Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem

The necessary and sufficient conditions for solvability of ISP under consideration are obtained.

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Опубліковано в: :Журнал математической физики, анализа, геометрии
Дата:2007
Автори: Zubkova, E.I., Rofe-Beketov, F.S.
Формат: Стаття
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Опубліковано: Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України 2007
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Цитувати:Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem / E.I. Zubkova, F.S. Rofe-Beketov // Журнал математической физики, анализа, геометрии. — 2007. — Т. 3, № 1. — С. 47-60. — Бібліогр.: 14 назв. — англ.

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Digital Library of Periodicals of National Academy of Sciences of Ukraine
id nasplib_isofts_kiev_ua-123456789-106437
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spelling Zubkova, E.I.
Rofe-Beketov, F.S.
2016-09-28T17:56:15Z
2016-09-28T17:56:15Z
2007
Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem / E.I. Zubkova, F.S. Rofe-Beketov // Журнал математической физики, анализа, геометрии. — 2007. — Т. 3, № 1. — С. 47-60. — Бібліогр.: 14 назв. — англ.
1812-9471
https://nasplib.isofts.kiev.ua/handle/123456789/106437
The necessary and sufficient conditions for solvability of ISP under consideration are obtained.
en
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
Журнал математической физики, анализа, геометрии
Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem
Article
published earlier
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection DSpace DC
title Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem
spellingShingle Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem
Zubkova, E.I.
Rofe-Beketov, F.S.
title_short Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem
title_full Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem
title_fullStr Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem
title_full_unstemmed Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem
title_sort inverse scattering problem on the axis for the schrödinger operator with triangular 2 x 2 matrix potential. i. main theorem
author Zubkova, E.I.
Rofe-Beketov, F.S.
author_facet Zubkova, E.I.
Rofe-Beketov, F.S.
publishDate 2007
language English
container_title Журнал математической физики, анализа, геометрии
publisher Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
format Article
description The necessary and sufficient conditions for solvability of ISP under consideration are obtained.
issn 1812-9471
url https://nasplib.isofts.kiev.ua/handle/123456789/106437
citation_txt Inverse Scattering Problem on the Axis for the Schrödinger Operator with Triangular 2 x 2 Matrix Potential. I. Main Theorem / E.I. Zubkova, F.S. Rofe-Beketov // Журнал математической физики, анализа, геометрии. — 2007. — Т. 3, № 1. — С. 47-60. — Бібліогр.: 14 назв. — англ.
