Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2009, Article ID 314656, 18 pages
doi:10.1155/2009/314656
Research Article
The Existence of Positive Solution to
Three-Point Singular Boundary Value Problem of
Fractional Differential Equation
Yuansheng Tian1 and Anping Chen1, 2
1
2
Department of Mathematics, Xiangnan University, Chenzhou 423000, China
School of Mathematics and Computational Science, Xiangtan University, Xiangtan 411105, China
Correspondence should be addressed to Anping Chen, chenap@263.net
Received 13 May 2009; Accepted 23 June 2009
Recommended by Yong Zhou
We investigate the existence of positive solution to nonlinear fractional differential equation
three-point singular boundary value problem: Dq ut ft, ut 0, 0 < t < 1, u0 0,
u1 αDq−1/2 ut|tξ , where 1 < q ≤ 2 is a real number, ξ ∈ 0, 1/2, α ∈ 0, ∞ and
αΓqξ q−1/2 < Γq 1/2, Dq is the standard Riemann-Liouville fractional derivative, and
f ∈ C0, 1 × 0, ∞, 0, ∞, limt → 0 ft, · ∞ i.e., f is singular at t 0. By using the
fixed-point index theory, the existence result of positive solutions is obtained.
Copyright q 2009 Y. Tian and A. Chen. 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
Fractional differential equations have been of great interest recently. It is caused both by the
intensive development of the theory of fractional calculus itself and the applications of such
constructions in various sciences such as physics, mechanics, chemistry, and engineering. For
details, see 1–10 and the reference therein. However, up to our knowledge, most of those
papers have studied the existence and multiplicity of solution or positive solution to the
initial value problem of nonlinear fractional differential equations; see 1, 4, 10, 11.
Recently, there are a few paper considering the Dirichlet-type boundary value problem
for nonlinear ordinary differential equations of fractional order; see 12–14. Delbosco 13
has investigated the nonlinear Dirichlet-type problem
uq−1 Dq ut fut,
0 < t < 1, 1 < q < 2,
u0 u1 0.
1.1
2
Abstract and Applied Analysis
He has proved that if fu is Lipschitizan function, then the problem has at least one solution
ut in a certain subspace of C0, 1 in which fractional derivative has a Holder property.
When ft, u is continuous on 0, 1 × 0, ∞, by the use of some fixed-point theorem on
cones, Bai and Lü 12 and Zhang 14 have given the existence of positive solutions to the
equation
Dq ut ft, ut 0,
0 < t < 1,
1.2
with boundary condition
u0 u1 0,
1.3
u0 u 0 u1 u 1,
respectively.
This paper is to study the existence of positive solution for the three-point singular
boundary value problem of nonlinear fractional differential equation
Dq ut ft, ut 0,
u0 0,
0 < t < 1,
u1 αDq−1/2 ut .
1.4
tξ
By using the fixed-point index theory, where 1 < q ≤ 2 is a real number, ξ ∈ 0, 1/2,
α ∈ 0, ∞ satisfy that αΓqξq−1/2 < Γq 1/2, Dq is the standard Riemann-Liouville
fractional derivative, and the function f satisfies the following condition:
H1 f ∈ C0, 1 × 0, ∞, 0, ∞, limt → 0 ft, · ∞, there exists a constant b : 0 <
b < 1 such that tb ft, ut is continuous function on 0, 1 × 0, ∞.
The organization of this paper is as follows. In Section 2, we present some necessary
definitions and Preliminary results that will be used to prove our main results. The proof of
our main result is given in Section 3. In Section 4, we will give an example to ensure our main
result.
2. Preliminaries
The material in this section is basic in some sense. For the reader’s convenience, we present
some necessary definitions from fractional calculus theory and preliminary results.
Definition 2.1. The fractional integral of order q > 0 of a function x : 0, ∞ → R is given by
1
I xt Γ q
q
t
t − sq−1 xsds,
0
provided that the right side is pointwise defined on 0, ∞.
2.1
Abstract and Applied Analysis
3
Definition 2.2. The fractional derivative of order q > 0 of a continuous function x : 0, ∞ →
R is given by
n t
d
xs
1
ds,
Dq xt q−n1
Γ n − q dt
0 t − s
2.2
where n q 1, provided that the right side is pointwise defined on 0, ∞.
