# Unbounded Solutions of a Boundary Value  Problem for Abstractnth-Order Differential Equations on an Infinite Interval

Authors: Zhenbin Liu; Lishan Liu; Yonghong Wu; Jing Zhao
Journal: Journal of Applied Mathematics and Stochastic Analysis
 (2008)

---

## Abstract

The existence of unbounded nonnegative solutions of a boundary value problem fornth-order differential equations defined on an infinite interval is obtained by means of the Mönch fixed-point theorem. An example is then presented to demonstrate the application of our results.


---

## Body

## 1. Introduction

Over the last few years, have focused their research
on the study of boundary value problems for nonlinear differential and integral
equations defined on an infinite interval, and various theoretical results have
been obtained [1–8]. In [4], the existence of multiple positive solutions of a
boundary value problem (BVP) for nth-order
nonlinear impulsive integral-differential equations defined on an infinite
interval in a Banach space is obtained by means of the fixed-point index theory
of completely continuous operators. However, the result requires the use of the
measures of noncompactness condition α
(
f
(
t
,
P
r
,
…
,
P
r
)
)
=
0 (where P
r
=
{
x
∈
P
:
∥
x
∥
≤
r
}) and the
normal and solid cone P in a real
Banach space. In [6], by using the Mönch fixed-point theorem, a class of
infinite boundary value problems for first-order impulsive differential
equations in a Banach space is considered and the existence of positive
solutions is obtained, but the solutions are limited to bounded solutions only.To generalize and further develop the existing results
in this field, in this paper we discuss the existence of unbounded solutions
for a class ofnth-order
nonlinear differential equations defined on an infinite interval in a Banach
space by using the Mönch fixed-point theorem under certain conditions weaker
than those in [4]. The boundary value problem in question is as follows:(1.1)
u
(
n
)
(
t
)
=
f
(
t
,
u
(
t
)
,
u
′
(
t
)
,
…
,
u
(
n
−
1
)
(
t
)
)
,
u
(
i
)
(
0
)
=
β
i
u
(
i
+
1
)
(
0
)
,
i
=
0
,
1
,
…
,
n
−
2,
u
(
n
−
1
)
(
+
∞
)
=
β
u
(
n
−
1
)
(
0
)
,
where J
=
[
0
,
+
∞
), u
(
n
−
1
)
(
+
∞
)
=
lim
t
→
+
∞
u
(
n
−
1
)
(
t
), β
>
1, β
i
>
0
(
i
=
0
,
1
,
…
,
n
−
2
), f
∈
C
[
J
×
E
×
⋯
×
E
,
E
], in which (
E
,
∥
⋅
∥
) is a real
Banach space.LetC
[
J
,
E
] be the space of
all continuous functions u
:
J
→
E, and let L
[
J
,
E
] be the Banach
space of all strongly measurable functions u
:
J
→
E with ∫
J
∥
u
(
t
)
∥
d
t
<
+
∞, equipped with the norm ∥
u
∥
1
=
∫
J
∥
u
(
t
)
∥
d
t. Let C
e
[
J
,
E
]
=
{
u
∈
C
[
J
,
E
]
:
e
−
t
∥
u
(
t
)
∥
→
0
f
o
r
t
→
+
∞
}, then it is clear that C
e
[
J
,
E
] is a Banach
space with norm
(1.2)
∥
u
∥
e
=
sup
t
∈
J
{
e
−
t
∥
u
(
t
)
∥
}
.
Let D
e
n
−
1
[
J
,
E
]
=
{
u
∈
C
n
−
1
[
J
,
E
]
:
lim
t
→
+
∞
e
−
t
∥
u
(
i
)
(
t
)
∥
→
0,
i
=
0
,
1
,
…
,
n
−
1
}, then it is also easy to see that D
e
n
−
1
[
J
,
E
] is a Banach
space with norm
(1.3)
∥
u
∥
D
=
max
{
∥
u
∥
e
,
∥
u
′
∥
e
,
…
,
∥
u
(
n
−
1
)
∥
e
}
.
Let P be a cone of
the Banach space E, C
e
[
J
,
P
]
=
{
u
∈
C
e
[
J
,
E
]
:
u
(
t
)
≥
θ
,
t
∈
J
}
, and D
e
n
−
1
[
J
,
P
]
=
{
u
∈
D
e
n
−
1
[
J
,
E
]
:
u
(
i
)
(
t
)
≥
θ
,
t
∈
J
,
i
=
0
,
1
,
…
,
n
−
1
}, where θ denotes the
zero element of E. Then, it is obvious that C
e
[
J
,
P
] is a cone in
space C
e
[
J
,
E
], and D
e
n
−
1
[
J
,
P
] is a cone in
space D
e
n
−
1
[
J
,
E
].Deffinition 1.1.
A
functionu
∈
D
e
n
−
1
[
J
,
P
]
∩
C
n
[
J
,
E
] is called a
nonnegative solution of BVP(1.1) if u
(
t
)
satisfies (1.1)
for t
∈
J.The rest of the paper is organized as follows. In
Section2, we give some lemmas which provide a theoretical basis for the proof
of our main results. The main theorem is presented and proved in Section 3. In
Section 4, an example is given to demonstrate the application of our results.


