, whi
h is similar to the
ontinues
ase
2
mentioned in [MO℄ and [Ko℄. For this purpose, we introdu
e the
onformal mapping (the
κ : Λ K
quasi-momentum) → by
cosκ(λ) = ( 1)N∆(λ,q), λ Λ, and κ(it) i as t .
− ∈ → ± ∞ → ±∞ (1.7)
Λ = C N−1γ K = κ : 0 6 Reκ 6 Nπ N−1κ(γ )
1 n 1 n
Here and below, \∪ is the domain, { }\∪ is
Γ = (πn+i ψ ,πn i ψ )
n n n

alled the quasi-momentum domain and | | − | | is an ex
ised verti
al
slit.
ψ R2N−2, q H 2
Theorem 1.2. i) For ea
h ∈ there is a unique point ∈ and a unique
κ : Λ K

onformal mapping → su
h that the following identities hold true
κ(λ (q) i0) = πn i ψ (q) , κ(ν (q) i0) = πn iψ (q), n Z .
n n n 1,n N−1
± ± | | ± ± ∈ (1.8)
ii) For (ψ1,n)n∈ZN−1, there is a standard algorithm to re
over a,b.
iii) The following two-sided estimates hold true
1 1
e2max|ψn| < (λ− λ+)2 < b2 +2a2 < N(λ− λ+)2 < 16Ne2max|ψn|.
4 4 N − 0 N − 0 (1.9)
A
ru
ial argument in the proof of Theorem 1.1 is the fa
t that the gradients of the
Diri
hlet eigenvalues and of the norming
onstants are linear independent. More pre
isely,
we de(cid:28)ne the symple
ti
form ∧ by
N
f g = (f g f g ) (f g f g )
∧ X 1,n 2,n − 2,n 1,n − 1,n−1 2,n − 2,n 1,n−1 (1.10)
n=1
for f = (f1,δn,f2,n)n∈ZN, g = (g1,n,g2,n)n∈ZN ∈ C2N withn,fm1,0 ≡Zf1,N and g1,0 ≡ g1,N. Note
n,m
that below stands for the Krone
ker symbol for all ∈ . Then we show
n,m Z ,
N−1
Theorem 1.3. For all ∈ it holds true
d ν d ν = 0,
q n q m
∧ (1.11)
d ψ d ψ = 0,
q 1,n q 1,m
∧ (1.12)
d ψ d ν = 2δ .
q 1,n q m n,m
∧ (1.13)
d ν , d ψ , n Z , H 2.
q n q 1,n N−1
In parti
ular, ∈ is a basis of
Pös
hel and Trubowitz [PT℄ proved an analogue of Theorem 1.3 for the Sturm-Liouville
[0,1].
problem on the interval We use their arguments in our proof. Note that van Moerbeke
µ
n
[vM℄ proved (using another approa
h) that the gradients of the Neumann eigenvalues and
of the norming
onstants are linear independent. Namely, van Moerbeke used the Ja
obi
matri
es with removed rows and
olumns. Remark that our proof also
an be applied for
µ .
n
the
ase of
There are di(cid:27)erent approa
hes to the inverse spe
tral problem for the periodi
Ja
obi
operator. The investigation on this topi
started in 1976 by van Moerbeke [vM℄ and by Date
3
andTanaka [DT℄. Bothworks obtainedthe re
onstru
tion, but notthe
hara
terization: van
Moerbeke did it using the Stieltjes inverse spe
tral method from [Ah℄ or [GK℄, and Date and
Tanaka did it applying the su(cid:30)x shifting by a
onstant. The nonlinear Toda latti
e turned
out an important appli
ation of these methods (see [To℄). The (cid:28)rst for us known work on
the
hara
terization is the paper [Pe℄ by Perkolab, where some analogue of Theorem 1.2 is
showed using [MO℄. Further, Korotyaev and Kutsenko [KoKu℄ showed Theorems 1.1 and 1.2
in terms of the Neumann eigenvalues applying approa
h [KaKo℄. That is, [KoKu℄ extended
the result of Mar
henko and Ostrovski about the height-slit mapping for the Hill operator
(see [MO℄, [Ko1℄) to the
ase of the periodi
Ja
obi matrix using the Neumann eigenvalues.
