# Multiple Positive Solutions of Third-Order BVP with Advanced Arguments and Stieltjes Integral Conditions

Authors: Jian Chang; Jian-Ping Sun; Ya-Hong Zhao
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

We consider the following third-order boundary value problem with advanced arguments and Stieltjes integral boundary conditions:u
′
′
′
t
+
f
t
,
u
α
t
=
0
,
t
∈
0
,
1
,
u
0
=
γ
u
η
1
+
λ
1
u and u
′
′
0
=
0
,
u
1
=
β
u
η
2
+
λ
2
u, where 0
<
η
1
<
η
2
<
1, 0
≤
γ
,
β
≤
1, α
:
[
0,1
]
→
[
0,1
] is continuous, α
(
t
)
≥
t for t
∈
[
0,1
], and α
(
t
)
≤
η
2 for t
∈
[
η
1
,
η
2
]. Under some suitable conditions, by applying a fixed point theorem due to Avery and Peterson, we obtain the existence of multiple positive solutions to the above problem. An example is also included to illustrate the main results obtained.


---

## Body

## 1. Introduction

Third-order differential equations arise from a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross-section, a three-layer beam, electromagnetic waves or gravity driven flows, and so on [1].Recently, third-order boundary value problems (BVPs for short) have received much attention from many authors; see [2–20] and the references therein. However, it is necessary to point out that all the unknown functions in the above-mentioned papers do not depend on advanced arguments.In 2012, Jankowski [21] studied the existence of multiple positive solutions to the BVP(1)
u
′
′
′
t
+
h
t
f
t
,
u
α
t
=
0
,
t
∈
0,1
,
u
0
=
u
′
′
0
=
0
,
u
1
=
β
u
η
+
λ
u
,where the unknown function u depended on an advanced argument α satisfying the following condition:(
A
0
)
α
:
[
0,1
]
→
[
0,1
] was continuous and α
(
t
)
≥
t for t
∈
[
0,1
].λ denoted a linear functional on C
[
0,1
] given by(2)
λ
u
=
∫
0
1
u
t
d
Λ
tinvolving a Stieltjes integral with a suitable function Λ of bounded variation. The measure d
Λ could be a signed one. The situation with a signed measure d
Λ was first discussed in [22, 23] for second-order differential equations; it was also discussed in [24, 25] for second-order impulsive differential equations.Among the boundary conditions in (1), only u
(
1
) was related to u
(
η
) and a Stieltjes integral. When u
(
0
) was also related to u
(
η
) and the Stieltjes integral, the authors in [26, 27] obtained the existence and multiplicity of positive solutions to the BVP(3)
u
′
′
′
t
+
f
t
,
u
α
t
=
0
,
t
∈
0,1
,
u
0
=
γ
u
η
+
λ
u
,
u
′
′
0
=
0
,
u
1
=
β
u
η
+
λ
u
,where λ
[
u
] was defined as in (2) and the condition (
A
0
) was imposed on the advanced argument α. The main tools used were the Guo-Krasnoselskii fixed point theorem [28, 29] and a fixed point theorem due to Avery and Peterson [30].In this paper, we are concerned with the following third-order BVP with advanced arguments and Stieltjes integral boundary conditions:(4)
u
′
′
′
t
+
f
t
,
u
α
t
=
0
,
t
∈
0,1
,
u
0
=
γ
u
η
1
+
λ
1
u
,
u
′
′
0
=
0
,
u
1
=
β
u
η
2
+
λ
2
u
,where(5)
λ
i
u
=
∫
0
1
u
t
d
Λ
i
t
,
i
=
1,2
,here Λ
1 and Λ
2 are suitable functions of bounded variation. It is important to indicate that it is not assumed that λ
i
u
(
i
=
1,2
) is positive to all positive u. Throughout this paper, we always assume that 0
<
η
1
<
η
2
<
1, 0
≤
γ
,
β
≤
1, f
:
[
0,1
]
×
[
0
,
+
∞
)
→
[
0
,
+
∞
) is continuous, and the advanced argument α satisfies the following condition:(
A
0
′
)
α
:
[
0,1
]
→
[
0,1
] is continuous, α
(
t
)
≥
t for t
∈
[
0,1
] and α
(
t
)
≤
η
2 for t
∈
[
η
1
,
η
2
].In order to obtain our main results, we need the following concepts and fixed point theorem [30].LetE be a real Banach space and let K be a cone in E.A mapΘ is said to be a nonnegative continuous convex functional on K if Θ
:
K
→
[
0
,
∞
) is continuous and Θ
(
t
u
+
(
1
-
t
)
v
)
≤
t
Θ
(
u
)
+
(
1
-
t
)
Θ
(
v
) for all u
,
v
∈
K and t
∈
[
0,1
].Similarly, a mapΦ is said to be a nonnegative continuous concave functional on K if Φ
:
K
→
[
0
,
∞
) is continuous and Φ
(
t
u
+
(
1
-
t
)
v
)
≥
t
Φ
(
u
)
+
(
1
-
t
)
Φ
(
v
) for all u
,
v
∈
K and t
∈
[
0,1
].Letφ and Θ be nonnegative continuous convex functionals on K, let Φ be a nonnegative continuous concave functional on K, and let Ψ be a nonnegative continuous functional on K. For positive numbers a
,
b
,
c
,
d, we define the following sets:(6)
K
φ
,
d
=
u
∈
K
:
φ
u
<
d
,
K
φ
,
Φ
,
b
,
d
=
u
∈
K
:
b
≤
Φ
u
,
φ
u
≤
d
,
K
φ
,
Θ
,
Φ
,
b
,
c
,
d
=
u
∈
K
:
b
≤
Φ
u
,
Θ
u
≤
c
,
φ
u
≤
d
,
R
φ
,
Ψ
,
a
,
d
=
u
∈
K
:
a
≤
Ψ
u
,
φ
u
≤
d
.Theorem 1 (Avery and Peterson fixed point theorem).
