Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum
The theorem of the necessary and sufficient conditions for the solvability of ISP under consideration is proved. The method of addition of the discrete spectrum to the considered matrix not self-adjoint case is developed.
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irk-123456789-1064442016-09-29T03:02:15Z Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum Zubkova, E.I. Rofe-Beketov, F.S. The theorem of the necessary and sufficient conditions for the solvability of ISP under consideration is proved. The method of addition of the discrete spectrum to the considered matrix not self-adjoint case is developed. 2007 Article Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum / E.I. Zubkova, F.S. Rofe-Beketov // Журнал математической физики, анализа, геометрии. — 2007. — Т. 3, № 2. — С. 176-195. — Бібліогр.: 15 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106444 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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The theorem of the necessary and sufficient conditions for the solvability of ISP under consideration is proved. The method of addition of the discrete spectrum to the considered matrix not self-adjoint case is developed. |
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Article |
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Zubkova, E.I. Rofe-Beketov, F.S. |
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Zubkova, E.I. Rofe-Beketov, F.S. Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum Журнал математической физики, анализа, геометрии |
| author_facet |
Zubkova, E.I. Rofe-Beketov, F.S. |
| author_sort |
Zubkova, E.I. |
| title |
Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum |
| title_short |
Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum |
| title_full |
Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum |
| title_fullStr |
Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum |
| title_full_unstemmed |
Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum |
| title_sort |
inverse scattering problem on the axis for the schrödinger operator with triangular 2 x 2 matrix potential. ii. addition of the discrete spectrum |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| publishDate |
2007 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/106444 |
| citation_txt |
Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. II. Addition of the Discrete Spectrum / E.I. Zubkova, F.S. Rofe-Beketov // Журнал математической физики, анализа, геометрии. — 2007. — Т. 3, № 2. — С. 176-195. — Бібліогр.: 15 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
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| first_indexed |
2025-07-07T18:30:01Z |
| last_indexed |
2025-07-07T18:30:01Z |
| _version_ |
1837013930694672384 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2007, vol. 3, No. 2, pp. 176�195
Inverse Scattering Problem on the Axis
for the Schr�odinger Operator with Triangular 2� 2
Matrix Potential. II. Addition of the Discrete Spectrum
E.I. Zubkova
Ukrainian State Academy of Railway Transport
7 Feyerbakh Sq., Kharkiv, 61050, Ukraine
E-mail:bond@kart.edu.ua
F.S. Rofe-Beketov
Mathematical Division, B. Verkin Institute for Low Temperature Physics & Engineering
National Academy of Sciences of Ukraine
47 Lenin Ave., Kharkiv, 61103, Ukraine
E-mail:rofebeketov@ilt.kharkov.ua
Received February 14, 2006
The theorem of the necessary and su�cient conditions for the solvability
of ISP under consideration is proved. The method of addition of the discrete
spectrum to the considered matrix not self-adjoint case is developed.
Key words: scattering on the axis, inverse problem, triangular matrix
potential, discrete spectrum.
Mathematics Subject Classi�cation 2000: 47A40, 81U40.
This work constitutes Part II of [15]. Notations, de�nitions, numerations of
the statements, formulas, etc., extend those of [15]. The formulas (1)�(49) are
contained in [15]. References [1�5] repeat those of [15], but references [6�14] of
[15] are omitted here. Correction to Part I see at the end of the present Part II.
Theorem 1. If a set of values fR+(k); k 2 R; k2j < 0; Z+
j (t); j = 1; pg forms
the right SD for the scattering problem on the axis for the Schr�odinger operator
(1) with an upper triangular 2�2 matrix potential having the second moment ((2)
with m = 2) and with the real diagonal and without virtual level, conditions 1)�6)
should be satis�ed:
1) R+(k) is continuous in k 2 R : r+ll (k) = r+ll (�k); jr
+
ll (k)j � 1� Clk
2
1+k2
; l =
1; 2; R+(0) = �I; I � R+(�k)R+(k) = O(k2) as k ! 0 and R+(k) = O(k�1)
with k ! �1 (note that replacing the last condition by R+(k) = o(k�1) we obtain
a necessary condition too).
c
E.I. Zubkova and F.S. Rofe-Beketov, 2007
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
2) The function
F+
R (x) =
1
2�
1Z
�1
R+(k)eikxdk (50)
is absolutely continuous, and with a > �1 one has
+1R
a
�
1 + x2
� �� d
dx
F+
R (x)
�� dx
<1.
3) The functions zall(z), l = 1; 2, given by (48), are continuously di�erentiable
in the closed upper half-plane.
4) The function
F�R (x) = �
1
2�
1Z
�1
C(k)�1R+(�k)C(�k)e�ikxdk (51)
is absolutely continuous, and with a < +1 one has
aR
�1
�
1 + x2
� �� d
dx
F�R (x)
�� dx
< 1. Here c12(z) is given by (49), cll(z) � all(z) is determined by (48) (one
can show* that condition 4 is also necessary in version 4a, namely, if c12(z) is
constructed as in (42), (43), and cll(z) � all(z) is constructed as in (39), which
correspond to the absence of discrete spectrum).
5) degZ+
j (t) �
2P
l=1
sign z
[j]+
ll
� 1; j = 1; p, z
[j]+
ll
� 0 and z
[j]+
ll
are constant.
6) rgZ+
j (t) = rg diagZ+
j (t) = rg diagZ+
j (0); j = 1; p.
The necessary conditions 1)�6) listed above (with condition 4 being replaced
by its version 4a) become su�cient together with the following assumption:
H) The function ka11(�k)a22(k)fr
+
11(�k)r
+
12(k) + r+12(�k)r
+
22(k)g satis�es
the H�older condition in the �nite points as well as at in�nity.
(The claims of the theorem related solely to the diagonal matrix elements,
are direct consequences of [1, 2].)
Remark 3. In the case when the discrete spectrum is absent, conditions 5 and
6 of Th. 1 become inapplicable, and conditions 4 and 4a become the same.
P r o o f of Theorem 1. The necessity. Similarly to [1], under condition
(2) one has that R+(k) is a continuous function of k 2 R. In this context,
since the upper triangular potential of the scattering problem (1) has its principal
diagonal formed by real functions, the following relations hold: r+ll (k) = r+ll (�k)
and
��r+ll (k)�� � 1� Clk
2
1+k2
, l = 1; 2 (see [1]).
*E.g., using the procedure of subsequent eliminating eigenvalues or the properties of
the Fourier transform, which we omit here.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 177
E.I. Zubkova and F.S. Rofe-Beketov
Furthermore, (9) and (11) imply with k 2 R that there exist the limits
kE�(x; k) = fE+(x; k)[R
+(k) + I] +E+(x;�k)�E+(x; k)gkC(k): (52)
Since the scattering problem in (1), (2) is assumed to have no virtual
level, the de�nition (10) implies the existence of the limits
lim
k!0
kA(k) = C1; lim
k!0
kC(k) = C2;
det C1 6= 0; det C2 6= 0:
(53)
Thus, passage to a limit as k ! 0 in (52) yields 0 = lim
k!0
fE+(x; k)[R
+(k)
+I]gkC(k) = E+(x; 0) lim
k!0
[R+(k) + I]C2. By a continuity of R+(k) one has
R+(0) = �I. Also by (53), we deduce from (14) that I �R+(�k)R+(k) = O(k2)
as k ! 0.
