Existence and Uniqueness of Mittag-Leffler

EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 8, No. 4, 2015, 478-498
ISSN 1307-5543 – www.ejpam.com
Existence and Uniqueness of Mittag-Leffler-Ulam Stable
Solution for Fractional Integrodifferential Equations with
Nonlocal Initial Conditions
Mohamed I. Abbas
Department of Mathematics and Computer Science, Faculty of Science, Alexandria University, Alexandria 21511, Egypt
Abstract. In this paper, the existence and uniqueness of mild solution for fractional integrodifferential
equations with nonlocal initial conditions are investigated by using Hölder’s inequality, p−mean continuity and Schauder’s fixed point theorem in Banach spaces. The Mittag-Leffler-Ulam stability results
are also obtained by using generalized singular Gronwall’s inequality.
2010 Mathematics Subject Classifications: 26A33, 34A08,34D20, 45N05.
Key Words and Phrases: Mild solutions, Hölder’s inequality, Schauder’s fixed point theorem, The
Mittag-Leffler-Ulam stability, Generalized singular Gronwall’s inequality.
1. Introduction
During the past decades, fractional differential equations have attracted many authors
(see for instance [10, 11, 14] and [15]). This is mostly because they efficiently describe many
phenomena arising in engineering, physics, economy and science. There has been a significant
development in nonlocal problems for fractional differential equations or inclusions (see for
instance [1, 7, 12] and [22]).
As we all know, the main difficulty to study the fractional evolution equations is how to
obtain the suitable fractional resolvent family generated by the infinitesimal generator A in
Banach space. In order to solve this problem, some authors introduced an α-resolvent family
under the Riemann-Liouville fractional derivative and some constraints, (see, for example, [2,
6]), and the others introduced suitable operator families with the Caputo fractional derivative
in terms of some probability density functions and operator semigroup (see, for example, [17,
22] and [23]). For the latter, a pioneering work has been reported by El-Borai [4, 5].
On the other hand, in the theory of functional equations there are some special kind of data
dependence: Ulam-Hyers, Ulam-Hyers-Rassias, Ulam-Hyers-Bourgin and Aoki-Rassias (see [3,
Email address: [email protected]
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M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
479
8] and [9]). Recently, J. Wang et al. [18, 19] discussed four type Mittag- leffler-Ulam stability
of fractional differential equations and obtained some new and interesting stability results.
Motivated by the above work, we consider the nonlocal Cauchy problem of the following form
¨
C α
D x(t) = Ax(t) + f (t, x(t), I β x(t)), t ∈ J := [0, b]
(1)
x(0) + g(x) = x 0 ,
where C Dα is the Caputo fractional derivative of order 0 < α < 1, I β is the Riemann-Liouville
fractional integration of order 0 < β < 1, b > 0, A is the infinitesimal generator of a C0
semigroup {Q(t)} t≥0 of operators on E, f : J × E × E → E, g : C(J, E) → E are given functions
satisfying some assumptions and x 0 is an element of the Banach space E.
For any strongly continuous semigroup (i.e. C0 semigroup) {Q(t)} t≥0 on E, we define the
generator
Q(t)u − u
Au = lim+
in E.
t→0
t
The domain D(A) of this linear operator is the set of all u ∈ E for which the limit above exists.
Then D(A) is dense in E and A is closed, meaning that for un ∈ D(A), if un → u and Aun → v in
E, then u ∈ D(A) and Au = v. For more details, we refer the reader to [13].
2. Preliminaries
In this section, we introduce preliminary facts which are used throughout this paper. We
assume that E is a Banach space with the norm | · |. Let C(J, E) be the Banach space of
continuous functions from J into E with the norm kxk = sup t∈J |x(t)|, where x ∈ C(J, E).
Let B(E) be the space of all bounded linear operators from E to E with the norm
kQkB(E) = sup{|Q(u)| : |u| = 1}, where Q(u) ∈ B(E) and u ∈ E. Throughout this paper, let A
be the infinitesimal generator of a C0 semigroup {Q(t)} t≥0 of operators on E. Clearly
M := sup kQkB(E) < ∞.
(2)
t∈[0,b]
We need some basic definitions and properties of the fractional calculus theory which are used
further in this paper. For more details, see I. Podlubny [15].
Definition 1. The fractional integral of order α with the lower limit zero for a function
h ∈ AC[0, ∞) is defined as
Z t
h(s)
1
α
ds, t > 0, 0 < α < 1,
I h(t) =
Γ(α) 0 (t − s)1−α
provided the right side is point-wise defined on [0, ∞), where Γ(·) is the gamma function.
Definition 2. Riemann-Liouville derivative of order α with the lower limit zero for a function
h ∈ AC[0, ∞) can be written as
Z t
h(s)
1
d
L α
D h(t) =
ds, t > 0, 0 < α < 1.
Γ(1 − α) d t 0 (t − s)α
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
480
Definition 3. The Caputo derivative of order α with the lower limit zero for a function
h ∈ AC[0, ∞) can be written as
C
Remark 1.
Dα h(t) = L Dα (h(t) − h(0)), t > 0, 0 < α < 1.
• If h(t) ∈ C 1 [0, ∞), then
Z t
′
′
h (s)
1
C α
D h(t) =
ds = I 1−α h (t), t > 0, 0 < α < 1.
α
Γ(1 − α) 0 (t − s)
• The Caputo derivative of a constant is equal to zero.
• If h is an abstract function with values in E, then integrals which appear in Definitions 1-3
are taken in Bochner’s sense.
For measurable functions m : J → R, define the norm
(R
1
( J |m(t)| p d t) p , 1 ≤ p < ∞,
kmk L p (J) =
infµ(J̄)=0 {sup t∈J−J̄ |m(t)|}, p = ∞,
where µ(J̄) is the Lebesgue measure on J̄. Let L p (J, R) be the Banach space of all Lebesgue
measurable functions m : J → R with kmk L p (J) < ∞.
Lemma 1 (Hölder’s inequality). Assume that q, p ≥ 1 and 1q + 1p = 1. If l ∈ L q (J, R) and
m ∈ L p (J, R), then for 1 ≤ p ≤ ∞, lm ∈ L 1 (J, R) and klmk L 1 (J) ≤ klk L q (J) .kmk L p (J) .
Lemma 2 ([21] p-mean continuity). For each ψ ∈ L p (J, E) with 1 ≤ p < +∞, we have
Rb
lim r→0 0 kψ(t + r) − ψ(t)k p d t = 0, where ψ(s) = 0 for s not in J.
Lemma 3 (Bochner’s Theorem). A measurable function H : [0, b] → R is Bochner’s integrable
if |H| is Lebesgue integrable.
Lemma 4 (Schauder Fixed Point Theorem). If B is a closed bounded and convex subset of a
Banach space E and F : B → B is completely continuous, then F has a fixed point in B.
We end this section with an important singular type Gronwall inequality.
Theorem 1 ([16] Theorem 1.4). For any t ∈ [0, b), if
Z t
n
X
u(t) ≤ a(t) +
(t − s)βi −1 u(s)ds,
bi (t)
i=1
0
where all the functions are not negative and continuous. The constants βi > 0.
bi (i = 1, 2, . . . , n) are the bounded and monotonic increasing functions on [0, b), then


