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J. Nonlinear Sci. Appl. 8 (2015), 99–109
Research Article
Positive solutions for Caputo fractional differential
equations involving integral boundary conditions
Yong Wang∗, Yang Yang
School of Science, Jiangnan University, Wuxi 214122, China.
Communicated by J. J. Nieto
Abstract
In this work we study integral boundary value problem involving Caputo differentiation
 q
c

 Dt u(t) = f (t, u(t)), 0 < t < 1,
Z 1
Z 1
0
0

h(t)u(t)dt, γu (0) − δu (1) =
g(t)u(t)dt,
 αu(0) − βu(1) =
0
0
where α, β, γ, δ are constants with α > β > 0, γ > δ > 0, f ∈ C([0, 1] × R+ , R), g, h ∈ C([0, 1], R+ ) and c Dtq
is the standard Caputo fractional derivative of fractional order q(1 < q < 2). By using some fixed point
c
theorems we prove the existence of positive solutions. 2015
All rights reserved.
Keywords: Caputo fractional boundary value problem, fixed point theorem, positive solution.
2010 MSC: 34B15, 34B25.
1. Introduction
In recent years, fractional differential equations have been widely used in diffusion and transport theory,
chaos and turbulence, viscoelastic mechanics, non-newtonian fluid mechanics etc. It has received highly
attention and becomes one of the hottest issues in the international research field. For instance, Westerlund
[14] utilized fractional differential equations to depict the transmission of electromagnetic wave, the one
dimensional model is
∂ 2 E(x, t)
∂ 2 E(x, t)
ν
+
µ
ε
x
D
E(x,
t)
+
= 0,
µ 0 ε0
0
0
00
t
∂t2
∂t2
∗
Corresponding author
Email addresses: yongwangjn@163.com (Yong Wang), yynjnu@126.com (Yang Yang)
Received 2014-12-14
Y. Wang, Y. Yang, J. Nonlinear Sci. Appl. 8 (2015), 99–109
100
ν
where µ0 , ε0 , x0 are constants, 0 Dtν E(x, t) = ∂ E(x,t)
is a fractional derivative.
∂tν
As an excellent tool, fixed point method is used for investigating nonlinear boundary value problems
and there are a lot of papers devoted to this direction. We refer the reader to some papers involving
fractional differential equations [1, 2, 4, 5, 11, 12, 13, 15, 16, 17, 18] and the references therein. For example,
Leggett-Williams fixed point theorem is used to study the existence of multiple positive solutions for some
integral boundary value problems [4, 11, 16]. However, all these works were done under assumption that the
nonlinear term is nonnegative. Therefore, it is natural to discuss the existence of positive solutions while
the nonlinear term is sign-changing, for instance, see [15, 17, 18].
In [17] the author obtained the existence of positive solutions for the coupled integral boundary value
problem for systems of nonlinear fractional q-difference equations
Dqα u(t) + λf (t, u(t), v(t)) = 0, Dqβ v(t) + λg(t, u(t), v(t)) = 0, t ∈ (0, 1), λ > 0,
Z 1
Z 1
Dqj u(0) = Dqj v(0) = 0, 0 ≤ j ≤ n − 2, u(1) = µ
v(s)dq s, v(1) = ν
u(s)dq s.
0
(1.1)
0
Under the semipositone nonlinearities, by applying the nonlinear alternative of Leray-Schauder type and
Krasnoselskii’s fixed point theorems, several existence theorems for (1.1) had been established.
Motivated by the above works, we investigate the existence of positive solutions for integral boundary
value problems involving Caputo differentiation
 q
c

