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J. Nonlinear Sci. Appl. 8 (2015), 110–129
Research Article
Positive solutions for singular coupled integral
boundary value problems of nonlinear Hadamard
fractional differential equations
Wengui Yang
Ministry of Public Education, Sanmenxia Polytechnic, Sanmenxia, Henan 472000, China.
Communicated by J. J. Nieto
Abstract
In this paper, we study the existence of positive solutions for a class of coupled integral boundary value
problems of nonlinear semipositone Hadamard fractional differential equations
Dα u(t) + λf (t, u(t), v(t)) = 0,
u(j) (1) = v (j) (1) = 0,
Dβ v(t) + λg(t, u(t), v(t)) = 0, t ∈ (1, e), λ > 0,
Z e
Z e
ds
ds
0 ≤ j ≤ n − 2, u(e) = µ
v(s) , v(e) = ν
u(s) ,
s
s
1
1
where λ, µ, ν are three parameters with 0 < µ < β and 0 < ν < α, α, β ∈ (n − 1, n] are two real numbers
and n ≥ 3, Dα , Dβ are the Hadamard fractional derivative of fractional order, and f, g are sign-changing
continuous functions and may be singular at t = 1 or/and t = e. First of all, we obtain the corresponding
Green’s function for the boundary value problem and some of its properties. Furthermore, by means of the
nonlinear alternative of Leray-Schauder type and Krasnoselskii’s fixed point theorems, we derive an interval
of λ such that the semipositone boundary value problem has one or multiple positive solutions for any λ
c
lying in this interval. At last, several illustrative examples were given to illustrate the main results. 2015
All rights reserved.
Keywords: Hadamard fractional differential equations, coupled integral boundary conditions, positive
solutions, Green’s function, fixed point theorems.
2010 MSC: 34A08, 34B16, 34B18.
Email address: wgyang0617@yahoo.com (Wengui Yang)
Received 2014-11-4
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
111
1. Introduction
We consider the following coupled integral boundary value problem for systems of nonlinear semipositone
Hadamard fractional differential equations
Dα u(t) + λf (t, u(t), v(t)) = 0,
u(j) (1) = v (j) (1) = 0,
Dβ v(t) + λg(t, u(t), v(t)) = 0, t ∈ (1, e), λ > 0,
Z e
Z e
ds
ds
0 ≤ j ≤ n − 2, u(e) = µ
v(s) , v(e) = ν
u(s) ,
s
s
1
1
(1.1)
where λ, µ, ν are three parameters with 0 < µ < β and 0 < ν < α, α, β ∈ (n − 1, n] are two real numbers
and n ≥ 3, Dα , Dβ are the Hadamard fractional derivative of fractional order, and f, g are sign-changing
continuous functions and may be singular at t = 1 or/and t = e. To the best knowledge of the author,
there are few papers which deal with the coupled integral boundary value problems for systems of nonlinear
Hadamard fractional differential equations.
Coupled boundary value problems have wide applications in various fields of sciences and engineering,
for example, the Sturm-Liouville problems, heat equation, reaction-diffusion equations, mathematical biology and so on. In recent years, there have been some significant developments in the study of ordinary
differential equations and partial differential equations involving fractional derivatives with coupled boundary conditions, as shown by the papers [26, 27, 32, 38, 42, 43] and the references therein. For example, by
mixed monotone method, Cui et al. [15] established sufficient conditions for the existence and uniqueness
of positive solutions to a singular differential system with integral boundary value conditions. By using
the properties of the Green’s function and the Guo-Krasnosel’skii fixed point theorem, Wang et al. [35]
obtained some existence results of positive solutions for higher-order singular semipositone fractional differential systems with coupled integral boundary conditions and parameters under some conditions concerning
the nonlinear functions.
Due to the fact that fractional-order models are more accurate than integer-order models (that is, there
are more degrees of freedom in the fractional-order models), the subject of fractional differential equations
has recently developed into a interesting topic for many researchers in view of its numerous applications in
the field of physics, engineering, mechanics, chemistry, and so forth. For some recent work on the topic, see
[1, 4, 6, 11, 16, 19, 28, 30, 31, 37]. Specially, the study of coupled systems of fractional order differential
equations has been addressed extensively by several researchers, see [3, 5, 18, 20, 21, 29, 33, 36, 40] and the
references cited therein. For instance, By applying some standard fixed point theorems, Jiang et al. [23]
and Yuan et al. [41] considered the existence of positive solutions to the four-point coupled boundary value
problems for systems of nonlinear semipositone fractional differential equations under different conditions,
respectively. In [20], Hao and Zhai studied the existence of at least one positive solution to a coupled system
of fractional boundary value problems by using Schauder fixed point theorem.
However, we should point out that most of the work on the topic is based on Riemann-Liouville and
Caputo type fractional differential equations in the last few years. In 1892, Hadamard introduced another
kind of fractional derivatives, i.e., Hadamard type fractional differential equations, which differs from the
preceding ones in the sense that the kernel of the integral and derivative contain logarithmic function of
arbitrary exponent. Details and properties of Hadamard fractional derivative and integral can be found
in [12, 13, 14, 17, 22, 24]. Recently, there are some results on Hadamard type fractional differential equations/inclusions, see [9, 10] and the references cited therein. For example, by applying some standard fixed
point theorems, Ahmad and Ntouyas [7, 8] studied the existence and uniqueness of solutions for fractional
integral boundary value problem involving Hadamard type fractional differential equations/systems with integral boundary conditions, respectively. In [34], based on some classical fixed point theorems, Thiramanus
et al. investigated the existence and uniqueness of solutions for a fractional boundary value problem involving Hadamard-type fractional differential equations and nonlocal fractional integral boundary conditions.
In [39], by applying some inequalities associated with Green’s function and Guo-Krasnosel’skii fixed point
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
112
theorems, the author showed the existence of positive solutions for a class of singular four-point coupled
boundary value problem of nonlinear semipositone Hadamard fractional differential equations.
Motivated by the results mentioned above and wide applications of coupled boundary value conditions, we
consider the existence of positive solutions for singular Hadamard fractional differential equations boundary
value problem (1.1). In Section 2, we present some preliminaries and lemmas that will be used to prove our
main results. And we obtain the corresponding Green’s function for boundary value problem (1.1) and some
of its properties. The main theorems are formulated and proved in Section 3. At last, several illustrative
examples were given to illustrate the main results in Section 4.
2. Preliminaries
For the convenience of the reader, we firstly present some basic concepts of Hadamard type fractional
calculus to facilitate analysis of problem (1.1).
Definition 2.1. [24] The Hadamard derivative of fractional order q for a function g : [1, ∞) → R is defined
as
n Z t 1
d
t n−q−1
ds
q
D g(t) =
t
log
g(s) , n − 1 < q < n,
Γ(n − q) dt
s
s
1
where n = [q] + 1, [q] denotes the integer part of the real number q and log(·) = loge (·).
Definition 2.2. [24] The Hadamard fractional integral of order q for a function g : [1, ∞) → R is defined as
Z t
1
t q−1
ds
q
I g(t) =
log
g(s) , q > 0,
Γ(q) 1
s
s
provided the integral exists.
Now we derive the corresponding Green’s function for boundary value problem (1.1), and obtain some
properties of the Green’s function.
Lemma 2.3. Let x, y ∈ C[0, 1] be given functions. Then the boundary value problem
Dα u(t) + x(t) = 0,
u(j) (1) = v (j) (1) = 0,
Dqβ v(t) + y(t) = 0,
0 ≤ j ≤ n − 2,
t ∈ (1, e),
Z e
ds
u(e) = µ
v(s) ,
s
1
Z
v(e) = ν
e
u(s)
1
ds
,
s
has an integral representation
Z e
Z e

ds
ds


H1 (t, s)y(s) ,
u(t)
=
G
(t,
s)x(s)
+
1

s
s
Z 1e
Z 1e
ds
ds


 v(t) =
G2 (t, s)y(s) +
H2 (t, s)x(s) ,
s
s
1
1
(2.1)
(2.2)
where

(log t)α−1 (1 − log s)α−1 (αβ − µν + µν log s) (log(t/s))α−1


−
,

(αβ
−
µν)Γ(α)
Γ(α)
G1 (t, s) =
(log t)α−1 (1 − log s)α−1 (αβ − µν + µν log s)



,
(αβ − µν)Γ(α)