Lemma 6. The coe�cients A(k) and B(k) given by (10), admit representa-
tions as follows:
A(k) = I �
1
2ik
8<:
1Z
�1
V (x)dx+
0Z
�1
A1(t)e
�iktdt
9=; ;
B(k) =
1
2ik
1Z
�1
B1(t)e
�iktdt;
with A1(t) being a summable matrix function whose �rst moment with m = 2 is
on (�1; 0];
B1(t) is a summable matrix function whose �rst moment with m = 2 is on
(�1;1).
P r o o f of Lemma 6 coincides with that of the lemma by V.A. Marchenko
[1, Lemma 3.5.1.], if one takes into account that under condition (2) the kernel
K(x; t) of the transformation operator is a summable function, which has (when
m = 2) the �rst moment with respect to t 2 [x;1).
Lemma 6 and the de�nition of re�ection coe�cient (11), (24) imply R+(k) =
o( 1
k
) as k ! �1. Condition 1) of the theorem is proved completely.
Condition 2) of the theorem follows from the arguments, with the help of
which the Marchenko equation is derived for the given right SD (see [1, 4, 5]).
The fact that zall(z) (48) as well as the same function zcll(z) are continuous
in the closed upper half-plane is proved in [1] by an application of Lemma 3.5.1
from [1]. The continuous di�erentiability of these functions in the closed upper
half-plane is a direct consequence of Lemma 6.
178 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
Use Remark 2 to Lemma 5 [15], the relation (29) between the left and the right
re�ection coe�cients, and the argument that derives the Marchenko equation by
a contour integration for the given left SD (see [1], the text that starts at (3.5.14)
and ends at (3.5.190)) to prove that condition 4) holds.
The necessity of conditions 5) and 6) of the theorem follows from claims a)
and b) of Lemma 1 [15].
The necessity of assumptions of Theorem 1 is proved.
Prove a su�ciency of assumptions of the theorem.
1. The Case of the Absence of Discrete Spectrum
First, reconstruct the problem (1), (2) with m = 2, given R+(k), without
eigenvalues and normalizing polynomials. In this case, use (29) and the formulas
(39), (40) of Lemma 5 [15], to construct the function R�0 (k) as follows:
R�0 (k) = �C0(k)
�1R+(�k)C0(�k); C0(k) =
�
a011(k) c012(k)
0 a022(k)
�
; (54)
where zero indices indicate the absence of eigenvalues.
Prove that R+(k) and R�0 (k) are the right and the left re�ection coe�cients
of the same di�erential equation (1), whose potential is triangular, summable,
and has the second moment on the real axis. Since R+(k) and R�0 (k) are up-
per triangular and the diagonal elements r+ll (k) and r
[0]�
ll (k), l = 1; 2, satisfy the
assumptions of the Marchenko lemma [1, Lemma 3.5.3], one deduces that the
Marchenko equations constructed respectively to R+(k) and R�0 (k) have unique
solutions K0
+(x; y) and K0
�
(x; y), and analogously eK0
+(x; y) and eK0
�
(x; y). (In
fact, the equations for diagonal elements are solvable unambiguously by [1], and
the equations for k012+ and k012� di�er from those ones for the diagonal ele-
ments only by a free term). By the same Lemma 3.5.3 of [1], the functions
E0
�
(x; k) = e�ikxI �
�1R
x
K0
�
(x; t)e�iktdt are the Jost solutions of the Schr�odinger
equations on the entire axis, in which the potentials V �0 (x) possess the prop-
erty (2) with m = 2 (i.e., have the second moment), and similarly eE0
�
(x; k) =
e�ikxI �
�1R
x
eK0
�
(x; t)e�iktdt are the Jost tilde-solutions.
To prove that R+(k) and R�0 (k) are the right and the left re�ection coe�cients
of the same equation, it su�ces to demonstrate that
E0
�
(x; k)C0(k)
�1 = E0
+(x;�k) +E0
+(x; k)R
+(k);
E0
+(x; k)A0(k)
�1 = E0
�
(x;�k) +E0
�
(x; k)R�0 (k); k 2 R:
(55)
We follow the ideas of [1, 3] to prove (55).
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 179
E.I. Zubkova and F.S. Rofe-Beketov
De�ne a function
�+(x; y) := F+
R (x+ y) +
1Z
x
K0
+(x; t)F
+
R (t+ y)dt;
with F+
R given by (50). It is quite plausible from the above that at every �xed x,
the function �+(x; y) is in L(�1;1) since F+
R (y) 2 L(�1;1). Furthermore,
by virtue of (50) one has
1Z
�1
�+(x; y)e
�ikydy = E0
+(x; k)R
+(k):
By the Marchenko equation �+(x; y) = �K0
+(x; y) with x < y <1.
Since
1R
x
K0
+(x; y)e
�ikydy = E0
+(x;�k) � e�ikxI, one has
1R
�1
�+(x; y)e
�ikydy
=
xR
�1
�+(x; y)e
�ikydy + e�ikxI �E0
+(x;�k). Thus,
E0
+(x; k)R
+(k) +E0
+(x;�k) = H�(x; k)C0(k)
�1; (56)
where
H�(k) = e�ikx
8<:I + lim
N!1
xZ
�N
�+(x; y)e
�ik(y�x)dy
9=;C0(k):
It su�ces to show that
H�(x; k) = E0
�
(x; k) (57)
and this proves (55). In fact, consider the system
E0
+(x; k)R
+(k) +E0
+(x;�k) = H�(x; k)C0(k)
�1;
E0
+(x; k) +E0
+(x;�k)R
+(�k) = H�(x;�k)C0(�k)
�1
with respect to E0
+(x;�k) to deduce from (14) that
H�(x; k)R
�
0 (k) +H�(x;�k) = E0
+(x; k)A0(k)
�1; (58)
which by virtue of (57) yields (55).
Similarly to the proof of Theorem 6.5.1 of [3] it is possible to establish the
following three properties of the function H�(x; k):
1. H�(x; k) admits an analytic continuation into the upper half-plane, and
for large z one has the estimate jH�(x; z) � e�ixzIj = O
�
exIm z
jzj
I
�
:
180 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
2. zH�(x; z) is continuous in the closed upper half-plane, and zH�(x; z) =
o(I) as z ! 0 (uniformly in x).