 Dt u(t) = f (t, u(t)), 0 < t < 1,
Z 1
Z 1
(1.2)
0
0

h(t)u(t)dt, γu (0) − δu (1) =
g(t)u(t)dt,
 αu(0) − βu(1) =
0
0
where α, β, γ, δ are real constants with α > β > 0, γ > δ > 0, f ∈ C([0, 1] × R+ , R), g, h ∈ C([0, 1], R+ )
and c Dtq is the standard Caputo fractional derivative of fractional order q(1 < q < 2). We consider the two
cases:
(1) The nonlinearity is asymptotically linear at infinity, maybe it is negative and unbounded.
(2) The nonlinearity is bounded from below, including sign-changing.
2. Preliminaries
We first offer some basic definitions and facts used throughout this paper. For more details, see [7, 9, 10].
Definition 2.1. For a function f given on the interval [a, b], the Caputo derivative of fractional order q is
defined as
Z t
1
c q
D f (t) =
(t − s)n−q−1 f (n) (s)ds, n = [q] + 1,
Γ(n − q) 0
where [q] denotes the integer part of q.
Definition 2.2. The Riemann-Liouville fractional integral of order q for a function f is defined as
Z t
1
I q f (t) =
(t − s)q−1 f (s)ds, q > 0,
Γ(q) 0
provided that such integral exists.
Lemma 2.3. Let q > 0. Then the differential equation c Dq u(t) = 0 has solutions
u(t) = c0 + c1 t + c2 t2 + · · · + cn−1 tn−1 ,
where ci ∈ R, i = 0, 1, 2, . . . , n, n = [q] + 1.
Y. Wang, Y. Yang, J. Nonlinear Sci. Appl. 8 (2015), 99–109
101
Lemma 2.4. Let q > 0. Then
I q (c Dq u)(t) = u(t) + c0 + c1 t + c2 t2 + · · · + cn−1 tn−1 ,
where ci ∈ R, i = 0, 1, 2, . . . , n, n = [q] + 1.
Lemma 2.5. Let q ∈ (1, 2) and y ∈ C[0, 1]. Then boundary value problem
( q
c
Dt u(t) = y(t), 0 < t < 1,
αu(0) − βu(1) = 0, γu0 (0) − δu0 (1) = 0,
has a unique solution u in the form
Z
1
G(t, s)y(s)ds,
u(t) =
0
where
G(t, s) =

(t−s)q−1

 Γ(q) +
β(1−s)q−1
(α−β)Γ(q)

 β(1−s)q−1 +
(α−β)Γ(q)
+
βδ(q−1)(1−s)q−2
(α−β)(γ−δ)Γ(q)
βδ(q−1)(1−s)q−2
(α−β)(γ−δ)Γ(q)
+
+
δ(q−1)t(1−s)q−2
,
(γ−δ)Γ(q)
δ(q−1)t(1−s)q−2
,
(γ−δ)Γ(q)
0 ≤ s ≤ t ≤ 1,
0 ≤ t ≤ s ≤ 1.
Proof. By Definitions 2.1, 2.2 and Lemmas 2.3, 2.4, we have
Z t
1
u(t) =
(t − s)q−1 y(s)ds + c1 + c2 t
Γ(q) 0
for c1 , c2 ∈ R. Then
(2.1)
Z 1
1
u(0) = c1 , u(1) =
(1 − s)q−1 y(s)ds + c1 + c2 ,
Γ(q) 0
Z
q−1 1
0
0
(1 − s)q−2 y(s)ds + c2 .
u (0) = c2 , u (1) =
Γ(q) 0
In view of the boundary conditions αu(0) − βu(1) = 0, γu0 (0) − δu0 (1) = 0, we obtain
Z 1
Z 1
β
βδ(q − 1)
q−1
c1 =
(1 − s) y(s)ds +
(1 − s)q−2 y(s)ds,
(α − β)Γ(q) 0
(α − β)(γ − δ)Γ(q) 0
Z 1
δ(q − 1)
c2 =
(1 − s)q−2 y(s)ds.
(γ − δ)Γ(q) 0
Substituting c1 , c2 into the equation (2.1), we find
Z t
Z 1
1
δ(q − 1)t
q−1
u(t) =
(t − s) y(s)ds +
(1 − s)q−2 y(s)ds
Γ(q) 0
(γ − δ)Γ(q) 0
Z 1
Z 1
β
βδ(q − 1)
q−1
+
(1 − s) y(s)ds +
(1 − s)q−2 y(s)ds
(α − β)Γ(q) 0
(α − β)(γ − δ)Γ(q) 0
Z 1
=
G(t, s)y(s)ds.
0
This completes the proof.
Lemma 2.6. Let q ∈ (1, 2). Then boundary value problem
 q
c

