# Positive Solutions for a Class of Fourth-Orderp-Laplacian Boundary Value Problem Involving Integral Conditions

Authors: Yan Sun
Journal: Discrete Dynamics in Nature and Society
 (2015)

---

## Abstract

Under some conditions concerning the first eigenvalues corresponding to the relevant linear operator, we obtain sharp optimal criteria for the existence of positive solutions forp-Laplacian problems with integral boundary conditions. The main methods in the paper are constructing an available integral operator and combining fixed point index theory. The interesting point of the results is that the nonlinear term contains all lower-order derivatives explicitly. Finally, we give some examples to demonstrate the main results.


---

## Body

## 1. Introduction

In this paper, we discuss the existence of positive solutions for the followingp-Laplacian integral boundary value problems:(1)ϕpy′′t′′=atft,yt,y′t,y′′t,y′′′t,0<t<1,y0=y1=∫01bsysds,y′′0=y′′1=ϕq∫01csϕpy′′sds,where ϕ
p
(
t
)
=
|
t
|
p
-
2
t, p
>
1, ϕ
q
=
ϕ
p
-
1, 1
/
p
+
1
/
q
=
1, a
∈
L
1
[
0,1
] is symmetric on the interval [
0,1
] (i.e., a
(
1
-
t
)
=
a
(
t
) and a
(
t
)
≥
0 for t
∈
[
0,1
]), f
:
[
0,1
]
×
[
0
,
+
∞
)
×
(
-
∞
,
+
∞
)
×
(
-
∞
,
+
∞
)
×
(
-
∞
,
+
∞
)
→
[
0
,
+
∞
) is continuous, and b
,
c
∈
L
1
[
0,1
] are nonnegative symmetric on [
0,1
].Boundary value problems of ordinary differential equations have become an important research field in recent years. Fourth-orderp-Laplacian boundary value problems arise in applied mathematics, physics, gas diffusion through porous media, engineering, elastic mechanics, electromagnetic waves of gravity driven flows, and the various areas of adiabatic tubular reactor processes, as well as biological problems; see [1–5] and the references therein.In [6], by using the fixed point theorem for strict set contraction operator, Zhang et al. considered the existence of positive solutions for the following nth-order impulsive boundary value problems with integral boundary conditions in Banach spaces:(2)znt+gt,zt,z′t,…,zn-2t=θ,t∈J,t≠tk,k=1,…,m,Δzn-1t=tk=-Ikzn-2tk,k=1,…,m,zi0=θ,i=0,1,…,n-3,zn-20=zn-21=∫01htzn-2tdt,where J
=
[
0,1
], 0
<
t
1
<
t
2
<
⋯
<
t
k
<
⋯
<
t
m
<
1, g
∈
C
[
J
×
P
n
-
1
,
P
], I
k
∈
C
[
P
,
P
], k
=
1
,
…
,
m, h
∈
L
1
[
0,1
], and Δ
z
(
n
-
1
)
t
=
t
k
=
z
(
n
-
1
)
(
t
k
+
)
-
z
(
n
-
1
)
(
t
k
-
) denotes the jump of z
(
n
-
1
)
(
t
) at t
=
t
k. They also gave the upper and lower bounds for these positive solutions.In [7], by means of the fixed point index theory, Sun and Zhang studied the following singular nonlinear Sturm-Liouville problems:(3)-Lϕx=axgϕx,0<x<1,R1ϕ=α1ϕ0+β1ϕ′0=0,R2ϕ=α2ϕ1+β2ϕ′1=0,-Lϕx=fx,ϕx,0<x<1,R1ϕ=α1ϕ0+β1pxϕ′0=0,R2ϕ=α2ϕ1+β2pxϕ′1=0,where (
L
ϕ
)
(
x
)
=
p
x
ϕ
′
x
′
+
q
(
x
)
ϕ
(
x
) and a
(
x
) is allowed to be singular at x
=
0 and x
=
1.In [2], by making use of the theory of the spectrum, Sergejeva studied the regions of solvability for three-point boundary value problem:(4)-y′′t=μy+-λy-+ht,y,y′′,t∈0,1,y0=0,y1=cy12,where y
+
=
max
⁡
{
y
,
0
}, y
-
=
max
⁡
{
-
y
,
0
}, and h is a bounded function. The solvability results are established for the problem with h
≡
0.Motivated and inspired greatly by the above mentioned works, the aim of the paper is to improve, generalize, and supplement the previous results. We obtain the existence results on nonlinear fourth-order integral boundary value problems with one-dimensionalp-Laplacian integral boundary value problem (1). The novelty of our results is in exploring some optimal criteria for the existence of positive solutions of problem (1). The methods used in our work will depend on an application of fixed point index theory together with the first eigenvalues corresponding to the relevant linear operator. It can also be seen that the nonlinear term involves all lower-order derivatives explicitly.The paper is organized as follows. In Section2, we state some lemmas and several preliminary results. The main results are formulated and proved in Section 3. Examples are given in Section 4.