## 2. Some Lemmas

Here we first list some assumptions to be used
throughout the rest of the paper.(
H
1
) There exist a
(
t
) and b
i
(
t
)
∈
C
[
J
,
ℝ
+
] such that a
(
t
), e
t
b
i
(
t
)
∈
L
[
J
,
ℝ
+
]
(
i
=
0
,
1
,
2
,
…
,
n
−
1
) and
(2.1)
∥
f
(
t
,
x
0
,
x
1
,
…
,
x
n
−
1
)
∥
≤
a
(
t
)
+
∑
i
=
0
n
−
1
b
i
(
t
)
∥
x
i
(
t
)
∥
∀
t
∈
J
,
x
i
∈
E
,
∑
i
=
0
n
−
1
b
i
*
<
β
−
1
β
*
+
β
,
β
*
=
max
0
≤
j
≤
n
−
2
{
∏
i
=
j
n
−
2
β
i
}
,
b
i
*
=
∫
0
+
∞
e
t
b
i
(
t
)
d
t
<
+
∞
,
i
=
0
,
1
,
2
,
…
,
n
−
1.(
H
2
) There exists l
i
(
t
)
∈
C
[
J
,
ℝ
+
] such that e
t
l
i
(
t
)
∈
L
[
J
,
ℝ
+
]
(
i
=
0
,
1
,
…
,
n
−
1
)
,
(2.2)
α
(
f
(
t
,
D
0
,
D
1
,
…
,
D
n
−
1
)
)
≤
∑
i
=
0
n
−
1
l
i
(
t
)
α
(
D
i
)
∀
t
∈
J
,
D
i
⊂
E
,
2
(
β
*
+
β
)
β
−
1
∑
i
=
0
n
−
1
l
i
<
1
,
l
i
=
∫
0
+
∞
e
t
l
i
(
t
)
d
t
<
+
∞
,
i
=
0
,
1
,
…
,
n
−
1,
where α
(
⋅
) denotes the
Kuratowski measure of noncompactness in E. For details on the definition and properties of the
measure of noncompactness, the reader is referred to [9, 10].In the following, we give various lemmas which are to
be used for the proof of the main results to be presented in Section3.Lemma 2.1 ( See [4]).
Let (H
1) be satisfied. Then u
∈
D
e
n
−
1
[
J
,
P
]
∩
C
n
[
J
,
E
] is a solution
of BVP(1.1) if and only if u
∈
D
e
n
−
1
[
J
,
P
] is a solution
of the following integral equation:
(2.3)
u
(
t
)
=
1
β
−
1
[
∑
j
=
0
n
−
2
(
∏
i
=
j
n
−
2
β
i
)
t
j
j
!
+
t
n
−
1
(
n
−
1
)
!
]
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
+
1
(
n
−
1
)
!
∫
0
t
(
t
−
s
)
n
−
1
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
.Proof.
Ifu
∈
D
e
n
−
1
[
J
,
P
]
∩
C
n
[
J
,
E
] is a solution
of BVP(1.1), then by condition (
H
1
) we have the
convergence of the infinite integral
(2.4)
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
.
Integrating the
first equation in (1.1) from 0 to t, we have
(2.5)
u
(
n
−
1
)
(
t
)
−
u
(
n
−
1
)
(
0
)
=
∫
0
t
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
.
By virtue of u
(
n
−
1
)
(
+
∞
)
=
β
u
(
n
−
1
)
(
0
), let t
→
+
∞ in (2.5), we
get
(2.6)
u
(
n
−
1
)
(
0
)
=
1
β
−
1
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
.
From  the second equation in (1.1), we have(2.7)
