# Existence of Positive Solution to System of Nonlinear Third-Order Three-Point BVPs

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

---

## Abstract

We are concerned with the following system of third-order three-point boundary value problems:u
′
′
′
(
t
)
+
f
(
t
,
v
(
t
)
)
=
0, t
∈
(
0,1
), v
′
′
′
(
t
)
+
g
(
t
,
u
(
t
)
)
=
0, t
∈
(
0,1
), u
(
0
)
=
u
′
′
(
0
)
=
0, u
′
(
1
)
=
α
u
(
η
), v
(
0
)
=
v
′
′
(
0
)
=
0, and v
′
(
1
)
=
α
v
(
η
), where 0
<
η
<
1 and 0
<
α
<
1
/
η. By imposing some suitable conditions on f and g, we obtain the existence of at least one positive solution to the above system. The main tool used is the theory of the fixed-point index.


---

## 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, there are a lot of papers concerning the existence of positive solutions to third-order three-point boundary value problems (BVPs for short); see [2–13] and the references therein. Yet, only in a few papers has the problem of existence of positive solutions to systems of third-order three-point BVPs been considered.It is worth mentioning that there are some excellent works on systems of second-order or higher-order multipoint BVPs; see [14–18].In this paper, we study the existence of positive solution for the following system of nonlinear third-order three-point BVPs:(1)
u
′
′
′
t
+
f
t
,
v
t
=
0
,
t
∈
0,1
,
v
′
′
′
t
+
g
t
,
u
t
=
0
,
t
∈
0,1
,
u
0
=
u
′
′
0
=
0
,
u
′
1
=
α
u
η
,
v
0
=
v
′
′
0
=
0
,
v
′
1
=
α
v
η
.Throughout this paper, we always assume that the following conditions are fulfilled:(
H
1
)
0
<
η
<
1 and 0
<
α
<
1
/
η.
(
H
2
)
f
,
g
∈
C
[
0,1
]
×
[
0
,
∞
)
,
[
0
,
∞
) and f
(
t
,
0
)
≡
0, g
(
t
,
0
)
≡
0, t
∈
[
0,1
].If(
u
,
v
)
∈
C
3
(
[
0,1
]
,
[
0
,
∞
)
)
×
C
3
(
[
0,1
]
,
[
0
,
∞
)
) satisfies the differential equations and boundary conditions in system (1), then (
u
,
v
) is said to be a solution of system (1). If (
u
,
v
) is a solution of system (1) and u
(
t
)
>
0, v
(
t
)
>
0, t
∈
(
0,1
), then (
u
,
v
) is said to be a positive solution of system (1).To end this section, we state the following results on the theory of the fixed-point index [19].LetE be a real Banach space, P
⊂
E a cone, and θ the zero element in E. For ρ
>
0, we denote(2)
B
ρ
=
u
∈
E
:
u
<
ρ
.Theorem 1.
LetA
:
B
ρ
¯
∩
P
→
P be a completely continuous operator. If there exists u
0
∈
P
∖
{
θ
} such that (3)
u
-
A
u
≠
λ
u
0
,
u
∈
∂
B
ρ
∩
P
,
λ
≥
0
,then i
(
A
,
B
ρ
∩
P
,
P
)
=
0.Theorem 2.
LetA
:
B
ρ
¯
∩
P
→
P be a completely continuous operator which has no fixed point on ∂
B
ρ
∩
P. If A
u
≤
u for all u
∈
∂
B
ρ
∩
P, then i
(
A
,
B
ρ
∩
P
,
P
)
=
1.


