# About Positivity of Green's Functions for Nonlocal Boundary Value Problems with Impulsive Delay Equations

Authors: Alexander Domoshnitsky; Irina Volinsky
Journal: The Scientific World Journal
 (2014)

---

## Abstract

The impulsive delay differential equation is considered(
L
x
)
(
t
)
=
x
′
(
t
)
+
∑
i
=
1
m
p
i
(
t
)
x
(
t
-
τ
i
(
t
)
)
=
f
(
t
)
,
t
∈
[
a
,
b
]
,
x
(
t
j
)
=
β
j
x
(
t
j
-
0
)
,
j
=
1
,
…
,
k
,
a
=
t
0
<
t
1
<
t
2
<
⋯
<
t
k
<
t
k
+
1
=
b
,
x
(
ζ
)
=
0
,
ζ
∉
[
a
,
b
]
, with nonlocal boundary condition l
x
=
∫
a
b
φ
s
x
′
s
d
s
+
θ
x
a
=
c
,
φ
∈
L
∞
a
,
b
;
θ
,
c
∈
R
. Various results on existence and uniqueness of solutions and on positivity/negativity of the Green's functions for this equation are obtained.


---

## Body

## 1. Introduction

Mathematical models with impulsive differential equations attract the topic attention of many researchers (see [1–7]); many important results on boundary value problems and stability of these equations have been obtained. One of possible approaches to study impulsive equations is the theory of generalized differential equations allowing researchers to consider systems with continuous solutions as well as systems with discontinuous solutions and discrete systems in the frame of the same theory [8–15]. In this paper we use the approach proposed in the monograph [1] and developed in [16–20].Various comparison theorems for solutions of the Cauchy and periodic problems for ordinary differential equations with impulses have been obtained in [17, 21–24]. On the basis of the comparison theorems, tests of stability are proposed in [25, 26]. Theory of impulsive differential equations and inclusions was presented in the book [27].Equations with nonlocal boundary conditions are applied in modelling different processes (see, e.g., the recent work of Skubachevskii [28]). Nonlocal problems for nonimpulsive functional differential equations were considered in ([25], Chapter 15). Existence results for nonlocal boundary value problems with impulsive equations are presented in [29–35]. There are almost no results on sign constancy of Green's function for impulsive boundary value problems. Concerning nonlocal impulsive boundary value problems, we know, there are no results about positivity/negativity of Green's function. In this paper we propose results of this sort.In this paper we consider the following impulsive equation:(1)
(
L
x
)
(
t
)
=
x
′
(
t
)
+
∑
i
=
1
m
p
i
(
t
)
x
(
t
-
τ
i
(
t
)
)
=
f
(
t
)
,
t
∈
[
a
,
b
]
,
(2)
x
(
t
j
)
=
β
j
x
(
t
j
-
0
)
,
j
=
1
,
…
,
k
,
(3)
a
=
t
0
<
t
1
<
t
2
<
⋯
<
t
k
<
t
k
+
1
=
b
,
(4)
x
(
ζ
)
=
0
,
ζ
∉
[
a
,
b
]
,
with different types of boundary conditions:
(5)
l
x
=
∫
a
b
‍
φ
(
s
)
x
′
(
s
)
d
s
+
θ
x
(
a
)
=
c
,
φ
∈
L
∞
[
a
,
b
]
;
θ
,
c
∈
R
,
θ
≠
0
,
(6)
x
(
a
)
=
c
,
(7)
x
(
a
)
=
γ
x
(
b
)
,
γ
∈
R
,
γ
≠
1
.We develop the ideas presented in [17] and we have obtained various results on the existence and uniqueness of solutions for impulsive boundary value problems. The main contribution of the presented paper is the formulation and proof of positivity/negativity conditions for Green's functions for the following impulsive functional differential boundary value problems: (1), (2), (4), and (5); (1), (2), (4), and (6); (1), (2), (4), (7).


