# Monotone Iterative Technique and Symmetric Positive Solutions to Fourth-Order Boundary Value Problem with Integral Boundary Conditions

Authors: Huihui Pang; Chen Cai
Journal: Discrete Dynamics in Nature and Society
 (2014)

---

## Abstract

The purpose of this paper is to investigate the existence of symmetric positive solutions for a class of fourth-order boundary value problem:u
4
(
t
)
+
β
u
′
′
(
t
)
=
f
(
t
,
u
(
t
)
,
u
′
′
(
t
)
), 0
<
t
<
1, u
(
0
)
=
u
(
1
)
=
∫
0
1
‍
p
(
s
)
u
(
s
)
d
s, u
′
′
(
0
)
=
u
′
′
(
1
)
=
∫
0
1
‍
q
s
u
′
′
(
s
)
d
s, where p
,
q
∈
L
1
[
0,1
], f
∈
C
(
[
0,1
]
×
[
0
,
∞
)
×
(
-
∞
,
0
]
,
[
0
,
∞
)
). By using a monotone iterative technique, we prove that the above boundary value problem has symmetric positive solutions under certain conditions. In particular, these solutions are obtained via the iteration procedures.


---

## Body

## 1. Introduction

The deformation of an elastic beam in equilibrium state, whose two ends are simply supported, can be described by a fourth-order ordinary differential equation BVP (short for boundary value problem). At present, two-point situation of fourth-order BVP has been studied by many authors, generally using the nonlinear alternatives of Leray-Schauder, the fixed point index theory, and the method of upper and lower solutions, monotone iteration; see [1–6].Recently, problems with integral boundary value conditions arise naturally in thermal conduction problems [7], semiconductor problems [8], and hydrodynamic problems [9]. Hence, the existence results of positive solutions to this kind of problems have received a great deal of attentions. We refer the readers to [10–15].In [13], Ma studied the following problem:
(1)
u
4
t
=
h
t
f
t
,
u
,
0
<
t
<
1
,
(2)
u
0
=
u
1
=
∫
0
1
p
s
u
s
d
s
,
u
′′
0
=
u
′′
1
=
∫
0
1
q
s
u
′′
s
d
s
,
where p
,
q
∈
L
1
[
0,1
] and h and f are continuous. The existence of at least one symmetric positive solution is obtained by the application of the fixed point index in cones.In [14], authors study the existence and nonexistence of symmetric positive solutions of the following fourth-order BVP:
(3)
ϕ
p
x
′′
t
′′
=
w
t
f
t
,
x
t
,
0
<
t
<
1
,
x
0
=
x
1
=
∫
0
1
g
s
x
s
d
s
,
ϕ
p
x
′′
0
=
ϕ
p
x
′′
1
=
∫
0
1
h
s
ϕ
p
x
′′
s
d
s
.
The argument was based on the fixed point theory in cones.For fourth-order differential equation(4)
u
4
t
+
β
u
′′
t
=
λ
f
t
,
u
t
,
u
′′
t
,
0
<
t
<
1
,
subject to boundary value conditions (2), author in [15] established the existence of positive solutions by the use of the Krasnoseliis fixed point theorem in cone.The existing literature indicates that researches of fourth-order two point BVPs are excellent and methods are developed to be various. However, as to fourth-order BVPs with integral boundary value conditions, methods applied are relatively limited. Most of results are obtained by the use of fixed point theory in the cone or the fix point index theorem.In this paper, we will apply the monotone iterative technique to the following fourth-order BVP with integral boundary conditions:(5)
u
4
t
+
β
u
′′
t
=
f
t
,
u
t
,
u
′′
t
,
0
<
t
<
1
,
(6)
u
0
=
u
1
=
∫
0
1
p
s
u
s
d
s
,
u
′′
0
=
u
′′
1
=
∫
0
1
q
s
u
′′
s
d
s
.
We do not assume that the upper and lower solutions to the boundary value problem should exist but construct the specific form of the symmetric upper and lower solutions. And we will construct successive iterative schemes for approximating solutions. In addition, it is worth stating that the first term of our iterative scheme is a simple function or a constant function. Therefore, the iterative scheme is feasible. Under the appropriate assumptions on nonlinear term, a new and general result to the existence of symmetric positive solution of BVP (5) and (6) is obtained.We assume that the following conditions hold throughout the paper:(S
1)
π
2
/
4
⩽
β
<
π
2;
(S
2)
∫
0
1
p
(
s
)
d
s
<
1
,
∫
0
1
q
(
s
)
sin
β
s
d
s
+
∫
0
1
q
(
s
)
sin
β
(
1
-
s
)
d
s
<
sin
β;
(S
3)
f
∈
C
(
[
0,1
]
×
[
0
,
+
∞
)
×
(
-
∞
,
0
]
,
[
0
,
+
∞
)
),  p
,
q
∈
L
1
[
0,1
],  p
(
s
)
⩾
0,  q
(
s
)
⩾
0.


