Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential
The characteristic properties of the scattering data for the Schr¨odinger operator on the axis with a triangular 2 × 2 matrix potential are obtained. A difference between the necessary and sufficient conditions for solvability of ISPunder consideration, contained in the previous works of the authors...
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
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nasplib_isofts_kiev_ua-123456789-1065452025-02-23T17:33:00Z Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential Zubkova, E.I. Rofe-Beketov, F.S. The characteristic properties of the scattering data for the Schr¨odinger operator on the axis with a triangular 2 × 2 matrix potential are obtained. A difference between the necessary and sufficient conditions for solvability of ISPunder consideration, contained in the previous works of the authors, is eliminated. The authors are deeply grateful to D.G. Shepel’sky who attracted their attention to the question about possibility to omit condition H in the main theorem of [4, 5]. 2009 Article Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential / E.I. Zubkova, F.S. Rofe-Beketov // Журнал математической физики, анализа, геометрии. — 2009. — Т. 5, № 3. — С. 296-309. — Бібліогр.: 10 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106545 en Журнал математической физики, анализа, геометрии application/pdf Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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The characteristic properties of the scattering data for the Schr¨odinger operator on the axis with a triangular 2 × 2 matrix potential are obtained. A difference between the necessary and sufficient conditions for solvability of ISPunder consideration, contained in the previous works of the authors, is eliminated. |
| format |
Article |
| author |
Zubkova, E.I. Rofe-Beketov, F.S. |
| spellingShingle |
Zubkova, E.I. Rofe-Beketov, F.S. Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential Журнал математической физики, анализа, геометрии |
| author_facet |
Zubkova, E.I. Rofe-Beketov, F.S. |
| author_sort |
Zubkova, E.I. |
| title |
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential |
| title_short |
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential |
| title_full |
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential |
| title_fullStr |
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential |
| title_full_unstemmed |
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential |
| title_sort |
necessary and sufficient conditions in inverse scattering problem on the axis for the triangular 2 x 2 matrix potential |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| publishDate |
2009 |
| url |
https://nasplib.isofts.kiev.ua/handle/123456789/106545 |
| citation_txt |
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis for the Triangular 2 x 2 Matrix Potential / E.I. Zubkova, F.S. Rofe-Beketov // Журнал математической физики, анализа, геометрии. — 2009. — Т. 5, № 3. — С. 296-309. — Бібліогр.: 10 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT zubkovaei necessaryandsufficientconditionsininversescatteringproblemontheaxisforthetriangular2x2matrixpotential AT rofebeketovfs necessaryandsufficientconditionsininversescatteringproblemontheaxisforthetriangular2x2matrixpotential |
| first_indexed |
2025-11-24T02:16:01Z |
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2025-11-24T02:16:01Z |
| _version_ |
1849636223710134272 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2009, v. 5, No. 3, pp. 296–309
Necessary and Sufficient Conditions in Inverse Scattering
Problem on the Axis for the Triangular 2 × 2 Matrix
Potential
E.I. Zubkova
Ukrainian State Academy of Railway Transport
7 Feyerbakh Sq., Kharkiv, 61050, Ukraine
E-mail:zubkova elena@list.ru
F.S. Rofe-Beketov
Mathematical Division, B. Verkin Institute for Low Temperature Physics and Engineering
National Academy of Sciences of Ukraine
47 Lenin Ave., Kharkiv 61103, Ukraine
E-mail:rofebeketov@ilt.kharkov.ua
Received October 1, 2008
The characteristic properties of the scattering data for the Schrödinger
operator on the axis with a triangular 2× 2 matrix potential are obtained.
A difference between the necessary and sufficient conditions for solvability
of ISP under consideration, contained in the previous works of the authors,
is eliminated.
Key words: scattering on the axis, inverse problem, triangular matrix
potential.
Mathematics Subject Classification 2000: 47A40, 81U40.
To the blessed memory of Alexander Ya. Povzner (27.06.1915–22.04.2008)
The monograph by V.A. Marchenko [1, Chapter 3] contains a complete
solution of the inverse scattering problem (ISP) on the axis for the Schrödinger
equation with a real scalar potential having the first moment. A solution of ISP
on the axis for the potential having the second moment is considered in [2], [3,
Ch. VI]. In the works of the authors [4, 5] for ISP on the axis for the Schrödinger
system of equations with an upper triangular 2×2 matrix potential V (x), having
the second moment and not having a virtual level (that is, when there was no
bounded solution on the axis for k = 0), there were obtained some necessary and
c© E.I. Zubkova and F.S. Rofe-Beketov, 2009
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis...
close, but somewhat different, sufficient conditions on the scattering data (SD)
for the problem
−Y ′′ + V (x)Y = k2Y, −∞ < x <∞. (1)
In the present paper the necessary and sufficient conditions on SD to the
problem above are given in four versions (see Theorems 1 and 2 below, each
in two versions), where the definitions and notations from [4, 5] are used. In
particular, R±(k), k ∈ R; k2
j < 0, and Z±
j (t), j = 1, p, stand respectively for
right and left upper triangular matrical reflection coefficients, negative eigen-
values of the problem, and corresponding right and left upper triangular nor-
malizing polynomials determined by (15) in [4]. Normalizing polynomials for
a scalar nonselfadjoint scattering problem on the semiaxis were introduced by
V.E. Lyantse [6]. A more exhaustive list of references is available in [1]–[7].
