Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 130, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
UNIQUENESS AND EXISTENCE OF POSITIVE SOLUTIONS
FOR SINGULAR FRACTIONAL DIFFERENTIAL EQUATIONS
NEMAT NYAMORADI, TAHEREH BASHIRI,
S. MANSOUR VAEZPOUR, DUMITRU BALEANU
Abstract. In this article, we study the existence of positive solutions for the
singular fractional boundary value problem
−Dα u(t) = Af (t, u(t)) +
k
X
Bi I βi gi (t, u(t)),
t ∈ (0, 1),
i=1
Dδ u(0) = 0,
Dδ u(1) = aD
α−δ−1
2
˛
(Dδ u(t))˛t=ξ
where 1 < α ≤ 2, 0 < ξ ≤ 1/2, a ∈ [0, ∞), 1 < α − δ < 2, 0 < βi < 1, A, Bi ,
1 ≤ i ≤ k, are real constant, Dα is the Riemann-Liouville fractional derivative
of order α. By using the Banach’s fixed point theorem and Leray-Schauder’s
alternative, the existence of positive solutions is obtained. At last, an example
is given for illustration.
1. Introduction
Fractional calculus is the field of mathematical analysis which deals with the
investigation and applications of integrals and derivatives of arbitrary order, the
fractional calculus may be considered an old and yet novel topic.
Recently, fractional differential equations have been of great interest. This is
because of both the intensive development of the theory of fractional calculus itself
and its applications in various sciences, such as physics, mechanics, chemistry, engineering, etc. For example, for fractional initial value problems, the existence and
multiplicity of solutions were discussed in [2, 4, 10, 11], moreover, fractional derivative arises from many physical processes,such as a charge transport in amorphous
semiconductors [18], electrochemistry and material science are also described by
differential equations of fractional order [5, 6, 7, 14, 15]. Bai and Lü [3] considered
the boundary value problem of fractional order differential equation
D0α+ u(t) + f (t, u(t)) = 0,
t ∈ (0, 1),
u(0) = u(1) = 0,
2000 Mathematics Subject Classification. 34A08, 74H20,30E25.
Key words and phrases. Existence of solutions; Banachs fixed point theorem;
Leray-Schauders alternative.
c
2014
Texas State University - San Marcos.
Submitted August 21, 2013. Published June 6, 2014.
1
2
N. NYAMORADI, T. BASHIRI, S. M. VAEZPOUR, D. BALEANU
EJDE-2014/??
where D0α+ is the standard Riemann-Liouville fractional derivative of order 1 < α ≤
2 and
f : [0, 1] × [0, ∞) → [0, ∞) is continuous.
Hussein [9], considered the following nonlinear m-point boundary value problem
of fractional type
D0α+ x(t) + q(t)f (t, x(t)) = 0,
0
00
a.e. on [0, 1], α ∈ (n − 1, n], n ≥ 2,
x(0) = x (0) = x (0) = · · · = x
(n−2)
(0) = 0,
x(1) =
m−2
X
ξi x(ηi ),
i=1
Pm−2
where 0 < η1 < · · · < ηm−2 < 1, ξi > 0 with i=1 ξi ηiα−1 < 1, q is a real-valued
continuous function and f is a nonlinear Pettis integrable function.
Motivated by the above works, the purpose of this paper is to discuss the following singular fractional boundary value problem:
−Dα u(t) = Af (t, u(t)) +
k
X
Bi I βi gi (t, u(t)),
t ∈ (0, 1),
i=1
Dδ u(0) = 0,
Dδ u(1) = aD
α−δ−1
2
(1.1)
(Dδ u(t))t=ξ ,
where 1 < α ≤ 2, 0 < ξ ≤ 12 , a ∈ (0, ∞), 1 < α − δ < 2, 0 < βi < 1, A, Bi ,
1 ≤ i ≤ k, are real constant, Dα is the Riemann-Liouville fractional derivative of
order α.
The rest of this paper is organized as follows. In Section 2, we give some preliminaries. In Sections 3 and 4, we study the existence and uniqueness of solutions
for system (1.1) by Banach’s fixed point theorem and Leray-Schauder’s alternative,
respectively. At last, in Section 5, an example is also given to illustrate our theory.
2. Preliminaries
In this section, we present notation and some preliminary lemmas that will be
used in the proofs of the main results.
Definition 2.1 ([16, 17]). The Riemann-Liouville fractional integral operator of
order α > 0, of function f ∈ L1 (R+ ) is defined as
Z t
1
I0α+ f (t) =
(t − s)α−1 f (s)ds,
Γ(α) 0
where Γ(·) is the Euler gamma function.
