Lyapunov functions for periodic matrix-valued Jacobi operators

0.3 MB·English

by  Evgeny Korotyaev

Lyapunov fun
tions for periodi
matrix-valued Ja
obi
operators
7
0
0
∗ †
2 Evgeny Korotyaev Anton Kutsenko

n
a February 2, 2008
J

6
1

] Abstra
t
P
We
onsider periodi
matrix-valued Ja
obi operators. The spe
trum of this operator
S
is absolutely
ontinuous and
onsists of intervals separated by gaps. We de(cid:28)ne the
.
h
Lyapunov fun
tion, whi
h is analyti
on an asso
iated Riemann surfa
e. On ea
h sheet
t
a
the Lyapunov fun
tion has the standard properties of the Lyapunov fun
tion for the
m
s
alar
ase. We show that this fun
tion has (real or
omplex) bran
h points, whi
h we
[

all resonan
es. We prove that there exist two types of gaps: i) stable gaps, i.e., the

1 endpoints are periodi
and anti-periodi
eigenvalues, ii) unstable (resonan
e) gaps, i.e.,
v
the endpoints are resonan
es (real bran
h points). We show that some spe
tral data
8
4 determine the spe
trum (
ounting multipli
ity) of the Ja
obi operator.
4
1
0
7 1 Introdu
tion and main results
0
ℓ2(Z)m
/
h Consider self-adjoint matrix-valued Ja
obi operators J a
ting on and are given by
mat (Jy)n = anyn+1+bnyn+a⊤n−1yn−1, n ∈ Z, yn ∈ Cm,y = (yn)n∈Z ∈ ℓ2(Z)m,m > 1, (1.1)
a ,b = b⊤ m m deta = 0
v: wnherZe n n n are p-periodi
sequen
es oσf(th)e × real matri
es and n 6 for all
i ∈ . It is well known that the spe
trum J of J is absolutely
ontinuous and
onsists
X [λ+ ,λ−],λ+ < λ− 6 λ+,n = 1,...,N <
of non-degenerated intervals n−1 n n−1 n n ∞. These intervals
r γ = (λ−,λ+) > 0
a n n n
are separated by the gaps with the length . Introdu
e the fundamental
m m ϕ = (ϕn(z))n∈Z,ϑ = (ϑn(z))n∈Z
× matrix-valued solutions of the equation
a y +b y +a⊤ y = zy , ϕ ϑ 0, ϕ ϑ I , (z,n) C Z,
n n+1 n n n−1 n−1 n 0 ≡ 1 ≡ 1 ≡ 0 ≡ m ∈ × (1.2)
I m m 2m 2m M
m p
where is the identity × matrix. We de(cid:28)ne the monodromy × matrix and
T ,n > 1
n
the tra
e by
ϑ (z) ϕ (z)
M (z) = p p , T (z) = TrMn(z).
p (cid:18) ϑ (z) ϕ (z) (cid:19) n p (1.3)
p+1 p+1
∗
Institut fu(cid:4)r Mathematik, Humboldt Universit(cid:4)at zu Berlin, Rudower Chaussee 25, 12489, Berlin, Ger-
many, e-mail: evgenymath.hu-berlin.de
†
Department of Mathemati
s of Sankt-Petersburg State University, Russia e-mail: ku
enkoarambler.u
1
M D
Introdu
e the modi(cid:28)ed monodromy matrix and the determinant by
a 0
M = P M P−1, P = 0 , D(z,τ) = det(M(z) τI ), τ,z C.
0 p 0 0 (cid:18) 0 I (cid:19) − 2m ∈ (1.4)
m
τ = τ (z),j N M(z) N = 1,,..,m
j j 2m m
Let ∈ be the eigenvalues of , where { }. An eigenvalue
M(z)
of is
alled a multiplier.
Theorem (Lyapunov-Poin
ar(cid:1)e) i) The following identities hold
0 I
M⊤JM = J, J = m ,
(cid:18) I 0 (cid:19) (1.5)
m
−
D(z,τ) = τ2mD(z,τ−1), all z,τ C,τ = 0.
∈ 6 (1.6)
τ(z) d > 1 z C z R τ−1(z)
ii) Iτf(z) is a multiplier of multipli
ityd for some ∈M(z)(,ozr C∈ ), then 2m
(or ) isτ±a1(mz)u,ljtipliNer of muσlt(ipl)i
=ity m. MzoreCov:erτ,(eza)
h= 1 ∈ , has exa
tly
multipτli(ezr)s j ∈ m and J ∪j=z1{ C∈ | τj(z)| = 1}. τ′(z) = 0
iii) If is a simple multiplier for some ∈ and | | , then 6 .
m = 1
It is well knoσw(n t)ha=t thpe[λs+pe
,tλru−m] of the sλ
+ala<r (λ− 6 λ)+Ja<
o.b..i <opeλr−at6orλJ+is a<bsλo−lutely
λ
o±ntinuous and J ∪1 n−1 n , where [λ0+ ,λ−1],n 1N p p−1 pγan=d
n are 2-periodi
eigenvalues. The intervals n−1 n ∈ p are separated by gaps n
(λ−,λ−) γ > 0 γ γ = 0
n n of lengths | n| . If a gap n is degenerate, i.e. | n| , then the
orresponding
σ σ ∆(z) = ϕp+1(z)+ϑp(z)
segments σn(, )n=+1 mλergeC. :Re∆
a(zll)tha[t t1h,e1]Lyapunov fun(
tio1n)p−n∆(λ±) = 1 2 nand Nthe
spe
trum J { ∈ ∈ − }. Note that − n for all ∈ p.
We re
all well-known fa
ts (see e.g. [P℄ or [KKu℄).
F p
Theorem 1.1. A real polynoamia>l 0,bis thRe,LnyapZunovFfu(nz)
t=ionczfpo+r sOo(mzpe−s1)
alarz -periodi
n n
Ja
obico>pe0rator Fwi′t(hz n)u=m0b,ers( 1)p−jF(z∈) > 1 ∈ i(cid:27)j N zas< ..→. <∞z for
j j p−1 1 p−1
some and − for all ∈ and for some .
∆ t t [ 1,1] c > 0 ∆
Remark. 1) Zeros of − for any (cid:28)xed ∈ − and a
onstant determine .
σ( ) = σ( ) , ∆ = ∆ ∆,∆
2) J J forsomeJa
obioperatorsJ J i(cid:27) , where arethe
orresponding
Lyapuano=v 1fu,nb
et=ion0s. n Z ∆ = Te(z) e T e
3) If n n for all ∈ , then p 2 , where p is the Chebyshev polynomial.
Before we des
ribe the
ontent of our paper, we brie(cid:29)y
omment on ba
kground litera-
ture for matrix-valued Ja
obi operators. Inverse spe
tral theory for s
alar periodi
Ja
obi
operators is well understood, see a book [T℄ and papers [BGGK℄, [K℄, [KKu℄, [P℄, [vM℄ and
referen
es therein. The
orrespondingtheoryforperiodi
Ja
obimatri
eswithmatrix-valued

