Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 214042, 17 pages
doi:10.1155/2012/214042
Research Article
Positive Solutions for a Fractional Boundary Value
Problem with Changing Sign Nonlinearity
Yongqing Wang,1 Lishan Liu,1, 2 and Yonghong Wu2
1
2
School of Mathematical Sciences, Qufu Normal University, Shandong, Qufu 273165, China
Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia
Correspondence should be addressed to Yongqing Wang, wyqing9801@163.com
Received 14 April 2012; Accepted 25 June 2012
Academic Editor: Bashir Ahmad
Copyright q 2012 Yongqing Wang et al. 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.
We discuss the existence of positive solutions to the following fractional m-point boundary value
β
α
problem with changing sign nonlinearity D0
ut λft, ut 0, 0 < t < 1, u0 0, D0 u1 m−2
β
i1 ηi D0 uξi , where λ is a positive parameter, 1 < α ≤ 2, 0 < β < α − 1, 0 < ξ1 < · · · < ξm−2 < 1
m−2
α−β−1
α
< 1, D0
is the standard Riemann-Liouville derivative, f and may be singular
with i1 ηi ξi
at t 0 and/or t 1 and also may change sign. The work improves and generalizes some previous
results.
1. Introduction
In this paper, we consider the following fractional differential equation with m-point boundary conditions:
α
D0
ut λft, ut 0,
u0 0,
β
D0 u1 0 < t < 1,
m−2
β
ηi D0 uξi ,
1.1
i1
α−β−1
<
where 1 < α ≤ 2, λ > 0 is a parameter, 0 < β < α − 1, 0 < ξ1 < · · · < ξm−2 < 1 with m−2
i1 ηi ξi
α
is the standard Riemann-Liouville derivative, and f ∈ C0, 1 × 0, ∞ → −∞, ∞
1, D0
may be singular at t 0 and/or t 1 and also may change sign. In this paper, by a positive
solution to 1.1, we mean a function u ∈ C0, 1 which is positive on 0, 1 and satisfies 1.1.
2
Abstract and Applied Analysis
In recent years, great efforts have been made worldwide to study the existence of
solutions for nonlinear fractional differential equations by using nonlinear analysis methods
1–24. Fractional-order multipoint boundary value problems BVP have particularly
attracted a great deal of attention see, e.g., 13–19. In 10, the authors discussed some
properties of the Green function for the Direchlet-type BVP of nonlinear fractional differential
equations
α
ut ft, ut 0,
D0
u0 0,
0 < t < 1,
u1 0,
1.2
α
is the standard Riemann-Liouville derivative and f : C0, 1×0, ∞ →
where 1 < α < 2, D0
0, ∞ is continuous. By using the Krasnosel’skii fixed-point theorem, the existence of
positive solutions was obtained under some suitable conditions on f.
In 14, the authors investigated the existence and multiplicity of positive solutions by
using some fixed-point theorems for the fractional differential equation given by
α
ut ft, ut 0,
D0
u0 0,
0 < t < 1,
β
β
D0 u1 aD0 uξ,
1.3
where 1 < α ≤ 2, 0 ≤ β ≤ 1, 0 < ξ < 1, 0 ≤ a ≤ 1 with aξα−β−2 < 1 − β, 0 ≤ α − β − 1, f is
nonnegative.
It should be noted that in most of the works in literature the nonlinearity needs to
be nonnegative in order to establish positive solutions. As far as we know, semipositone
fractional nonlocal boundary value problems with 1 < α ≤ 2 have been seldom studied due
to the difficulties in finding and analyzing the corresponding Green function.
