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Hindawi Publishing Corporation
Boundary Value Problems
Volume 2009, Article ID 937064, 19 pages
doi:10.1155/2009/937064
Research Article
Positive Solutions to Singular and
Delay Higher-Order Differential Equations on
Time Scales
Liang-Gen Hu,
1
Ti-Jun Xiao,
2
and Jin Liang
3
1
Department of Mathematics, University of Science and Technology of China, Hefei 230026, China
2
School of Mathematical Sciences, Fudan University, Shanghai 200433, China
3
Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China
Correspondence should be addressed to Jin Liang,
Received 21 March 2009; Accepted 1 July 2009
Recommended by Juan Jos
´
e Nieto
We are concerned with singular three-point boundary value problems for delay higher-order
dynamic equations on time scales. Theorems on the existence of positive solutions are obtained
by utilizing the fixed point theorem of cone expansion and compression type. An example is given
to illustrate our main result.
Copyright q 2009 Liang-Gen Hu 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.
1. Introduction
In this paper, we are concerned with the following singular three-point boundary value
problem BVP for short for delay higher-order dynamic equations on time scales:
−1
n
u
Δ
2n
t
w
t
f
t, u
t − c
,t∈
a, b
,
u
t
ψ
t
,t∈
a − c, a
,
u
Δ
2i
a
− β
i1
u
Δ
2i1
a
α
i1
u
Δ
2i
,
γ
i1
u
Δ
2i
u
Δ
2i
b
, 0 ≤ i ≤ n − 1,
1.1
where c ∈ 0, b − a/2, ∈ a, b, β
i
≥ 0, 1 <γ
i
< b − a β
i
/ − a β
i
,0≤ α
i
<
b − γ
i
γ
i
− 1a − β
i
/b − , i 1, 2, ,n and ψ ∈ Ca − c, a. The functional
w : a, b → 0, ∞ is continuous and f : a, b × 0, ∞ → 0, ∞ is continuous. Our
2 Boundary Value Problems
nonlinearity w may have singularity at t a and/or t b, and f may have singularity at
u 0.
To understand the notations used in 1.1, we recall the following definitions which
can be found in 1, 2.
a A time scale T is a nonempty closed subset of the real numbers R. T has the topology
that it inherits from the real numbers with the standard topology. It follows that the
jump operators σ, ρ : T → T,
σ
t
inf
{
τ ∈ T : τ>t
}
,ρ
t
sup
{
τ ∈ T : τ<t
}
1.2
supplemented by inf ∅ : sup T and sup ∅ : inf T are well defined. The point
t ∈ T is left-dense, left-scattered, right-dense, right-scattered if ρtt, ρ
t <t,
σtt, σt <t, respectively. If T has a left-scattered maximum t
1
right-scattered
minimum t
2
, define T
k
T −{t
1
} T
k
T −{t
2
}; otherwise, set T
k
T T
k
T.
By an interval a, b we always mean the intersection of the real interval a, b with
the given time scale, that is, a, b ∩T. Other types of intervals are defined similarly.
b For a function f : T → R and t ∈ T
k
,theΔ-derivative of f at t, denoted by f
Δ
t,
is the number provided it exists with the property that, given any ε>0, there is a
neighborhood U ⊂ T of t such that
f
σ
t
− f
s
− f
Δ
t
σ
t
− s
≤ ε
|
σ
t
− s
|
, ∀s ∈ U. 1.3
c For a function f : T → R and t ∈ T
k
,the∇-derivative of f at t, denoted by f
∇
t,
is the number provided it exists with the property that, given any ε>0, there is a
neighborhood U ⊂ T of t such that
f
ρ
t
− f
s
− f
∇
t
ρ
t
− s
≤ ε
ρ
t
− s
, ∀s ∈ U. 1.4
d If F
Δ
tftΦ
∇
tgt, then we define the integral
t
a
f
Δ F
t
− F
a
t
a
g
∇ Φ
t
− Φ
a
. 1.5
Theoretically, dynamic equations on time scales can build bridges between continuous
and discrete mathematics. Practically, dynamic equations have been proposed as models in
the study of insect population models, neural networks, and many physical phenomena
which include gas diffusion through porous media, nonlinear diffusion generated by
nonlinear sources, chemically reacting systems as well as concentration in chemical of
biological problems 2. Hence, two-point and multipoint boundary value problems for
dynamic equations on time scales have attracted many researchers’ attention see, e.g., 1–19
and references therein. Moreover, singular boundary value problems have also been treated
in many papers see, e.g., 4, 5, 12–14, 18 and references therein.
