On the Abstract Inverse Scattering Problem for Trace Class Perturbations
The scattering problem for a pair of selfadjoint operators {L₀, L}, where L - L₀ is of trace-class, is studied. The explicit form of the scattering matrix and its properties are defined. The equation for the inverse problem is obtained.
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irk-123456789-1405632018-07-11T01:23:18Z On the Abstract Inverse Scattering Problem for Trace Class Perturbations Hatamleh, R. Zolotarev, V.A. The scattering problem for a pair of selfadjoint operators {L₀, L}, where L - L₀ is of trace-class, is studied. The explicit form of the scattering matrix and its properties are defined. The equation for the inverse problem is obtained. 2017 Article On the Abstract Inverse Scattering Problem for Trace Class Perturbations / R. Hatamleh, V.A. Zolotarev // Журнал математической физики, анализа, геометрии. — 2017. — Т. 13, № 1. — С. 3-34. — Бібліогр.: 21 назв. — англ. 1812-9471 DOI: doi.org/10.15407/mag13.01.003 Mathematics Subject Classification 2000: 47A45 http://dspace.nbuv.gov.ua/handle/123456789/140563 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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The scattering problem for a pair of selfadjoint operators {L₀, L}, where L - L₀ is of trace-class, is studied. The explicit form of the scattering matrix and its properties are defined. The equation for the inverse problem is obtained. |
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Hatamleh, R. Zolotarev, V.A. |
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Hatamleh, R. Zolotarev, V.A. On the Abstract Inverse Scattering Problem for Trace Class Perturbations Журнал математической физики, анализа, геометрии |
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Hatamleh, R. Zolotarev, V.A. |
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Hatamleh, R. |
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On the Abstract Inverse Scattering Problem for Trace Class Perturbations |
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On the Abstract Inverse Scattering Problem for Trace Class Perturbations |
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On the Abstract Inverse Scattering Problem for Trace Class Perturbations |
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On the Abstract Inverse Scattering Problem for Trace Class Perturbations |
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On the Abstract Inverse Scattering Problem for Trace Class Perturbations |
| title_sort |
on the abstract inverse scattering problem for trace class perturbations |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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On the Abstract Inverse Scattering Problem for Trace Class Perturbations / R. Hatamleh, V.A. Zolotarev // Журнал математической физики, анализа, геометрии. — 2017. — Т. 13, № 1. — С. 3-34. — Бібліогр.: 21 назв. — англ. |
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Журнал математической физики, анализа, геометрии |
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AT hatamlehr ontheabstractinversescatteringproblemfortraceclassperturbations AT zolotarevva ontheabstractinversescatteringproblemfortraceclassperturbations |
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2025-07-10T10:44:47Z |
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2025-07-10T10:44:47Z |
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1837256454994657280 |
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Journal of Mathematical Physics, Analysis, Geometry
2017, vol. 13, No. 1, pp. 3�34
On the Abstract Inverse Scattering Problem for Trace
Class Perturbations
R. Hatamleh
Department of Mathematics, Jadara University, Irbid-Jordan
E-mail: raedhat@yahoo.com
V.A. Zolotarev
B. Verkin Institute for Low Temperature Physics and Engineering
National Academy of Sciences of Ukraine
47 Nauky Ave., Kharkiv 61103, Ukraine
E-mail: vazolotarev@gmail.com
Received July 15, 2015, revised November 3, 2015
The scattering problem for a pair of selfadjoint operators {L0, L}, where
L − L0 is of trace-class, is studied. The explicit form of the scattering
matrix and its properties are de�ned. The equation for the inverse problem
is obtained.
Key words: scattering matrix, Friedrichs�Faddeev model, inverse scat-
tering problem.
Mathematics Subject Classi�cation 2010: 47A45.
Introduction
The scattering theory is not only the source of new non-traditional mathe-
matical methods in the realm of perturbations of linear operators but it is also
an important instrument for solving some problems of mathematical physics and
performing calculus [1, 2]. Along with the problems of scattering theory, such as
the existence of wave operators and their completeness, the properties of scatter-
ing matrix etc. (the Kato�Birman theory), the inverse problem, contained in the
reconstruction of the initial operator system by the physically measured scattering
data, plays an important, if not the main, role [1�6].
In this paper, the scattering problem for the pair of selfadjoint operators
{L0, L} is solved under the assumption that L−L0 is a trace-class operator. Sim-
ilar problems were also studied in [8�10]. For simplicity sake (although this does
not complicate signi�cantly the considerations mentioned below), the operator
c© R. Hatamleh and V.A. Zolotarev, 2017
R. Hatamleh and V.A. Zolotarev
L0 is assumed to be bounded. The main goal of the paper is to construct the
scattering operator S for the given pair {L0, L} and solve the inverse problem.
The paper consists of 5 sections, in the �rst of which the wave operators W±
and the scattering operator S are calculated using the methods of [2]. In Section
2, for the operator system {L0, L} under study, the Friedrichs�Faddeev model is
constructed and the scattering matrix S(x) is calculated. The main properties of
S(x) are proven and the link between S(x) and the function of spectral shift ξ(x)
is determined (the Birman�Krein Theorem). Section 3 is dedicated to the study
of the Friedrichs�Faddeev model {Q,L}. The eigenfunctions of the operator L
are described and the type of spectral projectors of absolutely continuous part of
the spectrum of the operator L is de�ned. In Section 4, the main integral equa-
tion of the inverse problem (an analogue of the Marchenko equation) is obtained.
The last section is dedicated to the solution of the inverse Riemann scattering
problem for the case where dimE = 1. Similar problems for one-dimensional
perturbations are studied in [19�21]. Notice that the statement of the problem is
not traditional: the Friedrichs�Faddeev model is unambiguously restored by the
scattering data. The solution of the inverse problem for the case dimE > 1 will
be the subject of further publications.
1. Scattering Operator
I. In a separable Hilbert space H we consider a pair of bounded linear selfad-
joint operators {L0, L} such that
L
def= L0 − ϕ∗Rϕ, (1)
where ϕ : H → E (E is a separable Hilbert space); R = R∗ is an operator in E.
It follows from (1) that
R0(λ) = R(λ)−R0(λ)ϕ∗RϕR(λ) (λ 6∈ σ (L0) ∪ σ(L)),
where R0(λ) = (L0 − λI)−1 and R(λ) = (L− λI)−1, and thus
ϕR(λ) = (I − V (λ)R)−1ϕR0(λ),
where V (λ) is a Nevanlinna [9] operator-function in E,
V (λ) def= ϕR0(λ)ϕ∗. (2)
Thus
R(λ) = R0(λ) + R0(λ)ϕ∗N(λ)ϕR0(λ), (3)
where
N(λ) def= R(I − V (λ)R)−1. (4)
4 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
Theorem 1. The function N(λ) (4) is Nevanlinna [12], (N(λ) − R) is
holomorphic at the in�nitely distant point and the following statements are true:
1)N(λ) = (I −RV (λ))−1R;
2)N∗(λ) = N
(
λ̄
)
;
3)N(λ)−N(w) = N(λ)[V (λ)− V (w)]N(w) (5)
= (λ− w)N(λ)ϕR0(λ)R0(w)ϕ∗N(w);
4) the representation
N(λ) = R +
b∫
a
dσ(t)
t− λ
is true, where λ, w belong to the analyticity domain of N(λ); V (λ) is given by
(2); R0(λ) = (L0 − λI)−1; σ(t) is a nondecreasing operator-function of bounded
variation in E; −∞ < a < b < ∞.
P r o o f. Relation 1) follows from the identity (I−RV (λ))R = R(I−V (λ)R).
Equality 2) follows from 1) and from the fact that R = R∗, V ∗(λ) = V
(
λ̄
)
. Since
N(λ)−N(w) = (I −RV (λ))−1R[V (λ)− V (w)]R(I − V (w)R)−1
= N(λ)[V (λ)− V (w)]N(w),
3) is a corollary of V (λ)− V (w) = (λ− w)ϕR0(λ)R0(w)ϕ∗. Taking into account
that
N(λ) = R + RV (λ)R(I − V (λ)R)−1, sup
y>1
‖yV (iy)‖ < ∞,
we obtain 4) in view of the well-known [12] representation for the Nevanlinna
functions.
