Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 58, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SOLVABILITY OF FRACTIONAL MULTI-POINT
BOUNDARY-VALUE PROBLEMS WITH p-LAPLACIAN
OPERATOR AT RESONANCE
TENGFEI SHEN, WENBIN LIU, TAIYONG CHEN, XIAOHUI SHEN
Abstract. In this article, we consider the multi-point boundary-value problem for nonlinear fractional differential equations with p-Laplacian operator:
α−1
α
D0β+ ϕp (D0α+ u(t)) = f (t, u(t), D0α−2
+ u(t), D0+ u(t), D0+ u(t)),
u(0) = u0 (0) = D0α+ u(0) = 0,
D0α−1
+ u(1) =
m
X
t ∈ (0, 1),
σi D0α−1
+ u(ηi ),
i=1
P
α
where 2 < α ≤ 3, 0 < β ≤ 1, 3 < α + β ≤ 4, m
i=1 σi = 1, D0+ is the standard
p−2
Riemann-Liouville fractional derivative. ϕp (s) = |s|
s is p-Laplacians operator. The existence of solutions for above fractional boundary value problem
is obtained by using the extension of Mawhin’s continuation theorem due to
Ge, which enrich konwn results. An example is given to illustrate the main
result.
1. Introduction
In recent years, fractional differential equations play a important role in many
fields such as physics, engineering, biology, control theory, etc., see [1, 12, 15, 17,
18]. It has been studied extensively by scholars have obtained many results, see
[2, 5, 10, 11, 14, 19, 22].
However, the existence of solutions for fractional boundary value problems at
resonance is less studied, see [3, 4, 7, 9, 20, 21]. There are few articles which consider
the boundary value problems (BVPs for shorts) at resonance for nonlinear fractional
differential equation with p-Laplacian operator. In 2012, Chen, Liu and Hu [6]
considered existence of solutions of boundary value problems for a Caputo fractional
differential equation with p-Laplacian operator at resonance by coincidence degree
theory by Mawhin:
D0β+ ϕp (D0α+ u(t)) = f (t, u(t), D0α+ u(t)),
t ∈ (0, 1),
D0α+ u(0)
= D0α+ u(1) = 0,
≤ 2, D0α+ is a Caputo
2
(1.1)
where 0 < α, β < 1, 1 < α + β
fractional derivative,
ϕp (s) = |s|p−2 s is a p-Laplacian operator, f : [0, 1] × R → R is continuous.
2000 Mathematics Subject Classification. 34A08, 34B15.
Key words and phrases. Fractional differential equation; boundary value problem;
p-Laplacian operator; Coincidence degree theory; Resonance.
c
2014
Texas State University - San Marcos.
Submitted January 13, 2013. Published February 28, 2014.
1
2
T. SHEN, W. LIU, T. CHEN, X. SHEN
EJDE-2014/58
In this article, we study fractional multi-point boundary value problem with pLaplacian operator at resonance by using the extension of Mawhin’s continuation
theorem due to Ge,
α−1
α
D0β+ ϕp (D0α+ u(t)) = f (t, u(t), D0α−2
+ u(t), D0+ u(t), D0+ u(t)),
u(0) = u0 (0) = D0α+ u(0) = 0,
D0α−1
+ u(1) =
m
X
t ∈ (0, 1),
σi D0α−1
+ u(ηi ),
(1.2)
i=1
Pm
where 2 < α ≤ 3, 0 < β ≤ 1, 3 < α + β ≤ 4, ηi ∈ (0, 1), σi ∈ R, i=1 σi = 1,
1 < m, m ∈ N , ϕp (s) = |s|p−2 s, 1 < p, 1/p + 1/q = 1, ϕp is invertible and its
inverse operator is ϕq , D0α+ is Riemann-Liouville standard fractional derivative,
f : [0, 1] × R4 → R is continuous.
There are few articles to investigate fractional multi-point boundary value problem with p-Laplacian operator at resonance. By constructing suitable continuous
linear projectors and using the extension of Mawhin’s continuation theorem due
to Ge, the existence of solutions were obtained. Our paper perfect and generalize
some known results.
To investigate the problem, we use the condition
m
X
1
σi ηiqβ−β+1 ) 6= 0.