 Dt u(t) = 0, 0 < t < 1,
Z 1
Z
0
0

αu(0)
−
βu(1)
=
h(t)u(t)dt,
γu
(0)
−
δu
(1)
=

0
0
1
g(t)u(t)dt
Y. Wang, Y. Yang, J. Nonlinear Sci. Appl. 8 (2015), 99–109
102
can be expressed in the form
where φ(t) :=
β+(α−β)t
(α−β)(γ−δ)
1
Z
1
u(t) =
α−β
1
Z
g(t)u(t)dt,
h(t)u(t)dt + φ(t)
0
0
for t ∈ [0, 1].
Proof. By Lemma 2.3 we see
u(t) = c3 + c4 t, where c3 , c4 ∈ R.
Consequently,
Z
1
Z
h(t)u(t)dt, (γ − δ)c4 =
(α − β)c3 − βc4 =
u(t) =
t
γ−δ
1
Z
g(t)u(t)dt +
0
1
α−β
g(t)u(t)dt.
0
0
Hence
1
Z
1
h(t)u(t)dt +
0
β
(α − β)(γ − δ)
Z
1
g(t)u(t)dt.
0
This completes the proof.
Let q ∈ (1, 2) and y ∈ C[0, 1]. Then from Lemmas 2.5, 2.6 we can obtain the boundary value problem
 q
c

