# Solvability for a Discrete Fractional Three-Point Boundary Value Problem at Resonance

Authors: Weidong Lv
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

This paper is concerned with the existence of solutions to a discrete three-point boundary value problem at resonance involving the Riemann-Liouville fractional difference of orderα
∈
(
0,1
]. Under certain suitable nonlinear growth conditions imposed on the nonlinear term, the existence result is established by using the coincidence degree continuation theorem. Additionally, a representative example is presented to illustrate the effectiveness of the main result.


---

## Body

## 1. Introduction

For any numbera
∈
R, we denote N
a
=
{
a
,
a
+
1
,
a
+
2
,
…
} and [
c
,
d
]
N
a
=
{
c
,
c
+
1
,
…
,
d
-
1
,
d
}, for any c
,
d
∈
N
a with c
<
d, throughout this paper. It is also worth noting that, in what follows, we appeal to the conventions ∑
s
=
k
1
k
2
‍
u
(
s
)
=
0 and ∏
s
=
k
1
k
2
‍
u
(
s
)
=
1 for a given function u defined on N
a and k
1
,
k
2
∈
N
a with k
1
>
k
2.In this paper, we will consider the existence of solutions for the following discrete fractional three-point boundary value problem:(1)
Δ
α
u
(
t
)
=
f
(
t
+
α
-
1
,
u
(
t
+
α
-
1
)
)
,
t
∈
[
0
,
b
]
N
0
,
u
(
α
-
1
)
-
β
u
(
α
+
η
)
=
γ
u
(
α
+
b
)
,
where 0
<
α
≤
1 is a real number, Δ
α denotes the Riemann-Liouville fractional difference of order α, f
:
[
α
-
1
,
α
+
b
-
1
]
N
α
-
1
×
R
→
R, η
∈
[
0
,
b
-
1
]
N
0, β
,
γ
>
0, and
(2)
β
∏
i
=
1
η
+
1
‍
α
+
i
-
1
i
+
γ
∏
i
=
1
b
+
1
‍
α
+
i
-
1
i
=
1
,
which implies that the problem (1) is at resonance. We note that the problem (1) happens to be at resonance in the sense that the associated linear homogeneous boundary value problem
(3)
Δ
α
u
(
t
)
=
0
,
t
∈
[
0
,
b
]
N
0
,
u
(
α
-
1
)
-
β
u
(
α
+
η
)
=
γ
u
(
α
+
b
)
,
has u
(
t
)
=
c
t
α
-
1
_
,
t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1, c
∈
R, as a nontrivial solution.The continuous fractional calculus has received increasing attention within the last ten years or so and the theory of fractional differential equations has been a new important mathematical branch due to its demonstrated applications in various fields of science and engineering. For more details, see [1–14] and references therein. Significantly less is known, however, about the discrete fractional calculus, but in recent several years, a lot of papers have appeared; see [15–36]. For example, in [19], Atıcı and Eloe explored a discrete fractional conjugate boundary value problem with the Riemann-Liouville fractional difference. To the best of our knowledge, this is a pioneering work on discussing boundary value problems in discrete fractional calculus. After that, Goodrich studied discrete fractional boundary value problems involving the Riemann-Liouville fractional difference intensively and obtained a series of excellent results; see [20–26]. Bastos et al. in [28, 29] considered the discrete fractional calculus of variations and established the necessary conditions for fractional difference variational problems. Abdeljawad introduced the Caputo fractional difference and developed some useful properties of it in [30]. Ferreira [35] investigated the existence and uniqueness of solutions for some discrete fractional nonresonance boundary value problems of order less than one by using the Banach fixed point theorem.Although the solvability of fractional boundary value problems has been studied extensively, there are few papers dealing with it under resonance conditions, besides [37–41]. Additionally, as far as we know, the existence of solutions to discrete fractional boundary value problems at resonance has not been studied.Motivated by the aforementioned results, we will investigate the discrete fractional boundary value problem (1) at resonance and establish some sufficient conditions for the existence of solutions to it by using the coincidence degree theory.For the sake of convenience, we will always assume that the following conditions hold in this paper:(
H
1
)
α
∈
(
0,1
]
,
η
∈
[
0
,
b
-
1
]
N
0, β
,
γ
>
0, and β
∏
i
=
1
η
+
1
‍
(
α
+
i
-
1
)
/
i
+
γ
∏
i
=
1
b
+
1
‍
(
α
+
i
-
1
)
/
i
=
1;
(
H
2
)
f
:
[
α
-
1
,
α
+
b
-
1
]
N
α
-
1
×
R
→
R is continuous.The remainder of this paper is organized as follows. Section2 preliminarily provides some necessary basic knowledge for the theory of discrete fractional calculus and the coincidence degree continuation theorem. In Section 3, the existence result of solutions for problem (1) will be established with the help of the coincidence degree theory. Finally in Section 4, a concrete example is provided to illustrate the possible application of the established analytical result.


