KdV Flow on Generalized Reflectionless Potentials
The purpose of this article is to construct KdV fow on a space of generalized reflectionless potentials by applying Sato's Grassmannian approach The point is that the base space contains not only rapidly decreasing potentials but also oscillating ones such as periodic ones, which makes it poss...
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Kotani, S. 2016-09-29T20:23:44Z 2016-09-29T20:23:44Z 2008 KdV Flow on Generalized Reflectionless Potentials / S. Kotani // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 4. — С. 490-528. — Бібліогр.: 16 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106520 The purpose of this article is to construct KdV fow on a space of generalized reflectionless potentials by applying Sato's Grassmannian approach The point is that the base space contains not only rapidly decreasing potentials but also oscillating ones such as periodic ones, which makes it possible for us to discuss the shift invariant probability measures on it. en Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України Журнал математической физики, анализа, геометрии KdV Flow on Generalized Reflectionless Potentials Article published earlier |
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KdV Flow on Generalized Reflectionless Potentials |
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KdV Flow on Generalized Reflectionless Potentials Kotani, S. |
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KdV Flow on Generalized Reflectionless Potentials |
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KdV Flow on Generalized Reflectionless Potentials |
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KdV Flow on Generalized Reflectionless Potentials |
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KdV Flow on Generalized Reflectionless Potentials |
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kdv flow on generalized reflectionless potentials |
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Kotani, S. |
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Kotani, S. |
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2008 |
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Журнал математической физики, анализа, геометрии |
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The purpose of this article is to construct KdV fow on a space of generalized reflectionless potentials by applying Sato's Grassmannian approach The point is that the base space contains not only rapidly decreasing potentials but also oscillating ones such as periodic ones, which makes it possible for us to discuss the shift invariant probability measures on it.
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1812-9471 |
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KdV Flow on Generalized Reflectionless Potentials / S. Kotani // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 4. — С. 490-528. — Бібліогр.: 16 назв. — англ. |
| work_keys_str_mv |
AT kotanis kdvflowongeneralizedreflectionlesspotentials |
| first_indexed |
2025-11-25T12:07:53Z |
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2025-11-25T12:07:53Z |
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1850514431130206208 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2008, vol. 4, No. 4, pp. 490�528
KdV Flow on Generalized Re�ectionless Potentials
S. Kotani
Kwansei Gakuen University School of Science and Technology Hyogo
2-1 Gakuen, Sanda-shi, Hyogo 669-1337, Japan
E-mail:kotani@kwansei.ac.jp
Received April 14, 2008
The purpose of this article is to construct KdV �ow on a space of gene-
ralized re�ectionless potentials by applying Sato's Grassmannian approach.
The point is that the base space contains not only rapidly decreasing poten-
tials but also oscillating ones such as periodic ones, which makes it possible
for us to discuss the shift invariant probability measures on it.
Key words: Korteweg de Vries equation, inverse spectral methods, re�ec-
tionless potentials.
Mathematics Subject Classi�cation 2000: 35Q53, 37K15.
1. Introduction
The KdV equation is
@u
@t
= �
@3u
@x3
+ 6u
@u
@x
;
and this describes the dynamics of shallow waters. As is well known, n-soliton
solutions for the KdV equation are given by
u(t; x) = �2
@2
@x2
log det(I +A(t; x));
where
A(t; x) =
�p
mimj
�i + �j
e�(�i+�j)x+4(�3i+�
3
j )t
�
1�i;j�n
with mi; �i > 0: For each �xed t 2 R; u(t; �) is a re�ectionless potential which ap-
pears in 1-D scattering theory. V.A. Marchenko [13, 14] considered the compact
uniform closure of re�ectionless potentials, which we call the space of genera-
lized re�ectionless potentials, and made an attempt to solve the KdV equation
starting from an element of this closure. However, he had to impose the solvabil-
ity condition on an integral equation, which made it impossible to solve the KdV
c
S. Kotani, 2008
KdV Flow on Generalized Re�ectionless Potentials
equation in its full generality. On the other hand, M. Sato and Y. Sato established
a uni�ed approach for a large class of completely integrable systems. They con-
structed solutions based on dynamics (�ows) on in�nite dimensional Grassmann
manifold, and it was rewritten from an analytic point of view by G. Segal and
G. Wilson [16]. R.A. Johnson [9] mentioned the applicability of their approach
to this space of generalized re�ectionless potentials, which admits a certain class
of oscillating functions. However, to apply this method we have to prove the
transversality, which is equivalent to the solvability of the integral equation con-
sidered by Marchenko. The �rst purpose of this paper is to construct a KdV �ow
on this space by showing the transversality. In the case when the base space is
a set of rapidly decreasing smooth functions, H.P. McKean [15] applied Sato's
theory to construct the KdV �ow on it.
For �0 < 0 let
�0 be the compact uniform closure of all re�ectionless
potentials whose associated Schr�odinger operators have their spectrum in [�0;1).
Set
� =
�
g; g(z) is holomorphic on D; g(0) = 1, g(z) 6= 0 for 8z 2 D;
takes real values on R and g(�z) = g(z)�1 for 8z 2 D
�
;
where D is the closed unit disc. We construct a homomorphism K between the
group � and the group of all homeomorphisms on
�0 : This K induces the shift
operation if we choose
gx(z) = e�xz 2 �
and solutions for the KdV equation if we choose
gx;t(z) = e�xz+4tz3 2 �:
Any other higher order KdV equation can be solved in this way on
�0 : We also
discuss the isospectral property under K:
The motivation of this paper is to construct a nice solution for the KdV equa-
tion starting from a certain random initial data. This problem was raised by
V.E. Zakharov and the author was taught it by S.A. Molchanov. We would like
to construct a solution as a typical random �eld fu(t; x)gt;x2R which is shift in-
variant with respect to t and x: In this respect, there are already solutions which
are quasiperiodic in time and space, which is a special case of shift invariant ran-
dom �elds. However our aim is to give a very random solution. The construction
of the KdV �ow is a starting point in solving this problem. Since
�0 is compact
and the KdV �ow fK(g)gg2� is commutative, the space of all probability mea-
sures on
�0 invariant with respect to fK(g)gg2� is a non-empty compact convex
set. Therefore we have many ergodic K(g)�invariant probability measures on
�0 : It is interesting to study the spectral property for the associated Schr�odinger
operators under these probability measures. The problem of V.E. Zakharov is just
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 491
S. Kotani
the problem on �nding such a probability measure under which K(g)q behaves as
random as possible, especially the spectrum of the Schr�odinger operators should
have a dense point spectrum on [�0; 0], whereas the spectrum in (0;1) is always
purely absolutely continuous for any potential from
�0 : Since a KdV-�ow invari-
ant probability measure is automatically shift invariant, hence we can de�ne the
Floquet exponent. We discuss the relationship between the KdV-�ow and the
Floquet exponent, although it is still unsatisfactory.
We try to give a self-contained explanation of this subject as far as possible,
since it may be di�cult to obtain a complete view only by citing necessary facts.
2. Spectral Theory of 1-D Schr�odinger Operators
and Dyson Formula
Let us consider a one-dimensional Schr�odinger operator
L = �
d2
dx2
+ q(x)
with potential q; which is a real valued function of L1
loc(R): In this section we
introduce the Gelfand�Levitan inverse spectral theory and the Dyson formula
which solves the inverse spectral problem by the Fredholm determinants of the
integral operators associated with the spectral measures.
Suppose q(x) satis�es
q(x) � �cx2
for every su�ciently large jxj with a constant c. Then it is known that L has
a unique selfadjoint extension on L2(R): Under this condition, for � 2 CnR there
exist unique solutions f�(x; �) of
Lf = �f; f(0) = 1 and f 2 L2(R�);
where R+ = [0;1); R� = (�1; 0]. Set
m�(�) = m�(�; q) = �f 0�(0; �):
These functions become the Herglotz ones which are holomorphic on the upper
half-plane with positive imaginary parts. We call these functions as H-functions,
see Appendix for the properties of H-functions. Let g�(x; y) be the Green function
for L� �, that is
(L� �)�1 (x; y) = g�(x; y):
It is well known that
g�(x; y) = g�(y; x) = �
f+(x; �)f�(y; �)
m+(�) +m�(�)
for x � y (1)
492 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
is valid. The Gelfand�Levitan inverse spectral theory says that the potential
q on R+(resp.R�) can be recovered from m+(resp.m�) by solving the integral
equation(2) of Fredholm type. Let
�+(x; y) =
Z
R
�
1� cos
p
�x
� �
1� cos
p
�y
�
�2
�+(d�)� x ^ y
with a measure �+ representing m+. De�ne
F+(x; y) =
@2
@x@y
�+(x; y)
and consider
K(x; y) + F+(x; y) +
Z x
0
F+(y; t)K(x; t)dt = 0: (2)
Then F+ is continuous and the integral equation (2) is uniquely solvable in the
space of continuous functions on [0; x] for each �xed x > 0. Then the potential q
is given by
q(x) = 2
d
dx
K(x; x) for x > 0: (3)
For details see V.A. Marchenko [12].
For later purpose we give another representation of q by a determinant.
