# Spectrums of Solvable Pantograph Differential-Operators for First Order

Authors: Z. I. Ismailov; P. Ipek
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

By using the methods of operator theory, all solvable extensions of minimal operator generated by first order pantograph-type delay differential-operator expression in the Hilbert space of vector-functions on finite interval have been considered. As a result, the exact formula for the spectrums of these extensions is presented. Applications of obtained results to the concrete models are illustrated.


---

## Body

## 1. Introduction

The quantitative and qualitative theory of linear pantograph differential equations, sometimes known as pantograph-type delay differential equations, was first studied in detail by T. Kato and J. B. McLeod [1], L. Fox et al. [2], and A. Iserles [3] in the nineteen seventies.These equations arose as a mathematical model of an industrial problem involving wave motion in the overhead supply line to an electrified railway system, so they are often called pantograph equations.In industrial applications in works [2, 4] and in studies on biology and economics, control and electrodynamics in works [5–7] have been researched (for more comprehensive list of features see [3]).Since an analytical computation of solutions, eigenvalues, and eigenfunctions of corresponding problems is very difficult theoretically and technically, then in this theory methods of numerical analysis play a significant role (for more information see [8–13]).Let us remember that an operatorS
:
D
(
S
)
⊂
H
→
H in Hilbert space H is called solvable, if S is one-to-one, S
D
(
S
)
=
H, and S
-
1
∈
L
(
H
).In this work, by using methods of operator theory all solvable extensions of minimal operator generated by pantograph-type delay differential-operator expression for first order in the Hilbert space of vector-functions at a finite interval have been described in terms of boundary values in Section2. Consequently, the resolvent operators of these extensions can be written clearly.The exact formula for the spectrums of these extensions has been given in Section3. Applications of obtained results to the concrete models have been illustrated in Section 4.


