Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 9 (2016), 717–726
Research Article
Positive solutions of p-Laplacian fractional
differential equations with integral boundary value
conditions
Yunhong Li∗, Guogang Li
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018, Hebei, P. R. China.
Communicated by R. Saadati
Abstract
In this work, we investigate the existence of solutions of p-Laplacian fractional differential equations with
integral boundary value conditions. Using the five functionals fixed point theorem, the existence of multiple
positive solutions is obtained for the boundary value problems. An example is also given to illustrate the
c
effectiveness of our main result. 2016
All rights reserved.
Keywords: Multiple positive solutions, p-Laplacian, the five functionals fixed point theorem.
2010 MSC: 34B15, 34B18.
1. Introduction
The equation with p-Laplacian operator arises in the modeling of different physical and natural phenomena, non-Newtonian mechanics, nonlinear flow laws and many other branches of engineering. The study of
differential equations with p-Laplacian operator was initiated by many authors, one may see [1]–[8], [11],
[12]–[14] and references therein.
In [6], Guo et al. discussed the existences of solution for the following boundary value problems for the
p-Laplacian equation:
(φp (u0 (t)))0 + a(t)f (t, u(t)) = 0, t ∈ (0, 1),
subject to one of the following boundary conditions:
∗
Corresponding author
Email addresses: mathhong@126.com (Yunhong Li), lgg-2000@163.com (Guogang Li)
Received 2015-07-11
Y. Li, G. Li, J. Nonlinear Sci. Appl. 9 (2016), 717–726
0
φp (u (0)) =
m−2
X
718
0
ai φp (u (ξi )), u(1) =
i=1
m−2
X
bi u(ξi ),
i=1
or
u(0) =
m−2
X
ai u(ξi ), φp (u0 (1)) =
i=1
m−2
X
bi φp (u0 (ξi )).
i=1
Using the five functionals fixed point theorem, they obtained the existence of multiple (at least three)
positive solutions for above boundary value problems.
Chen et al. [3] showed the existence solutions by coincidence degree for the Caputo fractional p-Laplacian
equations:
D0β+ φp (D0α+ x(t)) = f (t, x(t), D0α+ x(t)), 0 < t < 1,
D0α+ x(0) = D0α+ x(1) = 0,
where 0 < α, β ≤ 1 and 1 < α + β ≤ 2, φp (s) = |s|p−2 s, p > 1, f : [0, 1] × R2 → R is continuous, D0β+ and
D0α+ are the Caputo fractional derivatives. They used Lu = D0β+ φp (D0α+ x(t)) with D0α+ x(0) = D0α+ x(1) = 0
and obtained dim kerL = 1.
Zhang et al. [14] discussed the eigenvalue problem for a class of singular p-Laplacian fractional differential
equations involving the Riemann-Stieltjes integral boundary conditions
− Dtβ φp (Dtα x) (t) = λf (t, x(t)), t ∈ (0, 1),
Z 1
α
x(s)dA(s),
x(0) = 0, Dt x(0) = 0, x(1) =
0
Dtβ
where
and Dtα are the standard Riemann-Liouville fractional derivatives with 1 < α ≤ 2, 0 < β ≤ 1,
R1
A is a function of bounded variation and 0 x(s) dA(s) denotes the Riemann-Stieltjes integral of x with
respect to A, f (t, x) : (0, 1) × (0, ∞) → [0, ∞) is continuous and may be singular at t = 0, 1 and x = 0. their