In what follows the matrices are denoted by upper-case letters and their elements
by the corresponding lower-case ones with two subscripts. The subscript ‘0’ indi-
cates that the corresponding values were built with a discrete spectrum not being
taken into account.
Theorem 1. Consider the problem (1) with an upper triangular 2× 2 matrix
potential having the second moment. The potential is assumed to have a real
diagonal and to be such that the problem (1) has no virtual level. The items 1)–
7) listed below provide necessary and sufficient conditions for a set of the values{
R+(k), k ∈ R; k2
j < 0, Z+
j (t), j = 1, . . . , p <∞
}
(2)
to be right SD for (1). Here R+(k) and Z+
j (t) (j = 1, p) are upper triangular 2×2
matrix functions. This theorem is valid in two versions: either under condition
4 or 4a.
1) R+(k) is continuous in k ∈ R; r+
ll (k) = r+
ll (−k),
∣∣r+
ll (k)
∣∣ ≤ 1 − Clk
2
1+k2 , with
Cl > 0, l = 1, 2, R+(0) = −I; I − R+(−k)R+(k) = O(k2) as k → 0 and
R+(k) = O(k−1) as k → ±∞ (here the last two conditions can be enhanced
as the necessary ones up to the requirements of continuity for the function
{I − R+(−k)R+(k)}k−2 for k ∈ R and the estimate R+(k) = o(k−1) for
k → ±∞).
2) The function
F+
R (x) =
1
2π
∞∫
−∞
R+(k)eikxdk
is absolutely continuous, and for every a > −∞ one has
+∞∫
a
(1 + x2)
∣∣ d
dxF
+
R (x)
∣∣ dx <∞.
Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3 297
E.I. Zubkova and F.S. Rofe-Beketov
3) The functions zc0ll(z) ≡ za0
ll(z), l = 1, 2, given by
zc0ll(z) ≡ za0
ll(z) := ze
− 1
2πi
∞∫
−∞
ln(1−|r+
ll
(k)|2)
k−z
dk
, Im z > 0, (3)
are continuously differentiable in the closed upper half-plane. Hence they
are well defined on the real axis by continuity, and one has lim
z→0
(za0
ll(z)) �= 0
due to the absence of a virtual level.
4) The function
F−
R (x) ≡ − 1
2π
∞∫
−∞
C(k)−1R+(−k)C(−k)e−ikxdk (4)
is absolutely continuous, and for every a < +∞ one has
a∫
−∞
(
1 + x2
) ∣∣ d
dxF
−
R (x)
∣∣ dx <∞. (5)
Here the matrix C(k) is defined as follows. For l = 1, 2 its elements cll(k)
are given by
zcll(z) ≡ zall(z) := zc0ll(z)
p∏
j=1
(
z−kj
z+kj
)sl
j
, Im z > 0, (6)
where Im kj > 0, slj = sign z[j]+
ll ≥ 0. Furthermore, c21(k) ≡ 0, c12(k) ≡
c12(k + i0) with
zc12(z) =
−ψ+(0)
p∏
j=1
k
κj
j +a1z+...+aκzκ
p∏
j=1
(z+kj)
κj
+ ψ+(z)
a0
11(z), Im z > 0. (7)
The constants a1, . . . , aκ are uniquely derivable� by the given polynomials
Zj(t) and by all(z) (6), κ =
p∑
j=1
κj =
p∑
j=1
(sign z
[j]+
11 + sign z[j]+
22 ). Besides,
zc12(z) is bounded and continuous in the closed upper half-plane
ψ±(z) =
1
2πi
∞∫
−∞
h(k)
k − z
p∏
j=1
(
k − kj
k + kj
)s1j
dk, ±Im z > 0, (8)
From the equation system (20) in [4] in which one has to substitute c12(kj) (7) (or (49) in
[4]), a12(kj) ((49) in [4]), and all(kj), ȧll(kj) (6) (cf. Remark 2 in [4] corrected in [5]).
298 Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis...
h(k) = ka11(−k)a22(k){r+
11(−k)r+
12(k) + r+
12(−k)r+
22(k)}, h(0) = 0, (9)
ψ+(0) =
1
2πi
∞∫
−∞
k−1h(k)
p∏
j=1
(
k − kj
k + kj
)s1j
dk. (10)
4a) (5) holds when in (4) one substitutes C(k) by the matrix C0(k), where
c0ll(z) ≡ a0
ll(z), l = 1, 2, are given by (3), and c012(z) is determined as
follows:
zc012(z) = [ψ+
0 (z)− ψ+
0 (0)]a
0
11(z), Im z > 0, (11)
ψ+
0 (z) =
1
2πi
∞∫
−∞
h0(k)
k − z
dk, (12)
h0(k) = ka0
11(−k)a0
22(k){r+
11(−k)r+
12(k) + r+
12(−k)r+
22(k)}, h0(0) = 0,
(13)
ψ+
0 (0) =
1
2πi
∞∫
−∞
k−1h0(k)dk. (14)
5) degZ+
j (t) ≤
2∑
l=1
sign z[j]+
ll − 1, j = 1, p, with z
[j]+
ll being nonnegative and
constant.
