Inverse problem for the interior spectral data of the equation of hydrogen atom
We consider the inverse problem for the second-order differential operators with regular singularity and show that the potential function can be uniquely determined by the set of values of eigenfunctions at a certain interior point and parts of two spectra. Розглянуто обернену задачу для диференцiал...
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Український математичний журнал
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| Цитувати: | Inverse problem for the interior spectral data of the equation of hydrogen atom / E.S. Panakhov, M. Sat // Український математичний журнал. — 2012. — Т. 64, № 11. — С. 1516-1525. — Бібліогр.: 21 назв. — англ. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1860041540284448768 |
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| author | Panakhov, E.S. Sat, M. |
| author_facet | Panakhov, E.S. Sat, M. |
| citation_txt | Inverse problem for the interior spectral data of the equation of hydrogen atom / E.S. Panakhov, M. Sat // Український математичний журнал. — 2012. — Т. 64, № 11. — С. 1516-1525. — Бібліогр.: 21 назв. — англ. |
| collection | DSpace DC |
| container_title | Український математичний журнал |
| description | We consider the inverse problem for the second-order differential operators with regular singularity and show that the potential function can be uniquely determined by the set of values of eigenfunctions at a certain interior point and parts of two spectra.
Розглянуто обернену задачу для диференцiальних операторiв другого порядку з регулярною сингулярнiстю та показано, що потенцiальна функцiя однозначно визначається множиною значень власних функцiй у деякiй внутрiшнiй точцi та частинами двох спектрiв.
|
| first_indexed | 2025-12-07T16:55:43Z |
| format | Article |
| fulltext |
UDC 517.9
E. S. Panakhov (Firat Univ., Elazig, Turkey),
M. Sat (Erzincan Univ., Turkey)
INVERSE PROBLEM FOR INTERIOR SPECTRAL DATA
OF THE HYDROGEN ATOM EQUATION
ОБЕРНЕНА ЗАДАЧА ДЛЯ ВНУТРIШНIХ СПЕКТРАЛЬНИХ ДАНИХ
РIВНЯННЯ АТОМА ВОДНЮ
We consider the inverse problem for second-order differential operators with regular singularity and show that the potential
function can be uniquely determined by the set of values of eigenfunctions at some interior point and parts of two spectra.
Розглянуто обернену задачу для диференцiальних операторiв другого порядку з регулярною сингулярнiстю та пока-
зано, що потенцiальна функцiя однозначно визначається множиною значень власних функцiй у деякiй внутрiшнiй
точцi та частинами двох спектрiв.
1. Introduction. The inverse Sturm – Liouville problem is primarily a model problem. Typically,
in an inverse eigenvalue problem, one measures the frequencies of a vibration system and tries
to infer some physical properties of the system. Inverse problems of spectral analysis involve the
reconstruction of a linear operator from its spectral characteristics [1, 2]. A problem of this kind was
first investigated by Ambarzumyan in 1929 [7]. Later, inverse problems for a regular and singular
Sturm – Liouville operator appeared in various versions [3 – 14].
The inverse problem for interior spectral data of the differential operator consists in reconstruction
of this operator from the known eigenvalues and some information on eigenfunctions at some internal
point. The technique employed is similar to those used in [9]. Similar problems for the Sturm –
Liouville operator and Dirac operator were formulated and studied in [10].
The main goal of the present work is to study the inverse problem of reconstructing the singular
Sturm – Liouville operator on the basis of spectral data of a kind: one spectrum and some information
on eigenfunctions at the internal point.
Consider the following singular Sturm – Liouville operator L satisfying
Ly = −y′′ +
[
` (`+ 1)
x2
− 2
x
+ q(x)
]
y = λy, 0 < x < π, (1.1)
with boundary conditions,
y(0) = 0, (1.2)
y′(π, λ) +Hy(π, λ) = 0, (1.3)
where q(x) is assumed to be real valued and square integrable, λ spectral parameter, ` ∈ N0, and H
finite real number. The operator L is self adjoint on the L2 (0, π) and (1.2), (1.3) boundary conditions
has a discrete spectrum {λn} .