oe(cid:30)
ients is still largely a wide open (cid:28)eld. Some new parti
ular results were re
ently ob-
tained by Gesztesy and
oauthors [CG℄,[CGR℄,[GKM℄ and see referen
es therein. Note that
for (cid:28)nite Ja
obi matri
es with matrix-valued
oe(cid:30)
ients the
omplete solution of the inverse
problem was given re
ently by Chelkak and Korotyaev [CK℄.
M(z) D(τ,z) = 0
The eigenvalues of are the zeros of the equation . This is an algebrai
τ 2m z
equation in of degree , where thτe(
zo)e,(cid:30)j
ieNnts are polynomials of . It is well known (see
j 2m
e.g. Chapter 8, [Fo℄) that the roots ∈
onstitute one or several bran
hes of one
2
C
or several analyti
fun
tions that have only algebrai
singularities in . Thus the number
M(z) N z
e
of eigenvalues of is a
onstant with the ex
eption of some spe
ial values of (see
below the de(cid:28)niτtio(nz),ojf a rNesonan
e). There is a (cid:28)niNte n=um2mber of su
h points on the plane.
j 2m e
If the fun
tions ∈ are all distin
t, then . If some of them are identi
al,
N < 2m M(z)
e
then and is permanently degenerate.
m = 1
If , then the Riemann surfa
e for the multipliers has 2 sheets, but the Lyapunov
m > 2
fun
tion is entire. Similarly, in the
ase it is mor∆e
(ozn)v=eni1e(nτt(fzo)r+usτ−to1(
zo)n),stjru
tNthe
Riemann surfa
e for the Lyapunov fun
tions given by j 2 j j ∈ m,
m
whi
hσh(as) = shmeetsz(:se∆e t(hz)e eq[ua1t,io1n] (1.7))#. ANote that the Lyapunov-Poin
ar(cid:1)e TAheorem
gives J j=1{ j ∈ − }. Let be a number of elements of a set .
S
∆ ,s = 1,..,p 6 m p
s 0 s
TheoreRm,1p.2.>T1hereexistanalyti
fun
tions onthe -sheetedRiemann
surfa
e s s having the folωlowingNpro,pseretieNs: , ω = N
i) There exist disjoint subsets s ⊂ m ∈ p0 s m su
h that all bran
hes of
∆ ,s N ∆ (z) = 1(τ (z)+τ−1(z))S, j ω
s ∈ p0 have the form j 2 j j ∈ s and satisfy
e D(z,τ) p0 τ +τ−1
= Φ (z,ν), Φ (z,ν) = (ν ∆ (z)), ν = , τ = 0,
s s j
(2τ)m − 2 6 (1.7)
Y1 jY∈ωs
z,τ C Φ (z,ν) ν,z C
s
for any ∈ , where the fun
tions are some polynomials with respe
t to ∈ .
∆ = ∆ i ω ,j ω Φ = Φ ∆ = ∆
i j k s k s k s
If for some ∆∈,j ∈N , then and . Y = (α,β) R
ii) Let some bran
h j ∈ m be real analyti
on seome ienterval ⊂ and
1 < ∆ (z) < 1 z Y ∆′(z) = 0 z Y
− j foρr,ρan,sy ∈N ,.#Tωhe>n 2 j 6 for ea
h ∈ .
iii) All fun
tions s ∈ p0 s given by (1.8) are polynomials and
N0
ρ = ρ , ρ (z) = (∆ (z) ∆ (z))2, z C, where ρ = 1, if #ω = 1.
s s i j s s
− ∈ (1.8)
Y1 i<jY,i,j∈ωs
(λ−,λ+)
n n
iv) Ea
h endpo∆int,ojf aNgap is a periodρi
(or anti-periodi
)eigenvalue or a real bran
h
j m
point oτf0s,ojmeN ∈ , whi
h is a zero ofA(b=elo(awasu.
.ah )p−o1ints are
alled resonan
es).
v) Let j ∈ m be eigenvalues of a matrix p 1 2 p . Then following asymptoti
s
hold zp
∆j(z) = 2 (τj0 +O(z−m1 )), τj(z) = zpτj0 +O(z−m1 )), j ∈ Nm, (1.9)
ρs(z) = zp|ωs|(|ωs|−1)(cs +O(z−m1 )), cs = 2|ωs|(1−|ωs|) (τk0 −τj0)2, s ∈ Np0 (1.10)
Y
k<j,k,j∈ωs
z τ0 = τ0 j,k ω ,j = k c = 0 ρ (z) =
as | | → ∞ Moreover, if j 6 k for all ∈ s 6 , then s 6 and s
z|ωs|(|ωs|−1)(c +O(z−1)) z
s
as | | → ∞.
z z ρ
0 0
De(cid:28)nition. The number is a resonan
e of J, if is a zero of given by (1.8).
3
a = diag a ,a ,...,a , b = diag b ,b ,...,b , n Z
n 1,n 2,n m,n n 1,n 2,n m,n
Example. Leat > 0,b { R,(j,n) N }Zand { = m } for all ∈
and for some j,n j,n ∈ ℓ2(∈Z) m × . Then the operator J ⊕1 Jj, where Jj is
a s
alar Ja
obi operator, a
ting in and is given by
( jy)n = aj,nyn+1 +aj,n−1yn−1 +bj,nyn, n Z, y = (yn)n∈Z ℓ2(Z).
J ∈ ∈
∆ ∆
j j j
In this
ase ea
h is the standard Lyapunov fun
tion for J , and the properties of is
ρ = 1 ω = j
j
well known, see Theorem 1.1 and [T℄. Thus , sin
e { }.
Below we show the following identity (see (3.2))
mp
D(z,τ) = cτm (z λ (τ)), z,τ C, τ = 0, c = ( 1)mdetA ,
j p
− ∈ 6 − (1.11)
Yj=1
λ (τ) D(z,τ) = 0 τ Λ(τ) =
j
wλhe(rτe),n Nare ,zeτrosCof the polynomialσ( ) = mp σfor (cid:28)xed σ. =Deλ(cid:28)n(Se1)t,hSe1s=et τ C :
{ n ∈ mp} ∈ . Then we obtain J j=1 j, where j j { ∈
τ = 1 N0 = 0,.S..,m m > 0
| | } is the bounded
lose set. Ler m { ∆} f,ojr N . We des
ribe minimum
j m
spe
tral data whi
h determine all Lyapunov fun
tions ∈ for some Ja
obi operator.
κ R,j N0 cosκ = cosκ j = n
TheorΛem=1Λ.3(e.iκL0e)t k ∈ Λ ∈Λ(meiκsaj)t,is#fyΛ = (mn 6 j)p+1j,fjor aNll 6 .
0 j j m
i) Let andlet Λ⊂,j N0 − ∈ forDso(m,e)Ja
obi operator
J. Then∆th,ej speN
tral data j ∈ σm(, d)etermine the polynomial · · , all Lyapunov
j m
fun
tions ∈ and the spe
trum J (
ounted a