## 2. Preliminaries

Throughout the paper, we make the following assumptions:(H
1) f
∈
C
(
[
0,1
]
×
[
0
,
+
∞
)
×
(
-
∞
,
+
∞
)
×
(
-
∞
,
+
∞
)
×
(
-
∞
,
+
∞
)
,
[
0
,
+
∞
)
) and h
(
t
,
y
)
≤
f
t
,
y
,
y
′
,
y
′
′
,
y
′
′
′
≤
g
(
t
,
y
), h
,
g
:
[
0,1
]
×
[
0
,
+
∞
)
→
[
0
,
+
∞
), are continuous and also h
(
·
,
y
) and g
(
·
,
y
) are symmetric on [
0,1
] for all y
≥
0 (i.e., h
(
1
-
t
,
y
)
=
h
(
t
,
y
), g
(
1
-
t
,
y
)
=
g
(
t
,
y
)).(H
2) a
,
b
,
c
∈
L
1
[
0,1
] are all nonnegative and symmetric on [
0,1
], and a
(
t
)
≢
0 is continuous on (
0,1
), and a
(
t
) may be singular at t
=
0 and/or t
=
1, and μ
∈
(
0,1
) and ν
∈
(
0,1
), where(5)μ=∫01bsds,ν=∫01csds.In what follows, we will consider the Banach spaceE
=
C
[
0,1
], C
+
[
0,1
]
=
{
x
∈
C
[
0,1
]
∣
x
(
t
)
≥
0
}, equipped with the maximum norm z
=
max
⁡
0
≤
t
≤
1
⁡
z
t. Denote cone P by(6)P=z∈C+0,1∣ztis  concave  fort∈0,1;then, P is a positive cone in C
[
0,1
].LetB
r
=
z
∈
C
+
[
0,1
]
∣
z
<
r for r (r
>
0), and(7)Ht,s=Gt,s+11-μ∫01Gs,τbτdτ,H1t,s=Gt,s+11-ν∫01Gs,υcυdυ,Gt,s=s1-t,0≤s≤t≤1,t1-s,0≤t≤s≤1.Here G
(
t
,
s
), H
(
t
,
s
), and H
1
(
t
,
s
) are Green’s functions of problem (1).Lemma 1 (see [1]).
Suppose that (H
1) and (H
2) hold; then, for t
,
s
∈
[
0,1
], we have(I)
G
(
t
,
s
)
≥
0, H
(
t
,
s
)
≥
0, H
1
(
t
,
s
)
≥
0;
(II)
e
(
t
)
e
(
s
)
≤
G
(
t
,
s
)
≤
e
(
t
)
≤
1
/
4, where e
(
t
)
=
t
(
1
-
t
);
(III)
G
(
1
-
t
,
1
-
s
)
=
G
(
t
,
s
), H
(
1
-
t
,
1
-
s
)
=
H
(
t
,
s
), and H
1
(
1
-
t
,
1
-
s
)
=
H
1
(
t
,
s
);
(IV)
ρ
e
(
s
)
≤
H
(
t
,
s
)
≤
γ
s
(
1
-
s
)
=
γ
e
(
s
)
≤
1
/
4
γ; ρ
1
e
(
s
)
≤
H
1
(
t
,
s
)
≤
γ
1
s
(
1
-
s
)
=
γ
1
e
(
s
)
≤
1
/
4
γ
1, where γ
=
1
/
1
-
μ, ρ
=
∫
0
1
e
(
τ
)
b
(
τ
)
d
τ
/
1
-
μ, γ
1
=
1
/
1
-
ν, ρ
1
=
∫
0
1
e
(
τ
)
c
(
τ
)
d
τ
/
1
-
ν.Define an integral operatorA
:
C
+
[
0,1
]
→
C
+
[
0,1
] by(8)
A
y
t
=
∫
0
1
H
t
,
s
ϕ
q
∫
0
1
H
1
s
,
τ
a
τ
·
f
τ
,
y
τ
,
y
′
τ
,
y
′
′
τ
,
y
′
′
′
τ
d
τ
d
s
.It is well known thaty
∈
C
+
[
0,1
] is a solution of problem (1) if and only if y is a fixed point of operator equation y
=
A
y.Lemma 2.
Suppose that (H
1) and (H
2) hold, if y
∈
E is a solution of the following integral equation: (9)
