# Existence and Monotone Iteration of Positive Pseudosymmetric Solutions for a Third-Order Four-Point BVP

Authors: Xuezhe Lv; Libo Wang; Minghe Pei
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We study the existence and monotone iteration of solutions for a third-order four-point boundary value problem. An existence result of positive, concave, and pseudosymmetric solutions and its monotone iterative scheme are established by using the monotone iterative technique. Meanwhile, as an application of our results, an example is given.


---

## Body

## 1. Introduction

The third-order equations arise in many areas of applied mathematics and physics, such as the deflection of a curved beam having a constant or varying cross section, three-layer beam, electromagnetic waves, or gravity-driven flows [1], and thus have been studied extensively in the literature; see [1–29] and references therein. Recently, wide attention has been paid to the third-order boundary value problems with nonlocal boundary conditions; see [4, 6–9, 11, 12, 15, 16, 20, 23–30] and references therein.In 2006, using the monotone iterative technique, Zhou and Ma [30] obtained the existence of positive solutions and established a corresponding iterative scheme for the following third-order p-Laplacian problem:
(1)
(
ϕ
p
(
u
′′
(
t
)
)
)
′
=
q
(
t
)
f
(
t
,
u
(
t
)
)
,
t
∈
(
0,1
)
,
u
(
0
)
=
∑
i
=
1
m
‍
α
i
u
(
ξ
i
)
,
u
′
(
η
)
=
0
,
u
′′
(
1
)
=
∑
i
=
1
n
‍
β
i
u
′′
(
θ
i
)
.In 2009, Sun et al. [23] studied the existence of positive solutions for the following third-order p-Laplacian problem:
(2)
(
ϕ
p
(
u
′′
(
t
)
)
)
′
=
q
(
t
)
f
(
t
,
u
(
t
)
,
u
′
(
t
)
,
u
′′
(
t
)
)
,
t
∈
(
0,1
)
,
u
(
0
)
=
∑
i
=
1
m
‍
α
i
u
(
ξ
i
)
,
u
′
(
η
)
=
0
,
u
′′
(
1
)
=
∑
i
=
1
n
‍
β
i
u
′′
(
θ
i
)
.
By applying a monotone iterative method, the authors obtained the existence of positive solutions for the problem and established iterative schemes for approximating the solutions.In 2011, Zhang [29] considered the following singular third-order three-point boundary value problem:
(3)
u
′′′
(
t
)
+
f
(
t
,
u
(
t
)
)
=
0
,
t
∈
(
0,1
)
,
u
(
0
)
=
u
′
(
0
)
=
0
,
u
′
(
1
)
=
α
u
′
(
η
)
.
The existence and uniqueness of solutions and corresponding iterative scheme to the problem are obtained by applying the cone theory and the Banach contraction mapping principle.In 2013, Li et al. [7] studied third-order four-point boundary value problem with p-Laplacian of the form
(4)
(
ϕ
p
(
u
′′
(
t
)
)
)
′
+
q
(
t
)
f
(
t
,
u
(
t
)
,
u
′
(
t
)
)
=
0
,
t
∈
(
0,1
)
,
u
(
0
)
=
0
,
u
(
1
)
=
u
(
η
)
,
u
′′
(
1
+
η
2
)
=
0
.
By using the monotone iterative technique, the existence result of positive pseudosymmetric solutions and its monotone iterative scheme are established for the problem.Motivated by above works and [31], in this paper, we consider the existence and monotone iteration of positive pseudosymmetric solutions of the following third-order four-point boundary value problem:
(5)
u
′′′
(
t
)
+
q
(
t
)
f
(
t
,
u
(
t
)
,
u
′
(
t
)
,
u
′′
(
t
)
)
=
0
,
t
∈
(
0,1
)
,
subject to boundary conditions
(6)
α
u
(
0
)
-
β
u
′
(
0
)
=
0
,
u
′
(
η
)
+
u
′
(
1
)
=
0
,
u
′′
(
1
+
η
2
)
=
0
,
where η
∈
(
0,1
) and α
>
0, β
≥
0. Here we say u
*
(
t
) is positive solution of BVP (5), (6), if u
*
(
t
) is the solution of BVP (5), (6) and satisfies u
*
(
t
)
>
0 for t
∈
(
0,1
].To the best of our knowledge, the pseudosymmetric solutions for the second-order boundary value problem have been studied by some authors, see [31–33]. And [7] is the only one concerned with the third-order boundary value problem. We note that the nonlinearity of f in our problem contains explicitly t and every derivatives of u up to order two.This work is organized as follows. In Section2, some notations and preliminaries are introduced. The main results are discussed in Section 3. As an application of our results, an example is given in the last section.