Definition 2.2 ([16, 17]). The Riemann-Liouville fractional derivative of order
α > 0 of a continuous function f : (0, ∞) → R is defined as
d n Z t
1
(t − s)n−α−1 f (s)ds,
D0α+ f (t) =
Γ(n − α) dt
0
where n = [α] + 1.
Lemma 2.3 ([12]). The equality D0γ+ I0γ+ f (t) = f (t) with γ > 0 holds for f ∈
L1 (0, 1).
Lemma 2.4 ([12]). Let α > 0. Then the differential equation
D0α+ u = 0
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UNIQUENESS AND EXISTENCE OF POSITIVE SOLUTIONS
3
has a unique solution u(t) = c1 tα−1 + c2 tα−2 + · · · + cn tα−n , ci ∈ R, i = 1, . . . , n,
there n − 1 < α ≤ n.
Lemma 2.5 ([12]). Let α > 0. Then the following equality holds for u ∈ L1 (0, 1),
D0α+ u ∈ L1 (0, 1);
I0α+ D0α+ u(t) = u(t) + c1 tα−1 + c2 tα−2 + · · · + cn tα−n ,
ci ∈ R, i = 1, . . . , n, there n − 1 < α ≤ n.
Lemma 2.6 ([3]). For λ > −1 and α > 0,
D0α+ tγ =
Γ(γ + 1) γ−α
t
.
Γ(γ − α + 1)
Lemma 2.7 ([13]). Suppose that g ∈ L1 (0, 1) and α, β be two constant such that
0 ≤ β ≤ 1 < α, then
Z t
Z t
Γ(α)
(t − s)α−β−1 g(s) ds.
D0β+
(t − s)α−1 g(s)ds =
Γ(α
−
β)
0
0
Now, we consider (1.1). By the substitution u(t) = I δ y(t) = D−δ y(t), problem
(1.1) is turned into
−Dα−δ y(t) = Af (t, I δ y(t)) +
k
X
Bi I βi gi (t, I δ y(t)),
t ∈ (0, 1),
(2.1)
i=1
y(0) = 0,
y(1) = aD
α−δ−1
2
y(t))
t=ξ
.
Lemma 2.8. For any h ∈ C[0, 1] ∩ L(0, 1), the unique solution of the boundary
value problem
−Dα−δ y(t) = h(t), t ∈ (0, 1),
(2.2)
α−δ−1
y(0) = 0, y(1) = aD 2 y(t))t=ξ
is
y(t) = −I α−δ h(t) +
tα−δ−1 Γ( α−δ+1
)
2
Γ( α−δ+1
) − aΓ(α − δ)ξ
2
α−δ−1
2
I α−δ h(1) − aI
α−δ+1
2
h(ξ) .
Proof. By applying Lemma 2.5, equation (2.2) is equivalent to the integral equation
y(t) = −I α−δ h(t) − c1 tα−δ−1 − c2 tα−δ−2 ,
(2.3)
for some arbitrary constants c1 , c2 ∈ R.
By the boundary condition y(0) = 0, we conclude that c2 = 0. Then, we have
y(1) = −I α−δ h(1) − c1 ,
and it follows from lemma (2.6) that
D
α−δ−1
2
y(t) = −D
= −I
α−δ−1
2
α−δ+1
2
I α−δ h(t) − c1 D
h(t) − c1
α−δ−1
2
tα−δ−1
Γ(α − δ) α−δ−1
t 2 .
Γ( α−δ+1
)
2
Therefore,
D
α−δ−1
2
y(t)
t=ξ
Z
=−
0
ξ
α−δ−1
(ξ − s) 2
Γ(α − δ) α−δ−1
h(s)ds − c1 α−δ+1 t 2 .
α−δ+1
Γ( 2 )
Γ( 2 )
4
N. NYAMORADI, T. BASHIRI, S. M. VAEZPOUR, D. BALEANU
EJDE-2014/??
α−δ−1
2
y(t))t=ξ , we obtain that
h Z 1 (1 − s)α−δ−1
Γ( α−δ+1
)
2
c1 =
−
h(s)ds
α−δ−1
Γ(α − δ)
Γ( α−δ+1
) − aΓ(α − δ)ξ 2
0
2
Z ξ
α−δ−1
i
(ξ − s) 2
h(s)ds
.
+a
)
Γ( α−δ+1
0
2
So, by the boundary condition y(1) = aD
Therefore, the unique solution of equation (2.2) is
y(t) = −I α−δ h(t) +
ξ
Z
−a
0
tα−δ−1 Γ( α−δ+1
)
2
hZ
α−δ−1
2
)
Γ( α−δ+1
2
α−δ−1
2
(ξ − s)
Γ( α−δ+1
)
2
− aΓ(α − δ)ξ
i
h(s)ds .