Boundary Value Problems 3
In 2004, J. J. DaCunha et al. 13 considered singular second-order three-point
boundary value problems on time scales
u
ΔΔ
t
f
t, u
t
0,
0, 1
∩ T,
u
0
0,u
p
u
σ
2
1
1.6
and obtained the existence of positive solutions by using a fixed point theorem due to Gatica
et al. 14, where f : 0, 1 × 0, ∞ → 0, ∞ is decreasing in u for every t ∈ 0, 1 and may
have singularity at u 0.
In 2006, Boey and Wong 11 were concerned with higher-order differential equation
on time scales of the form
−1
n−1
y
Δ
n
t
−1
p1
F
t, y
σ
n−1
t
,t∈
a, b
,
y
Δ
i
a
0, 0 ≤ i ≤ p − 1,
y
Δ
i
σ
b
0,p≤ i ≤ n − 1,
1.7
where p, n are fixed integers satisfying n ≥ 2, 1 ≤ p ≤ n − 1. They obtained some existence
theorems of positive solutions by using Krasnosel’skii fixed point theorem.
Recently, Anderson and Karaca 8 studied higher-order three-point boundary value
problems on time scales and obtained criteria for the existence of positive solutions.
The purpose of this paper is to investigate further the singular BVP for delay higher-
order dynamic equation 1.1. By the use of the fixed point theorem of cone expansion
and compression type, results on the existence of positive solutions to the BVP 1.1 are
established.
The paper is organized as follows. In Section 2, we give some lemmas, which will be
required in the proof of our main theorem. In Section 3, we prove some theorems on the
existence of positive solutions for BVP 1.1. Moreover, we give an example to illustrate our
main result.
2. Lemmas
For 1 ≤ i ≤ n,letG
i
t, s be Green’s function of the following three-point boundary value
problem:
−u
ΔΔ
t
0,t∈
a, b
,
u
a
− β
i
u
Δ
a
α
i
u
,γ
i
u
u
b
,
2.1
where ∈ a, b and α
i
,β
i
,γ
i
satisfy the following condition:
C
β
i
≥ 0, 1 <γ
i
<
b − a β
i
− a β
i
, 0 ≤ α
i
<
b − γ
i
γ
i
− 1
a − β
i
b −
. 2.2
4 Boundary Value Problems
Throughout the paper, we assume that σbb.
From 8, we know that for any t, s ∈ a, b × a, b and 1 ≤ i ≤ n,
G
i
t, s
⎧
⎨
⎩
G
i
1
t, s
,s∈
a,
,
G
i
2
t, s
,s∈
, b
,
2.3
where
G
i
1
t, s
1
d
i
⎧
⎨
⎩
γ
i
t −
b − t
σ
s
β
i
− a
,
σ
s
≤ t,
γ
i
σ
s
−
ω
b − σ
s
t β
i
− a
α
i
− b
t − σ
s
,t≤ s,
G
i
2
t, s
1
d
i
⎧
⎨
⎩
σ
s
1 − α
i
α
i
β
i
− a
b − t
γ
i
− a β
i
t − σ
s
,σ
s
≤ t,
t
1 − α
i
α
i
β
i
− a
b − σ
s
,t≤ s,
d
i
γ
i
− 1
a − β
i
1 − α
i
b
α
i
− γ
i
.
2.4
The following four lemmas can be found in 8.
Lemma 2.1. Suppose that the condition (C) holds. Then the Green function of G
i
t, s in 2.3 satisfies
G
i
t, s
> 0,
t, s
∈
a, b
×
a, b
. 2.5
Lemma 2.2. Assume that the condition (C) holds. Then Green’s function G
i
t, s in 2.3 satisfies
G
i
t, s
≤ max
{
G
i
b, s
,G
i
σ
s
,s
}
,
t, s
∈
a, b
×
a, b
. 2.6
Remark 2.3. 1 If s ∈ γ
i
− a β
i
− α
i
− β
i
a/1 − α
i
,b,s≤ t, we know that G
i
t, s is
nonincreasing in t and
G
i
b, s
G
i
σ
s
,s
γ
i
− a β
i
b − σ
s
σ
s
1 − α
i
α
i
β
i
− a
b − σ
s
≥
γ
i
− a β
i
b
1 − α
i
α
i
β
i
− a
> 0.