ording to its multipli
ity).
Λ Λ #Λ = #Λ 1
1 1 1 1
ii) Let ⊂ and let − . Then there exist in(cid:28)nitely many Ja
obi operators,
Λ Λ Λ j = 2,...m
having tehe same spe
tral deata 0, 1, j, , but di(cid:27)erent determinants.
eikκj =e1 k N λ (eiκj),n N
n mp
Example. Note that if m = 2for somκe =∈0,κ, t=heπn Λ ∈ are k-periodi
0 1 0
eigenvalues. Consider the
ase . Let . Then is a set of all periodi
Λ #Λ = 2p,#Λ = p+1
1 0 1
eigenvaluesaΛnd= λisasetofsλomeσa(nti)-periodi
eigenvaluessu
h that κ = π .
Let the set 2 { }, where ∈ ΛJ ,Λis ,aΛsome four-periodi
eigenvalued,eit.e(.M, (2z) 2.τIDu)e
to Theorem 1.3, thee spe
traledata 0 1 2 determine the polynomial p − m ,
σ( ) ∆ ,∆
1 2
the spe
trum J and the Lyapunov fun
tions .
p > 3 mp λs(τ),s N
Tτ heCorem 1.4. i) Let . Then ea
h sum Pj=1 j ∈ p−1 does not depend on
∈ . In parti
ular, the following identities hold
mp p mp p
λ (τ) = Trb , λ2(τ) = Tr(b2 +2a a⊤), all τ C.
n n n n n n ∈ (1.12)
Xn=1 Xn=1 Xn=1 Xn=1
ii) The following estimate is ful(cid:28)lled:
pm
λ2(τ) > 2pm(detA )p2m,
n p
(1.13)
Xn=1
where the identityκholdRs,tjrueNi(cid:27) bn ≡ 0, aconas⊤nκ==(dceotsAκp)p2mIm fojr=alnl n ∈ Np.
j m n j
iλii) Le(teniκujm) =ber2scos 1∈(κ +∈2πn) satisfy(j,n,k)6 N0 Nfo0r all N6 .bTh=en0,thaeae⊤ig=enIvalues
nm+k p j for all ∈ m × p−1 × p i(cid:27) n n n m for
n N p a = I
all ∈ pm and n=1 n m.
Q
4
We des
ribe a priori estimates
= max a (j,k) b (j,k) a = a (j,k) ,
∞ n n n n
Proposition1.5. LetkJk {| |, | |}, wherethematri
es { }
b = b (j,k)
n n
{ }. Then the following estimates hold
6 max λ+ , λ− = 6 (4m 1) ,
kJk∞ {| 0| | N|} kJk − kJk∞ (1.14)
λ− +λ+ λ− λ+ p
+ | N 0| 6 N − 0 6 (4m 1) , if Trb = 0.
∞ ∞ j
kJk 2 2 − kJk (1.15)
Xj=1
Remark. A priori estimates for s
alar Ja
obi operators were obtained in [BGGK℄, [K℄,
[KKr℄.
The plan of our paper is as follows. In Se
t. 2 we prove Theorem 1.2. In Se
t. 3 we
prove Theorem 1.3, 1.4 and Proposition 1.5. In Se
tion 4 we
onsider examples for the
ase
m = p = 2
, where we
onstru
t the
omplex and real resonan
es. In Se
t. 5 we shortly re
all
the well known results about the properties of point spe
trum and the absen
e of singular