y
t
=
∫
0
1
H
t
,
s
ϕ
q
∫
0
1
H
1
s
,
τ
a
τ
·
f
τ
,
y
τ
,
y
′
τ
,
y
′
′
τ
,
y
′
′
′
τ
d
τ
d
s
;then, y is a solution of problem (1).Lemma 3.
Suppose that (H
1) and (H
2) hold; then, A
:
P
→
P is a completely continuous operator.Proof.
First, we prove thatA
:
P
→
P
. For every y
∈
P, we have(10)
A
y
′
′
t
=
-
ϕ
q
∫
0
1
H
1
t
,
s
a
s
·
f
s
,
y
s
,
y
′
s
,
y
′
′
s
,
y
′
′
′
s
d
s
≤
0
,
A
y
0
=
∫
0
1
H
0
,
s
ϕ
q
∫
0
1
H
1
s
,
τ
a
τ
·
f
τ
,
y
τ
,
y
′
τ
,
y
′
′
τ
,
y
′
′
′
τ
d
τ
d
s
≥
0
,
A
y
1
=
∫
0
1
H
1
,
s
ϕ
q
∫
0
1
H
1
s
,
τ
a
τ
·
f
τ
,
y
τ
,
y
′
τ
,
y
′
′
τ
,
y
′
′
′
τ
d
τ
d
s
≥
0
.Thus, (
A
y
)
(
t
)
≥
0, for all t
∈
[
0,1
]
. Hence, (
A
y
)
(
t
) is nonnegative concave functional on [
0,1
].
By Lemma1, for t
∈
[
0,1
], we obtain(11)
A
y
t
=
∫
0
1
H
t
,
s
ϕ
q
∫
0
1
H
1
s
,
τ
a
τ
·
f
τ
,
y
τ
,
y
′
τ
,
y
′
′
τ
,
y
′
′
′
τ
d
τ
d
s
≤
∫
0
1
H
t
,
s
ϕ
q
∫
0
1
H
1
s
,
τ
a
τ
·
g
τ
,
y
τ
d
τ
d
s
≤
γ
γ
1
q
-
1
∫
0
1
e
s
·
ϕ
q
∫
0
1
e
τ
a
τ
g
τ
,
y
τ
d
τ
d
s
<
+
∞
.Hence, A
y
∈
P, and A
(
P
)
⊂
P, which implies A
:
P
→
P
.
Next, we show thatA
:
P
→
P is completely continuous.
For any natural numbern
≥
2, set(12)
a
n
t
=
inf
t
≤
s
≤
1
/
n
⁡
a
s
,
0
≤
t
≤
1
n
,
a
t
,
1
n
≤
t
≤
n
-
1
n
,
inf
n
-
1
/
n
≤
s
≤
t
⁡
a
s
,
n
-
1
n
≤
t
≤
1
.Then, a
n
:
[
0,1
]
→
[
0
,
+
∞
) is continuous, and a
n
(
t
)
≤
a
(
t
), t
∈
[
0,1
]. Let(13)
A
n
y
t
=
∫
0
1
H
t
,
s
ϕ
q
∫
0
1
H
1
s
,
τ
a
n
τ
·
f
τ
,
y
τ
,
y
′
τ
,
y
′
′
τ
,
y
′
′
′
τ
d
τ
d
s
.
Obviously,A
n
:
P
→
P is a completely continuous operator. From Lemma 1, we obtain(14)limn→∞⁡Any-Ay≤limn→∞⁡max0≤t≤1⁡∫01Ht,s·ϕq∫01H1s,τaτ-anτgτ,yτdτds≤γγ1q-1limn→∞⁡∫01es·ϕq∫01eτaτgτ,yτdτds=γγ1q-1·limn→∞⁡∫enesϕq∫01eτaτgτ,yτdτds=0,where e
(
n
)
=
[
0
,
1
/
n
]
∪
[
n
-
1
/
n
,
1
]. Therefore, A
:
P
→
P is a completely continuous operator. The proof is completed.Now we define linear integral operator as follows:(15)Tyt=∫01Ht,sϕq∫01H1s,τaτyτdτds,L=maxt∈0,1⁡∫01Ht,sϕq∫01H1s,τaτdτds.It is easy to see thatT
:
C
+
[
0,1
]
→
C
+
[
0,1
] is a completely continuous linear operator, and T
(
P
)
⊂
P
.Lemma 4.
Assume that (H
1) and (H
2) hold. Then, the operator T
:
P
→
P defined by (15) is a completely continuous linear operator and T
