# Boundary Value Problems for Fractional Differential Equations with Fractional Multiterm Integral Conditions

Authors: Jessada Tariboon; Sotiris K. Ntouyas; Arisa Singubol
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We discuss the existence and uniqueness of solutions for boundary value problems involving multiterm fractional integral boundary conditions.
Our study relies on standard fixed point theorems. Illustrative examples are also presented.


---

## Body

## 1. Introduction

In this paper, we study the existence and uniqueness of solutions for the following fractional differential equation:(1)
D
c
q
x
(
t
)
=
f
(
t
,
x
(
t
)
)
,
1
<
q
≤
2
,
0
<
t
<
T
,
subject to nonlocal fractional integral boundary conditions
(2)
∑
i
=
1
m
‍
λ
i
I
α
i
x
(
η
i
)
=
ω
1
,
∑
j
=
1
n
‍
μ
j
(
I
β
j
x
(
T
)
-
I
β
j
x
(
ξ
j
)
)
=
ω
2
,
where D
c
q denotes the Caputo fractional derivative of order q, f
:
[
0
,
T
]
×
ℝ
→
ℝ is a continuous function, η
i
,
ξ
j
∈
(
0
,
T
), λ
i
,
μ
j
∈
ℝ, for all i
=
1,2
,
…
,
m
,
j
=
1,2
,
…
,
n
,
ω
1
,
ω
2
∈
ℝ, and I
ϕ is the Riemann-Liouville fractional integral of order ϕ
>
0
(
ϕ
=
α
i
,
β
j
,
i
=
1,2
,
…
,
m
,
j
=
1,2
,
…
,
n
).The significance of studying problem (1)-(2) is that condition (2) is very general and includes many conditions as special cases. In particular, if α
i
=
β
j
=
1, for all i
=
1,2
,
…
,
m, j
=
1,2
,
…
,
n, then condition (2) reduces to
(3)
λ
1
∫
0
η
1
‍
x
(
s
)
d
s
+
λ
2
∫
0
η
2
‍
x
(
s
)
d
s
+
⋯
+
λ
m
∫
0
η
m
‍
x
(
s
)
d
s
=
ω
1
,
μ
1
∫
ξ
1
T
‍
x
(
s
)
d
s
+
μ
2
∫
ξ
2
T
‍
x
(
s
)
d
s
+
⋯
+
μ
n
∫
ξ
n
T
‍
x
(
s
)
d
s
=
ω
2
.
Note that condition (3) does not contain values of unknown function x at the left-side and right-side of boundary points t
=
0 and t
=
T, respectively.In recent years, the boundary value problems of fractional order differential equations have emerged as an important area of research, since these problems have applications in various disciplines of science and engineering such as mechanics, electricity, chemistry, biology, economics, control theory, signal and image processing, polymer rheology, regular variation in thermodynamics, biophysics, aerodynamics, viscoelasticity and damping, electrodynamics of complex medium, wave propagation, and blood flow phenomena (see [1–5]). Many researchers have studied the existence theory for nonlinear fractional differential equations with a variety of boundary conditions; for instance, see the papers [6–16] and the references therein.The main objective of this paper is to develop some existence and uniqueness results for the boundary value problem (1)-(2) by using standard fixed point theorems. The paper is organized as follows. In Section 2, we recall some preliminary facts that we need in the sequel, and Section 3 contains our main results. Finally, illustrative examples are presented in Section 4.


## 2. Preliminaries

In this section, we introduce some notations and definitions of fractional calculus [2, 3] and present preliminary results needed in our proofs later.Definition 1.
For an at leastn-times differentiable function g
:
[
0
,
∞
)
→
ℝ, the Caputo derivative of fractional order q is defined as
(4)
D
c
q
g
(
t
)
=
1
Γ
(
n
-
q
)
∫
0
t
‍
(
t
-
s
)
n
-
q
-
1
g
(
n
)
(
s
)
d
s
,
h
h
h
h
h
h
n
-
1
<
q
<
n
,
n
=
[
q
]
+
1
,
where [
q
] denotes the integer part of the real number q.Definition 2.
The Riemann-Liouville fractional integral of orderq is defined as
(5)
I
q
g
(
t
)
=
1
Γ
(
q
)
∫
0
t
‍
g
(
s
)
(
t
-
s
)
1
-
q
d
s
,
q
>
0
,
provided the integral exists.Lemma 3.
Forq
>
0, the general solution of the fractional differential equation c
D
q
x
(
t
)
=
0 is given by
(6)
x
(
t
)
=
c
0
+
c
1
t
+
⋯
+
c
n
-
1
t
n
-
1
,
where c
i
∈
ℝ, i
=
1,2
,
…
,
n
-
1
(
n
=
[
q
]
+
1
).In view of Lemma3, it follows that
(7)
I
q
D
c
q
x
(
t
)
=
x
(
t
)
+
c
0
+
c
1
t
+
⋯
+
c
n
-
1
t
n
-
1
,
for some c
i
∈
ℝ, i
=
1,2
,
…
,
n
-
1
(
n
=
[
q
]
+
1
).For convenience, we set(8)
Ω
1
=
∑
i
=
1
m
‍
λ
i
η
i
α
i
+
1
Γ
(
α
i
+
2
)
,
Ω
2
=
∑
i
=
1
m
‍
λ
i
η
i
α
i
Γ
(
α
i
+
1
)
,
Ω
3
=
∑
j
=
1
n
‍
μ
j
T
β
j
+
1
-
ξ
j
β
j
+
1
Γ
(
β
j
+
2
)
,
Ω
4
=
∑
j
=
1
n
‍
μ
j
T
β
j
-
ξ
j
β
j
Γ
(
β
j
+
1
)
,
Δ
=
Ω
1
Ω
4
-
Ω
2
Ω
3
.Lemma 4.
LetΔ
≠
0, 1
<
q
≤
2, α
i
,
β
j
>
0, η
i
,
ξ
j
∈
(
0
,
T
) for i
=
1,2
,
…
,
m, j
=
1,2
,
…
,
n, and h
∈
C
(
[
0
,
T
]
,
ℝ
). Then, the problem
(9)
D
c
q
x
(
t
)
=
h
(
t
)
,
t
∈
(
0
,
T
)
(10)
∑
i
=
1
m
‍
λ
i
I
α
i
x
(
η
i
)
=
ω
1
,
∑
j
=
1
n
‍
μ
j
(
I
β
j
x
(
T
)
-
I
β
j
x
(
ξ
j
)
)
=
ω
2
,
has a unique solution given by
(11)
x
(
t
)
=
I
q
h
(
t
)
+
Ω
4
t
-
Ω
3
Δ
(
ω
1
-
∑
i
=
1
m
‍
λ
i
I
α
i
+
q
h
(
η
i
)
)
+
Ω
1
-
Ω
2
t
Δ
(
ω
2
-
∑
j
=
1
n
‍
μ
j
(
I
β
j
+
q
h
(
T
)
-
I
β
j
+
q
h
(
ξ
j
)
)
)
.Proof.
Using Lemma3, (9) can be expressed as an equivalent integral equation:
(12)
x
(
t
)
=
I
q
h
(
t
)
+
c
1
t
+
c
2
.
Taking the Riemann-Liouville fractional integral of order p
>
0 for (12), we have
(13)
I
p
x
(
t
)
=
I
p
+
q
h
(
t
)
+
c
1
t
p
+
1
Γ
(
p
+
2
)
+
c
2
t
p
Γ
(
p
+
1
)
.
From the first condition of (10) and the equation (13) with p
=
α
i, it follows that
(14)
c
1
∑
i
=
1
m
‍
λ
i
η
i
α
i
+
1
Γ
(
α
i
+
2
)
+
c
2
∑
i
=
1
m
‍
λ
i
η
i
α
i
Γ
(
α
i
+
1
)
=
ω
1
-
∑
i
=
1
m
‍
λ
i
I
α
i
+
q
h
(
η
i
)
.
According to the above process, the second conditions of (10) and (13) with p
=
β
j imply that
(15)
c
1
∑
j
=
1
n
‍
μ
j
T
β
j
+
1
-
ξ
j
β
j
+
1
Γ
(
β
j
+
2
)
+
c
2
∑
j
=
1
n
‍
μ
j
T
β
j
-
ξ
j
β
j
Γ
(
β
j
+
1
)
=
ω
2
-
∑
j
=
1
n
‍
μ
j
(
I
β
j
+
q
h
(
T
)
-
I
β
j
+
q
h
(
ξ
j
)
)
.
Solving the system of linear equations for constants c
1 and c
2, we have
(16)
c
1
=
Ω
4
Δ
(
ω
1
-
∑
i
=
1
m
‍
λ
i
I
α
i
+
q
h
(
η
i
)
)
-
Ω
2
Δ
(
ω
2
-
∑
j
=
1
n
‍
μ
j
(
I
β
j
+
q
h
(
T
)
-
I
β
j
+
q
h
(
ξ
j
)
)
)
,
c
2
=
Ω
1
Δ
(
ω
2
-
∑
j
=
1
n
‍
μ
j
(
I
β
j
+
q
h
(
T
)
-
I
β
j
+
q
h
(
ξ
j
)
)
)
-
Ω
3
Δ
(
ω
1
-
∑
i
=
1
m
‍
λ
i
I
α
i
+
q
h
(
η
i
)
)
.
Substituting constants c
1 and c
2 into (12), we obtain (11) as required.