0
1
(1 − s)α−δ−1
h(s)ds
Γ(α − δ)
The proof is complete.
Thus, the solution of the problem (1.1) can be written as
u(t) = I δ y(t)
h
= I δ − I α−δ h(t) +
= −I α h(t) +
Z
×
0
t
α−δ−1
2
Γ( α−δ+1
) − aΓ(α − δ)ξ
2
Γ( α−δ+1 )
2
) − aΓ(α − δ)ξ
Γ( α−δ+1
2
α−δ−1
2
I α−δ h(1) − aI
α−δ+1
2
I α−δ h(1) − aI
α−δ+1
2
h(ξ)
I α−δ h(1) − aI
α−δ+1
2
h(ξ)
h(ξ)
i
(t − s)δ−1 α−δ−1
s
ds
Γ(δ)
= −I α h(t) +
×
)
tα−δ−1 Γ( α−δ+1
2
n tα−1
Z
Γ( α−δ+1
)
2
Γ( α−δ+1
) − aΓ(α − δ)ξ
2
1
o
(1 − ν)δ−1 ν α−δ−1 dν ,
α−δ−1
2
Γ(δ) 0
where we have used the substitution s = tν in the integral of the last term. Using
the relation for the Beta function B(·, ·),
Z 1
Γ(α)Γ(β)
,
B(α, β) =
(1 − u)α−1 uβ−1 du =
Γ(α + β)
0
one has
)
tα−1 Γ(α − δ)Γ( α−δ+1
α−δ+1
α−δ
2
2
u(t) = −I α h(t) +
I
h(1)
−
aI
h(ξ)
.
α−δ−1
Γ(α) Γ( α−δ+1
) − aΓ(α − δ)ξ 2
2
(2.4)
The solution of the original nonlinear problem (1.1) can be obtained by replacing
h with the right hand side of the fractional equation of (1.1) in (2.4).
The basic space used in this paper is the real Banach space C = C([0, 1], R) of all
continuous functions from [0, 1] → R endowed with the norm kuk = supt∈[0,1] |u(t)|.
In relation to problem (1.1), we define an operator T : C → C as
Z t
Z t
k
α−1
X
(t − s)α+βi −1
(t − s)
f (s, u(s))ds −
Bi
gi (s, u(s))ds
(T u)(t) = −A
Γ(α)
Γ(α + βi )
0
0
i=1
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UNIQUENESS AND EXISTENCE OF POSITIVE SOLUTIONS
α−1
+t
1
h Z
A0 A
0
+
k
X
Z
1
Bi
0
i=1
5
α−δ−1
(1 − s)
Γ(α − δ)
f (s, u(s))ds
(1 − s)α−δ+βi −1
gi (s, u(s))ds
Γ(α − δ + βi )
α−δ−1
ξ
(ξ − s) 2
f (s, u(s))ds
)
Γ( α−δ+1
0
2
Z ξ
α−δ−1
k
i
X
(ξ − s) 2 +βi
−a
gi (s, u(s))ds ,
Bi
α−δ+1
+ βi )
0 Γ(
2
i=1
Z
− aA
where
A0 =
Γ(α − δ)Γ( α−δ+1
)
2
) − aΓ(α − δ)ξ
Γ(α) Γ( α−δ+1
2
α−δ−1
2
.
It is clear that the existence of a positive solution for the system (1.1) is equivalent
to the existence of a nontrivial fixed point of T on C.
For convenience of the reader, we set
α−δ+1
∆ = sup
n h
|A|
t∈[0,1]
+
k
X
h
|Bi |
i=1
+a
ξ
i
tα
1
ξ 2
+ |A0 |tα−1
+ a α−δ+1
Γ(α + 1)
Γ(α − δ + 1)
Γ( 2 + 1)
tα+βi
1
+ |A0 |tα−1
Γ(α + βi + 1)
Γ(α − δ + βi + 1)
α−δ+1
+βi
2
Γ( α−δ+1
+ βi + 1)
2
(2.5)
io
.
3. Existence results via Banach’s fixed point theorem
In this section, by using Banach’s fixed point theorem, we will establish find a
unique solution of problem (1.1). Now, we state our results.
Theorem 3.1. Assume that f, gi : [0, 1] × R → R, i = 1, . . . , k, are continuous
functions satisfying the condition
(A1) |f (t, u) − f (t, v)| ≤ L1 |u − v|, |gi (t, u) − gi (t, v)| ≤ Li+1 |u − v|, for i =
1, . . . , k, t ∈ [0, 1], Li > 0, (i = 1, . . . , k + 1), u, v ∈ R.
