Hindawi Publishing Corporation
Boundary Value Problems
Volume 2010, Article ID 862079, 23 pages
doi:10.1155/2010/862079
Research Article
Positive Solutions for Fourth-Order Singular
p-Laplacian Differential Equations with Integral
Boundary Conditions
Xingqiu Zhang1, 2 and Yujun Cui3
1
Department of Mathematics, Huazhong University of Science and Technology, Wuhan,
Hubei 430074, China
2
Department of Mathematics, Liaocheng University, Liaocheng, Shandong 252059, China
3
Department of Applied Mathematics, Shandong University of Science and Technology,
Qingdao 266510, China
Correspondence should be addressed to Xingqiu Zhang, zhxq197508@163.com
Received 7 April 2010; Accepted 12 August 2010
Academic Editor: Claudianor O. Alves
Copyright q 2010 X. Zhang and Y. Cui. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
By employing upper and lower solutions method together with maximal principle, we establish a
necessary and sufficient condition for the existence of pseudo-C3 0, 1 as well as C2 0, 1 positive
solutions for fourth-order singular p-Laplacian differential equations with integral boundary
conditions. Our nonlinearity f may be singular at t 0, t 1, and u 0. The dual results for
the other integral boundary condition are also given.
1. Introduction
In this paper, we consider the existence of positive solutions for the following nonlinear
fourth-order singular p-Laplacian differential equations with integral boundary conditions:
ϕp x t
ft, xt, xt, 0 < t < 1,
1
x0 gsxsds, x1 0,
0
ϕp x 0 ϕp x 1 1
0
hsϕp x s ds,
1.1
2
Boundary Value Problems
where ϕp t |t|p−2 · t, p ≥ 2, ϕq ϕ−1
p , 1/p 1/q 1, f ∈ CJ × R × R , R , J 0, 1,
1
R 0, ∞, R 0, ∞, I 0, 1, and g, h ∈ L1 0, 1 is nonnegative. Let σ1 0 1 −
1
1
sgsds, σ2 0 hsds. Throughout this paper, we always assume that 0 ≤ 0 gsds < 1,
1
0 < 0 hsds < 1 and nonlinear term f satisfies the following hypothesis:
H ft, u, v : J × R × R → R is continuous, nondecreasing on u and nonincreasing
on v for each fixed t ∈ J, and there exists a real number b ∈ R such that, for any
r ∈ J,
ft, u, rv ≤ r −b ft, u, v,
∀t, u, v ∈ J × R × R ,
1.2
there exists a function ξ : 1, ∞ → R , ξl < l and ξl/l2 is integrable on 1, ∞
such that
ft, lu, v ≤ ξlft, u, v,
∀t, u, v ∈ J × R × R , l ∈ 1, ∞.
1.3
Remark 1.1. Condition H is used to discuss the existence and uniqueness of smooth positive
solutions in 1.
i Inequality 1.2 implies that
ft, u, cv ≥ c−b ft, u, v,
if c ≥ 1.
1.4
Conversely, 1.4 implies 1.2.
ii Inequality 1.3 implies that
−1
ft, cu, v ≥ ξ c−1
ft, u, v,
if 0 < c < 1.
1.5
Conversely, 1.5 implies 1.3.
Remark 1.2. Typical functions that satisfy condition H are those taking the form ft, u, v
−μj
ni1 ai tuλi m
, where ai , bj ∈ C0, 1, 0 < λi < 1, μj > 0 i 1, 2, . . . , m;
j1 bj tu
j 1, 2, . . . , m.
Remark 1.3. It follows from 1.2 and 1.3 that
ft, u, u ≤
⎧ u
⎪
⎪
⎪ξ v ft, v, v,
⎨
if u ≥ v ≥ 0,
⎪ v b
⎪
⎪
⎩
ft, v, v,
u
if v ≥ u ≥ 0.
1.6
Boundary Value Problems
3
Boundary value problems with integral boundary conditions arise in variety of
different areas of applied mathematics and physics. For example, heat conduction, chemical
engineering, underground water flow, thermoelasticity, and plasma physics can be reduced
to nonlocal problems with integral boundary conditions. They include two point, three point,
and nonlocal boundary value problems see 2–5 as special cases and have attracted much
attention of many researchers, such as Gallardo, Karakostas, Tsamatos, Lomtatidze, Malaguti,
Yang, Zhang, and Feng see 6–13, e.g.. For more information about the general theory of
integral equations and their relation to boundary value problems, the reader is referred to the
book by Corduneanu 14 and Agarwal and O’Regan 15.
Recently, Zhang et al. 13 studied the existence and nonexistence of symmetric
positive solutions for the following nonlinear fourth-order boundary value problems:
ωtft, xt, 0 < t < 1,
ϕp x t
1
gsxsds,
x0 x1 0
ϕp x 0 ϕp x 1 1
1.7
hsϕp x s ds,
0
where ϕp t |t|p−2 · t, p > 1, ϕq φp−1 , 1/p 1/q 1, ω ∈ L0, 1 is nonnegative, symmetric
on the interval 0, 1, f : 0, 1 × 0, ∞ → 0, ∞ is continuous, and g, h ∈ L1 0, 1 are
nonnegative, symmetric on 0, 1.