 Dt u(t) = y(t), 0 < t < 1,
Z 1
Z 1
(2.2)
0
0

αu(0)
−
βu(1)
=
h(t)u(t)dt,
γu
(0)
−
δu
(1)
=
g(t)u(t)dt

0
0
is equivalent to
Z
u(t) =
0
1
1
G(t, s)y(s)ds +
α−β
Z
1
Z
h(t)u(t)dt + φ(t)
0
1
g(t)u(t)dt.
0
Throughout this paper we always assume that the following condition holds:
(H1) κ = κ1 κ4 − κ2 κ3 > 0, κ1 ≥ 0, κ4 ≥ 0, where
Z 1
Z 1
1
κ1 = 1 −
h(t)dt, κ2 =
h(t)φ(t)dt,
α−β 0
0
Z 1
Z 1
1
κ3 =
g(t)dt, κ4 = 1 −
g(t)φ(t)dt.
α−β 0
0
Lemma 2.7. Suppose (H1) holds. Then (2.3) is equivalent to
Z 1
u(t) =
H(t, s)y(s)ds,
0
where
Z 1
Z 1
1
H(t, s) = G(t, s) +
κ4
h(t)G(t, s)dt + κ2
g(t)G(t, s)dt
κ(α − β)
0
0
Z 1
Z 1
φ(t)
+
κ3
h(t)G(t, s)dt + κ1
g(t)G(t, s)dt .
κ
0
0
Proof. Multiplying h(t) on both sides of (2.3) and integrating over [0, 1], we find
Z 1
h(t)u(t)dt
0
Z 1
Z 1
Z 1
Z 1
Z 1
Z 1
1
=
h(t)
G(t, s)y(s)dsdt +
h(t)dt
h(t)u(t)dt +
h(t)φ(t)dt
g(t)u(t)dt.
α−β 0
0
0
0
0
0
(2.3)
Y. Wang, Y. Yang, J. Nonlinear Sci. Appl. 8 (2015), 99–109
103
Similarly,
Z 1
g(t)u(t)dt
Z 1
Z 1
=
g(t)
G(t, s)y(s)dsdt +
0
0
0
1
α−β
1
Z
Z
g(t)dt
0
1
Z
h(t)u(t)dt +
0
1
Z
g(t)φ(t)dt
0
1
g(t)u(t)dt.
0
Consequently, we get
κ1 −κ2
−κ3 κ4
# " R
#
R1
" R1
1
h(t)u(t)dt
h(t)
G(t,
s)y(s)dsdt
R01
R 01
= R01
.
0 g(t)u(t)dt
0 g(t) 0 G(t, s)y(s)dsdt
Therefore,
#
" R
#
R1
" R1
1
1
κ
κ
h(t)
G(t,
s)y(s)dsdt
h(t)u(t)dt
4
2
R01
R 01
R01
.
=
κ κ3 κ1
0 g(t) 0 G(t, s)y(s)dsdt
0 g(t)u(t)dt
As a result, we have
1
Z
u(t) =
G(t, s)y(s)ds
Z 1
Z 1
Z 1
Z 1
1
+
κ4
h(t)
G(t, s)y(s)dsdt + κ2
g(t)
G(t, s)y(s)dsdt
κ(α − β)
0
0
0
0
Z 1
Z 1
Z 1
Z 1
φ(t)
+
κ3
h(t)
G(t, s)y(s)dsdt + κ1
g(t)
G(t, s)y(s)dsdt .
κ
0
0
0
0
0
This completes the proof.
Lemma 2.8. Let
Z 1
Z 1
φ(0)
κ4
h(t)dt + κ2
g(t)dt +
κ3
h(t)dt + κ1
g(t)dt ,
κ
0
0
0
0
Z 1
Z 1
Z 1
Z 1
φ(1)
1
κ4
h(t)dt + κ2
g(t)dt +
κ3
h(t)dt + κ1
g(t)dt .
K2 := 1 +
κ(α − β)
κ
0
0
0
0
1
K1 := 1 +
κ(α − β)
Z
1
Z
1
Then the following inequalities hold:
K1 M(s) ≤ H(t, s) ≤
α
K2 M(s), ∀t ∈ [0, 1], s ∈ (0, 1).
β
Proof. Let
g1 (t, s) =
(t − s)q−1
β(1 − s)q−1
βδ(q − 1)(1 − s)q−2 δ(q − 1)t(1 − s)q−2
+
+
+
, 0 ≤ s ≤ t ≤ 1,
Γ(q)
(α − β)Γ(q)
(α − β)(γ − δ)Γ(q)
(γ − δ)Γ(q)
and
g2 (t, s) =
β(1 − s)q−1
βδ(q − 1)(1 − s)q−2 δ(q − 1)t(1 − s)q−2
+
+
, 0 ≤ t ≤ s ≤ 1.
(α − β)Γ(q)
(α − β)(γ − δ)Γ(q)
(γ − δ)Γ(q)
For given s ∈ (0, 1), g1 , g2 are increasing with respect to t for t ∈ [0, 1]. Hence,
min G(t, s) = min{ min g1 (t, s), min g2 (t, s)} = min{g1 (s, s), g2 (0, s)} = g2 (0, s)
t∈[0,1]
t∈[s,1]
=
s)q−1
t∈[0,s]
βδ(q − 1)(1 − s)q−2
β(1 −
+
:= M(s),
(α − β)Γ(q)
(α − β)(γ − δ)Γ(q)
Y. Wang, Y. Yang, J. Nonlinear Sci. Appl. 8 (2015), 99–109
104
max G(t, s) = max{ max g1 (t, s), max g2 (t, s)} = max{g1 (1, s), g2 (s, s)} = g1 (1, s)
t∈[0,1]
t∈[s,1]
=
s)q−1
(1 −
Γ(q)
t∈[0,s]
+
β(1 − s)q−1
