Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction
In the paper, the spectral problem generated by the Sturm-Liouville equation -y'' + q(x)y = (λ² - ip(x)λ)y, where q(x) is a real L₂(0, a)-function and p(x) is a peace-wise constant, is considered with the Dirichlet boundary conditions at the ends of the interval (0, a). The spectrum of the...
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2012
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Kobyakova, L. 2016-10-03T16:14:03Z 2016-10-03T16:14:03Z 2012 Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction / L. Kobyakova // Журнал математической физики, анализа, геометрии. — 2012. — Т. 8, № 3. — С. 280-295. — Бібліогр.: 19 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106724 In the paper, the spectral problem generated by the Sturm-Liouville equation -y'' + q(x)y = (λ² - ip(x)λ)y, where q(x) is a real L₂(0, a)-function and p(x) is a peace-wise constant, is considered with the Dirichlet boundary conditions at the ends of the interval (0, a). The spectrum of the problem is compared with the spectra of auxiliary problems with the Dirichlet-Dirichlet and the Dirichlet-Neumann boundary conditions on the halves of the interval. Asymptotic formulas are obtained for the eigenvalues of this problem. В статье рассматривается спектральная задача, порожденная уравнением Штурма-Лиувилля -y'' + q(x)y = (λ² - ip(x)λ)y, где q(x) - вещественная L₂(0, a)-функция, а p(x) является кусочно-постоянной, с краевыми условиями Дирихле на концах интервала (0, a). Спектр данной задачи сравнивается со спектром вспомогательной задачи с краевыми условиями Дирихле-Дирихле и Дирихле-Неймана на полуинтервалах. Получены асимптотические формулы для собственных значений задачи. The author is thankful to professor V.N. Pivovarchik for the attention to the work during the time of its writing. en Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України Журнал математической физики, анализа, геометрии Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction Article published earlier |
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Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction |
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Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction Kobyakova, L. |
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Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction |
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Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction |
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Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction |
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Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction |
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spectral problem generated by the equation of smooth string with piece-wise constant friction |
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Kobyakova, L. |
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Kobyakova, L. |
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2012 |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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In the paper, the spectral problem generated by the Sturm-Liouville equation -y'' + q(x)y = (λ² - ip(x)λ)y, where q(x) is a real L₂(0, a)-function and p(x) is a peace-wise constant, is considered with the Dirichlet boundary conditions at the ends of the interval (0, a). The spectrum of the problem is compared with the spectra of auxiliary problems with the Dirichlet-Dirichlet and the Dirichlet-Neumann boundary conditions on the halves of the interval. Asymptotic formulas are obtained for the eigenvalues of this problem.
В статье рассматривается спектральная задача, порожденная уравнением Штурма-Лиувилля -y'' + q(x)y = (λ² - ip(x)λ)y, где q(x) - вещественная L₂(0, a)-функция, а p(x) является кусочно-постоянной, с краевыми условиями Дирихле на концах интервала (0, a). Спектр данной задачи сравнивается со спектром вспомогательной задачи с краевыми условиями Дирихле-Дирихле и Дирихле-Неймана на полуинтервалах. Получены асимптотические формулы для собственных значений задачи.
|
| issn |
1812-9471 |
| url |
https://nasplib.isofts.kiev.ua/handle/123456789/106724 |
| citation_txt |
Spectral Problem Generated by the Equation of Smooth String with Piece-Wise Constant Friction / L. Kobyakova // Журнал математической физики, анализа, геометрии. — 2012. — Т. 8, № 3. — С. 280-295. — Бібліогр.: 19 назв. — англ. |
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2025-11-25T21:42:07Z |
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1850560159852527616 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2012, vol. 8, No. 3, pp. 280–295
Spectral Problem Generated by the Equation of Smooth
String with Piece-Wise Constant Friction
L. Kobyakova
South-Ukrainian National Pedagogical University
26 Staroportofrankivska St., Odesa, 65020, Ukraine
E-mail: Luda Kobyakova@rambler.ru
Received October 3, 2011
In the paper, the spectral problem generated by the Sturm–Liouville
equation
−y′′ + q(x)y = (λ2 − ip(x)λ)y,
where q(x) is a real L2(0, a)-function and p(x) is a peace-wise constant, is
considered with the Dirichlet boundary conditions at the ends of the interval
(0, a). The spectrum of the problem is compared with the spectra of auxiliary
problems with the Dirichlet–Dirichlet and the Dirichlet–Neumann boundary
conditions on the halves of the interval. Asymptotic formulas are obtained
for the eigenvalues of this problem.
Key words: spectral problem, Sturm–Liouville equation, eigenvalues.
Mathematics Subject Classification 2010: 34B08, 47A75.
1. Introduction
The equation for the transverse displacement u(s, t) of an inhomogeneous
string is of the form [1],
d2u(s, t)
ds2
− ρ(s)
d2u(s, t)
dt2
= 0, (1.1)
where ρ(s) is the linear density of the string, s is a spatial coordinate, t is time.
It should be noted that the most general case of the string is considered in [2].