Then the boundary-value problem (1.1) has a unique solution if L <
L = max{Li : i = 1, . . . , k + 1} and ∆ is given by (2.5).
1
∆,
where
Proof. Assume that M = max{Mi : i = 1, . . . , k + 1}, where Mi are finite numbers
∆M
given by supt∈[0,1] |f (t, 0)| = M1 , supt∈[0,1] |gi (t, 0)| = Mi+1 . Selecting r > 1−L∆
,
we show that T Br ⊂ Br , where Br = {u ∈ C : kuk ≤ r}. Using that |f (s, u(s))| ≤
|f (s, u(s)) − f (s, 0)| + |f (s, 0)| ≤ L1 r + M1 , |gi (s, u(s))| ≤ |gi (s, u(s)) − gi (s, 0)| +
|gi (s, 0)| ≤ Li+1 r + Mi+1 , i = 1, . . . , k, for u ∈ Br , and (2.5) we can show that
k(T u)k
α−δ+1
≤ (Lr + M ) sup
t∈[0,1]
n h
|A|
i
tα
1
ξ 2
+ |A0 |tα−1
+ a α−δ+1
Γ(α + 1)
Γ(α − δ + 1)
Γ( 2 + 1)
6
N. NYAMORADI, T. BASHIRI, S. M. VAEZPOUR, D. BALEANU
+
k
X
h
|Bi |
i=1
+a
ξ
EJDE-2014/??
1
tα+βi
+ |A0 |tα−1
Γ(α + βi + 1)
Γ(α − δ + βi + 1)
α−δ+1
+βi
2
io
Γ( α−δ+1
2
+ βi + 1)
≤ (Lr + M )∆ < r,
which implies that T Br ⊂ Br . Now, for u, v ∈ C we obtain
kT u − T vk
n Z
≤ sup |A|
t∈[0,1]
k
X
t
0
α−1
(t − s)
Γ(α)
|f (s, u(s)) − f (s, v(s))|ds
t
(t − s)α+βi −1
|gi (s, u(s)) − gi (s, v(s))|ds
Γ(α + βi )
0
i=1
h Z 1 (1 − s)α−δ−1
+ |A0 |tα−1 |A|
|f (s, u(s)) − f (s, v(s))|ds
Γ(α − δ)
0
Z 1
k
X
(1 − s)α−δ+βi −1
+
|Bi |
|gi (s, u(s)) − gi (s, v(s))|ds
Γ(α − δ + βi )
0
i=1
Z ξ
α−δ−1
(ξ − s) 2
|f (s, u(s)) − f (s, v(s))|ds
+ |A|a
Γ( α−δ+1
)
0
2
Z ξ
α−δ−1
k
io
X
(ξ − s) 2 +βi −1
+a
|Bi |
|g
(s,
u(s))
−
g
(s,
v(s))|ds
i
i
Γ( α−δ−1
+ βi )
0
2
i=1
α−δ+1
n h
i
tα
1
ξ 2
≤ L sup |A|
+ |A0 |tα−1
+ a α−δ+1
Γ(α + 1)
Γ(α − δ + 1)
Γ( 2 + 1)
t∈[0,1]
+
+
k
X
Z
|Bi |
h
|Bi |
i=1
+a
ξ
1
tα+βi
+ |A0 |tα−1
Γ(α + βi + 1)
Γ(α − δ + βi + 1)
α−δ+1
+βi
2
Γ( α−δ+1
2
+ βi + 1)
io
ku − vk
= L∆ku − vk.
By the assumption, L < 1/∆. Therefore, T is a contraction. Thus, by the contraction mapping principle (Banach’s fixed point theorem) the proof is complete. Now we present another variant of existence uniqueness result based on the
Hölder inequality.
Theorem 3.2. Suppose that the continuous functions f, gi satisfy the following
conditions:
(A2) |f (t, u(t))−f (t, v(t))| ≤ m(t)|u−v|, |gi (t, u(t))−gi (t, v(t))| ≤ ni (t)|u−v|, for
1
t ∈ [0, 1], u, v ∈ R, m, ni ∈ L γ ([0, 1], R+ ), i = 1, . . . , k, and γ ∈ (0, α−δ−2).