6) rg Z+
j (t) = rg diag Z+
j (t) = rg diag Z+
j (0), j = 1, p.
7) The function h0(k) given by (13) satisfies the Hölder condition on the
real axis, that is, there exist constants α and µ, 0 < µ ≤ 1, such that
|h0(k1) − h0(k2)| ≤ α|k1 − k2|µ for all −∞ < k1 < k2 < ∞. (Notice
that within the ‘only if ’ part this condition can be enhanced up to µ = 1.)
Moreover, h0(k) = o( 1
k ) as k → ±∞ (the last estimate is a consequence of
the conditions 1 and 3 of this theorem). Obviously, one gets an equivalent
condition replacing in the present condition 7 h0(k) (13) with h(k) (9).
Remark 1. The conditions of the theorem related only to the diagonal matrix
elements are direct consequences of [1, Ch. 3], [2], [3, Ch. VI].
Remark 2. Condition 7 of Theorem 1 corresponds to the condition H of the
main theorem in [4, 5] taken in somewhat reduced form containing no Hölder con-
dition for h(k) (13) in the neighborhood of infinity, that is, without the condition
|h(k1)− h(k2)| ≤ α| 1
k1
− 1
k2
|µ for |k1|, |k2| ≥ 1, α = const > 0, 0 < µ ≤ 1. How-
ever, right at the point ∞ the Hölder condition |h(k)− h(∞)| ≤ α 1
|k|µ holds with
Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3 299
E.I. Zubkova and F.S. Rofe-Beketov
h(∞) = 0 and µ = 1 due to our condition 7. (Notice that the notations h0(k)
and h(k) for the expressions (13) and (9) are not found in [4, 5]. Furthermore, to
avoid a confusion, we call Theorem 1 from [4, 5] the ‘main theorem’ owing to the
fact that the term ‘main theorem’ is in the title of [4], while ‘Theorem 1’ refers
to the present work.)
P r o o f of Theorem 1. Prove that condition 7 is necessary. Recall that the
condition H in [4], [5] was among sufficient conditions of the main theorem and
there was no claim that this condition was necessary.
Now notice that by condition 3 of Theorem 1, the functions ka0
ll(k), l = 1, 2,
defined by (3), and hence also kall(k) (6), have continuous derivatives on the real
axis. Demonstrate that
a0
ll(k) = 1 +O(1/k), da0
ll(k)/dk = o(1/k), k → ±∞; (15)
all(k) = 1 +O(1/k), dall(k)/dk = o(1/k), k → ±∞. (16)
These facts are easily deducible as conditions 1 – 5 of Theorem 1, applied to the
diagonal elements of the matrix values in (2), are necessary and sufficient for the
above elements to be SD for problem (1) corresponding to the diagonal part of
the potential V (x). (By Theorem 3.5.1 from [1] under assumption that there is
a finite second moment for the potential.) The above allows one to establish (15),
(16) using the properties of a direct scattering problem, in particular, by Lemma
6 from [5]�. In the lemma the matrix functions A(k), B(k), defined by formulas
(10) from [4] as matrix Wronskians, divided by 2ik, for several Jost solutions of
the problem (1), are represented in the form
A(k) = I − 1
2ik
{
∞∫
−∞
V (x)dx +
0∫
−∞
A1(t)e−iktdt
}
= I +O( 1
k ),
B(k) = 1
2ik
∞∫
−∞
B1(t)e−iktdt = o( 1
k ), k → ±∞,
so that all(k) (6) appear to be the diagonal elements of A(k). These expressions
provide differentiability for kA(k) and kB(k) as well as for their asymptotics and
the asymptotics of their derivatives, since A1(t), B1(t) are summable matrix func-
tions on the left semiaxis (correspondingly, on the axis) having the first moment,
which allows one to differentiate in k under the integral sign in the right-hand
sides of the above relations. For this, one has to use also formula (53) from [5]:
lim
k→0
kA(k) = C1, where detC1 �= 0 due to the absence of a virtual level.
Notice that Lemma 6 from [5] is an analog of Lemma 3.5.1 from [1] with regard to the
considered case of matrix potential, which has the second moment on the axis.
300 Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis...
Thus we have proved not only asymptotics (15), (16), but also, using the
definition R+(k) = −A(k)−1B(−k) (see (24) from [4]), established the following
lemma.
Lemma 1. Suppose a matrix potential V (x) has the second moment on the
axis. The matrix reflection coefficient R+(k) for the problem (1), as in Theo-
rem 1, is a bounded function of k on the axis with a continuous derivative and
such that d
dkR
+(k) = o( 1
k ) as k → ±∞.
Use conditions 1 and 3 of Theorem 1, the estimates (15), (16), and Lemma 1
to deduce from (13) that h0(k) = o( 1
k ) as k → ±∞, h0(k) = O(k) as k → 0 and
that there exists a bounded on the axis derivative of h0(k): |dh0(k)/dk| < const,
−∞ < k <∞ providing the Hölder condition for h0(k) (even with µ = 1). Thus
it is proved that condition 7 of Theorem 1 is necessary for h0(k), hence also for
h(k).