Substituting u(s, t) = v(λ, s)eiλt into equation (1.1), we obtain the equation
for the amplitude function v(λ, s),
d2v
ds2
+ λ2ρ(s)v(λ, s) = 0, (1.2)
where λ is the spectral parameter.
c© L. Kobyakova, 2012
Spectral Problem Generated by the Equation of Smooth String
It is known that if the density ρ(s) is twice differentiable (let, e.g., ρ(s) ∈
W 2
2 (0, l)) and ρ(s) ≥ ε > 0 for all s ∈ [0, l], l is the length of the string, then
applying the Liouville transformation [3]
x(s) =
s∫
0
ρ(s′)1/2 ds′,
y(λ, x) = ρ[x]1/2v(λ, s(x)), ρ[x] := ρ(s(x)),
equation (1.1) can be reduced to the Sturm–Liouville equation
−y′′ + q(x)y = λ2y, x ∈ (0, a), (1.3)
where q(x) = ρ[x]−1/4 d2
dx2 (ρ[x]1/4), a =
l∫
0
ρ(s)1/2 ds.
Numerous amounts of literature is devoted to boundary problems generated
by equation (1.3) (see, e.g., [1, 4–6], etc.).
Physically motivated is the consideration of the problem on the vibrations of
a string subject to viscous friction (damping). Pioneering papers [7–9] dealt with
the problem describing pointwise friction. Further the theory of such problems
was developed in [10, 11].
In the case of a distributed friction we have
d2u
ds2
+ (λ2ρ(s)− iλp(s))u = 0 (1.4)
instead of (1.2) with real ρ(s) and nonnegative p(s).
In this case, if ρ(s) ∈ W 2
2 (0, l), ρ(s) ≥ ε > 0 and p(s) ∈ L2(0, l), the Liouville
transformation is used to obtain
−y′′ + q(x)y = (λ2 − iλp(x))y. (1.5)
The problem generated by equation (1.5) on the semi-axis was considered in
[12]. The corresponding inverse problem, namely, the problem on finding a pair
q and p by using the scattering data, appeared to be more complicated than the
classical inverse Sturm–Liouville problem on the semi-axis. It is natural, since one
of the operators included in the corresponding operator pencil is antisymmetric
(skew-symmetric).
In parallel, there was developed the theory of direct and inverse problems
generated by the diffusion equation
−y′′ + q(x)y = (λ2 − λp(x))y, (1.6)
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3 281
L. Kobyakova
where p(x) and q(x) are real. In this case the corresponding quadratic operator
pencil contains only self-adjoint and symmetric operators and the problem turns
out to be easier. Here the papers [13–15] should be mentioned.
The direct problems generated by equation (1.5) on the finite interval were
considered in [16], where the separability of the spectrum from the real axis was
analyzed. However, not putting essential restrictions on p(x), we can obtain only
rough asymptotics of eigenvalues.
The present paper studies the direct boundary value problem generated by
equation (1.5) with
p(x) =
{
p1 = const, x ∈ [0, a/2];
p2 = const, x ∈ [a/2, a],
where p1 > 0, p2 > 0, a is the length of the interval, and by the Dirichlet–
Dirichlet boundary conditions
y(λ, 0) = 0, y(λ, a) = 0.
In Sec. 2, the boundary value problem generated by equation (1.6) with
Dirichlet–Dirichlet conditions at both ends (the main problem) is given. It is
natural to consider two pairs of the auxiliary problems on half-intervals with the
Dirichlet–Neumann and Dirichlet–Dirichlet boundary conditions at the ends to
compare their spectra with the spectrum of the main problem.
In Sec. 3, the operator theoretical approach to the problem is given. It is
proved that in certain domains the eigenvalues of the main problem interlace
with the elements of the union of spectra of the auxiliary problems.
In Sec. 4, we derive asymptomatic formulas for the eigenvalues of the main
problem.
2. Statement of the Problems
For the sake of convenience, let us divide the interval [0, a] into two parts and
measure the distance on the left side of the interval from the left to the right, and
on the right side from the right to the left. Then our boundary value problem
takes the form
−y′′1 + q1(x)y1 = (λ2 − ip1λ)y1, for x ∈ [0, a/2]; (2.1)
−y′′2 + q2(x)y2 = (λ2 − ip2λ)y2, for x ∈ [0, a/2]; (2.2)
y1(λ, 0) = 0, y2(λ, 0) = 0, (2.3)
where the matching conditions at the midpoint a/2 of the interval are
y1(λ, a/2) = y2(λ, a/2), (2.4)
y′1(λ, a/2) = −y′2(λ, a/2), (2.5)
282 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3
Spectral Problem Generated by the Equation of Smooth String
and qj(x) ∈ L2[0, a/2], pj > 0 for j = 1, 2. For definiteness, we will assume that
p1 < p2.
