Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 9 (2016), 827–835
Research Article
Existence and uniqueness of mild and classical
solutions to fractional order Hadamard-type Cauchy
problem
Qutaibeh Katatbeha,∗, Ahmad Al-Omarib
a
Department of Mathematics and Statistics, Faculty of Science and Arts, Jordan University of Science and Technology, Irbid-Jordan
22110.
b
Al al-Bayt University, Faculty of Sciences, Department of Mathematics P.O. Box 130095, Mafraq 25113, Jordan.
Communicated by R. Saadati
Abstract
We consider the existence and uniqueness of a mild and classical solution to impulsive nonlocal conditions
fractional-order Hadamard-type Cauchy problem. The results are obtained by means of fixed point methods. Finally, we illustrate our results by an example of fractional-order Hadamard-type Cauchy problem.
c
2016
All rights reserved.
Keywords: Hadamard fractional derivative, integral boundary conditions, fixed point theorems, impulsive
equations.
2010 MSC: 34A08, 34N05, 34A12.
1. Introduction and preliminaries
In recent studies, the theory of fractional differential equations and inclusions has been into the focus of
many of them. This is due to its extensive applications in numerous branches of applied sciences such as,
physics, economics, and engineering sciences [2, 3, 5, 8, 10, 12, 13]. Fractional differential equations and their
applications make it possible to find appropriate models for describing real-world problems which cannot be
described by using classical integral-order differential equations. Some recent contributions to the subject
can be found in [1] and references therein. It has been noticed that most of the work on the topic is based
∗
Corresponding author
Email addresses: qutaibeh@just.edu.jo (Qutaibeh Katatbeh), omarimutah1@yahoo.com (Ahmad Al-Omari)
Received 2015-08-17
Q. Katatbeh, A. Al-Omari, J. Nonlinear Sci. Appl. 9 (2016), 827–835
828
on Riemann–Liouville and Caputo-type fractional differential equations. However, there are other kinds
of fractional derivatives that appears side by side with Riemann–Liouville and Caputo derivatives. The
fractional derivative is due to Hadamard who introduced it in 1892 [7], but which differs from the preceding
ones, in the sense that the kernel of the integral (in the definition of the Hadamard derivative) contains a
logarithmic function of arbitrary exponent. The details and properties of the Hadamard fractional derivative
and integral can be found in [4, 6, 9]. In this paper, we study a boundary value problem of Hadamard-type
fractional differential inclusions.
Our aim is to study the existence and uniqueness of the mild and classical solutions to impulsive nonlocalconditions fractional Cauchy problem on Hadamard-type fractional differential inclusions. Throughout the
paper we shall use the notation:
Ω = {(t, s) : 1 ≤ s ≤ t ≤ T },
M = sup{kS(t)k , t ∈ [1, T ]},
and
X = C([1, T ], R).
Dα u(t) + A u(t) = f (t, u(t), u(b1 (t)), . . . , u(br (t))) +
Z
t
Z
1+
f1 (t, s, u(s)) ds +
1
f2 (t, s, u(s)) ds,
(1.1)
1
∆u|tk = Ik (u(t−
k )), k = 1, 2, . . . m.
u(1) + g(u) = u0 ,
where t ∈ J \ {t1 , t2 , . . . , tm } ⊆ [1, T ] = J and 1 < t1 < · · · < tm < T where 0 < α ≤ 1. The operator
−A generates an analytic compact semigroup (S(t))t≥0 of uniformly bounded linear operators on a Banach
space R. Dα is a Hadamard-type fractional derivative operator and f : C([1, T ] × Rr+1 → R, fi : Ω × R → R
(i = 1, 2), g : X → R, bi : J → J (i = 1, . . . , r) are function satisfying some assumptions, and u0 ∈ R and
−
+
−
∆u|t=tk = u(t+
k ) − u(tk ), where u(tk ) and u(tk ), represent the right and left limits of u(t) at t = tk ; Ik
(k = 1, 2, . . . , m) are functions to be specified later.
