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Applied Mathematics Letters 23 (2010) 1129–1132
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Applied Mathematics Letters
journal homepage: www.elsevier.com/locate/aml
An existence result for a superlinear fractional differential equation
Dumitru Băleanu a,∗ , Octavian G. Mustafa b , Ravi P. Agarwal c
a
Çankaya University, Department of Mathematics & Computer Science, Ögretmenler Cad. 14 06530, Balgat – Ankara, Turkey
b
University of Craiova, DAL, Department of Mathematics & Computer Science, Tudor Vladimirescu 26, 200534 Craiova, Romania
c
Florida Institute of Technology, Department of Mathematical Sciences, Melbourne, FL 32901, USA
article
abstract
info
Article history:
Received 18 January 2010
Received in revised form 7 April 2010
Accepted 30 April 2010
We establish the existence and uniqueness of solution for the boundary value problem
α 0
λ
0
α
0 Dt (x ) + a(t )x = 0, t > 0, x (0) = 0, limt →+∞ x(t ) = 1, where 0 Dt designates the
Riemann–Liouville derivative of order α ∈ (0, 1) and λ > 1. Our result might be useful
for establishing a non-integer variant of the Atkinson classical theorem on the oscillation
of Emden–Fowler equations.
© 2010 Elsevier Ltd. All rights reserved.
Keywords:
Sequential differential equation
Emden–Fowler equation
Nonlinear oscillation
1. Introduction
00
λ
In a seminal paper [1] from 1955, Atkinson established
R +∞ that the equation x +a(t )x = 0, where a : [t0 , +∞) → (0, +∞)
is continuous and λ > 1, oscillates if and only if t
ta(t )dt = +∞. It was demonstrated there that the boundary value
0
problem (BVP)
x00 + a(t )xλ = 0,
lim x(t ) = 1,
lim x0 (t ) = 0
t →+∞
(1)
t →+∞
R +∞
ta(t )dt < 1; cf. [1, p. 646].
has at least one solution when the Atkinson hypothesis (AH) holds, namely η = λ t
0
In the present note, we shall formulate a fractional variant of (1) and establish the existence and uniqueness of its solution
under some AH-type restriction. During the last few years the fractional calculus has been applied successfully to a variety
of applied problems; see [2–9].
As a fractional counterpart of (1), we introduce the two-point BVP
α
0 Dt
(x0 ) + a(t )xλ = 0,
where (0 Dαt f )(t ) =
1
Γ (1−α)
lim t 1−α x0 (t ) = 0,
t &0
·
d
dt
hR
t
λ
f (s)
0 (t −s)α
lim x(t ) = 1,
t →+∞
(2)
i
ds stands for the Riemann–Liouville derivative of order α of some function f ;
λ
cf. [10, p. 68]. Here, α ∈ (0, 1), x = |x | · sign x and Γ stands for Euler’s function Gamma.
The regularity of the solution is standard, see [11,10,3]; we ask that x ∈ C ([0, +∞), R) ∩ C 1 ((0, +∞), R) and
limt &0 [t 1−α x0 (t )] = 0. Thus, the problem can be recast as
x0 (t ) = −
∗
1
Γ (α)
t
Z
0
a(s)
· [x(s)]λ ds,
(t − s)1−α
t > 0.
Corresponding author.
E-mail addresses: [email protected] (D. Băleanu), [email protected] (O.G. Mustafa), [email protected] (R.P. Agarwal).
0893-9659/$ – see front matter © 2010 Elsevier Ltd. All rights reserved.
doi:10.1016/j.aml.2010.04.049
(3)
1130
D. Băleanu et al. / Applied Mathematics Letters 23 (2010) 1129–1132
To study (3), we shall employ a technique from [12]. Precisely, we ask that the singular integral equation
y(t ) =
A(s)
t
Z
(t − s)1−α
0
+∞
Z
y(τ )dτ
1−
λ
ds,
t ≥ 0,
(4)
s
where A : [0, +∞) → R is continuous, has a solution in C ∩ L∞ ∩ L1 . It is obvious that (4) leads to (3) for A = −a · [Γ (α)]−1
and y = x0 .
The present work is organized as follows. Section 2 contains our existence result regarding the problem (2). Section 3
gives an illustrative example of coefficient A(t ).
2. Main result
1
Introduce the function w : [1, +∞) → [0, +∞) with the formula w(ζ ) =
λ−λ (λ − 1)λ−1 for all ζ ≥ 1, where ζ0 = λλ (λ − 1)−λ .
Notice that λ[1 + ζ0 w(ζ0 )]λ−1 = w(ζ1 ) and |uλ − v λ | ≤
0
1
w(ζ0 )
ζ λ −1
.
ζ
Then, 0 ≤ w(ζ ) ≤ w(ζ0 ) =
· |u − v| for all u, v such that |u|, |v| ≤ 1 + ζ0 w(ζ0 ).
Assume that the next AH restriction holds true
+∞
Z
0
Z
A(s)
t
0
(t − s)
t ≥0
A(s)
t
Z
ds dt + sup 1−α
0
( t − s)
ds = ε · w(ζ0 ) < 1
1−α
(5)
for some ε ∈ (0, 1).
Theorem 1. The problem (2) has a solution x(t ) = 1 −
problem (4).