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fulltext Journal of Mathematical Physics, Analysis, Geometry 2007, v. 3, No. 1, pp. 47�60 Inverse Scattering Problem on the Axis for the Schr�odinger Operator with Triangular 2� 2 Matrix Potential. I. Main Theorem 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 necessary and su�cient conditions for solvability of ISP under con- sideration are obtained. Key words: scattering on the axis, inverse problem, triangular matrix potential. Mathematics Subject Classi�cation 2000: 47A40, 81U40. The paper deals with the scattering problem on the axis for the Schr�odinger equation system with the upper triangular matrix potential V (x) = (vrd(x)) 2 1 : v21 � 0 whose principal diagonal is real: Imvrr(x) � 0, r = 1; 2, l[Y ] � �Y 00 + V (x)Y = k2Y; �1 < x <1: (1) The scalar selfadjoint case of ISP for (1) was considered in [1] under the condition 1Z �1 (1 + jxjm)jV (x)jdx <1 (2) with m = 1; in [2] and [3] essentially for m = 2. The problem on the semi-axis for selfadjoint systems with the condition (2) for m = 1 was considered in [4], c E.I. Zubkova and F.S. Rofe-Beketov, 2007 E.I. Zubkova and F.S. Rofe-Beketov for triangular n � n systems in [5]. The scalar not selfadjoint case on the semi- axis with an exponentially small potential at in�nity was considered in [9], and a similar case of the axis in [6]. We also mention the monographs [13, 14]. Consider the case m = 2. Besides (1), we also consider the tilde(�)-equation el[ eZ] � � eZ 00 + eZV (x) = k2 eZ; �1 < x <1: (3) The solutions E�(x; k), eE�(x; k) of equations (1), (3) with asymptotics E�(x; k) � e�ikxI; eE�(x; k) � e�ikxI; x! �1; Im k � 0; (4) where I is a unit matrix, are called the Jost solutions. For them, the representa- tions by B.Ya. Levin are known [7] (see also [1, 3, 4]) E�(x; k) = Ie�ikx� �1R x K�(x; t)e �iktdt; eE�(x; k) = Ie�ikx� �1R x eK�(x; t)e �iktdt; Im k � 0; (5) in terms of transformation operators, with V (x) = �2dK�(x; x)=dx = �2d eK�(x; x)=dx: (6) In addition to the Jost solutions (4), we need the solutions E^ �(x; k) � e�ikxI; eE^ �(x; k) � e�ikxI; x! �1; Im k � 0; k 6= 0; (7) which form fundamental systems together with E�(x; k) (respectively, together with eE�(x; k)). The matrix solutions E^ �(x; k) were constructed and studied in [4]; we form the solutions eE^ �(x; k) in a similar way. However, unlike the Jost solutions, the solutions (7) are not determined unambiguously by their asymp- totics for Im k > 0. Nevertheless, if one of the solutions (7), (e.g., E^ +(x; k)) is �xed, then the associated solution eE^ +(x; k) is determined unambiguously by the additional condition Wf eE^ +(x; k); E ^ +(x; k)g � eE^ +(x; k) d dx E^ +(x; k) � d dx eE^ +(x; k)E ^ +(x; k) = 0; Im k � 0; k 6= 0: (8) For every �0 > 0 solutions (7) can be chosen as analytic in k for jkj > �0, Imk > 0, and we will assume this in what follows (for the scalar case see also [8, 9].) Since with real k 6= 0 the pairs of functions E+(x;�k), E�(x;�k) (respec- tively, eE+(x;�k), eE�(x;�k)) form the fundamental systems of solutions of (1) 48 Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 Inverse Scattering Problem on the Axis for the Schr�odinger Operator (respectively, of (3)), and their Wronski determinants do not depend on x, one has the following relations (see, for example, [10]): E+(x; k) = E�(x;�k)A(k) +E�(x; k)B(k); E�(x; k) = E+(x;�k)C(k) +E+(x; k)D(k);eE+(x; k) = C(k) eE�(x;�k) �D(�k) eE�(x; k);eE�(x; k) = A(k) eE+(x;�k)�B(�k) eE+(x; k); k 2 Rnf0g; (9) with A(k) = 1 2ik Wf eE�(x; k); E+(x; k)g;C(k) = � 1 2ik Wf eE+(x; k); E�(x; k)g; B(k) = � 1 2ik Wf eE�(x;�k); E+(x; k)g;D(k) = 1 2ik Wf eE+(x;�k); E�(x; k)g: (10) The values R+(k) = D(k)C(k)�1; R�(k) = B(k)A(k)�1 (11) are called the right and the left re�ection coe�cients, and T+(k) = C(k)�1; T�(k) = A(k)�1; (12) the right and the left transmission coe�cients, respectively.