Qk
Z t
n
∞
X
X
Pk
′
′
i=1 [bi (t)Γ(βi )]

u(t) ≤ a(t) +
(t − s) i=1 βi −1 a(s)ds .
Pk
Γ( i=1 βi ′ )
0
k=1 1′ ,2′ ,··· ,k′ =1
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
481
Remark 2. For n = 2, if the constants b1 , b2 ≥ 0, β1 , β2 > 0, a(t) is nonnegative and locally
integrable on 0 ≤ t < b and u(t) is nonnegative and locally integrable on 0 ≤ t < b with
t
Z
u(t) ≤ a(t) + b1
t
Z
(t − s)
β1 −1
(t − s)β2 −1 u(s)ds,
u(s)ds + b2
0
0
then

Z t
Z t
∞
X
(b1 Γ(β1 ))k
(b2 Γ(β2 ))k
kβ1 −1
kβ2 −1
(t − s)
a(s)ds +
(t − s)
a(s)ds .
u(t) ≤ a(t) +
Γ(kβ
)
Γ(kβ
)
1
2
0
0
k=1
Remark 3. Under the hypotheses of Remark 2, let a(t) is a nondecreasing function on 0 ≤ t < b.
Then we have
u(t) ≤ a(t) Eβ1 b1 Γ(β1 )t β1 + Eβ2 b2 Γ(β2 )t β2 ,
P∞
zk
where Eα is the Mittag-Leffler function [15] defined by Eα [z] = k=0 Γ(kα+1)
, z ∈ C.
3. Existence and Uniqueness of Mild Solutions
We recall the following definition of a mild solution for the nonlocal Cauchy problem (1).
For more details, one can see [20, 24].
Definition 4. By the mild solution of the nonlocal Cauchy problem (1), we mean that the function
x ∈ C(J, E) which satisfies
t
Z
(t − s)α−1 T (t − s) f (s, x(s), I β x(s))ds, t ∈ [0, b],
x(t) = S(t)(x 0 − g(x)) +
0
where
Z
∞
ξα (θ )Q(t α θ )dθ ,
S(t) =
0
Z
∞
θ ξα (θ )Q(t α θ )dθ ,
T (t) =α
0
1
1
1
ξα (θ ) = θ −1− α ρα (θ − α ),
α
∞
1X
Γ(nα + 1)
ρα (θ ) =
sin(nπα), θ ∈ (0, ∞),
(−1)n−1 θ −αn−1
π n=1
n!
where ξα is the probability density function defined on (0, ∞), which has properties ξα (θ ) ≥ 0
for all θ ∈ (0, ∞) and
Z∞
Z∞
1
.
(3)
ξα (θ )dθ = 1,
θ ξα (θ )dθ =
Γ(1 + α)
0
0
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
482
Before stating and proving the main results, we introduce the following hypotheses.
(H1) Q(t) is a compact operator for every t > 0.
(H2) The function f : J × E × E → E satisfies that f (., x, y) : J → E is measurable for all
x, y ∈ C(J, E) and f (t, ., .) : E × E → E is continuous for all t ∈ J and there exists a
positive function µ(·) ∈ L p (J, R+ ) for some p with 1 < p < ∞ such that
| f (t, x, y)| ≤ µ(t)(kxk + k yk), x, y ∈ C(J, E), t ∈ J.
(H3) The function g : C(J, E) → E is completely continuous and there exist constants L > 0,
L ′ > 0 such that:
′
|g(x)| ≤ Lkxk + L , x ∈ C(J, E).
For each positive constant k, let Bk = {x ∈ C(J, E) : kxk ≤ k}. Then Bk is clearly a bounded
closed and convex subset in C(J, E).
The following existence result for nonlocal Cauchy problem (1) is based on Schauder fixed
point theorem.
Theorem 2. If assumptions (H1) − (H3) are satisfied, then the nonlocal Cauchy problem (1) has
a mild solution provided that



p−1
p−1
β+ 2
α− 1p
p
p
p−1
M kb
kµk L p (J,R+ ) + b
M L −
kµk L p2 (J,R+ )  < 1.
Γ(1 + α) p + (α − 1)p − 1
Γ(1 + β)
Proof. For any positive constant k and x ∈ Bk , according to (2) and (H3), it follows that
Z
∞
′
ξα (θ )Q(t α θ )(x 0 − g(x))dθ ≤ M (|x 0 | + Lk + L ).
(4)
0
R∞
Therefore, the function 0 ξα (θ )Q(t α θ )(x 0 − g(x))dθ exists.
In view of (2), (3) and (H2), we get
Z t Z ∞
θ (t − s)α−1 ξα (θ )Q((t − s)α θ ) f (s, x(s), I β x(s))dθ ds
0
0
Z tZ ∞
≤M
θ ξ (θ )(t − s)α−1 f (s, x(s), I β x(s)) dθ ds
α
0
0
M
≤
Γ(1 + α)
Z
M
≤
Γ(1 + α)
Z
t
(t − s)α−1 µ(s)(kxk + kI β xk)ds
0
t
(t − s)
0
α−1
k
µ(s)(k +
Γ(β)
Z
s
(s − τ)β−1 dτ)ds
0
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
t
sβ
ds
(t − s) µ(s) 1 +
βΓ(β)
0