Pk
(A3) |A|kmkZ1 + i=1 |Bi |kni kZi+1 < 1, where
1 1 − γ 1−γ
|A0 | 1 − γ 1−γ
Z1 =
+
Γ(α) α − γ
Γ(α − δ) α − δ − γ
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UNIQUENESS AND EXISTENCE OF POSITIVE SOLUTIONS
+
a|A0 | 1 − γ 1−γ α−δ+1 −γ
ξ 2
,
) α−δ+1
−γ
Γ( α−δ+1
2
2
1−γ
1 − γ 1−γ
1
|A0 |
1−γ
)
+(
)
Γ(α + βi ) α + βi − γ
Γ(α − δ + βi ) α − δ + βi − γ
1−γ α−δ+1
a|A0 |
1−γ
+ ( α−δ+1
) α−δ+1
ξ 2 +βi −γ , (i = 1, . . . , k),
Γ( 2 + βi )
+
β
−
γ
i
2
R1
1
γ
and kµk = ( 0 |µ(s)| γ ds) , µ = m, n.
Then, the boundary value problem (1.1) has a unique solution.
Zi+1 = (
Proof. For u, v ∈ R and t ∈ [0, 1], by Hölder inequality, we have
kT u − T vk
n Z
≤ sup |A|
t∈[0,1]
+
k
X
t
Z
|Bi |
0
h
+ |A0 | |A|
Z
1
|Bi |
0
i=1
ξ
m(s)|u(s) − v(s)|ds
(t − s)α+βi −1
ni (s)|u(s) − v(s))|ds
Γ(α + βi )
1
Z
0
k
X
(t − s)
Γ(α)
0
i=1
+
α−1
t
α−δ−1
(1 − s)
Γ(α − δ)
m(s)|u(s) − v(s)|ds
(1 − s)α−δ+βi −1
ni (s)|u(s) − v(s))|ds
Γ(α − δ + βi )
α−δ−1
(ξ − s) 2
m(s)|u(s) − v(s)|ds
Γ( α−δ+1
)
0
2
Z ξ
α−δ−1
k
io
X
(ξ − s) 2 +βi
+a
|Bi |
n
(s)|u(s)
−
v(s))|ds
i
α−δ+1
+ βi )
0 Γ(
2
i=1
Z
+ a|A|
≤ sup
t∈[0,1]
n |A|kmk 1 − γ 1−γ
Γ(α)
α−γ
tα−γ +
k
X
|Bi |kni k 1 − γ 1−γ α+βi −γ
(
t
)
Γ(α + βi ) α + βi − γ
i=1
k
h |A|kmk 1 − γ 1−γ X
1−γ
|Bi |kni k
1−γ
+ |A0 |
+
(
)
Γ(α − δ) α − δ − γ
Γ(α − δ + βi ) α − δ + βi − γ
i=1
a|A|kmk 1 − γ 1−γ α−δ+1 −γ
+ α−δ+1 α−δ+1
ξ 2
Γ( 2 )
−γ
2
k
1−γ α−δ+1
io
X
|Bi |kni k
1−γ
+a
( α−δ+1
) α−δ+1
ξ 2 +βi −γ ku − vk
Γ( 2 + βi )
+ βi − γ
2
i=1
n
h 1 1 − γ 1−γ
|A0 | 1 − γ 1−γ
≤ A|kmk
+
Γ(α) α − γ
Γ(α − δ) α − δ − γ
i
1−γ
α−δ+1
a|A0 |
1−γ
ξ 2 −γ
+ α−δ+1 α−δ+1
Γ( 2 )
−γ
2
k
h
1 − γ 1−γ
X
1
+
|Bi |kni k (
)
Γ(α + βi ) α + βi − γ
i=1
7
8
N. NYAMORADI, T. BASHIRI, S. M. VAEZPOUR, D. BALEANU
EJDE-2014/??
1−γ
|A0 |
1−γ
)
Γ(α − δ + βi ) α − δ + βi − γ
1−γ α−δ+1
io
a|A0 |
1−γ
+ ( α−δ+1
) α−δ+1
ξ 2 +βi −γ ku − vk
Γ( 2 + βi )
+ βi − γ
2
+(
= [|A|kmkZ1 +
k
X
|Bi |kni kZi+1 ]ku − vk.
i=1
By the condition (A3), it follows that T is a contraction mapping. Hence, by
the Banach’s fixed point theorem T has a unique fixed point which is the unique
solution of the problem (1.1). Then, the proof is complete.
4. Existence result via Leray-Schauder’s alternative
In this section, by using the Leray-schauder’s alternative, we will find at least one
solution to problem (1.1). The proof of the main result in this section is based on the
Leray-schauder’s alternative [1, 8] that we recall here for the reader’s convenience.
