# Eigenvalue Problem for Nonlinear Fractional Differential Equations with Integral Boundary Conditions

Authors: Guotao Wang; Sanyang Liu; Lihong Zhang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

By employing known Guo-Krasnoselskii fixed point theorem, we investigate the eigenvalue interval for the existence and nonexistence of at least one positive solution of nonlinear fractional differential equation with integral boundary conditions.


---

## Body

## 1. Introduction

Fractional calculus has been receiving more and more attention in view of its extensive applications in the mathematical modelling coming from physical and other applied sciences; see books [1–5]. Recently, the existence of solutions (or positive solutions) of nonlinear fractional differential equation has been investigated in many papers (see [6–28] and references cited therein). However, in terms of the eigenvalue problem of fractional differential equation, there are only a few results [29–33].To the best of author’s knowledge, no paper has considered the eigenvalue problem of the following nonlinear fractional differential equation with integral boundary conditions:(1)
C
D
α
u
(
t
)
+
λ
f
(
t
,
u
(
t
)
)
=
0
,
0
<
t
<
1
,
n
<
α
≤
n
+
1
,
n
≥
2
,
n
∈
N
.
u
(
0
)
=
u
′′
(
0
)
=
u
′′′
(
0
)
=
⋯
=
u
(
n
)
(
0
)
=
0
,
u
(
1
)
=
ξ
∫
0
1
‍
u
(
s
)
d
s
,
where 0
<
ξ
<
2, C
D
α is the Caputo fractional derivative, and f
:
[
0,1
]
×
[
0
,
∞
)
→
[
0
,
∞
) is a continuous function.Our proof is based upon the properties of the Green function and Guo-Krasnoselskii’s fixed point theorem given in [34]. Our purpose here is to give the eigenvalue interval for nonlinear fractional differential equation with integral boundary conditions. Moreover, according to the range of the eigenvalue λ, we establish some sufficient conditions for the existence and nonexistence of at least one positive solution of the problem (1).