ontinuous spe
trum. In the proof we use [CK℄, [BBK℄.
2 The Lyapunov fun
tions
y = (yn)n∈Z anyn+1 +bnyn +
We re
all well-known results. For any solution of the equation
a⊤ y = zy
n−1 n−1 n
we de(cid:28)ne
y 0 I
f = n , and = m , n Z.
n (cid:18) y (cid:19) Tn (cid:18) a−1a⊤ a−1(z b ) (cid:19) ∈ (2.1)
n+1 − n n−1 n − n
ϑ ϕ
f f = f M = n n
Then n satis(cid:28)es n Tn n−1. Thus the matrix-valued fun
tion n (cid:18) ϑ ϕ (cid:19)
n+1 n+1
M = M ,M = I M = n
sna>tis1(cid:28),eswherne njT=n1Xjn−=1Xn.0..X1 f2omr.maTthrii
sesgiXvejs. tWhee rmewonriotderTonmyinmthaetrfioxrm n Qj=1Tj,
Q
I 0 0 I a⊤ 0
= m m n−1 = P−1R 0P ,
Tn (cid:18) 0 a−n1 (cid:19)(cid:18) −Im z −bn (cid:19)(cid:18) 0 Im (cid:19) n nTn n−1 (2.2)
where
I 0 0 I
P = a⊤ I , R = P m = a⊤ a−1, 0 = m .
n n ⊕ m n n(cid:18) 0 a−n1 (cid:19) n ⊕ n Tn (cid:18) −Im z −bn (cid:19) (2.3)
Then using (2.2), we get
n n n
M = P−1R 0P = P−1 R 0 P = P−1M P , M = R 0.
n j jTj j−1 n jTj 0 n n 0 n jTj (2.4)
Yj=1 (cid:16)Yj=1 (cid:17) Yj=1
R , j > 1
j j
Matri
es T , satisfy
0 I
R⊤JR = J, ( 0)⊤J 0 = J, J = m ,
j j Tj Tj (cid:18) I 0 (cid:19) (2.5)
m
−
5
whi
h yields M⊤JM = J, M−1 = JM⊤J, n Z.
n n n − n ∈ (2.6)
Using (2.4) we obtain
a⊤ϑ (a⊤)−1 a⊤ϕ a−1ϑ⊤a a−1ϑ⊤
M = P M P−1 = n n 0 n n , M⊤ = 0 n n 0 n+1
n n n 0 (cid:18) ϑ (a⊤)−1 ϕ (cid:19) n (cid:18) ϕ⊤a ϕ⊤ (cid:19) (2.7)
n+1 0 n+1 n n n+1
and ϕ⊤ ϕ⊤a
M−1 = JM⊤J = n+1 − n n .
n − n (cid:18) −a−01ϑ⊤n+1 a−01ϑ⊤nan (cid:19) (2.8)
M M−1 = I M−1M = I
n n 2m n n 2m
Due to and we dedu
e that
a⊤ϑ (a⊤)−1ϕ⊤ a⊤ϕ (a⊤)−1ϑ⊤ = I , ϑ (a⊤)−1ϕ⊤ = ϕ (a⊤)−1ϑ⊤,
n n 0 n+1 − n n 0 n+1 m n 0 n n 0 n (2.9)
ϑ⊤a ϕ ϑ⊤ a ϕ = a , ϕ⊤ a⊤ϕ = ϕ⊤a⊤ϕ , ϑ⊤ a⊤ϑ = ϑ⊤a⊤ϑ .
n n n+1 − n+1 n n 0 n+1 n n n n n+1 n+1 n n n n n+1 (2.10)
Re
all the asymptoti
s of fundamental solutions
ϕ = znA +O(λn−1), ϑ = O(zn−1), n > 1, A = (a a ...a )−1
n+1 n n+1 n 1 2 n
(2.11)
z M
p
as → ∞. Substituting last asymptoti
s into we obtain
0 0
M (z) = zp +O(zp−1).
p (cid:18) 0 A (cid:19) (2.12)
p
D(z,τ)
It is well known that the determinant satis(cid:28)es