As for the claim that conditions 1–6 of Theorem 1 are necessary, it was es-
tablished in [4], [5]. However, when proving the necessity of condition 4, we used
Remark 2 from [4] corrected in [5], providing the expression (7)–(10) for zc12(z) as
a bounded solution (together with −za12(−z)) of the Riemann–Hilbert problem
with the factorized coefficient a11(k)
a22(−k) for the half-plane
kc12(k)
a11(k)
=
−ka12(−k)
a22(−k) + h(k), −∞ < k <∞. (17)
Here h(k) satisfies not only condition 7, but also the Hölder condition in the
neighborhood of infinity, as required in [8, Chapter II], [9, Chapter 2]. Now we
abandon the last restriction adhering h(k) only to condition 7. It is easy to see
that in this case Remark 2 from [4] corrected in [5] is also valid. In particular,
(7)–(10) of this paper hold for zc12(z) as one can see from the Sokhotski–Plemelj
formulas. Furthermore, a bounded in both half-planes Im z > 0 and Im z < 0
solution of the problem (17), given by formulas (7)–(10) and Remark 2 of [4]
corrected in [5], is unique, because a bounded solution of the homogeneous equa-
tion corresponding to (17) is identically zero. In fact, it turns both sides of this
homogeneous equation into a single constant by the Liouville theorem. On the
other hand, with k → 0 both sides of this homogeneous equation vanish due to
condition 3. So, a bounded solution of the homogeneous equation corresponding
to (17) is trivial.
It is established that conditions 1–7 of Theorem 1, except condition 4a, are
necessary.
Now notice that condition 7 of Theorem 1, as well as the condition H in the
main theorem of [4, 5], is used only in constructing c12(z) in the formulation of
condition 4 and in constructing c012(z) in the formulation of condition 4a. In both
Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3 301
E.I. Zubkova and F.S. Rofe-Beketov
theorems these constructions use the same argument based on solving either the
Riemann–Hilbert problem (17) or a similar problem with all functions from (17)
but with subscript ‘0’. Thereby conditions 1–7 of Theorem 1 (with condition 4a)
appear to be sufficient in parallel with the sufficiency of conditions 1–6 and H of
the main theorem (with condition 4 used in the form of 4a) proved in [4, 5]. Thus
we have
Claim 1. The ‘only if ’ part of Theorem 1 with condition 4 and its ‘if ’ part
with condition 4a are proved.
Lemma 2. Claim 1 implies that condition 4a is necessary.�
First, we sketch a proof of Lemma 2 and then we will be able to claim that
Theorem 1 is proved completely in both versions, that is, either with condition
4 or with condition 4a. If one proves that conditions 1–7 (with condition 4
used) imply conditions 1–7 (with condition 4a used, the sufficiency of which has
been shown earlier), one also proves that conditions 1–7 (with condition 4) are
sufficient, while their necessity has already been established.
So, let (2) be an SD for the problem considered, hence it satisfies condition 4
of Theorem 1. By omitting an eigenvalue (either simple or multiple), prove that
the values {
R+(k), k ∈ R; k2
j < 0, Z+
j (t), j = 1, . . . , p− 1 <∞
}
(18)
also satisfy condition 4 of Theorem 1, along with conditions 1–3 and 5–7, which
are obviously valid. This procedure is to be repeated p times to prove that
condition 4a is necessary for SD (2).
Begin with considering the diagonal elements of the values (18). To simplify
our notation, we omit the indices ll of the diagonal elements. Prove that the
functions of the form f
[p−1]−
R (x) = 1
2π
∞∫
−∞
r−p (k)
(
k−kp
k+kp
)2
e−ikxdk = f
[p]−
R (x) −
4µ2
p
x∫
−∞
f
[p]−
R (t)(x−t)e−µp(x−t)dt+4µp
x∫
−∞
f
[p]−
R (t)e−µp(x−t)dt, with µj ≡ −ikj > 0,
r−j (k) ≡ − r+(−k)aj(−k)
aj(k)
, f [j]−
R (x) ≡ 1
2π
∞∫
−∞
r−j (k)e
−ikxdk, j = 0, . . . , p, derived from
the diagonal elements (2), satisfy condition 4 of Theorem 1. Obviously, f [p−1]−
R (x)
is absolutely continuous.
Prove that d
dxf
[p−1]−
R (x) satisfies (5). After simple computations, which in-
clude integration by parts and changing integration order in some multiple inte-
This fact was mentioned without proof in the formulation of the main theorem in [4] and
[5], with a footnote on a possible method of proving it in [5].
302 Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis...
grals, we get
a∫
−∞
(1 + x2)
∣∣∣ ddxf [p−1]−
R (x)
∣∣∣ dx
≤ 9
a∫
−∞
(1 + x2)
∣∣∣ ddxf [p]−
R (x)
∣∣∣ dx+ 24
µp
a∫
−∞
t
∣∣∣ ddtf [p]−
R (t)
∣∣∣ dt
+ 32
µ2
p
a∫
−∞
∣∣∣ ddtf [p]−
R (t)
∣∣∣ dt − 4
(
µp + µpa
2 + 2a+ 2
µp
)
e−µpa
a∫
−∞
t
∣∣∣ ddtf [p]−
R (t)
∣∣∣ eµptdt
− 4
µ2
p
(µ3
pa(1 + a2) + 4(aµp + 1)2 + 2µ2
p + 4)e
−µpa
a∫
−∞
∣∣∣ ddtf [p]−
R (t)
∣∣∣ eµptdt.