Lemma 4.1. (Nonlinear alternative for single valued maps [1, 8]). Let E be a
Banach space, C a closed, convex subset of E, U an open subset of C and 0 ∈ U .
Suppose that F :Ū → C is a completely continuous operator. Then, either
(i) F has a fixed point in Ū, or
(ii) there is a u ∈ ∂U (the boundary of U in C) and λ ∈ (0, 1) with u = λF (u).
We now state our main result in this section.
Theorem 4.2. Suppose that f, gi : [0, 1] × R → R, i = 1, . . . , k, are continuous
functions. Assume that:
(H1) There exist functions p, pi ∈ L1 ([0, 1], R+ ), i = 1, . . . , k, and nondecreasing
functions ψ, ψi : R+ → R+ , i = 1, . . . , k, such that
|f (t, x)| ≤ p(t)ψ(kxk), |gi (t, x)| ≤ pi (t)ψi (kxk),
for all (t, x) ∈ [0, 1] × R and i = 1, . . . , k.
(H2) There exists a constant M > 0 such that
|A|ψ(M )kpkL1 Ω +
M
Pk
i=1
|Bi |Ωi ψi (M )kpi kL1
>1
where
α−δ+1
1
ξ 2
1
+ |A0 |
+ a α−δ+1
,
Ω=
Γ(α + 1)
Γ(α − δ + 1)
Γ( 2 + 1)
and
α−δ+1
1
1
ξ 2 +βi
Ωi =
+ |A0 |
+ a α−δ+1
,
Γ(α + βi + 1)
Γ(α − δ + βi + 1)
Γ( 2 + βi + 1)
(i = 1, . . . , k).
Then, the boundary-value problem (1.1) has at least one solution on [0, 1].
Proof. Consider the operator T : C → C with
(T u)(t)
EJDE-2014/130
UNIQUENESS AND EXISTENCE OF POSITIVE SOLUTIONS
9
Z t
k
α−1
X
(t − s)
(t − s)α+βi −1
f (s, u(s))ds −
gi (s, u(s))ds
Bi
Γ(α)
Γ(α + βi )
0
0
i=1
h Z 1 (1 − s)α−δ−1
+ tα−1 A0 A
f (s, u(s))ds
Γ(α − δ)
0
Z 1
k
X
(1 − s)α−δ+βi −1
+
gi (s, u(s))ds
Bi
Γ(α − δ + βi )
0
i=1
Z ξ
Z ξ
α−δ−1
α−δ−1
k
i
X
(ξ − s) 2
(ξ − s) 2 +βi
f
(s,
u(s))ds
−
a
g
(s,
u(s))ds
.
− aA
B
i
i
α−δ+1
)
+ βi )
Γ( α−δ+1
0
0 Γ(
2
2
i=1
Z
t
= −A
We show that T maps bounded sets into bounded sets in C([0, 1], R). for a positive
number r, let Br = {u ∈ C([0, 1], R) : kuk ≤ r} be a bounded set in C([0, 1], R).
Then, we have
|(T u)(t)|
Z t
Z t
k
α−1
X
(t − s)
(t − s)α+βi −1
≤ |A|
p(s)ψ(kuk)ds +
|Bi |
pi (s)ψi (kuk)ds
Γ(α)
Γ(α + βi )
0
0
i=1
h Z 1 (1 − s)α−δ−1
p(s)ψ(kuk)ds
+ |A0 |tα−1 |A|
Γ(α − δ)
0
Z 1
k
X
(1 − s)α−δ+βi −1
+
|Bi |
pi (s)ψi (kuk)ds
Γ(α − δ + βi )
0
i=1
Z ξ
α−δ−1
(ξ − s) 2
p(s)ψ(kuk)ds
+ |A|a
)
Γ( α−δ+1
0
2
Z ξ
α−δ−1
k
i
X
(ξ − s) 2 +βi −1
|Bi |
+a
p
(s)ψ
(kuk)ds
i
i
Γ( α−δ−1
+ βi )
0
2
i=1
α−δ+1
i
h
ξ 2
1
tα
α−1
1
+ |A0 |t
+ a α−δ+1
≤ |A|ψ(r)kpkL
Γ(α + 1)
Γ(α − δ + 1)
Γ( 2 + 1)
k
h
X
tα+βi
1
+
|Bi |ψi (r)kpi kL1
+ |A0 |tα−1
Γ(α + βi + 1)
Γ(α − δ + βi + 1)
i=1
+a
ξ
α−δ+1
+βi
2
Γ( α−δ+1
2
+ βi + 1)
i
.
