R E S E A R C H
Open Access
Nontrivial solutions for an integral boundary
value problem involving Riemann–Liouville
fractional derivatives
Zhengqing Fu
1, Shikun Bai
2, Donal O’Regan
3and Jiafa Xu
2**Correspondence: xujiafa292@sina.com
2School of Mathematical Sciences,
Chongqing Normal University, Chongqing, China
Full list of author information is available at the end of the article
Abstract
In this paper using topological degree we study the existence of nontrivial solutions
for a fractional differential equation involving integral boundary conditions. Here, the
nonlinear term may be sign-changing and may also depend on the derivatives of the
unknown function.
Keywords:
Fractional integral boundary value problems; Nontrivial solutions;
Topological degree
1 Introduction
We study the existence of nontrivial solutions for the following integral boundary value
problem involving Riemann–Liouville fractional derivatives:
⎧
⎪
⎪
⎨
⎪
⎪
⎩
D
β0+D
α0+
u
(
t
) =
f
(
t
,
u
(
t
),
u
(
t
), –
D
α0+u
(
t
)),
0 <
t
< 1,
u
(0) =
u
(0) = 0,
u
(1) =
01g
(
t
)
u
(
t
)
dt
,
D
α0+
u
(0) =
D
α0++1u
(0) = 0,
D
α0++1u
(1) =
10
h
(
t
)
D
α+1 0+
u
(
t
)
dt
,
(1.1)
where
α,
β
∈
(2, 3] are two real numbers,
D
α0+
,
D
β
0+
are the Riemann–Liouville fractional
derivatives, and
f
∈
C
([0, 1]
×
R
3,
R
) (
R
= (–
∞
, +
∞
)). Moreover, the functions
g
,
h
are
defined on [0, 1] and satisfy the condition:
(H0)
g
,
h
≥
0
with
01g
(
t
)
t
α–2dt
∈
[0, 1)
, and
1 0h
(
t
)
t
β–2
dt
∈
[0, 1)
.
Fractional-order problems arise naturally in engineering and scientific disciplines such
as physics, biophysics, chemistry, control theory, signal and image processing, and
aero-dynamics; we refer the reader to [
1
–
3
]. For example, in [
4
,
5
] the authors introduced a
fractional-order model of infection of CD4
+T-cells, and the system takes the following
form:
⎧
⎪
⎪
⎨
⎪
⎪
⎩
D
α1(
T
) =
s
–
KVT
–
dT
+
bI
,
D
α2(
I
) =
KVT
– (
b
+
δ)
I
,
D
α3(
I
) =
N
δ
I
–
cV
,
where
D
αiare fractional derivatives,
i
= 1, 2, 3. Many results on the existence and
multiplic-ity of solutions (or positive solutions) of nonlinear fractional differential equations can be
found for example in [
6
–
37
] and the references therein. In [
6
–
11
,
19
,
20
,
26
], the authors
used the fixed point index theory to study the existence of (positive) solutions for various
boundary value problems of fractional differential equations, for example, Bai in [
6
]
ob-tained positive solutions for the nonlocal fractional-order differential equation boundary
value problem
⎧
⎨
⎩
D
α0+
u
(
t
) +
f
(
t
,
u
(
t
)) = 0,
0 <
t
< 1,
u
(0) = 0,
β
u
(η) =
u
(1),
where 1 <
α
≤
2, 0 <
βη
α–1< 1, 0 <
η
< 1,
f
is continuous on [0, 1]
×
R
+and satisfies the
conditions
(H)
B1lim inf
u→0+
f
(
t
,
u
)
u
>
λ
1,
lim sup
u→+∞f
(
t
,
u
)
u
<
λ
1,
uniformly on
t
∈
[0, 1],
and
(H)
B2lim inf
u→+∞
f
(
t
,
u
)
u
>
λ
1,
lim sup
u→0+f
(
t
,
u
)
u
<
λ
1,
uniformly on
t
∈
[0, 1],
where
λ
1is the first eigenvalue corresponding of the relevant linear operator. These
con-ditions can also be found in some integer-order differential equations; we refer the reader
to [
38
–
52
] and the references therein. Integral boundary conditions arise in thermal
con-duction problems, semiconductor problems and hydrodynamic problems (see [
11
]) and
we refer the reader to [
10
,
11
,
18
,
19
,
26
,
28
–
30
,
33
,
34
,
39
,
43
,
50
–
52
] and the references
therein. In [
26
] the authors studied the integral boundary value problem of the nonlinear
Hadamard fractional differential equation
⎧
⎨
⎩
D
β(ϕ
p(
D
αu
(
t
))) =
f
(
t
,
u
(
t
)),
1 <
t
<
e
,
u
(1) =
u
(1) =
u
(
e
) = 0,
D
αu
(1) = 0,
ϕ
p
(
D
αu
(
e
)) =
μ
e1
ϕ
p(
D
α
u
(
t
))
dt t,
where
α
∈
(2, 3],
β
∈
(1, 2] and
D
α,
D
βare Hadamard fractional derivatives.
Many papers in the literature also considered sign-changing nonlinearity problems; see
[
4
,
5
,
7
–
9
,
13
–
19
,
22
–
30
,
34
,
35
,
38
–
52
] and the references therein. In [
15
], the authors
studied the fractional differential equation with a singular decreasing nonlinearity and a
p
-Laplacian operator:
⎧
⎨
⎩
–
D
α0+
(ϕ
p(–
D
γ0+z
))(
x
) =
f
(
x
,
z
(
x
)),
0 <
x
< 1,
z
(0) = 0,
D
γ0+z
(0) =
D
γ0+z
(1) = 0,
z
(1) =
01z
(
x
)
d
χ(
x
).
Using a double iterative technique, they showed that the above problem has a unique
pos-itive solution, and from an iterative technique, they established an appropriate sequence,
which converges uniformly to the unique positive solution.
function
u
and its integer-order, fractional-order derivatives
u
, –
D
α0+
u
, (2) the nonlinearity
can be unbounded on [0, 1]
×
R
3, which improves results on semipositone problems (i.e.,
boundedness from below); see [
5
,
21
,
42
,
43
,
45
].
2 Preliminaries
We present some definitions and notations from fractional calculus theory involving
Riemann–Liouville fractional derivatives; for details see the books [
1
–
3
].
