Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2012, Article ID 841349, 18 pages
doi:10.1155/2012/841349
Research Article
The Existence of Solutions for a Fractional
2m-Point Boundary Value Problems
Gang Wang,1 Wenbin Liu,1 Jinyun Yang,2
Sinian Zhu,1 and Ting Zheng1
1
2
Department of Mathematics, China University of Mining and Technology, Xuzhou 221008, China
School of Mathematics and Physical Sciences, Xuzhou Institute of Technology, Xuzhou 221008, China
Correspondence should be addressed to Gang Wang, wangg0824@163.com
Received 24 June 2011; Revised 9 October 2011; Accepted 11 October 2011
Academic Editor: Yongkun Li
Copyright q 2012 Gang 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.
By using the coincidence degree theory, we consider the following 2m-point boundary value
α−1
α−2
α
problem for fractional differential equation D0
ut ft, ut, D0
ut, D0
ut et, 0 < t < 1,
m−2
m−2
3−α
α−2
α−2
α
α
and I0
are
I0 ut|t0 0, D0 u1 i1 ai D0 uξi , u1 i1 bi uηi , where 2 < α ≤ 3, D0
the standard Riemann-Liouville fractional derivative and fractional integral, respectively. A new
result on the existence of solutions for above fractional boundary value problem is obtained.
1. Introduction
Fractional differential equations have been of great interest recently. This is because of the
intensive development of the theory of fractional calculus itself as well as its applications.
Apart from diverse areas of mathematics, fractional differential equations arise in a variety
of different areas such as rheology, fluid flows, electrical networks, viscoelasticity, chemical
physics, and many other branches of science see 1–4 and references cited therein. The
research of fractional differential equations on boundary value problems, as one of the focal
topics has attained a great deal of attention from many researchers see 5–13.
However, there are few papers which consider the boundary value problem at
resonance for nonlinear ordinary differential equations of fractional order. In 14, Hu and
Liu studied the following BVP of fractional equation at resonance:
α
xt f t, xt, x t, x t ,
D0
x0 x1,
0 ≤ t ≤ 1,
x 0 x 0 0,
where 1 < α ≤ 2, D0α is the standard Caputo fractional derivative.
1.1
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In 15, Zhang and Bai investigated the nonlinear nonlocal problem
α
ut ft, ut, 0 < t < 1,
D0
u0 0,
βu η u1,
1.2
where 1 < α ≤ 2, they consider the case βηα−1 1, that is, the resonance case.
In 16, Bai investigated the boundary value problem at resonance
α
α−1
D0
ut f t, ut, D0
ut et,
2−α
I0
ut|t0 0,
α−1
D0
u1 0 < t < 1,
1.3
m−2
α−1
βi D0
u ηi
i0
is considered, where 1 < α ≤ 2 is a real number, D0α and I0α are the standard RiemannLiouville fractional derivative and fractional integral, respectively, and f : 0, 1 × R2 → R
is continuous and et ∈ L1 0, 1, m ≥ 2, 0 < ξi < 1, βi ∈ R, i 1, 2, . . . , m − 2 are given
constants such that m−2
i1 βi 1.
In this paper, we study the 2m-point boundary value problem
α
α−1
α−2
D0
ut f t, ut, D0
ut, D0
ut et,
3−α
I0
ut|t0 0,
α−2
D0
u1 m−2
0 < t < 1,
u1 α−2
ai D0
uξi ,
m−2
i1
bi u ηi ,
1.4
1.5
i1
where 2 < α ≤ 3, m ≥ 2, 0 < ξ1 < · · · < ξm < 1, 0 < η1 < · · · < ηm < 1, ai , bi ∈ R, f :
0, 1 × R3 → R, f satisfies Carathéodory conditions, D0α and I0α are the standard RiemannLiouville fractional derivative and fractional integral, respectively.
Setting
m−2
m−2
1
1
α1
α
,
Λ2 1−
1−
ai ξi
ai ξi ,
Λ1 αα 1
αα − 1
i1
i1
m−2
m−2
ΓαΓα − 1
Γα2
2α−1
2α−2
Λ3 ,
Λ4 .