2.7
Therefore, we have
G
i
b, s
≤ G
i
t, s
≤ G
i
σ
s
,s
≤ δ
i
G
i
b, s
, 2.8
where
δ
i
b
1 − α
i
α
i
β
i
− a
γ
i
− a β
i
> 1.
2.9
Boundary Value Problems 5
2 If t and s satisfy the other cases, then we get that G
i
t, s is nondecreasing in t and
G
i
t, s
≤ G
i
b, s
. 2.10
Lemma 2.4. Assume that (C) holds. Then Green’s function G
i
t, s in 2.3 verifies the following
inequality:
G
i
t, s
≥ min
t − a
b − a
,
b − t
γ
i
b − a
G
i
b, s
≥ min
t − a
δ
i
b − a
,
b − t
γ
i
b − a
max
{
G
i
b, s
,G
i
σ
s
,s
}
.
2.11
Remark 2.5. If s ∈ , γ
i
− a β
i
− α
i
− β
i
a/1 − α
i
,s≤ t, then we find
γ
i
− 1
a − β
i
1 − α
i
σ
s
α
i
− γ
i
< 0. 2.12
So there exists a misprint on 8, Page 2431, line 23.From2.3, it follows that
G
i
t, s
G
i
b, s
σ
s
1 − α
i
α
i
β
i
− a
b − t
γ
i
− a β
i
t − σ
s
γ
i
− a β
i
b − σ
s
≥
β
i
− a
b − t
γ
i
− a β
i
t − σ
s
γ
i
− a β
i
b − a
≥
b − t
γ
i
b − a
.
2.13
Consequently, we get
G
i
t, s
≥
b − t
γ
i
b − a
G
i
b, s
.
2.14
If s ∈ γ
i
− a β
i
− α
i
− β
i
a/1 − α
i
,b, s ≤ t, then, from 2.8,weobtain
G
i
t, s
≥
t − a
b − a
G
i
b, s
≥
t − a
δ
i
b − a
G
i
σ
s
,s
.
2.15
6 Boundary Value Problems
Remark 2.6. If we set h
i
t : min{t − a/δ
i
b − a, b − t/γ
i
b − a}, then we have
G
i
t, s
≥ h
i
t
max
{
G
i
b, s
,G
i
σ
s
,s
}
,
t, s
∈
a, b
×
a, b
. 2.16
Denote
G
i
·,s
max
t∈a,b
|
G
i
t, s
|
,s∈
a, b
.
2.17
Thus we have
G
i
t, s
≥ h
i
t
G
i
·,s
,
t, s
∈
a, b
×
a, b
. 2.18
Lemma 2.7. Assume that condition (C) is satisfied. For G
i
t, s as in 2.3, put H
1
t, s : G
1
t, s
and recursively define
H
j
t, s
b
a
H
j−1
t, r
G
j
r, s
Δr
2.19
for 2 ≤ j ≤ n.ThenH
n
t, s is Green’s function for the homogeneous problem
−1
n
u
Δ
2n
t
0,t∈
a, b
,
u
Δ
2i
a
− β
i1
u
Δ
2i1
a
α
i1
u
Δ
2i
,
γ
i1
u
Δ
2i
u
Δ
2i
b
, 0 ≤ i ≤ n − 1.
2.20
Lemma 2.8. Assume that (C) holds. Denote
K :
n−1
j1
k
j
,L:
n−1
j1
l
j
,
2.21
then Green’s function H
n
t, s in Lemma 2.7 satisfies
h
1
t
L
G
n
·,s
≤ H
n
t, s
≤ K
G
n
·,s
,
t, s
∈
a, b
×
a, b
, 2.22
where
k
j
b
a
G
j
·,s
Δs>0,l
j
b
a
G
j
·,s
h
j1
s
Δs, 1 ≤ j ≤ n − 1.
2.23
Boundary Value Problems 7
Proof. We proceed by induction on n ≥ 2. We denote the statement by P n.FromLemma 2.7,
it follows that
H
2
t, s
b
a
H
1
t, r
G
2
r, s
Δr
≤
b
a
G
1
·,r
G
2
·,s
Δr k
1
G
2
·,s
,
2.24
and from 2.18 , we have
H
2
t, s
b
a
H
1
t, r
G
2
r, s
Δr
≥
b
a
h
1
t
G
1
·,r
× h
2
r
G
2
·,s
Δr
h
1
t
l
1
G
2
·,s
.
2.25
So P2 is true.