In 23, the authors investigated the following fractional differential equation with
three-point boundary conditions:
α
ut ft, ut et 0,
D0
u0 0,
β
0 < t < 1,
β
D0 u1 aD0 uξ,
1.4
where 1 < α ≤ 2, 0 < β ≤ 1, 0 < ξ < 1, 0 ≤ a ≤ 1,0 ≤ α − β − 1, et ∈ L0, 1, and f satisfies the
Caratheodory conditions. The authors obtained the properties of the Green function for 1.4
as follows:
βtα−1 s1 − sα−β−1
tα−1 1 − sα−β−1
≤ Gt, s ≤
.
Γα
Γα 1 − aξα−β−1
1.5
By using the Schauder fixed-point theorem, the authors obtained the existence of positive
solution of 1.4 with the following assumptions:
A1 for each L > 0, there exists a function φL 0 such that ft, tα−1 x ≥ φL t for a.e.
t ∈ 0, 1, for all x ∈ 0, L;
Abstract and Applied Analysis
3
A2 there exist gx, hx, and kt 0, such that
0 ≤ ft, x ≤ kt gx hx ,
∀x ∈ 0, ∞, a.e. t ∈ 0, 1,
1.6
here g : 0, ∞ → 0, ∞ is continuous and nonincreasing, h : 0, ∞ → 0, ∞ is continuous,
and h/g is nondecreasing;
A3 There exist two positive constants R > r > 0 such that
R > ΦR1 γ∗ ≥ r > 0,
1
ksg rsα−1 ds < ∞,
0
R≥
1
hR
gR
1
0
1.7
1 − sα−β−1
α−1
ds γ∗ .
ksg
rs
Γα 1 − aξα−β−1
Here
ΦR1 1
0
βs1 − sα−β−1
φR sds.
Γα
1.8
The assumptions on nonlinearity are not suitable for frequently used conditions, such as
superlinear or some sublinear. For instance, ft, x xμ , μ > 0, obviously, f does not satisfy
A1 .
Inspired by the previous work, the aim of this paper is to establish conditions for the
existence of positive solutions of the more general BVP 1.1. Our work presented in this
paper has the following new features. Firstly, we consider few cases of 1 < α ≤ 2 which
has been studied before, and in dealing with the difficulties related to the Green function
for this case, some new properties of the Green function have been discovered. Secondly, the
BVP 1.1 possesses singularity; that is, f may be singular at t 0 and/or t 1. Thirdly,
the nonlinearity f may change sign and may be unbounded from below. Finally, we impose
weaker positivity conditions on the nonlocal boundary term; that is, some of the coefficients
ηi may be negative.
The rest of the paper is organized as follows. In Section 2, we present some
preliminaries and lemmas that are to be used to prove our main results. We also discover
some new positive properties of the corresponding Green function. In Section 3, we discuss
the existence of positive solutions of the semipositone BVP 1.1. In Section 4, we give an
example to demonstrate the application of our theoretical results.
2. Basic Definitions and Preliminaries
For the convenience of the reader, we present here the necessary definitions from fractional
calculus theory. These definitions can also be found in the recent literature.
4
Abstract and Applied Analysis
Definition 2.1. The fractional integral of order α > 0 of a function u : 0, ∞ → R is given by
α
ut I0
1
Γα
t
2.1
t − sα−1 usds,
0
provided that the right-hand side is pointwisely defined on 0, ∞.
Definition 2.2. The Riemann-Liouville fractional derivative of order α > 0 of a function u :
0, ∞ → R is given by
α
D0
ut n t
d
1
t − sn−α−1 usds,
Γn − α dt
0
2.2
where n α 1 and α denotes the integer part of the number α, provided that the righthand side is pointwisely defined on 0, ∞.
α
Lemma 2.3 see 3. Let α > 0. Then the following equality holds for u ∈ L0, 1, D0
u ∈ L0, 1;
2.3
α
α
D0
ut ut c1 tα−1 c2 tα−2 · · · cn tα−n ,
I0
where ci ∈ R, i 1, 2, . . . , n, n − 1 < α ≤ n.