1−
1−
bi ηi
bi ηi
Γ2α
Γ2α − 1
i1
i1
1.6
In this paper, we will always suppose that the following conditions hold:
C1:
m−2
m−2
m−2
m−2
i1
i1
i1
i1
ai ξi ai 1,
bi ηiα−1 bi ηiα−2 1,
1.7
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3
C2:
Λ Λ1 Λ4 − Λ2 Λ3 /
0.
1.8
We say that boundary value problem 1.4 and 1.5 is at resonance, if BVP
α
ut 0,
D0
3−α
I0
ut|t0 0,
Dα−2
0 u1 m−2
α−2
ai D0
uξi ,
i1
u1 m−2
bi u ηi
1.9
i1
has ut atα−1 btα−2 , a, b ∈ R as a nontrivial solution.
The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions, and lemmas. In Section 3, we establish a theorem on existence of solutions
for BVP 1.4-1.5 under nonlinear growth restriction of f, basing on the coincidence degree
theory due to Mawhin see 17.
Now, we will briefly recall some notation and an abstract existence result.
Let Y, Z be real Banach spaces, L : dom L ⊂ Y → Z a Fredholm map of index zero,s
and P : Y → Y, Q : Z → Z continuous projectors such that
Y Ker L ⊕ Ker P,
Z Im L ⊕ Q,
Im P Ker L,
Ker Q Im L.
1.10
It follows that L|dom L∩Ker P : dom L ∩ Ker P → Im L is invertible. We denote the inverse of the
map by Kp . If Ω is an open-bounded subset of Y such that dom L ∩ Ω / ∅, the map N : Y → Z
will be called L-compact on Ω if QNΩ is bounded and Kp I − QN : Ω → Y is compact.
The lemma that we used is 17, Theorem 2.4.
Lemma 1.1. Let L be a Fredholm operator of index zero and let N be L-compact on Ω. Assume that
the following conditions are satisfied:
i Lx /
λNx, for all x, λ ∈ domL \ KerL ∩ ∂Ω × 0, 1;
ii Nx ∈
/ ImL, for all x ∈ KerL ∩ ∂Ω;
iii degJQN|KerL , KerL ∩ Ω, 0 /
0,
where Q : Z → Z is a projection as above with KerQ Im L and J : ImQ → KerL is any
isomorphism. Then the equation Lx Nx has at least one solution in domL ∩ Ω.
2. Preliminaries
For the convenience of the reader, we present here some necessary basic knowledge and
definitions about fractional calculus theory. These definitions can be found in the recent
literature 1–16, 18.
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Journal of Applied Mathematics
Definition 2.1. The fractional integral of order α > 0 of a function y : 0, ∞ → R is given by
α
yt I0
1
Γα
t
t − sα−1 ysds,
2.1
0
provided the right side is pointwise defined on 0, ∞, where Γ· is the Gamma function.
Definition 2.2. The fractional derivative of order α > 0 of a function y : 0, ∞ → R is given
by
α
yt
D0
n t
ys
d
1
ds,
α−n1
Γn − α dt
0 t − s
2.2
where n α 1, provided the right side is pointwise defined on 0, ∞.
Definition 2.3. We say that the map f : 0, 1 × Rn → R satisfies Carathéodory conditions with
respect to L1 0, 1 if the following conditions are satisfied:
i for each z ∈ Rn , the mapping t → ft, z is Lebesgue measurable;
ii for almost every t ∈ 0, 1, the mapping t → ft, z is continuous on Rn ;
iii for each r > 0, there exists ρr ∈ L1 0, 1, R such that for a.e. t ∈ 0, 1 and every
|z| ≤ r, we have ft, z ≤ ρr t.
Lemma 2.4 see 15. Assume that u ∈ C0, 1 ∩ L1 0, 1 with a fractional derivative of order α > 0
that belongs to C0, 1 ∩ L1 0, 1. Then
α
α
D0
ut ut c1 tα−1 c1 tα−2 · · · cN tα−N
I0
2.3
for some ci ∈ R, i 1, 2, . . . , N, where N is the smallest integer greater than or equal to α.
We use the classical Banach space C0, 1 with the norm
max |xt|,
x∞ t∈0,1
2.4
L0, 1 with the norm
x1 1
|xt|dt.