(log t)β−1 (1 − log s)β−1 (αβ − µν + µν log s) (log(t/s))β−1


−
,

(αβ − µν)Γ(β)
Γ(β)
G2 (t, s) =
(log t)β−1 (1 − log s)β−1 (αβ − µν + µν log s)



,
(αβ − µν)Γ(β)
H1 (t, s) =
µα(log t)α−1 (1 − log s)β−1 log s
,
(αβ − µν)Γ(β)
H2 (t, s) =
1 ≤ s ≤ t ≤ e,
(2.3)
1 ≤ t ≤ s ≤ e,
1 ≤ s ≤ t ≤ e,
(2.4)
1 ≤ t ≤ s ≤ e,
νβ(log t)β−1 (1 − log s)α−1 log s
.
αΓ(α)
(2.5)
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
113
Proof. As argued in [24], the solution of Hadamard differential system in (2.1) can be written the following
equivalent integral equations
Z t
ds
1
t α−1
α−1
α−2
α−n
u(t) = c11 (log t)
+ c12 (log t)
+ · · · + c1n (log t)
−
log
x(s) ,
Γ(α) 1
s
s
(2.6)
β−1
Z t
t
ds
1
log
y(s) .
v(t) = c21 (log t)β−1 + c22 (log t)β−2 + · · · + c2n (log t)β−n −
Γ(β) 1
s
s
From Dqj u(0) = Dqj v(0) = 0, 0 ≤ j ≤ n − 2, we have cin = ci(n−1) = · · · = ci2 = 0 (i = 1, 2). Thus, (2.6)
reduces to
Z t
t α−1
1
ds
α−1
log
u(t) = c11 (log t)
−
x(s) ,
Γ(α) 1
s
s
(2.7)
β−1
Z t
t
1
ds
β−1
log
v(t) = c21 (log t)
−
y(s) .
Γ(β) 1
s
s
Re
Re
ds
Using the boundary conditions u(e) = µ 1 v(s) ds
s and v(e) = ν 1 u(s) s , from (2.7), we obtain
Z e
Z e
ds
(1 − log s)α−1
ds
v(s) +
c11 = µ
x(s) ,
s
Γ(α)
s
1
1
(2.8)
Z e
Z e
β−1
(1 − log s)
ds
ds
y(s) .
c21 = ν
u(s) +
s
Γ(β)
s
1
1
Combining (2.7) and (2.8), we have
Z e
Z e
Z t
ds
(1 − log s)α−1
ds
1
t α−1
ds
α−1
u(t) = (log t)
µ
v(s) +
x(s)
−
log
x(s) ,
s
Γ(α)
s
Γ(α) 1
s
s
1
1
Z e
β−1
Z e
Z t
β−1
ds
(1 − log s)
ds
1
t
ds
v(t) = (log t)β−1 ν
u(s) +
y(s)
−
log
y(s) .
s
Γ(β)
s
Γ(β) 1
s
s
1
1
Integrating the above equations (2.9) from 0 to 1, we obtain
Z e
Z e
Z e
Z e
ds
(1 − log s)α−1
ds
ds dt
α−1
(log t)
µ
v(s) +
x(s)
u(s) =
s
s
Γ(α)
s
t
1
1
1
1
!
α−1
Z t
Z e
1
t
ds dt
−
log
x(s)
Γ(α)
s
s
t
1
1
Z e
Z e
Z e
µ
ds
(1 − log s)α−1
ds
(1 − log s)α
=
v(s) +
x(s) −
x(s)ds
α 1
s
αΓ(α)
s
αΓ(α)
1
1
Z
Z e
ds
µ e
(1 − log s)α−1 log s
ds
=
v(s) +
x(s) ,
α 1
s
αΓ(α)
s
1
and
Z
1
e
Z e
ds dt
ds
(1 − log s)β−1
(log t)
u(s) +
y(s)
s
Γ(β)
s
t
1
1
1
!
β−1
Z e
Z t
1
t
ds dt
−
log
y(s)
Γ(β) 1
s
s
t
1
Z e
Z e
Z e
ν
ds
(1 − log s)β−1
ds
(1 − log s)β
=
u(s) +
y(s) −
y(s)ds
β 1
s
βΓ(β)
s
βΓ(β)
1
1
Z
Z e
ν e
ds
(1 − log s)β−1 log s
ds
=
y(s) .
u(s) +
β 1
s
βΓ(β)
s
1
ds
v(s) =
s
Z
e
β−1
Z
ν
e
(2.9)
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
Re
Re
ds
Solving for 1 u(s) ds
s and 1 v(s) s , we have
Z e
Z e
Z e
µ
(1 − log s)β−1 log s
(1 − log s)α−1 log s
ds
αβ
ds
ds
,
u(s) =
y(s) +
x(s)
s
αβ − µν α 1
βΓ(β)
s
αΓ(α)
s
1
1
Z e
Z e
Z e
ds
ν
(1 − log s)α−1 log s
αβ
ds
(1 − log s)β−1 log s
ds
v(s) =
.
x(s) +
y(s)
s
αβ − µν β 1
αΓ(α)
s
βΓ(β)
s
1
1
114
(2.10)
Combining (2.7), (2.8) and (2.10), we get
Z
Z e
µαβ(log t)α−1 ν e (1 − log s)α−1 log s
ds
(1 − log s)β−1 log s
ds
u(t) =
x(s) +
y(s)
αβ − µν
β 1
αΓ(α)
s
βΓ(β)
s
1
α−1
Z e
Z t
α−1
α−1
(log t)
(1 − log s)
t
ds
1
ds
+
log
x(s) −
x(s)
Γ(α)
s
Γ(α) 1
s
s
1
Z e
Z e
α−1
α−1
α−1
µν(log t)
(1 − log s)
log s
(log t)
(1 − log s)α−1
ds
ds
=
x(s) +
x(s)
(αβ − µν)Γ(α)
s
Γ(α)
s
1
1
α−1
Z t
Z e
α−1
β−1
1
t
ds
µα(log t)
(1 − log s)
log s
ds
−
log
x(s) +
y(s) ,
Γ(α) 1
s
s
(αβ − µν)Γ(β)
s
1
and
Z
Z e
ναβ(log t)β−1 µ e (1 − log s)β−1 log s
ds
(1 − log s)α−1 log s
ds
v(t) =
y(s) +
x(s)
αβ − µν
α 1
βΓ(β)
s
αΓ(α)
s
1
β−1
Z t
Z e
β−1
β−1
ds
1
t
ds
(log t)
(1 − log s)
y(s) −
log
y(s)
+
Γ(β)
s
Γ(β) 1
s
s
1
Z e
Z e
β−1
β−1
β−1
µν(log t)
(1 − log s)
log s
ds
(log t)
(1 − log s)β−1
ds
=
y(s) +
y(s)
(αβ − µν)Γ(β)
s
Γ(β)
s
1
1
β−1
Z t
Z e
β−1
α−1
νβ(log t)
(1 − log s)
log s
1
t
ds
ds
−
log
y(s) +
x(s) .
Γ(β) 1
s
s
αΓ(α)
s
1
Hence, we have
Z e
Z e

ds
ds


+
H1 (t, s)y(s) ,
u(t)
=
G
(t,
s)x(s)
1

s
s
Z1e
Z 1e
ds
ds


 v(t) =
G2 (t, s)y(s) +
H2 (t, s)x(s) .
s
s
1
1
This completes the proof of the lemma.
From Lemma 2.3, the system (1.1) can be expressed the following integral form
Z e
Z e

ds
ds


G1 (t, s)f (s, u(s), v(s)) +
H1 (t, s)g(s, u(s), v(s))
,
 u(t) = λ
s
s
Z 1 e
Z 1e