## 2. Description of Solvable Extensions

In the Hilbert spaceL
2
(
H
,
(
0,1
)
) of vector-functions consider a linear pantograph differential-operator expression for first order in the form
(1)
l
(
u
)
=
u
′
(
t
)
+
∑
m
=
1
n
‍
A
m
(
t
)
u
(
α
m
t
)
,
where (1)
H is a separable Hilbert space with inner product (
·
,
·
)
H and norm ∥
·
∥
H,
(2)
operator-functionA
m
(
·
)
:
[
0,1
]
→
L
(
H
), m
=
1,2
,
3
,
…
,
n, is continuous on the uniform operator topology,
(3)
m
=
1,2
,
3
,
…
,
n
-
1, 0
<
α
m
<
1, α
n
=
1.On the other hand, here the following differential expression will be considered:(2)
m
(
u
)
=
u
′
(
t
)
in the Hilbert space L
2
(
H
,
(
0,1
)
) corresponding to (1).It is clear that the formally adjoint expression of (2) is of the form
(3)
m
+
(
v
)
=
-
v
′
(
t
)
.
Now define operator M
0
′ on the dense in L
2
(
H
,
(
0,1
)
) set of vector-functions D
0
′,
(4)
D
0
′
≔
{
u
(
t
)
∈
L
2
(
H
,
(
0,1
)
)
:
u
(
t
)
=
∑
k
=
1
n
‍
φ
k
(
t
)
f
k
,
φ
k
(
t
)
∈
C
0
∞
(
0,1
)
,
f
k
∈
H
,
k
=
1,2
,
…
,
n
,
n
∈
N
∑
k
=
1
n
‍
}
,
as M
0
′
u
=
m
(
u
).The closure ofM
0
′ in L
2
(
H
,
(
0,1
)
) is called the minimal operator generated by differential-operator expression (2) and is denoted by M
0.In a similar way, the minimal operatorM
0
+ in L
2
(
H
,
(
0,1
)
) corresponding to differential expression (3) can be defined.The adjoint operator ofM
0
+
(
M
0
) in L
2
(
H
,
(
0,1
)
) is called the maximal operator generated by (2)((3)) and is denoted by M
(
M
+
).Now define an operatorP
α
m in L
2
(
H
,
(
0,1
)
) in the form
(5)
P
α
m
u
(
t
)
=
u
(
α
m
t
)
,
u
∈
L
2
(
H
,
(
0,1
)
)
,
m
=
1,2
,
3
,
…
,
n
.
Then for u
∈
L
2
(
H
,
(
0,1
)
) and for m
=
1,2
,
3
,
…
,
n it is obtained that
(6)
∥
P
α
m
u
∥
L
2
(
H
,
(
0,1
)
)
2
=
∫
0
1
(
u
(
α
m
t
)
,
u
(
α
m
t
)
)
H
d
t
=
1
α
m
∫
0
α
m
(
u
(
x
)
,
u
(
x
)
)
H
d
x
≤
1
α
m
∫
0
1
∥
u
(
x
)
∥
H
2
d
x
=
1
α
m
∥
u
∥
L
2
(
H
,
(
0,1
)
)
2
.
Therefore we have P
α
m
∈
L
(
L
2
(
H
,
(
0,1
)
)
) and ∥
P
α
m
∥
≤
1
/
α
m, m
=
1,2
,
3
,
…
,
n.In this situation the following defined operator(7)
A
α
(
t
)
=
∑
m
=
1
n
‍
A
m
(
t
)
P
α
m
,
0
<
α
<
1
,
is a linear bounded operator in L
2
(
H
,
(
0,1
)
).Throughout this work the following defined operators(8)
L
0
:
=
M
0
+
A
α
(
t
)
,
L
0
:
W
o
2
1
(
H
,
(
0,1
)
)
⊂
L
2
(
H
,
(
0,1
)
)
⟶
L
2
(
H
,
(
0,1
)
)
,
L
:
=
M
+
A
α
(
t
)
,
L
:
W
2
1
(
H
,
(
0,1
)
)
⊂
L
2
(
H
,
(
0,1
)
)
⟶
L
2
(
H
,
(
0,1
)
)
will be called the minimal and maximal operators corresponding to differential expression (1) in L
2
(
H
,
(
0,1
)
), respectively.Now letU
(
t
,
s
), t
,
s
∈
[
0,1
], be the family of evolution operators corresponding to the homogeneous differential equation
(9)
U
t
(
t
,
s
)
f
+
A
α
(
t
)
U
(
t
,
s
)
f
=
0
,
t
,
s
∈
[
0,1
]
,
U
(
s
,
s
)
f
=
f
,
f