## 2. Preliminary

In this section, we give a definition and some lemmas which help to simplify the presentation of our main result.Definition 1 (see [33]).
Letu
∈
C
[
0,1
], η
∈
(
0,1
). We say u is pseudosymmetric about η on [
0,1
], if u is symmetric on [
η
,
1
]; that is,
(7)
u
(
t
)
=
u
(
1
+
η
-
t
)
,
∀
t
∈
[
η
,
1
]
.
Let the Banach space E
=
C
2
[
0,1
] be endowed with the norm
(8)
∥
u
∥
=
max
⁡
{
∥
u
∥
∞
,
∥
u
′
∥
∞
,
∥
u
′′
∥
∞
}
,
where ∥
u
(
i
)
∥
∞
=
max
⁡
0
≤
t
≤
1
|
u
(
i
)
(
t
)
|,  i
=
0,1
,
2. Define a cone P
⊂
E by
(9)
P
=
{
u
∈
E
∣
u
(
t
)
is
nonnegative
,
concave
,
and
pseudosymmetric
about
η
on
⁡
[
0,1
]
}
,
and by “≤” denote the induced partial ordering via cone P; that is, for u
1,  u
2
∈
E, u
1
≤
u
2 if and only if u
2
-
u
1
∈
P.
For convenience, we denote the following.(
H
0
)
q
(
t
) is a nonnegative continuous function defined on (
0,1
),  q
(
t
)
≢
0 on any subinterval of (
0,1
). In addition, ∫
0
1
‍
q
(
t
)
d
t
<
+
∞ and
(10)
q
(
t
)
=
q
(
1
+
η
-
t
)
on
⁡
(
η
,
1
)
.
(
H
1
)  f
(
t
,
u
,
v
,
w
)
:
[
0,1
]
×
[
0
,
+
∞
)
×
ℝ
×
(
-
∞
,
0
]
→
ℝ is continuous,
(11)
f
(
t
,
u
,
v
,
w
)
≤
0
on
⁡
[
0
,
1
+
η
2
]
×
[
0
,
+
∞
)
2
×
(
-
∞
,
0
]
,
and, for all(
t
,
u
,
v
,
w
)
∈
[
η
,
1
]
×
[
0
,
+
∞
)
×
ℝ
×
(
-
∞
,
0
],
(12)
f
(
t
,
u
,
v
,
w
)
=
-
f
(
1
+
η
-
t
,
u
,
-
v
,
w
)
.
(
H
2
)  f
(
t
,
u
,
v
,
w
) is nonincreasing in u, nonincreasing in v, and nondecreasing in w on [
0
,
(
1
+
η
)
/
2
]
×
[
0
,
+
∞
)
2
×
(
-
∞
,
0
].
(
H
3
)  f
(
t
,
0,0
,
0
)
≢
0 on [
0,1
].
Now, we define an operatorT
:
P
→
C
2
[
0,1
] as follows: for u
∈
P,
(13)
(
T
u
)
(
t
)
=
{
∫
0
t
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
,
0
≤
t
≤
1
+
η
2
,
∫
t
1
‍
∫
(
1
+
η
)
/
2
τ
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
∫
0
η
‍
∫
τ
(
1
+
η
)
/
2
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
,
1
+
η
2
≤
t
≤
1
.
Obviously, under assumptions (
H
0
) and (
H
1
), the operator T is well defined.Lemma 2.
Assume that(
H
0
) and (
H
1
) hold. Then u
∈
P is a solution of BVP (5), (6) if and only if u
∈
P is a fixed point of T.Proof.
At first we show the necessity. Supposeu
∈
P is a solution of BVP (5), (6). Then, integrating (5) and using (6) we infer that
(14)
u
′′
(
t
)
+
∫
(
1
+
η
)
/
2
t
‍
q
(
r
)
f
(
r
,
u
(
r
)
,
u
′
(
r
)
,
u
′′
(
r
)
)
d
r
=
0
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
∈
[
0,1
]
,
(15)
α
u
(
0
)
-
β
u
′
(
0
)
=
0
,
u
′
(
η
)
+
u
′
(
1
)
=
0
.
For t
∈
[
0
,
(
1
+
η
)
/
2
], integrating (14) on [
t
,
(
1
+
η
)
/
2
] and taking into account u
′
(
(
1
+
η
)
/
2
)
=
0, we get
(16)
u
′
(
t
)
=
∫
t
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′′
(
s
)
)
d
s
d
r
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
∈
[
0
,
1
+
η
2
]
.