To seek necessary and sufficient conditions for the existence of solutions to the
ordinary differential equations is important and interesting, but difficult. Professors Wei
16, 17, Du and Zhao 18, Graef and Kong 19, Zhang and Liu 20, and others have
done much excellent work under some suitable conditions in this direction. To the author’s
knowledge, there are no necessary and sufficient conditions available in the literature for the
existence of solutions for integral boundary value problem 1.1. Motivated by above papers,
the purpose of this paper is to fill this gap. It is worth pointing out that the nonlinearity
ft, u, v permits singularity not only at t 0, 1 but also at v 0. By singularity, we mean that
the function f is allowed to be unbounded at the points t 0, 1 and v 0.
2. Preliminaries and Several Lemmas
A function xt ∈ C2 0, 1 and ϕp x t ∈ C2 0, 1 is called a C2 0, 1 positive solution of
BVP 1.1 if it satisfies 1.1 xt > 0 for t ∈ 0, 1. A C2 0, 1 positive solution of 1.1 is
called a psuedo-C3 0, 1 positive solution if ϕp x t ∈ C1 0, 1 xt > 0, −x t > 0 for
t ∈ 0, 1. Denote that
E x : x ∈ C2 0, 1, and ϕp x t ∈ C2 0, 1 .
2.1
4
Boundary Value Problems
Definition 2.1. A function αt ∈ E is called a lower solution of BVP 1.1 if αt satisfies
≤ ft, αt, αt, 0 < t < 1,
ϕp α t
1
α0 − gsαsds ≤ 0, α1 ≤ 0,
0
− ϕp α 0 −
1
2.2
hsϕp α s ds ≤ 0,
0
− ϕp α 1 −
1
hsϕp α s ds ≤ 0.
0
Definition 2.2. A function βt ∈ E is called an upper solution of BVP 1.1 if βt satisfies
ϕp β t
≥ f t, βt, βt , 0 < t < 1,
1
β0 − gsβsds ≥ 0, β1 ≥ 0,
0
− ϕp β 0 −
1
2.3
hsϕp β s ds ≥ 0,
0
− ϕp β 1 −
1
hsϕp β s ds ≥ 0.
0
Suppose that 0 < ak < bk < 1, and
Fk 2
x : x ∈ C ak , bk , ϕp x t ∈ C ak , bk , xak −
− ϕp x ak −
bk
ak
2
bk
ak
gsxsds ≥ 0, xbk ≥ 0,
hsϕp x s ds ≥ 0, − ϕp x bk −
bk
ak
hsϕp x s ds ≥ 0 .
2.4
To prove the main results, we need the following maximum principle.
Lemma 2.3 Maximum principle. If x ∈ Fk , such that ϕp x t ≥ 0, t ∈ ak , bk , then
xt ≥ 0, −x t ≥ 0, t ∈ ak , bk .
Boundary Value Problems
5
Proof. Set
−x t yt, t ∈ ak , bk ,
σt, t ∈ ak , bk ,
ϕp x t
xak −
bk
ak
gsxsds r1 ,
bk
− ϕp x ak −
bk
ak
2.6
2.7
hsϕp x s ds r3 ,
ak
− ϕp x bk −
xbk r2 ,
2.5
2.8
hsϕp x s ds r4 ,
2.9
then ri ≥ 0, i 1, 2, 3, 4, σt ≥ 0, t ∈ ak , bk and
−ϕp y t σt,
ϕp y ak −
ϕp y bk −
bk
ak
bk
ak
t ∈ ak , bk ,
hsϕp ys ds r3 ,
2.10
hsϕp ys ds r4 .
Let
ϕp y t zt,
2.11
then
−z t σt,
zak −
bk
ak
hszsds r3 ,
t ∈ ak , bk ,
zbk −
bk
ak
2.12
hszsds r4 .
2.13
By integration of 2.12, we have
z t z ak −
t
ak
σsds.
2.14
Integrating again, we get
zt zak z ak t − ak −
t
ak
t − sσsds.
2.15
6
Boundary Value Problems
Let t bk in 2.15, we obtain that
z ak 1
r4 − r3
bk − ak bk − ak
bk
ak
bk − sσsds.
2.16
Substituting 2.13 and 2.16 into 2.15, we obtain that
r4 − r3
zt r3 t − ak bk − ak
bk
ak
Gk t, sσsds bk
ak
hszsds,
2.17
where
⎧
1 ⎨bk − ts − ak ,
Gk t, s bk − ak ⎩b − st − a ,
k
k
0 ≤ s ≤ t ≤ 1,
2.18
0 ≤ t ≤ s ≤ 1.
