Accepted Manuscript
Local and global existence of mild solution to impulsive fractional semilinear integro-differential equation with noncompact semigroup Haide Gou, Baolin Li PII: DOI: Reference:
S1007-5704(16)30174-5 10.1016/j.cnsns.2016.05.021 CNSNS 3877
To appear in:
Communications in Nonlinear Science and Numerical Simulation
Received date: Revised date: Accepted date:
3 February 2016 17 May 2016 23 May 2016
Please cite this article as: Haide Gou, Baolin Li, Local and global existence of mild solution to impulsive fractional semilinear integro-differential equation with noncompact semigroup, Communications in Nonlinear Science and Numerical Simulation (2016), doi: 10.1016/j.cnsns.2016.05.021
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Highlights • We establish a general framework to find the mild solutions for impulsive fractional integro-differential equations, which will provide an effective way to deal with such
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problems. • Due to the latest development regarding the mild solution for the considered problem
of fractional order with impulsive conditions the definition of mild solution defined by the authors of [10] is not appropriate, because classical solutions of the impulsive fractional differential equations do not satisfy the definition of a mild solution given by the author and the semigroup property T (t + s) = T (t)T (s) for thesystem is not
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used correctly,
• The motivations of this paper are two aspect : on the one hand, we observed that
most of the existing articles are only devoted to study the local existence of mild solutions for fractional evolution equations, up until now the existence of mild solution for an impulsive fractional functional integro differential equations with noncompact
AC
CE
PT
ED
M
semigroup in Banach spaces has not been considered in the literature.
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Local and global existence of mild solution to
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impulsive fractional semilinear integro-differential equation with noncompact semigroup Haide Gou, Baolin Li†,
∗
College of Mathematics and Statistics, Northwest Normal University
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Lanzhou, 730070, PR China
Abstract: In this paper, we study local and global existence of mild solution for an impulsive fractional functional integro differential equation with noncompact
M
semigroup in Banach spaces. We establish a general framework to find the mild solutions for impulsive fractional integro-differential equations, which will
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provide an effective way to deal with such problems. The theorems proved in this paper improve and extend some related conclusions on this topic. Finally, two applications are given to illustrate that our results are valuable.
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Keywords: Mild solutions; Impulsive fractional integro-differential equations; Equicontinuous semigroup
CE
AMS(2000) Subject Classification: 26A33, 34A37, 34G20, 47H10
Introduction
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1
Fractional differential equations arise in many engineering and scientific disciplines as
the mathematical modelling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, and so forth, involves derivatives of fractional order. In consequence, the subject of fractional differential ∗ †
Supported by the National Natural Science Foundation of China (Grant No.11061031). Corresponding author.
E-mail address:
[email protected]
1
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equations is gaining much importance and attention. For more details, see [1-7] and the references therein. Impulsive differential equations have become important in recent years as mathematical models of phenomena in both physical and social sciences. There has been a significant
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development in impulsive theory especially in the area of impulsive differential equations with fixed moments. In [8], Heard and Rakin considered the following integro-differential equation in a Banach space X: (
u0 + A(t)u(t) = f (t, u(t)) + u(t0 ) = u0 ,
Rt
q(s − t)g(s, u(s))ds,
t0
t > t0 ≥ 0,
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where for each t ≥ 0, the linear operator −A(t) is the infinitesimal generator of an analytic
semigroup in X, the nonlinear operator f is defined from [0, ∞) × X into X and satisfies a H¨ older condition of the form
kf (t, y1 ) − f (s, y2 )k ≤ C[|t − s|v + ky1 − y2 kγk ], 0 < v, k, γ < 1, k · k is the norm on X and k · kk is the graph norm on Xk = D(Ak (0)), the
M
nonlinear map g is assumed to satisfy a local Lipschitz condition with respect to the norm
of X. Also, the uniqueness of solution is proved under the restriction that the space X is a
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Hilbert space and γ = 1. In paper [9], Xiaobao Shu study the existence of mild solutions to a class of fractional semilinear integro-differential equation of order 1 < α < 2 in a Banach space X
Dtα u(t) = Au(t) + f (t, u(t)) +
PT
(
u(0) + m(u) = u0 ∈ X,
Rt
u0 (0)
0
q(t − s)g(s, u(s))ds, t ∈ [0, T ],
+ n(u) = u1 ∈ X,
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where −A is a sectorial operator of type (M, θ, α, µ), defined from the domain D(A) ⊂ X
into X, the nonlinear map f, g are defined from [0, T ] × X → X are continuous functions.