For a selfadjoint operator L in H and for all ε > 0,
∫
R+
eiλte−itLe−εtdt = −iR(λ + iε),
∫
R+
e−iλteitLe−εtdt = iR(λ− iε) (6)
take place, where λ ∈ R and R(λ) = (L− λI)−1. Consider an operator-function
δ(λ, ε) def=
1
2πi
{R(λ + iε)−R(λ− iε)} =
ε
π
R(λ + iε)R(λ− iε) (7)
(λ ∈ R, ε > 0), then [2] ∫
R
δ(λ, ε)dλ = I. (8)
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 5
R. Hatamleh and V.A. Zolotarev
II. We de�ne the wave operators [1 � 3],
W±
def= s− lim
t→∓∞ exp{itL} · exp {−itL0}P a
0 , (9)
where P a
0 is an orthoprojector on an absolutely continuous subspace of the oper-
ator L0. The Kato�Rosenblum Theorem [1, 2] yields that if R is of trace-class,
then the limits (9) exist and W± are complete, i. e., W±H = P aH, where P a is an
orthoprojector on an absolutely continuous subspace of the operator L. The oper-
ators W± (9) are partial isometries from P a
0 H on P aH [1, 2] and they intertwine
the absolutely continuous parts of the operators L0 and L,
LaW± = W±La
0, (10)
where La
0 = L0P
a
0 , La = LP a. The existence of W± (9) yields that the limits
W± exist in the weak convergence sense and these limits coincide. Use the weak
convergence by Abel [2].
Theorem 2. If the limits W± (9) exist, then
〈W±f0, f〉 = lim
ε→+0
∫
R+
2εe−2εt 〈exp{∓itL} exp {±itL0} f0, f〉 dt, (11)
∀f0 ∈ P a
0 H, ∀f ∈ P aH.
P r o o f. Let us show that A → 0 as ε → +0, where
A
def= 〈W−f0, f〉 − 2ε
∫
R+
e−2εt
〈
eitLe−itL0f0, f
〉
dt =
= 2ε
∫
R+
e−2εt
〈(
W− − eitLe−itL0
)
f0, f
〉
dt.
For all δ > 0 there exists N = N (δ, f0, f) such that
∣∣〈(W−eitLe−itL0
)
f0, f
〉∣∣ < δ
takes place for t > N , therefore
∣∣∣∣∣∣
2ε
∞∫
N
e−2εt
〈(
W− − eitLe−itL0
)
f0, f
〉
dt
∣∣∣∣∣∣
< δe−2εN .
Taking into account the unitarity of eitL, eitL0 and the isometricity of W−, we
obtain
∣∣∣∣∣∣
2ε
N∫
0
e−2εt
〈(
W− − eitLe−itL0
)
f0, f
〉
dt
∣∣∣∣∣∣
≤ 2 ‖f0‖ · ‖f‖
(
1− e−2εN
)
.
6 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
Using the obvious inequality 1 − e−δ ≤ δ, which is true for all δ ∈ R+, we have
|A| < δe−2εN + 4εN ‖f0‖ · ‖f‖. Let ε be such that εN < δ. Then |A| < Cδ
(C = max {1, 4 ‖f0‖ · ‖f‖}).
Theorem 3. Let R in (1) be of trace-class, then for the wave operators W±
(9), the following representations are true:
〈W±f0, f〉 = 〈f0, f〉+ lim
ε→+0
∫
R
〈R0(λ± iε)ϕ∗N(λ± iε)ϕδ0(λ, ε)f0, f〉 dt, (12)
where f0 ∈ P a
0 H, f ∈ P aH; the functions N(λ) and δ(λ, ε) are given by (4) and
(7), respectively; R0(λ) = (L0 − λI)−1.
P r o o f. Since
〈W−f0, f〉 = lim
ε→+0
∫
R+
e−2εt
〈
e−itL0f0, e
−itLf
〉
dt
in view of (11), then using the Parseval inequality for the Fourier transform and
(6), we obtain
〈W−f0, f〉 = lim
ε→+0
ε
π
∫
R
〈R0(λ + iε)f0, R(λ + iε)f〉 dλ.
Taking into account (3) and (7), we have
〈W−f0, f〉 =
= lim
ε→+0
∫
R
〈δ0(λ, ε)f0, f〉 dλ +
∫
R
〈R0(λ− iε)ϕ∗N(λ− iε)ϕδ0(λ, ε)f0, f〉 dλ
,
which, in view of (8), gives us (12). For W+, formula (12) is proved in a similar
way.
Formula (12) implies
〈
W ∗
∓f, f0
〉
= 〈f, f0〉+ lim
ε→+0
∫
R
〈δ0(µ, ε)ϕ∗N(µ± iε)ϕR0(µ± iε)f, f0〉 dµ. (13)
De�ne the scattering operator [1�3],
S
def= W ∗
+W−. (14)
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 7
R. Hatamleh and V.A. Zolotarev
The operator S maps P a
0 H onto itself isometrically, and
SLa
0 = La
0S (15)
in view of (10), where La
0 is an absolutely continuous part of L0.
Theorem 4. The scattering operator S (14) of the pair {L0, L} (L is given
by (1), R is of trace-class) is given by the formula
〈(S − I)f0, g0〉 = −2πi lim
ε→+0
∫
R
〈δ0(λ, ε)ϕ∗N(λ− iε)ϕδ0(λ, ε)f0, g0〉 dλ, (16)
where f0, g0 ∈ P a
0 H; N(λ) and δ0(λ, ε) are given by (4) and (7), respectively.
P r o o f. Formulas (12), (13) imply
〈Sf0, g0〉 = 〈W−f0,W+g0〉 = 〈f0, W+g0〉+ lim
ε→+0
∫
R
〈R0(λ− iε)ϕ∗N(λ− iε)ϕ
×δ0(λ, ε)f0,W+g0〉 dλ = 〈f0, g0〉+ lim
ε1→+0
∫
R
〈δ0 (µ, ε1) ϕ∗N (µ− iε1) ϕ
×R0 (µ− iε1) f0, g0〉 dµ + lim
ε→+0
∫
R
〈R0(λ− iε)ϕ∗N(λ− iε)ϕδ0(λ, ε)f0, g0〉 dλ
+ lim
ε1→+0
lim
ε→+0
∫
R
dλ
∫
R
dµ 〈δ0 (µ, ε1) ϕ∗N (µ− iε1) ϕR0 (µ− iε1) R0(λ− iε)ϕ∗
×N(λ− iε)ϕδ0(λ, ε)f0, g0〉 .
The double integral, in view of 3) (5), is equal to
B
def=
∫
R
dλ
∫
R
dµ
〈
δ0 (µ, ε1) ϕ∗
N (µ− iε1)−N(λ− iε)
µ− λ + i (ε− ε1)
ϕδ0(λ, ε)f0, g0
〉
.
Calculate the expression
∫
R
dλ
µ− λ + i (ε− ε1)
δ0(λ, ε) =
1
2πi
∫
R
dλ
µ− λ + i (ε1 − ε)
∫
R
dEx
x− λ− iε
−
∫
R
dEx
x− λ + iε
=
1
2πi
∫
R
dEx
x− µ + i (ε1 − 2ε)
∫
R
[
1
λ− x + iε
8 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
− 1
λ− µ + i (ε1 − ε))
]
dλ
}
− 1
2πi
∫
R
dEx
x− µ + iε1
∫
R
[
1
λ− x− iε
− 1
µ− λ + i (ε1 − ε)
]
dλ
}
.
If ε > ε1, then
∫
R
[
1
λ− x + iε
− 1
λ− µ + i (ε1 − ε)
]
dλ = −2πi,
∫
R
[
1
λ− x− iε
− 1
µ− λ + i (ε1 − ε)
]
dλ = 0,
and ∫
R
dλ
µ− λ + i (ε− ε1)
δ0(µ, ε) = −R0 (µ− i (2ε− ε1)) (ε > ε1) .
Analogously,
∫
R
dµ
µ− λ + i (ε− ε1)
δ0 (µ, ε1) = R0 (λ1 − ε) (ε > ε1) .
Calculate the double integral
B = −
∫
R
dµ 〈δ0 (µ, ε1) ϕ∗R0 (µ− i (2ε− ε1)) f0, g0〉
−
∫
R
dλ 〈R0(λ− iε)ϕ∗N(λ− iε)ϕδ0(λ, ε)f0, g0〉 .
Substituting this expression into the sum for 〈Sf0, g0〉, we obtain
〈(S − I)f0, g0〉 = lim
ε,ε1→+0
∫
R
dµ 〈δ (µ, ε1) ϕ∗N (µ− iε1) ϕ×
×{R0 (µ− iε1)−R0 (µ + i (2ε− ε1))} f0, g0〉 .