Definition 2.1
(see [
1
–
3
]) The Riemann–Liouville fractional derivative of order
α
> 0 of
a continuous function
f
: (0, +
∞
)
→
(–
∞
, +
∞
) is given by
D
α0+f
(
t
) =
1
Γ
(
n
–
α)
d
d
t
n t0
(
t
–
s
)
n–α–1f
(
s
)
ds
,
where
n
= [α] + 1, [α] denotes the integer part of number
α, provided that the right side is
pointwise defined on (0, +
∞
).
Definition 2.2
(see [
1
–
3
]) The Riemann–Liouville fractional integral of order
α
> 0 of a
function
f
: (0, +
∞
)
→
(–
∞
, +
∞
) is given by
I
α0+
f
(
t
) =
1
Γ
(α)
t
0
(
t
–
s
)
α–1f
(
s
)
ds
,
provided that the right side is pointwise defined on (0, +
∞
).
Lemma 2.3
(see [
5
, Lemma 2.3])
Let
α
> 0,
then
,
for u
,
D
α0+
u
∈
C
(0, 1)
∩
L
(0, 1),
we have
I
0+αD
α0+u
(
t
) =
u
(
t
) +
c
1t
α–1+
c
2t
α–2+
· · ·
+
cN
t
α–N,
for some ci
∈
R
,
i
= 1, 2, . . . ,
N
,
where N is the smallest integer greater than or equal to
α.
Lemma 2.4
Suppose that
(H0)
holds
.
If let
–
D
α0+
u
:=
v
,
then the fractional boundary value
problem
⎧
⎨
⎩
–
D
α0+
u
(
t
) =
v
(
t
),
0 <
t
< 1,
u
(0) =
u
(0) = 0,
u
(1) =
01g
(
t
)
u
(
t
)
dt
,
(2.1)
can be transformed into its equivalent Hammerstein integral equation
,
which takes the
form
u
(
t
) =
10
G
1(t
,
s
)
v
(
s
)
ds
,
(2.2)
where
G
1(t
,
s
) =
G
1(t
,
s
) +
t
α–1
1 –
01g
(
t
)
t
α–2dt
1
0
G
1(t
,
s
) =
1
Γ
(α)
⎧
⎨
⎩
t
α–1(1 –
s
)
α–2– (
t
–
s
)
α–1,
0
≤
s
≤
t
≤
1,
t
α–1(1 –
s
)
α–2,
0
≤
t
≤
s
≤
1,
(2.3)
G
2(t
,
s
) =
1
Γ
(α)
⎧
⎨
⎩
t
α–2(1 –
s
)
α–2– (
t
–
s
)
α–2,
0
≤
s
≤
t
≤
1,
t
α–2(1 –
s
)
α–2,
0
≤
t
≤
s
≤
1.
Proof
From Lemma
2.3
we have
u
(
t
) = –
I
α0+
v
(
t
) +
c
1t
α–1+
c
2t
α–2+
c
3t
α–3,
for
c
i∈
R
,
i
= 1, 2, 3.
Note that
u
(0) =
u
(0) = 0, and we obtain
c
2=
c
3= 0. Then we have
u
(
t
) = –
I
α0+
v
(
t
) +
c
1t
α–1= –
t
0
(
t
–
s
)
α–1Γ
(α)
v
(
s
)
ds
+
c
1t
α–1
.
Therefore, from the condition
u
(1) =
01g
(
t
)
u
(
t
)
dt
, we have the equation
–
10
(1 –
s
)
α–2Γ
(α)
v
(
s
)
ds
+
c
1= –
10
g
(
t
)
t
0
(
t
–
s
)
α–2Γ
(α)
v
(
s
)
ds dt
+
c
1 10
g
(
t
)
t
α–2dt
.
Solving this, we have
c
1=
1
1 –
01g
(
t
)
t
α–2dt
1
0
(1 –
s
)
α–2Γ
(α)
v
(
s
)
ds
–
10
g
(
t
)
t
0
(
t
–
s
)
α–2Γ
(α)
v
(
s
)
ds dt
=
1
1 –
01g
(
t
)
t
α–2dt
1
0
(1 –
s
)
α–2Γ
(α)
v
(
s
)
ds
–
10
g
(
t
)
1s
(
t
–
s
)
α–2Γ
(α)
v
(
s
)
dt ds
.
As a result, we obtain
u
(
t
) = –
t
0
(
t
–
s
)
α–1Γ
(α)
v
(
s
)
ds
+
1
1 –
01g
(
t
)
t
α–2dt
1
0
t
α–1(1 –
s
)
α–2Γ
(α)
v
(
s
)
ds
–
t
α–1
1 –
01g
(
t
)
t
α–2dt
1
0
g
(
t
)
1s
(
t
–
s
)
α–2Γ
(α)
v
(
s
)
dt ds
= –
t
0
(
t
–
s
)
α–1Γ
(α)
v
(
s
)
ds
+
10
t
α–1(1 –
s
)
α–2Γ
(α)
v
(
s
)
ds
–
10
t
α–1(1 –
s
)
α–2Γ
(α)
v
(
s
)
ds
+
1
1 –
01g
(
t
)
t
α–2dt
1
0
t
α–1(1 –
s
)
α–2Γ
(α)
v
(
s
)
ds
–
t
α–1
1 –
01g
(
t
)
t
α–2dt
1
0
g
(
t
)
1s
(
t
–
s
)
α–2Γ
(α)
v
(
s
)
dt ds
=
10
G
1(t
,
s
)
v
(
s
)
ds
+
t
α–1
1 –
01g
(
t
)
t
α–2dt
1
0 1
0
g
(
t
)
G
2(t
,
s
)
dtv
(
s
)
ds
=
10
G
1(
t
,
s
)
v
(
s
)
ds
.
For convenience, let
G
2(t
,
s
) =
∂
∂
t
G
1(t
,
s
)
= (α
– 1)
G
2(t
,
s
) +
t
α–2
1 –
01g
(
t
)
t
α–2dt
+
10
g
(
t
)
G
2(t
,
s
)
dt
,
for
t
,
s
∈
[0, 1].
Lemma 2.5
Suppose that
(H0)
holds
.