## 2. Preliminaries

LetE
=
C
[
0,1
] be equipped with the maximum norm. Then E is a Banach space.Lemma 3 (see [7]).
For anyy
∈
E, the BVP(4)
u
′
′
′
t
+
y
t
=
0
,
t
∈
0,1
,
u
0
=
u
′
′
0
=
0
,
u
′
1
=
α
u
ηhas a unique solution (5)
u
t
=
∫
0
1
G
t
,
s
y
s
d
s
,
t
∈
0,1
,where (6)
G
t
,
s
=
1
2
1
-
α
η
2
t
-
1
-
α
η
t
2
+
s
2
-
α
t
η
2
+
s
2
,
s
≤
min
⁡
η
,
t
,
2
t
-
t
2
-
s
2
+
α
η
t
-
s
2
,
η
≤
s
≤
t
,
2
t
-
2
t
s
1
-
α
η
-
α
t
η
2
+
s
2
,
t
≤
s
≤
η
,
2
t
1
-
s
,
s
≥
max
⁡
η
,
t
.For convenience, we denote(7)
a
=
1
+
α
η
1
-
α
η
,
γ
=
η
2
2
1
+
α
η
min
⁡
1
,
α
.Obviously, a
>
0 and 0
<
γ
<
1.Lemma 4 (see [7]).
For anyt
,
s
∈
0,1
×
0,1, 0
≤
G
(
t
,
s
)
≤
a
(
1
-
s
).Lemma 5.
For anyt
,
s
∈
η
,
1
×
0,1, G
(
t
,
s
)
≥
γ
a
(
1
-
s
).Proof.
SinceG
(
t
,
1
)
=
0, we only need to consider t
,
s
∈
η
,
1
×
0,1.
Ifs
≤
η
≤
t, then (8)
G
t
,
s
a
1
-
s
=
2
t
-
1
-
α
η
t
2
+
s
2
-
α
t
η
2
+
s
2
2
1
+
α
η
1
-
s
>
2
t
-
1
-
α
η
t
+
t
s
-
α
t
η
+
η
s
2
1
+
α
η
1
-
s
=
t
2
1
+
α
η
≥
η
2
1
+
α
η
>
η
2
2
1
+
α
η
≥
γ
.
Ifη
≤
s
≤
t, then (9)
G
t
,
s
a
1
-
s
=
2
t
-
t
2
-
s
2
+
α
η
t
-
s
2
2
1
+
α
η
1
-
s
≥
2
t
-
2
t
1
-
α
η
-
2
α
η
t
s
2
1
+
α
η
1
-
s
≥
α
η
2
1
+
α
η
>
α
η
2
2
1
+
α
η
≥
γ
.
Ifη
≤
t
≤
s, then (10)
G
t
,
s
a
1
-
s
=
t
1
+
α
η
≥
η
1
+
α
η
>
η
2
2
1
+
α
η
≥
γ
.Therefore, for any t
,
s
∈
η
,
1
×
0,1, G
(
t
,
s
)
≥
γ
a
(
1
-
s
).Lemma 6.
Lety
∈
E and y
(
t
)
≥
0, t
∈
0,1. Then the unique solution u of BVP (4) satisfies (11)
u
t
≥
0
,
t
∈
0,1
,
m
i
n
t
∈
η
,
1
u
t
≥
η
u
.Proof.
Sincey
(
t
)
≥
0, t
∈
[
0,1
], it follows from Lemma 4 that u
(
t
)
≥
0 for t
∈
[
0,1
]. In view of u
′
′
′
(
t
)
=
-
y
(
t
)
≤
0, t
∈
0,1, we have u
′
′
(
t
)
≤
u
′
′
(
0
)
=
0, t
∈
0,1. This shows that u
′
(
t
)
≥
u
′
(
1
)
=
α
u
(
η
)
≥
0, t
∈
0,1, which indicates that(12)
m
i
n
t
∈
η
,
1
u
t
=
u
η
,
u
=
u
1
.On the other hand, it follows from the fact u
′
′
(
t
)
≤
0 for t
∈
0,1 that u is concave down on [
0,1
], which together with u
(
0
)
=
