Directory UMM :Journals:Journal_of_mathematics:EJQTDE:
Electronic Journal of Qualitative Theory of Differential Equations
2011, No. 27, 1-16; http://www.math.u-szeged.hu/ejqtde/
Existence of Solutions for Fractional Differential Equations with
Multi-point Boundary Conditions at Resonance on a Half-line
Huacheng Zhou, Chunhai Kou∗and Feng Xie†
Department of Applied Mathematics, Donghua University
Shanghai 201620, P.R. China
Abstract
In this paper, we investigate the existence of solutions for multi-point boundary
value problems at resonance concerning fractional differential equation on a half-line.
Our analysis relies on the coincidence degree of Mawhin. As an application, an example
is presented to illustrate the main results.
2010 Mathematics Subject Classification. 34A08; 34B10; 34B40.
Key words and phrases. Fractional differential equations; multi-point boundary
conditions; resonance; coincidence degree theorem; half-line
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, 9, 10, 11, 16, 17, 25] 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
[18, 19, 20, 21, 29, 34]).
In this paper, we study the existence of solutions for the fractional differential equation
at resonance with multi-point boundary value problem on a half-line:
α
α−1
D0+
x(t) = f (t, x(t), D0+
x(t)), t ∈ (0, +∞),
α−1
x(t) = βx(η),
x(0) = 0, lim D0+
t→+∞
(1.1)
(1.2)
α
where 1 < α ≤ 2, η > 0, f : [0, +∞) × R × R → R is an S-Carath´eodory function, and D0+
is the standard Riemann-Liouville fractional derivative. Moreover, we need the following
condition
Γ(α) = βη α−1 .
(1.3)
∗
†
E-mail: [email protected].
Corresponding author, E-mail: [email protected].
EJQTDE, 2011 No. 27, p. 1
The condition (1.3) is critical since the fractional differential operator in (1.1) has a
nontrivial kernel. Boundary value problems with such critical conditions are so-called problems at resonance. Boundary value problems for differential equations of integer order at
resonance have been studied by many authors (see [3, 4, 5, 6, 7, 12, 13, 14, 22, 23, 24, 26, 27]
and references cited therein).
More recently, various types of multi-point boundary value problems for fractional differential equations at resonance on a bounded domain have been analyzed by Kosmatov [8],
Jiang [35], Bai [30], Bai and Zhang [31, 32].
However, to our knowledge, it is rare for work to be done on the solutions of fractional
differential equations at resonance on a half-line. In this paper, our goal is to fill this gap
in the literature.
The layout of this paper is as follow. In Section 2, we provide some necessary background.
In particular, we shall introduce some lemmas and definitions related with problem (1.1)
and (1.2). In Section 3, the main results of problem (1.1) and (1.2) will be stated and
proved. Finally, one example is also included to illustrate the main results.
2
Background materials and preliminaries
In this section, to establish the existence of solutions, we present some necessary background
and lemmas. These definitions and lemmas can be found in the recent literature [1, 2, 15,
28, 36].
Definition 2.1. The Riemann-Liouville fractional integral of order α > 0 of a function
f : (0, +∞) → R is given by
Z t
1
α
(t − s)α−1 f (s)ds
I0+ f (t) =
Γ(α) 0
provided the right side is pointwise defined on (0, +∞), and we have
α µ
I0+
t =
Γ(µ + 1) µ+α
t , α > 0, µ > −1.
Γ(µ + α + 1)
Definition 2.2. The Riemann-Liouville fractional derivative of order α > 0 of a function
f : (0, +∞) → R is given by
α
n−α
D0+
f (t) = D n I0+
f (t)
where n = [α] + 1, D n =
dn
,t
dtn
> 0, and we have for λ > −1
α λ
D0+
t =
Γ(λ + 1) λ−α
t .
Γ(λ − α + 1)
Lemma 2.1. Assume that u ∈ C(0, +∞) ∩ Lloc (0, +∞) with a fractional derivative of
order α > 0 that belongs to C(0, +∞) ∩ Lloc (0, +∞). Then
α
α
I0+
D0+
x(t) = x(t) + c1 tα−1 + c2 tα−2 + · · · + cn tα−N ,
for some ci ∈ R, i = 1, . . . , N, where N is the smallest integer greater than or equal to α.
EJQTDE, 2011 No. 27, p. 2
We now give the background from the coincidence degree theory.
Definition 2.3. Let X and Y be normed spaces. A linear operator L : dom(L) ⊂ X →
Y is said to be a Fredholm operator of index zero provided that
(i) ImL is closed subset of Y , and
(ii) dim KerL=codim ImL < +∞.
From definition2.3 it follows that there exist continuous projects P : X → X and
Q : Y → Y such that ImP = KerL, KerQ = ImL, X = KerL ⊕ KerP, Y = ImL ⊕ ImQ and
L|DomL∩KerP : DomL ∩ KerP → ImL is invertible. We denote the inverse that map
by Kp : ImL → DomL ∩ KerP . The generalized inverse of L denoted by KP,Q : Y →
DomL ∩ KerP is defined by KP,Q = KP (I − Q).
Definition 2.4. Let L : Dom(L) ⊂ X → Y be a Fredholm operator, E be a metric
space, and N : E → Y be an operator. We say that N is L-compact on E if QN : E → Y
and KP,Q N : E → X are compact on E.
Definition 2.5. f : [0, +∞) × R2 → R is said to be an S-Carath´eodory function if and
only if
(i) for each (u, v) ∈ R2 , t 7→ f (t, u, v) is measurable on [0, +∞);
(ii) for a.e. t ∈ [0, +∞), (u, v) 7→ f (t, u, v) is continuous on R2 ;
(iii) for each r > 0, there exists ϕr (t) ∈ L1 [0, +∞) ∩ C[0, +∞) satisfying supt≥0 |ϕr (t)| <
+∞, ϕr (t) > 0, t ∈ (0, +∞) such that
max{|u|, |v|} ≤ r impiles |f (t, (1 + tα−1 )u, v)| ≤ ϕr (t), a.e. t ∈ [0, +∞).
Since the Arzela-Ascoli theorem fails to work in the space C∞ , we need a modified compactness criterion to prove that N is L-compact(C∞ = {x ∈ C[0, +∞), limt→+∞ x(t) exists}).
Lemma 2.6.[28] Let M ⊂ C∞ = {x ∈ C[0, +∞), limt→+∞ x(t) exists}. Then M is
relatively compact if the following conditions hold:
(i) all functions from M are uniformly bounded;
(ii) all functions from M are equicontinuous on any compact interval of [0, +∞);
(iii) all functions from M are equiconvergent at infinity, that is, for any given ε > 0,
there exists a T = T (ε) > 0 such that |f (t1 ) − f (t2 )| < ε, for all t1 , t2 > T and f ∈ M.
In this paper, we use the space X, Y defined by
α−1
X = {x ∈ C[0, +∞) : lim x(t)/(1 + tα−1 ) exists, lim D0+
x(t) exists},
t→+∞
t→+∞
Y = {y ∈ C[0, +∞) : y ∈ L1 [0, +∞), sup |y(t)| < +∞},
t≥0
α−1
with the norm kxkX = max{kxk0 , kD0+
xk∞ } and kykY = max{kykL1 , kyk∞ } respectively,
where k · k∞ is the supremum norm on [0, +∞) and kxk0 = supt≥0 |x(t)|/(1 + tα−1 ), kykL1 =
R +∞
|y(s)|ds. By the standard arguments, we can prove that (X, k · kX ), (Y, k · kY ) are
0
Banach spaces.
Define L be the linear operator from DomL ∩ X to Y with
α
DomL = {x ∈ X : D0+
x ∈ L1 [0, +∞) ∩ C[0, +∞), x satisfies (1.2)}
and
α
Lx = D0+
x, x ∈ DomL.
EJQTDE, 2011 No. 27, p. 3
We define N : X → Y by setting
α−1
(Nx)(t) = f (t, x(t), D0+
x(t)).
Lemma 2.7. The operator L : Dom(L) ⊂ X → Y is a Fredholm operator of index zero.
Furthermore, the linear projector operator Q : Y → Y can be defined by
Z +∞
Z η
β
α−1
(Qy)(t) = w(t)
y(s)ds −
(η − s) y(s)ds
Γ(α) 0
0
Rη
R +∞
β
(η − s)α−1 w(s)ds = 1 and the
where w(t) ∈ Y satisfy w(t) > 0 and 0 w(s)ds − Γ(α)
0
linear operator Kp : ImL → DomL ∩ KerP can be written by
Z t
1
KP y(t) =
(t − s)α−1 y(s)ds, t ∈ [0, +∞),
Γ(α) 0
also
kKP ykX ≤
1
kykL1 , for all y ∈ ImL.
Γ(α)
Proof. It is clear that KerL = {x = ctα−1 : c ∈ R}. Now we show that
Z +∞
Z η
β
ImL = {y ∈ Y :
(η − s)α−1 y(s)ds = 0}.
y(s)ds −
Γ(α)
0
0
(2.1)
α
If y ∈ ImL, then there exists a function x ∈ DomL such that y(t) = D0+
x(t). By Lemma2.1,
α
I0+
y(t) = x(t) + c1 tα−1 + c2 tα−2 .