## 3. Main Results

Let𝒞
=
C
(
[
0
,
T
]
,
ℝ
) denote the Banach space of all continuous functions from [
0
,
T
] to ℝ endowed with the norm defined by ∥
x
∥
=
sup
⁡
t
∈
[
0
,
T
]
|
x
(
t
)
|. As in Lemma 4, we define an operator 𝒬
:
𝒞
→
𝒞 by
(17)
𝒬
x
(
t
)
=
I
q
f
(
s
,
x
(
s
)
)
(
t
)
+
Ω
4
t
-
Ω
3
Δ
(
ω
1
-
∑
i
=
1
m
‍
λ
i
I
α
i
+
q
f
(
s
,
x
(
s
)
)
(
η
i
)
)
+
Ω
1
-
Ω
2
t
Δ
×
(
ω
2
-
∑
j
=
1
n
‍
μ
j
(
-
I
β
j
+
q
f
(
s
,
x
(
s
)
)
(
ξ
j
)
I
β
j
+
q
f
(
s
,
x
(
s
)
)
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
-
I
β
j
+
q
f
(
s
,
x
(
s
)
)
(
ξ
j
)
)
ω
2
-
∑
j
=
1
n
‍
μ
j
(
I
β
j
+
q
f
(
s
,
x
(
s
)
)
(
T
)
)
.
It should be noticed that problem (1)-(2) has solutions if and only if the operator 𝒬 has fixed points.We are in a position to establish our main results. In the following subsections, we prove existence as well as existence and uniqueness results for the BVP (1)-(2) by using a variety of fixed point theorems.


### 3.1. Existence and Uniqueness Result via Banach’s Fixed Point Theorem

Theorem 5.
Assume that(
H
1
)
there exists a constantL
>
0 such that |
f
(
t
,
u
)
-
f
(
t
,
v
)
|
≤
L
|
u
-
v
|, for each t
∈
[
0
,
T
] and u
,
v
∈
ℝ.If
(18)
L
Λ
<
1
,
where Λ is defined by
(19)
Λ
=
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
,
then problem (1)-(2) has a unique solution in [
0
,
T
].Proof.
We transform problem (1)-(2) into a fixed point problem, x
=
𝒬
x, where the operator 𝒬 is defined as in (17). Observe that the fixed points of the operator 𝒬 are solutions of problem (1)-(2). Applying the Banach contraction mapping principle, we will show that 𝒬 has a unique fixed point.
We letsup
⁡
t
∈
[
0
,
T
]
|
f
(
t
,
0
)
|
=
M
<
∞ and choose
(20)
r
≥
Λ
M
+
Φ
1
-
L
Λ
,
where a constant Φ is defined by
(21)
Φ
=
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
.
Now, we show that 𝒬
B
r
⊂
B
r, where B
r
=
{
x
∈
𝒞
:
∥
x
∥
≤
r
}. For any x
∈
B
r, we have
(22)
|
𝒬
x
(
t
)
|
≤
sup
⁡
t
∈
[
0
,
T
]
{
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
t
|
Ω
4
|
|
Δ
|
|
ω
1
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
t
|
Ω
2
|
|
Δ
|
|
ω
2
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
3
|
+
t
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
t
|
Ω
2
|
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
t
|
Ω
4
|
|
Δ
|
|
ω
1
|
}
≤
I
q
(
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
0
)
|
+
|
f
(
s
,
0
)
|
)
(
T
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
×
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
(
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
0
)
|
+
|
f
(
s
,
0
)
|
)
(
η
i
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
(
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
0
)
|
+
|
f
(
s
,
0
)
|
)
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
(
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
0
)
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
f
(
s
,
0
)
|
)
(
ξ
j
)
)
≤
(
L
r
+
M
)
I
q
(
1
)
(
T
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
(
L
r
+
M
)
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
(
1
)
(
η
i
)
+
(
L
r
+
M
)
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
I
β
j
+
q
(
1
)
(
T
)
+
I
β
j
+
q
(
1
)
(
ξ
j
)
)
≤
(
L
r
+
M
)
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
(
L
r
+
M
)
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
(
L
r
+
M
)
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
=
(
L
r
+
M
)
Λ
+
Φ
≤
r
,
which implies that 𝒬
B
r
⊂
B
r.
Next, we letx
,
y
∈
𝒞. Then, for t
∈
[
0
,
T
], we have
(23)
|
𝒬
x
(
t
)
-
𝒬
y
(
t
)
|
≤
I
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
×
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
η
i
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
ξ
j
)
)
≤
I
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
T
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
×
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
η
i
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
ξ
j
)
)
≤
T
q
L
∥
x
-
y
∥
Γ
(
q
+
1
)
+
(
|
Ω
3
|
+
T
|
Ω
4
|
)
L
∥
x
-
y
∥
|
Δ
|
×
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
(
|
Ω
1
|
+
T
|
Ω
2
|
)
L
∥
x
-
y
∥
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
=
L
Λ
∥
x
-
y
∥
,
which implies that ∥
𝒬
x
-
𝒬
y
∥
≤
L
Λ
∥
x
-
y
∥. As L
Λ
<
1, 𝒬 is a contraction. Therefore, we deduce by Banach's contraction mapping principle that 𝒬 has a fixed point which is the unique solution of problem (1)-(2). The proof is completed.