Again integrating (16) on [
0
,
t
]
⊂
[
0
,
(
1
+
η
)
/
2
] one can obtain
(17)
u
(
t
)
=
∫
0
t
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
u
(
0
)
.
But, from (15) and (16), it follows that
(18)
u
(
0
)
=
β
α
∫
0
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
.
Therefore, for t
∈
[
0
,
(
1
+
η
)
/
2
], one has
(19)
u
(
t
)
=
∫
0
t
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
.
For t
∈
[
(
1
+
η
)
/
2
,
1
], integrating (14) on [
(
1
+
η
)
/
2
,
t
], we get
(20)
u
′
(
t
)
=
-
∫
(
1
+
η
)
/
2
t
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
∈
[
1
+
η
2
,
1
]
.
Again integrating (20) on [
t
,
1
]
⊂
[
(
1
+
η
)
/
2
,
1
] one obtains
(21)
u
(
t
)
=
∫
t
1
‍
∫
(
1
+
η
)
/
2
τ
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
u
(
1
)
,
t
∈
[
1
+
η
2
,
1
]
.
In particular, we have
(22)
u
(
1
+
η
2
)
=
∫
(
1
+
η
)
/
2
1
‍
∫
(
1
+
η
)
/
2
τ
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
u
(
1
)
.
In (19), we take t
=
(
1
+
η
)
/
2, and then
(23)
u
(
1
+
η
2
)
=
∫
0
(
1
+
η
)
/
2
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
.
From (22) and (23) one has
(24)
u
(
1
)
=
∫
0
1
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
.
Hence from (21) it follows that
(25)
u
(
t
)
=
∫
t
1
‍
∫
(
1
+
η
)
/
2
τ
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
∫
0
1
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
=
∫
t
1
‍
∫
(
1
+
η
)
/
2
τ
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
(
∫
0
η
‍
+
∫
η
(
1
+
η
)
/
2
‍
+
∫
(
1
+
η
)
/
2
1
‍
)
×
∫
τ
(
1
+
η
)
/
2
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
‍
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
∈
[
1
+
η
2
,
1
]
.
Notice that from (
H
0
), (
H
1
), and the fact that u
∈
P, we have
(26)
(
∫
η
(
1
+
η
)
/
2
‍
+
∫
(
1
+
η
)
/
2
1
‍
)
×
∫
τ
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
=
0
.
Therefore
(27)
u
(
t
)
=
∫
t
1
‍
∫
(
1
+
η
)
/
2
τ
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
∫
0
η
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
∈
[
1
+
η
2
,
1
]
.
This together with (19) implies that u
(
t
) is fixed point of T.
The sufficiency, by direct computation and using the fact thatu
∈
P, follows immediately.The following lemmas are some properties of the operatorT.Lemma 3.
Assume that(
H
0
) and (
H
1
) hold. Then T
P
⊂
P.Proof.
From the definition ofT, it is easy to check that T
u is nonnegative on [
0,1
] and satisfies (6) for all u
∈
P. Furthermore, since
(28)
(
T
u
)
′
′
(
t
)
=
-
∫
(
1
+
η
)
/
2
t
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
∈
[
0,1
]
,
it follows that T
u is concave on [
0,1
].
Next we prove thatT
u is pseudosymmetric about η on [
0,1
]. In fact, if t
∈
[
η
,
(
1
+
η
)
/
2
], then 1
+
η
-
t
∈
[
(
1
+
η
)
/
2
,
1
]; it follows that
(29)
(
T
u
)
(
1
+
η
-
t
)
=
∫
1
+
η
-
t
1
‍
∫
(
1
+
η
)
/
2
τ
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
∫
0
η
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
.
Note that u is pseudosymmetric about η on [
0,1
]; that is, u
(
t
)
=
u
(
1
+
η
-
t
) for t
∈
[
η
,
1
], and then
(30)
u
′
(
t
)
=
-
u
′
(
1
+
η
-
t
)
,
u
′′
(
t
)
=
u
′′
(
1
+
η
-
t
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
∈
[
η
,
1
]
.
Thus, for all r
∈
[
η
,
1
], from (
H
0
) and (
H
1
), we have
(31)
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
=
-
∫
(
1
+
η
)
/
2
r
‍
q
(
1
+
η
-
s
)
×
f
(
1
+
η
-
s
,
u
(
s
)
,
-
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
=
-
∫
(
1
+
η
)
/
2
r
‍
q
(
1
+
η
-
s
)
×
f
(
u
′
′
1
+
η
-
s
,
u
(
1
+
η
-
s
)
,
u
′
(
1
+
η
-
s
)
,
u
′
′
(
1
+
η
-
s
)
)
d
s
=
∫
(
1
+
η
)
/
2
1
+
η
-
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
.