This kind of representation was remarked �rst by F. Dyson [3] in the scattering
case (in which the potential q is decaying su�ciently fast at �1; see Th. 4 below)
and by K. Iwasaki [8] in the case of boundary value problems on �nite intervals.
For the sake of completeness we give a proof of this formula here. Let F x
+ be the
integral operator on C([0; x]) with kernel F+.
Theorem 1. (Dyson formula)
q(x) = �2
d2
dx2
log det(I + F x
+): (4)
P r o f. Set
Fx(t; s) = xF+(xt; xs)
and consider the integral operator Fx with kernel Fx(t; s) on L
2([0; 1]): Then it is
easy to see that det(I + F x
+) = det(I + Fx) holds. Therefore
d
dx
log det(I + F x
+) =
d
dx
log det(I + Fx)
= tr
�
(I + Fx)
�1 @Fx
@x
�
=
Z 1
0
@Fx
@x
(t; t)dt+
ZZ
[0;1]2
�x(t; s)
@Fx
@x
(s; t)dtds;
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 493
S. Kotani
where �x = (I + Fx)
�1 � I and �x(t; s) is the kernel for �x: Observing
x
@Fx
@x
(t; s) = Fx(t; s) + t
@Fx
@t
(t; s) + s
@Fx
@s
(t; s);
we have
x
d
dx
log det(I + Fx) = tr (Fx + �xFx)
+
Z 1
0
t
�
@Fx
@t
(t; t) +
Z 1
0
@Fx
@t
(t; s)�x(s; t)ds
�
dt
+
Z 1
0
t
�
@Fx
@s
(t; t) +
Z 1
0
�x(t; s)
@Fx
@s
(s; t)ds
�
dt
= �
Z 1
0
�x(t; t)dt�
Z 1
0
t
@�x
@t
(t; t)dt�
Z 1
0
t
@�x
@s
(t; t)dt
= �
Z 1
0
�x(t; t)dt�
Z 1
0
t
d
dt
(�x(t; t)) dt
= ��x(1; 1);
where we have used the identity Fx+�x+Fx�x = Fx+�x+�xFx. On the other
hand, from (2) we see
xK(x; xt) = �xF+(x; xt)� x
Z 1
0
�x(t; s)F+(x; xs)ds
= �Fx(t; 1) �
Z 1
0
�x(t; s)Fx(s; 1)ds
= �x(t; 1);
which implies
d
dx
log det(I + F x
+) = �K(x; x):
Consequently, the proposition can be proved from (3).
The formula (4) may be called Dyson formula.
3. Inverse Scattering Problem and Re�ectionless Potentials
If the potential satis�es
jq(x)j (1 + jxj) 2 L1(R);
one can de�ne two linearly independent solutions e�(x; k) for k 2 C+(= fk 2 C;
Im k � 0g) of Le = k2e satisfying the following asymptotic behaviour�
e+(x; k) ' eikx as x! +1;
e�(x;�k) ' e�ikx as x! �1 .
494 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
Since the pairs fe+(x; k); e+(x;�k)g ; fe�(x; k); e�(x;�k)g form a fundamental
system of solutions of Le = k2e for nonzero real k, we can introduce a(k); b(k) as�
e+(x; k) = a(k)e�(x; k) + b(k)e�(x;�k);
e�(x;�k) = a(k)e+(x;�k)� b(�k)e+(x; k):
The real valuedness of the potential V implies8<:
a(k) = a(�k);
b(k) = b(�k);
ja(k)j2 = 1 + jb(k)j2 ;
(5)
for nonzero real k: Set 8>>>>><>>>>>:
r+(k) = �
b(�k)
a(k)
;
r�(k) =
b(k)
a(k)
;
t(k) =
1
a(k)
:
r+(k)(resp. r�(k)) is called the right re�ection coe�cient(resp. the left
re�ection coe�cient) and t(k) is called the transmission coe�cient (see
V.A. Marchenko [12]). From (5) we see that
0 < jt(k)j � 1 and
��r+(k)�� = ��r�(k)�� � 1
for every k 2 Rn f0g : It is known that the single r+(k)(equivalently r�(k)) de-
termines the other fr�(k); t(k)g. It is also known that a(k) is holomorphic on
C+ and has only �nitely many simple poles fi�jgnj=1
on the pure imaginary axis.
At k = i�j , e
+(x; i�j) and e
�(x;�i�j) are linearly dependent and belong to L2(R):
Therefore
n
��2j
on
j=1
are eigenvalues of L: Set
�
m�
j
��2
=
Z
R
��e�(x; i�j)��2 dx:
Then it is not di�cult to see that�
m�
j
��2
= �
�
m+
j
�2
a0(i�j)
2
holds. The triple
n
r+(k); i�j ;m
+
j ; 1 � j � n
o
is called the right scattering
data. The inverse scattering problem is to obtain the potential V from the
right (or left) scattering data, and the basic part of the problem was solved by
V.A. Marchenko. His procedure is as follows. Since r+(k) = O(jkj�1) as jkj ! 1;
R+(x) =
1
�
Z
R
r+(k)e2ikxdk
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 495
S. Kotani
is well-de�ned. It is known that R+(x) is locally absolutely continuous andZ 1
0
��R+(x)
�� dx+
Z 1
0
(1 + jxj)
��R+0(x)
�� dx <1: (6)
De�ne
F+(x) = R+(x) + 2
nX
j=1
e�2�jxm+
j :
Then (5) makes it possible to consider an integral equation on L2([0;1)) for each
�xed x 2 R
K(t) + F+(x+ t) +
Z 1
0
F+(x+ t+ s)K(s)ds = 0:
It is also known that this equation is uniquely solvable and we denote its solution
by K+(x; t): Then the following theorem is valid.
Theorem 2. It holds that
q(x) = �
@
@x
K+(x; 0):
F. Dyson [3] discovered a compact expression of V: Let
F+
x f(t) =
Z 1
0
F+(x+ t+ s)f(s)ds:
The property (6) implies the operator F+
x de�nes a trace class operator
on L2([0;1)) for each �xed x 2 R:
Theorem 3. (Dyson formula) It holds that
q(x) = �2
d2
dx2
log det(I + F+
x ): (7)
A similar expression is possible by using the left scattering data fr�(k); i�j ;
m�
j ; 1 � j � ng. A potential q is called re�ectionless if
r+(k) � 0 (and hence r�(k) � 0):
Now it is easy to see from Th. 3 that q is re�ectionless if and only if
V (x) = �2
d2
dx2
log det(I +A+(x)) (8)
496 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
with
A+(x) =
0@
q
m+
i m
+
j
�i + �j
e�(�i+�j)x
1A
1�i;j�n
:
In this case the potential q is decaying exponentially fast and analytic on R:
This re�ectionless property can be interpreted by m�(�) as follows. Since the
de�nition of m� implies
m+(�) =
e0+(0;
p
�)
e+(0;
p
�)
; m�(�) = �
e0�(0;�
p
�)
e�(0;�
p
�)
for � 2 C+;
and for � > 0; e�(0;�
p
� + i0) exist �nitely , we see
m+(� + i0) +m�(� + i0) =
2i
p
�b(
p
�)
e+(0;
p
� + i0)e�(0;�
p
� + i0)
:
Hence we see that q is re�ectionless if and only if
m+(� + i0) = �m�(� + i0) for all � > 0 (9)
holds.
4. Generalized Re�ectionless Potentials
In this section we give the closure of the class of all re�ectionless potentials.
To this end we characterize H-functions m� satisfying the property (9). We pre-
pare a lemma.
Lemma 4. An H-function m satis�es
Rem(� + i0) = 0 a:e: on (0;1) (10)
if and only if there exists a measure � on (�1; 0] satisfyingZ 0
�1
�(d�)
1 + j�j
<1;
and
� 0 such that
m(�) = �i
p
�
� i
p
�
Z 0
�1
�(d�)
� � �
: (11)
Moreover, setting
�(�) =
+
Z 0
�1
�(d�)
� � �
;
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 497
S. Kotani
for some real � we have
m(�) = �+
Z 0
�1
�
1
� � �
�
�
1 + �2
�p
���(d�)
�
1
�
Z 1
0
�
1
� � �
�
�
1 + �2
�p
��(�)d�: (12)
P r o f. Suppose the characteristics ofm are f�; �; �g : Set �(�) = arg (m(� + i0)) :
The identity (10) implies �(�) =
�
2
a.e. on (0;1): Therefore Appendix implies
m(�) = exp
�
c+
1
�
Z
R
�
1
� � �
�
�
1 + �2
�
�(�)d�
�
=
1
�i
p
�
exp
c+
1
�
Z
(�1;0]
�
1
� � �
�
�
1 + �2
�
�(�)d�
!