## 2. Preliminaries

Since the theory of discrete fractional calculus is in its infancy to some extent, in order to make this paper self-contained, we begin by presenting here some necessary basic definitions and lemmas about it. For more details, see [15, 16, 19, 34].Definition 1 (see [15]).
For anyt and ν, the falling factorial function is defined as
(4)
t
ν
_
=
Γ
(
t
+
1
)
Γ
(
t
+
1
-
ν
)
,
provided that the right-hand side is well defined. We appeal to the convention that if t
+
1
-
ν is a pole of the Gamma function and t
+
1 is not a pole, then t
ν
_
=
0.Definition 2 (see [42]).
Theνth discrete fractional sum of a function f
:
N
a
→
R, for ν
>
0, is defined by
(5)
Δ
a
-
ν
f
(
t
)
=
1
Γ
(
ν
)
∑
s
=
a
t
-
ν
‍
(
t
-
s
-
1
)
ν
-
1
_
f
(
s
)
,
t
∈
N
a
+
ν
.
Also, we define the trivial sum Δ
a
0
f
(
t
)
=
f
(
t
), t
∈
N
a.Definition 3 (see [15]).
Theνth discrete Riemann-Liouville fractional difference of a function f
:
N
a
→
R, for ν
>
0, is defined by
(6)
Δ
a
ν
f
(
t
)
=
Δ
n
Δ
a
-
(
n
-
ν
)
f
(
t
)
,
t
∈
N
a
+
n
-
ν
,
where n is the smallest integer greater than or equal to ν and Δ
n is the nth order forward difference operator. If ν
=
n
∈
N
1, then Δ
a
n
f
(
t
)
=
Δ
n
f
(
t
).Remark 4.
From the Definitions2 and 3, it is easy to see that Δ
a
-
ν maps functions defined on N
a to functions defined on N
a
+
ν and Δ
a
ν maps functions defined on N
a to functions defined on N
a
+
n
-
ν, where n is the smallest integer greater than or equal to ν. Also, it is worth reminding the reader that the t in Δ
a
ν
f
(
t
) (or Δ
a
-
ν
f
(
t
)) represents an input for the function Δ
a
ν
f (or Δ
a
-
ν
f) and not for the function f. For ease of notation, we throughout this paper omit the subscript a in Δ
a
ν
f
(
t
) and Δ
a
-
ν
f
(
t
) when it does not to lead to domains confusion and general ambiguity.Lemma 5 (see [19]).
Letν
>
0. Then Δ
-
ν
Δ
ν
f
(
t
)
=
f
(
t
)
+
c
1
t
ν
-
1
_
+
c
2
t
ν
-
2
_
+
⋯
+
c
n
t
ν
-
n
_, where c
i
∈
R, i
=
1,2
,
…
,
n, and n is the smallest integer greater than or equal to ν.Lemma 6 (see [21]).