Z t
Z t
Mk
1
=
(t − s)α−1 sβ µ(s)ds
(t − s)α−1 µ(s)ds +
Γ(1 + α)
Γ(1
+
β)
0
0

Z t
p−1
1p

Z
t
p
(α−1)p
Mk 
(t − s) p−1 ds
≤
(µ(s)) p ds
Γ(1 + α)
0
0

p−1
1p
Z t
Z t
p
(α−1)p
1
(sβ p µ(s)) p ds 
+
(t − s) p−1 ds
Γ(1 + β)
0
0

p−1
p
p+(α−1)p−1
p−1
Mk 
p
kµk L p (J,R+ )
t
≤
Γ(1 + α)
p + (α − 1)p − 1
Mk
=
Γ(1 + α)
+
Z
483
α−1
1
Γ(1 + β)
p−1
p + (α − 1)p − 1
p−1
p
t
p+(α−1)p−1
p
t
Z
β p2
p−1
Z
2
t
p
2
Mk
≤
Γ(1 + α)
×t
≤
β p2 +p−1
p2
α− 1p
M kb
Γ(1 + α)
p−1
p + (α − 1)p − 1
t

 p−1
2

p+(α−1)p−1
p

0

p−1
p
1
p2
(µ(s)) p ds
s p−1 ds
0

p

1
kµk p
 1 
L (J,R+ ) +

Γ(1 + β) 1 + β p2
p−1
kµk L p2 (J,R+ )
p−1
p + (α − 1)p − 1
p−1
p

β+
p−1
p2

kµk L p (J,R+ ) + b
kµk L p2 (J,R+ )  ,
Γ(1 + β)
(5)
R
∞
for all t ∈ J. Thus, 0 θ (t − s)α−1 ξα (θ )Q((t − s)α θ ) f (s, x(s), I β x(s))dθ is Lebesgue integrable with
respect to s ∈ [0, t] for all t ∈ [0, b]. From Lemma 3 (Bochner’s theorem), it
R∞
follows that 0 θ (t −s)α−1 ξα (θ )Q((t −s)α θ ) f (s, x(s), I β x(s))dθ is Bochner’s integrable with
respect to s ∈ [0, t] for all t ∈ J.
For each positive constant k, define an operator F on Bk by the formula
t
Z
(t − s)α−1 T (t − s) f (s, x(s), I β x(s))ds, t ∈ [0, b],
(F x)(t) = S(t)(x 0 − g(x)) + α
0
(6)
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
484
where x ∈ Bk . Let
k=

1−M L−
α− 1
M kb p
Γ(1+α)
M |x 0 | + L ′

p−1
p
p−1
kµk L p (J,R+ ) +
p+(α−1)p−1
β+
.
p−1
(7)
p2
b
2
Γ(1+β) kµk L p (J,R+ )
In the following, we will prove that F has a fixed point on Bk . Our proof will be divided into
two steps.
Step I. kF xk ≤ k whenever x ∈ Bk .
For each x ∈ Bk and t ∈ J, by using the similar method as we did in (4) and (5), we have
Z t
|(F x)(t)| = S(t)(x 0 − g(x)) + α
(t − s)α−1 T (t − s) f (s, x(s), I β x(s))ds
0
Z
∞
α
ξα (θ )Q(t θ )(x 0 − g(x))dθ ≤
0
Z Z
t ∞
θ (t − s)α−1 ξα (θ )Q((t − s)α θ ) f (s, x(s), I β x(s))dθ ds
+ α
0 0
′
≤M (|x 0 | + Lk + L )
α− 1p
+
M kb
Γ(1 + α)
p−1
p + (α − 1)p − 1
p−1
p

β+
p−1
p2
kµk L p (J,R+ ) + b
kµk L p2 (J,R+ ) 
Γ(1 + β)
=k.
(8)
Hence kF xk ≤ k for each x ∈ Bk .
Step II. F is a completely continuous operator.
Firstly, we will prove that F is continuous on Bk . For any x n , x ⊆ Bk , n = 1, 2, . . .
with limn→∞ kx n − xk = 0, we get
lim x n (t) = x(t), for t ∈ J.
n→∞
Thus by condition (H2), we have
lim f (t, x n (t), I β x n (t)) = f (t, x(t), I β x(t)),
n→∞
for t ∈ J.
So, we can conclude that
sup | f (s, x n (s), I β x n (s)) − f (t, x(s), I β x(s))| → 0, as n → ∞.
s∈[0,b]
On the other hand, for t ∈ J
|F x n (t) − F x(t)|