(1
−
∆=
Γ(β + 1)q−1 (qβ − β + 1)
i=1
The rest of this article is organized as follows: In Section 2, we give some notations, definitions and Lemmas. In Section 3, basing on the extension of Mawhin’s
continuation theorem due to Ge, we establish a theorem of existence result for BVP
(1.2).
2. Preliminaries
For the convenience of the reader, we present here some basic knowledge and
definitions for fractional calculus theory, that can be found in [2, 8, 12].
Let X and Y be two Banach spaces with norms k · kX and kukY , respectively.
A continuous operator
M |dom M ∩X : X ∩ dom M → Y
is said to be quasi-linear if
(i) Im M := M (X ∩ dom M ) is a closed subset of Y ,
(ii) ker M := {u ∈ X ∩ dom M : M u = 0} is is linearly homeomorphic to Rn ,
n < ∞.
Let X1 = ker M and X2 be the complement space of X1 in X, then X = X1 ⊕X2 .
On the other hand, suppose Y1 is a subspace of Y and Y2 is the complement space
of Y1 in Y so that Y = Y1 ⊕ Y2 . Let P : X → X1 be a projector and Q : Y → Y1 a
semi-projector, and Ω ⊂ X an open and bounded set with origin θ ∈ Ω. Where θ
is the origin of a linear space.
Suppose Nλ : Ω → Y , λ ∈ [0, 1] is a continuous operator. Denote N1 by N . Let
Σλ = {u ∈ Ω : M u = Nλ u}. Nλ is said to be M -compact in Ω if there is a Y1 ⊂ Y
with dim Y1 = dim X1 and an operator R : Ω × [0, 1] → X continuous and compact
such that for λ ∈ [0, 1],
(I − Q)Nλ (Ω) ⊂ Im M ⊂ (I − Q)Y,
(2.1)
QNλ x = θ, λ ∈ (0, 1) ⇔ QN x = θ,
(2.2)
EJDE-2014/58
FRACTIONAL MULTI-POINT BOUNDARY-VALUE PROBLEMS
R(·, λ) |Σλ = (I − P ) |Σλ
3
(2.3)
and R(·, 0) is the zero operator,
M [P + R(·, λ)] = (I − Q)Nλ .
(2.4)
Lemma 2.1 ([8]). Let (X, k · kX ) and (Y, k · kY ) be two Banach spaces, and Ω ⊂ X
an open and bounded nonempty set. Suppose M : X ∩ dom M → Y is a quasi-linear
operator Nλ : Ω → Y , λ ∈ [0, 1] is M -compact in Ω. In addition, if:
(i) M u 6= Nλ u for all (u, λ) ∈ (dom M ∩ ∂Ω) × (0, 1),
(ii) QN u 6= 0 for all u ∈ ∂Ω ∩ ker M ,
(iii) deg(JQN, ker M ∩ Ω, 0) 6= 0, where J : Im Q → ker M is a homeomorphism
with J(θ) = θ and N = N1 ,
then the equation M u = N u has at least one solution in dom M ∩ Ω̄.
Definition 2.2. The Riemann-Liouville fractional integral of order α > 0 of a
function u is given by
Z t
1
I0α+ u(t) =
(t − s)α−1 u(s)ds,
Γ(α) 0
provided that the right side integral is pointwise defined on (0, +∞).
Definition 2.3. The Riemann-Liouville fractional derivative of order α > 0 of a
function u is given by
Z
d n t
u(s)
1
α
( )
ds,
D0+ u(t) =
α−n+1
Γ(n − α) dt
0 (t − s)
provided that the right side integral is pointwise defined on (0, +∞).
Lemma 2.4. Assume that u ∈ C(0, 1) ∩ L1 (0, 1) with a fractional derivative of
order α > 0 that belongs to C(0, 1) ∩ L1 (0, 1). Then
α
α
I0+
D0+
u(t) = u(t) + c1 tα−1 + c2 tα−2 + · · · + cN tα−N ,
for some ci ∈ R, i = 1, 2, . . . , N , where N is the smallest integer grater than or
equal to α.
Lemma 2.5. Assume u(t) ∈ C[0, 1] and 0 ≤ β ≤ α, then D0β+ I0α+ u(t) = I0α−β
u(t).
+
And, for all α ≥ 0, β > −1, we have
D0α+ tβ =
Γ(β + 1) β−α
t
,
Γ(β − α + 1)
giving in particular D0α+ tα−m = 0, m = 1, 2, . . . , N , where N is the smallest integer
grater than or equal to α.