results are derived based on the method of upper and lower solutions and the Schauder fixed point theorem.
Motivated by the aforementioned works, this paper is concerned with the existence of positive solutions
for the coupled system of p-Laplacian fractional differential equations with integral boundary value conditions
D0β+ (φp (D0α+ u(t))) + f (t, u(t)) = 0, t ∈ (0, 1),
(1.1)
φp (D0α+ u(0))(i)
= 0, i = 1, 2, . . . , l − 1,
Z 1
α
φp (D0+ u(1)) =
h(t)φp (D0α+ u(t))dt,
(1.2)
u(j) (0) = 0, j = 1, 2, . . . , n − 1,
Z 1
u(0) =
k(t)u(t)dt,
(1.4)
(1.3)
0
(1.5)
0
where φp (u) = |u|p−2 u, p > 1. D0β+ and D0α+ are the Caputo fractional derivatives, l − 1 < β ≤ l, n − 1 <
α ≤ n, l ≥ 1, n ≥ 1 and l + n − 1 < α + β ≤ l + n. Using the five functionals fixed point theorem, the
existence of multiple positive solutions is obtained for the aforementioned boundary value problems.
We will suppose the following conditions are satisfied:
R1
R1
(H1 ) k(t), h(t) ∈ L1 [0, 1], k(t) ≥ 0, h(t) ≥ 0, 0 < 0 k(t)dt < 1 and 0 < 0 h(t)dt < 1;
(H2 ) f (t, u) : [0, 1] × [0, ∞) → [0, ∞) is continuous.
2. Background and definitions
For the convenience, we present some necessary basic knowledge and definitions about fractional calculus
theory, which can be found in [9, 10].
Y. Li, G. Li, J. Nonlinear Sci. Appl. 9 (2016), 717–726
719
Definition 2.1. The fractional integral of order α > 0 of a function y : (0, ∞) → R is given by
α
I0+
y(t) =
1
Γ(α)
Z
t
(t − s)α−1 y(s)ds,
0
provided that the right side is pointwise defined on (0, ∞).
Definition 2.2. For a continuous function y : (0, ∞) → R, the Caputo derivative of fractional order α > 0
is defined as
Z t
1
α
D0+
y(t) =
(t − s)n−α−1 y (n) (s)ds,
Γ(n − α) 0
where n = [α] + 1, provided that the right side is pointwise defined on (0, ∞).
α u(t) = 0 has
Lemma 2.3. Let α > 0 and u ∈ AC N −1 [0, 1]. Then the fractional differential equation D0+
u(t) = c0 + c1 t + c2 t2 + . . . + cN −1 tN −1 , ci ∈ R, i = 1, 2, . . . , N
as the unique solution, where N is the smallest integer greater than or equal to α.
Let K be a cone in real Banach space E and γ, ϕ, θ be nonnegative continuous convex functional on K
and let ω, ψ be nonnegative continuous concave functional on K. Then for nonnegative numbers h, a, b, d
and c, we define the following convex sets
P (γ, c) = {x ∈ K|γ(x) < c},
P (γ, ω, a, c) = {x ∈ K|a ≤ ω(x), γ(x) ≤ c},
Q(γ, ϕ, d, c) = {x ∈ K|ϕ(x) ≤ d, γ(x) ≤ c},
P (γ, θ, ω, a, b, c) = {x ∈ K|a ≤ ω(x), θ(x) ≤ b, γ(x) ≤ c},
Q(γ, ϕ, ψ, h, d, c) = {x ∈ K|h ≤ ψ(x), ϕ(x) ≤ d, γ(x) ≤ c}.
Theorem 2.4. Let K be a cone in real Banach space E. Suppose there exist nonnegative continuous concave
functionals ω and ψ on K, and nonnegative continuous convex functionals γ, ϕ, and θ on K such that for
some positive numbers c and m,
ω(x) ≤ ϕ(x), and kxk ≤ mγ(x) for all x ∈ P (γ, c).
Suppose further that T : P (γ, c) → P (γ, c) is completely continuous and there exist h, d, a, b ≥ 0 with