:
Hence �i
p
�m(�) is again an H-function which takes real values on (0;1) and is
analytic there. Therefore
�i
p
�m(�) = �1 + �1�+
Z 0
�1
�
1
� � �
�
�
1 + �2
�p
���(d�);
which implies Z
(�1;�1]
j�j�
3
2 �(d�) <1: (13)
On the other hand, an H-function �m�1 also satis�es (10), hence m(�)=
�
�i
p
�
�
is an H-function such that
m(�)
�i
p
�
= �2 + �2�+
Z 0�
�1
�
1
� � �
�
�
1 + �2
�
�(d�)
p
��
+
c
��
with some c � 0; which impliesZ
[�1;0)
j�j�
1
2 �(d�) <1:
Moreover, we see �2 = 0: Now, from (13) it follows
m(�)
�i
p
�
=
+
Z 0+
�1
�(d�)
� � �
;
with 8>><>>:
= �2 +
Z 0�
�1
�
��
1 + �2
�
�(d�)
p
��
�(d�) = I(�1;0)(�)
�(d�)
p
��
+ cÆf0g(d�)
:
498 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
Conversely assume m is given by (11). Then all we have to show is that m is
an H-function. To see this, we note m(�) = �i
p
��(�), and � is an H-function
taking real values on (0;1); from which the conclusion follows.
Proposition 5. Herglotz functions m� satisfy the property (9) if and only if
there exist measures �� on (�1; 0] satisfyingZ 0
�1
�+(d�) + ��(d�)
1 + j�j
<1; (14)
and � 2 R;
� 0 such that
m�(�) = ���
1
2
Z 0
�1
�
1
� � �
�
�
1 + �2
�p
�� (�+(d�) � ��(d�)) (15)
�
i
p
�
2
�
i
p
�
2
Z 0+
�1
�+(d�) + ��(d�)
� � �
:
P r o f. Let the characteristics of m� be f��; ��; ��g : First note m+(�) +
m�(�) is an H-function satisfying
Re (m+(� + i0) +m�(� + i0)) = 0 a:e: on (0;1);
since we have the condition (9). Hence we immediately see8><>:
�+(d�) = ��(d�) =
��(�)
2�
p
�d� on (0,1)
with �(�) =
+
Z
(�1;0)
�+(d�) + ��(d�)
(� � �)
p
��
+
c
��
:
Introduce
��(d�) = I(�1;0)(�)
��(d�)p
��
+
1
2
cÆf0g(d�):
Then m� can be represented as
m�(�) = �� + ���+
Z 0
�1
�
1
� � �
�
�
1 + �2
�p
����(d�)
+
Z 1
0
�
1
� � �
�
�
1 + �2
�
��(�)
2�
p
�d�:
On the other hand, we know from Lem. 4 that
m+(�) +m�(�) = �i
p
�
� i
p
�
Z 0
�1
�+(d�) + ��(d�)
� � �
; (16)
hence �� = 0; which concludes (15).
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 499
S. Kotani
Now, for �0 < 0 we introduce
�0 a class of potentials q as all compact uni-
form limit on R of some re�ectionless potentials whose associated Schr�odinger
operators have their spectrum in [�0;1): An element of
�0 is called a general-
ized re�ectionless potential. A potential of the form (8) is called the classical
re�ectionless potential and the set of all these potentials is denoted by
cl
�0
.
We try to parametrize the set by measures on [�
p
��0;
p
��0] de�ned by
��0 =
8<:
�; a measure on [�
p
��0;
p
��0] satisfyingZ
[�
p
��0;
p
��0]
�(d�)
��0 � �2
� 1
9=; :
V.A. Marchenko [14] showed the following result, which we prove again here.
Theorem 6. Then two spaces
�0 and ��0 are homeomorphic and m� are
given by
m�(�z2) = �z �
Z
[�
p
��0;
p
��0]
�(d�)
�� � z
; (17)
or the characteristic measures �� of m� are
��(d�) =
( p
���(d�) on (�1; 0] with � = �
p
��
�
1
2�
p
��(�)d� on (0,1)
;
where
�(�) = �2 + 2
Z
[�
p
��0;
p
��0]
�(d�)
��2 � �
:
P r o f. Choose a q 2
cl
�0
whose associated Schr�odinger operator has its
spectrum in [�0;1): Since q is rapidly decreasing, �� have �nitely many points
in their supports in [�0; 0): It is easy to see that
m�(�) = i
p
�+O
�
1
p
�
�
; as �!1; (18)
hence
= �2: We introduce a measure � on [�
p
��0;
p
��0] byZ p
��0
�
p
��0
f(�)� (d�) =
1
2
Z 0
�0
f(
p
��)�+ (d�) +
1
2
Z 0
�0
f(�
p
��)�� (d�) :
Since they satisfy the property (9), we can apply Prop. 5 and the formula (17)
follows. On the other hand, the condition that the spectrum of the operator is
contained in [�0;1) implies
g�(0; 0) = � (m+(�) +m�(�))
�1 > 0 for � < �0:
500 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
Hence (16) shows Z
[�0;0]
�+(d�) + ��(d�)
2 (� � �)
< 1; for � < �0;
which implies the condition in the de�nition of ��0 : Now the expression (17) is
easily deduced from (15). Then it is routine that for a compact uniform limit of
some re�ectionless potentials the associated m� also have the representation (17)
by choosing a suitable � from ��0 : In the next section we will prove that any
element � from ��0 with at most �nitely many points as its support gives a clas-
sical re�ectionless potential. This together with the theorem below completes the
proof.
For q; de�ne a shift by
Txq(�) = q(�+ x) for x 2 R:
Introducing another pair of linearly independent solutions f'�; �g for
Lf = �f satisfying
�
f(0) = 1; f 0(0) = 0 =) '�(x)
f(0) = 0; f 0(0) = 1 =) �(x):
(19)
Then the uniqueness of f+ implies
f+(y; �;Txq) =
f+(y + x; �; q)
f+(x; �; q)
;
hence
m+ (�;Txq) = f 0+(0; �;Txq)
=
f 0+(x; �; q)
f+(x; �; q)
=
'0�(x; q) +m+(�; q)
0
�(x)
'�(x; q) +m+(�; q) �(x)
: (20)
m�(�; q) also has a similar expression, therefore it is easy to check that an identity
m+ (� + i0;Txq) +m�(� + i0;Txq) = 0 a.e. on [0;1)
holds if so is the case x = 0; which implies
�0 is a shift-invariant space.
Moreover, an asymptotic expansion of the Green function shows that8>>><>>>:
g�(x; x; q) =
1
�2i
p
�
�
q(x)
4i�
p
�
+ o
�
j�j�
3
2
�
,
@2g�(x; y; q)
@x@y
�����x=y = i
p
�
2
�
q(x)
4i
p
�
+ o
�
j�j�
1
2
�
as �!1
: (21)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 501
S. Kotani
On the other hand, we have another expression for the Green function by
using f'�; �g : Let
M(�) =
0BB@ �
1
m+(�) +m�(�)
�
m+(�)
m+(�) +m�(�)
+
1
2
�
m+(�)
m+(�) +m�(�)
+
1
2
m+(�)m�(�)
m+(�) +m�(�)
1CCA ;
which is a matrix valued Herglotz function. Set ��(x) = ('�(x); �(x))
T .
Then (1) shows
g�(x; y) = (M(�)��(x); ��(y)) : (22)
Here the inner product on C2 is de�ned without taking the complex conjugate.
For a potential q 2
�0 ; it is easy to see that for z 2 C
2 an H-function (M(�)z; z)
satis�es the identity (9) a.e. on [0;1): Therefore, applying Lem. 4, we have
M(�) = �
i
p
�
2
�
0 0
0 �1
�
�
i
p
�
2
Z 0
�0
1
� � �
�(d�);
where �(d�) is a real matrix valued non-negative de�nite measure on
[�0; 0]. Here we have used the asymptotics (21). Now it follows from (22) that
g�(x; y)
�
i
p
�
2
= � �(x) �(y) +
Z 0
�0
1
� � �
(�(d�)��(x); ��(y))
= � �(x) �(y) +
Z 0
�0
1
� � �
(�(d�) (��(x)� ��(x)) ; ��(y)� ��(y))
+
Z 0
�0
1
� � �
(�(d�)��(x); ��(y)) :
Since �rst two terms are holomorphic on C as a function of �; the asymptotics of
the Green function shows that the above right-hand side behaves like O(��1) as
�!1; which implies the sum of �rst two terms becomes zero. Hence we have
g�(x; y) = �
i
p
�
2
Z 0
�0
1
� � �
(�(d�)��(x); ��(y)) : (23)
This combined with (21) shows8>>>>>>>><>>>>>>>>:
Z 0
�0
(�(d�)��(x); ��(x)) = 1Z 0
�0
� (�(d�)��(x); ��(x)) =
V (x)
2Z 0
�0
�
�(d�)�0�(x); �
0
�(x)
�
= �
V (x)
2
: (24)
502 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
This, in particular, implies
2�0 � V (x) � 0
for any q 2
�0 : If we observe the identity
��00� (x) + q(x)��(x) = ���(x);
the repeating use of (24) shows that q is in�nitely di�erentiable and their deriva-
tives have bounds depending only on �0; which was proved by D.S. Lundina [11]
through a di�erent argument. We state the above argument as a theorem together
with the re�nements by V.A. Marchenko [14].
Theorem 7. The followings hold:
(i) The shift acts on
�0 :
(ii) Any element of
�0 is in�nitely di�erentiable and all its derivatives have
bounds ���q(n)(x)��� � 2
�p
��0
�n+2
(n+ 1)!
for n = 0; 1; 2; : : : : In particular,
�0 becomes compact in the compact uniform
metric.