Lemma 2.3 see 7. 1 If x ∈ L0, 1, ρ > σ > 0, then Dσ I ρ xt I ρ−σ xt.
2 If ρ > 0, λ > 0, then Dρ tλ−1 Γλ/Γλ − ρtλ−ρ−1 .
Lemma 2.4 see 12. Assume that x ∈ C0, 1 ∩ L0, 1 with a fractional derivative of order q > 0
that belongs to C0, 1 ∩ L0, 1. Then
I q Dq xt xt A1 tq−1 A2 tq−2 · · · AN tq−N ,
2.3
Ai ∈ R, i 1, 2, . . . , N, where N is the smallest integer greater than or equal to q.
Lemma 2.5. If y ∈ C0, 1 ∩ L0, 1 and 1 < q ≤ 2, ξ ∈ 0, 1, α ∈ R satisfy that
αΓqξq−1/2 /
Γq 1/2, then the problems
Dq ut yt 0,
u0 0,
0 < t < 1,
u1 αDq−1/2 ut
2.4
tξ
have the unique solution
t
t − sq−1
ysds
Γ q
0
1
ξ
tq−1 Γ q 1 /2
1 − sq−1
ξ − sq−1/2
ysds − α
ysds .
Γ q
Γ q 1 /2 − αΓ q ξq−1/2
0
0 Γ q 1 /2
2.5
ut −
Proof. By applying Lemma 2.4, we may reduce 2.4 to an equivalent integral equation
ut −I q yt A1 tq−1 A2 tq−2
2.6
for some A1 , A2 ∈ R. Consequently the general solution of 2.4 is
ut −
t
t − sq−1 ys
ds A1 tq−1 A2 tq−2 .
Γ
q
0
2.7
4
Abstract and Applied Analysis
Note that u0 0, we have A2 0 and
u1 −
1
1 − sq−1 ys
ds A1 .
Γ q
0
2.8
On the other hand, by 2.6 and Lemma 2.3, we have
Dq−1/2 ut −Dq−1/2 I q yt A1 Dq−1/2 tq−1
Γ q
q1/2
−I
yt A1 tq−1/2 .
Γ q 1 /2
2.9
Therefore
Dq−1/2 ut
tξ
−
Γ q
ξ − sq−1/2
ysds A1 ξq−1/2 .
Γ
q
1
/2
Γ
q
1
/2
0
ξ
2.10
By u1 αDp ut|tξ , combine with 2.8 and 2.10, we obtain
1
ξ
Γ q 1 /2
1 − sq−1
ξ − sq−1/2
A1 ysds − α
ysds . 2.11
Γ q
Γ q 1 /2 − αΓ q ξq−1/2
0
0 Γ q 1 /2
So, the unique solution of problem 2.4 is
t
t − sq−1
ysds
Γ q
0
1
ξ
tq−1 Γ q 1 /2
1 − sq−1
ξ − sq−1/2
ysds − α
ysds .
Γ q
Γ q 1 /2 − αΓ q ξq−1/2
0
0 Γ q 1 /2
2.12
ut −
The proof is completed.
Abstract and Applied Analysis
5
Lemma 2.6. If y ∈ C0, 1, 0, ∞ ∩ L0, 1 and 1 < q ≤ 2, ξ ∈ 0, 1/2, α ∈ 0, ∞ satisfy that
αΓqξq−1/2 < Γq 1/2, then the unique solution of the problem 2.4
t
t − sq−1
ysds
Γ q
0
1
ξ
tq−1 Γ q 1 /2
1 − sq−1
ξ − sq−1/2
ysds − α
ysds
Γ q
Γ q 1 /2 − αΓ q ξq−1/2
0
0 Γ q 1 /2
ut −
1
αtq−1
Gt, sysds Γ q 1 /2 − αΓ q ξq−1/2
0
ξ
1
q−1 q−1/2
q−1/2
q−1 q−1/2
1 − s ξ
ysds 1 − s ξ
×
− ξ − s
ysds .