2.5
0
Definition 2.5. For n ∈ N, we denote by ACn 0, 1 the space of functions ut which have
continuous derivatives up to order n − 1 on 0, 1 such that un−1 t is absolutely continuous:
ACn 0, 1{u | 0, 1 → R and Dn−1 ut is absolutely continuous in 0, 1}.
Lemma 2.6 see 15. Given μ > 0 and N μ 1 we can define a linear space
Cμ 0, 1 ut | ut I0α xt c1 tμ−1 c2 tμ−2 · · · cN tμ−N−1 , t ∈ 0, 1 ,
2.6
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5
where x ∈ 0, 1, ci ∈ R, i 1, 2, . . . , N − 1. By means of the linear functional analysis theory, we
can prove that with the
μ μ−N−1 u u∞ ,
uCμ D0 u · · · D0
∞
∞
2.7
Cμ 0, 1 is a Banach space.
Remark 2.7. If μ is a natural number, then Cμ 0, 1 is in accordance with the classical Banach
space Cn 0, 1.
Lemma 2.8 see 15. f ⊂ Cμ 0, 1 is a sequentially compact set if and only if f is uniformly
bounded and equicontinuous. Here uniformly bounded means there exists M > 0, such that for every
u∈f
μ μ−N−1 u u∞ < M,
uCμ D0 u · · · D0
∞
∞
2.8
and equicontinuous means that ∀ε > 0, ∃δ > 0, such that
∀t1 , t2 ∈ 0, 1, |t1 − t2 | < δ, ∀u ∈ f ,
|ut1 − ut2 | < ε,
α−i
α−i
∀t1 , t2 ∈ 0, 1, |t1 − t2 | < δ, ∀u ∈ f, ∀i 1, 2, . . . , N − 1 .
ut2 < ε,
D0 ut1 − D0
2.9
Lemma 2.9 see 1. Let α > 0, n α 1. Assume that u ∈ L1 0, 1 with a fractional integration
of order n − α that belongs to ACn 0, 1. Then the equality
n
α
α
I0
D0
u t ut −
i1
n−i
n−α
u t
|t0 α−i
I0
t
Γα − i 1
2.10
holds almost everywhere on 0, 1.
α
Definition 2.10 see 16. Let I0
L1 0, 1, α > 0 denote the space of functions ut,
α
v, v ∈ L1 0, 1.
represented by fractional integral of order α of a summable function: u I0
1
Let Z L1 0, 1, with the norm y 0 |ys|ds, Y Cα−1 0, 1 defined by Lemma 2.6,
α−2
with the norm uCα−1 D0α−1
u∞ D0 u∞ u∞ , where Y is a Banach space.
Define L to be the linear operator from dom L ⊂ Y to Z with
α
u ∈ L1 0, 1, u satisfies1.5 ,
dom L u ∈ Cα−1 0, 1 | D0
Lu D0α u,
u ∈ dom L,
2.11
2.12
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Journal of Applied Mathematics
we define N : Y → Z by setting
α−1
α−2
utD0
ut et.
Nut f t, ut, D0
2.13
Then boundary value problem 1.4 and 1.5 can be written as Lu Nu.
3. Main Results
Lemma 3.1. Let L be defined by 2.12, then
KerL atα−1 btα−2 | a, b ∈ R ∼
R2 ,
ImL y∈Z|
1
1 − sysds −
0
1
α−1
1 − s
ysds −
0
m−2
ξi
i1
0
ai
m−2
ηi
i1
0
bi
ξi − sysds 0,
ηi − s
α−1
3.1
ysds 0 .
Proof. In the following lemma, we use the unified notation of both for fractional integrals
−α
α
and fractional derivatives assuming that I0
D0
for α < 0.
α
α
Let Lu D0 u, by Lemma 2.9, D0 ut 0 has solution
ut 3
i1
3−i
3−α
u t
|t0 α−i
I0
t
Γα − i 1
3−α 3−α 3−α
u t |t0 α−1
I0
I0 u t |t0 α−2
I0 u t |t0 α−3
t t t
Γα
Γα − 1
Γα − 2
3−α α−2
ut|t0 α−2
Dα−1 ut|t0 α−1 D0
I0 u t |t0 α−3
t t t .
