Electronic Journal of Differential Equations, Vol. 2008(2008), No. 146, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE OF POSITIVE SOLUTIONS FOR SINGULAR
FRACTIONAL DIFFERENTIAL EQUATIONS
TINGTING QIU, ZHANBING BAI
Abstract. In this article, we establish the existence of a positive solution to
a singular boundary-value problem of nonlinear fractional differential equation. Our analysis rely on nonlinear alternative of Leray-Schauder type and
Krasnoselskii’s fixed point theorem in a cone.
1. Introduction
Many papers and books on fractional calculus differential equation have appeared recently. Most of them are devoted to the solvability of the linear fractional equation in terms of a special function and to problems of analyticity in
the complex domain(see, for example [2, 8]). Moreover, Delbosco and Rodino [3]
considered the existence of a solution for the nonlinear fractional differential equation D0α+ u = f (t, u), where 0 < α < 1, and f : [0, a] × R → R, 0 < a ≤ +∞ is a
given function,continuous in (0, a)×R. They obtained results for solutions by using
the Schauder fixed point theorem and the Banach contraction principle. Recently,
Zhang [11] considered the existence of positive solution for equation D0α+ u = f (t, u),
where 0 < α < 1, and f : [0, 1] × [0, +∞) → [0, +∞), is a given continuous function,
by using the sub-and super-solution method.
In this article, we discuss the existence of a positive solution to boundary-value
problems of the nonlinear fractional differential equation
D0α+ u(t) + f (t, u(t)) = 0,
0
0 < t < 1,
00
u(0) = u (1) = u (0) = 0 ,
(1.1)
where 2 < α ≤ 3, D0α+ is the Caputo’s differentiation, and f : (0, 1]×[0, ∞) → [0, ∞)
with lim f (t, ·) = +∞ (that is f is singular at t = 0). We obtain two results
t→0+
about this boundary-value problem, by using Krasnoselskii’s fixed point theorem
and nonlinear alternative of Leray-Schauder type in a cone.
For existence theorems for fractional differential equation and applications, we
refer the reader to the survey by Kilbas and Trujillo [6]. Concerning the definitions
2000 Mathematics Subject Classification. 34B15.
Key words and phrases. Boundary value problem; positive solution;
singular fractional differential equation; fixed-point theorem.
c
2008
Texas State University - San Marcos.
Submitted April 4, 2008. Published October 24, 2008.
Supported by grants 10371006 from the National Nature Science Foundation of China,
and 10626033 from the Mathematics Tianyuan Foundation of China.
1
2
T. QIU, Z. BAI
EJDE-2008/146
of fractional integral and derivative and related basic properties, we refer the reader
to Samko, Kilbas, and Marichev [5] and Delbosco and Rodino [3].
2. Preliminaries
For the convenience of the reader, we present here the necessary definitions from
fractional calculus theory. These definitions and properties can be found in the
literature.
Definition 2.1. The Riemann-Liouville fractional integral of order α > 0 of a
function f : (0, ∞) → R is given by
Z t
1
α
(t − s)α−1 f (s)ds,
I0+
f (t) =
Γ(α) 0
provided that the right-hand side is pointwise defined on (0, ∞).
Definition 2.2. The Caputo fractional derivative of order α > 0 of a continuous
function f : (0, ∞) → R is given by
Z t
1
f (n) (s)
α
D0+ f (t) =
ds,
Γ(n − α) 0 (t − s)α−n+1
where n − 1 < α ≤ n, provided that the right-hand side is pointwise defined on
(0, ∞).
Lemma 2.3 ([10]). Let n − 1 < α ≤ n, u ∈ C n [0, 1]. Then
α
α
I0+
D0+
u(t) = u(t) − C1 − C2 t − · · · − Cn tn−1 ,
where Ci ∈ R, i = 1, 2, . . . n.
Lemma 2.4 ([10]). The relation
α+β
α β
Ia+
Ia+ ϕ = Ia+
ϕ
is valid in following case
Reβ > 0, Re(α + β) > 0, ϕ(x) ∈ L1 (a, b).