α−1
α
In this article, we take X = {u|u, D0α−2
+ u, D0+ u, D0+ u ∈ C[0, 1]} with the norm
α−2
α−1
α
kukX = max{kuk∞ , kD0+ uk∞ kD0+ uk∞ , kD0+ uk∞ }, where kuk∞ = maxt∈[0,1]
|u(t)|, and Y = C[0, 1] with the norm kykY = kyk∞ . By means of the linear
functional analysis theory, it is easy to prove that X and Y are Banach spaces. so,
we omit it.
Define the operator M : dom M ⊂ X → Y by
M u = D0β+ ϕp (D0α+ u(t)),
(2.5)
4
T. SHEN, W. LIU, T. CHEN, X. SHEN
EJDE-2014/58
n
dom M = u ∈ X : D0β+ ϕp (D0α+ u) ∈ Y, u(0) = u0 (0) = D0α+ u(0) = 0,
D0α−1
+ u(1) =
m
X
o
σi D0α−1
+ u(ηi ) .
(2.6)
i=1
Define the operator Nλ : X → Y , λ ∈ [0, 1],
α−1
α
Nλ u(t) = f (t, u(t), D0α−2
+ u(t), D0+ u(t), D0+ u(t)), t ∈ [0, 1],
then (1.2) is equivalent to the operator equation M u = N u, where N = N1 .
3. Main result
In this section, we show existence of solutions for BVP (1.2). Let us make some
assumptions which will be used throughout this article.
(H1) There exist nonnegative functions a, b, c, d, e ∈ Y such that
|f (t, u, v, w, z))| ≤ a(t) + b(t)|u|p−1 + c(t)|v|p−1 + d(t)|w|p−1 + e(t)|z|p−1 ,
for all t ∈ [0, 1], (u, v, w, z) ∈ R4 .
(H2) There exists a constant A > 0 such that
Z s
Z 1 1
(s − τ )β−1 f (τ, u, v, w, z)dτ ds
ϕq
Γ(β) 0
0
Z s
Z ηi
m
X
1
(s − τ )β−1 f (τ, u, v, w, z)dτ )ds 6= 0,
−
σi
ϕq (
Γ(β)
0
0
i=1
for all t ∈ [0, 1], (u, v, w, z) ∈ R4 , |v| + |w| > A.
(H3) There exists a constant B > 0 such that
Z
Z s
1
1 1
ϕq (
(s − τ )β−1 f (τ, cτ α−1 , cΓ(α)τ, cΓ(α), 0)dτ )ds
0 6= Λ := c
∆ 0
Γ(β) 0
Z s
Z ηi
m
X
1
−
σi
(s − τ )β−1 f (τ, cτ α−1 , cΓ(α)τ, cΓ(α), 0)dτ )ds ,
ϕq (
Γ(β) 0
0
i=1
for all |c| > B, c ∈ R.
Theorem 3.1. Let f : [0, 1] × R4 → R be continuous and the condition (H1)–(H3)
hold. Then BVP (1.2) has at least one solution provided that
Dkbk
1
∞
+
Dkck
+
Dkdk
+
kek
< 1.
(3.1)
∞
∞
∞
Γ(β + 1) Γ(α)p−1
Lemma 3.2. The operator M : dom M ∩ X → Y is a quasi-linear, and
ker M = {u ∈ X : u(t) = ctα−1 , ∀t ∈ [0, 1], c ∈ R}
Z s
Z 1
n
1
Im M = y ∈ Y :
ϕq (
(s − τ )β−1 y(τ )dτ )ds
Γ(β)
0
0
Z ηi
Z s
m
o
X
1
−
σi
ϕq (
(s − τ )β−1 y(τ )dτ )ds = 0
Γ(β) 0
0
i=1
(3.2)
(3.3)
EJDE-2014/58
FRACTIONAL MULTI-POINT BOUNDARY-VALUE PROBLEMS
5
Proof. By Lemma 2.4 and D0β+ ϕp (D0α+ u(t)) = 0, we have
D0α+ u(t) = ϕq (c0 tβ−1 ).
From condition D0α+ u(0) = 0, we obtain that c0 = 0. Thus,
u(t) = c1 tα−1 + c2 tα−2 + c3 tα−3 .