We now assume that P m is true for some positive integer m ≥ 2. From Lemma 2.7,it
follows that
H
m1
t, s
b
a
H
m
t, r
G
m1
r, s
Δr
≤
b
a
H
m
t, r
G
m1
r, s
Δr
≤
⎛
⎝
b
a
m−1
j1
k
j
×
G
m
·,r
Δr
⎞
⎠
G
m1
·,s
m
j1
k
j
G
m1
·,s
,
H
m1
t, s
b
a
H
m
t, r
G
m1
r, s
Δr
≥
b
a
h
1
t
×
m−1
j1
l
j
G
m
·,r
h
m1
r
G
m1
·,s
Δr
h
1
t
m
j1
l
j
G
m1
·,s
.
2.26
So Pm 1 holds. Thus Pn is true by induction.
8 Boundary Value Problems
Lemma 2.9 see 20. Let E, · be a real Banach space and P ⊂ E a cone. Assume that T :
P
ζ,η
→ P is completely continuous operator such that
i Tu u for u ∈ ∂P
ζ
and Tu u for u ∈ ∂P
η
,
ii Tu u for u ∈ ∂P
ζ
and Tu u for u ∈ ∂P
η
.
Then T has a fixed point u
∗
∈ P with ζ ≤u
∗
≤η.
3. Main Results
We assume that {a
m
}
m≥1
and {b
m
}
m≥1
are strictly decreasing and strictly increasing sequences,
respectively, with lim
m →∞
a
m
a, lim
m →∞
b
m
b and a
1
<b
1
. A Banach space E Ca, b is
the set of real-valued continuous in the topology of T functions ut defined on a, b with
the norm
u
max
t∈a,b
|
u
t
|
.
3.1
Define a cone by
P
u ∈ E : u
t
≥
h
1
t
L
K
u
,t∈
a, b
.
3.2
Set
P
ξ
{
u ∈ P :
u
<ξ
}
,∂P
ξ
{
u ∈ P :
u
ξ
}
,ξ>0,
P
ζ,η
u ∈ P : ζ<u <η
, 0 <ζ<η,
Y
1
{
t ∈
a, b
: t − c<a
}
,Y
2
{
t ∈
a, b
: t − c ≥ a
}
,
Y
m
{
t ∈ Y
2
: t − c ∈
a, a
m
∪
b
m
,b
}
.
3.3
Assume that
C1 ψ : a − c, a → 0, ∞ is continuous;
C2 we have
0 <K
q
p
G
n
·,s
w
s
Δs, K
b
a
G
n
·,s
w
s
Δs<∞,
3.4
for constants p and q with a c<p<q<b;
C3 the function f : a, b × 0, ∞ → R
is continuous and w : a, b → R
is
continuous satisfying
lim
m →∞
sup
u∈P
ζ,η
K
Y
m
G
n
·,s
w
s
f
s, u
s − c
Δs 0, ∀0 <ζ<η.
3.5
Boundary Value Problems 9
We seek positive solutions u : a, b → R
, satisfying 1.1. For this end, we transform
1.1 into an integral equation involving the appropriate Green function and seek fixed points
of the following integral operator.
Define an operator T : C
a, b → Ca, b by
Tu
t
b
a
H
n
t, s
w
s
f
s, u
s − c
Δs, ∀u ∈ C
a, b
,
3.6
where C
a, b{u ∈ Ca, b | ut ≥ 0,t∈ a, b}.
Proposition 3.1. Let (C1), (C2), and (C3) hold, and let ζ, η be fixed constants with 0 <ζ<η.Then
T : P
ζ,η
→ P is completely continuous.
Proof. We separate the proof into four steps.
Step 1. For each u ∈ P
ζ,η
, Tu is bounded.
By condition C3, there exists some positive integer m
0
satisfying
sup
u∈P
ζ,η
K
Y
m
0
G
n
·,s
w
s
f
s, u
s − c
Δs ≤ 1,
3.7
where
Y
m
0
{
t ∈ Y
2
: t − c ∈
a, a
m
0
∪
b
m
0
,b
}
; 3.8
here, we used the fact that for each u ∈ P
ζ,η
and t ∈ a
m
0
c, b
m
0
c ∩ a, b,
η ≥ u
t − c
≥
h
1
t − c
L
K
u
≥ ζ min
h
1
a
m
0
L
K
,
h
1
b
m
0
L
K
,
h
1
b − c
L
K
ζh > 0,
3.9
where
h min
h
1
a
m
0
L
K
,
h
1
b
m
0
L
K
,
h
1
b − c
L
K
.