M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
485
Z t
α−1
β
β
= S(t)(g(x n ) − g(x)) + α
(t − s)
T (t − s) f (s, x n (s), I x n (s)) − f (s, x(s), I x(s)) ds
0
Z
∞
ξα (θ )Q(t α θ )(g(x n ) − g(x))dθ ≤
0
Z Z
t ∞
α−1
α
β
β
θ (t − s) ξα (θ )Q((t − s) θ ) f (s, x n (s), I x n (s)) − f (s, x(s), I x(s)) dθ ds
+ α
0 0
Z t
αM
≤M kg(x n ) − g(x)k +
(t − s)α−1 | f (s, x n (s), I β x n (s)) − f (s, x(s), I β x(s))|ds
Γ(α + 1) 0
M bα
sup | f (s, x n (s), I β x n (s)) − f (t, x(s), I β x(s))|,
≤M kg(x n ) − g(x)k +
Γ(α + 1) s∈[0,b]
which implies
kF x n − F xk ≤ M kg(x n ) − g(x)k +
M bα
sup | f (s, x n (s), I β x n (s)) − f (t, x(s), I β x(s))|.
Γ(α + 1) s∈[0,b]
Hence, by condition (H3) we get kF x n −F xk → 0, as n → ∞. This means that F is continuous.
Next, we will show that {F x, x ∈ Bk } is equicontinuous. For any x ∈ Bk and
0 ≤ t 1 ≤ t 2 ≤ b, we get
|(F x)(t 2 ) − (F x)(t 1 )|
= [S(t 2 ) − S(t 1 )](x 0 − g(x))
Z t2
(t 2 − s)α−1 T (t 2 − s) f (s, x(s), I β x(s))ds
+α
0
−α
(t 1 − s)α−1 T (t 1 − s) f (s, x(s), I β x(s))ds
0
Z
∞
α
α
ξα (θ )[Q(t 2 θ ) − Q(t 1 θ )](x 0 − g(x))dθ ≤
0
Z Z
t2 ∞
+ α
θ (t 2 − s)α−1 ξα (θ )Q((t 2 − s)α θ ) f (s, x(s), I β x(s))dθ ds
0 0
Z t1 Z ∞
α−1
α
β
−
θ (t 1 − s) ξα (θ )Q((t 1 − s) θ ) f (s, x(s), I x(s))dθ ds
0
0
Z
∞
ξα (θ )[Q(t 2α θ ) − Q(t 1α θ )](x 0 − g(x))dθ ≤
0
Z Z
t2 ∞
α−1
α
β
θ (t 2 − s) ξα (θ )Q((t 2 − s) θ ) f (s, x(s), I x(s))dθ ds
+ α
t
0
Z
t1
1
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
486
Z Z
t1 ∞
α−1
α−1
α
β
+ α
θ [(t 2 − s)
− (t 1 − s) ]ξα (θ )Q((t 2 − s) θ ) f (s, x(s), I x(s))dθ ds
0 0
Z Z
t1 ∞
θ (t 1 − s)α−1 ξα (θ )[Q((t 2 − s)α θ ) − Q((t 1 − s)α θ )] f (s, x(s), I β x(s))dθ ds
+ α
0 0
Z
∞
α
α
ξα (θ )[Q(t 2 θ ) − Q(t 1 θ )](x 0 − g(x))dθ + α(I1 + I2 + I3 ),
=
0
where
Z Z
t2 ∞
θ (t 2 − s)α−1 ξα (θ )Q((t 2 − s)α θ ) f (s, x(s), I β x(s))dθ ds ,
I1 =α t
0
Z 1 Z
t1 ∞
α−1
α−1
α
β
I2 =α θ [(t 2 − s)
− (t 1 − s) ]ξα (θ )Q((t 2 − s) θ ) f (s, x(s), I x(s))dθ ds ,
0 0
Z Z
t1 ∞
θ (t 1 − s)α−1 ξα (θ )[Q((t 2 − s)α θ ) − Q((t 1 − s)α θ )] f (s, x(s), I β x(s))dθ ds .
I3 =α 0 0
By using analogous argument performed in (4) and (5), we can conclude that
Z t2
αM k
1
I1 ≤
sβ )ds
(t 2 − s)α−1 µ(s)(1 +
Γ(α + 1) t
Γ(β + 1)
1
Z t
p−1
Z t
1p
p
2
2
(α−1)p
Mk 
p
≤
(t 2 − s) p−1 ds
(µ(s)) ds
Γ(1 + α)
t1
t1

Z t
p−1
Z t
1p
p
2
2
(α−1)p
1
(t − s) p−1 ds
(sβ p µ(s)) p ds 
+
Γ(1 + β)
t
t
1
1
p−1
p
p+(α−1)p−1
p−1
p
(t 2 − t 1 )
p + (α − 1)p − 1

Z t
Z t
12
p−1
2
2
2
p
p
2
βp
2
1

s p−1 ds
(µ(s)) p ds
× kµk L p (J,R+ ) +
Γ(1 + β)
t
t
Mk
≤
Γ(1 + α)

1
Mk
≤
Γ(1 + α)

1
p−1
p
1
p−1
(t 2 − t 1 )α− p
p + (α − 1)p − 1

 p−1
2
p

1
 1 
×
kµk L p (J,R+ ) + Γ(1 + β)
β p2
1+
p−1

β p2
p−1 +1
(t 2
β p2
p−1 +1
− t1
)
p−1
p2

kµk L p2 (J,R+ ) 
.
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
487
One can deduce that lim t 1 →t 2 I1 = 0. Also, note that
αM k
I2 ≤
Γ(α + 1)
Z
t1
1
sβ )ds
Γ(β + 1)
1p Z t
[(t 2 − s)α−1 − (t 1 − s)α−1 ]µ(s)(1 +
0
Z
αM k 
≤
Γ(α + 1)
t1
1
[(t 2 − s)α−1 − (t 1 − s)α−1 ] p ds
(µ(s))
p
p−1
p−1
p
ds
0
0
1p Z
t1
Z
t1
p
1
(sβ µ(s)) p−1 ds
[(t 2 − s)α−1 − (t 1 − s)α−1 ] p ds
Γ(β + 1)
0
0

Z t
1p
1
αM k 
p
≤
[(t 2 − s)α−1 − (t 1 − s)α−1 ] p ds kµk p−1
L
(J,R+ )
Γ(α + 1)
0
p−1
p
+
+
1
Γ(β + 1)
Z t
1
×
1p Z
t1
Z
t1
[(t 2 − s)α−1 − (t 1 − s)α−1 ] p ds
s
p2