## 2. Solution's Representation Formulas

Define the spaceD
(
t
1
,
…
,
t
k
) of piecewise continuous functions x
:
[
a
,
b
]
→
R, isomorphic to the topological product L
×
R, where L is the space of measurable essential bounded functions z
:
[
a
,
b
]
→
R, by the following equality:
(8)
x
(
t
)
=
∫
a
t
‍
Ω
(
t
,
s
)
z
(
s
)
d
s
+
ω
(
t
)
α
,
ω
(
t
)
=
∑
i
=
1
k
+
1
‍
χ
[
t
i
-
1
,
t
i
)
(
t
)
∏
j
=
1
i
‍
β
i
-
j
,
Ω
(
t
,
s
)
=
∑
i
=
1
k
+
1
‍
χ
[
t
i
-
1
,
t
i
)
(
t
)
χ
[
t
i
-
1
,
t
i
)
(
s
)
β
0
+
∑
i
=
2
k
+
1
‍
∑
r
=
1
i
-
1
‍
χ
[
t
i
-
1
,
t
i
)
(
t
)
χ
[
t
i
-
1
,
t
i
)
(
s
)
∏
j
=
1
i
-
r
‍
β
i
-
j
,
where a
=
t
0
<
t
1
<
t
2
<
⋯
<
t
k
<
t
k
+
1
=
b.It is clear thatx
(
t
) is absolutely continuous in (
t
i
-
1
,
t
i
), i
=
1
,
…
,
k
+
1, satisfying the equality x
(
t
i
)
=
β
i
x
(
t
i
-
0
). It is clear that x
(
t
) has discontinuity of the first kind and is continuous from the right at points t
i, i
=
1
,
…
,
k.By [17], the general solution of (1)–(4) has the following representation:
(9)
x
(
t
)
=
C
(
t
,
a
)
x
(
a
)
+
∫
a
t
C
(
t
,
s
)
f
(
s
)
d
s
.
‍Theorem 1.
If the boundary value problem (1)–(5) is uniquely solvable in the space D
(
t
1
,
…
,
t
k
) for every essential bounded f and c
∈
R, then its solution can be represented in the form
(10)
x
(
t
)
=
∫
a
b
‍
G
(
t
,
s
)
f
(
s
)
d
s
+
C
(
t
,
a
)
c
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
,
where Green's function G
(
t
,
s
) of this problem is
(11)
G
(
t
,
s
)
=
C
(
t
,
s
)
-
C
(
t
,
a
)
∫
s
b
‍
φ
(
w
)
C
w
′
(
w
,
s
)
d
w
+
φ
(
s
)
θ
+
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
d
w
and C
(
t
,
s
)
=
0  for t
<
s
.Proof.
From the boundary condition (5), we obtain
(12)
∫
a
b
‍
φ
(
s
)
x
′
(
s
)
d
s
+
θ
x
(
a
)
=
c
⟹
x
(
a
)
=
c
θ
-
1
θ
∫
a
b
‍
φ
(
s
)
x
′
(
s
)
d
s
.
Substituting (12) into (9), we obtain
(13)
x
(
t
)
=
C
(
t
,
a
)
θ
(
c
-
∫
a
b
‍
φ
(
s
)
x
′
(
s
)
d
s
)
+
∫
a
t
‍
C
(
t
,
s
)
f
(
s
)
d
s
.
The Cauchy functionC
(
t
,
s
) can be represented in the following form:
(14)
C
(
t
,
s
)
=
∑
i
=
0
k
‍
∑
j
=
0
k
‍
C
i
j
(
t
,
s
)
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
×
[
H
t
j
(
s
)
-
H
t
j
+
1
(
s
)
]
,
where C
i
j
(
t
,
s
)
=
C
(
t
,
s
) in the rectangle t
i
≤
t
<
t
i
+
1, s
i
≤
s
<
s
i
+
1.
Let us take derivative in (9):
(15)
x
′
(
t
)
=
C
t
′
(
t
,
a
)
x
(
a
)
+
d
d
t
(
∫
a
t
‍
C
(
t
,
s
)
f
(
s
)
d
s
)
=
C
t
′
(
t
,
a
)
x
(
a
)
+
∫
a
t
‍
C
t
′
(
t
,
s
)
f
(
s
)
d
s
+
C
(
t
,
t
)
f
(
t
)
.
Since, according to the definition of the Cauchy function,C
(
t
,
t
)
=
1, we obtain(16)
x
′
(
t
)
=
C
t
′
(
t
,
a
)
x
(
a
)
+
∫
a
t
‍
C
t
′
(
t
,
s
)
f
(
s
)
d
s
+
f
(
t
)
.
Let us substitute (16) into (13) as follows:
(17)
x
(
t
)
=
C
(
t
,
a
)
θ
×
(
c
-
∫
a
b
‍
φ
(
w
)
[
∫
a
w
‍
C
w
′
(
w
,
τ
)
f
(
τ
)
d
τ
+
f
(
w
)
C
w
′
(
w
,
a
)
x
(
a
)
+
∫
a
w
‍
C
w
′
(
w
,
τ
)
f
(
τ
)
d
τ
+
f
(
w
)
]
d
w
)
+
∫
a
t
‍
C
(
t
,
s
)
f
(
s
)
d
s
=
c
θ
C
(
t
,
a
)
-
C
(
t
,
a
)
θ
×
∫
a
b
‍
φ
(
w
)
[
∫
a
w
‍
C
w
′
(
w
,
τ
)
C
w
′
(
w
,
a
)
x
(
a
)
+
∫
a
w
‍
C
w
′
(
w
,
τ
)
f
(
τ
)
d
τ
+
f
(
w
)
]
d
w
+
∫
a
t
‍
C
(
t
,
s
)
f
(
s
)
d
s
=
c
θ
C
(
t
,
a
)
+
∫
a
t
‍
C
(
t
,
s
)
f
(
s
)
d
s
-
C
(
t
,
a
)
θ
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
x
(
a
)
d
w
-
C
(
t
,
a
)
θ
∫
a
b
‍
φ
(
w
)
(
∫
a
w
‍
C
w
′
(
w
,
τ
)
f
(
τ
)
d
τ
)
d
w
-
C
(
t
,
a
)
θ
∫
a
b
‍
φ
(
w
)
f
(
w
)
d
w
.
SinceC
(
t
,
s
)
=
0, s
>
t, it follows that ∫
a
t
C
(
t
,
s
)
d
s
=
∫
a
t
C
(
t
,
s
)
d
s
+
∫
t
b
C
(
t
,
s
)
d
s
=
∫
a
b
C
(
t
,
s
)
d
s.
Changing the order of integration in the third double integral, we get(18)
x
(
t
)
=
c
θ
C
(
t
,
a
)
+
∫
a
b
‍
C
(
t
,
w
)
f
(
w
)
d
w
-
C
(
t
,
a
)
θ
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
x
(
a
)
d
w
-
C
(
t
,
a
)
θ
∫
a
b
‍
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
f
(
w
)
d
s
d
w
-
C
(
t
,
a
)
θ
∫
a
b
‍
φ
(
w
)
f
(
w
)
d
w
=
c
θ
C
(
t
,
a
)
+
∫
a
b
‍
[
C
(
t
,
w
)
f
(
w
)
-
C
(
t
,
a
)
θ
φ
(
w
)
C
w
′
(
w
,
a
)
x
(
a
)
-
C
(
t
,
a
)
θ
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
f
(
w
)
d
s
-
C
(
t
,
a
)
θ
φ
(
w
)
f
(
w
)
]
d
w