0 implies that u
(
η
)
≥
η
u
(
1
). In view of (12), we get that m
i
n
t
∈
η
,
1
u
(
t
)
≥
η
u.Lemma 7.
There existsx
0
∈
0
,
4
3
/
9
π such that(13)
x
0
e
-
x
0
+
α
e
-
η
x
0
+
2
x
0
e
1
/
2
x
0
cos
⁡
3
2
x
0
-
2
α
e
1
/
2
η
x
0
sin
⁡
π
6
+
3
2
η
x
0
=
0
.Proof.
Let(14)
φ
x
=
x
e
-
x
+
α
e
-
η
x
+
2
x
e
1
/
2
x
cos
⁡
3
2
x
-
2
α
e
1
/
2
η
x
sin
⁡
π
6
+
3
2
η
x
,
x
∈
0
,
4
3
9
π
.Then it is obvious that (15)
φ
0
=
0
,
φ
′
0
=
3
1
-
α
η
>
0
.This indicates that there exists x
∗
∈
0
,
4
3
/
9
π such that φ
(
x
∗
)
>
0, which together with the fact that (16)
φ
4
3
9
π
=
4
3
9
π
e
-
4
3
/
9
π
+
α
e
-
4
3
/
9
π
η
-
4
3
9
π
e
2
3
/
9
π
-
2
α
e
2
3
/
9
π
η
sin
⁡
π
6
+
2
π
3
η
<
4
3
9
π
e
-
4
3
/
9
π
+
α
e
-
4
3
/
9
π
η
-
4
3
9
π
e
2
3
/
9
π
-
α
e
2
3
/
9
π
η
=
-
4
3
9
π
e
2
3
/
9
π
-
e
-
4
3
/
9
π
+
α
e
2
3
/
9
π
η
-
e
-
4
3
/
9
π
η
<
0implies that there exists x
0
∈
x
∗
,
4
3
/
9
π
⊂
0
,
4
3
/
9
π such that φ
(
x
0
)
=
0; that is, (13) is satisfied.In the remainder of this paper, we always assume thatx
0 is defined as in Lemma 7 and u
0
(
t
)
=
-
e
-
x
0
t
+
2
e
1
/
2
x
0
t
s
i
n
π
/
6
+
3
/
2
x
0
t, t
∈
[
0,1
].Corollary 8.
x
0
3 is an eigenvalue of the eigenvalue problem (17)
u
′
′
′
t
+
λ
u
t
=
0
,
t
∈
0,1
,
u
0
=
u
′
′
0
=
0
,
u
′
1
=
α
u
ηand u
0 is an eigenfunction corresponding to the eigenvalue x
0
3.Proof.
By Lemma7, it is obvious.Let(18)
P
=
u
∈
E
:
u
t
≥
0
,
t
∈
0,1
,
P
0
=
u
∈
P
:
m
i
n
t
∈
η
,
1
u
t
≥
η
u
.Then P and P
0 are cones in E. Now, we define a linear operator B as follows: (19)
B
u
t
=
∫
0
1
G
t
,
s
u
s
d
s
,
u
∈
P
,
t
∈
0,1
.Lemma 9.
ConsiderB
:
P
→
P
0.Proof.
In view of Lemmas3, 4, and 6, it is not difficult to verify that B
(
P
)
⊂
P
0.Lemma 10.
Consideru
0
∈
P and(20)
u
0
t
=
x
0
3
B
u
0
t
,
t
∈
0,1
.Proof.
Obviously,u
0 is continuous and u
0
(
t
)
≥
e
-
x
0
t
(
e
3
/
2
x
0
t
-
1
)
≥
0, t
∈
0,1, which indicates that u
0
∈
P. And it follows from Lemma 3 and Corollary 8 that (20) is satisfied.