Proceeding to the limit as ε → ε1 + 0 (ε > ε1) and taking into account (7),
we obtain (16). For ε1 > ε, the considerations are similar and we again obtain
formula (16).
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 9
R. Hatamleh and V.A. Zolotarev
2. The Friedrichs�Faddeev Model
III. Consider the subspace H1 (⊆ H),
H1
def= span {R0(λ)ϕ∗E : λ ∈ C\σ (L0)} , (17)
where R0(λ) = (L0 − λI)−1. The subspace H1 reduces the operators L0 and L1
(1). Since H⊥
1 ⊂ Kerϕ, then L|H⊥
1
= L0|H⊥
0
and the wave operators W± (9) cor-
responding to these restrictions coincide with the projector P a
1 (P a
1 is a projector
on the absolutely continuous subspace of L0|H⊥
1
). The scattering operator S (14)
on H⊥
1 is a trivial one; S|H⊥
1
= P a
1 . Therefore the supposition that H = H1 (17)
is reasonable, and thus ϕ∗E is the generating subspace of the operator L0 [13].
In E, specify a non-descending operator-function, F (x) = ϕExϕ∗, where Ex
is the resolution of identity of the operator L0. The operator L0 being bounded,
the support of the measure dF (x) is �nite. De�ne the Hilbert space
L2
R(E, dF (x)) def=
f(x) ∈ E :
∫
R
〈dF (x)f(x), f(x)〉 < ∞
(18)
obtained as a result of the closure of the class of continuous vector-functions f(x)
from E. If H1 = H, then the operator U : L2
R(E, dF (x)) → H, given by the
formula
f = Uf(x), f
def=
∫
R
dExϕ∗f(x) (19)
(f ∈ H, f(x) ∈ L2
R(E, dF (x))), speci�es the unitary isomorphism between H
(= H1) and L2
R(E, dF (x)) (18).
Theorem 5. Let H = H1 (17) and U : L2
R(E, dF (x)) → H be given by (19),
then the operators Q = U∗L0U and L = U∗LU are
(Qf)(x) def= xf(x), (Lf) def= xf(x)−R
∫
R
dF (t)f(t), (20)
where f(x) ∈ L2
R(E, dF (x)).
P r o o f. The representation of Q (20) follows from the equality
L0Uf(x) = L0
∫
R
dExϕ∗f(x) =
∫
R
dExϕ∗(xf(x)) = UQf(x).
Since ϕ̃ = ϕU is given by
ϕ̃f(x) =
∫
R
dF (x)f(x),
10 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
then, taking into account that ϕ̃∗ is the operator of enclosure of E into L2
R(E, dF (x)),
we obtain the representation of L in view of (1).
Formulas (20), well-known in the scattering theory [2, 4], are called the
Friedrichs�Faddeev model.
The scattering operator S (14) corresponds to the operator of multiplication by
the operator-function S(x) = U∗SU in E (U is given by (19)), which is commonly
[1, 2] named a scattering matrix. The above follows from the commutativity
of U∗SU and Q (20) in view of (15).
Further, we will suppose that the operator ϕ from (1) belongs to the Hilbert�
Shmidt class. Denote the function F (x) = ϕExϕ∗ which also belongs to this
class for all x ∈ R. Moreover, we suppose that the function F (x) is absolutely
continuous in the sense that there exists a nonnegative integrable (by Bochner)
operator-function a(·) : R → B(E), (B(E) is the space of bounded operators in
E), such that
F (∆) =
∫
∆
a(x)dx (a(x) ≥ 0)
for any Borel set ∆ from R. Formally, we write this condition of absolute conti-
nuity in the form
dF (x) = a(x)dx. (21)
Theorem 6. Let H = H1 (17), and the function F (x) = ϕExϕ∗ be absolutely
continuous, moreover, we suppose that R in (1) is of trace-class and ϕ belongs to
the Hilbert�Schmidt class. Then the scattering matrix S(x) = U∗SU correspond-
ing to S (14) is given by
S(x) = I − 2πiN−(x)a(x) (x ∈ R), (22)
where N−(x) = N(x− i0) are the boundary values on R from C− of the function
N(λ) (4).
P r o o f. Let
f0 =
∫
R
dExϕ∗f0(x) (f0 ∈ L2
R(E, dF (x))). (23)
Then (16) implies
〈(S − I)f0, g0〉
= −2πi lim
ε→+0
∫
R
〈
δ0(λ, ε)ϕ∗N(λ− iε)ϕδ0(λ, ε)
∫
R
dExϕ∗f0(x), g0
〉
dλ.
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 11
R. Hatamleh and V.A. Zolotarev
Since
ϕδ0(λ, ε)
∫
R
dExϕ∗f0(x) =
ε
π
ϕ
∫
R
dExϕ∗f0(x)
=
ε
π
∫
R
dF (x)f0(x)
(x− λ)2 + ε2
=
1
π
∫
R
ε
(x− λ)2 + ε2
a(x)f0(x)dx,
then, using the properties of the Poisson kernel [2, 14, 15], we obtain
ϕδ0(λ, x)
∫
R
dExϕ∗f0(x) → a(λ)f0(λ) (ε → +0).
Taking into account [13, 14],
〈δ0(λ, ε)f, g〉 → d
dλ
〈Eλf, g〉 (ε → +0),
and the existence of the boundary values N(λ − iε) → N−(λ) (= N(λ − i0)) as
ε → +0 [11, 12], we obtain
〈(S − I)f0, g0〉 = −2πi
∫
R
〈dEλϕ∗N−(λ)a(λ)f0(λ), g0〉 .
Hence,
Sf0 =
∫
R
dEλϕ∗ {I − 2πiN−(λ)a(λ)} f0(λ) = U{I − 2πiN−(λ)a(λ)}f0(λ),
which gives us (22).
Let
S+ = S∗ def= W ∗
−W+. (24)
Then, similarly to (16),
〈(
S+ − I
)
f0, g0
〉
= 2πi lim
ε→+0
∫
R
〈δ0(λ, ε)ϕ∗N(λ + iε)ϕδ0(λ, ε)f0, g0〉 dλ,
and thus the scattering matrix S+(x) = U∗S+U can be written in the form
S+(x) = I + 2πiN+(x)a(x) (x ∈ R), (25)
where N+(x) = N(x + i0) are the boundary values on R from C+ of the function
N(λ) (4).
12 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
It is obvious that as x /∈ supp a(x) the matrices S(x) (22) and S+(x) (25) are
equal to the identity operator, moreover,
a(x)S+(x) = S∗(x)a(x) (x ∈ R). (26)
Hence, S+(x) is adjoint to S(x) in the metric of L2
R(E, dF (x)).
Let us �nd the functional realizations of the wave operators W± (12). Substi-
tute f0 (23) into representation (12) for W−, then
〈(W− − I) f0, f〉
= lim
ε→+0
∫
R
〈
R0(λ− iε)ϕ∗N(λ− iε)ϕδ0(λ, ε)
∫
R
dExϕ∗f0(x), f
〉
dλ
= lim
ε→+0
∫
R
〈
dEyϕ
∗
∫
R
N(λ− iε)
y − λ + iε
ϕδ0(λ, ε)
∫
R
dExϕ∗f0(x), f
〉
dλ,
which gives
〈(W− − I) f0, f〉 = −
∫
R
〈
dExϕ∗
∫
R
N−(λ)
λ− x− i0
a(λ)f(λ)dλ, f
〉
.
Theorem 7. If H = H1 (17), the function F (x) = ϕExϕ∗ is absolutely
continuous (21), the operator R from (1) is of trace-class, and ϕ is a Hilbert�
Schmidt operator, then the wave operators W̃± = U∗W±U are
(
W̃∓f
)
(x) = f(x)−
∫
R
N∓(λ)
λ− x∓ i0
a(λ)f(λ)dλ, (27)
where f(x) ∈ L2
R(E, dF (x)).
Using (22), (24), we obtain
(
W̃−f
)
(x) = f(x)− 1
2πi
∫
R
I − S(λ)
λ− x− i0
f(λ)dλ,
(
W̃+f
)
(x) = f(x)− 1
2πi
∫
R
S+(λ)− I
λ− x + i0
f(λ)dλ,
(28)
which, by the Sokhotsky formulas [16], gives us the representations
(
W̃−f
)
(x) =
1
2
(I + S(λ))f(λ)− 1
2πi
6
∫
R
I − S(λ)
λ− x
f(λ)dλ,
(
W̃+f
)
(x) =
1
2
(
S+(λ) + I
)
f(λ)− 1
2πi
6
∫
R
S+(λ)− I
λ− x
dλ.