* Use the following well-known relations for coe�cients A(�k)C(k) = I �B(k)D(k); C(�k)A(k) = I �D(k)B(k); B(�k)C(k) +A(k)D(k) = D(�k)A(k) + C(k)B(k) = 0; (13) which follow from (9), (10), to deduce that (see [1, 10]) (I �R�(�k)R�(k))�1 = A(k)C(�k); (I �R+(�k)R+(k))�1 = C(k)A(�k): (14) The eigenvalues of problem (1) k2 j , j = 1; p, coincide with the set of eigenvalues for the scalar scattering problems with the real diagonal elements of the matrix potential V (x), since they are the roots of the equation detA(k) = a11(k)a22(k) = 0; Imk > 0: Therefore the number of eigenvalues is �nite, and k2 j < 0. Besides that, we assume the absence of a virtual level, i.e., the absence of a bounded on the x-axis nontrivial solution of (1) at k = 0. *These were derived from the �rst two relations in (9). In a similar way, the last two relations in (9) can be used to determine the tilde-coe�cients eR�(k) ø eT�(k) corresponding to (3) which coincide with R�(k) and T�(k) (see Lemma 2 below). Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 49 E.I. Zubkova and F.S. Rofe-Beketov We call the polynomials Z+ j (t) = �ie�ikj tReskjfW +(k)C(k)�1eiktg; Z� j (t) = �ieikj tReskjfW �(k)A(k)�1e�iktg; j = 1; p; t 2 R; (15) with W�(k) = � 1 2ik Wf eE^ �(x; k); E�(x; k)g; (16) respectively, the right and the left normalizing polynomials (cf. [8, 9, 6] in the scalar case). Normalizing polynomials do not depend on a choice of eE^ � in the expression for W�(k) (16) and possess the properties: Lemma 1. a) One has the following inequalities that involve degrees of nor- malizing polynomials: degZ� j (t) � 2X l=1 sign z [j]� ll � 1 � 1; j = 1; p; (17) with the diagonal elements z [j]� ll being nonnegative and independent of t. (The degree of the identically zero polynomial is assumed to be negative.) b) The ranks of normalizing polynomials satisfy the following relations: rgZ� j (t) = rg diagZ� j (t) = rg diagZ� j (0); j = 1; p: (18) c) The matrix elements of normalizing polynomials and those of the matrices C(k), A(k) are related as follows: a11(kj)z [j]� 12 (0) + c12(kj)z [j]� 22 + i _a11(kj)(z [j]� 12 )0(0) = 0; z [j]� 11 a12(kj) + z [j]� 12 (0)a22(kj) + i(z [j]� 12 )0(0) _a22(kj) = 0; (19) z [j]+ 11 c12(kj) + z [j]+ 12 (0)a22(kj)� i(z [j]+ 12 )0(0) _a22(kj) = 0; a11(kj)z [j]+ 12 (0) + a12(kj)z [j]+ 22 � i _a11(kj)(z [j]+ 12 )0(0) = 0; (20) a11(kj)(z [j]� 12 )0(0) = a22(kj)(z [j]� 12 )0(0) = 0; all(kj)z [j]� ll = 0; l = 1; 2; j = 1; p: (21) P r o o f. The claims a) and b) can be proved similarly to [5]. Prove (20), (21) for Z+ j (t). Use (17) to deduce from the de�nition (15) that Z+ j (0) = �i d dk (W+(k)C(k)�1(k � kj) 2)kj ; (Z+ j )0(0) = (W+(k)C(k)�1(k � kj) 2)kj , that is (Z+ j )0(0)C(kj) = (W+(k)C(k)�1C(k)(k � kj) 2)kj = (W+(k)(k � kj) 2)kj = 0; 50 Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 Inverse Scattering Problem on the Axis for the Schr�odinger Operator Z+ j (0)C(kj)� i(Z+ j )0(0) _C(kj) = �i d dk (W+(k)C(k)�1(k � kj) 2)kjC(kj)� i(W+(k)C(k)�1(k � kj) 2)kj _C(kj) = �i d dk (W+(k)C(k)�1C(k)(k � kj) 2)kj = �i d dk (W+(k)(k � kj) 2)kj = 0. Now write down the latter relations separately for the matrix elements to get (20) and (21) for Z+ j (t), j = 1; p. The validity of (19), (21) for Z� j (t) can be proved in a similar way. Lemma 1 is proved. The set of values fR+(k); k 2 R; k2 j < 0; Z+ j (t); j = 1; pg, respectively, fR�(k); k 2 R; k2 j < 0; Z� j (t); j = 1; pg are called the right and the left scattering data (SD) for equation (1). In this context, as in the scalar case [1�3], the right