∈
H
.
The operator U
(
t
,
s
), t
,
s
∈
[
0,1
], is linear continuous, boundedly invertible in H and
(10)
U
-
1
(
t
,
s
)
=
U
(
s
,
t
)
,
s
,
t
∈
[
0,1
]
(for more detailed analysis of this concept see [14]).Let us introduce the operator(11)
U
z
(
t
)
:
=
U
(
t
,
0
)
z
(
t
)
,
U
:
L
2
(
H
,
(
0,1
)
)
⟶
L
2
(
H
,
(
0,1
)
)
.
In this case it is easy to see that, for the differentiable vector-function z
∈
L
2
(
H
,
(
0,1
)
), z
:
[
0,1
]
→
H satisfies the following relation:
(12)
l
(
U
z
)
=
(
U
z
)
′
(
t
)
+
A
α
(
t
)
U
z
(
t
)
=
U
z
′
(
t
)
+
(
U
t
′
+
A
α
(
t
)
U
)
z
(
t
)
=
U
m
(
z
)
.
From this U
-
1
l
(
U
z
)
=
m
(
z
). Hence it is clear that if L
~ is some extension of the minimal operator L
0, that is, L
0
⊂
L
~
⊂
L, then
(13)
U
-
1
L
0
U
=
M
0
,
M
0
⊂
U
-
1
L
U
=
M
~
⊂
M
,
U
-
1
L
U
=
M
.
For example, prove the validity of the last relation. It is known that
(14)
D
(
M
0
)
=
W
o
2
1
(
H
,
(
0,1
)
)
,
D
(
M
)
=
W
2
1
(
H
,
(
0,1
)
)
.
If u
∈
D
(
M
), then l
(
U
z
)
=
U
m
(
z
)
∈
L
2
(
H
,
(
0,1
)
); that is, U
u
∈
D
(
L
). From the last relation M
⊂
U
-
1
L
U. Contrarily, if a vector-function u
∈
D
(
L
), then
(15)
m
(
U
-
1
v
)
=
U
-
1
l
(
v
)
∈
L
2
(
H
,
(
0,1
)
)
;
that is, U
-
1
v
∈
D
(
M
). From last relation U
-
1
L
⊂
M
U; that is, U
-
1
L
U
⊂
M. Hence, U
-
1
L
U
=
M.Theorem 1.
Each solvable extensionL
~ of the minimal operator L
0 in L
2
(
H
,
(
0,1
)
) is generated by the pantograph differential-operator expression (1) and boundary condition
(16)
(
K
+
E
)
u
(
0
)
=
K
U
(
0,1
)
u
(
1
)
,
where K
∈
L
(
H
) and E is an identity operator in H. The operator K is determined uniquely by the extension L
~; that is, L
~
=
L
K.
On the contrary, the restriction of the maximal operatorL
0 to the manifold of vector-functions satisfies condition (16) for some bounded operator K
∈
L
(
H
) is a solvable extension of the minimal operator L
0 in the L
2
(
H
,
(
0,1
)
).Proof.
Firstly, all solvable extensionsM
~ of the minimal operator M
0 in L
2
(
H
,
(
0,1
)
) in terms of boundary values are described.
Consider the following so-called Cauchy extensionM
c
(17)
M
c
u
=
u
′
(
t
)
,
u
(
0
)
=
0
,
M
c
:
D
(
M
c
)
=
{
u
∈
W
2
1
(
H
,
(
0,1
)
)
:
u
(
0
)
=
0
}
⊂
L
2
(
H
,
(
0,1
)
)
⟶
L
2
(
H
,
(
0,1
)
)
of the minimal operator M
0. It is clear that M
c is a solvable extension of M
0 and
(18)
M
c
-
1
f
(
t
)
=
∫
0
t
f
(
x
)
d
x
,
f
∈
L
2
(
H
,
(
0,1
)
)
,
M
c
-
1
:
L
2
(
H
,
(
0,1
)
)
⟶
L
2
(
H
,
(
0,1
)
)
.
Now assume that M
~ is a solvable extension of the minimal operator M
0 in L
2
(
H
,
(
0,1
)
). In this case it is known that the domain of M
~ can be written in direct sum in the form
(19)
D
(
M
~
)
=
D
(
M
0
)
⊕
(
M
c
-
1
+
K
)
V
,
where V
=
Ker
⁡
M
=
H
,
K
∈
L
(
H
) [15, 16]. Therefore for each u
(
t
)
∈
D
(
M