3.10
Set
D : max
f
t, ψ
t − c
: t ∈ Y
1
,
Q : max
f
t, u
t − c
: t ∈ Y
2
,ζh≤ u
t − c
≤ η
.
3.11
10 Boundary Value Problems
Then we obtain
Tu
t
≤ sup
t∈a,b
sup
u∈P
ζ,η
b
a
H
n
t, s
w
s
f
s, u
s − c
Δs
≤ K sup
u∈P
ζ,η
Y
1
G
n
·,s
w
s
f
s, u
s − c
Δs
sup
u∈P
ζ,η
K
Y
m
0
G
n
·,s
w
s
f
s, u
s − c
Δs
sup
u∈P
ζ,η
K
Y
2
\Y
m
0
G
n
·,s
w
s
f
s, u
s − c
Δs
≤ 1 max
{
D, Q
}
K
b
a
G
n
·,s
w
s
Δs<∞.
3.12
Consequently, Tuis bounded and well defined.
Step 2. T : P
ζ,η
→ P. For every u ∈ P
ζ,η
,wegetfrom2.22
Tu
sup
t∈a,b
b
a
H
n
t, s
w
s
f
s, u
s − c
Δs
≤ K
b
a
G
n
·,s
w
s
f
s, u
s − c
Δs.
3.13
Then by the above inequality
Tu
t
b
a
H
n
t, s
w
s
f
s, u
s − c
Δs
≥
b
a
h
1
t
L
G
n
·,s
w
s
f
s, u
s − c
Δs
≥
h
1
t
L
K
Tu
.
3.14
This leads to Tu ∈ P.
Boundary Value Problems 11
Step 3. We will show that T : P
ζ,η
→ P is continuous. Let {u
m
}
m≥1
be any sequence in P
ζ,η
such that lim
m →∞
u
m
u ∈ P
ζ,η
. Notice also t hat as m →∞,
φ
m
s
f
s, u
m
s − c
− f
s, u
s − c
w
s
−→ 0, for s ∈
a c, b
,
f
s, u
m
s − c
− f
s, u
s − c
w
s
f
s, ψ
s − c
− f
s, ψ
s − c
w
s
0, for s ∈
a, a c
,
Y
2
H
n
t, s
φ
m
s
Δs ≤ sup
x∈P
ζ,η
2K
Y
2
G
n
·,s
w
s
f
s, x
s
Δs<∞.
3.15
Now these together with C2 and the Lebesgue dominated convergence theorem 10 yield
that as m →∞,
Tu
m
− Tu
sup
t∈
a,b
b
a
H
n
t, s
w
s
f
s, u
m
s − c
− f
s, u
s − c
Δs −→ 0.
3.16
Step 4. T : P
ζ,η
→ P is compact.
Define
w
m
t
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
min
{
w
t
,w
a
m
}
,a≤ t ≤ a
m
,
w
t
,a
m
≤ t ≤ b
m
,
min
{
w
t
,w
b
m
}
,b
m
≤ t ≤ b,
f
m
t, u
t − c
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
f
t, ψ
t − c
,a≤ t<a c,
min
f
t, u
t − c
,f
t, u
a
m
,a c ≤ t ≤ a
m
c,
f
t, u
t − c
,t∈
a
m
c, b
m
c
∩
a, b
,
min
f
t, u
t − c
,f
t, u
b
m
,t∈
b
m
c, b
∩
a, b
,
3.17
and an operator sequence {T
m
} for a fixed m by
T
m
ut
b
a
H
n
t, sw
m
sf
m
s, us − cΔs, ∀t ∈ a, b. 3.18
Clearly, the operator sequence {T
m
} is compact by using the Arzela-Ascoli theorem
3, for each m ∈ N. We will prove that T
m
converges uniformly to T on P
ζ,η
. For any u ∈ P
ζ,η
,
12 Boundary Value Problems
we obtain
T
m
u − Tu
sup
t∈a,b
b
a
H
n
t, s
w
m
s
f
m
s, u
s − c
− w
s
f
s, u
s − c
Δs
≤ K
b
a
G
n
·,s
w
m
s
f
m
s, u
s − c
− w
s
f
s, u
s − c
Δs
≤ K
Y
1
G
n
·,s
|
w
m
s
− w
s
|
f
s, ψ
s − c
Δs
K
Y
2
G
n
·,s
w
m
s
f
m
s, u
s − c
− w
s
f
s, u
s − c
Δs.