Combined with u(0) = u0 (0) = 0 , we have c2 = c3 = 0, u(t) = c1 tα−1 , c1 ∈ R.
Thus, (3.2) is satisfied.
If y ∈ Im M , then there exists a function u ∈ dom M such that
y(t) = D0β+ ϕp (D0α+ u(t)).
Then by Lemma 2.4 and boundary value condition, we have
u(t) = I0α+ ϕq (I0β+ y(s)) + c1 tα−1 ,
β
α−1 α
D0α−1
+ u(t) = D0+ I0+ ϕq (I0+ y(s)) + c1 Γ(α).
Pm
Combing this with i=1 σi = 1, we obtain
Z s
Z 1
1
(s − τ )β−1 y(τ )dτ )ds
ϕq (
Γ(β)
0
0
Z ηi
Z s
m
X
1
−
σi
ϕq (
(s − τ )β−1 y(τ )dτ )ds = 0.
Γ(β)
0
0
i=1
On the other hand, suppose y ∈ Y and satisfies (3.3). Let u(t) = I0α+ ϕq (I0β+ y(t)),
then u ∈ dom M and M u(t) = D0β+ ϕp (D0α+ u(t)) = y(t). so y ∈ Im M and Im M :=
M (dom M ) is a closed subset of Y . Thus, M is a quasi-linear operator.
Lemma 3.3. Let Ω ⊂ X be an open and bounded set, then Nλ is M -compact in Ω.
Proof. Define the continuous projectors P : X → X1 and Q : Y → Y1 by
1
α−1
Dα−1
, t ∈ [0, 1],
P u(t) =
+ u(0)t
Γ(α) 0
Z 1
Z s
1
1
Qy(t) = ϕp ( (
ϕq (
(s − τ )β−1 y(τ )dτ )ds
∆ 0
Γ(β) 0
Z ηi
Z s
m
X
1
σi
ϕq (
−
(s − τ )β−1 y(τ )dτ )ds)), t ∈ [0, 1].
Γ(β)
0
0
i=1
Obviously, X1 = ker M = Im P and Y1 = ImQ. Thus, we have dim Y1 = dim X1 =
1. For any y ∈ Y , we have
Z s
1 Z 1
1
Q2 y = Q(Qy) = Qyϕp
ϕq (
(s − τ )β−1 dτ )ds
∆ 0
Γ(β) 0
Z ηi Z s
m
X
1
(s − τ )β−1 dτ ds = Qy.
−
σi
ϕq
Γ(β) 0
0
i=1
Hence, Q2 = Q, Q is a semi-projector. Based on the definition of M and Q, it is easy
to see that ker Q = Im M . Let Ω ⊂ X be an open and bounded set with θ ∈ Ω. For
each u ∈ Ω, we can get Q[(I − Q)Nλ (u)] = 0. Thus, (I − Q)Nλ (u) ∈ Im M = ker Q.
Taking any y ∈ ImM and noting Qy = 0 , we can get y ∈ (I − Q)Y . So (2.1) holds.
It is easy to verify (2.2).
6
T. SHEN, W. LIU, T. CHEN, X. SHEN
EJDE-2014/58
Define R : Ω × [0, 1] → X2 by
Z t
1 Z s
1
R(u, λ)(t) =
(t − s)α−1 ϕq
(s − τ )β−1 ((I − Q)Nλ u(τ ))dτ ds.
Γ(α) 0
Γ(β) 0
By the continuity of f , it is easy to get that R(u, λ) is continuous on Ω×[0, 1]. Moreover, for all u ∈ Ω, there exists a constant L > 0 such that |I0β+ (I − Q)Nλ u(τ ))| ≤ L,
α−1
α
so we can easily obtain that R(Ω, λ), D0α−2
+ R(Ω, λ), D0+ R(Ω, λ) and D0+ R(Ω, λ)
are uniformly bounded. By Arzela-Ascoli theorem, we just need to prove that
R : Ω × [0, 1] → X2 is equicontinuous.