If
α,
β
,
f are as in
(
1.1
),
then the fractional boundary
value problem
(
1.1
)
is equivalent to the Hammerstein integral equation
v
(
t
) =
10
H
1(t
,
s
)
f
s
,
10
G
1(s
,
τ
)
v
(τ
)
d
τ
,
10
G
2(s
,
τ
)
v
(τ
)
d
τ
,
v
(
s
)
ds
,
(2.4)
where
H
1(
t
,
s
) =
H
1(
t
,
s
) +
t
β–11 –
01h
(
t
)
t
β–2dt
1
0
h
(
t
)
H
2(
t
,
s
)
dt
,
H
1(t
,
s
) =
1
Γ
(β)
⎧
⎨
⎩
t
β–1(1 –
s
)
β–2– (
t
–
s
)
β–1,
0
≤
s
≤
t
≤
1,
t
β–1(1 –
s
)
β–2,
0
≤
t
≤
s
≤
1,
H
2(
t
,
s
) =
1
Γ
(β)
⎧
⎨
⎩
t
β–2(1 –
s
)
β–2– (
t
–
s
)
β–2,
0
≤
s
≤
t
≤
1,
t
β–2(1 –
s
)
β–2,
0
≤
t
≤
s
≤
1.
(2.5)
Proof
Substituting –
D
α0+
u
=
v
into (
1.1
), we have
⎧
⎨
⎩
–
D
β0+v
(
t
) =
f
(
t
,
01G
1(t
,
s
)
v
(
s
)
ds
,
01G
2(t
,
s
)
v
(
s
)
ds
,
v
(
t
)),
0 <
t
< 1,
v
(0) =
v
(0) = 0,
v
(1) =
01h
(
t
)
v
(
t
)
dt
.
(2.6)
Using
f
(
t
) to replace
f
(
t
,
·
,
·
,
·
), and from Lemma
2.3
we obtain
v
(
t
) = –
I
0+βf
(
t
) +
c
1t
β–1+
c
2t
β–2+
c
3t
β–3,
for
ci
∈
R
,
i
= 1, 2, 3.
Note that
v
(0) =
v
(0) = 0 implies
c
2=
c
3= 0. Hence,
v
(
t
) = –
t
0
(
t
–
s
)
β–1Γ
(β)
f
(
s
)
ds
+
c
1t
β–1
.
From the condition
v
(1) =
01h
(
t
)
v
(
t
)
dt
, we get
–
10
(1 –
s
)
β–2Γ
(β
)
f
(
s
)
ds
+
c
1= –
10
h
(
t
)
t
0
(
t
–
s
)
β–2Γ
(β)
f
(
s
)
ds dt
+
c
1 10
h
(
t
)
t
β–2dt
.
Solving this equation, we obtain
c
1=
1
1 –
01h
(
t
)
t
β–2dt
1
0
(1 –
s
)
β–2Γ
(β)
f
(
s
)
ds
–
1
1 –
01h
(
t
)
t
β–2dt
1
0
h
(
t
)
t
0
(
t
–
s
)
β–2=
1
1 –
01h
(
t
)
t
β–2dt
1
0
(1 –
s
)
β–2Γ
(β)
f
(
s
)
ds
–
1
1 –
01h
(
t
)
t
β–2dt
1
0
h
(
t
)
1s
(
t
–
s
)
β–2Γ
(β
)
f
(
s
)
dt ds
.
Consequently, we have
v
(
t
) = –
t
0
(
t
–
s
)
β–1Γ
(β)
f
(
s
)
ds
+
1
1 –
01h
(
t
)
t
β–2dt
1
0
t
β–1(1 –
s
)
β–2Γ
(β
)
f
(
s
)
ds
–
t
β–1
1 –
01h
(
t
)
t
β–2dt
1
0
h
(
t
)
1s
(
t
–
s
)
β–2Γ
(β)
f
(
s
)
dt ds
= –
t
0
(
t
–
s
)
β–1Γ
(β)
f
(
s
)
ds
+
10
t
β–1(1 –
s
)
β–2Γ
(β
)
f
(
s
)
ds
–
10
t
β–1(1 –
s
)
β–2Γ
(β)
f
(
s
)
ds
+
1
1 –
01h
(
t
)
t
β–2dt
1
0
t
β–1(1 –
s
)
β–2Γ
(β)
f
(
s
)
ds
–
t
β–1
1 –
01h
(
t
)
t
β–2dt
1
0
h
(
t
)
1s
(
t
–
s
)
β–2Γ
(β)
f
(
s
)
dt ds
=
10
H
1(
t
,
s
)
f
(
s
)
ds
+
t
β–11 –
01h
(
t
)
t
β–2dt
1
0 1
0
h
(
t
)
H
2(
t
,
s
)
dt
f
(
s
)
ds
=
10
H
1(t
,
s
)
f
(
s
)
ds
.
This completes the proof.
Lemma 2.6
The functions Gi
,
H
1(
i
= 1, 2)
satisfy the properties
:
(i)
Gi
(
t
,
s
),
H
1(t
,
s
)
≥
0
for
t
,
s
∈
[0, 1]
×
[0, 1]
,
(ii)
t
β–1φ
β
(
s
)
≤
H
1(t
,
s
)
≤
φ
β(
s
)
for
t
,
s
∈
[0, 1]
×
[0, 1]
,
where
φ
β(
s
) =
s(1–s)β–2
Γ(β)
+
1
0h(t)H2(t,s)dt
1–01h(t)tβ–2dt
,
s
∈
[0, 1]
.
Proof
We only prove (ii). From [
20
] we have
t
β–11
Γ
(β)
s
(1 –
s
)
β–2
≤
H
1(t
,
s
)
≤
1
Γ
(β
)
s
(1 –
s
)
β–2
,
for
t
,
s
∈
[0, 1]
×
[0, 1].
Combining this with (
2.5
), we easily obtain the inequalities in (ii). This completes the
proof.