Let us introduce the second singular Sturm – Liouville operator L̃ satisfying
L̃y = −y′′ +
[
` (`+ 1)
x2
− 2
x
+ q̃(x)
]
y = λy, 0 < x < π, (1.4)
c© E. S. PANAKHOV, M. SAT, 2012
1516 ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
INVERSE PROBLEM FOR INTERIOR SPECTRAL DATA OF THE HYDROGEN ATOM EQUATION 1517
subject to the same boundary conditions (1.2), (1.3), where q̃(x) is assumed to be real valued and
square integrable. The operator L̃ is self adjoint on the L2 (0, π) and (1.2), (1.3) boundary conditions
has a discrete spectrum
{
λ̃n
}
.
2. Main results. Before giving main results of this article, we mention some known results. We
will consider the equation
R′′ +
2
x
R′ − ` (`+ 1)
x2
R+
(
E +
2
x
)
R = 0, 0 < x <∞. (2.1)
In quantum mechanics, the study of the energy levels of the hydrogen atom leads to this equation
[16]. The substitution R =
y
x
reduces this equation to the form
y′′ +
[
E +
2
x
− ` (`+ 1)
x2
]
y = 0. (2.2)
As known [17 – 19] the solution of (2.2) is bounded at zero, one has the following asymptotic formula
for λ→∞
y (x) =
eπ/2
√
λ∣∣∣∣Γ(`+ 1 +
i√
λ
)∣∣∣∣
1√
λ
cos
[√
λx+
1√
λ
ln
√
λx− (`+ 1)
π
2
+ α
]
+ o (1),
where
α = arg Γ
(
`+ 1 +
i√
λ
)
.
Eigenvalues of the problem (1.1) – (1.3) are the roots of the (1.3). These spectral characteristics
and eigenfunctions satisfy the following asymptotic expression, respectively [18]:
ρn =
√
λn = n+
`
2
+O
(
lnn
n
)
, (2.3)
ϕ (x, λn) = cos
[(
n+
`
2
)
x− `π
2
]
+O
(
lnn
n
)
, (2.4)
ϕ′ (x, λn) = −
(
n+
`
2
)
sin
[(
n+
`
2
)
x− `π
2
]
+O
(
lnn
n
)
. (2.5)
Next, we present the main results in this article. When b =
π
2
, we get the following uniqueness
theorem.
Theorem 2.1. If for every n ∈ N we have
λn = λ̃n,
y′n
(π
2
)
yn
(π
2
) =
ỹ′n
(π
2
)
ỹn
(π
2
) , (2.6)
then
q(x) = q̃(x) a.e. on the interval (0, π).
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
1518 E. S. PANAKHOV, M. SAT
In the case b 6= π
2
, the uniqueness of q(x) can be proved if we require the knowledge of a part
of the second spectrum.
Let m(n) be a sequence of natural numbers with a property
m(n) =
n
σ
(1 + εn), 0 < σ ≤ 1, εn → 0. (2.7)
Lemma 2.1. Let m(n) be a sequence of natural numbers satisfying (2.7) and b ∈
(
0,
π
2
)
are
so chosen that σ >
2b
π
. If for any n ∈ N
λm(n) = λ̃m(n),
y′m(n)(b)
ym(n)(b)
=
ỹ′m(n)(b)
ỹm(n)(b)
, (2.8)
then
q(x) = q̃(x) a.e. on (0, b].
Let l(n) and r(n) be a sequence of natural numbers such that
l(n) =
n
σ1
(1 + ε1,n), 0 < σ1 ≤ 1, ε1,n → 0, (2.9)
r(n) =
n
σ2
(1 + ε2,n), 0 < σ2 ≤ 1, ε2,n → 0, (2.10)
and let µn be the eigenvalues of the problem (1.1), (1.2 ) and (2.11) and µ̃n be the eigenvalues of the
problem (1.4), (1.2) and (2.11)
y′(π, λ) +H1y(π, λ) = 0, H 6= H1. (2.11)
Using Mochizuki and Trooshin’s method from Lemma 2.1 and Theorem 2.1, we will prove that the
following theorem holds.