By virtue of the boundary condition (1.2), we have
α−1 α
α
lim D0+
I0+ y(t) = βI0+
y(η)
t→+∞
and therefore
Z
+∞
0
β
y(s)ds −
Γ(α)
Z
0
η
(η − s)α−1 y(s)ds = 0.
(2.2)
α
On the other hand, suppose y ∈ Y and satisfies (2.2), let x(t) = I0+
y(t), then x(0) =
R
t
α−1
α−1
α
0, D0+ x(t) = 0 y(s)ds. Thus limt→+∞ D0+ x(t) = βx(η). Then x ∈ DomL and D0+
x(t) =
y(t). That is to say, (2.1) holds.
For y ∈ Y , taking the projector
Z +∞
Z η
β
α−1
(η − s) y(s)ds .
(Qy)(t) = w(t)
y(s)ds −
Γ(α) 0
0
It is easy to
verify that the operator
R +∞
R η Q is a projector. Letting y1 = y − Qy. Then y1 ∈
ImL(since 0 y1 (s)ds − (β/Γ(α)) 0 (η − s)α−1 y1 (s)ds = 0). Hence Y = ImL + ImQ. From
y ∈ ImQ, there exists a constant c ∈ R, such that y = cw(t), and from y ∈ ImL, we obtain
Z +∞
Z η
β
c=
cw(s)ds −
(η − s)α−1 cw(s)ds = 0,
Γ(α)
0
0
EJQTDE, 2011 No. 27, p. 4
which implies ImL ∩ ImQ = {0} and Y = ImL ⊕ ImQ. Thus we have
dim KerL = dim ImQ = codim ImL = 1.
This implies that L is a Fredholm operator of index zero.
Taking P : X → X as follows:
(P x)(t) =
1
D α−1 x(0)tα−1 ,
Γ(α) 0+
then the generalized inverse KP : ImL → DomL ∩ KerP can be written by
Z t
1
KP y(t) =
(t − s)α−1 y(s)ds, t ∈ [0, +∞).
Γ(α) 0
In fact, for y ∈ ImL, we have
α
(LKP )y(t) = D0+
((KP y)(t)) = y(t),
and for x ∈ DomL ∩ KerP , we know
α
(KP L)x(t) = (KP )D0+
x(t) = x(t) + c1 tα−1 + c2 tα−2 ,
α−1
for some c1 , c2 ∈ R. In view of x ∈ DomL ∩ KerP , D0+
x(0) = 0 and (1.2), we obtain
c1 = c2 = 0. Therefore
(KP L)x(t) = x(t).
This shows that KP = (L|DomL∩KerP )−1 .
Again from the definition of KP , we have
Z
1 t (t − s)α−1
≤ 1 kykL1 ,
kKP yk0 = sup
y(s)ds
Γ(α)
α−1
1+t
t≥0 Γ(α)
0
(2.3)
and
α−1
kD0+
KP yk∞
=
α−1 1
sup |D0+
Γ(α)
t≥0
Z
0
t
(t − s)
α−1
y(s)ds| = sup |
t≥0
Z
0
t
y(s)ds| ≤ kykL1 .
(2.4)
It follows from (2.3) and (2.4) that
kKp ykX ≤
1
kykL1 .
Γ(α)
This completes the proof of Lemma 2.7.
Lemma 2.8. Let f be an S-Carath´eodory function, then N is L-compact.
Proof. Obviously, QN and KP (I − Q)N are continuous. So we only need to prove the
compactness, i.e. QN and KP (I − Q)N maps bounded sets into relatively compact ones.
Suppose U ⊂ X is a bounded set. Then there exists r > 0 such that kxkX ≤ r, for all
x ∈ U. Because f is an S-Carath´eodory function, there exists ϕr (t) ∈ L1 [0, +∞)∩C[0, +∞)
satisfying supt≥0 |ϕr (t)| < +∞, ϕr (t) > 0, t ∈ (0, +∞) such that
α−1
|f (t, x(t), D0+
x(t))| = |f (t, (1 + tα−1 )
x(t)
, D α−1 x(t))| ≤ ϕr (t), a.e. t ∈ [0, +∞).
1 + tα−1 0+
EJQTDE, 2011 No. 27, p. 5
Then for any x ∈ U,
kQNxkL1 =
Z
+∞
0
Z
w(t)(
+∞
α−1
f (s, x(s), D0+
x(s)) ds
0
Z η
β
α−1
α−1
(η − s) f (s, x(s), D0+ x(s))ds) dt
−
Γ(α) 0
Z +∞
Z +∞
Z
βη α−1 η
≤
|w(t)|dt
ϕr (s)ds +
ϕr (s)ds
Γ(α) 0
0
0
βη α−1
kwkL1 kϕr kL1 ,
≤ 1+
Γ(α)
(2.5)
and
kQNxk∞
Z
= sup w(t)(
t≥0
0
+∞
α−1
f (s, x(s), D0+
x(s)) ds
Z η
α−1
α−1
(η − s) f (s, x(s), D0+ x(s))dt)
β
Γ(α) 0
α−1
βη
kwk∞ kϕr kL1 .
≤ 1+
Γ(α)
−
(2.6)
It follows from (2.5) and (2.6) that kQNxkY = max{kQNxkL1 , kQNxk∞ } ≤ (1+ βηΓ(α) )kwkY kϕr kL1 .
Noting that ImQ ≃ R, we have QN is compact.
Furthermore, for any x ∈ U we have
Z t
(KP,Qx)(t)
(t − s)α−1
α−1
= 1
f (s, x(s), D0+
x(s))
1 + tα−1 Γ(α)
α−1
1
+
t
0
Z +∞
α−1
− w(s)
f (τ, x(τ ), D0+
x(τ )) dτ
0
Z η
β
α−1
α−1
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
−
Γ(α) 0
(2.7)
Z +∞
Z t
Z t
1
1
≤
ϕr (s)ds +
w(s)ds
ϕr (τ )dτ
Γ(α) 0
Γ(α) 0
0
Z η
β
α−1
(η − τ ) ϕr (τ )dτ
+
Γ(α) 0
βη α−1
1
1
1+
kwkL1 kϕr kL1 ,
kϕr kL1 +
≤
Γ(α)
Γ(α)
Γ(α)
α−1
and
α−1
|D0+
(KP,Q x)(t)|
Z t
Z +∞
α−1
α−1
= (f (s, x(s), D0+ x(s)) − w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
0
0
Z η
β
α−1
α−1
(2.8)
−
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
Γ(α) 0
βη α−1
kwkL1 kϕr kL1 .
≤ kϕr kL1 + 1 +
Γ(α)
EJQTDE, 2011 No. 27, p. 6
It follows from (2.7) and (2.8) that KP,Q U is uniformly bounded. Meanwhile, for any
t1 , t2 ∈ [0, T ] with T is a positive constant
(KP,Qx)(t1 ) (KP,Q x)(t2 )
1 + tα−1 − 1 + tα−1
1
2
Z t1
α−1
α−1
1
(t
−
s)
(t
−
s)
2
1
α−1
≤
α−1 −
α−1 f (s, x(s), D0+ x(s))
Γ(α) 0
1 + t1
1 + t2
Z +∞
α−1
− w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
0
Z η
β
α−1
α−1
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
−
Γ(α) 0
Z t2
1
(t2 − s)α−1
α−1
+
α−1 f (s, x(s), D0+ x(s))
Γ(α) t1 1 + t2
Z +∞
α−1
− w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
0
Z η
β
α−1
α−1
−
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
Γ(α) 0
(2.9)
Z +∞
Z t1
α−1
α−1
(t2 − s)
1
(t1 − s)
|ϕr (s) + w(s)
ϕr (τ )dτ
≤
α−1 −
Γ(α) 0
1 + t1
1 + tα−1
0
2
Z η
β
α−1
(η − τ ) ϕr (τ )dτ ) ds
−
Γ(α) 0
Z +∞
Z t2
1
(t2 − s)α−1
+
ϕr (τ )dτ
ϕr (s) + w(s)
Γ(α) t1 1 + tα−1
0
2
Z η
β
α−1
(η − τ ) ϕr (τ )dτ ) ds
−
Γ(α) 0
Z t1
α−1
(t1 − s)α−1 (t2 − s)α−1
1
βη
1
≤
) ds
−
α−1 ϕr (s) + w(s)kϕr kL (1 +
Γ(α) 0 1 + tα−1
Γ(α)
1
+
t
1
2
Z t2
1
(t2 − s)α−1
βη α−1
+
ϕr (s) + w(s)kϕr kL1 (1 + Γ(α) ) ds
Γ(α) t1 1 + tα−1
2
→ 0 as t1 → t2
and
α−1
α−1
|D0+
(KP,Q x)(t1 ) − D0+
(KP,Q x)(t2 )|
Z t2
Z +∞
α−1
f (s, x(s), D0+ x(s)) − w(s)
=
ϕr (τ )dτ
t1
0
Z η
β
α−1
(η − τ ) ϕr (τ )dτ ) ds
−
Γ(α) 0
Z t2
α−1
βη
)ds
≤
ϕr (s) + w(s)kϕr kL1 (1 +
Γ(α)
t1
→ 0 as t1 → t2 .