=
c
θ
C
(
t
,
a
)
+
∫
a
b
‍
[
[
(
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
+
φ
(
w
)
)
C
(
t
,
w
)
-
C
(
t
,
a
)
θ
×
(
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
+
φ
(
w
)
)
]
f
(
w
)
]
d
w
-
∫
a
b
‍
C
(
t
,
a
)
θ
φ
(
w
)
C
w
′
(
w
,
a
)
x
(
a
)
d
w
=
∫
a
b
‍
[
(
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
+
φ
(
w
)
)
C
(
t
,
w
)
-
C
(
t
,
a
)
θ
×
(
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
+
φ
(
w
)
)
]
f
(
w
)
d
w
+
C
(
t
,
a
)
θ
(
c
-
x
(
a
)
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
d
w
)
.
Substituting now (16) into (12), we obtain
(19)
x
(
a
)
=
c
θ
-
1
θ
∫
a
b
‍
φ
(
s
)
×
[
C
s
′
(
s
,
a
)
x
(
a
)
+
∫
a
s
‍
C
s
′
(
s
,
w
)
f
(
w
)
d
w
+
f
(
s
)
]
d
s
=
c
θ
-
x
(
a
)
1
θ
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
-
1
θ
∫
a
b
‍
φ
(
s
)
[
∫
a
s
‍
C
s
′
(
s
,
w
)
f
(
w
)
d
w
+
f
(
s
)
]
d
s
,
x
(
a
)
[
1
θ
(
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
)
]
=
1
θ
(
c
-
∫
a
b
‍
φ
(
s
)
[
∫
a
s
‍
C
s
′
(
s
,
w
)
f
(
w
)
d
w
+
f
(
s
)
]
d
s
)
,
x
(
a
)
=
c
-
∫
a
b
‍
φ
(
s
)
[
∫
a
s
‍
C
s
′
(
s
,
w
)
f
(
w
)
d
w
+
f
(
s
)
]
d
s
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
=
-
∫
a
b
‍
φ
(
s
)
f
(
s
)
d
s
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
+
c
-
∫
a
b
‍
φ
(
s
)
[
∫
a
s
‍
C
s
′
(
s
,
w
)
f
(
w
)
d
w
]
d
s
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
=
-
∫
a
b
‍
φ
(
s
)
f
(
s
)
d
s
+
∫
a
b
‍
f
(
w
)
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
d
w
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
+
c
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
.
Substituting (19) into (18), we obtain
(20)
x
(
t
)
=
∫
a
b
‍
[
(
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
+
φ
(
w
)
)
C
(
t
,
w
)
-
C
(
t
,
a
)
θ
×
(
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
+
φ
(
w
)
)
]
f
(
w
)
d
w
+
C
(
t
,
a
)
θ
(
c
+
[
(
∫
a
b
‍
φ
(
s
)
f
(
s
)
d
s
+
∫
a
b
‍
f
(
w
)
×
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
d
w
-
c
)
×
(
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
)
-
1
]
×
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
d
w
)
,
x
(
t
)
=
∫
a
b
[
C
(
t
,
w
)
-
C
(
t
,
a
)
θ
(
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
+
φ
(
w
)
)
+
(
C
(
t
,
a
)
/
θ
)
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
×
[
φ
(
w
)
+
∫
w
b
‍
φ
(
s
)
C
s
′
(
s
,
w
)
d
s
]
]
f
(
w
)
d
w
+
C
(
t
,
a
)
θ
(
c
-
c
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
)
=
∫
a
b
‍
G
(
t
,
w
)
f
(
w
)
d
w
+
g
(
t
)
.
It is clear that Green's function G
(
t
,
s
) of this problem is of the following form:
(21)
G
(
t
,
s
)
=
C
(
t
,
s
)
-
C
(
t
,
a
)
θ
(
∫
s
b
‍
φ
(
τ
)
C
τ
′
(
τ
,
s
)
d
τ
+
φ
(
s
)
)
+
(
C
(
t
,
a
)
/
θ
)
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
d
w
θ
+
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
d
w
×
[
φ
(
s
)
+
∫
s
b
‍
φ
(
w
)
C
w
′
(
w
,
s
)
d
w
]
=
C
(
t
,
s
)
-
C
(
t
,
a
)
θ
(
∫
s
b
‍
φ
(
τ
)
C
τ
′
(
τ
,
s
)
d
τ
+
φ
(
s
)
)
×
[
1
-
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
d
w
θ
+
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
d
w
]
=
C
(
t
,
s
)
-
C
(
t
,
a
)
∫
s
b
‍
φ
(
w
)
C
w
′
(
w
,
s
)
d
w
+
φ
(
s
)
θ
+
∫
a
b
‍
φ
(
w
)
C
w
′
(
w
,
a
)
d
w
,
g
(
t
)
=
C
(
t
,
a
)
θ
(
c
-
c
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
)
=
C
(
t
,
a
)
c
θ
+
∫
a
b
‍
φ
(
s
)
C
s
′
(
s
,
a
)
d
s
,
and C
(
t
,
s
)
=
0, for t
<
s.
One has(22)
x
(
t
)
=
∫
a
b
‍
G
(
t
,
s
)
f
(
s
)
d
s
+
g
(
t
)
.
Theorem 1 has been proven.Theorem1 demonstrates the importance to know exactly or approximately the Cauchy function C
(
t
,
s
) of impulsive equations (1)–(4).Substitutingc
=
0, φ
(
s
)
≡
γ, θ
=
γ
-
1 we obtain the Corollary.Corollary 2.
If generalized periodic problem (1)–(4), (7) is uniquely solvable, then its solution can be represented in the following form:
(23)
x
(
t
)
=
∫
a
b
‍
W
(
t
,
s
)
f
(
s
)
d
s
,
where the Green's function W
(
t
,
s
) is as follows:
(24)
W
(
t
,
s
)
=
C
(
t
,
s
)
-
C
(
t
,
a
)
γ
∫
s
b
‍
C
w
′
(
w
,
s
)
d
w
+
γ
(
γ
-
1
)
+
γ
∫
a
b
‍
C
w
′
(
w
,
a
)
d
w
.