Hence, for t
∈
[
η
,
(
1
+
η
)
/
2
],
(32)
∫
1
+
η
-
t
1
‍
∫
(
1
+
η
)
/
2
τ
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
=
-
∫
1
+
η
-
t
1
‍
∫
(
1
+
η
)
/
2
1
+
η
-
τ
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
=
-
∫
t
η
‍
∫
(
1
+
η
)
/
2
τ
¯
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
(
-
τ
¯
)
=
∫
t
η
‍
∫
(
1
+
η
)
/
2
τ
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
.
From (29) and (32), it follows that
(33)
(
T
u
)
(
1
+
η
-
t
)
=
∫
t
η
‍
∫
(
1
+
η
)
/
2
τ
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
∫
0
η
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
=
∫
0
t
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
=
(
T
u
)
(
t
)
,
t
∈
[
η
,
1
+
η
2
]
.
If t
∈
[
(
1
+
η
)
/
2
,
1
], then 1
+
η
-
t
∈
[
η
,
(
1
+
η
)
/
2
]. From (33), we have
(34)
(
T
u
)
(
1
+
η
-
t
)
=
(
T
u
)
(
1
+
η
-
(
1
+
η
-
t
)
)
=
(
T
u
)
(
t
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
∈
[
1
+
η
2
,
1
]
.
This together with (33) implies
(35)
(
T
u
)
(
t
)
=
(
T
u
)
(
1
+
η
-
t
)
,
t
∈
[
η
,
1
]
.
In summary, T
u
∈
P, and then T
P
⊂
P.The following lemma can be easily verified by a standard argument.Lemma 4.
Assume that(
H
0
) and (
H
1
) hold. Then T
:
P
→
P is completely continuous.Lemma 5.
Assume that(
H
0
), (
H
1
), and (
H
2
) hold. Then T is nondecreasing on P; that is, T
u
1
≤
T
u
2 for u
1,  u
2
∈
P with u
1
≤
u
2.Proof.
Letu
1,  u
2
∈
P with u
1
≤
u
2. Then u
2
-
u
1
∈
P. By the definition of P, u
2
(
t
)
-
u
1
(
t
) is nonnegative, concave, and pseudosymmetric about η on [
0,1
]. Therefore
(36)
u
2
′
(
t
)
≥
u
1
′
(
t
)
,
0
≤
t
≤
1
+
η
2
;
u
2
′
(
t
)
≤
u
1
′
(
t
)
,
1
+
η
2
≤
t
≤
1
,
u
2
′′
(
t
)
≤
u
1
′′
(
t
)
,
t
∈
[
0,1
]
.
From (
H
2
) and the definition of T it follows that
(37)
(
T
u
2
)
(
t
)
-
(
T
u
1
)
(
t
)
=
∫
0
t
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
[
f
(
s
,
u
2
(
s
)
,
u
2
′
(
s
)
,
u
2
′′
(
s
)
)
-
f
(
s
,
u
1
(
s
)
,
u
1
′
(
s
)
,
u
1
′′
(
s
)
)
]
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
[
f
(
s
,
u
2
(
s
)
,
u
2
′
(
s
)
,
u
2
′′
(
s
)
)
-
f
(
s
,
u
1
(
s
)
,
u
1
′
(
s
)
,
u
1
′′
(
s
)
)
]
d
s
d
r
≥
0
,
t
∈
[
0
,
1
+
η
2
]
,
(38)
(
T
u
2
)
′
′
(
t
)
-
(
T
u
1
)
′
′
(
t
)
=
∫
t
(
1
+
η
)
/
2
‍
q
(
s
)
[
f
(
s
,
u
2
(
s
)
,
u
2
′
(
s
)
,
u
2
′′
(
s
)
)
-
f
(
s
,
u
1
(
s
)
,
u
1
′
(
s
)
,
u
1
′′
(
s
)
)
]
d
s
≤
0
,
t
∈
[
0
,
1
+
η
2
]
.