(2.10)
It follows from (2.9) and (2.10) that KP,Q U is equicontinuous. From Lemma 2.6, we can see
α−1
that if KP,Q U/(1+tα−1 ), D0+
KP,QU are equiconvergent at infinity, then KP,Q U is relatively
EJQTDE, 2011 No. 27, p. 7
compact in X. In fact, considering that the following estimate
Z +∞
βη α−1
ε
1
ϕr (s) + w(s)kϕr kL1 (1 +
)ds < ,
Γ(α) L
Γ(α)
3
(2.11)
holding for ε > 0 and some L > 0, we have
lim sup |g(t, s) − 1| ≤ lim g(t, L) = 0,
t→∞ s∈[0,L]
where g(t, s) =
t1 , t2 ≥ T ,
(t−s)α−1
,
1+tα−1
t→∞
s ∈ [0, L], t ∈ [L, +∞). Thus, there exists T > L such that for
sup |g(t1 , s) − g(t2 , s)|
s∈[0,L]
≤ sup |g(t1 , s) − 1| + sup |g(t2 , s) − 1|
s∈[0,L]
s∈[0,L]
(2.12)
−1
βη α−1
ε
kϕr kL1 (1 + kwkL1 (1 +
))
.
<
3
Γ(α)
Therefore, it follows from (2.11) and (2.12) that for t1 , t2 ≥ T , we get
(KP,Q x)(t1 ) (KP,Q x)(t2 )
1 + tα−1 − 1 + tα−1
1
2
Z L
(t1 − s)α−1 (t2 − s)α−1
1
α−1
−
≤
α−1 f (s, x(s), D0+ x(s))
Γ(α) 0 1 + tα−1
1
+
t
1
2
Z +∞
α−1
− w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
0
Z η
β
α−1
α−1
−
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
Γ(α) 0
Z +∞
Z t1
(t1 − s)α−1
1
α−1
α−1
f (τ, x(τ ), D0+
x(τ ))dτ
+
f (s, x(s), D0+ x(s)) − w(s)
Γ(α) L
1 + tα−1
0
1
Z η
β
α−1
α−1
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
−
Γ(α) 0
Z +∞
Z t2
1
(t2 − s)α−1
α−1
α−1
+
x(τ ))dτ
f (τ, x(τ ), D0+
α−1 f (s, x(s), D0+ x(s)) − w(s)
Γ(α) L
1 + t2
0
Z η
β
α−1
α−1
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
−
Γ(α) 0
Z +∞
Z L
ε
α−1
α−1
|f (s, x(s), D0+ x(s))| + w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
≤
3
0
0
−1
Z η
β
βη α−1
α−1
α−1
−
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
))
kϕr kL1 (1 + kwkL1 (1 +
Γ(α) 0
Γ(α)
Z t1
βη α−1
1
ϕr (s) + w(s)kϕr kL1 (1 +
)ds
+
Γ(α) L
Γ(α)
Z t2
1
βη α−1
+
ϕr (s) + w(s)kϕr kL1 (1 +
)ds
Γ(α) L
Γ(α)
≤ε
EJQTDE, 2011 No. 27, p. 8
and
α−1
α−1
|D0+
(KP,Q x)(t1 ) − D0+
(KP,Q x)(t2 )|
Z +∞
Z max{t1 ,t2 }
α−1
f (s, x(s), D x(s)) − w(s)
≤
ϕr (τ )dτ
0+
0
min{t1 ,t2 }
Z η
β
α−1
(η − τ ) ϕr (τ )dτ ) ds
−
Γ(α) 0
Z +∞
βη α−1
1
)ds
≤
ϕr (s) + w(s)kϕr kL (1 +
Γ(α)
L
≤ ε.
So we complete the proof.
The following fixed point theorem due to Mawhin is fundamental in the proofs of our
main results.
Lemma 2.9.[36] Let Ω ⊂ X be open and bounded, L be a Fredholm mapping of index
zero and N be L-compact on Ω. Assume that the following conditions are satisfied:
(i) Lx 6= λNx for every (x, λ) ∈ (DomL\KerL) ∩ ∂Ω) × (0, 1);
(ii) Nx ∈
/ImL for every x ∈KerL ∩ ∂Ω;
(iii) deg(JQN|KerL∩∂Ω , Ω ∩ KerL, 0)6= 0, with Q : Y → Y a continuous projector such
that KerQ=ImL and J :ImQ →KerL is an isomorphism.
Then the equation Lx = Nx has at least one solution in DomL ∩ Ω.
3
Main results
In this section, we present our main results and prove them.
Theorem 3.1. Assume that f is an S-Carath´eodory function and the following conditions are satisfied:
(H1 ) There exist functions a, b, c ∈ L1 [0, +∞) such that
|f (t, u, v)| ≤ a(t)|u| + b(t)|v| + c(t), a.e. t ∈ [0, +∞) and all (u, v) ∈ R2
and
kak1 =
Z
(3.1)
+∞
a(s)(1 + tα−1 )ds < +∞;
0
α−1
(H2 ) There exists constant A > 0 such that for x ∈ DomL, if |D0+
x(t)| > A for all
t ∈ [0, +∞), then
Z η
Z +∞
β
α−1
α−1
(η − s)α−1 f (s, x(s), D0+
x(s))ds 6= 0;
f (s, x(s), D0+ x(s))ds −
Γ(α)
0
0
(H3 ) There exists constant B > 0 such that for all c ∈ R with |c| > B, either
Z +∞
Z η
β
α−1
α−1
α−1
c
(η − s) f (s, cs , cΓ(α))ds < 0
f (s, cs , cΓ(α))ds −
Γ(α) 0
0
(3.2)
EJQTDE, 2011 No. 27, p. 9
or else
Z
c
+∞
f (s, cs
α−1
0
β
, cΓ(α))ds −
Γ(α)
Z
η
(η − s)
0
α−1
f (s, cs
α−1
, cΓ(α))ds
> 0.
(3.3)
Then the BVP (1.1), (1.2) has at least one solution provided that
2
2
kak1 +
kbkL1 < 1.
Γ(α)
Γ(α)
Proof. We construct an open bounded set Ω ⊂ X that satisfies the assumption of
Lemma 2.9. Let
Ω1 = {x ∈ DomL\KerL : Lx = λNx, for some λ ∈ [0, 1]}.
For x ∈ Ω1 , we have x ∈
/KerL, λ 6= 0 and Nx ∈ ImL. Note that KerQ = ImL and, thus,
Z +∞
Z η
β
α−1
α−1
f (s, x(s), D0+ x(s))ds −
(η − s)α−1 f (s, x(s), D0+
x(s))ds = 0
Γ(α) 0
0
α−1
since QNx = 0. It follows from (H2 ) that there exists t0 ∈ [0, +∞) such that |D0+
x(t0 )| ≤
R
t
0
α−1
α−1
α
A. In view of D0+ x(0) = D0+ x(t0 ) − 0 D0+ x(t)dt, we have
α−1
α
|D0+
x(0)| ≤ A + kD0+
xkL1 = A + kLxkL1 ≤ A + kNxkL1 .
(3.4)
Again for x ∈ Ω1 , x ∈ DomL\KerL, then (I − P )x ∈ DomL ∩ KerP and LP x = 0, thus
from Lemma 2.7, we have
k(I − P )xkX = kKP L(I − P )xkX ≤
1
kL(I − P )xkL1
Γ(α)
1
1
kLxkL1 ≤
kNxkL1 .
=
Γ(α)
Γ(α)
(3.5)
From (3.4), (3.5), we have
kxkX ≤ kP xkX + k(I − P )xkX ≤
A
2
+
kNxkL1 .
Γ(α) Γ(α)
(3.6)
If (3.6) holds, from (3.1), we have
kxkX ≤
2
A
α−1
xk∞ + kckL1 ).
+
(kak1 kxk0 + kbkL1 kD0+
Γ(α) Γ(α)
(3.7)
Thus from kxk0 ≤ kxkX and (3.7), we have
kxk0 ≤
A
1
1−
(
2
kak1 Γ(α)
Γ(α)
+
2
α−1
xk∞ + kckL1 )).
(kbkL1 kD0+
Γ(α)
(3.8)
α−1
Again from (3.7), (3.8) and kD0+
xk∞ ≤ kxkX we obtain
α−1
kD0+
xk∞ ≤
1
1−
2
kak1
Γ(α)
A
−
(
2
1 Γ(α)
kbk
L
Γ(α)
+
2
kckL1 ).
Γ(α)
(3.9)
EJQTDE, 2011 No. 27, p. 10
It follows from (3.8) and (3.9) that, there exists an M > 0 such that kxkX ≤ M for all
x ∈ Ω1 , that is, Ω1 is bounded.
Let
Ω2 = {x ∈ KerL : Nx ∈ ImL}.
For x ∈ Ω2 , x ∈ KerL implies that x can be expressed by x = ctα−1 , where c is an arbitrary
constant, and QNx = 0, thus
Z +∞
Z η
1
α−1
α−1
α−1
(η − s) f (s, cs , cΓ(α))ds = 0.
w(t)
f (s, cs , cΓ(α))ds −
Γ(α) 0
0
1
1
It follows from (H2 ) that, we get kxkX ≤ Γ(α)
|c| ≤ Γ(α)
A. So Ω2 is bounded too.
We define the isomorphism J : ImQ → KerL by
J(cw(t)) = ctα−1 , c ∈ R.