~
) it is true that
(20)
u
(
t
)
=
u
0
(
t
)
+
M
c
-
1
f
+
K
f
,
u
0
∈
D
(
M
0
)
,
f
∈
H
.
That is, u
(
t
)
=
u
0
(
t
)
+
t
f
+
K
f, u
0
∈
D
(
M
0
), f
∈
H. Hence u
(
0
)
=
K
f, u
(
1
)
=
f
+
K
f
=
(
K
+
E
)
f. Hence u
(
0
)
=
K
f, u
(
1
)
=
f
+
K
f
=
(
K
+
E
)
f and from these relations it is obtained that
(21)
(
K
+
E
)
u
(
0
)
=
K
u
(
1
)
.
On the other hand, uniqueness of operator K
∈
L
(
H
) follows from [15]. Therefore, M
~
=
M
K. This completes the necessary part of this assertion.
On the contrary, ifM
K is an operator generated by differential expression (2) and boundary condition (21), then M
K is bounded, boundedly invertible, and
(22)
M
K
-
1
:
L
2
(
H
,
(
0,1
)
)
⟶
L
2
(
H
,
(
0,1
)
)
,
M
K
-
1
f
(
t
)
=
∫
0
t
f
(
x
)
d
x
+
K
∫
0
1
f
(
x
)
d
x
,
hhhhhhhhhhhhhhhhi
f
∈
L
2
(
H
,
(
0,1
)
)
.
Consequently, all solvable extensions of the minimal operator M
0 in L
2
(
H
,
(
0,1
)
) are generated by differential expression (2) and boundary condition (21) with any linear bounded operator K.
Now consider the general case. For this in theL
2
(
H
,
(
0,1
)
) introduce an operator in the form
(23)
U
:
L
2
(
H
,
(
0,1
)
)
⟶
L
2
(
H
,
(
0,1
)
)
,
(
U
z
)
(
t
)
≔
U
(
t
,
0
)
z
(
t
)
,
z
∈
L
2
(
H
,
(
0,1
)
)
.
From the properties of the family of evolution operators U
(
t
,
s
), t
,
s
∈
[
0,1
], it is implied that an operator U is linear bounded and has a bounded inverse and
(24)
(
U
-
1
z
)
(
t
)
=
U
(
0
,
t
)
z
(
t
)
.
On the other hand, from the relations
(25)
U
-
1
L
0
U
=
M
0
,
U
-
1
L
~
U
=
M
~
,
U
-
1
L
U
=
M
,
it is implied that an operator U is one-to-one between sets of solvable extensions of minimal operators L
0 and M
0 in L
2
(
H
,
(
0,1
)
).
The extensionL
~ of the minimal operator L
0 is solvable in L
2
(
H
,
(
0,1
)
) if and only if the operator M
~
=
U
-
1
L
~
U is an extension of the minimal M
0 in L
2
(
H
,
(
0,1
)
). Then, u
∈
D
(
L
~
) if and only if
(26)
(
K
+
E
)
U
(
0,0
)
u
(
0
)
=
K
U
(
0,1
)
u
(
1
)
;
that is, (
K
+
E
)
u
(
0
)
=
K
U
(
0,1
)
u
(
1
). This proves the validity of the claims in the theorem.Corollary 2.
In particular the resolvent operatorR
λ
(
L
K
), λ
∈
ρ
(
L
K
), of any solvable extension L
K of the minimal operator L
0, generated by pantograph-type delay differential expression
(27)
l
(
u
)
=
u
′
(
t
)
+
A
(
t
)
u
(
α
t
)
,
0
<
α
<
1
,
with boundary condition in L
2
(
H
,
(
0,1
)
),
(28)
(
K
+
E
)
u
(
0
)
=
K
U
(
0,1
)
u
(
1
)
,
is of the form
(29)
R
λ
(
L
K
)
f
(
t
)
=
U
(
t
,
0
)
[
(
E
+
K
(
1
-
e
λ
)
)
-
1
K
∫
0
1
e
λ
(
1
-
s
)
U
(
0
,
s
)
f
(
s
)
d
s
k
+
∫
0
t
e
λ
(
t
-
s
)
U
(
0
,
s
)
f
(
s
)
d
s
]
,
f
∈
L
2
(
H
,
(
0,1
)
).Remark 3.
Note that in the general caseA
P
α
≠
P
α
A, for any A
∈
L
(
H
).
Indeed, if(30)
(
A
f
)
(
t
)
=
t
f
(
t
)
,
f
∈
L
2
(
H
,
(
0,1
)