### 3.2. Existence Result via Krasnoselskii’s Fixed Point Theorem

Lemma 6 (Krasnoselskii’s fixed point theorem, [17]).
LetM be a closed, bounded, convex, and nonempty subset of a Banach space X. Let A and B be the operators such that (a) A
x
+
B
y
∈
M whenever x
,
y
∈
M; (b) A is compact and continuous; and (c) B is a contraction mapping. Then, there exists z
∈
M such that z
=
A
z
+
B
z.Theorem 7.
Letf
:
[
0
,
T
]
×
ℝ
→
ℝ be a continuous function satisfying (
H
1
). Assume that(
H
2
)
|
f
(
t
,
u
)
|
≤
θ
(
t
),for all
(
t
,
u
)
∈
[
0
,
T
]
×
ℝ, and θ
∈
C
(
[
0
,
T
]
,
ℝ
+
).Then, the boundary value problem (1)-(2) has at least one solution on [
0
,
T
] provided
(24)
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
<
1
.Proof.
Settingsup
⁡
t
∈
[
0
,
T
]
|
θ
(
t
)
|
=
∥
θ
∥ and choosing
(25)
ρ
≥
∥
θ
∥
Λ
+
Φ
,
(Λ and Φ are defined in (19) and (21), resp.), we consider B
ρ
=
{
x
∈
𝒞
(
[
0
,
T
]
,
ℝ
)
:
∥
x
∥
≤
ρ
}. We define the operators 𝒬
1 and 𝒬
2 on B
ρ by
(26)
𝒬
1
x
(
t
)
=
I
q
f
(
s
,
x
(
s
)
)
(
t
)
,
𝒬
2
x
(
t
)
=
Ω
3
-
Ω
4
t
Δ
ω
1
+
Ω
2
t
-
Ω
1
Δ
ω
2
+
Ω
3
-
Ω
4
t
Δ
∑
i
=
1
m
‍
λ
i
I
α
i
+
q
f
(
s
,
x
(
s
)
)
(
η
i
)
+
Ω
2
t
-
Ω
1
Δ
∑
j
=
1
n
‍
μ
j
(
(
ξ
j
)
I
β
j
+
q
f
(
s
,
x
(
s
)
)
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
-
I
β
j
+
q
f
(
s
,
x
(
s
)
)
(
ξ
j
)
)
.
For any x
,
y
∈
B
ρ, we have
(27)
|
𝒬
1
x
(
t
)
+
𝒬
2
y
(
t
)
|
≤
sup
⁡
t
∈
[
0
,
T
]
{
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
i
i
i
i
i
i
i
i
i
i
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
y
(
s
)
)
|
(
η
i
)
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
y
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
y
(
s
)
)
|
(
ξ
j
)
)
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
}
≤
∥
θ
∥
[
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
]
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
=
∥
θ
∥
Λ
+
Φ
≤
ρ
.
This shows that 𝒬
1
x
+
𝒬
2
y
∈
B
ρ. It is easy to see using (24) that 𝒬
2 is a contraction mapping.
Continuity off implies that the operator 𝒬
1 is continuous. Also, 𝒬
1 is uniformly bounded on B
ρ as
(28)
∥
𝒬
1
x
∥
≤
T
q
Γ
(
q
+
1
)
∥
θ
∥
.
Now, we prove the compactness of the operator 𝒬
1.
We definesup
⁡
(
t
,
x
)
∈
[
0
,
T
]
×
B
ρ
|
f
(
t
,
x
)
|
=
f
¯
<
∞, and consequently we have
(29)
|
(
𝒬
1
x
)
(
t
2
)
-
(
𝒬
1
x
)
(
t
1
)
|
=
1
Γ
(
q
)
|
∫
0
t
1
‍
[
(
t
2
-
s
)
q
-
1
-
(
t
1
-
s
)
q
-
1
]
f
(
s
,
x
(
s
)
)
d
s
i
i
i
i
i
i
i
i
i
i
i
i
+
∫
t
1
t
2
‍
(
t
2
-
s
)
q
-
1
f
(
s
,
x
(
s
)
)
d
s
∫
0
t
1
‍
|
≤
f
¯
Γ
(
q
+
1
)
|
t
1
q
-
t
2
q
|
,
which is independent of x and tends to zero as t
2
-
t
2
→
0. Thus, 𝒬
1 is equicontinuous. So 𝒬
1 is relatively compact on B
ρ. Hence, by the Arzelá-Ascoli theorem, 𝒬
1 is compact on B
ρ. Thus, all the assumptions of Lemma 6 are satisfied. So the conclusion of Lemma 6 implies that the boundary value problem (1)-(2) has at least one solution on [
0
,
T
].