## 3. Main Results

Obviously,(
u
,
v
) is a solution of system (1) if and only if (
u
,
v
)
∈
C
(
[
0,1
]
,
[
0
,
∞
)
)
×
C
(
[
0,1
]
,
[
0
,
∞
)
) is a solution of the following system:(21)
u
t
=
∫
0
1
G
t
,
s
f
s
,
v
s
d
s
,
t
∈
0,1
,
v
t
=
∫
0
1
G
t
,
s
g
s
,
u
s
d
s
,
t
∈
0,1
.Moreover, system (21) can be written as the integral equation: (22)
u
t
=
∫
0
1
G
t
,
s
f
s
,
∫
0
1
G
s
,
τ
g
τ
,
u
τ
d
τ
d
s
,
t
∈
0,1
.If we define an operator A on P by (23)
A
u
t
=
∫
0
1
G
t
,
s
f
s
,
∫
0
1
G
s
,
τ
g
τ
,
u
τ
d
τ
d
s
,
t
∈
0,1
,then it is easy to verify that A
:
P
→
P is completely continuous. Moreover, if u
∈
P is a fixed point of A and v
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
g
(
s
,
u
(
s
)
)
d
s, t
∈
0,1, then (
u
,
v
) is a solution of system (1).Theorem 11.
Assume that the following conditions are fulfilled:(
A
1
)
There exists a constantp
∈
0,1 such that (24)
1
l
i
m
x
→
∞
f
t
,
x
x
p
>
0
,
2
l
i
m
x
→
∞
g
t
,
x
x
1
/
p
=
∞uniformly on 0,1.
(
A
2
)
There exists a constantq
∈
0
,
∞ such that (25)
1
l
i
m
x
→
0
+
f
t
,
x
x
q
<
∞
,
2
l
i
m
x
→
0
+
g
t
,
x
x
1
/
q
=
0uniformly on 0,1.
Then system (1) has at least one positive solution.Proof.
First, let(26)
M
=
u
∈
P
:
u
-
A
u
=
λ
u
0
,
λ
≥
0
.Then we may assert that M is a bounded subset of E.
In fact, ifu
∈
M, then there exists λ
≥
0 such that u
=
A
u
+
λ
u
0, which together with Lemma 10 implies that(27)
u
t
=
A
u
t
+
λ
u
0
t
=
∫
0
1
G
t
,
s
f
s
,
∫
0
1
G
s
,
τ
g
τ
,
u
τ
d
τ
d
s
+
λ
x
0
3
B
u
0
t
=
∫
0
1
G
t
,
s
F
u
s
d
s
+
λ
x
0
3
B
u
0
t
=
B
F
u
t
+
λ
x
0
3
u
0
t
,
t
∈
0,1
,where F
:
P
→
P is defined as follows: (28)
F
u
t
=
f
t
,
∫
0
1
G
t
,
s
g
s
,
u
s
d
s
,
t
∈
0,1
.In view of (27) and Lemma 9, we have u
∈
P
0. This implies that(29)
u
≤
1
η
m
i
n
t
∈
η
,
1
u
t
.On the other hand, from (1) of (
A
1
), we know that there exist positive constants C
1 and C
2 such that (30)
f
t
,
x
≥
C
1
x
p
-
C
2
,
t
,
x
∈
0,1
×
0
,
∞
,which together with Jensen inequality and Lemma 4 implies that(31)
A
u
t
=
∫
0
1
G
t
,
s
f
s
,
∫
0
1
G
s
,
τ
g
τ
,
u
τ
d
τ
d
s
≥
∫
0
1
G
t
,
s
C
1
∫
0
1
G
s
,
τ
g
τ
,
u
τ
d
τ
p
-
C
2
d
s
≥
C
1
∫
0
1
G
t
,
s
∫
0
1
G
s
,
τ
p
g
τ
,
u
τ
p
d