ds
ds

 v(t) = λ
G2 (t, s)g(s, u(s), v(s)) +
H2 (t, s)f (s, u(s), v(s))
.
s
s
1
1
(2.11)
Lemma 2.4. For t, s ∈ [1, e], the functions G1 (t, s) and H1 (t, s) defined by (2.3) and (2.5) satisfy
µν
max{(αβ − µν)(α − 1) + µν, αβ}
(log t)α−1 ρ1 (s) ≤ G1 (t, s) ≤
ρ1 (s),
(αβ − µν)Γ(α)
(αβ − µν)Γ(α)
(2.12)
µν
αβ
(log t)α−1 ρ2 (s) ≤ H1 (t, s) ≤
ρ2 (s),
(αβ − µν)Γ(β)
(αβ − µν)Γ(β)
(2.13)
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
G1 (t, s) ≤
max{(αβ − µν)(α − 1) + µν, αβ}
(log t)α−1 ,
(αβ − µν)Γ(α)
115
H1 (t, s) ≤
αβ
(log t)α−1 ,
(αβ − µν)Γ(β)
(2.14)
where
ρ1 (s) = (1 − log s)α−1 log s,
ρ2 (s) = (1 − log s)β−1 log s.
(2.15)
Proof. First, we will show that (2.12) is true. On the one hand, when 1 ≤ s ≤ t ≤ e, we have
(log t)α−1 (1 − log s)α−1 (αβ − µν + µν log s) − (log t − log s)α−1 (αβ − µν)
(αβ − µν)Γ(α)
α−1
α−1
(log t)
(1 − log s)
(αβ − µν + µν log s) − (log t)α−1 (1 − log s/ log t)α−1 (αβ − µν)
=
(αβ − µν)Γ(α)
α−1
α−1
(log t)
(1 − log s)
(αβ − µν + µν log s) − (log t)α−1 (1 − log s)α−1 (αβ − µν)
≥
(αβ − µν)Γ(α)
α−1
α−1
µν
µν(log t)
(1 − log s)
log s
=
(log t)α−1 ρ1 (s), t, s ∈ [1, e],
=
(αβ − µν)Γ(α)
(αβ − µν)Γ(α)
G1 (t, s) =
and
G1 (t, s) =
≤
(αβ − µν)[(log t − log t log s)α−1 − (log t − log s)α−1 ] + µν(log t)α−1 (1 − log s)α−1 log s
(αβ − µν)Γ(α)
R log t−log t log s α−2
x
dx + µν(log t)α−1 (1 − log s)α−1 log s
(αβ − µν)(α − 1) log t−log s
(αβ − µν)Γ(α)
(αβ − µν)(α − 1)(log t − log t log s)α−2 [(log t − log t log s) − (log t − log s)]
≤
(αβ − µν)Γ(α)
α−1
α−1
µν(log t)
(1 − log s)
log s
+
(αβ − µν)Γ(α)
(αβ − µν)(α − 1)(log t)α−2 (1 − log s)α−2 (1 − log t) log s + µν(log t)α−1 (1 − log s)α−1 log s
=
(αβ − µν)Γ(α)
α−2
(αβ − µν)(α − 1)(log t)
(1 − log s)α−2 (1 − log s) log s + µν(log t)α−1 (1 − log s)α−1 log s
=
(αβ − µν)Γ(α)
α−1
[(αβ − µν)(α − 1) + µν](1 − log s)
log s
≤
(αβ − µν)Γ(α)
max{(αβ − µν)(α − 1) + µν, αβ}
≤
ρ1 (s), t, s ∈ [1, e],
(αβ − µν)Γ(α)
On the other hand, when 1 ≤ t ≤ s ≤ e, since 0 < µ < β and 0 < ν < α, we also have
G1 (t, s) ≥
µν(log t)α−1 (1 − log s)α−1 log s
µν
=
(log t)α−1 ρ1 (s),
(αβ − µν)Γ(α)
(αβ − µν)Γ(α)
t, s ∈ [1, e],
and
(log t)α−1 (1 − log s)α−1 (αβ − µν + µν log s)
αβ(log s)α−1 (1 − log s)α−1
≤
(αβ − µν)Γ(α)
(αβ − µν)Γ(α)
α−1
αβ(1 − log s)
log s
[(αβ − µν)(α − 1) + µν](1 − log s)α−1 log s
≤
≤
(αβ − µν)Γ(α)
(αβ − µν)Γ(α)
max{(αβ − µν)(α − 1) + µν, αβ}
≤
ρ1 (s), t, s ∈ [1, e],
(αβ − µν)Γ(α)
G1 (t, s) =
Next we show that (2.13) holds for t, s ∈ [1, e]. In fact, since 0 < µ < β and 0 < ν < α, we get
H1 (t, s) ≥
µν(log t)α−1 (1 − log s)β−1 log s
µν
=
(log t)α−1 ρ2 (s),
(αβ − µν)Γ(β)
(αβ − µν)Γ(β)
t, s ∈ [1, e],
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
116
and
H1 (t, s) ≤
αβ(1 − log s)β−1 log s
αβ
=
ρ2 (s),
(αβ − µν)Γ(β)
(αβ − µν)Γ(β)
t, s ∈ [1, e].
Finally, we will prove (2.14) is valid for any t, s ∈ [1, e]. Noticing (1 − log s)α−2 (1 − log t) ≤ 1, (1 −
log s)α−1 log s ≤ 1, and (1 − log s)β−1 log s ≤ 1, when 1 ≤ s ≤ t ≤ e, we have
(αβ − µν)(α − 1)(log t)α−2 (1 − log s)α−2 (1 − log t) log s + µν(log t)α−1 (1 − log s)α−1 log s
(αβ − µν)Γ(α)
α−2
(αβ − µν)(α − 1)(log t)
(1 − log s)α−2 (1 − log t) log t + µν(log t)α−1 (1 − log s)α−1 log s
≤
(αβ − µν)Γ(α)
α−1
[(αβ − µν)(α − 1) + µν](log t)
≤
(αβ − µν)Γ(α)
max{(αβ − µν)(α − 1) + µν, αβ}
(log t)α−1 , t, s ∈ [1, e],
≤
(αβ − µν)Γ(α)
G1 (t, s) ≤
and when 1 ≤ t ≤ s ≤ e, since 0 < µ < β and 0 < ν < α, we also have
G1 (t, s) ≤
αβ(log t)α−1
max{(αβ − µν)(α − 1) + µν, αβ}
≤
(log t)α−1 ,
(αβ − µν)Γ(α)
(αβ − µν)Γ(α)
t, s ∈ [1, e].
And we have
H1 (t, s) =
µα(log t)α−1 (1 − log s)β−1 log s
αβ
≤
(log t)α−1 ,
(αβ − µν)Γ(β)
(αβ − µν)Γ(β)
t, s ∈ [1, e].
This completes the proof of the lemma.
Similarly, we have
Lemma 2.5. For t, s ∈ [1, e], the functions G1 (t, s) and H1 (t, s) defined by (2.4) and (2.5) satisfy
µν
max{(αβ − µν)(β − 1) + µν, αβ}
(log t)β−1 ρ2 (s) ≤ G2 (t, s) ≤
ρ2 (s),
(αβ − µν)Γ(β)
(αβ − µν)Γ(β)
µν
αβ
(log t)β−1 ρ1 (s) ≤ H2 (t, s) ≤
ρ1 (s),
(αβ − µν)Γ(α)
(αβ − µν)Γ(α)
max{(αβ − µν)(β − 1) + µν, αβ}
αβ
G2 (t, s) ≤
(log t)β−1 , H2 (t, s) ≤
(log t)β−1 ,
(αβ − µν)Γ(β)
(αβ − µν)Γ(α)
where ρ1 (s) and ρ2 (s) are defined in (2.15).
Remark 2.6. From Lemmas 2.4 and 2.5, for t, s ∈ [1, e], we have
a(log t)α−1 ρ1 (s) ≤ G1 (t, s) ≤ bρ1 (s),
H1 (t, s) ≤ b(log t)α−1 ,
G1 (t, s) ≤ b(log t)α−1 ,
a(log t)β−1 ρ2 (s) ≤ G2 (t, s) ≤ bρ2 (s),
a(log t)β−1 ρ1 (s) ≤ H2 (t, s) ≤ bρ1 (s),
a(log t)α−1 ρ2 (s) ≤ H1 (t, s) ≤ bρ2 (s),
G2 (t, s) ≤ b(log t)β−1 ,
H2 (t, s) ≤ b(log t)β−1 ,
where
µν
,
(αβ − µν) max{Γ(α), Γ(β)}
max{(αβ − µν)(α − 1) + µν, αβ} max{(αβ − µν)(β − 1) + µν, αβ}
,
.
b = max
(αβ − µν)Γ(α)
(αβ − µν)Γ(β)
a=
In the rest of the paper, we always suppose the following assumptions hold:
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
117
(H1 ) f (t, u, v), g(t, u, v) ∈ C([1, e] × [0, +∞) × [0, +∞), (−∞, +∞)), moreover there exists two functions
q1 (t), q2 (t) ∈ L1 ([1, e], (0, +∞)) such that f (t, u, v) ≥ −q1 (t) and g(t, u, v) ≥ −q2 (t) for any t ∈ [1, e],
u, v ∈ [0, +∞).
∗
(H1 ) f (t, u, v), g(t, u, v) ∈ C((1, e) × [0, +∞) × (0, +∞), (−∞, +∞)), f, g may be singular at t = 1, e,
moreover there exists two functions q1 (t), q2 (t) ∈ L1 ((1, e), (0, +∞)) such that f (t, u, v) ≥ −q1 (t) and
g(t, u, v) ≥ −q2 (t) for any t ∈ (1, e), u, v ∈ [0, +∞).
(H2 ) f (t, 0, 0) > 0 and g(t, 0, 0) > 0 for t ∈ [1, e].
(H3 ) There exists [θ1 , θ2 ] ⊂ (1, e) such that lim inf min
u↑+∞ t∈[θ1 ,θ2 ]
f (t,u,v)
u
= +∞ and lim inf min
v↑+∞ t∈[θ1 ,θ2 ]
g(t,u,v)
v
= +∞.
(H∗3 ) There exists [θ1 , θ2 ] ⊂ (1, e) such that lim inf min f (t,u,v)
= +∞ and lim inf min g(t,u,v)
= +∞.
v
u
v↑+∞ t∈[θ1 ,θ2 ]
u↑+∞ t∈[θ1 ,θ2 ]
R
R
Re
e
e
ds
ds
(H4 ) 1 ρi (s)qi (s) ds
s < +∞ (i = 1, 2), 1 ρ1 (s)f (s, u, v) s < +∞, 1 ρ2 (s)g(s, u, v) s < +∞ for any u, v ∈
[0, m], m > 0 is any constant.
Lemma 2.7. Assume the condition (H1 ) or (H∗1 ) holds, then the boundary value problem
−Dα ω1 (t) = λq1 (t),
(j)
(j)
ω1 (1) = ω2 (1) = 0,
−Dβ ω2 (t) = λq2 (t),
0 ≤ j ≤ n − 2,
t ∈ (1, e), λ > 0,
Z e
ds
ω2 (s) ,
ω1 (e) = µ
s
1
Z
e
ω1 (s)
ω2 (e) = ν
1
have an unique solution
Z e
Z e