If the first part of (H3 ) is fulfilled, we set
Ω3 = {x ∈ KerL : −λJ −1 x + (1 − λ)QNx = 0}.
For every x = ctα−1 ∈ Ω3 , one has λJ −1 x = (1 − λ)QNx,
Z +∞
Z η
β
α−1
α−1
α−1
λcw(t) = (1−λ)w(t)
f (s, cs , cΓ(α))ds −
(η − s) f (s, cs , cΓ(α))ds .
Γ(α) 0
0
If λ = 1, then c = 0 and, if |c| > B, in view of (3.2), one has
Z +∞
Z η
β
α−1
α−1
2
α−1
(η − s) f (s, cs , cΓ(α))ds < 0,
λc w(t) = (1−λ)w(t)c
f (s, cs , cΓ(α))ds −
Γ(α) 0
0
which contradicts λc2 w(t) > 0. If the other part of (H3 ) is satisfied, we take
Ω3 = {x ∈ KerL : λJ −1 x + (1 − λ)QNx = 0}
1
1
and, again, obtain a contradiction. Thus, in ether case kxkX ≤ ( Γ(α)
+ 1)|c| ≤ ( Γ(α)
+ 1)B,
that is, Ω3 is bounded.
In what follows, we shall prove that all conditions of Lemma 2.9 are satisfied. Set Ω be
a bounded open subset of X such that ∪3i=1 Ωi ⊂ Ω. we know that L is a Fredholm operator
of index zero and N is L-compact on Ω. By the definition of Ω, we have
(i)Lx 6= λNx for every (x, λ) ∈ (DomL\KerL) ∩ ∂Ω) × (0, 1);
(ii) Nx ∈
/ImL, for every x ∈ KerL ∩ ∂Ω.
At last we will prove that (iii) of Lemma 2.9 is satisfied. To this end, let
H(x, λ) = ±λIdx + (1 − λ)JQNx,
where Id is the identical operator. By virtue of the definition of Ω, we know Ω ⊃ Ω3 , thus
H(x, λ) 6= 0 for x ∈ KerL ∩ ∂Ω, then by homotopy property of degree, we get
deg(JQN|KerL∩∂Ω , Ω ∩ KerL, 0) = deg(H(·, 0), Ω ∩ KerL, 0)
= deg(H(·, 1), Ω ∩ KerL, 0)
= deg(±Id, Ω ∩ KerL, 0)
= ±1 6= 0.
EJQTDE, 2011 No. 27, p. 11
So, the third assumption of Lemma 2.9 is fulfilled and Lx = Nx has at least one solution
in DomL ∩ Ω. The proof is complete.
Corollary 3.2. Assume that f is an S-Carath´eodory function and the conditions (H1 )
and (H3 ) in Theorem 3.1 are satisfied with (H2 ) replaced with
′
(H2 ): There exist functions l, m ∈ L1 [0, +∞) with l(t), m(t) ≥ 0 and l(t) ≡
/0 such that
f (t, u, v) ≥ l(t)|v| − m(t).
Then the BVP (1.1) and (1.2) has at least one solution provided that
2
2
kak1 +
kbkL1 < 1.
Γ(α)
Γ(α)
′
Proof. We only need to prove that the hypothesis (H2 ) implies the hypothesis (H2 ) in
Theorem 3.1.
In fact, setting
Z η
−1
Z +∞
β
α−1
(1 −
A = 2kmkL1
(η − s) )l(s)ds +
l(s)ds
,
Γ(α)
0
η
we have
Z +∞
β
−
Γ(α)
Z
η
α−1
(η − s)α−1 f (s, x(s), D0+
x(s))ds
Z η
Z +∞
β
α−1
α−1
α−1
=
(1 −
(η − s) )f (s, x(s), D0+ x(s))ds +
f (s, x(s), D0+
x(s))ds
Γ(α)
0
η
Z η
Z +∞
β
α−1
>
(1 −
(η − s) )(l(s)A − m(s))ds +
(l(s)A − m(s))ds
Γ(α)
0
η
Z η
Z +∞
β
α−1
= 2kmkL1 −
(η − s) )(m(s)ds −
(1 −
m(s)ds ≥ 0.
Γ(α)
0
η
0
α−1
f (s, x(s), D0+
x(s))ds
0
Therefore, the hypothesis (H2 ) in Theorem 3.1 is satisfied and the conclusion of the corollary
follows from Theorem 3.1.
To illustrate how our main results can be used in practice we present an example.
Example 3.1. Consider the following boundary value problem
√
1
3
1
π −t
2
2
2
x(t)) + D0+
x(t)),
(3.10)
D0+ x(t) =
e (3 sin(x2 (t) + D0+
8
1
√
1
2
x(0) = 0, lim D0+
(3.11)
x(t) = πx( ).
t→+∞
4
Conclusion. The BVP (3.10), (3.11) has at least one solution in X = {x ∈ C[0, +∞) :
1
√
2
limt→+∞ x(t)/(1 + t) exists, limt→+∞ D0+
x(t) exists}.
√
1
3
1
Proof. Let α = 2 , β = π, η = 4 , it is easily to see√Γ(α) = βη 2 , that is, the BVP
(3.10), (3.11) is a resonance problem. And let f (t, u, v) = 8π e−t (3 sin(u2 + v) + v), then we
have
√
π −t
|f (t, u, v)| ≤
e (3 + |v|).
8
EJQTDE, 2011 No. 27, p. 12
Setting a(t) = 0, b(t) =
conditions
√
π −t
e , c(t)
8
=
√
3 π −t
e .
8
Obviously (H2 ) is fulfilled as well as the
1
2
2
1
< 1.
3 kak1 +
3 kbkL =
2
Γ( 2 )
Γ( 2 )
1
2
x(t)| > 4 holds for t ∈ [0, +∞), from
Now taking A = 4, for any x ∈ X, assuming |D0+
1
1
2
2
x(t) > A or D0+
x(t) < −A holds for t ∈ [0, +∞).
the condition of (H2 ), we have either D0+
1
2
x(t) > A holds, for t ∈ [0, +∞), then
If D0+
Z +∞ √
1
1
π −s
2
2
x(s)) + D0+
x(s))ds
QNx = w(t)
e (3 sin(x2 (s) + D0+
8
0
!
√
Z 1
1
1
4 1
1
π −s
2
2
−2
e (3 sin(x2 (s) + D0+
x(s)) + D0+
x(s))ds
( − s) 2
4
8
0
Z +∞
8e−s ds(A − 3) > 0,
> w(t)
1
4
1
2
and if D0+
x(t) < −A holds, for t ∈ [0, +∞), then
Z +∞ √
1
1
π −s
2
2
e (3 sin(x2 (s) + D0+
x(s)) + D0+
x(s))ds
QNx = w(t)
8
0
!
√
Z 1
1
1
4 1
1
π −s
2
2
x(s)) + D0+
x(s))ds
−2
( − s) 2
e (3 sin(x2 (s) + D0+
4
8
0
Z +∞
< w(t)
8e−s ds(3 − A) < 0.
1
4
Thus the condition (H1 ) holds.
It is easy to see that for all c > 0, we have
Z +∞ √
√
√
π −s
π
π
2
e (3 sin(c s + c
)+c
)ds
c
8
2
2
0
!
√
√
√
Z 1
4 1
1
π −s
π
π
−2
( − s) 2
e (3 sin(c2 s + c
)+c
)ds
4
8
2
2
0
Z +∞ √
√
π
π −s
c−3
e ds
>c
1
2
8
4
and for all c < 0, we have
Z +∞ √
√
√
π −s
π
π
2
c
e (3 sin(c s + c
)+c
)ds
8
2
2
0
!
√
√
√
Z 1
4 1
1
π −s
π
π
e (3 sin(c2 s + c
)+c
)ds
( − s) 2
−2
4
8
2
2
0
√
Z +∞ √
π
π −s
>c
c+3
e ds.
1
2
8
4
EJQTDE, 2011 No. 27, p. 13
Hence the second inequality in (H3 ) becomes
Z +∞ √
√
√
π −s
π
π
2
e (3 sin(c s + c
)+c
)ds
c
8
2
2
0
!
√
√
√
Z 1
4 1
1
π −s
π
π
−2
( − s) 2
e (3 sin(c2 s + c
)+c
)ds
4
8
2
2
0
√
Z +∞ √
√
π
π
π −s
c −3 ,c
c+3 }
e ds > 0
> min{c
1
2
2
8
4
and is satisfied for all c with |c| > √6π . So the condition (H3 ) holds. Thus, Theorem
3.1 implies BVP (3.10) and (3.11) has at least one solution in X = {x ∈ C[0, +∞) :
1
√
2
x(t) exists}.
limt→+∞ x(t)/(1 + t) exists, limt→+∞ D0+
Remark 1. Example implies that there is a large number of functions that satisfy the
conditions of Theorem 3.1. In addition, the conditions of Theorem 3.1 also easy to check.
Acknowledgment
This work is supported by the National Natural Science Foundation of China (Nos. 10971221
and 10701023) and the Natural Science Foundation of Shanghai (No. 10ZR1400100).