(29)
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 13
R. Hatamleh and V.A. Zolotarev
The intertwining conditions follow directly from (10),
LaW̃± = W̃±Qa. (30)
Construct a direct proof. Using (27), we get
(
W̃−Qaf
)
(x) = xf(x)− πiN−(x)a(x)xf(x)−
∫
R
N−(λ)
λ− x
a(λ)λf(λ)
= x
(
W̃−f
)
(x)− 6
∫
R
N−(λ)a(λ)f(λ)dλ.
Taking into account the identity N(λ) = R + RV (λ)N(λ), we obtain
(
W̃−Qaf
)
(x) = x
(
W̃−f
)
(x)−R
∫
R
a(λ)f(λ)−R
∫
R
−πia(λ)+ 6
∫
R
a(t)dt
t− λ
×N−(λ)a(λ)f(λ)dλ = x
(
W̃−f
)
(x)−R
∫
R
a(λ) {f(λ)− πiN−(λ)a(λ)f(λ)
− 6
∫
R
N−(x)
x− λ
a(x)f(x)dx
dλ = La
(
W̃−f
)
(x).
Equalities (30) show that W̃± (27) are the transformation operators [4, 5].
IV. The absolute continuity of F (x) (21) and (7) imply
a(x) = lim
ε→+0
ϕδ0(λ, ε)ϕ =
1
2πi
lim
ε→+0
[V (x + iε)− V (x− iε)],
thus
V+(x)− V−(x) = 2πia(x), (31)
where V±(x) are the boundary values on R from C± of the function V (λ) (2).
Equality (31) coincides with the Stieltjes�Perron inversion formula [2, 12] and
may also follow from the Sokhotsky formula [16] for V±(x).
Using 3) (5) and (31), we obtain
N+(x)−N−(x) = 2πiN+(x)a(x)N−(x),
N+(x)−N−(x) = 2πiN−(x)a(x)N+(x),
(32)
where N±(x) are the boundary values on R from C± of the function N(λ) (4).
14 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
Theorem 8. If the representing measure F (x) of the function V (λ) (2) is
absolutely continuous (21) and the suppositions of Theorem 6 take place, then
1) S∗(x)a(x)S(x) = a(x), (S+(x))∗ a(x)S+(x) = a(x),
2) S+(x)S(x) = S(x)S+(x) = I,
(33)
where S(x) and S+(x) are given by (22) and (25).
P r o o f. The de�nition of S(x) from (22) implies
S∗(x)a(x)S(x) = a(x)− 2πia(x)N−(x)a(x) + 2πia(x)N+(x)a(x)−
−(2πi)2a(x)N+(x)a(x)N−(x)a(x) = a(x)
in view of (32). We prove the second relation from 1) in a similar way. Taking
into account (22), (25), and (32), we obtain
S+(x)S(x) = I−2πiN−(x)a(x)+2πiN+(x)a(x)−(2πi)2N+(x)a(x)N−(x)a(x) = I
in view of (32).
Corollary 1. The equalities 1) (33) prove that the reciprocal matrix-functions
S(x) and S+(x) are unitary in the metric of the space L2
R(E, dF (x)) (18) when
dF (x) = a(x)dx.
Using (22) and (31), we obtain
S(x) = I − (I −RV−(x))−1 R (V+(x)− V−(x)) = (I −RV−(x))−1 (I −RV+(x)) ,
therefore
S(x) = (I −RV−(x))−1 (I −RV+(x)) .
Thus,
S(x) (I −RV+(x))−1 = (I −RV−(x))−1 , (34)
which, in view of 1) (5), gives us
S(x)N+(x) = N−(x). (35)
Theorem 9. Let the suppositions of Theorem 6 be true. Then the Nevanlinna
function N(λ) (4) is the solution of the Riemann boundary value problem [16],
(35), the coe�cient of which is the scattering matrix S(x) (22).
The equalities 2) (33) are dual to (34),
S+(x)N−(x) = N+(x), (36)
where S+(x) is given by (25).
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 15
R. Hatamleh and V.A. Zolotarev
V. From (3) and the fact that N(λ) (4) is of trace-class we have
tr {R(λ)−R0(λ)} = tr
{
ϕR2
0(λ)ϕ∗N(λ)
}
= tr
{
V ′(λ)N(λ)
}
.
Using (4) and the formula from [17],
d
dλ
ln det(I −A(λ)) = − tr
{
A′(λ)(I −A(λ))−1
}
,
we obtain
tr
{
V ′(λ)N(λ)
}
= − d
dλ
D(λ),
where
D(λ) def= det(I −RV (λ)) (37)
is the perturbation operator [17] of the pair of operators Q, L (20). The
existence of (37) is determined by the fact that R is of trace-class.
The following statement belongs to M. G. Krein [2, 18].
Theorem 10. Let R be of trace-class. Then for D(λ) (37),
lnD(λ) = −
∫
R
ξ(x)
x− λ
dx (Imλ 6= 0) (38)
takes place, where ξ(x) is the spectral shift function [2, 18],
ξ(x) def= − 1
π
lim
ε→+0
arg D(x + iε). (39)
The limit (38) exists almost everywhere for x ∈ R, and
∫
R
|ξ(x)|dx ≤ ‖R‖1;
∫
R
ξ(x)dx = trR
(‖R‖1 is the trace-class norm of R). Moreover, ξ(x) ≤ k+ (ξ(x) ≥ k−) almost
everywhere if R has exactly k+ positive (k− negative) eigenvalues.
Using this theorem, we obtain
tr {R(λ)−R0(λ)} =
∫
R
ξ(x)
(x− λ)2
dx (Imλ 6= 0),
which implies the well-known Krein trace formula [2, 18].
Theorem 11 (M. G. Krein). Let R be of trace-class and f(λ) be a holo-
morphic function as λ ∈ σ (L0) ∪ σ (L1), then
tr {f(L)− f (L0)} =
∫
R
ξ(x)f ′(x)dx.
16 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
The link between a scattering matrix and a function of spectral shift was estab-
lished by M. Sh. Birman and M. G. Krein [2].
Theorem 12 (M. Sh. Birman, M. G. Krein). If R is of trace-class, then
for almost all x ∈ R the formula
detS(x) = exp{−2πiξ(x)} (40)
is true, where S(x) is the scattering matrix (22), and ξ(x) is the function of
spectral shift (39).
P r o o f. Equality (34) yields
detS(x) = D(x + i0)D−1(x− i0),
where D(λ) is given by (37). To conclude the proof it remains to use de�nition
(39) of the function ξ(x).
3. Scattering Matrix
VI.We proceed to the analysis of the Friedrichs�Faddeev model (20) assuming
that F (x) is absolutely continuous (21). To �nd the resolvent RL(λ) = (L−λI)−1,
we denote (L − λI)−1f(x) = g(x). Then
(x− λ)g(x)−R
∫
R
a(t)g(t)dt = f(x).
In terms of G(x) = (x− λ)g(x) this equation takes the form
G(x)−R
∫
R
a(t)
G(t)
t− λ
dt = f(x).
Multiplying this equality by a(x)(x− λ)−1 and integrating it, we obtain
∫
R
a(x)
G(x)
x− λ
dx− V (λ)R
∫
R
a(t)
G(t)
t− λ
dt =
∫
R
a(x)
f(x)
x− λ
dx.
Hence, ∫
R
a(x)
G(x)
x− λ
dx = (I − V (λ)R)−1
∫
R
a(x)
f(x)
x− λ
dx.
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 17
R. Hatamleh and V.A. Zolotarev
Theorem 13. The resolvent RL(λ) = (L − λI)−1 of the operator L (20) is
given by
(RL(λ)f) (x) =
f(x)
x− λ
+
N(λ)
x− λ
∫
R
a(t)
f(t)
t− λ
dt, (41)
where f(x) ∈ L2
R(E, dF (x)), the function F (x) is absolutely continuous (21), and
N(λ) is given by formula (4).
The description of the discrete spectrum σp(L) of the operator L (20) lies in
the following.
Theorem 14. Suppose that the function F (x) = ϕExϕ∗ is absolutely con-
tinuous and R is of trace-class. Then the eigenfunctions Lfλ(x) = λfλ(x) of the
operator L (20) are given by
fλ(x) =
hλ
x− λ
, (42)
where hλ ∈ Ker(I − RV (λ)) and it does not depend on x, besides, dimKer(I −
RV (λ)) < ∞; the eigenvalues λ lie outside the supp a(x) and are zeroes of the
function D(λ) (36), D(λ) = D
(
λ̄
)
; the order of the values λ is dimKer(I −
RV (λ)); the eigenvalues λ ∈ σp(L) have the limiting points of the �absolutely
continuous closure� [8] of the set σ0 :=
{
x ∈ R : x ∈ supp a(x)
}
as their accumu-
lation points. The operator R on the subspace Ker(I −RV (λ)) is invertible.
