# Existence and Nonexistence of Positive Solutions for Coupled Riemann-Liouville Fractional Boundary Value Problems

Authors: Johnny Henderson; Rodica Luca; Alexandru Tudorache
Journal: Discrete Dynamics in Nature and Society
 (2016)

---

## Abstract

We investigate the existence and nonexistence of positive solutions for a system of nonlinear Riemann-Liouville fractional differential equations with two parameters, subject to coupled integral boundary conditions.


---

## Body

## 1. Introduction

Fractional differential equations describe many phenomena in various fields of engineering and scientific disciplines such as physics, biophysics, chemistry, biology, economics, control theory, signal and image processing, aerodynamics, viscoelasticity, and electromagnetics (see [1–5]). Integral boundary conditions arise in thermal conduction problems, semiconductor problems, and hydrodynamic problems (see, e.g., [6–9]).We consider the system of nonlinear fractional differential equations with parametersS
D
0
+
α
u
t
+
λ
f
t
,
u
t
,
v
t
=
0
,
t
∈
0,1
,
D
0
+
β
v
t
+
μ
g
t
,
u
t
,
v
t
=
0
,
t
∈
0,1
,with the coupled integral boundary conditionsB
C
u
0
=
u
′
0
=
⋯
=
u
n
-
2
0
=
0
,
u
′
1
=
∫
0
1
v
s
d
H
s
,
v
0
=
v
′
0
=
⋯
=
v
m
-
2
0
=
0
,
v
′
1
=
∫
0
1
u
s
d
K
s
,where α
∈
(
n
-
1
,
n
], β
∈
(
m
-
1
,
m
], n
,
m
∈
N, n
,
m
≥
3, D
0
+
α and D
0
+
β denote the Riemann-Liouville derivatives of orders α and β, respectively, and the integrals from B
C are Riemann-Stieltjes integrals. The boundary conditions B
C include multipoint and integral boundary conditions, as well as the sum of these in a single framework.Under some assumptions on the nonnegative functionsf and g, we present intervals for the parameters λ and μ such that positive solutions of S-B
C exist. By a positive solution of problem S-B
C we mean a pair of functions (
u
,
v
)
∈
C
(
0,1
;
R
+
)
×
C
(
0,1
;
R
+
) satisfying S and B
C with u
(
t
)
>
0 for all t
∈
(
0,1
] or v
(
t
)
>
0 for all t
∈
(
0,1
]. The nonexistence of positive solutions for S-B
C is also investigated. The existence, multiplicity, and nonexistence of positive solutions (u
(
t
)
,
v
(
t
)
≥
0 for all t
∈
[
0,1
] and (
u
,
v
)
≠
(
0,0
)) for system S with different coupled boundary conditions, namely,B
C
1
u
0
=
u
′
0
=
⋯
=
u
n
-
2
0
=
0
,
u
1
=
∫
0
1
v
s
d
H
s
,
v
0
=
v
′
0
=
⋯
=
v
m
-
2
0
=
0
,
v
1
=
∫
0
1
u
s
d
K
s
,were investigated in [10, 11] (where f and g are nonnegative and nonsingular functions) and in [12] (where λ
=
μ
=
1 and f
(
t
,
u
,
v
) and g
(
t
,
u
,
v
) are replaced by f
~
(
t
,
v
) and g
~
(
t
,
u
), respectively, with f
~ and g
~ nonnegative functions, singular or not). In this paper, Green’s functions associated with problem S-B
C, the inequalities satisfied by these functions, and the cone defined in the proof of the main results are different than the corresponding ones that the authors used in [10–12] for problem S-B
C
1. Existence results for the positive solutions of problem S-B
C, where f and g are sign-changing functions which may be singular at t
=
0 or t
=
1 and satisfy some different assumptions than those used in this paper, were obtained in [13]. We also mention the paper [14], where the authors studied the existence and multiplicity of positive solutions for system S with α
=
β, λ
=
μ and the boundary conditions u
(
i
)
(
0
)
=
v
(
i
)
(
0
)
=
0, i
=
0
,
…
,
n
-
2, u
(
1
)
=
a
v
(
ξ
), and v
(
1
)
=
b
u
(
η
), with ξ
,
η
∈
(
0,1
), 0
<
a
b
ξ
η
<
1, and f and g are sign-changing nonsingular or singular functions. The results obtained in [14] are relying on a nonlinear alternative of Leray-Schauder type and the Krasnosel’skii’s fixed point theorem. For other recent results concerning the coupled fractional boundary value problems we refer the reader to [15–17].The paper is organized as follows. Section2 contains some auxiliary results which investigate a nonlocal boundary value problem for fractional differential equations and presents the properties of Green’s functions associated to our problem S-B
C. In Section 3, we prove the main existence theorems for the positive solutions with respect to a cone for S-B
C which are based on the Guo-Krasnosel’skii fixed point theorem, and then the nonexistence of positive solutions is studied in Section 4. Finally, in Section 5, two examples are given to illustrate our main results.