3.19
From C1, C2, and the Lebesgue dominated convergence theorem 10, we see that the
right-hand side 3.19 can be sufficiently small for mbeing big enough. Hence the sequence
{T
m
} of compact operators converges uniformly to T on P
ζ,η
so that operator T is compact.
Consequently, T : P
ζ,η
→ P is completely continuous by using the Arzela-Ascoli theorem
3.
Proposition 3.2. It holds that v ∈ P
ζ,η
is a solution of 1.1 if and only if Tv v.
Proof. If v ∈ P
ζ,η
and Tv v, then we have
−1
n
v
Δ
2n
t
−1
n
Tv
Δ
2n
t
w
t
f
t, v
t − c
,
3.20
and for any 0 ≤ i ≤ n − 1,
v
Δ
2i
a
− β
i1
v
Δ
2i1
a
α
i1
v
Δ
2i
,γ
i1
v
Δ
2i
v
Δ
2i
b
.
3.21
From 8, Lemma 3.1, we know that vt ≥ 0ona, b. So we conclude that v is the solution
of BVP 1.1.
For convenience, we list the following notations and assumptions:
R
μK
q
p
G
n
·,s
wsΔs
−1
,μ min
h
1
p
L
K
,
h
1
q
L
K
;
κ
K
b
a
G
n
·,s
wsΔs
−1
;
3.22
Boundary Value Problems 13
f
ξ
μξ
:
f
t, u
t − c
u
t
,t∈
p, q
,u∈
μξ, ξ
; 3.23
f
ζ
ρ
:
f
t, u
t − c
u
t
,t∈ Y
2
,u∈
ρ, ζ
; 3.24
S
ρ
sup
u∈∂P
ρ
K
Y
2
G
n
·,s
w
s
f
s, u
s − c
Δs, ρ > 0. 3.25
From condition C2 and 3.12, we have Sρ < ∞.
Theorem 3.3. Assume that there exist positive constants ρ,ζ, ξ, r with ζ<μξ, r<κand ζ ≥
κSρ/κ − r such that
i f
ξ
μξ
>Rand f
ζ
ρ
<r;
ii ft, ψt − c/ut <r, for all t ∈ Y
1
and u ∈ ρ, ζ.
If (C1), (C2), and (C3) hold, then the boundary value problem 1.1 has at least one positive solution
u such that
u
t
⎧
⎨
⎩
ψ
t
, if t ∈
a − c, a
,
u
∗
t
, if t ∈
a, b
,
ζ ≤
u
∗
≤ ξ.
3.26
Proof. Define the operator T : P
ζ,ξ
→ P by 3.6.Fromi and 3.23, it follows that there
exists ε
1
> 0 such that
f
t, u
t − c
≥
R ε
1
u
t
, for t ∈
p, q
,u∈
μξ, ξ
. 3.27
We claim that
Tu u, ∀u ∈ ∂P
ξ
. 3.28
If it is false, then there exists some u
1
∈ ∂P
ξ
with Tu
1
≤ u
1
,thatis,u
1
− Tu
1
∈ P which implies
that u
1
t ≥ Tu
1
t for t ∈ a, b.
Set
λ min
u
1
t
: t ∈
p, q
≥ min
h
1
p
L
K
,
h
1
q
L
K
u
1
μξ. 3.29
14 Boundary Value Problems
We know from 2.22 and 3.27 that for t ∈ p, q,
u
1
t
≥ Tu
1
t
b
a
H
n
t, s
w
s
f
s, u
1
s − c
Δs
Y
1
H
n
t, s
w
s
f
s, u
1
s − c
Δs
Y
2
H
n
t, s
w
s
f
s, u
1
s − c
Δs
≥
q
p
H
n
t, s
w
s
f
s, u
1
s − c
Δs
≥ min
h
1
p
,h
1
q
L
q
p
G
n
·,s
w
s
f
s, u
1
s − c
Δs
≥
R ε
1
min
t∈p,q
u
1
t
μK
q
p
G
n
·,s
w
s
Δs
≥ λR
μK
q
p
G
n
·,s
w
s
Δs
λε
1
μK
q
p
G
n
·,s
w
s
Δs
λ λε
1
μK
q
p
G
n
·,s
w
s
Δs,
3.30
the first inequality of C2 implies that
u
1
t
>λ, ∀t ∈
p, q
. 3.31
Clearly, 3.31 contradicts 3.29. This means that 3.28 holds.