ds
ds


H1 (t, s)q2 (s)
,
G1 (t, s)q1 (s) +
 ω1 (t) = λ
s
s
1
1
Z e
Z e

ds
ds

 ω2 (t) = λ
G2 (t, s)q2 (s) +
H2 (t, s)q1 (s)
,
s
s
1
1
which satisfy
Z e

ds
α−1


(q1 (s) + q2 (s)) ,
 ω1 (t) ≤ λb(log t)
s
Z1 e
ds


 ω2 (t) ≤ λb(log t)β−1
(q1 (s) + q2 (s)) ,
s
1
ds
,
s
(2.16)
t ∈ [1, e],
(2.17)
t ∈ [1, e].
Proof. It follows from Lemma 2.3, Remark 2.6 and the condition (H1 ) or (H∗1 ) that (2.16) and (2.17) hold.
Let E = [1, e] × [1, e], then E is a Banach space with the norm
k(u, v)k1 = kuk + kvk,
kuk = max |u(t)|,
t∈[1,e]
kvk = max |v(t)|
t∈[1,e]
for any (u, v) ∈ E. Let
P = {(u, v) ∈ E : u(t) ≥ ω(log t)α−1 kuk, v(t) ≥ ω(log t)β−1 kvk for t ∈ [1, e]},
where 0 < ω = a/b < 1. Then P is a cone of E.
Next we only consider the following singular boundary value problem
Dα x(t) + λ(f (t, [x(t) − ω1 (t)]∗ , [y(t) − ω2 (t)]∗ ) + q1 (t)) = 0,
t ∈ (1, e),
λ > 0,
Dβ y(t) + λ(g(t, [x(t) − ω1 (t)]∗ , [y(t) − ω2 (t)]∗ ) + q2 (t)) = 0, t ∈ (1, e), λ > 0,
Z e
Z e
ds
ds
(j)
(j)
x (1) = y (1) = 0, 0 ≤ j ≤ n − 2, x(e) = µ
y(s) , y(e) = ν
x(s) ,
s
s
1
1
(2.18)
where a modified function [z(t)]∗ for any z ∈ C[1, e] by [z(t)]∗ = z(t), if z(t) ≥ 0, and [z(t)]∗ = 0, if z(t) < 0.
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
118
Lemma 2.8. If (x, y) ∈ C[1, e] × C[1, e] with x(t) > ω1 (t) and y(t) > ω2 (t) for any t ∈ (1, e) is a positive
solution of the singular system (2.18), then (x − ω1 , y − ω2 ) is a positive solution of the singular system
(1.1).
Proof. In fact, if (x, y) ∈ C[1, e] × C[1, e] is a positive solution of the singular system (2.18) such that
x(t) > ω1 (t) and y(t) > ω2 (t) for any t ∈ (1, e], then from (2.18) and the definition of [·]∗ , we have
Dα x(t) + λ(f (t, x(t) − ω1 (t), y(t) − ω2 (t)) + q1 (t)) = 0,
t ∈ (1, e),
λ > 0,
Dβ y(t) + λ(g(t, x(t) − ω1 (t), y(t) − ω2 (t)) + q2 (t)) = 0, t ∈ (1, e), λ > 0,
Z e
Z e
ds
ds
(j)
(j)
y(s) , y(e) = ν
x (1) = y (1) = 0, 0 ≤ j ≤ n − 2, x(e) = µ
x(s) .
s
s
1
1
(2.19)
Let u = x − ω1 and v = y − ω2 , then Dα u(t) = Dα x(t) − Dα ω1 (t) and Dβ v(t) = Dβ y(t) − Dβ ω2 (t) for
t ∈ (1, e), which imply that
− Dα u(t) = −Dα x(t) + Dα ω1 (t) = −Dα x(t) − λq1 (t),
β
β
β
β
− D v(t) = −D y(t) + D ω2 (t) = −D y(t) − λq2 (t),
t ∈ (1, e),
t ∈ (1, e).
Thus (2.19) becomes
Dα u(t) + λf (t, u(t), v(t)) = 0,
u(j) (1) = v (j) (1) = 0,
Dβ v(t) + λg(t, u(t), v(t)) = 0, t ∈ (1, e), λ > 0,
Z e
Z e
ds
ds
0 ≤ j ≤ n − 2, u(e) = µ
v(s) , v(e) = ν
u(s) ,
s
s
1
1
i.e., (x − ω1 , y − ω2 ) is a positive solution of the singular system (1.1). This proves Lemma 2.8.
Employing Lemma 2.3, the singular system (2.18) can be expressed as
Z e

ds


u(t) =λ
G1 (t, s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s))


s

1

Z e



ds


+λ
H1 (t, s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) ,

s
Z e 1
ds



v(t) =λ
G2 (t, s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s))