For u ∈ Ω, 0 < t1 < t2 ≤ 1, 2 < α ≤ 3, 0 < β ≤ 1, 3 < α + β ≤ 4, we have
|R(u, λ)(t2 ) − R(u, λ)(t1 )|
Z t2
1
=
|
(t2 − s)α−1 ϕq (I0β+ ((I − Q)Nλ u(τ )))ds
Γ(α) 0
Z t1
−
(t1 − s)α−1 ϕq (I0β+ ((I − Q)Nλ u(τ )))ds|
0
Z t1
Z t2
ϕq (L)
α−1
α−1
≤
(
((t2 − s)
− (t1 − s)
)ds +
(t2 − s)α−1 ds)
Γ(α) 0
t1
ϕq (L) α
(t − tα
=
1 ),
Γ(α + 1) 2
α−2
|D0α−2
+ R(u, λ)(t2 ) − D0+ R(u, λ)(t1 )|
Z t2
Z t1
=|
(t − s)ϕq (I0β+ ((I − Q)Nλ u(τ )))ds −
(t − s)ϕq (I0β+ ((I − Q)Nλ u(τ )))ds|
0
0
Z t1
Z t2
≤ ϕq (L)(
(t2 − s) − (t1 − s)ds +
(t2 − s)ds)
0
t1
ϕq (L) 2
(t2 − t21 )
=
2
and
α−1
|D0α−1
+ R(u, λ)(t2 ) − D0+ R(u, λ)(t1 )|
Z t2
Z
=|
ϕq (I0β+ ((I − Q)Nλ u(τ )))ds −
0
0
t1
ϕq (I0β+ ((I − Q)Nλ u(τ )))ds|
≤ ϕq (L)(t2 − t1 ).
Since tα is uniformly continuous on [0, 1], it follows that R(Ω, λ), D0α−2
+ R(Ω, λ)
β
and D0α−1
R(Ω,
λ)
are
equicontinuous.
Similarly,
we
can
get
I
((I
−
Q)N
+
λ u(τ )) ⊂
0+
C[0, 1] is equicontinuous, Considering of ϕq (s) is uniformly continuous on [−L, L],
we have D0α+ R(Ω, λ) = ϕq (I0β+ ((I − Q)Nλ (Ω))) is also equicontinuous. So, we can
obtain that R : Ω × [0, 1] → X2 is compact.
For each u ∈ Σλ , we have D0β+ ϕp (D0α+ u(t)) = Nλ (u(t)) ∈ Im M . Thus,
Z t
Z s
1
1
R(u, λ)(t) =
(t − s)α−1 ϕq (
(s − τ )β−1 ((I − Q)Nλ u(τ ))dτ )ds
Γ(α) 0
Γ(β) 0
Z t
Z s
1
1
α−1
=
(t − s)
ϕq (
(s − τ )β−1 D0β+ ϕp (D0α+ u(τ ))dτ )ds,
Γ(α) 0
Γ(β) 0
EJDE-2014/58
FRACTIONAL MULTI-POINT BOUNDARY-VALUE PROBLEMS
7
which together with u(0) = u0 (0) = D0α+ u(0) = 0 yields
R(u, λ)(t) = u(t) −
1
α−1
Dα−1
= (I − P )u(t).
+ u(0)t
Γ(α) 0
It is easy to verify that R(u, 0)(t) is the zero operator. So (2.3) holds. Besides, for
any u ∈ Ω,
M [P u + R(u, λ)](t)
Z t
Z s
1
1
α−1
= M[
(t − s)
ϕq (
(s − τ )β−1 ((I − Q)Nλ u(τ ))dτ )ds
Γ(α) 0
Γ(β) 0
1
α−1
Dα−1
+
+ u(0)t
Γ(α) 0
= (I − Q)Nλ u(t),
which implies (2.4). So Nλ is M -compact in Ω.
Lemma 3.4. Suppose (H1), (H2) hold, Then the set
Ω1 = u ∈ dom M \ ker M : M u = λN u, λ ∈ (0, 1)}
is bounded.
Proof. By lemma 2.4, for each u ∈ dom M , D0α−1
+ u ∈ C[0, 1], we have
α−2
u(t) = I0α−1
D0α−1
+ c2 tα−3 .
+
+ u(t) + c1 t
Combining this with u(0) = u0 (0) = 0, we get c1 = c2 = 0. Thus,
Z t
1
α−1
kuk∞ = kI0α−1
D
uk
≤
|
(t − s)α−2 ds|kD0α−1
+
+ uk∞
∞
0+
Γ(α − 1) 0
1
kDα−1
≤
+ uk∞ .