2
(p−1)
2
p
ds
0
0
(µ(s)) p−1 ds
β p2
(p−1)2

p−1
2

p

0
2
 (p−1)
2

≤
(p−1)2
β+ 2
p

b
1
αM k 
kµk p


+

p−1
+
(J,R
)
L
Γ(α + 1)
Γ(β + 1) 1 + β p2 2
(p−1)
[(t 2 − s)α−1 − (t 1 − s)α−1 ] p ds
×
kµk
L
p2
p−1 (J,R+ )



1p
b
Z

p
.
0
Using Lagrange mean value theorem, one can obtain (t 2 − s)α−1 − (t 1 − s)α−1 → 0 as t 1 → t 2 ,
Rb
for s ∈ J. By Lemma 2, we can deduce that 0 [(t 2 − s)α−1 − (t 1 − s)α−1 ] p ds → 0 as t 1 → t 2 .
Thus we deduce that lim t 1 →t 2 I2 = 0.
For t 1 = 0, 0 < t 2 ≤ b, it is easy to see that I3 = 0. For t 1 > 0 and ǫ > 0 be enough small,
we have
Z t 1 −ǫ Z ∞
θ (t 1 − s)α−1 ξα (θ )kQ((t 2 − s)α θ ) − Q((t 1 − s)α θ )kB(E) | f (s, x(s), I β x(s))|dθ ds
I3 ≤α
0
0
Z
t1
∞
Z
θ (t 1 − s)α−1 ξα (θ )kQ((t 2 − s)α θ ) − Q((t 1 − s)α θ )kB(E) | f (s, x(s), I β x(s))|dθ ds
+α
t 1 −ǫ
≤
αk
Γ(α + 1)
Z
0
t 1 −ǫ
(t 1 − s)α−1 µ(s)(1 +
0
sβ
) sup kQ((t 2 − s)α θ ) − Q((t 1 − s)α θ )kB(E) ds
Γ(β + 1) s∈[0,t 1 −ǫ]
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
488
Z t1
2αM k
sβ
+
(t 1 − s)α−1 µ(s)(1 +
)ds
Γ(α + 1) t −ǫ
Γ(β + 1)
 1
Z t −ǫ
p−1
Z t −ǫ
1p
p
1
1
(α−1)p
αk

≤
(t 1 − s) p−1 ds
(µ(s)) p ds
Γ(α + 1)
0
0
+
×
1
Γ(1 + β)
sup
s∈[0,t 1 −ǫ]
t 1 −ǫ
Z
(t 1 − s)
(α−1)p
p−1
p−1
Z
p
(sβ p µ(s)) p ds
ds
0

kQ((t 2 − s)α θ ) − Q((t 1 − s)α θ )kB(E)
2αM k 
+
Γ(α + 1)
t1
Z
(t 1 − s)
(α−1)p
p−1
Z
p−1
p
(µ(s)) p ds
t 1 −ǫ

(t 1 − s)
(α−1)p
p−1
1p
t1
ds
t 1 −ǫ
t1
×

0

Z
1p
t 1 −ǫ
p−1
p
Z
1p
t1
(sβ p µ(s)) p ds
ds
t 1 −ǫ
+
1
Γ(1 + β)


t 1 −ǫ
p−1
p−1
p+(α−1)p−1
p
p
p+(α−1)p−1
p−1
1
αk
p−1
− ǫ p−1
t1
kµk L p (J,R+ ) +
≤
Γ(α + 1) p + (α − 1)p − 1
Γ(1 + β)

Z t −ǫ
Z t −ǫ
12
p−1
2
1
1
p
p
β p2
2
 sup kQ((t 2 − s)α θ ) − Q((t 1 − s)α θ )kB(E)
×
s p−1 ds
(µ(s)) p ds
0
2αM k
+
Γ(α + 1)
Z
s∈[0,t 1 −ǫ]
0
t1
p−1
p + (α − 1)p − 1

12
2
(µ(s)) p ds
×

p−1
p
ǫ
p+(α−1)p−1
p
1
kµk L p (J,R+ ) +
Γ(1 + β)
Z
t1
s
β p2
p−1
p−1
2
p
ds
t 1 −ǫ
p

t 1 −ǫ
p−1
p+(α−1)p−1
p−1
p
p
p+(α−1)p−1
p−1
1
p−1
p−1
t1
kµk L p (J,R+ ) +
−ǫ
p + (α − 1)p − 1
Γ(1 + β)

p−1
 2
p

β p2 +p−1
1
 sup kQ((t − s)α θ ) − Q((t − s)α θ )k
p2


2
kµk
t
−
ǫ
×
p
+
2
1
B(E)
1
2
L (J,R ) 
βp
s∈[0,t 1 −ǫ]
1 + p−1
αk
≤
Γ(α + 1)

2αM k
+
Γ(α + 1)
p−1
p + (α − 1)p − 1
p−1
p
ǫ
p+(α−1)p−1
p
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498


 p−1
2


×
kµk L p (J,R+ ) +
489
p
1
 1 
Γ(1 + β) 1 + β p2
ǫ
β p2 +p−1
p2

kµk L p2 (J,R+ ) 

p−1
Since (H1) implies the continuity of {Q(t)} t≥0 in t in the uniform operator topology, it is
easy to see that I3 tends to zero independently of x ∈ Bk as t 2 − t 1 → 0, ǫ → 0. Thus
(F x)(t 2 ) − (F x)(t 1 ) tends to zero independently of x ∈ Bk as t 2 − t 1 → 0, which means that
{F x, x ∈ Bk } is equicontinuous.
It remains to prove that for t ∈ [0, b], the set V (t) = {(F x)(t), x ∈ Bk } is relatively compact
in E. Obviously, V (0) is relatively compact in E. Let 0 < t ≤ b be fixed. For ∀ǫ ∈ (0, t) and
∀δ > 0, define an operator Fǫ,δ on Bk by the formula
∞
Z
ξα (θ )Q(t α θ )(x 0 − g(x))dθ
(Fǫ,δ x)(t) =
δ
t−ǫ Z ∞
Z
θ (t − s)α−1 ξα (θ )Q((t − s)α θ ) f (s, x(s), I β x(s))dθ ds
+α
0
Z
δ
∞
ξα (θ )Q(t α θ − ǫ α δ)(x 0 − g(x))dθ
=Q(ǫ α δ)
δ
t−ǫ Z ∞
Z