s

1

Z e


ds



+λ
H2 (t, s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) ,
s
1
t ∈ [1, e],
(2.20)
t ∈ [1, e].
By a solution of the singular system (2.18), we mean a solution of the corresponding system of integral
equation (2.20). Defined an operator T : P → P by
T (x, y) = (T1 (x, y), T2 (x, y)),
where operators Ti : P → C[1, e] (i = 1, 2) are defined by
Z e
ds
T1 (x, y)(t) =λ
G1 (t, s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s))
s
1
Z e
ds
+λ
H1 (t, s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) ,
s
Z e 1
ds
T2 (x, y)(t) =λ
G2 (t, s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s))
s
1
Z e
ds
H2 (t, s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) ,
+λ
s
1
t ∈ [1, e],
(2.21)
t ∈ [1, e].
Clearly, if (x, y) ∈ P is a fixed point of T , then (x, y) is a solution of the singular system (2.18).
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
119
Lemma 2.9. Assume the condition (H1 ) or (H∗1 ) holds, then
T :P →P
is a completely continuous operator.
Proof. For any fixed (x, y) ∈ P , there exists a constant L > 0 such that k(x, y)k1 ≤ L. And then,
[x(s) − ω1 (s)]∗ ≤ x(s) ≤ kxk ≤ k(x, y)k1 ≤ L, [y(s) − ω2 (s)]∗ ≤ y(s) ≤ kyk ≤ k(x, y)k1 ≤ L, s ∈ [1, e].
For any t ∈ [1, e], it follows from (2.20) and Remark 2.6 that
Z e
ds
T1 (x, y)(t) =λ
G1 (t, s) f (s, [x(s) − w1 (s)]∗ , [y(s) − w2 (s)]∗ ) + q1 (s)
s
1
Z e
ds
H1 (t, s) g(s, [x(s) − w1 (s)]∗ , [y(s) − w2 (s)]∗ ) + q2 (s)
+λ
s
Z e 1
ds
bρ1 (s) f (s, [x(s) − w1 (s)]∗ , [y(s) − w2 (s)]∗ ) + q1 (s)
≤λ
s
1
Z e
ds
+λ
bρ2 (s) g(s, [x(s) − w1 (s)]∗ , [y(s) − w2 (s)]∗ ) + q2 (s)
s
Z1 e
Z e
ds
ds
(ρ1 (s) + ρ1 (s)) + λM µ
≤λM b
(ρ1 (s)q1 (s) + ρ2 (s)q2 (s))
s
s
1
1
Z e
ds
≤2λM b + λM µ
(q1 (s) + q2 (s)) < +∞,
s
1
where
M = max
max
f (t, u, v),
t∈[1,e],u,v∈[0,L]
g(t, u, v) + 1.
max
t∈[1,e],u,v∈[0,L]
Similarly, we have
Z
|T2 (x, y)(t)| ≤ 2λM b + λM µ
e
(q1 (s) + q2 (s))
1
ds
< +∞,
s
Thus T : P → E is well defined.
Next, we show that T : P → P . For any fixed (x, y) ∈ P , t ∈ [1, e], by (2.21) and Remark 2.6, we have
Z e
ds
T1 (x, y)(t) ≤λb
ρ1 (s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s))
s
1
Z e
ds
+ λb
ρ2 (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) ,
s
Z e 1
ds
T2 (x, y)(t) ≤λb
ρ2 (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s))
s
1
Z e
ds
+ λb
ρ1 (s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) ,
s
1
which implies that
Z
e
ds
ρ1 (s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s))
s
1
Z e
ds
+ λb
ρ2 (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) ,
s
Z e 1
ds
kT2 (x, y)k ≤λb
ρ2 (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s))
s
1
Z e
ds
+ λb
ρ1 (s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) .
s
1
kT1 (x, y)k ≤λb
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
120
On the other hand, from (2.20) and Remark 2.6, we also obtain
Z e
ds
α−1
ρ1 (s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s))
T1 (x, y)(t) ≥λa(log t)
s
1
Z e
ds
ρ2 (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s)) ,
+ λa(log t)β−1
s
Z e 1
ds
ρ2 (s)(g(s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q2 (s))
T2 (x, y)(t) ≥λa(log t)β−1
s
1
Z e
ds
ρ1 (s)(f (s, [x(s) − ω1 (s)]∗ , [y(s) − ω2 (s)]∗ ) + q1 (s)) .
+ λa(log t)α−1
s
1
So we have
T1 (x, y) ≥ ω(log t)α−1 kT1 (x, y)k,
T2 (x, y) ≥ ω(log t)β−1 kT2 (x, y)k,
t ∈ [1, e].
This implies that T (P ) ⊂ P . According to the Ascoli-Arzela theorem, we can easily get that T : P → P is
completely continuous. This completes the proof of the lemma.
The following nonlinear alternative of Leray-Schauder type and Krasnoselskii’s fixed point theorem will
play major role in our next analysis.
Theorem 2.10 (Nonlinear alternative of Leray-Schauder type, see [2]). Let X be a Banach space with
Ω ∈ X closed and convex. Assume U is a relatively open subset of with 0 ∈ U , and let
S:U →Ω
be a compact, continuous map. Then either
(a) S has a fixed point in U , or
(b) there exists u ∈ ∂U and v ∈ (0, 1), with u = vSu.
Theorem 2.11 (Krasnoselskii’s fixed point theorem, see [25]). Let X be a Banach space, and let P ⊂ X
be a cone in X. Assume Ω1 , Ω2 are open subsets of X with 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2 , and let S : P → P be a
completely continuous operator such that, either
(a) kSwk ≤ kwk, w ∈ P ∩ ∂Ω1 , kSwk ≥ kwk, w ∈ P ∩ ∂Ω2 , or
(b) kSwk ≥ kwk, w ∈ P ∩ ∂Ω1 , kSwk ≤ kwk, w ∈ P ∩ ∂Ω2 .
Then S has a fixed point in P ∩ (Ω2 \Ω1 ).
3. Main results
Theorem 3.1. Suppose that (H1 ) and (H2 ) hold. Then there exists a constant λ∗ > 0 such that the boundary
value problem (1.1) has at least one positive solution for any 0 < λ ≤ λ∗ .
Proof. Fix δ ∈ (0, 1). From (H2 ), let 0 < ε < 1 be such that
f (t, u, v) ≥ δf (t, 0, 0) and g(t, u, v) ≥ δg(t, 0, 0),
Let f (ε) =
max
1≤t≤e,0≤u,v≤ε
{f (t, u, v) + q1 (t)}, g(ε) =
for 1 ≤ t ≤ e, 0 ≤ u, v ≤ ε.
max
1≤t≤e,0≤u,v≤ε
{g(t, u, v) + q2 (t)}, and ci =
(i = 1, 2), we have
lim
z↓0
f (z)
g(z)
= +∞ and lim
= +∞.
z↓0 z
z
Suppose 0 < λ < ε/(8ch(ε)) := λ∗ , where c = max(c1 , c2 ) and h(ε) = max(f (ε), g(ε)). Since
lim
z↓0
h(z)
h(ε)
1
= +∞ and
<
,
z
ε
8cλ
(3.1)
Re
1
bρi (s) ds
s
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
121
then exists a R0 ∈ (0, ε) such that
h(R0 )
1
=
.
R0
8cλ
Let U = {(u, v) ∈ P |k(u, v)k1 < R0 }, (u, v) ∈ ∂U and θ ∈ (0, 1) be such that (u, v) = θT (u, v), i.e.,