Letν
∈
R and t
,
s
∈
R such that (
t
-
s
)
ν
_ is well defined. Then
(7)
Δ
s
(
t
-
s
)
ν
_
=
-
ν
(
t
-
s
-
1
)
ν
-
1
_
.Next, we will briefly recall some notations in the frame of Mawhin’s coincidence degree continuous theorem. For more details, see [43].LetX and Y be two real Banach spaces. Consider an operator equation L
u
=
N
u, where L
:
Dom
⁡
L
⊂
X
→
Y is a linear operator and N
:
X
→
Y is a nonlinear operator. The operator L will be called a Fredholm operator of index zero if dim
⁡
Ker
⁡
L
=
codim
Im
⁡
L
<
∞ and Im
⁡
L is closed in Y. If L is a Fredholm operator of index zero, then there exist continuous projectors P
:
X
→
X and Q
:
Y
→
Y such that Ker
⁡
L
=
Im
⁡
P, Im
⁡
L
=
Ker
⁡
Q and X
=
Ker
⁡
L
⊕
Ker
⁡
P, Y
=
Im
⁡
L
⊕
Im
⁡
Q. It follows that L
|
Dom
⁡
L
∩
Ker
⁡
P
:
Dom
⁡
L
∩
Ker
⁡
P
→
Im
⁡
L is invertible and its inverse is denoted by K
p.IfΩ is an open bounded subset of X and Dom
⁡
L
∩
Ω
¯
≠
∅, the operator N
:
X
→
Y will be called L-compact on Ω
¯ if Q
N
(
Ω
¯
) is bounded and K
p
(
I
-
Q
)
N
:
Ω
¯
→
X is compact.Now, we present the coincidence degree continuation theorem as follows, which will be used in the sequel to establish the existence of solutions for problem (1).Theorem 7 (see [43]).
LetL
:
Dom
L
⊂
X
→
Y be a Fredholm operator of index zero and let N
:
X
→
Y be L-compact on Ω
¯. Assume that the following conditions are satisfied:(i)
L
x
≠
λ
N
x for every (
x
,
λ
)
∈
[
(
Dom
⁡
L
∖
Ker
⁡
L
)
∩
∂
Ω
]
×
(
0,1
);
(ii)
N
x
∉
Im
⁡
L for every x
∈
Ker
⁡
L
∩
∂
Ω;
(iii)
deg
⁡
(
Q
N
|
Ker
⁡
L
,
Ω
∩
Ker
⁡
L
,
0
)
≠
0, where Q is a projection such that Im
⁡
L
=
Ker
⁡
Q.Then the operator equation L
x
=
N
x has at least one solution in Dom
⁡
L
∩
Ω
¯.Finally, we wish to fix our framework for the study of problem (1). First of all, we denote X
=
{
u
:
[
α
-
1
,
α
+
b
]
N
α
-
1
→
R
} and Y
=
{
y
:
[
0
,
b
]
N
0
→
R
}, and it is clear that X and Y are two Banach spaces when equipped with the usual maximum norm; that is, for any u
∈
X and y
∈
Y, ∥
u
∥
=
max
⁡
{
|
u
(
t
)
|
:
t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1
} and ∥
y
∥
=
max
⁡
{
|
y
(
t
)
|
:
t
∈
[
0
,
b
]
N
0
}. Next, we define the linear operator L
:
Dom
⁡
L
⊂
X
→
Y by
(8)
L
u
=
Δ
α
u
,
with
(9)
Dom
⁡
L
=
{
u
:
u
∈
X
,
u
(
α
-
1
)
-
β
u
(
α
+
η
)
=
γ
u
(
α
+
b
)
}
,
and the nonlinear operator N
:
X
→
Y as
(10)
N
u
(
t
)
=
f
(
t
+
α
-
1
,
u
(
t
+
α
-
1
)
)
,
for
t
∈
[
0
,
b
]
N
0
.
Then the problem (1) is equivalent to an operator equation
(11)
L
u
=
N
u
,
u
∈
Dom
⁡
L
.


## 3. Main Results

In this section, we will establish the existence of at least one solution for the problem (1). To accomplish this, we firstly present here several lemmas which will be used in the sequel.For convenience, we define the operatorQ
1
:
Y
→
Y by
(12)
(
Q
1
y
)
(
t
)
=
β
∑
s
=
0
η
‍
(
α
+
η
-
s
-
1
)
α
-
1
_
y
(
s
)
+
γ
∑
s
=
0
b
‍
(
α
+
b
-
s
-
1
)
α
-
1
_
y
(
s
)
,
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
,
b
]
N
0
,
and by Lemma 6, we can find that Q
1
(
1
)
=
(
β
(
α
+
η
)
α
_
+
γ
(
α
+
b
)
α
_
)
/
α
>
0.Lemma 8.
If(
H
1
) holds, then
(13)
Ker
⁡
L
=
{
u
∈
X
:
u
=
c
t
α
-
1
_
,
c
∈
R
}
,
(14)
Im
⁡
L
=
{
y
∈
Y
:
Q
1
y
=
0
}
,
where L and Q
1 are defined by (8) and (12), respectively.Proof .
