Hindawi Publishing Corporation
International Journal of Differential Equations
Volume 2011, Article ID 401803, 14 pages
doi:10.1155/2011/401803
Research Article
The Existence of Solutions for
a Nonlinear Fractional Multi-Point Boundary
Value Problem at Resonance
Xiaoling Han and Ting Wang
Department of Mathematics, Northwest Normal University, Lanzhou 730070, China
Correspondence should be addressed to Xiaoling Han, hanxiaoling@nwnu.edu.cn
Received 16 May 2011; Accepted 16 June 2011
Academic Editor: Nikolai Leonenko
Copyright q 2011 X. Han and T. Wang. 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 solution for a multipoint boundary value problem of fractional differential equation. An existence result is obtained with the use of the coincidence degree theory.
1. Introduction
In this paper, we study the multipoint boundary value problem
α
α−1
α−2
D0
ut f t, ut, D0
ut, D0
ut et,
3−α
I0
u0 0,
α−2
D0
u0 n
α−2
βj D0
u ξj ,
0 < t < 1,
u1 j1
m
αi u ηi ,
1.1
1.2
i1
where 2 < α ≤ 3, 0 < ξ1 < ξ2 < · · · < ξn < 1, n ≥ 1, 0 < η1 < · · · < ηm < 1, m ≥ 2, αi , βj ∈ R,
m
i1
αi ηiα−1 m
αi ηiα−2 1,
n
βj ξj 0,
n
i1
j1
j1
βj 1,
1.3
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α
α
and I0
are the
f : 0, 1 × R3 → R satisfying the Carathéodory conditions, e ∈ L1 0, 1. D0
standard Riemann-Liouville derivative and integral, respectively. We assume, in addition,
that
n
m
Γα2 Γα − 1 α
2α−1
βj ξ 1 − αi ηi
R
Γ2αΓα 1 j1 j
i1
n
m
Γα2 Γα − 1 α1
2α−2
1 − αi ηi
−
βj ξ
Γα 2Γ2α − 1 j1 j
i1
1.4
/ 0,
where Γ is the Gamma function. Due to condition 1.3, the fractional differential operator in
1.1, 1.2 is not invertible.
Fractional differential equation can describe many phenomena in various fields
of science and engineering. Many methods have been introduced for solving fractional
differential equations, such as the popular Laplace transform method, the iteration method.
For details, see 1, 2 and the references therein.
Recently, there are some papers dealing with the solvability of nonlinear boundary
value problems of fractional differential equation, by use of techniques of nonlinear analysis
fixed-point theorems, Leray-Schauder theory, etc., see, for example, 3–6. But there are few
papers that consider the fractional-order boundary problems at resonance. Very recently 7,
Y. H. Zhang and Z. B. Bai considered the existence of solutions for the fractional ordinary
differential equation
α−n−1
α
α−1
ut f t, ut, D0
ut, . . . , D0
ut et,
D0
0 < t < 1,
1.5
subject to the following boundary value conditions:
α−n−1
n−α
I0
u0 D0
α−2
u0 · · · D0
u0 0,
u1 σu η ,
1.6
where n > 2 is a natural number, n − 1 < α ≤ n is a real number, f : 0, 1 × Rn → R is continuous, and e ∈ L1 0, 1, σ ∈ 0, ∞, and η ∈ 0, 1 are given constants such that σηα−1 1.
α
α
and I0
are the standard Riemann-Liouville derivative and integral, respectively. By the
D0
conditions, the kernel of the linear operator is one dimensional.
Motivated by the above work and recent studies on fractional differential equations
8–18, in this paper, we consider the existence of solutions for multipoint boundary value
problem 1.1, 1.2 at resonance. Note that under condition 1.3, the kernel of the linear
operator in 1.1, 1.2 is two dimensional. Our method is based upon the coincidence degree
theory of Mawhin 18.
Now, we will briefly recall some notation and abstract existence result.