Notice that
bk
ak
hszsds bk
ak
r3
r4 − r3
hs r3 s − ak bk − ak
bk
ak
bk
ak
r4 − r3
hsds bk − ak
hsds ·
bk
ak
bk
ak
bk
ak
Gk s, τστdτ s − ak hsds bk
ak
bk
ak
b
hs
k
ak
hszsds ds
Gk s, τστdτ ds
hszsds,
2.19
therefore,
bk
ak
b
k
r4 − r3 bk
r3
hszsds hsds s − ak hsds
b
bk − ak ak
ak
1 − k hsds
1
ak
bk
ak
hs
b
k
ak
Gk s, τστdτ ds .
2.20
Boundary Value Problems
7
Substituting 2.20 into 2.17, we have
b
bk
k
r4 − r3
1
hs
Gk s, τστdτ ds
zt r3 t − ak b
bk − ak
ak
1 − akk hsds ak
b
b
k
k
1
r4 − r3 bk
r3
hsds Gk t, sσsds
s − ak hsds bk
bk − ak ak
ak
ak
1−
hsds
ak
bk − t
t − ak
1
r3 r4 bk − ak
bk − ak
1 − σ2k
bk
ak
ak
ak
bk − s
hsdsr3 bk − ak
bk
ak
s − ak
hsdsr4
bk − ak
Kk t, sσsds
1
bk − t
bk − ak 1 − σ2k
bk
b
k
bk
ak
bk
bk − s
s − ak
t − ak
1
hsds r3 hsds r4
bk − ak
bk − ak 1 − σ2k ak bk − ak
Kk t, sσsds,
2.21
where
1
Kk t, s Gk t, s 1 − σ2k
bk
ak
Gk s, τhτdτ,
σ2k bk
ak
hsds.
2.22
Obviously, Gk t, s ≥ 0, Kk t, s ≥ 0, σ2k ≥ 0. From 2.21, it is easily seen that zt ≥ 0 for
t ∈ ak , bk . By 2.11, we know that ϕp yt ≥ 0, that is, yt ≥ 0. Thus, we have proved that
−x t ≥ 0, t ∈ ak , bk . Similarly, the solution of 2.5 and 2.7 can be expressed by
bk
bk − s
bk − t
bk − t
r1
xt gs
bk − ak 1 − σ1k bk − ak ak
bk − ak
bk
bk
s − ak
t − ak
bk − t
r2 gs
Hk t, sysds,
bk − ak 1 − σ1k bk − ak ak
bk − ak
ak
2.23
where
bk − t
Hk t, s Gk t, s 1 − σ1k
bk
ak
Gk s, τgτdτ,
By 2.23, we can get that xt ≥ 0, t ∈ ak , bk .
σ1k
1
bk − ak
bk
ak
gsbk − sds.
2.24
8
Boundary Value Problems
Lemma 2.4. Suppose that H holds. Let xt be a C2 0, 1 positive solution of BVP 1.1. Then there
exist two constants 0 < I1 < I2 such that
I1 1 − t ≤ xt ≤ I2 1 − t,
t ∈ 0, 1.
2.25
Proof. Assume that xt is a C2 0, 1 positive solution of BVP 1.1. Then xt can be stated as
xt 1
0
1−t
Gt, s −x s ds 1 − σ1
1
Gτ, sgτ −x s dτ ds,
2.26
0
where
Gt, s ⎧
⎨t1 − s,
0 ≤ t ≤ s ≤ 1,
⎩s1 − t,
0 ≤ s ≤ t ≤ 1.
2.27
It is easy to see that
1
Gτ, sgτ −x s dτ ds > 0.
2.28
Gτ, sgτ −x s dτ ds 1 − tx0 ≥ 0.
2.29
1
x0 1 − σ1
0
By 2.26, for 0 ≤ t ≤ 1, we have that
1−t
xt ≥
1 − σ1
1
0
From 2.26 and 2.27, we get that
1
0 ≤ xt ≤ 1 − t
s −x s ds 0
1
1 − σ1
1
Gτ, sgτ −x s dτ ds .
2.30
Gτ, sgτ −x s dτ ds,
2.31
0
Setting
I1 x0,
I2 1
0
s −x s ds 1
1 − σ1
1
0
then from 2.29 and 2.30, we have 2.25.
Lemma 2.5. Suppose that H holds. And assume that there exist lower and upper solutions of BVP
1.1, respectively, αt and βt, such that αt, βt ∈ E, 0 ≤ αt ≤ βt for t ∈ 0, 1, α1 β1 0. Then BVP 1.1 has at least one C2 0, 1 positive solution xt such that αt ≤ xt ≤ βt,
t ∈ 0, 1. If, in addition, there exists Ft ∈ L1 0, 1 such that
ft, x, x ≤ Ft,
for αt ≤ xt ≤ βt,
2.32
Boundary Value Problems
9
then the solution xt of BVP 1.1 is a pseudo-C3 0, 1 positive solution.