Let
E
:=
C
[0, 1],
v
:=
max
t∈[0,1]|
v
(
t
)
|
(here
v
∈
E
),
P
:=
{
v
∈
E
:
v
(
t
)
≥
0,
∀
t
∈
[0, 1]
}
. Then
(
E
,
·
) is a real Banach space, and
P
is a cone on
E
. Now, we define an operator
A
:
E
→
E
as follows:
(
Av
)(
t
) :=
10
H
1(t
,
s
)
f
s
,
10
G
1(s
,
τ
)
v
(τ
)
d
τ
,
10
G
2(s
,
τ
)
v
(τ
)
d
τ
,
v
(
s
)
ds
,
(2.7)
for all
v
∈
E
. Moreover, we note that the continuity of
G
1,
G
2,H
1and
f
implies that
A
:
equivalent to that of fixed points of
A
, and then from (
1.1
) and (
2.6
) (–
D
α0+
u
=
v
), we see
that if there exists a
y
∈
E
such that
Ay
=
y
, then
y
is a solution for (
1.1
). Therefore, in what
follows we study the existence of fixed points of
A
. For this purpose we need to define a
linear operator
La
,b,c:
E
→
E
as follows:
(
La
,b,cv)(
t
) :=
10
Ha
,b,c(
t
,
s
)
v
(
s
)
ds
,
∀
v
∈
E
,
(2.8)
where
H
a,b,c(
t
,
s
) :=
aH
3(
t
,
s
) +
bH
2(
t
,
s
) +
cH
1(
t
,
s
),
∀
t
,
s
∈
[0, 1], with
a
,
b
,
c
≥
0 and
a
2+
b
2+
c
2= 0; here
H
2(t
,
s
) =
10
H
1(t
,
τ
)
G
2(τ,
s
)
d
τ
,
H
3(t
,
s
) =
10
H
1(t
,
τ
)
G
1(τ,
s
)
d
τ
,
∀
t
,
s
∈
[0, 1].
Moreover, we know that the continuity of
Hi
(
i
= 1, 2, 3) implies that
La
,b,cis a completely
continuous operator and
La
,b,c(
P
)
⊂
P
. Let
r
(
La
,b,c) denote the spectral radius of
La
,b,c, and
from Gelfand’s theorem we see that
r
(
La
,b,c) > 0 (the proof is standard; see [
20
, Lemma 5]).
Let
P
0=
{
v
∈
P
:
v
(
t
)
≥
t
β–1v
,
∀
t
∈
[0, 1]
}
. Then if we define an operator (
A
1v
)(
t
) =
10
H
1(t
,
s
)
v
(
s
)
ds
, where
H
1is defined by (
2.5
), and from Lemma
2.6
(ii) we have
A
1(P
)
⊂
P
0.(2.9)
Indeed, if
v
∈
P
, Lemma
2.6
(ii) implies that
1
0
t
β–1φ
β(
s
)
v
(
s
)
ds
≤
1
0
H
1(t
,
s
)
v
(
s
)
ds
≤
1
0
φ
β(
s
)
v
(
s
)
ds
,
and
(
A
1v
)(
t
)
≥
t
β–1A
1v
.
Lemma 2.7
(see [
53
, Theorem 19.3])
Let P be a reproducing cone in a real Banach space
E and let L
:
E
→
E be a compact linear operator with L
(
P
)
⊂
P
.
Let r
(
L
)
be the spectral
radius of L
.
If r
(
L
) > 0,
then there exists
ϕ
∈
P
\ {
0
}
such that L
ϕ
=
r
(
L
)ϕ.
Therefore, from Lemma
2.7
we see that there exists
ϕ
a,b,c∈
P
\ {
0
}
such that
L
a,b,cϕ
a,b,c=
r
(
L
a,b,c)ϕ
a,b,c.
(2.10)
In what follows, we prove that
ϕ
a,b,c∈
P
0.(2.11)
Indeed, from (
2.10
) we have
ϕ
a,b,c(
t
) =
1
r
(
La
,b,c)
(
La
,b,cϕ
a,b,c)(
t
) =
1
r
(
La
,b,c)
1
0
Ha
,b,c(
t
,
s
)ϕ
a,b,c(
s
)
ds
=
1
r
(
La
,b,c)
1
0
aH
3(
t
,
s
) +
bH
2(
t
,
s
) +
cH
1(
t
,
s
)
Using Lemma
2.6
(ii) and the definitions of
Hi
(
i
= 1, 2, 3), we have
ϕ
a,b,c≤
1
r
(
L
a,b,c)
1
0
a
1
0
φ
β(τ
)
G
1(τ,
s
)
d
τ
+
b
1
0
φ
β(τ
)
G
2(τ,
s
)
d
τ
+
c
φ
β(
s
)
×
ϕ
a,b,c(
s
)
ds
and
ϕ
a,b,c(
t
)
≥
1
r
(
La
,b,c)
1
0
a
1
0
t
β–1φ
β
(τ
)
G
1(τ
,
s
)
d
τ
+
b
10
t
β–1φ
β
(τ
)
G
2(τ
,
s
)
d
τ
+
ct
β–1φ
β(
s
)
ϕ
a,b,c(
s
)
ds
≥
t
β–1ϕ
a,b,c
.
Therefore, (
2.11
) is true.
Lemma 2.8
(see [
54
])
Let E be a Banach space and
Ω
a bounded open set in E
.
Suppose
that A
:
Ω
→
E is a continuous compact operator
.
If there exists u
0∈
E
\ {
0
}
such that
u
–
Au
=
μ
u
0,
∀
u
∈
∂Ω
,
μ
≥
0,
then the topological degree
deg
(
I
–
A
,
Ω, 0) = 0.
Lemma 2.9
(see [
54
])
Let E be a Banach space and
Ω
a bounded open set in E with
0
∈
Ω
.
Suppose that A
:
Ω
→
E is a continuous compact operator
.
If
Au
=
μ
u
,
∀
u
∈
∂Ω,
μ
≥
1,
then the topological degree
deg
(
I
–
A
,
Ω, 0) = 1.
3 Main results
Let
α
i,
β
i,
γ
i≥
0 (
i
= 1, 2) with
α
12+
β
12+
γ
12= 0,
α
22+
β
22+
γ
22= 0, and
r
–1(
L
αi,βi,γi) =
λ
αi,βi,γifor
i
= 1, 2. Now, we list our assumptions for
f
as follows:
(H1)
f
∈
C
([0, 1]
×
R
3,
R
)
.
(H2) There exist two nonnegative functions
b
(
t
),
c
(
t
)
∈
C
[0, 1]
with
c
(
t
)
≡
0
and a
function
K
(
x
1,
x
2,
x
3)
∈
C
[
R
3,
R
+]
such that
f
(
t
,
x
1,
x
2,
x
3)
≥
–
b
(
t
) –
c
(
t
)
K
(
x
1,
x
2,
x
3),
∀
x
i∈
R
,
t
∈
[0, 1],
i
= 1, 2, 3.