At first, in view of Lemma5 and (
H
1
), we can easily verify that (13) holds. Next, we prove that (14) also holds.
For anyy
∈
Im
⁡
L, then there exists a function u
∈
Dom
⁡
L such that y
=
Δ
α
u. Based on Lemma 5, we have
(15)
u
(
t
)
=
1
Γ
(
α
)
∑
s
=
0
t
-
α
‍
(
t
-
s
-
1
)
α
-
1
_
y
(
s
)
+
c
t
α
-
1
_
,
where t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1, c
∈
R. From conditions u
(
α
-
1
)
-
β
u
(
α
+
η
)
=
γ
u
(
α
+
b
) and (
H
1
), we can easily obtain that Q
1
y
=
0.
Conversely, for anyy
∈
Y with Q
1
y
=
0, if we set u
(
t
)
=
Δ
-
α
y
(
t
)
,
t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1, then u
(
α
-
1
)
=
0, and it is easy to verify that u
∈
Dom
⁡
L. Moreover, by the relation Δ
α
Δ
-
α
y
=
y, we have L
u
=
y, which lead to y
∈
Im
⁡
L. So we get that (14) holds. The proof is complete.Lemma 9.
If(
H
1
) holds, then L defined by (8) is a Fredholm operator of index zero.Proof .
In order to show thatL is a Fredholm operator of index zero, firstly, we consider the following operator Q
:
Y
→
Y defined by
(16)
Q
y
=
α
(
Q
1
y
)
β
(
α
+
η
)
α
_
+
γ
(
α
+
b
)
α
_
=
Q
1
y
Q
1
(
1
)
,
where Q
1 is defined by (12). Evidently, Im
⁡
Q
=
R and Q is a continuous linear projector. In fact, for any y
∈
Y, we have
(17)
Q
1
(
Q
y
)
=
Q
1
(
Q
1
y
Q
1
(
1
)
)
=
Q
1
y
Q
1
(
1
)
Q
1
(
1
)
=
Q
1
y
,
Q
2
y
=
Q
(
Q
y
)
=
Q
1
(
Q
y
)
Q
1
(
1
)
=
Q
1
y
Q
1
(
1
)
=
Q
y
;
that is to say, Q
:
Y
→
Y is idempotent. Hence, Q is a projector.
From the definition ofQ and (14), it is easy to see that y
∈
Im
⁡
L leads to Q
y
=
0, and if y
∈
Ker
⁡
Q, we can get Q
1
y
=
0, which implies that y
∈
Im
⁡
L. So, we derive Ker
⁡
Q
=
Im
⁡
L.
For anyy
∈
Y, set y
=
(
y
-
Q
y
)
+
Q
y. Since Q
y
∈
Im
⁡
Q and (
I
-
Q
)
y
∈
Ker
⁡
Q, we have Y
=
Im
⁡
Q
+
Ker
⁡
Q. Moreover, take y
0
∈
Ker
⁡
Q
∩
Im
⁡
Q. Then y
0 can be written as y
0
(
t
)
=
c
,
t
∈
[
0
,
b
]
N
0, c
∈
R, for y
0
∈
Im
⁡
Q. On the other hand, since y
0
∈
Ker
⁡
Q
=
Im
⁡
L, by (14), we can get Q
1
y
0
=
Q
1
(
c
)
=
c
Q
1
(
1
)
=
0, which implies that y
0
=
c
=
0. So, we have Im
⁡
Q
∩
Ker
⁡
Q
=
{
0
} and Y
=
Im
⁡
Q
⊕
Ker
⁡
Q
=
Im
⁡
Q
⊕
Im
⁡
L.
Now, sincedim
⁡
Ker
⁡
L
=
codim
Im
⁡
L
=
dim
⁡
Im
⁡
Q
=
1 and Im
⁡
L is closed in Y, L is a Fredholm operator of index zero. The proof is complete.LetP
:
X
→
X be defined by
(18)
P
u
(
t
)
=
1
Γ
(
α
)
u
(
α
-
1
)
t
α
-
1
_
,
t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1
.
It is clear that P
:
X
→
X is a linear continuous projector and
(19)
Im
⁡
P
=
{
u
∈
X
:
u
=
c
t
α
-
1
_
,
c
∈
R
}
=
Ker
⁡
L
.