## 2. Preliminaries

Givenm
(
x
)
∈
L
1
[
0,1
], let
(7)
δ
1
=
1
-
∫
0
1
m
x
d
x
;
δ
2
=
sin
β
-
∫
0
1
m
x
sin
β
x
d
x
-
∫
0
1
m
x
sin
β
1
-
x
d
x
and δ
i
≠
0,  i
=
1,2. Denoted by H
i
(
t
,
s
),  i
=
1,2, the Green’s function of the following problem:
(8)
-
u
′′
t
+
λ
i
u
t
=
0
,
0
<
t
<
1
,
u
0
=
u
1
=
∫
0
1
m
s
u
s
d
s
.
Then, careful calculation yields
(9)
H
1
t
,
s
=
G
1
t
,
s
+
1
δ
1
∫
0
1
G
1
s
,
x
m
x
d
x
,
G
1
t
,
s
=
t
1
-
s
,
0
⩽
t
⩽
s
⩽
1
,
s
1
-
t
,
0
⩽
s
⩽
t
⩽
1
,
H
2
t
,
s
=
G
2
t
,
s
+
sin
β
t
+
sin
β
1
-
t
δ
2
∫
0
1
G
2
s
,
x
m
x
d
x
,
G
2
t
,
s
=
sin
β
t
sin
β
1
-
s
β
sin
β
,
0
⩽
t
⩽
s
⩽
1
,
sin
β
s
sin
β
1
-
t
β
sin
β
,
0
⩽
s
⩽
t
⩽
1
.Denote(10)
ρ
1
=
1
1
-
∫
0
1
p
x
d
x
;
ρ
2
=
sin
β
-
∫
0
1
q
x
sin
β
x
d
x
-
∫
0
1
q
x
sin
β
1
-
x
d
x
;
ρ
2
t
=
1
ρ
2
sin
β
t
+
sin
β
1
-
t
.Lemma 1 (see [15]).
Suppose that(
S
1
)
–
(
S
3
) hold. Then, for any g
∈
C
[
0,1
], u solves the problem
(11)
u
4
t
+
β
u
′′
t
=
g
t
,
0
<
t
<
1
,
u
0
=
u
1
=
∫
0
1
p
s
u
s
d
s
,
u
′′
0
=
u
′′
1
=
∫
0
1
q
s
u
′′
s
d
s
if and only if u
(
t
)
=
∫
0
1
‍
∫
0
1
‍
K
1
t
,
s
K
2
s
,
τ
g
τ
d
τ
d
s, where
(12)
K
1
t
,
s
=
G
1
t
,
s
+
ρ
1
∫
0
1
G
1
s
,
x
p
x
d
x
,
K
2
s
,
τ
=
G
2
s
,
τ
+
ρ
2
s
∫
0
1
G
2
τ
,
x
q
x
d
x
.
During the process of getting the above solution, we can also know(13)
u
′′
t
=
-
∫
0
1
K
2
t
,
s
g
s
d
s
.Lemma 2.
If(
S
2
) is satisfied, the following results are true:(1)
K
i
(
t
,
s
)
⩾
0, for t
,
s
∈
[
0,1
],  i
=
1,2;
(2)
G
i
(
1
-
t
,
1
-
s
)
=
G
i
(
t
,
s
)
,
G
i
(
t
,
s
)
⩽
G
i
(
t
,
t
), for t
,
s
∈
[
0,1
],  i
=
1,2.Denote(14)
l
1
=
∫
0
1
G
1
x
,
x
p
x
d
x
;
l
2
=
∫
0
1
G
2
x
,
x
q
x
d
x
;
h
t
=
G
2
t
,
t
+
ρ
2
t
l
2
;
l
3
=
∫
0
1
h
t
d
t
.
As β
∈
[
π
/
2
,
π
), it is easy to check that h
(
t
)
=
h
(
1
-
t
) and h
′′
(
t
)
⩽
0, for t
∈
[
0,1
]. Hence, from the symmetry and concavity of h
(
t
), we have
(15)
h
t
⩽
h
1
2
⩽
2
∫
0
1
h
t
d
t
=
2
l
3
.
In addition, for t
,
s
∈
[
0,1
], the following results hold:
(16)
K
1
t
,
s
=
G
1
t
,
s
+
ρ
1
∫
0
1
G
1
s
,
x
p
x
d
x
⩽
G
1
t
,
t
+
ρ
1
∫
0
1
G
1
x
,
x
p
x
d
x
=
t
1
-
t
+
ρ
1
l
1
,
(17)
K
1
1
2
,
s
⩽
1
4
+
ρ
1
l
1
.
Further,
(18)
K
2
s
,
τ
=
G