3. H�(x; k) � e�ikxI 2 L2(�1;1) in k.
Use the above properties of H�(x; k) to prove (57). For x < y, consider
an analytic in the upper half-plane function [H�(x; z) � e�ixzI]eiyz . Use the
method of contour integration to obtain, in view of properties 1�3,
lim
R!1
RZ
�R
h
H�(x; k) � e�ixkI
i
eiykdk = 0; x < y:
Hence,
H�(x; k) = e�ikxI +
xZ
�1
G�(x; y)e
�ikydy; (59)
for some G�(x; y) 2 L2(�1; x).
From (58) and (59) one has E0
+(x; k)A0(k)
�1 � eikxI =
xR
�1
G�(x; y)e
ikydy
+e�ikxR�0 (k) +
xR
�1
G�(x; y)e
�ikydyR�0 (k).
By a construction, A0(z) and A
�1
0 (z) are regular in the open upper half-plane,
hence with t < x one has
0 = 1
2�
1R
�1
(E0
+(x; k)A0(k)
�1 � eikxI)e�iktdt
= G�(x; t) + F�R0
(x+ t) +
xR
�1
G�(x; y)F
�
R0
(t+ y)dt:
That is, G�(x; y) satis�es the Marchenko equation. It follows from the unam-
biguous solvability of the Marchenko equation that K0
�
(x; t) � G�(x; t), whence
one deduces (57) in view of (59).
Thus R+(k) and R�0 (k) are the right and the left re�ection coe�cients for the
problem (1), (2) with m = 2 under the absence of discrete spectrum, so in this
special case Theorem 1 is proved.
2. The Addition of Discrete Spectrum
Now consider a general case when the problem might have a �nite number p
of di�erent eigenvalues (for the considered 2� 2 triangular matrix potential they
can be either simple or of multiplicity two, and respectively the ranks of normal-
izing polynomials Z+
j (t) will be either 1 or 2). We proceed by induction (cf., for
example, [3] in the scalar case). Suppose that for the data
fR+(k); k21; : : : k
2
p; Z
+
1 (t); : : : Z+
p (t)g (60)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 181
E.I. Zubkova and F.S. Rofe-Beketov
the inverse problem on the axis is solved, that is those values from the right SD for
a problem of the form (1), (2) with m = 2 and a potential V0(x) (do not confuse
index 0 with the previous notation of Sect. 1!). We are about to show how in this
case to obtain a solution of the inverse problem with the p+1 di�erent eigenvalues
and the normalizing polynomials, that is, with the right SD of the form
fR+(k); k21 ; : : : k
2
p; k
2
p+1; Z
+
1 (t); : : : Z+
p (t); Z+
p+1(t)g: (61)
Denote by E0
+(x; k) and eE0
+(x; k) the Jost solutions for the equations, respectively,
�Y 00 + V0(x)Y = k2Y; �1 < x < +1; (62)
� eZ 00 + eZV0(x) = k2 eZ; �1 < x < +1; (63)
with asymptotics E0
+(x; k) � eikxI, eE0
+(x; k) � eikxI as x ! +1, Im k � 0.
Since k2p+1 is not an eigenvalue of the equations (62) and (63), one has that
E0
+(x; kp+1) and eE0
+(x; kp+1) decay exponentially as x! +1, and increase
exponentially as x! �1.
We generalize the procedure of attaching the discrete spectrum expounded in
[3, Ch. VI, � 6] to the considered matrix not self-adjoint case.
Set
F (x; y) =
E0
+(x; kp+1)Z
+
p+1(0)
eE0
+(y; kp+1)� i
d
dk
n
E0
+(x; k)(Z
+
p+1)
0(0) eE0
+(y; k)
o
kp+1
;
and consider the degenerate integral equation about B(x; y)
B(x; y) + F (x; y) +
1Z
x
B(x; t)F (t; y)dt = 0; x < y: (64)
Solve it to obtain
B(x; y) = �E0
+(x; kp+1)Z
+
p+1(0)
h
I +
1R
x
eE0
+(t; kp+1)E
0
+(t; kp+1)dtZ
+
p+1(0)
i
�1
� eE0
+(y; kp+1) + ib(x; y)(Z+
p+1)
0(0); (65)
with
b(x; y) =
_e011+(x;kp+1)e
0
22+(y;kp+1)
1+z
[p+1]+
22
1R
x
e022+(t;kp+1)2dt
+
e011+(x;kp+1) _e
0
22+(y;kp+1)
1+z
[p+1]+
11
1R
x
e011+(t;kp+1)2dt
�
e011+(x;kp+1)e
0
22+(y;kp+1)z
[p+1]+
11
1R
x
_e011+(t;kp+1)e
0
11+(t;kp+1)dt
(1+z
[p+1]+
11
1R
x
e011+(t;kp+1)2dt)(1+z
[p+1]+
22
1R
x
e022+(t;kp+1)2dt)
�
e011+(x;kp+1)e
0
22+(y;kp+1)z
[p+1]+
22
1R
x
_e022+(t;kp+1)e
0
22+(t;kp+1)dt
(1+z
[p+1]+
11
1R
x
e011+(t;kp+1)2dt)(1+z
[p+1]+
22
1R
x
e022+(t;kp+1)2dt)
:
(66)
182 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
Set
4 V (x) = �2
d
dx
B(x; x); V (x) = V0(x) +4 V (x): (67)
Lemma 7. The matrix function 4V (x) given by (67) possesses the property
+1Z
�1
(1 + jxjm)j 4 V (x)jdx <1; (68)
if V0(x) satis�es condition (2) with m � 1.
P r o o f. Since E0
+(x; kp+1) and eE0
+(x; kp+1) decay exponentially as x! +1,
it follows from (65)�(67) that 4V (x) also decays exponentially as x! +1.
Thus it remains to show that 4V (x) has m moments at �1 if V0(x) satis�es
(2). For this we introduce the notations
�(x; kp+1) = e�ikp+1xE0
+(x; kp+1);e�(x; kp+1) = e�ikp+1x eE0
+(x; kp+1);
�(x; kp+1) = e�2ikp+1x
1R
x
eE0
+(t; kp+1)E
0
+(t; kp+1)dt
(69)
and generalize the techniques of [3] to obtain the following three Lemmas 8�10
which are similar to Lemmas 6.6.1�6.6.3 of [3].
Lemma 8. The matrix functions �(x; kp+1), e�(x; kp+1), given by (69), and
_'ll(x; kp+1) = d
dk
�
e�ikxe0ll+(x; k)
�
kp+1
(where 'll are the diagonal elements of
�; e�, l = 1; 2), are bounded in the neighborhood of x = �1.
A p r o o f results immediately from the representations (5) for the Jost
solutions and the inequalities for transformation operators (see [4]):
jK0
+(x; t)j � C
1Z
x+t
2
jV0(s)jds; j eK0
+(x; t)j � C
1Z
x+t
2
jV0(s)jds; (70)
with some constant C and also with the use of exponentially rising solutions as
x! �1 with asymptotics of (7).