Lastly, the inverse problem in terms of the gap lengths was solved in [BGGK℄ based on the
approa
h from [GT℄ and in [Ko1℄ based on the approa
h from [KaKo℄.
Our note isorganizedas follows: Se
tion 2 displays somepreliminarystatements interms
ν .
n
of In Se
tion 3 we prove Theorem 1.3; this is te
hni
ally the most di(cid:30)
ult part of this
note. In Se
tion 4, we show Theorem 1.1 using the argument from [Ko℄ and [KoKu℄, where
µ
n
this theorem in terms of is proved. Then Theorem 1.1 together with [KoKr℄ and [KoKu℄
dire
tly implies Theorem 1.2.
2 Preliminaries
q
Inthisse
tion, we willdeterminethegradients(withrespe
t to )oftheDiri
hleteigenvalues
andthe norming
onstants interms ofthe fundamentalsolutions. Forthis purpose, we de(cid:28)ne
the Wronskian by f,g = a (f g f g ), n Z,
n n n n+1 n+1 n
{ } − ∈ (2.1)
f = (fn)n∈Z, g = (gn)n∈Z fn,gn C.
for the sequen
es with ∈ Below, we use the notation
∂ = ∂q = (∂xk,∂bk)k∈ZN.
ν , ψ , n Z , H 2
n 1,n N−1
Lemma 2.1. Ea
h from the fun
tions ∈ is real analyti
on and
satis(cid:28)es
∂ϕ (ν (q),q)
N n
d ν = ,
q n − ϕ′ (ν (q),q) (2.2)
N n
ϕ′ (ν (q),q)d ν +∂ϕ (ν (q),q)
d ψ = N+1 n q n N+1 n .
q 1,n
ϕN+1(νn(q),q) (2.3)
Proof. This proof is similar to the
ontinues
ase [Ko℄ (see also [KoKu℄).
ϕˆ = ϕ(ν (q),q),ϑˆ = ϑ(ν (q),q) (n,q) Z H 2.
n n N−1
LkemZma 2.2. Let for all ∈ × Then for all
N
∈ , the following identities hold
(2a ϕˆ ϕˆ ,ϕˆ2)
d ν = k k k+1 k ,
qk n − a ϕˆ ϕˆ′ (2.4)
N N+1 N
′ ˆ ′ ˆ
d ψ = B +(ϕˆ ϑ ϕˆ ϑ )d ν ,
qk 1,n n,k N+1 N − N N+1 q n (2.5)
1
ˆ ˆ ˆ
B = (a (ϕˆ ϑ +ϕˆ ϑ ),ϕˆ ϑ ).
n,k k k+1 k k k+1 k k
a (2.6)
N
4
k,j Z
N
Proof. We assume ∈ .
i) We want to show (2.4) applying (2.2). That is, we have to determine the derivation
∂ ϕˆ . ∂ ϕˆ ϕ
qk N Firstly, we
al
ulate xk N using the equation (1.2) for j
a ϕ +(b λ)ϕ +a ϕ = 0,
j−1 j−1 j j j j+1
−
x
k
and its derivation with respe
t to
a ∂ ϕ +(b λ)∂ ϕ +a ∂ ϕ = a (δ ϕ +δ ϕ +δ δ ϕ ).
j−1 xk j−1 j − xk j j xk j+1 − k j,k k+1 j,k+1 k k,N j,1 0
χ χ = 1 k < N χ = 0
k<N k<N k<N
Below, stands for the
hara
teristi
fun
tion, i.e. for and
k N N ∂ ϕ
for ≥ϕ (re
alling that is (cid:28)xed). Multiplyingthe (cid:28)rst equatjionZby , xk j and the se
ond
j N
one by and taking the di(cid:27)eren
e, we sum the result over all ∈ that is,
2a ϕ ϕ = a (χ 2ϕ ϕ +δ (ϕ ϕ +ϕ ϕ ))
k k k+1 k k<N k k+1 k,N 0 1 N N+1
N
= a (ϕ ∂ ϕ ϕ ∂ ϕ )+a (ϕ ∂ ϕ ϕ ∂ ϕ )
X j−1 j−1 xk j − j xk j−1 j j+1 xk j − j xk j+1
j=1
N
= ∂ ϕ,ϕ ∂ ϕ,ϕ = ∂ ϕ,ϕ ∂ ϕ,ϕ = ∂ ϕ,ϕ
X{ xk }j −{ xk }j−1 { xk }N −{ xk }0 { xk }N
j=1
ϕ = 0. λ = ν ϕˆ = 0,
0 n N
sin
e Next, setting and re
alling it gives
2a ϕˆ ϕˆ
k k k+1
∂ ϕˆ = .