## 2. Auxiliary Results

In this section, we present some auxiliary results that will be used to prove our main results.We consider the fractional differential system(1)
D
0
+
α
u
t
+
x
t
=
0
,
t
∈
0,1
,
D
0
+
β
v
t
+
y
t
=
0
,
t
∈
0,1
,with the coupled integral boundary conditions(2)
u
0
=
u
′
0
=
⋯
=
u
n
-
2
0
=
0
,
u
′
1
=
∫
0
1
v
s
d
H
s
,
v
0
=
v
′
0
=
⋯
=
v
m
-
2
0
=
0
,
v
′
1
=
∫
0
1
u
s
d
K
s
,where α
∈
(
n
-
1
,
n
], β
∈
(
m
-
1
,
m
], n
,
m
∈
N, n
,
m
≥
3, and H
,
K
:
[
0,1
]
→
R are functions of bounded variation.Lemma 1 (see [13]).
IfH
,
K
:
[
0,1
]
→
R are functions of bounded variation, Δ
=
(
α
-
1
)
(
β
-
1
)
-
∫
0
1
‍
τ
α
-
1
d
K
(
τ
)
∫
0
1
‍
τ
β
-
1
d
H
(
τ
)
≠
0 and x
,
y
∈
C
(
0,1
)
∩
L
1
(
0,1
), then the pair of functions (
u
,
v
)
∈
C
(
[
0,1
]
)
×
C
(
[
0,1
]
) given by(3)
u
t
=
∫
0
1
G
1
t
,
s
x
s
d
s
+
∫
0
1
G
2
t
,
s
y
s
d
s
,
t
∈
0,1
,
v
t
=
∫
0
1
G
3
t
,
s
y
s
d
s
+
∫
0
1
G
4
t
,
s
x
s
d
s
,
t
∈
0,1
,is solution of problem (1)-(2), where(4)
G
1
t
,
s
=
g
1
t
,
s
+
t
α
-
1
Δ
∫
0
1
τ
β
-
1
d
H
τ
∫
0
1
g
1
τ
,
s
d
K
τ
,
∀
t
,
s
∈
0,1
,
G
2
t
,
s
=
β
-
1
t
α
-
1
Δ
∫
0
1
g
2
τ
,
s
d
H
τ
,
∀
t
,
s
∈
0,1
,
G
3
t
,
s
=
g
2
t
,
s
+
t
β
-
1
Δ
∫
0
1
τ
α
-
1
d
K
τ
∫
0
1
g
2
τ
,
s
d
H
τ
,
∀
t
,
s
∈
0,1
,
G
4
t
,
s
=
α
-
1
t
β
-
1
Δ
∫
0
1
g
1
τ
,
s
d
K
τ
,
∀
t
,
s
∈
0,1
,
(5)
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
,
g
2
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
.Lemma 2 (see [13]).
The functionsg
1, g
2 given by (5) have the following properties:
(a)g
1
,
g
2
:
[
0,1
]
×
[
0,1
]
→
R
+ are continuous functions, and g
1
(
t
,
s
)
>
0, g
2
(
t
,
s
)
>
0 for all (
t
,
s
)
∈
(
0,1
]
×
(
0,1
).
(b)g
1
(
t
,
s
)
≤
h
1
(
s
), g
2
(
t
,
s
)
≤
h
2
(
s
) for all (
t
,
s
)
∈
[
0,1
]
×
[
0,1
], where h
1
(
s
)
=
s
(
1
-
s
)
α
-
2
/
Γ
(
α
) and h
2
(
s
)
=
s
(
1
-
s
)
β
-
2
/
Γ
(
β
) for all s
∈
[
0,1
].
(c)g
1
(
t
,
s
)
≥
t
α
-
1
h
1
(
s
), g
2
(
t
,
s
)
≥
t
β
-
1
h
2
(
s
) for all (
t
,
s
)
∈
[
0,1
]
×
[
0,1
].Lemma 3 (see [13]).
IfH
,
K
:
[
0,1
]
→
R are nondecreasing functions, and Δ
>
0, then G
i
,
(
i
=
1
,
…
,
4
) given by (4) are continuous functions on [
0,1
]
×
[
0,1
] and satisfy G
i
(
t
,
s
)
≥
0 for all (
t
,
s
)
∈
[
0,1
]
×
[
0,1
], (i
=
1
,
…
,
4). Moreover, if x
,
y
∈
C
(
0,1
)
∩
L
1
(
0,1
) satisfy x
(
t
)
≥
0, y
(
t
)
≥
0 for all t
∈
(
0,1
), then the solution (
u
,
v
) of problem (1)-(2) given by (3) satisfies u
(
t
)
≥
0
,
v
(
t
)
≥
0 for all t
∈
[
0,1
].Lemma 4.
Assume thatH
,
K
:
[
0,1
]
→
R are nondecreasing functions and Δ
>
0. Then the functions G
i
,
(
i
=
1
,
…
,
4
) satisfy for all (
t
,
s
)
∈
[
0,1
]
×
[
0,1
] the following relations:(
a
1
)
G
1
(
t
,
s
)
≤
J
1
(
s
), where J
1
s
=
h
1
s
+
1
/
Δ
∫
0
1
‍
τ
β
-
1
d
H
τ
∫
0
1
‍
g
1
τ
,
s
d
K
τ.
(
a
2
)
G
1
(
t
,
s
)
≥
t
α
-
1
J
1
(
s
).
(
b
1
)
G
2
(
t
,
s
)
≤
J
2
(
s
), where J
2
s
=
β
-
1
/
Δ
∫
0
1
‍
g
2
τ
,
s
d
H
τ.
(
b
2
)
G
2
(
t
,
s
)
=
t
α
-
1
J
2
(
s
).
(
c
1
)
G
3
(
t
,
s
)
≤
J
3
(
s
), where J
3
s
=
h
2
s
+
1
/
Δ
∫
0
1
‍
τ
α
-
1
d
K
τ
∫
0
1
‍
g
2
τ
,
s
d
H
τ.
(
c
2
)
G
3
(
t
,
s
)
≥
t
β
-
1
J
3
(
s
).
(
d
1
)
G
4
(
t
,
s
)
≤
J
4
(
s
), where J
4
s
=
α
-
1
/
Δ
∫
0
1
‍
g
1
τ
,
s
d
K
τ.
(
d
2
)
G
4
(
t
,
s
)
=
t
β
-
1
J
4
(
s
).Proof.
The above inequalities follow from the properties of the functionsg
i, (i
=
1
,
…
,
4) from Lemma 2.Lemma 5.
Assume thatH
,
K
:
[
0,1
]
→
R are nondecreasing functions, Δ
>
0, and x
,
y
∈
C
(
0,1
)
∩
L
1
(
0,1
), x
(
t
)
≥
0, and y
(
t
)
≥
0 for all t
∈
(
0,1
). Then the solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] of problem (1)-(2) given by (3) satisfies the inequalities u
(
t
)
≥
t
α
-
1
u
(
t
′
), v
(
t
)
≥
t
β
-
1
v
(
t
′
), for all t
,
t
′
∈
[
0,1
].Proof.
By using Lemma4, we obtain for all t
,
t
′
∈
[
0,1
] the following inequalities:(6)
u
t
=
∫
0
1
G
1
t
,
s
x
s
d
s
+
∫
0
1
G
2
t
,
s
y
s
d
s
≥
t
α
-
1
∫
0
1
J
1
s
x
s
d
s
+
∫
0
1
J
2
s
y
s
d
s
≥
t
α
-
1
∫
0
1
G
1
t
′
,
s
x
s
d
s
+
∫
0
1
G
2
t
′
,
s
y
s
d
s
=
t
α
-
1
u
t
′
,
v
t
=
∫
0
1
G
3
t
,
s
y
s
d
s
+
∫
0
1
G
4
t
,
s
x
s
d
s
≥
t
β
-
1
∫
0
1
J
3
s
y
s
d
s
+
∫
0
1
J
4
s
x
s
d
s
≥
t
β
-
1
∫
0
1
G
3
t
′
,
s
y
s
d
s
+
∫
0
1
G
4
t
′
,
s
x
s
d
s
=
t
β
-
1
v
t
′
.In the proof of our main existence results we will use the Guo-Krasnosel’skii fixed point theorem presented below (see [18]).Theorem 6.
LetX be a Banach space and let C
⊂
X be a cone in X. Assume Ω
1 and Ω
2 are bounded open subsets of X with 0
∈
Ω
1
⊂
Ω
1
¯
⊂
Ω
2 and let A
:
C
∩
(
Ω
2
¯
∖
Ω
1
)
→
C be a completely continuous operator such that either(i)
A
u
≤
u, u
∈
C
∩
∂
Ω
1, and A
u
≥
u, u
∈
C
∩
∂
Ω
2, or
(ii)
A
u
≥
u, u
∈
C
∩
∂
Ω
1, and A
u
≤
u, u
∈
C
∩
∂
Ω
2.Then A has a fixed point in C
∩
(
Ω
2
¯
∖
Ω
1
).