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In particular, the following fractional order integro-differential equation in a Banach
space X: Rt α u (t) + Au(t) = f (t, u(t)) + 0 q(t − s)g(s, u(s))ds, t > 0, α ∈ (0, 1], u(0) = u0 ∈ X, ∆u| − t=tk = Ik (u(tk )),
k = 1, . . . , m,
was mentioned and established the local and global existence of mild solutions using the fixed point technique by Rashid and Al-Omari in [10]. Here −A is assumed to be an infinitesimal
generator of a compact semigroup T (t), t ≥ 0. Due to the latest development regarding the 2
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mild solution for the considered problem of fractional order with impulsive conditions the definition of mild solution defined by the authors of [10] is not appropriate, because classical solutions of the impulsive fractional differential equations do not satisfy the definition of a system is not used correctly, for more details (see [11,12]).
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mild solution given by the author and the semigroup property T (t + s) = T (t)T (s) for the However, to the best of our knowledge, local and global existence of mild solution for an impulsive fractional functional integro differential equation with noncompact semigroup in Banach spaces has not been investigated yet. Motivated by this consideration, in this paper, we investigate local and global existence of mild solution for an impulsive fractional
∆u|t=tk = Ik (u(t− k )), u(0) = u ∈ X, 0
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functional integro differential equation with noncompact semigroup in Banach spaces X : Rt c α Dt u(t) + Au(t) = f (t, u(t)) + 0 q(t − s)g(s, u(s))ds, t ≥ 0, t 6= tk , k = 1, . . . , m,
(1)
where c Dtα is the Caputo fractional derivative of order α ∈ (0, 1], A : D(A) ⊂ X → X is
a closed linear operator and−A generates a uniformly bounded C0 -semigroup T (t)(t ≥ 0)
M
in X, the nonlinear maps f, g : [0, ∞) × X → X, and q : I → X, are continuous. I = [0, T ), 0 < T ≤ ∞, u0 ∈ X. Ik : X → X, 0 = t0 < t1 < · · · < tm < tm+1 = T, ∆u|t=tk =
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− + − u(t+ k ) − u(tk ), u(tk ) = limm→0 u(tk + m) and u(tk ) = limm→0 u(tk − m) represent the right
and left limits of u(t) at t = tk , respectively.
The motivations of this paper are two aspect : on the one hand, we observed that
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most of the existing articles (see,[13-15,17-19]) are only devoted to study the local existence of mild solutions for fractional evolution equations, up until now the existence of mild solution for an impulsive fractional functional integro differential equation with noncompact
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semigroup in Banach spaces has not been considered in the literature. In order to fill this gap, we are concerned with the existence of local mild solutions and global mild solutions
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for the problem (1) in this paper; on the other hand, we find that the fractional evolution equations have been extensively studied in recent years by using various fixed point theorems when the corresponding semigroup T (t)(t ≥ 0) is compact, this is very convenient to the
equations with compact resolvent. But for the case that the corresponding semigroup T (t)(t ≥ 0) is noncompact, there are very few papers which studied these equations, only
Wang et al. [20] discussed the local existence of mild solutions for nonlocal problems of fractional evolution equations under the situation that T (t)(t ≥ 0) is an analytic semigroup
of uniformly bounded linear operators. In this article, by using the famous Sadovskiis fixed point theorem (see Lemma 2.5) and a new estimation technique of the measure of 3
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noncompactness (see Lemma 2.6), we study local and global existence of mild solution for the problem (1). Our results can be considered as a contribution to these nascent fields. The paper is organized as follows: In Section 2, we introduce some notations and recall some basic known results. In Section 3 we present the existence of local solutions to the
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problem (1) in Banach space. In Section 4 we discuss the existence theorem of global solutions for the problem (1). In Section 5, we give two examples to illustrate our results.