2
s
,
τ
+
ρ
2
s
∫
0
1
G
2
τ
,
x
q
x
d
x
⩽
G
2
s
,
s
+
ρ
2
s
∫
0
1
G
2
x
,
x
q
x
d
x
=
G
2
s
,
s
+
ρ
2
s
l
2
=
h
s
and therefore
(19)
K
2
1
2
,
τ
⩽
h
1
2
.We consider Banach spaceE
=
C
2
[
0,1
] equipped with the norm u
=
max
⁡
u
∞
,
u
′′
∞, where u
∞
=
sup
⁡
t
∈
[
0,1
]
|
u
(
t
)
|. In this paper, a symmetric positive solution u
* of (5) means a function which is symmetric and positive on (
0,1
) and satisfies (5) as well as the boundary conditions (6).In this paper, we always suppose that the following assumptions hold:(H1)
f
(
t
,
u
,
v
)
=
f
(
1
-
t
,
u
,
v
) for t
∈
[
0,1
],  u
∈
[
0
,
∞
),  v
∈
(
-
∞
,
0
];
(H2)
f
(
t
,
u
1
,
v
1
)
⩽
f
(
t
,
u
2
,
v
2
), for 0
⩽
u
1
⩽
u
2
<
+
∞,  0
⩽
|
v
1
|
⩽
|
v
2
|
<
+
∞;
(H3)
f
(
t
,
0,0
)
≢
0, for t
∈
[
0,1
].Denote(20)
P
=
u
∈
E
:
u
t
⩾
0
,
u
′′
t
⩽
0
,
u
t
=
u
1
-
t
,
t
∈
0,1
u
∈
E
:
u
t
⩾
0
,
u
′′
.
It is easy to see that P is a cone in E.We define the operatorT as follows:
(21)
T
u
t
=
∬
0
1
K
1
t
,
s
K
2
s
,
τ
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
,
00000000000000000000000000000000000000
i
t
∈
0,1
.
By the above argument, we know that, for any u
∈
E, T
u
∈
E and
(22)
T
u
′′
t
=
-
∫
0
1
K
2
t
,
τ
f
τ
,
u
τ
,
u
′′
τ
d
τ
,
0000000000000000000000000000000
t
∈
0,1
.Lemma 3.
If(
S
1
)
–
(
S
3
) are satisfied, T
:
P
→
P is completely continuous; that is, T is continuous and compact.Proof.
For anyu
∈
P, from (21) and (22), combining Lemma 2 and (S
3), we know that (
T
u
)
(
t
)
⩾
0 and T
u
′′
(
t
)
⩽
0 for t
∈
[
0,1
]. We now prove that T
u is symmetric about 1
/
2.
Fort
∈
[
0,1
],
(23)
T
u
1
-
t
=
∬
0
1
K
1
1
-
t
,
s
K
2
s
,
τ
·
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
=
∬
1
0
G
1
1
-
t
,
1
-
s
G
2
1
-
s
,
1
-
τ
·
f
1
-
τ
,
u
1
-
τ
,
u
′′
1
-
τ
·
d
1
-
τ
d
1
-
s
+
∫
1
0
∫
0
1
G
1
1
-
t
,
1
-
s
·
ρ
2
1
-
s
∫
0
1
G
2
τ
,
x
q
x
d
x
·
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
1
-
s
+
∬
0
1
ρ
1
∫
0
1
G
1
s
,
x
p
x
d
x
·
G
2
s
,
τ
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
+
∬
0
1
ρ
1
∫
0
1
G
1
s
,
x
p
x
d
x
·
G
2
s
,
τ
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
=
∬
0
1
G
1
t
,
s
G
2
s
,
τ
·