u
(
i
)
(
0
)
=
β
i
u
(
i
+
1
)
(
0
)
=
β
i
β
i
+
1
u
(
i
+
2
)
(
0
)
=
β
i
β
i
+
1
,
…
,
β
n
−
2
u
(
n
−
1
)
(
0
)
=
∏
j
=
i
n
−
2
β
j
u
(
n
−
1
)
(
0
)
.
Substituting
(2.6) into (2.7) yields for i
=
0
,
1
,
…
,
n
−
2,
(2.8)
u
(
i
)
(
0
)
=
1
β
−
1
∏
j
=
i
n
−
2
β
j
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
.
From (2.5) and
(2.6), we have
(2.9)
u
(
n
−
1
)
(
t
)
=
1
β
−
1
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
+
∫
0
t
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
.
Integrating (2.9) from 0 to t and using (2.8)
for i
=
n
−
2, we have
(2.10)
u
(
n
−
2
)
(
t
)
=
u
(
n
−
2
)
(
0
)
+
t
β
−
1
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
+
∫
0
t
d
s
∫
0
s
f
(
w
,
u
(
w
)
,
…
,
u
(
n
−
1
)
(
w
)
)
d
w
=
(
β
n
−
2
β
−
1
+
t
β
−
1
)
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
+
∫
0
t
(
t
−
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
.
It is not difficult to show by mathematical induction
that u satisfies
(2.3). Conversely, if u
∈
D
e
n
−
1
[
J
,
P
] is a solution
of (2.3), then direct differentiation of (2.3) gives
(2.11)
u
(
i
)
(
t
)
=
1
β
−
1
[
∑
j
=
i
n
−
2
(
∏
s
=
j
n
−
2
β
s
)
t
j
−
i
(
j
−
i
)
!
+
t
n
−
i
−
1
(
n
−
i
−
1
)
!
]
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
+
1
(
n
−
i
−
1
)
!
∫
0
t
(
t
−
s
)
n
−
i
−
1
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
,
i
=
0
,
1
,
…
,
n
−
2,
(2.12)
u
(
n
−
1
)
(
t
)
=
1
β
−
1
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
+
∫
0
t
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
,
u
(
n
)
(
t
)
=
f
(
t
,
u
(
t
)
,
u
′
(
t
)
,
…
,
u
(
n
)
(
t
)
)
∀
t
∈
J
.
Consequently, u
∈
C
n
[
J
,
E
], and by (2.11) and (2.12), it is easy to see that u
(
t
) satisfies (1.1). The proof of Lemma 2.1 is completed.We now
consider an operatorA defined by
(2.13)
(
A
u
)
(
t
)
=
1
β
−
1
[
∑
j
=
0
n
−
2
(
∏
s
=
j
n
−
2
β
s
)
t
j
j
!
+
t
n
−
1
(
n
−
1
)
!
]
∫
0
+
∞
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
+
1
(
n
−
1
)
!
∫
0
t
(
t
−
s
)
n
−
1
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
,
∀
t
∈
J
.
By Lemma 2.1, u
∈
D
e
n
−
1
[
J
,
P
]
∩
C
n
[
J
,
E
] is a solution
of BVP(1.1) if and only if u is a fixed