Γ(α) 0
Take any u ∈ Ω1 , then N u ∈ ImM = ker Q. Thus, QN u = 0 for all t ∈ [0, 1]. It
α−1
follows from (H2) that there exists t0 ∈ [0, 1] such that |D0α−2
+ u(t0 )|+|D0+ u(t0 )| ≤
A. Thus
Z t
α−1
D0α−1
u(t)
=
D
u(t
)
+
D0α+ u(t)dt,
+
0
0+
α−2
D0α−2
+ u(t) = D0+ u(t0 ) +
Z
t0
t
t0
D0α−1
+ u(t)dt,
α
kD0α−1
+ uk∞ ≤ A + kD0+ uk∞ ,
α−1
α
kD0α−2
+ uk∞ ≤ A + kD0+ uk∞ ≤ 2A + kD0+ uk∞ ,
1
kuk∞ ≤
(A + kD0α+ uk∞ ).
Γ(α)
Combined with M u = λN u and D0α+ u(0) = 0, we obtain
ϕp (D0α+ u(t)) = λI0β+ N u(t).
From (H1) and λ ∈ (0, 1), we have
Z t
1
α−1
α
α
(t − s)β−1 |f (s, u(s), D0α−2
|ϕp (D0+ u(t))| ≤
+ u(s), D0+ u(s), D0+ u(s))|ds
Γ(β) 0
8
T. SHEN, W. LIU, T. CHEN, X. SHEN
≤
1
Γ(β)
Z
EJDE-2014/58
t
0
p−1
(t − s)β−1 (a(s) + b(s)|u(s)|p−1 + c(s)|D0α−2
+ u(s)|
p−1
+ d(s)|D0α−1
+ e(s)|D0α+ u(s)|p−1 )ds
+ u(s)|
1
p−1
p−1
≤
(kak∞ + kbk∞ kuk∞
+ kck∞ kD0α−2
+ uk∞
Γ(β + 1)
p−1
+ kdk∞ kD0α−1
),
+ kek∞ kD0α+ ukp−1
+ uk∞
∞
∀t ∈ [0, 1],
which together with |ϕp (D0α+ u(t))| = |D0α+ u(t)|p−1 , and the basic inequality (|a| +
|b|)p ≤ Cp (|a|p + |b|p ), where Cp = 2p−1 when p > 1 and where Cp = 1 when
0 < p ≤ 1, a, b ∈ R (see [13]). We can get
kD0α+ ukp−1
∞ ≤
1
α−2
p−1
(kak∞ + kbk∞ kukp−1
∞ + kck∞ kD0+ uk∞
Γ(β + 1)
p−1
+ kdk∞ kD0α−1
+ kek∞ kD0α+ ukp−1
+ uk∞
∞ )
≤
1
Ap−1
1
α
p−1
(kak∞ + kbk∞ D(
kD
uk
+
)
+
∞
Γ(β + 1)
Γ(α)p−1 0
Γ(α)p−1
p−1
+ kck∞ D(2p−1 Ap−1 + kD0α+ uk∞
) + kdk∞ D(Ap−1 + kD0α+ ukp−1
∞ )
p−1
+ kek∞ kD0α+ uk∞
).
where D = max{1, 2p−2 }. From (3.1), we can see that there exists a constant
M1 > 0 such that
kD0α−1
+ uk∞ ≤ A + M1 := M2 ,
A
1
M1 +
:= M4 .
≤ 2A + M1 := M3 , kuk∞ ≤
Γ(α)
Γ(α)
kD0α+ uk∞ ≤ M1 ,
kD0α−2
+ uk∞
Thus
α−1
α
kukX = max kuk∞ , kD0α−2
+ uk∞ , kD0+ uk∞ , kD0+ uk∞
≤ max{M1 , M2 , M3 , M4 } := M.
Therefore, Ω1 is bounded.
Lemma 3.5. Suppose (H2) holds, then the set Ω2 = {u ∈ ker M : N u ∈ Im M } is
bounded.
Proof. For each u ∈ Ω2 , we can have that u(t) = ctα−1 for all c ∈ R and QN u =
0. It follow from (H2) that there exists a t0 ∈ [0, 1] such that |D0α−1
+ u(t0 )| +
α−2
A
|D0+ u(t0 )| ≤ A, which implies |c| ≤ Γ(α)(1+t0 ) . Therefore, Ω2 is bounded.