0
Γα
Γα − 1
Γα − 2
3.2
Combine with 1.5, So,
Ker L atα−1 btα−2 | a, b ∈ R ∼
R2 .
3.3
Let y ∈ Z and let
1
ut Γα
t
0
t − sα−1 ysds c1 tα−1 c2 tα−2 c3 tα−3 .
3.4
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7
α
ut yt a.e. t ∈ 0, 1 and, if
Then D0
1
1 − sysds −
1 − s
α−1
ξi
i1
0
m−2
ηi
i1
0
ai
0
1
m−2
ysds −
bi
0
ξi − sysds 0,
α−1
ηi − s
3.5
ysds 0
hold, then ut satisfies the boundary conditions 1.5. That is, u ∈ dom L and we have
y ∈ Z | y satisfies 3.4 ⊆ Im L.
3.6
α
u ∈ Im L, we have
Let u ∈ dom L. Then for D0
3.7
α
yt ut − c1 tα−1 − c2 tα−2 − c3 tα−3 ,
I0
where
c1 α−1
ut|t0
D0
,
Γα
c2 α−2
ut|t0
D0
,
Γα − 1
c3 3−α
ut|t0
I0
,
Γα − 2
3.8
which, due to the boundary value condition 1.5, implies that satisfies 3.5. In fact, from
m−2
3−α
α−2
α−2
I0
ut|t0 0 we have c3 0, from D0
u1 m−2
i1 ai D0 uξi , u1 i1 bi uηi , we
have
1
1 − sysds −
0
1
0
1 − sα−1 ysds −
m−2
ξi
i1
0
m−2
ηi
i1
0
ai
bi
ξi − sysds 0,
ηi − s
α−1
3.9
ysds 0.
Hence,
y ∈ Z | y satisfies 3.4 ⊇ Im L.
3.10
y ∈ Z | y satisfies 3.4 Im L.
3.11
Therefore,
The proof is complete.
Lemma 3.2. The mapping L : domL ⊂ Y → Z is a Fredholm operator of index zero, and
Qyt T1 yt tα−1 T2 yt tα−2 ,
3.12
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Journal of Applied Mathematics
where
1
Λ4 Q1 y − Λ2 Q2 y ,
Λ
T1 y 1
Λ3 Q1 y − Λ1 Q2 y ,
Λ
T2 y 3.13
define by Kp : ImL → domL ∩ KerP by
Kp yt 1
Γα
t
0
α
yt,
t − sα−1 ysds I0
y ∈ ImL,
3.14
and for all y ∈ ImL, Kp yCα−1 ≤ 1/Γα 2y1 .
Proof. Consider the continuous linear mapping Q1 : Z → Z and Q2 : Z → Z defined by
Q1 y 1
1 − sysds −
0
Q2 y 1
1 − sα−1 ysds −
0
m−2
ξi
i1
0
m−2
ηi
i1
0
ai
bi
ξi − sysds,
ηi − s
α−1
3.15
ysds.
Using the above definitions, we construct the following auxiliary maps T1 : Z → Z and
T2 : Z → Z:
T1 y 1
Λ4 Q1 y − Λ2 Q2 y ,
Λ
1
T2 y Λ3 Q1 y − Λ1 Q2 y .
Λ
3.16
Since the condition C2 holds, the mapping defined by
Qyt T1 yt tα−1 T2 yt tα−2
3.17
is well defined.
Recall C2 and note that
1
T1 T1 ytα−1 Λ4 Q1 T1 ytα−1 − Λ2 Q2 T1 ytα−1
Λ
Λ1 Λ2
Λ2 Λ3
1
Λ4 Λ1
Λ4 Λ3
Λ4
Q1 y −
Q2 y − Λ2
Q1 y −
Q2 y
Λ
Λ
Λ
Λ
Λ
T1 y,
3.18
Journal of Applied Mathematics
9
and similarly we can derive that
T1 T2 ytα−2 0,
T2 T1 ytα−1 0,
T2 T2 ytα−2 T2 y.