)
,
hhhhhhhhh
A
:
L
2
(
0,1
)
⟶
L
2
(
0,1
)
,
then for 0
<
α
<
1 and f
∈
L
2
(
0,1
)
(31)
(
A
P
α
)
f
(
t
)
=
A
(
P
α
f
(
t
)
)
=
A
(
f
(
α
t
)
)
=
t
f
(
α
t
)
,
0
<
t
<
1
,
(
P
α
A
)
f
(
t
)
=
P
α
(
A
f
(
t
)
)
=
P
α
(
t
f
(
t
)
)
=
(
α
t
)
f
(
α
t
)
,
0
<
t
<
1
.Corollary 4.
Assume that for anyt
∈
(
0,1
) and any u
∈
W
2
1
(
H
,
(
0,1
)
)
(32)
(
A
α
u
)
(
α
t
)
=
A
α
u
(
α
t
)
.
In this case, all solvable extensions of minimal operator L
0 are generated by the following differential expression
(33)
l
(
u
)
=
u
′
(
t
)
+
A
(
t
)
u
(
α
t
)
,
0
<
α
<
1
,
and boundary condition
(34)
(
K
+
E
)
u
(
0
)
=
K
∑
n
=
0
∞
‍
(
-
1
)
n
n
!
A
n
u
(
α
n
)
,
K
∈
B
(
H
)
in the Hilbert L
2
(
H
,
(
0,1
)
) and vice versa.
Note that the series in the right side of the last equality is convergent, because for anyu
∈
W
2
1
(
H
,
(
0,1
)
)
(35)
∑
n
=
0
∞
‍
∥
(
-
1
)
n
n
!
A
n
u
(
α
n
)
∥
H
≤
∑
n
=
0
∞
‍
∥
A
∥
n
n
!
max
⁡
[
0,1
]
∥
u
∥
H
<
+
∞
.Corollary 5.
All solvable extensionsL
K of the minimal operator L
0 generated by pantograph differential expression l
(
u
)
=
u
′
(
t
)
+
u
(
α
t
), 0
<
α
<
1, are described with boundary conditions
(36)
(
K
+
E
)
u
(
0
)
=
K
[
u
(
1
)
-
u
(
α
)
1
!
-
u
(
α
2
)
2
!
+
⋯
]
=
K
∑
n
=
0
∞
‍
1
n
!
u
(
α
n
)
in the Hilbert space L
2
(
H
,
(
0,1
)
).Corollary 6.
It can be proved that all the solvable extensions of the minimal operator are generated by pantograph-type differential-operator expressions for first order(37)
l
(
u
)
=
u
′
(
t
)
+
u
(
α
1
t
)
+
u
(
α
2
t
)
in L
2
(
H
,
(
0,1
)
) generated by l
(
·
) and boundary condition
(38)
(
K
+
E
)
u
(
0
)
=
K
[
(
α
2
3
)
u
(
1
)
-
(
u
(
α
1
)
+
u
(
α
2
)
)
+
1
2
!
(
u
(
α
1
2
)
+
2
u
(
α
1
α
2
)
+
u
(
α
2
2
)
)
-
1
3
!
(
u
(
α
1
3
)
+
u
(
α
1
α
2
2
)
+
2
u
(
α
1
2
α
2
)
+
2
u
(
α
1
α
2
2
)
+
u
(
α
1
2
α
2
)
+
u
(
α
2
3
)
)
+
⋯
]
in L
2
(
H
,
(
0,1
)
).Remark 7.
Theorem1 can be generalized in the differential expression
(39)
l
φ
(
u
)
:
=
u
′
(
t
)
+
∑
m
=
1
n
‍
A
m
(
t
)
u
(
φ
m
(
t
)
)
,
where φ
m
:
[
0,1
]
→
[
0,1
], φ
m
>
0
(
<
0
), φ
m
∈
C
1
[
0,1
], P
φ
m
:
L
2
(
H
,
(
0,1
)
)
→
L
2
(
H
,
(
0,1
)
), P
φ
m
u
(
t
)
=
u
(
φ
m
(
t
)
), m
=
1,2
,
…
,
n, A
φ
(
t
)
:
=
∑
m
=
1
n
A
m
(
t
)
P
φ
m.Theorem 8.
All solvable extensions of minimal operator corresponding to pantograph-type delay differential-operator expressionl
φ
(
·
) in Hilbert space L
2
(
H
,
(
0,1
)
) are described by l
φ
(
·
) and boundary condition
(40)
(
K
+
E
)
u
(
0
)
=
K
U
φ
(
0,1
)
u
(
1
)
,
where K
∈
B
(
H
) and U
φ
(
t
,
s
), t
,
s
∈
[
0,1
], is a family of evolution operators corresponding to the homogeneous differential equation
(41)
(
U
φ
)
t
′
(
t
,
s
)
+
A
φ
(
t
)
U
(
t
,
s
)
=
0
,
t
,
s
∈
[
0,1
]
,
with boundary condition U
φ
(
s
,
s
)
f
=
f
,
f
∈
H and vice versa.