Lemma 3.6. Suppose (H3) holds, then the set
Ω3 = {u ∈ ker M : (−1)m λJ −1 u + (1 − λ)QN u = 0, λ ∈ [0, 1]}
is bounded, where m = 1 when Λ < 0 and m = 2 when Λ > 0.
Proof. Case 1, suppose Λ < 0, for each u ∈ Ω3 , we can get that u(t) = ctα−1 for all
c ∈ R. We define the isomorphism J : Im Q → ker M by J(c) = ctα−1 , c ∈ R, t ∈
EJDE-2014/58
FRACTIONAL MULTI-POINT BOUNDARY-VALUE PROBLEMS
9
[0, 1]. So, we have
Z s
1 Z 1
1
λc = (1 − λ)ϕp
ϕq (
(s − τ )β−1 f (τ, cτ α−1 , cΓ(α)τ, cΓ(α), 0)dτ )ds
∆ 0
Γ(β) 0
Z s
Z ηi m
X
1
−
(s − τ )β−1 f (τ, cτ α−1 , cΓ(α)τ, cΓ(α), 0)dτ ds .
σi
ϕq
Γ(β) 0
0
i=1
(3.4)
If λ = 0, then |c| ≤ B because of the first part of (H3). If λ ∈ (0, 1], we can also
obtain |c| ≤ B. Otherwise, if |c| > B, in view of the first part of (H3), one has
Z s
1 Z 1
1
ϕq (
(s − τ )β−1 f (τ, cτ α−1 , cΓ(α)τ, cΓ(α), 0)dτ )ds
c(1 − λ)ϕp
∆ 0
Γ(β) 0
Z ηi Z s
m
X
1
ϕq
−
σi
(s − τ )β−1 f (τ, cτ α−1 , cΓ(α)τ, cΓ(α), 0)dτ ds ≤ 0.
Γ(β) 0
0
i=1
(3.5)
On the other hand, λc2 > 0 which contradicts to (3.4). Therefore, Ω3 is bounded.
Case 2, suppose Λ > 0, it is similar to case 1 to proof Ω3 is bounded. So, we
omit it.
Proof of Theorem 3.1. Assume that Ω is a bounded open set of X with ∪3i=1 Ωi ⊂ Ω.
By Lemma 3.3, we obtain that N is M -compact on Ω. Then by Lemmas 3.4 and
3.5, we have
(i) M x 6= Nλ x for each (u, λ) ∈ (dom M \ ker M ) × (0, 1),
(ii) QN u 6= 0, for all u ∈ ∂Ω ∩ ker M .
Thus, we need to prove that (iii) of Lemma 2.1 is true, Let I be the identity operator
in the Banach space X, and H(u, λ) = (−1)m λJ −1 (u) + (1 − λ)QN (u). According
to Lemma 3.6 we know that for each u ∈ ∂Ω ∩ ker M , H(u, λ) 6= 0. Thus, by the
homotopic property of degree, we have
deg(JQN |ker M , Ω ∩ ker M, 0) = deg(H(·, 0), Ω ∩ ker M, 0)
= deg(H(·, 1), Ω ∩ ker M, 0)
= deg(±I, Ω ∩ ker M, 0) 6= 0.
which means (iii) of Lemma 2.1 is satisfied. Consequently, by Lemma 2.1, the
equation M u = N u has at least one solution in dom M ∩ Ω. Namely, BVP (1.2)
have at least one solution in the space X.
Acknowledgements. This research as supported by grant 11271364 from the
NNSF of China. The authors are grateful to those who give valuable suggestions
about the original manuscript.
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Tengfei Shen
College of Sciences, China University of Mining and Technology, Xuzhou 221008, China
E-mail address: shentengfei1987@126.com
Wenbin Liu (corresponding author)
College of Sciences, China University of Mining and Technology, Xuzhou 221008, China
E-mail address: wblium@163.com
Taiyong Chen
College of Sciences, China University of Mining and Technology, Xuzhou 221008, China
E-mail address: taiyongchen@cumt.edu.cn
Xiaohui Shen
College of Sciences, China University of Mining and Technology, Xuzhou 221008, China
E-mail address: shenxiaohuicool@163.com