Proof. For each k, for all xt ∈ Ek {x : x ∈ C2 ak , bk , and ϕp x t ∈ C2 ak , bk }, we
defined an auxiliary function
⎧
⎪
ft, αt, αt,
⎪
⎪
⎨
Fk xt ft, xt, xt,
⎪
⎪
⎪
⎩ f t, βt, βt ,
if xt ≤ αt,
2.33
if αt ≤ xt ≤ βt,
if xt ≥ βt.
By condition H, we have that Fk : Ek → 0, ∞ is continuous.
Let {ak }, {bk } be sequences satisfying 0 < · · · < ak1 < ak < · · · < a1 < b1 < · · · < bk <
bk1 < · · · < 1, ak → 0 and bk → 1 as k → ∞, and let {rki }, i 1, 2, 3, be sequences satisfying
αak −
bk
ak
gsαsds ≤ rk1 ≤ βak −
bk
ak
gsβsds,
αbk ≤ rk2 ≤ βbk ,
rk1 −→ 0, rk2 −→ 0, as k −→ ∞,
bk
bk
− ϕp α ak −
hsϕp α s ds ≤ rk3 ≤ − ϕp β ak −
hsϕp β s ds , 2.34
ak
− ϕp α bk −
1
ak
hsϕp α s ds ≤ rk4 ≤ − ϕp β bk −
0
bk
ak
hsϕp β s ds ,
rk3 −→ 0, rk4 −→ 0, as k −→ ∞.
For each k, consider the following nonsingular problem:
Fk xt,
ϕp x t
xak −
bk
ak
gsxsds rk1 ,
− ϕp x ak −
t ∈ ak , bk ,
− ϕp x bk −
bk
ak
bk
ak
xbk rk2 ,
hsϕp x s ds rk3 ,
hsϕp x s ds rk4 .
2.35
10
Boundary Value Problems
For convenience, we define linear operators as follows:
bk
bk − s
bk − t
1
hsds r3
Bk xt bk − ak 1 − σ2k ak bk − ak
bk
bk
s − ak
t
1
hsds r4 Kk t, sxsds,
bk − ak 1 − σ2k ak bk − ak
ak
bk
bk − s
bk − t
bk − t
r1
gs
Ak xt bk − ak 1 − σ1k bk − ak ak
bk − ak
bk
bk
s − ak
t − ak
bk − t
r2 gs
Hk t, sxsds.
bk − ak 1 − σ1k bk − ak ak
bk − ak
ak
2.36
By the proof of Lemma 2.3, xt is a solution of problem 2.35 if and only if it is the
fixed point of the following operator equation:
xt.
F
xt Ak ϕ−1
B
k
k
p
2.37
By 2.33, it is easy to verify that Ak ϕ−1
p Bk Fk : Ek → Ek is continuous and Fk Ek is
a bounded set. Moreover, by the continuity of Gk t, s, we can show that Ak ϕ−1
p Bk is a
−1
−1
compact operator and Ak ϕp Bk Ek is a relatively compact set. So, Ak ϕp Bk Fk : Ek →
Ek is a completely continuous operator. In addition, x ∈ Ek is a solution of 2.35 if and
only if x is a fixed point of operator Ak ϕ−1
p Bk Fk x x. Using the Shauder’s fixed point
−1
theorem, we assert that Ak ϕp Bk Fk has at least one fixed point xk ∈ C2 ak , bk , by xk t 2
Ak ϕ−1
p Bk Fk xk t, we can get ϕp xk ∈ C ak , bk .
We claim that
αt ≤ xk t ≤ βt,
t ∈ ak , bk .
2.38
From this it follows that
ft, xk t, xk t,
ϕp xk t
t ∈ ak , bk .
2.39
≤βt on ak , bk . By the definition of Fk , we have
Indeed, suppose by contradiction that xk t/
Fk xk t f t, βt, βt ,
t ∈ ak , bk .
2.40
Therefore,
f t, βt, βt ,
ϕp xk t
t ∈ ak , bk .
2.41
Boundary Value Problems
11
On the other hand, since βt is an upper solution of 1.1, we also have
≥ f t, βt, βt ,
ϕp β t
t ∈ ak , bk .
2.42
Then setting
zt ϕp −β t − ϕp −xk t ,
t ∈ ak , bk .
2.43
By 2.41 and 2.42, we obtain that
−z t ≥ 0,
zak −
bk
ak
t ∈ ak , bk , x ∈ C2 ak , bk ,
hszsds ≥ 0,
zbk −
bk
ak
hszsds ≥ 0.
2.44
By Lemma 2.3, we can conclude that
zt ≥ 0,
t ∈ ak , bk .
2.45
Hence,
− β t − xk t ≥ 0,
t ∈ ak , bk .
2.46
Set
ut βt − xk t,
t ∈ ak , bk .
2.47
Then
−u t ≥ 0,
uak −
bk
ak
t ∈ ak , bk , x ∈ C2 ak , bk ,
gsusds ≥ 0,
ubk ≥ 0.