Also, proceeding as the proof of Lemma 9, we can show that
(20)
X
=
Ker
⁡
P
⊕
Im
⁡
P
=
Ker
⁡
P
⊕
Ker
⁡
L
.Define operatorK
p
:
Im
⁡
L
→
Dom
⁡
L
∩
Ker
⁡
P by
(21)
K
p
y
(
t
)
=
Δ
-
α
y
(
t
)
=
1
Γ
(
α
)
∑
s
=
0
t
-
α
‍
(
t
-
s
-
1
)
α
-
1
_
y
(
s
)
,
i
i
i
i
i
i
i
i
i
i
i
t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1
,
where y
∈
Im
⁡
L.From the definitions ofP and K
p, it is easy to see that the inverse of L
|
Dom
⁡
L
∩
Ker
⁡
P is K
p. In fact, if y
∈
Im
⁡
L, then we have
(22)
L
K
p
y
=
Δ
α
Δ
-
α
y
=
y
.
Also, if u
∈
Dom
⁡
L
∩
Ker
⁡
P, by Lemma 5, we have
(23)
K
p
L
u
(
t
)
=
u
(
t
)
+
c
t
α
-
1
_
,
where t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1
,
c
∈
R. Then it follows from K
p
L
u
∈
Ker
⁡
P and u
∈
Ker
⁡
P that
(24)
K
p
L
u
(
α
-
1
)
=
u
(
α
-
1
)
+
c
(
α
-
1
)
α
-
1
_
=
Γ
(
α
)
c
=
0
,
which implies that c
=
0. Consequently, we have K
p
L
u
=
u for u
∈
Dom
⁡
L
∩
Ker
⁡
P. So, K
p
=
(
L
|
Dom
⁡
L
∩
Ker
⁡
P
)
-
1.Lemma 10.
Suppose that(
H
2
) holds. If Ω
⊂
X is an open bounded subset and Dom
⁡
L
∩
Ω
¯
≠
∅, then N is L-compact on Ω
¯.Proof.
By the continuity off, we can verify that Q
N
(
Ω
¯
) and K
p
(
I
-
Q
)
N
(
Ω
¯
) are bounded. So we get that K
p
(
I
-
Q
)
N
(
Ω
¯
) is compact. Therefore N is L-compact on Ω
¯. The proof is complete.To establish the main result, we need the following conditions.(
H
3
)
There exist two nonnegative functionsp
,
q
∈
X with {
∏
i
=
1
b
‍
(
(
α
+
i
)
/
i
)
+
(
(
b
+
1
)
/
α
)
}
∥
p
∥
<
1 such that
(25)
|
f
(
t
,
u
)
|
≤
p
(
t
)
|
u
|
+
q
(
t
)
,
for
(
t
,
u
)
∈
[
α
-
1
,
α
+
b
-
1
]
N
α
-
1
×
(
-
∞
,
+
∞
)
.
(
H
4
)
There exists a constantM
>
0 such that
(26)
Q
1
N
u
≠
0
,
for each u
∈
X satisfying |
u
(
t
)
|
>
M, t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1.
(
H
5
)
There exists a constantM
*
>
0 such that for any u
(
t
)
=
c
t
α
-
1
_, t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1
,
c
∈
R, if |
c
|
>
M
*, then either
(27)
c
Q
1
N
u
>
0
or
(28)
c
Q
1
N
u
<
0
.Theorem 11.
If(
H
1
)
–
(
H
5
) hold, then the problem (1) has at least one solution in X.Proof.
This proof will be divided into four main steps. Now let us prove the steps one by one.
Step
1. Set Ω
1
=
{
u
∈
Dom
⁡
L
∖
Ker
⁡
L
:
L
u
=
λ
N
u
,
λ
∈
(
0,1
)
} and prove that Ω
1 is bounded in X.
Foru
∈
Ω
1, then u
∈
Dom
⁡
L
∖
Ker
⁡
L and L
u
=
λ
N
u, so N
u
∈
Im
⁡
L. By (14), we have Q
1
N
u
=
0. From (
H
4
), there exists a constant t
0
∈
[
α
-
1
,
α
+
b
]
N
α
-
1 such that |
u
(
t
0
)
|
≤
M. Since L
u
=
λ
N
u, by Lemma 5, we have
(29)
u
(
t
)
=
λ
Γ
(
α
)
∑
s
=
0
t
-
α
‍
(
t
-
s
-
1
)
α
-
1
_
N
u
(
s
)
+
c
t
α
-
1
_
,
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1
,
c
∈
R
.