xk N aNϕˆN+1 (2.7)
∂ ϕˆ ϕ
Se
ondly, we
al
ulate bk N using the equation (1.2) for j
a ϕ +(b λ)ϕ +a ϕ = 0,
j−1 j−1 j j j j+1
−
b
k
and its derivation with respe
t to
a ∂ ϕ +(b λ)∂ ϕ +a ∂ ϕ = δ ϕ .
j−1 bk j−1 j − bk j j bk j+1 − j,k k
∂ ϕ ϕ
Multiplying the (cid:28)rst equatjion bZy ,bk j and the se
ond one by j and taking the di(cid:27)eren
e,
N
we sum the result over all ∈ that is,
N
2
ϕ = a (ϕ ∂ ϕ ϕ ∂ ϕ )+a (ϕ ∂ ϕ ϕ ∂ ϕ )
k X j−1 j−1 bk j − j bk j−1 j j+1 bk j − j bk j+1
j=1
N
= ∂ ϕ,ϕ ∂ ϕ,ϕ = ∂ ϕ,ϕ ∂ ϕ,ϕ = ∂ ϕ,ϕ
X{ bk }j −{ bk }j−1 { bk }N −{ bk }0 { bk }N
j=1
ϕˆ2
ϕ = 0. λ = ν , ϕˆ = 0 ∂ ϕˆ = k
sin
e 0 Setting n we see that N implies bk N aNϕˆN+1 and by (2.7), we
get (2.4).
5
∂ ϕˆ
ii) In order to prove (2.5) and (2.6), we determine qk N+1 and then substitute it into
∂ ϕˆ ϕ
the identity (2.3). Firstly, we
al
ulate xk N+1 using the equation (1.2) for j
a ϕ +(b λ)ϕ +a ϕ = 0,
j−1 j−1 j j j j+1
−
ϑ x
j k
and the derivation of the equation (1.2) for with respe
t to
a ∂ ϑ +(b λ)∂ ϑ +a ∂ ϑ = a (δ ϑ +δ ϑ +δ δ ϑ ).
j−1 xk j−1 j − xk j j xk j+1 − k j,k k+1 j,k+1 k k,N j,1 0
∂ ϑ ϕ
Multiplying the (cid:28)rst equatjion bZy ,xk j and the se
ond one by j and taking the di(cid:27)eren
e,
N
we sum the result over all ∈ that is,
a (χ (ϕ ϑ +ϕ ϑ )+δ (ϕ ϑ +ϕ ϑ ))
k k<N k k+1 k+1 k k,N 1 0 N+1 N
N
= a (ϕ ∂ ϑ ϕ ∂ ϑ )+a (ϕ ∂ ϑ ϕ ∂ ϑ )
X j−1 j−1 xk j − j xk j−1 j j+1 xk j − j xk j+1
j=1
N
= ∂ ϑ,ϕ ∂ ϑ,ϕ = ∂ ϑ,ϕ ∂ ϑ,ϕ = ∂ ϑ,ϕ
X{ xk }j −{ xk }j−1 { xk }N −{ xk }0 { xk }N
j=1
∂ ϑ = ∂ ϑ = 0. λ = ν ,
sin
e xk 0 xk 1 Setting n we get
ˆ ˆ ˆ
a (ϕˆ ϑ +ϕˆ ϑ ) = ∂ ϑ,ϕˆ .
k k k+1 k+1 k { xk }N
ϕˆ = 0
N
We observe that implies
ˆ ˆ ˆ ˆ ˆ
a (ϕˆ ϑ +ϕˆ ϑ ) = a ϕˆ ∂ ϑ = a ϑ ∂ ϕˆ +a ϑ ∂ ϕˆ
k k k+1 k+1 k N N+1 xk N − N N xk N+1 N N+1 xk N
∂ ϑ,ϕ = 0
sin
e xk{ }N and hen
e
a
k ˆ ˆ ˆ ˆ
∂ ϕˆ = (ϑ (ϑ ϕˆ +ϕˆ ϑ )+2ϑ ϕˆ ϕˆ ).