Lemma 2.5. Given f ∈ C[0, 1], and 2 < α ≤ 3, the unique solution of
0 < t < 1,
D0α+ u(t) + f (t) = 0,
u(0) = u0 (1) = u00 (0) = 0.
is
Z
u(t) =
(2.1)
1
G(t, s)f (s)ds
0
where
(
G(t, s) =
(α−1)t(1−s)α−2 −(t−s)α−1
,
Γ(α)
t(1−s)α−2
Γ(α−1) ,
0 ≤ s ≤ t ≤ 1,
0 ≤ t ≤ s ≤ 1.
(2.2)
Proof. We may apply Lemma2.3 to reduce Eq.(2.1) to an equivalent integral equation
α
u(t) = −I0+
f (t) + C1 + C2 t + C3 t2
for some Ci ∈ R, i = 1, 2, 3. By Lemma2.4 we have
α−1
1
α
1
1
u0 (t) = −D0+
I0+
f (t) + C2 + 2C3 t = −D0+
I0+
I0+
f (t) + C2 + 2C3 t
α−1
= −I0+
f (t) + C2 + 2C3 t
EJDE-2008/146
EXISTENCE OF POSITIVE SOLUTIONS
3
α−1
α−2
α−2
1
1
1
u00 (t) = −D0+
I0+
f (t) + 2C3 = −D0+
I0+
I0+
f (t) + 2C3 = −I0+
f (t) + 2C3 .
From u(0) = u0 (1) = u00 (0) = 0, one has
Z 1
1
C1 = 0, C2 =
(1 − s)α−2 f (s)ds, C3 = 0.
Γ(α − 1) 0
Therefore, the unique solution of problem (2.1) is
Z t
Z 1
1
1
(t − s)α−1 f (s)ds +
t(1 − s)α−2 f (s)ds
u(t) = −
Γ(α) 0
Γ(α − 1) 0
Z t
Z 1
t(1 − s)α−2
t(1 − s)α−2
(t − s)α−1
=
−
f (s)ds +
f (s)ds
Γ(α − 1)
Γ(α)
Γ(α − 1)
0
t
Z 1
=
G(t, s)f (s)ds
0
For G(t, s), since 2 < α ≤ 3, 0 ≤ s ≤ t ≤ 1 we can obtain
(α − 1)t(1 − s)α−2 ≥ t(1 − s)α−2 ≥ t(t − s)α−2 ≥ (t − s)α−1
obviously,we get G(t, s) > 0. The proof is complete.
Lemma 2.6 ([7]). Let E be a Banach space, P ⊆ E a cone, and Ω1 , Ω2 are
two bounded open balls of E centered at the origin with Ω1 ⊂ Ω2 . Suppose that
A : P ∩ (Ω2 \ Ω1 ) → P is a completely continuous operator such that either
(i) kAxk ≤ kxk, x ∈ P ∩ ∂Ω1 and kAxk ≥ kxk, x ∈ P ∩ ∂Ω2 , or
(ii) kAxk ≥ kxk, x ∈ P ∩ ∂Ω1 and kAxk ≤ kxk, x ∈ P ∩ ∂Ω2
holds. Then A has a fixed point in P ∩ (Ω2 \ Ω1 ).
Lemma 2.7 ([4]). Let E be a Banach space with C ⊆ E closed and convex. Assume
U is a relatively open subset of C with 0 ∈ U and A : U → C is a continuous
compact map. Then either
(1) A has a fixed point in U ; or
(2) there exists u ∈ ∂U and λ ∈ (0, 1) with u = λAu.
3. Main Results
For our construction, we let E = C[0, 1] and kuk = max |u(t)| which is a Banach
0≤t≤1
space. We seek solutions of (1.1) that lie in the cone
P = {u ∈ E : u(t) ≥ 0, 0 ≤ t ≤ 1} .
Define operator T : P → P , by
Z
T u(t) =
1
G(t, s)f (s, u(s))ds .
0
Lemma 3.1. Let 0 < σ < 1, 2 < α ≤ 3, F : (0, 1] → R is continuous and
lim+ F (t) = ∞. Suppose that tσ F (t) is continuous function on [0, 1]. Then the
t→0
function
Z
H(t) =
G(t, s)F (s)ds
0
is continuous on [0, 1].
t
4
T. QIU, Z. BAI
EJDE-2008/146
Rt
Proof. By the continuity of tσ F (t) and H(t) = 0 G(t, s)s−σ sσ F (s)ds It is easily
to check that H(0) = 0. The proof is divided into three cases:
σ
Case 1: t0 = 0, ∀t ∈ (0, 1].