2m 2m
D(z,τ) = det(M(z) τI) = τ2m + τ2m−jξ (z) = (τ τ (z)), z,τ C,
j j
− − ∈ (2.13)
Xj=1 Yj=1
s−1
1 1
ξ = 1, ξ = T , ξ = (T +T ξ ), ...,ξ = T ξ ,..,
0 1 1 2 2 1 1 s s−j j
− −2 −s (2.14)
X0
ξ = ξ j N
j 2m−j m
see p.331-333 [RS℄. TheTid,ennt>ity1(1.6) gives for all ∈ . Re
all that the
n
Chebyshev polynomials satisfy:
[n]
τn +τ−n 2 (n j 1)!
T (ν) = = 2m−1 c νn−2j, c = ( 1)jn − − 2n−2j−m,
n n,j n,j
2 − (n 2j)!j! (2.15)
X0 −
ν = τ+τ−1 τn+τ−n = T (ν)
2 , see [AS℄. Then (2.13) and the identity 2 n yield
D(τ,z) (τm +τ−m) (τm−1 +τ1−m) m ξ
= +ξ +...+2−mξ = j T (ν),
(2τ)m 2m 1 2m m 2m−1 j (2.16)
Xj=0
6
1 m τn +τ−n m
TrMn(z) = j j = T (∆ (z)), z C.
n j
2 2 ∈ (2.17)
X1 Xj=1
The substitution of (2.15) into the equality (2.16) gives
m
D(τ,z)
Φ(ν,z) = = φ (z)νm−j,
(2τ)m j (2.18)
X0
ξ
1
φ = 1, φ = c ξ = , φ = c +c ξ , φ = c ξ +c ξ ,...,
0 1 m−1,0 1 2 m,1 m−2,0 2 3 m−1,1 1 m−3,0 3
2 (2.19)
φ = c +c ξ +c ξ +...+c ξ ,
2n m,n m−2,n−1 2 m−4,n−2 4 m−2n,0 2n
(2.20)
φ = c ξ +c ξ +c ξ +...+c ξ .
2n+1 m−1,n 1 m−3,n−1 3 m−5,n−2 5 m−2n−1,0 2n+1
(2.21)
a = I ,b = 0 n Z 0
n m n
LDe0t for all ∈ and denote the
orreDsp0(oτn,dzi)ng=J(aτ
2o+bi1oper2aτtTor(bzy))Jm . If
is the
orresponding determinant, then we obtain − p 2 and
Φ0 = D0(τ,z) = (ν T (z))m = mCm( T (z))jνm−j, CN = N!
(2τ)m − p 2 0 j − p 2 where m (N−m)!m!.
P Φ(ν,z) = D(τ,z) = mφ (z)νm−j
Proof of Theorem 1.2. i) We have the identity (2τ)m 0 j , see
φ ∆ (z),..,∆ P(z)
j 1 m
(Φ2(.z1,8ν),)w=he0re the polyznomCials are given by (2.19)-(2.21). are the roots of
for (cid:28)xed ∈mν.m−Tjhwen(zt)he statemwent i) follows from thze well-known about the
zeros of a polynomial 0 j , where j is a polynomials in , see [Fo℄.
∆′(z)P= 1(1 τ−2(z))τ′(z) = 0,z Y
ii) We have j 2 − 6 ∈ , sin
e by the Lyapunov-Poin
ar(cid:1)e
τ′(z) = 0 z Y
Theorem, 6 for all ∈ .
f = τn + α τn−1 + .. + α , g = b τs +
1 n 0
iii) Re
all that the resultant for polynomials
β τs−1 +..+β
1 s
is given by
1 α .. α 0 0 .. 0
1 n
 0 1 α .. α 0 .. 0  
1 n n
. . . . . . . .  lines
 