Definition 1.1 ([9]). The Hadamard derivative of fractional order α for a function g : [1, ∞) → R is defined
by
n Z t 1
d
t n−α−1 g(s)
Dα g(t) =
t
log
ds,
Γ(n − α) dt
s
s
1
α ∈ (n − 1, n), n = [α] + 1, where [α] denotes the integer part of the real number α and log(.) = loge (.).
Definition 1.2 ([9]). The Hadamard fractional integral of order α for a function g : [1, ∞) → R is defined
as
1
I g(t) =
Γ(α)
for α > 0, provided the integral exists.
α
Z t
1
t
log
s
α−1
g(s)
ds
s
Let us recall the generalized Gronwall inequality, which can be found in [11],
Lemma 1.3. Suppose α > 0, a(t) and u(t) are nonnegative function and locally integrable on 1 ≤ t < T and
g(t) is a nonnegative, nondecreasing continuous function defined on 1 ≤ t < T , and g(t) ≤ M (constant). If
Z t
ds
t α−1
u(t) ≤ a(t) + g(t)
log
u(s)
on 1 ≤ t < T,
s
s
1
then
#
Z t "X
∞
[g(t)Γ(α)]n
t n α−1
ds
u(t) ≤ a(t) +
log
a(s)
.
Γ(n α)
s
s
1
n=1
Q. Katatbeh, A. Al-Omari, J. Nonlinear Sci. Appl. 9 (2016), 827–835
829
Definition 1.4. A function u ∈ C([1, T ], R) satisfying the integral equation
u(t) = S(t − 1)u0 − S(t − 1)g(u)
Z t
t α−1 h
1
S(t − s) log
+
f (s, u(s), u(b1 (s)), . . . , u(br (s)))
Γ(α) 1
s
Z s
Z 1+
ds
+
f1 (s, τ, u(τ )) dτ +
f2 (s, τ, u(τ )) dτ
s
1
1
m
X
+
S(t − tk )Ik (u(t−
t∈J
k )),
k=1
is said to be a mild solution to the integrodifferential evolution impulsive nonlocal Cauchy problem (1.1)
with fractional-order Hadamard-type.
2. Main result
Theorem 2.1. The integrodifferential evolution impulsive nonlocal Cauchy problem (1.1) has a unique mild
solution u, if the following conditions are satisfied:
(i) f : J × Rm+1 → R is a continuous function with respect to the first variable in J, fi : Ω × R → R
(i = 1, 2) are continuous functions with respect to the first and second variables in Ω; g : X → R,
and bi : J → J (i = 1, . . . , m) are continuous functions on J and there exist positive constants L, Li
(i = 1, 2) and K such that:
!
m
X
(2.1)
kf (s, z0 , z1 , . . . , zm ) − f (s, ze0 , ze1 , . . . , zem )k ≤ L
kzi − zei k
i=0
for s ∈ J, zi , zei ∈ R (i = 1, 2, . . . , m),
kfi (s, τ, z) − fi (s, τ, ze)k ≤ Li (kz − zek) (i = 1, 2)
(2.2)
for (s, τ ) ∈ Ω, z, ze ∈ R and
kg(u) − g(e
u)k ≤ K ku − u
ek∞ for u, u
e ∈ X.
(ii)
(2.3)
M logα T
(mL + K)Γ(α + 1) log−α T + L(m + 1) + L1 T + L2 T
< 1.
Γ(α + 1)
(iii) The functions Ik : R → R are continuous and there exists a ρ1 such that kIk (x)k ≤ ρ1 for all x ∈ R
and k = 1, . . . , m and u0 ∈ R.