0 < d < a such that each of the following is satisfied:
(C1 ) {x ∈ P (γ, θ, ω, a, b, c)|ω(x) > a} =
6 ∅ and ω(T x) > a for x ∈ P (γ, θ, ω, a, b, c),
(C2 ) {x ∈ Q(γ, ϕ, ψ, h, d, c)|ϕ(x) < d} =
6 ∅ and ϕ(T x) < d for x ∈ Q(γ, ϕ, ψ, h, d, c),
(C3 ) ω(T x) > a provided x ∈ P (γ, ω, a, c) with θ(T x) > b,
(C4 ) ϕ(T x) < d provided x ∈ Q(γ, ϕ, d, c) with ψ(T x) < h.
Then T has at least three fixed points x1 , x2 , x3 ∈ P (γ, c) such that
ϕ(x1 ) < d, a < ω(x2 ) and d < ϕ(x3 ) with ω(x3 ) < a.
3. Preliminary lemmas
Lemma 3.1. The boundary value problems (1.1)–(1.5) has a solution u(t) if and only if u(t) solves the
equation
Z t
1
u(t) =
(t − s)α−1 r(s)ds + b0 ,
(3.1)
Γ(α) 0
Y. Li, G. Li, J. Nonlinear Sci. Appl. 9 (2016), 717–726
720
where
r(s) = φq a0 −
1
Γ(β)
s
Z
(s − τ )
β−1
f dτ ,
hR
i
R1
R
t
β−1 f dτ − 1 h(t)
β−1 f dτ dt
(1
−
τ
)
(t
−
τ
)
0
0
0
h
i
a0 =
,
R1
Γ(β) 1 − 0 h(t)dt
hR
i
R1
t
α−1 r(s)ds dt
k(t)
(t
−
s)
0
0
h
i
b0 =
.
R1
Γ(α) 1 − 0 k(t)dt
(3.2)
0
φq (s) is the inverse function of φp (s), a.e., φq (s) = |s|q−2 s,
1
p
+
1
q
(3.3)
(3.4)
= 1.
Proof. From (1.1), we get
D0β+ (φp (D0α+ u(t))) = −f (t, u(t)).
For t ∈ [0, 1], integrate from 0 to t, in view of Lemma 2.3, we have
Z t
1
α
(t − τ )β−1 f (τ, u(τ ))dτ + a0 + a1 t + · · · + al−1 tl−1 .
φp (D0+ u(t)) = −
Γ(β) 0
By (1.2), we obtain a1 = · · · = al−1 = 0. i.e.,
φp (D0α+ u(t))
t
Z
1
=−
Γ(β)
(t − τ )β−1 f (τ, u(τ ))dτ + a0 .
0
From (1.3), we get
hR
i
R
t
β−1 f dτ dt
β−1 f dτ − 1 h(t)
(t
−
τ
)
(1
−
τ
)
0
0
0
i
h
a0 =
.
R1
Γ(β) 1 − 0 h(t)dt
R1
In view of (3.5), we have
D0α+ u(t)
= φq a0 −
1
Γ(β)
Z
t
β−1
(t − τ )
f dτ .
0
For t ∈ [0, 1], integrate from 0 to t, in view of Lemma 2.3, we get
Z t
1
u(t) =
(t − s)α−1 r(s)ds + b0 + b1 t + · · · + bn−1 tn−1 .
Γ(α) 0
By (1.4), we get b1 = · · · = bn−1 = 0. i.e.,
1
u(t) =
Γ(α)
Z
t
(t − s)α−1 r(s)ds + b0 .
0
From (1.5), we get
R1
b0 =
0
i
− s)α−1 r(s)ds dt
h
i
.
R1
Γ(α) 1 − 0 k(t)dt
k(t)
hR
t
0 (t
So we have
1
u(t) =
Γ(α)
The proof is complete.
Z
0
t
(t − s)α−1 r(s)ds + b0 .
(3.5)
Y. Li, G. Li, J. Nonlinear Sci. Appl. 9 (2016), 717–726
721
Lemma 3.2. Every solution satisfied in Lemma 3.1 is non-negative and non-decreasing function for all
t ∈ [0, 1].
R1
Proof. For f (t, u(t)) ≥ 0, h(t) ≥ 0 and 0 < 0 h(t)dt < 1, in view of (3.2), we get
Z s
1
β−1
r(s) = φq a0 −
(s − τ )
f dτ ,
Γ(β) 0
h
i
R 1 (1 − τ )β−1 f dτ − R 1 h(t) R t (t − τ )β−1 f dτ dt R s
(s − τ )β−1 f dτ
0
0
0
0
h
i
= φq
−
R1
Γ(β)
Γ(β) 1 − 0 h(t)dt
h
i
R 1 (1 − τ )β−1 f dτ − R 1 h(t) R 1 (1 − τ )β−1 f dτ dt R 1
(1 − τ )β−1 f dτ
0
0
0
0
h
i
≥ φq
−
R1
Γ(β)
Γ(β) 1 − h(t)dt
0
R 1
= φq
0
τ )β−1 f dτ
(1 −
Γ(β)
R1
−
0
(1 − τ )β−1 f dτ
Γ(β)
= 0,
thus,
1
u(t) =
Γ(α)
t
Z
(t − s)α−1 r(s)ds + b0
0
is nondecreasing. u(0) is minimum of u(t) for t ∈ [0, 1]. In view of k(t) ≥ 0, 0 <
hR
i
R1
t
α−1 r(s)ds dt
(t
−
s)
k(t)
0
0
i
h
,
u(0) = b0 =
R1
Γ(α) 1 − 0 k(t)dt
R1
0
k(t)dt < 1, and
so u(0) ≥ 0. Thus, we obtain u(t) ≥ 0 for t ∈ [0, 1]. The proof is complete.
4. Main results