θ (t − s)
+
0
α−1
α
α
β
ξα (θ )Q((t − s) θ − ǫ δ) f (s, x(s), I x(s))dθ ds ,
δ
where x ∈ Bk . Then from the compactness of Q(ǫ α δ)(ǫ α δ > 0), we obtain that the set Vǫ,δ (t) =
{(Fǫ,δ x)(t), x ∈ Bk } is relatively compact in E. Obviously, V (0) is relatively compact in E for
∀ǫ ∈ (0, t) and ∀δ > 0. Moreover, for every x ∈ Bk , we have
Z
δ
α
ξ (θ )Q(t θ )(x 0 − g(x))dθ |(F x)(t) − (Fǫ,δ x)(t)| ≤ 0 α
Z Z
t δ
+ α
θ (t − s)α−1 ξα (θ )Q((t − s)α θ ) f (s, x(s), I β x(s))dθ ds
0 0
Z Z
t ∞
+
θ (t − s)α−1 ξα (θ )Q((t − s)α θ ) f (s, x(s), I β x(s))dθ ds
0 δ
Z t−ǫ Z ∞
−
θ (t − s)α−1 ξα (θ )Q((t − s)α θ ) f (s, x(s), I β x(s))dθ ds
0
δ
Zδ
′
≤M (x 0 + Lk + L )
ξα (θ )dθ
0
Z tZ
δ
θ (t − s)α−1 ξα (θ )µ(s)(1 +
+ αM k
0
0
sβ
)dθ ds
Γ(β + 1)
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
Z
t
490
∞
Z
θ (t − s)α−1 ξα (θ )µ(s)(1 +
+ αM k
δ
t−ǫ
sβ
)dθ ds
Γ(β + 1)
δ
Z
′
≤M (x 0 + Lk + L )
ξα (θ )dθ
0
Z δ
β
s
(t − s)α−1 µ(s)(1 +
+ αM k
θ ξα (θ )dθ
)ds
Γ(β + 1)
0
0
Z ∞
Z t
β
s
)
θ ξα (θ )dθ ds
+ αM k
(t − s)α−1 µ(s)(1 +
Γ(β + 1)
0
t−ǫ
Zδ
Z
t
′
≤ M (x 0 + Lk + L )
ξα (θ )dθ
0
Z δ
sβ
(t − s) µ(s)(1 +
+ αM k
θ ξα (θ )dθ
)ds
Γ(β + 1)
0
0

Z t
sβ
α−1
) .
(t − s) µ(s)(1 +
+ αM k
Γ(β + 1)
t−ǫ
Z
t
α−1
Using again the same analogous performed in (5), we have
t
Z
(t − s)
0
α−1
p−1
p
p+(α−1)p−1
p−1
sβ
p
t
µ(s)(1 +
)ds ≤
Γ(β + 1)
p + (α − 1)p − 1

2
× kµk L p (J,R+ ) +
β p +p−1
p2

t
kµk L p2 (J,R+ )  ,
Γ(1 + β)
and
Z
p−1
β
p
p+(α−1)p−1
p
−
1
s
p
)ds ≤
ǫ
(t − s)α−1 µ(s)(1 +
Γ(β + 1)
p + (α − 1)p − 1
t−ǫ

2
t
× kµk L p (J,R+ ) +
β p +p−1
p2

ǫ
kµk L p2 (J,R+ )  ,
Γ(1 + β)
we obtain
Z
′
δ
|(F x)(t) − (Fǫ,δ x)(t)| ≤M (x 0 + Lk + L )
ξα (θ )dθ
0
+ αM k
p−1
p + (α − 1)p − 1

p−1
p
t
p+(α−1)p−1
p
β p2 +p−1
p2

kµk L p (J,R+ ) + t
kµk L p2 (J,R+ ) 
Γ(1 + β)
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
Z
δ
×
0
×ǫ
491
p−1
θ ξα (θ )dθ + αM k
p + (α − 1)p − 1

2
p+(α−1)p−1
p
β p +p−1
p2
p−1
p

kµk L p (J,R+ ) + ǫ
kµk L p2 (J,R+ )  .
Γ(1 + β)
Therefore, there are relatively compact sets {(Fǫ,δ x)(t), x ∈ Bk } arbitrarily close to the set
{(F x)(t), x ∈ Bk } for t ∈ (0, b]. Hence, {(F x)(t), x ∈ Bk } is relatively compact in E. Moreover,
{(F x)(t), x ∈ Bk } is uniformly bounded by (8). Therefore, {(F x)(t), x ∈ Bk } is relatively
compact by Ascoli-Arzèla Theorem.
Also, Since F is continuous on BK . Then F is a completely continuous operator. Obviously
F maps Bk into itself. Hence, Schauder fixed point theorem shows that F has a fixed point
x ∈ Bk , which means that the nonlocal Cauchy problem (1) has at least one mild solution on
J. The proof is complete.
The following existence and uniqueness result for the nonlocal Cauchy problem (1) is based
on Banach contraction principle. We will need the following assumption.
(H4) There exists a positive constant L f such that for any x, x ∗ , y, y ∗ ∈ C(J, Bk ), we have
| f (t, x, x ∗ ) − f (t, y, y ∗ )| ≤ L f (kx − yk + kx ∗ − y ∗ )k,
for t ∈ J, where k is defined as in (7).
Theorem 3. If assumptions (H2) − (H4) are satisfied, then the nonlocal Cauchy problem (1) has
a unique mild solution provided that
α
αM L f
b
Γ(α)bα+β
ML +
< 1.
(9)
+
Γ(α + 1) α
Γ(α + β)
Proof. It is easy to see that
Z tZ
R∞
0
ξα (θ )Q(t α θ )(x 0 − g(x))dθ exists and
∞
θ (t − s)α−1 ξα (θ )Q((t − s)α θ ) f (s, x(s), I β x(s))dθ ds
0
0
Bochner’s integrable with respect to s ∈ [0, t] for all t ∈ J. For x ∈ Bk , Consider the operator
F on Bk which is given by (6).
Obviously, it is sufficient to proof that F has a unique fixed point on Bk .
According to (8), we know that F is an operator from Bk into itself. For any x, y ∈ Bk and
t ∈ J, according to (H3), (H4) and (2), we have
Z
∞
α
|(F x)(t) − (F y)(t)| ≤ ξα (θ )Q(t θ )(g( y) − g(x))dθ 0
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
492
Z Z
t ∞
α−1
α
β
β
+ α
θ (t − s) ξα (θ )Q((t − s) θ )[ f (s, x(s), I x(s)) − f (s, y(s), I y(s))]dθ ds
0 0
≤M Lkx − yk
+
αM L f
t
Z
(t − s)
Γ(α + 1)
t
Z
α−1
|x(s) − y(s)|ds +
(t − s)
0
s
Z
α−1
0
0