point of the operator A in u
∈
D
e
n
−
1
[
J
,
P
].Lemma 2.2 (See [10, 11,
13]).
LetE be a Banach
space, let J
0
=
[
a
,
b
]
be a finite
interval, and let H
⊂
C
[
J
0
,
E
]
be a countable
set. Assume that there exists ρ
∈
L
[
J
0
,
ℝ
+
] such that ∥
u
(
t
)
∥
≤
ρ
(
t
)
,
t
∈
J
0
,
u
∈
H. Then α
(
{
u
(
t
)
:
u
∈
H
}
)
∈
L
[
J
0
,
ℝ
+
] and
(2.14)
α
(
{
∫
J
0
u
(
t
)
d
t
:
u
∈
H
}
)
≤
2
∫
J
0
α
(
{
u
(
t
)
:
u
∈
H
}
)
d
t
.Lemma 2.3.
LetE be a Banach
space, and H
⊂
C
[
J
,
E
]. If H
is a countable
set and there exists ρ
∈
L
[
J
,
ℝ
+
] such that ∥
u
(
t
)
∥
≤
ρ
(
t
)
,
t
∈
J
,
u
∈
H, then α
(
{
u
(
t
)
:
u
∈
H
}
)
is integrable
on J, and
(2.15)
α
(
{
∫
0
+
∞
u
(
t
)
d
t
:
u
∈
H
}
)
≤
2
∫
0
+
∞
α
(
{
u
(
t
)
:
u
∈
H
}
)
d
t
.Proof.
By∥
u
(
t
)
∥
≤
ρ
(
t
) for any u
∈
H and t
∈
J, we get u
(
t
)
∈
L
[
J
,
E
] for all u
∈
H and α
(
{
u
(
t
)
:
u
∈
H
}
)
≤
2
ρ
(
t
). As ρ
(
t
)
∈
L
[
J
,
ℝ
+
], α
(
{
u
(
t
)
:
u
∈
H
}
) is integrable
in J. For any ε
>
0, from ρ
(
t
)
∈
L
[
J
,
ℝ
+
], there exists T
0
>
0 such that ∫
t
+
∞
ρ
(
s
)
d
s
<
ε
f
o
r
a
n
y
t
>
T
0. So for any t
>
T
0 and u
∈
H, we have
(2.16)
∥
∫
0
t
u
(
s
)
d
s
−
∫
0
+
∞
u
(
s
)
d
s
∥
=
∥
∫
t
+
∞
u
(
s
)
d
s
∥
<
∫
t
+
∞
ρ
(
s
)
d
s
<
ε
.
Let
(2.17)
S
(
t
)
=
{
∫
0
t
u
(
s
)
d
s
:
u
∈
H
}
,
T
=
{
∫
0
+
∞
u
(
s
)
d
s
:
u
∈
H
}
.
By (2.16) for any t
>
T
0 and u
∈
H, we obtain
(2.18)
d
(
∫
0
t
u
(
s
)
d
s
,
T
)
=
inf
u
∈
H
{
∥
∫
0
t
u
(
s
)
d
s
−
∫
0
+
∞
u
(
s
)
d
s
∥
}
≤
∥
∫
0
t
u
(
s
)
d
s
−
∫
0
+
∞
u
(
s
)
d
s
∥
<
ε
,
where d denotes the
distance of a point to a set. Using the same method, we have
(2.19)
d
(
∫
0
+
∞
u
(
s
)
d
s
,
S
(
t
)
)
<
ε
,
t
>
T
0
,
u
∈
H
.
So
(2.20)
sup
u
∈
H
d
(
∫
0
t
u
(
s
)
d
s
,
T
)
≤
ε
,
sup
u
∈
H
d
(
∫
0
+
∞
u
(
s
)
d
s
,
S
(
t
)
)
≤
ε
.
Hence, for any t
>
T
0, we have
(2.21)
d
h
(
{
∫
0
t
u
(
s
)
d
s
:
u
∈
H
}
,
{
∫
0
+
∞
u
(
s
)
d
s
:
u
∈
H
}
)
=
max
{
sup
u
∈
H
d
(
∫
0
t
u
(
s
)
d
s
,
T
)
,
sup
u
∈
H
d
(
∫
0
+
∞
u
(
s
)
d
s
,
S
(
t
)
)
}
≤
ε
,
where d
h
(
D
1
,
D
2
) denotes the
Hausdorff distance of the sets D
1 and D
2, that is,
(2.22)
d
h
(
D
1
,
D
2
)
=
max
{
sup
x
∈
D
1
d
(
x
,
D
2
)
,
sup
x
∈
D
2
d
(
x
,
D
1
)
}
.