## 3. Auxiliary Results

Let us construct the Cauchy functionC
(
t
,
s
) for several simple equations.Consider now the following auxiliary equation:(25)
x
′
(
t
)
+
p
(
t
)
x
(
t
)
=
f
(
t
)
.Let us denote the Cauchy function of the nonimpulsive equation (25) by K
(
t
,
s
). It is known that K
(
t
,
s
)
=
e
-
∫
s
t
‍
p
(
ζ
)
d
ζ. For every s this function is absolutely continuous function with respect to t, and K
(
t
,
s
)
=
0 for s
>
t.Using the fact that the Cauchy functionC
(
t
,
s
) for every fixed s as a function of t is a solution of
(26)
x
′
(
t
)
+
p
(
t
)
x
(
t
)
=
0
,
t
∈
[
s
,
b
]
satisfying the condition C
(
s
,
s
)
=
1, we obtain the following theorem.Theorem 3.
The Cauchy functionC
(
t
,
s
) of (25) and (2) can be represented in the following form:
(27)
C
(
t
,
s
)
=
∑
i
=
1
k
‍
∑
j
=
0
i
-
1
‍
∏
r
=
j
+
1
i
‍
β
r
K
(
t
,
s
)
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
×
[
H
t
j
(
s
)
-
H
t
j
+
1
(
s
)
]
+
∑
i
=
0
k
‍
H
0
(
t
-
s
)
K
(
t
,
s
)
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
×
[
H
t
i
(
s
)
-
H
t
i
+
1
(
s
)
]
,
where H
a
(
t
), Heaviside function, is defined by the following equality:
(28)
H
a
(
t
)
=
{
1
,
t
≥
a
,
0
,
t
<
a
.Now let us consider the following auxiliary equation:(29)
x
′
(
t
)
=
f
(
t
)
,
t
∈
[
a
,
b
]
,
x
(
t
j
)
=
β
j
x
(
t
j
-
0
)
,
j
=
1
,
…
,
k
,
a
=
t
0
<
t
1
<
t
2
<
⋯
<
t
k
<
t
k
+
1
=
b
.Figure1 describes the Cauchy function C
0
(
t
,
s
) of the problem (29), (2), and (4) in the case k
=
3.Figure 1In the case ofk impulses we obtain the following theorem.Theorem 4.
The Cauchy functionC
0
(
t
,
s
) of the problem (29), (2) and its derivatives has the following representations:
(30)
C
0
(
t
,
s
)
=
∑
i
=
1
k
‍
∑
j
=
0
i
-
1
‍
∏
r
=
j
+
1
i
‍
β
r
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
×
[
H
t
j
(
s
)
-
H
t
j
+
1
(
s
)
]
+
∑
i
=
0
k
‍
H
0
(
t
-
s
)
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
×
[
H
t
i
(
s
)
-
H
t
i
+
1
(
s
)
]
,
C
0
(
t
,
a
)
=
∑
i
=
1
k
‍
∏
r
=
1
i
‍
β
r
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
+
H
0
(
t
-
a
)
[
H
t
0
(
t
)
-
H
t
1
(
t
)
]
,
(
C
0
(
t
,
s
)
)
t
′
=
∑
i
=
1
k
‍
∑
j
=
0
i
-
1
‍
∏
r
=
j
+
1
i
‍
β
r
[
H
t
j
(
s
)
-
H
t
j
+
1
(
s
)
]
×
[
δ
t
i
(
t
)
-
δ
t
i
+
1
(
t
)
]
+
∑
i
=
0
k
‍
[
H
t
i
(
s
)
-
H
t
i
+
1
(
s
)
]
×
(
δ
0
(
t
-
s
)
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
+
H
0
(
t
-
s
)
[
δ
t
i
(
t
)
-
δ
t
i
+
1
(
t
)
]
)
,
(
C
0
(
t
,
a
)
)
t
′
=
∑
i
=
1
k
‍
∏
r
=
1
i
‍
β
r
[
δ
t
i
(
t
)
-
δ
t
i
+
1
(
t
)
]
+
δ
0
(
t
-
a
)
[
H
t
0
(
t
)
-
H
t
1
(
t
)
]
+
H
0
(
t
-
a
)
[
δ
t
0
(
t
)
-
δ
t
1
(
t
)
]
.Denote byG
0
(
t
,
s
) Green's function of problem (29), (5), and by W
0
(
t
,
s
) Green's function of generalized periodic problem (29), (7).Let us denoteI
(
φ
,
θ
,
β
)
=
θ
+
∫
a
b
φ
(
t
)
(
C
0
(
t
,
a
)
)
t
′
d
t.Letβ
=
C
o
l
{
β
1
,
…
,
β
k
}. The computation of integrals in formula (11) leads us to the following equality:
(31)
I
(
φ
,
θ
,
β
)
=
θ
+
∑
i
=
1
k
-
1
‍
∏
r
=
1
i
‍
β
r
×
[
φ
(
t
i
+
)
+
φ
(
t
i
-
)
2
-
φ
(
t
i
+
1
+
)
+
φ
(
t
i
+
1
-
)
2
]
+
β
1
β
2
⋯
β
k
[
φ
(
t
k
+
)
+
φ
(
t
k
-
)
2
-
φ
(
t
k
+
1
-
)
2
]
+
φ
(
t
0
+
)
2
-
φ
(
t
1
+
)
+
φ
(
t
1
-
)
2
+
φ
(
t
0
+
)
2
.DenotingI
1
(
φ
,
β
)
=
φ
(
s
)
+
∫
s
b
φ
(
t
)
(
C
0
(
t
,
s
)
)
t
′
d
t, we obtain for a
≤
s
<
t
1
(32)
I
1
(
φ
,
β
)
=
φ
(
s
)
+
∑
i
=
1
k
-
1
‍
∏
r
=
1
i
‍
β
r
×
[
φ
(
t
i
+
)
+
φ
(
t
i
-
)
2
-
φ
(
t
i
+
1
+
)
+
φ
(
t
i
+
1
-
)
2
]
+
β
1
β
2
⋯
β
k
[
φ
(
t
k
+
)
+
φ
(
t
k
-
)
2
-
φ
(
t
k
+
1
-
)
2
]
+
φ
(
s
+
)
2
+
H
0
(
t
0
-
s
)
φ
(
t
0
+
)
2
-
φ
(
t
1
+
)
+
φ
(
t
1
-
)
2
.DenotingI
2
(
φ
,
β
)
=
φ
(
s
)
+
∫
s
b
φ
(
t
)
(
C
0
(
t
,
s
)
)
t
′
d
t, we obtain for t
1
≤
s
<
t
2
(33)
I
2
(
φ
,
β
)
=
φ
(
s
)
+
∑
i
=
2
k
-
1
‍
∏
r
=
2
i
‍
β
r
×
[
φ
(
t
i
+
)
+
φ
(
t
i
-
)
2
-
φ
(
t
i
+
1
+
)
+
φ
(
t
i
+
1
-
)
2
]
+
β
2
β
3
⋯
β
k
[
φ
(
t
k
+
)
+
φ
(
t
k
-
)
2
-
φ
(
t
k
+
1
-
)
2
]
+
φ
(
s
+
)
2
+
H
0
(
t
1
-
s
)
φ
(
t
1
+
)
2
-
φ
(
t
2
+
)
+
φ
(
t
2
-
)
2
⋮DenotingI
k
+
1
(
φ
,
β
)
=
φ
(
s
)
+
∫
s
b
φ
(
t
)
(
C
0
(
t
,
s
)
)
t
′
d
t, we obtain for t
k
≤
s
<
b
(34)
I
k
+
1
(
φ
,
β
)
=
φ
(
s
)
+
φ
(
s
+
)
2
+
H
0
(
t
k
-
s
)
φ
(
t
k
+
)
2
-
φ
(
t
k
+
1
-
)
2
.Let us denoteJ
(
β
,
γ
)
=
(
γ
-
1
)
+
γ
∫
a
b
(
C
0
(
t
,
a
)
)
t
′
d
t.The computation of integrals in formula (24) leads us to the following equality:
(35)
J
(
β
,
γ
)
=
(
γ
-
1
)
+
γ
(
∑
i
=
1
k
‍
∏
r
=
1
i
‍
β
r
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
+
1
)
.DenotingJ
1
(
β
,
γ
)
=
γ
∫
s
b
(
C
0
(
t
,
s
)
)
t
′
d
t
+
γ, we obtain for a
≤
s
<
t
1
(36)
J
1
(
β
,
γ
)
=
γ
(
∑
i
=
1
k
‍
∏
r
=
1
i
‍
β
r
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
+
H
t
0
(
s
)
+
H
0
(
t
0
-
s
)
-
1
∑
i
=
1
k
‍
)
+
γ
.DenotingJ
2
(
β
,
γ
)
=
γ
∫
s
b
(
C
0
(
t
,
s
)
)
t
′
d
t
+
γ, we obtain for t
1
≤
s
<
t
2
(37)
J
2
(
β
,
γ
)
=
γ
(
∑
i
=
2
k
‍
∏
r
=
1
i
‍
β
r
[
H
t
i
(
t
)
-
H
t
i
+
1
(
t
)
]
+
H
t
1
(
s
)
+
H
0
(
t
1
-
s
)
∑
i
=
2
k
‍
)
+
γ
⋮DenotingJ
k
+
1
(
β
)
=
γ
∫
s
b
(
C
0
(
t
,
s
)
)
t
′
d
t
+
γ, we obtain for t
k
≤
s
<
b
(38)
J
k
+
1
(
β
)
=
γ
(
H
t
k
(
s
)
-
H
t
k
+
1
(
s
)
+
H
0
(
t
k
-
s
)
-
H
0
(
t
k
+
1
-
s
)
H
t
k
(
s
)
-
H
t
k
+
1
(
s
)
+
H
0
(
t
k
-
s
)
)
+
γ
=
γ
(
H
t
k
(
s
)
+
H
0
(
t
k
-
s
)
)
+
γ
.Corollary 5.
Let us assume thatI
(
φ
,
θ
,
β
)
≠
0; then Green's function G
0
(
t
,
s
) of problem (29), (5) in the case of three impulsive points (i.e., k
=
3
) is described by Figure 2.Figure 2Corollary 6.
Let us assume thatJ
(
β
,
γ
)
≠
0; then Green's function W
0
(
t
,
s
) of generalized periodic problem (29), (7) in the case of three impulsive points (i.e. k
=
3) is described by Figure 3.Figure 3