u = θT1 (u, v) and v = θT2 (u, v). we claim that k(u, v)k1 6= R0 . In fact, for (x, y) ∈ ∂U and k(u, v)k1 = R0 ,
we have
Z e
ds
u(t) =θT1 (u, v)(t) ≤ λ
G1 (t, s) f (s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q1 (s)
s
1
Z e
ds
H1 (t, s) g(s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q2 (s)
+λ
s
1
Z e
Z e
Z e
Z e
(3.2)
ds
ds
ds
ds
≤λ
G1 (t, s)f (R0 ) + λ
H1 (t, s)g(R0 ) ≤ λ
bρ1 (s)f (R0 ) + λ
bρ2 (s)g(R0 )
s
s
s
s
1
1
Z e1
Z1 e
ds
ds
bρ2 (s) g(R0 ) ≤ 2cλh(R0 ),
bρ1 (s) f (R0 ) + λ
≤λ
s
s
1
1
and similarly, we also have
v(t) = θT2 (u, v)(t) ≤ 2cλh(R0 ).
(3.3)
It follows that R0 = k(u, v)k1 ≤ 4cλh(R0 ), that is
1
1
h(R0 )
h(R0 )
≥
>
=
,
R0
4λc
8cλ
R0
which implies that k(u, v)k1 6= R0 . By the nonlinear alternative of Leray-Schauder type, T has a fixed point
(u, v) ∈ U . Moreover, combining (3.1)-(3.3) and the fact that R0 < ε, we obtain
Z e
ds
u(t) =λ
G1 (t, s) f (s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q1 (s)
s
1
Z e
ds
+λ
H1 (t, s) g(s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q2 (s)
s
Z e
Z e 1
ds
ds
H1 (t, s) δg(s, 0, 0) + q2 (s)
+λ
≥λ
G1 (t, s) δf (s, 0, 0) + q1 (s)
s
s
1
Z1 e
Z e
ds
≥λ
G1 (t, s)q1 (s)dq s + λ
H1 (t, s)q2 (s) = w1 (t), for t ∈ (1, e),
s
1
1
and similarly, we also have
v(t) ≥ w2 (t), for t ∈ (1, e).
Then T has a positive fixed point (x, y) and k(u, v)k1 ≤ R0 < 1. Namely, (u, v) is positive solution of the
boundary value problem (3.1) with u(t) ≥ w1 (t) and v(t) ≥ w2 (t), for t ∈ (1, e).
Let x(t) = u(t) − w1 (t) ≥ 0 and y(t) = u(t) − w2 (t) ≥ 0. Then (x, y) is a nonnegative solution (positive
on (1, e)) of the boundary value problem (1.1).
Theorem 3.2. Suppose that (H∗1 ) and (H3 )-(H4 ) hold. Then there exists a constant λ∗ > 0 such that the
boundary value problem (1.1) has at least one positive solution for any 0 < λ ≤ λ∗ .
2 Re
Proof. Let Ω1 = {(u, v) ∈ E × E : kuk < R1 , kvk < R1 }, where R1 = max(1, r), r = ba 1 q1 (s) + q2 (s) ds
s .
Choose
R1
∗
−1 R1
λ = min 1,
(R + 1) ,
,
2
2r
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
where R =
Re
1
bρ1 (s)
max
0≤u,v≤R1
122
Re
ds
f (s, u, v) + q1 (s) s + 1 bρ2 (s)
max
0≤u,v≤R1
g(s, u, v) + q2 (s) ds
s and R1 ≥ 0.
Then, for any (u, v) ∈ P ∩ ∂Ω1 , we have kuk = R1 or kvk = R1 . Moreover u(t) − w1 (t) ≤ u(t) ≤ kuk ≤ R1 ,
v(t) − w2 (t) ≤ v(t) ≤ kvk ≤ R1 , and it follows that
Z e
ds
kT1 (u, v)(t)k ≤λ
bρ1 (s) f (s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q1 (s)
s
1
Z e
ds
bρ2 (s) g(s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q2 (s)
+λ
s
Z e 1
ds
bρ1 (s)
max f (s, u, v) + q1 (s)
≤λ
0≤u,v≤R1
s
1
Z e
ds
R1
bρ2 (s)
max g(s, u, v) + q2 (s)
+λ
= λR ≤
,
0≤u,v≤R
s
2
1
1
and similarly, we also have kT2 (u, v)(t)k ≤ R1 /2. This implies
kT (u, v)k1 = kT1 (u, v)k + kT2 (u, v)k ≤ R1 ≤ k(u, v)k1 , for (u, v) ∈ P \∂Ω1 .
On the other hand, choose two constants N1 , N2 > 1 such that
Z θ2
Z θ2
a2
ds
a2
α−1 ds
ρ1 (s)(log s)
≥ 1, λN2 γ
ρ2 (s)(log s)β−1
≥ 1,
λN1 γ
2b θ1
s
2b θ1
s
where γ = min {(log t)α−1 }. By assumptions (H3 ) and (H4 ), there exists a constant B > R1 such that
t∈[θ1 ,θ2 ]
f (t, u, v)
> N1 ,
u
namely
f (t, u, v) > N1 u,
for t ∈ [θ1 , θ2 ], u > B, v > 0,
(3.4)
and
g(t, u, v)
> N2 , namely g(t, u, v) > N2 v, for t ∈ [θ1 , θ2 ], u > 0, v > B.
(3.5)
v
Choose R2 , let Ω1 = {(u, v) ∈ E × E : kuk < R2 , kvk < R2 }. Then for any (u, v) ∈ (P1 × P2 ) ∩ ∂Ω2 , we have
kuk = R2 or kvk = R2 . If kuk = R2 , we can state that
Z e
Z e
ds
ds
u(t) − w1 (t) =u(t) − λ
G1 (t, s)q1 (s) + λ
H1 (t, s)q2 (s)
s
s
1
Z1 e
Z e
ds
ds
α−1
α−1
≥u(t) − λ
b(log t)
q1 (s) + λ
b(log t)
q2 (s)
s
s
1
1
Z e
ds
=u(t) − λb(log t)α−1
q1 (s) + q2 (s)
s
1
2 Z e
ds
a
a
b
=u(t) − λ (log t)α−1
q1 (s) + q2 (s)
= u(t) − λ (log t)α−1 r
b
a 1
s
b
u(t)
λr
1
≥u(t) − λr
= 1−
u(t) ≥ u(t) ≥ 0, t ∈ [1, e],
kuk
R2
2
and then
∗
1
u(t)
2
min {[u(t) − w1 (t)] } = min {u(t) − w1 (t)} ≥ min
t∈[θ1 ,θ2 ]
t∈[θ1 ,θ2 ]
n ω
o
a
α−1
≥ min
(log t)
kuk = R2 min (log t)α−1 ≥ B + 1 > B.
2b t∈[θ1 ,θ2 ]
t∈[θ1 ,θ2 ] 2∆
t∈[θ1 ,θ2 ]
Since B > R1 ≥ r, from (3.4), we have
f (t, [u(t) − w1 (t)]∗ , [v(t) − w2 (t)]∗ ) ≥ N1 [u(t) − w1 (t)]∗ ≥
N1
u(t),
2
for t ∈ [θ1 , θ2 ].
(3.6)
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
123
It follows from (3.6) that
Z e
ds
G1 (t, s) f (s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q1 (s)
T1 (u, v)(t) =λ
s
1
Z e
ds
H1 (t, s) g(s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q2 (s)
+λ
s
Z e 1
ds
≥λ
G1 (t, s) f (s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q1 (s)
s
1
Z θ2
ds
G1 (t, s)f (s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ )
≥λ
s
θ1
Z θ2
Z θ2
N1
a
ds
N1
ds
α−1
α−1
a(log t)
ρ1 (s) u(s) ≥ λa (log t)
≥λ
ρ1 (s) (log s)α−1 kuk
2
s
2
b
s
θ1
θ1
Z θ2
2
ds
a
ρ1 (s)(log s)α−1 R2
≥λN1
min {(log t)α−1 }
2b t∈[θ1 ,θ2 ]
s
θ1
Z θ2
2
a
ds
≥λN1 γ
ρ1 (s)(log s)α−1 R2 ≥ R2 , for t ∈ [θ1 , θ2 ].
2b θ1
s
If kvk = R2 , we obtain
Z
v(t) − w2 (t) = v(t) − λ
1
e
ds
G2 (t, s)q2 (s) + λ
s
Z
1
e
ds
H2 (t, s)q1 (s)
s
1
≥ v(t) ≥ 0,
2
t ∈ [1, e],
and then
∗
min {[v(t) − w2 (t)] } =
t∈[θ1 ,θ2 ]
1
v(t)
2
min {v(t) − w2 (t)} ≥ min
t∈[θ1 ,θ2 ]
o
na
o
n
a
≥
min
(log t)β−1 kvk = R2 min (log t)β−1 ≥ B + 1 > B.
2b t∈[θ1 ,θ2 ]
t∈[θ1 ,θ2 ] 2b
t∈[θ1 ,θ2 ]
Since B > R1 ≥ r, from (3.5), one verifies that
g(t, [u(t) − w1 (t)]∗ , [v(t) − w2 (t)]∗ ) ≥ N2 [v(t) − w2 (t)]∗ ≥
N2
v(t),
2
for t ∈ [θ1 , θ2 ].