So, by (2.21), we obtain for any t
>
T
0,
(2.23)
|
α
(
{
∫
0
t
u
(
s
)
d
s
:
u
∈
H
}
)
−
α
(
{
∫
0
+
∞
u
(
s
)
d
s
:
u
∈
H
}
)
|
≤
2
d
h
(
{
∫
0
t
u
(
s
)
d
s
:
u
∈
H
}
,
{
∫
0
+
∞
u
(
s
)
d
s
:
u
∈
H
}
)
≤
2
ε
.Hence, we have
(2.24)
lim
t
→
+
∞
α
(
{
∫
0
t
u
(
s
)
d
s
:
u
∈
H
}
)
=
α
(
{
∫
0
+
∞
u
(
s
)
d
s
:
u
∈
H
}
)
.
By Lemma 2.3 for any t
∈
[
0
,
+
∞
), we have
(2.25)
α
(
{
∫
0
t
u
(
s
)
d
s
:
u
∈
H
}
)
≤
2
∫
0
t
α
(
{
u
(
s
)
:
u
∈
H
}
)
d
s
.
Hence, from (2.24) and (2.25), we have
(2.26)
α
(
{
∫
0
+
∞
u
(
s
)
d
s
:
u
∈
H
}
)
=
lim
t
→
+
∞
α
(
{
∫
0
t
u
(
s
)
d
s
:
u
∈
H
}
)
≤
2
lim
t
→
+
∞
∫
0
t
α
(
{
u
(
s
)
:
u
∈
H
}
)
d
s
=
2
∫
0
+
∞
α
(
{
u
(
s
)
:
u
∈
H
}
)
d
s
.
Therefore, (2.15) is satisfied. The proof of Lemma 2.3
is completed.Remark 2.4.
Lemma2.3 generalizes
Lemma 2.2 from a finite
interval to an infinite interval, and it plays an important role in studying
the differential equations defined on an infinite interval. It should be
emphasized that Lemma 2.3 has no counterpart in the existing literature.Lemma 2.5 (See [2, Lemma 7]).
Let(
H
1
)
be satisfied,
and let V
=
{
u
m
}
⊂
D
e
n
−
1
[
J
,
E
] be a countable
bounded set. Then
(2.27)
α
D
(
A
V
)
=
max
{
sup
t
∈
J
[
e
−
t
α
(
(
A
V
)
(
i
)
(
t
)
)
]
:
i
=
0
,
1
,
…
,
n
−
1
}
,
where α
D
(
⋅
) denotes the
Kuratowski measure of noncompactness in D
e
n
−
1
[
J
,
E
], (
A
V
)
(
i
)
=
{
(
A
u
m
)
(
i
)
(
t
)
:
u
m
∈
V
,
m
=
1
,
2
,
…
}
(
i
=
0
,
1
,
2
,
…
,
n
−
1
).Lemma 2.6.
If condition(
H
1
)
is satisfied,
then the operator A
is continuous
and
(2.28)
A
(
D
e
n
−
1
[
J
,
P
]
)
⊂
D
e
n
−
1
[
J
,
P
]
.Proof.
For anyu
∈
D
e
n
−
1
[
J
,
P
], we have
(2.29)
(
A
u
)
(
t
)
=
1
β
−
1
[
∑
j
=
0
n
−
2
(
∏
s
=
j
n
−
2
β
s
)
t
j
j
!
+
t
n
−
1
(
n
−
1
)
!
]
∫
0
+
∞
f
(
(
s
)
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
+
1
(
n
−
1
)
!
∫
0
t
(
t
−
s
)
n
−
1
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
,
(2.30)
(
A
u
)
(
i
)
(
t
)
=
1
β
−
1
[
∑
j
=
i
n
−
2
(
∏
s
=
j
n
−
2
β
s
)
t
j
−
i
(
j
−
i
)
!
+
t
n
−
i
−
1
(
n
−
i
−
1
)
!
]
∫
0
+
∞
f
(
s
,
u
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
+
1
(
n
−
i
−
1
)
!
∫
0
t
(
t
−
s
)
n
−
i
−
1
f
(
(
s
)
,
u