(iii) Any element of
�0 is holomorphic on the strip
n
jIm zj <
p
��0
�1
o
and satis�es
jq(z)j � �2�0
�
1�
p
��0 jIm zj
��2
:
Among potentials in
�0 ; the potentials having �nite band structure are of
particular interest. We say a potential has a �nite band structure if there exists
a �nite number of nonoverlapping intervals [�i; �i], i = 1; 2; : : : ; n, in [�0; 0] on
which
m+(� + i0) = �m�(� + i0) for all � 2 [�i; �i] (i = 1; 2; : : : ; n)
holds. To compute �� in this case, �rst we consider
g(�) = � (m+(�) +m�(�))
�1 (= g�(0; 0)):
Taking log; we see
g(�) = exp
�
+
1
�
Z
R
�
1
� � �
�
�
1 + �2
�
arg g(� + i0)d�
�
=
1
�2i
p
�
exp
1
�
nX
i=1
Z �i
�i
�
2
� � �
d� +
1
�
nX
i=1
Z �i+1
�i
arg g(� + i0)
� � �
d�
!
;
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 503
S. Kotani
where we set �n+1 = 0: The constant factor is determined from the behaviour
(18) of g(�): Let �i be a unique zero of g(�) in [�i; �i+1] for i = 1; 2; : : : ; n: ThenZ �i+1
�i
arg g(� + i0)
� � �
d� = �
Z �i
�i
1
� � �
d� = � log
�� �i
�� �i
;
hence
g(�) =
1
�2i
p
�
exp
1
2
nX
i=1
log
�� �i
�� �i
+
nX
i=1
log
�� �i
�� �i
!
=
1
�2i
p
�
vuut nY
i=1
(�� �i)
2
(�� �i) (�� �i)
:
��i is a pole of m+(�) + m�(�); hence �+ (f�ig) + �� (f�ig) > 0: However if
�� (f�ig) > 0; then the Schr�odinger operator L has two non-trivial solutions f�
satisfying
Lf� = �if�; f�(0) = 0 and f� 2 L2(R�);
which means f� are linearly dependent, hence �i is an eigenvalue of L on L2(R):
This contradicts the fact that L has no spectrum outside the set
S =
[
1�i�n
[�i; �i] [ [0;1):
Therefore �+ (f�ig)�� (f�ig) = 0: Let
"+i =
�
1 if �+ (f�ig) > 0
0 otherwise
; "�i = 1� "+i :
Noting �+ = �� on S; we have
��(d�) = �(�)IS(�)d� +
nX
i=1
"�i �iÆ(f�ig)(d�); (25)
with
�(�) =
8>>>>><>>>>>:
1
�
p
��
p
j(� � �i) (� � �i)j
j� � �ij
sY
j:j 6=i
(� � �j) (� � �j)
(� � �j)
2
if � 2 [�i; �i]
1
�
p
�
vuut nY
i=1
(� � �i) (� � �i)
(� � �i)
2
if � � 0;
and
�i = 2
p
��i (�i � �i) (�i � �i)
vuutY
j:j 6=i
(�i � �j) (�i � �j)
(�i � �j)
2
:
504 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
A measure � on [�
p
��0;
p
��0] is de�ned by
�(d�) =
8><>:
1
p
��
�+(d�); if � =
p
�� > 0
1
p
��
��(d�); if � = �
p
�� < 0
:
In this case we can determine q by using the Theta function on the compact
Riemann surface for a hyperelliptic curve
w2 = �
nY
i=1
(�� �i) (�� �i) : (26)
We choose a homology basis f�i; �jg1�i;j�n on the surface and a basis of diffe-
rential forms of the �rst kind f!ig1�i�n satisfying
1
2�i
I
�i
!j = Æij :
Set
Bij =
I
�i
!j:
This matrix is called a period matrix for the surface, and it is known that B is
symmetric and its real part is negative de�nite. If all the points f��i;��jg lie on
the real line, the matrix B becomes real. Hence in this case B is a real symmetric
negative de�nite matrix. De�ne the Theta function
� (z1; z2; : : : ; zn) =
X
m2Zn
exp
�
1
2
(Bm;m) + (z;m)
�
:
Then it is known (see A.R. Its�V.B. Matveev [7]) that
Proposition 8. There exist c 2 R; a;b 2 Rn such that
q(x) = c� 2
d2
dx2
log � (xa+ b) : (27)
5. Characterization of Classical Re�ectionless Potentials
In this section we characterize classical re�ectionless potentials in terms of �
and give a concrete description of the measure
m(d�) =
nX
i=1
m+
i Æf�ig(d�)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 505
S. Kotani
associated with the scattering data.
For � from ��0 , �rst we compute F+ of Sect. 2. Set
�+(x; y) = �
1
2�
Z 1
0
(1� cos
p
�x)(1� cos
p
�y)
�2
p
�� (�) d�
+
Z
[�0;0]
(1� cos
p
�x)(1� cos
p
�y)
�2
�+ (d�)� x ^ y:
De�ne
F+(x; y) =
@2�+(x; y)
@x@y
:
Here we remark the positive-de�niteness of F+; which will be useful when applying
Sato theory.
Lemma 9. F+ is positive de�nite.
P r o f. A routine calculation shows
F+(x; y) =
�1
2�
Z 1
0
sin
p
�x sin
p
�y
p
�
(� (�) + 2) d�
+
Z 0
�0
sinh
p
��x sinh
p
��y
��
�+ (d�) ;
which implies the positive-de�niteness of F+:
Further calculation shows
F+(x; y) =
1
4
Z
[�0;0]
e
p
��(x+y) � e
p
��jx�yj
��
�+ (d�)
+
1
4
Z
[�0;0]
e�
p
��jx�yj � e�
p
��(x+y)
��
�� (d�) :
Replacing f�+; ��g with �; we have
F+(x; y) =
Z
[�
p
��0;
p
��0]
e�jx+yj � e�jx�yj
2�
�(d�): (28)
Now we compute the Fredholm determinant of the integral operator F+ in
L2([0; a]; dx) with kernel F+(x; y): For later purpose we decompose F+ into two
parts:
F+ = �V +B;
506 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
with
V (x; y) =
8<:
Z p
��0
�
p
��0
sinh �(x� y)
�
�(d�) for x � y
0 for x < y
;
B(x; y) =
Z �
p
�0
�
p
��0
sinh �x
�
e�y�(d�):
To compute the inverse (I � V )�1 we set
m(�) =
Z 1
0
e
p
��xdx
Z p
��0
�
p
��0
sinh �x
�
�(d�) =
Z p
��0
�
p
��0
�(d�)
��2 � �
: (29)
We note that m is a function of Herglotz type. Further, in this case m takes
negative values on (0;1). Therefore
em(�) �
m(�)
1�m(�)
= �1 +
1
1�m(�)
is an H-function as well and takes negative values on (0;1): Here we use the
condition Z
[�
p
��0;
p
��0]
�(d�)
��0 � �2
� 1;
which implies em(�) has no singularity on (�1; �0); hence there exists a unique
measure e� on R such that
em(�) =
Z
[0;
p
��0]
e�(d�)
��2 � �
: (30)
De�ning
eV (x; y) =
8<:
Z
[0;
p
��0]
sinh �(x� y)
�
e�(d�) for x � y
0 for x < y
;
we see
(I � V )�1 = I + eV :
Now we employ the method used by Ikeda�Kusuoka�Manabe [6]
I + F+ = I � V +B
= (I � V )(I + (I � V )�1B)
= (I � V )(I + (I + eV )B);
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 507
S. Kotani
which leads us to�eV B� (x; y) = Z x
0
eV (x; z)B(z; y)dz
=
Z p
��0
�
p
��0
Z p
��0
0
1
�2 � �2
�
sinh �x
�
�
sinh �x
�
�
e�ye� (d�) �(d�):
Now set
g(x; �) �
sinh �x
�
+
Z p
��0
0
1
�2 � �2
�
sinh �x
�
�
sinh �x
�
�e�(d�)
=
�
1 + em(��2)
� sinh �x
�
+
Z
[0;1)
1
�2 � �2
sinh �x
�
e�(d�):
We assume � has its support only on a �nite set f�ig1�i�n of [�
p
��0;
p
��0]
and
�2i 6= �2j if i; j 2 f1; 2; : : : ; ng , i 6= j: (31)
We try to compute the determinant keeping its generalization to � with in�nite
support in mind. (30) shows that e� has a �nite support f�ig1�i�n in (0;
p
��0]:
(30) implies also
� or � � 2 supp� =) em(��2) + 1 = 0: (32)
Therefore, for � or �� 2supp�
g(x; �) =
Z
[0;1)
1
�2 � �2
sinh �x
�
e�(d�);
and we have
eB(x; y) � B(x; y) +
�eV B� (x; y)
=
Z p
��0
�
p
��0
g(x; �)e�y� (d�)
=
Z
[0;
p
��0]
sinh �x
�
e�(d�)Z p
��0
�
p
��0
1
�2 � �2
e�y�(d�): (33)
To compute the determinant further we remark here a duality relation for
determinants. Let (X;F ; �); (�;M; �) be the measure spaces and
K(x; �); L(x; �) 2 L2(X � �; �� �):
508 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
De�ne 8><>:
F (x; y) =
Z
�
K(x; �)L(y; �)�(d�)
bF (�; �) =
Z
X
K(x; �)L(x; �)�(dx)
:
Lemma 10. F and F̂ de�ne trace class operators on L2(X;�) and L2(�; �)
respectively and it holds that
det(I + F ) = det(I + bF ):
P r o f. This is an in�nite dimensional version of the identity
det(I +AB) = det(I +BA)
for any n�m matrix A and m� n matrix B: We omit the proof.