Proof. Introduce the operator F : X → X by
(F ω)(t) = S(t − 1)u0 − S(t − 1)g(ω)
Z t
h
t
1
+
S(t − s)(log )α−1 f (s, ω(s), ω(b1 (s)), . . . , ω(br (s)))
Γ(α) 1
s
Z s
Z 1+
ds
+
f1 (s, τ, ω(τ )) dτ +
f2 (s, τ, ω(τ )) dτ
s
1
1
m
X
+
S(t − tk )Ik (ω(t−
t ∈ J.
k )),
k=1
(2.4)
Q. Katatbeh, A. Al-Omari, J. Nonlinear Sci. Appl. 9 (2016), 827–835
830
We have
||(F ω)(t) − (F ω̂)(t)|| ≤ ||S(t − 1)||||g(ω) − g(ω̂)||
Z t
t α−1 h
1
S(t − s) log
+
f (s, ω(s), ω(b1 (s)), . . . , ω̂(br (s)))
Γ(α) 1
s
i ds
− f (s, ω̂(s), ω̂(b1 (s)), . . . , ω̂(br (s)))
s
α−1 Z s
Z t
1
t
ds
+
S(t − s) log
||f1 (s, τ, ω(τ )) − f1 (s, τ, ω̂(τ ))|| dτ
Γ(α) 1
s
s
1
α−1 Z 1+
Z t
1
t
ds
+
S(t − s) log
||f2 (s, τ, ω(τ )) − f2 (s, τ, ω̂(τ ))|| dτ
Γ(α) 1
s
s
1
m
X
−
+
||S(t − tk )|| ||Ik (ω(t−
k )) − Ik (ω̂(tk ))||
k=1
"
#
Z m
X
t (α−1)
ML t
ds
log
≤ M K||ω − ω̂||∞ +
||ω(s) − ω̂(s)|| +
||ω(bi (s)) − ω̂(bi (s))||
Γ(α) 1
s
s
i=1
α−1 Z s
Z t
M L1
t
ds
+
log
||ω(τ ) − ω̂(τ )|| dτ
Γ(α) 1
s
s
1
α−1 Z 1+
Z t
t
ds
M L2
log
||ω(τ ) − ω̂(τ )|| dτ
+
Γ(α) 1
s
s
1
+ M mC||ω − ω̂||∞
Z M L(m + 1)||ω − ω̂||∞ t
t (α−1) ds
≤ M K||ω − ω̂||∞ +
log
Γ(α)
s
s
1
α−1
Z t
t
ds
M L1 T ||ω − ω̂||∞
log
+
Γ(α)
s
s
1
α−1
Z t
M L2 T ||ω − ω̂||∞
t
ds
+
log
+ M mC||ω − ω̂||∞
Γ(α)
s
s
1
M L(m + 1)||ω − ω̂||∞ logα T
≤ M K||ω − ω̂||∞ +
Γ(α)
α
α
M L1 T ||ω − ω̂||∞ log T
M L2 T ||ω − ω̂||∞ logα T
+
+
+ M mC||ω − ω̂||∞
Γ(α)
α
Γ(α)
α
M logα T
≤ (mL + K)Γ(α + 1) log−α T + L(m + 1) + L1 T + L2 T
||ω − ω̂||∞ .
Γ(α + 1)
Using the assumption, we get
(mL + K)Γ(α + 1) log−α T + L(m + 1) + L1 T + L2 T
M logα T
< 1.
Γ(α + 1)
This implies that
||(F ω)(t) − (F ω̂)(t)|| ≤ ||ω − ω̂||∞
and hence F is a contraction on X.
The following theorem gives the conditions for the existence of a unique classical solution for fractional
implicit nonlocal conditions with fractional order Hadamard-type.
Q. Katatbeh, A. Al-Omari, J. Nonlinear Sci. Appl. 9 (2016), 827–835
831
Theorem 2.2. Assume that:
(i) R is a reflexive Banach space and u0 ∈ R.