2.48
By Lemma 2.3, we can conclude that
ut ≥ 0,
t ∈ ak , bk ,
2.49
which contradicts the assumption that xk t/
≤βt. Therefore, xk t/
≤βt is impossible.
Similarly, we can show that αt ≤ xk t. So, we have shown that 2.38 holds.
Using the method of 21 and Theorem 3.2 in 22, we can obtain that there is a C2 0, 1
positive solution ωt of 1.1 such that αt ≤ ωt ≤ βt, and a subsequence of {xk t}
converging to ωt on any compact subintervals of 0, 1.
In addition, if 2.32 holds, then |ϕp x t | ≤ Ft. Hence, ϕp x t is absolutely
integrable on 0, 1. This implies that xt is a pseudo-C3 0, 1 positive solution of 1.1.
12
Boundary Value Problems
3. The Main Results
Theorem 3.1. Suppose that H holds, then a necessary and sufficient condition for BVP 1.1 to
have a pseudo-C3 0, 1 positive solution is that the following integral condition holds:
0<
1
fs, 1 − s, 1 − sds < ∞.
3.1
0
Proof. The proof is divided into two parts, necessity and suffeciency.
Necessity. Suppose that xt is a pseudo-C3 0, 1 positive solution of 1.1. Then both
ϕp x 0 and ϕp x 1 exist. By Lemma 2.4, there exist two constants 0 < I1 < I2 such that
I1 1 − t < xt < I2 1 − t,
t ∈ 0, 1.
3.2
Without loss of generality, we may assume that 0 < I1 < 1 < I2 . This together with condition
H implies that
1
0
1 1
1
1
1
f s, xs, xs ≤ ξ
I2b fs, xs, xsds
I
I
I
1
2
1
0
0
1
I b · ϕp x 1 − ϕp x 0 < ∞.
ξ
I1 2
fs, 1 − s, 1 − sds ≤
3.3
On the other hand, since xt is a pseudo-C3 0, 1 positive solution of 1.1, we have
≡ 0,
ft, xt, xt /
t ∈ 0, 1.
3.4
Otherwise, let zt ϕp x t. By the proof of Lemma 2.3, we have that zt ≡ 0, t ∈ 0, 1,
that is, x t ≡ 0 which contradicts that xt is a pseudo-C3 0, 1 positive solution. Therefore,
there exists a positive t0 ∈ 0, 1 such that ft0 , xt0 , xt0 > 0. Obviously, xt0 > 0. By 1.6
we have
0 < ft0 , xt0 , xt0 ≤
⎧ xt0 ⎪
⎪
ft0 , 1 − t0 , 1 − t0 ,
ξ
⎪
⎪
⎨ 1 − t0
if xt0 ≥ 1 − t0 ,
⎪
⎪
1 − t0 b
⎪
⎪
⎩
ft0 , 1 − t0 , 1 − t0 ,
xt0 if xt0 ≤ 1 − t0 .
3.5
Consequently, ft0 , 1 − t0 , 1 − t0 > 0, which implies that
1
0
fs, 1 − s, 1 − sds > 0.
3.6
Boundary Value Problems
13
It follows from 3.3 and 3.6 that
0<
1
fs, 1 − s, 1 − sds < ∞,
3.7
0
which is the desired inequality.
Sufficiency. First, we prove the existence of a pair of upper and lower solutions. Since ξl/l2
is integrable on 1, ∞, we have
lim inf
l → ∞
ξl
0.
l
3.8
Otherwise, if liml → ∞ infξl/l m0 > 0, then there exists a real number N > 0 such that
ξl/l2 > m0 /2l when l > N, which contradicts the condition that ξl/l2 is integrable on
1, ∞. In view of condition H and 3.8, we obtain that
ft, ru, v ≥ hrft, u, v,
r ∈ 0, 1,
ξ p
p−1
r
lim sup
lim sup −1 lim inf
0,
p → ∞
r → 0
hr p → ∞
p
h p
3.9
3.10
where hr ξr −1 −1 .
Suppose that 3.1 holds. Firstly, we define the linear operators A and B as follows:
Bxt 1
0
Axt 1
1
Gt, sxsds 1 − σ2
Gt, sxsds 0
1−t
1 − σ1
1
Gs, τhτxsdτ ds,
3.11
0
1
Gs, τgτxsdτ ds,
3.12
0
where Gt, s is given by 2.27. Let
b1 t Aϕ−1
p Bft, 1 − t, 1 − t,
t ∈ 0, 1.
3.13
It is easy to know from 3.11 and 3.12 that ϕp −b t ∈ C1 0, 1. By Lemma 2.4, we know
that there exists a positive number k < 1 such that
k1 1 − t ≤ b1 t ≤
1
1 − t,
k1
t ∈ 0, 1.
3.14
Take 0 < l1 < k1 sufficiently small, then by 3.10, we get that l1 k1 /hl1 k1 ≤ k1 , that is,
hl1 k1 − l1 ≥ 0,
1
ξ
l1 k1
−
1
≤ 0.
l1
3.15
14
Boundary Value Problems
Let
αt l1 b1 t,
βt 1
b1 t,
l1
t ∈ 0, 1.