Lemma 9. One has the following inequalities (m � 1) for the matrix functions
�(x; kp+1), e�(x; kp+1) (69) and _'ll(x; kp+1):
0R
�1
(1 + jtjm)j d
dt
�(t; kp+1)jdt <1;
0R
�1
(1 + jtjm)j d
dt
e�(t; kp+1)jdt <1;
0R
�1
(1 + jtjm)j d
dt
_'ll(t; kp+1)jdt <1; l = 1; 2:
(71)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 183
E.I. Zubkova and F.S. Rofe-Beketov
A p r o o f is similar to that of Lemma 6.6.2 [3] in view of the fact that V0(x)
satis�es (2) with m � 1.
Lemma 10. The following statements are valid for the matrix function
�(x; kp+1) (69):
a) j�(x; kp+1)j and j�
�1(x; kp+1)j are bounded as x! �1;
b)
0Z
�1
(1 + jtjm)j
d
dt
�(t; kp+1)jdt <1; m � 1; (72)
c) j _
ll(x; kp+1)j are bounded as x! �1, l = 1; 2, and
0Z
�1
(1 + jtjm)j
d
dt
_
ll(t; kp+1)jdt <1; m � 1; l = 1; 2; (73)
with
ll, l = 1; 2, being the diagonal elements of the matrix function �(x; kp+1).
P r o o f s of propositions a) and b) of Lemma 10 are similar to that of
Lemma 6.6.3 [3] taking into account that V0(x) satis�es (2) with m � 1. Also
note that the boundedness of j��1(x; kp+1)j as x ! �1 follows from the fact
that j�(x; kp+1)j is bounded and j
ll(x; kp+1)j � al > 0, l = 1; 2, as x! �1.
Prove the proposition c) of Lemma 10.
From (69) one has
_
ll(x; kp+1) = 2
1R
x
_'ll(t; kp+1)'ll(t; kp+1)e
�2ikp+1(x�t)dt
�2i
1R
x
'2
ll(t; kp+1)(x� t)e�2ikp+1(x�t)dt
= 2
0R
�1
_'ll(x� z; kp+1)'ll(x� z; kp+1)e
�2ikp+1zdz
�2i
0R
�1
'2
ll(x� z; kp+1)ze
�2ikp+1zdz:
(74)
Thus, in view of Lemma 8, we obtain that
j _
ll(x; kp+1)j � Cll(
0Z
�1
e�2ikp+1zdz +
0Z
�1
jzje�2ikp+1zdz) <1; l = 1; 2;
as x! �1.
184 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
Since
1R
0
e0ll+(x; kp+1)dx <1, with the notation
0ll(x; kp+1) = e�2ikp+1x
0Z
x
e0ll+(t; kp+1)
2dt; �ll(kp+1) =
1Z
0
e0ll+(t; kp+1)
2dt;
we obtain the following representation:
ll(x; kp+1) = �ll(kp+1)e
�2ikp+1x +
0ll(x; kp+1);
and hence
_
ll(x; kp+1) = �2ixe�2ikp+1x�ll(kp+1) + e�2ikp+1x _�ll(kp+1) + _
0ll(x; kp+1):
Thus it su�ces to prove (73) for the function d
dt
_
0ll(t; kp+1).
So,
_
0ll(x; kp+1) = �2ix
0Z
x
'll(t; kp+1)
2e�2ikp+1(x�t)dt+ 2
0Z
x
[ _'ll(t; kp+1)
+ it'll(t; kp+1)]'ll(t; kp+1)e
�2ikp+1(x�t)dt
= 2
0Z
x
_'ll(x� z; kp+1)'ll(x� z; kp+1)e
�2ikp+1zdz
� 2i
0Z
x
'll(x� z; kp+1)
2ze�2ikp+1zdz:
Hence
d
dt
_
0ll(t; kp+1) = �2 _'ll(0; kp+1)'ll(0; kp+1)e
�2ikp+1t
+2i'll(0; kp+1)
2te�2ikp+1t + 2
0Z
t
�
d
dt
_'ll(t� z; kp+1)'ll(t� z; kp+1)
+ _'ll(t� z; kp+1)
d
dt
'll(t� z; kp+1)
�
e�2ikp+1zdz
� 4i
0Z
t
d
dt
'll(t� z; kp+1)'ll(t� z; kp+1)ze
�2ikp+1zdz:
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 185
E.I. Zubkova and F.S. Rofe-Beketov
Thus, in view of Lemmas 8 and 9, one has
0Z
�1
jtjm
���� ddt _
0ll(t; kp+1)
���� dt � C0 + C1
0Z
�1
� 0Z
t
���� ddt _'ll(t� z; kp+1)
���� e�2ikp+1zdz
+
0Z
t
���� ddt'll(t� z; kp+1)
���� e�2ikp+1zdz
�
jtjmdt
+C2
0Z
�1
jtjm
0Z
t
���� ddt'll(t� z; kp+1)
���� jzje�2ikp+1zdzdt
= C0 + C1
0Z
�1
e�2ikp+1zdz
zZ
�1
jtjm
����� ddt _'ll(t� z; kp+1)
����+ ���� ddt'll(t� z; kp+1)
�����dt
+C2
0Z
�1
jzje�2ikp+1zdz
zZ
�1
jtjm
���� ddt'll(t� z; kp+1)
���� dt <1;
and Lemma 10 is proved.