Next we will show that
Tu u, ∀u ∈ P
ζ
. 3.32
Suppose on the contrary that there exists some u
2
∈ ∂P
ζ
with u
2
≤ Tu
2
for all t ∈ a, b.
For t, u ∈ Y
2
× ρ, ζ,fromi and 3.24, there exists ε
2
> 0 such that
f
t, u
t − c
≤
r − ε
2
u
t
. 3.33
and for t, u ∈ Y
1
× ρ, ζ, there exists ε
2
> 0, from ii, such that
f
t, ψ
t − c
≤
r − ε
2
u
t
. 3.34
Put
Y
3
:
t ∈ Y
2
: u
2
t
>ρ
, u
2
t
⎧
⎨
⎩
min
u
2
t
,ρ
,t∈ Y
2
,
ρ, t ∈ Y
1
.
3.35
Boundary Value Problems 15
If Y
3
∅, then we take u
2
tρ.Itiseasytoseethath
1
t − cLζ/K ≤ u
2
t − c ≤u
2
ζ
for t ∈ Y
2
and u
2
t ∈ C
a, b, u
2
ρ,thatis,u
2
∈ ∂P
ρ
.From3.33 and 3.34,wefindthat
Tu
2
sup
t∈a,b
b
a
H
n
t, s
w
s
f
s, u
2
s − c
Δs
≤ K
b
a
G
n
·,s
w
s
f
s, u
2
s − c
Δs
K
Y
1
G
n
·,s
w
s
f
s, ψ
s − c
Δs K
Y
3
G
n
·,s
w
s
f
s, u
2
s − c
Δs
K
Y
2
\Y
3
G
n
·,s
w
s
f
s, u
2
s − c
Δs
≤
r − ε
2
max
t∈Y
1
u
2
t
Y
1
G
n
·,s
w
s
Δs
sup
t,u
2
∈Y
3
×ρ,ζ
f
t, u
2
t − c
K
Y
3
G
n
·,s
w
s
Δs
sup
u
2
∈∂P
ρ
K
Y
2
G
n
·,s
w
s
f
s, u
2
s − c
Δs
≤ ζrK
b
a
G
n
·,s
w
s
Δs S
ρ
− ζε
2
K
b
a
G
n
·,s
w
s
Δs
ζrκ
−1
− ζε
2
κ
−1
S
ρ
<ζ
u
2
3.36
yielding a contradiction with u
2
≤ Tu
2
for all t ∈ a, b. This means that 3.32 holds.
Therefore, from 3.28, 3.32 and Lemma 2.9, we conclude that the operator T has at least
one fixed point u
∗
∈ P
ζ,ξ
. From the definition of the cone P and 2.18,weseethatu
∗
t > 0
for all t ∈ a, b.Thus,Proposition 3.2 implies that u
∗
is a solution of BVP 1.1.Soweobtain
the desired result.
Adopting the same argument as in Theorem 3.3 , we obtain the following results.
Corollary 3.4. Let ρ, ζ, r, f
ζ
ρ
be as in Theorem 3.3. Suppose that (ii) of Theorem 3.3 holds and
lim
ξ →∞
f
ξ
μξ
∞. If (C1), (C2), and (C3) holds , then boundary value problem 1.1 has at least
one positive solution u ∈ P
ζ,η
such that
u
t
⎧
⎨
⎩
ψ
t
, if t ∈
a − c, a
,
u
∗∗
t
, if t ∈
a, b
,
ζ ≤
u
∗∗
≤ η, ζ < μη.
3.37
16 Boundary Value Problems
Theorem 3.5. Assume that there exist positive constants ρ
i
,ζ
i
,ξ
i
,r with ζ
i
<μξ
i
, r<κand ζ
i
≥
κSρ
i
/κ − r, i 1, 2, ,msuch that
iii f
ξ
i
μξ
i
>Rand f
ζ
i
ρ
i
<r;
iv ft, ψt − c/ut <r, for all t ∈ Y
1
and u ∈ ρ
i
,ζ
i
.
If (C1), (C2), and (C3) hold, then boundary value problem 1.1 has at least m positive solutions
u
i
∈ P
ζ
i
,ξ
i
such that for i 1, 2, ,m
u
i
t
⎧
⎨
⎩
ψ
t
, if t ∈
a − c, a
,
u
∗
i
t
, if t ∈
a, b
ζ ≤
u
∗
i
≤ ξ.