 0 ... 0 1 α .. α 
R(f,g) = det 1 n  .

 β0 β1 .. βs 0 .. 0   (2.22)
 
 0 β β .. β 0 .. 0  
 0 1 s  s
 . . . . . . . .   lines
 
 0 ... 0 β β .. β 
0 1 s
 



f τ ,..,τ
1 n
The dis
riminant of the polynomial with zeros is given by
Disf = (τi τj)2 = ( 1)n(n2−1)R(f,f′).
− −
Yi<j
DisΦ (λ,τ) = (∆ (λ) ∆ (λ))2 = ( 1)Nj(Nj−1)R(Φ ,Φ′)
Thus we have j i<s,i,s∈ωj i − s − 2 j j is polyno-
Φ (λ,τ) Q ρ = N0Dis Φ
miailv,)siEn
ae
h jgap hasistthheefpoormlynγonm=ial.(λT−nh,eλn+nt)h=e fu∩nmj=
t1iγonn,j, whQere1 γn,j =j(iλs−np,jo,λly+nn,jo)m⊂ialR. is
∆ (z) / [ 1,1] z γ ∆ (λ− ) =
some (cid:28)nite interval su
h that j ∈ − for all ∈ n,j. Note that j n,j ± or
∆ (λ+ ) = 1 λ± ∆ (z)
j n,j ± or n,j is the bran
h point of j , otherwise we have a
ontradi
tion.
7
F
v) Using (2.12) we de(cid:28)ne by
F(t) = tpM = F +tF (t), F = 0 A , t = z−1,
p 0 1 0 m p
⊕
F F (t) = const+O(t) t 0 τ (t),j N
1 1 j 2m
where is a some maFtr(itx)polynomial anτd0,j N as → A. Let τ (∈t)
be the eigenvalues of . Re
all that j ∈ m are eigenvalues of p. Theen, j are
τ1/mj
given by
onvergent Puiseux series (the Taylor series in ) e
∞ k
k
τ (t) = τ0 + α tmj, t < r, m = m, m N, j w ,s N
j j j,k | | j j ∈ ∈ s ∈ k
Xk=1 X1
e
r > 0 w = N ,w = N w ,...,w = N N
for some , see p=4, [RS℄, where 1 m1 2 m1+m2 \ 1 k m \ m−mk
τ0 = τ0 j,j′ w τ (t),j w
and j j′ for all ∈ s. Here j ∈ s are the bran
hes of one or more multivalued
τ = 0
analyti
fun
tions with at worst algeebrai
bran
h points at and satisfy
τj(z) = zpτj(t) = zp(τj0 +O(z−m1s)) = τj0zp +O(zp−m1s), j ∈ ws.
τ0 e τ (t),j N t m = 1
If ea
h eigenvaluτe(zj)i=s sτim0zpple+, Oth(eznp−e1a)
,hj j N ∈ m is anal2y∆ti
f=orτs(mza)l+l , i1.e.,= τs0zp +.
Thus we obtain j j e∈ m, whi
h yields j j τj(z) j
O(zp−1) ρ
s
and using (1.8), we get asymptoti
s for .
3 Inverse Problems
In order to prove Theorem 1.3 we need the Lemma 3.1-3.3.
W = W s s s W = cosjκ
Lemmκar 3.C1,.sL>et1 { cor,sj}κj,rr==0 cboesaκr′ × martr=ixrw′ ith
omdpeotnWent=s 0 r,j r for
some ∈ , where 6 for all 6 . Then 6 .
cosjz = T (cosz) T
j j
Proof. UsWing = cosjκ = T (co,swκhe)re theWCyhe=by0shev polynomialsy = (ayre)gsiveCnsb+y1,(y2.=150),
we obtain r,j r j Pr(x.)L=et s y Tfo(rx)s,oPme=v0e
tor n 0 ∈ cosκ ,6 j .
Then we dedu
e that a polynomial j=0 j j 6 has distin
tzeros j ∈
N0 degPP6 degT = s W
s s
, whi
h gives a
ontradi
tion, sin
e . Thus is invertible.
c = ( 1)mdetA ξ (z),j N
p j m
Lemma 3.2. Let − . The polynomials ∈ , given by (2.13), satisfy