## 3. Existence Results for the Positive Solutions

In this section, we will give sufficient conditions onλ
,
μ
,
f, and g such that positive solutions with respect to a cone for our problem S-B
C exist.We present now the assumptions that we will use in the sequel:(H1)
H
,
K
:
[
0,1
]
→
R are nondecreasing functions and Δ
=
(
α
-
1
)
(
β
-
1
)
-
∫
0
1
‍
τ
α
-
1
d
K
(
τ
)
×
∫
0
1
‍
τ
β
-
1
d
H
(
τ
)
>
0.
(H2)
The functionsf
,
g
:
[
0,1
]
×
[
0
,
∞
)
×
[
0
,
∞
)
→
[
0
,
∞
) are continuous.Forc
∈
(
0,1
), we introduce the following extreme limits:(7)
f
0
s
=
limsup
u
+
v
→
0
+
⁡
max
t
∈
0,1
⁡
f
t
,
u
,
v
u
+
v
,
g
0
s
=
limsup
u
+
v
→
0
+
⁡
max
t
∈
0,1
⁡
g
t
,
u
,
v
u
+
v
,
f
0
i
=
liminf
u
+
v
→
0
+
⁡
min
t
∈
c
,
1
⁡
f
t
,
u
,
v
u
+
v
,
g
0
i
=
liminf
u
+
v
→
0
+
⁡
min
t
∈
c
,
1
⁡
g
t
,
u
,
v
u
+
v
,
f
∞
s
=
limsup
u
+
v
→
∞
⁡
max
t
∈
0,1
⁡
f
t
,
u
,
v
u
+
v
,
g
∞
s
=
limsup
u
+
v
→
∞
⁡
max
t
∈
0,1
⁡
g
t
,
u
,
v
u
+
v
,
f
∞
i
=
liminf
u
+
v
→
∞
⁡
min
t
∈
c
,
1
⁡
f
t
,
u
,
v
u
+
v
,
g
∞
i
=
liminf
u
+
v
→
∞
⁡
min
t
∈
c
,
1
⁡
g
t
,
u
,
v
u
+
v
.In the definitions of the extreme limits above, the variables u and v are nonnegative.By using Lemma1, a solution of the following nonlinear system of integral equations(8)
u
t
=
λ
∫
0
1
G
1
t
,
s
f
s
,
u
s
,
v
s
d
s
+
μ
∫
0
1
G
2
t
,
s
g
s
,
u
s
,
v
s
d
s
,
t
∈
0,1
,
v
t
=
μ
∫
0
1
G
3
t
,
s
g
s
,
u
s
,
v
s
d
s
+
λ
∫
0
1
G
4
t
,
s
f
s
,
u
s
,
v
s
d
s
,
t
∈
0,1
,is a solution for problem S-B
C.We consider the Banach spaceX
=
C
(
[
0,1
]
) with the supremum norm · and the Banach space Y
=
X
×
X with the norm u
,
v
Y
=
u
+
v. We define the cone P
⊂
Y by(9)
P
=
u
,
v
∈
Y
;
u
t
≥
t
α
-
1
u
,
v
t
≥
t
β
-
1
v
,
∀
t
∈
0,1
.Forλ
,
μ
>
0, we introduce the operators T
1
,
T
2
:
Y
→
X and T
:
Y
→
Y defined by(10)
T
1
u
,
v
t
=
λ
∫
0
1
G
1
t
,
s
f
s
,
u
s
,
v
s
d
s
+
μ
∫
0
1
G
2
t
,
s
g
s
,
u
s
,
v
s
d
s
,
0
≤
t
≤
1
,
T
2
u
,
v
t
=
μ
∫
0
1
G
3
t
,
s
g
s
,
u
s
,
v
s
d
s
+
λ
∫
0
1
G
4
t
,
s
f
s
,
u
s
,
v
s
d
s
,
0
≤
t
≤
1
,and T
(
u
,
v
)
=
(
T
1
(
u
,
v
)
,
T
2
(
u
,
v
)
), (
u
,
v
)
∈
Y. It is clear that if (
u
,
v
) is a fixed point of operator T, then (
u
,
v
) is a positive solution of problem S-B
C.Lemma 7.
If(
H
1
)-(
H
2
) hold, then T
:
P
→
P is a completely continuous operator.Proof.
Let(
u
,
v
)
∈
P be an arbitrary element. Because T
1
(
u
,
v
) and T
2
(
u
,
v
) satisfy problem (1)-(2) for x
(
t
)
=
λ
f
(
t
,
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
], and y
(
t
)
=
μ
g
(
t
,
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
], then by Lemma 5, we obtain(11)
T
1
u
,
v
t
≥
t
α
-
1
T
1
u
,
v
t
′
,
T
2
u
,
v
t
≥
t
β
-
1
T
2
u
,
v
t
′
,
∀
t
,
t
′
∈
0,1
,and so(12)
T
1
u
,
v
t
≥
t
α
-
1
T
1
u
,
v
,
T
2
u
,
v
t
≥
t
β
-
1
T
2
u
,
v
,
∀
t
∈
0,1
.
By(
H
2
) and the above inequalities, we deduce that T
(
u
,
v
)
∈
P. Hence, we get T
(
P
)
⊂
P. By using standard arguments, we can easily show that T
1 and T
2 are completely continuous, and then T is a completely continuous operator.Forc
∈
(
0,1
), we denote A
=
∫
0
1
‍
J
1
(
s
)
d
s, B
=
∫
0
1
‍
J
2
(
s
)
d
s, C
=
∫
0
1
‍
J
3
(
s
)
d
s, D
=
∫
0
1
‍
J
4
(
s
)
d
s, A
~
=
∫
c
1
‍
J
1
(
s
)
d
s, B
~
=
∫
c
1
‍
J
2
(
s
)
d
s, C
~
=
∫
c
1
‍
J
3
(
s
)
d
s, and D
~
=
∫
c
1
‍
J
4
(
s
)
d
s, where J
i
,
i
=
1
,
…
,
4, are defined in Section 2 (Lemma 4).Forf
0
s
,
g
0
s
,
f
∞
i
,
g
∞
i
∈
(
0
,
∞
) and numbers α
1
,
α
2
∈
[
0,1
], α
3
,
α
4
∈
(
0,1
), a
∈
[
0,1
], and b
∈
(
0,1
), we define the numbers(13)
L
1
=
max
⁡
a
α
1
γ
f
∞
i
A
~
,
1
-
a
α
2
γ
f
∞
i
D
~
,
L
2
=
min
⁡
b
α
3
f
0
s
A
,
1
-
b
α
4
f
0
s
D
,
L
3
=
max
⁡
a
1
-
α
1
γ
g
∞
i
B
~
,
1
-
a
1
-
α
2
γ
g
∞
i
C
~
,
L
4
=
min
⁡
b
1
-
α
3
g
0
s
B
,
1
-
b
1
-
α
4
g
0
s
C
,
L
2
′
=
min
⁡
b
f
0
s
A
,
1
-
b
f
0
s
D
,
L
4
′
=
min
⁡
b
g
0
s
B
,
1
-
b
g
0
s
C
,where γ
=
m
i
n
{
c
α
-
1
,
c
β
-
1
}.Theorem 8.
Assume that(
H
1
) and (
H
2
) hold, c
∈
(
0,1
), α
1
,
α
2
∈
[
0,1
], α
3
,
α
4
∈
(
0,1
), and a
∈
[
0,1
] and b
∈
(
0,1
).
(1) Iff
0
s
,
g
0
s
,