3.19
So, for y ∈ Z, it follows from the four relations above that
Q2 y Q
T1
T1 y tα−1 T2 y tα−2
T1 y tα−1 T2 y tα−2 tα−1 T2 T1 y tα−1 T2 y tα−2 tα−2
T1 y t
α−1
T2 y t
3.20
α−2
Qy,
that is, the map Q is idempotent. In fact Q is a continuous linear projector.
Note that y ∈ Im L implies Qy 0. Conversely, if Qy 0, so
Λ4 Q1 y − Λ2 Q2 y 0,
Λ1 Q2 y − Λ3 Q1 y 0,
3.21
but
Λ4 −Λ2 Λ4 Λ1 − Λ2 Λ3 / 0,
−Λ3 Λ1 3.22
then we must have Q1 y Q2 y 0; since the condition C2 holds, this can only be the case if
Q1 y Q2 y 0, that is, y ∈ Im L. In fact Ker Q Im L, take y ∈ Z in the form y y−QyQy
so that y − Qy ∈ Ker Q Im L, Qy ∈ Im Q, thus, Z Im L Im Q, Let y ∈ Im L ∩ Im Q and
assume that y atα−1 btα−2 is not identically zero on 0, 1. Then, since y ∈ Im L, from 3.5
and the condition C2, we have
Q1 y 1
0
Q2 y 1
0
1 − s as
α−1
bs
α−2
ds −
m−2
ξi
i1
0
ai
ξi − s asα−1 bsα−2 ds 0,
m−2
ηi α−1 α−1
α−1
α−1
α−2
ηi − s
as bs
ds −
as bsα−2 ds 0.
bi
1 − s
i1
3.23
0
So
aΛ1 bΛ2 0,
aΛ3 bΛ4 0,
3.24
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Journal of Applied Mathematics
but
Λ1 Λ2 Λ1 Λ4 − Λ2 Λ3 / 0,
Λ3 Λ4 3.25
we derive a b 0, which is a contradiction. Hence, Im L ∩ Im Q {0}; thus Z Im L ⊕ Im Q.
Now, dimKerL 2 codimImL and so L is a Fredholm operator of index zero.
Let P : Y → Y be defined by
P ut 1
1
α−1
D0
Dα−2 ut|t0 tα−2 ,
ut|t0 tα−1 Γα
Γα − 1 0
t ∈ 0, 1.
3.26
Note that P is a continuous linear projector and
α−1
α−2
u0 D0
u0 0 .
Ker P u ∈ Y | D0
3.27
It is clear that Y Ker P ⊕ Ker L.
Note that the projectors P and Q are exact. Define by Kp : Im L → dom L ∩ Ker P by
Kp yt 1
Γα
t
0
α
yt,
t − sα−1 ysds I0
y ∈ Im L.
3.28
t − sysds,
3.29
Hence we have
α−1
D0
Kp y t t
α−2
D0
ysds,
0
Kp y t t
0
then
Kp y ≤
∞
1 y ,
1
Γα
α−1 D0 Kp y ≤ y1 ,
∞
α−2 D0 Kp y ≤ y1 ,
∞
3.30
and thus
1
2 y1 .
Γα
3.31
α α
I0 yt yt.
LKp yt D0
3.32
α
α
D0
ut ut c1 tα−1 c2 tα−2 c3 tα−3 ,
Kp L ut I0
3.33
Kp y α−1 ≤
C
In fact, if y ∈ Im L, then
Also, if u ∈ dom L ∩ Ker P , then
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11
where
c1 α−1
ut|t0
D0
,
Γα
c2 α−2
ut|t0
D0
,
Γα − 1
c3 3−α
ut|t0
I0
,
Γα − 2
3.34
and from the boundary value condition 1.5 and the fact that u ∈ dom L ∩ Ker P , P u 0,
α−1
α−2
3−α
D0
ut|t0 D0
ut|t0 I0
ut|t0 0, we have c1 c2 c3 0, thus
Kp L ut ut.
3.35
This shows that Kp L|dom L∩Ker P −1 . The proof is complete. Using 3.16, we write
QNut T1 Nutα−1 T2 Nutα−2 ,
t
1
Kp I − QNut t − sα−1 Nus − QNusds.