xk N+1 −a N k k+1 k k+1 N+1 k k+1 (2.8)
N
∂ ϕˆ ϕ
Se
ondly, we
al
ulate bk N+1 using the equation (1.2) for j
a ϕ +(b λ)ϕ +a ϕ = 0,
j−1 j−1 j j j j+1
−
ϑ b
j k
and the derivation of the equation (1.2) for with respe
t to
a ∂ ϑ +(b λ)∂ ϑ +a ∂ ϑ = δ ϑ .
j−1 bk j−1 j − bk j j bk j+1 − j,k k
∂ ϑ ϕ
Multiplying the (cid:28)rst equatjion bZy ,bk j and the se
ond one by j and taking the di(cid:27)eren
e,
N
we sum the result over all ∈ that is,
N
ϕ ϑ = a (ϕ ∂ ϑ ϕ ∂ ϑ )+a (ϕ ∂ ϑ ϕ ∂ ϑ )
k k X j−1 j−1 bk j − j bk j−1 j j+1 bk j − j bk j+1
j=1
6
N
= ∂ ϑ,ϕ ∂ ϑ,ϕ = ∂ ϑ,ϕ ∂ ϑ,ϕ = ∂ ϑ,ϕ
X{ bk }j −{ bk }j−1 { bk }N −{ bk }0 { bk }N
j=1
∂ ϑ = ∂ ϑ = 0. λ = ν , ϕˆ = 0
sin
e bk 0 bk 1 We set n then N implies
ˆ ˆ ˆ ˆ
ϕˆ ϑ = a ϕˆ ∂ ϑ = a ϑ ∂ ϕˆ +a ϑ ∂ ϕˆ ,
k k N N+1 bk N − N N bk N+1 N N+1 bk N
∂ ϑ,ϕ = 0. ∂ ϕˆ = a−1(ϕˆ ϕˆ ϑˆ ϑˆ ϕˆ2)
by bk{ }N Therefore, bk N+1 N N+1 k k − N+1 k and (2.8) yield
1
ˆ ˆ ˆ ˆ 2
∂ ϕˆ = ϕˆ (a (ϕˆ ϑ +ϕˆ ϑ ),ϕˆ ϑ ) ϑ (2a ϕˆ ϕˆ ,ϕˆ ) .
qk N+1 a (cid:16) N+1 k k+1 k k k+1 k k − N+1 k k k+1 k (cid:17)
N
Substituting this expression into the identity (2.3), one obtains (2.5) and (2.6).
3 The proof of Theorem 1.3.
d ν , d ψ , n Z , H 2
q n q 1,n N−1
In this Se
tion, we will show that ∈ is a basis of using the
symple
ti
form de(cid:28)ned in (1.10). First of all, we display some identities for the Wronskian
de(cid:28)ned in (2.1).
ˆ ˜
ϕˆ = ϕ(ν (q),q),ϑ = ϑ(ν (q),q) ϕ˜ = ϕ(ν (q),q),ϑ = ϑ(ν (q),q), n,m
n n m m
LZemm,qa3H.1.2L.et k Z , and ∈
N−1 N
∈ Then for all ∈ the following identities hold true
N k
ϕˆ ϕ˜ = 0, ϕ˜,ϕˆ = (ν ν ) ϕˆ ϕ˜ ,
X j j { }k n − m X i i (3.1)
j=1 i=1
N ˜ ˆ k
ϑ,ϑ
ˆ ˜ N ˜ ˆ ˆ ˜
ϑ ϑ = { } , ϑ,ϑ = (ν ν ) ϑ ϑ ,
X j j (ν ν ) { }k n − m X i i (3.2)
j=1 n − m i=1
N ˆ k
a (1 ϕ˜ ϑ )
ˆ 0 N+1 N ˆ ˆ
ϑ ϕ˜ = − , ϕ˜,ϑ = a +(ν ν ) ϑ ϕ˜ ,
X j j (ν ν ) { }k − 0 n − m X i i (3.3)
j=1 n − m i=1
N ˜ k
a (ϑ ϕˆ 1)
˜ 0 N N+1 ˜ ˜
ϕˆ ϑ = − , ϑ,ϕˆ = a +(ν ν ) ϕˆ ϑ ,
X j j (ν ν ) { }k 0 n − m X i i (3.4)
j=1 n − m i=1
N
2 ′
ϕˆ = a ϕˆ ϕˆ .