Let Y, Z be real Banach spaces, let L : domL ⊂ Y → Z be a Fredholm map of
index zero, and let P : Y → Y, Q : Z → Z be continuous projectors such that ImP KerP , KerQ ImL, and Y KerL ⊕ KerP , Z ImL ⊕ ImQ. It follows that
L|domL∩KerP : domL ∩ KerP → ImL is invertible. We denote the inverse of the map by
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3
∅, the map N : Y → Z will be
KP . If Ω is an open-bounded subset of Y such that domL∩Ω /
called L-compact on Ω if QNΩ is bounded and KP I − QN : Ω → Y is compact.
The theorem that we used is Theorem 2.4 of 18.
Theorem 1.1. Let L be a Fredholm operator of index zero and N be L-compact on Ω. Assume that the
following conditions are satisfied:
i Lx / λNx for every x, λ ∈ domL \ KerL ∩ ∂Ω × 0, 1,
ii Nx ∈
/ ImL for every x ∈ KerL ∩ ∂Ω,
iii degJQN|KerL , Ω ∩ KerL, 0 /
0, where Q : Z → Z is a projection as above with
ImL KerQ, and J : ImQ → KerL is any isomorphism,
then the equation Lx Nx has at least one solution in domL ∩ Ω.
The rest of this paper is organized as follows. In Section 2, we give some notation and
Lemmas. In Section 3, we establish an existence theorem of a solution for the problem 1.1,
1.2.
2. Background Materials and Preliminaries
For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory, and these definitions can be found in the recent literature
1, 2.
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
0
provided the right side is pointwise defined on 0, ∞. And we let I0
yt yt for every
continuous y : 0, ∞ → R.
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, ∞.
Lemma 2.3 see 3. 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
I0
D0
ut ut C1 tα−1 C2 tα−2 · · · CN tα−N ,
for some Ci ∈ R, i 1, 2, . . . , N, where N is the smallest integer greater than or equal to α.
2.3
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We use the classical space C0, 1 with the norm x∞ maxt∈0,1 |xt|. Given μ > 0
and N μ 1, one can define a linear space
μ
Cμ 0, 1 : u | ut I0 xt c1 tμ−1 c2 tμ−2 · · · cN−1 tμ−N−1 , t ∈ 0, 1 ,
2.4
where x ∈ C0, 1 and ci ∈ R, i 1, 2, . . . , N − 1. By means of the linear function analysis
μ
μ−N−1
u∞ u∞ , Cμ 0, 1
theory, one can prove that with the norm uCμ D0 u∞ · · ·D0
is a Banach space.
Lemma 2.4 see 7. F ⊂ Cμ 0, 1 is a sequentially compact set if and only if F is uniformly bounded
and equicontinuous. Here, uniformly bounded means that there exists M > 0 such that for every
u ∈ F,
μ μ−N−1 u u∞ < M,
2.5
uCμ D0 u · · · D0
∞
∞
and equicontinuous means that for all ε > 0, ∃δ > 0 such that
|ut1 − ut2 | < ε, ∀t1 , t2 ∈ 0, 1, |t1 − t2 | < δ, ∀u ∈ F,
2.6
α−i
α−i
ut2 < ε, t1 , t2 ∈ 0, 1, |t1 − t2 | < δ, ∀u ∈ F, ∀i ∈ {0, . . . , N − 1}.
D0 ut1 − D0
1
α−1
xt Let Z L1 0, 1 with the norm g1 0 |gs|ds. Y Cα−1 0, 1 {u | ut I0
α−1
α−2
α−2
ct , t ∈ 0, 1}, where x ∈ C0, 1, c ∈ R, with the norm uCα−1 D0 u∞ D0 u∞ u∞ , and Y is a Banach space.
Definition 2.5. By a solution of the boundary value problem 1.1, 1.2, we understand a
α−1
u is absolutely continuous on 0, 1 and satisfies 1.1,
function u ∈ Cα−1 0, 1 such that D0
1.2.
Definition 2.6. We say that the map f : 0, 1 × R → R satisfies the Carathéodory conditions
with respect to L1 0, 1 if the following conditions are satisfied:
i for each z ∈ R, the mapping t → ft, z is Lebesgue measurable,
ii for almost every t ∈ 0, 1, the mapping z → ft, z is continuous on R,
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.