On the other hand, since condition 4 of Theorem 1 holds for f [p]−
R (x), the inequal-
ity of the form (5) holds for d
dxf
[p−1]−
R (x), hence f [p−1]−
R (x) satisfies condition 4
of Theorem 1.
Now consider a nondiagonal element of the values (18) and prove that this ele-
ment satisfies condition 4 of Theorem 1. Here three cases are possible, depending
on the specific form of the matrix polynomial to be omitted.
Case I. Let s1p = sign z[p]+
11 = 1, s2p = sign z[p]+
22 = 0, that is Z+
p (t) =(
z
[p]+
11 z
[p]+
12 (t)
0 0
)
. Then cp11(k) = cp−1
11 (k)k−kp
k+kp
, cp22(k) = cp−1
22 (k), cp12(k) =
cp−1
12 (k)+Pκ−1(k)
Qκ(k) c
p−1
11 (k), where κ =
p∑
j=1
κj =
p∑
j=1
(
sign z[j]+
11 + sign z[j]+
22
)
, Pκ−1(k)
= ψ+(0)
p−1∏
j=1
k
κj
j +a
p
1−a1kp+k(a
p
2−a1−a2kp)+· · ·+kκ−2(apκ−1−aκ−2−aκ−1kp)+
kκ−1(apκ − aκ−1), degPκ−1(k) ≤ κ− 1; Qκ(k) = (k + kp)
p−1∏
j=1
(k + kj)s
2
j (k − kj)s
1
j ,
degQκ(k) = κ. (Here κ = κ(p).)
Thus we have
r
[p−1]−
12 (k) =
k − kp
k + kp
r
[p]−
12 (k)− r
[p]−
11 (k)
Pκ−1(−k)
Qκ(−k)
(
k − kp
k + kp
)2
+ r
[p]−
22 (k)
Pκ−1(k)
Qκ(k)
.
Therefore, using the specific form of the Fourier transform of a ratio of polyno-
mials (see Sect. 2, Ch. I, Ex. 5 [10]), one can deduce that the function
f
[p−1]−
12 (x) ≡ 1
2π
∞∫
−∞
r
[p−1]−
12 (k)e−ikxdk
= f
[p]−
12 (x) + 2µpe−µpx
x∫
−∞
f
[p]−
12 (t)eµptdt − α̃pe
−µpx
x∫
−∞
f
[p]−
11 (t)(x− t)eµptdt
Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3 303
E.I. Zubkova and F.S. Rofe-Beketov
−
p∑
j=1
e−µjx
x∫
−∞
{
s1jαjf
[p]−
11 (t)− s2jγjf
[p]−
22 (t)
}
eµj tdt
−
p−1∑
j=1
eµjx
+∞∫
x
{
s2jβjf
[p]−
11 (t)− s1jδjf
[p]−
22 (t)
}
e−µj tdt
is absolutely continuous. Here αj , βj , γj, δj are constants with α̃p �= 0, γp �= 0.
Assume βp = δp = 0 to deduce that for all a <∞
a∫
−∞
(1 + x2)
∣∣∣ ddxf [p−1]−
12 (x)
∣∣∣ dx
≤ 3
a∫
−∞
(1 + x2)
∣∣∣ ddxf [p]−
12 (x)
∣∣∣ dx+ 4
µp
a∫
−∞
t
∣∣∣ ddtf [p]−
12 (t)
∣∣∣ dt
+ 4
µ2
p
a∫
−∞
∣∣∣ ddtf [p]−
12 (t)
∣∣∣ dt+ a∫
−∞
(1 + t2)
{(
p∑
j=1
s1j |αj |+s2j |βj|
µj
+ |α̃p|
µ2
p
)∣∣∣ ddtf [p]−
11 (t)
∣∣∣
+
p∑
j=1
s2j |γj |+s1j |δj |
µj
∣∣∣ ddtf [p]−
22 (t)
∣∣∣} dt+ 2 a∫
−∞
t
{(
p∑
j=1
s1j |αj |−s2j |βj|
µ2
j
+ 2|α̃p|
µ3
p
)∣∣∣ ddtf [p]−
11 (t)
∣∣∣
+
p∑
j=1
s2j |γj |−s1j |δj |
µ2
j
∣∣∣ ddtf [p]−
22 (t)
∣∣∣} dt+ 2 a∫
−∞
{(
p∑
j=1
s1j |αj |+s2j |βj |
µ3
j
+ 3|α̃p|
µ4
p
)∣∣∣ ddtf [p]−
11 (t)
∣∣∣
+
p∑
j=1
s2j |γj |+s1j |δj |
µ3
j
∣∣∣ ddtf [p]−
22 (t)
∣∣∣} dt+Mp(a) <∞,
where Mp(a) ≡ −2e−µpa[1 + a2 + 2a
µp
+ 2
µ2
p
]
a∫
−∞
| ddtf
[p]−
12 (t)|eµptdt −
p∑
j=1
e−µja
µj
[1 +
a2 + 2a
µp
+ 2
µ2
p
]
a∫
−∞
(s1j |αj || ddtf
[p]−
11 (t)| + s2j |γj || ddtf
[p]−
22 (t)|)eµj tdt +
p−1∑
j=1
eµja
µj
[1 + a2 −
2a
µp
+ 2
µ2
p
]
+∞∫
a
(s2j |βj | · | ddtf
[p]−
11 (t)|+ s1j |δj || ddtf
[p]−
22 (t)|)e−µj tdt− |α̃p|e−µpa
µp
[(1 + a2)a−
3a2+1
µp
+ 6a
µ2
p
+ 6
µ3
p
] ·
a∫
−∞
| ddtf
[p]−
11 (t)|eµptdt + |α̃p|e−µpa
µp
[1 + a2 − 2a
µp
+ 2
µ2
p
]
a∫
−∞
t| ddtf
[p]−
11 (t)|eµptdt <∞.