(t − s)
α+β−1
ds kx − yk
0
bα Γ(α)bα+β
+
α
Γ(α + β)
kx − yk.
kx − yk.
which means that F is a contraction according to (9). By applying Banach contraction principle, we know that F has a unique fixed point on Bk . The proof is complete.
4. Mittag-Leffler-Ulam Stabilities
In this section, we consider the Mittag-Leffler-Ulam stability of the nonlocal Cauchy problem (1). Let ε be a positive real number and ϕ : J → R+ be a continuous function. We consider
the following inequalities
C Dα y(t) − Ay(t) − f t, y(t), I β y(t) ≤ε, t ∈ J
(10)
C Dα y(t) − Ay(t) − f t, y(t), I β y(t) ≤ϕ(t), t ∈ J
(11)
C
α
β
D y(t) − Ay(t) − f t, y(t), I y(t) ≤εϕ(t), t ∈ J
(12)
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
493
Definition 5. Eq. (1) is Mittag-Leffler-Ulam-Hyers stable, with respect to Eα if there exists a real
number c > 0 such that for each ε > 0 and for each solution y ∈ C 1 (J, E) of the inequality (10),
there exists a mild solution x ∈ C(J, E) of Eq. (1) with
y(t) − x(t) ≤ cεE [t], t ∈ J.
α
Definition 6. Eq. (1) is generalized Mittag-Leffler-Ulam-Hyers stable, with respect to Eα if there
exists θ ∈ C(R+ , R+ ), θ (0) = 0 such that for each solution y ∈ C 1 (J, E) of the inequality (10),
there exists a mild solution x ∈ C(J, E) of Eq. (1) with
y(t) − x(t) ≤ θ (ε)E [t], t ∈ J.
α
Definition 7. Eq. (1) is Mittag-Leffler-Ulam-Hyers-Rassias stable with respect to ϕEα if there
exists Cϕ > 0 such that for each ε > 0 and for each solution y ∈ C 1 (J, E) of the inequality (12),
there exists a mild solution x ∈ C(J, E) of Eq.(1) with
y(t) − x(t) ≤ C εϕ(t)E [t], t ∈ J.
ϕ
α
Definition 8. Eq. (1) is generalized Mittag-Leffler-Ulam-Hyers-Rassias stable with respect to ϕEα
if there exists Cϕ > 0 such that for each solution y ∈ C 1 (J, E) of the inequality (11), there exists
a mild solution x ∈ C(J, E) of Eq.(1) with
y(t) − x(t) ≤ C ϕ(t)E [t], t ∈ J.
ϕ
α
Remark 4. It is clear that: (i) Definition 5 =⇒ Definition 6; (ii) Definition 7 =⇒ Definition 8.
Remark 5. A function y ∈ C 1 (J, E) is a solution of the inequality (10) if and only if there exist
a function h ∈ C(J, E) (which depend on y) such that
(i) |h(t)| ≤ ε, t ∈ J,
(ii)
C
Dα y(t) = Ay(t) + f t, y(t), I β y(t) + h(t), t ∈ J.
One can have similar remarks for the inequalities (11) and (12).
Remark 6. If y ∈ C 1 (J, E) i a solution of the inequality (10), then y is a solution of the following
integral inequality
Z t
M bα
ε.
(t − s)α−1 T (t − s) f (s, y(s), I β y(s))ds ≤
y(t) − S(t)( y0 − g( y)) −
Γ(α + 1)
0
We have similar remarks for the solutions of inequalities (11) and (12).
Theorem 4. If assumptions (H3) and (H4) are satisfied, then the nonlocal Cauchy problem (1)
is Mittag-Leffler-Ulam-Hyers stable.
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
494
Proof. Let y ∈ C 1 (J, E) is a solution of the inequality (10). Let us denote by x ∈ C(J, E)
the unique mild solution of the nonlocal Cauchy problem
¨
C α
D x(t) = Ax(t) + f (t, x(t), I β x(t)), t ∈ J
(13)
x(0) = y(0).
We have
t
Z
(t − s)α−1 T (t − s) f (s, y(s), I β y(s))ds.
x(t) = S(t)( y0 − g( y)) +
0
Then we get
Z t
α−1
β
y(t) − S(t)( y − g( y)) −
(t − s)
T (t − s) f (s, y(s), I y(s))ds
0
0
Z
t
≤ (t − s)α−1 T (t − s)h(s)ds
0
Z tZ ∞
θ (t − s)α−1 ξα (θ )Q((t − s)α−1 θ )|h(s)|dθ ds
≤α
0
0
Z t
αM
≤
ε
(t − s)α−1 ds
Γ(α + 1)
0
M bα
ε.
≤
Γ(α + 1)
From these relations, we have
Z t
α−1
β
(t − s)
T (t − s) f (s, x(s), I x(s))ds
| y(t) − x(t)| = y(t) − S(t)( y0 − g( y)) −
0
Z t
(t − s)α−1 T (t − s) f (s, y(s), I β y(s))ds
≤ y(t) − S(t)( y0 − g( y)) −
0
Z
t
α−1
β
β
+ (t − s)
T (t − s)[ f (s, y(s), I y(s)) − f (s, x(s), I x(s))]ds
0
≤
≤
M bα
ε
Γ(α + 1)
Z Z
t ∞
+ α
θ (t − s)α−1 ξα (θ )Q((t − s)α−1 θ )[ f (s, y(s), I β y(s))−
0 0
− f (s, x(s), I β x(s))]dθ ds
M bα
ε+
Γ(α + 1)
Z t
αM
(t − s)α−1 | f (s, y(s), I β y(s)) − f (s, x(s), I β x(s))|ds
+
Γ(α + 1) 0
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
495
Z t
αM L f
M bα
ε+
(t − s)α−1 |x(s) − y(s)|ds
≤
Γ(α + 1)
Γ(α + 1)
0