In parallel with the main problem we will consider the pairs of the boundary
value problem related to it:
I. Dirichlet–Neumann problem (the Dirichlet condition on the left end and
the Neumann condition on the right end) on the half-intervals
−y′′1 + q1(x)y1 = (λ2 − ip1λ)y1, (2.6)
y1(λ, 0) = 0, y′1(λ, a/2) = 0; (2.7)
and
−y′′2 + q2(x)y2 = (λ2 − ip2λ)y2, (2.8)
y2(λ, 0) = 0, y′2(λ, a/2) = 0. (2.9)
II. Dirichlet–Dirichlet problem (the Dirichlet conditions at both ends) on the
half-intervals
−y′′1 + q1(x)y1 = (λ2 − ip1λ)y1, (2.10)
y1(λ, 0) = 0, y1(λ, a/2) = 0; (2.11)
and
−y′′2 + q2(x)y2 = (λ2 − ip2λ)y2, (2.12)
y2(λ, 0) = 0, y2(λ, a/2) = 0. (2.13)
Let us look for the solution of problem (2.1)–(2.5) in the form yj(λ, x) =
MjSj(
√
λ2 − ipjλ, x), where Mj are constants, Sj(
√
λ2 − ipjλ, x) are the solu-
tions of equations (2.1) and (2.2) satisfying the following initial conditions:
Sj(
√
λ2 − ipjλ, 0) = 0, S′j(
√
λ2 − ipjλ, 0) = 1, j = 1, 2. (2.14)
Matching conditions (2.4), (2.5) at the midpoint of the interval imply
M1S1(
√
λ2 − ip1λ, a/2) = M2S2(
√
λ2 − ip2λ, a/2),
M1S
′
1(
√
λ2 − ip1λ, a/2) = −M2S
′
2(
√
λ2 − ip2λ, a/2).
(2.15)
The system of equations (2.15) with respect to the unknown M1 and M2 pos-
sesses a nontrivial solution if its determinant, which is said to be the characteristic
function ϕ(λ) of problem (2.1)–(2.5), equals zero
ϕ(λ) = S1(
√
λ2 − ip1λ, a/2) · S′2(
√
λ2 − ip2λ, a/2) +
+S2(
√
λ2 − ip2λ, a/2) · S′1(
√
λ2 − ip1λ, a/2) = 0. (2.16)
It is clear that the set of zeros of the function ϕ(λ) is the spectrum of problem
(2.1)–(2.5).
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3 283
L. Kobyakova
3. Operator Theoretical Approach to the Problem
Here and further in the paper we will use the following definitions:
Definition 1 (see, e.g., [17]).
1) A number λ0 ∈ C is said to be an eigenvalue of the pencil L(λ) if there
exists a vector y0 ∈ D(L) (called an eigenvector of L(λ)) such that y0 6= 0 and
L(λ0)y0 = 0.
2) The vectors y1, y2, . . . , ym−1 are called associated to y0 if
k∑
s=1
1
s!
dsL(λ)
dλs
∣∣∣
λ=λ0
yk−s = 0, k = 1,m− 1.
The number m is said to be the length of the chain composed of the eigen- and
associated vectors.
3) The geometric multiplicity of an eigenvalue is defined to be the number of
the corresponding linearly independent eigenvectors.
4) The algebraic multiplicity of an eigenvalue is defined to be the greatest
value of the sum of the lengths of chains corresponding to linearly independent
eigenvectors.
5) If algebraic and geometric multiplicity of an eigenvalue coincide, we call it
semisimple.
6) An eigenvalue is said to be isolated if it has some deleted neighbourhood
contained in the resolvent set. An isolated eigenvalue λ0 of finite algebraic mul-
tiplicity is said to be normal if the image ImL(λ0) is closed.
Let us introduce the operators A and K acting in the Hilbert space
H = L2(0, a/2)⊕ L2(0, a/2),
A
( y1
y2
)
=
( −y′′1 + q1(x)y1
−y′′2 + q2(x)y2
)
,
D(A) =
{( y1
y2
)
:
y1, y2 ∈ W 2
2 (0, a/2), y1(0) = y2(0) = 0;
y1(a/2) = y2(a/2), y′1(a/2) = −y′2(a/2),
}
,
where W 2
2 (0, a/2) is the corresponding Sobolev space, and
K =
(
p1I 0
0 p2I
)
, D(K) = H,
where I is the identity operator in L2(0, a/2).
It is obvious that the operator K is strictly positive, i.e., K ≥ p1I > 0.
Let us denote
I =
(
I 0
0 I
)
284 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3
Spectral Problem Generated by the Equation of Smooth String
and consider the operator pencil
L(λ) = λ2I− iλK −A,
acting in the space H with the domain
D(L) = D(I) ∩D(K) ∩D(A) = D(A)
not depending on spectral parameter. It is natural to identify the spectrum of
the main problem with the spectrum of L(λ).
It is known that the operator A is self-adjoint and bounded below, i.e., there
exists a number β > 0 such that the inequality A + βI ≥ 0 is true and for some
β1 > β the operator A+β1I possesses a compact resolvent. This implies that the
number of negative eigenvalues of the operator A is finite. Since the operators K
and I are bounded, the spectrum of the pencil L consists of normal eigenvalues
which accumulate only to +∞.
Theorem 1.
1) The eigenvalues of problem (2.1)–(2.5) lie in the strip Imλ ∈
[
p1
2 ; p2
2
]
and
on the intervals (−i∞; ip1
2 ) or ( ip2
2 ; +i∞) of the imaginary axis.
2) Non-pure imaginary eigenvalues of problem (2.1)–(2.5) are symmetrical
with respect to the imaginary axis, i.e., λk = −λk, and algebraic multiplicities of
the symmetrically located eigenvalues coincide.