We return to the proof of Lemma 7. Use the notation of (69) to rewrite
4V (x):
4V (x) = �2
d
dx
B(x; x)
= �2
d
dx
[��(x; kp+1)Z
+
p+1(0)fIe
�2ikp+1x + �(x; kp+1)Z
+
p+1(0)g
�1e�(x; kp+1)
+i
�
e
�2ikp+1x d
dk
('11(x;k)'22(x;k))kp+1
+2ix'11(x;kp+1)'22(x;kp+1)e
�2ikp+1x
(e
�2ikp+1x+z
[p+1]+
11
11(x;kp+1))(e
�2ikp+1x+z
[p+1]+
22
22(x;kp+1))
+
_'11(x;kp+1)'22(x;kp+1)z
[p+1]+
11
11(x;kp+1)+'11(x;kp+1) _'22(x;kp+1)z
[p+1]+
22
22(x;kp+1)
(e
�2ikp+1x+z
[p+1]+
11
11(x;kp+1))(e
�2ikp+1x+z
[p+1]+
22
22(x;kp+1))
�
'11(x;kp+1)'22(x;kp+1)
2
(z
[p+1]+
11 _
11(x;kp+1)+z
[p+1]+
22 _
22(x;kp+1))
(e
�2ikp+1x+z
[p+1]+
11
11(x;kp+1))(e
�2ikp+1x+z
[p+1]+
22
22(x;kp+1))
�
(Z+
p+1)
0(0)]:
Now consider the possible (with the assumption 6 of the theorem being taken
into account) Cases I�III:
I) z
[p+1]+
11 = 0. In view of the assumption 5) of the theorem one has (Z+
p+1)
0(t)
� 0, Z+
p+1(t) � Z+
p+1, that is
186 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
4V (x) = 2 1
e
�2ikp+1x+z
[p+1]+
22
22
�f�0(x; kp+1)Z
+
p+1
e�(x; kp+1) + �(x; kp+1)Z
+
p+1
e�0(x; kp+1)g
�2
�2ikp+1e
�2ikp+1x+z
[p+1]+
22
022
(e
�2ikp+1x+z
[p+1]+
22
22)2
�(x; kp+1)Z
+
p+1
e�(x; kp+1):
(75)
II) z
[p+1]+
22 = 0. In view of the assumption 5) of the theorem one has (Z+
p+1)
0(t)
� 0, Z+
p+1(t) � Z+
p+1, that is
4V (x) = 2 1
e
�2ikp+1x+z
[p+1]+
11
11
�f�0(x; kp+1)Z
+
p+1
e�(x; kp+1) + �(x; kp+1)Z
+
p+1
e�0(x; kp+1)g
�2
�2ikp+1e
�2ikp+1x+z
[p+1]+
11
011
(e
�2ikp+1x+z
[p+1]+
11
11)2
�(x; kp+1)Z
+
p+1
e�(x; kp+1):
(76)
III) z
[p+1]+
ll > 0, l = 1; 2, then
4V (x) = 2�0(x; kp+1)fe
�2ikp+1xZ+
p+1(0)
�1 + �(x; kp+1)g
�1e�(x; kp+1)
+2�(x; kp+1)fe
�2ikp+1xZ+
p+1(0)
�1 + �(x; kp+1)g
�1e�0(x; kp+1)
�2�(x; kp+1)fe
�2ikp+1xZ+
p+1(0)
�1 +�(x; kp+1)g
�1[�2ikp+1e
�2ikp+1xZ+
p+1(0)
�1
+�0(x; kp+1)]fe
�2ikp+1xZ+
p+1(0)
�1 + �(x; kp+1)g
�1e�(x; kp+1)
�2i
�
�2ikp+1e
�2ikp+1x[ d
dk
('11'22)+2ix'11'22]
(e
�2ikp+1x+z
[p+1]+
11
11)(e
�2ikp+1x+z
[p+1]+
22
22)
+
e
�2ikp+1x[ d2
dkdx
('11'22)+2i'11'22+2ix('11'22)
0]
(e
�2ikp+1x+z
[p+1]+
11
11)(e
�2ikp+1x+z
[p+1]+
22
22)
+
z
[p+1]+
11 [( _'11'22)
0
11+ _'11'22
0
11�
('11'22)
0
2
_
11�
'11'22
2
_
011]
(e
�2ikp+1x+z
[p+1]+
11
11)(e
�2ikp+1x+z
[p+1]+
22
22)
+
z
[p+1]+
22 [('11 _'22)
0
22+'11 _'22
0
22�
('11'22)
0
2
_
22�
'11'22
2
_
022]
(e
�2ikp+1x+z
[p+1]+
11
11)(e
�2ikp+1x+z
[p+1]+
22
22)
�
�
�4ikp+1e
�4ikp+1x�2ikp+1e
�2ikp+1x(z
[p+1]+
11
11+z
[p+1]+
22
22)
(e
�2ikp+1x+z
[p+1]+
11
11)2(e
�2ikp+1x+z
[p+1]+
22
22)2
+
e
�2ikp+1x(z
[p+1]+
11
011+z
[p+1]+
22
022)+z
[p+1]+
11 z
[p+1]+
22 (
11
22)
0
(e
�2ikp+1x+z
[p+1]+
11
11)2(e
�2ikp+1x+z
[p+1]+
22
22)2
�
�
n
e�2ikp+1x
h
d
dk
('11'22) + 2ix'11'22
i
+ _'11'22z
[p+1]+
11
11
+'11 _'22z
[p+1]+
22
22 �
'11'22
2
(z
[p+1]+
11 _
11 + z
[p+1]+
22 _
22)
o�
(Zp+1)
0(0).
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 187
E.I. Zubkova and F.S. Rofe-Beketov
It follows from Lemmas 8�10 that in each of three cases the matrix function
4V (x) satis�es (68) if V0(x) satis�es (2) with m � 1. Lemma 7 is proved.
It is possible to deduce from (64) the following di�erential equation for B(x; y)
(compare, for instance, [3]):
@2B(x; y)
@x2
� V (x)B(x; y) =
@2B(x; y)
@y2
�B(x; y)V0(y); (77)
with B(x; x) = 1
2
1R
x
4V (t)dt.
It follows from (77), in view of the fact that B(x; y) tends to zero as y ! +1
(see (65)), that the function
E+(x; k) = E0
+(x; k) +
1Z
x
B(x; y)E0
+(y; k)dy; Im k � 0; (78)
is the Jost solution for the equation
�Y 00 + V (x)Y = k2Y; �1 < x < +1; (79)
with the asymptotics E+(x; k) � eikxI as x! +1.
Lemma 11. The right SD of the problems (79), (2) with m � 1 coincide with
the values given in (61).
A p r o o f of the lemma is based on the computation of the coe�cients
A(k) and B(k) (10) for the equation (79).
In view of the assumptions 5) and 6) of the theorem one has three possible
cases again:
Case I). Z+
p+1(t) � Z+
p+1 =
0 z
[p+1]+
12
0 z
[p+1]+
22
!
: z
[p+1]+
22 > 0, then we obtain
from (65) and (78), in view of (9) and (27) as x! �1
E+(x; k) = E0
+(x; k) �E0
+(x; kp+1)Z
+
p+1[I +
1R
x
eE0
+(t; kp+1)E
0
+(t; kp+1)dtZ
+
p+1]
�1
�
1R
x
eE0
+(y; kp+1)E
0
+(y; k)dy
� eikx
�
I �
2kp+1
(k+kp+1)z
[p+1]+
22 a022(kp+1)
A0(kp+1)Z
+
p+1
�
A0(k)
+e�ikx
�
I +
2kp+1
(k�kp+1)z
[p+1]+
22 a022(kp+1)
A0(kp+1)Z
+
p+1
�
B0(k),
188 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
hence
A(k) = �(k)A0(k); B(k) = �(�k)B0(k); k 2 R; (80)
with
�(k) = I �
2kp+1
(k+kp+1)z
[p+1]+
22 a022(kp+1)
A0(kp+1)Z
+
p+1
=
0B@ 1 �
2kp+1(a
0
11(kp+1)z
[p+1]+
12 +a012(kp+1)z
[p+1]+
22 )
(k+kp+1)z
[p+1]+
22 a022(kp+1)
0
k�kp+1
k+kp+1
1CA : (81)
It is clear from the �rst relation in (80) that the equation (79) has the same
eigenvalues k21 ; k
2
2 ; : : : ; k
2
p as the initial equation (62), and one more eigenvalue
k2p+1.