Considering |
u
(
t
0
)
|
≤
M, we get
(30)
|
c
t
0
α
-
1
_
|
≤
M
+
1
Γ
(
α
)
∑
s
=
0
t
0
-
α
‍
(
t
0
-
s
-
1
)
α
-
1
_
|
N
u
(
s
)
|
.
Consequently, by (29), (30), Lemma 6, (
H
3
), and the monotonicity of functions t
α
_ and t
α
-
1
_ on [
α
-
1
,
α
+
b
]
N
α
-
1, we have
(31)
|
u
(
t
)
|
≤
1
Γ
(
α
)
∑
s
=
0
t
-
α
‍
(
t
-
s
-
1
)
α
-
1
_
|
N
u
(
s
)
|
+
t
α
-
1
_
t
0
α
-
1
_
|
c
t
0
α
-
1
_
|
≤
1
Γ
(
α
)
∑
s
=
0
t
-
α
‍
(
t
-
s
-
1
)
α
-
1
_
|
N
u
(
s
)
|
+
(
α
-
1
)
α
-
1
_
(
α
+
b
)
α
-
1
_
×
{
M
+
1
Γ
(
α
)
∑
s
=
0
t
0
-
α
‍
(
t
0
-
s
-
1
)
α
-
1
_
|
N
u
(
s
)
|
}
=
1
Γ
(
α
)
∑
s
=
0
t
-
α
‍
(
t
-
s
-
1
)
α
-
1
_
|
f
(
s
+
α
-
1
,
u
(
s
+
α
-
1
)
)
|
+
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
{
M
+
1
Γ
(
α
)
∑
s
=
0
t
0
-
α
‍
(
t
0
-
s
-
1
)
α
-
1
_
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
+
α
-
1
,
u
(
s
+
α
-
1
)
)
|
{
M
+
1
Γ
(
α
)
∑
s
=
0
t
0
-
α
‍
(
t
0
-
s
-
1
)
α
-
1
_
}
≤
1
Γ
(
α
)
∑
s
=
0
t
-
α
‍
(
t
-
s
-
1
)
α
-
1
_
×
{
p
(
s
+
α
-
1
)
|
u
(
s
+
α
-
1
)
|
+
q
(
s
+
α
-
1
)
}
+
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
×
{
M
+
1
Γ
(
α
)
∑
s
=
0
t
0
-
α
‍
(
t
0
-
s
-
1
)
α
-
1
_
iiiiiiii
×
{
p
(
s
+
α
-
1
)
|
u
(
s
+
α
-
1
)
|
+
q
(
s
+
α
-
1
)
}
{
M
+
1
Γ
(
α
)
∑
s
=
0
t
0
-
α
‍
(
t
0
-
s
-
1
)
α
-
1
_
}
≤
∥
p
∥
∥
u
∥
+
∥
q
∥
Γ
(
α
)
{
∑
s
=
0
t
-
α
‍
(
t
-
s
-
1
)
α
-
1
_
+
(
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
)
i
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
×
∑
s
=
0
t
0
-
α
‍
(
t
0
-
s
-
1
)
α
-
1
_
}
+
M
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
=
∥
p
∥
∥
u
∥
+
∥
q
∥
Γ
(
α
+
1
)
{
t
α
_
+
(
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
)
t
0
α
_
}
+
M
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
≤
∥
p
∥
∥
u
∥
+
∥
q
∥
Γ
(
α
+
1
)
{
(
α
+
b
)
α
_
+
(
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
)
iiiiiiiiiiiiiiiiiiiiiiiiiiiii
×
(
α
+
b
)
α
_
{
(
α
+
b
)
α
_
+
(
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
)
}
+
M
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
≤
{
∏
i
=
1
b
‍
α
+
i
i
+
b
+
1
α
}
(
∥
p
∥
∥
u
∥
+
∥
q
∥
)
+
M
∏
i
=
1
b
+
1
‍
i
α
+
i
-
1
.