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(Received February 23, 2011)
EJQTDE, 2011 No. 27, p. 16
2011, No. 27, 1-16; http://www.math.u-szeged.hu/ejqtde/
Existence of Solutions for Fractional Differential Equations with
Multi-point Boundary Conditions at Resonance on a Half-line
Huacheng Zhou, Chunhai Kou∗and Feng Xie†
Department of Applied Mathematics, Donghua University
Shanghai 201620, P.R. China
Abstract
In this paper, we investigate the existence of solutions for multi-point boundary
value problems at resonance concerning fractional differential equation on a half-line.
Our analysis relies on the coincidence degree of Mawhin. As an application, an example
is presented to illustrate the main results.
2010 Mathematics Subject Classification. 34A08; 34B10; 34B40.
Key words and phrases. Fractional differential equations; multi-point boundary
conditions; resonance; coincidence degree theorem; half-line
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, 9, 10, 11, 16, 17, 25] 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
[18, 19, 20, 21, 29, 34]).
In this paper, we study the existence of solutions for the fractional differential equation
at resonance with multi-point boundary value problem on a half-line:
α
α−1
D0+
x(t) = f (t, x(t), D0+
x(t)), t ∈ (0, +∞),
α−1
x(t) = βx(η),
x(0) = 0, lim D0+
t→+∞
(1.1)
(1.2)
α
where 1 < α ≤ 2, η > 0, f : [0, +∞) × R × R → R is an S-Carath´eodory function, and D0+
is the standard Riemann-Liouville fractional derivative. Moreover, we need the following
condition
Γ(α) = βη α−1 .
(1.3)
∗
†
E-mail: [email protected].
Corresponding author, E-mail: [email protected].
EJQTDE, 2011 No. 27, p. 1
The condition (1.3) is critical since the fractional differential operator in (1.1) has a
nontrivial kernel. Boundary value problems with such critical conditions are so-called problems at resonance. Boundary value problems for differential equations of integer order at
resonance have been studied by many authors (see [3, 4, 5, 6, 7, 12, 13, 14, 22, 23, 24, 26, 27]
and references cited therein).
More recently, various types of multi-point boundary value problems for fractional differential equations at resonance on a bounded domain have been analyzed by Kosmatov [8],
Jiang [35], Bai [30], Bai and Zhang [31, 32].
However, to our knowledge, it is rare for work to be done on the solutions of fractional
differential equations at resonance on a half-line. In this paper, our goal is to fill this gap
in the literature.
The layout of this paper is as follow. In Section 2, we provide some necessary background.
In particular, we shall introduce some lemmas and definitions related with problem (1.1)
and (1.2). In Section 3, the main results of problem (1.1) and (1.2) will be stated and
proved. Finally, one example is also included to illustrate the main results.
2
Background materials and preliminaries
In this section, to establish the existence of solutions, we present some necessary background
and lemmas. These definitions and lemmas can be found in the recent literature [1, 2, 15,
28, 36].
Definition 2.1. The Riemann-Liouville fractional integral of order α > 0 of a function
f : (0, +∞) → R is given by
Z t
1
α
(t − s)α−1 f (s)ds
I0+ f (t) =
Γ(α) 0
provided the right side is pointwise defined on (0, +∞), and we have
α µ
I0+
t =
Γ(µ + 1) µ+α
t , α > 0, µ > −1.
Γ(µ + α + 1)
Definition 2.2. The Riemann-Liouville fractional derivative of order α > 0 of a function
f : (0, +∞) → R is given by
α
n−α
D0+
f (t) = D n I0+
f (t)
where n = [α] + 1, D n =
dn
,t
dtn
> 0, and we have for λ > −1
α λ
D0+
t =
Γ(λ + 1) λ−α
t .
Γ(λ − α + 1)
Lemma 2.1. Assume that u ∈ C(0, +∞) ∩ Lloc (0, +∞) with a fractional derivative of
order α > 0 that belongs to C(0, +∞) ∩ Lloc (0, +∞). Then
α
α
I0+
D0+
x(t) = x(t) + c1 tα−1 + c2 tα−2 + · · · + cn tα−N ,
for some ci ∈ R, i = 1, . . . , N, where N is the smallest integer greater than or equal to α.
EJQTDE, 2011 No. 27, p. 2
We now give the background from the coincidence degree theory.
Definition 2.3. Let X and Y be normed spaces. A linear operator L : dom(L) ⊂ X →
Y is said to be a Fredholm operator of index zero provided that
(i) ImL is closed subset of Y , and
(ii) dim KerL=codim ImL < +∞.
From definition2.3 it follows that there exist continuous projects P : X → X and
Q : Y → Y such that ImP = KerL, KerQ = ImL, X = KerL ⊕ KerP, Y = ImL ⊕ ImQ and
L|DomL∩KerP : DomL ∩ KerP → ImL is invertible. We denote the inverse that map
by Kp : ImL → DomL ∩ KerP . The generalized inverse of L denoted by KP,Q : Y →
DomL ∩ KerP is defined by KP,Q = KP (I − Q).
Definition 2.4. Let L : Dom(L) ⊂ X → Y be a Fredholm operator, E be a metric
space, and N : E → Y be an operator. We say that N is L-compact on E if QN : E → Y
and KP,Q N : E → X are compact on E.
Definition 2.5. f : [0, +∞) × R2 → R is said to be an S-Carath´eodory function if and
only if
(i) for each (u, v) ∈ R2 , t 7→ f (t, u, v) is measurable on [0, +∞);
(ii) for a.e. t ∈ [0, +∞), (u, v) 7→ f (t, u, v) is continuous on R2 ;
(iii) for each r > 0, there exists ϕr (t) ∈ L1 [0, +∞) ∩ C[0, +∞) satisfying supt≥0 |ϕr (t)| <
+∞, ϕr (t) > 0, t ∈ (0, +∞) such that
max{|u|, |v|} ≤ r impiles |f (t, (1 + tα−1 )u, v)| ≤ ϕr (t), a.e. t ∈ [0, +∞).
Since the Arzela-Ascoli theorem fails to work in the space C∞ , we need a modified compactness criterion to prove that N is L-compact(C∞ = {x ∈ C[0, +∞), limt→+∞ x(t) exists}).
Lemma 2.6.[28] Let M ⊂ C∞ = {x ∈ C[0, +∞), limt→+∞ x(t) exists}. Then M is
relatively compact if the following conditions hold:
(i) all functions from M are uniformly bounded;
(ii) all functions from M are equicontinuous on any compact interval of [0, +∞);
(iii) all functions from M are equiconvergent at infinity, that is, for any given ε > 0,
there exists a T = T (ε) > 0 such that |f (t1 ) − f (t2 )| < ε, for all t1 , t2 > T and f ∈ M.
In this paper, we use the space X, Y defined by
α−1
X = {x ∈ C[0, +∞) : lim x(t)/(1 + tα−1 ) exists, lim D0+
x(t) exists},
t→+∞
t→+∞
Y = {y ∈ C[0, +∞) : y ∈ L1 [0, +∞), sup |y(t)| < +∞},
t≥0
α−1
with the norm kxkX = max{kxk0 , kD0+
xk∞ } and kykY = max{kykL1 , kyk∞ } respectively,
where k · k∞ is the supremum norm on [0, +∞) and kxk0 = supt≥0 |x(t)|/(1 + tα−1 ), kykL1 =
R +∞
|y(s)|ds. By the standard arguments, we can prove that (X, k · kX ), (Y, k · kY ) are
0
Banach spaces.
Define L be the linear operator from DomL ∩ X to Y with
α
DomL = {x ∈ X : D0+
x ∈ L1 [0, +∞) ∩ C[0, +∞), x satisfies (1.2)}
and
α
Lx = D0+
x, x ∈ DomL.
EJQTDE, 2011 No. 27, p. 3
We define N : X → Y by setting
α−1
(Nx)(t) = f (t, x(t), D0+
x(t)).
Lemma 2.7. The operator L : Dom(L) ⊂ X → Y is a Fredholm operator of index zero.
Furthermore, the linear projector operator Q : Y → Y can be defined by
Z +∞
Z η
β
α−1
(Qy)(t) = w(t)
y(s)ds −
(η − s) y(s)ds
Γ(α) 0
0
Rη
R +∞
β
(η − s)α−1 w(s)ds = 1 and the
where w(t) ∈ Y satisfy w(t) > 0 and 0 w(s)ds − Γ(α)
0
linear operator Kp : ImL → DomL ∩ KerP can be written by
Z t
1
KP y(t) =
(t − s)α−1 y(s)ds, t ∈ [0, +∞),
Γ(α) 0
also
kKP ykX ≤
1
kykL1 , for all y ∈ ImL.
Γ(α)
Proof. It is clear that KerL = {x = ctα−1 : c ∈ R}. Now we show that
Z +∞
Z η
β
ImL = {y ∈ Y :
(η − s)α−1 y(s)ds = 0}.
y(s)ds −
Γ(α)
0
0
(2.1)
α
If y ∈ ImL, then there exists a function x ∈ DomL such that y(t) = D0+
x(t). By Lemma2.1,
α
I0+
y(t) = x(t) + c1 tα−1 + c2 tα−2 .