P r o o f. Equality Lfλ(x) = λfλ(x) implies
(x− λ)fλ(x)−R
∫
R
a(t)fλ(t)dt = 0,
and for hλ(x) = (x− λ)fλ(x) we obtain
hλ(x)−R
∫
R
a(t)
hλ(t)
t− λ
dt = 0.
It is seen from the relation that hλ(x) = hλ does not depend on x and [I −
RV (λ)]hλ = 0. The �nite dimensionality of Ker(I − RV (λ)) provides the total
continuity of the operator R. Taking into account that (I−RV (λ))∗ = I−V
(
λ̄
)
R
(in view of R = R∗, V ∗(λ) = V
(
λ̄
)
) and using det(I − AB) = det(I −BA) [17],
we can prove that D(λ) = D
(
λ̄
)
. There are no complex roots of D(λ) (37) as L
is selfadjoint.
It remains to prove that R is not degenerate on Ker(I − RV (λ)) in view
of dimKer(I − RV (λ)) < ∞. If hλ ∈ Ker(I − RV (λ)), then hλ = RV (λ)hλ.
Assuming the contrary, Rhλ = 0, we obtain
18 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
‖hλ‖2 = 〈hλ, hλ〉 = 〈hλ, RV (λ)hλ〉 = 〈Rhλ, V (λ)hλ〉 = 0,
and hence hλ = 0.
The singularities of N(λ) (4) consist of an absolutely continuous part coincid-
ing with σ0 :=
{
x ∈ R : x ∈ supp a(x)
}
(32), (σ0 ⊆ [a, b], −∞ < a < b < ∞, in
view of the boundedness of L0), and a discrete part which is the eigenvalues of L
that are zeroes of D(λ) (37) and lie outside
{
x ∈ R : supp a(x)
}
. Thus,
σ(L) = σ0 ∪ σp, (43)
where
σp
def= σp(L) = {κn ∈ R : D (κn) = 0} . (44)
The sets σ0 and σp (44) do not intersect with each other except the limit points.
Notice that if σ0 consists of the �nite (denumerable) number of not-intersecting
segments from [a, b], then the point spectrum (44) of the semiaxes (−∞, a) and
(b,∞) is in the lacunas between these segments and probably has some limit
points of these intervals as its accumulation points.
If κn ∈ σp (44), then the subspace
Fn
def= Ker (I −RV (κn)) (45)
is nontrivial and �nite-dimensional, mn
def= dimFn < ∞.
Lemma 1. Suppose that
1) supp a(x) = [a, b]; 2)
⋂
x∈[a,b]
Ker a(x) = {0}. (46)
Then the subspaces Fn (45) do not intersect each other, Fn ∩ Fk = {0} (k 6= n).
P r o o f. Let there be a nonzero vector h ∈ Fk ∩ Fn (k 6= n). Then
R−1h = V (κn) h = V (κk) h, in view of the invertibility of R on â Fn (Theorem
14). Therefore, 〈{V (κn)− V (κk)}h, h〉 = 0, i. e.,
(κn − κk)
∫
R
〈a(x)h, h〉
(x− κn) (x− κk)
dx = 0.
Since κn 6= κk (n 6= k) and the integrand has a constant sign (κn, κk 6∈ [a, b]),
then 〈a(x)h, h〉 = 0 for all x ∈ [a, b]. Taking into account that a(x) ≥ 0, we obtain
a(x)h = 0 for all x ∈ [a, b], which contradicts condition 2) in (46).
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 19
R. Hatamleh and V.A. Zolotarev
Corollary 2. If condition 1) (46) does not hold, then the statement of the
lemma is not true. Let for dimE = 1 and supp a(x) = [a, c] ∪ [d, b] (a < c < d <
b), the equation 1 = RV (λ) always have two roots: κ1 ∈ (−∞, a) and κ2 ∈ (c, d)
as R > 0, or κ1 ∈ (c, d) and κ2 ∈ (b,∞) as R < 0. Then the equality
c∫
a
a(x)dx
(x− κ1) (x− κ2)
+
b∫
d
a(x)dx
(x− κ1) (x− κ2)
= 0
does not imply a(x) = 0.
Let κn ∈ σp (44). Then the eigenfunctions of the operator L (Theorem 14)
have the form
fs
κn
(x) =
hs
n
x− κn
(1 ≤ s ≤ mn, n ∈ N),
where {hs
n}mn
1 is the orthonormal basis in Fn (45), and D (κn) = 0 (D(λ) is given
by (36)). It is obvious that
〈
fs
κn
(x), fp
κn
(x)
〉
L2 =
〈
V ′ (κn) hs
n, hp
n
〉
E
(1 ≤ s, p ≤ mn),
where V ′ (κn) is the value of the derivative of the function V (λ) as λ = κn. Show
that V ′ (κn) is non-degenerate on Fn (45). Assuming the contrary, we suppose
that there is a vector h ∈ Fn such that V ′ (κn) h = 0. Since R−1h = V (κn) h,
then for all f ∈ E we have
∣∣〈R−1h, f
〉∣∣2 = |〈V (κn) h, f〉|2 =
∣∣∣∣∣∣
∫
R
〈a(x)h, f〉
x− κn
dx
∣∣∣∣∣∣
2
≤
∫
R
〈a(x)h, h〉
(x− κn)2
dx ·
∫
R
〈a(x)f, f〉dx =
〈
V ′ (κn) h, h
〉 ∫
R
〈a(x)f, f〉dx = 0.
The arbitrariness of f ∈ E implies R−1h = 0. Consequently, h = 0 (R is invertible
on Fn by Theorem 14).
Hence, the operator PnV ′ (κn) Pn is positive and invertible (Pn is an ortho-
projector on Fn (45)). Therefore there exist
Mn
def=
{
PnV ′ (κn) Pn
}− 1
2 (n ∈ N), (47)
where Mn > 0 and rankMn = mn. Then the vector-functions
fs (κn, x) =
Mnhs
n
x− κn
(1 ≤ s ≤ mn, n ∈ N) (48)
20 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
form the orthonormal basis in L2
R(E, a(x)dx) corresponding to the eigenvalue κn
(for all n ∈ N).
The functions fs (κn, x) (48) are orthogonal for di�erent κn as the eigenfunc-
tions of the operator L. Therefore,
∫
R
1
(x− κn) (x− κl)
〈
a(x)Mnhs
n,Mlh
k
l
〉
dx = 0,
where {hs
n}mn
1 and
{
hk
l
}ml
1
are orthonormal bases in Fn and Fl (44), respectively.
Hence,
Ml {V (κn)− V (κl)}Ml = 0 (∀κn 6= κl).
Show that this equality holds automatically. Formula (45) implies Pn = RV (κn) Pn
or R−1Mn = V (κn) Mn in view of the invertibility of R on Fn. Taking into ac-
count MlR
−1 = MlV (κl), we have
Ml {V (κn)− V (κl)}Mn = MlR
−1Mn −MlR
−1Mn = 0.
Theorem 15. Suppose that the operator R is of trace-class; the representing
measure dF (x) of the function V (λ) (2) is absolutely continuous (21); {κn}∞1 are
the set of zeroes of the function D(λ) (37), D (κn) = 0, and κn 6∈ σ0.
Then the function N(λ) (4) is given by
N(λ) = R +
∞∑
n=1
M2
n
κn − λ
+
∫
σ0
b(x)
x− λ
dx, (49)
where b(x) ≥ 0, supp b(x) = σ0, and Mn are given by (47). Moreover,
∫
σ0
b(x)dx < ∞;
∞∑
n=1
M2
n < ∞. (50)
P r o o f. Let us turn back to representation 4) (5) of the function N(λ)
(4). Then the absolute continuity of dσ(t) on σ0 follows from (32) and the Kato�
Rosenblum theorem [2]. It is necessary to prove that discontinuities of the measure
dσ(t) at the points κn coinside with M2
n.
Write the Laurent expansion of N(λ) (4) at the point κn ∈ σp (44),
N(λ) =
σn
κn − λ
+ R + ...,
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 21
R. Hatamleh and V.A. Zolotarev
where σn ≥ 0, and substitute the obtained equality into the identity N(λ) =
R + RV (λ)N(λ). Then
σn
κn − λ
+ R + . . . = R + RV (λ)
(
σn
κn − λ
+ R + ...