0
ξ
2.13
is nonnegative on 0, 1, where
Gt, s ⎧
⎪
t1 − sq−1 − t − sq−1
⎪
⎪
,
⎪
⎪
⎪
Γ q
⎪
⎨
0 ≤ s ≤ t ≤ 1,
⎪
⎪
⎪
q−1
⎪
⎪
⎪ t1 − s
⎪
,
⎩
Γ q
0 ≤ t ≤ s ≤ 1.
2.14
Proof. By Lemma 2.5, the unique solution of problem 2.4 is
t
t − sq−1
ysds
Γ q
0
1
ξ
tq−1 Γ q 1 /2
1 − sq−1
ξ − sq−1/2
ysds − α
ysds
Γ q
Γ q 1 /2 − αΓ q ξq−1/2
0
0 Γ q 1 /2
ut −
t
t − sq−1
ysds
Γ q
0
1
tq−1 Γ q 1 /2
1 − sq−1
ysds
q−1/2
Γ q
Γ q 1 /2 − αΓ q ξ
0
ξ
αtq−1 Γ q 1 /2
ξ − sq−1/2
− ysds
q−1/2
Γ q 1 /2 − αΓ q ξ
0 Γ q 1 /2
−
6
Abstract and Applied Analysis
t
t − sq−1
ysds
Γ q
0
1
αΓ q ξq−1/2
1 − sq−1
q−1
1 ysds
q−1/2 t
Γ q
Γ q 1 /2 − αΓ q ξ
0
−
ξ
αtq−1
− ξ − sq−1/2 ysds
Γ q 1 /2 − αΓ q ξq−1/2 0
t
1
t − sq−1
1 − sq−1 tq−1
−
ysds
ysds Γ q
Γ q
0
0
1
αtq−1 ξq−1/2
1 − sq−1 ysds
Γ q 1 /2 − αΓ q ξq−1/2 0
ξ
αtq−1
− ξ − sq−1/2 ysds
Γ q 1 /2 − αΓ q ξq−1/2 0
1
t
t1 − sq−1 − t − sq−1
t1 − sq−1
ysds ysds
Γ q
Γ q
0
t
1
ξ
αtq−1
q−1 q−1/2
q−1/2
1 − s ξ
ysds − ξ − s
ysds
Γ q 1 /2 − αΓ q ξq−1/2
0
0
1
αtq−1
Gt, sysds Γ q 1 /2 − αΓ q ξq−1/2
0
ξ
1
q−1 q−1/2
q−1/2
q−1 q−1/2
1 − s ξ
ysds 1 − s ξ
×
− ξ − s
ysds .
0
ξ
2.15
Observing the expression of Gt, s, it is clear that Gt, s > 0 for s, t ∈ 0, 1. On the
other hand, by ξ ∈ 0, 1/2, we have
s
1 − s ≥ 1 −
,
ξ
2
∀s ∈ 0, ξ.
2.16
Hence
q−1
1 − s
≥
s
1−
ξ
q−1/2
.
2.17
It implies that
1 − sq−1 ξq−1/2 ≥ ξ − sq−1/2 .
Therefore ut is nonnegative on 0, 1.
2.18
Abstract and Applied Analysis
7
Lemma 2.7. Let 1 < q ≤ 2, ξ ∈ 0, 1/2, α ∈ 0, ∞ satisfy that αΓqξq−1/2 < Γq 1/2 and
y : 0, 1 → 0, ∞ is continuous, and limt → 0 yt ∞. Suppose that there exists a constant
b : 0 < b < 1 such that tb yt is continuous function on 0, 1. Then the unique solution of 2.4
t
t − sq−1
ysds
Γ q
0
1
ξ
tq−1 Γ q 1 /2
1 − sq−1
ξ − sq−1/2
ysds − α
ysds .
Γ q
Γ q 1 /2 − αΓ q ξq−1/2
0
0 Γ q 1 /2
2.19
ut −
is continuous on 0, 1.
Proof. Since tb yt is continuous in 0, 1, thus there exists a constant L > 0 such that |tb yt| ≤
L for all t ∈ 0, 1 and
1
ut − Γ q
t
t − sq−1 s−b · sb ysds
0
1
tq−1 Γ q 1 /2
1 − sq−1 s−b · sb ysds
Γ q 1 /2 − αΓ q ξq−1/2 Γ q 0
ξ
αtq−1
− ξ − sq−1/2 s−b · sb ysds.