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
+
∬
0
1
G
1
t
,
s
ρ
2
s
∫
0
1
G
2
τ
,
x
q
x
d
x
·
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
+
∬
0
1
ρ
1
∫
0
1
G
1
s
,
x
p
x
d
x
·
G
2
s
,
τ
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
+
∬
0
1
ρ
1
∫
0
1
G
1
s
,
x
p
x
d
x
·
G
2
s
,
τ
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
=
T
u
t
.
So, T
P
⊂
P. The continuity of T is obvious. We now prove that T is compact. Let Ω
⊂
P be a bounded set. Then, there exists R such that
(24)
Ω
=
u
∈
P
:
u
⩽
R
.
For any u
∈
Ω, we have
(25)
0
⩽
f
t
,
u
t
,
u
′′
t
⩽
max
⁡
f
t
,
u
,
u
′′
∣
t
∈
0,1
,
u
∈
0
,
R
,
u
′′
∈
-
R
,
0
=
:
M
.
Therefore, from (17) and (18), we have
(26)
T
u
∞
=
T
u
1
2
=
∬
0
1
K
1
1
2
,
s
K
2
s
,
τ
·
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
⩽
M
∬
0
1
1
4
+
ρ
1
l
1
h
s
d
τ
d
s
=
1
4
+
ρ
1
l
1
l
3
M
,
and from (19), we have
(27)
T
u
′′
∞
=
-
T
u
′′
1
2
=
∫
0
1
K
2
1
2
,
τ
f
τ
,
u
τ
,
u
′′
τ
d
τ
⩽
M
∫
0
1
h
1
2
d
τ
=
M
h
1
2
.
So, T
u is uniformly bounded. Next we prove that T
u is equicontinuous.
For0
⩽
t
1
⩽
t
2
⩽
s
⩽
1, we have
(28)
T
u
t
2
-
T
u
t
1
∞
=
∬
0
1
K
1
t
2
,
s
-
K
1
t
1
,
s
·
K
2
s
,
τ
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
⩽
M
∬
0
1
K
1
t
2
,
s
-
K
1
t
1
,
s
·
K
2
s
,
τ
d
τ
d
s
=
M
∬
0
1
G
1
t
2
,
s
-
G
1
t
1
,
s
·
K
2
s
,
τ
d
τ
d
s
⩽
M
∬
0
1
t
2
-
t
1
K
2
s
,
τ
d
τ
d
s
=
N
1
t
2
-
t
1
,
where N
1
=
M
∬
0
1
K
2
s
,
τ
d
τ
d
s and
(29)
T
u
′′
t
2
-
T
u
′′
t
1
∞
⩽
M
∫
0
1
K
2
t
2
,
τ
-
K
2
t
1
,
τ
d
τ
⩽
M
∫
0
1
G
2
t
2
,
τ
-
G
2
t
1
,
τ
d
τ
+
M
ρ
2
t
2
-
ρ
2
t
1
·
∬
0
1
G
2
τ
,
x
q
x
d
x
d
τ
.
According to the Lagrange mean value theorem, we obtain that
(30)
G
2
t
2
,
τ
-
G
2
t
1
,
τ
=
sin
β
1
-
τ
β
sin
β
sin
β
t
2
-
sin
β
t
1
⩽
1
β
sin
β
β
t
2
-
β
t
1
=
1
sin
β
t
2
-
t
1
.
Similarly, we have
(31)
ρ
2
t
2
-
ρ
2
t
1
⩽
2
β
ρ
2
t
2
-
t
1
.
Hence, there exists a positive constant N
2 such that
(32)
T
u
′′
t
2
-
T
u
′′
t
1
∞
⩽
N
2
t
2
-
t
1
.
And the similar results can be obtained for 0
⩽
s
⩽
t
1
⩽
t
2
⩽
1 and 0
⩽
t
1
⩽
s
⩽
t
2
⩽
1.
The Arzelà-Ascoli theorem guarantees thatT
Ω is relatively compact which means that T is compact.