Let E be the real Banach space C[0, 1] with the maximum norm and define the cone K ⊂ E by
K = u |u ∈ E and u(t) are non-negative, nondecreasing function on [0, 1] .
Define the operator T on K by
1
T u(t) =
Γ(α)
Z
t
(t − s)α−1 r(s)ds + b0 ,
(4.1)
0
where r(s), b0 , is defined by (3.2) and (3.4). Obviously, u is a solution of equation (1.1)–(1.5) if and only if
u is a fixed point of operator T . In order to obtain the result, we define the nonnegative continuous concave
functionals ω, ψ and the nonnegative continuous convex functionals θ, ϕ, γ on K by
ω(u) = min u(t) = u(t1 ),
t∈[t1 ,t2 ]
1
ψ(u) = min u(t) = u
,
1
λ
t∈[ λ
,1]
(4.2)
θ(u) = max u(t) = u(t2 ),
(4.4)
ϕ(u) = max u(t) = u(1),
(4.5)
γ(u) = max u(t) = u(1),
(4.6)
(4.3)
t∈[t1 ,t2 ]
1
t∈[ λ
,1]
t∈[0,1]
where 0 < t1 < t2 < 1, 0 <
1
λ
< 1. It is easy to see that, for each u ∈ K,
Y. Li, G. Li, J. Nonlinear Sci. Appl. 9 (2016), 717–726
722
ω(u) = u(t1 ) ≤ u(1) = ϕ(u) and kuk = γ(u).
Now, for the convenience, we introduce the following notations. Let
1
h
i,
R1
Γ(α + 1) 1 − 0 k(t)dt
R1
α
tα1
0 hk(t)t dt
i,
e3 =
+
R
Γ(α + 1) Γ(α + 1) 1 − 1 k(t)dt
0
R t2
h(t)((1 − t)β − (1 − t2 )β )dt
h
i
e4 = t1
,
R1
Γ(β + 1) 1 − 0 h(t)dt
α
t1
e7 =
,
t2
e1 =
e2 =
1
h
Γ(β + 1) 1 −
R1
0
h(t)dt
i,
e5 =
1 − (1 − λ1 )β
h
i,
R1
Γ(β + 1) 1 − 0 h(t)dt
e6 =
(1 − λ1 )β
h
i,
R1
Γ(β + 1) 1 − 0 h(t)dt
e8 = λα .
The following theorem is main result in this paper.
Theorem 4.1. Suppose that the conditions (H1 ) and (H2 ) hold. In addition, assume there exist nonnegative
numbers d, a and c such that 0 < h < e8 h ≤ d < a < ea7 ≤ b ≤ c, and f (t, u) satisfies the following growth
conditions:
(H3 ) f (t, u) ≤ e12 φp ec1 , for t ∈ [0, 1] and u ∈ [0, c],
(H4 ) f (t, u) > e14 φp ea3 , for t ∈ [t1 , t2 ] and u ∈ [a, b],
φp
d
e1
e
− e5 φp ec
2
1
(H5 ) f (t, u) <
, for t ∈ [ λ1 , 1] and u ∈ [h, d].
e6
Then the boundary value problems (1.1)–(1.5) have at least three positive solutions u1 , u2 and u3 such that
kui k < c, for i = 1, 2, 3,
ku1 k < d, a < ω(u2 ), and ku3 k > d, ω(u3 ) < a.
Proof. First, we show that T : P (γ, c) → P (γ, c) is a completely continuous operator.
Let u ∈ K, in view of (4.1), we have T u(t) are nonnegative and nondecreasing, consequently, T : K → K.
Applying the Arzela-Ascoli Theorem and standard arguments, we conclude that T is a completely continuous
operator. If u ∈ P (γ, c), in view of (4.1), we have
1
T u(t) =
Γ(α)
Z
t
(t − s)α−1 r(s)ds + b0 ,
0
so k T u k= T u(1), where r(s) and b0 are defined in (3.2) and (3.4). By (3.2) and condition (H3 ), we have
R1
(1 − τ )β−1 f dτ
h
i
(s − τ )
f dτ ≤ φq
R1
0
Γ(β) 1 − 0 h(t)dt
R1
− τ )β−1 dτ
1
c
1
c
1
c
0 (1
h
i = φq
h
i = .
≤ φq
φp
φp
R
R
1
1
e2
e1 Γ(β) 1 − h(t)dt
e2
e1 Γ(β + 1) 1 − h(t)dt
e1
0
0
r(s) = φq a0 −
1
Γ(β)
Z
s
β−1
0
By (4.1), we have
T u(1) =
1
Γ(α)
Z
0
1
(1 − s)α−1 r(s)ds + b0
Y. Li, G. Li, J. Nonlinear Sci. Appl. 9 (2016), 717–726
723
i
α−1 r(s)ds dt
−
s)
(1 −
0
h
i
= 0
+
R1
Γ(α)
Γ(α) 1 − 0 k(t)dt