Γα 0
3.36
By Lemma 2.8 and a standard method, we obtain the following lemma.
Lemma 3.3 see 16. For every given e ∈ L1 0, 1, Kp I − QN : Y → Y is completely continuous.
Assume that the following conditions on the function ft, x, y, z are satisfied.
H1 There exist functions at, bt, ct, dt, rt ∈ L1 0, 1, and a constant θ ∈ 0, 1
such that for all x, y, z ∈ R3 , t ∈ 0, 1, one of the following inequalities is satisfied:
f t, x, y, z ≤ at|x| bty ct|z| dt|x|θ rt,
3.37
f t, x, y, z ≤ at|x| bty ct|z| dtyθ rt,
3.38
f t, x, y, z ≤ at|x| bty ct|z| dt|z|θ rt.
3.39
α−1
H2 There exists a constant A > 0, such that for x ∈ dom L \ Ker L satisfying |D0
xt| α−2
|D0 xt| > A for all t ∈ 0, 1, we have
0,
Q1 Nxt /
or Q2 Nxt /
0.
3.40
H3 There exists a constant B > 0 such that for every a, b ∈ R satisfying a2 b2 > B then
either
aT1 N atα−1 btα−2 bT2 N atα−1 btα−2 < 0,
3.41
12
Journal of Applied Mathematics
or
aT1 N atα−1 btα−2 bT2 N atα−1 btα−2 > 0.
3.42
Remark 3.4. T1 Natα−1 btα−2 and T2 Natα−1 btα−2 from H3 stand for the images of ut atα−1 btα−2 under the maps T1 N and T2 N, respectively.
Lemma 3.5. Suppose H1-H2 hold, then the set
Ω1 {x ∈ domL \ KerL : Lx λNx, λ ∈ 0, 1}
3.43
Ω1 {x ∈ dom L \ Ker L : Lx λNx, λ ∈ 0, 1}.
3.44
is bounded.
Proof. Take
Then for x ∈ Ω1 , Lx λNx thus λ / 0, Nx ∈ Im L Ker Q, and hence QNxt for all
t ∈ 0, 1. By the definition of Q, we have Q1 Nxt Q2 Nxt 0. It follows from H2 that
α−1
α−2
ut0 | |D0
ut0 | ≤ A.
there exists t0 ∈ 0, 1, such that |D0
Now
α−1
D0
xt
α−2
D0
xt
α−1
D0
xt0 α−2
D0
xt0 t
t0
t
t0
α
D0
xsds,
3.45
α−1
D0
xsds,
and so
α−1
α−1
xt
D0 x0 ≤ D0
∞
α−1 α−1
≤ D0
xt0 D0
x
1
≤ A Lx1 ≤ A Nx1 ,
α−2
α−2
xt
D0 x0 ≤ D0
∞
α−1 α−2
≤ D0
xt0 D0
x
∞
α−1
α−2
α ≤ D0
xt0 D0
xt0 D0
x 1
≤ A Lx1 ≤ A Nx1 .
3.46
Journal of Applied Mathematics
13
Therefore, we have
1
1
α−1
α−1
α−2
α−2 D
D
x0t
x0t
P xCα−1 0
Γα 0
α−1
Γα − 1
C
1
1
α−1
α−1
α−2
α−2 Γα D0 x0t Γα − 1 D0 x0t ∞
α−1
α−1
α−2
D0
x0 D0
x0t D0
x0
∞
3.47
∞
1
1
α−1
α−2
D0 x0 1 D0 x0
Γα
Γα − 1
1
1
≤ 2
A Nx1 1 A Nx1 .
