# Existence and Multiple Positive Solutions for Boundary Value Problem of Fractional Differential Equation withp-Laplacian Operator

Authors: Min Jiang; Shouming Zhong
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

This paper investigates the existence, multiplicity, nonexistence, and uniqueness of positive solutions to a kind of two-point boundary value problem for nonlinear fractional differential equations withp-Laplacian operator. By using fixed point techniques combining with partially ordered structure of Banach space, we establish some criteria for existence and uniqueness of positive solution of fractional differential equations with p-Laplacian operator in terms of different value of parameter. In particular, the dependence of positive solution on the parameter was obtained. Finally, several illustrative examples are given to support the obtained new results. The study of illustrative examples shows that the obtained results are applicable.


---

## Body

## 1. Introduction

In this paper, we consider boundary value problem for a fractional differential equation withp-Laplacian operator
(1)
(
ϕ
p
(
D
0
+
α
x
(
t
)
)
)
′
+
λ
ω
(
t
)
f
(
t
,
x
(
t
)
)
=
0
,
0
<
t
<
1
,
x
′
(
0
)
=
x
′′
(
0
)
=
0
,
x
(
1
)
=
η
∫
0
1
‍
g
(
s
)
x
(
s
)
d
s
,
where 2
<
α
≤
3
,
0
<
η
<
1 is a constant and λ
>
0 is a parameter. p
>
1, ϕ
p
(
s
) is the p-Laplacian operator; that is, ϕ
p
(
s
)
=
|
s
|
p
-
2
s
,
ϕ
p
-
1
(
s
)
=
ϕ
q
(
s
)
,
1
/
p
+
1
/
q
=
1.D denotes the Caputo fractional derivative.Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various fields of sciences and engineering such as physics, chemistry, aerodynamics, electrodynamics of complex medium, electrical circuits, and biology (see [1–7] and their references). Differential equations with p-Laplacian arise naturally in non-Newtonian mechanics, nonlinear elasticity, glaciology, population biology, combustion theory, and nonlinear flow laws. Since the p-Laplacian operator and fractional calculus arise from so many applied fields, the fractional p-Laplacian differential equations are worth studying. Recently, there have appeared a very large number of papers which are devoted to the existence of solutions of boundary value problems and initial value problems for the fractional differential equations (see [8–12]), and the existence of solutions of boundary value problems for the fractional p-Laplacian differential equations has just begun in recent years (see [13–22]). On the other hand, there are few papers that consider the eigenvalue intervals of fractional boundary value problems (see [23, 24]). In [23], the author discussed the following system:
(2)
D
0
+
α
x
(
t
)
+
λ
h
(
t
)
f
(
x
(
t
)
)
=
0
,
0
<
t
<
1
,
mmmmmmmmmmmmmmmn
2
<
α
≤
3
,
x
(
0
)
=
x
′
(
0
)
=
x
′′
(
0
)
=
x
′′
(
1
)
=
0
.
By the use of appropriate conditions with respect to lim
⁡
t
→
0
(
f
(
t
)
/
t
)
=
f
0 and lim
⁡
t
→
∞
(
f
(
t
)
/
t
)
=
f
∞, the author proved that the above problem has at least one or two positive solutions for some λ, where f
0
=
0
,
l
,
∞
,
f
∞
=
0
,
l
,
∞
,
0
<
l
<
∞. But almost all the results which the author obtained depend on both f
0 and f
∞; the case depends on one of f
0 and f
∞; the author only discussed f
0
=
0
,
∞ and f
∞
=
0
,
∞; the case f
0
=
l or f
∞
=
l has not been discussed. On the other hand, there exist several results on the existence of one solution to fractional p-Laplacian boundary value problems (BVPs); there are, to the best of our knowledge, relatively few results on the nonexistence and the uniqueness of positive solutions to fractional p-Laplacian differential equation with parameter.Motivated by the above questions, in this paper, we will establish several sufficient conditions for the existence of positive solutions of (1) by using fixed point theorem and fixed point index theory.The work is organized in the following fashion. In Section2, we provide some necessary background. In particular, we will introduce some lemmas and definitions associated with fixed point index theory. The main results will be stated and proved in Section 3. Two examples are given in Section 4.


## 2. Preliminaries

In this section, we introduce definitions and preliminary facts which are used throughout this paper.Definition 1 (see [1]).
The Riemann-Liouville fractional integral of orderα
>
0 of a function x
:
(
0
,
+
∞
)
→
R is given by
(3)
I
0
+
α
x
(
t
)
=
1
Γ
(
α
)
∫
0
t
‍
(
t
-
s
)
α
-
1
x
(
s
)
d
s
,
provided that the right side is point-wise defined on (
0
,
+
∞
) and Γ is the Gamma function.Definition 2 (see [1]).
The Caputo fractional derivative of orderα
>
0 of a continuous function x
:
(
0
,
+
∞
)
→
R is given by
(4)
D
0
+
α
x
(
t
)
=
1
Γ
(
n
-
α
)
∫
0
t
‍
(
t
-
s
)
n
-
α
-
1
x
(
n
)
(
s
)
d
s
,
where n
=
[
α
]
+
1, provided that the right side is point-wise defined on (
0
,
+
∞
).Lemma 3 (see [1]).
Letα
>
0, and assume that x
∈
C
[
0,1
], and then the fractional differential equation D
0
α
x
(
t
)
=
0 has unique solutions:
(5)
p
p
p
p
p
u
(
t
)
=
c
0
+
c
1
t
+
c
2
t
2
+
⋯
+
c
n
-
1
t
n
-
1
,
c
i
∈
R
,
i
=
0,1
,
2
,
…
,
n
-
1
,
n
-
1
<
α
≤
n
.Lemma 4 (see [1]).
Letα
>
0. Then,
(6)
I
0
+
α
D
0
+
α
x
(
t
)
=
x
(
t
)
+
c
0
+
c
1
t
+
c
2
t
2
+
⋯
+
c
n
-
1
t
n
-
1
,
where c
i
∈
R
,
i
=
0,1
,
2
,
…
,
n
-
1
,
n
-
1
<
α
≤
n.Lemma 5 (see [25]).
LetP be a cone in a real Banach space E, and let Ω be a bounded open set of E. Assume that operator A
:
P
∩
Ω
-
→
P is completely continuous. If there exists a x
0
>
0 such that
(7)
x
-
A
x
≠
t
x
0
,
∀
x
∈
P
∩
∂
Ω
,
t
≥
0
,
then i
(
A
,
P
∩
Ω
,
P
)
=
0.Lemma 6 (see [26]).
LetE be a Banach space and P
⊆
E a cone, and Ω
1 and Ω
2 are open set with 0
∈
Ω
1
,
Ω
-
1
⊆
Ω
2, and let T
:
P
∩
(
Ω
-
2
∖
Ω
1
)
→
P be completely continuous operator such that either(i)
∥
T
u
∥
≤
∥
u
∥
,
u
∈
P
∩
∂
Ω
1, and ∥
T
u
∥
≥
∥
u
∥
,
u
∈
P
∩
∂
Ω
2or(ii)
∥
T
u
∥
≥
∥
u
∥
,
u
∈
P
∩
∂
Ω
1, and ∥
T
u
∥
≤
∥
u
∥
,
u
∈
P
∩
∂
Ω
2holds. Then T has a fixed point in P
∩
(
Ω
-
2
∖
Ω
1
).