Consequently,
kT uk
α−δ+1
i
1
ξ 2
1
+ |A0 |
+ a α−δ+1
Γ(α + 1)
Γ(α − δ + 1)
Γ( 2 + 1)
k
h
X
1
1
+
|Bi |ψi (r)kpi kL1
+ |A0 |
Γ(α
+
β
+
1)
Γ(α
−
δ
+ βi + 1)
i
i=1
≤ |A|ψ(r)kpkL1
+a
ξ
h
α−δ+1
+βi
2
Γ( α−δ+1
+ βi + 1)
2
i
10
N. NYAMORADI, T. BASHIRI, S. M. VAEZPOUR, D. BALEANU
= |A|ψ(r)kpkL1 Ω +
k
X
EJDE-2014/??
|Bi |ψi (r)kpi kL1 Ωi ,
i=1
Pk
let K = |A|ψ(r)kpkL1 Ω + i=1 |Bi |ψi (r)kpi kL1 Ωi , therefore, we conclude that
kT uk ≤ K. Thus, T maps bounded sets into bounded sets in C([0, 1], R).
Next, we show that T maps bounded sets into equicontinuous sets of C([0, 1], R).
Let t1 , t2 ∈ [0, 1] with t1 < t2 and u ∈ Br ; thus, we have
|(T u)(t2 ) − (T u)(t1 )|
Z t1
|A|
[(t2 − s)α−1 − (t1 − s)α−1 ]|f (s, u(s))|ds
≤
Γ(α) 0
Z t2
|A|
α−1
+
(t2 − s)
|f (s, u(s))|ds
Γ(α) t1
Z t1
k
X
|Bi |
[(t2 − s)α+βi −1 − (t1 − s)α+βi −1 ]|gi (s, u(s))|ds
+
Γ(α
+
β
)
i
0
i=1
Z t2
k
X
|Bi |
(t2 − s)α+βi −1 |gi (s, u(s))|ds
+
Γ(α
+
β
)
i
t1
i=1
h Z 1 (1 − s)α−δ−1
α−1
α−1
+ [(t2 )
− (t1 )
]|A0 | |A|
|f (s, u(s))|ds
Γ(α − δ)
0
Z 1
k
X
(1 − s)α−δ+βi −1
+
|Bi |
|gi (s, u(s))|ds
Γ(α − δ + βi )
0
i=1
Z ξ
α−δ−1
(ξ − s) 2
|f (s, u(s))|ds
+ a|A|
)
Γ( α−δ+1
0
2
Z ξ
α−δ−1
k
i
X
(ξ − s) 2 +βi
|Bi |
+a
|gi (s, u(s))|ds
α−δ+1
+ βi )
0 Γ(
2
i=1
Z t1
|A|
≤
[(t2 − s)α−1 − (t1 − s)α−1 ]p(s)ψ(r)ds
Γ(α) 0
Z t2
|A|
α−1
+
(t2 − s)
p(s)ψ(r)ds
Γ(α) t1
Z t1
k
X
|Bi |
+
[(t2 − s)α+βi −1 − (t1 − s)α+βi −1 ]pi (s)ψi (r)ds
Γ(α
+
β
)
i
0
i=1
Z t2
k
X
|Bi |
+
(t2 − s)α+βi −1 pi (s)ψi (r)ds
Γ(α + βi ) t1
i=1
h Z 1 (1 − s)α−δ−1
α−1
α−1
+ [(t2 )
− (t1 )
]|A0 | |A|
p(s)ψ(r)ds
Γ(α − δ)
0
Z 1
k
X
(1 − s)α−δ+βi −1
+
|Bi |
pi (s)ψi (r)ds
Γ(α − δ + βi )
0
i=1
EJDE-2014/130
Z
+ a|A|
0
ξ
UNIQUENESS AND EXISTENCE OF POSITIVE SOLUTIONS
11
Z ξ
α−δ−1
α−δ−1
k
i
X
(ξ − s) 2
(ξ − s) 2 +βi
p(s)ψ(r)ds
+
a
p
(s)ψ
(r)ds
|B
|
i
i
i
α−δ+1
Γ( α−δ+1
)
+ βi )
0 Γ(
2
2
i=1
Obviously, the right-hand side of the above inequality tends to zero independently
of u ∈ Br as t2 − t1 → 0. Therefore, T : C([0, 1], R) → C([0, 1], R) is completely
continuous by application of the Arzelá-Ascoli theorem.