Set
⎧
α−β−1
,
1 ⎨tα−1 1 − s
G0 t, s α−β−1
Γα ⎩tα−1 1 − s
− t − sα−1 ,
ps 1 −
0 ≤ t ≤ s ≤ 1,
0 ≤ s ≤ t ≤ 1,
ξi − s α−β−1
ηi
,
1−s
s≤ξ
2.4
2.5
i
2.6
Gt, s G0 t, s qstα−1 ,
where
qs ps − p0
1 − sα−β−1 ,
Γαp0
p0 1 −
m−2
α−β−1
ηi ξi
.
2.7
i1
For convenience in presentation, we here list the assumption to be used throughout the paper.
H1 p0 > 0, qs ≥ 0 on 0, 1.
Remark 2.4. If ηi 0 i 1, . . . , m−2, we have p0 1 and qs ≡ 0. If ηi ≥ 0 i 1, . . . , m−2
α−β−1
< 1, we have qs ≥ 0 on 0, 1.
and m−2
i1 ηi ξi
Abstract and Applied Analysis
5
Lemma 2.5 see 14. Assume that gt ∈ L0, 1 and α > β > 0. Then
β
t
D0
Γα
t − sα−1 gsds Γ
α−β
0
t
t − sα−β−1 gsds.
2.8
0
Lemma 2.6. Assume H1 holds, and yt ∈ L0, 1. Then the unique solution of the problem
α
D0
ut yt 0,
u0 0,
β
D0 u1 0 < t < 1,
m−2
β
ηi D0 uξi 2.9
i1
is
ut 1
Gt, sysds,
2.10
0
where Gt, s is the Green function of the boundary value problem 2.9.
Proof. From Lemma 2.3, the solution of 2.9 is
α
yt c1 tα−1 c2 tα−2 .
ut −I0
2.11
Consequently,
1
ut −
Γα
t
t − sα−1 ysds c1 tα−1 c2 tα−2 .
2.12
0
From u0 0, we have c2 0.
By Lemma 2.5, we have
1
β
D0 ut − Γ α−β
t
c1 Γα α−β−1
.
t
t − sα−β−1 ysds Γ α−β
0
2.13
Therefore,
1
1
c1 Γα
β
D0 u1 − ,
1 − sα−β−1 ysds Γ α−β 0
Γ α−β
ξi
c1 Γα α−β−1
1
β
D0 uξi − .
ξ
ξi − sα−β−1 ysds Γ α−β 0
Γ α−β i
2.14
6
Abstract and Applied Analysis
β
By D0 u1 m−2
i1
β
ηi D0 uξi , we have
1
c1 0
1
0
1 − sα−1 ysds −
m−2
i1
ηi
ξi
Γαp0
0
ξi − sα−β−1 ysds
1 − sα−β−1 psysds
.
Γαp0
2.15
Therefore, the solution of 2.9 is
ut c1 t
α−1
1
−
Γα
1
t
t − sα−1 ysds
0
2.16
Gt, sysds.
0
Lemma 2.7. The function G0 t, s has the following properties:
1 G0 t, s > 0, for t, s ∈ 0, 1;
2 ΓαG0 t, s ≤ tα−1 , for t, s ∈ 0, 1;
3 βtα−1 hs ≤ ΓαG0 t, s ≤ hstα−2 , for t, s ∈ 0, 1,
where
hs s1 − sα−β−1 .
2.17
Proof. 1 When 0 < t ≤ s < 1, it is clear that
G0 t, s 1 α−1
t 1 − sα−β−1 > 0.
Γα
2.18
When 0 < s ≤ t < 1, we have
tα−1 1 − sα−β−1 − t − sα−1 ≥ tα−1 1 − sα−β−1 − tα−1 1 − sα−1
tα−1 1 − sα−β−1 1 − 1 − sβ > 0.
2.19
2 By 2.4, for any t, s ∈ 0, 1, we have
ΓαG0 t, s ≤ tα−1 1 − sα−β−1 ≤ tα−1 .