σ Since
t F (t) is continuous in [0, 1], there exists a
constant M > 0, such that t F (t) ≤ M , for t ∈ [0, 1]. Hence
Z
t (α − 1)t(1 − s)α−2 − (t − s)α−1 −σ σ
H(t) − H(0) = s s F (s)ds
Γ(α)
0
Z 1
t(1 − s)α−2 −σ σ
+
s s F (s)ds
Γ(α
−
1)
t
Z t
Z 1 t(1 − s)α−2
(t − s)α−1 −σ σ
−σ σ
=
s s F (s)ds −
s s F (s)ds
Γ(α
−
1)
Γ(α)
0
0
Z 1 t(1 − s)α−2
Z t (t − s)α−1
≤
s−σ sσ F (s)ds + s−σ sσ F (s)ds
Γ(α − 1)
Γ(α)
0
0
Z 1
Z t
t(1 − s)α−2 −σ
(t − s)α−1 −σ
≤M
s ds + M
s ds
Γ(α − 1)
Γ(α)
0
0
M α−σ
Mt
B(1 − σ, α − 1) +
t
B(1 − σ, α)
=
Γ(α − 1)
Γ(α)
Γ(1 − σ)M t Γ(1 − σ)M tα−σ
=
+
→ 0 (as t → 0)
Γ(α − σ)
Γ(1 + α − σ)
where B denotes the beta function.
Case 2: t0 ∈ (0, 1), for all t ∈ (t0 , 1]
H(t) − H(t0 )
Z t (α − 1)t(1 − s)α−2 − (t − s)α−1
=
s−σ sσ F (s)ds
Γ(α)
0
Z 1
Z 1
t(1 − s)α−2 −σ σ
t0 (1 − s)α−2 −σ σ
+
s s F (s)ds −
s s F (s)ds
Γ(α − 1)
Γ(α − 1)
t
t0
Z t0
(α − 1)t0 (1 − s)α−2 − (t0 − s)α−1 −σ σ
−
s s F (s)ds
Γ(α)
0
Z t
Z 1 t(1 − s)α−2
(t − s)α−1 −σ σ
=
s−σ sσ F (s)ds −
s s F (s)ds
Γ(α − 1)
Γ(α)
0
0
Z 1
Z t0
t0 (1 − s)α−2 −σ σ
(t0 − s)α−1 −σ σ
−
s s F (s)ds +
s s F (s)ds
Γ(α
−
1)
Γ(α)
0
0
Z 1 (t − t )(1 − s)α−2
0
−σ σ
=
s s F (s)ds
Γ(α − 1)
0
Z t0
Z t
(t − s)α−1 − (t0 − s)α−1 −σ σ
(t − s)α−1 −σ σ
−
s s F (s)ds −
s s F (s)ds
Γ(α)
Γ(α)
0
t0
Z
Z t0 h
i
M (t − t0 ) 1
M
≤
(1 − s)α−2 s−σ ds +
(t − s)α−1 − (t0 − s)α−1 s−σ ds
Γ(α − 1) 0
Γ(α) 0
Z t
M
−
(t − s)α−1 s−σ ds
Γ(α) t0
≤
M (t − t0 )
M tα−σ
M tα−σ
0
B(1 − σ, α − 1) +
B(1 − σ, α) −
B(1 − σ, α)
Γ(α − 1)
Γ(α)
Γ(α)
EJDE-2008/146
=
EXISTENCE OF POSITIVE SOLUTIONS
Γ(1 − σ)M tα−σ
Γ(1 − σ)M (t − t0 ) Γ(1 − σ)M tα−σ
0
+
−
→0
Γ(α − σ)
Γ(1 + α − σ)
Γ(1 + α − σ)
5
(as t → t0 ).
Case 3: t0 ∈ (0, 1], for all t ∈ [0, t0 ). The proof is similar to that of Case 2; we
omitted it.
Lemma 3.2. Let 0 < σ < 1, 2 < α ≤ 3, f : (0, 1] × [0, +∞) → [0, +∞) is
continuous and lim+ f (t, ·) = +∞, tσ f (t, u(t)) is continuous function on [0, 1] ×
t→0
[0, +∞), then the operator T : P → P is completely continuous.