We now prove that (37) and (38) hold for t
∈
[
(
1
+
η
)
/
2
,
1
]. In fact, if t
∈
[
(
1
+
η
)
/
2
,
1
], then 1
+
η
-
t
∈
[
η
,
(
1
+
η
)
/
2
]
⊂
[
0
,
(
1
+
η
)
/
2
], and hence, from the fact that  T
u
1 and T
u
2 are pseudosymmetric about η on [
0,1
], it follows that, for t
∈
[
(
1
+
η
)
/
2
,
1
],
(39)
(
T
u
2
)
(
t
)
-
(
T
u
1
)
(
t
)
=
(
T
u
2
)
(
1
+
η
-
t
)
-
(
T
u
1
)
(
1
+
η
-
t
)
≥
0
,
(
T
u
2
)
′′
(
t
)
-
(
T
u
1
)
′′
(
t
)
=
(
T
u
2
)
′′
(
1
+
η
-
t
)
-
(
T
u
1
)
′′
(
1
+
η
-
t
)
≤
0
.
So
(40)
(
T
u
2
)
(
t
)
-
(
T
u
1
)
(
t
)
≥
0
,
t
∈
[
0,1
]
,
and (
T
u
2
)
(
t
)
-
(
T
u
1
)
(
t
) is concave on [
0,1
].
Finally, we show that(
T
u
2
)
(
t
)
-
(
T
u
1
)
(
t
) is pseudosymmetric about η on [
0,1
]. To do this we let S
(
t
)
=
(
T
u
2
)
(
t
)
-
(
T
u
1
)
(
t
). We note that T
u
1,  T
u
2
∈
P, (
T
u
1
)
(
t
) and (
T
u
2
)
(
t
) are pseudosymmetric about η on [
0,1
], and thus
(41)
S
(
1
+
η
-
t
)
=
(
T
u
2
)
(
1
+
η
-
t
)
-
(
T
u
1
)
(
1
+
η
-
t
)
=
(
T
u
2
)
(
t
)
-
(
T
u
1
)
(
t
)
=
S
(
t
)
,
aaaaaaaaaaaaaaaaaaaaaaaaaa
t
∈
[
η
,
1
]
.
In summary, T
u
2
-
T
u
1
∈
P; that is, T
u
1
≤
T
u
2.


## 3. Main Result

Now we establish existence result of positive, concave, and pseudosymmetric solutions and its monotone iterative scheme for BVP (5), (6).Theorem 6.
Assume that(
H
0
), (
H
1
), (
H
2
), and (
H
3
) hold. Suppose also that there exist two positive numbers a and b with a
>
b such that
(42)
inf
⁡
t
∈
(
0
,
(
1
+
η
)
/
2
]
q
(
t
)
f
(
t
,
a
,
a
,
-
a
)
≥
-
b
,
where a
,
b satisfy
(43)
a
≥
max
⁡
{
β
(
1
+
η
)
2
4
α
(
7
8
-
1
2
(
1
+
η
)
2
)
-
1
,
(
1
+
η
)
3
8
+
β
(
1
+
η
)
2
4
α
,
(
1
+
η
)
2
}
b
.
Then BVP (5), (6) has positive, concave, and pseudosymmetric solutions w
*,  v
*
∈
P with
(44)
∥
w
*
∥
∞
≤
a
,
lim
⁡
n
→
∞
T
n
w
0
=
w
*
,
w
h
e
r
e
w
0
(
t
)
=
a
t
(
1
+
η
-
t
)
2
(
1
+
η
)
2
+
7
8
a
;
∥
v
*
∥
∞
≤
a
,
lim
⁡
n
→
∞
T
n
v
0
=
v
*
,
w
h
e
r
e
v
0
(
t
)
≡
0
.Proof.
We denoteP
¯
a
=
{
u
∈
P
:
∥
u
∥
≤
a
}. In what follows, we first show that T
P
¯
a
⊂
P
¯
a. To do this, let u
∈
P
¯
a; then obviously
(45)
0
≤
u
(
t
)
≤
max
⁡
t
∈
[
0,1
]
u
(
t
)
=
∥
u
∥
∞
≤
a
,
t
∈
[
0,1
]
,
max
⁡
t
∈
[
0,1
]
|
u
′′
(
t
)
|
=
-
min
⁡
t
∈
[
0,1
]
u
′′
(
t
)
≤
a
.
Also since u
(
t
) is concave and pseudosymmetric about η on [
0,1
], then u
′
(
t
) is nonincreasing on [
0,1
], u
′
(
(
1
+
η
)
/
2
)
=
0, and u
′
(
t
)
=
-
u
′
(
1
+
η
-
t
) for t
∈
[
(
1
+
η
)
/
2
,
1
]. Hence u
′
(
t
)
≥
0 for t
∈
[
0
,
(
1
+
η
)
/
2
] and |
u
′
(
t
)
| achieve the maximum at t
=
0. Consequently
(46)
max
⁡
t
∈
[
0,1
]
|
u
′
(
t
)
|
=
u
′
(
0
)
≤
a
.