## 2. Preliminaries

For the convenience of the readers, we first present some background materials.Definition 1.
For a functionf
:
[
0
,
∞
)
→
ℝ, the Caputo derivative of fractional order α is defined as
(2)
C
D
α
f
(
t
)
=
1
Γ
(
n
-
α
)
∫
0
t
‍
(
t
-
s
)
n
-
α
-
1
f
(
n
)
(
s
)
d
s
,
111111111111111
11
1111
n
=
[
α
]
+
1
,
where [
α
] denotes the integer part of the real number α.Definition 2.
The Riemann-Liouville fractional integral of orderα for a function f is defined as
(3)
I
α
f
(
t
)
=
1
Γ
(
α
)
∫
0
t
‍
(
t
-
s
)
α
-
1
f
(
s
)
d
s
,
α
>
0
,
provided that such integral exists.Lemma 3.
Letα
>
0; then
(4)
I
α
C
D
α
u
(
t
)
=
u
(
t
)
+
C
0
+
C
1
t
+
C
2
t
2
+
⋯
+
C
n
-
1
t
n
-
1
,
for some C
i
∈
ℝ,  i
=
1,2
,
…
,
n, n
=
[
α
]
+
1.Lemma 4 (see [34]).
LetE be a Banach space, and let P
⊂
E be a cone. Assume that Ω
1, Ω
2 are open subsets of E with 0
∈
Ω
1, Ω
¯
1
⊂
Ω
2, and let T
:
P
∩
(
Ω
¯
2
∖
Ω
1
)
→
P be a completely continuous operator such that(i)
∥
T
u
∥
≥
∥
u
∥, u
∈
P
∩
∂
Ω
1, and ∥
T
u
∥
≤
∥
u
∥, u
∈
P
∩
∂
Ω
2, or
(ii)
∥
T
u
∥
≤
∥
u
∥, u
∈
P
∩
∂
Ω
1, and ∥
T
u
∥
≥
∥
u
∥, u
∈
P
∩
∂
Ω
2.
ThenT has a fixed point in P
∩
(
Ω
¯
2
∖
Ω
1
).Lemma 5.
Letn
<
α
≤
n
+
1, n
≥
2, n
∈
N, and ξ
≠
2. Assume y
∈
C
[
0,1
]; then the unique solution of the problem
(5)
C
D
α
u
(
t
)
+
y
(
t
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
u
′′
(
0
)
=
u
′′′
(
0
)
=
⋯
=
u
(
n
)
(
0
)
=
0
,
u
(
1
)
=
ξ
∫
0
1
‍
u
(
s
)
d
s
,
is given by the expression
(6)
u
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
y
(
s
)
d
s
,
where
(7)
G
(
t
,
s
)
=
{
2
t
(
1
-
s
)
α
-
1
(
α
-
ξ
+
ξ
s
)
-
(
2
-
ξ
)
α
(
t
-
s
)
α
-
1
(
2
-
ξ
)
Γ
(
α
+
1
)
,
111111111111111111111
1
1111
0
≤
s
≤
t
≤
1
,
2
t
(
1
-
s
)
α
-
1
(
α
-
ξ
+
ξ
s
)
(
2
-
ξ
)
Γ
(
α
+
1
)
,
0
≤
t
≤
s
≤
1
.Proof.
It is well known that the equationC
D
α
u
(
t
)
+
y
(
t
)
=
0 can be reduced to an equivalent integral equation:
(8)
u
(
t
)
=
-
I
α
y
(
t
)
-
∑
i
=
0
n
‍
b
i
t
i
=
-
∫
0
t
‍
(
t
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
-
∑
i
=
0
n
‍
b
i
t
i
,
for some b
i
∈
ℝ  (
i
=
0,1
,
2
,
…
,
n
).
By the conditionsu
(
0
)
=
u
′′
(
0
)
=
u
′′′
(
0
)
=
⋯
=
u
(
n
)
(
0
)
=
0 and u
(
1
)
=
ξ
∫
0
1
‍
u
(
s
)
d
s, we can get that b
0
=
b
2
=
b
3
=
⋯
=
b
n
=
0 and
(9)
b
1
=
-
∫
0
1
‍
(
1
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
-
ξ
∫
0
1
‍
u
(
s
)
d
s
.
Hence, we have
(10)
u
(
t
)
=
-
∫
0
t
‍
(
t
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
+
t
∫
0
1
‍
(
1
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
+
ξ
t
∫
0
1
‍
u
(
s
)
d
s
.
Put∫
0
1
‍
u
(
s
)
d
s
=
A; then, from (10), we deduce that