3) The number of pure imaginary eigenvalues is finite.
P r o o f. 1) Let us transform the spectral parameter τ = λ− ip1
2 . Then we
obtain the quadratic operator pencil
L1(τ) = τ2I− iτK1 −A1,
where
K1 =
(
0 0
0 p2 − p1
)
, A1 = A−
(
p2
1
4 I 0
0 (p2
1
4 − p1(p1−p2)
2 )I
)
.
It is clear that K1 ≥ 0, and A1 is a self-adjoint operator bounded below. We
apply Lemma 2.2 from [17] to the pencil L1(τ) and conclude that the spectrum
of the pencil L1(τ) can lie only in the half-plane Imτ ≥ 0 and on the imaginary
axis. As far as τ = λ − ip1
2 , the spectrum of the pencil L(λ) can lie only in the
half-plane Imλ ≥ p1/2 and on the imaginary axis.
To prove that non-pure imaginary eigenvalues lie in the half-plane Imλ ≤ p2
2 ,
we apply the transformation of the spectral parameter τ = −λ + ip2
2 . Then we
obtain the quadratic pencil
L2(τ) = τ2I− iτK2 −A2,
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3 285
L. Kobyakova
where
K2 =
(
p2 − p1 0
0 0
)
, A2 = A−
(
(−p2
2
4 + p2(p2−p1)
2 )I 0
0 −p2
2
4 I
)
.
Evidently, K2 ≥ 0 and A2 is a self-adjoint operator bounded below. We apply
Lemma 2.2 from [17] to the pencil L2(τ) and conclude that the spectrum of this
pencil L2(τ) can lie only in the half-plane Imτ ≥ 0 and on the imaginary axis.
Since τ = −λ+ ip2
2 , the spectrum of the pencil L(λ) can lie only in the half-plane
Imλ ≤ p2/2 and on the imaginary axis.
We have proved that the spectrum of the pencil L(λ) can lie only in the strip
p1/2 ≤ Imλ ≤ p2/2 and on the imaginary axis.
Let us prove assertion 2).
Since the functions Sj(
√
λ2 − ipjλ, x) and S′j(
√
λ2 − ipjλ, x) are even func-
tions of the first argument, i.e., they are entire functions of λ2 − ipjλ, and the
identity (−λ)2 − i(−λ)pj = λ2 − iλpj is true, then the function ϕ(λ) possesses
the symmetry property ϕ(−λ) = ϕ(λ). This implies assertion 2).
Let us prove assertion 3).
As it was mentioned above, the number of negative eigenvalues of the operator
A is finite and thereby the number of negative eigenvalues of the operator A1 is
also finite. Then, applying Theorem 2.2 from [17], we obtain that the number of
eigenvalues of the pencil L1(τ) lying on the negative imaginary half-axis (−i∞, 0)
is finite and coincides with the number of negative eigenvalues of the operator A1.
Similarly, the number of eigenvalues of the pencil L2(τ), which lie on the negative
imaginary axis (−i∞, 0), is finite and equals the number of negative eigenvalues
of the operator A2. This implies assertion 3).
Consequently, the pure imaginary eigenvalues of problem (2.1)–(2.5) are on
the finite interval and do not accumulate to a point of the imaginary axis. The
theorem is proved.
In the sequel we will use
Definition 2. The function f(z), z ∈ C\R, is said to be a Nevalinna function
if it maps the open upper half-plane into the closed upper half-plane and takes
complex conjugate values at non-real points symmetric with respect to the real
axis (f(z) = f(z)).
Now we consider the pure imaginary eigenvalues of problem (2.1)–(2.5). For
conveniency, we compare their location with the location of pure imaginary eigen-
values of problems (2.6)–(2.7), (2.8)–(2.9).
Let us denote by {λk}, {µ(1)
k } and {µ(2)
k } the eigenvalues of problems (2.1)–
(2.5), (2.6)–(2.7) and (2.8)–(2.9), respectively, lying on the union of the intervals
286 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3
Spectral Problem Generated by the Equation of Smooth String
(−i∞; ip1/2)∪ (ip2/2;+i∞); and let {ζk} = {µ(1)
k }∪{µ(2)
k }. We enumerate them
in the following way:
on the interval (−i∞; ip1/2): |ζ−1 − ip1/2| ≥ |ζ−2 − ip1/2| ≥ . . .,
|λ−1 − ip1/2| ≥ |λ−2 − ip1/2| ≥ . . .,
and on the interval (ip2/2;+i∞): |ζ1| ≥ |ζ2| ≥ . . .,
|λ1| ≥ |λ2| ≥ . . ..
Theorem 2.
1) The elements of the sequences {ζk} and {λk}, which lie on the intervals
(−i∞; ip1/2) and (ip2/2;+i∞), interlace as follows:
|ζ1| ≥ |λ1| ≥ |ζ2| ≥ |λ2| ≥ ...
|ζ−1 − ip1/2| ≥ |λ−1 − ip1/2| ≥ |ζ−2 − ip1/2| ≥ |λ−2 − ip1/2| ≥ ...
2) The multiplicities of ζk do not exceed 2.
3) The equality λk = ζk for k > 0 is valid if and only if λk = ζk+1, and for
k < 0, if and only if λk = ζk−1.