R1
Proof. For each u ∈ P , let T u(t) = 0 G(t, s)f (s, u(s))ds. By Lemma3.1 and the
fact that f, G(t, s) are non-negative, we have T : P → P .
Let u0 ∈ P and ku0 k = C0 , if u ∈ P and ku − u0 k < 1, then kuk < 1 + C0 = C.
By the continuity of tσ f (t, u(t)), we know that tσ f (t, u(t)) is uniformly continuous
on [0, 1] × [0, C].
Thus for all > 0, there exists δ > 0(δ < 1), such that |tσ f (t, u2 )−tσ f (t, u1 )| < ,
for all t ∈ [0, 1], and u1 , u2 ∈ [0, C] with |u2 − u1 | < δ. Obviously, if ku − u0 k < δ,
then u(t), u0 (t) ∈ [0, C] and ku(t) − u0 (t)k < δ, for all t ∈ [0, 1]. Hence,
|tσ f (t, u(t)) − tσ f (t, u0 (t))| < ,
for all t ∈ [0, 1].
(3.1)
u ∈ P , with ku − u0 k < δ. It follows from (3.1) that
kT u − T u0 k = max T u(t) − T u0 (t)
0≤t≤1
Z
≤ max
0≤t≤1
1
G(t, s)s−σ sσ f (s, u(s)) − sσ f (s, u0 (s))ds
0
1
Z
G(t, s)s−σ ds
<
0
1
(α − 1)(1 − s)α−2 − (1 − s)α−1 −σ
s ds
Γ(α)
0
Z 1
≤
(1 − s)α−2 s−σ ds
Γ(α − 1) 0
Γ(1 − σ)
B(1 − σ, α − 1) =
.
=
Γ(α − 1)
Γ(α − σ)
By the arbitrariness of u0 , T : P → P is continuous. Let M ⊂ P be bounded; i.e.,
there exists a positive constant b such that kuk ≤ b, for all u ∈ p.
Since tσ f (t, u) is continuous in [0, 1] × [0, +∞), let
Z
=
L=
Then
Z
T u(t) ≤
0
thus
max
0≤t≤1,u∈M
tσ f (t, u) + 1,
1
−σ
G(t, s)s
σ
Z
|s f (s, u(s))|ds ≤ L
0
∀u ∈ M.
1
G(1, s)s−σ ds =
Γ(1 − σ)L
;
Γ(α − σ)
Γ(1 − σ)L
kT uk = max T u(t) ≤
.
0≤t≤1
Γ(α − σ)
So, T (M ) is equicontinuous. For > 0 set
n
o
Γ(α − σ)
δ = min Γ(1−α)L
,
, Γ(1−α)L Γ(1−α)L2α .
Γ(1−α)L
2LΓ(1 − σ)
Γ(α−σ) + Γ(1+α−σ)
Γ(α−σ) + Γ(1+α−σ)
6
T. QIU, Z. BAI
EJDE-2008/146
For u ∈ M , t1 , t2 ∈ [0, 1], with t1 < t2 , for 0 < t2 − t1 < δ, we have
T u(t2 ) − T u(t1 )
Z 1
Z 1
G(t1 , s)f (s, u(s))ds
G(t2 , s)f (s, u(s))ds −
=
0
0
Z 1
G(t2 , s) − G(t1 , s) s−σ sσ f (s, u(s))ds
=
0
Z 1
≤L
G(t2 , s) − G(t1 , s)s−σ ds
0
Z 1
Z t2 (α − 1)t (1 − s)α−2 − (t − s)α−1
t2 (1 − s)α−2 −σ
2
2
≤ L
s−σ ds +
s ds
Γ(α)
Γ(α − 1)
0
t2
Z 1
Z t1
t1 (1 − s)α−2 −σ (α − 1)t1 (1 − s)α−2 − (t1 − s)α−1 −σ
s ds −
s ds
−
Γ(α)
Γ(α − 1)
t1
0
Z 1
Z t2
h
(1 − s)α−2 −σ
(t2 − s)α−1 −σ
≤ L (t2 − t1 )
s ds +
s ds
Γ(α − 1)
Γ(α)
0
0
Z t1
(t1 − s)α−1 −σ i
s ds
−
Γ(α)
0
Z
Z t2
(t2 − t1 ) 1 −σ
L
=L
s−σ (t2 − s)α−1 ds
s (1 − s)α−2 ds +
Γ(α − 1) 0
Γ(α) 0
Z t1
L
−
s−σ (t1 − s)α−1 ds
Γ(α) 0
LΓ(1 − σ) α−σ
(t2 − t1 )Γ(1 − σ)
≤L
+
(t
− t1α−σ ).