Γα
Γα − 1
≤
2
Note that I − P x ∈ dom L ∩ Ker P for all x ∈ Ω1 . Then, by Lemma 3.2, we have
I − P xCα−1 Kp LI − P xCα−1 ≤
2
1
Nx1 ,
Γα
3.48
so, we have
xCα−1 ≤ I − P xCα−1 P xCα−1
1
1
≤ 2
A Nx1 1 A Nx1 Γα
Γα − 1
1
2
Nx1
Γα
1
2
5 Nx1
Γα Γα − 1
1
1
1
3 A
Γα Γα − 1 Γα − 2
3.49
≤ mNx1 nA,
where m 2/Γα 1/Γα − 1 5, n 1/Γα 1/Γα − 1 1/Γα − 2 3, A
is a constant. This is for all x ∈ Ω1 . If the first condition of H1 is satisfied, then, we have
α−1 α−2 x , D0 x
xCα−1 max x∞ , D0
∞
∞
α−1 α−2 x c1 D0
x
≤ m a1 x∞ b1 D0
∞
α−2 θ
d1 D0 x D ,
∞
∞
3.50
14
Journal of Applied Mathematics
where D r1 e1 n/m, and consequently, for
x∞ ≤ xCα−1 ,
α−1 D0 x ≤ xCα−1 ,
α−2 D0 x ≤ xCα−1 ,
∞
∞
3.51
so
m
α−1 α−2 α−2 θ
x c1 D0
x d1 D0
x D ,
b1 D0
∞
∞
∞
1 − ma1
m
α−2 α−2 θ
α−1 x
≤
x
x
D
,
d
c
D0 1 D0
1 D0
∞
∞
∞
1 − ma1 − mb1
m
α−2 θ
α−2 d1 D0 x D .
D0 x ≤
∞
∞
1 − ma1 − mb1 − mc1
x∞ ≤
3.52
But θ ∈ 0, 1 and a1 b1 c1 ≤ 1/m, so there exists A1 , A2 , A3 > 0 such that
α−2 D0 x ≤ A1 ,
∞
α−1 D0 x ≤ A2 ,
∞
x∞ ≤ A3 .
3.53
Therefore, for all x ∈ Ω1 ,
α−1 α−2 ≤ max{A1 , A2 , A3 },
x , D0 x
xCα−1 max x∞ , D0
∞
∞
3.54
we can prove that Ω1 is also bounded.
If 3.38 or 3.39 holds, similar to the above argument, we can prove that Ω1 is bounded too.
Lemma 3.6. Suppose H3 holds, then the set
Ω2 {x ∈ KerL : Nx ∈ ImL}
3.55
Ω2 {x ∈ Ker L : Nx ∈ Im L},
3.56
is bounded.
Proof. Let
for x ∈ Ω2 , x ∈ Ker L {x ∈ dom L : x atα−1 btα−2 , a, b ∈ R, t ∈ 0, 1} and QNxt 0;
thus T1 Natα−1 btα−2 T2 Natα−1 btα−2 0. By H3, a2 b2 ≤ B, that is, Ω2 is bounded.
Lemma 3.7. Suppose H3 holds, then the set
Ω3 {x ∈ KerL : −λJx 1 − λQNx 0,
is bounded.
λ ∈ 0, 1}
3.57
Journal of Applied Mathematics
15
Proof. We define the isomorphism J : Ker L → Im Q by
J atα−1 btα−2 atα−1 btα−2 .
3.58
If the first part of H3 is satisfied, let
Ω3 {x ∈ Ker L : −λJx 1 − λQNx 0, λ ∈ 0, 1}.
3.59
For every x atα−1 btα−2 ∈ Ω3 ,
λ atα−1 btα−2 1 − λ T1 N atα−1 btα−2 tα−1 T2 N atα−1 btα−2 tα−1 .
3.60
If λ 1, then a b 0, and if a2 b2 > B, then by H3
λ a2 b2 1 − λ aT1 N atα−1 btα−2 bT2 N atα−1 btα−2 < 0,
3.61
which, in either case, is a contradiction. If the other part of H3 is satisfied, then we take
Ω3 {x ∈ Ker L : λJx 1 − λQNx 0, λ ∈ 0, 1},
3.62
and, again, obtain a contradiction. Thus, in either case
xCα−1 atα−1 btα−2 Cα−1
atα−1 btα−2 aΓα∞ aΓαt bΓα − 1∞
∞
3.63
≤ 1 2Γα|a| 1 Γα − 1|b|
≤ 2 2Γα Γα − 1B,
for all x ∈ Ω3 , that is, Ω3 is bounded.
Remark 3.8. Suppose the second part of H3 holds, then the set
Ω3 {x ∈ Ker L : λJx 1 − λQNx 0, λ ∈ 0, 1}
3.64
is bounded.