Case II. Now let s1p = 0, s2p = 1, that is Z+
p (t) =
(
0 z
[p]+
12 (t)
0 z
[p]+
22
)
. Then
cp11(k) = cp−1
11 (k), cp22(k) = cp−1
22 (k)k−kp
k+kp
, cp12(k) = cp−1
12 (k)k−kp
k+kp
+ Pκ−1(k)
Qκ(k) c
p−1
11 (k),
where Pκ−1(k) = ψ+(0)
p−1∏
j=1
k
κj
j + ap1+ a1kp + k(a
p
2 − a1+ a2kp) + · · ·+ kκ−1(apκ −
aκ−1)+
2kpψ−(−kp)
k [
p−1∏
j=1
(k+kj)κj −
p−1∏
j=1
k
κj
j ], degPκ−1(k) ≤ κ−1; Qκ(k) =
p∏
j=1
(k+
kj)s
2
j (k − kj)s
1
j , degQκ(k) = κ.
304 Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis...
This implies
r
[p−1]−
12 (k) =
k − kp
k + kp
r
[p]−
12 (k)− r
[p]−
11 (k)
Pκ−1(−k)
Qκ(−k)
k − kp
k + kp
+ r
[p]−
22 (k)
Pκ−1(k)
Qκ(k)
k − kp
k + kp
.
Using again the quoted above Example 5 to Ch. I, Section 2 of [10], one can
deduce that the function
f
[p−1]−
12 (x) ≡ 1
2π
∞∫
−∞
r
[p−1]−
12 (k)e−ikxdk = f
[p]−
12 (x) + 2µpe−µpx
x∫
−∞
f
[p]−
12 (t)eµptdt
−
p∑
j=1
e−µjx
x∫
−∞
{
s1jαjf
[p]−
11 (t)− s2jγjf
[p]−
22 (t)
}
eµjtdt
−
p−1∑
j=1
eµjx
+∞∫
x
{
s2jβjf
[p]−
11 (t)− s1jδjf
[p]−
22 (t)
}
e−µjtdt
+γ̃pe−µpx
x∫
−∞
f
[p]−
22 (t)(x− t)eµptdt
is absolutely continuous. Here αj, βj , γj , δj are constants with αp �= 0, γ̃p �= 0.
Again set βp = δp = 0 to get for all a <∞
a∫
−∞
(1+x2)
∣∣∣∣ ddxf [p−1]−
12 (x)
∣∣∣∣ dx ≤ 3
a∫
−∞
(1+x2)
∣∣∣∣ ddxf [p]−
12 (x)
∣∣∣∣ dx+ 4
µp
a∫
−∞
t
∣∣∣∣ ddtf [p]−
12 (t)
∣∣∣∣ dt
+
4
µ2
p
a∫
−∞
∣∣∣∣ ddtf [p]−
12 (t)
∣∣∣∣ dt +
a∫
−∞
(1 + t2)
p∑
j=1
s1j |αj |+ s2j |βj |
µj
∣∣∣∣ ddtf [p]−
11 (t)
∣∣∣∣
+
p∑
j=1
s2j |γj |+ s1j |δj |
µj
+
|γ̃p|
µ2
p
∣∣∣∣ ddtf [p]−
22 (t)
∣∣∣∣
dt
+2
a∫
−∞
t
p∑
j=1
s1j |αj | − s2j |βj |
µ2
j
∣∣∣∣ ddtf [p]−
11 (t)
∣∣∣∣
+
p∑
j=1
s2j |γj | − s1j |δj |
µ2
j
− 2|γ̃p|
µ3
p
∣∣∣∣ ddtf [p]−
22 (t)
∣∣∣∣
dt
+2
a∫
−∞
p∑
j=1
s1j |αj |+ s2j |βj |
µ3
j
∣∣∣∣ ddtf [p]−
11 (t)
∣∣∣∣
p∑
j=1
s2j |γj |+ s1j |δj |
µ3
j
+
3|γ̃p|
µ4
p
∣∣∣∣ ddtf [p]−
22 (t)
∣∣∣∣
dt+Mp(a) <∞,
Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3 305
E.I. Zubkova and F.S. Rofe-Beketov
where Mp(a) ≡ −2e−µpa[1 + a2 + 2a
µp
+ 2
µ2
p
]
a∫
−∞
| ddtf
[p]−
12 (t)|eµptdt −
p∑
j=1
e−µja
µj
[1 +
a2 + 2a
µp
+ 2
µ2
p
]
a∫
−∞
(s1j |αj || ddtf
[p]−
11 (t)| + s2j |γj || ddtf
[p]−
22 (t)|)eµj tdt +
p−1∑
j=1
eµja
µj
[1 + a2 −
2a
µp
+ 2
µ2
p
]
+∞∫
a
(s2j |βj | · | ddtf
[p]−
11 (t)|+ s1j |δj || ddtf
[p]−
22 (t)|)e−µj tdt− |γ̃p|e−µpa
µp
[(1 + a2)a−
3a2+1
µp
+ 6a
µ2
p
+ 6
µ3
p
] ·
a∫
−∞
| ddtf
[p]−
22 (t)|eµptdt + |γ̃p|e−µpa
µp
[1 + a2 − 2a
µp
+ 2
µ2
p
]
a∫
−∞
t| ddtf
[p]−
22 (t)|eµptdt <∞.