## 3. Main Results

In this section, we present some new results on the existence, multiplicity, nonexistence, and the uniqueness of positive solution of problem (1) and dependence of the positive solution x
λ
(
t
) on the parameter λ.LetE
=
C
[
0,1
]; then E is a real Banach space with the norm ∥
·
∥ defined by ∥
x
∥
=
max
⁡
0
≤
t
≤
1
|
x
(
t
)
|.Lemma 7.
Lety
(
t
)
∈
C
(
0,1
)
∩
L
1
(
0,1
) and y
(
t
)
≥
0; then the solution of the problem
(8)
(
ϕ
p
(
D
0
+
α
x
(
t
)
)
)
′
+
λ
y
(
t
)
=
0
,
0
<
t
<
1
,
x
′
(
0
)
=
x
′′
(
0
)
=
0
,
x
(
1
)
=
η
∫
0
1
‍
g
(
s
)
x
(
s
)
d
s
is given by
(9)
x
(
t
)
=
1
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
m
m
m
-
∫
0
t
‍
(
t
-
s
)
α
-
1
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
ϕ
q
(
λ
∫
0
τ
‍
y
(
μ
)
d
μ
)
d
τ
m
m
m
n
m
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
y
(
μ
)
d
μ
)
d
τ
)
d
s
}
,
(10)
x
(
t
)
≥
0
,
min
⁡
0
≤
t
≤
1
x
(
t
)
≥
σ
∥
x
∥
,
where 2
<
α
≤
3
,
0
<
η
<
1
,
1
-
η
∫
0
1
‍
g
(
s
)
d
s
>
0 and σ
=
η
∫
0
1
‍
(
1
-
s
)
g
(
s
)
d
s
/
(
1
-
η
∫
0
1
‍
s
g
(
s
)
d
s
).Proof.
It is easy to see by integration of (8) that
(11)
ϕ
p
(
D
0
+
α
x
(
t
)
)
-
ϕ
p
(
D
0
+
α
x
(
0
)
)
=
-
λ
∫
0
t
‍
y
(
s
)
d
s
.
By the boundary condition x
′
(
0
)
=
x
′′
(
0
)
=
0, we can easily get that D
0
+
α
x
(
0
)
=
0, and we obtain that
(12)
ϕ
p
(
D
0
+
α
x
(
t
)
)
=
-
λ
∫
0
t
‍
y
(
s
)
d
s
;
that is
(13)
D
0
+
α
x
(
t
)
=
-
ϕ
q
(
λ
∫
0
t
‍
y
(
s
)
d
s
)
.
By Lemmas 3 and 4, we get that
(14)
x
(
t
)
=
-
I
α
ϕ
q
(
λ
∫
0
t
‍
y
(
s
)
d
s
)
+
c
0
+
c
1
t
+
c
2
t
2
.
Using the boundary condition x
′
(
0
)
=
x
′′
(
0
)
=
0, x
(
1
)
=
η
∫
0
1
‍
g
(
s
)
x
(
s
)
d
s, we get c
1
=
c
2
=
0 and
(15)
c
0
=
1
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
-
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
m
×
∫
0
1
‍
g
(
t
)
(
∫
0
t
‍
(
t
-
s
)
α
-
1
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
)
d
t
.
Thus,
(16)
x
(
t
)
=
-
1
Γ
(
α
)
∫
0
t
‍
(
t
-
s
)
α
-
1
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
+
1
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
-
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
g
(
t
)
(
∫
0
t
‍
(
t
-
s
)
α
-
1
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
)
d
t
=
1
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
m
m
m
m
m
-
∫
0
t
‍
(
t
-
s
)
α
-
1
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
ϕ
q
(
λ
∫
0
τ
‍
y
(
μ
)
d
μ
)
d
τ
mmmmmmnm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
y
(
μ
)
d
μ
)
d
τ
)
d
s
}
.
Direct differentiation of (9) implies
(17)
x
′
(
t
)
=
-
1
Γ
(
α
-
1
)
∫
0
t
‍
(
t
-
s
)
α
-
2
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
≤
0
.
By differentiation of (17), we get
(18)
x
′′
(
t
)
=
-
1
Γ
(
α
-
2
)
∫
0
t
‍
(
t
-
s
)
α
-
3
ϕ
q
(
λ
∫
0
s
‍
y
(
τ
)
d
τ
)
d
s
≤
0
.
Thus the solution of problem (8) is nonincreasing and concave on [
0,1
], and
(19)
∥
x
∥
=
x
(
0
)
,
min
⁡
0
≤
t
≤
1
x
(
t
)
=
x
(
1
)
.
On the other hand, as we assume that y
(
t
)
≥
0, we see that
(20)
x
(
1
)
=
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
ϕ
q
(
λ
∫
0
τ
‍
y
(
μ
)
d
μ
)
d
τ
m
m
m
m
m
m
m
m
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
y
(
μ
)
d
μ
)
d
τ
)
d
s
}
=
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
(
∫
0
s
‍
[
(
1
-
τ
)
α
-
1
-
(
s
-
τ
)
α
-
1
]
m
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
y
(
μ
)
d
μ
)
d
τ
m
m
n
n
+
∫
s
1
‍
(
1
-
τ
)
α
-
1
ϕ
q
(
λ
∫
0
τ
‍
y
(
μ
)
d
μ
)
d
τ
)
d
s
}
≥
0
.
Therefore, x
(
t
)
≥
0 and is concave for t
∈
[
0,1
]. So for every t
∈
[
0,1
],
(21)
x
(
t
)
≥
t
x
(
1
)
+
(
1
-
t
)
x
(
0
)
.
Therefore,
(22)
η
∫
0
1
‍
x
(
t
)
g
(
t
)
d
t
≥
η
x
(
1
)
∫
0
1
‍
t
g
(
t
)
d
t
+
η
x
(
0
)
∫
0
1
‍
(
1
-
t
)
g
(
t
)
d
t
.
Since x
(
1
)
=
η
∫
0
1
‍
x
(
t
)
g
(
t
)
d
t
,
1
-
η
∫
0
1
‍
g
(
t
)
d
t
>
0, we have
(23)
x
(
1
)
≥
η
∫
0
1
‍
(
1
-
t
)
g
(
t
)
d
t
1
-
η
∫
0
1
‍
t
g
(
t
)
d
t
x
(
0
)
;
that is,
(24)
min
⁡
0
≤
t
≤
1
x
(
t
)
=
x
(
1
)
≥
η
∫
0
1
‍
(
1
-
t
)
g
(
t
)
d
t
1
-
η
∫
0
1
‍
t
g
(
t
)
d
t
x
(
0
)
=
σ
∥
x
∥
.
The Lemma is proved.We construct a coneP in E by
(25)
P
=
{
x
∈
E
:
x
≥
0
,
min
⁡
0
≤
t
≤
1
x
(
t
)
≥
σ
∥
x
∥
}
,
where σ is defined in Lemma 7. It is easy to see P is a closed convex cone of E.DefineT
:
P
→
E by
(26)
(
T
x
)
(
t
)
=
1
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
m
m
m
m
m
m
-
∫
0
t
‍
(
t
-
s
)
α
-
1
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
n
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
m
m
m
m
m
n
m
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
m
m
×
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
.It is clear that the fixed points of the operatorϕ
q
(
λ
)
T are the solutions of the boundary value problems (1).We make the following hypotheses:(H1)
f
:
[
0,1
]
×
[
0
,
∞
)
→
[
0
,
∞
) is continuous;
(H2)
ω
:
(
0,1
)
→
[
0
,
∞
) is continuous and not identical zero on any closed subinterval of (
0,1
) with 0
<
∫
0
1
‍
ω
(
t
)
d
t
<
+
∞;
(H3)
1
-
η
∫
0
1
‍
g
(
s
)
d
s
>
0;
(H4)
f
ϕ
0
∈
[
0
,
∞
], where f
ϕ
0
:
=
lim
⁡
x
→
0
(
f
(
t
,
x
)
/
ϕ
p
(
x
)
) uniformly for t
∈
[
0,1
];
(H5)
f
ϕ
∞
∈
[
0
,
∞
], where f
ϕ
∞
:
=
lim
⁡
x
→
∞
(
f
(
t
,
x
)
/
ϕ
p
(
x
)
) uniformly for t
∈
[
0,1
];
(H6)
f
(
t
,
x
)
>
0 for any t
∈
[
0,1
] and x
>
0.Lemma 8.
Assume that (H1)–(H3) hold. ThenT
:
P
→
E is completely continuous.Proof.
First, we show thatT is continuous. It is easy to see T
(
P
)
⊂
P. For x
n
(
t
)
∈
E and x
n
(
t
)
→
x
(
t
), as n
→
∞, by the continuous of f
(
t
,
x
(
t
)
), we get f
(
t
,
x
n
(
t
)
)
→
f
(
t
,
x
(
t
)
), as n
→
∞. This implies that
(27)
ϕ
q
(
∫
0
t
‍
w
(
s
)
f
(
s
,
x
n
(
s
)
)
d
s
)
⟶
ϕ
q
(
∫
0
t
‍
w
(
s
)
f
(
s
,
x
(
s
)
)
d
s
)
.
So we have
(28)
sup
⁡
t
∈
[
0,1
]
|
ϕ
q
(
∫
0
t
‍
w
(
s
)
f
(
s
,
x
n
(
s
)
)
d
s
)
mmm
-
ϕ
q
(
∫
0
t
‍
w
(
s
)
f
(
s
,
x
(
s
)
)
d
s
)
|
⟶
0
,
n
⟶
∞
.
Denote Π
=
|
(
T
x
n
)
(
t
)
-
(
T
x
)
(
t
)
|; then
(29)
Π
≤
1
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
|
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
n
(
τ
)
)
d
τ
)
m
m
m
m
m
m
m
m
m
m
m
-
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
|
d
s
mmm
+