Setting
K(x; �2) =
sinh �x
�
; L(y; �2) =
Z
R
1
�2 � �2
e�y�(d�);
we have eB(x; y) =
Z
[0;1)
K(x; �2)L(y; �2)e�(d�):
Hence the corresponding bB becomes
bB(�2; �2) =
Z a
0
K(x; �2)L(x; �2)dx;
we see from Lem. 10
det(I + eB) = det(I + bB): (34)
Now we compute bB(�2; �2):
bB(�2; �2) =
Z
R
1
�2 � �2
�(d�)
Z a
0
sinh �x
�
e�xdx
=
Z
R
1
�2 � �2
�(d�)
1
2�
e(�+�)a � 1
� + �
�
e(���)a � 1
� � �
!
=
Z
R
e�a
2� (�2 � �2)
�(d�)
�
e�a
� + �
�
e��a
� � �
�
+
Z
R
�(d�)
(�2 � �2) (�2 � �2)
:
(35)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 509
S. Kotani
HoweverZ
R
�(d�)
(�2 � �2) (�2 � �2)
=
8<:
m(��2)�m(��2)
�2 � �2
= 0 if �2 6= �2
�~�(f�g)�1 if �2 = �2
; (36)
hence setting �k = �(f�kg); ~�j = ~�(f�jg) and8>><>>:
aik =
1
�2i � �2k
;
bjk =
1
�j + �k
+
e�2�ja
�j � �k
;
(37)
we have
det(I + bB) = det
X
k
aike
(�j+�j)a
e�j
2�j
bjk
!!
= exp
0@a nX
j=1
(�j + �j)
1A0@Y
j
e�j�j
2�j
1Adet ((aij)) det((bij)) : (38)
To compute det ((bij)) we set8<: C+
ij =
1
�i + �j
; C�
ij =
1
�i � �j
�ij(a) = e�2�iaÆij
: (39)
Then
bij =
�
C+ +�(a)C��
ij
:
Introduce
H(�) = 1�
nY
i=1
�� �i
�+ �i
and eH(�) =
H(�)
1�H(�)
= �1 +
nY
i=1
�+ �i
�� �i
:
Then we can show that there exist real numbers f�ig1�i�n ; f�ig1�i�n such that
H(�) =
nX
i=1
�i
�+ �i
and eH(�) =
nX
i=1
�i
�� �i
:
Lemma 11. We have the identities:
(i)
nX
i=1
�i
�j + �i
= 1;
nX
i=1
�i
�i + �j
= 1 for 8j = 1; 2; : : : ; n;
(ii) �i =
0@ nX
j=1
�j
(�j + �i)
2
1A�1
; �i = (�i + �i)
nY
j:j 6=i
�i + �j
�i � �j
:
510 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
P r o f. We omit the proof, since the computation is elementary.
Lemma 12. (C+)
�1
=
�
�i�j
�j + �i
�
and C� (C+)
�1
=
�
�j (1�H(��i))
�i + �j
�
:
P r o f. Set
Pij =
�i
�j + �i
; Qij =
�i
�i + �j
and S = (�iÆij) :
Then Lem. 11 shows
(PQ)ij =
nX
k=1
�i
�k + �i
�k
�k + �j
=
8>>>><>>>>:
�i
�i � �j
nX
k=1
�
�k
�k � �i
�
�k
�k � �j
�
= 0 if i 6= j
�i
nX
k=1
�k
(�k + �i)
2
= 1 if i = j
= Æij :
Therefore �
C+
��1
= Q�1S = PS;
which implies
C� �C+
��1
ij
=
nX
k=1
1
�i � �k
�k
�j + �k
�j
=
�j
�i + �j
nX
k=1
�
�k
�i � �k
+
�k
�j + �k
�
=
�j (1�H(��i))
�i + �j
:
Lemma 13. �i (1�H(��i)) =
(1�H(��i))2
2�i
e�i > 0 for i = 1; 2; : : : ; n:
P r o f. Observe �1 � em(��) has zeroes
�
�2i
1�i�n and poles
�
�2i
1�i�n ;
hence
�em(��) =
nX
i=1
e�i
�2i � �
= 1�
nY
i=1
�� �2i
�� �2i
:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 511
S. Kotani
On the other hand, similarly as we obtained �i in Lemma 11, we have
e�i = ��2i � �2i
� nY
j:j 6=i
�2j � �2i
�2j � �2i
= (�i � �i) (�i + �i)
nY
j:j 6=i
(�i � �j) (�i + �j)
(�i � �j) (�i + �j)
= �i (�i � �i)
nY
j:j 6=i
�i � �j
�j + �i
=
2�i�i
1�H(��i)
;
hence
�i (1�H(��i)) =
(1�H(��i))2
2�i
e�i > 0:
Setting
mi = �i (1�H(��i)) > 0;
we see
Proposition 14. V 2
cl
�0
if and only if the associated � has a �nite support.
Moreover, the Fredholm determinant is given by
det(I + F+) = const:� det
�
Æij +
p
mimj
�i + �j
e�a(�i+�j)
�
;
with
const: = exp
0@aX
j
(�j + �j)
1A0@Y
j
�e�j�j
2�j
�1Adet
�
C+
�
det ((aij)) ;
and the measure m(d�) =
nX
i=1
miÆf�ig(d�) can be represented as
m(d�) =
(1�H(��))2
2�
e�(d�):
512 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
P r o f. Summing up the above argument, we have
det(I + F+) = exp
0@aX
j
(�j + �j)
1AY
j
�e�j�j
2�j
�
det ((aij))
� det
�
C+ +�(a)C��
= const:det
�
I +�(a)C� �C+
��1�
= const:det
�
Æij + e�2a�i
�j (1�H(��i))
�i + �j
�
= const:det
�
Æij + e�2a�i
�i (1�H(��i))
�i + �j
�
= const:det
�
Æij +
p
mimj
�i + �j
e�a(�i+�j)
�
:
Remark 15. If we replace � with b� constructed by re�ection from �,
that is, b�(d�) = �(�d�): Then e� remains unchanged and C+(resp.C�) turns to
C�(resp.C+); hence
det(I + bF+)(a) = det(I + F+)(�a):
Remark 16. The condition (31) can be removed if we approximate �
by a sequence of measures satisfying (31).
6. Construction of KdV-Flow
In this section we construct the KdV-�ow on
�0 by applying the theory of
M. Sato�Y. Sato. They gave a very transparent view for a class of integrable
systems including the KdV equation and later it was developed by Date�Jimbo�
Kashiwara�Miwa [1]. However, their original argument is quite algebraic. So we
imply here a more analytic version by S. Segal�G. Wilson [15] and give a complete
proof by calculating the � -functions for classical re�ectionless potentials.
Let S1 be the unit circle in C; and H = L2(S1): Introduce two orthogonal
subspaces H� of H8<: H+ =
n
f 2 H; f(z) =
X
n�0
fnz
n with
X
n�0
jfnj2 <1
o
H� =
n
f 2 H; f(z) =
X
n��1
fnz
n with
X
n��1
jfnj2 <1
o
:
Then it is easy to see that
H = H+ �H� (orthogonal sum). (40)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 513
S. Kotani
Let PH�
be the orthogonal projections to H�, respectively. Let W be a closed
subspace of H satisfying:
(i) PH+
: W �! H+ is a Fredholm operator (i.e., has �nite dimensional
kernel and cokernel) with index 0; that is dimKer = dimCoKer:
(ii) PH�
: W �! H� is a trace class operator;
(iii) f 2W �! z2f 2W ;
(iv) H� \W = f0g(transversality).
We denote by Gr(2)(H) the set of all closed subspaces W satisfying the condi-
tions (i), (ii), (iii) and (iv). The properties (i) and (iv) assure the unique existence
of a bounded operator A from H+ to H� such that
W = ff +Af ; f 2 H+g :
This is because (iv) implies dimKer = 0, and hence (i) implies PH+
(W ) = H+:
Conversely, if such an operator A exists, then (iv) holds. Introduce
� =
�
g; g(z) is holomorphic on D; g(0) = 1 , g(z) 6= 0 for 8z 2 D;
takes real values on R and g(�z) = g(z)�1 for 8z 2 D
�
;
where D = fz 2 C; jzj � 1g : Apparently � is a commutative group acting on
Gr(2)(H) by multiplication but for the condition (iv): For g 2 �; we represent it
as
g�1 =
�
a b
0 d
�
corresponding to the decomposition (40). Now, for g 2 � and W 2 Gr(2)(H)
we de�ne
�W (g) = det(I + a�1bA): (41)
Although the following lemma was proved in [16], we give a proof for the sake of
completeness.
Lemma 17. �W (g) 6= 0 if and only if g�1W is transverse to H�:
P r o f. Suppose for f 2 H+; f + a�1bAf = 0. Set
f1 = g�1Af � bAf 2 H�:
Then
gf1 = Af � a�1bAf = Af + f 2W;
which completes the proof.