scattering data is determined unambiguously by the left data, and conversely (see (29) and Lemma 3 below). SD of the scalar problem that corresponds to some diagonal entry of the triangular matrix potential coincides with the system of corresponding diagonal entries of the matrix SD. Lemma 2. The associated right scattering data (SD) and the tilde-scattering data fR+(k); k 2 R; k2j < 0; Z+ j (t); j = 1; pg (22) for problems (1) and (3) of the above form coincide. Analogously, the left data for problems (1) and (3) coincide. P r o o f. We present a proof for the case of the right SD. De�ne the re�ection coe�cient for problem (3) byeR+(k) = �A�1(k)B(�k); k 2 R; (23) with A(k) and B(k) determined by (10). The third equality in (13), together with (23), implies R+(k) = D(k)C�1(k) = �A�1(k)B(�k) = eR+(k); k 2 R: (24) The coincidence of the eigenvalues k2 j , Imkj > 0, for problems (1) and (3) follows from the upper-triangular form of the problems and the fact that the diagonal elements of the matrix potential V (x) are real, that is detA(k) = detC(k) = a11(k)a22(k), Imk � 0. By the de�nition of a normalizing polynomial for problem (3) one has eZ+ j (t) = �ie�ikj tReskjfA �1(k)fW+(k)eiktg; (25) with fW+(k) = � 1 2ik Wf eE�(x; k);E^ +(x; k)g: (26) In the upper half-plane, similarly to (9), one has the following representations: E+(x; k) = E�(x; k)W �(k) +E^ �(x; k)A(k); E^ +(x; k) = E�(x; k)W ^(k)�E^ �(x; k)fW+(k); E�(x; k) = E+(x; k)W +(k) +E^ +(x; k)C(k); Im k > 0; (27) Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 51 E.I. Zubkova and F.S. Rofe-Beketov with W�(k) de�ned by (16) and fW+(k) by (26), W^(k) = � 1 2ik Wf eE^ �(x; k); E^ +(x; k)g. Now substitute the initial two relations of (27) into the third one to obtain E�(x; k) = E�(x; k)W �(k)W+(k)+E^ �(x; k)A(k)W+(k)+ E�(x; k)W ^(k)C(k)� E^ �(x; k)fW+(k)C(k). Group the summands to get E�(x; k)(I�W �(k)W+(k)�W^(k)C(k)) = E^ �(x; k)(A(k)W+(k)�fW+(k)C(k)), with Imk > 0. Since the solutions E�(x; k) and E^ �(x; k) form a fundamental system with Im k > 0, the following relations are valid: A(k)W+(k) = fW+(k)C(k), I =W�(k)W+(k) +W^(k)C(k), Im k > 0. So, A(k)�1fW+(k) = W+(k)C(k)�1, that is Z+ j (t) = eZ+ j (t), which was to be proved. Lemma 2 is proved. Remark 1. Similarly to (24), one has for the left re�ection coe�cient R�(k) := B(k)A(k)�1 = �C(k)�1D(�k) =: eR�(k); k 2 R; (28) and for the left normalizing polynomial Z� j (t) = �ieikjtReskjfW �(k)A(k)�1e�iktg = �ieikj tReskjfC(k)�1fW�(k)e�iktg, j = 1; p, with fW�(k) = 1 2ik Wf eE+(x; k); E ^ �(x; k)g. (28) and (24) allow to write down a relationship between the right and the left re�ection coe�cients: R�(k) = �A(�k)R+(�k)A(k)�1 = �C(k)�1R+(�k)C(�k); k 2 R; (29) A relationship between the right and the left normalizing polynomials is set up by the following Lemma 3. One has the relations as follows between the right and the left normalizing polynomials for any � 2 R: Z� j (t) = �Cj(t� �)[Z+ j (�) +Qj ] �1A <kj> �1 ; Z+ j (t) = �Aj(t� �)[Z� j (�) +Qj ] �1C <kj> �1 ; (30) Z� j (t) = �C <kj> �1 [Z+ j (�) +Qj] �1Aj(� � t); Z+ j (t) = �A <kj> �1 [Z� j (�) +Qj] �1Cj(� � t); (31) with Cj(t) = eikjtReskjfC(k)�1e�iktg = C <kj> �1 + (�it)C <kj> �2 , Aj(t) = e�ikj t� ReskjfA(k)�1eiktg = A <kj> �1 + itA <kj> �2 , Qj being the arbitrary upper triangular matrices with the property q [j] ll = 0 if z [j]� ll 6= 0 and q [j] ll 6= 0 if z [j]� ll = 0, l = 1; 2, rgQj = 2� rgZ� j (t). 