## 3. Spectrum of Solvable Extensions

In this section, the spectrum structure of solvable extensions of minimal operatorL
0 in L
2
(
H
,
(
0,1
)
) will be investigated.Firstly, prove the following fact.Theorem 9.
IfL
~ is a solvable extension of a minimal operator L
0 and M
~
=
U
-
1
L
~
U corresponds to the solvable extension of a minimal operator M
0, then the spectrum of these extensions is true σ
(
L
~
)
=
σ
(
M
~
).Proof.
Consider a problem to the spectrum for a solvable extensionL
K of a minimal operator L
0 generated by pantograph differential-operator expression (1); that is,
(42)
L
K
u
=
λ
u
+
f
,
λ
∈
C
,
f
∈
L
2
(
H
,
(
0,1
)
)
.
From this it is obtained that
(43)
(
L
K
-
λ
E
)
u
=
f
or (
U
M
K
U
-
1
-
λ
E
)
u
=
f. Hence U
(
M
K
-
λ
)
(
U
-
1
u
)
=
f.
Therefore, the validity of the theorem is clear.Now prove the following result for the spectrum of solvable extension.Theorem 10.
IfL
K is a solvable extension of the minimal operator L
0 in the space L
2
(
H
,
(
0,1
)
), then spectrum of L
K has the form
(44)
σ
(
L
K
)
=
{
λ
∈
C
:
λ
=
ln
⁡
|
μ
+
1
μ
|
+
i
arg
(
μ
+
1
μ
)
+
2
n
π
i
,
μ
∈
σ
(
K
)
∖
{
0
,
-
1
}
,
n
∈
Z
|
μ
+
1
μ
|
}
.Proof.
Firstly, the spectrum of the solvable extensionM
K
=
U
-
1
L
K
U of the minimal operator M
0 in L
2
(
H
,
(
0,1
)
) will be investigated.
Consequently, consider the following problem for the spectrum; that is,M
K
u
=
λ
u
+
f, λ
∈
C, f
∈
L
2
(
H
,
(
0,1
)
). Then
(45)
u
′
=
λ
u
+
f
,
(
K
+
E
)
u
(
0
)
=
K
u
(
1
)
,
λ
∈
C
,
hi
f
∈
L
2
(
H
,
(
0,1
)
)
,
K
∈
L
(
H
)
.
It is clear that a general solution of the above differential equation in L
2
(
H
,
(
0,1
)
) has the form
(46)
u
λ
(
t
)
=
e
λ
t
f
0
+
∫
0
t
e
λ
(
t
-
s
)
f
(
s
)
d
s
,
f
0
∈
H
.
Therefore, from the boundary condition (
K
+
E
)
u
λ
(
0
)
=
K
u
λ
(
1
) it is obtained that
(47)
(
E
+
K
(
1
-
e
λ
)
)
f
0
=
K
∫
0
1
e
λ
(
1
-
s
)
f
(
s
)
d
s
.
For λ
m
=
2
m
π, m
∈
Z, from the last relation it is established that
(48)
f
0
(
m
)
=
K
∫
0
1
e
λ
m
(
1
-
s
)
f
(
s
)
d
s
,
m
∈
Z
.
Consequently, in this case the resolvent operator of M
K is in the form
(49)
R
λ
m
(
M
K
)
f
(
t
)
=
K
e
λ
m
t
∫
0
1
e
λ
m
(
1
-
s
)
f
(
s
)
d
s
+
∫
0
t
e
λ
(
t
-
s
)
f
(
s
)
d
s
,
hhhhhhhhhhhhhh
f
∈
L
2
(
H
,
(
0,1
)
)
,
m
∈
Z
.
On the other hand, it is clear that R
λ
m
(
M
K
)
∈
B
(
L
2
(
H
,
(
0,1
)
)
), m
∈
Z.
Now assume thatλ
≠
2
m
π, m
∈
Z, λ
∈
C. Then using (47) we have
(50)
(
K
-
1
e
λ
-
1
E
)
f
0
=
1
1
-
e
λ
K
∫
0
1
e
λ
(
1
-
s
)
f
(
s
)
d
s
,
hhhhhhhhhhhhhh
f
0
∈
H
,
f
∈
L
2
(
H
,
(
0,1
)
)
.
Therefore, λ
∈
σ
(
M
K
) if and only if
(51)
μ
=
1
e
λ
-
1
∈
σ
(
K
)
.
In this case since μ
≠
0,
(52)
e
λ
=
μ
+
1
μ
,
μ
∈
σ
(
K
)
,
μ
≠
-
1
.
Then
(53)
λ
n
=
ln
⁡
|
μ
+
1
μ
|
+
i
arg
(
μ
+
1
μ
)
+
2
n
π
i
,
n
∈
Z
.
Later on, using the last relation and Theorem 9 the validity of the claim in the theorem is proved.Now one result on the asymptotic behavior of eigenvalues of solvable extensions in special cases will be proved.Theorem 11.
LetK
∈
L
(
H
), K
≠
0, and σ
(
K
)
=
σ
P
(
K
). In addition, assume that there exists α
,
β
>
0 such that for any μ
ϵ
σ
P
(
K
) are true |
μ
|
≥
α
>
0 and |
μ
+
1
|
≥
β
>
0. Then λ
n
(
L
K
)
~
2
n
π as n
→
∞. (i.e., there exist lim
⁡
n
→
∞
⁡
(
|
λ
n
(
L
k
)
|
/
2
n
π
) and 0
<
lim
⁡
n
→
∞
⁡
(
|
λ
n
(
L
k
)
|
/
2
n
π
)
<
∞).Proof.
In this case forn
≥
1
(54)
|
λ
n
(
M
K
)
|
2
=
ln
⁡
2
|
μ
+
1
μ
|
+
|
arg
(
μ
+
1
μ
)
+
2
n
π
|
2
.
Since for any μ
∈
σ
p
(
K
)
(55)
|
μ
+
1
μ
|
≥
β
|
μ
|
≥
β
∥
K
∥
>
0
,
|
μ
+
1
μ
|
≤
1
+
1
|
μ
|
≤
1
+
1
α
,
then
(56)
ln
⁡
β
∥
K
∥
≤
ln
⁡
|
μ
+
1
μ
|
≤
ln
⁡
(
1
+
1
α
)
.
Therefore, for any μ
∈
σ
p
(
K
) is true
(57)
min
⁡
{
|
ln
⁡
(
β
∥
K
∥
)
|
,
|
ln
⁡
(
1
+
1
α
)
|
}
≤
|
ln
⁡
⁡
|
μ
+
1
μ
|
|
≤
max
⁡
{
|
ln
⁡
(
β
∥
K
∥
)
|
,
|
ln
⁡
(
1
+
1
α
)
|
}
.
On the other hand, for any n
∈
Z
(58)
(
2
n
π
)
2
≤
|
arg
(
μ
+
1
μ
)
+
2
n
π
|
2
≤
(
2
(
n
+
1
)
π
)
2
.
Consequently, for any n
∈
N
(59)
(
2
n
π
)
2
(
1
+
1
4
n
2
π
2
min
⁡
2
{
|
ln
⁡
(
β
∥
K
∥
)
|
,
|
ln
⁡
(
1
+
1
α
)
|
}
)
≤
|
λ
n
(
M
K
)
|
2
≤
(
2
n
π
)
2
(
(
2
(
n
+
1
)
π
2
n
π
)
2
+
1
(
2
n
π
)
2
max
⁡
2
{
|
ln
⁡
(
β
∥
K
∥
)
|
,
|
ln
⁡
(
1
+
1
α
)
|
}
(
2
(
n
+
1
)
π
2
n
π
)
2
)
.
This means that λ
n
(
M
K
)
~
2
n
π as n
→
∞.