Γ(α − σ)
Γ(1 + α − σ) 2
Case 1: t1 = 0, t2 < δ.
T u(t2 ) − T u(t1 ) = L t2 Γ(1 − σ) + LΓ(1 − σ) tα−σ
Γ(α − σ)
Γ(1 + α − σ) 2
δΓ(1 − σ)
LΓ(1 − σ)
<L
+
δ < .
Γ(α − σ)
Γ(1 + α − σ)
Case 2: δ ≤ t1 < t2 < 1.
T u(t2 ) − T u(t1 ) < L δΓ(1 − σ) + LΓ(1 − σ) δ α−σ
Γ(α − σ)
Γ(1 + α − σ)
LδΓ(1 − σ) + LΓ(1 − σ)δ α−σ
=
Γ(α − σ)
2LδΓ(1 − σ)
<
< .
Γ(α − σ)
Case 3: 0 < t1 < δ, t2 < 2δ.
T u(t2 ) − T u(t1 ) < L δΓ(1 − σ) + LΓ(1 − σ) 2α δ < .
Γ(α − σ)
Γ(1 + α − σ)
Therefore, T (M ) is equicontinuous. The Arzela-Ascoli theorem implies that T (M )
is compact. Thus, the operator T : P → P is completely continuous.
EJDE-2008/146
EXISTENCE OF POSITIVE SOLUTIONS
7
Theorem 3.3. Let 0 < σ < 1, 2 < α ≤ 3, f : (0, 1] × [0, +∞) → [0, +∞)
is continuous and lim+ f (t, ·) = +∞, tσ f (t, y) is continuous function on [0, 1] ×
t→0
[0, +∞). Assume that there exist two distinct positive constant ρ, µ(ρ > µ) such
that
(H1) tσ f (t, ω) ≤ ρ Γ(α−σ)
Γ(1−σ) , for (t, ω) ∈ [0, 1] × [0, ρ];
(H2) tσ f (t, ω) ≥ µ Γ(α−σ)
Γ(1−σ) , for (t, ω) ∈ [0, 1] × [0, µ].
Then (1.1) has at least one positive solution.
Proof. From Lemma 3.2 we have T : P → P is completely continuous. We divide
the proof into the following two steps.
Step1: Let Ω1 = {u ∈ P : kuk < α−σ−1
α−σ µ}, for u ∈ K ∩ ∂Ω1 and all t ∈ [0, 1], we
µ.
It
follows
from (H2) that
have 0 ≤ u(t) ≤ α−σ−1
α−σ
Z 1
Z 1
T u(1) =
G(1, s)f (s, u(s))ds =
G(1, s)s−σ sσ f (s, u(s))ds
0
≥µ
0
Γ(α − σ)
Γ(1 − σ)
Z
1
G(1, s)s−σ ds
0
Z
Γ(α − σ) h 1 (α − 1)(1 − s)α−2 − (1 − s)α−1 −σ i
=µ
s ds
Γ(1 − σ) 0
Γ(α)
Z
Z 1
Γ(α − σ) h 1 (1 − s)α−1 −σ
(1 − s)α−1 −σ i
=µ
s ds −
s ds
Γ(1 − σ) 0 Γ(α − 1)
Γ(α)
0
Γ(α − σ) B(1 − σ, α − 1) B(1 − σ, α − 1) =µ
−
Γ(1 − σ)
Γ(α − 1)
Γ(α)
α−σ−1
≥
µ = kuk .
α−σ
Hence,
kT uk = max |T u(t)| ≥
0≤t≤1
α−σ−1
µ = kuk,
α−σ
for u ∈ P ∩ ∂Ω1 .