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
d
s
,
i
=
1
,
…
,
n
−
1
,
(
A
u
)
(
n
)
(
t
)
=
f
(
t
,
u
(
t
)
,
u
′
(
t
)
,
…
,
u
(
n
−
1
)
(
t
)
)
.
So, by (2.30) and condition (
H
1
) for i
=
0
,
1
,
…
,
n
−
1, we have
(2.31)
e
−
t
∥
(
A
u
)
(
i
)
(
t
)
∥
≤
β
*
+
β
β
−
1
e
−
t
∑
j
=
i
n
−
1
t
j
−
i
(
j
−
i
)
!
∫
0
+
∞
∥
f
(
(
s
)
,
u
(
s
)
,
u
′
(
s
)
,
…
,
u
(
n
−
1
)
(
s
)
)
∥
d
s
≤
β
*
+
β
β
−
1
e
−
t
∑
j
=
i
n
−
1
t
j
−
i
(
j
−
i
)
!
(
a
*
+
∑
i
=
0
n
−
1
b
i
*
∥
u
∥
D
)
,
where a
*
=
∫
0
+
∞
a
(
t
)
d
t. Hence, lim
t
→
+
∞
e
−
t
∥
(
A
u
)
(
i
)
(
t
)
∥
=
0. Therefore, (2.28) is satisfied.
Now we show that the operatorA is continuous.
Let u
m
,
u
¯
∈
D
e
n
−
1
[
J
,
P
], ∥
u
m
−
u
¯
∥
D
→
0. Then r
=
sup
m
∥
u
m
∥
D
<
+
∞ and ∥
u
¯
∥
≤
r. By the integrability of a
(
t
) and e
t
b
i
(
t
)
(
i
=
0
,
1
,
…
,
n
−
1
) for any ε
>
0, there exists T
>
0 such that
(2.32)
β
*
+
β
β
−
1
[
2
∫
T
+
∞
a
(
t
)
d
t
+
2
r
n
∫
T
+
∞
e
t
b
i
(
t
)
d
t
]
<
ε
2
.
On the other
hand, by the continuity of f, it is easy to see that
(2.33)
∫
0
T
∥
f
(
s
,
u
m
(
s
)
,
…
,
u
m
(
n
−
1
)
(
s
)
)
−
f
(
s
,
u
¯
(
s
)
,
…
,
u
¯
(
n
−
1
)
(
s
)
)
∥
d
s
→
0,
m
→
+
∞
.
Hence, for the above ε
>
0, there exists a natural number m
0 such that for
any m
>
m
0,
(2.34)
β
*
+
β
β
−
1
∫
0
T
∥
f
(
s
,
u
m
(
s
)
,
…
,
u
m
(
n
−
1
)
(
s
)
)
−
f
(
s
,
u
¯
(
s
)
,
…
,
u
¯
(
n
−
1
)
(
s
)
)
∥
d
s
<
ε
2
.
Thus, from
(2.13), (2.32), (2.34), and (
H
1
) for any m
>
m
0, we obtain
(2.35)
∥
A
u
m
−
A
u
¯
∥
D
≤
β
*
+
β
β
−
1
∫
0
+
∞
∥
f
(
s
,
u
m
(
s
)
,
…
,
u
m
(
n
−
1
)
(
s
)
)
−
f
(
s
,
u
¯
(
s
)
,
…
,
u
¯
(
n
−
1
)
(
s
)
)
∥
d
s
≤
ε
2
+
β
*
+
β
β
−
1
[
2
∫
T
+
∞
a
(
t
)
d
t
+
∑
i
=
0
n
−
1
(
∥
u
m
∥
D
+
∥
u
¯
∥
D
)
∫
T
+
∞
e
t
b
i
(
t
)
d
t
]
≤
ε
2
+
β
*
+
β
β
−
1
[
2
∫
T
+
∞
a
(
t
)
d
t
+
2
r
n
∫
T
+
∞
e
t
b
i
(
t
)
d
t
]
<
ε
.
Therefore, the continuity of A is proved. So,
the proof of Lemma 2.6 is completed.Lemma 2.7 ([10, 12]
(Mönch)).
LetV
be a closed and
convex subset of E and x
0
∈
V. Assume that operator A
:
V
→
V
has the
following property:
(2.36)
C
⊂
V
c
o
u
n
t
a
b
l
e
,
C
⊂
c
o
¯
(
{
x
0
}
∪
A
(
V
)
)
⇒
V
i
s
r
e
l
a
t
i
v
e
l
y
c
o
m
p
a
c
t
.
Then A has a fixed
point in V.