3.16
Thus, from 3.14 and 3.16, we have
l1 k1 1 − t ≤ αt ≤ 1 − t ≤ βt ≤
1
1 − t,
l1 k1
t ∈ 0, 1.
3.17
Considering p ≥ 2, it follows from 3.15, 3.17, and condition H that
ft, αt, αt ≥ ft, l1 k1 1 − t, 1 − t ≥ hl1 k1 ft, 1 − t, 1 − t
p−1
≥ l1 ft, 1 − t, 1 − t ≥ l1 ft, 1 − t, 1 − t
ϕp α t , t ∈ 0, 1,
1
1
ft, 1 − t, 1 − t
f t, βt, βt ≤ f t,
1 − t, 1 − t ≤ ξ
l1 k1
l1 k1
1
≤
ft, 1 − t, 1 − t
l1
p−1
1
≤
ft, 1 − t, 1 − t
l1
ϕp β t , t ∈ 0, 1.
3.18
From 3.13 and 3.16, it follows that
α0 1
gtαtdt,
β0 0
1
gtβtdt,
0
ϕp α 0 ϕp α 1 1
α1 0,
β1 0,
hsϕp α s ds,
3.19
0
ϕp β 0 ϕp β 1 1
hsϕp β s ds.
0
Thus, we have shown that αt and βt are lower and upper solutions of BVP 1.1,
respectively.
Additionally, when αt ≤ xt ≤ βt, t ∈ 0, 1, by 3.17 and condition H, we have
1
0 ≤ ft, xt, xt ≤ f t,
1 − t, l1 k1 1 − t
l1 k1
1
1
ft, 1 − t, l1 k1 1 − t ≤ ξ
≤ξ
l1 k1 −b ft, 1 − t, 1 − t Ft.
l1 k1
l1 k1
3.20
Boundary Value Problems
15
1
From 3.1, we have 0 Ftdt < ∞. So, it follows from Lemma 2.5 that BVP 1.1 admits a
pseudo-C3 0, 1 positive solution such that αt ≤ xt ≤ βt.
Remark 3.2. Lin et al. 23, 24 considered the existence and uniqueness of solutions for some
fourth-order and k, n − k conjugate boundary value problems when ft, u, v qtgu hv, where
g : 0, ∞ −→ 0, ∞ is continuous and nondecreasing,
h : 0, ∞ −→ 0, ∞ is continuous and nonincreasing,
3.21
under the following condition:
P1 for t ∈ 0, 1 and u, v > 0, there exists α ∈ 0, 1 such that
gtu ≥ tα gu,
h t−1 v ≥ tα hv.
3.22
Lei et al. 25 and Liu and Yu 26 investigated the existence and uniqueness of positive
solutions to singular boundary value problems under the following condition:
P2 ft, λu, 1/λv ≥ λα ft, u, v, for all u, v > 0, λ ∈ 0, 1, where α ∈ 0, 1 and
ft, u, v is nondecreasing on u and nonincreasing on v.
Obviously, 3.21-3.22 imply condition P2 and condition P2 implies condition
H. So, condition H is weaker than conditions P1 and P2 . Thus, functions considered in
this paper are wider than those in 23–26.
In the following, when ft, u, u admits the form ft, u, that is, nonlinear term f is not
mixed monotone on u, but monotone with respect u, BVP 1.1 becomes
ft, xt, 0 < t < 1,
ϕp x t
1
gsxsds, x1 0,
x0 0
ϕp x 0 ϕp x 1 1
3.23
hsϕp x s ds.
0
If ft, u satisfies one of the following:
H∗ ft, u : J × R → R is continuous, nondecreasing on u, for each fixed t ∈ 0, 1,
there exists a function ξ : 1, ∞ → R , ξl < l and ξl/l2 is integrable on 1, ∞
such that
ft, lu ≤ ξlft, u,
∀t, u ∈ J × R , l ∈ 1, ∞.
3.24
16
Boundary Value Problems
Theorem 3.3. Suppose that H∗ holds, then a necessary and sufficient condition for BVP 3.23 to
have a pseudo-C3 0, 1 positive solution is that the following integral condition holds
0<
1
fs, 1 − sds < ∞.
3.25
0
Proof. The proof is similar to that of Theorem 3.1; we omit the details.
Theorem 3.4. Suppose that H∗ holds, then a necessary and sufficient condition for problem 3.23
to have a C2 0, 1 positive solution is that the following integral condition holds
0<
1
s1 − sfs, 1 − sds < ∞.
3.26
0
Proof. The proof is divided into two parts, necessity and suffeciency.
Necessity. Assume that xt is a C2 0, 1 positive solution of BVP 3.23. By Lemma 2.4, there
exist two constants I1 and I2 , 0 < I1 < I2 , such that
I1 1 − t ≤ xt ≤ I2 1 − t,
t ∈ 0, 1.