LetE be a real Banach space and let K be a cone in E. Let φ and Θ be nonnegative continuous convex functionals on K, let Φ be a nonnegative continuous concave functional on K, and let Ψ be a nonnegative continuous functional on K satisfying Ψ
(
k
u
)
≤
k
Ψ
(
u
) for 0
≤
k
≤
1, such that, for some positive numbers M and d,(7)
Φ
u
≤
Ψ
u
,
u
≤
M
φ
ufor all u
∈
K
(
φ
,
d
)
¯. Suppose S
:
K
(
φ
,
d
)
¯
→
K
(
φ
,
d
)
¯ is completely continuous and there exist positive numbers a
,
b
,
c with a
<
b, such that(C1) {
u
∈
K
(
φ
,
Θ
,
Φ
,
b
,
c
,
d
)
:
Φ
(
u
)
>
b
}
≠
ϕ and Φ
(
S
u
)
>
b for u
∈
K
(
φ
,
Θ
,
Φ
,
b
,
c
,
d
);
(C2) Φ
(
S
u
)
>
b for u
∈
K
(
φ
,
Φ
,
b
,
d
) with Θ
(
S
u
)
>
c;
(C3) θ
∉
R
(
φ
,
Ψ
,
a
,
d
) and Ψ
(
S
u
)
<
a for u
∈
R
(
φ
,
Ψ
,
a
,
d
) with Ψ
(
u
)
=
a.Then S has at least three fixed points u
1
,
u
2
,
u
3
∈
K
(
φ
,
d
)
¯, such that(8)
b
<
Φ
u
1
,
a
<
Ψ
u
2
w
i
t
h
Φ
u
2
<
b
,
Ψ
u
3
<
a
.


## 2. Main Results

For convenience, we denote(9)
Δ
=
1
-
γ
1
-
η
2
β
+
1
-
β
η
1
γ
,
k
t
,
s
=
1
2
1
-
t
t
-
s
2
,
0
≤
s
≤
t
≤
1
,
t
1
-
s
2
,
0
≤
t
≤
s
≤
1
,
ρ
i
=
1
-
η
2
β
∫
0
1
d
Λ
i
t
-
1
-
β
∫
0
1
t
d
Λ
i
t
,
i
=
1,2
,
τ
i
=
η
1
γ
∫
0
1
d
Λ
i
t
+
1
-
γ
∫
0
1
t
d
Λ
i
t
,
i
=
1,2
.In the remainder of this paper, we always assume thatΔ
-
ρ
1
>
0, Δ
-
τ
2
>
0, and (
Δ
-
ρ
1
)
(
Δ
-
τ
2
)
>
ρ
2
τ
1, and for Λ
i
(
i
=
1,2
), the following conditions are fulfilled: (10)
∫
0
1
d
Λ
i
t
≥
∫
0
1
t
d
Λ
i
t
≥
0
,
κ
i
s
=
∫
0
1
k
t
,
s
d
Λ
i
t
≥
0
,
s
∈
0,1
,
i
=
1,2
.Then, ρ
i
≥
0
,
τ
i
≥
0
(
i
=
1,2
) and Δ
>
0.Lemma 2 (see [21]).
One has0
≤
k
(
t
,
s
)
≤
(
1
/
2
)
(
1
+
s
)
(
1
-
s
)
2, (
t
,
s
)
∈
[
0,1
]
×
[
0,1
].Lemma 3.
For anyy
∈
C
[
0,1
], the BVP(11)
u
′
′
′
t
=
-
y
t
,
t
∈
0,1
,
u
0
=
γ
u
η
1
+
λ
1
u
,
u
′
′
0
=
0
,
u
1
=
β
u
η
2
+
λ
2
uhas the unique solution(12)
u
t
=
1
-
η
2
β
-
t
1
-
β
Δ
λ
1
u
+
η
1
γ
+
t
1
-
γ
Δ
λ
2
u
+
1
-
η
2
β
γ
-
t
γ
1
-
β
Δ
∫
0
1
k
η
1
,
s
y
s
d
s
+
η
1
β
γ
+
t
β
1
-
γ
Δ
∫
0
1
k
η
2
,
s
y
s
d
s
+
∫
0
1
k
t
,
s
y
s
d
s
,
t
∈
0,1
.Proof.