∫
0
t
‍
(
t
-
s
)
α
-
1
|
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
n
(
τ
)
)
d
τ
)

-

ϕ
q
m
m
m
m
m
m
m
m
m
m
m
×
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
|
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
mmmim
m
×
|
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
n
(
μ
)
)
d
μ
)
m
m
m
m
m
m
m
-
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
|
d
τ
mmmim
+
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
m
×
|
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
n
(
μ
)
)
d
μ
)
-
ϕ
q
m
n
m
m
m
m
m
i
m
×
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
|
d
τ
)
d
s
}
≤
sup
⁡
s
∈
[
0,1
]
|
ϕ
q
(
∫
0
s
‍
w
(
τ
)
f
(
τ
,
x
n
(
τ
)
)
d
τ
)
m
m
m
m
m
-
ϕ
q
(
∫
0
s
‍
w
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
|
×
2
Γ
(
α
+
1
)
(
1
+
η
∫
0
1
‍
g
(
s
)
d
s
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
⟶
0
,
mmmmmmmmmmmmmmmmmn
n
⟶
∞
.
Hence,T is continuous.
Second, we show thatT is compact.
Forx
(
t
)
∈
B
r
=
{
x
∈
P
:
∥
x
∥
≤
r
}, by condition (H1), f
(
t
,
x
(
t
)
)
<
∞. Denote M
=
max
⁡
0
≤
t
≤
1
,
∥
x
∥
≤
r
f
(
t
,
x
(
t
)
).
Then(30)
|
T
x
(
t
)
|
≤
1
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
|
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
|
d
s
m
m
m
m
m
m
+
∫
0
t
‍
(
t
-
s
)
α
-
1
|
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
|
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
i
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
i
×
|
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
|
d
τ
mim
+
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
n
m
i
i
×
|
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
|
d
τ
)
d
s
}
≤
1
Γ
(
α
)
×
{
∫
0
1
‍
(
1
-
s
)
α
-
1
|
ϕ
q
(
M
∫
0
s
‍
ω
(
τ
)
d
τ
)
|
d
s
+
∫
0
t
‍
(
t
-
s
)
α
-
1
|
ϕ
q
(
M
∫
0
s
‍
ω
(
τ
)
d
τ
)
|
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
×
|
ϕ
q
(
M
∫
0
τ
‍
ω
(
μ
)
d
μ
)
d
τ
|
n
+
∫
0
s
‍
(
s
-
τ
)
α
-
1
n
×
|
ϕ
q
(
M
∫
0
τ
‍
ω
(
μ
)
d
μ
)
|
d
τ
)
d
s
}
≤
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
ϕ
q
(
M
∫
0
1
‍
ω
(
τ
)
d
τ
)
×
∫
0
1
‍
g
(
s
)
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
d
τ
+
∫
0
s
‍
(
s
-
τ
)
α
-
1
d
τ
)
d
s
+
1
Γ
(
α
)
ϕ
q
(
M
∫
0
1
‍
ω
(
τ
)
d
τ
)
×
{
∫
0
1
‍
(
1
-
s
)
α
-
1
d
s
+
∫
0
t
‍
(
t
-
s
)
α
-
1
d
s
}
≤
2
ϕ
q
(
M
∫
0
1
‍
ω
(
τ
)
d
τ
)
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
τ
)
d
τ
)
<
∞
;
that is, T maps bounded sets into bounded sets in P.
Forx
(
t
)
∈
B
r, 0
<
t
1
<
t
2
<
1,
(31)
|
T
x
(
t
2
)
-
T
x
(
t
1
)
|
=
1
Γ
(
α
)
|
∫
0
t
2
‍
(
t
2
-
s
)
α
-
1
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
m
m
m
m
m
i
-
∫
0
t
1
‍
(
t
1
-
s
)
α
-
1
m
m
m
m
m
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
|
≤
1
Γ
(
α
)
ϕ
q
(
M
∫
0
1
‍
ω
(
τ
)
d
τ
)
m
m
×
|
∫
t
1
t
2
‍
∫
0
t
1
‍
(
(
t
2
-
s
)
α
-
1
-
(
t
1
-
s
)
α
-
1
)
d
s
m
m
n
m
+
∫
t
1
t
2
‍
(
t
2
-
s
)
α
-
1
d
s
|
.
By mean value theorem, we obtain that
(32)
∫
0
t
1
‍
(
(
t
2
-
s
)
α
-
1
-
(
t
1
-
s
)
α
-
1
)
d
s
≤
∫
0
t
1
‍
(
α
-
1
)
(
1
-
s
)
α
-
2
(
t
2
-
t
1
)
d
s
≤
∫
0
1
‍
(
α
-
1
)
(
1
-
s
)
α
-
2
(
t
2
-
t
1
)
d
s
=
t
2
-
t
1
.
Thus
(33)
|
T
x
(
t
2
)
-
T
x
(
t
1
)
|
≤
1
Γ
(
α
)
ϕ
q
(
M
∫
0
1
‍
ω
(
τ
)
d
τ
)
[
(
t
2
-
t
1
)
+
1
α
(
t
2
-
t
1
)
α
]
.
This shows that |
T
x
(
t
2
)
-
T
x
(
t
1
)
|
→
0, as t
2
-
t
1
→
0, so {
T
x
:
x
∈
B
r
} is equicontinuous. Therefore, the operator T
:
P
→
P is completely continuous by the Arzela-Ascoli theorem.Now for convenience we introduce the following notations. Let(34)
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
d
τ
)
d
s
=
γ
1
,
ϕ
q
(
∫
0
1
‍
ω
(
τ
)
d
τ
)
=
γ
2
,
m
(
r
~
)
=
min
⁡
{
min
⁡
t
f
(
t
,
x
)
ϕ
p
(
r
~
)
,
x
∈
[
σ
r
~
,
r
~
]
}
.Theorem 9.
Assume that (H1)–(H5) hold.(i)
If0
<
f
ϕ
0
<
+
∞, then there exists ξ
0
>
0 such that, for every 0
<
r
<
ξ
0, problem (1) has a positive solution x
r
(
t
) satisfying ∥
x
r
∥
=
r associated with
(35)
λ
=
λ
r
∈
[
λ
-
0
,
λ
0
]
,
where λ
-
0 and λ
0 are two positive finite numbers.
(ii)
If0
<
f
ϕ
∞
<
+
∞, then there exists ξ
0
*
>
0 such that, for every R
>
ξ
0
*, problem (1) has a positive solution x
R
(
t
) satisfying ∥
x
R
∥
=
R for any
(36)
λ
=
λ
R
∈
[
λ
-
0
*
,
λ
0
*
]
,
where λ
-
0
* and λ
0
* are two positive finite numbers.
(iii)
Iff
ϕ
0
=
+
∞, then there exists ξ
1
>
0 such that, for any 0
<
r
*
<
ξ
1, problem (1) has a positive solution x
r
*
(
t
) satisfying ∥
x
r
*
∥
=
r
* for any
(37)
λ
=
λ
r
*
∈
(
0
,
λ
*
]
,
where λ
* is a positive finite number.
(iv)
Iff
ϕ
∞
=
+
∞, then there exists ξ
-
1
>
0 such that, for every R
*
>
ξ
-
1, problem (1) has a positive solution x
R
*
(
t
) satisfying ∥
x
R
*
∥
=
R
* for any
(38)
λ
=
λ
R
*
∈
(
0
,
λ
*
]
,
where λ
* is a positive finite number.
(v)
If there existsξ
>
0 such that m
(
r
~
)
>
ξ, then problem (1) has positive solution x
r
~
(
t
) satisfying ∥
x
r
~
∥
=
r
~ for any
(39)
λ
=
λ
r
~
∈
(
0
,
λ
~
]
,
where λ
~ is a positive finite number.Proof.
(i) It follows from0
<
f
ϕ
0
<
+
∞ that there exists l
1
,
l
2
,
ρ
>
0, such that
(40)
l
1
ϕ
p
(
x
)
≤
f
(
t
,
x
)
≤
l
2
ϕ
p
(
x
)
,
ppip
(
∀
t
∈
[
0,1
]
,
0
≤
x
≤
ρ
)
.
Letξ
0
=
ρ
/
σ; we show that ξ
0 is required. When x
∈
P
∩
∂
Ω
r, we have
(41)
x
(
t
)
≥
σ
∥
x
∥
=
σ
r
;
we need x
(
t
)
<
ρ; this implies that σ
r
<
ρ, as r
<
ξ
0, so σ
r
<
σ
ξ
0
=
ρ.
Letλ
0
=
(
1
/
l
1
)
(
Γ
(
α
)
/
σ
γ
1
)
p
-
1. Then we may assume that
(42)
x
-
ϕ
q
(
λ
0
)
T
x
≠
0
,
(
∀
x
∈
P
∩
∂
Ω
r
)
;
if not, then there exists x
r
∈
P
∩
∂
Ω
r such that ϕ
q
(
λ
0
)
T
x
r
=
x
r, and then (35) already holds for λ
r
=
λ
0.
Defineψ
(
t
)
≡
1
,
∀
t
∈
[
0,1
]; then ψ
(
t
)
∈
P, and ∥
ψ
∥
≡
1. We now show that
(43)
x
-
ϕ
q
(
λ
0
)
T
x
≠
κ
ψ
(
∀
x
∈
P
∩
∂
Ω
r
,
κ
≥
0
)
.
In fact, if there exists x
1
∈
P
∩
∂
Ω
r
,
κ
1
≥
0 such that x
1
-
ϕ
q
(
λ
0
)
T
x
1
=
κ
1
ψ, then (42) implies that κ
1
>
0. On the other hand x
1
=
ϕ
q
(
λ
0
)
T
x
1
+