βδ(q − 1)(1 − s)q−2 δ(q − 1)(1 − s)q−2
α
+
+
= M(s).
(α − β)Γ(q)
(α − β)(γ − δ)Γ(q)
(γ − δ)Γ(q)
β
Therefore,
M(s) ≤ G(t, s) ≤
α
M(s), ∀t ∈ [0, 1], s ∈ (0, 1).
β
This yields
Z 1
Z 1
1
g(t)G(t, s)dt
h(t)G(t, s)dt + κ2
κ4
K1 M(s) ≤ G(t, s) +
κ(α − β)
0
0
Z 1
Z 1
φ(t)
+
g(t)G(t, s)dt
h(t)G(t, s)dt + κ1
κ3
κ
0
0
α
≤ K2 M(s).
β
This completes the proof.
Let E := C[0, 1], kuk := maxt∈[0,1] |u(t)|, P := {u ∈ E : u(t) ≥ 0, ∀t ∈ [0, 1]}. Then (E, k · k) becomes
a real Banach space and P is a cone on E.
Definition 2.9. Given a cone P in a real Banach space E, a functional α : P → [0, ∞) is said to be
nonnegative continuous concave on P , provided α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y), for all x, y ∈ P with
t ∈ [0, 1].
Let a, b, r > 0 be constants and α as defined above, we denote Pr = {y ∈ P : kyk < r}, P {α, a, b} =
{y ∈ P : α(y) ≥ a, kyk ≤ b}.
Lemma 2.10. (Leggett-Williams fixed point theorem, see [3, 8]) Assume E is a real Banach space, P ⊂ E
is a cone. Let A : P c → P c be completely continuous and α be a nonnegative continuous concave functional
on P such that α(y) ≤ kyk, for y ∈ P c . Suppose that there exist 0 < a < b < d ≤ c such that
(1) {y ∈ P (α, b, d)| α(y) > b} =
6 ∅ and α(Ay) > b, for all y ∈ P (α, b, d),
(2) kAyk < a, for all kyk ≤ a,
(3) α(Ay) > b for all y ∈ P (α, b, c) with kAyk > d.
Then A has at least three fixed points y1 , y2 , y3 satisfying
ky1 k < a,
b < α(y2 ),
ky3 k > a,
α(y3 ) < b.
Lemma 2.11. (see [6]) Let E be a Banach space, and A : E → E be a completely continuous operator.
Assume that T : E → E is a bounded linear operator such that 1 is not an eigenvalue of T and
kAu − T uk
= 0.
kuk
kuk→∞
lim
Then A has a fixed point in E.
Define A : E → E
Z
(Au)(t) =
1
H(t, s)f (s, u(s))ds.
0
Then, by Lemmas 2.5, 2.6 and 2.7, the existence of solutions for (1.2) is equivalent to the existence of fixed
points for the operator A. Furthermore, in view of the continuity H and f , we can adopt the Ascoli-Arzela
theorem to prove A is a completely continuous operator.
Y. Wang, Y. Yang, J. Nonlinear Sci. Appl. 8 (2015), 99–109
105
3. Main results
For convenience, we set
Z 1
α
β
K1 β
α
βδ
, θ=
ξ=
K2 M(s)ds = K2
+
,
β
q(α − β)Γ(q) (α − β)(γ − δ)Γ(q)
K2 α
0 β
α2 K22
β
βδ
D1 =
,
+
K1 β 2 q(α − β)Γ(q) (α − β)(γ − δ)Γ(q)
Z 1
β
βδ
M(s)ds = K1
l = K1
.
+
q(α − β)Γ(q) (α − β)(γ − δ)Γ(q)
0
Theorem 3.1. Let (H1) hold true. Moreover, suppose that
(H2) f (t, 0) 6≡ 0 for all t ∈ [0, 1],
(H3) limu→+∞ f (t,u)
= λ, uniformly for t ∈ [0, 1], where |λ| < ξ −1 .
u
Then (1.2) has a positive solution.
Proof. Define T : P → P
Z
1
H(t, s)u(s)ds.
(T u)(t) = λ
(3.1)
0
Clearly, T is a bounded linear operator, and by Lemmas 2.5, 2.6 and 2.7 we know that (3.1) is equivalent to
 q
c