So, by the fact that {
∏
i
=
1
b
‍
(
(
α
+
i
)
/
i
)
+
(
(
b
+
1
)
/
α
)
}
∥
p
∥
<
1 in (
H
3
) and (31), we can derive that Ω
1 is bounded.
Step
2. Set Ω
2
=
{
u
∈
Ker
⁡
L
:
N
u
∈
Im
⁡
L
} and prove that Ω
2 is bounded in X.
For anyu
∈
Ω
2, there exists a constant c
∈
R such that u
(
t
)
=
c
t
α
-
1
_, t
∈
[
α
-
1
,
α
+
b
]
N
α
-
1, and N
u
∈
Im
⁡
L. So it follows from (14) that Q
1
N
u
=
0. By virtue of (
H
5
) and the fact that t
α
-
1
_ is decreasing for t on [
α
-
1
,
α
+
b
]
N
α
-
1, we can derive that |
u
(
t
)
|
≤
|
c
|
(
α
-
1
)
α
-
1
_
≤
M
*
Γ
(
α
), which implies that Ω
2 is bounded in X.
Step
3. Set Ω
3
=
{
u
∈
Ker
⁡
L
:
λ
J
u
+
(
1
-
λ
)
θ
Q
N
u
=
0
,
λ
∈
[
0,1
]
} and prove that Ω
3 is bounded in X, where J
:
Ker
⁡
L
→
Im
⁡
Q is a linear isomorphism defined by
(32)
J
(
c
t
α
-
1
_
)
=
c
,
θ
=
{
1
,
if
(
H
5
)
(
27
)
holds
,
-
1
,
if
(
H
5
)
(
28
)
holds
.
For any c
t
α
-
1
_
∈
Ω
3, there exists λ
∈
[
0,1
] such that
(33)
λ
c
=
-
(
1
-
λ
)
θ
Q
N
(
c
t
α
-
1
_
)
.
If λ
=
0, then Q
N
(
c
t
α
-
1
_
)
=
0. Hence Q
1
N
(
c
t
α
-
1
_
)
=
0. By (
H
5
), we get |
c
|
≤
M
*. If λ
=
1, then c
=
0. For λ
∈
(
0,1
), if |
c
|
>
M
*, then, from (
H
5
), we can obtain that
(34)
θ
c
Q
1
N
(
c
t
α
-
1
_
)
>
0
.
Therefore, we have
(35)
λ
c
2
=
-
(
1
-
λ
)
θ
c
Q
N
(
c
t
α
-
1
_
)
<
0
,
which is a contradiction. So, Ω
3
⊂
{
u
∈
Ker
⁡
L
:
u
=
c
t
α
-
1
_
,
|
c
|
≤
M
*
} is bounded in X.
Step
4. Let Ω be a bounded open set such that Ω
⊃
∪
i
=
1
3
Ω
¯
i and prove that
(36)
deg
⁡
(
Q
N
|
Ker
⁡
L
,
Ω
∩
Ker
⁡
L
,
0
)
≠
0
.
It follows from Lemma 10 that N is L-compact on Ω
¯. Then by Steps 1 and 2, we have (i)
L
u
≠
λ
N
u for every (
u
,
λ
)
∈
[
(
Dom
⁡
L
∖
Ker
⁡
L
)
∩
∂
Ω
]
×
(
0,1
);
(ii)
N
u
∉
Im
⁡
L for every u
∈
Ker
⁡
L
∩
∂
Ω.
At last, we prove that condition (iii) of Theorem7 is satisfied. Let
(37)
H
(
u
,
λ
)
=
λ
J
u
+
(
1
-
λ
)
θ
Q
N
u
.
According to the arguments in Step 3, we have
(38)
H
(
u
,
λ
)
≠
0
,
∀
u
∈
Ker
⁡
L
∩
∂
Ω
,
and therefore, via the homotopy property of degree, we get that
(39)
deg
⁡
(
Q
N
|
Ker
⁡
L
,
Ω
∩
Ker
⁡
L
,
0
)
=
deg
⁡
(
θ
H
(
·
,
0
)
,
Ω
∩
Ker
⁡
L
,
0
)
=
deg
⁡
(
θ
H
(
·
,
1
)
,
Ω
∩
Ker
⁡
L
,
0
)
=
deg
⁡
(
θ
J
,
Ω
∩
Ker
⁡
L
,
0
)
≠
0
,
which implies that condition (iii) of Theorem 7 is satisfied. Then by Theorem 7, we can conclude that L
u
=
N
u has at least one solution in Dom
⁡
L
∩
Ω
¯; that is, (1) has at least one solution in X. The proof is completed.