κ
1
ψ
≥
κ
1
ψ; we may choose κ
*
=
max
⁡
{
κ
∣
x
1
≥
κ
ψ
}
}; then κ
1
≤
κ
*
<
+
∞
,
x
1
≥
κ
*
ψ. Therefore
(44)
κ
*
=
κ
*
∥
ψ
∥
≤
∥
x
1
∥
=
r
.
Consequently, for any t
∈
[
0,1
], (26) and (44) imply that
(45)
x
1
(
t
)
=
ϕ
q
(
λ
0
)
1
Γ
(
α
)
×
{
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
1
(
τ
)
)
d
τ
)
d
s
m
-
∫
0
t
‍
(
t
-
s
)
α
-
1
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
1
(
τ
)
)
d
τ
)
d
s
}
+
η
ϕ
q
(
λ
0
)
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
mmmm
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
mmmmmmim
×
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
1
(
μ
)
)
d
μ
)
d
τ
mmmmmm
m
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
m
m
×
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
1
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
+
κ
1
ψ
≥
η
σ
ϕ
q
(
λ
0
)
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
mmmmmmnm
×
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
1
(
μ
)
)
d
μ
)
d
τ
mmmmmnm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
ϕ
q
m
m
m
m
m
n
m
m
m
×
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
1
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
+
σ
ϕ
q
(
λ
0
)
Γ
(
α
)
m
m
×
{
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
1
(
τ
)
)
d
τ
)
d
s
}
mm
+
κ
1
ψ
≥
σ
ϕ
q
(
λ
0
)
Γ
(
α
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
1
(
τ
)
)
d
τ
)
d
s
+
κ
1
ψ
≥
σ
ϕ
q
(
λ
0
)
Γ
(
α
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
l
1
∫
0
s
‍
ω
(
τ
)
ϕ
p
(
x
1
(
τ
)
)
d
τ
)
d
s
+
κ
1
ψ
≥
σ
ϕ
q
(
λ
0
)
Γ
(
α
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
l
1
∫
0
s
‍
ω
(
τ
)
ϕ
p
(
κ
*
ψ
(
τ
)
)
d
τ
)
d
s
+
κ
1
ψ
=
σ
κ
*
l
1
q
-
1
λ
0
q
-
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
d
τ
)
d
s
+
κ
1
ψ
=
σ
κ
*
l
1
q
-
1
λ
0
q
-
1
γ
1
Γ
(
α
)
+
κ
1
ψ
=
κ
*
+
κ
ψ
,
which implies that x
1
(
t
)
≥
(
κ
*
+
κ
)
ψ
(
t
)
,
t
∈
[
0,1
], which is a contradiction to the definition of κ
*. Thus, (42) holds and, by Lemma 5, the fixed point index
(46)
i
(
ϕ
q
(
λ
0
)
T
,
P
∩
∂
Ω
r
,
P
)
=
0
.
On the other hand, by the fact that the fixed point index of constants operator is 1, so
(47)
i
(
ϑ
,
P
∩
∂
Ω
r
,
P
)
=
1
,
where ϑ is the zero operator. It follows therefore from (46) and (47) and the homotopy invariance property that there exist x
r
∈
P
∩
∂
Ω
r and 0
<
ν
0
<
1 such that ν
0
ϕ
q
(
λ
0
)
T
x
r
=
x
r, but x
r
=
ϕ
q
(
λ
r
)
T
x
r, which implies that ν
0
ϕ
q
(
λ
0
)
T
x
r
=
ϕ
q
(
λ
r
)
T
x
r; it follows that we get ν
0
ϕ
q
(
λ
0
)
=
ϕ
q
(
λ
r
)
); that is,
(48)
ν
0
λ
0
q
-
1
=
λ
r
q
-
1
,
and then
(49)
λ
r
=
ν
0
1
/
(
q
-
1
)
λ
0
=
ν
0
p
-
1
λ
0
<
λ
0
.
From the proof above, for any r
<
ξ
0, there exists a positive solution x
r
∈
P
∩
∂
Ω
r associated with λ
=
λ
r
>
0.
Thus(50)
x
r
(
t
)
=
ϕ
q
(
λ
r
)
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
r
(
τ
)
)
d
τ
)
d
s
mmmmmmm
-
∫
0
t
‍
(
t
-
s
)
α
-
1
m
m
m
m
m
m
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
r
(
τ
)
)
d
τ
)
d
s
}
m
+
η
ϕ
q
(
λ
r
)
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
m
×
{
∫
0
1
‍
g
(
s
)
m
m
m
n
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
mmnmm
×
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
r
(
μ
)
)
d
μ
)
d
τ
mmmnm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
i
m
×
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
r
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
with ∥
x
r
∥
=
r.
Next, we show thatλ
r
≥
λ
-
0. In fact,
(51)
x
r
(
t
)
=
ϕ
q
(
λ
r
)
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
r
(
τ
)
)
d
τ
)
d
s
mmmmmmm
-
∫
0
t
‍
(
t
-
s
)
α
-
1
m
m
m
n
m
m
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
r
(
τ
)
)
d
τ
)
d
s
}
+
η
ϕ
q
(
λ
r
)
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
mmmmmmnm
×
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
r
(
μ
)
)
d
μ
)
d
τ
mmmmnmm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
m
n
×
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
r
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
≤
η
ϕ
q
(
λ
r
)
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
g
(
s
)
∫
0
1
‍
(
1
-
τ
)
α
-
1
mmmmmmmm
×
ϕ
q
(
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
r
(
μ
)
)
d
μ
)
d
τ
d
s
+
ϕ
q
(
λ
r
)
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
r
(
τ
)
)
d
τ
)
d
s
=
ϕ
q
(
λ
r
)
η
∫
0
1
‍
g
(
s
)
d
s
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
r
(
τ
)
)
d
τ
)
d
s
+
ϕ
q
(
λ
r
)
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
r
(
τ
)
)
d
τ
)
d
s
≤
ϕ
q
(
λ
r
)
η
∫
0
1
‍
g
(
s
)
d
s
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
×
ϕ
q
(
l
2
∫
0
s
‍
ω
(
τ
)
x
r
(
τ
)
d
τ
)
d
s
+
ϕ
q
(
λ
r
)
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
n
m
×
ϕ
q
(
l
2
λ
r
∫
0
s
‍
ω
(
τ
)
x
r
(
τ
)
d
τ
)
d
s
=
ϕ
q
(
λ
r
)
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
×
ϕ
q
(
l
2
∫
0
s
‍
ω
(
τ
)
x
r
(
τ
)
d
τ
)
d
s
≤
∥
x
r
∥
ϕ
q
(
λ
r
l
2
)
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
×
ϕ
q
(
∫
0
1
‍
ω
(
τ
)
d
τ
)
d
s
=
∥
x
r
∥
ϕ
q
(
λ
r
l
2
)
γ
2
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
,
which implies that
(52)
∥
x
r
∥
≤
∥
x
r
∥
ϕ
q
(
λ
r
l
2
)
γ
2
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
;
that is,
(53)
ϕ
q
(
λ
r
l
2
)
γ
2
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
≥
1
;
it follows that we get
(54)
λ
r
≥
1
l
2
(
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
γ
2
)
p
-
1
:
=
λ
-
0
.
In conclusion, λ
r
∈
[
λ
-
0
,
λ
0
].
(ii) It follows from0
<
f
ϕ
∞
<
∞ that there exist l
1
*
,
l
2
*
,
ρ
*
>
0 such that
(55)
l
1
*
ϕ
p
(
x
)
≤
f
(
t
,
x
)
≤
l
2
*
ϕ
p
(
x
)
p
p
p
p
p
(
∀
t
∈
[
0,1
]
,
x
≥
ρ
*
)
.
Let ξ
0
*
=
ρ
*
/
σ, the following proof are similar to (i).
(iii) It follows fromf
ϕ
0
=
+
∞, there exists l
*
>
0
,
ρ
*
>
0, such that
(56)
f
ϕ
0
≥
l
*
ϕ
p
(
x
)
,
(
∀
t
∈
[
0,1
]
,
0
≤
x
≤
ρ
*
)
.
Let ξ
1
=
ρ
*