By virtue of the boundary condition (1.2), we have
α−1 α
α
lim D0+
I0+ y(t) = βI0+
y(η)
t→+∞
and therefore
Z
+∞
0
β
y(s)ds −
Γ(α)
Z
0
η
(η − s)α−1 y(s)ds = 0.
(2.2)
α
On the other hand, suppose y ∈ Y and satisfies (2.2), let x(t) = I0+
y(t), then x(0) =
R
t
α−1
α−1
α
0, D0+ x(t) = 0 y(s)ds. Thus limt→+∞ D0+ x(t) = βx(η). Then x ∈ DomL and D0+
x(t) =
y(t). That is to say, (2.1) holds.
For y ∈ Y , taking the projector
Z +∞
Z η
β
α−1
(η − s) y(s)ds .
(Qy)(t) = w(t)
y(s)ds −
Γ(α) 0
0
It is easy to
verify that the operator
R +∞
R η Q is a projector. Letting y1 = y − Qy. Then y1 ∈
ImL(since 0 y1 (s)ds − (β/Γ(α)) 0 (η − s)α−1 y1 (s)ds = 0). Hence Y = ImL + ImQ. From
y ∈ ImQ, there exists a constant c ∈ R, such that y = cw(t), and from y ∈ ImL, we obtain
Z +∞
Z η
β
c=
cw(s)ds −
(η − s)α−1 cw(s)ds = 0,
Γ(α)
0
0
EJQTDE, 2011 No. 27, p. 4
which implies ImL ∩ ImQ = {0} and Y = ImL ⊕ ImQ. Thus we have
dim KerL = dim ImQ = codim ImL = 1.
This implies that L is a Fredholm operator of index zero.
Taking P : X → X as follows:
(P x)(t) =
1
D α−1 x(0)tα−1 ,
Γ(α) 0+
then the generalized inverse KP : ImL → DomL ∩ KerP can be written by
Z t
1
KP y(t) =
(t − s)α−1 y(s)ds, t ∈ [0, +∞).
Γ(α) 0
In fact, for y ∈ ImL, we have
α
(LKP )y(t) = D0+
((KP y)(t)) = y(t),
and for x ∈ DomL ∩ KerP , we know
α
(KP L)x(t) = (KP )D0+
x(t) = x(t) + c1 tα−1 + c2 tα−2 ,
α−1
for some c1 , c2 ∈ R. In view of x ∈ DomL ∩ KerP , D0+
x(0) = 0 and (1.2), we obtain
c1 = c2 = 0. Therefore
(KP L)x(t) = x(t).
This shows that KP = (L|DomL∩KerP )−1 .
Again from the definition of KP , we have
Z
1 t (t − s)α−1
≤ 1 kykL1 ,
kKP yk0 = sup
y(s)ds
Γ(α)
α−1
1+t
t≥0 Γ(α)
0
(2.3)
and
α−1
kD0+
KP yk∞
=
α−1 1
sup |D0+
Γ(α)
t≥0
Z
0
t
(t − s)
α−1
y(s)ds| = sup |
t≥0
Z
0
t
y(s)ds| ≤ kykL1 .
(2.4)
It follows from (2.3) and (2.4) that
kKp ykX ≤
1
kykL1 .
Γ(α)
This completes the proof of Lemma 2.7.
Lemma 2.8. Let f be an S-Carath´eodory function, then N is L-compact.
Proof. Obviously, QN and KP (I − Q)N are continuous. So we only need to prove the
compactness, i.e. QN and KP (I − Q)N maps bounded sets into relatively compact ones.
Suppose U ⊂ X is a bounded set. Then there exists r > 0 such that kxkX ≤ r, for all
x ∈ U. Because f is an S-Carath´eodory function, there exists ϕr (t) ∈ L1 [0, +∞)∩C[0, +∞)
satisfying supt≥0 |ϕr (t)| < +∞, ϕr (t) > 0, t ∈ (0, +∞) such that
α−1
|f (t, x(t), D0+
x(t))| = |f (t, (1 + tα−1 )
x(t)
, D α−1 x(t))| ≤ ϕr (t), a.e. t ∈ [0, +∞).
1 + tα−1 0+
EJQTDE, 2011 No. 27, p. 5
Then for any x ∈ U,
kQNxkL1 =
Z
+∞
0
Z
w(t)(
+∞
α−1
f (s, x(s), D0+
x(s)) ds
0
Z η
β
α−1
α−1
(η − s) f (s, x(s), D0+ x(s))ds) dt
−
Γ(α) 0
Z +∞
Z +∞
Z
βη α−1 η
≤
|w(t)|dt
ϕr (s)ds +
ϕr (s)ds
Γ(α) 0
0
0
βη α−1
kwkL1 kϕr kL1 ,
≤ 1+
Γ(α)
(2.5)
and
kQNxk∞
Z
= sup w(t)(
t≥0
0
+∞
α−1
f (s, x(s), D0+
x(s)) ds
Z η
α−1
α−1
(η − s) f (s, x(s), D0+ x(s))dt)
β
Γ(α) 0
α−1
βη
kwk∞ kϕr kL1 .
≤ 1+
Γ(α)
−
(2.6)
It follows from (2.5) and (2.6) that kQNxkY = max{kQNxkL1 , kQNxk∞ } ≤ (1+ βηΓ(α) )kwkY kϕr kL1 .
Noting that ImQ ≃ R, we have QN is compact.
Furthermore, for any x ∈ U we have
Z t
(KP,Qx)(t)
(t − s)α−1
α−1
= 1
f (s, x(s), D0+
x(s))
1 + tα−1 Γ(α)
α−1
1
+
t
0
Z +∞
α−1
− w(s)
f (τ, x(τ ), D0+
x(τ )) dτ
0
Z η
β
α−1
α−1
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
−
Γ(α) 0
(2.7)
Z +∞
Z t
Z t
1
1
≤
ϕr (s)ds +
w(s)ds
ϕr (τ )dτ
Γ(α) 0
Γ(α) 0
0
Z η
β
α−1
(η − τ ) ϕr (τ )dτ
+
Γ(α) 0
βη α−1
1
1
1+
kwkL1 kϕr kL1 ,
kϕr kL1 +
≤
Γ(α)
Γ(α)
Γ(α)
α−1
and
α−1
|D0+
(KP,Q x)(t)|
Z t
Z +∞
α−1
α−1
= (f (s, x(s), D0+ x(s)) − w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
0
0
Z η
β
α−1
α−1
(2.8)
−
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
Γ(α) 0
βη α−1
kwkL1 kϕr kL1 .
≤ kϕr kL1 + 1 +
Γ(α)
EJQTDE, 2011 No. 27, p. 6
It follows from (2.7) and (2.8) that KP,Q U is uniformly bounded. Meanwhile, for any
t1 , t2 ∈ [0, T ] with T is a positive constant
(KP,Qx)(t1 ) (KP,Q x)(t2 )
1 + tα−1 − 1 + tα−1
1
2
Z t1
α−1
α−1
1
(t
−
s)
(t
−
s)
2
1
α−1
≤
α−1 −
α−1 f (s, x(s), D0+ x(s))
Γ(α) 0
1 + t1
1 + t2
Z +∞
α−1
− w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
0
Z η
β
α−1
α−1
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
−
Γ(α) 0
Z t2
1
(t2 − s)α−1
α−1
+
α−1 f (s, x(s), D0+ x(s))
Γ(α) t1 1 + t2
Z +∞
α−1
− w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
0
Z η
β
α−1
α−1
−
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
Γ(α) 0
(2.9)
Z +∞
Z t1
α−1
α−1
(t2 − s)
1
(t1 − s)
|ϕr (s) + w(s)
ϕr (τ )dτ
≤
α−1 −
Γ(α) 0
1 + t1
1 + tα−1
0
2
Z η
β
α−1
(η − τ ) ϕr (τ )dτ ) ds
−
Γ(α) 0
Z +∞
Z t2
1
(t2 − s)α−1
+
ϕr (τ )dτ
ϕr (s) + w(s)
Γ(α) t1 1 + tα−1
0
2
Z η
β
α−1
(η − τ ) ϕr (τ )dτ ) ds
−
Γ(α) 0
Z t1
α−1
(t1 − s)α−1 (t2 − s)α−1
1
βη
1
≤
) ds
−
α−1 ϕr (s) + w(s)kϕr kL (1 +
Γ(α) 0 1 + tα−1
Γ(α)
1
+
t
1
2
Z t2
1
(t2 − s)α−1
βη α−1
+
ϕr (s) + w(s)kϕr kL1 (1 + Γ(α) ) ds
Γ(α) t1 1 + tα−1
2
→ 0 as t1 → t2
and
α−1
α−1
|D0+
(KP,Q x)(t1 ) − D0+
(KP,Q x)(t2 )|
Z t2
Z +∞
α−1
f (s, x(s), D0+ x(s)) − w(s)
=
ϕr (τ )dτ
t1
0
Z η
β
α−1
(η − τ ) ϕr (τ )dτ ) ds
−
Γ(α) 0
Z t2
α−1
βη
)ds
≤
ϕr (s) + w(s)kϕr kL1 (1 +
Γ(α)
t1
→ 0 as t1 → t2 .