Γ q 1 /2 − αΓ q ξq−1/2 0
2.20
For any t0 ∈ 0, 1, we will prove ut → ut0 t → t0 , t ∈ 0, 1. For the convenience, the
proof is divided into three cases.
Case 1 t0 0. It is easy to know that u0 0. For all t ∈ 0, 1, we have
1 t
q−1 −b
b
|ut − ut0 | ≤ t − s s · s ysds
Γ q 0
1
tq−1 Γ q 1 /2
q−1/2 1 − sq−1 s−b · sb ysds
Γ q 1 /2 − αΓ q ξ
Γ q 0
ξ
αtq−1
q−1/2 −b
b
s · s ysds
q−1/2 ξ − s
Γ q 1 /2 − αΓ q ξ
0
8
Abstract and Applied Analysis
L
≤ Γ q
t
t − sq−1 s−b ds
0
1
Ltq−1 Γ q 1 /2
q−1/2 1 − sq−1 s−b ds
Γ q 1 /2 − αΓ q ξ
Γ q 0
ξ
αLtq−1
q−1/2 ξ − sq−1/2 s−b ds
Γ q 1 /2 − αΓ q ξ
0
Ltq−1 Γ q 1 /2
Ltq−b B 1 − b, q q−1/2 B 1 − b, q
Γ q
Γ q 1 /2 − αΓ q ξ
Γ q
αLξ1−bq−1/2 tq−1
q−1/2 B 1 − b, q 1 /2
Γ q 1 /2 − αΓ q ξ
LΓ1 − bΓ q 1 /2
LΓ1 − b q−b
t tq−1
Γ 1−bq
Γ q 1 /2 − αΓ q ξq−1/2 Γ 1 − b q
αLΓ1 − bΓ q 1 /2 ξ1−bq−1/2
tq−1 −→ 0 t −→ 0,
Γ q 1 /2 − αΓ q ξq−1/2 Γ 1 − b q 1 /2
2.21
where B denotes beta function.
Case 2 t0 ∈ 0, 1, for all t ∈ t0 , 1. We have
|ut − ut0 |
t
t0
1
1
− t − sq−1 ysds t0 − sq−1 ysds
Γ q 0
Γ q 0
q−1
ξ
q−1
q−1/2
1
tq−1 − t0 Γ q 1 /2
1 − s
ξ − s
ysds − α
ysds q−1/2
Γ q
Γ q 1 /2 − αΓ q ξ
0
0 Γ q 1 /2
t
1 t0 1
q−1
q−1
≤ t − s − t0 − sq−1 s−b · sb ysds t − s s−b · sb ysds
Γ q 0
Γ q t0
q−1
1
tq−1 − t0 Γ q 1 /2
q−1 −b
b
q−1/2 1 − s s · s ysds
Γ q 1 /2 − αΓ q ξ
Γ
q
0
q−1
ξ
α tq−1 − t0
q−1/2 −b
s · sb ysds
q−1/2 ξ − s
Γ q 1 /2 − αΓ q ξ
0
Abstract and Applied Analysis
9
t
L
t − sq−1 − t0 − sq−1 s−b ds t − sq−1 s−b ds
Γ q t0
0
q−1
1
tq−1 − t0 Γ q 1 /2 L
1 − sq−1 s−b ds
Γ q 1 /2 − αΓ q ξq−1/2 Γ q 0
q−1
ξ
α tq−1 − t0 L
q−1/2 ξ − sq−1/2 s−b ds
Γ q 1 /2 − αΓ q ξ
0
q−1
q−1
t
Γ q 1 /2 L
−
t
0
L
q−b
q−b
t − t0 B 1 − b, q B 1 − b, q
Γ q
Γ q 1 /2 − αΓ q ξq−1/2 Γ q
q−1
α tq−1 − t0 Lξ1−bq−1/2 q1
B 1 − b,
2
Γ q 1 /2 − αΓ q ξq−1/2
LΓ1 − bΓ q 1 /2
LΓ1 − b q−b q−b q−1
t − t0
q−1/2 tq−1 − t0
Γ 1−bq
Γ q 1 /2 − αΓ q ξ
Γ 1−bq
L
≤ Γ q
t0 αLΓ1 − bΓ q 1 /2 ξ1−bq−1/2
q−1
−→ 0 t −→ t0 .
q−1/2 tq−1 − t0
Γ q 1 /2 − αΓ q ξ
Γ 1 − b q 1 /2
2.22
Case 3 t0 ∈ 0, 1, for all t ∈ 0, t0 . The proof is similar to the step 2. Here, we omit it.