)
.
Multiplying the last equality by (κn − λ) and passing to limit as λ → κn, we
obtain that σn = RV (κn) σn (V (κn) exist since κn 6∈ σ0). Show that σn :
Fn → Fn and σnF⊥
n = 0. The de�nition of Fn (45) implies Pn = RV (κn) Pn (Pn
is an orthoprojector on Fn). Then
σn = RV (κn) σn = RV (κn)
[
Pn + P⊥
n
]
σn = Pnσn + RV (κn) P⊥
n σn,
where P⊥
n = I − Pn. Therefore, P⊥
n σn = RV (κn) P⊥
n σn, or (I −RV (κn))P⊥
n σn
= 0. Thus, P⊥
n σnf ∈ Fn for all f ∈ Fn and P⊥
n σnPn = 0, which proves that σn :
Fn → Fn and σn|F⊥n = 0, in view of the self-adjointness of σn.
Now we have to prove that σn = M2
n (47). First, we multiply the equality
R = (I − RV (λ))N(λ) from the right by Pn. Then, taking into account the
invertibility of R on Fn, we obtain
R−1 − V (λ)
κn − λ
(κn − λ) N(λ)Pn = Pn.
Using the Laurent expansion of N(λ) at the point λ = κn and passing to limit as
λ → κn, we have V ′ (κn)σnPn = Pn, which gives us M−2
n σn = Pn (σnPn = Pnσn).
The convergence of the integral and the series in (50) follows from the �niteness
of variation of the measure dσ(t) at representation 4) (5) of the function N(λ).
VII. Find the resolution of the identity Ea
λ corresponding to the absolutely
continuous part of La of the operator L. Then
d
dλ
〈Ea
λf, f〉 = lim
ε→+0
〈δ(λ, ε)f, f〉,
where δ(λ, ε) (7) corresponds to the operator L. Using (41), we obtain
〈δ(λ, ε)f, f〉 =
1
2πi
{〈RL(λ + iε)f, f〉 − 〈RL(λ− iε)f, f〉} =
1
2πi
〈
f(x)
x− λ− iε
+
N(λ + iε)
x− λ− iε
∫
R
a(t)
t− λ− iε
f(t)dt, f(x)
〉
− 1
2πi
〈
f(x)
x− λ + iε
− N(λ− iε)
x− λ + iε
×
∫
R
a(t)
t− λ + iε
f(t)dt, f(x)
〉
= Aε + Bε + Bε,
22 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
where,
Aε
def=
1
π
∫
R
ε
(x− λ)2 + ε2
〈a(x)f(x), f(x)〉dx,
Bε
def=
1
π
∫
R
〈
N(λ + iε)
∫
R
a(t)
t− λ− iε
f(t)dt,
∫
R
a(x)
x− λ + iε
f(x)dx
〉
.
Lemma 2. For all λ ∈ R, the formula
A0 + B0 + B0 = 〈a(λ)f(λ), f(λ)〉 − π2 〈a(λ)N−(λ)a(λ)f(λ), N−(λ)a(λ)f(λ)〉
+Re
〈
[N1(λ) + N−(λ)] a(λ)f(λ), 6
∫
R
a(x)
t− λ
f(t)dt
〉
(51)
+
〈
a(λ)N−(λ) 6
∫
R
a(t)
t− λ
f(t)dt,N−(λ) 6
∫
R
a(t)
t− λ
f(t)dt
〉
is true, where A0 (B0) = Aε (Bε)|ε=+0 , and N±(λ) are the boundary values on R
from C± of the function N(λ) (4).
P r o o f. The property of the Poisson kernel [14, 15] yields
Aε → A0 = 〈a(λ)f(λ), f(λ)〉 (ε → +0).
Using the Sokhotsky formulas [16], we have
B0 = −2πi
〈
N+(λ)
1
2
a(λ)f(λ) +
1
2πi
6
∫
R
a(t)
t− λ
f(t)dt
,
−
1
2
a(λ)f(λ) +
1
2πi
6
∫
R
a(t)
t− λ
f(t)dt
〉
.
Finally, after elementary transformations, taking into account (32), we obtain
(51).
Theorem 16. Suppose that the suppositions of Theorem 6 take place. Then
for the identity resolution Ea
λ of the absolutely continuous part La of the operator
L (20), the representation
d
dλ
〈Ea
λf, f〉 =
〈
a(λ)
1
2
(
I + S+(λ)
)
f(λ) + N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt
,
1
2
(
I + S+(λ)
)
f(λ) + N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt
〉 (52)
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 23
R. Hatamleh and V.A. Zolotarev
is true, where S+(λ) is the scattering matrix (25), and N+(λ) is the boundary
value on R from C+ of the function N(λ) (4).
P r o o f. Since N−(λ) = N+(λ)− 2πiN+(λ)a(λ)N−(λ) (32), then the third
summand from (51) is given by
1
2
〈
[N+(λ) + N−(λ)] 6
∫
R
a(t)
t− λ
f(t)dt, a(λ)f(λ)
〉
=
〈
N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt,
a(λ)f(λ)〉+
〈
a(λ)N−(λ) 6
∫
R
a(λ)
t− λ
f(t)dt, πiN−(λ)a(λ)f(λ)
〉
.
Substituting this expression in (51) and taking into account that S(λ)N+(λ) =
N−(λ) (35) and S∗(λ)a(λ)S(λ) = a(λ) 1) (33), we obtain
A0 + B0 + B0 = 〈a(λ)f(λ), f(λ)〉 − 〈a(λ)πiN+(λ)a(λ)f(λ), πiN+(λ)a(λ)f(λ)〉
+
〈
N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt, a(λ)f(λ)
〉
+
〈
a(λ)N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt,
πiN+(λ)a(λ)f(λ)〉+
〈
a(λ)f(λ), N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt
〉
+
〈
a(λ)πiN+(λ)a(λ)f(λ), N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt
〉
+
〈
a(λ)N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt,N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt
〉
.
Hence, after elementary transformations we obtain
A0 + B0 + B0 =
〈
a(λ)
{
f(λ) + πiN+(λ)a(λ)f(λ) + N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt
}
,
f(λ) + πiN+(λ)a(λ)f(λ) + N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt
〉
,
which gives us (52) in view of (25).
24 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
4. Equation of the Inverse Scattering Problem
VIII. Associate every function f(x) ∈ L2
R(E, dF (x)) (18) with the Fourier
coe�cients,
fs (κn) def= 〈f(x), f s (κn, x)〉 (1 ≤ s ≤ mn, n ∈ N), (53)
corresponding to the discrete spectrum σp(L) (44) of the operator L, where
fs (κn, x) are given by (48). Then
fs (κn) =
〈∫
R
a(x)
x− κn
f(x)dx,Mnhs
n
〉
, (54)
where {hs
n}mn
1 are the orthonormal basis in Fn (45), and Mn are given by (47).
In the case of absolutely continuous spectrum of the operator L,
f̃(λ) def=
1
2
(
I + S+(λ)
)
f(λ) + N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt (55)
are the �Fourier coe�cients� of the function f(x) ∈ L2
R(E, dF (x)) by Theorem
16. The space L2
R(E, dF (x)) is the orthogonal sum of two subspaces, the �rst of
which is the eigenfunctions (associated states) of the operator L, and the second
subspace corresponds to the absolutely continuous spectrum of L. Write the
Parseval equality [4],
〈f(x), g(x)〉L2
R(E,dF (x)) =
∞∑
n=1
mn∑
s=1
fs (κn) gs (κn) +
〈
f̃(λ), g̃(λ)
〉
L2
R(E,dF (λ))
, (56)
where f , g ∈ L2
R(E, dF (x)). (54) implies
mn∑
s=1
fs (κn) gs (κn) =
∫
R
dx
∫
R
dy
〈
M2
n
a(x)
x− κn
f(x),
a(y)
y − κn
g(y)
〉
since {hs
n〉mn
1 is the orthonormal basis in Fn. Using
1
(x− κn) (y − κn)
=
1
x− y
[
1
y − κn
− 1
x− κn
]
,
we obtain
∞∑
n=1
mn∑
s=1
fs (κn) gs (κn) = −
∫
R
dx
∫
R
dy
〈
P (x)− P (y)
x− y
a(x)f(x), a(y)f(y)
〉
,
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 25
R. Hatamleh and V.A. Zolotarev
where
P (x) def=
∞∑
n=1
M2
n
x− κn
. (57)
By the condition of (50), the series of (57) converges.