Theorem 2.2. Let l(n) and r(n) be a sequence of natural numbers satisfying (2.9) and (2.10),
and
π
2
< b < π are so chosen that σ1 >
2b
π
− 1, σ2 > 2− 2b
π
. If for any n ∈ N we have
λn = λ̃n, µl(n) = µ̃l(n) and
y′r(n)(b)
yr(n)(b)
=
ỹ′r(n)(b)
ỹr(n)(b)
, (2.12)
then
q(x) = q̃(x) a.e. on (0, π).
3. Proof of the main results. In this section we present the proofs of main results in this paper.
Proof of Theorem 2.1. Before proving the Theorem 2.1, we will mention some results, which
will be needed later. We get the initial value problems
−y′′ +
[
` (`+ 1)
x2
− 2
x
+ q(x)
]
y = λy, 0 < x < π, (3.1)
y(0) = 0, (3.2)
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
INVERSE PROBLEM FOR INTERIOR SPECTRAL DATA OF THE HYDROGEN ATOM EQUATION 1519
and
−ỹ′′ +
[
` (`+ 1)
x2
− 2
x
+ q̃(x)
]
ỹ = λỹ, 0 < x < π, (3.3)
ỹ(0) = 0. (3.4)
It can be shown [19] that there exists a kernel K (x, t)
(
K̃ (x, t)
)
continuous on (0, π)× (0, π)
such that by using the transformation operator every solution of equations (3.1), (3.2 ) and (3.3), (3.4)
can be expressed in the form
y(x, λ) = cos
[(
n+
`
2
)
x− `π
2
]
+
x∫
0
K (x, t) cos
[(
n+
`
2
)
t− `π
2
]
dt, (3.5)
ỹ(x, λ) = cos
[(
n+
`
2
)
x− `π
2
]
+
x∫
0
K̃ (x, t) cos
[(
n+
`
2
)
t− `π
2
]
dt, (3.6)
respectively, where the kernel K (x, t)
(
K̃ (x, t)
)
is the solution of the problem
∂2K (x, t)
∂x2
−
(
2
x
− ` (`+ 1)
x2
+ q̃(x)
)
K (x, t) =
∂2K (x, t)
∂t2
−
(
2
t
− ` (`+ 1)
t2
+ q(t)
)
K (x, t)
subject to the boundary conditions
K (x, x) =
1
2
x∫
0
[
q̃(t)− q(t)
]
,
K (x, 0) = 0.
After the transformations
z =
1
4
(x+ t)2 , w =
1
4
(x− t)2 , K (x, t) = (z − w)−v+(1/2) u (z, w) ,
we obtain the following problem
(
α = −v +
1
2
)
:
∂2u
∂z∂w
− α
z − w
∂u
∂z
+
α
z − w
∂u
∂w
=
(q̃ − q)u
4
√
zw
− u√
z (z − w)
,
u(z, z − δ) = 0,
∂u
∂z
+
α
z
u =
1
4
[
q̃
(√
z
)
− q
(√
z
)]
zv−1,
for a constant δ. This problem can be solved by using the Riemann method [20, 21].
Multiplying (3.1) by ỹ (x, λ) and ( 3.3) by y (x, λ) , subtracting and integrating from 0 to
π
2
, we
obtain
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
1520 E. S. PANAKHOV, M. SAT
π/2∫
0
(q (x)− q̃ (x)) y (x, λ) ỹ(x, λ)dx =
[
ỹ(x, λ)y′(x, λ)− y(x, λ)ỹ′(x, λ)
] ∣∣∣π/2
0
. (3.7)
The functions y (x, λ) and ỹ (x, λ) satisfy the same initial conditions (3.2) and ( 3.4), i.e.,
ỹ(0, λ)y′(0, λ)− y(0, λ)ỹ′(0, λ) = 0.