### 3.3. Existence Result via Leray-Schauder’s Nonlinear Alternative

Theorem 8 (nonlinear alternative for single valued maps, [18]).
LetE be a Banach space, C a closed, convex subset of E, U an open subset of C, and 0
∈
U. Suppose that F
:
U
¯
→
C is a continuous, compact (i.e., F
(
U
¯
) is a relatively compact subset of C) map. Then, either(i)
F has a fixed point in U
¯, or
(ii)
there isu
∈
∂
U (the boundary of U in C) and λ
∈
(
0,1
) with u
=
λ
F
(
u
).Theorem 9.
Assume that(
H
3
)
there exist a continuous nondecreasing functionψ
:
[
0
,
∞
)
→
(
0
,
∞
) and a function p
∈
C
(
[
0
,
T
]
,
ℝ
+
) such that
(30)
|
f
(
t
,
x
)
|
≤
p
(
t
)
ψ
(
∥
x
∥
)
f
o
r
e
a
c
h
(
t
,
x
)
∈
[
0
,
T
]
×
ℝ
;
(
H
4
)
there exists a constantM
>
0 such that
(31)
M
ψ
(
M
)
∥
p
∥
Λ
+
Φ
>
1
,
whereΛ and Φ are defined in (19) and (21), respectively.Then, the boundary value problem (1)-(2) has at least one solution on [
0
,
T
].Proof.
Let the operator𝒬 be defined by (17). Firstly, we will show that 𝒬
maps bounded sets (balls) into bounded sets in
C
(
[
0
,
T
]
,
ℝ
). For a number r
>
0, let B
r
=
{
x
∈
C
(
[
0
,
T
]
,
ℝ
)
:
∥
x
∥
≤
r
} be a bounded ball in C
(
[
0
,
T
]
,
ℝ
). Then, for t
∈
[
0
,
T
] we have
(32)
|
𝒬
x
(
t
)
|
≤
sup
⁡
t
∈
[
0
,
T
]
{
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
i
i
i
i
i
i
i
i
i
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
}
≤
ψ
(
∥
x
∥
)
I
q
p
(
s
)
(
T
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
ψ
(
∥
x
∥
)
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
p
(
s
)
(
η
i
)
+
ψ
(
∥
x
∥
)
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
I
β
j
+
q
p
(
s
)
(
T
)
+
I
β
j
+
q
p
(
s
)
(
ξ
j
)
)
≤
ψ
(
∥
x
∥
)
∥
p
∥
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
ψ
(
∥
x
∥
)
∥
p
∥
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
ψ
(
∥
x
∥
)
∥
p
∥
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
-
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
=
ψ
(
∥
x
∥
)
∥
p
∥
{
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
}
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
,
and, consequently,
(33)
∥
𝒬
x
∥
≤
ψ
(
r
)
∥
p
∥
Λ
+
Φ
,
where Λ and Φ are defined by (19) and (21), respectively.
Next, we will show that𝒬
maps bounded sets into equicontinuous sets of
C
(
[
0
,
T
]
,
ℝ
). Let τ
1
,
τ
2
∈
[
0
,
T
] with τ
1
<
τ
2 and x
∈
B
r. Then, we have
(34)
|
(
𝒬
x
)
(
τ
2
)
-
(
𝒬
x
)
(
τ
1
)
|
≤
1
Γ
(
q
)
|
∫
0
τ
1
‍
[
(
τ
2
-
s
)
q
-
1
-
(
τ
1
-
s
)
q
-
1
]
f
(
s
,
x
(
s
)
)
d
s
iiiiiiiii
+
∫
τ
1
τ
2
‍
(
τ
2
-
s
)
q
-
1
f
(
s
,
x
(
s
)
)
d
s
∫
0
τ
1
‍
|
+
|
ω
1
Ω
4
|
+
|
ω
2
Ω
2
|
|
Δ
|
|
τ
2
-
τ
1
|
+
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
(
τ
2
-
τ
1
)
+
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
(
τ
2
-
τ
1
)
≤
ψ
(
r
)
Γ
(
q
)
|
∫
0
τ
1
‍
[
(
τ
2
-
s
)
q
-
1
-
(
τ
1
-
s
)
q
-
1
]
p
(
s
)
d
s
iiiiiiiii
+
∫
τ
1
τ
2
‍
(
τ
2
-
s
)
q
-
1
p
(
s
)
d
s
∫
0
τ
1
‍
|
+
|
ω
1
Ω
4
|
+
|
ω
2
Ω
2
|
|
Δ
|
(
τ
2
-
τ
1
)
+
|
Ω
4
|
|
Δ
|
ψ
(
r
)
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
p
(
s
)
(
η
i
)
(
τ
2
-
τ
1
)
+
|
Ω
2
|
|
Δ
|
ψ
(
r
)
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
p
(
s
)
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
p
(
s
)
(
ξ
j
)
)
(
τ
2
-
τ
1
)
.
As τ
2
-
τ
1
→
0, the right-hand side of the above inequality tends to zero independently of x
∈
B
r. Therefore, by the Arzelá-Ascoli theorem, the operator 𝒬
:
C
(
[
0
,
T
]
,
ℝ
)
→
C
(
[
0
,
T
]
,
ℝ
) is completely continuous.
Letx be a solution. Then, for t
∈
[
0
,
T
], following the similar computations as in the first step, we have
(35)
|
x
(
t
)
|
≤
ψ
(
∥
x
∥
)
∥
p
∥
Λ
+
Φ
,
which leads to
(36)
∥
x
∥
ψ
(
∥
x
∥
)
∥
p
∥
Λ
+
Φ
≤
1
.
In view of (
H
4
), there exists M such that ∥
x
∥
≠
M. Let us set
(37)
U
=
{
x
∈
C
(
[
0
,
T
]
,
ℝ
)
:
∥
x
∥
<
M
}
.
We see that the operator 𝒬
:
U
¯
→
C
(
[
0
,
T
]
,
ℝ
) is continuous and completely continuous. From the choice of U, there is no x
∈
∂
U such that x
=
ν
𝒬
x for some ν
∈
(
0,1
). Consequently, by the nonlinear alternative of Leray-Schauder type, we deduce that 𝒬 has a fixed point x
∈
U
¯ which is a solution of problem (1)-(2). This completes the proof.


### 3.4. Existence Result via Leray-Schauder Degree

Theorem 10.
Letf
:
[
0
,
T
]
×
ℝ
→
ℝ be a continuous function. Assume that(
H
5
)
there exist constants0
≤
κ
<
Λ
-
1, where Λ is given by (19) and N
>
0 such that
(38)
|
f
(
t
,
x
)
|
≤
κ
|
x
|
+
N
f
o
r
e
a
c
h
(
t
,
x
)
∈
[
0
,
T
]
×
ℝ
.Then, the boundary value problem (1)-(2) has at least one solution on [
0
,
T
].Proof.
Let us define an operator𝒬
:
𝒞
→
𝒞 as in (17) and consider the fixed point problem
(39)
x
=
𝒬
x
.
In view of the fixed point problem (39), we are going to prove the existence of at least one solution x
∈
𝒞 satisfying (39). Define a ball B
R
⊂
𝒞 with a constant radius R
>
0 given by
(40)
B
R
=
{
x
∈
𝒞
:
max
⁡
t
∈
[
0,1
]
|
x
(
t
)
|
<
R
,
R
>
0
}
.
Hence, it is sufficient to show that 𝒬
:
B
¯
R
→
𝒞 satisfies
(41)
x
≠
λ
𝒬
x
,
∀
x
∈
∂
B
R
,
∀
λ
∈
[
0,1
]
.
We define
(42)
H
(
λ
,
x
)
=
λ
𝒬
x
,
x
∈
C
,
λ
∈
[
0,1
]
.
As shown in Theorem 9, we deduce that the operator 𝒬 is continuous, uniformly bounded, and equicontinuous. Then, by the Arzelá-Ascoli theorem, a continuous map h
λ defined by h
λ
(
x
)
=
x
-
H
(
λ
,
x
)
=
x
-
λ
𝒬
x is completely continuous. If (41) holds, then the following Leray-Schauder degree is well defined, and by the homotopy invariance of topological degree, it follows that
(43)
deg
⁡
(
h
λ
,
B
R
,
0
)
=
deg
⁡
(
I
-
λ
𝒬
,
B
R
,
0
)
=
deg
⁡
(
h
1
,
B
R
,
0
)
=
deg
⁡
(
h
0
,
B
R
,
0
)
=
deg
⁡
(
I
,
B
R
,
0
)
=
1
≠
0
,
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
0
∈
B
R
,
where I denotes the unit operator. By the nonzero property of Leray-Schauder degree, h
1
(
x
)
=
x
-
λ
𝒬
x
=
0 for at least one x
∈
B
R. In order to prove (41), we assume that x
=
λ
𝒬
x for some λ
∈
[
0,1
] and for all t
∈
[
0
,
T
]. Then,
(44)
|
x
(
t
)
|
=
|
λ
(
𝒬
x
)
(
t
)
|
≤
sup
⁡
t
∈
[
0
,
T
]
{
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
|
Ω
4
|
t
|
Δ
|
|
ω
1
|
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
|
Ω
2
|
t
|
Δ
|
|
ω
2
|
+
|
Ω
3
|
+
|
Ω
4
|
t
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
×
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
|
Ω
2
|
t
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
|
Ω
4
|
t
|
Δ
|
|
ω
1
|
}
≤
I
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
|
ω
2
|
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
≤
(
κ
|
x
|
+
N
)
I
q
(
1
)
(
T
)
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
|
ω
2
|
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
(
κ
|
x
|
+
N
)
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
(
1
)
(
η
i
)
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
(
κ
|
x
|
+
N
)
×
∑
j
=
1
n
‍
|
μ
j
|
(
I
β
j
+
q
(
1
)
(
T
)
+
I
β
j
+
q
(
1
)
(
ξ
j
)
)
≤
(
κ
|
x
|
+
N
)
[
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
)
Γ
(
β
j
+
q
+
1
)
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
)
Γ
(
β
j
+
q
+
1
)
]
+
Φ
=
(
κ
|
x
|
+
N
)
Λ
+
Φ
.
Computing directly for ∥
x
∥
=
sup
⁡
t
∈
[
0
,
T
]
|
x
(
t
)
|, we have
(45)
∥
x
∥
≤
N
Λ
+
Φ
1
-
κ
Λ
.
Let R
=
(
N
Λ
+
Φ
)
/
(
1
-
κ
Λ
)
+
1; then (41) holds. This completes the proof.