Step 2: Let Ω2 = {u ∈ P : kuk < ρ}, for u ∈ K ∩ ∂Ω2 and all t ∈ [0, 1], we have
0 ≤ u(t) ≤ ρ. By assumption (H1),
Z 1
T u(t) =
G(t, s)f (s, u(s))ds
0
Z
1
G(t, s)s−σ sσ f (s, u(s))ds
Z
Γ(α − σ) h t (α − 1)t(1 − s)α−2 − (t − s)α−1 −σ
≤ρ
s ds
Γ(1 − σ) 0
Γ(α)
Z 1
t(1 − s)α−2 −σ i
+
s ds
Γ(α − 1)
t
Z
Z t
Γ(α − σ) h 1 t(1 − s)α−2 −σ
(t − s)α−1 −σ i
≤ρ
s ds −
s ds
Γ(1 − σ) 0 Γ(α − 1)
Γ(α)
0
Z 1
i
Γ(α − σ) h
t
≤ρ
s−σ (1 − s)α−2 ds
Γ(1 − σ) Γ(α − 1) 0
=
0
8
T. QIU, Z. BAI
EJDE-2008/146
Γ(α − σ) B(1 − σ, α − 1)
Γ(1 − σ)
Γ(α − 1)
Γ(α − σ) Γ(1 − σ)
=ρ
= ρ.
Γ(1 − σ) Γ(α − σ)
≤ρ
So kT u(t)k ≤ kuk, for u ∈ P ∩ ∂Ω2 . Therefore, by (ii) of Lemma 2.6, we complete
the proof.
Theorem 3.4. Let 0 < σ < 1, 2 < α ≤ 3, f : (0, 1] × [0, +∞) → [0, +∞)
is continuous and lim+ f (t, ·) = +∞, tσ f (t, y) is continuous function on [0, 1] ×
t→0
[0, +∞). Suppose the following conditions are satisfied:
(H3) there exists a continuous, nondecreasing function ϕ : [0, +∞) → (0, ∞)
with tσ f (t, ω) ≤ ϕ(ω), for (t, ω) ∈ [0, 1] × [0, +∞)
r
(H4) there exists r > 0, with ϕ(r)
> Γ(α−σ)
Γ(1−σ)
Then (1.1) has one positive solution.
Proof. Let U = {u ∈ P : kuk < r}, we have U ⊂ P . From Lemma 3.2, we know
T : U → P is completely continuous. If there exists u ∈ ∂U , λ ∈ (0, 1) such that
u = λT u,
(3.2)
By (H3) and (3.2), for t ∈ [0, 1] we have
Z
1
Z
0
Z
≤
1
G(t, s)f (s, u(s))ds ≤
u(t) = λT u(t) = λ
G(t, s)s−σ sσ f (s, u(s))ds
0
1
G(t, s)s−σ ϕ(u(s))ds
0
Z
1
G(t, s)s−σ ds
≤ ϕ(kuk)
0
Z
1
G(1, s)s−σ ds
≤ ϕ(kuk)
0
1
(α − 1)t(1 − s)α−2 − (t − s)α−1 −σ
s ds
Γ(α)
0
tB(1 − σ, α − 1)
≤ ϕ(kuk)
Γ(α − 1)
Γ(1 − σ)
≤ ϕ(kuk)
.
Γ(α − σ)
Z
= ϕ(kuk)
Γ(1−σ)
Consequently, kuk ≤ ϕ(kuk) Γ(α−σ)
; namely,
kuk
Γ(1 − σ)
≤
.
ϕ(kuk)
Γ(α − σ)
Combining (H4) and the above inequality, we have kuk =
6 r, which is contradiction
with u ∈ ∂U . According to Lemma 2.7, T has a fixed point u ∈ U , therefore, (1.1)
has a positive solution.
EJDE-2008/146
EXISTENCE OF POSITIVE SOLUTIONS
9
As an example, consider the fractional differential equation
(t − 12 )2 ln(2 + u)
= 0, 0 < t < 1
tσ
0
00
u(0) = u (1) = u (0) = 0,
D0α+ u(t) +
(3.3)
where 0 < σ < 1, 2 < α ≤ 3. Then (3.3) has a positive solution.
References
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[7] M. A. Krasnoselskii; Positive Solution of Operator Equation, Noordhoff,Groningen, 1964.
[8] Y. Ling, S. Ding; A Class of Analytic Function Defined by Fractive Derivative, J. Math.
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Tingting Qiu
College of Information Science and Engineering, Shandong University of Science and
Technology, Qingdao, 266510, China
E-mail address: qiutingting19833@163.com
Zhanbing Bai
College of Information Science and Engineering, Shandong University of Science and
Technology, Qingdao, 266510, China
E-mail address: zhanbingbai@163.com