(H3)
lim
α1|x1|+β1|x2|+γ1|x3|→+∞α1|x1K|+(x1β1,|x2x2,|x3+γ)1|x3|= 0
.
(H4)
lim inf
α1|x1|+β1|x2|+γ1|x3|→+∞f(t,x1,x2,x3)
α1|x1|+β1|x2|+γ1|x3|
>
λ
α1,β1,γ1, uniformly for
t
∈
[0, 1]
.
(H5)
lim sup
α2|x1|+β2|x2|+γ2|x3|→0α |f(t,x1,x2,x3)|2|x1|+β2|x2|+γ2|x3|
<
λ
α2,β2,γ2, uniformly for
t
∈
[0, 1]
.
We now present our main result.
Proof
From (H4) there exist
ε
0> 0 and
X
0> 0 such that
f
(
t
,
x
1,x
2,x
3)≥
(λ
α1,β1,γ1+
ε
0)α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
,
∀
t
∈
[0, 1],
α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
>
X
0.(3.1)
For any given
ε
with
ε
0–
c
ε
> 0, and from (H3) there exists
X
1>
X
0such that
K
(
x
1,
x
2,
x
3)
≤
ε
α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
,
∀
α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
>
X
1.
(3.2)
It follows from (H2), (
3.1
), (
3.2
) that
f
(
t
,
x
1,x
2,x
3)≥
(λ
α1,β1,γ1+
ε
0)α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
–
b
(
t
) –
c
(
t
)
K
(
x
1,x
2,x
3)≥
(λ
α1,β1,γ1+
ε0)
α1
|
x
1|
+
β1
|
x
2|
+
γ1
|
x
3|
–
b
(
t
) –
ε
c
(
t
)
α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
≥
λ
α1,β1,γ1+
ε
0–
ε
c
(
t
)
α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
–
b
(
t
)
≥
λ
α1,β1,γ1+
ε
0–
ε
c
α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
–
b
,
∀
α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
>
X
1.(3.3)
Let
CX1
= (λ
α1,β1,γ1+
ε
0–
ε
c
)
X
1+
max
0≤t≤1,α1|x1|+β1|x2|+γ1|x3|≤X1|
f
(
t
,
x
1,x
2,x
3)|
,
K
∗=
max
α1|x1|+β1|x2|+γ1|x3|≤X1K
(
x
1,x
2,x
3). Then it easy to see thatf
(
t
,
x
1,x
2,x
3)≥
λ
α1,β1,γ1+
ε
0–
ε
c
α
1|
x
1|
+
β
1|
x
2|
+
γ
1|
x
3|
–
b
(
t
) –
CX1
,
∀
(
t
,
x
1,x
2,x
3)∈
[0, 1]
×
R
3.
(3.4)
Note that
ε
can be chosen arbitrarily small, and we let
R
>
max
(
b
+
c
K
∗+
CX1
)
01φ
h(
s
)
ds
1 –
ε
M
α1,β1,γ1c
,
(λ
α1,β1,γ1+ 2ε
0– 2
c
ε)(
b
+
c
K
∗+
CX1
)
1 0φ
h(
s
)
ds
ε
0–
c
ε
–
ε
c
M
α1,β1,γ1(λ
α1,β1,γ1+ 2ε
0– 2
c
ε)
,
where
M
α1,β1,γ1=
10
φ
h(
s
)(α
1 10
G
1(
s
,
τ
)
d
τ
+
β
1 10
G
2(
s
,
τ
)
d
τ
+
γ
1)
ds
, and
φ
h(
s
) =
(1–s)β–2Γ(β)
+
1
0h(t)H2(t,s)dt
1–01h(t)tβ–2dt
,
s
∈
[0, 1].
Now we prove that
v
–
Av
=
μϕ
α1,β1,γ1,
∀
v
∈
∂
BR
,
μ
≥
0,
(3.5)
where
ϕ
α1,β1,γ1is the positive eigenfunction of
L
α1,β1,γ1corresponding to the eigenvalue
λ
α1,β1,γ1, and then
ϕ
α1,β1,γ1=
λ
α1,β1,γ1L
α1,β1,γ1ϕ
α1,β1,γ1and
ϕ
α1,β1,γ1∈
P
0by (
2.11
).
Suppose (
3.5
) is not true. Then there exists
v
0∈
∂
BR
and
μ
0> 0 such that
Let
v
(
t
) =
10
H
1(
t
,
s
)
b
(
s
) +
c
(
s
)
K
1
0
G
1(
s
,
τ
)
v
0(τ
)
d
τ
,
10
G
2(
s
,
τ
)
v
0(τ
)
d
τ
,
v
0(
s
)
+
CX1
ds
,
∀
t
∈
[0, 1].
(3.7)
Then we have
v
(
t
)
≤
10
H
1(
t
,
s
)
b
(
s
) +
c
(
s
)
ε
α
1 1 0G
1(
s
,
τ
)
v
0(τ
)
d
τ
+
β1
10
G
2(s
,
τ
)
v
0(τ)
d
τ
+
γ1
v
0(s
)
+
K
∗+
C
X1ds
≤
1 0t
β–1φ
h(
s
)
b
(
s
) +
c
(
s
)
ε
α1
1 0G
1(s
,
τ
)
v
0(τ)
d
τ
+
β
1 10
G
2(s
,
τ
)
v
0(τ)
d
τ
+
γ
1v
0(s
)
+
K
∗+
CX1
ds
≤
t
β–11
0
φ
h(
s
)
b
+
c
ε
α
1 1 0G
1(s
,
τ
)
d
τ
+
β
1 10
G
2(s
,
τ
)
d
τ
+
γ
1v
0+
K
∗+
CX1
ds
≤
t
β–1b
+
c
K
∗+
CX1
1
0
φ
h(
s
)
ds
+
t
β–1ε
c
1
0
φ
h(
s
)
α
11
0
G
1(s
,
τ
)
d
τ
+
β
1 10
G
2(s
,
τ
)
d
τ
+
γ
1v
0ds
≤
b
+
c
K
∗+
CX1
1
0
φ
h(
s
)
ds
+
ε
c
R
1
0
φ
h(
s
)
α1
1
0
G
1(s
,
τ
)
d
τ
+
β1
10
G
2(s
,
τ
)
d
τ
+
γ1
ds
.