## 3.1. Existence and Uniqueness Result via Banach’s Fixed Point Theorem

Theorem 5.
Assume that(
H
1
)
there exists a constantL
>
0 such that |
f
(
t
,
u
)
-
f
(
t
,
v
)
|
≤
L
|
u
-
v
|, for each t
∈
[
0
,
T
] and u
,
v
∈
ℝ.If
(18)
L
Λ
<
1
,
where Λ is defined by
(19)
Λ
=
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
,
then problem (1)-(2) has a unique solution in [
0
,
T
].Proof.
We transform problem (1)-(2) into a fixed point problem, x
=
𝒬
x, where the operator 𝒬 is defined as in (17). Observe that the fixed points of the operator 𝒬 are solutions of problem (1)-(2). Applying the Banach contraction mapping principle, we will show that 𝒬 has a unique fixed point.
We letsup
⁡
t
∈
[
0
,
T
]
|
f
(
t
,
0
)
|
=
M
<
∞ and choose
(20)
r
≥
Λ
M
+
Φ
1
-
L
Λ
,
where a constant Φ is defined by
(21)
Φ
=
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
.
Now, we show that 𝒬
B
r
⊂
B
r, where B
r
=
{
x
∈
𝒞
:
∥
x
∥
≤
r
}. For any x
∈
B
r, we have
(22)
|
𝒬
x
(
t
)
|
≤
sup
⁡
t
∈
[
0
,
T
]
{
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
t
|
Ω
4
|
|
Δ
|
|
ω
1
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
t
|
Ω
2
|
|
Δ
|
|
ω
2
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
3
|
+
t
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
t
|
Ω
2
|
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
t
|
Ω
4
|
|
Δ
|
|
ω
1
|
}
≤
I
q
(
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
0
)
|
+
|
f
(
s
,
0
)
|
)
(
T
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
×
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
(
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
0
)
|
+
|
f
(
s
,
0
)
|
)
(
η
i
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
(
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
0
)
|
+
|
f
(
s
,
0
)
|
)
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
(
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
0
)
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
f
(
s
,
0
)
|
)
(
ξ
j
)
)
≤
(
L
r
+
M
)
I
q
(
1
)
(
T
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
(
L
r
+
M
)
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
(
1
)
(
η
i
)
+
(
L
r
+
M
)
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
I
β
j
+
q
(
1
)
(
T
)
+
I
β
j
+
q
(
1
)
(
ξ
j
)
)
≤
(
L
r
+
M
)
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
(
L
r
+
M
)
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
(
L
r
+
M
)
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
=
(
L
r
+
M
)
Λ
+
Φ
≤
r
,
which implies that 𝒬
B
r
⊂
B
r.
Next, we letx
,
y
∈
𝒞. Then, for t
∈
[
0
,
T
], we have
(23)
|
𝒬
x
(
t
)
-
𝒬
y
(
t
)
|
≤
I
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
×
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
η
i
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
ξ
j
)
)
≤
I
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
T
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
×
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
η
i
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
-
f
(
s
,
y
(
s
)
)
|
(
ξ
j
)
)
≤
T
q
L
∥
x
-
y
∥
Γ
(
q
+
1
)
+
(
|
Ω
3
|
+
T
|
Ω
4
|
)
L
∥
x
-
y
∥
|
Δ
|
×
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
(
|
Ω
1
|
+
T
|
Ω
2
|
)
L
∥
x
-
y
∥
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
=
L
Λ
∥
x
-
y
∥
,
which implies that ∥
𝒬
x
-
𝒬
y
∥
≤
L
Λ
∥
x
-
y
∥. As L
Λ
<
1, 𝒬 is a contraction. Therefore, we deduce by Banach's contraction mapping principle that 𝒬 has a fixed point which is the unique solution of problem (1)-(2). The proof is completed.