P r o o f. Let us introduce the notation ξj = λ2 − ipjλ and consider the
functions Φj(
√
ξj) =
Sj(
√
ξj ,a/2)
S′j(
√
ξj ,a/2)
and Φj(
√
λ2 − ipjλ) = Sj(
√
λ2−ipjλ,a/2)
S′j(
√
λ2−ipjλ,a/2)
,
j = 1, 2.
Since Sj(
√
ξj , a/2) and S′j(
√
ξj , a/2) are entire functions, Φj(
√
ξj) is mero-
morphic.
Using the standard method (see, e.g., [2, p. 661–662], we can show that
Im
Sj(
√
ξj , a/2)
S′j(
√
ξj , a/2)
= Im ξj ·
a/2∫
0
|S2
j (
√
ξj , x)| dx
|S′j(
√
ξj , a/2)|2 .
It is clear that if Im ξj > 0, then Im
Sj(
√
ξj ,a/2)
S′j(
√
ξj ,a/2)
> 0. Consequently, the
function Φj(
√
ξj) maps the open upper half-plane into the open upper half-plane,
and Φj(
√
ξj) = Φj(
√
ξj). Thus they are Nevalinna functions.
From meromorphic and Nevalinna properties of the function Φj(
√
ξj) it fol-
lows (see [18, p. 399]) that in the half-planes Im ξj > 0 and Im ξj < 0, Φj(
√
ξj) it
does not have zeros and poles, i.e., all its zeros and poles lie on the real axis and
strictly interlace. Moreover, the function Φj(
√
ξj) is monotonically increasing on
its intervals of continuity.
Using the integral representations of the functions Sj(
√
ξj , a/2) and
S′j(
√
ξj , a/2) (see (1.2.11) in [19, p. 18]), we can state that Φj(
√
ξj) −→ +0
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3 287
L. Kobyakova
as ξj −→ −∞, i.e., the most left element of the interlaced sequence of zeros and
poles of the function Φj(
√
ξj) is its pole. Hence, the number of its negative zeros
and poles is finite.
Since Im ξj = Reλ(2Imλ− pj), the function Φj(
√
λ2 − ipjλ) (j = 1, 2) maps
the quadrants D
(j)
1 = {λ : Reλ > 0, Im λ > pj/2} and D
(j)
2 = {λ : Reλ < 0,
Imλ < pj/2} into the open upper half-plane, and the quadrants D
(j)
3 = {λ :
Reλ < 0, Imλ > pj/2} and D
(j)
4 = {λ : Reλ > 0, Imλ < pj/2} into the open
lower half-plane.
Consequently, the function Φj(
√
λ2 − ipjλ) (j = 1, 2) does not have zeros and
poles in the quadrants D
(j)
1 , D
(j)
2 , D
(j)
3 and D
(j)
4 because its zeros and poles are
on the imaginary axis and on the straight line Imλ = pj/2 and are interlaced.
The function Φ1(
√
λ2 − ip1λ) + Φ2(
√
λ2 − ip2λ) maps the quadrants D+
1 =
{λ : Reλ > 0, Imλ > p2/2} and D+
2 = {λ : Reλ < 0, Imλ < p1/2} into the
upper half-plane, that is, for all λ ∈ D+
1 ∪ D+
2 : 0 < arg(Φ1(
√
λ2 − ip1λ)+
+Φ2(
√
λ2 − ip2λ)) < π, and the quadrants D+
3 = {Reλ > 0, Imλ < p1/2}
and D+
4 = {Reλ < 0, Imλ > p2/2} into the lower half-plane, that is, for all
λ ∈ D+
3 ∪D+
4 : −π < arg(Φ1(
√
λ2 − ip1λ) + Φ2(
√
λ2 − ip2λ)) < 0.
The functions Φ1(
√
λ2 − ip1λ) and Φ2(
√
λ2 − ip2λ) do not have poles in these
quadrants, therefore the sum of these functions does not have them as well.
Let us show that the function Φ1(
√
λ2 − ip1λ) + Φ2(
√
λ2 − ip2λ) also does
not have zeros in the quadrants D+
1 , D+
2 and D+
3 , D+
4 .
Let us assume that the function Φ1(
√
λ2 − ip1λ) + Φ2(
√
λ2 − ip2λ) has a
zero, for example, in the quadrant D+
1 , and let Γ be a simple closed contour
completely located in the quadrant D+
1 circling this zero. Then the increment
of the argument of the function Φ1(
√
λ2 − ip1λ) + Φ2(
√
λ2 − ip2λ) along the
contour Γ is not less than 2π, which is impossible because the whole contour lies
in the open half-plane.
It is obvious that Φj(
√
λ2 − ipjλ) −→ +0 as λ −→ ±i∞. If λ moves along
the imaginary axis towards ipj/2, then the function Φj(
√
λ2 − ipjλ) increases
monotonically on each of the intervals of its continuity.
To prove that the zeros and the poles of Φ1(
√
λ2 − ip1λ)+Φ2(
√
λ2 − ip2λ) on
the intervals (−i∞; ip1/2) and (ip2/2;+i∞) are interlaced, we consider a contour
surrounding exactly two poles which lie both on (−i∞; ip1/2) or on (ip2, i∞). If
there is no zero between them, then the increment of the argument of the function
Φ1(
√
λ2 − ip1λ) + Φ2(
√
λ2 − ip2λ) along the contour is 4π, which is impossible
because Imξj · ImΦj(ξj) > 0.