## 4. Applications

Example 12.
Let(60)
(
H
,
∥
·
∥
H
)
=
(
C
,
|
·
|
)
,
A
(
t
)
=
a
(
t
)
∈
C
(
R
)
.
By Theorem 1, all solvable extensions L
k of minimal operator L
0 generated by l
(
u
)
=
u
′
(
t
)
+
a
(
t
)
u
(
α
t
), 0
<
α
<
t, in L
2
(
0,1
) are described with l
(
·
) and boundary condition
(61)
(
k
+
1
)
u
(
0
)
=
k
exp
⁡
(
-
∫
0
1
a
(
t
)
d
t
)
u
(
1
)
,
k
∈
C
.
In addition, the resolvent operator of these extensions is in the form
(62)
R
λ
(
L
k
)
f
(
t
)
=
exp
⁡
(
-
∫
0
t
a
(
x
)
P
α
d
x
)
×
[
(
λ
(
t
-
s
)
+
∫
0
s
a
(
x
)
P
α
d
x
)
(
1
+
k
(
1
-
e
λ
)
-
1
)
k
×
∫
0
1
exp
⁡
⁡
(
λ
(
1
-
s
)
+
∫
0
s
a
(
x
)
P
α
d
x
)
f
(
s
)
d
s
⁡
+
∫
0
t
exp
⁡
(
λ
(
t
-
s
)
+
∫
0
s
a
(
x
)
P
α
d
x
)
f
(
s
)
d
s
]
,
λ
∈
σ
(
L
k
), f
∈
L
2
(
0,1
) and for k
≠
0
,
-
1 spectrum of this extension L
k is in the form
(63)
σ
(
L
k
)
=
{
λ
∈
C
:
λ
=
ln
⁡
|
k
+
1
k
|
+
i
arg
(
k
+
1
k
)
+
2
n
π
i
,
n
∈
Z
}
.Example 13.
Let(64)
(
H
,
∥
·
∥
H
)
=
(
C
,
|
·
|
)
,
a
,
b
∈
C
(
C
)
.
Consider the following pantograph functional-differential expression in the form
(65)
l
(
u
)
=
u
′
(
t
)
+
a
(
t
)
u
(
t
)
+
b
(
t
)
u
(
α
t
)
,
0
<
α
<
1
,
in L
2
(
0,1
). Then by Theorem 1, all solvable extensions L
k of minimal operator L
0 are generated by l
(
·
) and boundary condition
(66)
(
k
+
1
)
u
(
0
)
=
k
exp
⁡
(
∫
0
1
(
a
(
s
)
+
b
(
s
)
P
α
)
d
s
)
u
(
1
)
,
hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
k
∈
C
and vice versa.
Moreover, the resolvent operator of these extensions is(67)
R
λ
(
L
k
)
f
(
t
)
=
exp
⁡

(
-
∫
0
t
(
a
(
s
)
+
b
(
s
)
P
α
)
d
s
)
×
[
(
1
+
k
(
1
-
e
λ
)
-
1
)
k
∫
0
1
exp
⁡