Case III. Finally, let s1p = 1, s2p = 1, that is Z+
p (t) =
(
z
[p]+
11 z
[p]+
12 (t)
0 z
[p]+
22
)
.
Then cp11(k) = cp−1
11 (k)k−kp
k+kp
, cp22(k) = cp−1
22 (k)k−kp
k+kp
, cp12(k) = cp−1
12 (k)k−kp
k+kp
+
Pκ−1(k)
Qκ(k) c
p−1
11 (k), where κ =
p∑
j=1
κj =
p∑
j=1
(
sign z[j]+
11 + sign z[j]+
22
)
, Pκ−1(k) = ap1 +
a1k
2
p+k(ψ+(0)
p−1∏
j=1
k
κj
j +ap2+a2k
2
p)+ · · ·+kκ−1(apκ−aκ−2)+2kpψ−(−kp)
p−1∏
j=1
(k+
kj)κj + 2k2
pψ
−(−kp)
k [
p−1∏
j=1
(k + kj)κj −
p−1∏
j=1
k
κj
j ], degPκ−1(k) ≤ κ − 1; Qκ(k) = (k +
kp)2
p−1∏
j=1
(k + kj)s
2
j (k − kj)s
1
j , degQκ(k) = κ.
Thus we obtain
r
[p−1]−
12 (k) = r
[p]−
12 (k)− 4kp
k+kp
r
[p]−
12 (k) + 4k2
p
(k+kp)2
r
[p]−
12 (k)
r
[p]−
11 (k) Rκ(k)
(k+kp)3
p−1∏
j=1
(k−kj)
s2
j (k+kj)
s1
j
+ r
[p]−
22 (k) Tκ(k)
(k+kp)3
p−1∏
j=1
(k+kj)
s2
j (k−kj)
s1
j
,
where Rκ(k) ≡ (−1)κPκ−1(−k)(k−kp), degRκ(k) ≤ κ; Tκ(k) ≡ Pκ−1(k)(k−kp),
deg Tκ(k) ≤ κ.
Use again Example 5 to Chapter I, Section 2 of [10] to conclude that the
function
f
[p−1]−
12 (x) ≡ 1
2π
∞∫
−∞
r
[p−1]−
12 (k)e−ikxdk = f
[p]−
12 (x) + 4µpe−µpx
x∫
−∞
f
[p]−
12 (t)eµptdt
−4µ2
pe
−µpx
x∫
−∞
f
[p]−
12 (t)(x− t)eµptdt
−
p∑
j=1
e−µjx
x∫
−∞
{s1jαjf [p]−
11 (t)− s2jγjf
[p]−
22 (t)}eµj tdt
306 Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3
Necessary and Sufficient Conditions in Inverse Scattering Problem on the Axis...
−
p−1∑
j=1
eµjx
+∞∫
x
{s2jβjf [p]−
11 (t)− s1jδjf
[p]−
22 (t)}e−µj tdt
−e−µpx
x∫
−∞
(α[1]
p f
[p]−
11 (t)− γ
[1]
p f
[p]−
22 (t))(x − t)eµptdt
−e−µpx
x∫
−∞
(α[2]
p f
[p]−
11 (t)− γ
[2]
p f
[p]−
22 (t))(x − t)2eµptdt
is absolutely continuous. Here αj, βj , γj , δj are constants with α
[l]
p �= 0 �= γ
[l]
p ,
l = 1, 2. Set βp = δp = 0 to deduce that for all a <∞ one has
a∫
−∞
(1 + x2)| ddxf
[p−1]−
12 (x)|dx ≤ 9
a∫
−∞
(1 + x2)| ddxf
[p]−
12 (x)|dx + 24
µp
a∫
−∞
t| ddtf
[p]−
12 (t)|dt
+ 32
µ2
p
a∫
−∞
| ddtf
[p]−
12 (t)|dt +
a∫
−∞
(1 + t2){(
p∑
j=1
s1j |αj |+s2j |βj |
µj
+ |α[1]
p |
µ2
p
+ 2|α[2]
p |
µ3
p
)| ddtf
[p]−
11 (t)|
+(
p∑
j=1
s2j |γj |+s1j |δj |
µj
+ |γ[1]
p |
µ2
p
+ 2|γ[2]
p |
µ3
p
)| ddtf
[p]−
22 (t)|}dt
+2
a∫
−∞
t{(
p∑
j=1
s1j |αj |−s2j |βj |
µ2
j
+ 2|α[1]
p |
µ3
p
+ 6|α[2]
p |
µ4
p
)| ddtf
[p]−
11 (t)|
+(
p∑
j=1
s2j |γj |−s1j |δj |
µ2
j
+ 2|γ[1]
p |
µ3
p
+ 6|γ[2]
p |
µ4
p
)| ddtf
[p]−
22 (t)|}dt + 2
a∫
−∞
{(
p∑
j=1
s1j |αj |+s2j |βj |
µ3
j
+
3|α[1]
p |
µ4
p
+ 12|α[2]
p |
µ5
p
)| ddtf
[p]−
11 (t)|+ (
p∑
j=1
s2j |γj |+s1j |δj |
µ3
j
+ 3|γ[1]
p |
µ4
p
+ 12|γ[2]
p |
µ5
p
)| ddtf
[p]−
22 (t)|}dt
+Mp(a) <∞,
where Mp(a) ≡ −4e−µpa[µpa(1 + a2) − 2a2 + 8a
µp
+ 8
µ2
p
]
a∫
−∞
| ddtf
[p]−
12 (t)|eµptdt +
4µpe−µpa[1+a2− 2a
µp
+ 2
µ2
p
]
a∫
−∞
t| ddtf
[p]−
12 (t)|eµptdt−
p∑
j=1
e−µja
µj