Since the monotonically increasing functions Φj(
√
ξj) −→ +0 (j = 1, 2) as
ξj −→ −∞, their sum Φ1(
√
ξ1)+Φ2(
√
ξ2) −→ +0 as ξ1 −→ −∞ and ξ2 −→ −∞.
288 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3
Spectral Problem Generated by the Equation of Smooth String
When solving the equation λ2 − ipjλ = ξj with respect to λ, we obtain
λ1,2 =
−ipj±
√
4ξj−p2
j
2 . It is clear that λ −→ ±i∞, as ξj −→ −∞, that is, the
pure imaginary zeros and poles of the functions Φj(
√
λ2 − ipjλ) lie on the finite
interval of an imaginary axis and their number is also finite.
Let us divide both sides of (2.16) by S′1(
√
λ2 − ip1λ, a/2)·S′2(
√
λ2 − ip1λ, a/2):
ϕ(λ)
S′1(
√
λ2 − ip1λ, a/2) · S′2(
√
λ2 − ip2λ, a/2)
=
=
S1(
√
λ2 − ip1λ, a/2)
S′1(
√
λ2 − ip1λ, a/2)
+
S2(
√
λ2 − ip2λ, a/2)
S′2(
√
λ2 − ip1λ, a/2)
.
It was shown above that the zeros of each of the summands in the right-hand
side of the last equation are strictly interlaced with its poles on the intervals
(−i∞; ip1/2) and (ip2/2;+i∞). This implies that the zeros of the numerator in
the left-hand side interlace with the zeros of the denominator on these intervals.
Concluding from the above, the zeros of the characteristic function ϕ(λ)
are the eigenvalues λk of problem (2.1)–(2.5), and the zeros of the functions
S′1(
√
λ2 − ip1λ, a/2) and S′2(
√
λ2 − ip2λ, a/2) are the eigenvalues {µ(1)
k } and
{µ(2)
k } of problems (2.6)–(2.7) and (2.8)–(2.9), respectively. Whence assertions 1)
follows.
2) The eigenvalues {µ(1)
k } and {µ(2)
k } of problems (2.6)–(2.7) and (2.8)–(2.9)
are simple, that is, their multiplicities equal 1. Thus the multiplicity of each ζk ∈
∈ {µ(1)
k } ∪ {µ(2)
k } does not exceed 2.
3) If λk = ζk = µ
(1)
p , then ϕ(λk) = 0 and S′1(
√
µ
(1)
p
2 − ip1µ
(1)
p , a/2) = 0.
Consequently, the second term of (2.16) is
S′2(
√
µ
(1)
p
2 − ip1µ
(1)
p , a/2) · S1(
√
µ
(1)
p
2 − ip1µ
(1)
p , a/2) = 0.
The function S1(
√
µ
(1)
p
2 − ip1µ
(1)
p , x) and its derivative S′1(
√
µ
(1)
p
2 − ip1µ
(1)
p , x)
do not vanish simultaneously, consequently, S1(
√
µ
(1)
p
2 − ip1µ
(1)
p , a/2) 6= 0.
It means that S′2(
√
µ
(1)
p
2 − ip1µ
(1)
p , a/2) = 0. So, µ
(1)
p is an eigenvalue of
problem (2.8)–(2.9).
If eigenvalues of the auxiliary problems (2.6)–(2.7) and (2.8)–(2.9) on the in-
terval (ip2/2;+i∞) (on the interval (−i∞; ip1/2) ) coincide, then they are num-
bered as ζk and ζk+1 (ζk−1 and ζk). The theorem is proved.
Now let us compare the location of the pure imaginary eigenvalues of problem
(2.1)–(2.5) with the location of the pure imaginary eigenvalues of problems (2.10)–
(2.11), (2.12)–(2.13).
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3 289
L. Kobyakova
We denote by {λk}, {ν(1)
k } and {ν(2)
k } the eigenvalues of problems (2.1)–
(2.5), (2.10)–(2.11) and (2.12)–(2.13), respectively, lying on the intervals of the
imaginary axis (−i∞; ip1/2) and (ip2/2;+i∞); let {ξk} = {ν(1)
k } ∪ {ν(2)
k }, i.e.,
λk, ζk ∈ (−i∞; ip1/2) ∪ (ip2/2;+i∞).
Let us enumerate them in the following way:
on the interval (−i∞; ip1/2): |ξ−1 − ip1/2| ≤ |ξ−2 − ip1/2| ≤ . . .,
|λ−1 − ip1/2| ≤ |λ−2 − ip1/2| ≤ . . .,
and on the interval (ip2/2;+i∞): |ξ1| ≥ |ξ2| ≥ . . .,
|λ1| ≥ |λ2| ≥ . . ..
Theorem 3.
1) The sequences {ξk} and {λk} are interlaced on the intervals (−i∞; ip1/2)
and (ip2/2;+i∞),
|λ1| ≥ |ξ1| ≥ |λ2| ≥ ... ,
|λ−1 − ip1/2| ≥ |ξ−2 − ip1/2| ≥ |λ−2 − ip1/2| ≥ ...