Theorem 3.9. If C1-C2 and H1–H3 hold, then the boundary value problem 1.4-1.5 has
at least one solution.
Proof. Set Ω to be a bounded open set of Y such that ∪3i1 Ω ⊂ Ω. It follows from Lemmas 3.2
and 3.3 that L is a Fredholm operator of index zero, and the operator Kp I − QN : Ω → Y is
16
Journal of Applied Mathematics
compact N, thus, is L-compact on Ω. By Lemmas 3.5 and 3.6, we get that the following two
conditions are satisfied:
i Lx /
λNx for every x, λ ∈ dom L \ Ker L ∩ ∂Ω × 0, 1;
ii Nx ∈
/ Im L, for every x ∈ Ker L ∩ ∂Ω.
Finally, we will prove that iii of Lemma 1.1 is satisfied. Let Hx, λ ±λJx 1 −
λQNx, where I is the identity operator in the Banach space Y . According to Lemma 3.7 or
Remark 3.8, we know that Hx, λ /
0, for all x ∈ ∂Ω ∩ Ker L, and thus, by the homotopy
property of degree,
degQN|Ker L , Ker L ∩ Ω, 0 degH·, 0, Ker L ∩ Ω, 0
degH·, 1, Ker L ∩ Ω, 0
deg±I, Ker L ∩ Ω, 0
⎤
⎡ Λ
4 −Λ2 Λ1 Λ4 − Λ2 Λ3
⎢ Λ
⎥
Λ
sgn⎣±
⎦ sgn ±
−Λ3 Λ1 Λ
Λ
Λ
3.65
±1 /
0.
Then by Lemma 1.1, Lx Nx has at least one solution in dom L ∩ Ω, so the boundary value
problem 1.4 and 1.5 has at least one solution in the space Cα−1 0, 1. The proof is finished.
4. An Example
Let us consider the following boundary value problem:
5/2
α−1
α−2
D0
ut f t, ut, D0
ut, D0
ut et,
3−α
I0
ut|t0 0,
2
1
1/2
− D0
,
u
3
3
1/2
1/2
D0
u1 2D0
u
0 < t < 1,
u1 64
1
81
1
u
− u
,
5
4
5
9
4.1
where
1
1 3/2
1 1/2
α−1
α−2
f t, ut, D0
sinut D0
ut, D0
ut ut D0
ut
40
20
20
1/5
1/2
5 cos D0
ut
.
4.2
Journal of Applied Mathematics
17
Corresponding to the problem 1.4-1.5, we have that et 1 3sin2 t, α 5/2, a1 −1,
a2 2, ξ1 1/3, ξ2 2/3, b1 64/5, b2 −81/5, η1 1/4, η2 1/9 and
1
1
1
f t, x, y, z sin x y z 5 cos z1/5 ,
40
20
20
4.3
then there is
a1 a2 1,
a1 ξ1 a2 ξ2 1,
b1 η11/2 b2 η21/2 1,
b1 η13/2 b2 η23/2 1,
2
2
4
4
7/2
5/2
,
Λ2 ,
1 − ai ξi
1 − ai ξi
Λ1 35
15
i1
i1
2
2
1 Γ5/2Γ3/2
Γ5/22
4
3
Λ3 Λ4 1 − bi ηi ,
1 − bi ηi ,
24
6
24
i1
i1
4.4
Λ Λ1 Λ4 − Λ2 Λ3 /
0,
f t, x, y, z ≤ 1 |x| 1 y 1 |z| 5|z|1/5 .
40
20
20
Again, taking a 1/40, b c 1/20, then
a1 b1 c1 1/8,
1
1
≈ 0.131,
m 2/Γα 1/Γα − 1 5
4.5
therefore
a1 b1 c1 <
1
.
m
4.6
Take A 181, B 81. By simple calculation, we can get that C1-C2 and H1–H3 hold.
By Lemma 1.1, we obtain that 4.1 has at least one solution.
Acknowledgments
This work is sponsored by NNSF of China 10771212 and the Fundamental Research Funds
for the Central Universities 2010LKSX09.
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