f
∞
i
,
g
∞
i
∈
(
0
,
∞
), L
1
<
L
2, and L
3
<
L
4, then, for each λ
∈
(
L
1
,
L
2
) and μ
∈
(
L
3
,
L
4
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(2) Iff
0
s
=
0, g
0
s
,
f
∞
i
,
g
∞
i
∈
(
0
,
∞
) and L
3
<
L
4
′, then, for each λ
∈
(
L
1
,
∞
) and μ
∈
(
L
3
,
L
4
′
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(3) Ifg
0
s
=
0, f
0
s
,
f
∞
i
,
g
∞
i
∈
(
0
,
∞
), and L
1
<
L
2
′, then, for each λ
∈
(
L
1
,
L
2
′
) and μ
∈
(
L
3
,
∞
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(4) Iff
0
s
=
g
0
s
=
0, f
∞
i
,
g
∞
i
∈
(
0
,
∞
), then, for each λ
∈
(
L
1
,
∞
) and μ
∈
(
L
3
,
∞
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(5) If{
f
0
s
,
g
0
s
,
f
∞
i
∈
(
0
,
∞
)
,
g
∞
i
=
∞
} or {
f
0
s
,
g
0
s
,
g
∞
i
∈
(
0
,
∞
)
,
f
∞
i
=
∞
} or {
f
0
s
,
g
0
s
∈
(
0
,
∞
)
,
f
∞
i
=
g
∞
i
=
∞
}, then, for each λ
∈
(
0
,
L
2
) and μ
∈
(
0
,
L
4
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(6) If{
f
0
s
=
0
,
g
0
s
,
f
∞
i
∈
(
0
,
∞
)
,
g
∞
i
=
∞
} or {
f
0
s
=
0
,
f
∞
i
=
∞
,
g
0
s
,
g
∞
i
∈
(
0
,
∞
)
} or {
f
0
s
=
0
,
g
0
s
∈
(
0
,
∞
)
,
f
∞
i
=
g
∞
i
=
∞
}, then, for each λ
∈
(
0
,
∞
) and μ
∈
(
0
,
L
4
′
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(7) If{
f
0
s
,
f
∞
i
∈
(
0
,
∞
)
,
g
0
s
=
0
,
g
∞
i
=
∞
} or {
f
0
s
,
g
∞
i
∈
(
0
,
∞
)
,
g
0
s
=
0
,
f
∞
i
=
∞
}, or {
f
0
s
∈
(
0
,
∞
)
,
g
0
s
=
0
,
f
∞
i
=
g
∞
i
=
∞
}, then, for each λ
∈
(
0
,
L
2
′
) and μ
∈
(
0
,
∞
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(8) If{
f
0
s
=
g
0
s
=
0
,
f
∞
i
∈
(
0
,
∞
)
,
g
∞
i
=
∞
} or {
f
0
s
=
g
0
s
=
0
,
f
∞
i
=
∞
,
g
∞
i
∈
(
0
,
∞
)
} or {
f
0
s
=
g
0
s
=
0
,
f
∞
i
=
g
∞
i
=
∞
}, then, for each λ
∈
(
0
,
∞
) and μ
∈
(
0
,
∞
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.Proof.
We consider the above coneP
⊂
Y and the operators T
1, T
2, and T. Because the proofs of the above cases are similar, in what follows we will prove one of them, namely, Case (
2
). So, we suppose f
0
s
=
0 and g
0
s
,
f
∞
i
,
g
∞
i
∈
(
0
,
∞
). Let λ
∈
(
L
1
,
∞
) and μ
∈
(
L
3
,
L
4
′
). We choose α
~
3
∈
0,1
-
μ
g
0
s
B
/
b and α
~
4
∈
0,1
-
μ
g
0
s
C
/
1
-
b. Let ε
>
0 be a positive number such that ε
<
f
∞
i, ε
<
g
∞
i and(14)
a
α
1
γ
f
∞
i
-
ε
A
~
≤
λ
,
a
1
-
α
1
γ
g
∞
i
-
ε
B
~
≤
μ
,
1
-
a
α
2
γ
f
∞
i
-
ε
D
~
≤
λ
,
1
-
a
1
-
α
2
γ
g
∞
i
-
ε
C
~
≤
μ
,
b
α
~
3
ε
A
≥
λ
,
b
1
-
α
~
3
g
0
s
+
ε
B
≥
μ
,
1
-
b
α
~
4
ε
D
≥
λ
,
1
-
b
1
-
α
~
4
g
0
s
+
ε
C
≥
μ
.
By using(
H
2
) and the definitions of f
0
s and g
0
s, we deduce that there exists R
1
>
0 such that f
(
t
,
u
,
v
)
≤
ε
(
u
+
v
) and g
(
t
,
u
,
v
)
≤
(
g
0
s
+
ε
)
(
u
+
v
) for all t
∈
[
0,1
], u
,
v
∈
R
+ with 0
≤
u
+
v
≤
R
1. We define the set Ω
1
=
{
(
u
,
v
)
∈
Y
,
u
,
v
Y
<
R
1
}. Now let (
u
,
v
)
∈
P
∩
∂
Ω
1, that is, (
u
,
v
)
∈
P with u
,
v
Y
=
R
1 or equivalently u
+
v
=
R
1. Then u
(
t
)
+
v
(
t
)
≤
R
1 for all t
∈
[
0,1
], and by Lemma 4, we obtain(15)
T
1
u
,
v
t
=
λ
∫
0
1
G
1
t
,
s
f
s
,
u
s
,
v
s
d
s
+
μ
∫
0
1
G
2
t
,
s
g
s
,
u
s
,
v
s
d
s
≤
λ
∫
0
1
J
1
s
f
s
,
u
s
,
v
s
d
s
+
μ
∫
0
1
J
2
s
g
s
,
u
s
,
v
s
d
s
≤
λ
∫
0
1
J
1
s
ε
u
s
+
v
s
d
s
+
μ
∫
0
1
J
2
s
g
0
s
+
ε
u
s
+
v
s
d
s
≤
λ
ε
∫
0
1
J
1
s
u
+
v
d
s
+
μ
g
0
s
+
ε
∫
0
1
J
2
s
u
+
v
d
s
=
λ
ε
A
+
μ
g
0
s
+
ε
B
u
,
v
Y
≤
b
α
~
3
+
b
1
-
α
~
3
u
,
v
Y
=
b
u
,
v
Y
,
t
∈
0,1
.Therefore, T
1
(
u
,
v
)
≤
b
u
,
v
Y.
In a similar manner, we conclude(16)
T
2
u
,
v
t
=
μ
∫
0
1
G
3
t
,
s
g
s
,
u
s
,
v
s
d
s
+
λ
∫
0
1
G
4
t
,
s
f
s
,
u
s
,
v
s
d
s
≤
μ
∫
0
1
J
3
s
g
s
,
u
s
,
v
s
d
s
+
λ
∫
0
1
J
4
s
f
s
,
u
s
,
v
s
d
s
≤
μ
∫
0
1
J
3
s
g
0
s
+
ε
u
s
+
v
s
d
s
+
λ
∫
0
1
J
4
s
ε
u
s
+
v
s
d
s
≤
μ
g
0
s
+
ε
∫
0
1
J
3
s
u
+
v
d
s
+
λ
ε
∫
0
1
J
4
s
u
+
v
d
s
=
μ
g
0
s
+
ε
C
+
λ
ε
D
u
,
v
Y
≤
1
-
b
1
-
α
~
4
+
1
-
b
α
~
4
u
,
v
Y
=
1
-
b
u
,
v
Y
,
t
∈
0,1
.Hence, T
2
(
u
,
v
)
≤
1
-
b
u
,
v
Y.
Then, for(
u
,
v
)
∈
P
∩
∂
Ω
1, we deduce(17)
T
u
,
v
Y
=
T
1
u
,
v
+
T
2
u
,
v
≤
b
u
,
v
Y
+
1
-
b
u
,
v
Y
=
u
,
v
Y
.
By the definitions off
∞
i and g
∞