## 3. Existence and Iterative of Solutions for BVP (5) and (6)

Theorem 4.
Assume that(
H
1
)
–
(
H
3
) hold. If there exists two positive numbers a
1
<
a such that
(33)
sup
⁡
t
∈
0,1
f
t
,
a
,
a
⩽
a
1
,
where a and a
1 satisfy
(34)
a
⩾
max
⁡
1
4
+
ρ
1
l
1
l
3
,
2
l
3
,
h
1
2
a
1
.
Then, problem (5) and (6) has concave symmetric positive solution w
*
,
v
*
∈
P with
(35)
w
*
∞
⩽
a
,
lim
⁡
n
→
∞
T
n
w
0
=
w
*
,
where
(36)
w
0
t
=
a
1
l
3
t
1
-
t
+
ρ
1
l
1
,
v
*
∞
⩽
a
,
lim
⁡
n
→
∞
T
n
v
0
=
v
*
,
where
(37)
v
0
t
=
0
.Proof.
We denoteP
a
=
{
w
∈
P
:
w
⩽
a
}. In what follows, we first prove that T
P
a
⊂
P
a.
Letw
∈
P
a; then 0
⩽
w
(
t
)
⩽
sup
⁡
t
∈
0,1
w
(
t
)
=
w
∞
⩽
a, |
w
′′
(
t
)
|
⩽
sup
⁡
t
∈
0,1
|
w
′′
(
t
)
|
=
w
′′
∞
⩽
a.
By assumption(
H
2
) and (33), for t
∈
[
0,1
], we have
(38)
0
⩽
f
t
,
w
t
,
w
′′
t
⩽
a
1
.
For anyw
(
t
)
∈
P
a, by Lemma 3, we know that T
w
∈
P. According to (17), (18), and (33), we get
(39)
T
w
∞
=
T
w
1
2
=
∬
0
1
K
1
1
2
,
s
K
2
s
,
τ
·
f
τ
,
u
τ
,
u
′′
τ
d
τ
d
s
⩽
a
1
∬
0
1
1
4
+
ρ
1
l
1
h
s
d
τ
d
s
=
a
1
1
4
+
ρ
1
l
1
∫
0
1
h
s
d
s
=
a
1
l
3
1
4
+
ρ
1
l
1
⩽
a
,
and from (19) and (33), we get
(40)
T
w
′′
∞
=
-
T
w
′′
1
2
=
∫
0
1
K
2
1
2
,
τ
f
τ
,
u
τ
,
u
′′
τ
d
τ
⩽
a
1
∫
0
1
K
2
1
2
,
τ
d
τ
⩽
a
1
∫
0
1
h
1
2
d
τ
=
a
1
h
1
2
⩽
a
.
Hence,T
w
⩽
a. Thus, we get T
P
a
⊂
P
a. Let w
0
(
t
)
=
a
1
l
3
(
t
(
1
-
t
)
+
ρ
1
l
1
), for t
∈
[
0,1
]; then w
0
⩽
a and w
0
(
t
)
∈
P
a. Let w
1
(
t
)
=
T
w
0; then w
1
∈
P
a. We denote
(41)
w
n
+
1
=
T
w
n
=
T
n
+
1
w
0
n
=
0,1
,
2
,
…
.
From the definition ofT, (16), (18), and (38), it follows that
(42)
w
1
t
=
T
w
0
t
=
∬
0
1
K
1
t
,
s
K
2
s
,
τ
·
f
τ
,
w
0
τ
,
w
0
′′
τ
d
τ
d
s
⩽
a
1
∬
0
1
K
1
t
,
s
K
2
s
,
τ
d
τ
d
s
⩽
a
1
∬
0
1
t
1
-
t
+
ρ
1
l
1
h
s
d
τ
d
s
=
a
1
t
1
-
t
+
ρ
1
l
1
∫
0
1
h
s
d
s
=
a
1
l
3
t
1
-
t
+
ρ
1
l
1
=
w
0
t
.
On the other hand, from (15), (18), and (38), we have
(43)
w
1
′′
t
=
T
w
0
′′
t
=
∫
0
1
K
2
t
,
τ
f
τ
,
w
0
τ
,
w
0
′′
τ
d
τ
⩽
a
1
∫
0
1
K
2
t
,
τ
d
τ
⩽
a
1
∫
0
1
h
t
d
τ
=
a
1
h
t
⩽
2
a
1
l
3
=
w
0
′′
t
.
From (
H
2
), it follows that
(44)
w
2
t
=
T
w
1
t
⩽
T
w
0
t
=
w
1
t
,
0
⩽
t
⩽
1
,
w
2
′′
t
=
T
w
1
′′
t
⩽
T
w
0
′′
t
=
w
1
′′
t
,
0
⩽
t
⩽
1
.
By induction,
(45)
w
n
+
1
t
⩽
w
n
t
,
w
n
+
1
′′
t
⩽
w
n
′′
t
,
0
⩽
t
⩽
1
n
=
0,1
,
2
,
…
.
SinceT
P
a
⊂
P
a, we have w
n
∈
P
a, n
=
0,1
,
2
,