(3.8)
Consequently, we have
v
≤
b
+
c
K
∗+
CX1
1
0
φ
h(
s
)
ds
+
ε
c
R
1
0
φ
h(
s
)
α1
1
0
G
1(s
,
τ
)
d
τ
+
β1
10
G
2(s
,
τ
)
d
τ
+
γ1
ds
<
R
.
(3.9)
Note that
v
∈
P
0, and then from (2.9
),
ϕ
α1,β1,γ1∈
P
0, andv
0(t
) +
v
(
t
) =
Av
0(t
) +
μ
0ϕ
α1,β1,γ1(
t
) +
v
(
t
)
=
10
H
1(
t
,
s
)
f
s
,
1 0G
1(
s
,
τ
)
v
0(τ
)
d
τ
,
10
+
b
(
s
) +
c
(
s
)
K
1
0
G
1(
s
,
τ
)
v
0(τ
)
d
τ
,
10
G
2(
s
,
τ
)
v
0(τ
)
d
τ
,
v
0(
s
)
+
CX1
ds
+
μ
0ϕ
α1,β1,γ1(
t
),
we have
v
0+
v
∈
P
0,
(3.10)
using the fact that
f
s
,
10
G
1(s
,
τ
)
v
0(τ)
d
τ
,
10
G
2(s
,
τ
)
v
0(τ)
d
τ
,
v
0(s
)
+
b
(
s
)
+
c
(
s
)
K
1
0
G
1(s
,
τ
)
v
0(τ)
d
τ
,
10
G
2(s
,
τ
)
v
0(τ)
d
τ
,
v
0(s
)
+
CX1
∈
P
.
Therefore, (
2.10
), (
3.4
) and (
3.7
) enable us to obtain
Av
0(
t
) +
v
(
t
)
=
10
H
1(t
,
s
)
f
s
,
10
G
1(s
,
τ
)
v
0(τ)
d
τ
,
10
G
2(s
,
τ
)
v
0(τ)
d
τ
,
v
0(s
)
ds
+
10
H
1(t
,
s
)
b
(
s
) +
c
(
s
)
K
1
0
G
1(s
,
τ
)
v
0(τ)
d
τ
,
10
G
2(s
,
τ
)
v
0(τ)
d
τ
,
v
0(s
)
+
CX1
ds
≥
λ
α1,β1,γ1+
ε
0–
ε
c
1 0
H
1(t
,
s
)
α
1 01G
1(s
,
τ
)
v
0(τ)
d
τ
+
β
1 01G
2(s
,
τ
)
v
0(τ)
d
τ
+
γ
1v
0(s
)
ds
–
10
H
1(t
,
s
)
b
(
s
) +
CX1
ds
+
10
H
1(t
,
s
)
b
(
s
) +
c
(
s
)
K
1
0
G
1(s
,
τ
)
v
0(τ)
d
τ
,
10
G
2(s
,
τ
)
v
0(τ)
d
τ
,
v
0(s
)
+
CX1
ds
≥
λ
α1,β1,γ1+
ε
0–
ε
c
1
0
H
1(t
,
s
)
α
1 10
G
1(s
,
τ
)
v
0(τ)
d
τ
+
β
1 10
G
2(s
,
τ
)
v
0(τ)
d
τ
+
γ
1v
0(s
)
ds
≥
λ
α1,β1,γ1+
ε
0–
ε
c
α
11
0
H
1(
t
,
s
)
10
G
1(
s
,
τ
)
v
0(τ
)
d
τ
ds
+
β
1 10
H
1(
t
,
s
)
10
G
2(
s
,
τ
)
v
0(τ
)
d
τ
ds
+
γ
1 10
H
1(
t
,
s
)
v
0(
s
)
ds
≥
λ
α1,β1,γ1+
ε
0–
ε
c
α
1 10
H
3(
t
,
τ
)
v
0(τ
)
d
τ
+
β
1 10
+
γ
1 10
H
1(
t
,
s
)
v
0(
s
)
ds
=
λ
α1,β1,γ1+
ε
0–
ε
c
1
0
H
α1,β1,γ1(
t
,
s
)
v
0(
s
)
ds
≥
λ
α1,β1,γ1+
ε
0–
ε
c
1
0
H
α1,β1,γ1(
t
,
s
)
v
0(
s
)
ds
.
(3.11)
From the definition of operator
L
α1,β1,γ1, we get
λ
α1,β1,γ1+
ε
0–
c
ε
1
0
H
α1,β1,γ1(
t
,
s
)
v
0(
s
)
ds
=
λ
α1,β1,γ11
0
H
α1,β1,γ1(
t
,
s
)
v
0(s
) +
v
(
s
)
ds
+
ε
0–
c
ε
1
0
H
α1,β1,γ1(
t
,
s
)
v
0(s
)
ds
–
λ
α1,β1,γ11
0
H
α1,β1,γ1(
t
,
s
)
v
(
s
)
ds
=
λ
α1,β1,γ1L
α1,β1,γ1(
v
0+
v
)(
t
) +
ε0
–
c
ε
1
0
H
α1,β1,γ1(
t
,
s
)
v
0(s
) +
v
(
s
)
ds
–
λ
α1,β1,γ1+
ε
0–
c
ε
1
0
H
α1,β1,γ1(
t
,
s
)
v
(
s
)
ds
.
(3.12)
From (
3.10
), we get
v
0(t
) +
v
(
t
)
≥
t
β–1v
0
+
v
≥
t
β–1(
v
0–
v
),
t
∈
[0, 1], and hence from
(
3.8
) we have
ε0
–
c
ε
1
0
H
α1,β1,γ1(
t
,
s
)
v
0(s
) +
v
(
s
)
ds
–
λ
α1,β1,γ1+
ε
0–
c
ε
1
0
H
α1,β1,γ1(
t
,
s
)
v
(
s
)
ds
≥
ε
0–
c
ε
R
–
v
1
0
s
β–1H
α1,β1,γ1(
t
,
s
)
ds
–
λ
α1,β1,γ1+
ε
0–
c
ε
×
b
+
c
K
∗+
CX1
1
0
φ
h(
s
)
ds
+
ε
c
RM
α1,β1,γ11
0
s
β–1H
α1,β1,γ1(
t
,
s
)
ds
≥
0.