[1+a2+ 2a
µp
+ 2
µ2
p
]
a∫
−∞
(s1j |αj |
| ddtf
[p]−
11 (t)|+s2j |γj || ddtf
[p]−
22 (t)|)eµj tdt+
p−1∑
j=1
eµja
µj
[1+a2− 2a
µp
+ 2
µ2
p
]
+∞∫
a
(s2j |βj || ddtf
[p]−
11 (t)|
+s1j |δj |·| ddtf
[p]−
22 (t)|)e−µj tdt− e−µpa
µpa
[(1+a2)a− 3a2+1
µp
+ 6a
µ2
p
+ 6
µ3
p
]
a∫
−∞
(|α[1]
p || ddtf
[p]−
11 (t)|
+|γ[1]
p | · | ddtf
[p]−
22 (t)|)eµptdt + e−µpa
µpa
[1 + a2 − 2a
µp
+ 2
µ2
p
]
a∫
−∞
t(|α[1]
p || ddtf
[p]−
11 (t)| + |γ[1]
p |
| ddtf
[p]−
22 (t)|)eµptdt− e−µpa
µpa
[(1+a2)a2+2(a2+a(1+a2))
µp
− 2
µ2
p
+24a
µ3
p
+ 24
µ4
p
]
a∫
−∞
(|α[2]
p || ddtf
[p]−
11 (t)|
+|γ[2]
p || ddtf
[p]−
22 (t)|)eµptdt+ e−µpa
µpa
[2a(1+a2)+2(3a2+1)
µp
+ 4a
µ2
p
+ 12
µ3
p
]
a∫
−∞
t(|α[2]
p || ddtf
[p]−
11 (t)|
+|γ[2]
p || ddtf
[p]−
22 (t)|)eµptdt − e−µpa
µpa
[1 + a2 + 2a)
µp
+ 2
µ2
p
]
a∫
−∞
t2(|α[2]
p || ddtf
[p]−
11 (t)| + |γ[2]
p |
| ddtf
[p]−
22 (t)|)eµptdt <∞.
Journal of Mathematical Physics, Analysis, Geometry, 2009, v. 5, No. 3 307
E.I. Zubkova and F.S. Rofe-Beketov
This case completes the proving of the fact that condition 4a is necessary for SD
(2); this also completes the proof of Lemma 2 as well as of Theorem 1 in two
versions.
Theorem 2. Conditions 1–6 of Theorem 1 (either with condition 4 or 4a),
together with additional condition 1a:
1a) R+(k) has a continuous derivative which is bounded on the entire axis and
is such that d
dkR
+(k) = o( 1
k ), k → ±∞,
are necessary and sufficient to characterize the scattering data for the problem
(1) of the case considered in Theorem 1.
P r o o f of Theorem 2. Lemma 1 establishes the necessity of condition 1a of
Theorem 2, while the necessity of conditions 1–6 (either with condition 4 or 4a)
are established by Theorem 1. As shown in the proof of Theorem 1, its condition
7 follows from conditions 1–6 and Lemma 1 (i.e., Th. 2, cond. 1a). Thus the
conditions of Theorem 2 imply all conditions 1–7 of Theorem 1, therefore the
conditions of Theorem 2 are necessary and sufficient either under condition 4 or
4a. Theorem 2 is proved.
The authors are deeply grateful to D.G. Shepel’sky who attracted their atten-
tion to the question about possibility to omit condition H in the main theorem
of [4, 5].
References
[1] V.A. Marchenko, Sturm–Liouville Operators and Applications. Naukova Dumka,
Kiev, 1977. (Russian) (Engl. transl.: Basel, Birkhäuser, 1986.)
[2] L.D. Faddeev, Inverse Scattering Problem in Quantum Theory II. In: Modern
Probl. Math. VINITI, Moscow, 3 (1974), 93–180. (Russian) (Engl. Transl.: J. Soviet
Math. 5 (1976), 334–396.)
[3] B.M. Levitan, Inverse Sturm-Liouville Problems. Nauka, Moscow, 1984. (Russian)
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