By integrating the differential equation in (82) three times from 0 to t and using the boundary condition u
′
′
(
0
)
=
0, we know that(13)
u
t
=
u
0
+
u
′
0
t
-
1
2
∫
0
t
t
-
s
2
y
s
d
s
,
t
∈
0,1
.And so,(14)
u
′
0
=
u
1
-
u
0
+
1
2
∫
0
1
1
-
s
2
y
s
d
s
.In view of (13), (14), and the boundary conditions u
(
0
)
=
γ
u
(
η
1
)
+
λ
1
[
u
] and u
(
1
)
=
β
u
(
η
2
)
+
λ
2
[
u
], we have(15)
u
t
=
1
-
t
γ
u
η
1
+
t
β
u
η
2
+
1
-
t
λ
1
u
+
t
λ
2
u
+
∫
0
1
k
t
,
s
y
s
d
s
,
t
∈
0,1
. Therefore,(16)
u
η
1
=
1
-
η
1
+
η
1
β
-
η
2
β
Δ
λ
1
u
+
η
1
Δ
λ
2
u
+
1
-
η
2
β
Δ
∫
0
1
k
η
1
,
s
y
s
d
s
+
η
1
β
Δ
∫
0
1
k
η
2
,
s
y
s
d
s
,
u
η
2
=
1
-
η
2
Δ
λ
1
u
+
η
2
+
η
1
γ
-
η
2
γ
Δ
λ
2
u
+
γ
-
η
2
γ
Δ
∫
0
1
k
η
1
,
s
y
s
d
s
+
1
-
γ
+
η
1
γ
Δ
∫
0
1
k
η
2
,
s
y
s
d
s
.Substituting (16) into (15), we get(17)
u
t
=
1
-
η
2
β
-
t
1
-
β
Δ
λ
1
u
+
η
1
γ
+
t
1
-
γ
Δ
λ
2
u
+
1
-
η
2
β
γ
-
t
γ
1
-
β
Δ
∫
0
1
k
η
1
,
s
y
s
d
s
+
η
1
β
γ
+
t
β
1
-
γ
Δ
∫
0
1
k
η
2
,
s
y
s
d
s
+
∫
0
1
k
t
,
s
y
s
d
s
,
t
∈
0,1
.LetC
[
0,1
] be equipped with the maximum norm. Then C
[
0,1
] is a Banach space. If we let (18)
K
=
u
∈
C
0,1
:
u
t
≥
0
,
t
∈
0,1
,
m
i
n
t
∈
η
1
,
η
2
u
t
≥
Γ
u
,
λ
i
u
≥
0
,
i
=
1,2
,where(19)
Γ
=
min
⁡
η
1
1
-
γ
+
η
1
γ
,
1
-
η
2
1
-
η
2
β
,then K is a cone in C
[
0,1
]. Now, we define operators T and S on K by(20)
T
u
t
=
1
-
η
2
β
-
t
1
-
β
Δ
λ
1
u
+
η
1
γ
+
t
1
-
γ
Δ
λ
2
u
+
F
u
t
,
t
∈
0,1
,
S
u
t
=
1
-
η
2
β
-
t
1
-
β
Δ
-
τ
2
+
η
1
γ
+
t
1
-
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
1
F
u
+
1
-
η
2
β
-
t
1
-
β
τ
1
+
η
1
γ
+
t
1
-
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
2
F
u
+
F
u
t
,
t
∈
0,1
,where (21)
F
u
t
=
1
-
η
2
β
γ
-
t
γ
1
-
β
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
η
1
β
γ
+
t
β
1
-
γ
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
0
1
k
t
,
s
f
s
,
u
α
s
d
s
,
t
∈
0,1
.Lemma 4.
T, S
:
K
→
K.Proof.
Letu
∈
K. Then it is easy to know that (22)
T
u
′
′
t
=
-
∫
0
1
f
s
,
u
α
s
d
s
≤
0
,
t
∈
0,1
, which shows that T
u is concave down on [
0,1
]. In view of (23)
F
u
0
=
1
-
η
2
β
γ
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
η
1
β
γ
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
≥
0
,
F
u
1
=
1
-
η
2
β
γ
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
1
-
γ
+
η
1
γ
β
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
≥
0
,we have(24)
T
u
0
=
1
-
η
2
β
Δ
λ
1
u
+
η
1
γ
Δ
λ
2
u
+
F
u
0
≥
0
,
T
u
1
=
1
-
η
2
β
Δ
λ
1
u
+
1
-
γ
+
η
1
γ
Δ
λ
2
u
+
F
u
1
≥
0
.So, (
T
u
)
(
t
)
≥
0, t
∈
[
0,1
].
Now, we prove thatmin
t
∈
η
1
,
η
2
⁡
(
T
u
)
(
t
)
≥
Γ
T
u. To do it we consider two cases.
Case 1. Let (
T
u
)
(
η
1
)
≤
(
T
u
)
(
η
2
). Then min
t
∈
η
1
,
η
2
⁡
(
T
u
)
(
t
)
=
(
T
u
)
(
η
1
) and there exists t
¯
∈
η
1
,
1 such that T
u
=
(
T
u
)
(
t
¯
).
Ift
¯
∈
η
1
,
η
2, then (25)
T
u
t
¯
-
T
u
0
t
¯
≤
T
u
η
1
-
T
u
0
η
1
. So, (26)
T
u
≤
η
2
η
1
T
u
η
1
-
η
2
-
η
1
η
1
T
u
0
,which together with(27)
T
u
0
=
γ
T
u
η
1
+
λ
1
uimplies that(28)
T
u
≤
η
2
-
η
2
-
η
1
γ
η
1
T
u
η
1
;that is,(29)
m
i
n
t
∈
η
1
,
η
2
T
u
t
≥
η
1
η
2
-
η
2
-
η
1
γ
T
u
.
Ift
¯
∈
(
η
2
,
1
], then (30)
T
u
t
¯
-
T
u
η
1
t
¯
-
η
1
≤
T
u
η
2
-
T
u
η
1
η
2
-
η
1
. So,(31)
T
u
≤
1
-
η
1
η
2
-
η
1
T
u
η
2
-
1
-
η
2
η
2
-
η
1
T
u
η
1
.On the other hand, it follows from (32)
T
u
η
1
-
T
u
0
η
1
≥
T
u
η
2
-
T
u
0
η
2and (27) that (33)
T
u
η
2
≤
η
2
-
η
2
-
η
1
γ
η
1
T
u
η
1
,which together with (31) implies that (34)
T
u
≤
1
-
γ
+
η
1
γ
η
1
T
u
η
1
;that is,(35)
m
i
n
t
∈
η
1
,
η
2
T
u
t
≥
η
1
1
-
γ
+
η
1
γ
T
u
.
Case 2. Let (
T
u
)
(
η
1
)
>
(
T
u
)
(
η
2
). Then min
t
∈
η
1
,
η
2
⁡
(
T
u
)
(
t
)
=
(
T
u
)
(
η
2
) and there exists t
¯
∈
[
0
,
η
2
) such that T
u
=
(
T
u
)
(
t
¯
).