## 3. Main Result

In this section, we present the main results we
obtained.Theorem 3.1.
Suppose(
H
1
) and (
H
2
)
hold. Then
BVP(1.1) has at least one nonnegative solution.Proof.
Choose(3.1)
R
>
a
*
(
β
−
1
β
*
+
β
−
∑
i
=
0
n
−
1
b
i
*
)
−
1
,
and let V
=
{
u
∈
D
e
n
−
1
[
J
,
P
]
:
∥
u
∥
D
≤
R
}. Obviously, V is a bounded
convex closed set. For any u
∈
V, by (2.28), (
H
1
) and the
definition of the norm ∥
⋅
∥
e, we have
(3.2)
∥
(
A
u
)
(
i
)
∥
e
=
sup
t
∈
J
{
e
−
t
∥
(
A
u
)
(
i
)
(
t
)
∥
}
≤
β
*
+
β
β
−
1
(
a
*
+
∑
j
=
0
n
−
1
b
j
*
∥
u
∥
D
)
≤
R
,
i
=
0
,
1
,
…
,
n
−
1,
from which and the definition of the norm ∥
⋅
∥
D, we obtain
(3.3)
∥
A
u
∥
D
=
max
{
∥
(
A
u
)
(
i
)
∥
e
:
0
≤
i
≤
n
−
1
}
≤
R
.
Therefore, by Lemma 2.6, we conclude that A is a continuous
operator from V to V.
Next, we prove thatC is a relatively
compact set if C
⊂
V is a countable
set satisfying C
⊂
c
o
¯
(
{
u
}
∪
A
C
) for some u
∈
V. From this and the properties of the Kuratowski
measure of noncompactness in D
e
n
−
1
[
J
,
E
], we have
(3.4)
α
D
(
C
)
≤
α
D
(
A
C
)
.
On the other
hand, by (2.28), Lemmas 2.2−2.5, and (
H
2
) for any t
∈
J and 0
≤
i
≤
n
−
1, we have
(3.5)
e
−
t
α
(
(
A
C
)
(
i
)
(
t
)
)
≤
β
*
+
β
β
−
1
2
∫
0
+
∞
α
(
f
(
s
,
C
(
s
)
,
C
′
(
s
)
,
…
,
C
(
n
−
1
)
(
s
)
)
)
d
s
≤
2
(
β
*
+
β
)
β
−
1
∑
j
=
0
n
−
1
l
j
α
D
(
C
)
.
Thus, by (3.4),
(3.5), and Lemma 2.5, we get(3.6)
α
D
(
C
)
≤
α
D
(
A
C
)
≤
[
2
(
β
*
+
β
)
β
−
1
∑
i
=
0
n
−
1
l
i
]
α
D
(
C
)
.
Hence, by (
H
2
), we have α
D
(
C
)
=
0, that is, C is a relatively
compact set in D
e
n
−
1
[
J
,
E
]. Therefore, by Lemma 2.7, we conclude that A has at least
one fixed point in V
⊂
D
e
n
−
1
[
J
,
P
], that is, BVP(1.1) has at least
one solution in D
e
n
−
1
[
J
,
P
]
∩
C
n
[
J
,
E
].