Let
Q(x) = q (x)− q̃ (x) , (3.8)
H(λ) =
π/2∫
0
Q (x) y (x, λ) ỹ(x, λ)dx. (3.9)
If the properties of y (x, λ) and ỹ (x, λ) are considered, the function H(λ) is an entire function.
Therefore the condition of the Theorem 2.1 imply
ỹ
(π
2
, λn
)
y′
(π
2
, λn
)
− y
(π
2
, λn
)
ỹ′
(π
2
, λn
)
= 0,
and hence
H (λn) = 0, n ∈ N.
In addition, using (3.5) and (3.9) for 0 < x ≤ π,∣∣H(λ)
∣∣ ≤ M
λ
, (3.10)
where M is constant.
Introduce the function
ω(λ) = y′(π, λ) +Hy(π, λ). (3.11)
By using the asymptotic forms of ϕ and ϕ′, we obtain
ω(λ) = −
(
n+
`
2
)
sin
[(
n+
`
2
)
− `π
2
]
+O
(
lnn
n
)
. (3.12)
The zeros of ω(λ) are the eigenvalues of L and hence it has only simple zeros λn because of the
seperated boundary conditions. It is an entire function of order
1
2
of λ. From this and from the
asymptotics for ω(λ) and H(λ), it follows that the function
Ψ(λ) =
H(λ)
ω (λ)
(3.13)
is an entire function. Asymptotic form of ω(λ) and with equation (3.13), we get∣∣Ψ(λ)
∣∣ = O
(
π√
λ
)
.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
INVERSE PROBLEM FOR INTERIOR SPECTRAL DATA OF THE HYDROGEN ATOM EQUATION 1521
So, for all λ, from the Liouville theorem,
Ψ(λ) = 0,
or
H(λ) = 0.
It was proved in [19] that there exists absolutely continuous function ˜̃K (x, τ) such that, we have
y (x, λ) ỹ (x, λ) =
=
1
2
1 + cos 2
[(
n+
`
2
)
x− `π
2
]
+
x∫
0
˜̃
K (x, τ) cos 2
[(
n+
`
2
)
τ − `π
2
]
dτ
, (3.14)
where ˜̃
K (x, t) = 2
[
K (x, x− 2τ) + K̃ (x, x− 2τ)
]
+
+2
x∫
−x+2τ
K (x, s) K̃ (x, s− 2τ) ds+
x−2τ∫
−x
K (x, s) K̃ (x, s+ 2τ) ds
.
We are now going to show that Q(x) = 0 a.e. on
(
0,
π
2
]
. From (3.9), (3.14) we obtain
1
2
π/2∫
0
Q(x)
1 + cos 2
[(
n+
`
2
)
x− `π
2
]
+
x∫
0
˜̃
K (x, τ) cos 2
[(
n+
`
2
)
τ − `π
2
]
dτ
dx = 0.
This can be written as
π/2∫
0
Q(x)dx+
π/2∫
0
cos 2
[(
n+
`
2
)
τ − `π
2
]Q(τ) +
π/2∫
τ
Q(x)
˜̃
K (x, τ) dx
dτ = 0.
Let λ→∞ along the real axis, by the Riemann – Lebesgue lemma, we should have
π/2∫
0
Q(x)dx = 0, (3.15)
and
π/2∫
0
cos 2
[(
n+
`
2
)
τ − `π
2
]Q(τ) +
π/2∫
τ
Q(x)
˜̃
K (x, τ) dx
dτ = 0. (3.16)
Thus from the completeness of the functions cos, it follows that
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
1522 E. S. PANAKHOV, M. SAT
Q(τ) +
π/2∫
τ
Q(x)
˜̃
K (x, τ) dx = 0, 0 < x <
π
2
. (3.17)
But this equation is a homogeneous Volterra integral equation and has only the zero solution. Thus
we have obtained
Q (x) = q (x)− q̃ (x) = 0,
or
q̃ (x) = q (x) ,
almost everywhere on
(
0,
π
2
]
.