(11)
A
=
∫
0
1
‍
u
(
t
)
d
t
=
-
∫
0
1
‍
∫
0
t
‍
(
t
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
d
t
+
∬
0
1
t
(
1
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
d
t
+
∫
0
1
‍
ξ
A
t
d
t
=
-
∫
0
1
‍
(
1
-
s
)
α
α
Γ
(
α
)
y
(
s
)
d
s
+
1
2
∫
0
1
‍
(
1
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
+
1
2
ξ
A
,
which implies that
(12)
A
=
-
2
2
-
ξ
∫
0
1
‍
(
1
-
s
)
α
α
Γ
(
α
)
y
(
s
)
d
s
+
1
2
-
ξ
∫
0
1
‍
(
1
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
.
Replacing this value in (10), we obtain the following expression for function u
(
t
):
(13)
u
(
t
)
=
-
∫
0
t
‍
(
t
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
+
t
∫
0
1
‍
(
1
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
-
2
ξ
2
-
ξ
∫
0
1
‍
t
(
1
-
s
)
α
α
Γ
(
α
)
y
(
s
)
d
s
+
ξ
2
-
ξ
∫
0
1
‍
t
(
1
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
=
-
∫
0
t
‍
(
t
-
s
)
α
-
1
Γ
(
α
)
y
(
s
)
d
s
+
∫
0
1
‍
2
t
(
1
-
s
)
α
-
1
(
α
-
ξ
+
ξ
s
)
(
2
-
ξ
)
α
Γ
(
α
)
y
(
s
)
d
s
=
∫
0
t
‍
(
(
(
2
-
ξ
)
Γ
(
α
+
1
)
)
-
1
(
2
t
(
1
-
s
)
α
-
1
(
α
-
ξ
+
ξ
s
)
-
(
2
-
ξ
)
α
(
t
-
s
)
α
-
1
)
×
(
(
2
-
ξ
)
Γ
(
α
+
1
)
)
-
1
y
(
s
)
d
s
)
+
∫
t
1
‍
2
t
(
1
-
s
)
α
-
1
(
α
-
ξ
+
ξ
s
)
(
2
-
ξ
)
Γ
(
α
+
1
)
y
(
s
)
d
s
=
∫
0
1
‍
G
(
t
,
s
)
y
(
s
)
d
s
.
This completes the proof.Lemma 6.
LetG be the Green function, which is given by the expression (7). For 0
<
λ
<
2, the following property holds:
(14)
t
G
(
1
,
s
)
≤
G
(
t
,
s
)
≤
2
α
ξ
(
α
-
2
)
G
(
1
,
s
)
,
∀
t
,
s
∈
(
0,1
)
.The proof is similar to that of Lemma 2.4 in [7], so we omit it.Consider the Banach spaceX
=
C
[
0,1
] with general norm
(15)
∥
u
∥
=
sup
⁡
t
∈
[
0,1
]
|
u
(
t
)
|
.
Define the cone P
=
{
u
∈
X
:
u
(
t
)
≥
(
ξ
(
α
-
1
)
/
2
α
)
t
∥
u
∥
}.Supposeu is a solution of (1). It is clear from Lemma 5 that
(16)
u
(
t
)
=
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
,
∀
t
∈
[
0,1
]
.Define the operatorS
λ
:
P
→
X as follows:
(17)
(
S
λ
u
)
(
t
)
=
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
,
∀
t
∈
[
0,1
]
.Lemma 7.
S
λ
:
P
→
P is completely continuous.Proof.
Since0
<
ξ
<
2, it is obvious that G
(
t
,
s
)
≥
0. So we have
(18)
∥
S
λ
u
∥
=
sup
⁡
t
∈
[
0,1
]
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≤
λ
∫
0
1
‍
2
α
ξ
(
α
-
2
)
G
(
1
,
s
)
f
(
s
,
u
(
s
)
)
d
s
,
(
S
λ
u
)
(
t
)
=
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≥
ξ
(
α
-
2
)
2
α
t
λ
∫
0
1
‍
2
α
ξ
(
α
-
2
)
G
(
1
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≥
ξ
(
α
-
2
)
2
α
t
∥
S
λ
u
∥
.
Therefore, S
λ
(
P
)
⊂
P. The other proof is similar to that in [7], so we omit it.