/
σ; we show that ξ
1 is required. When x
∈
P
∩
∂
Ω
r
*, we have x
(
t
)
≥
σ
∥
x
∥
=
σ
r
*; since x
≤
ρ
*, then we need σ
r
*
<
ρ
*, as r
*
<
ξ
1, so σ
r
*
<
ρ
* holds.
Letλ
*
=
(
1
/
l
*
)
(
Γ
(
α
)
/
σ
γ
1
)
1
/
(
q
-
1
); we proceed in the same way as in the proof of (i): replacing (42) we may assume that x
-
ϕ
q
(
λ
*
)
T
x
≠
0
,
(
∀
x
∈
P
∩
∂
Ω
r
*
), and replacing (43) we can prove x
-
ϕ
q
(
λ
*
)
T
x
≠
κ
ψ
,
(
∀
x
∈
P
∩
∂
Ω
r
*
,
κ
≥
0
). It follows from Lemma 5 that i
(
ϕ
q
(
λ
*
)
T
,
P
∩
Ω
r
*
,
P
)
=
0. Note that i
(
θ
,
P
∩
Ω
r
*
,
P
)
=
1; we can easily show that there exists x
r
*
∈
P
∩
∂
Ω
r
* and 0
<
ν
r
*
<
1 such that ν
r
*
ϕ
q
(
λ
*
)
T
x
r
*
=
x
r
*. Hence (37) holds for λ
r
*
=
λ
*
ν
r
*
1
-
p
<
λ
*.
The proof of Theorem9(iv) follows by the method similar to Theorem 9(iii); we omit it here.
(v) It follows thatm
(
r
~
)
>
ξ; for any x
∈
P
∩
∂
Ω
r
~, we have σ
r
~
≤
x
≤
r
~ and f
(
t
,
x
)
/
ϕ
p
(
x
)
≥
f
(
t
,
x
)
/
ϕ
p
(
r
~
)
≥
m
(
r
~
)
≥
ξ; then
(57)
f
(
t
,
x
)
≥
ξ
ϕ
p
(
x
)
,
∀
x
∈
[
σ
r
~
,
r
~
]
,
t
∈
[
0,1
]
.
Let λ
~
=
(
1
/
ξ
)
(
Γ
(
α
)
/
σ
γ
1
)
p
-
1. The following proof is similar to that of (iv). This finished the proof of (v).Remark 10.
In Theorem9, all the criteria obtained depend on one of f
ϕ
0 and f
ϕ
∞.LetΨ
(
x
)
=
ϕ
q
(
λ
)
T
x; the following theorems give out the multiply, nonexistence, and the dependence of parameter.Theorem 11.
Assume that (H1)–(H5) hold.(i)
Iff
ϕ
0
=
0 and f
ϕ
∞
=
0, lim
⁡
_
x
→
∞
f
(
t
,
x
)
=
f
~
∞
∈
(
0
,
+
∞
] uniformly for t
∈
[
0,1
], then there exists λ
*
>
0 such that the problem (1) has two solutions for any λ
>
λ
*.
(ii)
Iff
ϕ
0
=
0 and f
ϕ
∞
=
0, then there exists λ
_
>
0 such that, for any λ
<
λ
_, problem (1) has no solution.Proof.
(i) It follows fromf
~
∞
∈
(
0
,
+
∞
] that, for any ζ
~
>
0, there exists ρ
~
>
0, such that
(58)
f
(
t
,
x
)
>
ζ
~
,
(
∀
t
∈
[
0,1
]
,
x
≥
ρ
~
)
.
Sincef
ϕ
0
=
0, there exists ε
1
>
0
,
ρ
-
1
>
0 such that
(59)
f
(
t
,
x
)
≤
ε
1
ϕ
p
(
x
)
,
(
∀
t
∈
[
0,1
]
,
0
≤
x
≤
ρ
-
1
)
,
where ε
1 satisfied ϕ
q
(
λ
ε
1
)
γ
2
/
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
≤
1.
For∀
x
∈
P
∩
∂
Ω
ρ
-
1, we have
(60)
Ψ
(
x
)
=
(
ϕ
q
(
λ
)
T
x
)
(
t
)
=
1
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
mmmmm
-
∫
0
t
‍
(
t
-
s
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmmmm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
i
n
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
≤
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
g
(
s
)
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
n
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
ε
1
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≤
η
∫
0
1
‍
g
(
s
)
d
s
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
×
ϕ
q
(
λ
∫
0
1
‍
ω
(
μ
)
ε
1
ϕ
p
(
x
(
μ
)
)
d
μ
)
d
τ
m
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
n
×
ϕ
q
(
λ
∫
0
1
‍
ω
(
τ
)
ε
1
ϕ
p
(
x
(
τ
)
)
d
τ
)
d
s
≤
1
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
ϕ
q
(
λ
∫
0
1
‍
ω
(
τ
)
ε
1
ϕ
p
(
∥
x
∥
)
d
τ
)
=
ϕ
q
(
λ
ε
1
)
γ
2
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
∥
x
∥
≤
∥
x
∥
.
This implied
(61)
∥
Ψ
x
∥
≤
∥
x
∥
,
∀
x
∈
P
∩
∂
Ω
ρ
-
1
.
Let ρ
-
3
>
max
⁡
{
ρ
~
/
σ
,
ρ
-
1
} and λ
*
=
(
1
/
ζ
~
)
(
ρ
-
3
Γ
(
α
)
/
σ
γ
1
)
p
-
1; then ∀
x
∈
P
∩
∂
Ω
ρ
-
3; we have σ
∥
x
∥
≤
x
≤
ρ
-
3 and
(62)
(
Ψ
x
)
(
t
)
=
1
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
n
i
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
mmmmmmmi
-
∫
0
t
‍
(
t
-
s
)
α
-
1
m
m
m
m
m
i
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
n
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmmmm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
≥
σ
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
n
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmmmm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
+
σ
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
n
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≥
σ
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
>
σ
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
i
n
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
ζ
~
d
τ
)
d
s
>
σ
ϕ
q
(
λ
ζ
~
)
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
n
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
d
τ
)
d
s
=
σ
ϕ
q
(
λ
ζ
~
)
γ
1
Γ
(
α
)
=
ρ
-
3
,
which implies
(63)
∥
Ψ
x
∥
>
∥
x
∥
,
∀
x
∈
P
∩
∂
Ω
ρ
-
3
.
Next, forf
ϕ
∞
=
0, there exist ε
2
>
0
,
ρ
-
2
>
ρ
-
3
>
0 such that
(64)
f
(
t
,
x
)
≤
ε
2
ϕ
p
(
x
)
,
(
∀
t
∈
[
0,1
]
,
x
≥
ρ
-
2
)
,
where ε
2 satisfied ϕ
q
(
λ
ε
2
)
γ
2
/
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
≤
1.
Similar to the above proof, we get(65)
∥
Ψ
x
∥
≤
∥
x
∥
,
∀
x
∈
P
∩
∂
Ω
ρ
-
2
.
Applying Lemma7 to (61), (65), and (63) yields that Ψ has two fixed points x
1
,
x
2 such that x
1
∈
P
∩
(
Ω
-
ρ
-
3
∖
Ω
ρ
-
1
) and x
2
∈
P
∩
(
Ω
-
ρ
-
2
∖
Ω
ρ
-
3
).
(ii) It follows fromf
ϕ
0
=
0 and f
ϕ
∞
=
0 that there exists ε
1
,
ε
2
,
ρ
-
1
,
ρ
-
2
>
0, such that
(66)
f
(
t
,
x
)
≤
ε
1
ϕ
p
(
x
)
,
(
∀
t
∈
[
0,1
]
,
0
≤
x
≤
ρ
-
1
)
,
f
(
t
,
x
)
≤
ε
2
ϕ
p
(
x
)
,
(
∀
t
∈
[
0,1
]
,
x
≥
ρ
-
2
)
;
then for x
∈
[
ρ
-
1
,
ρ
-
2
], by the continuous of f
(
t
,
x
(
t
)
), there exists x
-
∈
[
ρ
-
1
,
ρ
-
2
] such that
(67)
max
⁡
ρ
-
1
≤
x
≤
ρ
-
2
f
(
t
,
x
)
ϕ
p
(
x
)
=
f
(
t
,
x
-
)
ϕ
p
(
x
-
)
.
Let Υ
_
=
max
⁡
{
ε
1
,
ε
2
,
f
(
t
,
x
-
)
/
ϕ
p
(
x
-
)
}
>
0; then
(68)
f
(
t
,
x
)
≤
Υ
_
ϕ
p
(
x
)
,
(
∀
0
≤
t
≤
1
,
x
≥
0
)
.
Assuming x
(
t
) is a positive solution of problem (1), we will show that this leads to a contradiction for λ
<
λ
_, where λ
_
=
(
1
/
Υ
_
)
(
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
/
γ
2
)
p
-
1. It follows from (26) that
(69)
∥
x
∥
=
x
(
0
)
=
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
i
m
i
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmmnm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