2
Preliminaries
Throughout this paper, we assume that A : D(A) ⊂ X → X is a closed linear operator
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and −A generates a uniformly bounded C0 -semigroup T (t)(t > 0) on a Banach space (X, k · k). Let I denote a closed subset of the interval [0, +∞) and let M = supt∈[0,+∞) kT (t)kB(X) ,
where B(X) stands for the Banach space of all linear and bounded operators in X. We denote by C(I, X) the Banach space of all continuous X-value functions on interval I and + P C(I, X) = {u : I → X : u ∈ C((tk−1 , tk ], X), k = 1, 2, . . . , m and there exist u(t− k ) and u(tk ), k =
1, 2, . . . , m with u(t− k ) = u(tk )} the Banach space of all continuous X-valued functions on
M
interval I with the supnorm kuk = supt∈I ku(t)k.
Definition 2.1. The fractional integral of order α > 0 with the lower limit zero for a
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function u is defined as
1 = Γ(α)
Z
0
t
(t − s)α−1 u(s)ds,
t > 0,
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Itα u(t)
where Γ(·) is the Gammma function.
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Definition 2.2. The Caputo fractional derivative of order α > 0 with the lower limit zero
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for a function u is defined as Z t 1 c α Dt u(t) = (t − s)n−α−1 u(n) (s)ds, Γ(n − α) 0
t > 0, 0 ≤ n − 1 < α < n,
where the function u(t) has absolutely continuous derivatives up to order n − 1. Remark 2.1. If u is an abstract function with values in X, then the integrals which appear in Definitions 2.1 and 2.2 are taken in Bochner’s sense. Definition 2.3. By a mild solution of the initial value problem ( c D α u(t) + Au(t) = f (t, u(t)) t ≥ 0, t u(0) = u0 ,
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on [0, ∞), we mean a continuous function u defined from [0, ∞) into X satisfying 0
where Tα (t) =
Z
∞
t
(t − s)α−1 Sα (t − s)f (s, u(s))ds, t ∈ [0, ∞), α
θα (s)T (t s)ds,
Sα (t) = α
Z
∞
sθα (s)T (tα s)ds,
0
0 ∞
1 X Γ(nα + 1) θα (s) = (−s)n−1 sin(nπα), πα n! n=1
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u(t) = Tα (t)u0 +
Z
s ∈ (0, ∞)
are the functions of Wright type defined on (0, ∞) which satisfies Z ∞ θα (s)ds = 1 θα (s) ≥ 0, s ∈ (0, ∞), and
Z
∞
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0
sv θα (s)ds =
0
Γ(1 + v) , Γ(1 + αv)
v ∈ [0, 1].
The following lemma can be found in [26,27].
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Lemma 2.1. The operators Tα (t) and Sα (t)(t ≥ 0) have the following properties: (i) For any fixed t ≥ 0, Tα (t) and Sα (t) are linear and bounded operators, i.e., for any
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u ∈ X,
kTα (t)uk ≤ M kuk,
kSα (t)uk ≤
αM kuk. Γ(α + 1)
into X.
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(ii) For every u ∈ X, t → Tα (t)u and t → Sα (t)u are continuous functions from [0, ∞)
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(iii) The operators Tα (t)(t ≥ 0) and Sα (t)(t ≥ 0) are strongly continuous in [0, ∞). Next, we recall some properties of the measure of noncompactness that will be used in the proof of our main results. We denote by ω(·) the Kuratowski measure of noncompactness
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on both the bounded sets of X and C(I, E). For more details of the measure of noncom-
pactness, see ([21,22]). For any D ⊂ C(I, X) and t ∈ I, let D(t) = {u(t)|u ∈ D} ⊂ X. If
D ⊂ C(I, X) is bounded, then D(t) is bounded in X and ω(D(t)) ≤ ω(D).
Lemma 2.2. [23] Let X be a Banach space, and let D ⊂ X be bounded. Then there exists a countable set D0 ⊂ D, such that ω(D) ≤ 2ω(D0 ).