52 Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 Inverse Scattering Problem on the Axis for the Schr�odinger Operator P r o o f. It is easy to show that at kj , the eigenvalues of problems (1) and (3), one has the following relations:( E+(x; kj)A <kj> �2 = i2E�(x; kj)(Z � j )0(0); E+(x; kj)A <kj> �1 + _E+(x; kj)A <kj> �2 = iE�(x; kj)Z � j (0) + i2 _E�(x; kj)(Z � j )0(0); (32)( E�(x; kj)C <kj> �2 = �i2E+(x; kj)(Z + j )0(0); E�(x; kj)C <kj> �1 + _E�(x; kj)C <kj> �2 = iE+(x; kj)Z + j (0)� i2 _E+(x; kj)(Z + j )0(0); (33)( C <kj> �2 eE+(x; kj) = i2(Z� j )0(0) eE�(x; kj); C <kj> �1 eE+(x; kj) + C <kj> �2 _eE+(x; kj) = iZ� j (0) eE�(x; kj) + i2(Z� j )0(0) _eE�(x; kj); (34)( A <kj> �2 eE�(x; kj) = �i2(Z+ j )0(0) eE+(x; kj); A <kj> �1 eE�(x; kj) +A <kj> �2 _eE�(x; kj) = iZ+ j (0) eE+(x; kj)� i2(Z+ j )0(0) _eE+(x; kj): (35) Prove the �rst relation in (30). The second one of (30) and (31) can be proved in a similar way. Transform the system (32) as follows:( �iE+(x; kj)A 0 j (�t) = i2E�(x; kj)(Z � j )0(t); E+(x; kj)Aj(�t)� i _E+(x; kj)A 0 j (�t) = iE�(x; kj)Z � j (t) + i2 _E�(x; kj)(Z � j )0(t); to be rewritten in a block matrix form:� E+(x; kj) _E+(x; kj) 0 E+(x; kj) � � � Aj(�t) �iA0 j (�t) � = � E�(x; kj) _E�(x; kj) 0 E�(x; kj) � � � iZ� j (t) i2(Z� j )0(t): � : Furthermore, it follows from (33) that� E�(x; kj) _E�(x; kj) 0 E�(x; kj) � � C <kj> �1 C <kj> �2 ! (Z+ j (t) +Qj) �1A <kj> �1 = � E+(x; kj) _E+(x; kj) 0 E+(x; kj) � � � iZ+ j (0) �i2(Z+ j )0(0) � (Z+ j (t) +Qj) �1A <kj> �1 : Applying the relation Z+ j (�)[Z+ j (t) +Qj] �1A <kj> �1 = Aj(� � t); (36) Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 53 E.I. Zubkova and F.S. Rofe-Beketov to be proved below, we deduce that� E�(x; kj) _E�(x; kj) 0 E�(x; kj) � � C <kj> �1 C <kj> �2 ! (Z+ j (t) +Qj) �1A <kj> �1 = � E+(x; kj) _E+(x; kj) 0 E+(x; kj) � � � iAj(�t) �i2(Aj) 0(�t) � = i � E�(x; kj) _E�(x; kj) 0 E�(x; kj) � � � iZ� j (t) i2(Z� j )0(t) � : Compare the left- and the right-hand sides to deduce Z� j (t) = �C <kj> �1 (Z+ j (t) +Qj) �1A <kj> �1 ; (Z� j )0(t) = iC <kj> �2 (Z+ j (t) +Qj) �1A <kj> �1 : Multiply the second relation by (� � t) and add the result to the �rst one to get Z� j (�) = �Cj(� � t)(Z+ j (t) + Qj) �1A <kj> �1 , or, equivalently, Z� j (t) = �Cj(t � �)(Z+ j (�) +Qj) �1A <kj> �1 . Now prove (36). Consider the cases: 1) a11(kj) = a22(kj) = 0, then z [j]� ll 6= 0, hence Qj = 0, and so by a virtue of the second relation in (20): Z+ j (�)Z+ j (t)�1A <kj> �1 = 0@ 1 z [j]+ 12 (�)�z [j]+ 12 (t) z [j]+ 22 0 1 1A � 1 _a11(kj) � d dk ( a12(k)(k�kj) 2 a11(k)a22(k) )kj 0 1 _a22(kj) ! = A <kj> �1 + i(� � t)A <kj> �2 = Aj(� � t): 2) a11(kj) = 0; a22(kj) 6= 0, hence q [j] 11 = 0; q [j] 22 6= 0 and Z+ j (t) = Z+ j (�) = Z+ j ; Aj(� � t) = A�1. Thus (36) is equivalent to Qj(Z + j + Qj) �1A <kj> �1 = 0. So, in our case Qj(Z + j +Qj) �1A <kj> �1 = 0 q [j] 12 0 q [j] 22 ! � 0@ 1 z [j]+ 11 � z [j]+ 12 +q [j] 12 z [j]+ 11 q [j] 22 0 1 q [j] 22 1A � 1 _a11(kj) � a12(kj) _a11(kj)a22(kj) 0 0 ! = 0: 54 Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 Inverse Scattering Problem on the Axis for the Schr�odinger Operator 3) a11(kj) 6= 0; a22(kj) = 0, hence q [j] 11 6= 0; q [j] 22 = 0 and Z+ j (t) = Z+ j (�) = Z+ j ; Aj(� � t) = A <kj> �1 . Thus (36) is equivalent to Qj(Z + j +Qj) �1A <kj> �1 = 0. By a virtue of the second relation in (20) one has Qj(Z + j +Qj) �1A <kj> �1 = q [j] 11 q [j] 12 0 0 ! � 0@ 1 q [j] 11 � z [j]+ 12 +q [j] 12 q [j] 11 z [j]+ 22 0 1 z [j]+ 22 1A � 0 � a12(kj) a11(kj) _a22(kj) 0 1 _a22(kj) ! = 