## 4. An Illustrative Example

In this section, we will illustrate the possible application of the above established analytical result with a concrete example.Example 1.
Consider the following discrete fractional boundary value problem:(40)
Δ
1
/
2
u
(
t
)
=
f
(
t
-
1
2
,
u
(
t
-
1
2
)
)
,
t
∈
[
0,3
]
N
0
,
u
(
-
1
2
)
-
16
9
u
(
3
2
)
=
128
105
u
(
7
2
)
,
where
(41)
f
(
t
,
u
)
=
{
2
-
t
2
100000
[
sin
(
t
+
u
2
)
]
2
,
(
t
,
u
)
∈
[
-
1
2
,
5
2
]
N
-
1
/
2
×
(
62
3
,
+
∞
)
,
3
31
u
-
t
2
100000
[
sin
(
t
+
u
2
)
]
2
,
(
t
,
u
)
∈
[
-
1
2
,
5
2
]
N
-
1
/
2
×
(
-
62
3
,
+
62
3
)
,
-
2
-
t
2
100000
[
sin
(
t
+
u
2
)
]
2
,
(
t
,
u
)
∈
[
-
1
2
,
5
2
]
N
-
1
/
2
×
(
-
∞
,
-
62
3
)
.
It is obvious that f is continuous. Corresponding to problem (1), there exist α
=
1
/
2, β
=
16
/
9, γ
=
128
/
105, η
=
1, b
=
3, and
(42)
β
∏
i
=
1
η
+
1
‍
α
+
i
-
1
i
+
γ
∏
i
=
1
b
+
1
‍
α
+
i
-
1
i
=
16
9
∏
i
=
1
2
‍
1
/
2
+
i
-
1
i
+
128
105
∏
i
=
1
4
‍
1
/
2
+
i
-
1
i
=
1
.
Therefore, the problem (40) is at resonance.Choosingp
(
t
)
=
3
/
31, q
(
t
)
=
t
2
/
100000, t
∈
[
-
1
/
2,7
/
2
]
N
-
1
/
2, then we have
(43)
|
f
(
t
,
u
)
|
≤
p
(
t
)
|
u
|
+
q
(
t
)
,
for
(
t
,
u
)
∈
[
-
1
2
,
5
2
]
N
-
1
/
2
×
R
,
{
∏
i
=
1
3
‍
1
/
2
+
i
i
+
3
+
1
1
/
2
}
∥
p
∥
=
163
16
×
3
31
=
489
496
<
1
.
Hence, the condition (
H
3
) in Theorem 11 holds.LetM
=
21. If |
u
(
t
)
|
>
M holds for any t
∈
[
-
1
/
2,7
/
2
]
N
-
1
/
2, then we can easily verify that
(44)
Q
1
N
u
≠
0
,
which implies that condition (
H
4
) of Theorem 11 holds.Furthermore, we can chooseM
*
=
50, to show that the condition (
H
5
) of Theorem 11 holds. In fact, for any u
(
t
)
=
c
t
-
1
/
2
_
,
t
∈
[
-
1
/
2,7
/
2
]
N
-
1
/
2 satisfying |
c
|
>
M
*, we can get that
(45)
c
N
u
(
t
)
>
0
,
for
t
∈
[
0,3
]
N
0
.
So, by the fact that (
t
-
s
-
1
/
2
)
-
1
/
2
_
>
0 for (
t
,
s
)
∈
[
0,3
]
N
0
×
[
0
,
t
]
N
0 and (45), we can derive that, for any u
(
t
)
=
c
t
α
-
1
_
,
t
∈
[
-
1
/
2,7
/
2
]
N
-
1
/
2 satisfying |
c
|
>
M
*,
(46)
c
Q
1
N
u
=
Q
1
(
c
N
u
)
>
0
,
which implies that (27) in (
H
5
) of Theorem 11 holds. Therefore, all conditions of Theorem 11 hold. Hence, we can conclude that problem (40) has at least one solution.



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*Source: 601092-2014-07-08.xml*