## 3.2. Existence Result via Krasnoselskii’s Fixed Point Theorem

Lemma 6 (Krasnoselskii’s fixed point theorem, [17]).
LetM be a closed, bounded, convex, and nonempty subset of a Banach space X. Let A and B be the operators such that (a) A
x
+
B
y
∈
M whenever x
,
y
∈
M; (b) A is compact and continuous; and (c) B is a contraction mapping. Then, there exists z
∈
M such that z
=
A
z
+
B
z.Theorem 7.
Letf
:
[
0
,
T
]
×
ℝ
→
ℝ be a continuous function satisfying (
H
1
). Assume that(
H
2
)
|
f
(
t
,
u
)
|
≤
θ
(
t
),for all
(
t
,
u
)
∈
[
0
,
T
]
×
ℝ, and θ
∈
C
(
[
0
,
T
]
,
ℝ
+
).Then, the boundary value problem (1)-(2) has at least one solution on [
0
,
T
] provided
(24)
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
<
1
.Proof.
Settingsup
⁡
t
∈
[
0
,
T
]
|
θ
(
t
)
|
=
∥
θ
∥ and choosing
(25)
ρ
≥
∥
θ
∥
Λ
+
Φ
,
(Λ and Φ are defined in (19) and (21), resp.), we consider B
ρ
=
{
x
∈
𝒞
(
[
0
,
T
]
,
ℝ
)
:
∥
x
∥
≤
ρ
}. We define the operators 𝒬
1 and 𝒬
2 on B
ρ by
(26)
𝒬
1
x
(
t
)
=
I
q
f
(
s
,
x
(
s
)
)
(
t
)
,
𝒬
2
x
(
t
)
=
Ω
3
-
Ω
4
t
Δ
ω
1
+
Ω
2
t
-
Ω
1
Δ
ω
2
+
Ω
3
-
Ω
4
t
Δ
∑
i
=
1
m
‍
λ
i
I
α
i
+
q
f
(
s
,
x
(
s
)
)
(
η
i
)
+
Ω
2
t
-
Ω
1
Δ
∑
j
=
1
n
‍
μ
j
(
(
ξ
j
)
I
β
j
+
q
f
(
s
,
x
(
s
)
)
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
-
I
β
j
+
q
f
(
s
,
x
(
s
)
)
(
ξ
j
)
)
.
For any x
,
y
∈
B
ρ, we have
(27)
|
𝒬
1
x
(
t
)
+
𝒬
2
y
(
t
)
|
≤
sup
⁡
t
∈
[
0
,
T
]
{
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
i
i
i
i
i
i
i
i
i
i
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
y
(
s
)
)
|
(
η
i
)
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
y
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
y
(
s
)
)
|
(
ξ
j
)
)
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
}
≤
∥
θ
∥
[
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
]
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
=
∥
θ
∥
Λ
+
Φ
≤
ρ
.
This shows that 𝒬
1
x
+
𝒬
2
y
∈
B
ρ. It is easy to see using (24) that 𝒬
2 is a contraction mapping.
Continuity off implies that the operator 𝒬
1 is continuous. Also, 𝒬
1 is uniformly bounded on B
ρ as
(28)
∥
𝒬
1
x
∥
≤
T
q
Γ
(
q
+
1
)
∥
θ
∥
.
Now, we prove the compactness of the operator 𝒬
1.
We definesup
⁡
(
t
,
x
)
∈
[
0
,
T
]
×
B
ρ
|
f
(
t
,
x
)
|
=
f
¯
<
∞, and consequently we have
(29)
|
(
𝒬
1
x
)
(
t
2
)
-
(
𝒬
1
x
)
(
t
1
)
|
=
1
Γ
(
q
)
|
∫
0
t
1
‍
[
(
t
2
-
s
)
q
-
1
-
(
t
1
-
s
)
q
-
1
]
f
(
s
,
x
(
s
)
)
d
s
i
i
i
i
i
i
i
i
i
i
i
i
+
∫
t
1
t
2
‍
(
t
2
-
s
)
q
-
1
f
(
s
,
x
(
s
)
)
d
s
∫
0
t
1
‍
|
≤
f
¯
Γ
(
q
+
1
)
|
t
1
q
-
t
2
q
|
,
which is independent of x and tends to zero as t
2
-
t
2
→
0. Thus, 𝒬
1 is equicontinuous. So 𝒬
1 is relatively compact on B
ρ. Hence, by the Arzelá-Ascoli theorem, 𝒬
1 is compact on B
ρ. Thus, all the assumptions of Lemma 6 are satisfied. So the conclusion of Lemma 6 implies that the boundary value problem (1)-(2) has at least one solution on [
0
,
T
].


## 3.3. Existence Result via Leray-Schauder’s Nonlinear Alternative

Theorem 8 (nonlinear alternative for single valued maps, [18]).
LetE be a Banach space, C a closed, convex subset of E, U an open subset of C, and 0
∈
U. Suppose that F
:
U
¯
→
C is a continuous, compact (i.e., F
(
U
¯
) is a relatively compact subset of C) map. Then, either(i)
F has a fixed point in U
¯, or
(ii)
there isu
∈
∂
U (the boundary of U in C) and λ
∈
(
0,1
) with u
=
λ
F
(
u
).Theorem 9.
Assume that(
H
3
)
there exist a continuous nondecreasing functionψ
:
[
0
,
∞
)
→
(
0
,
∞
) and a function p
∈
C
(
[
0
,
T
]
,
ℝ
+
) such that
(30)
|
f
(
t
,
x
)
|
≤
p
(
t
)
ψ
(
∥
x
∥
)
f
o
r
e
a
c
h
(
t
,
x
)
∈
[
0
,
T
]
×
ℝ
;
(
H
4
)
there exists a constantM
>
0 such that
(31)
M
ψ
(
M
)
∥
p
∥
Λ
+
Φ
>
1
,
whereΛ and Φ are defined in (19) and (21), respectively.Then, the boundary value problem (1)-(2) has at least one solution on [
0
,
T
].Proof.
Let the operator𝒬 be defined by (17). Firstly, we will show that 𝒬
maps bounded sets (balls) into bounded sets in
C
(
[
0
,
T
]
,
ℝ
). For a number r
>
0, let B
r
=
{
x
∈
C
(
[
0
,
T
]
,
ℝ
)
:
∥
x
∥
≤
r
} be a bounded ball in C
(
[
0
,
T
]
,
ℝ
). Then, for t
∈
[
0
,
T
] we have
(32)
|
𝒬
x
(
t
)
|
≤
sup
⁡
t
∈
[
0
,
T
]
{
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
i
i
i
i
i
i
i
i
i
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
}
≤
ψ
(
∥
x
∥
)
I
q
p
(
s
)
(
T
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
ψ
(
∥
x
∥
)
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
p
(
s
)
(
η
i
)
+
ψ
(
∥
x
∥
)
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
×
∑
j
=
1
n
‍
|
μ
j
|
(
I
β
j
+
q
p
(
s
)
(
T
)
+
I
β
j
+
q
p
(
s
)
(
ξ
j
)
)
≤
ψ
(
∥
x
∥
)
∥
p
∥
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
+
ψ
(
∥
x
∥
)
∥
p
∥
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
ψ
(
∥
x
∥
)
∥
p
∥
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
-
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
=
ψ
(
∥
x
∥
)
∥
p
∥
{
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
}
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
,
and, consequently,
(33)
∥
𝒬
x
∥
≤
ψ
(
r
)
∥
p
∥
Λ
+
Φ
,
where Λ and Φ are defined by (19) and (21), respectively.
Next, we will show that𝒬
maps bounded sets into equicontinuous sets of
C
(
[
0
,
T
]
,
ℝ
). Let τ
1
,
τ
2
∈
[
0
,
T
] with τ
1
<
τ
2 and x
∈
B
r. Then, we have
(34)
|
(
𝒬
x
)
(
τ
2
)
-
(
𝒬
x
)
(
τ
1
)
|
≤
1
Γ
(
q
)
|
∫
0
τ
1
‍
[
(
τ
2
-
s
)
q
-
1
-
(
τ
1
-
s
)
q
-
1
]
f
(
s
,
x
(
s
)
)
d
s
iiiiiiiii
+
∫
τ
1
τ
2
‍
(
τ
2
-
s
)
q
-
1
f
(
s
,
x
(
s
)
)
d
s
∫
0
τ
1
‍
|
+
|
ω
1
Ω
4
|
+
|
ω
2
Ω
2
|
|
Δ
|
|
τ
2
-
τ
1
|
+
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
(
τ
2
-
τ
1
)
+
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
(
τ
2
-
τ
1
)
≤
ψ
(
r
)
Γ
(
q
)
|
∫
0
τ
1
‍
[
(
τ
2
-
s
)
q
-
1
-
(
τ
1
-
s
)
q
-
1
]
p
(
s
)
d
s
iiiiiiiii
+
∫
τ
1
τ
2
‍
(
τ
2
-
s
)
q
-
1
p
(
s
)
d
s
∫
0
τ
1
‍
|
+
|
ω
1
Ω
4
|
+
|
ω
2
Ω
2
|
|
Δ
|
(
τ
2
-
τ
1
)
+
|
Ω
4
|
|
Δ
|
ψ
(
r
)
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
p
(
s
)
(
η
i
)
(
τ
2
-
τ
1
)
+
|
Ω
2
|
|
Δ
|
ψ
(
r
)
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
p
(
s
)
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
p
(
s
)
(
ξ
j
)
)
(
τ
2
-
τ
1
)
.
As τ
2
-
τ
1
→
0, the right-hand side of the above inequality tends to zero independently of x
∈
B
r. Therefore, by the Arzelá-Ascoli theorem, the operator 𝒬
:
C
(
[
0
,
T
]
,
ℝ
)
→
C
(
[
0
,
T
]
,
ℝ
) is completely continuous.
Letx be a solution. Then, for t
∈
[
0
,
T
], following the similar computations as in the first step, we have
(35)
|
x
(
t
)
|
≤
ψ
(
∥
x
∥
)
∥
p
∥
Λ
+
Φ
,
which leads to
(36)
∥
x
∥
ψ
(
∥
x
∥
)
∥
p
∥
Λ
+
Φ
≤
1
.
In view of (
H
4
), there exists M such that ∥
x
∥
≠
M. Let us set
(37)
U
=
{
x
∈
C
(
[
0
,
T
]
,
ℝ
)
:
∥
x
∥
<
M
}
.
We see that the operator 𝒬
:
U
¯
→
C
(
[
0
,
T
]
,
ℝ
) is continuous and completely continuous. From the choice of U, there is no x
∈
∂
U such that x
=
ν
𝒬
x for some ν
∈
(
0,1
). Consequently, by the nonlinear alternative of Leray-Schauder type, we deduce that 𝒬 has a fixed point x
∈
U
¯ which is a solution of problem (1)-(2). This completes the proof.