X j 0 N+1 N (3.5)
j=1
n,m Z j,k Z
N−1 N
Proof. We assume ∈ and ∈ .
ϕˆ ϕ˜
j j
i) To show (3.1), we
onsider the equation (1.2) for and respe
tively
a ϕˆ +b ϕˆ +a ϕˆ = ν ϕˆ ,
j−1 j−1 j j j j+1 n j
a ϕ˜ +b ϕ˜ +a ϕ˜ = ν ϕ˜ .
j−1 j−1 j j j j+1 m j
7
ϕ˜ ϕˆ
j j
Multiplying the (cid:28)rst equation by and the se
ond one by and taking the di(cid:27)eren
e, we
get
a (ϕ˜ ϕˆ ϕ˜ ϕˆ )+a (ϕ˜ ϕˆ ϕ˜ ϕˆ ) = (ν ν )ϕˆ ϕ˜
j−1 j j−1 j−1 j j j j+1 j+1 j n m j j
− − −
or equivalently, by the de(cid:28)nition (2.1) of the Wronskian,
ϕ˜,ϕˆ ϕ˜,ϕˆ = (ν ν )ϕˆ ϕ˜ .
j j−1 n m j j
{ } −{ } − (3.6)
j Z ϕ˜ = ϕˆ = ϕ˜ = ϕˆ = 0,
N 0 0 N N
Summing (3.6) over all ∈ and using we get the (cid:28)rst identity
in (3.1)
N
(ν ν ) ϕˆ ϕ˜ = ϕ˜,ϕˆ ϕ˜,ϕˆ = 0.
n − m X j j { }N −{ }0
j=1
k. k = 1,
Next, we show the se
ond identity in (3.1) by indu
tion on For the identity holds
ϕ˜,ϕˆ = 0. k 1,
0
a

ording to (3.6) sin
e { } Assuming that the identity holds for − we will
k :
verify it for By (3.6) and the indu
tion hypothesis, it follows
k−1 k
ϕ˜,ϕˆ ϕ˜,ϕˆ
k k−1
{ } = { } +ϕˆ ϕ˜ = ϕˆ ϕ˜ +ϕˆ ϕ˜ = ϕˆ ϕ˜ .
(ν ν ) (ν ν ) k k X i i k k X i i
n − m n − m i=1 i=1
This establishes the se
ond identity in (3.1). ˜ ˆ
ϑ,ϑ = 0.
0
ii) The proof of (3.2) is similar to that of (3.1) sin
e { } ˆ
ϑ ϕ˜
j j
iii) To show (3.3), we
onsider again the equation (1.2) for and respe
tively
ˆ ˆ ˆ ˆ
a ϑ +b ϑ +a ϑ = ν ϑ ,
j−1 j−1 j j j j+1 n j
a ϕ˜ +b ϕ˜ +a ϕ˜ = ν ϕ˜ .
j−1 j−1 j j j j+1 m j
ˆ
ϕ˜ ϑ
j j
Multiplying the (cid:28)rst equation by and the se
ond one by and taking the di(cid:27)eren
e, we
get ˆ ˆ ˆ ˆ ˆ
a (ϕ˜ ϑ ϕ˜ ϑ )+a (ϕ˜ ϑ ϕ˜ ϑ ) = (ν ν )ϑ ϕ˜
j−1 j j−1 j−1 j j j j+1 j+1 j n m j j
− − −
or equivalently ˆ ˆ ˆ
ϕ˜,ϑ ϕ˜,ϑ = (ν ν )ϑ ϕ˜ .
j j−1 n m j j
{ } −{ } − (3.7)
j Z ϕ˜,ϑˆ = a , a = a , ϕˆ = 0,
N 0 0 0 N N
Next, summing (3.7) over all ∈ and using { } − the (cid:28)rst
statement in (3.2) follows
N
ˆ ˆ ˆ ˆ
(ν ν ) ϑ ϕ˜ = ϕ˜,ϑ ϕ˜,ϑ = a (1 ϕ˜ ϑ ).
n − m X j j { }N −{ }0 0 − N+1 N
j=1
k. k = 1,
We proveˆthe se
ond identityˆ (3.2) by indu
tion on For it holds a

ording to (3.7),
ϕ˜,ϑ = a ϕ˜ ϑ = 0. k 1, k.