i, there exists R
¯
2
>
0 such that f
(
t
,
u
,
v
)
≥
(
f
∞
i
-
ε
)
(
u
+
v
) and g
(
t
,
u
,
v
)
≥
(
g
∞
i
-
ε
)
(
u
+
v
) for all u
,
v
≥
0 with u
+
v
≥
R
¯
2 and t
∈
[
c
,
1
]. We consider R
2
=
m
a
x
{
2
R
1
,
R
¯
2
/
γ
} and we define Ω
2
=
{
(
u
,
v
)
∈
Y
,
(
u
,
v
)
Y
<
R
2
}. Then for (
u
,
v
)
∈
P with u
,
v
Y
=
R
2, we obtain(18)
u
t
+
v
t
≥
t
α
-
1
u
+
t
β
-
1
v
≥
c
α
-
1
u
+
c
β
-
1
v
≥
γ
u
+
v
=
γ
u
,
v
Y
=
γ
R
2
≥
R
-
2
,
∀
t
∈
c
,
1
.
Then, by Lemma4, we conclude(19)
T
1
u
,
v
1
≥
λ
∫
0
1
J
1
s
f
s
,
u
s
,
v
s
d
s
+
μ
∫
0
1
J
2
s
g
s
,
u
s
,
v
s
d
s
≥
λ
∫
c
1
J
1
s
f
s
,
u
s
,
v
s
d
s
+
μ
∫
c
1
J
2
s
g
s
,
u
s
,
v
s
d
s
≥
λ
∫
c
1
J
1
s
f
∞
i
-
ε
u
s
+
v
s
d
s
+
μ
∫
c
1
J
2
s
g
∞
i
-
ε
u
s
+
v
s
d
s
≥
λ
γ
f
∞
i
-
ε
∫
c
1
J
1
s
u
,
v
Y
d
s
+
μ
γ
g
∞
i
-
ε
∫
c
1
J
2
s
u
,
v
Y
d
s
=
λ
γ
f
∞
i
-
ε
A
~
+
μ
γ
g
∞
i
-
ε
B
~
u
,
v
Y
≥
a
α
1
+
a
1
-
α
1
u
,
v
Y
=
a
u
,
v
Y
.So, T
1
(
u
,
v
)
≥
T
1
(
u
,
v
)
(
1
)
≥
a
u
,
v
Y.
In a similar manner, we deduce(20)
T
2
u
,
v
1
≥
μ
∫
0
1
J
3
s
g
s
,
u
s
,
v
s
d
s
+
λ
∫
0
1
J
4
s
f
s
,
u
s
,
v
s
d
s
≥
μ
∫
c
1
J
3
s
g
s
,
u
s
,
v
s
d
s
+
λ
∫
c
1
J
4
s
f
s
,
u
s
,
v
s
d
s
≥
μ
∫
c
1
J
3
s
g
∞
i
-
ε
u
s
+
v
s
d
s
+
λ
∫
c
1
J
4
s
f
∞
i
-
ε
u
s
+
v
s
d
s
≥
μ
γ
g
∞
i
-
ε
∫
c
1
J
3
s
u
,
v
Y
d
s
+
λ
γ
f
∞
i
-
ε
∫
c
1
J
4
s
u
,
v
Y
d
s
=
μ
γ
g
∞
i
-
ε
C
~
+
λ
γ
f
∞
i
-
ε
D
~
u
,
v
Y
≥
1
-
a
1
-
α
2
+
1
-
a
α
2
u
,
v
Y
=
1
-
a
u
,
v
Y
.So, T
2
(
u
,
v
)
≥
T
2
(
u
,
v
)
(
1
)
≥
1
-
a
u
,
v
Y.
Hence, for(
u
,
v
)
∈
P
∩
∂
Ω
2, we obtain(21)
T
u
,
v
Y
=
T
1
u
,
v
+
T
2
u
,
v
≥
a
u
,
v
Y
+
1
-
a
u
,
v
Y
=
u
,
v
Y
.
By using (17), (21), Lemma 7, and Theorem 6(i), we conclude that T has a fixed point (
u
,
v
)
∈
P
∩
(
Ω
¯
2
∖
Ω
1
) such that R
1
≤
u
+
v
≤
R
2, u
(
t
)
≥
t
α
-
1
u, and v
(
t
)
≥
t
β
-
1
v for all t
∈
[
0,1
]. If u
>
0 then u
(
t
)
>
0 for all t
∈
(
0,1
], and if v
>
0 then v
(
t
)
>
0 for all t
∈
(
0,1
].In what follows, forf
0
i
,
g
0
i
,
f
∞
s
,
g
∞
s
∈
(
0
,
∞
) and numbers α
1
,
α
2
∈
[
0,1
], α
3
,
α
4
∈
(
0,1
), a
∈
[
0,1
], and b
∈
(
0,1
), we define the numbers(22)
L
~
1
=
max
⁡
a
α
1
γ
f
0
i
A
~
,
1
-
a
α
2
γ
f
0
i
D
~
,
L
~
2
=
min
⁡
b
α
3
f
∞
s
A
,
1
-
b
α
4
f
∞
s
D
,
L
~
3
=
max
⁡
a
1
-
α
1
γ
g
0
i
B
~
,
1
-
a
1
-
α
2
γ
g
0
i
C
~
,
L
~
4
=
min
⁡
b
1
-
α
3
g
∞
s
B
,
1
-
b
1
-
α
4
g
∞
s
C
,
L
~
2
′
=
min
⁡
b
f
∞
s
A
,
1
-
b
f
∞
s
D
,
L
~
4
′
=
min
⁡
b
g
∞
s
B
,
1
-
b
g
∞
s
C
.Theorem 9.
Assume that(
H
1
) and (
H
2
) hold, c
∈
(
0,1
), α
1
,
α
2
∈
[
0,1
], α
3
,
α
4
∈
(
0,1
), a
∈
[
0,1
], and b
∈
(
0,1
).
(1) Iff
0
i
,
g
0
i
,
f
∞
s
,
g
∞
s
∈
(
0
,
∞
) and L
~
1
<
L
~
2 and L
~
3
<
L
~
4, then, for each λ
∈
(
L
~
1
,
L
~
2
) and μ
∈
(
L
~
3
,
L
~
4
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(2) Iff
0
i
,
g
0
i
,
f
∞
s
∈
(
0
,
∞
), g
∞
s
=
0, and L
~
1
<
L
~
2
′, then, for each λ
∈
(
L
~
1
,
L
~
2
′
) and μ
∈
(
L
~
3
,
∞
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(3) Iff
0
i
,
g
0
i
,
g
∞
s
∈
(
0
,
∞
), f
∞
s
=
0, and L
~
3
<
L
~
4
′, then, for each λ
∈
(
L
~
1
,
∞
) and μ
∈
(
L
~
3
,
L
~
4
′
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(4) Iff
0
i
,
g
0
i
∈
(
0
,
∞
), f
∞
s
=
g
∞
s
=
0, then for each λ
∈
(
L
~
1
,
∞
) and μ
∈
(
L
~
3
,
∞
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(5) If{
f
0
i
=
∞
,
g
0
i
,
f
∞
s
,
g
∞
s
∈
(
0
,
∞
)
} or {
f
0
i
,
f
∞
s
,
g
∞
s
∈
(
0
,
∞
)
,
g
0
i
=
∞
} or {
f
0
i
=
g
0
i
=
∞
,
f
∞
s
,
g
∞
s
∈
(
0
,
∞
)
}, then, for each λ
∈
(
0
,
L
~
2
) and μ
∈
(
0
,
L
~
4
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(6) If{
f
0
i
=
∞
,
g
0
i
,
f
∞
s
∈
(
0
,
∞
)
,
g
∞
s
=
0
} or {
f
0
i
,
f
∞
s
∈
(
0
,
∞
)
,
g
0
i
=
∞
,
g
∞
s
=
0
} or {
f
0
i
=
g
0
i
=
∞
,
f
∞
s
∈
(
0
,
∞
)
,
g
∞
s
=
0
}, then, for each λ
∈
(
0
,
L
~
2
′
) and μ
∈
(
0
,
∞
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(7) If{
f
0
i
=
∞
,
g
0
i
,
g
∞
s
∈
(
0
,
∞
)
,
f
∞
s
=
0
} or {
f
0
i
,
g
∞
s
∈
(
0
,
∞
)
,
g
0
i
=
∞
,
f
∞
s
=
0
} or {
f
0
i
=
g
0
i
=
∞
,
f
∞
s
=
0
,
g
∞
s
∈
(
0
,