(
P
)
⊂
P, the spectral radius r
(
T
)
≠
0, and T has a positive eigenfunction corresponding to its first eigenvalue λ
1
=
r
-
1
(
T
).Proof.
A simple modification of the argument in Lemma3 yields that T
:
P
→
P is a completely continuous linear operator and T
(
P
)
⊂
P. By (H
1) and (H
2), there is s
0
∈
(
0,1
) such that H
1
(
s
0
,
s
0
)
a
(
s
0
)
>
0. Choose constants u and v such that s
0
∈
(
u
,
v
)
⊂
[
u
,
v
]
⊂
(
0,1
) and H
1
(
s
,
s
)
a
(
s
)
>
0, for all t
,
s
∈
[
u
,
v
]. Choose a nonnegative continuous functional g
∈
C
[
0,1
] such that g
(
t
)
>
0, for all t
∈
(
u
,
v
). Then, for any t
∈
(
u
,
v
), we have(16)Tgt=∫01Ht,sϕq∫01H1s,τaτgτdτds≥∫uvHt,sϕq∫uvH1s,τaτgτdτds≥0.
Thus, there exists a constantw
>
0 such that w
(
T
g
)
(
t
)
≥
g
(
t
), for all t
∈
[
0,1
]. By famous Krein-Rutman theorems, we know that the spectral radius r
(
T
)
≠
0 and T has a positive eigenfunction ψ corresponding to its first eigenvalue λ
1
=
r
-
1
(
T
); that is, ψ
=
λ
1
T
ψ. This completes the proof.Lemma 5 (see [8]).
LetP be a cone in Banach space E, and let Ω
(
P
) be a bounded open set in P. Suppose that A
:
Ω
¯
(
P
)
→
P is completely continuous. If there exists y
0
∈
P
∖
θ, such that y
-
A
y
≠
μ
y
0, for all y
∈
∂
Ω
(
P
), μ
≥
0
. Then, the fixed point index i
(
A
,
Ω
(
P
)
,
P
)
=
0
.Lemma 6 (see [8]).
LetP be a cone in Banach space E, and let Ω
(
P
) be a bounded open set in P with θ
∈
Ω
(
P
). Suppose that A
:
Ω
¯
(
P
)
→
P is a completely continuous operator. If A
y
≠
μ
y, for all y
∈
∂
Ω
(
P
), μ
≥
1
. Then, the fixed point index i
(
A
,
Ω
(
P
)
,
P
)
=
1
.Lemma 7 (see [9]).
LetP and P
1 be cones in Banach space E, and let Ω
(
P
) be a bounded open set in P. Suppose that A
:
Ω
¯
(
P
)
→
P is a completely continuous operator, and A has no fixed point on ∂
Ω
(
P
). If there exist linear operators L
1
,
L
2
,
L
3
:
E
→
E which satisfy L
1
(
P
)
⊂
P, L
2
(
P
)
⊂
P
1, L
3
(
P
1
)
⊂
P
1, and y
0
∈
P
∖
θ, such that(1)
L
2
L
1
n
y
0
≥
L
2
y
0, for certain natural number n,
(2)
L
2
L
1
y
=
L
3
L
2
y, for all y
∈
P,
(3)
L
2
A
y
≥
L
2
L
1
y, for all y
∈
∂
Ω
P,then the fixed point index i
(
A
,
Ω
(
P
)
,
P
)
=
0
.