 Dt u(t) = λu(t), 0 < t < 1,
Z 1
Z 1
(3.2)

h(t)u(t)dt, γu0 (0) − δu0 (1) =
g(t)u(t)dt.
 αu(0) − βu(1) =
0
0
Next we show 1 is not an eigenvalue of T . We divide two cases.
Case 1. λ = 0.
This implies c Dtq u(t) = 0, and by Lemma 2.6 we get
1
α−β
u(t) =
Z
1
1
Z
h(t)u(t)dt + φ(t)
g(t)u(t)dt.
0
0
Multiplying h(t), g(t) on both sides and integrating over [0, 1], we find
1
Z
0
Z
0
1
1
1
h(t)u(t)dt =
α−β
Z
1
g(t)u(t)dt =
α−β
Z
Z
1
h(t)dt
Z
h(t)u(t)dt +
0
Z
g(t)dt
0
g(t)u(t)dt,
0
1
Z
h(t)u(t)dt +
0
1
Z
h(t)φ(t)dt
0
1
1
0
1
Z
g(t)φ(t)dt
0
1
g(t)u(t)dt.
0
Consequently,
κ1 −κ2
−κ3 κ4
# " R1
h(t)u(t)dt
0
R01
=
.
0
g(t)u(t)dt
0
This, together with (H1), yields
Z
1
Z
h(t)u(t)dt = 0,
0
1
g(t)u(t)dt = 0.
0
Also, u(t) ≡ 0, t ∈ [0, 1] for the fact that g, h ≥ 0 and g, h 6≡ 0. This contradicts to the definition of
eigenvalue and eigenfunction.
Case 2. λ 6= 0.
Y. Wang, Y. Yang, J. Nonlinear Sci. Appl. 8 (2015), 99–109
106
We assume that 1 is an eigenvalue of T , i.e., T u = u. So,
Z 1
Z 1
α
kuk = kT uk = |λ| max
H(t, s)u(s)ds ≤ |λ|
K2 M(s)dskuk
t∈[0,1] 0
0 β
β
α
βδ
kuk < kuk.
= |λ| K2
+
β
q(α − β)Γ(q) (α − β)(γ − δ)Γ(q)
This is impossible.
Above all, 1 is not an eigenvalue of T , as required.
On the other hand, by (H3), for all ε > 0, there exists M1 > 0 such that |f (t, u) − λu| ≤ εu, for t ∈
[0, 1], u ≥ M1 . Moreover, if u ≤ M1 , then |f (t, u) − λu| is bounded for all t ∈ [0, 1]. Consequently, there
exists ζ > 0 such that
|f (t, u) − λu| ≤ εu + ζ, for t ∈ [0, 1], u ∈ R+ .
Hence
Z 1
kAu − T uk = max H(t, s)(f (s, u(s)) − λu(s))ds
t∈[0,1] 0
Z 1
≤ max
H(t, s)|(f (s, u(s)) − λu(s))|ds
t∈[0,1] 0
1
Z
≤ max
H(t, s)ds(εkuk + ζ),
t∈[0,1] 0
and
maxt∈[0,1]
kAu − T uk
lim
≤ lim
kuk
kuk→∞
kuk→∞
R1
0
H(t, s)ds(εkuk + ζ)
= 0.
kuk
So, A has a fixed point in E. Note that 0 is not a fixed point of A, and thus A has a positive fixed point,
i.e., (1.2) has a positive solution. This completes the proof.
Theorem 3.2. Let (H1) hold true. Moreover, suppose that
(H4) There exists M > 0 such that f (t, u) ≥ −M for (t, u) ∈ [0, 1] × R+ ,
There exist positive constants e, b, c, N with M D1 < e < e + M D1 θ < b < θ2 c,
(H5) f (t, u) < ξe − M for t ∈ [0, 1], 0 ≤ u ≤ e,
(H6) f (t, u) ≥ bl N − M for t ∈ [0, 1], b − M D1 θ ≤ u ≤ θb2 ,
(H7) f (t, u) ≤ ξc − M for t ∈ [0, 1], 0 ≤ u ≤ c.
Then (1.2) has at least two positive solutions.
Proof. Let ω be a solution of
 q
c

 Dt u(t) = 1, 0 < t < 1,
Z 1
Z
0
0

αu(0)
−
βu(1)
=
h(t)u(t)dt,
γu
(0)
−
δu
(1)
=

0
1
θ
<N <
1
g(t)u(t)dt,
0
and z = M ω. By Lemma 2.7 we have
1
α
K2 M(s)ds
z(t) = M ω(t) = M
H(t, s)ds ≤ M
0
0 β
α
β
βδ
= M K2
+
β
q(α − β)Γ(q) (α − β)(γ − δ)Γ(q)
= M D1 θ < eθ.
Z
1
Z
cl
bξ
such that
Y. Wang, Y. Yang, J. Nonlinear Sci. Appl. 8 (2015), 99–109
107
We easily obtain that (1.2) has a positive solution u if and only if u + z = u
e is a solution of the boundary
value problem