(2.10)
It follows from (2.9) and (2.10) that KP,Q U is equicontinuous. From Lemma 2.6, we can see
α−1
that if KP,Q U/(1+tα−1 ), D0+
KP,QU are equiconvergent at infinity, then KP,Q U is relatively
EJQTDE, 2011 No. 27, p. 7
compact in X. In fact, considering that the following estimate
Z +∞
βη α−1
ε
1
ϕr (s) + w(s)kϕr kL1 (1 +
)ds < ,
Γ(α) L
Γ(α)
3
(2.11)
holding for ε > 0 and some L > 0, we have
lim sup |g(t, s) − 1| ≤ lim g(t, L) = 0,
t→∞ s∈[0,L]
where g(t, s) =
t1 , t2 ≥ T ,
(t−s)α−1
,
1+tα−1
t→∞
s ∈ [0, L], t ∈ [L, +∞). Thus, there exists T > L such that for
sup |g(t1 , s) − g(t2 , s)|
s∈[0,L]
≤ sup |g(t1 , s) − 1| + sup |g(t2 , s) − 1|
s∈[0,L]
s∈[0,L]
(2.12)
−1
βη α−1
ε
kϕr kL1 (1 + kwkL1 (1 +
))
.
<
3
Γ(α)
Therefore, it follows from (2.11) and (2.12) that for t1 , t2 ≥ T , we get
(KP,Q x)(t1 ) (KP,Q x)(t2 )
1 + tα−1 − 1 + tα−1
1
2
Z L
(t1 − s)α−1 (t2 − s)α−1
1
α−1
−
≤
α−1 f (s, x(s), D0+ x(s))
Γ(α) 0 1 + tα−1
1
+
t
1
2
Z +∞
α−1
− w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
0
Z η
β
α−1
α−1
−
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
Γ(α) 0
Z +∞
Z t1
(t1 − s)α−1
1
α−1
α−1
f (τ, x(τ ), D0+
x(τ ))dτ
+
f (s, x(s), D0+ x(s)) − w(s)
Γ(α) L
1 + tα−1
0
1
Z η
β
α−1
α−1
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
−
Γ(α) 0
Z +∞
Z t2
1
(t2 − s)α−1
α−1
α−1
+
x(τ ))dτ
f (τ, x(τ ), D0+
α−1 f (s, x(s), D0+ x(s)) − w(s)
Γ(α) L
1 + t2
0
Z η
β
α−1
α−1
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
−
Γ(α) 0
Z +∞
Z L
ε
α−1
α−1
|f (s, x(s), D0+ x(s))| + w(s)
f (τ, x(τ ), D0+
x(τ ))dτ
≤
3
0
0
−1
Z η
β
βη α−1
α−1
α−1
−
(η − τ ) f (τ, x(τ ), D0+ x(τ ))dτ ) ds
))
kϕr kL1 (1 + kwkL1 (1 +
Γ(α) 0
Γ(α)
Z t1
βη α−1
1
ϕr (s) + w(s)kϕr kL1 (1 +
)ds
+
Γ(α) L
Γ(α)
Z t2
1
βη α−1
+
ϕr (s) + w(s)kϕr kL1 (1 +
)ds
Γ(α) L
Γ(α)
≤ε
EJQTDE, 2011 No. 27, p. 8
and
α−1
α−1
|D0+
(KP,Q x)(t1 ) − D0+
(KP,Q x)(t2 )|
Z +∞
Z max{t1 ,t2 }
α−1
f (s, x(s), D x(s)) − w(s)
≤
ϕr (τ )dτ
0+
0
min{t1 ,t2 }
Z η
β
α−1
(η − τ ) ϕr (τ )dτ ) ds
−
Γ(α) 0
Z +∞
βη α−1
1
)ds
≤
ϕr (s) + w(s)kϕr kL (1 +
Γ(α)
L
≤ ε.
So we complete the proof.
The following fixed point theorem due to Mawhin is fundamental in the proofs of our
main results.
Lemma 2.9.[36] Let Ω ⊂ X be open and bounded, L be a Fredholm mapping of index
zero and N be L-compact on Ω. Assume that the following conditions are satisfied:
(i) Lx 6= λNx for every (x, λ) ∈ (DomL\KerL) ∩ ∂Ω) × (0, 1);
(ii) Nx ∈
/ImL for every x ∈KerL ∩ ∂Ω;
(iii) deg(JQN|KerL∩∂Ω , Ω ∩ KerL, 0)6= 0, with Q : Y → Y a continuous projector such
that KerQ=ImL and J :ImQ →KerL is an isomorphism.
Then the equation Lx = Nx has at least one solution in DomL ∩ Ω.
3
Main results
In this section, we present our main results and prove them.
Theorem 3.1. Assume that f is an S-Carath´eodory function and the following conditions are satisfied:
(H1 ) There exist functions a, b, c ∈ L1 [0, +∞) such that
|f (t, u, v)| ≤ a(t)|u| + b(t)|v| + c(t), a.e. t ∈ [0, +∞) and all (u, v) ∈ R2
and
kak1 =
Z
(3.1)
+∞
a(s)(1 + tα−1 )ds < +∞;
0
α−1
(H2 ) There exists constant A > 0 such that for x ∈ DomL, if |D0+
x(t)| > A for all
t ∈ [0, +∞), then
Z η
Z +∞
β
α−1
α−1
(η − s)α−1 f (s, x(s), D0+
x(s))ds 6= 0;
f (s, x(s), D0+ x(s))ds −
Γ(α)
0
0
(H3 ) There exists constant B > 0 such that for all c ∈ R with |c| > B, either
Z +∞
Z η
β
α−1
α−1
α−1
c
(η − s) f (s, cs , cΓ(α))ds < 0
f (s, cs , cΓ(α))ds −
Γ(α) 0
0
(3.2)
EJQTDE, 2011 No. 27, p. 9
or else
Z
c
+∞
f (s, cs
α−1
0
β
, cΓ(α))ds −
Γ(α)
Z
η
(η − s)
0
α−1
f (s, cs
α−1
, cΓ(α))ds
> 0.
(3.3)
Then the BVP (1.1), (1.2) has at least one solution provided that
2
2
kak1 +
kbkL1 < 1.
Γ(α)
Γ(α)
Proof. We construct an open bounded set Ω ⊂ X that satisfies the assumption of
Lemma 2.9. Let
Ω1 = {x ∈ DomL\KerL : Lx = λNx, for some λ ∈ [0, 1]}.
For x ∈ Ω1 , we have x ∈
/KerL, λ 6= 0 and Nx ∈ ImL. Note that KerQ = ImL and, thus,
Z +∞
Z η
β
α−1
α−1
f (s, x(s), D0+ x(s))ds −
(η − s)α−1 f (s, x(s), D0+
x(s))ds = 0
Γ(α) 0
0
α−1
since QNx = 0. It follows from (H2 ) that there exists t0 ∈ [0, +∞) such that |D0+
x(t0 )| ≤
R
t
0
α−1
α−1
α
A. In view of D0+ x(0) = D0+ x(t0 ) − 0 D0+ x(t)dt, we have
α−1
α
|D0+
x(0)| ≤ A + kD0+
xkL1 = A + kLxkL1 ≤ A + kNxkL1 .
(3.4)
Again for x ∈ Ω1 , x ∈ DomL\KerL, then (I − P )x ∈ DomL ∩ KerP and LP x = 0, thus
from Lemma 2.7, we have
k(I − P )xkX = kKP L(I − P )xkX ≤
1
kL(I − P )xkL1
Γ(α)
1
1
kLxkL1 ≤
kNxkL1 .
=
Γ(α)
Γ(α)
(3.5)
From (3.4), (3.5), we have
kxkX ≤ kP xkX + k(I − P )xkX ≤
A
2
+
kNxkL1 .
Γ(α) Γ(α)
(3.6)
If (3.6) holds, from (3.1), we have
kxkX ≤
2
A
α−1
xk∞ + kckL1 ).
+
(kak1 kxk0 + kbkL1 kD0+
Γ(α) Γ(α)
(3.7)
Thus from kxk0 ≤ kxkX and (3.7), we have
kxk0 ≤
A
1
1−
(
2
kak1 Γ(α)
Γ(α)
+
2
α−1
xk∞ + kckL1 )).
(kbkL1 kD0+
Γ(α)
(3.8)
α−1
Again from (3.7), (3.8) and kD0+
xk∞ ≤ kxkX we obtain
α−1
kD0+
xk∞ ≤
1
1−
2
kak1
Γ(α)
A
−
(
2
1 Γ(α)
kbk
L
Γ(α)
+
2
kckL1 ).
Γ(α)
(3.9)
EJQTDE, 2011 No. 27, p. 10
It follows from (3.8) and (3.9) that, there exists an M > 0 such that kxkX ≤ M for all
x ∈ Ω1 , that is, Ω1 is bounded.
Let
Ω2 = {x ∈ KerL : Nx ∈ ImL}.
For x ∈ Ω2 , x ∈ KerL implies that x can be expressed by x = ctα−1 , where c is an arbitrary
constant, and QNx = 0, thus
Z +∞
Z η
1
α−1
α−1
α−1
(η − s) f (s, cs , cΓ(α))ds = 0.
w(t)
f (s, cs , cΓ(α))ds −
Γ(α) 0
0
1
1
It follows from (H2 ) that, we get kxkX ≤ Γ(α)
|c| ≤ Γ(α)
A. So Ω2 is bounded too.
We define the isomorphism J : ImQ → KerL by
J(cw(t)) = ctα−1 , c ∈ R.