The main tool of this paper is the following well-known fixed-point index theorem
see 15.
Lemma 2.8. Let E be a Banach space, P ⊂ E is a cone. For r > 0, define Ωr {u ∈ P | u < r}.
Assume that T : Ωr → P is a completely continuous such that T u / u for u ∈ ∂Ωr {u ∈ P | u r}.
1 If T u ≥ u for u ∈ ∂Ωr , then iT, Ωr , P 0.
2 If T u ≤ u for u ∈ ∂Ωr , then iT, Ωr , P 1.
3. Main Results
For the convenience we introduce the following notations:
Γ1 − b
C1 ,
Γ 1−bq
10
Abstract and Applied Analysis
Γ1 − bΓ q 1 /2
1
αξ1−bq−1/2
C2 ,
Γ 1 − b q 1 /2
Γ q 1 /2 − αΓ q ξq−1/2 Γ 1 − b q
Γ q 1 /2 ξ1−b
Γ q
αΓ1 − bξq−1/2
C3 − .
Γ 1 − b q 1 /2
Γ q 1 /2 − αΓ q ξq−1/2 Γ 1 − b q
3.1
Remark 3.1. If 1 < q ≤ 2, 0 < b < 1, 0 < ξ ≤ 1/2, α > 0, and Γq 1/2 > αΓqξq−1/2 , then
α
C3 Γ q 1 /2 − αΓ q ξq−1/2
ξ
1 − sq−1 ξq−1/2 − ξ − sq−1/2 · s−b ds
0
1
1 − s
q−1 q−1/2
ξ
−b
· s ds
3.2
ξ
αΓ1 − bξq−1/2
Γ q 1 /2 − αΓ q ξq−1/2
Γ q 1 /2 ξ1−b
− > 0.
Γ 1−bq
Γ 1 − b q 1 /2
Γ q
Let E C0, 1 be a Banach spaces with the maximum norm u max0≤t≤1 |ut|.
Define the cone P ⊂ E by
P {u ∈ E | ut ≥ 0, 0 ≤ t ≤ 1}.
3.3
The positive solution which we consider in this paper is the form u0 0, ut > 0,
0 < t ≤ 1, u ∈ E.
Define an operator T : P → E by
t
t − sq−1
fs, usds
Γ q
0
1
tq−1 Γ q 1 /2
1 − sq−1
fs, usds
q−1/2
Γ q
Γ q 1 /2 − αΓ q ξ
0
T ut −
3.4
ξ
ξ − sq−1/2
−α
fs, usds .
0 Γ q 1 /2
Lemma 3.2. Assume that condition H1 holds. Then the operator T : P → P is completely
continuous.
Proof. If condition H1 holds, by Lemmas 2.6 and 2.7, we have T P ⊂ P . Let u0 ∈ P and
u0 a0 ; if u ∈ P and u − u0 < 1, then u < 1 a0 : a. By the continuous of tb ft, x, we
know that tb ft, x is uniformly continuous on 0, 1 × 0, a. Thus, for all ε > 0, there exists a
δ > 0 δ < 1 such that
b
t ft, x1 − tb ft, x2 <
ε
C1 C2
3.5
Abstract and Applied Analysis
11
for all t ∈ 0, 1 and x1 , x2 ∈ 0, a with |x1 −x2 | < δ. Obviously, if u−u0 ≤ δ, then u0 t, ut ∈
0, a and |ut − u0 t| < δ for each t ∈ 0, 1. Hence, we have
b