To compute the right re�ection coe�cient R+
1 (k) and C(k) that correspond
to the constructed equation (79), we use (24) and (14):
R+
1 (k) = �A(k)�1B(�k) = �A0(k)
�1��1(k)�(k)B0(�k) = R+(k); k 2 R;
(82)
and
C(k) = (I �R+
1 (�k)R
+
1 (k))
�1A(�k)�1
= (I �R+(�k)R+(k))�1A0(�k)
�1��1(�k) = C0(k)�
�1(�k); k 2 R: (83)
It is clear from (81) that ��1(�k) = �(k), hence C(k) = C0(k)�(k).
Now prove that the initial p normalizing polynomials of the problem (79), (2)
with m � 1 coincide with the normalizing polynomials of problem (62), (2) with
m � 1.
It follows from the de�nition (15) of the normalizing polynomials Z�j<1>(t) for
the equation (79) and the relation (80) that
Z�j<1>(t) = �if _W�(kj)A
0<kj>
�2 �(�kj)�W�(kj)A
0<kj>
�2 _�(�kj)
+W�(kj)A
0<kj>
�1 �(�kj)g � tW�(kj)A
0<kj>
�2 �(�kj); j = 1; p; (84)
with A
0<kj>
�2 = (A0(k)
�1(k � kj)
2)kj ; A
0<kj>
�1 = d
dk
(A0(k)
�1(k � kj)
2)kj being
the Laurent coe�cients.
One can show, using (27) and (32) with x ! +1 and noting that kj is
an eigenvalue of the equation (62) with kj < kp+1, j = 1; p, that
W�(kj)A
0<kj>
�2 = ��(�kj)(Z
�
j )0(0);
_W�(kj)A
0<kj>
�2 +W�(kj)A
0<kj>
�1 = i�(�kj)(Z
�
j )(0) + _�(�kj)(Z
�
j )0(0);
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 189
E.I. Zubkova and F.S. Rofe-Beketov
hence (84) can be rewritten as follows:
Z�j<1>(t) = �(�kj)Z
�
j (t)�(�kj)� i
d
dk
(�(k)(Z�j )0(0)�(k))k=�kj :
Next, use the relation (30) with � = t for the right and the left normalizing
polynomials Z�j<1>(t) of the problem (79):
Z+
j<1>(t) = �A
<kj>
�1 (Z�j<1>(t) +Qj)
�1C
<kj>
�1
= �[A
0<kj>
�1 �(�kj)�A
0<kj>
�2 _�(�kj)]�(kj)
�fZ�j (t)� i�(kj)
d
dk
(�(k)(Z�j )0(0)�(k))k=�kj�(kj) + �(kj)Qj�(kj)g
�1
��(kj)[�(�kj)C
0<kj>
�1 � _�(�kj)C
0<kj>
�2 ],
with C
0<kj>
�1 = d
dk
(C0(k)
�1(k � kj)
2)kj ; C
0<kj>
�2 = (C0(k)
�1(k � kj)
2)kj being
the Laurent coe�cients.
In view of (81) one has
fZ�j (t) + �(kj)Qj�(kj)� i�(kj)
d
dk
(�(k)(Z�j )0(0)�(k))k=�kj�(kj)g
�1
= (Z�j (t) + �(kj)Qj�(kj))
�1 +
2ikp+1
(k2j�k
2
p+1)(z
[j]�
11 +q
[j]
11 )(z
[j]�
22 +(
kj�kp+1
kj+kp+1
)2q
[j]
22 )
(Z�j )0(0),
hence
Z+
j<1>(t) = �A
0<kj>
�1 (Z�j (t) + �(kj)Qj�(kj))
�1C
0<kj>
�1
+A
0<kj>
�1 (Z�j (t) + �(kj)Qj�(kj))
�1�(kj) _�(�kj)C
0<kj>
�2
+A
0<kj>
�2 _�(�kj)�(kj)(Z
�
j (t) + �(kj)Qj�(kj))
�1C
0<kj>
�1
�
2ikp+1
(k2j�k
2
p+1)(z
[j]�
11 +q
[j]
11 )(z
[j]�
22 +(
kj�kp+1
kj+kp+1
)2q
[j]
22 )
A
0<kj>
�1 (Z�j )0(0)C
0<kj>
�1 .
If z
[j]�
11 > 0 and z
[j]�
22 > 0, then q
[j]
11 = q
[j]
22 = 0 with the de�nitions of Qj of
Lemma 3 [15], being taken into account;
A
0<kj>
�2 =
0
�a012(kj)
_a011(kj) _a
0
22(kj)
0 0
!
;
hence
Z+
j<1>(t) = Z+
j (t)
+
2kp+1
(k2j�k
2
p+1)z
[j]�
22 _a022(kj) _a
0
11(kj)
0@ 0
�a012(kj)
_a022(kj)
� i
(z
[j]�
12 )0(0)
z
[j]�
11
0 0
1A = Z+
j (t), j = 1; p,
in view of (19) and (21).
190 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
On the other hand, if z
[j]�
11 = 0 or z
[j]�
22 = 0, then by the assumption 5) of
Theorem 1 (Z�j )0(0) = 0 = A
0<kj>
�2 , hence Z+
j<1>(t) = Z+
j (t), j = 1; p.
Thus in any variant of Case I, the initial p normalizing polynomials of the
equation (79) turn out to be the same as those of the equation (62).
Now prove that Z+
p+1 is a normalizing polynomial of (79).
Similarly to (84), use (80), (81) to obtain
Z�p+1<1>(t) = �i d
dk
(W�(k)A�1
0 (k)�(�k)(k � kp+1)
2)kp+1
�t(W�(k)A�1
0 (k)�(�k)(k � kp+1)
2)kp+1
= �iW�(kp+1)A
�1
0 (kp+1)
h
2kp+1
z
[p+1]+
22 a022(kp+1)
A0(kp+1)Z
+
p+1
i
= �iW�(kp+1)Z
+
p+1
2kp+1
z
[p+1]+
22 a022(kp+1)
=W
� eE^
�
(x; kp+1); E
0
+(x; kp+1)
�
1R
x
e0+22 (t;kp+1)
2dt
1+z
[p+1]+
22
1R
x
e
0+
22 (t;kp+1)2dt
E0
+(x; kp+1)Z
+
p+1
�
1
z
[p+1]+
22 a022(kp+1)
Z+
p+1
= 1
a022(kp+1)
W
� eE^
�
(x; kp+1); E
0
+(x; kp+1)
�
I �
1R
x
e
0+
22 (t;kp+1)
2dt
1+z
[p+1]+
22
1R
x
e0+22 (t;kp+1)2dt
Z+
p+1
��
.
Assume x! �1 and apply the asymptotics E0
+(x; kp+1) � eikp+1xA0(kp+1),
that is, as x! �1
Z�p+1<1>(t) � Z�p+1<1> = 1
a022(kp+1)
Wfeikp+1xI; eikp+1xA0(kp+1)
�
0B@ 1 �
z
[p+1]+
12
z
[p+1]+
22
0 �
2ikp+1e
�2ikp+1x
z
[p+1]+
22 a022(kp+1)2
1CAg = 1
a022(kp+1)
A0(kp+1) �
0 0
0
(2ikp+1)
2
z
[p+1]+
22 a022(kp+1)2
!