In the following, we will prove 3.
2.20
Abstract and Applied Analysis
7
i When 0 < s ≤ t < 1, noticing that 0 < β < α − 1 ≤ 1, we have
∂ α−2
t s1 − sα−β−1 − tα−1 1 − sα−β−1 tα−2 1 − sα−β−1 t − s ln1 − s ≤ 0.
∂β
2.21
Therefore,
tα−2 s1 − sα−β−1 − tα−1 1 − sα−β−1 − t − sα−1 ≥ tα−2 s − tα−1 t − sα−1
2.22
− tα−2 t − s t − sα−1 ≥ 0,
which implies
ΓαG0 t, s ≤ hstα−2 .
2.23
On the other hand, we have
d
βs 1 − sβ ≤ 0,
ds
s ∈ 0, 1.
2.24
Therefore, βs 1 − sβ ≤ 1, which implies
1 − 1 − sβ ≥ βs.
2.25
Then
ΓαG0 t, s tα−1 1 − sα−β−1 − t − sα−1
≥ tα−1 1 − sα−β−1 − t − sβ t − tsα−β−1
s β α−1
1− 1−
t 1 − sα−β−1
t
≥ 1 − 1 − sβ tα−1 1 − sα−β−1
2.26
≥ βtα−1 hs.
ii When 0 < t ≤ s < 1, we have
ΓαG0 t, s tα−1 1 − sα−β−1 tα−2 t1 − sα−β−1
≤ tα−2 s1 − sα−β−1 hstα−2 ,
2.27
8
Abstract and Applied Analysis
On the other hand, clearly we have
ΓαG0 t, s tα−1 1 − sα−β−1 ≥ βtα−1 hs.
2.28
The inequalities 2.23–2.28 imply that 3 holds.
By Lemma 2.7, we have the following results.
Lemma 2.8. Assume H1 holds, then the Green function defined by 2.6 satisfies
1 Gt, s > 0, for all t, s ∈ 0, 1;
2 Gt, s ≤ tα−1 1/Γα qs, for all t, s ∈ 0, 1;
3 βtα−1 Φs ≤ Gt, s ≤ tα−2 Φs, for all t, s ∈ 0, 1,
where
Φs hs
qs .
Γα
2.29
Lemma 2.9. Assume H1 holds, then the function G∗ t, s : t2−α Gt, s satisfies
1 G∗ t, s > 0, for all t, s ∈ 0, 1;
2 G∗ t, s ≤ t1/Γα qs, for all t, s ∈ 0, 1;
3 βtΦs ≤ G∗ t, s ≤ Φs, for all t, s ∈ 0, 1.
For convenience, we list here four more assumptions to be used later:
H2 f ∈ C0, 1 × 0, ∞, −∞, ∞ satisfies
ft, x ≥ −rt,
f t, tα−2 x ≤ ztgx,
t ∈ 0, 1, x ∈ 0, ∞,
2.30
where r, z ∈ C0, 1, 0, ∞, g ∈ C0, ∞, 0, ∞.
1
1
H3 0 rsds < ∞, 0 < 0 zsds < ∞.
H4 There exists a, b ⊂ 0, 1 such that
lim inf min
x → ∞ t∈a,b
ft, x
∞.
x
2.31
H5 There exists c, d ⊂ 0, 1 such that
lim inf min ft, x ∞,
x → ∞ t∈c,d
gx
0.
lim
x → ∞ x
2.32
Abstract and Applied Analysis
9
Remark 2.10. The second limit of H5 implies
g ∗ u
0,
u → ∞ u
2.33
g ∗ u max gx.
2.34
lim
where
x∈0,u
Proof. By limu → ∞ gu/u 0, for any > 0, there exists N1 > 0, such that for any u > N1
we have
2.35
0 ≤ gu < u.
Let N max{N1 , g ∗ N1 /
}, for any u > N we have
0 ≤ g ∗ u < u g ∗ N1 < 2
u.