∞
)
}, then, for each λ
∈
(
0
,
∞
) and μ
∈
(
0
,
L
~
4
′
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.
(8) If{
f
0
i
=
∞
,
g
0
i
∈
(
0
,
∞
)
,
f
∞
s
=
g
∞
s
=
0
} or {
f
0
i
∈
(
0
,
∞
)
,
g
0
i
=
∞
,
f
∞
s
=
g
∞
s
=
0
} or {
f
0
i
=
g
0
i
=
∞
,
f
∞
s
=
g
∞
s
=
0
}, then, for each λ
∈
(
0
,
∞
) and μ
∈
(
0
,
∞
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for S-B
C.Proof.
We consider again the above coneP
⊂
Y and the operators T
1, T
2, and T. Because the proofs of the above cases are similar, in what follows we will prove one of them, namely, the first case of (
6
). So, we suppose f
0
i
=
∞, g
0
i
,
f
∞
s
∈
(
0
,
∞
), and g
∞
s
=
0. Let λ
∈
(
0
,
L
~
2
′
) and μ
∈
(
0
,
∞
). We choose α
~
3
∈
λ
f
∞
s
A
/
b
,
1 and α
~
4
∈
λ
f
∞
s
D
/
1
-
b
,
1, and let ε
>
0 be a positive number such that ε
<
g
0
i, ε
≤
λ
γ
A
~ and(23)
b
α
~
3
f
∞
s
+
ε
A
≥
λ
,
b
1
-
α
~
3
ε
B
≥
μ
,
1
-
b
α
~
4
f
∞
s
+
ε
D
≥
λ
,
1
-
b
1
-
α
~
4
ε
C
≥
μ
.
By using(
H
2
) and the definitions of f
0
i and g
0
i, we deduce that there exists R
3
>
0 such that f
(
t
,
u
,
v
)
≥
1
/
ε
(
u
+
v
) and g
(
t
,
u
,
v
)
≥
(
g
0
i
-
ε
)
(
u
+
v
) for all u
,
v
≥
0 with 0
≤
u
+
v
≤
R
3 and t
∈
[
c
,
1
]. We denote Ω
3
=
{
(
u
,
v
)
∈
Y
,
u
,
v
Y
<
R
3
}. Let (
u
,
v
)
∈
P with u
,
v
Y
=
R
3, that is, u
+
v
=
R
3. Because u
(
t
)
+
v
(
t
)
≤
u
+
v
=
R
3 for all t
∈
[
0,1
], then, by using Lemma 4, we obtain(24)
T
1
u
,
v
1
≥
λ
∫
0
1
J
1
s
f
s
,
u
s
,
v
s
d
s
+
μ
∫
0
1
J
2
s
g
s
,
u
s
,
v
s
d
s
≥
λ
∫
0
1
J
1
s
f
s
,
u
s
,
v
s
d
s
≥
λ
∫
0
1
J
1
s
1
ε
u
s
+
v
s
d
s
≥
λ
ε
∫
c
1
J
1
s
c
α
-
1
u
+
c
β
-
1
v
d
s
≥
λ
γ
ε
A
~
u
,
v
Y
≥
u
,
v
Y
.
Therefore,T
1
(
u
,
v
)
≥
T
1
(
u
,
v
)
(
1
)
≥
u
,
v
Y.
Thus, for an arbitrary element(
u
,
v
)
∈
P
∩
∂
Ω
3, we deduce (25)
T
u
,
v
Y
=
T
1
u
,
v
+
T
2
u
,
v
≥
T
1
u
,
v
≥
u
,
v
Y
.
Now, we define the functionsf
∗
,
g
∗
:
[
0,1
]
×
R
+
→
R
+, f
∗
(
t
,
x
)
=
m
a
x
0
≤
u
+
v
≤
x
f
(
t
,
u
,
v
), g
∗
(
t
,
x
)
=
m
a
x
0
≤
u
+
v
≤
x
g
(
t
,
u
,
v
), t
∈
[
0,1
], x
∈
R
+. Then f
(
t
,
u
,
v
)
≤
f
∗
(
t
,
x
), g
(
t
,
u
,
v
)
≤
g
∗
(
t
,
x
) for all t
∈
[
0,1
], u
≥
0, v
≥
0, and u
+
v
≤
x. The functions f
∗
(
t
,
·
)
,
g
∗
(
t
,
·
) are nondecreasing for every t
∈
[
0,1
], and they satisfy the conditions(26)
limsup
x
→
∞
⁡
max
t
∈
0,1
⁡
f
∗
t
,
x
x
≤
f
∞
s
,
limsup
x
→
∞
⁡
max
t
∈
0,1
⁡
g
∗
t
,
x
x
=
0
.Therefore, for ε
>
0, there exists R
-
4
>
0 such that, for all x
≥
R
-
4 and t
∈
[
0,1
], we have(27)
f
∗
t
,
x
x
≤
limsup
x
→
∞
⁡
max
t
∈
0,1
⁡
f
∗
t
,
x
x
+
ε
≤
f
∞
s
+
ε
,
g
∗
t
,
x
x
≤
limsup
x
→
∞
⁡
max
t
∈
0,1
⁡
g
∗
t
,
x
x
+
ε
=
ε
,and so f
∗
(
t
,
x
)
≤
(
f
∞
s
+
ε
)
x and g
∗
(
t
,
x
)
≤
ε
x.
We considerR
4
=
m
a
x
{
2
R
3
,
R
-
4
} and we denote Ω
4
=
{
(
u
,
v
)
∈
Y
,
u
,
v
Y
<
R
4
}. Let (
u
,
v
)
∈
P
∩
∂
Ω
4. By the definitions of f
∗ and g
∗, we obtain(28)
f
t
,
u
t
,
v
t
≤
f
∗
t
,
u
,
v
Y
,
g
t
,
u
t
,
v
t
≤
g
∗
t
,
u
,
v
Y
,
∀
t
∈
0,1
.Then, for all t
∈
[
0,1
], we conclude(29)
T
1
u
,
v
t
≤
λ
∫
0
1
J
1
s
f
s
,
u
s
,
v
s
d
s
+
μ
∫
0
1
J
2
s
g
s
,
u
s
,
v
s
d
s
≤
λ
∫
0
1
J
1
s
f
∗
s
,
u
,
v
Y
d
s
+
μ
∫
0
1
J
2
s
g
∗
s
,
u
,
v
Y
d
s
≤
λ
f
∞
s
+
ε
∫
0
1
J
1
s
u
,
v
Y
d
s
+
μ
ε
∫
0
1
J
2
s
u
,
v
Y
d
s
=
λ
f
∞
s
+
ε
A
+
μ
ε
B
u
,
v
Y
≤
b
α
~
3
+
b
1
-
α
~
3
u
,
v
Y
=
b
u
,
v
Y
.Therefore, T
1
(
u
,
v
)
≤
b
u
,
v
Y.
In a similar manner, we deduce(30)
T
2
u
,
v
t
≤
μ
∫
0
1
J
3
s
g
s
,
u
s
,
v
s
d
s
+
λ
∫
0
1
J
4
s
f
s
,
u
s
,
v
s
d
s
≤
μ
∫
0
1
J
3
s
g
∗
s
,
u
,
v
Y
d
s
+
λ
∫
0
1
J
4
s
f
∗
s
,
u
,
v
Y
d
s
≤
μ
ε
∫
0
1
J
3
s
u
,
v
Y
d
s
+
λ
f
∞
s
+
ε
∫
0
1
J
4
s
u
,
v
Y
d
s
=
μ
ε
C
+
λ
f
∞
s
+
ε
D
u
,
v
Y
≤
1
-
b
1
-
α
~
4
+
1
-
b
α
~
4
u
,
v
Y
=
1
-
b
u
,
v
Y
.So, T
2
(
u
,
v
)
≤
1
-
b
u
,
v
Y.
Then, for(
u
,
v
)
∈
P
∩
∂
Ω
4, it follows that (31)
T
u
,
v
Y
=
T
1
u
,
v
+
T
2
u
,
v
≤
b
u
,
v
Y
+
1
-
b
u
,
v
Y
=
u
,
v
Y
.
By using (25), (31), Lemma 7, and Theorem 6(ii), we conclude that T has a fixed point (
u
,
v
)
∈
P
∩
(
Ω
¯
4
∖
Ω
3
) such that R
3
≤
(
u
,
v
)
Y
≤
R
4.