## 3. Main Results

In this section, we establish sharp optimal criteria for the existence of positive solutions to problem (1) under superlinear cases and sublinear cases, respectively.Theorem 8.
Suppose that (H
1) and (H
2) hold. In addition, assume that(17)liminfy→+∞⁡ht,yty>λ1,uniformlyont∈0,1,(18)limsupy→0+⁡gt,yty<λ1,uniformlyont∈0,1,where λ
1 is the first eigenvalue of T defined by (15); then, problem (1) has at least one positive solution.To prove our main result, we need some preliminary results.Lemma 9.
Suppose that (H
1) and (H
2) hold. Define(19)P1=y∈P∣∫01y~∗tytdt≥λ1-1δy,where y
~
∗
(
t
)
=
y
∗
(
t
)
a
(
t
), y
∗
∈
P, is the positive eigenfunction corresponding to its first eigenvalue λ
1
=
r
-
1
(
T
); that is, y
∗
=
λ
1
T
y
∗, and(20)δ≤∫01y∗tatytdt∫01Ht,sϕq∫01H1s,τaτyτdτds;then, P
1 is a cone in C
+
[
0,1
], and T
(
P
)
⊂
P
1
.Proof.
It is obvious thatP
1 is a cone in C
+
[
0,1
]. For any y
∈
P, we obtain(21)∫01y~∗tTytdt=∫01y~∗tdt·∫01Ht,sϕq∫01H1s,τaτyτdτds=λ1-1∫01y~∗tytdt=λ1-1∫01y∗tatytdt≥λ1-1δTy.Thus, T
(
P
)
⊂
P
1. The proof is completed.Proof of Theorem8.
It follows from (17) that there exists ɛ
>
0, such that h
(
t
,
y
(
t
)
)
≥
(
λ
1
+
ɛ
)
y, for y is sufficiently large. In view of (H
1), we see that there exists b
≥
0 such that(22)ht,yt≥λ1+ɛy-b,0≤y<+∞.Let R
>
L
b
λ
1
2
(
ɛ
δ
)
-
1
∫
0
1
y
~
∗
(
t
)
d
t, where y
~
∗ and δ are defined by Lemma 9.
In the following, we prove that(23)y-Ay≠μy∗,∀y∈∂BR∩P,μ≥0,where y
∗
∈
P is a positive eigenfunction corresponding to its first eigenvalue λ
1
=
r
-
1
(
T
).
Otherwise, there existy
1
∈
∂
B
R
∩
P and μ
0
≥
0, such that(24)y1-Ay1=μ0y∗.Since A
(
P
)
⊂
P
1 and T
(
P
)
⊂
P
1, we know from (24) that y
1
∈
P
1
. Therefore, it follows from (19) and (22) that(25)∫01y~∗tAy1tdt-∫01y~∗ty1tdt≥∫01dt·∫01y~∗tHt,s·ϕq∫01H1s,τaτhτ,y1τdτds-∫01y~∗ty1tdt≥λ1+ɛ∫01dt∫01y~∗t·Ht,sϕq∫01H1s,τaτy1τdτds-b∫01dt∫01y~∗tHt,s·ϕq∫01H1s,τaτdτds-∫01y~∗t·y1tdt=λ1+ɛλ1-1∫01y~∗ty1tdt-bL∫01y~∗tdt-∫01y~∗ty1tdt=ɛλ1-1∫01y~∗ty1tdt-bL∫01y~∗tdt≥ɛλ1-1λ1-1δy1-bL∫01y~∗tdt=ɛλ1-2δR-bL∫01y~∗tdt>0.On the other hand, it follows from (24) that(26)∫01y~∗ty1tdt-∫01y~∗tAy1tdt=μ0∫01y~∗ty∗tdt≥0,which is a contradiction; thus, by Lemma 5, we see(27)iA,BR∩P,P=0.
It follows from (18) that there exists 0
≤
r
≤
R, such that(28)gt,yt<λ1y,∀0≤y≤r.Suppose that there exist y