0: Lemma 3 is proved. Lemma 4. For the problem (1), (2) under consideration with m � 1 one has the following decomposition of the Æ-function: Æ(x � t)I = 1 2� 1R �1 E+(x; k)A(k)�1 eE�(t; k) dk + pP j=1 1P l=0 d l ildkl fE+(x; k)(Z + j )(l)(0) eE+(t; k)gkj ; (37) or, equivalently, Æ(x � t)I = 1 2� 1R �1 fE+(x; k) eE+(t;�k) +E+(x; k)R +(k) eE+(t; k)g dk + pP j=1 1P l=0 d l ildkl fE+(x; k)(Z + j )(l)(0) eE+(t; k)gkj ; (38) which is known to be equivalent to the Parceval equation. P r o o f (cf. [11]) elaborates the method of contour integration combined with the passage to weak limit for the resolvent zRz(L) ! E as z ! 1, with E being the identity operator generated by the kernel Æ(x � t) as an integral operator, along with (9), (11). Lemma 3 and (29) indicate that for determining the right scattering data by a given left SD or conversely, it is su�cient to retrieve simultaneously the matrices A(k) and C(k). Lemma 5. Suppose that an upper triangular 2�2 matrix R+(k) is continuous in k 2 R; R+(0) = �I, R+(k) = O(k�1) as k ! �1, and I � R+(�k)R+(k) = O(k2) as k ! 0. Assume that its diagonal elements are such that r+ ll (k) = r+ ll (�k), jr+ ll (k)j � 1� Clk 2 1+k2 , l = 1; 2, and the associated functions zall(z) � zcll(z) = zexp 8<:� 1 2�i +1Z �1 ln(1� jr+ ll (k)j2) k � z dk 9=; ; l = 1; 2; Im z > 0; (39) Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 55 E.I. Zubkova and F.S. Rofe-Beketov are continuous in the closed upper half-plane (see [1], [2]) such that the function ka11(�k)a22(k) � r+11(�k)r + 12(k) + r+12(�k)r + 22(k) satis�es the H�older condition on the real axis, with the in�nite point included *. Then the following Riemann problem is solvable uniquely with respect to c12(k) and a12(�k) (cf. (14)), which are regular in the upper and the lower half-planes: kc12(k) = a11(k) a22(�k) � [�ka12(�k)] + ka22(k)ja11(k)j 2 � � r+11(�k)r + 12(k) + r+12(�k)r + 22(k) ; k 2 R; (40) and it turns out that this solution satis�es the assumption R�(0) = �I; (41) with R�(k) produced as in (29) via the matrices A(k) and C(k), determined from (39), (40), I is an identity matrix. The above solution admits a representation in the form c12(z) = +(z)� +(0) z a11(z); a12(z) = �(�z)� +(0) z a22(z); Im z > 0; (42) with �(z) = 1 2�i 1R �1 ka11(�k)a22(k) � r+11(�k)r + 12(k) + r+12(�k)r + 22(k) dk k�z ; �Imz > 0; +(0) = 1 2�i 1R �1 a11(�k)a22(k) � r+11(�k)r + 12(k) + r+12(�k)r + 22(k) dk: (43) P r o o f. The Riemann problem (40) under the assumptions of the lemma has index � = �1, hence its solution exists and is unique (see [12, Sect. 14.7]). The related formulas get simpler since the coe�cient of equation (40) is already written in a factorized form a11(k) a22(�k) , so we obtain zc12(z) = fa0 + +(z)ga11(z); Im z > 0; �za12(�z) = fa0 + �(z)ga22(�z); Im z < 0; (44) where a0 is a constant determined by the requirement for the right hand sides to be continuous at z = 0, i.e., a0 = � +(0) = � �(0); (45) *The latter assumption for a function f(k) (see [12]) means the existence of constants A and �, 0 < � � 1, such that for any x1 and x2 whose modulus exceeds one, jf(x1)�f(x2)j � Aj 1 x1 � 1 x2 j�. 