514 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
Throughout this section we assume �1 < �0 < 0; which is not essential.
For � 2 ��0 ; de�ne m+(�z2) by (17). Set
W� =
�
A(z2) +m+(�z2)B(z2); A;B 2 H+
: (42)
Here it should be noted that the conditionZ p
��0
�
p
��0
�(d�)
1� �2
�
Z p
��0
�
p
��0
�(d�)
��0 � �2
� 1
assures
��m+(�z2)
�� � 3 if jzj = 1; hence W� becomes a closed subspace of H:
This space was considered by R. Johnson [9] as an application of Sato theory.
We compute the operator a�1bA for this space and identify its Fredholm deter-
minant with a Fredholm determinant of an F+, which makes it possible to show
the transversality of gW� .
Lemma 18. For f 2 H+; the equation
f(z2) = B(z2)�
Z p
�0
�
p
�0
B(z2)�B(�2)
z2 � �2
�(d�) (43)
is uniquely solvable in H+, and the solution is given by
B(z2) = f(z2) +
Z p
�0
0
f(z2)� f(�2)
z2 � �2
e�(d�): (44)
P r o f. Suppose the support � is �nite. Note �rst
B(z2)�
Z p
��0
�
p
��0
B(z2)�B(�2)
z2 � �2
�(d�) = B(z2)(1 �m(�z2)) +
Z p
��0
�
p
��0
B(�2)�(d�)
z2 � �2
f(z2) +
Z p
��0
0
f(z2)� f(�2)
z2 � �2
e�(d�) = f(z2)(1 + em(�z2))�
Z p
��0
0
f(�2)
z2 � �2
e�(d�);
where m was introduced in (29). Then, substituting (44) into (43), we see
B(z2)�
Z p
��0
�
p
��0
B(z2)�B(�2)
z2 � �2
�(d�)
= f(z2)(1�m(�z2))(1 + em(�z2))� (1�m(�z2))
Z p
��0
0
f(�2)
z2 � �2
e�(d�)
+
Z p
��0
�
p
��0
f(�2)(1 + em(��2))�(d�)
z2 � �2
�
Z p
��0
�
p
��0
�(d�)
z2 � �2
Z p
��0
0
f(�2)e�(d�)
�2 � �2
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 515
S. Kotani
= f(z2)� (1�m(�z2))
Z p
��0
0
f(�2)
z2 � �2
e�(d�)
�
Z p
��0
0
f(�2)e�(d�)Z p
��0
�
p
��0
�(d�)
(�2 � �2) (z2 � �2)
= f(z2):
In this calculation we have used the fact
1 + em(��2) = 0 if � 2 supp� and 1�m(��2) = 0 if � 2 suppe�:
Now the rest of the proof is easy if we approximate a general � 2 ��0 by a sequence
of �'s with �nite supports.
For f 2 H+; we de�ne8>><>>:
Pf(z) =
f(z)� f(�z)
2z
;
Kf(z2) = f(z2) +
Z p
��0
0
f(z2)� f(�2)
z2 � �2
e�(d�):
Lemma 19. For the space of (42) we have
�
a�1bA
�
f(z) =
Z p
��0
�
p
��0
1� g(z)g(�)�1
� � z
KPf(�2)�(d�):
P r o f. Let f 2 H+: Then the de�nition of the operator A implies
f(z) +Af(z) = A(z2) +m+(�z2)B(z2);
with A;B 2 H+: However
m+(�z2)B(z2) = �zB(z2)�
Z p
��0
�
p
��0
B(z2)�B(�2)
� � z
�(d�)�
Z p
��0
�
p
��0
B(�2)�(d�)
� � z
;
and �rst two terms are contained in H+ and the last term is contained in H�
since �0 > �1: Hence we have
f(z) = A(z2)� zB(z2)�
Z p
��0
�
p
��0
B(z2)�B(�2)
� � z
�(d�)
Af(z) = �
Z p
��0
�
p
��0
B(�2)
� � z
�(d�);
516 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
which implies
f(z)� f(�z)
2z
= �B(z2) +
Z p
��0
�
p
��0
B(z2)�B(�2)
z2 � �2
�(d�):
Then Lemma 18 shows
Af(z) = �
Z p
��0
�
p
��0
KPf(�2)
� � z
�(d�):
On the other hand, for f 2 H
PH+
f(z) =
1
2�i
I
jz0j=1
f(z0)
z0 � z
dz0;
hence
a�1bAf(z) =
g(z)
2�i
I
jz0j=1
g(z0)�1
z0 � z
dz0
Z p
��0
�
p
��0
KPf(�2)
� � z0
�(d�)
= g(z)
Z p
��0
�
p
��0
KPf(�2)�(d�)
1
2�i
I
jz0j=1
g(z0)�1
(z0 � z) (� � z0)
dz0
=
Z p
��0
�
p
��0
KPf(�2)
1� g(z)g(�)�1
� � z
�(d�);
which concludes the lemma.
Now we compute the Fredholm determinant when the support of � consists
of a �nite set f�ig1�i�n. First note
det(I + a�1bA) = det(I +B);
with B de�ned by
Bf(z2) �
Z p
��0
�
p
��0
Kf(�2)
�
1� g(z)g(�)�1
� � z
�
1� g(�z)g(�)�1
� + z
�
�(d�)
2z
;
for f 2 H+: Since � has a �nite support, the following calculation is possible:
Kf(�2) = f(�2) +
Z p
��0
0
f(�2)� f(�2)
�2 � �2
e�(d�)
=
�
1 + eh(��2)� f(�2)� Z p
��0
0
f(�2)
�2 � �2
e�(d�)
=
Z p
��0
0
f(�2)
�2 � �2
e�(d�);
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 517
S. Kotani
for 1 + em(��2) = 0 if � 2supp� as we saw in (32). Hence
Bf(z2)
=
Z p
��0
�
p
��0
�
g(�z)g(�)�1
2z (� + z)
�
g(z)g(�)�1
2z (� � z)
+
1
�2 � z2
�
�(d�)
Z p
��0
0
f(�2)e�(d�)
�2 � �2
=
Z p
��0
0
f(�2)e�(d�)Z p
��0
�
p
��0
�
g(�z)g(�)�1
2z (� + z)
�
g(z)g(�)�1
2z (� � z)
+
1
�2 � z2
�
�(d�)
�2 � �2
=
Z p
��0
0
f(�2)e�(d�)Z p
��0
�
p
��0
�
g(�z)g(�)�1
2z (� + z)
�
g(z)g(�)�1
2z (� � z)
�
�(d�)
�2 � �2
+
Z p
��0
0
f(�2)e�(d�)Z p
��0
�
p
��0
�(d�)
(�2 � �2) (�2 � z2)
:
Note here
det(I +B) = det(I +QB)
if we denote the restriction of f 2 H+ to L2(e�) by Q: Then (36) shows
(I +QB) f(�2)
=
Z p
��0
0
f(�2)e�(d�)Z p
��0
�
p
��0
�
g(��)g(�)�1
2� (� + �)
�
g(�)g(�)�1
2� (� � �)
�
�(d�)
�2 � �2
;
which is the same as (35) if we replace e��a with g. Now the computation is quite
analogous to that of Sect. 5, and we have
Lemma 20. Suppose the support of � is �nite f�ig1�i�n: Then
�W�(g) = const:� det
�
Æij +
p
mimj
�i + �j
g(��i)�1g(�j)
�
;
with
const: =
nY
j=1
�
g(��j)g(�j)�1�je�j
2�j
�
det
�
C+
�
det ((aij)) :
Now we de�ne the KdV �ow. For g 2 � and x 2 R, introduce gx 2 � by
gx(z) = e�xzg(z):
Lemma 21. For � 2 ��0 de�ne W� by (42). Then for any g 2 �; there exists
a �g 2 ��0 such that
�W�(g
x) = det(I + F x
+;�g
): (45)
518 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
P r o f. First suppose the support of � is �nite. Then Lemma 20 implies
�W�(g
x) = exp
0@aX
j
(�j + �j)
1A nY
j=1
(g(�j)g(�j))
�1
nY
j=1
�
�je�j
2�j
�
det(C+) det(aij)
�det(Æij +
p
mimj
�i + �j
e�x(�i+�j)g(�i)g(�j));
with some constants ec1(g);ec2(g): Hence, regarding mig(�i)
2 as a new weight,
we can construct a classical re�ectionless potential qg with the scattering data�
r+(k) = 0; i�j ;mjg(�j)
2
: Let �g 2 ��0 be its spectral measure. Then from
Prop. 14 the identity (45) follows. For a general � 2 ��0 ; approximate it by
a sequence of measures f�ng from ��0 with �nite supports. Then it is easy
to see that �W�n
(gx) and det(I + F x
+;�ng
) converge as n ! 1 to �W�(g
x) and
det(I + F x
+;�g
), respectively, if we choose a subsequence of
�
�ng
converging to a
�g 2 ��0 if necessary. Hence we have (45) for a general �.
This lemma combined with Lem. 9 has the following conclusion.