=
(2ikp+1)
2
z
[p+1]+
22 a022(kp+1)3
A0(kp+1)
�
0 0
0 1
�
=
�4k2p+1
z
[p+1]+
22 a022(kp+1)3
�
0 a012(kp+1)
0 a022(kp+1)
�
.
It follows from (30) with � = t and (83) that
Z+
p+1<1> = �A
<kp+1>
�1 (Z�p+1<1> +Qp+1)
�1C
<kp+1>
�1
= �A�1
0 (kp+1)f
2kp+1
z
[p+1]+
22 a022(kp+1)
A0(kp+1)Z
+
p+1g
z
[p+1]+
22 a022(kp+1)
3
�4k2p+1a
0
22(kp+1)
f
2kp+1
z
[p+1]+
22 a022(kp+1)
�A0(kp+1)Z
+
p+1gC
�1
0 (kp+1) = Z+
p+1
a022(kp+1)
z
[p+1]+
22
z
[p+1]+
22
1
a022(kp+1)
= Z+
p+1,
and Lemma 11 in the Case I is proved.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 191
E.I. Zubkova and F.S. Rofe-Beketov
Case II). Z+
p+1(t) � Z+
p+1 =
z
[p+1]+
11 z
[p+1]+
12
0 0
!
: z
[p+1]+
11 > 0, then simi-
larly to the Case I we obtain
A(k) = �(k)A0(k); B(k) = �(�k)B0(k); k 2 R; (85)
with
�(k) = I �
2ikp+1
(k+kp+1)z
[p+1]+
11 a011(kp+1)
Z+
p+1C0(kp+1)
=
0@ k�kp+1
k+kp+1
�
2ikp+1(z
[p+1]+
11 c012(kp+1)+z
[p+1]+
12 a022(kp+1))
(k+kp+1)z
[p+1]+
11 a011(kp+1)
0 1
1A : (86)
It is clear from the �rst relation in (85) that the equation (79) has the same
eigenvalues k21 ; k
2
2 ; : : : ; k
2
p as the original equation (62), and one more eigenvalue
k2p+1.
Just as in (83), we deduce from (24) and (14) that
C(k) = C0(k)�
�1(�k) = C0(k)�(k); (87)
and the re�ection coe�cient of the equation (79) coincides with that of (62):
R+
1 (k) = R+(k).
The coincidence of the p initial normalizing polynomials of the equation (79)
with those of (62) can be proved just as in Case I with the substitution of �(k)
(86) for �(k) (81).
Now prove that Z+
p+1 is a normalizing polynomial of (79). Similarly to (84),
use (85) and (86) to obtain
Z�p+1<1>(t) � Z�p+1<1> = �iW�(kp+1)A
�1
0 (kp+1)
2kp+1
z
[p+1]+
11 a011(kp+1)
Z+
p+1
�C0(kp+1) =
1
z
[p+1]+
11 a011(kp+1)2
W
n eE^
�
(x; kp+1); E
0
+(x; kp+1)
�
e
0+
11 (x;kp+1)
1+z
[p+1]+
11
1R
x
(e0+11 (t;kp+1))2dt
Z+
p+1
1R
x
eE0
+(y; kp+1)E
0
+(y; kp+1)dy
o
Z+
p+1C0(kp+1)
=
W
�
e^11(x;kp+1);
e
0+
11
(x;kp+1)
z
[p+1]+
11
1R
x
(e
0+
11
(t;kp+1))
2dt
z
[p+1]+
11 a011(kp+1)
Z+
p+1C0(kp+1).
Supposing x ! �1 and using the asymptotics e0+11 (x; kp+1) � eikp+1xa011(kp+1),
that is with x! �1, one has
Z�p+1<1> =
W
�
e
ikp+1x;e
�ikp+1xa011(kp+1)
2ikp+1
�z
[p+1]+
11
a0
11
(kp+1)
2
z
[p+1]+
11 a011(kp+1)2
Z+
p+1C0(kp+1)
=
�4k2p+1
(z
[p+1]+
11 )2a011(kp+1)3
Z+
p+1C0(kp+1).
192 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
It follows from the relation (30) with � = t and (87) that
Z+
p+1<1> = �A
<kp+1>
�1 (Z�p+1<1> +Qp+1)
�1C
<kp+1>
�1 = �
2kp+1
z
[p+1]+
11 a011(kp+1)
A�1
0 (kp+1)
�Z+
p+1C0(kp+1)
(z
[p+1]+
11 )2a011(kp+1)
3
�4k2p+1
�
Z+
p+1C0(kp+1) +Qp+1
z
[p+1]+
11 a011(kp+1)
3
�4k2p+1
�
�1
�
2kp+1
z
[p+1]+
11 a011(kp+1)
Z+
p+1 = 1
a011(kp+1)2
a011(kp+1)(z
[p+1]+
11 )2a011(kp+1)
3 1
z
[p+1]+
11 a011(kp+1)
� 1
z
[p+1]+
11 a011(kp+1)
Z+
p+1 = Z+
p+1,
and in Case II Lemma 11 is proved too.
Case III). Z+
p+1(t) =
z
[p+1]+
11 z
[p+1]+
12 (t)
0 z
[p+1]+
22
!
: z
[p+1]+
11 > 0, z
[p+1]+
22 > 0,
then, similarly to Case I we obtain
A(k) =
(k)A0(k); B(k) =
(�k)B0(k); k 2 R; (88)
with
(k) =
k�kp+1
k+kp+1
I + i
�
a011(kp+1)
(k+kp+1)z
[p+1]+
22 a022(kp+1)
+
a022(kp+1)
(k+kp+1)z
[p+1]+
11 a011(kp+1)
�
2kp+1a
0
11(kp+1)
(k+kp+1)2z
[p+1]+
11 a011(kp+1)
�
(Z+
p+1)
0(0): (89)
The eigenvalues k2j , j = 1; p, of the equations (79) and (62) coincide, and (79)
has one more eigenvalue k2p+1 by (88) and (89).
(24) and (14) imply that the re�ection coe�cients for the equations (79) and
(62) coincide R+
1 (k) = R+(k), and
C(k) = C0(k)
�1(�k): (90)
The coincidence of the initial p normalizing polynomials of (79) and (62) can
be deduced similarly to Case I.