ξ (z) = ξ (z) = O(zpj), ξ (z) = czpm +O(zpm−1) as z , ,
j 2m−j m
→ ∞ (3.1)
mp m−1
D(z,τ)
c (z λ (τ)) = = (τm +τ−m)+ξ (z)+ (τm−j +τj−m)ξ (z).
− n τm m j (3.2)
nY=1 Xj=1
Proof. Using (1.6), (2.13) and Viete's formulas, we get
j
ξ (z) = ξ (z) = ( 1)j τ (z), 1 6 j 6 m.
j 2m−j − is (3.3)
16i1<X..<ij62mYs=1
8
τ (z) = τ0zp + o(zp) j N
Asymptoti
s (1.9) give j j , ∈ m and the Lyapunov-Poin
ar(cid:1)e Theorem
τ (z) = 1 = O(z−p) z τ0 = 0
yields m+j τj(z) as → ∞, sin
e ea
h j 6 . Then, using (3.3), we obtain
m
ξ (z) = O(zpj), ξ (z) = ( 1)m τ0zpm +o(zpm) = ( 1)mdetA zpm +O(zpm−1).
j m − j − p
Yj=1
Identity (3.2) follows from last identity and (1.6), (2.13).
We need the following Lemma about polynomials.
(z )k Ck (c )s Cs+1 s,k > 1
Lemma 3.3. Let n 1 ∈ and j 0 ∈ for some integers . Then there exists
r(z) g
a unique polynomial satisfying for some polynomial the following relations
s k
r(z) = zk c zj+O(zk−1) as z and r(z) = h(z)g(z), h(z) = (z z ),
j n
→ ∞ − (3.4)
Xj=0 nY=1
= g : degg 6 s , dim = s +1
s s
Proof. Intgrodu
e the linear spa
e of polynomia:ls P C{s+1 } P .
s s
Note that ∈ P . De(cid:28)ne the linear operator A P → by
1 dn
( g) = h(z)g(z) , n = k +j 1, j N , g .
A j n!dzn |z=0 − ∈ s+1 ∈ Ps
(cid:16) (cid:17)
g ( g) = c , j N h(z)g(z) = zk s c zj+O(zk−1)
We rewriteA inthe form A j j−1 ∈ s+1, where j=0 j
z : Cs+1 diPm = dimCs+1 =
s s
as → ∞. We dedu
e that A P → is an isomorphism, sin
e P
s + 1 g = 0 g = 0 g = −1(c )s r(z) = h(z)g(z) r(z) =
zk s ancdzijf+AO(zk−1,)thenz . If A n 0 and r , then
j=0 j as → ∞. Moreover, the polynomial is unique, sin
e A is an
P
isomorphism.
c = ( 1)mdetA ζ (z) c−1
p 0
Pζ r(zo)of ocf−T1ξhe(zo)rejm 1N.3. Re
all that − degζ 6.pjDe(cid:28)ne the pζol(yzn)o=mialspj ζ ≡zn ,
sojmeζ≡j,n ∈ Rj. In,tro∈du
emt.heLseemtsmKas3=.2{ypi(emlds−s−1)j+1,...,pa(nmd−thse)n}, js ∈ N0mP−1,n=K0 mj,n= {0fo}r,
whi
h satisfy m K = N0 , K K = ,for all r = s.
∪s=j s p(m−j) r ∩ s ∅ 6 (3.5)
h = τ−j +τj
j
Using (3.2) and (3.5) and setting , we obtain
pm m m
D(z,τ)
q(z,τ) = = (z λ (τ)) = h ζ (z) = h ζ (z)
n m−j j j m−j
cτm − (3.6)
nY=1 Xj=0 Xj=0
m p(m−j) m m m m
= h ζ zn = h ζ zn = h ζ zn
j m−j,n j m−j,n j m−j,n
Xj=0 Xn=0 Xj=0 Xs=j nX∈Ks Xj=0 Xs=j nX∈Ks
m s m s m pm
= h ζ zn = h ζ zn = η (τ)zn = η (τ)zn,
j m−j,n j m−j,n n n
Xs=0 Xj=0 nX∈Ks Xs=0 nX∈KsXj=0 Xs=0 nX∈Ks Xn=0
9
where s
η (τ) = h ζ , n K , s N0 .
n j m−j,n ∈ s ∈ m (3.7)
Xj=0
τ = eiκr,r N0
Substituting ∈ s into the identity (3.7) we get
η = 2W y , η = (η (eiκr))s , y = (ζ )s , W = (cosjκ )s ,