If the first part of (H3 ) is fulfilled, we set
Ω3 = {x ∈ KerL : −λJ −1 x + (1 − λ)QNx = 0}.
For every x = ctα−1 ∈ Ω3 , one has λJ −1 x = (1 − λ)QNx,
Z +∞
Z η
β
α−1
α−1
α−1
λcw(t) = (1−λ)w(t)
f (s, cs , cΓ(α))ds −
(η − s) f (s, cs , cΓ(α))ds .
Γ(α) 0
0
If λ = 1, then c = 0 and, if |c| > B, in view of (3.2), one has
Z +∞
Z η
β
α−1
α−1
2
α−1
(η − s) f (s, cs , cΓ(α))ds < 0,
λc w(t) = (1−λ)w(t)c
f (s, cs , cΓ(α))ds −
Γ(α) 0
0
which contradicts λc2 w(t) > 0. If the other part of (H3 ) is satisfied, we take
Ω3 = {x ∈ KerL : λJ −1 x + (1 − λ)QNx = 0}
1
1
and, again, obtain a contradiction. Thus, in ether case kxkX ≤ ( Γ(α)
+ 1)|c| ≤ ( Γ(α)
+ 1)B,
that is, Ω3 is bounded.
In what follows, we shall prove that all conditions of Lemma 2.9 are satisfied. Set Ω be
a bounded open subset of X such that ∪3i=1 Ωi ⊂ Ω. we know that L is a Fredholm operator
of index zero and N is L-compact on Ω. By the definition of Ω, we have
(i)Lx 6= λNx for every (x, λ) ∈ (DomL\KerL) ∩ ∂Ω) × (0, 1);
(ii) Nx ∈
/ImL, for every x ∈ KerL ∩ ∂Ω.
At last we will prove that (iii) of Lemma 2.9 is satisfied. To this end, let
H(x, λ) = ±λIdx + (1 − λ)JQNx,
where Id is the identical operator. By virtue of the definition of Ω, we know Ω ⊃ Ω3 , thus
H(x, λ) 6= 0 for x ∈ KerL ∩ ∂Ω, then by homotopy property of degree, we get
deg(JQN|KerL∩∂Ω , Ω ∩ KerL, 0) = deg(H(·, 0), Ω ∩ KerL, 0)
= deg(H(·, 1), Ω ∩ KerL, 0)
= deg(±Id, Ω ∩ KerL, 0)
= ±1 6= 0.
EJQTDE, 2011 No. 27, p. 11
So, the third assumption of Lemma 2.9 is fulfilled and Lx = Nx has at least one solution
in DomL ∩ Ω. The proof is complete.
Corollary 3.2. Assume that f is an S-Carath´eodory function and the conditions (H1 )
and (H3 ) in Theorem 3.1 are satisfied with (H2 ) replaced with
′
(H2 ): There exist functions l, m ∈ L1 [0, +∞) with l(t), m(t) ≥ 0 and l(t) ≡
/0 such that
f (t, u, v) ≥ l(t)|v| − m(t).
Then the BVP (1.1) and (1.2) has at least one solution provided that
2
2
kak1 +
kbkL1 < 1.
Γ(α)
Γ(α)
′
Proof. We only need to prove that the hypothesis (H2 ) implies the hypothesis (H2 ) in
Theorem 3.1.
In fact, setting
Z η
−1
Z +∞
β
α−1
(1 −
A = 2kmkL1
(η − s) )l(s)ds +
l(s)ds
,
Γ(α)
0
η
we have
Z +∞
β
−
Γ(α)
Z
η
α−1
(η − s)α−1 f (s, x(s), D0+
x(s))ds
Z η
Z +∞
β
α−1
α−1
α−1
=
(1 −
(η − s) )f (s, x(s), D0+ x(s))ds +
f (s, x(s), D0+
x(s))ds
Γ(α)
0
η
Z η
Z +∞
β
α−1
>
(1 −
(η − s) )(l(s)A − m(s))ds +
(l(s)A − m(s))ds
Γ(α)
0
η
Z η
Z +∞
β
α−1
= 2kmkL1 −
(η − s) )(m(s)ds −
(1 −
m(s)ds ≥ 0.
Γ(α)
0
η
0
α−1
f (s, x(s), D0+
x(s))ds
0
Therefore, the hypothesis (H2 ) in Theorem 3.1 is satisfied and the conclusion of the corollary
follows from Theorem 3.1.
To illustrate how our main results can be used in practice we present an example.
Example 3.1. Consider the following boundary value problem
√
1
3
1
π −t
2
2
2
x(t)) + D0+
x(t)),
(3.10)
D0+ x(t) =
e (3 sin(x2 (t) + D0+
8
1
√
1
2
x(0) = 0, lim D0+
(3.11)
x(t) = πx( ).
t→+∞
4
Conclusion. The BVP (3.10), (3.11) has at least one solution in X = {x ∈ C[0, +∞) :
1
√
2
limt→+∞ x(t)/(1 + t) exists, limt→+∞ D0+
x(t) exists}.
√
1
3
1
Proof. Let α = 2 , β = π, η = 4 , it is easily to see√Γ(α) = βη 2 , that is, the BVP
(3.10), (3.11) is a resonance problem. And let f (t, u, v) = 8π e−t (3 sin(u2 + v) + v), then we
have
√
π −t
|f (t, u, v)| ≤
e (3 + |v|).
8
EJQTDE, 2011 No. 27, p. 12
Setting a(t) = 0, b(t) =
conditions
√
π −t
e , c(t)
8
=
√
3 π −t
e .
8
Obviously (H2 ) is fulfilled as well as the
1
2
2
1
< 1.
3 kak1 +
3 kbkL =
2
Γ( 2 )
Γ( 2 )
1
2
x(t)| > 4 holds for t ∈ [0, +∞), from
Now taking A = 4, for any x ∈ X, assuming |D0+
1
1
2
2
x(t) > A or D0+
x(t) < −A holds for t ∈ [0, +∞).
the condition of (H2 ), we have either D0+
1
2
x(t) > A holds, for t ∈ [0, +∞), then
If D0+
Z +∞ √
1
1
π −s
2
2
x(s)) + D0+
x(s))ds
QNx = w(t)
e (3 sin(x2 (s) + D0+
8
0
!
√
Z 1
1
1
4 1
1
π −s
2
2
−2
e (3 sin(x2 (s) + D0+
x(s)) + D0+
x(s))ds
( − s) 2
4
8
0
Z +∞
8e−s ds(A − 3) > 0,
> w(t)
1
4
1
2
and if D0+
x(t) < −A holds, for t ∈ [0, +∞), then
Z +∞ √
1
1
π −s
2
2
e (3 sin(x2 (s) + D0+
x(s)) + D0+
x(s))ds
QNx = w(t)
8
0
!
√
Z 1
1
1
4 1
1
π −s
2
2
x(s)) + D0+
x(s))ds
−2
( − s) 2
e (3 sin(x2 (s) + D0+
4
8
0
Z +∞
< w(t)
8e−s ds(3 − A) < 0.
1
4
Thus the condition (H1 ) holds.
It is easy to see that for all c > 0, we have
Z +∞ √
√
√
π −s
π
π
2
e (3 sin(c s + c
)+c
)ds
c
8
2
2
0
!
√
√
√
Z 1
4 1
1
π −s
π
π
−2
( − s) 2
e (3 sin(c2 s + c
)+c
)ds
4
8
2
2
0
Z +∞ √
√
π
π −s
c−3
e ds
>c
1
2
8
4
and for all c < 0, we have
Z +∞ √
√
√
π −s
π
π
2
c
e (3 sin(c s + c
)+c
)ds
8
2
2
0
!
√
√
√
Z 1
4 1
1
π −s
π
π
e (3 sin(c2 s + c
)+c
)ds
( − s) 2
−2
4
8
2
2
0
√
Z +∞ √
π
π −s
>c
c+3
e ds.
1
2
8
4
EJQTDE, 2011 No. 27, p. 13
Hence the second inequality in (H3 ) becomes
Z +∞ √
√
√
π −s
π
π
2
e (3 sin(c s + c
)+c
)ds
c
8
2
2
0
!
√
√
√
Z 1
4 1
1
π −s
π
π
−2
( − s) 2
e (3 sin(c2 s + c
)+c
)ds
4
8
2
2
0
√
Z +∞ √
√
π
π
π −s
c −3 ,c
c+3 }
e ds > 0
> min{c
1
2
2
8
4
and is satisfied for all c with |c| > √6π . So the condition (H3 ) holds. Thus, Theorem
3.1 implies BVP (3.10) and (3.11) has at least one solution in X = {x ∈ C[0, +∞) :
1
√
2
x(t) exists}.
limt→+∞ x(t)/(1 + t) exists, limt→+∞ D0+
Remark 1. Example implies that there is a large number of functions that satisfy the
conditions of Theorem 3.1. In addition, the conditions of Theorem 3.1 also easy to check.
Acknowledgment
This work is supported by the National Natural Science Foundation of China (Nos. 10971221
and 10701023) and the Natural Science Foundation of Shanghai (No. 10ZR1400100).
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(Received February 23, 2011)
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