R1
1
c 
0h k(t)dt
i
≤
+
e1 Γ(α + 1) Γ(α + 1) 1 − R 1 k(t)dt
0


c 
1
h
i  = c.
=
e1 Γ(α + 1) 1 − R 1 k(t)dt
0
R1
s)α−1 r(s)ds
R1
k(t)
hR
t
0 (t
Thus, kT uk ≤ c. Consequently, we show T : P (γ, c) → P (γ, c).
Next, we show that conditions (C1 ) − (C4 ) in Theorem 2.4 are satisfied for T . It is easy to see that
{u ∈ P (γ, θ, ω, a, b, c)|ω(u) > a} =
6 ∅,
{u ∈ Q(γ, ϕ, ψ, h, d, c)|ϕ(u) < d} =
6 ∅.
To prove that the second part of condition (C1 ) holds, let u ∈ P (γ, θ, ω, a, b, c), in view of (4.2) and
(4.4), we get
ω(u) = u(t1 ) ≥ a, θ(u) = u(t2 ) ≤ b.
It implies that a ≤ u(t) ≤ b for t ∈ [t1 , t2 ]. From (3.2) and condition (H4 ), we get
Z s
Z 1
1
1
β−1
β−1
r(s) = φq a0 −
(s − τ )
f dτ ≥ φq a0 −
(1 − τ )
f dτ
Γ(β) 0
Γ(β) 0
R

R
R
R1
1
β−1 f dτ − 1 h(t) t (t − τ )β−1 f dτ dt
(1
−
τ
)
h(t)dt
·
0
0

i0
h
= φq  0
R1
Γ(β) 1 − 0 h(t)dt
R
≥ φq 

R1
1
t (1
0 h(t) h
φp
> φq 

≥ φq 
−
i
R1
Γ(β) 1 − 0 h(t)dt
a
e3
e4
φp
τ )β−1 f dτ dt
a
e3
e4
R t2
≥ φq
t1

β−1 f dτ dt
(1
−
τ
)

i
ht
R1
Γ(β) 1 − 0 h(t)dt
h(t)
R t2

β−1 dτ dt
(1
−
τ
)
t1
i 
ht
R1
Γ(β) 1 − 0 h(t)dt
R t2
h(t)
R t2

h(t)((1 − t)β − (1 − t2 )β )dt
≥ a.
h
i
R1
e3
Γ(β + 1) 1 − 0 h(t)dt
t1
R t2
By (4.1), we have
Z t1
1
(t1 − s)α−1 r(s)ds + b0
Γ(α) 0
hR
i
R1
t
R t1
α−1 r(s)ds dt
α−1
k(t)
(t
−
s)
(t1 − s)
r(s)ds
0
0
h
i
= 0
+
R1
Γ(α)
Γ(α) 1 − 0 k(t)dt
hR
i 
R
R1
t
α−1 ds dt
t1
α−1
k(t)
(t
−
s)
ds
0
0
a  0 (t1 − s)

h
i
>
+
R1
e3
Γ(α)
Γ(α) 1 − k(t)dt
T u(t1 ) =
0
Y. Li, G. Li, J. Nonlinear Sci. Appl. 9 (2016), 717–726
724