…. From Lemma 3, T is completely continuous. We assert that {
w
n
}
n
=
1
∞ has a convergent subsequence {
w
n
k
}
k
=
1
∞ and there exists w
*
∈
P
a such that w
n
k
→
w
*.
Letv
0
=
0,  t
∈
[
0,1
]; then v
0
∈
P
a. Let v
1
=
T
v
0; then v
1
∈
P
a; we denote
(46)
v
n
+
1
=
T
v
n
=
T
n
+
1
v
0
,
n
=
0,1
,
2
,
…
.
Similarly to{
w
n
}
n
=
1
∞, we assert that {
v
n
}
n
=
1
∞ has a convergent subsequence {
v
n
k
}
k
=
1
∞ and there exists v
*
∈
P
a, such that v
n
k
→
v
*.
Sincev
1
(
t
)
=
(
T
v
0
)
(
t
)
=
(
T
0
)
(
t
)
∈
P
a, we have
(47)
v
1
t
=
T
v
0
t
=
T
0
t
⩾
0
,
0
⩽
t
⩽
1
,
v
1
′′
t
=
T
v
0
′′
t
=
T
0
′′
t
⩾
0
,
0
⩽
t
⩽
1
.
Hence,
(48)
v
2
t
=
T
v
1
t
⩾
T
0
t
=
v
1
t
,
0
⩽
t
⩽
1
,
v
2
′′
t
=
T
v
1
′′
t
⩾
T
0
′′
t
=
v
1
′′
t
,
0
⩽
t
⩽
1
.
By induction, v
n
+
1
⩾
v
n, |
v
′′
n
+
1
(
t
)
|
⩾
|
v
′′
n
(
t
)
|, 0
⩽
t
⩽
1, (n
=
0,1
,
2
,
…). Hence, we assert that v
n
→
v
*, T
v
*
=
v
*.
Iff
(
t
,
0,0
)
≢
0,  0
⩽
t
⩽
1, then the zero function is not the solution of BVP (5) and (6). Thus, max
⁡
0
⩽
t
⩽
1
|
v
*
(
t
)
|
>
0; we have
(49)
v
*
⩾
min
⁡
⁡
t
,
1
-
t
max
⁡
0
⩽
t
⩽
1
v
*
t
>
0
,
0
<
t
<
1
.
It is well known that the fixed point of operatorT is the solution of BVP (5) and (6). Therefore, w
* and v
* are two positive, concave, and symmetric solutions of BVP (5) and (6).Example 5.
Consider the following fourth-order boundary value problem with integral boundary conditions:(50)
u
4
t
+
π
2
4
u
′′
t
=
f
t
,
u
t
,
u
′′
t
,
0
<
t
<
1
,
u
0
=
u
1
=
∫
0
1
s
u
s
d
s
,
u
′′
0
=
u
′′
1
=
∫
0
1
s
u
′′
s
d
s
,
where
(51)
f
t
,
u
,
v
=
1
30
t
1
-
t
+
1
2
v
2
+
ln
⁡
⁡
u
+
2
.
The calculation yields(52)
ρ
1
=
2
,
ρ
2
=
π
-
2
π
,
l
1
=
1
12
,
l
2
=
1
π
2
,
l
3
=
2
π
π
-
2
,
ρ
2
t
=
sin
⁡
π
/
2
t
+
sin
⁡
π
/
2
1
-
t
ρ
2
,
h
t
=
2
π
sin
π
2
t
sin
π
2
1
-
t
+
sin
⁡
π
/
2
t
+
sin
⁡
π
/
2
1
-
t
π
π
-
2
,
h
1
2
=
π
-
2
+
2
π
π
-
2
.
It is easy to check that assumptions (
S
1
)
–
(
S
3
) hold. Set a
=
12, a
1
=
10. Then we can verify that conditions (
H
1
) and (
H
2
) and (33) are satisfied. Then applying Theorem 4, BVP (50) has two concave symmetric positive solutions w
*
,
v
*
∈
P with
(53)
w
*
∞
⩽
10
,
lim
⁡
n
→
∞
T
n
w
0
=
w
*
,
where
(54)
w
0
t
=
60
t
1
-
t
+
10
3
π
π
-
2
.
v
*
∞
⩽
10
,
lim
⁡
n
→
∞
T
n
v
0
=
v
*
,
where
(55)
v
0
t
=
0
.



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*Source: 583875-2014-08-18.xml*