In this context,
Z�j<1>(t) = �if _W�(kj)A
0<kj>
�2
�1(kj) +W�(kj)A
0<kj>
�2
d
dk
(
�1(k))kj
+W�(kj)A
0<kj>
�1
�1(kj)g � tW�(kj)A
0<kj>
�2
�1(kj): (91)
Using kj < kp+1, j = 1; p, as x! �1, we get
Z�j<1>(t) =
(�kj)Z
�
j (t)
�1(kj)�
4ikp+1(kj + kp+1)
(kj � kp+1)3
(Z�j )0(0):
Thus, (30) with � = t, (88) and (89) imply
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 193
E.I. Zubkova and F.S. Rofe-Beketov
Z+
j<1>(t) = �[A
0<kj>
�1
�1(kj) +A
0<kj>
�2
d
dk
(
�1(k))kj ]
(kj)
��
Z�j (t) +
(kj)Qj
�
�1(�kj)
�
�1
+
4ikp+1(kj+kp+1)
(kj�kp+1)3(z
[j]�
11 +(
kj�kp+1
kj+kp+1
)2q
[j]
11 )(z
[j]�
22 +(
kj�kp+1
kj+kp+1
)2q
[j]
22 )
�kj�kp+1
kj+kp+1
�2
�(Z�j )0(0)
�1(�kj)[
(�kj)C
0<kj>
�1 � _
(�kj)C
0<kj>
�2 ]
= Z+
j (t) +A
0<kj>
�1 (Z�j (t) +
(kj)Qj
�1(�kj))
�1
�1(�kj) _
(�kj)C
0<kj>
�2
�A
0<kj>
�2
d
dk
(
�1(k))kj
(kj)(Z
�
j (t) +
(kj)Qj
�1(�kj))
�1C
0<kj>
�1
�
4ikp+1
(k2j�k
2
p+1)(z
[j]�
11 +(
kj�kp+1
kj+kp+1
)2q
[j]
11 )(z
[j]�
22 +(
kj�kp+1
kj+kp+1
)2q
[j]
22 )
A
0<kj>
�1 (Z�j )0(0)C
0<kj>
�1 = Z+
j (t)
in view of (19) and (21) and since Qj = 0, j = 1; p, with the de�nitions of Qj of
Lemma 3 [15], being taken into account, as z
[j]�
11 > 0 and z
[j]�
22 > 0. On the other
hand, if z
[j]�
11 = 0 or z
[j]�
22 = 0, then by assumption 5) of the theorem (Z�j )0(0) = 0,
A
0<kj>
�2 = (A0(k)
�1(k � kj)
2)kj = 0; C
0<kj>
�2 = (C0(k)
�1(k � kj)
2)kj = 0, hence
Z+
j<1>(t) = Z+
j (t), j = 1; p.
Prove that Z+
p+1(t) is a normalizing polynomial of the problem (79), (2) with
m � 1.
Using (88) and (89), similarly to (91), one has
Z�p+1<1>(t) = �2ikp+1W
�(kp+1)A
�
0 (kp+1)�
2kp+1
z
[p+1]+
22 a022(kp+1)
_W�(kp+1)(Z
+
p+1)
0(0)
�
�
1
z
[p+1]+
22 a022(kp+1)
+
a22(kp+1)
z
[p+1]+
11 a011(kp+1)2
�
2kp+1 _a
0
11(kp+1)
a011(kp+1)z
[p+1]+
22 a022(kp+1)
�
W�(kp+1)(Z
+
p+1)
0(0)
+
2ikp+1t
z
[p+1]+
22 a022(kp+1)
W�(kp+1)(Z
+
p+1)
0(0).
It is easy to show using the asymptotics as x! �1
E0
+(x; kp+1) � eikp+1xA0(kp+1); eE0
+(x; kp+1) � eikp+1xC0(kp+1);
that with x! �1
Z�p+1<1>(t) = (2ikp+1)
2[A0(kp+1)Z
+
p+1(t)C0(kp+1)]
�1
+2ikp+1
�
1
z
[p+1]+
11 z
[p+1]+
22 a011(kp+1)a
0
22(kp+1)
�
2� 2kp+1
�
_a022(kp+1)
a022(kp+1)
+
_a011(kp+1)
a011(kp+1)
��
+
a011(kp+1)
(z
[p+1]+
22 )2a022(kp+1)3
+
a022(kp+1)
(z
[p+1]+
11 )2a011(kp+1)3
�
(Z+
p+1)
0(0):
(92)
(30) with � = t implies
Z+
p+1<1>(t) = �A
<kp+1>
�1 (Z�p+1<1>(t))
�1C
<kp+1>
�1 : (93)
194 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Inverse Scattering Problem on the Axis for the Schr�odinger Operator
Use (88)�(90) and the de�nition to get
A
<kp+1>
�1 = d
dk
(A�1(k)(k � kp+1)
2)kp+1
= 2kp+1A
�1
0 (kp+1)
�i
h
1
z
[p+1]+
22 a022(kp+1)
+
a22(kp+1)
z
[p+1]+
11 a011(kp+1)2
�
2kp+1 _a
0
11(kp+1)
z
[p+1]+
22 a022(kp+1)a
0
11(kp+1)
i
(Z+
p+1)
0(0); (94)
and in a similar way,
C
<kp+1>
�1 = 2kp+1C
�1
0 (kp+1)� i
h
1
z
[p+1]+
11 a011(kp+1)
+
a11(kp+1)
z
[p+1]+
22 a022(kp+1)2
�
2kp+1 _a
0
11(kp+1)
z
[p+1]+
11 a011(kp+1)a
0
22(kp+1)
i
(Z+
p+1)
0(0): (95)
Use (92), (94), and (95) to deduce from (93), after some obvious computations,
that Z+
p+1<1>(t) = Z+
p+1(t), and Lemma 11 is proved in Case III too and therefore
it is proved completely.
It remains to notice that the scattering problems constructed above do not
have virtual levels, since R+(0) = R�(0) = �I for them.
Theorem 1 is proved completely.
Correction to Part I. In Remark 2 [15] the functions �(z) and +(0)
must be determined by modi�ed formulas (43), that is by (43) with additional
multiplier
pQ
j=1
(
k�kj
k+kj
)s
1
j in the integrand.
References
[1] V.A. Marchenko, Sturm�Liouville Operators and Applications. Naukova Dumka,
Kiev, 1977. (Russian) (Engl. transl.: Birkh�auser, Basel, 1986.)
[2] L.D. Faddeyev, Inverse Scattering Problem in Quantum Theory. II. � In: Modern
Probl. Math. 3 VINITI, Moscow, 1974, 93�180. (Russian)
[3] B.M. Levitan, Inverse Sturm�Liouville Problems. Nauka, Moscow, 1984 (Russian).
(Engl. transl.: VSP, Zeist, 1987.)
[4] Z.S. Agranovich and V.A. Marchenko, The Inverse Problem of Scattering Theory.
Kharkov State Univ., Kharkov, 1960. (Russian) (Engl. transl.: Gordon & Breach,
New York�London, 1963.)
[5] E.I. Bondarenko and F.S. Rofe-Beketov, Inverse Scattering Problem on the Semi-
Axis for a System with a Triangular Matrix Potential. � Mat. phys., aanal., geom.
10 (2003), 412�424. (Russian)
[15] E.I. Zubkova and F.S. Rofe-Beketov, Inverse Scattering Problem on the axis for the
Schr�odinger Operator with Triangular 2� 2 Matrix Potential. I. Main Theorem. �
J. Math. Phys., Anal., Geom. 3 (2007), 47�60.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 195
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