0 0 1 1
sin
e { } − and Supposing the identity is truth for − prove it for
(3.7) and the indu
tion hypothesis together imply
k−1
ˆ ˆ ˆ ˆ ˆ
ϕ˜,ϑ = ϕ˜,ϑ +(ν ν )ϑ ϕ˜ = (ν ν ) ϑ ϕ˜ +(ν ν )ϑ ϕ˜ a =
{ }k { }k−1 n − m k k n − m X i i n − m k k − 0
i=1
8
k
ˆ
= (ν ν ) ϑ ϕ˜ a .
n − m X i i − 0
i=1
So the se
ond identity in (3.2) follows. ˜ ˜
ϑ,ϕˆ = a ϑ,ϕˆ =
0 0 N
˜iv)The proof˜of (3.4) issimilartothat of (3.3), sin
e we have { } and { }
a ϑ ϕˆ = a ϑ ϕˆ .
N N N+1 0 N N+1
ϕ (λ) λ
j
v) To verify (3.5), we use the equation (1.2) for and its gradient with respe
t to
a ϕ (λ)+(b λ)ϕ (λ)+a ϕ (λ) = 0,
j−1 j−1 j j j j+1
−
′ ′ ′
a ϕ (λ)+(b λ)ϕ (λ)+a ϕ (λ) = ϕ (λ).
j−1 j−1 j − j j j+1 j
′
ϕ (λ) ϕ (λ)
j j
Multiplyingthe(cid:28)rstequationby andthese
ondoneby andtakingthedi(cid:27)eren
e,
we get
′ ′ ′ ′ 2
a (ϕ (λ)ϕ (λ) ϕ (λ)ϕ (λ))+a (ϕ (λ)ϕ (λ) ϕ (λ)ϕ (λ)) = ϕ (λ)
j−1 j j−1 − j−1 j j j j+1 − j+1 j j
or equivalently ′ ′ 2
ϕ (λ),ϕ(λ) ϕ (λ),ϕ(λ) = ϕ (λ).
{ }j −{ }j−1 j (3.8)
j Z ϕ ϕ′ 0,
N 1 1
Next, summing (3.8) over all ∈ and using ≡ ≡ we obtain
N
2 ′ ′ ′
ϕ (λ) = ϕ (λ),ϕ(λ) ϕ (λ),ϕ(λ) = ϕ (λ),ϕ(λ) .
X j { }N −{ }0 { }N
j=1
λ = ν ϕˆ = 0, N ϕˆ2 = ϕˆ′,ϕˆ = a ϕˆ ϕˆ′ .
Setting n and re
alling N it be
omes Pj=1 j { }N N N N
n Z
N−1
Nowwe willapplythede(cid:28)nition(1.10)ofthesymple
ti
form: For ∈ , we
onsider
ddqννn = (ddqkννn)k∈ZNBgivenBin (2.4) andnBnZ= (Bn,k)k∈qZN=dqe(cid:28)n.ed in (2.6). We observe that
q0 n ≡ qN n and n,0 ≡ n,N for all ∈ N−1 sin
e 0 N So we
an de(cid:28)ne a symple
ti
d ν B
q n n
form for and .
n,m Z ,
N−1
Theorem 3.2. For all ∈ the following identities hold
d ν d ν = 0,
q n q m
∧ (3.9)
B B = 0,
n m
∧ (3.10)
B d ν = 2δ .
n q m n,m
∧ − (3.11)
n,m Z ϕˆ =
N−1
Pϕ(rνoo(qf.),qW),eϑˆa=ssuϑm(νe(q),q)∈ ϕ˜ =. ϕF(uνrth(qe)r,mqo),rϑe˜,=weϑ(uνse(qt)h,eq)followingq abHbre2v.iations
n n m m
and for all ∈
n = m n = m
i) To verify the identity (3.9), let 6 sin
e the
aseC = 2(a2isϕˆobviϕˆo′usϕ˜. Apϕp˜l′y)in−g1,the
N N+1 N N+1 N
de(cid:28)nition (1.10) of the symple
ti
form and introdu
ing we
have
N
2 2 2 2
d ν d ν = C a (ϕˆ ϕˆ ϕ˜ ϕˆ ϕ˜ ϕ˜ ) a (ϕˆ ϕˆ ϕ˜ ϕˆ ϕ˜ ϕ˜ ) .