R1
α dt
k(t)t
a 
tα1
0h
i
=
+
e3 Γ(α + 1) Γ(α + 1) 1 − R 1 k(t)dt
0
= a.
Thus, in view of (4.2), we get ω(T u) = T u(t1 ) > a.
To show that the second part of condition (C2 ) holds, let u ∈ Q(γ, ϕ, ψ, h, d, c), in view of (4.3) and
(4.5), we get
1
ψ(u) = u( ) ≥ h, ϕ(u) = u(1) ≤ d.
λ
1
It implies that h ≤ u(t) ≤ d for t ∈ [ λ , 1], and 0 ≤ u ≤ c for t ∈ [0, 1]. In view of conditions (H3 ), (H5 ) and
(3.2), we get
Z s
1
(s − τ )β−1 f dτ ≤ φq (a0 )
r(s) = φq a0 −
Γ(β) 0
hR
i 
R 1
R
t
β−1 f dτ − 1 h(t)
β−1 f dτ dt
(1
−
τ
)
(t
−
τ
)
0
0
0

h
i
= φq 
R1
Γ(β) 1 − 0 h(t)dt
 R

 1

R1
Rλ
β−1 f dτ
1
β−1 f dτ
β−1 f dτ
1 (1 − τ )
(1
−
τ
)
(1
−
τ
)
λ
i  ≤ φq  0 h
i+
i
h
≤ φq  0 h
R1
R1
R1
Γ(β) 1 − 0 h(t)dt
Γ(β) 1 − 0 h(t)dt
Γ(β) 1 − 0 h(t)dt
R


1
e5
d
c
R λ1
β−1 dτ
β−1
φ
−
φ
1 (1 − τ )
p
p
(1 − τ )
dτ
e1
e2
e1
1
c
0 h
λ
i+
i
h
< φq  φp
R
R1
1
e2
e1 Γ(β) 1 − h(t)dt
e6
Γ(β) 1 − 0 h(t)dt
0


e5
d
c
1 β
1 β
φ
−
φ
p
p
)
)
(1
−
1
−
(1
−
e
e
e
1
c
1
2
1
λ
i+
i
h
h λR
≤ φq  φp
1
e2
e1 Γ(β + 1) 1 − R 1 h(t)dt
e6
Γ(β + 1) 1 − h(t)dt
0
=
0
d
.
e1
By (4.1), we have
T u(1) =
=
<
=
=
Z 1
1
(1 − s)α−1 r(s)ds + b0
Γ(α) 0
i
hR
R1
t
R1
α−1 r(s)ds dt
α−1
k(t)
(t
−
s)
r(s)ds
0
0
0 (1 − s)
h
i
+
R1
Γ(α)
Γ(α) 1 − 0 k(t)dt
hR
i 
R
R1
t
α−1 ds dt
1
α−1
k(t)
(t
−
s)
ds
0
0
d  0 (1 − s)