R +∞
t
y(s)ds, t ≥ 0, where y ∈ (C ∩ L∞ ∩ L1 )([0, +∞), R) verifies the
R
t
Proof. Introduce the set X of all the functions y ∈ (C ∩ L∞ ∩ L1 )([0, +∞), R) such that |y(t )| ≤ ζ0 · A(s)
0 (t −s)1−α
ds, t ≥ 0,
and the metric d(y1 , y2 ) = ky1 − y2 k∞ + ky1 − y2 kL1 for any y1 , y2 ∈ X . The Lebesgue dominated convergence theorem
shows that (X , d) is a complete metric space; see [12]. In particular, d(y, 0) ≤ εζ0 w(ζ0 ), y ∈ X .
The operator T : X → (C ∩ L∞ ∩ L1 )([0, +∞), R) with the formula
(T y)(t ) =
A(s)
t
Z
0
(t − s)1−α
+∞
Z
y(τ )dτ
1−
λ
ds,
t ≥ 0,
s
satisfies the conditions
Z t
Z t
A(s)
A(s)
λ
ds ≤ [1 + εζ0 w(ζ0 )] ds
|(T y)(t )| ≤ (1 + kykL1 ) 1−α
1−α
(
t
−
s
)
(
t
−
s
)
0
0
Z t
A(s)
t ≥ 0,
≤ ζ0 ds ,
1−α
0 ( t − s)
λ
that is T X ⊆ X , and respectively
kT y1 − T y2 k∞ + kT y1 − T y2 kL1 ≤ εky1 − y2 kL1 ≤ ε · d(y1 , y2 ).
Since T is an ε -contraction of (X , d), it has a fixed point y0 . This function is a solution of our problem (4).
3. A set of conditions for A(t )
Assume that the function A has a unique zero T > 0 and
T
Z
A(s)ds = −
0
Z
T
+∞
A(s)ds,
+∞
Z
t |A(t )|dt < +∞.
(6)
0
Suppose that B(t ) = t α kA|[t ,+∞) k∞ belongs to (L∞ ∩ L1 )((T , +∞), R). Notice that a simple candidate for B is provided
by the restriction |A(t )| ≤ c · t −(2+α) , t ≥ T , when c > 0.
We claim that
+∞
Z
0
Z
t
0
A(s)
Z t
A(s)
ds dt + sup ds < +∞.
1−α
(t − s)1−α (
t
−
s
)
t ≥0
0
D. Băleanu et al. / Applied Mathematics Letters 23 (2010) 1129–1132
1131
To establish the claim, let us start with the next computation
Z 2t
Z t
ds
A(s)
ds ≤ ds +
· t −α B(t )
1−α
(2t − s)1−α (
2t
−
s
)
(
2t
−
s)1−α
t
0
Z t
1
A(s)
=
ds +
· B(t ), t ≥ T .
1−α
(
2t
−
s
)
α
0
A(s)
2t
Z
0
Further,
A(s)
t
Z
t
Z
Z
1
0
s
A(τ )dτ ds
·
(2t − s)1−α
0
Z s
Z t
Z t
1
1
A(τ )dτ ds
A(τ )dτ − (1 − α)
= 1−α
2−α
t
0 (2t − s)
0
0
Z s
Z t
1
A(τ )dτ ds + o t α−1
when t → +∞,
= −(1 − α)
2
−α
0 (2t − s)
0
R +∞
Rt
since 0
A(s)ds = limt →+∞ 0 A(s)ds = 0. This follows from (6).
(2t − s)1−α
0
ds =
0
Now, we have
t
Z
0
s
Z
1
(2t − s)2−α
0
Z s
Z t
1
A(τ )dτ ds ≤
A(τ )dτ ds
2−α (
2t
−
s
)
0
0
Z
Z t
Z +∞ Z s
1 s
A(τ )dτ ds ≤ 1
A(τ )dτ ds
≤
2−α
t 2−α 0
0 t
0
0
= O t α−2 when t → +∞,
since again – recall (6) –
Z
+∞
2T
Z s
Z +∞ Z T Z s A(τ )dτ ds =
+
A(τ )dτ ds
0
2T
0
T
Z +∞ Z +∞
ds
−
A
(τ
)
d
τ
=
s
2T
Z
+∞
+∞
Z
|A(τ )|dτ ds
=
2T
Z
s
+∞
=
(s − 2T ) |A(s)|ds < +∞.
2T
We obtained that limt →+∞
A(s)
0 (t −s)1−α
Rt
ds = 0. Since limt &0
A(s)
0 (t −s)1−α
Rt
ds = 0 and the application t 7→
continuous, see [11,10,3], we deduce that it is bounded over [0, +∞).
Finally,
Z
+∞
4T
Z
2t
0
A(s)
0 (t −s)1−α
Rt
ds is
A(s)
Z +∞
Z +∞ Z t
1
1
ds dt ≤
B(t )dt +
A(τ )dτ dt
1−α
(2t − s)1−α α 4T
(4T )
4T
0
Z +∞
Z +∞ Z s
dt
ds < +∞.
+ (1 − α)
·
A
(τ
)
d
τ
2−α
4T
t
0
0
The claim is established.
Acknowledgement
We are grateful to the referees for their most useful comments.
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