It follows from (3.7) that
Z e
ds
T1 (u, v)(t) =λ
G1 (t, s) f (s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q1 (s)
s
1
Z e
ds
+λ
H1 (t, s) g(s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q2 (s)
s
Z e 1
ds
≥λ
H1 (t, s) g(s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ ) + q2 (s)
s
1
Z θ2
ds
≥λ
H1 (t, s)g(s, [u(s) − w1 (s)]∗ , [v(s) − w2 (s)]∗ )
s
θ1
Z θ2
Z θ2
N2
ds
N2
a
ds
α−1
α−1
≥λ
a(log t)
ρ2 (s) v(s) ≥ λa (log t)
ρ2 (s) (log s)β−1 kvk
2
s
2
b
s
θ1
θ1
Z
θ2
a2
ds
≥λN2
min {(log t)α−1 }
ρ2 (s)(log s)β−1 R2
2b t∈[θ1 ,θ2 ]
s
θ1
Z θ2
2
a
ds
≥λN2 γ
ρ2 (s)(log s)β−1 R2 ≥ R2 , for t ∈ [θ1 , θ2 ].
2b θ1
s
Thus, for any (u, v) ∈ (P1 × P2 ) ∩ ∂Ω2 , we always have
T1 (u, v)(t) ≥ R2 ,
for t ∈ [θ1 , θ2 ].
(3.7)
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
124
Similarly, for any (u, v) ∈ (P1 × P2 ) ∩ ∂Ω2 , it also holds
T2 (u, v)(t) ≥ R2 ,
for t ∈ [θ1 , θ2 ].
This implies
kT (u, v)k1 = kT1 (u, v)k + kT2 (u, v)k ≥ 2R2 ≥ k(u, v)k1 , for (u, v) ∈ (P1 × P2 )\∂Ω2 .
Thus condition (b) of Krasnoeselskii’s fixed point theorem is satisfied. As a result T has a fixed point (u, v)
with r ≤ R1 < kuk < R2 and r ≤ R1 < kvk < R2 .
Since r ≤ R1 < kuk < R2 and r ≤ R1 < kvk < R2 , we get
Z e
Z e
ds
ds
u(t) − w1 (t) =u(t) − λ
G1 (t, s)q1 (s) + λ
H1 (t, s)q2 (s)
s
s
1
1
Z e
Z e
ds
ds
a α−1
α−1
α−1
b(log t)
q1 (s) + λ
b(log t)
q2 (s)
≥ t
kuk − λ
b
s
s
1
1
Z e
ds
a
q1 (s) + q2 (s)
= (log t)α−1 kuk − λb(log t)α−1
b
s
1
a
a
a
α−1
α−1
α−1
≥ (log t)
r − λ (log t)
r = (1 − λ) (log t)
r ≥ 0, t ∈ (1, e),
b
b
b
and
Z e
Z e
ds
ds
H2 (t, s)q1 (s)
v(t) − w2 (t) =v(t) − λ
G2 (t, s)q2 (s) + λ
s
s
1
1
Z e
Z e
a β−1
ds
ds
β−1
β−1
≥ t
kvk − λ
b(log t)
q2 (s) + λ
b(log t)
q1 (s)
b
s
s
1
1
Z e
a
ds
= tβ−1 kvk − λb(log t)β−1
q1 (s) + q2 (s)
b
s
1
a
a
a
β−1
β−1
≥ (log t)
r − λ (log t)
r = (1 − λ) (log t)β−1 r ≥ 0, t ∈ (1, e).
b
b
b
Thus, (u, v) is positive solution of the boundary value problem (3.1) with u(t) > w1 (t) and v(t) > w2 (t)
for t ∈ (1, e). Let x(t) = u(t) − w1 (t) ≥ 0 and y(t) = v(t) − w2 (t) ≥ 0. Then (x, y) is a nonnegative solution
(positive on (1, e)) of the boundary value problem (1.1). This concludes the proof.
From the proof of Theorem 3.2, clearly condition (H3 ) can be replaced by condition (H∗3 ). So we have
the following theorem.
Theorem 3.3. Suppose that (H∗1 ), (H∗3 ) and (H4 ) hold. Then there exists a constant λ∗ > 0 such that the
boundary value problem (1.1) has at least one positive solution for any 0 < λ ≤ λ∗ .
Since condition (H1 ) implies conditions (H∗1 ) and (H4 ), then from the proof of Theorem 3.1 and 3.2, we
immediately have the following theorem.
Theorem 3.4. Suppose that (H1 )-(H3 ) hold. Then the boundary value problem (1.1) has at least two positive
solutions for λ > 0 sufficiently small.
In fact, let 0 < λ < min{λ∗ , λ∗ }, then the boundary value problem (1.1) has at least two positive
solutions.
Similarly, we conclude
Theorem 3.5. Suppose that (H1 )-(H2 ) and (H∗3 ) hold. Then the boundary value problem (1.1) has at least
two positive solutions for λ > 0 sufficiently small.
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
125
4. Some examples
Example 4.1. Consider the following coupled integral boundary value problem
!
1
Dα u(t) + λ ua + p
cos(2πv) = 0, t ∈ (1, e), λ > 0,
(1 − log t) log t
!
1
Dβ v(t) + λ v b + p
sin(2πu) = 0, t ∈ (1, e), λ > 0,
(1 − log t) log t
Z e
Z e
ds
ds
v(s) , v(e) = ν
u(j) (1) = v (j) (1) = 0, 0 ≤ j ≤ n − 2, u(e) = µ
u(s) ,
s
s
1
1
(4.1)
where a, b > 1. Then, if λ > 0 is sufficiently small, (4.1) has a positive solution (u, v) with u > 0, v > 0 for
t ∈ (1, e).
Proof. From (4.1), then we have
1
f (t, u, v) = ua + p
cos(2πv),
(1 − log t) log t
2
, i = 1, 2.
qi (t) = q(t) = p
(1 − log t) log t
1
g(t, u, v) = v b + p
sin(2πv),
(1 − log t) log t
Clearly, for t ∈ (1, e), we get
f (t, u, v) + q(t) ≥ ua + p
1
> 0,
1
g(t, u, v) + q(t) ≥ v b + p
> 0,
(1 − log t) log t
(1 − log t) log t
g(t, u, v)
f (t, u, v)
= +∞, lim inf
= +∞, for t ∈ [θ1 , θ2 ] ⊂ (1, e),
lim inf
v↑+∞
u↑+∞
u
v
for u, v ≥ 0. Thus (H∗1 ) and (H3 )-(H4 ) hold.
2 Re
2
2b2 π
ds
Let r = ba 1 √
s = a and R1 = 1 + r. We have
(1−log t) log t
Z
e
!
ds
s
1
!
Z e
ds
2
+
bρ2 (s)
max g(s, u, v) + p
0≤u,v≤R1
(1 − log t) log t s
1
!
!
Z e
Z e
3
ds
3
ds
a
b
≤
b R1 + p
+
b R1 + p
= b R1a + R1b + 6π .
(1 − log t) log t s
(1 − log t) log t s
1
1
R=
bρ1 (s)
2
max
f (s, u, v) + p
0≤u,v≤R1
(1 − log t) log t
Let
R1
R1
(R + 1)−1 ,
λ = min 1,
2
2r
∗
,
Now, if λ < λ∗ , Theorem 3.2 guarantees that (4.1) has a positive solution (u, v) with kuk ≥ 1 and kvk ≥
1.
Example 4.2. Consider the following coupled integral boundary value problem
π 2
α
D u(t) + λ
(v − a)(v − b) + cos
u
= 0, t ∈ (1, e), λ > 0,
1 + log t
2a
π 2
Dβ v(t) + λ
(u − c)(u − d) + sin
v
= 0, t ∈ (1, e), λ > 0,
1 + log t
2c
Z e
Z e
ds
ds
(j)
(j)
u (1) = v (1) = 0, 0 ≤ j ≤ n − 2, u(e) = µ
v(s) , v(e) = ν
u(s) ,
s
s
1
1
(4.2)
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
126
where b > a > 0, d > c > 0. Then, if λ > 0 is sufficiently small, (4.2) has two solutions (u1 , v1 ) and (u2 , v2 )
with ui (t) > 0 and vi (t) > 0 for t ∈ (1, e), i = 1, 2.
Proof. From (4.2), then we can see
π 2
(v − a)(v − b) + cos
u ,
1 + log t
2a
f (t, u, v) =
g(t, u, v) =
π 2
(u − c)(u − d) + sin
v .
1 + log t
2c
Clearly, there exists a constant q1 (t) = q2 (t) = m0 > 0 such that