To prove that q (x) = 0 on
[π
2
, π
)
almost everywhere, we should repeat the above arguments for
the supplementary problem
Ly = −y′′ +
[
` (`+ 1)
(π − x)x2
− 2
π − x
+ q(π − x)
]
y, 0 < x < π,
subject to the boundary conditions
y(π) = 0,
y′(0, λ) +Hy(0, λ) = 0,
where
q0(x) =
` (`+ 1)
x2
− 2
x
+ q1(x).
Consequently
q(x) = q̃(x) a.e. on the interval (0, π).
Therefore, Theorem 2.1 is proved.
Next, we show that Lemma 2.1 holds.
Proof of Lemma 2.1. As in the proof of Theorem 2.1 we can show that
G(ρ) =
b∫
0
Q(x)y (x, λ) ỹ(x, λ)dx =
[
ỹ(x, λ)y′(x, λ)− y(x, λ)ỹ′(x, λ)
] ∣∣∣
x=b
, (3.18)
where ρ =
√
λ =reiθ and Q (x) = q (x)− q̃ (x) . From the assumption
y′m(n)(b)
ym(n)(b)
=
ỹ′m(n)(b)
ỹm(n)(b)
,
together with the initial condition at 0 it follows that,
G(ρm(n)) = 0, n ∈ N.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
INVERSE PROBLEM FOR INTERIOR SPECTRAL DATA OF THE HYDROGEN ATOM EQUATION 1523
Next, we will show that G(ρ) = 0 on the whole ρ plane. The asymptotics (2.4), (2.5) imply that the
entire function G(ρ) is a function of exponential type ≤ 2b.
Define the indicator of function G(ρ) by;
h(θ) = lim
r→∞
sup
ln
∣∣G(reiθ)
∣∣
r
. (3.19)
Since
∣∣Im√λ ∣∣ = r |sin θ| , θ = arg
√
λ from (2.4) and (2.5) it follows that
h(θ) ≤ 2b |sin θ| . (3.20)
Let us denote by n(r) the number of zeros of G(ρ) in the disk {|ρ| ≤ r}. According to the [15] set
of zeros of every entire function of the exponential type, not identically zero, satisfies the inequality
lim
r→∞
inf
n(r)
r
≤ 1
2π
2π∫
0
h(θ)dθ, (3.21)
where n(r) is the number of zeros of G(ρ) in the disk |ρ| ≤ r. By (3.20),
1
2π
2π∫
0
h(θ)dθ ≤ b
π
2π∫
0
|sin θ| dθ =
4b
π
.
From the assumption and the known asymptotic expression (2.3) of the eigenvalues
√
λn we obtain
n(r) ≥ 2
∑
n
σ
[
1+O
(
lnn
n
)]
<r
1 = 2σr
(
1 + o (1)
)
, r →∞.
For the case σ >
2b
π
,
lim
r→∞
n(r)
r
≥ 2σ >
4b
π
= 2b
2π∫
0
|sin θ| dθ ≥ 1
2π
2π∫
0
h(θ)dθ. (3.22)
The inequalities (3.21) and (3.22) imply that G(ρ) = 0 on the whole ρ plane.
Similar to the proof of the Theorem 2.1, we have
q(x) = q̃(x) a.e. on the interval (0, b].
Lemma 2.1 is proved.
Now we prove that Theorem 2.2 is valid.
Proof of Theorem 2.2. From
λr(n) = λ̃r(n),
y′r(n)(b)
yr(n)(b)
=
ỹ′r(n)(b)
ỹr(n)(b)
,
where r(n) satisfies (2.10) and σ2 > 2− 2b
π
according to Lemma 2.1, we get
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
1524 E. S. PANAKHOV, M. SAT
q(x) = q̃(x) a.e. on [b, π). (3.23)
Thus, it needs to be proved that q(x) = q̃(x) a.e. on (0, b]. The eigenfunctions yn(x, λn) and
ỹn(x, λn) satisfy the same boundary condition at π. It means that
yn(x, λn) = ξnỹn(x, λn) (3.24)
on [b, π) for any n ∈ N where ξn are constants.