h
i
+
R1
e1
Γ(α)
Γ(α) 1 − 0 k(t)dt


R1
k(t)dt
d 
1
0h
i
+
e1 Γ(α + 1) Γ(α + 1) 1 − R 1 k(t)dt
0


1
d 
h
i  = d.
e1 Γ(α + 1) 1 − R 1 k(t)dt
0
So, we have ϕ(T u) = T u(1) < d. To see that (C3 ) is satisfied, let u ∈ P (γ, ω, a, c) with θ(T u) > b, that is,
T u(t2 ) > b. By (4.1), we have
Y. Li, G. Li, J. Nonlinear Sci. Appl. 9 (2016), 717–726
R t1
T u(t1 ) =
0
725
(t1 − s)α−1 r(s)ds
+ b0 .
Γ(α)
For
R t1
(t1
R0t2
0 (t2
− s)α−1 r(s)ds
− s)α−1 r(s)ds
=
tα1
R1
R0
α 1
t2
0
(1 − v)α−1 r(t1 v)dv
(1 − v)α−1 r(t2 v)dv
≥
tα1
= e7 ,
tα2
so, we get
T u(t1 ) ≥
e7
R t2
0
(t2 − s)α−1 r(s)ds
+ b0 ≥ e7
Γ(α)
R t2
0
(t2 − s)α−1 r(s)ds
+ b0
Γ(α)
= e7 T u(t2 ).
Thus, we have
ω(T u) = T u(t1 ) ≥ e7 T u(t2 ) > e7 b ≥ a.
Finally, to show (C4 ), we take u ∈ Q(γ, ϕ, d, c) with ψ(T u) < h, that is T u
R1
T u(1) =
0
1
λ
< h. By (4.1), we have
(1 − s)α−1 r(s)ds
+ b0 .
Γ(α)
For
R1
α−1 r(s)ds
0 (1 − s)
1
λ
R
0
( λ1 − s)α−1 r(s)ds
R1
=
α−1 r(s)ds
0 (1 − s)
≤ λα = e8 ,
R
α
1
v
1
α−1
r( λ )dv
λ
0 (1 − v)
thus, we have
T u(1) ≤
e8
R
1
λ
0
( λ1 − s)α−1 r(s)ds
+ b0 ≤ e8
Γ(α)
R
1
λ
0
( λ1 − s)α−1 r(s)ds
+ b0
Γ(α)
1
≤ e8 T u
.
λ
Therefore, the hypotheses of Theorem 2.4 are satisfied. So the boundary value problems (1.1)–(1.5) have at
least three positive solutions u1 , u2 and u3 such that
kui k < c for i = 1, 2, 3,
ku1 k < d, a < ω(u2 ) and ku3 k > d, ω(u3 ) < a.
The proof is complete.
5. Example
In this section, we present a simple example to explain our result.
Consider the following boundary value problems

0.82
0.7
+ f (t, u(t)) = 0, t ∈ (0, 1),

 D0+ (φp (D0+ u(t)))
R
1
0.11 φ (D 0.7 u(t))dt,
φp (D00.7
+ u(1)) = 0 (1 − t)
p
0+

R 1 0.1

u(0) = 0 t u(t)dt,
where

0.2,



3(u − 105)
√ + 0.2,
f (t, u) =

 3.2 + 0.1√u − 106,

3.2 + 0.1 764,
t ∈ [0, 1],
t ∈ [0, 1],
t ∈ [0, 1],
t ∈ [0, 1],
0 ≤ u ≤ 105,
105 < u ≤ 106,
106 < u ≤ 870,
870 < u.
Y. Li, G. Li, J. Nonlinear Sci. Appl. 9 (2016), 717–726
726
And we notice that α = 0.7, β = 0.82. If we take t1 = 0.05, t2 = 0.99, λ = 10, then 0 < t1 < t2 < 1.
It follows from a direct calculation that
e1 = 12.106009, e2 = 10.771129, e3 = 6.860732, e4 = 4.845449,
e5 = 0.891513,
e6 = 9.879617, e7 = 0.123690, e8 = 5.001872.
In addition, if we take p = 2, h = 1, d = 105, a = 106, b = 870, and c = 920, if f (t, u) satisfies the following
growth conditions:
1
c
f (t, u) ≤ φp
= 7.055464,
t ∈ [0, 1],
u ∈ [0, 920],
e2
e1
a
1
= 3.188610,
t ∈ [0.05, 0.99], u ∈ [106, 870],
f (t, u) > φp
e4
e3
φp ed1 − ee52 φp ec1
f (t, u) <
= 0.241238, t ∈ [0.1, 1],
u ∈ [1, 105],
e6
then all conditions of Theorem 4.1 are satisfied. Therefore, by Theorem 4.1, we know that the boundary
value problems have at least three positive solutions u1 , u2 and u3 such that
kui k < 920 for i = 1, 2, 3,
ku1 k < 105, ω(u2 ) > 106, and ku3 k > 105, ω(u3 ) < 106.
Acknowledgements:
The project is supported by the Science Foundation of Hebei University of Science and Technology(XL201144).
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