Define L to be the linear operator from domL ∩ Y to Z with
α
domL u ∈ Cα−1 0, 1 | D0
u ∈ L1 0, 1, u satisfies 1.2 ,
α
Lu D0
u,
2.7
u ∈ domL.
We define N : Y → Z by setting
α−1
α−2
Nut f t, ut, D0
ut, D0
ut et.
2.8
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5
Then boundary value problem 1.1, 1.2 can be written as
Lu Nu.
2.9
Lemma 2.7. Let condition 1.3 and 1.4 hold, then L : domL ∩ Y → Z is a Fredholm map of
index zero.
Proof. It is clear that KerL {atα−1 btα−2 | a, b ∈ R} ∼
R2 .
Let g ∈ Z and
1
ut Γα
t
t − sα−1 gsds c1 tα−1 c2 tα−2 ,
2.10
0
α
then D0
ut gt a.e., t ∈ 0, 1 and, if
1
1 − sα−1 gsds −
0
m
ηi
αi
ξj
βj
ηi − s
α−1
gsds 0,
0
i1
n
2.11
ξj − s gsds 0
0
j1
hold. Then ut satisfies the boundary conditions 1.2, that is, u ∈ domL, and we have
g ∈ Z | g satisfies 2.11 ⊆ ImL.
2.12
α
Let u ∈ domL, then for D0
u ∈ ImL, we have
α
α
ut I0
D0
ut c1 tα−1 c2 tα−2 c3 tα−3 ,
2.13
α
which, due to the boundary value condition 1.2, implies that D0
u satisfies 2.11. In fact,
m
3−α
from I0 u0 0, we have c3 0, from u1 i1 αi uηi , we have
1
1 − s
0
α−2
and from D0
u0 α−1
α
D0
usds
−
m
ηi
αi
i1
n
j1
0
ηi − s
α−1
α
D0
usds 0,
2.14
α−2
βj D0
uξj , we have
ξj
n
α
βj
usds 0.
ξj − s D0
j1
2.15
0
Hence,
ImL ⊆ g ∈ Z | g satisfies 2.11 .
2.16
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International Journal of Differential Equations
Therefore,
ImL g ∈ Z | g satisfies 2.11 .
2.17
Consider the continuous linear mapping Q1 : Z → Z and Q2 : Z → Z defined by
Q1 g 1
1 − sα−1 gsds −
0
m
ηi
i1
α−1
ηi − s
αi
gsds,
0
ξj
n
βj
Q2 g ξj − s gsds.
2.18
0
j1
Using the above definitions, we construct the following auxiliary maps R1 , R2 : Z → Z:
⎤
⎡
n
m
1 ⎣ Γα − 1 ΓαΓα − 1
α
2α−2
Q2 gt⎦,
1 − αi ηi
βj ξ Q1 gt −
R1 g R Γα 1 j1 j
Γ2α − 1
i1
⎡
n
Γα 1
Γα2
R2 g − ⎣
βj ξjα1 Q1 gt −
R Γα 2 j1
Γ2α
⎤
m
2α−1
Q2 gt⎦.
1 − αi ηi
2.19
i1
Since the condition 1.4 holds, the mapping Q : Z → Z defined by
Qy t R1 gt tα−1 R2 gt tα−2
2.20
is well defined.
Recall 1.4 and note that
R1 R1 gt
α−1
⎡
n
1 ⎣ Γα − 1 βj ξjα Q1 R1 gtα−1
R Γα 1 j1
−
ΓαΓα − 1
Γ2α − 1
⎤
m
1 − αi ηi2α−2 Q2 R1 gtα−1 ⎦
i1
⎡
n
m
1 ⎣ Γα − 1Γ α2 α
2α−1
βj ξ 1 − αi ηi
R1 g
R Γα 1Γ2α j1 j
i1
⎤
m
n
Γα − 1Γ α2
1 − αi ηi2α−2
−
βj ξjα1 ⎦
Γ2α − 1Γα 2
i1
j1
R1 g,
2.21
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7
and similarly we can derive that
R1 R2 gtα−2 0,
R2 R1 gtα−1 0,
2.22
R2 R2 gtα−2 R2 g.