(ii) f : J × Rm+1 → R fi : Ω × R → R (i = 1, 2), are continuous function with respect to the second
variable in J; g : X → R, bi : J → J (i = 1, . . . , m) are continuous function on J and there exist
positive constants C, Ci (i = 1, 2) and K such that:
!
m
X
kf (s, z0 , z1 , . . . , zm ) − f (e
s, ze0 , ze1 , . . . , zem )k ≤ C |s − se| +
kzi − zei k
(2.5)
i=0
for s, se ∈ J, zi , zei ∈ R (i = 1, 2, . . . , m),
kfi (s, τ, z) − fi (e
s, τ, ze)k ≤ Ci (|s − se| + kz − zek) (i = 1, 2)
(2.6)
for (s, τ ), (e
s, τ ) ∈ Ω, z, ze ∈ R and
kg(u) − g(e
u)k ≤ K ku − u
ek∞ for u, u
e ∈ X.
(2.7)
M logα T
< 1.
(iii) (mC + K)Γ(α + 1) log−α T + C(m + 1) + T (C1 + C2 )
Γ(α + 1)
(iv) The functions Ik : R → R are continuous and there exists a ρ1 such that kIk (x)k ≤ ρ1 for all x ∈ R
and k = 1, . . . , m.
Then the integrodifferential evolution impulsive nonlocal Cauchy problem (1.1) has a unique mild solution
u. Moreover, if u0 ∈ D(A), g(u) ∈ D(A), and if there is a positive constant κ such that
ku(bi (s)) − u(bi (e
s))k ≤ κ ku(s) − u(e
s)k for s, se ∈ J
(i = 1, . . . , m),
(2.8)
then u is a unique classical solution to problem (1.1).
Proof. Since all the assumptions of Theorem 2.1 are satisfied, it is easy to see that the presented problem
possesses a unique mild solution u. Now, we shall show that u is a classical solution to the problem.
Introducing the notation
N := max ||f (s, u(s), u(b1 (s), . . . , u(bm (s)))||
s∈J
Ni := max ||fi (s, τ, u(τ ))||, (i = 1, 2)
(s,τ )∈Γ
we my write
u(t + θ) − u(t) = (S(t + θ − 1)u0 − S(t + θ − 1)g(u)) − (S(t − 1)u0 − S(t − 1)g(u))
Z t+θ
i
1
t + θ α−1 h
+
S(t + θ − s) log
f (s, u(s), u(b1 (s)), . . . , u(br (s)))
Γ(α) 1
s
Z s
Z 1+
ds
+
f1 (s, τ, u(τ )) dτ +
f2 (s, τ, u(τ )) dτ
s
1
1
α−1 h
Z t
i
1
t
−
S(t − s) log
f (s, u(s), u(b1 (s)), . . . , u(br (s)))
Γ(α) 1
s
Z s
Z 1+
ds
+
f1 (s, τ, u(τ )) dτ +
f2 (s, τ, u(τ )) dτ
s
1
1
Q. Katatbeh, A. Al-Omari, J. Nonlinear Sci. Appl. 9 (2016), 827–835
+
m
X
S(t + θ − tk )Ik (u(t−
k )) −
k=1
m
X
832
S(t − tk )Ik (u(t−
k ))
k=1
= S(t − 1) (S(θ) − I) u0 − S(t − 1) (S(θ) − I) g(u)
Z 1+θ
i
t + θ α−1 h
1
S(t + θ − s) log
f (s, u(s), u(b1 (s)), . . . , u(br (s)))
+
Γ(α) 1
s
Z s
Z 1+
ds
+
f1 (s, τ, u(τ )) dτ +
f2 (s, τ, u(τ )) dτ
s
1
1
α−1 h
Z t
1
t
+
S(t − s) log
f (s + θ, u(s + θ), u(b1 (s + θ)), . . . , u(br (s + θ)))
Γ(α) 1
s
i ds
− f (s, u(s), u(b1 (s)), . . . , u(br (s)))
s
Z s
Z t
t α−1
ds
1
S(t − s) log
(f1 (s + θ, τ, u(τ )) − f1 (s, τ, u(τ ))) dτ
−
Γ(α) 1
s
s
1
α−1 Z s+θ
Z t
1
t
ds
+
S(t − s) log
(f1 (s + θ, τ, u(τ ))) dτ
Γ(α) 1
s
s
s
α−1 Z 1+
Z t
1
t
ds
+
S(t − s) log
(f2 (s + θ, τ, u(τ )) − f2 (s, τ, u(τ ))) dτ
Γ(α) 1
s
s
1
m
X
+
S(θ)Ik (u(t−
k ));
k=1
therefore,
||u(t + θ) − u(t)|| ≤ θM ||Au0 || + θM ||Ag(u)||
+
+
+
+
Z
Z (M (N + T N1 + T N2 )) 1+θ
t + θ α−1 ds
(M Cθ) t
t α−1 ds
log
+
log
Γ(α)
s
s
Γ(α) 1
s
s
1
!