g(t, u, v) + m0 > 0, for ∀t ∈ (1, e).
R e
Re
ds
and ε = 14 min(1, a, c), c0 = max 1 bρ1 (s) ds
s , 1 bρ2 (s) s , we obtain
f (t, u, v) + m0 > 0,
1
16(ab+cd+1)
Let δ =
f (t, u, v) ≥ δf (t, 0, 0) ≥ δ(ab + 1),
g(t, u, v) ≥ δg(t, 0, 0) ≥ δcd, for t ∈ (1, e), 0 ≤ u, v ≤ ε.
Thus (H1 )-(H2 ) and (H∗3 ) hold. Since
f (ε) =
g(ε) =
max
{f (t, u, v) + e1 (t)} ≤ 2(ab + cd) + m0 + 1,
max
{g(t, u, v) + e2 (t)} ≤ 2(ab + cd) + m0 + 1,
1≤t≤e,0≤u,v≤ε
1≤t≤e,0≤u,v≤ε
h(ε) = max(f (ε), g(ε)) ≤ 2(ab + cd) + m0 + 1.
We can choose
λ∗ =
ε
.
8c0 (2(ab + cd) + m0 + 1)
Now, if λ < λ∗ , Theorem 3.1 guarantees that (4.2) has a positive solution (u1 , v1 ) with ku1 k ≤ 1/4 and
kv1 k ≤ 1/4.
On the other hand, we have
lim inf
u↑+∞
f (t, u, v)
= +∞,
u
lim inf
v↑+∞
g(t, u, v)
= +∞, for t ∈ [θ1 , θ2 ] ⊂ (1, e),
v
2
for u, v ≥ 0. Thus (H1 )-(H3 ) also hold. Let r = 2ma0 b and R1 = 1 + r. We have
Z e
Z e
ds
ds
R=
bρ1 (s)
max f (s, u, v) + m0
+
bρ2 (s)
max g(s, u, v) + m0
,
0≤u,v≤R
0≤u,v≤R
s
s
1
1
1
1
and
R1
R1
λ = min 1,
(R + 1)−1 ,
2
2r
∗
,
Now, if λ < λ∗ , Theorem 3.2 guarantees that (4.2) has a positive solution (u2 , v2 ) with ku2 k ≥ 1 and
kv2 k ≥ 1.
Since all the conditions of Theorem 3.4 are satisfied, if λ < min(λ∗ , λ∗ ), Theorem 3.4 guarantees that
(4.2) has two solutions (u1 , v1 ) and (u2 , v2 ) with ui (t) > 0 and vi (t) > 0 for t ∈ (1, e), i = 1, 2.
Example 4.3. Consider the following coupled integral boundary value problem
Dα u(t) + λ(v a + cos(2πu)) = 0,
t ∈ (1, e),
Dβ v(t) + λ(ub + cos(2πv)) = 0,
t ∈ (1, e),
(j)
u
(1) = v
(j)
(1) = 0,
0 ≤ j ≤ n − 2,
λ > 0,
λ > 0,
Z e
ds
u(e) = µ
v(s) ,
s
1
Z
v(e) = ν
1
e
(4.3)
ds
u(s) ,
s
where a, b > 1. Then, if λ > 0 is sufficiently small, (4.3) has two solutions (u1 , v1 ) and (u2 , v2 ) with ui (t) > 0
and vi (t) > 0 for t ∈ (1, e), i = 1, 2.
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
127
Proof. From (4.3), then we can state that
f (t, u, v) = v a + cos(2πu),
g(t, u, v) = ub + cos(2πv),
q1 (t) = q2 (t) = q(t) = 2.
Clearly, we get
f (t, u, v) + q(t) ≥ v a + 1 > 0, g(t, u, v) + q(t) ≥ ub + 1 > 0, for t ∈ (1, e),
f (t, u, v)
g(t, u, v)
lim inf
= +∞, lim inf
= +∞, for t ∈ [θ1 , θ2 ] ⊂ (1, e),
v↑+∞
u↑+∞
v
u
for u, v ≥ 0. And f (t, 0, 0) = g(t, 0, 0) =R1 > 0, for t ∈R [1, e]. Thus
(H1 )-(H2 ) hold.
e
e
ds
ds
Let δ = 1/2 and ε = 1/8, c0 = max 1 bρ1 (s) s , 1 bρ2 (s) s , we obtain h(ε) = max(f (ε), g(ε)), where
f (ε) =
g(ε) =
Then
ε
8c0 h(ε)
max
{f (t, u, v) + e1 (t)} ≤ 8−a + 3,
max
{g(t, u, v) + e2 (t)} ≤ 8−a + 3.
1≤t≤e,0≤u,v≤ε
1≤t≤e,0≤u,v≤ε
≥
1
64c0 (1+3)
=
1
256c0 .
Let λ∗ =
1
256c0 .
Now, if λ < λ∗ , Theorem 3.1 guarantees that (4.3) has a
positive solution (u1 , v1 ) with ku1 k ≤ 1/8 and kv1 k ≤ 1/8.
On the other hand, let r = 4b2 /a and R1 = 1 + r. We have
Z e
Z e
ds
ds
R=
+
,
bρ1 (s)
max f (s, u, v) + 2
bρ2 (s)
max g(s, u, v) + 2
0≤u,v≤R1
0≤u,v≤R1
s
s
1
1
and
R1
R1
λ = min 1,
(R + 1)−1 ,
2
2r
∗
,
Now, if λ < λ∗ , Theorem 3.3 guarantees that (4.3) has a positive solution (u2 , v2 ) with ku2 k ≥ 1 and
kv2 k ≥ 1.
Since all the conditions of Theorem 3.5 are satisfied, if λ < min(λ∗ , λ∗ ), Theorem 3.5 guarantees that
(4.3) has two solutions (u1 , v1 ) and (u2 , v2 ) with ui (t) > 0 and vi (t) > 0 for t ∈ (1, e), i = 1, 2.
Example 4.4. Consider the following coupled integral boundary value problem
Dα u(t) + λ eu + v 2 + 7 cos(2π(t − 1)u) = 0, t ∈ (1, e), λ > 0,
Dβ v(t) + λ ev + u2 + 7 cos(2π(t − 1)v) = 0, t ∈ (1, e), λ > 0,
Z e
Z e
ds
ds
j
j
D u(0) = D v(0) = 0, 0 ≤ j ≤ n − 2, u(1) = µ
v(s) , v(1) = ν
u(s) .
s
s
1
1
(4.4)
Then, if λ > 0 is sufficiently small, (4.4) has two solutions (u1 , v1 ) and (u2 , v2 ) with ui (t) > 0 and vi (t) > 0
for t ∈ (1, e), i = 1, 2.
Proof. From (4.4), then we can see
f (t, u, v) = eu + v 2 + 7 cos(2π(t − 1)u),
g(t, u, v) = ev + u2 + 7 cos(2π(t − 1)v).
Clearly, there exists a constant q1 (t) = q2 (t) = 8 > 0 such that
f (t, 0, 0) = g(t, 0, 0) = 8,
f (t, u, v) + 8 ≥ 1 > 0,
g(t, u, v) + 8 ≥ 1 > 0, for ∀t ∈ (1, e).
Let δ = 1/100 and ε = 1/8, we obtain
f (t, u, v) ≥ δf (t, 0, 0),
g(t, u, v) ≥ δg(t, 0, 0), for t ∈ (1, e), 0 ≤ u, v ≤ ε.
Thus (H1 ), (H2 ), and (H∗3 ) hold. Since
h(ε) = max
max
{f (t, u, v) + q1 (t)},
1≤t≤e,0≤u,v≤ε
max
1≤t≤e,0≤u,v≤ε
{g(t, u, v) + q2 (t)}
≤ e + 16.
W. Yang, J. Nonlinear Sci. Appl. 8 (2015), 110–129
Let c0 = max
λ∗ =
R e
1
bρ1 (s) ds
s ,
Re
1
128
bρ2 (s) ds
s . We can choose
ε
.
8c0 (e + 16)
Now, if λ < λ∗ , Theorem 3.1 guarantees that (4.4) has a positive solution (u1 , v1 ) with ku1 k ≤ 1/8 and
kv1 k ≤ 1/8.
On the other hand, we have
lim inf
u↑+∞
f (t, u, v)
= +∞,
u
lim inf
v↑+∞
g(t, u, v)
= +∞, for t ∈ [θ1 , θ2 ] ⊂ (0, 1),
v
2
for u, v ≥ 0. Thus (H1 ), (H2 ), and (H3 ) also hold. Let r = 16b
ω and R1 = 1 + r. We have
Z e
Z e
ds
ds
bρ2 (s)
max g(s, u, v) + m0
bρ1 (s)
max f (s, u, v) + m0
+
R=
0≤u,v≤R1
0≤u,v≤R1
s
s
1
Z1 e
Z e
ds
ds
≤
bρ1 (s) eR1 + R12 + 7 + 8
+
bρ2 (s) eR1 + R12 + 7 + 8
s
s
1
1
R1
2
=c0 e + R1 + 15 ,
and
R1
R1
λ = min 1,
(R + 1)−1 ,
2
2r
∗
,
Now, if λ < λ∗ , Theorem 3.2 guarantees that (4.4) has a positive solution (u2 , v2 ) with ku2 k ≥ 1 and
kv2 k ≥ 1.
Since all the conditions of Theorem 3.4 are satisfied, if λ < min(λ∗ , λ∗ ), Theorem 3.4 guarantees that
(4.4) has two solutions (u1 , v1 ) and (u2 , v2 ) with ui (t) > 0 and vi (t) > 0 for t ∈ (1, e), i = 1, 2.
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