Ift
¯
∈
[
0
,
η
1
], then (36)
T
u
η
2
-
T
u
t
¯
η
2
-
t
¯
≥
T
u
η
2
-
T
u
η
1
η
2
-
η
1
.So,(37)
T
u
≤
η
2
η
2
-
η
1
T
u
η
1
-
η
1
η
2
-
η
1
T
u
η
2
.At the same time, since (38)
T
u
η
2
-
T
u
η
1
η
2
-
η
1
≥
T
u
1
-
T
u
η
1
1
-
η
1
,we have (39)
T
u
η
1
≤
1
-
η
1
1
-
η
2
T
u
η
2
-
η
2
-
η
1
1
-
η
2
T
u
1
,which together with(40)
T
u
1
=
β
T
u
η
2
+
λ
2
uimplies that(41)
T
u
η
1
≤
1
-
η
1
-
η
2
-
η
1
β
1
-
η
2
T
u
η
2
.In view of (37) and (41), we have (42)
T
u
≤
1
-
η
2
β
1
-
η
2
T
u
η
2
;that is,(43)
m
i
n
t
∈
η
1
,
η
2
T
u
t
≥
1
-
η
2
1
-
η
2
β
T
u
.
Ift
¯
∈
(
η
1
,
η
2
), then (44)
T
u
1
-
T
u
t
¯
1
-
t
¯
≥
T
u
1
-
T
u
η
2
1
-
η
2
.So,(45)
T
u
≤
1
-
η
1
1
-
η
2
T
u
η
2
-
η
2
-
η
1
1
-
η
2
T
u
1
, which together with (40) implies that (46)
T
u
≤
1
-
η
1
-
η
2
-
η
1
β
1
-
η
2
T
u
η
2
;that is,(47)
m
i
n
t
∈
η
1
,
η
2
T
u
t
≥
1
-
η
2
1
-
η
1
-
η
2
-
η
1
β
T
u
.
It follows from (29), (35), (43), and (47) that (48)
m
i
n
t
∈
η
1
,
η
2
T
u
t
≥
Γ
T
u
.
Finally, we need to show thatλ
i
T
u
≥
0
,
i
=
1,2. Since(49)
λ
i
F
u
=
∫
0
1
1
-
η
2
β
γ
-
t
γ
1
-
β
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
d
Λ
i
t
+
∫
0
1
η
1
β
γ
+
t
β
1
-
γ
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
d
Λ
i
t
+
∫
0
1
∫
0
1
k
t
,
s
f
s
,
u
α
s
d
s
d
Λ
i
t
=
γ
ρ
i
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
β
τ
i
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
0
1
κ
i
s
f
s
,
u
α
s
d
s
≥
0
,
i
=
1,2
,we have(50)
λ
i
T
u
=
ρ
i
Δ
λ
1
u
+
τ
i
Δ
λ
2
u
+
λ
i
F
u
≥
0
,
i
=
1,2
.
Therefore,T
:
K
→
K. Similarly, we may prove that S
:
K
→
K.Lemma 5.
T and S have the same fixed points in K.Proof.
On the one hand, ifu
∈
K is a fixed point of S, that is, u
=
S
u, then(51)
λ
1
u
=
λ
1
S
u
=
∫
0
1
1
-
η
2
β
-
t
1
-
β
Δ
-
τ
2
+
η
1
γ
+
t
1
-
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
1
F
u
+
1
-
η
2
β
-
t
1
-
β
τ
1
+
η
1
γ
+
t
1
-
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
2
F
u
+
F
u
t
d
Λ
1
t
=
Δ
Δ
-
τ
2
λ
1
F
u
+
Δ
τ
1
λ
2
F
u
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
,
λ
2
u
=
λ
2
S
u
=
∫
0
1
1
-
η
2
β
-
t
1
-
β
Δ
-
τ
2
+
η
1
γ
+
t
1
-
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
1
F
u
+
1
-
η
2
β
-
t
1
-
β
τ
1
+
η
1
γ
+
t
1
-
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
2
F
u
+
F
u
t
d
Λ
2
t
=
Δ
ρ
2
λ
1
F
u
+
Δ
Δ
-
ρ
1
λ
2
F
u
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
,which shows that(52)
λ
1
F
u
=
Δ
-
ρ
1
λ
1
u
-
τ
1
λ
2
u
Δ
,
λ
2
F
u
=
Δ
-
τ
2
λ
2
u
-
ρ
2
λ
1
u
Δ
.So,(53)
u
t
=
S
u
t
=
1
-
η
2
β
-
t
1
-
β
Δ
-
τ
2
+
η
1
γ
+
t
1
-
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
1
F
u
+
1
-
η
2
β
-
t
1
-
β
τ
1
+
η
1
γ
+
t
1
-
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
2
F
u
+
F
u
t
=
1
-
η
2
β
-
t
1
-
β
Δ
λ
1
u
+
η
1
γ
+
t
1
-
γ
Δ
λ
2
u
+
F
u
t
=
T
u
t
,
t
∈
0,1
,which indicates that u is a fixed point of T.
On the other hand, ifu
∈
K is a fixed point of T, that is, u
=
T
u, then (54)
λ
i
u
=
λ
i
T
u
=
∫
0
1
1
-
η
2
β
-
t
1
-
β
Δ
λ
1
u
+
η
1
γ
+
t
1
-
γ
Δ
λ
2
u
+
F
u
t
d
Λ
i
t
=
ρ
i
Δ
λ
1
u
+
τ
i
Δ
λ
2
u
+
λ
i
F
u
,
i
=
1,2
,which shows that (55)
λ
1
u
=
Δ
Δ
-
τ
2
λ
1
F
u
+
Δ
τ
1
λ
2
F
u
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
,
λ
2
u
=
Δ
ρ
2
λ
1
F
u
+
Δ
Δ
-
ρ
1
λ
2
F
u
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
.So,(56)
u
t
=
T
u
t
=
1
-
η
2
β
-
t
1
-
β
Δ
λ
1
u
+
η
1
γ
+
t
1
-
γ
Δ
λ
2
u
+
F
u
t
=
1
-
η
2
β
-
t
1
-
β
Δ
-
τ
2
+
η
1
γ
+
t
1
-
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
1
F
u
+
1
-
η
2
β
-
t
1
-
β
τ
1
+
η
1
γ
+
t
1
-
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
2
F
u
+
F
u
t
=
S
u
t
,
t
∈
0,1
,which indicates that u is a fixed point of S.Lemma 6.