3.27
Let c1 be a constant such that c1 I2 ≤ 1, 1/c1 ≥ 1. By condition H, we have
1 c1 xt
c1 xt
ft, xt f t,
1 − t ≥ f t,
1 − t
c1 1 − t
1−t
c1 xt
1 − t −1
ft, 1 − t
ft, 1 − t ≥
≥ξ
c1 xt
1−t
≥ c1 I1 ft, 1 − t,
3.28
t ∈ 0, 1.
By virtue of 3.28, we obtain that
ft, 1 − t ≤ c1 I1 −1 ft, xt c1 I1 −1 ϕp x t ,
t ∈ 0, 1.
3.29
By boundary value condition, we know that there exists a t0 ∈ 0, 1 such that
ϕp x
t0 0.
3.30
For t ∈ t0 , 1, by integration of 3.29, we get
t
t0
fs, 1 − sds ≤ c1 I1 −1 ϕp x t,
t ∈ t0 , 1.
3.31
Boundary Value Problems
17
Integrating 3.31, we have
1 t
t0
t0
−1
fs, 1 − sds dt ≤ c1 I1 1
t0
ϕp x
tdt
c1 I1 −1 ϕp x 1 − ϕp x t0 < ∞.
3.32
Exchanging the order of integration, we obtain that
1 t
t0
t0
fs, 1 − sds dt 1
t0
1 − sfs, 1 − sds < ∞.
3.33
Similarly, by integration of 3.29, we get
t0
sfs, 1 − sds < ∞.
3.34
0
Equations 3.33 and 3.34 imply that
1
s1 − sfs, 1 − sds < ∞.
3.35
0
Since xt is a C2 0, 1 positive solution of BVP 1.1, there exists a positive t0 ∈ 0, 1 such
that ft0 , xt0 > 0. Obviously, xt0 > 0. On the other hand, choose c2 < min{1, I1 , 1/I2 },
then c2 I2 < 1. By condition H, we have
1 c2 xt0 0 < ft0 , xt0 f t0 ,
1 − t0 c 2 1 − t0
1
c2 xt0 1
≤ξ
f t0 ,
ft0 , 1 − t0 .
1 − t0 ≤ ξ
c2
1 − t0
c2
3.36
Consequently, ft0 , t0 > 0, which implies that
1
s1 − sfs, 1 − sds > 0.
3.37
0
It follows from 3.35 and 3.37 that
0<
1
0
which is the desired inequality.
s1 − sfs, 1 − sds < ∞,
3.38
18
Boundary Value Problems
Sufficiency. Suppose that 3.26 holds. Let
b2 t Aϕ−1
p Bft, 1 − t,
t ∈ 0, 1.
3.39
It is easy to know, from 3.11 and 3.26, that
Bft, 1 − t ≤
1
1 − σ2
1
Gs, sfs, 1 − sds < ∞.
3.40
0
Thus, 3.12, 3.39, and 3.40 imply that 0 ≤ b2 t < ∞. By Lemma 2.4, we know that there
exists a positive number k2 < 1 such that
k2 1 − t ≤ b2 t ≤
1
1 − t,
k2
t ∈ 0, 1.
3.41
Take 0 < l2 < k2 sufficiently small, then by 3.10, we get that l2 k2 /hl2 k2 ≤ k2 , that is,
hl2 k2 − l2 ≥ 0,
ξ
1
l2 k2
1
≤ 0.
l2
3.42
t ∈ 0, 1.
3.43
−
Let
αt l2 b2 t,
βt 1
b2 t,
l2
Thus, from 3.41 and 3.43, we have
l2 k2 1 − t ≤ αt ≤ 1 − t ≤ βt ≤
1
1 − t,
l2 k2
t ∈ 0, 1.
3.44
Notice that p ≥ 2, it follows from 3.42−3.44 and condition H that
ft, αt ≥ ft, l2 k2 1 − t ≥ hl2 k2 ft, 1 − t
p−1
≥ l2 ft, 1 − t ≥ l2 ft, 1 − t
ϕp α t , t ∈ 0, 1,
1
1
ft, 1 − t
f t, βt ≤ f t, 1 − t ≤ ξ
lk
l2 k2
p−1
1
1
≤
ft, 1 − t
ft, 1 − t ≤
l2
l2
ϕp β t , t ∈ 0, 1.
3.45
Boundary Value Problems
19
From 3.39 and 3.43, it follows that
α0 1
gtαtdt,
β0 0
1
gtβtdt,
0
ϕp α 0 ϕp α 1 1
α1 0,
β1 0,
hsϕp α s ds,
3.46
0
ϕp β 0 ϕp β 1 1
hsϕp β s ds.
0
Thus, we have shown that αt and βt are lower and upper solutions of BVP 1.1,
respectively.
From the first conclusion of Lemma 2.5, we conclude that problem 1.1 has at least
one C2 0, 1 positive solution xt.