τ
d
s
-
C
2
∫
0
1
G
t
,
s
d
s
≥
C
1
∫
0
1
G
t
,
s
∫
0
1
G
s
,
τ
p
g
τ
,
u
τ
p
d
τ
d
s
-
C
3
,
t
∈
0,1
,where C
3
=
1
/
2
a
C
2.
From (2) of(
A
1
), we know that there exists C
4
>
0 such that(32)
g
t
,
x
p
≥
2
C
1
γ
p
a
p
-
1
η
m
1
2
x
-
C
4
,
t
,
x
∈
0,1
×
0
,
∞
,where m
1
=
min
t
∈
η
,
1
⁡
∫
η
1
‍
G
(
t
,
s
)
d
s
>
0.
So, it follows from (31), (32), and Lemmas 4 and 5 that(33)
u
t
=
A
u
t
+
λ
u
0
t
≥
A
u
t
≥
C
1
∫
0
1
G
t
,
s
∫
0
1
G
s
,
τ
p
g
τ
,
u
τ
p
d
τ
d
s
-
C
3
≥
C
1
∫
η
1
G
t
,
s
∫
0
1
G
s
,
τ
p
g
τ
,
u
τ
p
d
τ
d
s
-
C
3
≥
C
1
γ
p
a
p
∫
η
1
G
t
,
s
∫
0
1
1
-
τ
p
g
τ
,
u
τ
p
d
τ
d
s
-
C
3
≥
C
1
γ
p
a
p
∫
η
1
G
t
,
s
∫
0
1
1
-
τ
g
τ
,
u
τ
p
d
τ
d
s
-
C
3
≥
C
1
γ
p
a
p
-
1
∫
η
1
G
t
,
s
∫
0
1
G
s
,
τ
g
τ
,
u
τ
p
d
τ
d
s
-
C
3
≥
C
1
γ
p
a
p
-
1
∫
η
1
G
t
,
s
∫
0
1
G
s
,
τ
2
C
1
γ
p
a
p
-
1
η
m
1
2
u
τ
-
C
4
d
τ
d
s
-
C
3
=
2
η
m
1
2
∫
η
1
G
t
,
s
∫
0
1
G
s
,
τ
u
τ
d
τ
d
s
-
C
1
C
4
γ
p
a
p
-
1
∫
η
1
G
t
,
s
∫
0
1
G
s
,
τ
d
τ
d
s
-
C
3
=
2
η
m
1
2
∫
η
1
G
t
,
s
B
u
s
d
s
-
C
1
C
4
γ
p
a
p
-
1
∫
η
1
G
t
,
s
∫
0
1
G
s
,
τ
d
τ
d
s
+
C
3
≥
2
η
m
1
2
m
i
n
s
∈
η
,
1
B
u
s
∫
η
1
G
t
,
s
d
s
-
C
1
C
4
γ
p
a
p
∫
η
1
1
-
s
∫
0
1
G
s
,
τ
d
τ
d
s
+
C
3
≥
2
η
m
1
m
i
n
s
∈
η
,
1
B
u
s
-
C
5
,
t
∈
η
,
1
,where C
5
=
C
1
C
4
γ
p
a
p
∫
η
1
‍
(
1
-
s
)
∫
0
1
‍
G
(
s
,
τ
)
d
τ
d
s
+
C
3
>
0.
SinceB
(
u
)
∈
P
0, we have(34)
m
i
n
s
∈
η
,
1
B
u
s
≥
η
B
u
≥
η
∫
0
1
G
η
,
τ
u
τ
d
τ
≥
η
∫
η
1
G
η
,
τ
u
τ
d
τ
≥
η
m
i
n
τ
∈
η
,
1
u
τ
∫
η
1
G
η
,
τ
d
τ
≥
η
m
1
m
i
n
τ
∈
η
,
1
u
τ
.In view of (33) and (34), we get (35)
u
t
≥
2
m
i
n
τ
∈
η
,
1
u
τ
-
C
5
,
t
∈
η
,
1
.So, (36)
m
i
n
t
∈
η
,
1
u
t
≥
2
m
i
n
t
∈
η
,
1
u
t
-
C
5
,and so, (37)
m
i
n
t
∈
η
,
1
u
t
≤
C
5
,which together with (29) indicates that(38)
u
≤
C
5
η
.This shows that M is a bounded subset of E. Therefore, there exists a sufficiently larger R
>
1 such that (39)
u
-
A
u
≠
λ
u
0
,
u
∈
∂
B
R
∩
P
,
λ
≥
0
.So, it follows from Theorem 1 that(40)
i
A
,
B
R
∩
P
,
P
=
0
.
Next, from (1) of(
A
2
), we have (41)
b
≔
sup
⁡
f
t
,
x
x
q
:
t
,
x
∈
0,1
×
0,1
<
∞
,which together with f
(
t
,
0
)
≡
0, t
∈
[
0,1
], implies that(42)
f
t
,
x
≤
b
x
q
,
t
,
x
∈
0,1
×
0,1
.Let (43)
ε
=
m
i
n
1
m
2
,