n
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
i
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≤
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
g
(
s
)
m
m
n
×
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
i
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
d
s
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
i
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
=
1
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
n
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≤
1
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
n
×
ϕ
q
(
λ
∫
0
1
‍
ω
(
τ
)
Υ
_
ϕ
p
(
x
(
τ
)
)
d
τ
)
d
s
=
ϕ
q
(
λ
Υ
_
)
γ
2
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
∥
x
∥
.
This implies that 1
≤
ϕ
q
(
λ
Υ
_
)
γ
2
/
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
=
(
λ
Υ
_
)
q
-
1
γ
2
/
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
<
(
λ
_
Υ
_
)
q
-
1
γ
2
/
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
=
1, which is a contradiction. This finishes the proof of (ii).Remark 12.
In (i), the conditionlim
⁡
_
x
→
∞
f
(
t
,
x
)
=
f
~
∞
∈
(
0
,
+
∞
] uniformly for t
∈
[
0,1
] can be replaced with condition (H6). The same methods can be used to prove the results.Remark 13.
From the above proof, the conditionf
~
∞
∈
(
0
,
+
∞
] is important to keep the problem (1) having at least two positive solutions, if this condition is not satisfied, then there exists λ
_
>
0 small enough such that the problem (1) has no positive solutions for all λ
<
λ
_.Theorem 14.
Assume that (H1)–(H6) hold.(i)
Iff
ϕ
0
=
+
∞ and f
ϕ
∞
=
+
∞, then there exists λ
*
0
>
0 such that problem (1) has at least two solutions for λ
∈
(
0
,
λ
*
0
).
(ii)
Iff
ϕ
0
=
+
∞ and f
ϕ
∞
=
+
∞, then there exists λ
-
>
0 such that problem (1) has no solution for all λ
>
λ
-.Proof.
(i) It follows fromf
ϕ
0
=
+
∞ that there exists r
1
>
0, such that
(70)
f
(
t
,
x
)
≥
L
*
1
ϕ
p
(
x
)
,
(
∀
t
∈
[
0,1
]
,
0
≤
x
≤
r
1
)
,
where L
*
1 satisfies σ
2
ϕ
q
(
λ
L
*
1
)
γ
1
/
Γ
(
α
)
≥
1.
Then forx
∈
P
∩
∂
Ω
r
1, we have σ
∥
x
∥
≤
x
≤
r
1 for t
∈
[
0,1
] and
(71)
(
Ψ
x
)
(
t
)
=
1
Γ
(
α
)
×
{
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
n
n
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
mmmmm
-
∫
0
t
‍
(
t
-
s
)
α
-
1
m
m
m
m
m
n
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
n
n
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
≥
σ
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
m
n
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmmmm
m
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
n
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
+
σ
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
n
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≥
σ
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
n
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≥
σ
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
n
n
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
L
*
1
ϕ
p
(
x
(
τ
)
)
d
τ
)
d
s
≥
σ
2
ϕ
q
(
λ
L
*
1
)
∥
x
∥
Γ
(
α
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
mmmm
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
d
τ
)
d
s
=
σ
2
ϕ
q
(
λ
L
*
1
)
γ
2
Γ
(
α
)
∥
x
∥
,
which implies that
(72)
∥
Ψ
x
∥
≥
∥
x
∥
,
∀
x
∈
P
∩
∂
Ω
r
1
.
Next, from f
ϕ
∞
=
+
∞, there exists r
-
2
>
0, such that
(73)
f
(
t
,
x
)
≥
L
*
2
ϕ
p
(
x
)
,
∀
x
≥
r
-
2
,
where L
*
2 satisfies
(74)
σ
2
ϕ
q
(
λ
L
*
2
)
γ
2
Γ
(
α
)
≥
1
.
Let r
2
>
max
⁡
{
r
1
,
r
-
2
/
σ
}; then for x
∈
P
∩
∂
Ω
r
2 we have
(75)
x
(
t
)
≥
σ
∥
x
∥
=
σ
r
2
>
r
-
2
.
Similar to the above proof, we can get
(76)
∥
Ψ
x
∥
≥
∥
x
∥
,
∀
x
∈
P
∩
∂
Ω
r
2
.
Let r
1
<
r
3
<
r
2, M
r
3
=
max
⁡
{
f
(
t
,
x
)
∣
t
∈
[
0,1
]
,
∥
x
∥
≤
r
3
}
+
1
>
0, and λ
*
0
=
(
1
/
M
r
3
)
(
Γ
(
α
+
1
)
r
3
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
/
γ
2
)
p
-
1. Then for x
∈
P
∩
∂
Ω
r
3, we have
(77)
f
(
t
,
x
)
≤
M
r
3
,
∀
t
∈
[
0,1
]
.
Hence,
(78)
(
Ψ
x
)
(
t
)
=
1
Γ
(
α
)
{
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
mmmmmmn
-
∫
0
t
‍
(
t
-
s
)
α
-
1
m
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
}
+
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
n
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
i
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmmmnm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
≤
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
g
(
s
)
m
m
m
m
×
∫
0
1
‍
(
1
-
τ
)
α
-
1
ϕ
q
m
m
m
m
m
m
m
×
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
d
s
mm
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≤
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
g
(
s
)
m
m
m
m
×
∫
0
1
‍
(
1
-
τ
)
α
-
1
ϕ
q
m
m
m
m
m
m
n
×
(
λ
∫
0
τ
‍
ω
(
μ
)
M
r
3
d
μ
)
d
τ
d
s
mm
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
M
r
3
d
τ
)
d
s
≤
η
ϕ
q
(
λ
M
r
3
)
∫
0
1
‍
g
(
s
)
d
s
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
τ
)
α
-
1
mmm
×
ϕ
q
(
∫
0
1
‍
ω
(
μ
)
d
μ
)
d
τ
+
ϕ
q
(
λ
M
r
3
)
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
mmmmmmmm
×
ϕ
q
(
∫
0
1
‍
ω
(
τ
)
d
τ
)
d
s
=
γ
2
ϕ
q
(
λ
M
r
3
)
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
<
γ
2
ϕ
q
(
λ
*
0
M
r
3
)
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
=
r
3
.
This implies that
(79)
∥
Ψ
x
∥
<
∥
x
∥
,
∀
x
∈
P
∩
∂
Ω
r
3
.
Applying Lemma 7 to (72), (76), and (79) yields that Ψ has two fixed points x
1
,
x
2 such that x
1
∈
P
∩
(
Ω
-
r
3
∖
Ω
r
1
) and x
2
∈
P
∩
(
Ω
-
r
2
∖
Ω
r
3
).
(ii) By the proof of (i) and the continuity off
(
t
,
x
), there exists x
~
∈
[
r
1
,
r
-
2
] such that min
⁡
x
∈
[
r
1
,
r
-
2
]
(
f
(
t
,
x
)
/
ϕ
p
(
x
)
)
=
f
(
t
,
x
~
)
/
ϕ
p
(
x
~
).
LetΥ
-
=
min
⁡
{
L
*
1
,
L
*
2
,
f
(
t
,
x
~
)
/
ϕ
p
(
x
~
)
}; since (H6) holds, then Υ
-
>
0 and
(80)
f
(
t
,
x
)
≥
Υ
-
ϕ
p
(
x
)
,
∀
x
≥
0
,
t
∈
[
0,1
]
.
Assuming x
(
t