## 3. Main Result

For convenience, we list the denotation:(19)
F
0
=
lim
⁡
u
→
0
+
sup
⁡
t
∈
[
0,1
]
f
(
t
,
u
(
t
)
)
u
,
F
∞
=
lim
⁡
u
→
+
∞
sup
⁡
t
∈
[
0,1
]
f
(
t
,
u
(
t
)
)
u
,
f
0
=
lim
⁡
u
→
0
+
inf
⁡
t
∈
[
0,1
]
f
(
t
,
u
(
t
)
)
u
,
f
∞
=
lim
⁡
u
→
+
∞
inf
⁡
t
∈
[
0,1
]
f
(
t
,
u
(
t
)
)
u
.Next, we will establish some sufficient conditions for the existence and nonexistence of positive solution for problem (1).Theorem 8.
Letl
∈
(
0,1
) be a constant. Then for each
(20)
λ
∈
(
(
ξ
(
α
-
2
)
l
f
∞
2
α
∫
0
1
‍
s
G
(
1
,
s
)
d
s
)
-
1
,
(
2
α
F
0
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
)
-
1
)
,
problem (1) has at least one positive solution.Proof.
First, for anyε
>
0, from (20) we have
(21)
(
ξ
(
α
-
2
)
l
(
f
∞
-
ε
)
2
α
∫
0
1
‍
s
G
(
1
,
s
)
d
s
)
-
1
≤
λ
≤
(
2
α
(
F
0
+
ε
)
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
)
-
1
.
On the one hand, by the definition ofF
0, there exists r
1
>
0 such that, for any u
∈
[
0
,
r
1
], we have
(22)
f
(
t
,
u
)
≤
(
F
0
+
ε
)
u
.
Choose Ω
1
=
{
u
∈
X
:
∥
u
∥
≤
r
1
}. For u
∈
P
∩
∂
Ω
1, we have
(23)
∥
S
λ
u
∥
=
sup
⁡
t
∈
[
0,1
]
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≤
λ
∫
0
1
‍
2
α
ξ
(
α
-
2
)
G
(
1
,
s
)
(
F
0
+
ε
)
u
(
s
)
d
s
≤
λ
2
α
(
F
0
+
ε
)
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
∥
u
∥
≤
∥
u
∥
.
On the other hand, by the definition ofF
∞, there exists r
2
>
r
1 such that, for any u
∈
[
r
2
,
+
∞
), we have
(24)
f
(
t
,
u
)
≥
(
f
∞
-
ε
)
u
.
Take Ω
2
=
{
u
∈
X
:
∥
u
∥
≤
r
2
}. For u
∈
P
∩
∂
Ω
2, we have
(25)
∥
S
λ
u
∥
≥
(
S
λ
u
)
(
l
)
≥
λ
∫
0
1
‍
l
G
(
1
,
s
)
(
f
∞
-
ε
)
u
(
s
)
d
s
≥
λ
l
ξ
(
α
-
2
)
f
∞
2
α
∫
0
1
‍
s
G
(
1
,
s
)
d
s
∥
u
∥
≥
∥
u
∥
.
According to (23),  (25), and Lemma 4, S
λ has at least one fixed point u
∈
P
∩
(
Ω
¯
2
∖
Ω
1
) with r
1
≤
∥
u
∥
≤
r
2, which is a positive solution of (1).Remark 9.
IfF
0
=
0 and f
∞
=
∞, then we can get
(26)
2
α
F
0
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
=
0
,
ξ
(
α
-
2
)
l
f
∞
2
α
∫
0
1
‍
s
G
(
1
,
s
)
d
s
=
+
∞
.
Theorem 8 implies that, for λ
∈
(
0
,
+
∞
), problem (1) has at least one positive solution.Theorem 10.
Letl
∈
(
0,1
) be a constant. Then for each
(27)
λ
∈
(
(
ξ
(
α
-
2
)
l
f
0
2
α
∫
0
1
‍
s
G
(
1
,
s
)
d
s
)
-
1
,
(
2
α
F
∞
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
)
-
1
)
,
problem (1) has at least one positive solution.Proof.
First, it follows from (27) that, for any ε
>
0,
(28)
(
ξ
(
α
-
2
)
l
(
f
0
-
ε
)
2
α
∫
0
1
‍
s
G
(
1
,
s
)
d
s
)
-
1
≤
λ
≤
(
2
α
(
F
∞
+
ε
)
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
)
-
1
.
By the definition off