## 4. Estimation of Solutions

Let us denoteβ
=
max
⁡
1
≤
i
≤
k
⁡
β
i, Ω
=
(
1
/
|
I
(
φ
,
θ
,
β
)
|
)
|
(
1
-
β
k
+
1
)
/
(
1
-
β
)
|
(
(
2
k
+
1
)
/
2
)
Δ where Δ
=
(
max
⁡
1
≤
i
≤
k
+
1
⁡
|
t
i
-
t
i
-
1
|
)
2,
(39)
Ω
~
=
Ω
[
|
θ
|
+
∥
φ
∥
(
5
+
4
∑
i
=
1
k
‍
β
i
)
+
(
max
⁡
1
≤
i
≤
k
+
1
β
i
-
1
)
∥
φ
∥
(
3
+
2
∑
i
=
1
k
‍
β
i
)
]
.Theorem 7.
Let0
<
β
i, 1
≤
i
≤
k,  I
(
φ
,
θ
,
β
)
≠
0 and
(40)
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
n
|
p
i
(
s
)
|
Ω
~
<
1
;
then there exists a unique solution of problem (1)–(5) and this solution can be estimated as follows:
(41)
∥
x
∥
≤

Ω
~
(
b
-
a
)
∥
f
∥
|
I
(
φ
,
θ
,
β
)
|

+
|
c
|
max
⁡

[
max
⁡
1
≤
j
≤
k
⁡
∏
i
=
1
j
‍
β
i
,
1
]
(
1
-
Ω
~
)
|
I
(
φ
,
θ
,
β
)
|
,
for every essentially bounded function f and real c.Proof.
By Theorem1, problem (1)–(5) is equivalent to the integral equation:
(42)
x
(
t
)
=
∫
a
b
‍
G
0
(
t
,
s
)
ϕ
(
s
)
d
s
+
C
0
(
t
,
a
)
c
θ
+
∫
a
b
‍
φ
(
s
)
(
C
0
(
s
,
a
)
)
s
′
d
s
=
∫
a
b
‍
G
0
(
t
,
s
)
f
(
s
)
d
s
+
C
0
(
t
,
a
)
c
θ
+
∫
a
b
‍
φ
(
s
)
(
C
0
(
s
,
a
)
)
s
′
d
s
-
∫
a
b
‍
G
0
(
t
,
s
)
∑
i
=
1
m
‍
p
i
(
s
)
x
(
s
-
τ
i
(
s
)
)
d
s
.
Let us denoteψ
(
t
)
=
∫
a
b
G
0
(
t
,
s
)
f
(
s
)
d
s
+
C
0
(
t
,
a
)
(
c
/
(
θ
+
∫
a
b
φ
(
s
)
(
C
0
(
s
,
a
)
)
s
′
d
s
)
)
=
∫
a
b
G
0
(
t
,
s
)
f
(
s
)
d
s
+
(
c
C
0
(
t
,
a
)
/
I
(
φ
,
θ
,
β
)
).
Define the operatorM
:
D
(
t
1
,
…
,
t
k
)
→
D
(
t
1
,
…
,
t
k
) by the equality (
M
x
)
(
t
)
=
-
∫
a
b
G
0
(
t
,
s
)
∑
i
=
1
m
p
i
(
s
)
x
(
s
-
τ
i
(
s
)
)
d
s.
We can write equation which is equivalent to problem (1)–(5):
(43)
x
(
t
)
=
(
M
x
)
(
t
)
+
ψ
(
t
)
.
If∥
M
∥
<
1, then there exists the unique solution of (1)–(5) which can be represented in the form: x
(
t
)
=
(
I
-
M
)
-
1
ψ
(
t
).
It is clear that(44)
∥
M
x
∥
≤
∥
G
0
∥
∑
i
=
1
m
‍
∫
a
b
‍
|
p
i
(
s
)
|
|
x
(
s
-
τ
i
(
s
)
)
|
χ
[
a
,
b
]
×
(
s
-
τ
i
(
s
)
)
d
s
≤
∥
G
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
∥
x
∥
,
where ∥
x
∥
=
max
⁡
a
≤
t
≤
b
⁡
|
x
(
t
)
|. Thus we have got the following condition for existence of unique solution of problem (1)–(5):
(45)
∥
M
∥
≤
∥
G
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
n
|
p
i
(
s
)
|
<
1
.
Denote(46)
g
(
t
)
=
c
C
0
(
t
,
a
)
I
(
φ
,
θ
,
β
)
,
where C
0
(
t
,
a
) can be written as
(47)
C
0
(
t
,
a
)
=
{
1
,
a
≤
t
<
t
1
,
β
1
,
t
1
≤
t
<
t
2
,
β
1
β
2
,
t
2
≤
t
<
t
3
,
⋮
β
1
·
…
·
β
k
,
t
k
≤
t
<
b
.
It is clear that(48)
∥
g
∥
≤
|
c
|
max
⁡
[
max
⁡
1
≤
j
≤
k
⁡
∏
i
=
1
j
‍
β
i
,
1
]
|
I
(
φ
,
θ
,
β
)
|
.
Estimation of the solution of the problem (1)–(5) can be done as follows:
(49)
|
x
(
t
)
|
≤
∥
G
0
∥
∫
a
b
‍
|
f
(
s
)
|
d
s
+
∥
g
∥
+
∥
M
x
∥
⟹
∥
x
∥
≤
∥
G
0
∥
(
b
-
a
)
ess
sup
⁡
t
∈
[
a
,
b
]
|
f
(
t
)
|
+
∥
g
∥
+
∥
G
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
n
|
p
i
(
s
)
|
∥
x
∥
⟹
∥
x
∥
(
1
-
∥
G
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
n
|
p
i
(
s
)
|
)
≤
∥
G
0
∥
(
b
-
a
)
∥
f
∥
+
∥
g
∥
.
Let us denote(50)
Q
=
∥
G
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
n
|
p
i
(
s
)
|
.