Now, we can conclude the result by using the Leray-Schauder’s nonlinear alternative theorem. Consider the equations x = λT x for λ ∈ (0, 1) and assume that
u be a solution. Then, using the computations in proving that T is bounded, we
have
kuk = kλ(T u)k
α−δ+1
i
1
ξ 2
1
+ |A0 |
+ a α−δ+1
Γ(α + 1)
Γ(α − δ + 1)
Γ( 2 + 1)
k
h
X
1
1
+
|Bi |ψi (kuk)kpi kL1
+ |A0 |
Γ(α
+
β
+
1)
Γ(α
−
δ
+ βi + 1)
i
i=1
≤ |A|ψ(kuk)kpkL1
+a
ξ
h
α−δ+1
+βi
2
i
+ βi + 1)
Γ( α−δ+1
2
;
therefore,
|A|ψ(kuk)kpkL1 Ω +
kuk
Pk
i=1
≤ 1.
|Bi |Ωi ψi (kuk)kpi kL1
In view of (H2), there exists M such that kuk =
6 M . Let us set
U = {x ∈ C([0, 1], R) : kxk < M }.
It is obvious that the operator T : Ū → C([0, 1], R) is continuous and completely
continuous. From the choice of U there is no u ∈ ∂U such that u = λT (u) for
some λ ∈ (0, 1). Therefore, by the Leray-Schauder’s nonlinear alternative theorem
(Lemma 4.1), we conclude that T has a fixed point u ∈ Ū which is a solution of
the problem (1.1). Thus, the proof is completed.
5. Application
Example 5.1. Consider the singular boundary value problem
−D
3/2
u(t) = Af (t, u(t)) +
3
X
Bi I βi gi (t, u(t)),
t ∈ (0, 1),
(5.1)
i=1
D
1/4
u(0) = 0,
D
1/4
u(1) = aD
1/8
(D
1/4
u(t))
t=1/2
Here,
1 = 1/2, β2 = 1/3, β3 = 2/3, a = 2, f (t, u) =
√ A = Bi = 1, (i = 1, 2,33), β−1
9
1
−
t(1
+
u),
g
(t,
u)
=
tan
u + cos(et ). With the given data, we obtain:
i
125
29
A0 =
Γ(α − δ)Γ( α−δ+1
)
2
Γ(α) Γ( α−δ+1
) − aΓ(α − δ)ξ
2
α−δ−1
2
= −1.3365,
α−δ+1
∆ = sup
t∈[0,1]
n h
|A|
i
tα
1
ξ 2
+ |A0 |tα−1
+ a α−δ+1
Γ(α + 1)
Γ(α − δ + 1)
Γ( 2 + 1)
12
N. NYAMORADI, T. BASHIRI, S. M. VAEZPOUR, D. BALEANU
+
k
X
|Bi |
i=1
+a
ξ
h
EJDE-2014/??
1
tα+βi
+ |A0 |tα−1
Γ(α + βi + 1)
Γ(α − δ + βi + 1)
α−δ+1
+βi
2
Γ( α−δ+1
2
+ βi + 1)
io
= 8.9046,
9
and L1 = 9/125, L2 = L3 = L4 = 3/29 as |f (t, u) − f (t, v)| ≤ 125
|u − v|, |gi (t, u) −
3
1
gi (t, v)| ≤ 29 |u − v|. Obviously, L = max{Li : i = 1, . . . , 4} = 3/29 and L < ∆
.
Hence, all the assumptions of Theorem 3.1 are satisfied. Thus, by the conclusion
of Theorem 3.1, problem (5.1) has a unique solution.
Acknowledgments. The authors would like to thank the anonymous referees for
their suggestions and helpful comments which improved the presentation of the
original manuscript.
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EJDE-2014/130
UNIQUENESS AND EXISTENCE OF POSITIVE SOLUTIONS
13
Nemat Nyamoradi
Department of Mathematics, Faculty of Sciences, Razi University, 67149 Kermanshah,
Iran
E-mail address: nyamoradi@razi.ac.ir
Tahereh Bashiri
Department of Mathematics and Computer Sciences, Amirkabir University of Technology, Tehran, Iran
E-mail address: t.bashiri@aut.ac.ir
S. Mansour Vaezpour
Department of Mathematics and Computer Sciences, Amirkabir University of Technology, Tehran, Iran
E-mail address: vaez@aut.ac.ir
Dumitru Baleanu
Department of Mathematics and Computer Sciences, Faculty of Art and Sciences,
Cankaya University, 06530 Ankara, Turkey.
Institute of Space Sciences, P.O.BOX, MG-23, R 76900,Magurele-Bucharest, Romania.
Department of Chemical and Materials Engineering, Faculty of Engineering, King Abdulaziz University, P.O. Box 80204, Jeddah 21589, Saudi Arabia
E-mail address: dumitru@cankaya.edu.tr