Theorem 22. For � 2 ��0 and g 2 �; �W�(g) > 0; hence g�1W� is transverse
to H�:
For g 2 � and q 2
�0 ; we de�ne
(K(g)q) (x) = �2
d2
dx2
log �W�(g
x) (46)
as an element of
�0 . Then we can prove the following
Theorem 23. K(g) is a homeomorphism on
�0satisfying
K(g1g2) = K(g1)K(g2) and K(1) = id
and K(gx)q(�) = q(�+x); K(gx;t)q satis�es the KdV equation, where gx(z) = e�xz
and gx;t(z) = e�xz+4tz3 :
P r o f. Since everything is valid if K(g) is restricted to the classical re�ec-
tionless potentials, all we have to do is to approximate � 2 ��0 by a sequence of
measures f�ng from ��0 with �nite supports. The multiplicativity of K comes
from the cocycle property of �W
�
g
�1
1
W
(g2) =
�W (g1g2)
�W (g1)
:
We call this K as KdV-�ow on
�0 : It is well known also that K(g)q satis�es
the higher order KdV equations if we choose suitable one-parameter groups on �.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 519
S. Kotani
7. Characterization of W�
We have introduced a space W�. It is natural to ask when W coincides with
a W�: We combine the two properties (i), (iv) of W and summarize them again.
A closed subspace W of H is in Gr(2)(H) if and only if:
(i) PH+
:W �! H+ is one-to-one and onto.
(ii) PH�
: W �! H� is a trace class operator.
(iii) f 2W =) z2f 2W:
We set new conditions. For f 2 H, de�ne f(z) = f(z):
(iv) f 2W =) f2W:
Let
gx(z) = e�xz:
(v) For any x 2 R; g�1x W satis�es (i).
Suppose the conditions (i)�(v). Then, for any x 2 R; there exists a unique
fW (x; �) 2W such that
fW (x; z) = e�xz
�
1 +
a1(x)
z
+
a2(x)
z2
+ � � �
�
: (47)
Since
fW (x; z) = e�xz
1 +
a1(x)
z
+
a2(x)
z2
+ � � �
!
;
and the property (iv) implies fW (x; �) 2 W; the uniqueness shows fW (x; z) =
fW (x; z); hence ai(x) 2 R:
Lemma 23 (Segal�Wilson): fW satis�es
�
d2
dx2
fW (x; z) � 2a01(x)fW (x; z) = �z2fW (x; z);
and fai(x)g1�i<1 satisfy
2a0i(x) = a00i�1(x) + 2a01(x)ai�1(x) for i = 2; 3; : : : : (48)
P r o f. Di�erentiating both sides of (47), we have
d2
dx2
fW (x; z)
= z2fW (x; z)� 2ze�xz
�
a01(x)
z
+
a02(x)
z2
+ � � �
�
+ e�xz
�
a001(x)
z
+
a002(x)
z2
+ � � �
�
= z2fW (x; z)� 2a01(x)fW (x; z) + e�xz
1X
i=1
a00i (x)� 2a0i+1(x) + 2a01(x)ai(x)
zi
:
520 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
The linearity of the space W and (iii) imply
d2
dx2
fW (x; z)� z2fW (x; z) + 2a01(x)fW (x; z) 2W:
Therefore from (vi) it follows that8<: d2
dx2
fW (x; z)� z2fW (x; z) + 2a01(x)fW (x; z) = 0;
2a0i(x) = a00i�1(x) + 2a01(x)ai�1(x) for i = 2; 3; : : : :
Suppose V (x) is a real valued in�nitely di�erentiable function on R: Consider
a Schr�odinger operator
L = �
d2
dx2
+ q
on L2(R): If the boundaries �1 are of the limit circle type, then we have to
impose suitable boundary conditions at �1: For � 2 C such that Im (�) 6= 0 let
f+(x; �) be the solution for
Lf+ = �f+ ; f+(0; �) = 1
and�
f+ 2 L2(R+) if +1 is a limit point ;
f+ satis�es the boundary condition at +1 if +1 is a limit circle.
Set
m+(�) = f 0+(0; �);
which is called the Weyl function.
Lemma 25. There exist real valued smooth functions fbi(x)g1�i<1 such that
for each �xed n � 0
f+(x;�z2) = e�xz
�
1 +
b1(x)
z
+
b2(x)
z2
+ � � �+
bn(x)
zn
+O
�
1
zn+1
��
(49)
holds as jzj ! 1 in a region �
�
2
+ " < arg z < �" for any " > 0: Moreover, they
satisfy �
2b0i(x) = b00i�1(x) + 2b01(x)bi�1(x) for i = 2; 3; : : : ;
bi(0) = 0 for i = 1; 2; 3; : : : :
(50)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 521
S. Kotani
P r o f. Let '; be the solutions for Lf = �z2f satisfying '(0) = 0(0) = 1;
'0(0) = (0) = 0: Then
f+(x;�z2) = '(x; z) +m+(�z2) (x; z):
F. Gesztesy�B. Simon [5] proved for each a > 0 there exists a unique function
A(�) on [0; a] such that
m+(�z2) = �z �
Z a
0
A(�)e�2�zd�+O
�
e�2(a�")z
�
holds as jzj ! 1 in a region �
�
2
+ " < arg z < �" for any " > 0: The function
A is determined from m on [0; a] and smooth if so is m: On the other hand, ';
can be represented as8>><>>:
'(x; z) = cosh zx+
Z x
0
K1(x; y) cosh zydy;
(x; z) =
sinh zx
z
+
Z x
0
K2(x; y)
sinh zy
z
dy;
with smooth functions K1;K2: With these identities we see that f+(x;�z2) has
an asymptotic expansion (49) together with its derivatives f
(k)
+ of any order.
Substituting this expansion to the equation Lf+ = �z2f+; we have
0 = exz
�
�f 00+(x;�z
2) + q(x)f+(x;�z2) + z2f+(x;�z2)
�
= 2b
0
1(x) + q(x) +
2b02(x)� b
00
1(x) + b1(x)q(x)
z
+
2b03(x)� b002(x) + b2(x)q(x)
z2
+ � � �+
2b0n+1(x)� b00n(x) + bn(x)q(x)
zn
+O
�
1
zn+1
�
;
hence �
2b
0
1(x) + q(x) = 0
2b0n+1(x)� b00n(x) + bn(x)q(x) = 0
holds for any n � 1; which implies (50).
We introduce one more condition:
(vi) 1 2W:
Now we can state
Theorem 26. Suppose a subspace W of H satis�es the conditions (i) � (vi).
Then there exists a unique � 2 ��1 such that W =W�:
522 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
P r o f. If a subspace W of H satis�es the conditions (i)�(vi); then we can
de�ne fW in (50): The condition (vi) shows fW (0; z) = 1; and hence
ai(0) = 0 for any i = 1; 2; : : : :
Then, setting q(x) = �2a01(x); from Lems. 24 and 25 we see fW (x; z) = f+(x;�z2)
identically. On the other hand, for this q we introduce f�(x;�z2) satisfying the
boundary condition at �1; and we can show fW (x;�z) = f�(x;�z2) identically.
Therefore we have
m+(�z2) = f 0W (0; z); m�(�z2) = �f 0W (0;�z): (51)
Now we consider the analytic continuation of fW : fW (x; z) is holomorphic on
fjzj > 1g and fW (x; z) = fW (x; z): Therefore m� are holomorphic on Cn[�1;1)
and take real values on (�1;�1): Therefore the representing measure �� has no
mass on (�1;�1): This combined with (51) shows that fW (x; z) is analytically
continuable to Cn ([� 1;1] [ [�i; i]) keeping the property fW (x; z) = fW (x; z):
Then it is immediate that
m+(� + i0) = f 0W (0; i
p
� + 0) = f 0W (0;�i
p
� + 0) = �m�(� + i0) for a.e. � > 0:
Then Proposition 6 shows that
m�(�z2) = �z �
Z 1
�1
�(d�)
�� � z
with a measure � on [� 1;1]: We have to proveZ 1
�1
�(d�)
1� �2
� 1: (52)
Suppose (52) is not valid. Then there exists �0 < �1 such thatZ 1
�1
�(d�)
��0 � �2
= 1: (53)
Set
f(z) = �
Z 1
�1
B(�2)�(d�)
� � z
with B(z2) =
1
z2 � �0
:
Then f 2 H� since m+(�z2) 2W: On the other hand, (53) implies
B(z2) =
Z 1
�1
B(z2)�B(�2)
z2 � �2
�(d�):
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 523
S. Kotani
Let
A(z2) = �
Z 1
�1
B(z2)�B(�2)
z2 � �2
��(d�):
Then
f(z) = A(z2) +m+(�z2)B(z2)
holds. However the Taylor expansions for A;B around the origin converge uni-
formly on fjzj � 1g ; and we know m+(�z2) 2W: Hence the property (iii) implies
f 2 W; hence W \ H� 6= f0g, which contradicts (i). Therefore (52) holds, thus
we have q 2
�1: It is easy to see that W� � W since 1;m+(�z2) 2 W; which
shows W� =W in view of the property (i). The uniqueness of � is trivial since �
is determined from fW ; which completes the proof.