(3.13)
Combining (
3.11
), (
3.12
) and (
3.13
), and we obtain
Av
0(t
) +
v
(
t
)
≥
λ
α1,β1,γ1L
α1,β1,γ1(
v
0+
v
)(
t
),
t
∈
[0, 1].
(3.14)
Therefore, using (
3.6
) and (
3.14
) we have
v
0+
v
=
Av
0+
μ
0ϕ
α1,β1,γ1+
v
≥
λ
α1,β1,γ1L
α1,β1,γ1(
v
0+
v
) +
μ
0ϕ
α1,β1,γ1≥
μ
0ϕ
α1,β1,γ1.
Define
It is easy to see that
μ
∗≥
μ
0and
v
0+
v
≥
μ
∗ϕ
α1,β1,γ1. From
ϕ
α1,β1,γ1=
λ
α1,β1,γ1L
α1,β1,γ1×
ϕ
α1,β1,γ1, we obtain
λ
α1,β1,γ1L
α1,β1,γ1(
v
0+
v
)
≥
λ
α1,β1,γ1L
α1,β1,γ1μ
∗ϕ
α1,β1,γ1
=
μ
∗ϕ
α1,β1,γ1
.
Hence
v
0+
v
≥
λ
α1,β1,γ1L
α1,β1,γ1(
v
0+
v
) +
μ
0ϕ
α1,β1,γ1≥
μ
0+
μ
∗ϕ
α1,β1,γ1,
which contradicts the definition of
μ
∗. Therefore, (
3.5
) holds, and from Lemma
2.8
we
obtain
deg
(
I
–
A
,
BR
, 0) = 0.
(3.15)
From (H5) there exist 0 <
ε1
<
λ
α2,β2,γ2and 0 <
r
<
R
such that
f
(
t
,
x
1,
x
2,
x
3)
≤
(λ
α2,β2,γ2–
ε
1)
α
2|
x
1|
+
β
2|
x
2|
+
γ
2|
x
3|
,
for all
xi
∈
R
,
i
= 1, 2, 3,
t
∈
[0, 1] with 0
≤
α
2|
x
1|
+
β
2|
x
2|
+
γ
2|
x
3|
<
r
. Consequently, we
obtain
(
Av
)(
t
)
≤
(λ
α2,β2,γ2–
ε
1)
1
0
H
1(
t
,
s
)
α
2 10
G
1(
s
,
τ
)
v
(τ
)
d
τ
+
β
2 01G
2(
s
,
τ
)
v
(τ
)
d
τ
+
γ
2v
(
s
)
ds
≤
(λ
α2,β2,γ2–
ε1)
1
0
H
1(t
,
s
)
α2
10
G
1(s
,
τ
)
v
(τ
)
d
τ
+
β
2 10
G
2(s
,
τ
)
v
(τ
)
d
τ
+
γ
2v
(
s
)
ds
= (λ
α2,β2,γ2–
ε
1)1
0
H
α2,β2,γ2(
t
,
s
)
v
(
s
)
ds
= (λ
α2,β2,γ2–
ε
1)L
α2,β2,γ2|
v
|
(
t
),
∀
t
∈
[0, 1],
v
∈
E
,
v
≤
r
.
Now for this
r
, we prove that
Av
=
λ
v
,
∀
v
∈
∂
Br
,
λ
≥
1.
(3.16)
Assume the contrary. Then there exist
v
0∈
∂
Br
and
λ
0≥
1 such that
Av
0=
λ
0v
0. Letω(
t
) =
|
v
0(t
)
|
. Then
ω
∈
∂
Br
∩
P
and
ω
≤
1
λ0
(λ
α2,β2,γ2–
ε
1)L
α2,β2,γ2ω
≤
(λ
α2,β2,γ2–
ε
1)L
α2,β2,γ2ω.
By induction, we have
ω
≤
(λ
α2,β2,γ2–
ε
1)
n
L
nα2,β2,γ2
ω, for
n
= 1, 2, . . . . As a result, we have
ω
≤
(λ
α2,β2,γ2–
ε
1) nL
nand thus
1
≤
(λ
α2,β2,γ2–
ε
1)
nL
nα2,β2,γ2
.
Therefore, by Gelfand’s theorem, we have
(λ
α2,β2,γ2–
ε
1)r
(
L
α2,β2,γ2) = (λ
α2,β2,γ2–
ε
1)lim
n→∞n
L
nα2,β2,γ2≥
1.
This contradicts
(λ
α2,β2,γ2–
ε
1)r
(
L
α2,β2,γ2) = 1 –
ε
1r
(
L
α2,β2,γ2) < 1.
Thus (
3.16
) holds and from Lemma
2.9
we have
deg
(
I
–
A
,
Br
, 0) = 1.
(3.17)
Now (
3.15
) and (
3.17
) imply that
deg
(
I
–
A
,
BR
\
Br
, 0) =
deg
(
I
–
A
,
BR
, 0) –
deg
(
I
–
A
,
Br
, 0) = –1.
Therefore the operator
A
has at least one fixed point in
BR
\
Br
. Equivalently, (
1.1
) has at
least one nontrivial solution. This completes the proof.
Acknowledgements Not applicable.
Funding
Research supported by the National Natural Science Foundation of China (Grant No. 11601048), Natural Science Foundation of Chongqing (Grant No. cstc2016jcyjA0181), the Science and Technology Research Program of Chongqing Municipal Education Commission (Grant No. KJQN201800533), Natural Science Foundation of Chongqing Normal University (Grant No. 16XYY24).
Availability of data and materials Not applicable.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors conceived of the study, drafted the manuscript, and approved the final manuscript.