0, there exists r
1
>
0 such that, for any u
∈
[
0
,
r
1
], we have
(29)
f
(
t
,
u
)
≥
(
f
0
+
ε
)
u
.
Choose Ω
1
=
{
u
∈
X
:
∥
u
∥
≤
r
1
}. For u
∈
P
∩
∂
Ω
1, we have ∥
u
∥
=
r
1. Similar to the proof in Theorem 8, it holds from (28) and (29) that
(30)
∥
S
λ
u
∥
≥
(
S
λ
u
)
(
l
)
≥
λ
l
ξ
(
α
-
2
)
f
0
2
α
∫
0
1
‍
s
G
(
1
,
s
)
d
s
∥
u
∥
≥
∥
u
∥
.
NoteF
∞
=
lim
⁡
u
→
+
∞
⁡
sup
⁡
t
∈
[
0,1
]
⁡
f
(
t
,
u
(
t
)
)
/
u. There exists r
3
>
r
1, such that
(31)
f
(
t
,
u
)
≤
(
F
∞
+
ε
)
u
,
u
∈
(
r
3
,
+
∞
)
.
We consider the problem on two cases. (I) Suppose f is bounded. There exists M
>
0, such that f
(
t
,
u
(
t
)
)
≤
M, ∀
u
∈
(
r
3
,
+
∞
). Choose r
4
=
max
⁡
{
r
3
,
M
λ
(
2
α
/
ξ
(
α
-
2
)
)
∫
0
1
‍
G
(
1
,
s
)
d
s
}. Let Ω
2
′
=
{
u
∈
X
:
∥
u
∥
≤
r
4
}. For u
∈
P
∩
∂
Ω
2
′, we have
(32)
∥
S
λ
u
∥
=
sup
⁡
t
∈
[
0,1
]
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≤
λ
∫
0
1
‍
2
α
ξ
(
α
-
2
)
G
(
1
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≤
λ
M
2
α
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
∥
u
∥
≤
r
4
=
∥
u
∥
.
(II) Supposef is unbounded. There exists r
5
>
r
3 such that
(33)
f
(
t
,
u
(
t
)
)
≤
u
,
u
∈
(
r
5
,
+
∞
)
.
LetΩ
2
′′
=
{
u
∈
X
:
∥
u
∥
≤
r
5
}. For u
∈
P
∩
∂
Ω
2
′′, we have
(34)
∥
S
λ
u
∥
≤
λ
∫
0
1
‍
2
α
ξ
(
α
-
2
)
G
(
1
,
s
)
f
(
s
,
u
(
s
)
)
d
s
≤
λ
2
α
(
F
∞
+
ε
)
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
∥
u
∥
≤
∥
u
∥
.
Combining (I) and (II), take Ω
2
=
{
u
∈
X
:
∥
u
∥
≤
r
2
}; here, r
2
≥
max
⁡
{
r
4
,
r
5
}. Then for u
∈
P
∩
∂
Ω
2, we have
(35)
∥
S
λ
u
∥
≤
∥
u
∥
.
Hence, (30) and (42) together with Lemma 4 imply that S
λ has at least one fixed point u
∈
P
∩
(
Ω
¯
2
∖
Ω
1
) with r
1
≤
∥
u
∥
≤
r
2, which is a positive solution of (1).Theorem 11.
AssumeF
0
<
+
∞ and F
∞
<
+
∞. Problem (1) has no positive solution provided
(36)
λ
<
(
2
α
k
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
)
-
1
,
where k is a constant defined in (38).Proof.
SinceF
0
<
+
∞ and F
∞
<
+
∞, together with the definitions of F
0 and F
∞, there exist positive constants k
1, k
2, r
1, and r
2 satisfying r
1
<
r
2 such that
(37)
f
(
t
,
u
)
≤
k
1
u
,
u
∈
[
0
,
r
1
]
,
f
(
t
,
u
)
≤
k
2
u
,
u
∈
[
r
2
,
+
∞
]
.
Take
(38)
k
=
max
⁡
{
k
1
,
k
2
,
sup
⁡
(
t
,
u
)
∈
(
0,1
)
×
(
k
1
,
k
2
)
f
(
t
,
u
)
u
}
.
It follows thatf
(
t
,
u
)
≤
k
u for any u
∈
(
0
,
+
∞
). Suppose that v
(
t
) is a positive solution of (1). That is,
(39)
(
S
λ
v
)
(
t
)
=
v
(
t
)
,
∀
t
∈
J
.
In sequence,
(40)
∥
v
∥
=
∥
S
λ
v
∥
=
sup
⁡
t
∈
[
0,1
]
λ
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
v
(
s
)
)
d
s
≤
λ
∫
0
1
‍
2
α
ξ
(