2) The multiplicity of each ξk does not exceed 2.
3) The equation λk = ξk is valid for k > 0 if and only if λk = ξk−1, and for
k < 0 if and only if λk = ξk+1.
The proof of Theorem 3 is similar to that of Theorem 2.
4. Asymptotics of Eigenvalues
Theorem 4. The eigenvalues λn of problem (2.1)–(2.5) behave asymptoti-
cally as follows:
λn =
πn
a
+
i(p1 + p2)a
4
+
Bn
n
+ o
( 1
n
)
, (4.1)
where Bn = 1
πa
[
(−1)n p2−p1
4 sh (p2−p1)a
4 −(Q1 +Q2)
]
, Qj =
1
2
a/2∫
0
qj(t) dt, j = 1, 2.
P r o o f. Let us introduce (τ (j))2 = λ2 − ipjλ. Under this transformation
equations (2.1), (2.2) can be rewritten as
−y′′j + qj(x)yj = (τ (j))2yj , j = 1, 2. (4.2)
There exist (see formula (1.2.11), [19, p. 18]) the solutions of equations (4.2)
satisfying the initial conditions Sj(τ (j), 0) = 0, S′j(τ
(j), 0) = 1, and these solutions
are of the form
Sj(τ (j), x) =
sin τ (j)x
τ (j)
+
x∫
0
Kj(x, t,∞)
sin τ (j)x
τ (j)
dt, (4.3)
290 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3
Spectral Problem Generated by the Equation of Smooth String
where Kj(x, t,∞) = Kj(x, t)−Kj(x,−t), Kj(x, t) is the solution of the integral
equation (see (1.2.18), [19, p. 22])
Kj(x, t) =
1
2
x+t
2∫
0
gj(s) ds +
x+t
2∫
0
ds
x−t
2∫
0
gj(s + p) Kj(s + p, s− p)dp,
and Kj(x, x) =
1
2
x∫
0
gj(t) dt, Kj(x,−x) = 0.
The functions qj(x) ∈ L2(0, a/2), consequently, there exist the derivatives
∂Kj(x,t,∞)
∂x , ∂Kj(x,t,∞)
∂t , which belong to L2(0, a/2) as functions of each of its va-
riables.
Integrating by parts, we modify the right-hand side of (4.3) to the form
Sj(τ (j), x) =
sin τ (j)x
τ (j)
− 1
2
x∫
0
qj(t)dt · cos τ (j)x
(τ (j))2
+
x∫
0
∂Kj(x, t,∞)
∂t
cos τ (j)t
(τ (j))2
dt.
(4.4)
For the derivatives of Sj(τ (j), x) with respect to x, we have
S′j(τ
(j), x) = cos τ (j)x + Kj(x, x,∞) · sin τ (j)x
τ (j)
+
x∫
0
∂Kj(x, t,∞)
∂t
sin τ (j)t
τ (j)
dt.
(4.5)
Substituting (4.4) and (4.5) into the equation
ϕ(λ) = S1(τ (1), a/2)S′2(τ
(2), a/2) + S2(τ (2), a/2)S′1(τ
(1), a/2) = 0,
we obtain
ϕ(λ) =
sin(τ (1) + τ (2)) a
2
τ (1)
+ Θ(λ), (4.6)
where
Θ(λ) =
τ (1) − τ (2)
τ (1)τ (2)
sin τ (2) a
2
cos τ (1) a
2
+ (Q1 + Q2)
sin τ (1) a
2 sin τ (1) a
2
τ (1)τ (2)
+
1
τ (1)τ (2)
(
sin τ (1) a
2
Ks(τ (2)) + sin τ (2) a
2
Ks(τ (1))
)
− cos τ (1) a
2
cos τ (2) a
2
( Q1
(τ (1))2
+
Q2
(τ (2))2
)
+
1
(τ (1))2
cos τ (2) a
2
Kc(τ (1))
+
1
(τ (2))2
cos τ (1) a
2
Kc(τ (2))−Q1Q2
( cos τ (1) a
2 sin τ (2) a
2
(τ (1))2τ (2)
+
cos τ (2) a
2 sin τ (1) a
2
τ (1)(τ (2))2
)
+
Q2Kc(τ (1)) sin τ (2) a
2
(τ (1))2τ (2)
+
Q1Kc(τ (2)) sin τ (1) a
2
τ (1)(τ (2))2
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3 291
L. Kobyakova
+
Kc(τ (1))Ks(τ (2))
(τ (1))2τ (2)
+
Ks(τ (1))Kc(τ (2))
τ (1)(τ (2))2
− Q1Ks(τ (2)) cos τ (1) a
2
(τ (1))2τ (2)
− Q2Ks(τ (1)) cos τ (2) a
2
τ (1)(τ (2))2
,
Kc(τ (j)) =
a/2∫
0
∂Kj( a
2 , t,∞)
∂t
cos τ (j)t dt, Ks(τ (j)) =
a/2∫
0
∂Kj( a
2 , t,∞)
∂t
sin τ (j)t dt.
To find the asymptotic formulas for eigenvalues, we use the Rouché theorem.