(
λ
(
1
-
s
)
+
∫
0
s
(
a
(
x
)
+
b
(
x
)
P
α
)
d
x
)
f
(
s
)
d
s
+
∫
0
t
exp
⁡

(
λ
(
t
-
s
)
+
∫
0
s
(
a
(
x
)
+
b
(
x
)
P
α
)
d
x
)
f
(
s
)
d
s
]
,
hhhhhhhhhhhhhhi
λ
∈
C
,
f
∈
L
2
(
0,1
)
.
On the other hand, by Theorem 9 for k
∈
C
∖
{
(
0,0
)
,
(
-
1,0
)
} the spectrum of solvable extension L
k is in the form
(68)
σ
(
L
k
)
=
{
λ
∈
C
:
λ
=
ln
⁡
|
τ
+
1
τ
|
+
i
arg
(
τ
+
1
τ
)
+
2
n
π
i
,
n
∈
Z
}
.
Now consider the following differential equation
(69)
u
′
(
t
)
=
a
(
t
)
u
(
α
t
)
+
b
(
t
)
u
(
t
)
+
f
(
t
)
with initial-boundary value problem
(70)
u
(
0
)
=
u
0
in the Hilbert space L
2
(
0,1
), where a
(
·
)
,
b
(
·
)
∈
C
[
0,1
], f
∈
L
2
(
0,1
), and u
0
∈
C.
In order to solve this problem change the functionu
(
t
) by
(71)
y
(
t
)
=
u
(
t
)
-
u
0
,
0
<
t
<
1
.
Then the considered problem transforms the following problem:(72)
y
′
(
t
)
=
a
(
t
)
y
(
α
t
)
+
b
(
t
)
y
(
t
)
+
g
(
t
)
,
y
(
0
)
=
0
,
where g
(
t
)
=
f
(
t
)
-
(
a
(
t
)
+
b
(
t
)
)
u
0.
The last problem can be written in the form(73)
y
′
(
t
)
+
A
α
(
t
)
y
(
t
)
=
g
(
t
)
,
y
(
0
)
=
0
,
where A
α
(
t
)
=
-
a
(
t
)
P
α
-
b
(
t
)
E.
Then solution of the above Cauchy problem by Corollary2 can be analytically expressed in the form (
K
=
0
)
(74)
y
(
t
)
=
R
0
(
L
c
)
g
(
t
)
=
L
c
-
1
g
(
t
)
=
U
(
t
,
0
)
∫
0
t
U
(
0
,
x
)
g
(
x
)
d
x
.
Another approach to this problem has been investigated in [17].Example 14.
Consider the following integrodifferential equation for first order in the form(75)
u
′
(
t
)
+
∫
0
t
u
(
α
x
)
d
x
=
f
(
t
)
,
0
<
α
<
1
,
u
(
0
)
=
u
0
in Hilbert space L
2
(
0,1
). Changing the unknown function u
(
t
) by
(76)
y
(
t
)
=
u
(
t
)
-
u
0
,
0
<
t
<
1
,
the following initial-value problem for integrodifferential equation is obtained:
(77)
y
′
(
t
)
+
∫
0
t
y
(
α
x
)
d
x
=
g
(
t
)
,
y
(
0
)
=
0
,
where g
(
t
)
=
f
(
t
)
-
u
0
t, in L
2
(
0,1
).
The last equation can be rewritten in the form(78)
y
′
(
t
)
+
P
α
y
(
t
)
=
g
(
t
)
,
y
(
0
)
=
0
.
It is easy to see that the analytical solution of this problem is in the form
(79)
y
(
t
)
=
∫
0
t
e
-
P
α
(
t
-
s
)
g
(
s
)
d
s
.
Consequently, for 0
<
t
<
1
(80)
u
(
t
)
=
u
0
+
∫
0
t
e
-
P
α
(
t
-
s
)
(
f
(
s
)
-
u
0
s
)
d
s
=
u
0
-
e
-
P
α
t
∫
0
t
s
e
P
α
s
d
s
u
0
+
e
-
P
α
t
∫
0
t
e
P
α
s
f
(
s
)
d
s
.Example 15.
All solvable extensions of minimal operator generated by differential expression(81)
l
(
u
)
=
∂
u
(
t
,
x
)
∂
t
+
x
u
(
α
t
,
x
)
,
x
∈
(
-
1,1
)
,
hhhhhhhhhhhhhhhi
0
<
t
<
1
,
0
<
α
<
1
,
in the Hilbert space L
2
(
(
-
1,1
)
×
(
0,1
)
) are described by this l
(
·
) and boundary condition
(82)
(
K
+
E
)
u
(
0
,
x
)
=
K
U
(
0,1
)
u
(
1
,
x
)
,
where K
∈
B
(
L
2
(
-
1,1
)
) and U
(
t
,
s
), t
,
s
∈
[
0,1
], is a solution of operator equation
(83)
U
t
′
(
t
,
s
)
+
x
P
α
U
(
t
,
s
)
=
0
,
t
,
s
∈
[
0,1
]
,
hhhhhhi
U
(
s
,
s
)
f
=
f
,
f
∈
L
2
(
-
1,1
)
,
where P
α
u
(
t
)
=
u
(
α
t
), P
α
:
L
2
(
0,1
)
→
L
2
(
0,1
).



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*Source: 837565-2014-08-14.xml*