2.36
Therefore, limu → ∞ g ∗ u/u 0.
Lemma 2.11. Assume H1 holds and rt ∈ C0, 1 ∩ L0, 1 is nonnegative, then the BVP
α
ut rt 0,
D0
u0 0,
β
D0 u1 0 < t < 1,
m−2
β
ηi D0 uξi 2.37
i1
has a unique solution ωt 1
0
Gt, srsds with ωt ≤ ktα−1 , where
k
1
0
Proof. By Lemma 2.6, ωt Lemma 2.8, we have
ωt 1
0
1
0
1
qs rsds,
Γα
t ∈ 0, 1.
2.38
Gt, srsds is the unique solution of 2.37. By 2 of
Gt, srsds ≤ tα−1
1
0
1
qs rsds.
Γα
2.39
Let E C0, 1 be endowed with the maximum norm u max0≤t≤1 |ut| and define
a cone P by
P ut ∈ E : there exists lu > 0 such that βtu ≤ ut ≤ lu t ,
and then set Br {ut ∈ E : u < r}, Pr P ∩ Br , ∂Pr P ∩ ∂Br .
2.40
10
Abstract and Applied Analysis
Next we consider the following singular nonlinear BVP:
α
ut λ f t, ut − λωt rt 0,
D0
u0 0,
β
D0 u1 m−2
0 < t < 1,
β
ηi D0 uξi ,
2.41
i1
where λ > 0, vt max{vt, 0}, ωt is defined in Lemma 2.11.
Let
T ut λ
1
rs ds.
G∗ t, s f s, sα−2 us − λωs
2.42
0
2.41.
Clearly, if ut ∈ P is a fixed point of T , then yt tα−2 ut is a positive solution of
Lemma 2.12. Suppose that H1 –H3 hold. Then T : P → P is a completely continuous operator.
Proof. It is clear that T is well defined on P . For any u ∈ P , Lemma 2.9 implies
T ut ≥ βtλ
1
rs ds.
Φs f s, sα−2 us − λωs
2.43
0
On the other hand,
T ut ≤ λ
1
rs ds.
Φs f s, sα−2 us − λωs
2.44
0
Therefore, T ut ≥ βt T u. Noticing that
T ut ≤ λt
1
0
1
rs ds,
qs f s, sα−2 us − λωs
Γα
2.45
we have T : P → P .
Using the Ascoli-Arzela theorem, we can then get that T : P → P is a completely
continuous operator.
Lemma 2.13 see 25. Let E be a real Banach space and let P ⊂ E be a cone. Assume that Ω1 and
Ω2 are two-bounded open subsets of E with θ ∈ Ω1 , Ω1 ⊂ Ω2 , T : P ∩ Ω2 \ Ω1 → P a completely
continuous operator such that either
1 T u ≤ u, u ∈ P ∩ ∂Ω1 and T u ≥ u, u ∈ P ∩ ∂Ω2 , or
2 T u ≥ u, u ∈ P ∩ ∂Ω1 and T u ≤ u, u ∈ P ∩ ∂Ω2 .
Then T has a fixed point in P ∩ Ω2 \ Ω1 .
Abstract and Applied Analysis
11
3. Existence of Positive Solutions
Theorem 3.1. Suppose that H1 –H4 hold. Then there exists λ∗ > 0 such that the BVP 1.1 has
at least one positive solution for any λ ∈ 0, λ∗ .
Proof . Choose r1 > kβ−1 . Let
⎫
⎬
r1
,
λ∗ min 1, 1
⎩
g ∗ r1 1 0 Φszs rsds ⎭
3.1
g ∗ r max gx.
3.2
⎧
⎨
where
x∈0,r
In the following of the proof, we suppose λ ∈ 0, λ∗ .