c q

 Dt u(t) = fe(t, u(t) − z(t)), 0 < t < 1,
Z 1
Z 1
(3.3)
0
0

g(t)u(t)dt,
h(t)u(t)dt, γu (0) − δu (1) =
 αu(0) − βu(1) =
0
0
R+
R+
and u
e ≥ z for t ∈ (0, 1), where fe : [0, 1] ×
→
is defined by
(
f (t, y) + M, (t, y) ∈ [0, 1] × [0, +∞),
fe(t, y) =
f (t, 0) + M, (t, y) ∈ [0, 1] × (−∞, 0).
For u ∈ P , we define
Z
1
H(t, s)fe(s, u(s) − z(s))ds.
T u(t) =
0
K1 β
Next we check T (P ) ⊆ P0 , where P0 = {u ∈ P : mint∈[0,1] u(t) ≥ θkuk}, θ = K
. Indeed, for u ∈ P , Lemma
2α
2.8 implies
Z 1
Z 1
α
K1 M(s)fe(s, u(s) − z(s))ds ≤ T u(t) ≤
K2 M(s)fe(s, u(s) − z(s))ds.
0
0 β
Hence,
T u(t) ≥
K1 β
K2 α
Z
0
1
α
K2 M(s)fe(s, u(s) − z(s))ds ≥ θkT uk.
β
In what follows, we show that all the conditions of Lemma 2.10 are satisfied. We first define the
nonnegative, continuous concave functional α : P → [0, ∞) by α(u) = mint∈[0,1] |u(t)|. For each u ∈ P , it is
easy to see α(u) ≤ kuk. We prove that T (P c ) ⊆ P c . Let u ∈ P c . Then
(i) if u(t) ≥ z(t), we have 0 ≤ u(t) − z(t) ≤ u(t) ≤ c and fe(t, u(t) − z(t)) = f (t, u(t) − z(t)) + M ≥ 0. By
(H7) we have fe(t, u(t) − z(t)) ≤ ξc .
(ii) if u(t) < z(t), we have u(t) − z(t) < 0 and fe(t, u(t) − z(t)) = f (t, 0) + M ≥ 0. By (H7) we have
e
f (t, u(t) − z(t)) ≤ c .
ξ
Therefore, we have proved that, if u ∈ P c , then fe(t, u(t) − z(t)) ≤
Z
kT uk = max
t∈[0,1] 0
1
Z
H(t, s)fe(s, u(s) − z(s))ds ≤
0
1
c
ξ
for t ∈ [0, 1]. Then,
α
c
K2 M(s)ds = c.
β
ξ
Therefore, we have T (P c ) ⊆ P c . Especially, if u ∈ P e , then (H5) yields fe(t, u(t) − z(t)) ≤ ξe for t ∈ [0, 1].
So, we have T : P e → Pe , i.e., the assumption (2) of Lemma 2.10 holds.
To verify condition (1) of Lemma 2.10, let u(t) = θb2 , then u ∈ P , α(u) = b/θ2 > b, i.e., {u ∈
P (α, b, θb2 ) : α(u) > b} 6= ∅. Moreover, if u ∈ P (α, b, θb2 ), then α(u) ≥ b, and b ≤ kuk ≤ θb2 . Thus,
0 < b − M D1 θ ≤ u(t) − z(t) ≤ u(t) ≤ θb2 , t ∈ [0, 1]. From (H6) we obtain fe(t, u(t) − z(t)) ≥ bl N for t ∈ [0, 1].
By the definition of α, we have
Z 1
Z 1
b
e
α(T u) = min T u(t) ≥ θkT uk ≥ θ max
H(t, s)f (s, u(s) − z(s))ds ≥ θ N
K1 M(s)ds
l
t∈[0,1]
t∈[0,1] 0
0
= θN b > b.
Therefore, condition (1) of Lemma 2.10 is satisfied with d = b/θ2 .
Finally, we show condition (3) of Lemma 2.10 is satisfied. For this we choose u ∈ P (α, b, c) with
kT uk > b/θ2 . Then we have α(T u) = mint∈[0,1] T u(t) ≥ θkT uk ≥ θb > b. Hence, condition (3) of Lemma
2.10 holds with kT uk > b/θ2 .
Y. Wang, Y. Yang, J. Nonlinear Sci. Appl. 8 (2015), 99–109
108
From now on, all the hypotheses of Lemma 2.10 are satisfied. Hence T has at least three positive fixed
points u
e1 , u
e2 and u
e3 such that
ke
u1 k < e,
b < α(e
u2 ),
ke
u3 k > e,
α(e
u3 ) < b.
Furthermore, u
ei = ui + z (i = 1, 2, 3) are solutions of (3.3). Moreover,
u
e2 (t) ≥ θke
u2 k ≥ θα(e
u2 ) > θb > θM D1 ≥ z(t),
u
e3 (t) ≥ θke
u3 k > θe > θM D1 ≥ z(t),
t ∈ [0, 1],
t ∈ [0, 1].
So u2 = u
e2 − z, u3 = u
e3 − z are two positive solutions of (1.2). This completes the proof.
4. Examples
We now present two simple examples to explain our results. Let q = 1.5, α = γ = 2, β R= δ = 1, h(t) = t3 ,
1
9
5
60
20
10
√
g(t) = t2 . Then φ(t) = 1 + t, κ1 = 34 , κ2 = 20
, κ3 = 13 , κ4 = 12
, κ = 13
80 , K1 = 13 , K2 = 3 , 0 M(s)ds = 3 π ,
ξ=
400
√
9 π
≈ 25.08, ξ −1 ≈ 0.04, θ =
9
26 ,
D1 =
10400
√
81 π
≈ 72.44, l =
200
√
13 π
≈ 8.68.
Example 4.1. Let f (t, u) = λu + ρuσ − et + η, ∀t ∈ [0, 1], u ∈ R+ , where σ ∈ (0, 1), |λ| < 0.04, ρ 6= 0,
η 6= 0. Then (H2) and (H3) hold, by Theorem 3.1, (1.2) has at least one positive solution.
Example 4.2. We choose M = 0.01, e = 0.85, b = 2, N = 4, c = 30, and