2
∈
∂
B
r
∩
P, μ
1
≥
1, such that A
y
2
=
μ
1
y
2
. Without loss of generality, we may suppose that μ
1
>
1 (otherwise, the proof is completed). Then,(29)μ1y2t≤∫01Ht,s·ϕq∫01H1s,τaτgτ,y2τdτds<λ1∫01Ht,s·ϕq∫01H1s,τaτy2τdτds.Multiplying y
~
∗ and then integrating by (29), we have(30)μ1∫01y~∗ty2tdt<λ1∫01dt∫01y~∗tHt,s·ϕq∫01H1s,τaτy2τdτds=λ1∫01dt·∫01Ht,sy~∗t·ϕq∫01H1s,τaτy2τdτds=∫01y~∗t·y2tdt.By maximum principle, y
∗
(
t
)
>
0 and y
2
t
>
0, t
∈
0,1; then ∫
0
1
y
~
∗
(
t
)
y
2
(
t
)
d
t
>
0
. Hence, (30) implies μ
1
<
1 which is a contradiction. So by Lemma 6, we have(31)iA,Br∩P,P=1.It follows from (27) and (31) that(32)iA,BR∩P∖B¯r∩P,P=iA,BR∩P,P-iA,Br∩P,P=0-1=-1.Then, A has at least one fixed point on (
B
R
∩
P
)
∖
(
B
¯
r
∩
P
). This means problem (1) has at least one positive solution. The proof is completed.Corollary 10.
Suppose that (H
1) and (H
2) hold. In addition, assume that(33)0≤limsupy→0+⁡gt,yty=g0<h∞=liminfy→+∞⁡ht,yty≤+∞.Then, for any(34)ξ∈λ1h∞,λ1g0,where λ
1 is the first eigenvalue of T defined by (15), the following p-Laplacian integral boundary value problem(35)ϕpy′′t′′=ξatft,yt,y′t,y′′t,y′′′t,0<t<1,y0=y1=∫01bsysds,ϕpy′′0=ϕpy′′1=∫01csϕpy′′sdshas at least one positive solution.Proof.
By (34), we obtain(36)liminfy→+∞⁡ξht,yty>λ1,limsupy→0+⁡ξgt,yty<λ1.Therefore, it follows from Theorem 8 that Corollary 10 holds.Theorem 11.
Suppose that (H
1) and (H
2) hold. In addition, assume that(37)liminfy→0+⁡ht,yty>λ1,uniformlyont∈0,1;(38)limsupy→+∞⁡gt,yty<λ1,uniformlyont∈0,1,where λ
1 is the first eigenvalue of T defined by (15); then, problem (1) has at least one positive solution.Proof.
It follows from (37) that there exists r
1
>
0 such that(39)gt,yt≥λ1y,∀0≤y≤r1.Let y
∗
∈
P be the positive eigenfunction corresponding to its first eigenvalue λ
1
=
r
-
1
(
T
). Then, y
∗
=
λ
1
T
y
∗; that is,(40)y∗t=λ1Ty∗t=λ1∫01Ht,sϕq∫01H1s,τaτy∗τdτds,y∗t′′=λ1Ty∗t′′=-λ1ϕq∫01H1t,sasysds≤0,t∈0,1,y∗0=λ1∫01H0,sϕq∫01H1s,τaτy∗τdτds≥0,y∗1=λ1∫01H1,sϕq∫01H1s,τaτy∗τdτds≥0.Hence, y
∗
(
t
)
≥
0, for all t
∈
[
0,1
]
. Then, y
∗
(
t
) is nonnegative concave on [
0,1
]. Thus, y
∗
∈
P
∖
θ
.
Let(
T
2
y
)
(
t
)
=
λ
1
(
T
y
)
(
t
), y
∈
C
+
[
0,1
]. Then, T
2
:
C
+
[
0,1
]
→
C
+
[
0,1
] is a completely continuous linear operator, and T
2
(
P
1
)
⊂
P, T