α−1
Z t
m
X
(M C)
ds
t
|θ| + ku(s + θ) − u(s)k +
ku(bi (s + θ)) − u(bi (s))k
log
Γ(α) 1
s
s
i=1
Z (M C1 θT ) t
t α−1 ds
log
Γ(α)
s
s
1
α−1
Z t
Z (M N1 θ)
t
ds
(M C2 θT ) t
t α−1 ds
+
+ M mρ1 .
log
log
Γ(α)
s
s
Γ(α)
s
s
1
1
Consequently,
||u(t + θ) − u(t)|| ≤ θM ||Au0 || + θM ||Ag(u)|| + M mρ1
Z
(M (N + T N1 + T N2 )) 1+θ
t + θ α−1 ds
+
log
Γ(α)
s
s
1
α−1
Z t
(M C(1 + mκ))
t
ds
+
log
ku(s + θ) − u(s)k
Γ(α)
s
s
1
α−1
Z t
(2M Cθ + M C1 θT + M N1 θ + M C2 θT )
t
ds
log
+
,
Γ(α)
s
s
1
||u(t + θ) − u(t)|| ≤ θM ||Au0 || + θM ||Ag(u)|| + M mρ1
T ∗ (M (N + T N1 + T N2 ))
+
Γ(α)
Q. Katatbeh, A. Al-Omari, J. Nonlinear Sci. Appl. 9 (2016), 827–835
833
Z (M C(1 + mκ)) t
t α−1
ds
+
log
ku(s + θ) − u(s)k
Γ(α)
s
s
1
α
log T (2M Cθ + M C1 θT + M N1 θ + M C2 θT )
+
,
Γ(α)
where
)
α−1
ds
t
+
θ
, t ∈ [1, T ]
T ∗ := M ax
log
s
s
1
Z t
t α−1
ds
log
||u(t + θ) − u(t)|| ≤ C∗ (t) + C∗∗
ku(s + θ) − u(s)k .
s
s
1
(Z
1+θ
By the generalized Gronwall inequality from [11], it follows that
#
Z t "X
∞
t n α−1
ds
[C∗∗ Γ(α)]n
log
C∗ (s)
||u(t + θ) − u(t)|| ≤ C∗ (t) +
Γ(n
α)
s
s
1
n=1
||u(t + θ) − u(t)|| ≤ C∗ (t) +
∞
X
[C∗∗ Γ(α)]n
n=1
Γ(n α + 1)
(log t)n α C∗
||u(t + θ) − u(t)|| ≤ C∗ (t) + C∗ (t)Eα (C∗∗ Γ(α) logα t),
tk
for t ∈ [1, T ), θ > 0 and t + θ ∈
k=0 Γ(kα + 1)
(1, T ]. Hence u is Lipschitz continuous on J. The Lipschitz continuity of u on J combined with the Lipschitz
continuity of f on J × Rm+1 and fi (i = 1, 2) in J imply that the function
Z t
Z 1+
J 3 t → f (t, u(t), u(b1 (t)), . . . , u(br (t))) +
f1 (t, s, u(s)) ds +
f2 (t, s, u(s)) ds
where Eα is the Mittag-Leffler function defined by Eα (t) =
1
∞
P
1
is Lipschitz continuous on J. This property of f , together with assumptions and by Theorem 2.1 implies
that u is a unique classical solution to the present problem.