Author details
1College of Mathematics and System Sciences, Shandong University of Science and Technology, Qingdao, China.2School
of Mathematical Sciences, Chongqing Normal University, Chongqing, China. 3School of Mathematics, Statistics and
Applied Mathematics, National University of Ireland, Galway, Ireland.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Received: 8 February 2019 Accepted: 8 April 2019
References
1. Kilbas, A., Srivastava, H., Trujillo, J.: Theory and Applications of Fractional Differential Equations. North-Holland Mathematics Studies, vol. 204. Elsevier, Amsterdam (2006)
2. Podlubny, I.: Fractional Differential Equations. Mathematics in Science and Engineering, vol. 198. Academic Press, San Diego (1999)
4. Jiang, J., Liu, L., Wu, Y.: Positive solutions to singular fractional differential system with coupled boundary conditions. Commun. Nonlinear Sci. Numer. Simul.18(11), 3061–3074 (2013)
5. Wang, Y., Liu, L., Zhang, X., Wu, Y.: Positive solutions of an abstract fractional semipositone differential system model for bioprocesses of HIV infection. Appl. Math. Comput.258, 312–324 (2015)
6. Bai, Z.: On positive solutions of a nonlocal fractional boundary value problem. Nonlinear Anal.72, 916–924 (2010) 7. Wang, Y., Liu, L., Wu, Y.: Positive solutions for a nonlocal fractional differential equation. Nonlinear Anal.74(11),
3599–3605 (2011)
8. Cui, Y.: Uniqueness of solution for boundary value problems for fractional differential equations. Appl. Math. Lett.51, 48–54 (2016)
9. Wang, Y., Liu, L.: Positive solutions for a class of fractional infinite-point boundary value problems. Bound. Value Probl.
2018, Article ID 118 (2018)
10. Liu, X., Liu, L., Wu, Y.: Existence of positive solutions for a singular nonlinear fractional differential equation with integral boundary conditions involving fractional derivatives. Bound. Value Probl.2018, Article ID 24 (2018) 11. Jiang, J., Liu, W., Wang, H.: Positive solutions for higher order nonlocal fractional differential equation with integral
boundary conditions. J. Funct. Spaces2018, Article ID 6598351 (2018)
12. Hao, X., Wang, H., Liu, L., Cui, Y.: Positive solutions for a system of nonlinear fractional nonlocal boundary value problems with parameters andp-Laplacian operator. Bound. Value Probl.2017, Article ID 182 (2017)
13. Zuo, M., Hao, X., Liu, L., Cui, Y.: Existence results for impulsive fractional integro-differential equation of mixed type with constant coefficient and antiperiodic boundary conditions. Bound. Value Probl.2017, Article ID 161 (2017) 14. Zhang, X., Liu, L., Zou, Y.: Fixed-point theorems for systems of operator equations and their applications to the
fractional differential equations. J. Funct. Spaces2018, Article ID 7469868 (2018)
15. Wu, J., Zhang, X., Liu, L., Wu, Y., Cui, Y.: The convergence analysis and error estimation for unique solution of a
p-Laplacian fractional differential equation with singular decreasing nonlinearity. Bound. Value Probl.2018, Article ID 82 (2018)
16. Zhang, X., Wu, J., Liu, L., Wu, Y., Cui, Y.: Convergence analysis of iterative scheme and error estimation of positive solution for a fractional differential equation. Math. Model. Anal.23(4), 611–626 (2018)
17. He, J., Zhang, X., Liu, L., Wu, Y., Cui, Y.: Existence and asymptotic analysis of positive solutions for a singular fractional differential equation with nonlocal boundary conditions. Bound. Value Probl.2018, Article ID 189 (2018)
18. Zhang, X., Liu, L., Wu, Y., Zou, Y.: Existence and uniqueness of solutions for systems of fractional differential equations with Riemann–Stieltjes integral boundary condition. Adv. Differ. Equ.2018, Article ID 204 (2018)
19. Liu, B., Li, J., Liu, L., Wang, Y.: Existence and uniqueness of nontrivial solutions to a system of fractional differential equations with Riemann–Stieltjes integral conditions. Adv. Differ. Equ.2018, Article ID 306 (2018)
20. Xu, J., Wei, Z.: Positive solutions for a class of fractional boundary value problems. Nonlinear Anal., Model. Control
21(1), 1–17 (2016)
21. Pu, R., Zhang, X., Cui, Y., Li, P., Wang, W.: Positive solutions for singular semipositone fractional differential equation subject to multipoint boundary conditions. J. Funct. Spaces2017, Article ID 5892616 (2017)
22. Guo, Y.: Nontrivial solutions for boundary-value problems of nonlinear fractional differential equations. Bull. Korean Math. Soc.47(1), 81–87 (2010)
23. Zhang, K.: On a sign-changing solution for some fractional differential equations. Bound. Value Probl.2017, Article ID 59 (2017)
24. Cui, Y., Ma, W., Sun, Q., Su, X.: New uniqueness results for boundary value problem of fractional differential equation. Nonlinear Anal., Model. Control23(1), 31–39 (2018)
25. Zou, Y., He, G.: On the uniqueness of solutions for a class of fractional differential equations. Appl. Math. Lett.74, 68–73 (2017)
26. Zhang, K., Wang, J., Ma, W.: Solutions for integral boundary value problems of nonlinear Hadamard fractional differential equations. J. Funct. Spaces2018, Article ID 2193234 (2018)
27. Zhang, Y.: Existence results for a coupled system of nonlinear fractional multi-point boundary value problems at resonance. J. Inequal. Appl.2018, Article ID 198 (2018)
28. Ma, W., Meng, S., Cui, Y.: Resonant integral boundary value problems for Caputo fractional differential equations. Math. Probl. Eng.2018, Article ID 5438592 (2018)
29. Ma, W., Cui, Y.: The eigenvalue problem for Caputo type fractional differential equation with Riemann–Stieltjes integral boundary conditions. J. Funct. Spaces2018, Article ID 2176809 (2018)
30. Sun, Q., Meng, S., Cui, Y.: Existence results for fractional order differential equation with nonlocal Erdélyi–Kober and generalized Riemann–Liouville type integral boundary conditions at resonance. Adv. Differ. Equ.2018, Article ID 243 (2018)
31. Zhong, Q., Zhang, X., Lu, X., Fu, Z.: Uniqueness of successive positive solution for nonlocal singular higher-order fractional differential equations involving arbitrary derivatives. J. Funct. Spaces2018, Article ID 6207682 (2018) 32. Dong, X., Bai, Z., Zhang, S.: Positive solutions to boundary value problems ofp-Laplacian with fractional derivative.
Bound. Value Probl.2017, Article ID 5 (2017)
33. Song, Q., Bai, Z.: Positive solutions of fractional differential equations involving the Riemann–Stieltjes integral boundary condition. Adv. Differ. Equ.2018, Article ID 183 (2018)
34. Zhang, X.