T
,
S
:
K
→
K is completely continuous.Proof.
First, by Lemma4, we know that T
(
K
)
⊂
K.
Next, we show thatT is compact.
LetD
⊂
K be a bounded set. Then there exists M
1
>
0 such that u
≤
M
1 for any u
∈
D. Since Λ
1 and Λ
2 are functions of bounded variation, there exists M
2
>
0 such that (57)
∑
j
=
1
n
Λ
i
t
j
-
Λ
i
t
j
-
1
≤
M
2
,
i
=
1,2for any partition Δ
′
:
0
=
t
0
<
t
1
<
⋯
<
t
n
-
1
<
t
n
=
1. Let (58)
M
3
=
s
u
p
f
t
,
u
:
t
,
u
∈
0,1
×
0
,
M
1
. Then for any u
∈
D, we have (59)
T
u
=
m
a
x
t
∈
0,1
T
u
t
≤
1
-
η
2
β
Δ
λ
1
u
+
1
-
γ
+
η
1
γ
Δ
λ
2
u
+
1
-
η
2
β
γ
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
1
-
γ
+
η
1
γ
β
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
1
2
∫
0
1
1
+
s
1
-
s
2
f
s
,
u
α
s
d
s
≤
1
-
η
2
β
Δ
M
1
M
2
+
1
-
γ
+
η
1
γ
Δ
M
1
M
2
+
1
-
η
2
β
γ
M
3
Δ
∫
0
1
k
η
1
,
s
d
s
+
1
-
γ
+
η
1
γ
β
M
3
Δ
∫
0
1
k
η
2
,
s
d
s
+
5
24
M
3
,which shows that T
(
D
) is uniformly bounded.
On the other hand, for anyε
>
0, since k
(
t
,
s
) is uniformly continuous on [
0,1
]
×
[
0,1
], there exists δ
1
(
ε
)
>
0 such that, for any t
1
,
t
2
∈
[
0,1
] with |
t
1
-
t
2
|
<
δ
1
(
ε
), (60)
k
t
1
,
s
-
k
t
2
,
s
<
ε
5
M
3
+
1
,
s
∈
0,1
.Let(61)
δ
=
min
⁡
δ
1
ε
,
ε
Δ
5
1
-
β
M
1
M
2
+
1
,
ε
Δ
5
1
-
γ
M
1
M
2
+
1
,
ε
Δ
5
1
-
β
γ
M
3
∫
0
1
k
η
1
,
s
d
s
+
1
,
ε
Δ
5
1
-
γ
β
M
3
∫
0
1
k
η
2
,
s
d
s
+
1
.Then for any u
∈
D,
t
1
,
t
2
∈
[
0,1
] with |
t
1
-
t
2
|
<
δ, we have (62)
T
u
t
1
-
T
u
t
2
=
t
1
-
t
2
β
-
1
Δ
λ
1
u
+
t
1
-
t
2
1
-
γ
Δ
λ
2
u
+
t
1
-
t
2
β
-
1
γ
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
t
1
-
t
2
1
-
γ
β
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
0
1
k
t
1
,
s
-
k
t
2
,
s
f
s
,
u
α
s
d
s
≤
1
-
β
t
1
-
t
2
Δ
λ
1
u
+
1
-
γ
t
1
-
t
2
Δ
λ
2
u
+
1
-
β
γ
t
1
-
t
2
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
1
-
γ
β
t
1
-
t
2
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
0
1
k
t
1
,
s
-
k
t
2
,
s
f
s
,
u
α
s
d
s
≤
1
-
β
t
1
-
t
2
M
1
M
2
Δ
+
1
-
γ
t
1
-
t
2
M
1
M
2
Δ
+
1
-
β
γ
t
1
-
t
2
M
3
Δ
∫
0
1
k
η
1
,
s
d
s
+
1
-
γ
β
t
1
-
t
2
M
3
Δ
∫
0
1
k
η
2
,
s
d
s
+
M
3
∫
0
1
k
t
1
,
s
-
k
t
2
,
s
d
s
<
ε
,which shows that T
(
D
) is equicontinuous. It follows from Arzela-Ascoli theorem that T
(
D
) is relatively compact. Thus, we have shown that T is a compact operator.
Finally, we prove thatT is continuous.
Assume thatu
n
,
u
∈
K and lim
n
→
∞
⁡
u
n
=
u. Then there exists M
4
>
0 such that u
≤
M
4 and u
n
≤
M
4
,
n
=
1
,
2
,
…. For any ε
>
0, since f
(
s
,
x
) is uniformly continuous on [
0,1
]
×
[
0
,
M
4
], there exists δ
>
0 such that, for any x
1
,
x
2
∈
[
0
,
M
4
] with |
x
1
-
x
2
|
<
δ,(63)
f
s
,
x
1
-
f
s
,
x
2
<
ε
3
2
-
η
2
β
-
β
γ
/
Δ
∫
0
1
k
η
1
,
s
d
s
+
3
1
-
γ
+
η
1
γ
β
/
Δ
∫
0
1
k
η
2
,
s
d
s
+
5
/
8
,
s
∈
0,1
.At the same time, since lim
n
→
∞
⁡
u
n
=
u, there exists positive integer N such that, for any n
>
N,(64)
u
n
-
u
<
min
⁡
δ
,
ε
Δ
3
2
-
η
2
β
-
β
M
2
,
ε
Δ
3
1
-
γ
+
η
1
γ
M
2
.It follows from (63) and (64) that, for any n
>
N,(65)
T
u
n
-
T
u
=
max
t
∈
0,1
⁡
T
u
n
t
-
T
u
t
≤
2
-
η
2
β
-
β
Δ
λ
1
u
n
-
λ
1
u
+
1
-
γ
+
η
1
γ
Δ
λ
2
u
n
-
λ
2
u
+
2
-
η
2
β
-
β
γ
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
n
α
s
-
f
s
,
u
α
s
d
s
+
1
-
γ
+
η
1
γ
β
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
n
α
s
-
f
s
,
u
α
s
d
s
+
1
2
∫
0
1
1
+
s
1
-
s
2
f
s
,
u
n
α
s
-
f
s
,
u
α
s
d
s
≤
2
-
η
2
β
-
β
Δ
u
n
-
u
M
2
+
1
-
γ
+
η
1
γ
Δ
u
n
-
u
M
2
+
∫
0
1
2
-
η
2
β
-
β
γ
Δ
k
η
1
,
s
+
1
-
γ
+
η
1
γ
β
Δ
k
η
2
,
s
+
1
2
1
+
s
1
-
s
2
f
s
,
u
n
α
s
-
f
s
,
u
α
s
d
s
<
ε
,which indicates that T is continuous.