) is a positive solution of problem (1), we will show that this leads to contradiction for all λ
>
λ
-, where λ
-
=
(
1
/
Υ
-
)
(
Γ
(
α
)
/
γ
1
σ
)
p
-
1. It follows form (26) that
(81)
∥
x
∥
=
x
(
0
)
=
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
n
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmmmm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
ϕ
q
m
m
m
m
m
m
m
×
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≥
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≥
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
Υ
-
ϕ
p
(
x
(
τ
)
)
d
τ
)
d
s
≥
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
Υ
-
ϕ
p
(
σ
∥
x
∥
d
τ
)
)
d
s
≥
ϕ
q
(
λ
Υ
-
)
σ
∥
x
∥
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
m
m
m
m
m
×
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
d
τ
)
d
s
=
ϕ
q
(
λ
Υ
-
)
γ
1
σ
Γ
(
α
)
∥
x
∥
.
This implies that 1
≥
ϕ
q
(
λ
Υ
-
)
γ
1
σ
/
Γ
(
α
)
=
(
λ
Υ
-
)
q
-
1
γ
1
σ
/
Γ
(
α
)
>
(
λ
-
Υ
-
)
q
-
1
γ
1
σ
/
Γ
(
α
)
=
1, which is a contradiction. This completes the proof of (ii).Corollary 15.
Assume that (H1)–(H5) holds.(i)
Iflim
⁡
_
x
→
∞
f
(
t
,
x
)
=
f
~
∞
∈
(
0
,
+
∞
] uniformly for t
∈
[
0,1
], and one of f
ϕ
0
=
0 and f
ϕ
∞
=
0 is satisfied, then there exists λ
*
>
0 such that the problem (1) has at least one positive solution for any λ
>
λ
*.
(ii)
If one off
ϕ
0
=
+
∞ and f
ϕ
∞
=
+
∞ holds, then there exists λ
*
0
>
0 such that problem (1) has at least one positive solutions for any λ
∈
(
0
,
λ
*
0
).Proof.
(i) The conclusion is a direct consequence of Theorem11(i).
(ii) The conclusion is a direct consequence of Theorem14(i).Theorem 16.
Assume that (H1)–(H5) hold. Then the following conclusions hold.(i)
Iff
ϕ
0
=
0 and f
ϕ
∞
=
∞, then the problem (1) has a positive solution x
λ
(
t
) for all λ
>
0.
(ii)
Iff
ϕ
0
=
∞ and f
ϕ
∞
=
0, then the problem (1) has a positive solution x
λ
(
t
) for all λ
>
0.Proof.
We only prove (i); the proof of (ii) is similar, so we omit it here.
Letλ
>
0; since f
ϕ
0
=
0, there exists r
>
0
,
l
>
0 such that
(82)
f
(
t
,
x
)
≤
l
ϕ
p
(
x
)
,
(
∀
t
∈
[
0,1
]
,
x
∈
[
0
,
r
]
)
,
where l satisfied
(83)
γ
2
λ
q
-
1
l
q
-
1
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
≤
1
.
Similar to the proof of Theorem 14, we obtain that
(84)
∥
Ψ
x
∥
≤
∥
x
∥
,
∀
x
∈
P
∩
∂
Ω
r
.
From f
ϕ
∞
=
∞, there exists R
>
r
>
0
,
L
>
0 such that
(85)
f
(
t
,
x
)
≥
L
ϕ
p
(
x
)
,
(
∀
t
∈
[
0,1
]
,
x
≥
R
)
,
where L satisfied
(86)
σ
2
λ
q
-
1
L
q
-
1
γ
1
Γ
(
α
)
≥
1
.
We get that
(87)
∥
Ψ
x
∥
≥
∥
x
∥
,
∀
x
∈
P
∩
∂
Ω
R
.
Applying Lemma 7 to (84), (87) yields that Ψ has a fixed point x such that x
∈
P
∩
(
Ω
-
R
∖
Ω
r
).Theorem 17.
Assume that (H1)–(H6) hold. Then the following conclusions hold.
If0
<
f
ϕ
0
<
∞ and 0
<
f
ϕ
∞
<
∞, then there exists λ
^
1
,
λ
^
2
>
0 such that problem (1) has no positive solution for any λ
>
λ
^
1 and 0
<
λ
<
λ
^
2.Proof.
It follows from0
<
f
ϕ
0
<
∞ and 0
<
f
ϕ
∞
<
∞ that there exist L
^
1
>
0
,
L
^
2
>
0
,
l
^
1
>
0
,
l
^
2
>
0
, and r
^
2
>
r
^
1
>
0 such that
(88)
l
^
1
ϕ
p
(
x
)
≤
f
(
t
,
x
)
≤
L
^
1
ϕ
p
(
x
)
,
m
m
n
(
∀
t
∈
[
0,1
]
,
0
≤
x
≤
r
^
1
)
,
l
^
2
ϕ
p
(
x
)
≤
f
(
t
,
x
)
≤
L
^
2
ϕ
p
(
x
)
,
m
m
m
m
m
(
∀
t
∈
[
0,1
]
,
x
≥
r
^
2
)
.
Let M
^
=
max
⁡
{
f
(
t
,
x
)
/
ϕ
p
(
x
)
,
r
^
1
≤
x
≤
r
^
2
,
0
≤
t
≤
1
},m
^
=
min
⁡
{
f
(
t
,
x
)
/
ϕ
p
(
x
)
,
r
^
1
≤
x
≤
r
^
2
,
0
≤
t
≤
1
}. By the condition (H6), m
^
>
0. Then ∀
x
≥
0; we have
(89)
l
^
ϕ
p
(
x
)
≤
f
(
t
,
x
)
≤
L
^
ϕ
p
(
x
)
,
where L
^
=
max
⁡
{
L
^
1
,
L
^
2
,
M
^
}
>
0 and l
^
=
min
⁡
{
l
^
1
,
l
^
2
,
m
^
}
>
0.
Letλ
^
1
=
(
1
/
l
^
)
(
Γ
(
α
)
/
γ
1
σ
)
p
-
1 and λ
^
2
=
(
1
/
L
^
)
(
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
/
γ
2
)
p
-
1. If problem (1) has positive solution x
(
t
) in P, then
(90)
∥
x
∥
=
x
(
0
)
=
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
m
m
m
i
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmmm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
m
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
n
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≥
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≥
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
m
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
l
^
ϕ
p
(
x
(
τ
)
)
d
τ
)
d
s
≥
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
m
m
m
m
n
×
ϕ
q
(
λ
∫
0
s
‍
ω
(
τ
)
l
^
ϕ
p
(
σ
∥
x
∥
d
τ
)
)
d
s
≥
ϕ
q
(
λ
l
^
)
σ
∥
x
∥
Γ
(
α
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
(
∫
0
s
‍
ω
(
τ
)
d
τ
)
d
s
=
ϕ
q
(
λ
l
^
)
γ
1
σ
Γ
(
α
)
∥
x
∥
>
ϕ
q
(
λ
^
1
l
^
)
γ
1
σ
Γ
(
α
)
∥
x
∥
=
∥
x
∥
;
∥
x
∥
=
x
(
0
)
=
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
{
∫
0
1
‍
g
(
s
)
m
m
m
m
i
×
(
∫
0
1
‍
(
1
-
τ
)
α
-
1
ϕ
q
m
m
m
m
m
i
m
n
×
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
mmmmmm
-
∫
0
s
‍
(
s
-
τ
)
α
-
1
ϕ
q
m
m
m
m
m
m
m
m
×
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
)
d
s
}
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
m
m
m
m
m
m
n
×
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≤
η
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
g
(
s
)
m
m
m
×
∫
0
1
‍
(
1
-
τ
)
α
-
1
m
m
m
n
m
i
×
ϕ
q
(
λ
∫
0
τ
‍
ω
(
μ
)
f
(
μ
,
x
(
μ
)
)
d
μ
)
d
τ
d
s
+
1
Γ
(
α
)
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
m
m
m
m
m
m
×
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
=
1
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
m
m
m
×
(
λ
∫
0
s
‍
ω
(
τ
)
f
(
τ
,
x
(
τ
)
)
d
τ
)
d
s
≤
1
Γ
(
α
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
×
∫
0
1
‍
(
1
-
s
)
α
-
1
ϕ
q
m
m
m
×
(
λ
∫
0
1
‍
ω
(
τ
)
L
^
ϕ
p
(
x
(
τ
)
)
d
τ
)
d
s
=
ϕ
q
(
λ
L
^
)
γ
2
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
∥
x
∥
<
ϕ
q
(
λ
^
2
L
^
)
γ
2
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
∥
x
∥
=
∥
x
∥
,
which is a contradiction. This finishes the proof of Theorem 17.