n s n n n r=0 n m−j,n j=0 s r r,j=0 (3.8)
for all e(n,s) ∈ Ks × Ne0m. Due to Lemma 3.1, ea
h matrix Ws,s ∈ N0m is invertible. Thus
(3.8), (3.7) give
η (τ) =< (τj +τ−j)s ,(2W )−1η >, (n,s) K N0
n j=0 s n ∈ s × m (3.9)
< , > Rs e ηn( ),n Ns
ηwhe=re(η (·ei·κr))iss a s
alar produ
t in . Then the fun
tions · ∈η ( ),anre dNetermined by
n n r=0. We will use this fa
t to determine all fun
tions n · ∈ m. In order
teo des
ribe re
overing we need the following simple fa
t.
0 6 k 6 m 1 η (eiκj),(n,j) N0 N0
Lemmηa 3n.4. Lekt K Λ− . Then numbers nκ j N0 ∈ pm × k, the fun
-
tions n, ∈ ∪j=0 j, the set k+1 and the numbers j, ∈ m determine the numbers
η (eiκj),(n,j) N0 N0 η n k+1K
n ∈ pm × k+1, and the fun
tions n, ∈ ∪j=0 j.
Proof. Let Sk+1 = (ηn(eiκk+1))n∈∪kj=0Kj = (ηn(eiκk+1))pnm=p(m−k−1)+1. Using (3.6), we see
Λ q(z,eiκk+1)
k+1
that the elements of are some zeros of the polynomial and re
all that
#Λ = p(m k 1) + 1 S
k+1 k+1
− − . Also, usiqn(gz,(e3iκ.6k)+,1)we see that the
omponentsΛof ve
tor S
k+1 k+1
are the
oe(cid:30)
ients of the polynomiqa(lz,eiκk+1) . Then, due to Lemma 3.3,(η (eiκakn+d1))pm
n n=0
uniquelydeterminethepolyno(mηia(eliκj)) ajnd,Nus0ing(3.6), wedetermine η (τ).
Substituting obtained values n n∈Kk+1, ∈ k+1 into the (3.9), we determine n ,
n K
k+1
∈ . q(z,τ) = D(z,τ)
Now we des
ribe theΛre
overing pro
edure. Re
all thqa(zt,eiκ0) (ηcτm(ei.κ0S))upbmstituting
the elements of the set 0 intηo t(hτe),(n3.6)K, we determine and n nk=0. NT0hen
(3.9) determineηs(aτll)funn
tionms Kn = N∈0 0. Step by step, usiqn(gz,Lτe)mma 3.4 for ∈ m−1,
we determine n , ∈ ∪j=0 j pm. Then (3.6) gives . Using (2.13) we get
cq(z,τ) = D(z,τ) = τ2m + O(τ2m−1) τ q(z,τ)
as → ∞, then the polynomial uniquely
c D(z,τ)
determine
onstant aa,nbd,n Z. a > 0,n Z = m
ii) Assume that n n ∈ are diagonal, and ea
h n ∈ . Then J ⊕1 Jj,
∆ D
j j j
where J are s
alar Ja
obi operators. Let be,jtheNLyapunov fun
tions and let be the

orresponding determinant for the operators Jj ∈ m. Then we dedu
e that e
m
D(z,τ) = D , D (z,τ) = τ(τ +τ−1 2∆ (z)).
j j j
− (3.10)
Yj=1
e e
∆ (z) ∆′ (z0) = 0 ( 1)p−j∆ (z0) =
B∆y T(zh0e)or>em1,1.1j, theNLyapunov funz
t0io<n...m< z0satis(cid:28)es m j andq (−z) = s(∆m (zj)
|cosmκ )j +| cosκ ,s∈> 1p−1 for some 1 p. Then the polynomial s m −
0 0
satis(cid:28)es
1
( 1)p−jq (z0) = s ∆ (z0) +( 1)p−j 1+ cosκ > 1, (q )′(z0) = 0, j N .
− s j | m j | − − s 0 s j ∈ p−1
(cid:16) (cid:16) (cid:17) (cid:17)
10