Therefore,T
:
K
→
K is completely continuous. Similarly, we can prove that S
:
K
→
K is also completely continuous.For convenience, we denote(66)
D
1
=
γ
ρ
1
Δ
∫
0
1
k
η
1
,
s
d
s
+
β
τ
1
Δ
∫
0
1
k
η
2
,
s
d
s
+
∫
0
1
κ
1
s
d
s
,
D
2
=
γ
ρ
2
Δ
∫
0
1
k
η
1
,
s
d
s
+
β
τ
2
Δ
∫
0
1
k
η
2
,
s
d
s
+
∫
0
1
κ
2
s
d
s
,
D
3
=
1
-
η
2
β
γ
Δ
∫
0
1
k
η
1
,
s
d
s
+
1
-
γ
+
η
1
γ
β
Δ
∫
0
1
k
η
2
,
s
d
s
+
5
24
,
D
4
=
γ
ρ
1
Δ
∫
η
1
η
2
k
η
1
,
s
d
s
+
β
τ
1
Δ
∫
η
1
η
2
k
η
2
,
s
d
s
+
∫
η
1
η
2
κ
1
s
d
s
,
D
5
=
γ
ρ
2
Δ
∫
η
1
η
2
k
η
1
,
s
d
s
+
β
τ
2
Δ
∫
η
1
η
2
k
η
2
,
s
d
s
+
∫
η
1
η
2
κ
2
s
d
s
,
D
6
=
γ
1
-
η
2
Δ
∫
η
1
η
2
k
η
1
,
s
d
s
+
1
-
γ
+
η
1
γ
Δ
∫
η
1
η
2
k
η
2
,
s
d
s
.Let(67)
μ
>
1
-
η
2
β
Δ
-
τ
2
+
1
-
γ
+
η
1
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
1
+
1
-
η
2
β
τ
1
+
1
-
γ
+
η
1
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
2
+
D
3
,
0
<
L
<
η
1
η
2
-
η
2
-
η
1
γ
1
-
η
2
Δ
-
τ
2
+
η
1
γ
+
η
2
-
η
2
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
4
+
1
-
η
2
τ
1
+
η
1
γ
+
η
2
-
η
2
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
5
+
D
6
.Theorem 7.
Suppose that there exist positive constantsa, b, and d with a
<
b
<
b
/
Γ
≤
d such that the following conditions are fulfilled:(
A
1
)
f
(
t
,
u
)
≤
d
/
μ, (
t
,
u
)
∈
[
0,1
]
×
[
0
,
d
],
(
A
2
)
f
(
t
,
u
)
≥
b
/
L, (
t
,
u
)
∈
[
η
1
,
η
2
]
×
[
b
,
b
/
Γ
],
(
A
3
)
f
(
t
,
u
)
≤
a
/
μ, (
t
,
u
)
∈
[
0,1
]
×
[
0
,
a
].Then the BVP (4) has at least three positive solutions u
1, u
2, u
3 satisfying u
i
≤
d
(
i
=
1,2
,
3
) and(68)
m
i
n
t
∈
η
1
,
η
2
u
1
t
>
b
,
u
2
>
a
w
i
t
h
m
i
n
t
∈
η
1
,
η
2
u
2
t
<
b
,
u
3
<
a
.Proof.
Foru
∈
K, we define(69)
Φ
u
=
m
i
n
t
∈
η
1
,
η
2
u
t
,
φ
u
=
Θ
u
=
Ψ
u
=
u
.Then it is easy to know that Φ is a nonnegative continuous concave functional on K and φ, Θ and Ψ are nonnegative continuous convex functionals on K. In order to apply Theorem 1 to prove our main results, we use the operator S and take c
=
b
/
Γ.
First, we assert thatS
:
K
(
φ
,
d
)
¯
→
K
(
φ
,
d
)
¯.
In fact, ifu
∈
K
(
φ
,
d
)
¯, then 0
≤
u
t
≤
d
,
t
∈
[
0,1
], which together with (
A
1
) implies that (70)
λ
i
F
u
=
γ
ρ
i
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
β
τ
i
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
0
1
κ
i
s
f
s
,
u
α
s
d
s
≤
γ
ρ
i
Δ
∫
0
1
k
η
1
,
s
d
s
+
β
τ
i
Δ
∫
0
1
k
η
2
,
s
d
s
+
∫
0
1
κ
i
s
d
s
d
μ
=
D
i
d
μ
,
i
=
1,2
,
F
u
=
m
a
x
t
∈
0,1
F
u
t
≤
1
-
η
2
β
γ
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
1
-
γ
+
η
1
γ
β
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
1
2
∫
0
1
1
+
s
1
-
s
2
f
s
,
u
α
s
d
s
≤
1
-
η
2
β
γ
Δ
∫
0
1
k
η
1
,
s
d
s
+
1
-
γ
+
η
1
γ
β
Δ
∫
0
1
k
η
2
,
s
d
s
+
5
24
d
μ
=
D
3
d
μ
.It follows from (70) that(71)
φ
S
u
=
S
u
≤
1
-
η
2
β
Δ
-
τ
2
+
1
-
γ
+
η
1
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
1
F
u
+
1
-
η
2
β
τ
1
+
1
-
γ
+
η
1
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
2
F
u
+
F
u
≤
1
-
η
2
β
Δ
-
τ
2
+
1
-
γ
+
η
1
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
1
+
1
-
η
2
β
τ
1
+
1
-
γ
+
η
1
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
2
+
D
3
d
μ
≤
d
.This shows that S
:
K
(
φ
,
d
)
¯
→
K
(
φ
,
d
)
¯.