Let us denote
Φ(λ) =
sin(λa− i(p1+p2)a
4 )
λ
, (4.7)
Ψ(λ) = ϕ(λ)− Φ(λ) =
sin(τ (1) + τ (2)) a
2
τ (1)
− sin(λa− i(p1+p2)a
4 )
λ
+ Θ(λ). (4.8)
Let us consider the contours Γn which are the circles centered at πn
a + i(p1+p2)
4
with the radii r, |r| < π
2a .
For λ ∈ Γn, we have λ = πn
a + i(p1+p2)
4 + reiθ.
Let us multiply (4.6) by λ and prove that for all λ ∈ Γn
|λΦ(λ)| =
∣∣∣sin(λa− i(p1 + p2)a
4
)
∣∣∣ ≥ C1(r), (4.9)
where C1 depends only on r.
The function |λΦ(λ)| is continuous on each Γn and therefore it is bounded
on Γn. Since this function is periodic, for all n we have
min
λ∈Γ1
|λΦ(λ)| = min
λ∈Γn
|λΦ(λ)| ≤ |λΦ(λ)| ≤ max
λ∈Γn
|λΦ(λ)| = max
λ∈Γ1
|λΦ(λ)|.
We denote C1(r) = min
λ∈Γ1
|λΦ(λ)| and obtain inequality (4.9).
Let us show now that there exists a constant C2(r) such that for n large
enough (n ≥ N(r)) the inequality
|λΨ(λ)| ≤ C2(r)
n
(4.10)
holds on the circles Γn.
The moduli of summands in (4.8), which are the products of the constants
Q1, Q2 and functions sin τ (j) a
2 , cos τ (j) a
2 , Ks(τ (j)) and Kc(τ (j)), are bounded in
the strip Imλ ∈ [ (p1+p2)
4 − r, (p1+p2)
4 + r]. The factors of the form λ
(τ (1))α(τ (2))β
292 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3
Spectral Problem Generated by the Equation of Smooth String
(α, β = 0, 1, 2, α + β ≥ 2) satisfy λ
(τ (1))α(τ (2))β = O
(
1
λ
)
. Therefore, inequality
(4.10) will be proved if we can show that
∣∣∣sin(τ (1) + τ (2))
a
2
− sin(λa− i(p1 + p2)a
4
)
∣∣∣ = O
( 1
n
)
. (4.11)
Since (τ (j))2 = λ2 − ipjλ, we have
τ (j) = λ
√
1− ipj
λ
= λ
(
1− ipj
2λ
+ O
( 1
λ2
))
= λ− ipj
2
+ O
( 1
λ
)
,
(τ (1) + τ (2))
a
2
= λa− ia(p1 + p2)
4
+ O
( 1
λ
)
.
For λ ∈ Γn,
∣∣∣sin(λa− ia(p1 + p2)
4
)− sin(τ (1) + τ (2))
a
2
∣∣∣
=
∣∣∣2 cos
λa− ia(p1+p2)
4 ) +
(
λa− ia(p1+p2)
4 ) + O
(
1
λ
))
2
× sin
λa− ia(p1+p2)
4 )−
(
λa− ia(p1+p2)
4 ) + O
(
1
λ
))
2
∣∣∣
= 2
∣∣∣ cos
(
λa− ia(p1 + p2)
4
+ O
( 1
λ
))∣∣∣
∣∣∣sinO
( 1
λ
)∣∣∣ ≤ C(r)
∣∣∣O
( 1
|λ|
)∣∣∣ ≤ C2(r)
n
.
Taking into account (4.9) and (4.10), we obtain
|λϕ(λ)| = |λ| (|Φ(λ) + Ψ(λ)|) ≥ |λ| (|Φ(λ)| − |Ψ(λ)|) ≤ C1(r)− C2(r)
n
.
It is clear that for n large enough the equality C1(r)− C2(r)
n > 0 holds.
The function λΦ(λ) has exactly 1 simple zero (λn = πn
a + ia(p1+p2)
4 ) inside each
of the contours Γn. Therefore, by the Rouche theorem, the function λϕ(λ) also
has exactly 1 simple zero inside each of the counters Γn for n large enough. Since
the radius r of the contours Γn can be taken arbitrarily small, for the eigenvalues
we have the following asymptomatic expansion:
λn =
πn
a
+
ia(p1 + p2)
4
+ ∆n, ∆n −→ 0 for n −→∞. (4.12)
Let us substitute (4.12) into (4.6). Taking into account that
Kc(τ (j)) =
a/2∫
0
∂Kj(a/2, t,∞)
∂t
cos τ (j)t dt =
λ−→±∞
0,
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 3 293
L. Kobyakova
Ks(τ (j)) =
a/2∫
0
∂Kj(a/2, t,∞)
∂t
sin τ (j)t dt =
λ−→±∞
0
(Lemma 1.4.3 [19, p. 61]) and, expanding in power series in n the right-hand side
of (4.6), we obtain the coefficient
Bn =
1
πa
[
(−1)n p2 − p1
4
sh
(p2 − p1)a
4
− (Q1 + Q2)
]
. (4.13)
The theorem is proved.
Acknowledgements. The author is thankful to professor V.N. Pivovarchik
for the attention to the work during the time of its writing.
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