Equality (55) implies
B
def=
〈
f̃(λ), g̃(λ)
〉
L2
R(E,dF (λ))
=
∫
R
〈
a(λ)
{
1
2
(
I + S+(λ)
)
f(λ)
+N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt
,
1
2
(
I + S+(λ)
)
g(λ) + N+(λ) 6
∫
R
a(t)
t− λ
g(t)dt
〉
=
1
4
∫
R
〈[
2a(λ) + a(λ)S+(λ) +
(
S+(λ)
)∗
a(λ)
]
f(λ), g(λ)
〉
dλ
+
1
2
∫
R
〈
a(λ)
(
I + S+(λ)
)
f(λ), N+(λ) 6
∫
R
a(t)
t− λ
g(t)dt
〉
+
1
2
∫
R
〈
a(λ)N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt,
(
I + S+(λ)
)
g(λ)
〉
dλ
+
∫
R
〈
N−(λ)a(λ)N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt, 6
∫
R
a(t)g(t)
t− λ
dt
〉
dλ.
Using (26), (32), (33), we obtain
B = 〈f(x), g(x)〉L2
R(E,dF (x)) +
1
4
∫
R
〈
a(λ)
[
S+(λ) + S(λ)− 2I
]
f(λ), g(λ)
〉
dλ
+
1
2
∫
R
〈
N−(λ) [I + S∗(λ)] a(λ)f(λ), 6
∫
R
a(t)
t− λ
g(t)dt
〉
dλ
+
1
2
∫
R
〈
a(λ)[I + S(λ)]N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt, g(λ)
〉
dλ
+
1
2πi
∫
R
〈
[N+(λ)−N−(λ)] 6
∫
R
a(t)
t− λ
f(t)dt, 6
∫
R
a(t)
t− λ
g(t)dt
〉
dλ.
26 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
Taking into account (22), (25), and (35), we have
B = 〈f(x), g(x)〉L2
R(E,dF (x)) +
πi
2
∫
R
〈a(λ)[I − S(λ)]N+(λ)a(λ)f(λ), g(λ)〉 dλ
−1
2
∫
R
〈
a(λ)
∫
R
(I + S(t))N+(t)− (I + S(λ))N(λ)
t− λ
a(t)f(t)dt, g(λ)
〉
dλ
+
1
2πi
∫
R
〈
[I − S(λ)]N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt, 6
∫
R
a(t)
t− λ
g(t)dt
〉
dλ.
The last integral from this sum may be written in the form
C
def=
1
2πi
∫
R
dx 6
∫
R
dλ
x− λ
〈
(I − S(λ))N+(λ) 6
∫
R
a(t)
t− λ
f(t)dt, a(x)g(x)
〉
.
Using the Poincare�Bertrand formula [16] for rearranging the order of integration
in improper integrals,
1
πi
6
∫
R
dτ
τ − t
6
∫
R
ϕ(τ, ξ)
ξ − τ
dξ = πiϕ(t, t)+ 6
∫
R
dξ 6
∫
R
ϕ(τ, ξ)
(τ − t)(ξ − τ)
dτ,
we obtain
1
πi
6
∫
R
dλ
λ− x
6
∫
R
(I − S(λ))N+(λ)
t− λ
a(t)f(t)dt = πi[I − S(x)]N+(x)a(x)f(x)
+
1
πi
6
∫
R
dt 6
∫
R
(I − S(λ))N+(λ)
(t− λ)(λ− x)
a(t)f(t)dλ.
Therefore,
C = −πi
2
∫
R
〈[I − S(x)]N+(x)a(x)f(x), a(x)g(x)〉dx
+
1
2πi
∫
R
dx
〈∫
R
M(x)−M(t)
x− t
a(t)f(t)dt, a(x)g(x)
〉
,
where
M(x) def= 6
∫
R
(I − S(λ))N+(λ)
λ− x
dλ. (58)
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 27
R. Hatamleh and V.A. Zolotarev
Finally, we have
〈
f̃(λ), g̃(λ)
〉
L2
R(E,dF (λ))
= 〈f(x), g(x)〉L2
R(E,dF (x))
−1
2
∫
R
〈
a(λ)
∫
R
[I + S(t)]N+(t)− [I + S(λ)]N+(λ)
t− λ
a(t)f(t)dt, g(λ)
〉
dλ
+
1
2πi
∫
R
〈
a(λ)
∫
R
M(t)−M(λ)
t− λ
a(t)f(t)dt, g(λ)
〉
dλ.
Substituting the obtained expressions in the Parseval equality (56), we have
−
∫
R
dλ
〈
a(λ)
∫
R
P (x)− P (λ)
x− λ
a(x)f(x)dx, g(λ)
〉
−1
2
∫
R
dλ
〈
a(λ)
∫
R
[I + S(x)]N+(x)− [I + S(λ)]N+(λ)
x− λ
a(x)f(x)dx, g(λ)
〉
+
1
2πi
∫
R
dλ
〈
a(λ)
∫
R
M(x)−M(λ)
x− λ
a(x)f(x)dx, g(λ)
〉
= 0.
Let
G(λ) def= −P (λ)− 1
2
[I + S(λ)]N+(λ) +
1
2πi
6
∫
R
[I − S(x)]N+(x)
x− λ
dx. (59)
Then the last equality takes the form
∫
R
〈
a(λ)
∫
R
G(x)−G(λ)
x− λ
a(x)f(x)dx, g(λ)
〉
dλ = 0,
that, in view of the arbitrariness of g(x) ∈ L2
R(E, a(x)dx), gives
∫
R
G(x)−G(λ)
x− λ
a(x)f(x)dx = 0
for all f(x) ∈ L2
R(E, a(x)dx). This implies
∫
R
〈
a(λ)f(λ),
G(x)−G(λ)
x− λ
h
〉
= 0
28 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
for an arbitrary h ∈ E, in view of the self-adjointness of G(λ) (59), since S(λ)N+(λ) =
N−(λ) (35). Thus,
G(x)−G(λ)
x− λ
h = 0 (∀x, λ),
and G(λ) is a constant operator
G(λ) = B (= B∗) .
Since
G(λ) = −P (λ)−N+(λ) +
1
2
[I − S(λ)]N+(λ) +
1
2πi
6
∫
R
[I − S(x)]N+(x)
x− λ
dx,
then
G(λ) = −P (λ)−N+(λ) +
1
2πi
∫
R
[I − S(x)]N+(x)
x− λ− i0
dx (λ ∈ R)
in view of the Sokhotsky formula [16], therefore
G(λ) = −P (λ)−N(λ) +
1
2πi
∫
R
[I − S(x)]N+(x)
x− λ
dx (λ ∈ C+). (60)
Let λ = iy (y > 0), then the �rst and the third members of this sum converge to
zero as y → +∞ and N(iy) → R (y →∞) in view of 4) (5), and we get B = −R.
Theorem 17. Let f , g ∈ L2
R(E, dF (x)), where F (x) is absolutely continuous
(21), R is of trace-class, ϕ is a Hilbert�Schmidt operator, and f s (κn), f̃(λ),
and gs (κn), g̃(λ), are the corresponding Fourier coe�cients (54), (55) of these
functions.
For the Parseval equality (56) to be ful�lled, it is necessary and su�cient that
the relation
R− P (λ)− 1
2
[I + S(λ)]N+(λ) +
1
2πi
6
∫
R
[I − S(x)]N+(x)
x− λ
dx = 0 (λ ∈ R) (61)
hold, where P (λ) is given by (57); κn are zeroes of D(λ) (37); Mn are the �nite-
dimensional positive invertible operators (47) such that series (50) converges;
N+(λ) are the boundary values on R from C+ of the function N(λ) (4); S(λ)
is the scattering matrix (22).
Corollary 3. Taking into account S(λ)N+(λ) = N−(λ) (35), we write (61) in
the form
R− F (λ)− 1
2
[N+(λ) + N−(λ)] +
1
2πi
6
∫
R
N+(x)−N−(x)
x− λ
dx = 0. (62)
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 29
R. Hatamleh and V.A. Zolotarev
In view of the Sokhotsky formula [16], (49) implies
N±(λ) = R +
∞∑
n=1
M2
n
κn − λ
± πib(λ)+ 6
∫
R
b(x)
x− λ
dx (λ ∈ R). (63)
Substituting N±(λ) (63) into (62), we obtain an identity.
Relation (61) should be considered as an equation with respect to N+(λ) which
is an analogue of the Marchenko equation for the inverse scattering problem [4�7].