α
-
2
)
G
(
1
,
s
)
f
(
s
,
v
(
s
)
)
d
s
≤
λ
k
2
α
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
∥
v
∥
<
∥
v
∥
,
which is a contradiction. Hence, (1) has no positive solution.Theorem 12.
Assumef
0
>
0 and f
∞
>
0. Problem (1) has no positive solution provided
(41)
λ
>
(
ξ
k
(
α
-
2
)
2
α
∫
0
1
‍
s
2
G
(
1
,
s
)
d
s
)
-
1
,
where k is a constant defined in (43).Proof.
Sincef
0
>
0 and f
∞
>
0, together with the definitions of f
0 and f
∞, there exist positive constants k
1, k
2, r
1, and r
2 satisfying r
1
<
r
2 such that
(42)
f
(
t
,
u
)
≥
k
1
u
,
u
∈
[
0
,
r
1
]
,
f
(
t
,
u
)
≥
k
2
u
,
u
∈
[
r
2
,
+
∞
]
.
Take
(43)
k
=
min
⁡
{
k
1
,
k
2
,
inf
⁡
(
t
,
u
)
∈
(
0,1
)
×
(
k
1
,
k
2
)
f
(
t
,
u
)
u
}
.
It follows that f
(
t
,
u
)
≥
k
u for any u
∈
(
0
,
+
∞
). Suppose that v
(
t
) is a positive solution of (1). That is,
(44)
(
S
λ
v
)
(
t
)
=
v
(
t
)
,
∀
t
∈
J
.
In sequence,
(45)
∥
v
∥
≥
λ
∫
0
1
‍
s
G
(
1
,
s
)
f
(
s
,
v
(
s
)
)
d
s
≥
λ
k
ξ
(
α
-
2
)
2
α
∫
0
1
‍
s
2
G
(
1
,
s
)
d
s
∥
v
∥
>
∥
v
∥
,
which is a contradiction. Hence, (1) has no positive solution.Example 13.
Consider the fractional differential equation(46)
C
D
5
/
2
u
(
t
)
+
λ
f
(
t
,
u
(
t
)
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
u
′′
(
0
)
=
0
,
u
(
1
)
=
∫
0
1
‍
u
(
s
)
d
s
.
In this example, take
(47)
f
(
t
,
u
(
t
)
)
=
(
500
u
2
+
u
)
(
7
-
t
2
)
u
+
7
.
Obviously, we have
(48)
F
0
=
lim
⁡
u
→
0
+
sup
⁡
t
∈
[
0,1
]
(
500
u
2
+
u
)
(
7
-
t
2
)
u
(
u
+
7
)
=
1
,
f
∞
=
lim
⁡
u
→
+
∞
inf
⁡
t
∈
[
0,1
]
(
500
u
2
+
u
)
(
7
-
t
2
)
u
(
u
+
7
)
=
3000
.
Sinceα
=
5
/
2 and ξ
=
1, through a computation, we can get
(49)
∫
0
1
‍
G
(
1
,
s
)
d
s
=
∫
0
1
‍
2
t
(
1
-
s
)
α
-
1
(
α
-
ξ
+
ξ
s
)
-
(
2
-
ξ
)
α
(
t
-
s
)
α
-
1
(
2
-
ξ
)
Γ
(
α
+
1
)
d
s
=
∫
0
1
‍
2
(
1
-
s
)
3
/
2
(
3
/
2
+
s
)
-
(
5
/
2
)
(
1
-
s
)
3
/
2
Γ
(
7
/
2
)
d
s
≤
1
Γ
(
7
/
2
)
,
∫
0
1
‍
s
G
(
1
,
s
)
d
s
=
∫
0
1
‍
2
s
(
1
-
s
)
3
/
2
(
3
/
2
+
s
)
-
(
5
/
2
)
s
(
1
-
s
)
3
/
2
Γ
(
7
/
2
)
d
s
=
∫
0
1
‍
s
(
1
-
s
)
3
/
2
2
Γ
(
7
/
2
)
d
s
≥
2
35
Γ
(
7
/
2
)
.
Choosel
=
2
/
3; we have
(50)
(
ξ
(
α
-
2
)
l
f
∞
2
α
∫
0
1
‍
s
G
(
1
,
s
)
d
s
)
-
1
≤
7
Γ
(
7
/
2
)
80
<
Γ
(
7
/
2
)
10
≤
(
2
α
F
0
ξ
(
α
-
2
)
∫
0
1
‍
G
(
1
,
s
)
d
s
)
-
1
.
Theorem 8 implies that, for λ
∈
(
7
Γ
(
7
/
2
)
/
80, Γ
(
7
/
2
)
/
10
), the problem (46) has at least one positive solution.Remark 14.
In particular, if we takef
(
t
,
u
(
t
)
)
=
u
2
(
1
+
t
) in Example 13, then F
0
=
0 and f
∞
=
∞. Remark 9 implies that problem (46) has at least one positive solution for λ
∈
(
0
,
+
∞
).



---

*Source: 916260-2014-03-24.xml*