0 ≤ u ≤ 1,
0.01u,
2
ϕ(u) = −0.08u + 1.54u − 1.45, 1 ≤ u ≤ 1.9,


1.1872,
1.9 ≤ u < +∞.
Then ϕ ∈ C(R+ , R+ ). Furthermore, let f (t, u) = ϕ(u) − 0.01 for all t ∈ [0, 1], u ∈ R+ . Then M D1 =
104
−1 ≈ 2.89 and cl = 135 ≈ 5.19. Clearly,
√
≈ 0.72, M D1 θ = 9√4 π ≈ 0.25, θ2 c = 1215
338 ≈ 3.59, θ
bξ
26
81 π
M D1 < e < e + M D1 θ < b < θ2 c, θ−1 < N <
cl
bξ and f (t, u) ≥ −0.01 = −M . On the
√
π
(i) f (t, u) < 0.02 < ξe − M = 153
8000 − 0.01 ≈√0.024 for t ∈ [0, 1], 0 ≤ u ≤ 0.85,
(ii) f (t, u) > f (1.74) ≈ 0.98 > bl N − M = 1325 π − 0.01 ≈ 0.91 for t ∈ [0, 1], 1.74 <
other hand,
2−
4
√
9 π
≤u≤
1325
81
≈
16.69,
√
(iii) f (t, u) ≤ 1.1772 < 1.186 < ξc − M = 2740 π − 0.01 for t ∈ [0, 1], 0 ≤ u ≤ 30.
Hence, (H4)-(H7) are satisfied, by Theorem 3.2, (1.2) has at least two positive solutions.
Acknowledgements
Research supported by the NNSF-China (No. 61375004 and 11202084), NSFC-Tian Yuan Special Foundation (No. 11226116), Natural Science Foundation of Jiangsu Province of China for Young Scholar (No.
BK2012109), the China Scholarship Council (No. 201208320435).
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