## 4. Examples

In this section, we give out same examples to illustrate our main results can be used in practice.Example 1.
Consider the following fractional differential equation:(91)
(
ϕ
p
(
D
0
+
5
/
2
x
(
t
)
)
)
′
+
λ
t
(
1
+
t
2
)
1
/
2
x
p
-
1
arctan
x
=
0
,
mmmmmmmmmmmmmmmmmmmm
0
<
t
<
1
,
x
′
(
0
)
=
x
′′
(
0
)
=
0
,
x
(
1
)
=
1
2
∫
0
1
‍
1
2
x
(
s
)
d
s
,
where α
=
5
/
2
,
p
=
3
/
2
,
q
=
3
,
ω
(
t
)
=
t
,
f
(
t
,
x
)
=
(
1
+
t
2
)
1
/
2
x
p
-
1
arctan
x
,
η
=
1
/
2, and g
(
t
)
=
1
/
2.By simple computation,l
1
*
=
π
/
2
,
l
2
*
=
2
π
/
2
,
γ
1
=
∫
0
1
‍
(
1
-
s
)
3
/
2
ϕ
3
(
∫
0
s
‍
τ
d
τ
)
d
s
≈
0.0043
,
γ
2
=
ϕ
3
(
∫
0
1
‍
τ
d
τ
)
=
0.25
,
σ
≈
0.1429
,
λ
0
*
=
(
1
/
l
1
*
)
(
Γ
(
α
)
/
σ
γ
1
)
p
-
1
≈
2.5149
,
λ
-
0
*
=
(
1
/
l
2
*
)
(
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
g
(
s
)
d
s
)
‍
/
γ
2
)
p
-
1
≈
1.4222. Then, it is easy to see all the conditions of Theorem 9(ii) are satisfied; thus ∀
λ
∈
[
1.4222
,
2.5149
]
]; the problem (91) has at least one positive solution.Example 2.
Consider the boundary value problem(92)
(
ϕ
p
(
D
0
+
5
/
2
x
(
t
)
)
)
′
+
λ
t
(
1
+
t
2
)
1
/
2
x
p
e
-
x
=
0
,
mmmmmmmmmmmmmmmn
0
<
t
<
1
,
x
′
(
0
)
=
x
′′
(
0
)
=
0
,
x
(
1
)
=
1
2
∫
0
1
‍
1
2
x
(
s
)
d
s
,
where α
=
5
/
2
,
p
=
3
/
2
,
q
=
3
,
ω
(
t
)
=
t
,
f
(
t
,
x
)
=
(
1
+
t
2
)
1
/
2
x
p
e
-
x
,
η
=
1
/
2, and g
(
t
)
=
1
/
2:
(93)
f
ϕ
0
=
lim
⁡
x
→
0
f
(
t
,
x
)
ϕ
p
(
x
)
=
lim
⁡
x
→
0
(
1
+
t
2
)
1
/
2
x
p
e
-
x
x
p
-
1
=
0
,
mmmmmmmmmmmmmmmmn
∀
t
∈
[
0,1
]
,
f
ϕ
0
=
lim
⁡
x
→
∞
f
(
t
,
x
)
ϕ
p
(
x
)
=
lim
⁡
x
→
∞
(
1
+
t
2
)
1
/
2
x
p
e
-
x
x
p
-
1
=
0
,
mmmmmmmmmmmmmmmmmn
∀
t
∈
[
0,1
]
.
Let ρ
-
1
=
1
/
2
,
ρ
-
2
=
4; then
(94)
f
(
t
,
x
)
ϕ
p
(
x
)
≤
(
1
+
t
2
)
1
/
2
1
2
e
≤
1
2
e
,
mmmmn
∀
t
∈
[
0,1
]
,
x
∈
[
0
,
1
2
]
,
f
(
t
,
x
)
ϕ
p
(
x
)
≤
(
1
+
t
2
)
1
/
2
4
e
4
≤
4
2
e
4
,
mmmmi
∀
t
∈
[
0,1
]
,
x
∈
[
4
,
∞
)
,
so ε
1
=
1
/
2
e
,
ε
2
=
4
2
/
e
4, and ∀
x
∈
[
1
/
2
,
4
]
,
max
⁡
1
/
2
≤
x
≤
4
(
f
(
t
,
x
)
/
ϕ
p
(
x
)
)
=
2
e
-
1. We haveΥ
_
=
max
⁡
{
1
/
2
e
,
4
2
/
e
4
,
2
e
-
1
}
=
2
e
-
1,γ
2
=
0.25, and λ
_
=
(
1
/
Υ
_
)
(
Γ
(
α
+
1
)
(
1
-
η
∫
0
1
‍
g
(
s
)
d
s
)
/
γ
2
)
p
-
1
=
3
5
π
1
/
4
e
/
4. By Theorem 11(ii), for any λ
<
λ
_
=
3
5
π
1
/
4
e
/
4, the problem has no positive solution.


## 5. Conclusions

Recently, differential equations withp-Laplacian operator were widely discussed by several authors. In this paper, by using fixed point techniques combining with partially ordered structure of Banach space, we obtained the existence, multiply, and the dependence of parameter. These new results we presented can be used in numerical computation and analyze mathematical models of physical phenomena, mechanics, nonlinear dynamics, and many other related fields.



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*Source: 512426-2014-05-15.xml*