From (
H
0
), (
H
2
), and (42), it follows that
(47)
q
(
t
)
f
(
t
,
u
(
t
)
,
u
′
(
t
)
,
u
′′
(
t
)
)
≥
q
(
t
)
f
(
t
,
a
,
a
,
-
a
)
≥
inf
⁡
t
∈
(
0
,
(
1
+
η
)
/
2
]
⁡
q
(
t
)
f
(
t
,
a
,
a
,
-
a
)
≥
-
b
,
t
∈
(
0
,
1
+
η
2
]
.
This together with Lemma 3 implies
(48)
∥
T
u
∥
∞
=
(
T
u
)
(
1
+
η
2
)
=
∫
0
(
1
+
η
)
/
2
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
≤
(
1
+
η
)
3
8
b
+
β
(
1
+
η
)
2
4
α
b
≤
a
,
∥
(
T
u
)
′
∥
∞
=
(
T
u
)
′
(
0
)
=
∫
0
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
d
r
≤
(
1
+
η
)
2
8
b
<
a
,
∥
(
T
u
)
′
′
∥
∞
=
max
⁡
t
∈
(
0
,
(
1
+
η
)
/
2
]
⁡
|
-
∫
(
1
+
η
)
/
2
t
‍
q
(
s
)
×
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
∫
(
1
+
η
)
/
2
t
|
=
-
∫
0
(
1
+
η
)
/
2
‍
q
(
s
)
f
(
s
,
u
(
s
)
,
u
′
(
s
)
,
u
′
′
(
s
)
)
d
s
≤
1
+
η
2
b
<
a
.
Hence ∥
T
u
∥
≤
a, and thus T
P
¯
a
⊂
P
¯
a.
Letw
0
(
t
)
=
(
a
t
(
1
+
η
-
t
)
/
2
(
1
+
η
)
2
)
+
(
7
/
8
)
a
,
t
∈
[
0,1
]. Then ∥
w
0
∥
=
a, and thus w
0
∈
P
¯
a. Let w
1
=
T
w
0; then w
1
∈
P
¯
a. Define iterative sequence {
w
n
} as follows:
(49)
w
n
+
1
=
T
w
n
=
T
n
+
1
w
0
,
n
=
0,1
,
2
,
…
.
Since T
P
¯
a
⊂
P
¯
a, we have w
n
∈
P
¯
a,  n
=
0,1
,
2
,
…. From Lemma 4, {
w
n
}
n
=
1
∞ has a convergent subsequence {
w
n
k
}
k
=
1
∞ and there exists w
*
∈
P
¯
a such that
(50)
w
n
k
(
i
)
(
t
)
⇉
w
*
(
i
)
(
t
)
(
k
⟶
∞
)
on
⁡
[
0,1
]
,
i
=
0,1
,
2
.
From the definition of T and (42), for t
∈
[
0
,
(
1
+
η
)
/
2
], we have
(51)
w
1
(
t
)
=
T
w
0
(
t
)
=
∫
0
t
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
×
f
(
s
,
w
0
(
s
)
,
w
0
′
(
s
)
,
w
0
′
′
(
s
)
)
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
×
∫
(
1
+
η
)
/
2
r
‍
q
(
s
)
f
(
s
,
w
0
(
s
)
,
w
0
′
(
s
)
,
w
0
′
′
(
s
)
)
d
s
d
r
≤
∫
0
t
‍
∫
τ
(
1
+
η
)
/
2
‍
×
∫
r
(
1
+
η
)
/
2
‍
b
d
s
d
r
d
τ
+
β
α
∫
0
(
1
+
η
)
/
2
‍
∫
r
(
1
+
η
)
/
2
‍
b
d
s
d
r
=
b
∫
0
t
‍
∫
τ
(
1
+
η
)
/
2
‍
(
1
+
η
2
-
r
)
d
r
d
τ
+
β
α
b
×
∫
0
(
1
+
η
)
/
2
‍
(
1
+
η
2
-
r
)
d
r
≤
b
∫
0
t
‍
1
+
η
2
d
τ
+
β
α
b
(
1
+
η
2
)
2
≤
a
t
2
(
1
+
η
)
-
a
t
2
2
(
1
+
η
)
2
+
7
8
a
=
a
t
(
1
+
η
-
t
)
2
(
1
+
η
)
2
+
7
8
a
=
w
0
(
t
)
,
and, for t
∈
[
(
1
+
η
)
/
2
,
1
], we have
(52)
w
1
(
t
)
=
w
1
(
1
+
η
-
t
)
≤
w
0
(
1
+
η
-
t
)
=
w
0
(
t
)
.