Then the problem (1)–(5) has unique solution if
(51)
∥
M
∥
≤
Q
<
1
.
It is clear that(52)
∥
x
∥
≤
∥
G
0
∥
(
b
-
a
)
∥
f
∥
+
∥
g
∥
1
-
Q
.
Estimating the integrals in formula (11), we get the following inequalities:
(53)
|
I
(
φ
,
θ
,
β
)
|
≤
|
θ
|
+
2
∥
φ
∥
+
2
∥
φ
∥
∑
i
=
1
k
‍
β
i
=
|
θ
|
+
∥
φ
∥
(
2
+
2
∑
i
=
1
k
‍
β
i
)
,
|
I
1
(
φ
,
β
)
|
≤
∥
φ
∥
+
2
∥
φ
∥
∑
i
=
1
k
‍
β
i
+
2
∥
φ
∥
=
3
∥
φ
∥
+
2
∥
φ
∥
∑
i
=
1
k
‍
β
i
=
∥
φ
∥
(
3
+
2
∑
i
=
1
k
‍
β
i
)
,
|
I
2
(
φ
,
β
)
|
≤
∥
φ
∥
+
2
∥
φ
∥
∑
i
=
2
k
‍
β
i
+
2
∥
φ
∥
=
3
∥
φ
∥
+
2
∥
φ
∥
∑
i
=
2
k
‍
β
i
=
∥
φ
∥
(
3
+
2
∑
i
=
2
k
‍
β
i
)
,
⋮
|
I
k
+
1
(
φ
,
β
)
|
≤
2
∥
φ
∥
.
Let us writeG
0
=
G
1
+
G
2, where G
1 is upper triangular and G
2 lower triangular parts of G
0 and estimate ∥
G
0
∥. One has
(54)
∥
G
0
∥
≤
∥
G
1
∥
+
∥
G
2
∥
≤
Ω
[
max
⁡
1
≤
i
≤
k
+
1
|
I
i
(
φ
,
β
)
|
+
max
⁡
1
≤
i
≤
k
+
1
|
I
(
φ
,
θ
,
β
)
-
β
i
-
1
I
i
(
φ
,
β
)
|
]
≤
Ω
[
max
⁡
1
≤
i
≤
k
+
1
|
I
i
(
φ
,
β
)
|
+
|
I
(
φ
,
θ
,
β
)
|
+
max
⁡
1
≤
i
≤
k
+
1
|
I
i
(
φ
,
β
)
|
β
i
-
1
]
=
Ω
[
|
I
(
φ
,
θ
,
β
)
|
+
max
⁡
1
≤
i
≤
k
+
1
|
I
i
(
φ
,
β
)
|
(
β
i
-
1
+
1
)
]
.
From estimation of the integrals, we obtain(55)
∥
G
0
∥
≤
Ω
[
|
θ
|
+
∥
φ
∥
(
2
+
2
∑
i
=
1
k
‍
β
i
)
+
(
1
+
max
⁡
1
≤
i
≤
k
+
1
β
i
-
1
)
∥
φ
∥
(
3
+
2
∑
i
=
1
k
‍
β
i
)
]
≤
Ω
[
|
θ
|
+
∥
φ
∥
(
5
+
4
∑
i
=
1
k
‍
β
i
)
+
(
max
⁡
1
≤
i
≤
k
+
1
β
i
-
1
)
∥
φ
∥
(
3
+
2
∑
i
=
1
k
‍
β
i
)
]
=
Ω
~
.
By (50)
(56)
Q
≤
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
n
|
p
i
(
s
)
|
Ω
×
[
|
θ
|
+
∥
φ
∥
(
5
+
4
∑
i
=
1
k
‍
β
i
)
+
(
max
⁡
1
≤
i
≤
k
+
1
β
i
-
1
)
∥
φ
∥
(
3
+
2
∑
i
=
1
k
‍
β
i
)
]
.
Substituting (56) into (51) and (52) we obtain
(57)
∥
M
∥
≤
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
n
|
p
i
(
s
)
|
Ω
×
[
|
θ
|
+
∥
φ
∥
(
5
+
4
∑
i
=
1
k
‍
β
i
)
+
(
max
⁡
1
≤
i
≤
k
+
1
β
i
-
1
)
∥
φ
∥
(
3
+
2
∑
i
=
1
k
‍
β
i
)
]
,
(58)
∥
x
∥
≤
(
|
c
|
max
⁡
[
max
⁡
1
≤
j
≤
k
⁡
∏
i
=
1
j
‍
β
i
,
1
]
|
I
(
φ
,
θ
,
β
)
|
Ω
~
(
b
-
a
)
∥
f
∥
+
|
c
|
max
⁡
[
max
⁡
1
≤
j
≤
k
⁡
∏
i
=
1
j
‍
β
i
,
1
]
|
I
(
φ
,
θ
,
β
)
|
)
(
1
-
Ω
~
)
-
1
=
(
[
max
⁡
1
≤
j
≤
k
∏
i
=
1
j
‍
β
i
,
1
]
Ω
~
(
b
-
a
)
∥
f
∥
|
I
(
φ
,
θ
,
β
)
|
+
|
c
|
max
⁡
[
max
⁡
1
≤
j
≤
k
∏
i
=
1
j
‍
β
i
,
1
]
)
(
(
1
-
Ω
~
)
|
I
(
φ
,
θ
,
β
)
|
)
-
1
.
Theorem7 has been proven.Let us denoteΩ
2
=
|
1
-
β
k
+
1
|
(
2
k
+
1
)
Δ
/
2
|
1
-
β
|
|
J
(
β
,
γ
)
|,
(59)
Ω
2
~
=
Ω
2
[
|
γ
|
(
5
+
2
∑
i
=
1
k
‍
β
i
)
+
max
⁡
1
≤
i
≤
k
+
1
β
i
-
1
|
γ
|
(
3
+
∑
i
=
1
k
‍
β
i
)
]
.Theorem 8.
Let0
<
β
i, 1
≤
i
≤
k,  J
(
β
,
γ
)
≠
0 and
(60)
Ω
2
~
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
<
1
;
then there exists a unique solution of the generalized periodic problem (1)–(4), (7) and this solution can be estimated as follows:
(61)
∥
x
∥
≤
Ω
2
~
(
b
-
a
)
∥
f
∥
1
-
Ω
2
~
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
for every essentially bounded function f.Proof.
By Corollary2, problem (1)–(4), (7) is equivalent to the following integral equation:
(62)
x
(
t
)
=
∫
a
b
‍
W
0
(
t
,
s
)
ϕ
(
s
)
d
s
=
∫
a
b
‍
W
0
(
t
,
s
)
f
(
s
)
d
s
-
∫
a
b
‍
W
0
(
t
,
s
)
∑
i
=