t ft, ut − tb ft, u0 t <
ε
C1 C2
3.6
for all t ∈ 0, 1, u ∈ P with u − u0 < δ. It follows from 3.6 that
T u − T u0 max|T ut − T u0 t|
0≤t≤1
≤ max
0≤t≤1
1
Γ q
t
t − sq−1 s−b sb fs, us − sb fs, u0 sds
0
tq−1 Γ q 1 /2
q−1/2 Γ q 1 /2 − αΓ q ξ
Γ q
1
× 1 − sq−1 s−b sb fs, us − sb fs, u0 sds
0
αtq−1
Γ q 1 /2 − αΓ q ξq−1/2
ξ
q−1/2 −b b
b
× ξ − s
s s fs, us − s fs, u0 sds
0
< max
0≤t≤1
ε
Γ q C1 C2 t
t − sq−1 s−b ds
0
1
εtq−1 Γ q 1 /2
1 − sq−1 s−b ds
Γ q 1 /2 − αΓ q ξq−1/2 C1 C2 Γ q 0
ξ
αεtq−1
q−1/2 −b
ξ − s
s ds
Γ q 1 /2 − αΓ q ξq−1/2 C1 C2 0
εtq−1 Γ q 1 /2 B 1 − b, q
ε
q−b
t B 1 − b, q max 0≤t≤1 Γ q C1 C2 Γ q 1 /2 − αΓ q ξq−1/2 C1 C2 Γ q
q1
αεtq−1 ξ1−bq−1/2
B 1 − b,
2
Γ q 1 /2 − αΓ q ξq−1/2 C1 C2 Γ q 1 /2 Γ1 − b
Γ1 − b
≤
Γ 1−bq
Γ q 1 /2 − αΓ q ξq−1/2 Γ 1 − b q
αΓ1 − bΓ q 1 /2 ξ1−bq−1/2
ε
ε.
Γ q 1 /2 − αΓ q ξq−1/2 Γ 1 − b q 1 /2 C1 C2 3.7
By the arbitrariness of u0 , we have that T : P → P is continuous.
12
Abstract and Applied Analysis
Let Ω ⊂ P be bounded; that is, there exists a positive constant M such that Ω ∈ {u ∈
P | u ≤ M}. Since tb ft, ut is continuous on 0, 1 × 0, ∞, we let
N
max
t,u∈0,1×0,M
tb ft, ut 1.
3.8
For all u ∈ Ω, we have
1 t
q−1 −b
b
|T ut| ≤ t − s s · s ft, utds
Γ q 0
1
tq−1 Γ q 1 /2
q−1 −b
b
q−1/2 1 − s s · s fs, usds
Γ q 1 /2 − αΓ q ξ
Γ q 0
ξ
αtq−1
q−1/2 −b
b
s · s fs, usds
q−1/2 ξ − s
Γ q 1 /2 − αΓ q ξ
0
1
Ntq−1 Γ q 1 /2
N t
≤ t − sq−1 s−b ds q−1/2 1 − sq−1 s−b ds
Γ q 0
Γ q 1 /2 − αΓ q ξ
Γ q 0
ξ
αNtq−1
q−1/2 ξ − sq−1/2 s−b ds
Γ q 1 /2 − αΓ q ξ
3.9
0
Ntq−1 Γ q 1 /2
N
tq−b B 1 − b, q q−1/2 B 1 − b, q
Γ q
Γ q 1 /2 − αΓ q ξ
Γ q
q1
αNξ1−bq−1/2 tq−1
B 1 − b,
2
Γ q 1 /2 − αΓ q ξq−1/2
NΓ q 1 /2 Γ1 − b
NΓ1 − b
≤ Γ 1−bq
Γ q 1 /2 − αΓ q ξq−1/2 Γ 1 − b q
αNΓ1 − bΓ q 1 /2 ξ1−bq−1/2
Γ q 1 /2 − αΓ q ξq−1/2 Γ 1 − b q 1 /2
C1 C2 N.
Hence T Ω is bounded.
On the other hand, given ε > 0, set
1/q−b 1/q−1
1
ε
1
ε
ε
.
,
,
δ min
2 4NC1
2 2NC2
4NC1
3.10
For each u ∈ Ω, we will prove that if t1 , t2 ∈ 0, 1 and 0 < t2 − t1 < δ, then
|T ut2 − T ut1 | < ε.