Next, we claim that{
u
∈
K
(
φ
,
Θ
,
Φ
,
b
,
c
,
d
)
:
Φ
(
u
)
>
b
}
≠
ϕ and Φ
(
S
u
)
>
b for u
∈
K
(
φ
,
Θ
,
Φ
,
b
,
c
,
d
).
Indeed, the constant functionb
+
c
/
2
∈
{
u
∈
K
(
φ
,
Θ
,
Φ
,
b
,
c
,
d
)
:
Φ
(
u
)
>
b
}. Moreover, if u
∈
K
(
φ
,
Θ
,
Φ
,
b
,
c
,
d
), then b
≤
u
(
t
)
≤
c
,
t
∈
[
η
1
,
η
2
], which together with η
1
≤
t
≤
α
(
t
)
≤
η
2 for t
∈
[
η
1
,
η
2
] implies that b
≤
u
(
α
(
t
)
)
≤
c, t
∈
[
η
1
,
η
2
]. In view of (
A
2
), we have (72)
λ
1
F
u
=
γ
ρ
1
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
β
τ
1
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
0
1
κ
1
s
f
s
,
u
α
s
d
s
≥
γ
ρ
1
Δ
∫
η
1
η
2
k
η
1
,
s
f
s
,
u
α
s
d
s
+
β
τ
1
Δ
∫
η
1
η
2
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
η
1
η
2
κ
1
s
f
s
,
u
α
s
d
s
≥
γ
ρ
1
Δ
∫
η
1
η
2
k
η
1
,
s
d
s
+
β
τ
1
Δ
∫
η
1
η
2
k
η
2
,
s
d
s
+
∫
η
1
η
2
κ
1
s
d
s
b
L
=
D
4
b
L
,
λ
2
F
u
=
γ
ρ
2
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
β
τ
2
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
0
1
κ
2
s
f
s
,
u
α
s
d
s
≥
γ
ρ
2
Δ
∫
η
1
η
2
k
η
1
,
s
f
s
,
u
α
s
d
s
+
β
τ
2
Δ
∫
η
1
η
2
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
η
1
η
2
κ
2
s
f
s
,
u
α
s
d
s
≥
γ
ρ
2
Δ
∫
η
1
η
2
k
η
1
,
s
d
s
+
β
τ
2
Δ
∫
η
1
η
2
k
η
2
,
s
d
s
+
∫
η
1
η
2
κ
2
s
d
s
b
L
=
D
5
b
L
,
F
u
η
2
=
γ
1
-
η
2
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
1
-
γ
+
η
1
γ
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
≥
γ
1
-
η
2
Δ
∫
η
1
η
2
k
η
1
,
s
f
s
,
u
α
s
d
s
+
1
-
γ
+
η
1
γ
Δ
∫
η
1
η
2
k
η
2
,
s
f
s
,
u
α
s
d
s
≥
γ
1
-
η
2
Δ
∫
η
1
η
2
k
η
1
,
s
d
s
+
1
-
γ
+
η
1
γ
Δ
∫
η
1
η
2
k
η
2
,
s
d
s
b
L
=
D
6
b
L
.Since S
u is concave down on [
0,1
], we have (73)
S
u
η
2
-
S
u
0
η
2
≤
S
u
η
1
-
S
u
0
η
1
.So, (74)
S
u
η
2
≤
η
2
η
1
S
u
η
1
-
η
2
-
η
1
η
1
S
u
0
, which together with (75)
S
u
0
=
γ
S
u
η
1
+
Δ
Δ
-
τ
2
λ
1
F
u
+
Δ
τ
1
λ
2
F
u
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1implies that(76)
S
u
η
1
≥
η
1
η
2
-
η
2
-
η
1
γ
S
u
η
2
.Therefore, it follows from (72) and (76) that(77)
Φ
S
u
=
m
i
n
t
∈
η
1
,
η
2
S
u
t
=
m
i
n
S
u
η
1
,
S
u
η
2
≥
m
i
n
η
1
η
2
-
η
2
-
η
1
γ
S
u
η
2
,
S
u
η
2
=
η
1
η
2
-
η
2
-
η
1
γ
S
u
η
2
=
η
1
η
2
-
η
2
-
η
1
γ
1
-
η
2
Δ
-
τ
2
+
η
1
γ
+
η
2
-
η
2
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
1
F
u
+
1
-
η
2
τ
1
+
η
1
γ
+
η
2
-
η
2
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
2
F
u
+
F
u
η
2
≥
η
1
η
2
-
η
2
-
η
1
γ
1
-
η
2
Δ
-
τ
2
+
η
1
γ
+
η
2
-
η
2
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
4
+
1
-
η
2
τ
1
+
η
1
γ
+
η
2
-
η
2
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
5
+
D
6
b
L
>
b
.Thirdly, we assert that Φ
(
S
u
)
>
b for u
∈
K
(
φ
,
Φ
,
b
,
d
) with Θ
(
S
u
)
>
c.
To see this, we supposeu
∈
K
(
φ
,
Φ
,
b
,
d
) and Θ
(
S
u
)
=
S
u
>
c. Then(78)
Φ
S
u
=
m
i
n
t
∈
η
1
,
η
2
S
u
t
≥
Γ
S
u
>
Γ
c
=
b
.