4. Dual Results
Consider the fourth-order singular p-Laplacian differential equations with integral conditions:
ft, xt, xt, 0 < t < 1,
ϕp x t
1
gsxsds, x1 0,
x0 0
ϕp x 0 1
hsϕp x s ds,
x 1 0,
0
ft, xt, 0 < t < 1,
ϕp x t
1
gsxsds, x1 0,
x0 0
ϕp x 0 4.1
1
hsϕp x s ds,
4.2
x 1 0.
0
Firstly, we define the linear operator B1 as follows:
B1 xt 1
Gt, sxsds 0
where Gt, s is given by 2.27.
1−
1
1−t
0 1
− shsds
1
0
Gs, τhτxsdτ ds,
4.3
20
Boundary Value Problems
By analogous methods, we have the following results.
Assume that xt is a C2 0, 1 positive solution of problem 4.1. Then xt can be
expressed by
xt Aϕ−1
p B1 ft, xt, xt.
4.4
Theorem 4.1. Suppose that H holds, then a necessary and sufficient condition for 4.1 to have a
pseudo-C3 0, 1 positive solution is that the following integral condition holds:
0<
1
fs, 1 − s, 1 − sds < ∞.
4.5
0
Theorem 4.2. Suppose that H∗ holds, then a necessary and sufficient condition for problem 4.2 to
have a pseudo-C3 0, 1 positive solution is that the following integral condition holds:
0<
1
fs, 1 − sds < ∞.
4.6
0
Theorem 4.3. Suppose that H∗ holds, then a necessary and sufficient condition for problem 4.2 to
have a C2 0, 1 positive solution is that the following integral condition holds:
0<
1
s1 − sfs, 1 − sds < ∞.
4.7
0
Consider the fourth-order singular p-Laplacian differential equations with integral conditions:
ft, xt, xt, 0 < t < 1,
ϕp x t
1
gsxsds,
x0 0, x1 ϕp x 0 1
0
hsϕp x s ds,
x 1 0,
0
ft, xt, 0 < t < 1,
ϕp x t
1
gsxsds,
x0 0, x1 ϕp x 0 4.8
1
0
hsϕp x s ds,
4.9
x 1 0.
0
Define the linear operator A1 as follows:
A1 xt 1
0
Gt, sxsds 1−
1
0
t
sgsds
1
0
Gs, τgτxsdτ ds.
4.10
Boundary Value Problems
21
If xt is a C2 0, 1 positive solution of problem 4.8. Then xt can be expressed by
xt A1 ϕ−1
p B1 ft, xt, xt.
4.11
Theorem 4.4. Suppose that H holds, then a necessary and sufficient condition for problem 4.8 to
have a pseudo-C3 0, 1 positive solution is that the following integral condition holds:
0<
1
fs, s, sds < ∞.
4.12
0
Theorem 4.5. Suppose that H∗ holds, then a necessary and sufficient condition for problem 4.9 to
have a pseudo-C3 0, 1 positive solution is that the following integral condition holds:
0<
1
fs, sds < ∞.
4.13
0
Theorem 4.6. Suppose that H∗ holds, then a necessary and sufficient condition for problem 4.9 to
have a C2 0, 1 positive solution is that the following integral condition holds:
0<
1
s1 − sfs, sds < ∞.
4.14
0
Consider the fourth-order singular p-Laplacian differential equations with integral conditions:
ft, xt, xt, 0 < t < 1,
ϕp x t
1
gsxsds,
x0 0, x1 0
ϕp x 0 0,
ϕp x 1 hsϕp x s ds,
0
ft, xt, 0 < t < 1,
ϕp x t
1
gsxsds,
x0 0, x1 0
1
ϕp x 0 0,
4.15
ϕp x 1 1
4.16
hsϕp x s ds.
0
Define the linear operator B2 as follows:
B2 xt 1
0
Gt, sxsds 1−
1
0
t
shsds
1
0
Gs, τhτxsdτ ds.
4.17
22
Boundary Value Problems
If xt is a C2 0, 1 positive solution of problem 4.15. Then xt can be expressed by
xt A1 ϕ−1
p B2 ft, xt, xt.
4.18
Theorem 4.7. Suppose that H holds, then a necessary and sufficient condition for problem 4.15
to have a pseudo-C3 0, 1 positive solution is that the following integral condition holds:
0<
1
fs, s, sds < ∞.
4.19
0
Theorem 4.8. Suppose that H∗ holds, then a necessary and sufficient condition for problem 4.16
to have a pseudo-C3 0, 1 positive solution is that the following integral condition holds:
0<
1
fs, sds < ∞.
4.20
0
Theorem 4.9. Suppose that H∗ holds, then a necessary and sufficient condition for problem 4.16
to have a C2 0, 1 positive solution is that the following integral condition holds:
0<
1
s1 − sfs, sds < ∞.
4.21
0
Acknowledgments
The project is supported financially by a Project of Shandong Province Higher Educational
Science and Technology Program no. J10LA53 and the National Natural Science Foundation
of China no. 10971179.
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