Besides, we ought to search solution (61) in the class of Nevanlinna functions 4)
(5).
Let N+(λ) be the solution of equation (61), and N(λ) be its analytic contin-
uation into C+ in view of (60). Continue N(λ) into C− in a natural way, then,
using the Sokhotsky formula and (61), we obtain
N±(λ) = R− F (λ)± 1
2
(I − S(λ))N+(λ) +
1
2πi
6
∫
R
[I − S(x)]N+(x)
x− λ
dx. (64)
This implies the statement.
Theorem 18. Suppose that the suppositions of Theorem 6 are true, N+(λ)
is the solution of equation (61), N(λ) is its holomorphic extension into C+, and
N−(λ) is a boundary value on R from C− after the natural extension of N(λ) into
C−. Then the Riemann boundary value problem S(λ)N+(λ) = N−(λ) (38) takes
place. If
1
2i
(I − S(λ))N+(λ) ≥ 0 (λ ∈ R), (65)
then N−(λ) = N∗
+(λ).
P r o o f. The Riemann problem follows from (64) after the subtraction of
N−(λ) − N+(λ). Condition (65) is natural in view of (32). The fact that (65)
implies N−(λ) = N∗
+(λ) is obvious in view of (64).
5. Inverse Problem
IX. Let dimE = 1 and σ0 be the segment σ0 = [a, b] (−∞ < a < b < ∞).
The discrete spectrum σp (44) is formed by the zeroes of the equation 1 = RV (λ).
Taking into account monotonicity of V (λ) as λ ∈ R\[a, b], we see that the root
1 = RV (κ) is unique since κ ∈ (−∞, a) for R > 0, and κ ∈ (b,∞) for R < 0.
Thus, σp = {κ 6∈ [a, b]}.
Corollary 4. The set σp can be empty depending on the number R and on the
behavior of the representing measure dF (x) = a(x)dx of the function V (λ) (2) at
30 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
the end points of the segment [a, b]. If, for example, a(x) = α(x − a) (α > 0),
then
V (λ) = α(b− a) + α(λ− a)
b∫
a
dx
x− λ
.
The equation 1 = RV (λ) has no roots when R > {α(b − a)}−1 as V (a − 0) =
α(b− a).
In this case, the Riemann boundary value problem (34) can be written as
S(λ)
λ + i
λ− i
λ− κ
λ + i
(1−RV+(λ))−1 =
λ− κ
λ− i
(1−RV−(λ))−1 . (66)
Each function λ− κ
λ± i
(1−RV±(λ))−1 is holomorphically extendable in C± and
has no zeroes and singularities in C±, besides, it tends to 1 at the in�nity. Using
the standard method [16] of the solution of the boundary problem (66) and the
Sokhotsky formulas [16], we obtain
λ− κ
λ + i
(1−RV (λ)) = exp
− 1
2πi
∫
R
ln
[
S(x)
x + i
x− i
]
x− λ
dx
(λ ∈ C+). (67)
Let
ψ(x) =
1
π
arctg
1
x
(x ∈ R,−1
2
≤ ψ(x) ≤ 1
2
). (68)
Then, taking into account (39), we have
S(x)
x + i
x− i
= exp{−2πi(ξ(x)− ψ(x))},
where ξ(x) is the function of spectral shift (39). Equality (67) can be written in
the form
(1−RV (λ))−1 =
λ + i
λ− κ exp
∫
R
ξ(x)− ψ(x)
x− λ
dx
(λ ∈ C+). (69)
The function (1 − RV (λ))−1 maps C+ into C+ as R > 0, and � C− into C− as
R < 0. For de�niteness sake, take R > 0. Then the argument on the right-hand
side of equality (69) belongs to [0, π] as λ ∈ C+. This argument is
arg
λ + i
λ− κ +
∫
R
y
(t− x)2 + y2
(ξ(t)− ψ(t))dt, (70)
Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1 31
R. Hatamleh and V.A. Zolotarev
where λ = x+ iy ∈ C+. Each of the summands in this sum is a harmonic function
in C+, in view of the maximum principle, and its maximum and minimum values
are achieved on the boundary λ = x+ i0 ∈ R. Taking into account the properties
of the Poisson kernel [12], we obtain that the boundary value on R from C+ of
sum (70) is πψ(x)+π(ξ(x)−ψ(x)) = πξ(x). And thus 0 ≤ ξ(x) ≤ 1 for all x ∈ R
when R > 0.
Theorem 19. Let dimE = 1, σ0 and σp (44) be given by σ0 = [a, b]
(−∞ < a < b < ∞) and σp = {κ ∈ R\[a, b]}, and the scattering matrix
S(x) = exp{−2πξ(x)}, where ξ(x) is the function of spectral shift (39).
Then V (λ) (2) is given by
V (λ) = R−1
1− λ− κ
λ + i
exp
∫
R
ψ(x)− ξ(x)
x− λ
dx
(λ ∈ C+), (71)
where
0 ≤ ξ(x) ≤ 1 (R > 0) (−1 ≤ ξ(x) ≤ 0, R < 0), (72)
the function ϕ(x) is given by (68), and R−1 = V (κ).
It should be noticed that in (66) we could choose ia (a > 0) instead of i.
Either for ia or i the �nal result is the same: formula (71) does not depend on a.
By the scattering data, the scattering matrix S(x) = exp{−2πiξ(x)}, and the
number κ ∈ σp, using (71), we construct the Nevanlinna function V (λ) where
the constant R = V (r). Next, knowing V (λ), we �nd its representing measure
dF (x) = a(x)dx, using which we de�ne the space L2
R(E, dF (x)) (18) and the pair
of operators {Q,L} (20). Thus, the Friedrichs�Faddeev model is unambiguously
de�ned by the scattering data {S(x),κ} (20).
X. Let dimE = 1 and σ0 be given by
σ0 = [a, c1] ∪ [c2, c3] ∪ . . . ∪ [cm, b] (73)
(−∞ < a < c1 < c2 < . . . < cm < b < ∞, m ∈ N). Then the discrete
spectrum σp (44) may consist of m points {κk}m
1 that are zeroes of the equation
1 = RV (λ) and are in the lacunas of the set σ0 (73). Besides, there are two
possibilities: either κ1 ∈ (−∞, a), κ2 ∈ (c1, c2),. . . , κm ∈ (cm−1, cm) as R > 0;
or κ1 ∈ (c1, c2), κ2 ∈ (c3, c4), . . . , κm ∈ (b,∞), when R < 0. The roots {κk}m
1 in
the lacunas of the set σ0 (73) depend on the behavior of the representing measure
dF (x) = a(x)dx of the function V (λ) (2) at the extremities of zones, i. e., at the
points a < c1 < . . . < cm < b, as well as on the value of number R.
32 Journal of Mathematical Physics, Analysis, Geometry, 2017, vol. 13, No. 1
On the Abstract Inverse Scattering Problem for Trace Class Perturbations
Similarly to (66), in this case, we write the Riemann boundary value problem
(34) as
S(λ)
(
λ + i
λ− i
)m m∏
k=1
λ− κk
λ + i
(1−RV+(λ))−1 =
m∏
k=1
λ− λk
λ− i
(1−RV−(λ))−1 . (74)
Using again the considerations of the previous section, we obtain the unique so-
lution for this boundary problem,
(1−RV (λ))−1 =
m∏
k=1
λ + i
λ− κk
exp
∫
R
ξ(x)−mϕ(x)
x− λ
dx
(λ ∈ C+), (75)
where ξ(x) is the function of spectral shift (39), and ψ(x) is given by (68). In this
case, Theorem 19 can be formulated as follows.
Theorem 20. Let dimE = 1 and σ0 be given by (73), and the set σp (44) be
σp = {κk ∈ R\σ0, 1 ≤ k ≤ m} . Then the function V (λ) (2) is de�ned unambigu-
ously by the collection {S(x, {κk}m
1 },
V (λ) = R−1
1−
m∏
k=1
λ− κk
λ + i
exp
∫
R
mψ(x)− ξ(x)
x− λ
dx
(λ ∈ C+), (76)
ξ(x) is the function of spectral shift (39) (S(x) = exp{−2πiξ(x)}) and
0 ≤ ξ(x) ≤ 1 (R > 0); (−1 ≤ ξ(x) ≤ 0, R < 0), (77)
where ϕ(x) is given by (68), and R−1 = V (κn) (1 ≤ k ≤ m).
We can conclude that the Friedrichs�Faddeev model {Q,L} (20) is de�ned un-
ambiguously by the scattering matrix S(x) and the collection of numbers {κk}m
1 .
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