56 Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 Inverse Scattering Problem on the Axis for the Schr�odinger Operator and the functions �(z) determined by �(z) = 1 2�i 1R �1 ka22(k)ja11(k)j 2fr + 11(�k)r + 12(k)+r + 12(�k)r + 22(k)g a11(k) dk k�z = 1 2�i 1R �1 ka22(k)a11(�k)fr + 11(�k)r + 12(k) + r+12(�k)r + 22(k)g dk k�z ; �Im z > 0; (46) where it is implicit that all(�k) = all(k), l = 1; 2, by a virtue of (39). The second equality in (45) follows from (46) and the assumptions of the lemma. The solution (44) is derived from (40), (46), and the Sokhotski�Plemelj formulas, which all together result in kc12(k) a11(k) � +(k) = � ka12(�k) a22(�k) � �(k) = a0; (47) where the second equality (with a constant as a right-hand side) follows from the Liouville theorem, since the left and the central parts of (47) appear to be analytic continuations of each other to the entire complex plane. Applying (45), one can deduce (42) from (47). Now use (29) to construct a function r�12(k) = � r+11(�k)c12(�k) a11(k) � a22(�k) a11(k) r+12(�k)� r�22(�k)c12(k) a11(k) : The application of the Sokhotskiy�Plemelj formulas for �(z) (46) allows one to deduce from (44) with k 6= 0 on the real axis r�12(k) = � a22(�k) a11(k) r+12(�k)� r�11(k) k fa0 + +(�k)g � r�22(k) k fa0 + +(k)g: It follows that as k ! 0 by (47), (45), one has r�12(0) = 0 since r� ll (0) = �1 due to the properties of a scalar inverse scattering problem, which yields (41). Lemma 5 is proved. Remark 2. Suppose that in Lemma 5, besides the matrix R+(k) that satis�es the assumptions of the lemma, we are given the numbers k2 j < 0 and polynomials Z+ j (t), j = 1; p, t 2 R, satisfying the assumptions a), b), and (21) of Lemma 1. Assume also that zall(z) are given not by (39), but as follows: zall(z) � zcll(z) := ze � 1 2�i 1R �1 ln(1�jr+ ll (k)j2) k�z dk pY j=1 � z � kj z + kj � s l j ; (48) Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 57 E.I. Zubkova and F.S. Rofe-Beketov with sl j = sign z [j]+ ll � 0, l = 1; 2. In this case the Riemann problem (40) is also solvable uniquely under the assumptions (20) so that zc12(z) = 8<:� +(0) pQ j=1 k �j j +a1z+:::+a�z � pQ j=1 (z+kj) �j + +(z) 9=; a11(z) pQ j=1 � z+kj z�kj �s1j ; Im z > 0 �za12(�z) = 8<:� +(0) pQ j=1 k �j j +a1z+:::+a�z � pQ j=1 (z+kj) �j + �(z) 9=; a22(�z) pQ j=1 � z+kj z�kj � s 1 j ; Im z < 0; (49) with a1; : : : ; a� being retrievable unambiguously due to the systems (20), (21), and the given normalizing polynomials, � = pP j=1 �j = pP j=1 � sign z [j]+ 11 + sign z [j]+ 22 � , and the functions �(z) and +(0) determined by (43). The solution we get this way appears to be such that (43) is valid if R�(k) is constructed as in (29), which involves matrices A(k) and C(k) determined by (48), (49), (40). Note that problem (40) in our case has index � = �� 1. The determinant of the system that de�nes a1; : : : ; a� is nonzero in the cases under consideration, as one can see from (21). Theorem 1. If a set of values fR+(k); k 2 R; k2 j < 0; Z+ j (t); j = 1; pg forms the right SD for the scattering problem (1), (2) with m = 2 and an upper triangular 2 � 2 matrix potential (with a real diagonal and without virtual level), the 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). 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. 58 Journal of Mathematical Physics, Analysis, Geometry, 2007, v. 3, No. 1 Inverse Scattering Problem on the Axis for the Schr�odinger Operator 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 the version 4a, namely, if c12(z) is constructed as in (42), (43), and cll(z) � all(z) is constructed as in (39)). 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, the conditions 5) and 6) of the Theorem 1 become inapplicable, and the conditions 4 and 4a become the same. 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