1
m
‍
p
i
(
s
)
x
(
s
-
τ
i
(
s
)
)
d
s
.
Let us denoteψ
(
t
)
=
∫
a
b
W
0
(
t
,
s
)
f
(
s
)
d
s.
Define the operatorT
:
D
(
t
1
,
…
,
t
k
)
→
D
(
t
1
,
…
,
t
k
) by the equality (
T
x
)
(
t
)
=
-
∫
a
b
W
0
(
t
,
s
)
∑
i
=
1
m
p
i
(
s
)
x
(
s
-
τ
i
(
s
)
)
d
s.
We can write equation which is equivalent to problem (1)–(4), (7) as follows:
(63)
x
(
t
)
=
(
T
x
)
(
t
)
+
ψ
(
t
)
.
If∥
T
∥
<
1, then there exists the unique solution of (1)–(4), (7) which can be represented in the form: x
(
t
)
=
(
I
-
M
)
-
1
ψ
(
t
).
It is clear that(64)
∥
T
x
∥
≤
∥
W
0
∥
∑
i
=
1
m
‍
∫
a
b
‍
|
p
i
(
s
)
|
|
x
(
s
-
τ
i
(
s
)
)
|
χ
[
a
,
b
]
×
(
s
-
τ
i
(
s
)
)
d
s
≤
∥
W
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
∥
x
∥
,
where ∥
x
∥
=
max
⁡
a
≤
t
≤
b
⁡
|
x
(
t
)
|. Thus we have got the following condition for existence of unique solution of problem (1)–(4),(7):
(65)
∥
T
∥
≤
∥
W
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
n
|
p
i
(
s
)
|
<
1
.
Estimation of the solution of the problem (1)–(4),(7) can be done as follows:
(66)
|
x
(
t
)
|
≤
∥
W
0
∥
∫
a
b
‍
|
f
(
s
)
|
d
s
+
∥
T
x
∥
⟹
∥
x
∥
≤
∥
W
0
∥
(
b
-
a
)
ess
sup
⁡
t
∈
[
a
,
b
]
|
f
(
t
)
|
+
∥
W
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
∥
x
∥
⟹
∥
x
∥
(
1
-
∥
W
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
)
≤
∥
W
0
∥
(
b
-
a
)
∥
f
∥
.
Let us denote(67)
Q
=
∥
W
0
∥
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
.
Then the problem (1)–(4), (7) has unique solution if
(68)
∥
T
∥
≤
Q
<
1
.
It is clear that(69)
∥
x
∥
≤
∥
W
0
∥
(
b
-
a
)
∥
f
∥
1
-
Q
.
Estimating the integrals in formula (24), we get the following inequalities:
(70)
|
J
(
β
,
γ
)
|
≤
|
γ
|
+
|
γ
|
(
1
+
∑
i
=
1
k
‍
β
i
)
=
|
γ
|
(
2
+
∑
i
=
1
k
‍
β
i
)
,
|
J
1
(
β
,
γ
)
|
≤
|
γ
|
(
∑
i
=
1
k
‍
β
i
+
1
)
+
|
γ
|
≤
|
γ
|
(
3
+
∑
i
=
1
k
‍
β
i
)
,
|
J
2
(
β
,
γ
)
|
≤
|
γ
|
(
∑
i
=
2
k
‍
β
i
+
2
)
+
|
γ
|
≤
|
γ
|
(
3
+
∑
i
=
2
k
‍
β
i
)
,
⋮
|
J
k
+
1
(
β
)
|
=
3
|
γ
|
,
sup
⁡
1
≤
i
≤
k
+
1
|
J
i
(
β
,
γ
)
|
≤
|
γ
|
(
3
+
∑
i
=
1
k
‍
β
i
)
.
Let us writeW
0
=
W
1
+
W
2, where W
1 is upper triangular and W
2, lower triangular parts, and estimate ∥
W
0
∥:
(71)
∥
W
0
∥
≤
∥
W
1
∥
+
∥
W
2
∥
≤
Ω
2
[
max
⁡
1
≤
i
≤
k
+
1
|
J
i
(
β
,
γ
)
|
+
max
⁡
1
≤
i
≤
k
+
1
|
J
(
β
,
γ
)
-
β
i
-
1
J
i
(
β
,
γ
)
|
]
≤
Ω
2
[
max
⁡
1
≤
i
≤
k
+
1
|
J
i
(
β
,
γ
)
|
+
|
J
(
β
,
γ
)
|
+
max
⁡
1
≤
i
≤
k
+
1
|
J
i
(
β
,
γ
)
|
β
i
-
1
]
≤
Ω
2
[
|
J
(
β
,
γ
)
|
+
max
⁡
1
≤
i
≤
k
+
1
|
J
i
(
β
,
γ
)
|
(
β
i
-
1
+
1
)
]
.
We obtain(72)
∥
W
0
∥
≤
Ω
2
[
|
γ
|
(
2
+
∑
i
=
1
k
‍
β
i
)
+
(
1
+
max
⁡
1
≤
i
≤
k
+
1
β
i
-
1
)
×
|
γ
|
(
3
+
∑
i
=
1
k
‍
β
i
)
]
.
It is clear that(73)
∥
T
∥
≤
Q
≤
Ω
2
[
|
γ
|
(
5
+
2
∑
i
=
1
k
‍
β
i
)
+
max
⁡
1
≤
i
≤
k
+
1
β
i
-
1
|
γ
|
(
3
+
∑
i
=
1
k
‍
β
i
)
]
×
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
=
Ω
2
~
m
(
b
-
a
)
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
.
We obtain the following solution estimation:(74)
∥
x
∥
≤
Ω
2
~
(
b
-
a
)
∥
f
∥
1
-
Ω
2
~
m
(
b
-
a
)
ess
sup
⁡
s
∈
[
a
,
b
]
1
≤
i
≤
m
|
p
i
(
s
)
|
.
Theorem8 has been proven.