3. Example
In this section we provide an example to illustrate our main result. Consider the following impulsive
nonlocal fractional differential equation
Z t
Z 2
1 −t
e−t |u(t)|
et−s
es |u(s)|
α
e |u(t)| =
+
ds
+
ds
D u(t) +
2
100
(9 + et )(1 + |u(t)|)
1 5 + |u(s)|
1 (t + 5) (1 + |u(s)|)
3
t ∈[1, 2],
t 6= ,
0 < α ≤ 1.
2
−
|u( 32 )|
∆u|t= 3 =
−
2
8 + |u( 32 )|
|u(t)|
u(1) +
= u0 (t)
3 + |u(t)|
e−t |u(t)|
et−s
f (t, u(t), u(b(t))) =
,
f
(t,
s,
u(s))
=
,
1
(9 + et )(1 + |u(t)|)
5 + |u(s)|
Q. Katatbeh, A. Al-Omari, J. Nonlinear Sci. Appl. 9 (2016), 827–835
f2 (t, s, u(s)) =
es |u(s)|
(t + 5)2 (1 + |u(s)|)
and Ik (u) =
834
u(t)
.
8 + u(t)
We have
e−t
kf (t, u(t), u(b(t))) − f (t, v(t), v(b(t)))k =
9 + et
u
v 1 + u − 1 + v e−t |u − v|
=
(9 + et )(1 + u)(1 + v)
e−t
|u − v|
≤
9 + et
1
≤ |u − v|.
10
Therefore, the first condition in Theorem 2.2 holds with C =
1
10 .
We also have
1 t−s 1
−
kf1 (t, s, u(s)) − f1 (t, s, v(s))k = e 5 + u 5 + v
et−s |u − v|
=
(5 + u)(5 + v)
e
≤ |u − v|.
25
u(s)
es
v(s) kf2 (t, s, u(s)) − f2 (t, s, v(s))k =
−
(t + 5)2 1 + u(s) 1 + v(s) es |u − v|
=
(t + 5)2 (1 + u)(1 + v)
e
≤ |u − v|.
25
e
Hence the second condition in Theorem 2.2 holds with Ci = 25
(i = 1, 2). Moreover, we have
u
v 3|u − v|
1
kg(u) − g(v)k = −
=
≤ |u − v|,
3+u 3+v
(3 + u)(3 + v)
3
and
u
v 8|u − v|
1
kIk (u) − Ik (v)k = −
≤ |u − v|,
=
8+u 8+v
(8 + u)(8 + v)
8
and
kIk (u)k = |
u
1
|≤ .
u+8
8
Hence the condition (iv) in Theorem 2.2 holds with ρ1 = 18 . Let t ∈ [1, 2]. We shall check whether the
condition
M (log 2)α
<1
(mC + K)Γ(α + 1)(log 2)−α + C(m + 1) + 2(C1 + C2 )
Γ(α + 1)
is satisfied with m = 1 and we can take M = 1. Indeed
(mC + K)Γ(α + 1)(log 2)−α + C(m + 1) + 2(C1 + C2 )
M (log 2)α
<1
Γ(α + 1)
44
, which is satisfied for an α ∈ (0, 1], such as α = 0.2 or α = 0.8. Then by
49
Theorem 2.2, the problem (1.1) has a unique solution in [1, 2].
if and only if Γ(α + 1) >
Q. Katatbeh, A. Al-Omari, J. Nonlinear Sci. Appl. 9 (2016), 827–835
835
Acknowledgement
The authors thank Jordan University of Science and Technology and Al-Bayt University for their support.
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