1
b
+
1
m
2
q
+
1
1
/
q
,where m
2
=
m
a
x
t
∈
0,1
∫
0
1
‍
G
(
t
,
s
)
d
s
>
0. Then from (2) of (
A
2
) and g
(
t
,
0
)
≡
0, t
∈
[
0,1
], we know that there exists r
∈
0,1 such that(44)
g
t
,
x
≤
ε
x
1
/
q
,
t
,
x
∈
0,1
×
0
,
r
.By (44), for any u
∈
∂
B
r
∩
P, we have (45)
∫
0
1
G
s
,
τ
g
τ
,
u
τ
d
τ
≤
ε
∫
0
1
G
s
,
τ
u
τ
1
/
q
d
τ
≤
ε
m
2
u
1
/
q
≤
1
,
s
∈
0,1
,which together with (42) shows that (46)
A
u
t
=
∫
0
1
G
t
,
s
f
s
,
∫
0
1
G
s
,
τ
g
τ
,
u
τ
d
τ
d
s
≤
b
∫
0
1
G
t
,
s
∫
0
1
G
s
,
τ
g
τ
,
u
τ
d
τ
q
d
s
≤
b
ε
q
m
2
q
u
∫
0
1
G
t
,
s
d
s
≤
b
ε
q
m
2
q
+
1
u
<
b
+
1
ε
q
m
2
q
+
1
u
≤
u
,
t
∈
0,1
.This indicates that A
u
<
u, u
∈
∂
B
r
∩
P. So, it follows from Theorem 2 that(47)
i
A
,
B
r
∩
P
,
P
=
1
.In view of (40) and (47), we get (48)
i
A
,
B
R
∖
B
r
¯
∩
P
,
P
=
i
A
,
B
R
∩
P
,
P
-
i
A
,
B
r
∩
P
,
P
=
-
1
.Therefore, A has at least one fixed point u
∈
P and r
<
u
<
R. Let (49)
v
t
=
∫
0
1
G
t
,
s
g
s
,
u
s
d
s
,
t
∈
0,1
.Then (
u
,
v
) is a solution of system (1).
Thirdly, we proveu
(
t
)
>
0 for t
∈
0,1.
Suppose on the contrary that there existst
0
∈
0,1 such that u
(
t
0
)
=
0. Since u
′
′
′
(
t
)
=
-
f
(
t
,
v
(
t
)
)
≤
0, t
∈
(
0,1
), we know that u
′
′
(
t
)
≤
u
′
′
(
0
)
=
0, t
∈
(
0,1
). So, u
(
t
) is concave down on [
0,1
], which together with u
(
0
)
=
u
(
t
0
)
=
0 implies that u
(
1
)
=
0. At the same time, it follows from u
′
′
(
t
)
≤
0, t
∈
(
0,1
), that u
′
(
t
)
≥
u
′
(
1
)
=
α
u
(
η
)
≥
0, t
∈
0,1. So, 0
=
u
(
0
)
≤
u
(
t
)
≤
u
(
1
)
=
0, t
∈
0,1; that is, u
(
t
)
≡
0, t
∈
0,1, which contradicts with the fact that u
>
r
>
0. Therefore, u
(
t
)
>
0, t
∈
0,1.
Finally, we verifyv
(
t
)
>
0 for t
∈
0,1.
First, we may assert thatv
>
0. Indeed, if v
(
t
)
≡
0, t
∈
0,1, in view of f
(
t
,
0
)
≡
0, t
∈
[
0,1
], then we get (50)
u
t
=
∫
0
1
G
t
,
s
f
s
,
v
s
d
s
=
∫
0
1
G
t
,
s
f
s
,
0
d
s
=
0
,
t
∈
0,1
,which is a contradiction. This shows that v
>
0. So, by using the similar method to prove u
(
t
)
>
0 for t
∈
0,1, we may obtain v
(
t
)
>
0, t
∈
0,1.
To sum it up,(
u
,
v
) is a positive solution of system (1).



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*Source: 6874643-2016-07-13.xml*