## 4. Nonexistence Results for the Positive Solutions

We present in this section intervals forλ and μ for which there exists no positive solution of problem S-B
C that can be viewed as fixed point of operator T.Theorem 10.
Assume that(
H
1
) and (
H
2
) hold, and c
∈
(
0,1
). If f
0
s
,
f
∞
s
,
g
0
s
,
g
∞
s
<
∞, then there exist positive constants λ
0, μ
0 such that, for every λ
∈
(
0
,
λ
0
) and μ
∈
(
0
,
μ
0
), the boundary value problem S-B
C has no positive solution.Proof.
In a similar manner as in the proof of Theorem3.1 from [11], we can show that λ
0
=
m
i
n
1
/
(
4
M
1
A
)
,
1
/
(
4
M
1
D
) and μ
0
=
m
i
n
1
/
(
4
M
2
B
)
,
1
/
(
4
M
2
C
), where A
=
∫
0
1
‍
J
1
(
s
)
d
s, B
=
∫
0
1
‍
J
2
(
s
)
d
s, C
=
∫
0
1
‍
J
3
(
s
)
d
s, and D
=
∫
0
1
‍
J
4
(
s
)
d
s, satisfy the conditions of our theorem.Theorem 11.
Assume that(
H
1
) and (
H
2
) hold, and c
∈
(
0,1
). If f
0
i
,
f
∞
i
>
0 and f
(
t
,
u
,
v
)
>
0 for all t
∈
[
c
,
1
], u
≥
0
,
v
≥
0, and u
+
v
>
0, then there exists a positive constant λ
~
0 such that, for every λ
>
λ
~
0 and μ
>
0, the boundary value problem S-B
C has no positive solution.Proof.
From the assumptions of the theorem, we deduce that there existsm
1
>
0 such that f
(
t
,
u
,
v
)
≥
m
1
(
u
+
v
) for all t
∈
[
c
,
1
] and u
,
v
≥
0. We define λ
~
0
=
m
i
n
1
/
(
γ
m
1
A
~
)
,
1
/
(
γ
m
1
D
~
), where A
~
=
∫
c
1
‍
J
1
(
s
)
d
s and D
~
=
∫
c
1
‍
J
4
(
s
)
d
s. We will show that, for every λ
>
λ
~
0 and μ
>
0, problem S-B
C has no positive solution.
Letλ
>
λ
~
0 and μ
>
0. We suppose that S-B
C has a positive solution (
u
(
t
)
,
v
(
t
)
)
,
t
∈
[
0,1
].
IfA
~
≥
D
~, then λ
~
0
=
1
/
(
γ
m
1
A
~
), and therefore, we obtain(32)
u
1
=
T
1
u
,
v
1
=
λ
∫
0
1
G
1
1
,
s
f
s
,
u
s
,
v
s
d
s
+
μ
∫
0
1
G
2
1
,
s
g
s
,
u
s
,
v
s
d
s
≥
λ
∫
0
1
G
1
1
,
s
f
s
,
u
s
,
v
s
d
s
≥
λ
∫
c
1
G
1
1
,
s
f
s
,
u
s
,
v
s
d
s
≥
λ
m
1
∫
c
1
G
1
1
,
s
u
s
+
v
s
d
s
≥
λ
m
1
∫
c
1
J
1
s
u
s
+
v
s
d
s
≥
λ
m
1
γ
∫
c
1
J
1
s
u
+
v
d
s
=
λ
m
1
γ
A
~
u
,
v
Y
.
Then we conclude(33)
u
≥
u
1
≥
λ
m
1
γ
A
~
u
,
v
Y
>
λ
~
0
m
1
γ
A
~
u
,
v
Y
=
u
,
v
Y
,and so, (
u
,
v
)
Y
=
u
+
v
≥
u
>
u
,
v
Y, which is a contradiction.
IfA
~
<
D
~, then λ
~
0
=
1
/
(
γ
m
1
D
~
), and therefore, we deduce(34)
v
1
=
T
2
u
,
v
1
=
μ
∫
0
1
G
3
1
,
s
g
s
,
u
s
,
v
s
d
s
+
λ
∫
0
1
G
4
1
,
s
f
s
,
u
s
,
v
s
d
s
≥
λ
∫
0
1
G
4
1
,
s
f
s
,
u
s
,
v
s
d
s
≥
λ
∫
c
1
G
4
1
,
s
f
s
,
u
s
,
v
s
d
s
≥
λ
m
1
∫
c
1
G
4
1
,
s
u
s
+
v
s
d
s
≥
λ
m
1
∫
c
1
J
4
s
u
s
+
v
s
d
s
≥
λ
m
1
γ
∫
c
1
J
4
s
u
+
v
d
s
=
λ
m
1
γ
D
~
u
,
v
Y
.
Then we conclude(35)
v
≥
v
1
≥
λ
m
1
γ
D
~
u
,
v
Y
>
λ
~
0
m
1
γ
D
~
u
,
v
Y
=
u
,
v
Y
,and so, u
,
v
Y
=
u
+
v
≥
v
>
(
u
,
v
)
Y, which is a contradiction.
Therefore, the boundary value problemS-B
C has no positive solution.Theorem 12.
Assume that(
H
1
) and (
H
2
) hold, and c
∈
(
0,1
). If g
0
i
,
g
∞
i
>
0 and g
(
t
,
u
,
v
)
>
0 for all t
∈
[
c
,
1
], u
≥
0
,
v
≥
0, and u
+
v
>
0, then there exists a positive constant μ
~
0 such that, for every μ
>
μ
~
0 and λ
>
0, the boundary value problem S-B
C has no positive solution.Proof.
From the assumptions of the theorem, we deduce that there existsm
2
>
0 such that g
(
t
,
u
,
v
)
≥
m
2
(
u
+
v
) for all t
∈
[
c
,
1
] and u
,
v
≥
0. We define μ
~
0
=
m
i
n
1
/
(
γ
m
2
B
~
)
,
1
/
(
γ
m
2
C
~
), where B
~
=
∫
c
1
‍
J
2
(
s
)
d
s and C
~
=
∫
c
1
‍
J
3
(
s
)
d
s. Using a similar approach as that used in the proof of Theorem 11, we can show that, for every μ
>
μ
~
0 and λ
>
0, problem S-B
C has no positive solution.Theorem 13.
Assume that(
H
1
) and (
H
2
) hold, and c
∈
(
0,1
). If f
0
i
,
f
∞
i
,
g
0
i
,
g
∞
i
>
0 and f
(
t
,
u
,
v
)
>
0
,
g
(
t
,
u
,
v
)
>
0 for all t
∈
[
c
,
1
], u
≥
0
,
v
≥
0, and u
+
v
>
0, then there exist positive constants λ
^
0 and μ
^
0 such that, for every λ
>
λ
^
0 and μ
>
μ
^
0, the boundary value problem S-B
C has no positive solution.Proof.
From the assumptions of the theorem, we deduce that there existm
1
,
m
2
>
0 such that f
(
t
,
u
,
v
)
≥
m
1
(
u
+
v
) and g
(
t
,
u
,
v
)
≥
m
2
(
u
+
v
), for all t
∈
[
c
,
1
] and u
,
v
≥
0.
We defineλ
^
0
=
1
/
(
2
γ
m
1
A
~
) and μ
^
0
=
1
/
(
2
γ
m
2
C
~
), where A
~
=
∫
c
1
‍
J
1
(
s
)
d
s and C
~
=
∫
c
1
‍
J
3
(
s
)
d
s. Then, for every λ
>
λ
^
0 and μ
>
μ
^
0, problem S-B
C has no positive solution. Indeed, let λ
>
λ
^
0 and μ
>
μ
^
0. We suppose that S-B
C has a positive solution (
u
(
t
)
,
v
(
t
)
)
,
t
∈
[
0,1
]. In a similar manner as that used in the proofs of Theorems 11 and 12, we obtain(36)
u
≥
u
1
≥
λ
m
1
γ
A
~
u
,
v
Y
,
v
≥
v
1
≥
μ
m
2
γ
C
~
u
,
v
Y
,and so(37)
u
,
v
Y
=
u
+
v
≥
λ
m
1
γ
A
~
u
,
v
Y
+
μ
m
2
γ
C
~
u
,
v
Y
>
λ
^
0
m
1
γ
A
~
u
,
v
Y
+
μ
^
0
m
2
γ
C
~
u
,
v
Y
=
1
2
u
,
v
Y
+
1
2
u
,
v
Y
=
u
,
v
Y
,which is a contradiction. Therefore, the boundary value problem S-B
C has no positive solution.
We can also defineλ
^
0
′
=
1
/
(
2
γ
m
1
D
~
) and μ
^
0
′
=
1
/
(
2
γ
m
2
B
~
), where B
~
=
∫
c
1
‍
J
2
(
s
)
d
s and D
~
=
∫
c
1
‍
J
4
(
s
)
d
s. Then, for every λ
>
λ
^
0
′ and μ
>
μ
^
0
′, problem S-B
C has no positive solution. Indeed, let λ
>
λ
^
0
′ and μ
>
μ
^
0
′. We suppose that S-B
C has a positive solution (
u
(
t
)
,
v
(
t
)
)
,
t
∈
[
0,1
]. In a similar manner as that used in the proofs of Theorems 11 and 12, we obtain(38)
v
≥
v
1
≥
λ
m
1
γ
D
~
u
,
v
Y
,
u
≥
u
1
≥
μ
m
2
γ
B
~
u
,
v
Y
,and so(39)
u
,
v
Y
=
u
+
v
≥
μ
m
2
γ
B
~
u
,
v
Y
+
λ
m
1
γ
D
~
u
,
v
Y
>
μ
^
0
′
m
2
γ
B
~
u
,
v
Y
+
λ
^
0
′
m
1
γ
D
~
u
,
v
Y
=
1
2
u
,
v
Y
+
1
2
u
,
v
Y
=
u
,
v
Y
,which is a contradiction. Therefore, the boundary value problem S-B
C has no positive solution.