So, for g ∈ Z, it follows from the four relations above that
Q2 g R1 R1 gtα−1 R2 gtα−2 tα−1 R2 R1 gtα−1 R2 gtα−2 tα−2
R1 R1 gtα−1 tα−1 R1 R2 gtα−2 tα−1 R2 R1 gtα−1 tα−2 R2 R2 gtα−2 tα−2
R1 gt
α−1
R2 gt
2.23
α−2
Qg,
that is, the map Q is idempotent. In fact, Q is a continuous linear projector.
Note that g ∈ ImL implies Qg 0. Conversely, if Qg 0, then we must have
R1 g R2 g 0; since the condition 1.4 holds, this can only be the case if Q1 g Q2 g 0, that
is, g ∈ ImL. In fact, ImL KerQ.
Take g ∈ Z in the form g g − Qg Qg, so that g − Qg ∈ ImL KerQ and
Qg ∈ ImQ. Thus, Z ImL ImQ. Let g ∈ ImL ∩ ImQ and assume that gs asα−1 bsα−2 is not identically zero on 0, 1, then, since g ∈ ImL, from 2.11 and the condition
1.4, we derive a b 0, which is a contradiction. Hence, ImL ∩ ImQ {0}; thus,
Z ImL ⊕ ImQ.
Now, dim KerL 2 co dim ImL, 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 u0tα−2 ,
u0tα−1 Γα
Γα − 1 0
t ∈ 0, 1.
2.24
Note that P is a continuous linear projector and
α−1
α−2
KerP u ∈ Y | D0
u0 D0
u0 0 .
2.25
It is clear that Y KerL ⊕ KerP .
Note that the projectors P and Q are exact. Define KP : ImL → domL ∩ KerP by
1
KP gt Γα
t
0
α
gt.
t − sα−1 gsds I0
2.26
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International Journal of Differential Equations
Hence, we have
α−1
D0
KP g t t
α−2
D0
KP g t gsds,
0
t
t − sgsds,
2.27
0
α−1
α−2
KP g∞ ≤ g1 , D0
KP g∞ ≤ g1 , and thus
then KP g∞ ≤ 1/Γαg1 , D0
KP g α−1 ≤
C
2
1
g .
1
Γα
2.28
α α
In fact, if g ∈ ImL, then LKP g D0
I0 g g. Also, if u ∈ domL ∩ KerP , then
α
α
D0
gt gt c1 tα−1 c2 tα−2 c3 tα−3 ,
KP Lg t I0
2.29
from boundary value condition 1.2 and the fact that u ∈ domL ∩ KerP , we have c1 c2 c3 0. Thus,
−1
KP L|domL∩KerP .
2.30
QNut R1 Nutα−1 R2 Nutα−2 ,
1
1
KP I − QNut t − sα−1 Nus − QNusds.
Γα 0
2.31
Using 2.19, we write
With arguments similar to those of 7, we obtain the following Lemma.
Lemma 2.8. KP I−Q N : Y → Y is completely continuous.
3. The Main Results
Assume that the following conditions on the function ft, x, y, z are satisfied:
H1 there exists a constant A > 0, such that for u ∈ domL \ KerL satisfying
α−1
α−2
ut| |D0
ut| > A for all t ∈ 0, 1, we have
|D0
Q1 Nut /
0
or Q2 Nut /
0,
3.1
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H2 there exist functions a, b, c, d, r ∈ L1 0, 1 and a constant θ ∈ 0, 1 such that for
all x, y, z ∈ R3 and a.e., t ∈ 0, 1, one of the following inequalities is satisfied:
f t, x, y, z ≤ at|x| bty ct|z| dt|z|θ rt,
f t, x, y, z ≤ at|x| bty ct|z| dtyθ rt,
f t, x, y, z ≤ at|x| bty ct|z| dt|x|θ rt,
3.2
H3 there exists a constant B > 0 such that for every a, b ∈ R satisfying a2 b2 > B,
then either
aR1 N atα−1 btα−2 bR2 N atα−1 btα−2 < 0,
3.3
aR1 N atα−1 btα−2 bR2 N atα−1 btα−2 > 0.