For any u ∈ ∂Pr1 , noticing ut ≥ βtr1 and Lemma 2.11, we have
tα−2 ut − λωt ≥ βr1 − λk tα−1 ≥ βr1 − k tα−1 ≥ 0,
r1 ≥ ut − λt2−α ωt ≥ βr1 − k t ≥ 0.
3.3
3.4
Therefore,
T ut λ
1
rs ds
G∗ t, s f s, sα−2 us − λωs
0
≤λ
1
rs ds
Φs zsg us − λs2−α ωs
0
≤ λ g ∗ r1 1
1
3.5
Φszs rsds
0
< λ∗ g ∗ r1 1
1
Φszs rsds ≤ r1 .
0
Thus,
T u ≤ u,
∀u ∈ ∂Pr1 .
3.6
Now choose a real number
L>
b
λβ2 a
2
Φssα−1 ds
.
3.7
12
Abstract and Applied Analysis
By H4 , there exists a constant N > 0 such that
ft, x > Lx,
for any t ∈ a, b, x ≥ N.
3.8
Let
r2 r1 2N
2k
α−1 .
β
βa
3.9
Then for any u ∈ ∂Pr2 , we have
βr2 α−1
tα−2 ut − λωt ≥ βr2 − k tα−1 ≥
t ,
2
∀t ∈ 0, 1.
3.10
Thus, for any t ∈ a, b, we have tα−2 ut − λωt > N. Hence, we get
T u max λ
1
t∈0,1
≥ max λ
0
b
t∈0,1
≥ max λL
t∈0,1
rs ds
G∗ t, s f s, sα−2 us − λωs
G∗ t, sf s, sα−2 us − λωs ds
a
b
G∗ t, s sα−2 us − λωs ds
a
3.11
b
βr2 α−1
≥ max λL G∗ t, s
s ds
t∈0,1
2
a
λLβ2 r2 b
t Φssα−1 ds
≥ max
t∈0,1
2
a
λLβ2 r2 b
Φssα−1 ds ≥ r2 .
2
a
Therefore,
T u ≥ u,
∀u ∈ ∂Pr2 .
3.12
By Lemma 2.13, T has a fixed point u ∈ P such that r1 ≤ u ≤ r2 . Let ut tα−2 ut − λωt.
Since u ≥ r1 , by 3.3 we have ut ≥ 0 on 0, 1 and limt → 0 tα−2 ut 0. Notice that ωt is
the solution of 2.37 and tα−2 ut is the solution of 2.41. Thus, ut is a positive solution of
the BVP 1.1.
Theorem 3.2. Suppose that H1 –H3 and H5 hold. Then there exists λ∗ > 0 such that the BVP
1.1 has at least one positive solution for any λ ∈ λ∗ , ∞.
Abstract and Applied Analysis
13
Proof. By the first limit of H5 , there exists N > 0 such that
ft, x ≥
d
β2 c
2k
Φsds
for any t ∈ c, d, x ≥ N.
3.13
N
.
kcα−1
3.14
,
Let
λ∗ In the following part of the proof, we suppose λ > λ∗ .
Let
R1 2λk
.
β
3.15
Then for any u ∈ ∂PR1 , we have
tα−2 ut − λωt ≥ βR1 − λk tα−1 λktα−1 ≥ λ∗ ktα−1 ,
∀t ∈ 0, 1.
3.16
Therefore, tα−2 ut − λωt ≥ N, for any t ∈ c, d and u ∈ ∂PR1 . Then
T ut λ
1
rs ds
G∗ t, s f s, sα−2 us − λωs
0
≥λ
d
G∗ t, sf s, sα−2 us − λωs ds
c
≥
≥
2λk
d
β2 c Φsds
2λkt
β
d
c
Φsds
d
G∗ t, sds
3.17
c
d
Φsds R1 t.
c
This implies
T u ≥ u,
∀u ∈ ∂PR1 .