Remark 27. It is interesting to remove the condition (vi). Without (vi) one
can proceed in parallel with the above argument up to a certain point. In this case
we have to consider a meromorphic function on 1 < jzj � 1
ffW (x; z) =
fW (x; z)
fW (0; z)
instead of fW (x; z) itself. Since fW (0;1) = 1; there exists R � 1 such that
fW (0; z) 6= 0 for all z such that jzj > R: ffW satis�es LffW = �z2ffW and has
an expansion ffW (x; z) = e�xz
�
1 +
ea1(x)
z
+
ea2(x)
z2
+ � � �
�
on jzj > R: Since feai(x)g1�i<1 satisfy the same equations (50), we see eai(x) =
bi(x) for all i = 1; 2; : : : : Hence
ffW (x; z) = f+(x;�z2) (54)
identically, which implies
f 0W (0; z)
fW (0; z)
= m+(�z2): (55)
Then we can show as above that the left-hand side of (55) is analytically continu-
able to Cn ([�R;R] [ [�iR; iR]) ; which shows
m+(� + i0) = �m�(� + i0) for a.e. � > 0:
Hence h+ has an expression
m+(�z2) = �z �
Z R
�R
�(d�)
� � z
:
524 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
Let �0 � �R2 be any number such thatZ R
�R
�(d�)
��0 � �2
� 1;
then 2�0 � q(x) � 0; which implies the boundaries �1 are of limit point type.
It is not clear if we can avoid the possibility that there exists a zero of fW (0; z)
in 1 < jzj < R:
8. Isospectral Property of KdV-Flow
We introduce a subgroup �0 of � by
�0 = fg 2 �; log g is a polynomial of odd degree with real coe�cientsg :
Let
Lq = �
d2
dx2
+ q:
Theorem 28. For q 2
�0 and g 2 �0; there exists a unitary operator U(g; q)
on L2(R; dx) satisfying
LK(g)q = U(g; q)�1LqU(g; q):
P r o f. We prove this theorem only for gx(z) = e�xz and gx;t(z) = e�xz+4tz3 :
For a general g 2 �0 the proof is analogous. For gx(z) = e�xz; the proof is trivial,
since K(gx)q(�) = q(� + x) and we have only to set U(gx; q) = Tx independently
of V: For gx;t(z) = e�xz+4tz3 ; we employ the Lax representation. Set
u(t; x) = (K(gt;0)q) (x) = (K(gt;x)q) (0);
and de�ne an antisymmetric operator
Aq = 4D3 � 6qD � 3Dq
with D =
d
dx
: Then it is easy to see
[Lq; Aq] = LqAq �AqLq = 6qqx � qxxx;
where the left-hand side is a multiplication operator. Since u(t; x) satis�es the
KdV equation, we see
d
dt
Lu(t) = [Lu(t); Au(t)];
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 525
S. Kotani
where u(t)(�) = u(t; �): This formula is called the Lax representation of the KdV
equation. De�ne a one-parameter family of unitary operators by solving an equa-
tion
dU(t)
dt
= U(t)Au(t); U(0) = I:
Then it is obvious that Lu(t) = U(t)�1LqU(t); which concludes the proof.
Remark 29. One can expect Theorem 28 holds for any g 2 �; however to show
this we have to construct a pseudodi�erential operator for which the commutator
with Lq reduces to a multiplication operator.
9. Floquet Exponent
Let
=
�0 be the space of generalized re�ectionless potentials and fK(g)gg2�
be the KdV �ow. We consider the set of all probability measures PK (
) on
which are invariant under fK(g)gg2� and denote by Pshift (
) the set of all shift
invariant probability measures on
: Since the set of all probability measures
P (
) on
is compact and the �ow fK(g)gg2� is commutative, we easily see that
PK (
) is a non-empty compact convex set. For � 2 Pshift (
), set
w(�) = w�(�) =
Z
m+(�; q)�(dq); (56)
which is called Floquet exponent for � 2 Pshift (
). This was �rst introduced
by Johson�Moser. Then it is well known that
w(�) =
Z
m�(�; q)�(dq) = �
1
2
Z
g�(0; 0; q)
�1�(dq); (57)
and
w0(�) =
Z
g�(0; 0; q)�(dq) (58)
hold (see S. Kotani [10]). Then the commutativity and the isospectral property
under K imply easily the following
Theorem 30. If � 2 Pshift (
), then for all g 2 � it holds K(g)� 2 Pshift (
)
and we have wK(g)� = w�:
Let WK be the set of all functions w satisfying
(i) w;�iw; w0 are Herglotz functions,
(ii) w(�) < 0 on (�1;��0];
(iii) Rew(� + i0) = 0 on [0;1);
(iv) w(�)=
p
��! �1 as �! �1:
Then from the compactness of
we have
526 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
KdV Flow on Generalized Re�ectionless Potentials
Theorem 31. For any � 2 PK (
) ; w� 2 WK is valid: Conversely, for any
w 2 WK there exists a � 2 PK (
) such that w = w�:
It is natural to ask
To what extent does w determine � 2 PK (
) ? (59)
If w comes from a �nite bands spectrum, then the characterization of all potentials
with �nite bands spectrum (see [7]) shows that w determines uniquely �.
Appendix. In this Appendix we collect several fundamental facts of
H-functions (Herglotz functions). For the proofs see P.L. Duren [2]. As we de�ned
in Sect. 2, a holomorphic function de�ned on C+ is called a Herglotz function if
it maps C+ into C+: It is well known that an H-function m has an expression
m(�) = �+ ��+
Z
R
�
1
� � �
�
�
1 + �2
�
�(d�)
with �; � 2 R; � � 0 and a nonnegative measure � on R satisfyingZ
R
(1 + �2)�1� (d�) <1:
The triple f�; �; �g is called the characteristics for m: The measure � can be
recovered from m by
�([a; b)) =
1
�
lim
"#0
Z b
a
Imm(� + i")d�
if � (fag) = � (fbg) = 0: It is also known that for a.e. � 2 R
lim
"#0
1
�
Imm(� + i") = �0(�)
holds, where �0(�) denotes the density of absolutely continuous part of �: m itself
has a �nite limit
lim
"#0
m(� + i") = m(� + i0) 6= 0 a.e.� 2 R
unless m � 0: It is of some use to note that if m is an H-function, so is �m�1 and
logm: Especially logm is useful when we have to factorize m in an appropriate
way. Since
Im logm(�) = argm(�) 2 [0; �];
applying the above representation for m, we easily see that
logm(�) =
+
1
�
Z
R
�
1
� � �
�
�
1 + �2
�
argm(� + i0)d�
with
2 R:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4 527
S. Kotani
References
[1] E. Date, M. Jimbo, M. Kashiwara, and T. Miwa, Transformation Groups for Soliton
Equations. VII. � Publ. RIMS, Kyoto Univ. 18 (1982), 1077�1110.
[2] P.L. Duren, Theory of Hp spaces. 38. Pure and Appl. Math., Acad. Press, 1970.
[3] F. Dyson, Fredholm Determinants and Inverse Scattering Problems. � Comm.
Math. Phys. 47 (1976), 171�183.
[4] F. Gesztesy, W. Karwowski, and Z. Zhao, Limit of Soliton Solutions. � Duke Math.
J. 68 (1992), 101�150.
[5] F. Gesztesy and B. Simon, A New Approach to Inverse Spectral Theory, II. General
Real Potentials and the Connection to the Spectral Measure. � Ann. Math. 152
(2000), 593�643.
[6] N. Ikeda, S. Kusuoka, and S. Manabe, L�evy's Stochastic Area Formula for Gaussian
Processes. � Comm. Pure and Appl. Math. XLVII (1994), 329�360.
[7] A.R. Its and V.B. Matveev, Hill's Operator with Finitely Many Gaps. � Funct.
Anal.Appl. 9 (1975), 65�66.
[8] K. Iwasaki, On the Inverse Sturm�Liouville Problem with Spatial Symmetry. �
Funkc. Ekvacioj 31 (1988), 25�74.
[9] R.A. Johnson, On the Sato�Segal�Wilson Solutions of the KdV Equation. � Paci�c
J. Math. 132 (1988), 343�355.
[10] S. Kotani, Lyapounov Indices Determine Absolutely Continuous Spectra of Station-
ary One-Dimensional Schr�odinger Operators. � Proc. Taniguchi Symp. SA. Katata
(1984), 225�248.
[11] D.S. Lundina, Compactness of the Set of Re�ectionless Potentials. � Teor. Funkts.,
Funkts. Anal. i Prilozh. 44 (1985), 55�66. (Russian)
[12] V.A. Marchenko, Sturm�Liouville Operators and Applications. Birkh�auser, Boston,
1986.
[13] V.A. Marchenko, Nonlinear Equations and Operator Algebras. D. Reidel, Dor-
drechet. 1987.
[14] V.A. Marchenko, The Cauchy Problem for the KdV Equation with Nondecreasing
Initial Data. Springer Ser. in Nonlinear Dynamics, What is Integrability? (V.E. Za-
kharov, Ed.) (1990), 273�318.
[15] H.P. McKean, Geometry of KdV (5): Scattering from the Grassmannian Viewpoint.
� Comm. Pure and Appl. Math. XLII (1989), 687�701.
[16] G. Segal and G. Wilson, Loop Groups and Equations of KdV Type. � Inst. Haute
�Etude Sci. Publ. Math. 61 (1985), 5�65.
528 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 4
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