Thus w
0
(
t
)
-
w
1
(
t
)
≥
0,  t
∈
[
0,1
].
On the other hand, since, fort
∈
[
0
,
(
1
+
η
)
/
2
], we have
(53)
w
0
′
′
(
t
)
-
w
1
′
′
(
t
)
=
-
a
(
1
+
η
)
2
+
∫
(
1
+
η
)
/
2
t
‍
q
(
s
)
f
(
s
,
w
0
(
s
)
,
w
0
′
(
s
)
,
w
0
′
′
(
s
)
)
d
s
≤
-
a
(
1
+
η
)
2
+
b
≤
0
,
and, for t
∈
[
(
1
+
η
)
/
2
,
1
], we have
(54)
w
0
′′
(
t
)
-
w
1
′′
(
t
)
=
w
0
′′
(
1
+
η
-
t
)
-
w
1
′′
(
1
+
η
-
t
)
≤
0
,
it follows that w
0
′′
(
t
)
-
w
1
′′
(
t
)
≤
0 on [
0,1
]. Hence w
0
(
t
)
-
w
1
(
t
) is concave on [
0,1
].
Also, since, fort
∈
[
η
,
1
],
(55)
w
0
(
t
)
-
w
1
(
t
)
=
w
0
(
1
+
η
-
t
)
-
w
1
(
1
+
η
-
t
)
,
then w
0
(
t
)
-
w
1
(
t
) is pseudosymmetric about η on [
0,1
]. So w
1
≤
w
0, and hence from Lemma 5 it follows that T
w
1
≤
T
w
0; that is, w
2
≤
w
1. By induction, we can show without any difficulty that
(56)
w
n
+
1
≤
w
n
,
n
=
0,1
,
2
,
…
;
that is,
(57)
w
n
-
w
n
+
1
∈
P
,
n
=
0,1
,
2
,
…
.
Thus w
n
(
t
)
-
w
n
+
1
(
t
)
(
n
=
0,1
,
2
,
…
) is concave on [
0,1
] and pseudosymmetric about η on [
0,1
]; consequently
(58)
w
n
′
(
t
)
≥
w
n
+
1
′
(
t
)
,
t
∈
[
0
,
1
+
η
2
]
,
n
=
0,1
,
2
,
…
,
(59)
w
n
′
(
t
)
≤
w
n
+
1
′
(
t
)
,
t
∈
[
1
+
η
2
,
1
]
,
n
=
0,1
,
2
,
…
,
(60)
w
n
′′
(
t
)
≤
w
n
+
1
′′
(
t
)
,
t
∈
[
0,1
]
,
n
=
0,1
,
2
,
…
.
From (50)–(60), it follows that
(61)
w
n
(
i
)
(
t
)
⇉
w
*
(
i
)
(
t
)
(
n
⟶
∞
)
on
⁡
[
0,1
]
,
i
=
0,1
,
2
;
that is, w
n
→
w
*
(
n
→
∞
). Let n
→
∞ in (49) to obtain
(62)
w
*
=
T
w
*
.
Also, from (
H
3
), we have w
*
(
(
1
+
η
)
/
2
)
=
max
⁡
t
∈
[
0,1
]
w
*
(
t
)
>
0. This together with the concavity of w
* implies that
(63)
w
*
(
t
)
≥
w
*
(
(
1
+
η
)
/
2
)
-
0
(
(
1
+
η
)
/
2
)
-
0
t
=
2
1
+
η
w
*
(
1
+
η
2
)
t
>
0
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
t
∈
(
0
,
1
+
η
2
]
.
Again using the fact that w
* is pseudosymmetric about η on [
0,1
], we have
(64)
w
*
(
t
)
>
0
,
t
∈
[
1
+
η
2
,
1
]
.
Hence w
*
(
t
)
>
0 on (
0,1
]. Therefore, from Lemma 2, w
* is a concave pseudosymmetric positive solution of BVP (5), (6).
Letv
0
(
t
)
≡
0 on [
0,1
]; then v
0
∈
P
¯
a. Set
(65)
v
n
+
1
=
T
v
n
,
n
=
0,1
,
2
,
…
.