3.18
On the other hand, gx is continuous on 0, ∞, and thus from the second limit of
H5 , we have
g ∗ x
0,
x → ∞ x
lim
3.19
14
Abstract and Applied Analysis
where g ∗ x is defined by 3.2. For
1
2λ
−1
Φszsds
,
3.20
0
there exists X0 > 0 such that gu ≤ εx for any x ≥ X0 and u ∈ 0, x.
Let
R2 X0 R1 2λ
1
Φsrsds.
3.21
0
For any u ∈ ∂PR2 , by 3.16 we can get R2 ≥ ut − λt2−α ωt ≥ 0, for all t ∈ 0, 1. Therefore,
T u ≤ λ
1
rs ds
Φs zsg us − λs2−α ωs
0
≤ λεR2
1
Φszsds λ
0
≤
1
Φsrsds
3.22
0
R2 R2
R2 .
2
2
Thus,
T u ≤ u,
∀u ∈ ∂PR2 .
3.23
By Lemma 2.13, T has a fixed point u ∈ P such that R1 ≤ u ≤ R2 . Let ut tα−2 ut − λωt.
Since u ≥ R1 , by 3.16 we have ut ≥ 0 on 0, 1 and limt → 0 tα−2 ut 0. Notice that ωt
is a solution of 2.37 and tα−2 ut is a solution of 2.41. Thus, ut is a positive solution of
the BVP 1.1.
By the proof of Theorem 3.2, we have the following corollary.
Corollary 3.3. The conclusion of Theorem 3.2 is valid if H5 is replaced by H5∗ . There exist c, d ⊂
0, 1 and N > 0 such that for any t ∈ c, d and x ≥ N,
ft, x ≥
β2
d
c
2k
Φsds
gx
0.
lim
x → ∞ x
,
3.24
Abstract and Applied Analysis
15
4. Example
Example 4.1 a 4-point BVP with coefficients of both signs. Consider the following problem:
7/4
D0
ut λft, ut 0, t ∈ 0, 1, u0 0,
1 1/4
1
4
1/4
1/4
− D0 u
,
D0 u1 D0 u
4
2
9
4.1
ft, x x2 ln t.
4.2
where
We have
⎧
1/2
3/4
⎪
⎪
⎨t 1 − s ,
1
G0 t, s Γ7/4 ⎪
⎪
⎩t3/4 1 − s1/2 − t − s3/4 ,
0 ≤ t ≤ s ≤ 1,
4.3
0 ≤ s ≤ t ≤ 1,
⎧
1/4 − s 1/2 1 4/9 − s 1/2
⎪
⎪
−
,
⎪
1
−
⎪
⎪
1−s
2
1−s
⎪
⎪
⎪
⎪
⎪
⎪
⎨
1/2
ps 1 − 1 4/9 − s
,
⎪
⎪
2
1−s
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩1,
0≤s≤
1
,
4
4
1
<s≤ ,
4
9
4.4
4
< s ≤ 1.
9
By direct calculations, we have p0 1/6 and qs ≥ 0, which implies that H1 holds.
Let rt − ln t, zt t−1/2 , gx x2 . It is easy to see that H2 and H3 hold.
Moreover,
ft, x
∞.
x → ∞ t∈1/4,3/4
x
lim inf
min
4.5
Therefore, the assumptions of Theorem 3.1 are satisfied. Thus, Theorem 3.1 ensures that there
exists λ∗ > 0 such that the BVP 4.1 has at least one positive solution for any λ ∈ 0, λ∗ .
Remark 4.2. Noticing that λx2 does not satisfy A1 , therefore, the work in the present paper
improves and generalizes the main results of 23.
Acknowledgments
The first and second authors were supported financially by the National Natural Science
Foundation of China 11071141, 11101237 and the Natural Science Foundation of Shandong
Province of China ZR2011AQ008, ZR2011AL018. The third author was supported
financially by the Australia Research Council through an ARC Discovery Project Grant.
16
Abstract and Applied Analysis
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