Z tZ s
1
α−1
β−1
+
(t − s) (s − τ)
|x(τ) − y(τ)|dτds .
Γ(β) 0 0
Using a similar manner of the second integral as in Proof of Theorem 3, we get
Z t
αM L f
M bα
| y(t) − x(t)| ≤
ε+
(t − s)α−1 |x(s) − y(s)|ds
Γ(α + 1)
Γ(α + 1) 0
Z t
M Lf
(t − s)α+β−1 |x(s) − y(s)|ds.
+
Γ(α + β) 0
αM L
f
An application of Remark 2 and Remark 3 (with b1 = Γ(α+1)
, b2 =
β2 = α + β) to the last inequality yields the desired estimation
| y(t) − x(t)| ≤
M bα
Γ(α + 1)
M Lf
Γ(α+β) ,
β1 = α and
Eα M L f t α + Eα+β M L f t α+β ε.
Thus, the conclusion of our theorem hold.
The following theorem gives generalized Mittag-Leffler-Ulam-Hyers stability.
Theorem 5. If assumptions (H3) and (H4) are satisfied. Suppose there exist λ > 0 such that
1
Γ(α)
t
Z
(t − s)α−1 ϕ(s)ds ≤ λϕ(t),
0
for all t ∈ J, where ϕ ∈ C(J, R+ ) is nondecreasing. Then the nonlocal Cauchy problem (1) is
generalized Mittag-Leffler-Ulam-Hyers stable with respect to ϕEα .
Proof. Let y ∈ C 1 (J, E) is a solution of the inequality (11). By Remark 5, we have for t ∈ J
that y is a solution of the following integral inequality
Z t
α−1
β
(t − s)
T (t − s) f (s, y(s), I y(s))ds ≤ M λ ϕ(t).
y(t) − S(t)( y0 − g( y)) −
0
Let us denote by x ∈ C(J, E) the unique mild solution of the nonlocal Cauchy problem
¨
C α
D x(t) = Ax(t) + f (t, x(t), I β x(t)), t ∈ J
(14)
x(0) = y(0).
We have
t
Z
(t − s)α−1 T (t − s) f (s, y(s), I β y(s))ds.
x(t) = S(t)( y0 − g( y)) +
0
M. Abbas / Eur. J. Pure Appl. Math, 8 (2015), 478-498
496
Then we get
Z t
α−1
β
y(t) − S(t)( y − g( y)) −
(t
−
s)
T
(t
−
s)
f
(s,
y(s),
I
y(s))ds
0
0
Z
t
α−1
T (t − s)ϕ(s)ds
≤ (t − s)
0
Z tZ ∞
θ (t − s)α−1 ξα (θ )Q((t − s)α−1 θ )ϕ(s)dθ ds
≤α
0
0
αM
≤
Γ(α + 1)
t
Z
(t − s)α−1 ϕ(s)ds
0
≤M λ ϕ(t).
Again, from these relations, we have
Z t
(t − s)α−1 T (t − s) f (s, x(s), I β x(s))ds
| y(t) − x(t)| = y(t) − S(t)( y0 − g( y)) −
0
Z t
α−1
β
(t − s)
T (t − s) f (s, y(s), I y(s))ds
≤ y(t) − S(t)( y0 − g( y)) −
0
Z
t
+ (t − s)α−1 T (t − s)[ f (s, y(s), I β y(s)) − f (s, x(s), I β x(s))]ds
0
Z Z
t ∞
θ (t − s)α−1 ξα (θ )Q((t − s)α−1 θ )
≤M λ ϕ(t) + α 0 0
×[ f (s, y(s), I β y(s)) − f (s, x(s), I β x(s))]dθ ds
Z t
αM
≤M λ ϕ(t) +
(t − s)α−1 | f (s, y(s), I β y(s)) − f (s, x(s), I β x(s))|ds
Γ(α + 1) 0
Z t
αM L f
≤M λ ϕ(t) +
(t − s)α−1 |x(s) − y(s)|ds
Γ(α + 1)
0

Z tZ s
1
+
(t − s)α−1 (s − τ)β−1 |x(τ) − y(τ)|dτds .
Γ(β) 0 0
Using a similar manner of the second integral as in Proof of Theorem 3, we get
Z t
αM L f
M bα
(t − s)α−1 |x(s) − y(s)|ds
ε+
| y(t) − x(t)| ≤
Γ(α + 1)
Γ(α + 1) 0
Z t
M Lf
(t − s)α+β−1 |x(s) − y(s)|ds.
+
Γ(α + β) 0
REFERENCES
497
According to Remark 2 and Remark 3, we obtain our required assertion
| y(t) − x(t)| ≤ M λ Eα M L f t α + Eα+β M L f t α+β ϕ(t).
The conclusion of our theorem hold.
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