3.4
or else
Remark 3.1. R1 Natα−1 btα−2 and R2 Natα−1 btα−2 from H3 stand for the images of
ut atα−1 btα−2 under the maps R1 N and R2 N, respectively.
Theorem 3.2. If (H1)–(H3) hold, then boundary value problem 1.1-1.2 has at least one solution
provided that
a1 b1 c1 <
1
,
τ
3.5
where τ 5 2/Γα 1/Γα − 1.
Proof. Set
Ω1 {u ∈ domL \ KerL | Lu λNu for some λ ∈ 0, 1},
3.6
0, Nu ∈ ImL KerQ, and hence QNut 0 for all
then for u ∈ Ω1 , Lu λNu; thus, λ /
t ∈ 0, 1. By the definition of Q, we have Q1 Nut Q2 Nut 0. It follows from H1 that
α−1
α−2
ut0 | |D0
ut0 | ≤ A. Now,
there exists t0 ∈ 0, 1 such that |D0
α−1
D0
ut
α−2
ut
D0
α−1
D0
ut0 α−2
D0
ut0 t
t0
t
t0
α
D0
usds,
3.7
α−1
D0
usds,
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so
α−1
α−1
ut
D0 u0 ≤ D0
∞
α−1
α ≤ D0
ut0 D0
u 1
≤ A Lu1
≤ A Nu1 ,
α−2
α−2
ut
D0 u0 ≤ D0
3.8
∞
α−1 α−2
≤ D0
ut0 D0
u
∞
α−1
α−2
α ≤ D0
ut0 D0
ut0 D0
u 1
≤ A Lu1
≤ A Nu1 .
Now by 3.8, we have
P uCα−1
1
1
α−1
α−1
α−2
α−2 D u0t D u0t Γα 0
Γα − 1 0
Cα−1
1
1
α−1
α−1
α−2
α−2 Γα D0 u0t Γα − 1 D0 u0t ∞
α−1
α−1
α−2
D0
u0 D0
u0t D0
u0
∞
∞
3.9
1
1
α−1
α−2
D0 u0 1 D0 u0
Γα
Γα − 1
1
1
≤ 2
A Nu1 1 A Nu1 .
Γα
Γα − 1
≤
2
Note that I − P u ∈ ImKP domL ∩ KerP for u ∈ Ω1 , then, by 2.28 and 2.30,
I − P uCα−1 KP LI − P Cα−1
1
≤ 2−
LI − P u1
Γα
1
2−
Lu1
Γα
1
≤ 2−
Nu1 .
Γα
3.10
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11
Using 3.9 and 3.10, we obtain
uCα−1 P u I − P uCα−1
≤ P uCα−1 I − P uCα−1
1
1
1
A Nu1 1 A Nu1 2 Nu1
Γα
Γα − 1
Γα
1
1
2
1
A
5
Nu1 3 Γα Γα − 1
Γα Γα − 1
≤
2
τNu1 C1 ,
3.11
where C1 3 1/Γα 1/Γα − 1A is a constant. This is for all u ∈ Ω1 ,
uCα−1 ≤ τNu1 C1 .
3.12
If the first condition of H2 is satisfied, then we have
α−1 α−2 u , D0 u
max u∞ , D0
∞
∞
α−1 α−2 ≤ uCα−1 ≤ τ a1 u∞ b1 D0
u c1 D0
u
∞
∞
3.13
α−2 θ
d1 D0
u r1 e1 C1 ,
∞
and consequently,
u∞ ≤
τ
α−1 α−2 α−2 θ
u c1 D0
u d1 D0
u r1 e1
b1 D0
∞
∞
∞
1 − a1 τ
C1
,
1 − a1 τ
α−1 D0 u ≤
∞
τ
α−2 α−2 θ
u d1 D0
u r1 e1
c1 D0
∞
∞
1 − a1 τ − b1 τ
C1
,
1 − a1 τ − b1 τ
α−1 D0 u ≤
∞
α−2 θ
τd1 D0
u∞
1 − a1 τ − b1 τ − c1 τ
τr1 e1 C1
.