Then, from Lemma 3, the sequence {
v
n
} is well defined. Since v
1
∈
P
¯
a
⊂
P, we have v
1
≥
0
=
v
0, and thus from Lemma 5 it follows that
(66)
v
2
=
T
v
1
≥
T
v
0
=
v
1
.
By induction we can show that
(67)
v
n
+
1
≥
v
n
,
n
=
0,1
,
2
,
…
.
Similarly to {
w
n
}, we can show that there exists v
*
∈
P
¯
a such that v
n
→
v
*
(
n
→
∞
). Taking limit in (65), we get v
*
=
T
v
*. Obviously, v
(
t
)
>
0 on (
0,1
]. Therefore, from Lemma 2, v
* is a concave pseudosymmetric positive solution of BVP (5), (6). This completes the proof of the theorem.


## 4. An Example

Consider the following third-order four-point boundary value problem:(68)
u
′′′
(
t
)
+
3
(
3
t
-
2
)
×
(
e
-
u
/
8
+
64
64
+
u
′
2
+
arctan
u
′′
16
+
π
2
)
=
0
,
t
∈
(
0,1
)
,
u
(
0
)
-
1
2
u
′
(
0
)
=
0
,
u
′
(
1
3
)
+
u
′
(
1
)
=
0
,
u
′′
(
2
3
)
=
0
.
Let
(69)
f
(
t
,
u
,
v
,
w
)
=
3
(
3
t
-
2
)
(
e
-
u
/
8
+
64
64
+
v
2
+
arctan
w
16
+
π
2
)
.
Then f
∈
C
(
[
0,1
]
×
[
0
,
+
∞
)
×
ℝ
×
(
-
∞
,
0
]
,
ℝ
). It is easy to see that BVP (68) corresponds to BVP (5), (6) when q
(
t
)
≡
1, α
=
1, β
=
1
/
2, and η
=
1
/
3.Next we verify that all conditions of Theorem6 are satisfied. In fact, obviously the conditions (
H
0
), (
H
2
), and (
H
3
) hold. In addition, by the definition of f, we have
(70)
f
(
1
+
η
-
t
,
u
,
-
v
,
w
)
=
3
(
2
-
3
t
)
(
e
-
u
/
8
+
64
64
+
(
-
v
)
2
+
arctan
w
16
+
π
2
)
=
-
3
(
3
t
-
2
)
(
e
-
u
/
8
+
64
64
+
v
2
+
arctan
w
16
+
π
2
)
=
-
f
(
t
,
u
,
v
,
w
)
,
t
∈
[
1
3
,
1
]
,
f
(
t
,
u
,
v
,
w
)
≤
0
on
⁡
[
0
,
2
3
]
×
[
0
,
+
∞
)
2
×
(
-
∞
,
0
]
.
Hence the condition (
H
1
) is also satisfied.Now, we takea
=
16,  b
=
9. Then
(71)
max
⁡
{
β
(
1
+
η
)
2
4
α
(
7
8
-
1
2
(
1
+
η
)
2
)
-
1
,
(
1
+
η
)
3
8
+
β
(
1
+
η
)
2
4
α
,
(
1
+
η
)
2
(
7
8
-
1
2
(
1
+
η
)
2
)
-
1
}
=
16
9
,
and thus
(72)
a
=
max
⁡
{
β
(
1
+
η
)
2
4
α
(
7
8
-
1
2
(
1
+
η
)
2
)
-
1
,
(
1
+
η
)
3
8
+
β
(
1
+
η
)
2
4
α
,
(
1
+
η
)
2
(
7
8
-
1
2
(
1
+
η
)
2
)
-
1
}
b
.
On the other hand, we also have
(73)
inf
⁡
t
∈
(
0
,
2
/
3
]
f
(
t
,
16,16
,
-
16
)
=
inf
⁡
t
∈
(
0
,
2
/
3
]
3
(
3
t
-
2
)
(
e
-
2
+
64
64
+
1
6
2
+
arctan
(
-
1
)
+
π
2
)
>
-
9
=
-
b
.
In summary, all conditions of Theorem 6 are satisfied. Hence, from Theorem 6, BVP (68) has concave pseudosymmetric positive solution w
*,  v
*
∈
P with
(74)
∥
w
*
∥
∞
≤
16
,
lim
⁡
n
→
∞
T
n
w
0
=
w
*
,
where
w
0
(
t
)
=
-
9
2
t
2
+
6
t
+
14
;
∥
v
*
∥
∞
≤
16
,
lim
⁡
n
→
∞
T
n
v
0
=
v
*
,
where
v
0
(
t
)
≡
0
.



---

*Source: 518238-2014-04-08.xml*