Finally, we prove thatθ
∉
R
(
φ
,
Ψ
,
a
,
d
) and Ψ
(
S
u
)
<
a for u
∈
R
(
φ
,
Ψ
,
a
,
d
) with Ψ
(
u
)
=
a.
Indeed, it follows fromΨ
(
θ
)
=
0
<
a that θ
∉
R
(
φ
,
Ψ
,
a
,
d
). Moreover, if u
∈
R
(
φ
,
Ψ
,
a
,
d
) and Ψ
(
u
)
=
a, then 0
≤
u
(
t
)
≤
a, t
∈
[
0,1
], which together with (
A
3
) implies that (79)
λ
i
F
u
=
γ
ρ
i
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
β
τ
i
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
∫
0
1
κ
i
s
f
s
,
u
α
s
d
s
≤
γ
ρ
i
Δ
∫
0
1
k
η
1
,
s
d
s
+
β
τ
i
Δ
∫
0
1
k
η
2
,
s
d
s
+
∫
0
1
κ
i
s
d
s
a
μ
=
D
i
a
μ
,
i
=
1,2
,
F
u
=
m
a
x
t
∈
0,1
F
u
t
≤
1
-
η
2
β
γ
Δ
∫
0
1
k
η
1
,
s
f
s
,
u
α
s
d
s
+
1
-
γ
+
η
1
γ
β
Δ
∫
0
1
k
η
2
,
s
f
s
,
u
α
s
d
s
+
1
2
∫
0
1
1
+
s
1
-
s
2
f
s
,
u
α
s
d
s
≤
1
-
η
2
β
γ
Δ
∫
0
1
k
η
1
,
s
d
s
+
1
-
γ
+
η
1
γ
β
Δ
∫
0
1
k
η
2
,
s
d
s
+
5
24
a
μ
=
D
3
a
μ
.In view of (79), we have (80)
Ψ
S
u
=
S
u
≤
1
-
η
2
β
Δ
-
τ
2
+
1
-
γ
+
η
1
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
1
F
u
+
1
-
η
2
β
τ
1
+
1
-
γ
+
η
1
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
λ
2
F
u
+
F
u
≤
1
-
η
2
β
Δ
-
τ
2
+
1
-
γ
+
η
1
γ
ρ
2
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
1
+
1
-
η
2
β
τ
1
+
1
-
γ
+
η
1
γ
Δ
-
ρ
1
Δ
-
ρ
1
Δ
-
τ
2
-
ρ
2
τ
1
D
2
+
D
3
a
μ
<
a
,as required.
To sum up, all the hypotheses of Theorem1 are satisfied. Hence, the BVP (4) has at least three positive solutions u
1, u
2, u
3 satisfying u
i
≤
d
(
i
=
1,2
,
3
) and(81)
m
i
n
t
∈
η
1
,
η
2
u
1
t
>
b
,
u
2
>
a
w
i
t
h
m
i
n
t
∈
η
1
,
η
2
u
2
t
<
b
,
u
3
<
a
.


## 3. An Example

Example 1.
Consider the following BVP:(82)
u
′
′
′
t
+
f
t
,
u
α
t
=
0
,
t
∈
0,1
,
u
0
=
1
2
u
1
4
+
∫
0
1
u
t
d
3
2
t
2
-
t
,
u
′
′
0
=
0
,
u
1
=
1
4
u
1
2
+
∫
0
1
u
t
d
1
4
t
2
,where(83)
f
t
,
u
=
20
u
2
+
1
20
-
u
t
1
-
t
,
t
,
u
∈
0,1
×
0
,
1
20
,
1
20
+
1980
u
-
1
20
2
+
u
-
1
20
1
10
-
u
t
1
-
t
,
t
,
u
∈
0,1
×
1
20
,
1
10
,
5
+
1
10
u
-
1
10
2
t
1
-
t
,
t
,
u
∈
0,1
×
1
10
,
+
∞
,
α
t
=
2
t
2
,
t
∈
0
,
1
2
,
2
t
-
1
+
1
2
,
t
∈
1
2
,
1
.
SinceΛ
1
t
=
3
/
2
t
2
-
t and Λ
2
t
=
1
/
4
t
2
,
t
∈
0,1, a simple calculation shows that(84)
∫
0
1
d
Λ
1
t
=
∫
0
1
t
d
Λ
1
t
=
1
2
,
∫
0
1
d
Λ
2
t
=
1
4
,
∫
0
1
t
d
Λ
2
t
=
1
6
,
κ
1
s
=
1
8
s
4
-
1
6
s
3
+
1
24
,
κ
2
s
=
1
48
1
-
s
2
2
,
s
∈
0,1
.At the same time, in view of η
1
=
β
=
1
/
4 and η
2
=
γ
=
1
/
2, we get(85)
Δ
=
17
32
,
ρ
1
=
1
16
,
τ
1
=
5
16
,
ρ
2
=
3
32
,
τ
2
=
11
96
,
Γ
=
2
5
,
D
1
=
397
10880
,
D
2
=
3511
195840
,
D
3
=
845
3264
,
D
4
=
26561
2088960
,
D
5
=
25487
4177920
,
D
6
=
13
408
.If we choose μ
=
1
,
L
=
1
/
50
,
a
=
1
/
20
,
b
=
1
/
10, and d
=
6, then all the conditions of Theorem 7 are fulfilled. Therefore, it follows from Theorem 7 that the BVP (82) has at least three positive solutions u
1, u
2, u
3 satisfying u
i
≤
6
(
i
=
1,2
,
3
) and(86)
m
i
n
t
∈
1
/
4
,
1
/
2
u
1
t
>
1
10
,
u
2
>
1
20
w
i
t
h
m
i
n
t
∈
1
/
4
,
1
/
2
u
2
t
<
1
10
,
u
3
<
1
20
.



---

*Source: 3805804-2016-08-22.xml*