## 3.4. Existence Result via Leray-Schauder Degree

Theorem 10.
Letf
:
[
0
,
T
]
×
ℝ
→
ℝ be a continuous function. Assume that(
H
5
)
there exist constants0
≤
κ
<
Λ
-
1, where Λ is given by (19) and N
>
0 such that
(38)
|
f
(
t
,
x
)
|
≤
κ
|
x
|
+
N
f
o
r
e
a
c
h
(
t
,
x
)
∈
[
0
,
T
]
×
ℝ
.Then, the boundary value problem (1)-(2) has at least one solution on [
0
,
T
].Proof.
Let us define an operator𝒬
:
𝒞
→
𝒞 as in (17) and consider the fixed point problem
(39)
x
=
𝒬
x
.
In view of the fixed point problem (39), we are going to prove the existence of at least one solution x
∈
𝒞 satisfying (39). Define a ball B
R
⊂
𝒞 with a constant radius R
>
0 given by
(40)
B
R
=
{
x
∈
𝒞
:
max
⁡
t
∈
[
0,1
]
|
x
(
t
)
|
<
R
,
R
>
0
}
.
Hence, it is sufficient to show that 𝒬
:
B
¯
R
→
𝒞 satisfies
(41)
x
≠
λ
𝒬
x
,
∀
x
∈
∂
B
R
,
∀
λ
∈
[
0,1
]
.
We define
(42)
H
(
λ
,
x
)
=
λ
𝒬
x
,
x
∈
C
,
λ
∈
[
0,1
]
.
As shown in Theorem 9, we deduce that the operator 𝒬 is continuous, uniformly bounded, and equicontinuous. Then, by the Arzelá-Ascoli theorem, a continuous map h
λ defined by h
λ
(
x
)
=
x
-
H
(
λ
,
x
)
=
x
-
λ
𝒬
x is completely continuous. If (41) holds, then the following Leray-Schauder degree is well defined, and by the homotopy invariance of topological degree, it follows that
(43)
deg
⁡
(
h
λ
,
B
R
,
0
)
=
deg
⁡
(
I
-
λ
𝒬
,
B
R
,
0
)
=
deg
⁡
(
h
1
,
B
R
,
0
)
=
deg
⁡
(
h
0
,
B
R
,
0
)
=
deg
⁡
(
I
,
B
R
,
0
)
=
1
≠
0
,
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
0
∈
B
R
,
where I denotes the unit operator. By the nonzero property of Leray-Schauder degree, h
1
(
x
)
=
x
-
λ
𝒬
x
=
0 for at least one x
∈
B
R. In order to prove (41), we assume that x
=
λ
𝒬
x for some λ
∈
[
0,1
] and for all t
∈
[
0
,
T
]. Then,
(44)
|
x
(
t
)
|
=
|
λ
(
𝒬
x
)
(
t
)
|
≤
sup
⁡
t
∈
[
0
,
T
]
{
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
|
Ω
4
|
t
|
Δ
|
|
ω
1
|
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
|
Ω
2
|
t
|
Δ
|
|
ω
2
|
+
|
Ω
3
|
+
|
Ω
4
|
t
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
×
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
|
Ω
2
|
t
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
I
q
|
f
(
s
,
x
(
s
)
)
|
(
t
)
+
|
Ω
3
|
+
|
Ω
4
|
t
|
Δ
|
|
ω
1
|
}
≤
I
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
|
ω
2
|
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
|
f
(
s
,
x
(
s
)
)
|
(
η
i
)
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
(
ξ
j
)
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
T
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
I
β
j
+
q
|
f
(
s
,
x
(
s
)
)
|
(
ξ
j
)
)
≤
(
κ
|
x
|
+
N
)
I
q
(
1
)
(
T
)
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
|
ω
2
|
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
(
κ
|
x
|
+
N
)
∑
i
=
1
m
‍
|
λ
i
|
I
α
i
+
q
(
1
)
(
η
i
)
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
(
κ
|
x
|
+
N
)
×
∑
j
=
1
n
‍
|
μ
j
|
(
I
β
j
+
q
(
1
)
(
T
)
+
I
β
j
+
q
(
1
)
(
ξ
j
)
)
≤
(
κ
|
x
|
+
N
)
[
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
)
Γ
(
β
j
+
q
+
1
)
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
|
Ω
4
|
T
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
|
Ω
1
|
+
|
Ω
2
|
T
|
Δ
|
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
×
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
)
Γ
(
β
j
+
q
+
1
)
]
+
Φ
=
(
κ
|
x
|
+
N
)
Λ
+
Φ
.
Computing directly for ∥
x
∥
=
sup
⁡
t
∈
[
0
,
T
]
|
x
(
t
)
|, we have
(45)
∥
x
∥
≤
N
Λ
+
Φ
1
-
κ
Λ
.
Let R
=
(
N
Λ
+
Φ
)
/
(
1
-
κ
Λ
)
+
1; then (41) holds. This completes the proof.