Lemma 2.3. [21] Let X be a Banach space, and let D ⊂ C(I, E) is equicontinuous and bounded, then ω(D(t)) is continuous on I, and ω(D) = maxt∈I ω(D(t)). 5
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Lemma 2.4. [25] Let E be a Banach space, and let D = {un } ⊂ C(I, X) be a bounded and countable set. Then ω(D(t)) is the Lebesgue integral on X, and Z o n Z un (t)dt|n ∈ N ≤ 2 ω(D(t))dt. ω I
I
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Lemma 2.5. [22] Let X be a Banach space. Assume that D ⊂ X is a bounded closed and
convex set on X, Q : D → D is condensing. Then Q has at least one fixed point in D.
Lemma 2.6. [19] Suppose that a > 0, b ≥ 0 and β > 0. If u(t) is nonnegative and locally integrable on 0 ≤ t < Φ (someΦ ≤ +∞) with Z t u(t) ≤ a + b (t − s)β−1 u(s)ds
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0
on this interval, then
u(t) ≤ aEβ (bΓ(β)tβ ), 0 ≤ t < Φ, where Eβ is the Mittag-Leffler function defined by ∞ X
Eβ (z) =
3
M
n=0
zn , z ∈ C. Γ(nβ + 1)
Main results
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In this section, we prove the existence of local solutions to the problem (1) in Banach space. First of all, let us start by defining what we mean by a solution of the problem (1).
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Definition 3.1. A function u ∈ P C(I, E) is said to be a mild solution of the problem (1)
CE
if u satisfies the equation X Z u(t) = Tα (t)u0 +
tk
0
+
Z
t
(t−s)
AC
tk
α−1
α−1
(tk − s)
Z h Sα (t − s) f (s, u(s)) +
s
0
Z h Sα (t−s) f (s, u(s))+
0
s
i q(s − τ )g(τ, u(τ ))dτ ds
i X q(s−τ )g(τ, u(τ ))dτ ds+ Tα (t−tk )Ik (u(t− k )). 0
To prove our main results, we state the following basic assumptions of this paper.
(H1) The function f : I × X → X is continuous and there exists a constant N1 such that kf (t, u) − f (t, v)k ≤ N1 ku − vk
for all
u, v ∈ X.
(H2) The function g : I × X → X is continuous and there exists a constant N2 such that kg(t, u) − g(t, v)k ≤ N2 ku − vk 6
for all
u, v ∈ X.
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(H3) The functions Ik : X → X are continuous and there exists a constant µ > 0 and ρ > 0 such that
kIk (t, u)−Ik (t, v)k ≤ µku−vk and kIk (u)k ≤ ρkuk for all u, v ∈ X and k = 1, 2, . . . , m.
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(H4) For any R > 0 and a > 0, there exists positive constant Li = Li (R, a), i = 1, 2 such that for any equicontinuous and countable set D ⊂ BR = {u ∈ X : kuk ≤ R}, ω(f (t, D)) ≤ L1 ω(D),
ω(g(t, D)) ≤ L2 ω(D), t ∈ [0, a].
Theorem 3.1. Let f, g : I × X → X be a continuous function and -A be the generator of
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a C0 -semigroup (T (t))(t ≥ 0). If u ∈ P C(I, X) is a mild solution of (1) in the sense of Definition 3.1, then for any t ∈ (tk−1 , tk ], k = 1, . . . , m. Z h X Z tk u(t) = Tα (t)u0 + (tk − s)α−1 Sα (t − s) f (s, u(s)) + 0
+
Z
t
0
Z h Sα (t − s) f (s, u(s)) +
s
0
i q(s − τ )g(τ, u(τ ))dτ ds
i X q(s − τ )g(τ, u(τ ))dτ ds + Tα (t − tk )Ik (u(t− k )) 0
M
tk
(t − s)
α−1
s
is a solution of (1). In other words u(t) is a mild solution of (1).