2
y
∗
=
λ
1
T
y
∗
=
y
∗
.
For anyy
∈
∂
B
r
1
∩
P, it follows from (39) that(41)Ayt≥∫01Ht,sϕq∫01H1s,τaτhτ,yτdτds≥λ1∫01Ht,sϕq∫01H1s,τaτyτdτds=T2yt,t∈0,1.
SettingΩ
(
P
)
=
B
r
1
∩
P, L
1
=
L
3
=
T
2, L
2
=
I denotes the identical operator, and n
=
1, in Lemma 7. We have(42)iA,Br1∩P,P=0.
It follows from (38) that there exist r
2
>
r
1 and 0
<
σ
<
1, such that(43)gt,yt≤σλ1y,∀y≥r2.
Set(
T
1
y
)
(
t
)
=
σ
λ
1
(
T
y
)
(
t
). Then, T
1
:
C
+
[
0,1
]
→
C
+
[
0,1
] is a completely continuous linear operator, and T
1
(
P
1
)
⊂
P
. Let r
4
>
r
2
>
r
1
.
Now we prove thati
(
A
,
B
r
4
∩
P
,
P
)
=
1
.
Suppose that there existy
2
∈
∂
B
r
4
∩
P, μ
0
≥
1, such that A
y
2
=
μ
0
y
2
. Let e
y
2
=
t
∈
0,1
∣
y
2
t
>
r
2; then, from (43) we have(44)Ay2t≤∫01Ht,s·ϕq∫01H1s,τaτgτ,y2τdτds=∫ey2Ht,s·ϕq∫01H1s,τaτgτ,y2τdτds+∫0,1∖ey2Ht,s·ϕq∫01H1s,τaτgτ,y2τdτds≤σλ1∫01Ht,s·ϕq∫01H1s,τaτy2τdτds+∫01Ht,s·ϕq∫01H1s,τaτgτ,y2τdτds≤T1y2t+M,where M
=
sup
y
∈
∂
B
r
2
∩
P
⁡
∫
0
1
‍
H
(
t
,
s
)
ϕ
q
[
∫
0
1
‍
H
1
(
s
,
τ
)
a
(
τ
)
g
(
τ
,
y
2
τ
)
d
τ
]
d
s
. Thus, 0
≤
μ
0
y
2
t
=
A
y
2
t
≤
T
1
y
2
t
+
M, t
∈
[
0,1
]
. Since T
1
(
C
+
[
0,1
]
)
⊂
C
+
[
0,1
], we have 0
≤
(
T
1
j
(
A
y
2
)
)
(
t
)
≤
(
T
1
j
(
T
1
y
2
+
M
)
)
(
t
), t
∈
[
0,1
], for j
=
1
,
…
,
N
-
1
. It means that T
1
j
(
A
y
2
)
≤
T
1
j
(
T
1
y
2
+
M
), j
=
1
,
…
,
N
-
1
. Let(45)ɛ=121-rT1.For any y
∈
C
+
[
0,1
], denote(46)y∗=∑i=1NrT1+ɛN-iT1i-1y.It is easy to see that(47)μ0y2∗<rT1+3ɛ2y2∗.Since μ
0
≥
1, we have from (47) that 1
≤
r
(
T
1
)
+
3
ɛ
/
2, which contradicts (45). Thus, for any y
∈
∂
B
r
4
∩
P, μ
≥
1, we have that A
y
≠
μ
y
. Thus, from Lemma 6, we know that(48)iA,Br4∩P,P=1.
From (42) and (48), we see that(49)iA,Br4∩P∖B¯r1∩P,P=iA,Br4∩P,P-iA,Br1∩P,P=1-0=1.Then, A has at least one fixed point on (
B
r
4
∩
P
)
∖
(
B
¯
r
1
∩
P
). It means problem (1) has at least one positive solution. The proof is completed.Corollary 12.
Suppose that (H
1) and (H
2) hold. In addition, assume that(50)0≤g∞=limsupy→+∞⁡gt,yty≤liminfy→0+⁡gt,yty=g0≤+∞.Then, for any(51)ξ∈λ1h0,λ1g∞,where λ
1 is the first eigenvalue of T defined by (15), problem (35) has at least one positive solution.Proof.
By (51), we obtain(52)liminfy→0+⁡ξht,yty>λ1,limsupy→+∞⁡ξgt,yty<λ1.It follows from Theorem 11 that Corollary 12 holds. The proof is completed.