From (3.18) and (3.24) we obtain that,
G(ρ) = 0, for ρ2 = λn, n ∈ N,
and
G(ρ) = 0, for ρ2 = µl(n), n ∈ N.
W are going to show that inequality (3.21) fails and consequently, the entire function of expo-
nential type G(ρ) vanishes on the whole ρ-plane. Let ρn =
√
λn , sn =
√
µn. The ρn and sn have
the same asymptotics (2.3). Counting the number of ρn and sn located inside the disc of radius r, we
have
1 + 2r
[
1 +O
(
lnn
n
)]
of ρn’s and
1 + 2rσ1
[
1 +O
(
lnn
n
)]
of sn’s.
This means that
n(r) = 2 + 2
[
r(σ1 + 1) +O
(
lnn
n
)]
and
lim
r→∞
n(r)
r
= 2(σ1 + 1).
Repeating the last part of the proof of Lemma 2.1, and considering the condition σ1 >
2b
π
− 1, we
can show that G(ρ) = 0 identically on the whole ρ-plane which implies that
q(x) = q̃(x) a.e. on (0, b]
and consequently
q(x) = q̃(x) a.e. on (0, π).
Theorem 2.2 is proved.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
INVERSE PROBLEM FOR INTERIOR SPECTRAL DATA OF THE HYDROGEN ATOM EQUATION 1525
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Received 06.08.12
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 11
|
| id | nasplib_isofts_kiev_ua-123456789-165404 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1027-3190 |
| language | English |
| last_indexed | 2025-12-07T16:55:43Z |
| publishDate | 2012 |
| publisher | Український математичний журнал |
| record_format | dspace |
| spelling | Panakhov, E.S. Sat, M. 2020-02-13T12:20:39Z 2020-02-13T12:20:39Z 2012 Inverse problem for the interior spectral data of the equation of hydrogen atom / E.S. Panakhov, M. Sat // Український математичний журнал. — 2012. — Т. 64, № 11. — С. 1516-1525. — Бібліогр.: 21 назв. — англ. 1027-3190 https://nasplib.isofts.kiev.ua/handle/123456789/165404 517.9 We consider the inverse problem for the second-order differential operators with regular singularity and show that the potential function can be uniquely determined by the set of values of eigenfunctions at a certain interior point and parts of two spectra. Розглянуто обернену задачу для диференцiальних операторiв другого порядку з регулярною сингулярнiстю та показано, що потенцiальна функцiя однозначно визначається множиною значень власних функцiй у деякiй внутрiшнiй точцi та частинами двох спектрiв. en Український математичний журнал Український математичний журнал Статті Inverse problem for the interior spectral data of the equation of hydrogen atom Обернена задача для внутрiшнiх спектральних даних рiвняння атома водню Article published earlier |
| spellingShingle | Inverse problem for the interior spectral data of the equation of hydrogen atom Panakhov, E.S. Sat, M. Статті |
| title | Inverse problem for the interior spectral data of the equation of hydrogen atom |
| title_alt | Обернена задача для внутрiшнiх спектральних даних рiвняння атома водню |
| title_full | Inverse problem for the interior spectral data of the equation of hydrogen atom |
| title_fullStr | Inverse problem for the interior spectral data of the equation of hydrogen atom |
| title_full_unstemmed | Inverse problem for the interior spectral data of the equation of hydrogen atom |
| title_short | Inverse problem for the interior spectral data of the equation of hydrogen atom |
| title_sort | inverse problem for the interior spectral data of the equation of hydrogen atom |
| topic | Статті |
| topic_facet | Статті |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/165404 |
| work_keys_str_mv | AT panakhoves inverseproblemfortheinteriorspectraldataoftheequationofhydrogenatom AT satm inverseproblemfortheinteriorspectraldataoftheequationofhydrogenatom AT panakhoves obernenazadačadlâvnutrišnihspektralʹnihdanihrivnânnâatomavodnû AT satm obernenazadačadlâvnutrišnihspektralʹnihdanihrivnânnâatomavodnû |