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Proof. Using Definition 2.3, we have Z t Z h u(t) = Tα (t)u0 + (t − s)α−1 Sα (t − s) f (s, u(s)) +
PT
0
0
which leads to
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u(t1 ) = Tα (t1 )u0 +
Z
t1
0
(t1 − s)
α−1
s
i q(s − τ )g(τ, u(τ ))dτ ds, t ∈ [0, t1 ],
Z h Sα (t1 − s) f (s, u(s)) +
s
0
− − − since u(t+ 1 ) = u(t1 ) + I1 (u(t1 )) = u(t1 ) + I1 (u(t1 )), then
Z = Tα (t1 )u0 +
AC
u(t+ 1)
t1
0
(t1 −s)
Moreover, for t ∈ (t1 , t2 ], u(t) = Tα (t −
t1 )u(t+ 1)
t
t1
Z h Sα (t1 −s) f (s, u(s))+
0
α−1
(t − s)
s
i q(s−τ )g(τ, u(τ ))dτ ds+I1 (u(t− 1 )).
Z h Sα (t − s) f (s, u(s)) +
0
s
i q(s − τ )g(τ, u(τ ))dτ ds
Z s h i = Tα (t)u0 + (t1 − s)α−1 Sα (t − s) f (s, u(s)) + q(s − τ )g(τ, u(τ ))dτ ds 0 0 Z t Z s h i + (t − s)α−1 Sα (t − s) f (s, u(s)) + q(s − τ )g(τ, u(τ ))dτ ds + Tα (t − t1 )I1 (u(t− 1 )). t1
Z
+
Z
α−1
i q(s − τ )g(τ, u(τ ))dτ ds,
t1
0
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For t = t2 and t = t+ 2, Z s i h q(s − τ )g(τ, u(τ ))dτ ds (t1 − s) Sα (t2 − s) f (s, u(s)) + u(t2 ) = Tα (t2 )u0 + 0 0 Z t2 Z s h i + (t2 − s)α−1 Sα (t2 − s) f (s, u(s)) + q(s − τ )g(τ, u(τ ))dτ ds + Tα (t2 − t1 )I1 (u(t− 1 )), t1
α−1
t1
0
− − − since u(t+ 2 ) = u(t2 ) + I2 (u(t2 )) = u(t2 ) + I2 (u(t2 )), then
u(t+ 2)
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Z
Z s i h q(s − τ )g(τ, u(τ ))dτ ds (t1 − s) Sα (t2 − s) f (s, u(s)) + = Tα (t2 )u0 + 0 0 Z t2 Z s h i + (t2 − s)α−1 Sα (t2 − s) f (s, u(s)) + q(s − τ )g(τ, u(τ ))dτ ds Z
t1
α−1
t1
0
Hence for any t ∈ (t2 , t3 ], u(t) = Tα (t − t2 )u(t+ 2)+
t
t2
Z h (t − s)α−1 Sα (t − s) f (s, u(s)) +
s
0
i q(s − τ )g(τ, u(τ ))dτ ds
Z s h i α−1 = Tα (t)u0 + (t1 − s) Sα (t − s) f (s, u(s)) + q(s − τ )g(τ, u(τ ))dτ ds 0 0 Z t2 Z s h i + (t2 − s)α−1 Sα (t − s) f (s, u(s)) + q(s − τ )g(τ, u(τ ))dτ ds +
t2
Z h (t − s)α−1 Sα (t − s) f (s, u(s)) + t1 )I1 (u(t− 1 ))
0
0
s
i q(s − τ )g(τ, u(τ ))dτ ds
+ Tα (t − t2 )I2 (u(t− 2 )).
PT
+ Tα (t −
ED
t1 t
Z
t1
M
Z
Z
+
I2 (u(t− 2 )).
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+ Tα (t2 −
t1 )I1 (u(t− 1 ))
CE
By repeating the same procedure, we can easily deduce that Z h X Z tk u(t) = Tα (t)u0 + (tk − s)α−1 Sα (t − s) f (s, u(s)) +
0
0
t
(t − s)
AC
+
Z
tk
α−1
s
Z h Sα (t − s) f (s, u(s)) +
s
0
holds for any t ∈ (tk−1 , tk ], k = 1, . . . , m.
i q(s − τ )g(τ, u(τ ))dτ ds
i X q(s − τ )g(τ, u(τ ))dτ ds + Tα (t − tk )Ik (u(t− k )) 0
2
Theorem 3.2. Let X be a Banach space. A : D(A) ⊂ X → X be a closed linear operator and −A generates an equicontinuous C0 -semigroup T (t)(t ≥ 0) of uniformly bounded
operators in X. Suppose that the conditions (H1)-(H4) are satisfied. Then for every u0 ∈ P C(I, X) there exists a τ1 = τ1 (u0 ), 0 < τ1 < T such that the problem (1) has a mild solution u ∈ P C([0, τ1 ], X).