3.11
Abstract and Applied Analysis
13
In fact, similar to the proof of Lemma 2.7, we have
NΓ1 − b q−b q−b t2 − t1
|T ut2 − T ut1 | ≤ Γ 1−bq
q−1 q−1 NΓ q 1 /2 Γ1 − b
q−1/2 t2 − t1
Γ q 1 /2 − αΓ q ξ
Γ 1−bq
q−1 q−1 αNΓ1 − bΓ q 1 /2 ξ1−bq−1/2
q−1/2 t2 − t 1
Γ q 1 /2 − αΓ q ξ
Γ 1 − b q 1 /2
q−b q−b q−1 q−1 NC1 t2 − t1
NC2 t2 − t1 .
3.12
In the following, the proof is divided into three cases.
Case 1 δ ≤ t1 < t2 < 1, q − b − 1 ≤ 0.
q−b q−b q−1 q−1 NC2 t2 − t1
|T ut2 − T ut1 | ≤ NC1 t2 − t1
≤ NC1 q − b δq−b−1 t2 − t1 NC2 q − 1 δq−2 t2 − t1 < 2NC1 δq−b NC2 δq−1
≤ 2NC1 ·
ε
ε
NC2 ·
2NC2 2q−1
4NC1 2q−b
3.13
ε
ε
q−b
2 × 2q−1
2×2
< ε.
Case 2 δ ≤ t1 < t2 < 1, q − b − 1 > 0.
q−b q−b q−1 q−1 NC2 t2 − t1
|T ut2 − T ut1 | ≤ NC1 t2 − t1
≤ NC1 q − b t2 − t1 NC2 q − 1 δq−2 t2 − t1 < 2NC1 δ NC2 δq−1
≤ 2NC1 ·
ε
ε
NC2 ·
4NC1
2NC2 2q−1
ε
ε
2 2 × 2q−1
< ε.
3.14
14
Abstract and Applied Analysis
Case 3 0 ≤ t1 < δ, t2 < 2δ.
q−b q−b q−1 q−1 NC2 t2 − t1
|T ut2 − T ut1 | ≤ NC1 t2 − t1
q−b
≤ NC1 t2
q−1
NC2 t2
≤ NC1 2δq−b NC2 2δq−1
≤ NC1 ·
3.15
ε
ε
NC2 ·
4NC1
2NC2
ε ε
4 2
< ε.
Therefore, T Ω is equicontinuous. The Arzela-Ascoli Theorem implies that T Ω is
compact. Thus, the operator T : P → P is completely continuous.
We obtain the following existence results of the positive solution for problem 1.4.
Theorem 3.3. If condition H1 holds and assume further that there exist two positive constants
R > r > 0 such that
H2 tb ft, u > r/C3 , for t, u ∈ 0, 1 × 0, r;
H3 tb ft, u < R/C1 C2 , for t, u ∈ 0, 1 × 0, R,
then problem 1.4 has at least one positive solution u such that r < u < R.
Proof. Problem 1.4 has a solution u ut if and only if u is a solution of the operator
equation u T u. In order to apply Lemma 2.8, we separate the proof into the following two
steps.
Step 1. Let Ωr : {u ∈ P | u < r}. For any u ∈ ∂Ωr , we have u r and 0 ≤ ut ≤ r for
all t ∈ 0, 1. Observing the expression of Gt, s see 2.14, it is clear that G1, s 0. By
assumption H2 , we have
T u1 1
G1, sfs, usds
0
α
Γ q 1 /2 − αΓ q ξq−1/2
ξ
1 − sq−1 ξq−1/2 − ξ − sq−1/2 fs, usds
0
1
ξ
q−1 q−1/2
1 − s
ξ
fs, usds
Abstract and Applied Analysis
15
α
Γ q 1 /2 − αΓ q ξq−1/2
ξ
1 − sq−1 ξq−1/2 − ξ − sq−1/2 t−b · tb ft, utds
0
1
1 − s
q−1 q−1/2
ξ
·t
−b
· t ft, utds
b
ξ
αr
> Γ q 1 /2 − αΓ q ξq−1/2 C3
ξ
1
q−1 q−1/2
q−1/2
q−1 q−1/2
−b
−b
1 − s ξ
· s ds 1 − s ξ
×
− ξ − s
· s ds
0
ξ
1
ξ
αr
q−1 −b
q−1/2 −b
q−1/2
1 − s s ds − ξ − s
s ds
ξ
Γ q 1 /2 − αΓ q ξq−1/2 C3
0
0
q1
αr
ξq−1/2 B 1 − b, q − ξ1−bq−1/2 B 1 − b,
q−1/2 2
Γ q 1 /2 − αΓ q ξ
C3
Γ q 1 /2 ξ1−b
Γ q
r
αΓ1 − bξq−1/2
r.