## 4. Examples

In this section, we present some examples to illustrate our results.Example 1.
Consider the following boundary value problem with multiterm fractional integral:(46)
D
c
3
/
2
x
(
t
)
=
e
-
t
14
(
1
+
e
t
)
|
x
|
1
+
|
x
|
,
t
∈
(
0,1
)
,
1
2
I
2
/
3
x
(
3
4
)
-
2
3
I
1
/
2
x
(
1
4
)
+
3
4
I
4
/
3
x
(
2
5
)
=
3
2
,
2
5
(
I
3
/
2
x
(
1
)
-
I
3
/
2
x
(
1
8
)
)
+
1
3
(
I
1
/
4
x
(
1
)
-
I
1
/
4
x
(
5
8
)
)
=
-
2
3
.Here,q
=
3
/
2, m
=
3, λ
1
=
1
/
2, λ
2
=
-
2
/
3, λ
3
=
3
/
4, α
1
=
2
/
3, α
2
=
1
/
2, α
3
=
4
/
3, η
1
=
3
/
4, η
2
=
1
/
4, η
3
=
2
/
5, n
=
2, μ
1
=
2
/
5, μ
2
=
1
/
3, β
1
=
3
/
2, β
2
=
1
/
4, ξ
1
=
1
/
8, ξ
2
=
5
/
8, T
=
1, ω
1
=
3
/
2, ω
2
=
-
2
/
3, and f
(
t
,
x
)
=
(
e
-
t
/
(
14
(
1
+
e
t
)
)
)
(
|
x
|
/
(
1
+
|
x
|
)
). Since |
f
(
t
,
x
)
-
f
(
t
,
y
)
|
≤
(
1
/
28
)
|
x
-
y
|, then, (
H
1
) is satisfied with L
=
1
/
28. We can show that
(47)
Λ
=
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
≈
27.028626
,
L
Λ
≈
0.965308
<
1
.
Hence, by Theorem 5, the boundary value problem (46) has a unique solution on [
0,1
].Example 2.
Consider the following boundary value problem with multiterm fractional integral:(48)
D
c
7
/
5
x
(
t
)
=
1
12
(
1
+
t
2
)
x
2
1
+
|
x
|
+
2
|
x
|
1
+
|
x
|
,
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
t
∈
(
0
,
9
10
)
,
2
I
2
/
3
x
(
1
2
)
+
10
3
I
3
/
2
x
(
5
6
)
-
7
9
I
1
/
4
x
(
2
3
)
-
1
3
I
2
/
9
x
(
1
9
)
=
-
1
2
,
2
5
(
I
5
/
2
x
(
9
10
)
-
I
5
/
2
x
(
1
7
)
)
-
3
4
(
I
1
/
2
x
(
9
10
)
-
I
1
/
2
x
(
1
5
)
)
=
-
3
4
.Here,q
=
7
/
5, m
=
4, λ
1
=
2, λ
2
=
10
/
3, λ
3
=
-
7
/
9, λ
4
=
-
1
/
3,   α
1
=
2
/
3, α
2
=
3
/
2, α
3
=
1
/
4, α
4
=
2
/
9, η
1
=
1
/
2, η
2
=
5
/
6, η
3
=
2
/
3, η
4
=
1
/
9, n
=
2, μ
1
=
2
/
5, μ
2
=
-
3
/
4, β
1
=
5
/
2, β
2
=
1
/
2, ξ
1
=
1
/
7, ξ
2
=
1
/
5, T
=
9
/
10, ω
1
=
-
1
/
2, ω
2
=
-
3
/
4, and f
(
t
,
x
)
=
(
1
/
12
)
(
1
+
t
2
)
(
x
2
/
(
1
+
|
x
|
)
)
+
2
(
|
x
|
/
(
1
+
|
x
|
)
). It is easy to verify that
(49)
Λ
=
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
≈
1.641942
,
Φ
=
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
|
ω
1
|
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
|
ω
2
|
≈
2.582519
.
Clearly,
(50)
|
f
(
t
,
x
)
|
=
|
1
12
(
1
+
t
2
)
x
2
1
+
|
x
|
+
2
|
x
|
1
+
|
x
|
|
≤
1
12
(
1
+
t
2
)
(
|
x
|
+
24
)
.
Choosing p
(
t
)
=
(
1
/
12
)
(
t
2
+
1
) and ψ
(
|
x
|
)
=
|
x
|
+
24, we can show that
(51)
M
ψ
(
M
)
∥
p
∥
Λ
+
Φ
>
1
,
which implies that M
>
16.736885. Hence, by Theorem 9, the boundary value problem (48) has at least one solution on [
0,9
/
10
].Example 3.
Consider the following boundary value problem with multiterm fractional integral:(52)
D
c
3
/
2
x
(
t
)
=
1
5
π
sin
π
2
x
+
|
x
|
1
+
|
x
|
,
t
∈
(
0,1
)
,
-
1
2
I
1
/
3
x
(
3
5
)
+
4
5
I
3
/
2
x
(
2
5
)
+
3
2
I
2
/
5
x
(
1
4
)
-
2
5
I
1
/
4
x
(
2
3
)
=
1
2
,
-
2
7
(
I
3
/
2
x
(
1
)
-
I
3
/
2
x
(
2
3
)
)
+
1
3
(
I
1
/
2
x
(
1
)
-
I
1
/
2
x
(
3
4
)
)
-
3
4
(
I
2
/
3
x
(
1
)
-
I
2
/
3
x
(
1
4
)
)
=
-
2
3
.Here,q
=
3
/
2, m
=
4, λ
1
=
-
1
/
2, λ
2
=
4
/
5, λ
3
=
3
/
2, λ
4
=
-
2
/
5, α
1
=
1
/
3, α
2
=
3
/
2, α
3
=
2
/
5, α
4
=
1
/
4, η
1
=
3
/
5, η
2
=
2
/
5, η
3
=
1
/
4, η
4
=
2
/
3, n
=
3, μ
1
=
-
2
/
7, μ
2
=
1
/
3, μ
3
=
-
3
/
4, β
1
=
3
/
2, β
2
=
1
/
2, β
3
=
2
/
3, ξ
1
=
2
/
3, ξ
2
=
3
/
4, ξ
3
=
1
/
4, T
=
1, ω
1
=
1
/
2, ω
2
=
-
2
/
3, and f
(
t
,
x
)
=
(
1
/
5
π
)
(
sin
(
π
x
/
2
)
)
+
(
|
x
|
/
(
1
+
|
x
|
)
). We can show that
(53)
Λ
=
T
q
Γ
(
q
+
1
)
+
|
Ω
3
|
+
T
|
Ω
4
|
|
Δ
|
∑
i
=
1
m
‍
|
λ
i
|
η
i
α
i
+
q
Γ
(
α
i
+
q
+
1
)
+
|
Ω
1
|
+
T
|
Ω
2
|
|
Δ
|
∑
j
=
1
n
‍
|
μ
j
|
(
T
β
j
+
q
+
ξ
j
β
j
+
q
Γ
(
β
j
+
q
+
1
)
)
≈
3.400339
.
Since
(54)
|
f
(
t
,
x
)
|
=
|
1
5
π
sin
π
2
x
+
|
x
|
1
+
|
x
|
|
≤
1
10
|
x
|
+
1
,
then, (
H
5
) is satisfied with κ
=
1
/
10 and N
=
1 such that
(55)
κ
=
1
10
<
1
Λ
≈
0.294084
.
Hence, by Theorem 10, the boundary value problem (52) has at least one solution on [
0,1
].



---

*Source: 806156-2014-05-11.xml*