## 5. Examples

Letα
=
5
/
2 (n
=
3), β
=
7
/
3 (m
=
3), H
(
t
)
=
t
3, K
t
=
0
,
t
∈
[
0,1
/
3
)
;
1
,
t
∈
[
1
/
3,2
/
3
)
;
2
,
t
∈
[
2
/
3,1
], for all t
∈
[
0,1
]. Then ∫
0
1
u
(
s
)
d
K
(
s
)
=
u
1
/
3
+
u
2
/
3 and ∫
0
1
‍
v
(
s
)
d
H
(
s
)
=
3
∫
0
1
‍
s
2
v
(
s
)
d
s.We consider the system of fractional differential equationsS
0
D
0
+
5
/
2
u
t
+
λ
f
t
,
u
t
,
v
t
=
0
,
t
∈
0,1
,
D
0
+
7
/
3
v
t
+
μ
g
t
,
u
t
,
v
t
=
0
,
t
∈
0,1
,with the boundary conditionsB
C
0
u
0
=
u
′
0
=
0
,
u
′
1
=
3
∫
0
1
s
2
v
s
d
s
,
v
0
=
v
′
0
=
0
,
v
′
1
=
u
1
3
+
u
2
3
.Then we obtainΔ
=
(
α
-
1
)
(
β
-
1
)
-
∫
0
1
‍
s
α
-
1
d
K
(
s
)
∫
0
1
‍
s
β
-
1
d
H
(
s
)
=
26
3
-
6
2
-
3
/
13
3
>
0. The functions H and K are nondecreasing, and so assumption (
H
1
) is satisfied. Besides, we deduce(40)
g
1
t
,
s
=
1
Γ
5
/
2
t
3
/
2
1
-
s
1
/
2
-
t
-
s
3
/
2
,
0
≤
s
≤
t
≤
1
,
t
3
/
2
1
-
s
1
/
2
,
0
≤
t
≤
s
≤
1
,
g
2
t
,
s
=
1
Γ
7
/
3
t
4
/
3
1
-
s
1
/
3
-
t
-
s
4
/
3
,
0
≤
s
≤
t
≤
1
,
t
4
/
3
1
-
s
1
/
3
,
0
≤
t
≤
s
≤
1
,
h
1
s
=
s
1
-
s
1
/
2
Γ
5
/
2
,
h
2
s
=
s
1
-
s
1
/
3
Γ
7
/
3
.We also obtain(41)
J
1
s
=
4
3
π
s
1
-
s
1
/
2
+
4
26
3
-
6
2
-
3
π
1
+
2
2
1
-
s
1
/
2
-
1
-
3
s
3
/
2
-
2
-
3
s
3
/
2
,
0
≤
s
<
1
3
,
4
3
π
s
1
-
s
1
/
2
+
4
26
3
-
6
2
-
3
π
1
+
2
2
1
-
s
1
/
2
-
2
-
3
s
3
/
2
,
1
3
≤
s
<
2
3
,
4
3
π
s
1
-
s
1
/
2
+
4
26
3
-
6
2
-
3
π
1
+
2
2
1
-
s
1
/
2
,
2
3
≤
s
≤
1
,
J
2
s
=
52
3
26
3
-
6
2
-
3
Γ
7
/
3
3
13
1
-
s
1
/
3
-
3
13
1
-
s
13
/
3
-
3
5
s
1
-
s
10
/
3
-
3
7
s
2
1
-
s
7
/
3
,
J
3
s
=
s
1
-
s
1
/
3
Γ
7
/
3
+
13
1
+
2
2
26
3
-
6
2
-
3
Γ
7
/
3
3
13
1
-
s
1
/
3
-
3
13
1
-
s
13
/
3
-
3
5
s
1
-
s
10
/
3
-
3
7
s
2
1
-
s
7
/
3
,
J
4
s
=
26
3
26
3
-
6
2
-
3
π
1
+
2
2
1
-
s
1
/
2
-
1
-
3
s
3
/
2
-
2
-
3
s
3
/
2
,
0
≤
s
<
1
3
,
1
+
2
2
1
-
s
1
/
2
-
2
-
3
s
3
/
2
,
1
3
≤
s
<
2
3
,
1
+
2
2
1
-
s
1
/
2
,
2
3
≤
s
≤
1
.Forc
=
1
/
2, we deduce γ
=
1
/
(
2
2
). After some computations, we conclude A
=
∫
0
1
‍
J
1
(
s
)
d
s
≈
0.31258448, A
~
=
∫
1
/
2
1
‍
J
1
(
s
)
d
s
≈
0.18323168, B
=
∫
0
1
‍
J
2
(
s
)
d
s
≈
0.20906868, B
~
=
∫
1
/
2
1
‍
J
2
(
s
)
d
s
≈
0.13212807, C
=
∫
0
1
‍
J
3
(
s
)
d
s
≈
0.38549139, C
~
=
∫
1
/
2
1
‍
J
3
(
s
)
d
s
≈
0.25157025, D
=
∫
0
1
‍
J
4
(
s
)
d
s
≈
0.24263144, and D
~
=
∫
1
/
2
1
‍
J
4
(
s
)
d
s
≈
0.12808534.Example 1.
We consider the functions(42)
f
t
,
u
,
v
=
t
p
1
u
+
v
+
1
u
+
v
q
1
+
sin
⁡
v
u
+
v
+
1
,
g
t
,
u
,
v
=
t
3
p
2
u
+
v
+
1
u
+
v
q
2
+
cos
⁡
u
u
+
v
+
1
,for t
∈
[
0,1
], u
,
v
≥
0, where p
1
,
p
2
>
0 and q
1
,
q
2
>
1.
We havef
0
s
=
q
1, g
0
s
=
q
2
+
1, f
∞
i
=
1
/
2
p
1
(
q
1
-
1
), and g
∞
i
=
1
/
2
3
p
2
(
q
2
-
1
). We take a
=
b
=
α
1
=
α
2
=
α
3
=
α
4
=
1
/
2; then we obtain L
1
=
1
/
(
p
1
q
1
-
1
D
~
), L
2
=
1
/
(
4
q
1
A
), L
3
=
1
/
(
2
1
/
6
p
2
(
q
2
-
1
)
B
~
), and L
4
=
1
/
(
4
q
2
+
1
C
). The conditions L
1
<
L
2 and L
3
<
L
4 become p
1
(
q
1
-
1
)
/
q
1
>
4
A
/
D
~ and p
2
(
q
2
-
1
)
/
q
2
+
1
>
2
11
/
6
C
/
B
~. If p
1
(
q
1
-
1
)
/
q
1
≥
10 and p
2
(
q
2
-
1
)
/
q
2
+
1
≥
11, then the above conditions are satisfied. Therefore, by Theorem 8
(
1
), for each λ
∈
(
L
1
,
L
2
) and μ
∈
(
L
3
,
L
4
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for problem S
0-B
C
0. For example, if q
1
=
2, q
2
=
3, p
1
=
20, and p
2
=
22, then we obtain L
1
≈
0.39036474, L
2
≈
0.39989189, L
3
≈
0.15324297, and L
4
≈
0.16213072.
Besides, becausef
0
s
=
q
1, g
0
s
=
q
2
+
1, f
∞
s
=
p
1
(
q
1
+
1
), and g
∞
s
=
p
2
(
q
2
+
1
), we can apply Theorem 10. So, we conclude that there exist λ
0
,
μ
0
>
0 such that, for every λ
∈
(
0
,
λ
0
) and μ
∈
(
0
,
μ
0
), the boundary value problem S
0-B
C
0 has no positive solution. By Theorem 10, the positive constants λ
0 and μ
0 are given by λ
0
=
m
i
n
1
/
(
4
M
1
A
)
,
1
/
(
4
M
1
D
)
=
1
/
(
4
M
1
A
) and μ
0
=
m
i
n
1
/
(
4
M
2
B
)
,
1
/
(
4
M
2
C
)
=
1
/
(
4
M
2
C
). For example, if p
1
=
20, p
2
=
22, q
1
=
2, and q
2
=
3, then we obtain M
1
=
60, M
2
=
88, λ
0
≈
13.33
·
1
0
-
3, and μ
0
≈
7.37
·
1
0
-
3.Example 2.
We consider the functions(43)
f
t
,
u
,
v
=
t
+
1
a
~
u
2
+
v
2
,
g
t
,
u
,
v
=
t
b
~
e
u
+
v
-
1
,
t
∈
0,1
,
u
,
v
≥
0
,where a
~
,
b
~
>
0.
We havef
0
s
=
0, g
0
s
=
1, and f
∞
i
=
g
∞
i
=
∞. For b
=
1
/
2, we obtain L
4
′
=
1
/
(
2
C
)
≈
1.2970458. Then, by Theorem 8, (
6
) we conclude that, for each λ
∈
(
0
,
∞
) and μ
∈
(
0
,
L
4
′
), there exists a positive solution (
u
(
t
)
,
v
(
t
)
), t
∈
[
0,1
] for problem S
0-B
C
0.
Becauseg
0
i
=
2
-
b
~ and g
∞
i
=
∞, we can apply Theorem 12. Then there exists μ
~
0 such that, for every μ
>
μ
~
0 and λ
>
0, problem S
0-B
C
0 has no positive solution. For example, if b
~
=
1, then we deduce m
2
=
1
/
2 and μ
~
0
=
1
/
(
γ
m
2
C
~
)
≈
22.486.



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*Source: 2823971-2016-07-31.xml*