− ·
q−1/2
C3
Γ 1−bq
Γ 1 − b q 1 /2
Γ q 1 /2 − αΓ q ξ
3.16
So
T u ≥ u,
∀u ∈ ∂Ωr .
3.17
By Lemma 2.8, we have
iT, Ωr , P 0.
3.18
Step 2. Let ΩR : {u ∈ P | u < R}. For any u ∈ ∂ΩR , we have u R and 0 ≤ ut ≤ R for
all t ∈ 0, 1. By assumption H3 , for t ∈ 0, 1, we get
1 t
q−1 −b
b
|T ut| ≤ t − s s · s ft, utds
Γ q 0
1
tq−1 Γ q 1 /2
q−1/2 1 − sq−1 s−b · sb fs, usds
Γ q 1 /2 − αΓ q ξ
Γ q 0
ξ
αtq−1
q−1/2 −b
b
s · s fs, usds
q−1/2 ξ − s
Γ q 1 /2 − αΓ q ξ
0
16
Abstract and Applied Analysis
R
< Γ q C1 C2 t
t − sq−1 s−b ds
0
1
Rtq−1 Γ q 1 /2
1 − sq−1 s−b ds
q−1/2 Γ q 1 /2 − αΓ q ξ
Γ q C1 C2 0
ξ
αRtq−1
ξ − sq−1/2 s−b ds
q−1/2
Γ q 1 /2 − αΓ q ξ
C1 C2 0
R
tq−b B 1 − b, q
Γ q C1 C2 Rtq−1 Γ q 1 /2
B 1 − b, q
q−1/2 Γ q 1 /2 − αΓ q ξ
Γ q C1 C2 q1
αNξ1−bq−1/2 tq−1
B 1 − b,
q−1/2
2
Γ q 1 /2 − αΓ q ξ
C1 C2 Γ q 1 /2 Γ1 − b
Γ1 − b
≤
Γ 1−bq
Γ q 1 /2 − αΓ q ξq−1/2 Γ 1 − b q
αΓ1 − bΓ q 1 /2 ξ1−bq−1/2
R
R.
q−1/2 ·
C
Γ q 1 /2 − αΓ q ξ
Γ 1 − b q 1 /2
1 C2
3.19
Therefore
T u ≤ u,
∀u ∈ ∂ΩR .
3.20
By Lemma 2.8, we have
iT, ΩR , P 1.
3.21
Combine with 3.18 and 3.21, we have
i T, ΩR \ Ωr , P iT, ΩR , P − iT, Ωr , P 1 − 0 1.
3.22
Therefore, T has a fixed point u ∈ ΩR \ Ωr . Then problem 1.4 has at least one positive
solution u such that r < u < R.
Abstract and Applied Analysis
17
4. Example
Let q 3/2, α ξ 1/2. We consider the following boundary value problem:
D3/2 ut ft, ut 0,
u0 0,
0 < t < 1,
1 1/4
,
u1 D ut
2
t1/2
4.1
where
ft, u 1 −1/2 1 2 4
t
u t 1 .
4
18
4.2
Let b 1/2. By simple computation, we have
C1 ≈ 1.7724,
C2 ≈ 3.9824,
C3 ≈ 0.2635.
4.3
Choosing R 6, r 1/16, we have
1 1 2 4
R
u t 1 ≤1<
≈ 1.0426, t, u ∈ 0, 1 × 0, 6,
4 18
C1 C2
r
1
1 1 2 4
1
1
u t 1 ≥ >
, t, u ∈ 0, 1 × 0,
.
t1/2 ft, u ≈
4 18
4 C3 4.216
16
t1/2 ft, u 4.4
By Theorem 3.3, problem 4.1 has at least one solution u such that 1/16 < u < 6.
Acknowledgments
This work was supported by the Scientific Research Foundation of Hunan Provincial
Education Department 05A057, 08C826 was also supported by the Aid Program for Science
and Technology Innovative Research Team in Higher Educational Institutions of Hunan
Province, and the Construct Program of the Key Discipline in Hunan Province.
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