## 4. Examples

Example 1.
Consider the existence of positive solutions for the followingp-Laplacian integral boundary value problems:(53)ϕpy′′t′′=1+tπ-1y2+52016ln⁡1+y,0<t<1,y0=y1=∫01bsysds,y′′0=y′′1=ϕq∫01csϕpy′′sds,where f
t
,
y
t
,
y
′
t
,
y
′
′
t
,
y
′
′
′
t
=
(
1
+
t
)
(
π
-
1
)
y
2
+
5
/
2016
ln
⁡
(
1
+
y
). ϕ
p
(
t
)
=
|
t
|
p
-
2
t, p
>
1, ϕ
q
=
ϕ
p
-
1, 1
/
p
+
1
/
q
=
1, a
∈
L
1
[
0,1
] is symmetric on the interval [
0,1
], a
(
t
)
=
1 for t
∈
[
0,1
] is continuous, and b
,
c
∈
L
1
[
0,1
] are nonnegative symmetric on [
0,1
].Leth
(
t
,
y
)
=
5
/
2016
ln
⁡
(
1
+
y
)
+
(
π
-
1
)
y
2 and let g
(
t
,
y
)
=
(
1
+
t
)
(
π
-
1
)
y
2
+
5
/
2016
ln
⁡
(
1
+
y
)
. Obviously, h
(
t
,
y
)
≤
f
t
,
y
t
,
y
′
t
,
y
′
′
t
,
y
′
′
′
t
≤
g
(
t
,
y
), and(54)liminfy→+∞⁡ht,yty=+∞,uniformly  ont∈0,1,limsupy→0+⁡gt,yty=52016,uniformly  ont∈0,1.All conditions of Theorem8 are satisfied. Consequently, following from Theorem 8, we know that the boundary value problem (53) has at least one positive solution.Example 2.
Consider the existence of positive solutions for the followingp-Laplacian integral boundary value problems:(55)ϕpy′′t′′=y3+exp⁡yt-y′t-y′′′t-y′′t-1-6sin⁡y-sin6⁡yt+y′t+y′′t+y′′′t-sin⁡πt1-tπt1-t,0<t<1,y0=y1=∫01bsysds,y′′0=y′′1=ϕq∫01csϕpy′′sds,where ϕ
p
(
t
)
=
|
t
|
p
-
2
t, p
>
1, ϕ
q
=
ϕ
p
-
1, 1
/
p
+
1
/
q
=
1, a
∈
L
1
[
0,1
] is symmetric on the interval [
0,1
], a
(
t
)
=
1 for t
∈
[
0,1
] is continuous, and b
,
c
∈
L
1
[
0,1
] are nonnegative symmetric on [
0,1
]. For t
,
y
,
y
′
,
y
′
′
,
y
′
′
′
∈
(
0,1
)
×
[
0
,
+
∞
)
×
(
-
∞
,
+
∞
)
×
(
-
∞
,
+
∞
)
×
(
-
∞
,
+
∞
),(56)ft,yt,y′t,y′′t,y′′′t=y3+exp⁡yt-y′t-y′′′t-y′′t-1-6sin⁡y-sin6⁡yt+y′t+y′′t+y′′′t-sin⁡πt1-tπt1-t,0<t<1.Let(57)gt,y=y3+ey-1-6sin⁡y,ht,y=y3-216-3sin⁡πt1-tπt1-t,0<t<1.Obviously,h
(
t
,
y
)
≤
f
t
,
y
t
,
y
′
t
,
y
′
′
t
,
y
′
′
′
t
≤
g
(
t
,
y
), and(58)liminfy→+∞⁡ht,yty=+∞,uniformly  ont∈0,1,limsupy→0+⁡gt,yty=-5≠λ1,λ1is  the  first  eigenvalue  ofT,uniformly  ont∈0,1.All conditions of Theorem8 are satisfied. Consequently, following from Theorem 8, we know that the boundary value problem (55) has at least one positive solution.



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*Source: 418410-2015-08-26.xml*