1 − a1 τ − b1 τ − c1 τ
3.14
3.15
3.16
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α−1
u∞ ≤
Note that θ ∈ 0, 1 and a1 b1 c1 < 1/τ, so there exists M1 > 0 such that D0
M1 for all u ∈ Ω1 . The inequalities 3.14 and 3.15 show that there exist M2 , M3 > 0 such
α−1
u∞ ≤ M2 , u∞ ≤ M3 for all u ∈ Ω1 . Therefore, for all u ∈ Ω1 , uCα−1 u∞ that D0
α−1
α−2
u∞ ≤ M1 M2 M3 , that is, Ω1 is bounded given the first condition of
D0 u∞ D0
H2. If the other conditions of H2 hold, by using an argument similar to the above, we can
prove that Ω1 is also bounded.
Let
Ω2 {u ∈ KerL | Nu ∈ ImL}.
3.17
For u ∈ Ω2 , u ∈ KerL {u ∈ domL | u atα−1 btα−2 , a, b ∈ R, t ∈ 0, 1}, and QNatα−1 btα−2 0; thus, R1 Natα−1 btα−2 R2 Natα−1 btα−2 0. By H3, a2 b2 ≤ B, that is, Ω2
is bounded.
We define the isomorphism J : ImQ → KerL by
J atα−1 btα−2 atα−1 btα−2 ,
a, b ∈ R.
3.18
If the first part of H3 is satisfied, let
Ω3 u ∈ Ker L : −λJ −1 u 1 − λQNu 0, λ ∈ 0, 1 .
3.19
For every atα−1 btα−2 ∈ Ω3 ,
λ atα−1 btα−2 1 − λ R1 N atα−1 btα−2 tα−1 R2 N atα−1 btα−2 tα−2 .
3.20
If λ 1, then a b 0, and if a2 b2 > B, then by H3,
λ a2 b2 1 − λ aR1 N atα−1 btα−2 bR2 N atα−1 btα−2 < 0,
3.21
which, in either case, obtain a contradiction. If the other part of H3 is satisfied, then we take
Ω3 u ∈ Ker L : λJ −1 u 1 − λQNu 0, λ ∈ 0, 1 ,
3.22
and, again, obtain a contradiction. Thus, in either case,
α−1 α−2 u D0
u
uCα−1 u∞ D0
atα−1 btα−2 ∞
∞
Cα−1
atα−1 btα−2 aΓα∞ aΓαt bΓα − 1∞
∞
≤ 1 2Γα|a| 1 Γα − 1|b|
≤ 2 2Γα Γα − 1|a|,
3.23
International Journal of Differential Equations
13
for all u ∈ Ω3 , that is, Ω3 is bounded.
In the following, we will prove that all the conditions of Theorem 1.1 are satisfied.
3
Set Ω to be a bounded open set of Y such that Ui1
Ωi ⊂ Ω. by Lemma 2.8, the operator
KP I − QN : Ω → Y is compact; thus, N is L-compact on Ω, then by the above argument,
we have
i Lu /
λNx, for every u, λ ∈ domL \ Ker L ∩ ∂Ω × 0, 1,
ii Nu ∈
/ ImL, for every u ∈ KerL ∩ ∂Ω.
Finally, we will prove that iii of Theorem 1.1 is satisfied. Let Hu, λ ±Iu 1 − λJQNu,
where I is the identity operator in the Banach space Y . According to the above argument, we
know that
Hu, λ / 0,
∀u ∈ ∂Ω ∩ KerL,
3.24
and thus, by the homotopy property of degree,
deg JQN|KerL , Ω ∩ KerL, 0
degH. . . , 0, Ω ∩ KerL, 0
degH. . . , 1, Ω ∩ KerL, 0
3.25
deg±I, Ω ∩ KerL, 0
±1 /
0,
then by Theorem 1.1, Lu Nu has at least one solution in domL ∩ Ω, so boundary problem
1.1, 1.2 has at least one solution in the space Cα−1 0, 1. The proof is finished.
Acknowledgments
This work is supported by the NSFC no. 11061030, no. 11026060, the Fundamental Research
Funds for the Gansu Universities, the nwnu-kjcxgc-03-69, nwnu-kjcxgc-03-61.
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