3.38
Example 3.6. Let T R. Consider the following singular three-point boundary value problems
for delay four-order dynamic equations:
u
4
t
f
t, u
t − 1
0,t∈
0, 4
,
u
0
1
2
u
1
, 2u
1
u
4
,
u
0
1
2
u
1
, 2u
1
u
4
,
u
t
e
t
,t∈
−1, 0
,
3.39
where, for any t ∈ 0, 4, ρ 1, ζ 1480, μ 0.112, ξ 13500, M
1
1andM
2
1/50
2
,
f
t, u
t − 1
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
2M
1
u
t
,
t, u
∈
1, 4
×
ξ, ∞
,
M
1
u
t
1sin
π
u
t
− ϑφ
2
ϑ − ϑφ
cos
π
u
t
− ϑφ
2
ϑ − ϑφ
,
t, u
∈
1, 4
×
μξ, ξ
,
1
2
M
2
u
t
cos
π
u
t
− μ
2
ϑφ − μ
2M
1
ϑφ sin
π
u
t
− μ
2
ϑφ − μ
,
t, u
∈
1, 4
×
ζ, μξ
,
1
2
M
2
u
t
2 − sin
π
u
t
−
2
μ −
− cos
π
u
t
−
2
μ −
,
t, u
∈
1, 4
×
ρ, ζ
,
ρu
t
−1/2
− u
t
1/2
1
2
M
2
ρ,
t, u
∈
1, 4
×
0,ρ
,
1
2
M
2
u
t
,
t, u
∈
0, 1
× R.
3.40
Boundary Value Problems 17
Clearly, we know that
α
1
2
,β 0,γ 2,η 1,δ
5
4
,d
1
2
,
p
3
2
,q
7
2
,h
i
t
min
t
5
,
4 − t
8
,i 1, 2,
G
4,s
12s
s ∈
0, 1
,G
4,s
4
4 − s
s ∈
1, 3
,
G
s, s
4 − s
1 s
s ∈
3, 4
.
3.41
Simple computations yield
K
4
0
G
1
·,s
ds
1
0
12sds
3
1
4
4 − s
ds
4
3
1 s
4 − s
ds 24.17,
L
4
0
G
1
·,s
h
2
s
ds
1
0
12s
s
5
ds
20/13
1
4
4 − s
s
5
ds
3
20/13
4
4 − s
4 − s
8
ds
4
3
4 − s
2
1 s
8
ds
4.695,
μ min
h
1
p
L
K
,
h
1
q
L
K
0.112,
R
μK
7/2
3/2
G
2
·,s
ds
−1
0.282,
κ
K
4
0
G
2
·,s
ds
−1
1
24.17
2
.
3.42
Obviously,
lim
m →∞
sup
u∈P
ζ,η
K
Y
m
G
2
·,s
f
s, u
s − c
ds 0, ∀ 0 <ζ<η.
3.43
If t, u ∈ 1, 4 × 0, 1, then we have
h
t
h
t
ρ ≤ u
t
≤ ρ 1. 3.44
Therefore, we get
f
t, u
t − 1
≤ h
t
−1/2
− h
t
1/2
1
2
M
2
, for
t, u
∈
1, 4
×
0, 1
.
3.45
18 Boundary Value Problems
From 3.25, it follows that
S
1
sup
u∈∂P
1
K
4
1
G
2
·,s
f
s, u
s − 1
ds
≤ K
20/13
1
12s
5
s
1/2
−
s
5
1/2
1
2
M
2
ds
K
3
20/13
4
4 − s
8
4 − s
1/2
−
4 − s
8
1/2
1
2
M
2
ds
K
4
3
1 s
4 − s
8
4 − s
1/2
−
4 − s
8
1/2
1
2
M
2
ds
≤ 1120.
3.46
Thus,
ζ 1480 ≥
κS
1
κ − r
≈ 1461.37,ξ 13500,ζ<μξ.
3.47
Therefore, by Theorem 3.3 ,theBVP3.39 has at least one positive solution u such that
u
t
⎧
⎨
⎩
e
t
, if t ∈
−1, 0
,
u
∗
t
, if t ∈
0, 4
,
1480 ≤
u
∗
≤ 13500.
3.48
Acknowledgments
The authors would like to thank the referees for helpful comments and suggestions. The work
was supported partly by the NSF of China 10771202, the Research Fund for Shanghai Key
Laboratory of Modern Applied Mathematics 08DZ2271900, and the Specialized Research
Fund for the D octoral Program of Higher Education of China 2007035805.
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