## 4. An Example

Example 4.1.
Consider the following
boundary value problem for a second-order differential equation defined on an
infinite interval(4.1)
u
n
′
′
(
t
)
=
e
−
t
+
e
−
14
t
u
n
(
t
)
+
e
−
16
t
u
n
+
1
′
(
t
)
,
t
∈
J
,
u
n
(
0
)
=
u
n
+
1
′
(
0
)
,
u
n
′
(
+
∞
)
=
2
u
n
′
(
0
)
,
n
=
1
,
2
,
3
,
…
.ConclusionBVP(4.1) has at least one nonnegative
solution.Proof.
Let(5.1)
E
=
l
1
=
{
u
=
(
u
1
,
u
2
,
…
,
u
n
,
…
)
:
∑
n
=
0
∞
∥
u
n
∥
<
∞
}
with norm ∥
u
∥
=
∑
n
=
0
∞
∥
u
n
∥. P
=
{
u
=
(
u
1
,
u
2
,
…
,
u
n
,
…
)
∈
l
1
:
u
n
≥
0,
n
=
1
,
2
,
…
} is a cone in a
Banach space E and BVP(4.1) is
in the form of BVP(1.1) in E with n
=
2. In this situation, J
=
[
0
,
∞
)
u
=
(
u
1
,
…
,
u
n
,
…
), v
=
(
v
1
,
…
,
v
n
,
…
), and f
=
(
f
1
,
…
,
f
n
,
…
) in which
(5.2)
f
n
(
t
,
u
,
v
)
=
e
−
t
+
e
−
14
t
u
n
(
t
)
+
e
−
16
t
u
n
+
1
′
(
t
)
.
It is clear that f
∈
C
[
J
×
E
×
E
,
E
]. For any t
∈
J, u
,
v
∈
P, we have
(5.3)
|
f
n
(
t
,
u
,
v
)
|
≤
e
−
t
+
e
−
14
t
u
n
(
t
)
+
e
−
16
t
u
n
+
1
′
(
t
)
.
So we get
(5.4)
∥
f
(
t
,
u
,
v
)
∥
≤
a
(
t
)
+
b
0
(
t
)
∥
u
∥
+
b
1
(
t
)
∥
v
∥
,
where a
0
(
t
)
=
e
−
t, b
0
(
t
)
=
e
−
14
t, b
1
(
t
)
=
e
−
16
t. From (4.1) and (5.3), we have
(5.5)
a
*
=
∫
0
+
∞
e
−
t
d
t
=
1
,
b
0
*
=
∫
0
+
∞
e
t
e
−
14
t
d
t
=
1
13
,
b
1
*
=
∫
0
+
∞
e
t
e
−
16
t
d
t
=
1
15
,
β
*
=
β
0
=
1.
So, we have
(5.6)
β
−
1
β
*
+
β
=
2
−
1
1
+
2
=
1
3
>
1
13
+
1
15
=
b
0
*
+
b
1
*
.
Hence, the condition (
H
1
) is satisfied.
By (5.3) for any bounded sets D
1
,
D
2
⊂
l
1, we get
(5.7)
α
(
f
(
t
,
D
1
,
D
2
)
)
≤
l
1
(
t
)
α
(
D
1
)
+
l
2
(
t
)
α
(
D
2
)
,
where l
1
(
t
)
=
e
−
14
t, l
2
(
t
)
=
e
−
16
t. So
(5.8)
l
1
=
∫
0
+
∞
e
t
l
1
(
t
)
d
t
=
1
13
<
+
∞
,
l
2
=
∫
0
+
∞
e
t
l
2
(
t
)
d
t
=
1
15
<
+
∞
,
2
(
β
*
+
β
)
β
−
1
(
l
1
+
l
2
)
=
(
1
13
+
1
15
)
×
6
<
1.
Hence, the condition (
H
2
) is satisfied.
Therefore, our conclusion follows from Theorem 3.1.



---

*Source: 589480-2008-03-09.xml*