## 5. Sign Constancy of Green's Functions

The proofs of the following two assertions follow from the construction of Green's functions.Theorem 9.
Let(75)
0
<
β
i
,
1
≤
i
≤
k
,
I
(
φ
,
θ
,
β
)
·
I
j
(
φ
,
θ
,
β
)
<
0
,
j
=
1
,
…
,
k
+
1
;
then Green's function G
0
(
t
,
s
) of problem (29), (5) is positive for t
,
s
∈
[
a
,
b
].Theorem 10.
Let(76)
0
<
β
i
,
1
≤
i
≤
k
,
I
1
(
φ
,
θ
,
β
)
I
(
φ
,
θ
,
β
)
>
1
,
I
j
(
φ
,
θ
,
β
)
I
(
φ
,
θ
,
β
)
>
1
∏
i
=
1
j
-
1
‍
β
i
2
≤
j
≤
k
+
1
;
then Green's function G
0
(
t
,
s
) of problem (29), (5) is negative for t
,
s
∈
[
a
,
b
].Estimation of Green's function (11) leads us to the assertion.Theorem 11.
Let conditions (40) and (75) be fulfilled, p
i
(
t
)
≤
0 for i
=
1
,
…
,
m; then problem (1)–(5) is unique solvable and its Green's function G
(
t
,
s
) is positive for t
,
s
∈
[
a
,
b
].Proof.
Without losing generality we assume thatc
=
0.
Solution of problem (1)–(5) can be represented in the following form:
(77)
x
(
t
)
=
(
M
x
)
(
t
)
+
ψ
(
t
)
,
where
(78)
(
M
x
)
(
t
)
=
-
∫
a
b
‍
G
0
(
t
,
s
)
∑
i
=
1
m
‍
p
i
(
s
)
x
(
s
-
τ
i
(
s
)
)
d
s
,
ψ
(
t
)
=
∫
a
b
‍
G
0
(
t
,
s
)
f
(
s
)
d
s
.
Using Theorem9, we obtain that the operator M is positive.
The conditions of Theorem7 imply that ∥
M
∥
<
1.
Now it is clear that(79)
x
(
t
)
=
(
I
-
M
)
-
1
ψ
(
t
)
=
ψ
(
t
)
+
(
M
ψ
)
(
t
)
+
(
M
2
ψ
)
(
t
)
+
⋯
It follows from the positivity of the operatorM that for every nonnegative f we get according to (78) ψ
(
t
)
≥
0 and x
(
t
)
≥
ψ
(
t
)
≥
0.
It is clear that(80)
0
≤
x
(
t
)
-
ψ
(
t
)
=
∫
a
b
‍
G
(
t
,
s
)
f
(
s
)
d
s
-
∫
a
b
‍
G
0
(
t
,
s
)
f
(
s
)
d
s
=
∫
a
b
‍
[
G
(
t
,
s
)
-
G
0
(
t
,
s
)
]
f
(
s
)
d
s
.
From nonnegativity ofx
(
t
) for every nonnegative f
(
t
), we obtain that G
(
t
,
s
)
≥
G
0
(
t
,
s
)
≥
0.
Theorem11 has been proven.Theorem 12.
Let conditions (40) and (76) be fulfilled, p
i
(
t
)
≥
0 for i
=
1
,
…
,
m then problem (1)–(5) is unique solvable and its Green's function G
(
t
,
s
) is negative for t
,
s
∈
[
a
,
b
].We prove this assertion analogously to the proof of Theorem11.The proofs of the following two assertions follow from the construction of Green's functions.Theorem 13.
Let(81)
0
<
β
i
,
1
≤
i
≤
k
,
J
1
(
β
,
γ
)
J
(
β
,
γ
)
>
1
,
J
j
(
β
,
γ
)
J
(
β
,
γ
)
>
1
∏
i
=
1
j
-
1
‍
β
i
2
≤
j
≤
k
+
1
;
then Green's function of problem (29), (7) W
0
(
t
,
s
) is negative for t
,
s
∈
[
a
,
b
].Theorem 14.
Let(82)
0
<
β
i
,
1
≤
i
≤
k
,
max
⁡
1
≤
i
≤
k
+
1
J
(
β
,
γ
)
·
J
i
(
β
,
γ
)
<
0
;
then Green's function of problem (29), (7) W
0
(
t
,
s
) is positive for t
,
s
∈
[
a
,
b
].The proof of the following two theorems can be obtained analogously from the proof of Theorem11.Theorem 15.
Let conditions (60) and (82) be fulfilled, p
i
(
t
)
≤
0 for i
=
1
,
…
,
m; then problem (1)–(4), (7) is unique solvable and its Green's function W
(
t
,
s
) is positive for t
,
s
∈
[
a
,
b
].Theorem 16.
Let conditions (60) and (81) be fulfilled, p
i
(
t
)
≥
0 for i
=
1
,
…
,
m; then problem (1)–(4), (7) is unique solvable and its Green's function W
(
t
,
s
) is negative for t
,
s
∈
[
a
,
b
].



---

*Source: 978519-2014-02-13.xml*