q n ∧ q m X(cid:16) k k k+1 k − k k k+1 − k−1 k−1 k k − k k−1 k (cid:17)
k=1
9
Using the de(cid:28)nition (2.1) of the Wronskian, this equals
N
d ν d ν = C ϕˆ ϕ˜ ( ϕ˜,ϕˆ + ϕ˜,ϕˆ ).
q n ∧ q m X k k { }k { }k−1
k=1
By (3.1), it follows
N k−1 k N k−1 N k
d ν d ν
q n q m
∧ = ϕˆ ϕ˜ ( ϕˆ ϕ˜ + ϕˆ ϕ˜ ) = ϕˆ ϕ˜ ϕˆ ϕ˜ + ϕˆ ϕ˜ ϕˆ ϕ˜ .
C(ν ν ) X k k X i i X i i X k kX i i X k kX i i
n − m k=1 i=1 i=1 k=1 i=1 k=1 i=1
Below, we need the following simple identities
N k N N N k−1 N−1 N
z w = w z , z w = w z
X kX i X kX i X kX i X k X i (3.12)
k=1 i=1 k=1 i=k k=2 i=1 k=1 i=k+1
for all z = (zk)k∈ZN,w = (wk)k∈ZN ∈ CN. Then we
an use (3.12) and (3.5) to obtain
N k−1 N N N N
d ν d ν
q n q m
∧ = ϕˆ ϕ˜ ϕˆ ϕ˜ + ϕˆ ϕ˜ ϕˆ ϕ˜ = ϕˆ ϕ˜ ϕˆ ϕ˜ = 0,
C(ν ν ) X k kX i i X k kX i i X k kX i i
n − m k=1 i=1 k=1 i=k k=1 i=1
whi
h vanishes by the sum identity in (3.1).
n = m,
ii)We verify the identity (3.10) in the same manner like (3.9). Again, we
onsider 6
n = m
sin
e the
ase is obvious, and get
N
1
ˆ ˆ ˜ ˆ ˜ ˜
B B = a (ϕˆ ϑ +ϕˆ ϑ )ϕ˜ ϑ ϕˆ ϑ (ϕ˜ ϑ +ϕ˜ ϑ )
n ∧ m a2 X k(cid:16) k+1 k k k+1 k k − k k k+1 k k k+1 (cid:17)
N k=1
ˆ ˆ ˜ ˆ ˜ ˜
a (ϕˆ ϑ +ϕˆ ϑ )ϕ˜ ϑ ϕˆ ϑ (ϕ˜ ϑ +ϕ˜ ϑ )
k−1(cid:16) k k−1 k−1 k k k k k k k−1 k−1 k (cid:17)
− −
N
1
ˆ ˜ ˆ ˜ ˆ ˜
= a ϕ˜ ϕ˜ (ϑ ϑ ϑ ϑ )+ϑ ϑ (ϕˆ ϕ˜ ϕˆ ϕ˜ )
a2 X k(cid:16) k k k+1 k − k k+1 k k k+1 k − k k+1 (cid:17)
N k=1
ˆ ˜ ˆ ˜ ˆ ˜
+a ϑ ϑ (ϕˆ ϕ˜ ϕˆ ϕ˜ )+ϕˆ ϕ˜ (ϑ ϑ ϑ ϑ )
k−1(cid:16) k k k k−1 k−1 k k k k k−1 k−1 k (cid:17)
− −
N N
1
ˆ ˜ ˜ ˆ ˜ ˆ
= ( ϑ ϑ ( ϕ˜,ϕˆ + ϕ˜,ϕˆ )+ ϕˆ ϕ˜ ( ϑ,ϑ + ϑ,ϑ )).
a2 X k k { }k { }k−1 X k k { }k { }k−1
N k=1 n=1
Using(cid:28)rstly representations of theWronskian in(3.1)and(3.2) andapplying3.12, we obtain
a2 (B B ) N k−1 k N k−1 k
N n ∧ m = ϑˆ ϑ˜ ( ϕˆ ϕ˜ + ϕˆ ϕ˜ )+ ϕˆ ϕ˜ ( ϑˆ ϑ˜ + ϑˆ ϑ˜)
(ν ν ) X k k X i i X i i X k k X i i X i i
n − m k=1 i=1 i=1 k=1 i=1 i=1
10