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Proof. Since we are interested here only in local solutions, we may assume that T < ∞.
By using our assumption H1, H2 and H3 and let t0 > 0, R = M (2ku0 k + 1) > 0 be such that BR (u0 ) = {u : ku − u0 k ≤ R} and kf (t, u)k ≤ N1 , kg(t, u)k ≤ N2 for 0 ≤ t ≤ t0 and u ∈ BR (u0 ) and let choose
where q ∗ = sup0≤t≤T
Rt 0
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n h Γ(α + 1)(kx k + 1 − mρ ) i 1 h i1o Γ(α + 1) α α 0 1 0 τ0 = min t , T, , , (m + 1)(N1 + N2 q ∗ ) 4M (m + 1)(L1 + L2 q ∗ ) + 1
(2)
kq(t − s)kds. Set Ω = {u ∈ P C([0, τ1 ], X) : ku(t)k ≤ R, t ∈ [0, τ1 ]},
then Ω is a closed ball in P C([0, τ1 ], X) with center θ and radius R. Consider the operator Q : Ω → P C([0, τ1 ], X) defined by tk
0
+
Z
t
tk
α−1
(tk − s)
Z h (t − s)α−1 Sα (t − s) f (s, u(s)) +
0
Z h Sα (t − s) f (s, u(s)) +
0
s
s
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(Qu)(t) = Tα (t)u0 +
X Z
i q(s − τ )g(τ, u(τ ))dτ ds
i X q(s − τ )g(τ, u(τ ))dτ ds + Tα (t − tk )Ik (u(t− k )) 0
For any u ∈ Ω and t ∈ [0, τ1 ], by Lemma 2.1(i), we have
ED
M
Z tk αM (N1 + N2 q ∗ ) X (tk − s)α−1 ds k(Qu)(t)k ≤ M ku0 k + Γ(α + 1) 0
+ M mρ1 ≤ R.
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≤ M ku0 k +
0
q ∗ )τ α
Therefore, Qu ∈ Ω. Now we show that Q is continuous from Ω into Ω. To show this, we first observe that since f and g are continuous in [0, T ] × X, it follows that for any > 0
CE
and for a fixed u ∈ BR (u0 ) there exists δ1 (u), δ2 (u) > 0 such that for any v ∈ BR (u0 ) and
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let δ(u) = min{δ1 (u), δ2 (u)}. Then for any v ∈ Ω, ku − vk < δ(u) and choose M (m + 1)τ α N
1
Γ(α + 1)
+
M (m + 1)τ α N2 q ∗ + M mµ < . Γ(α + 1) δ(u)
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Then we have k(Qu)(t) − (Qv)(t)k ≤
+ +
tk Z t
(t − s)
X
(tk − s)α−1 kSα (t − s)kkf (s, u(s)) − f (s, v(s))kds
kSα (t − s)k
Z
s
0
|q(s − τ )|kg(τ, u(τ )) − g(τ, v(τ ))kdτ ds
kSα (t − s)kkf (s, u(s)) − f (s, v(s))kds
(t − s)α−1 kSα (t − s)k kTα (t −
0
≤
α−1
(tk − s)
0
+
tk
0
Z
s
0
tk )kkIk (u(t− k ))
M (m + 1)τ α N
1
Γ(α + 1)
|q(s − τ )|kg(τ, u(τ )) − g(τ, v(τ ))kdτ ds
− Ik (v(t− k ))k
M (m + 1)τ α N2 q ∗ + M mµ ku − vk ≤ . Γ(α + 1)
+
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+
X Z
tk
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X Z
Thus, we proved that Q : Ω → Ω is a continuous operator.
Now, we demonstrate that the operator Q : Ω → Ω is equicontinuous. For any u ∈ Ω
and 0 ≤ t1 < t2 ≤ τ1 , we get that
Z
M
k(Qu)(t2 ) − (Qu)(t1 )k ≤ kTα (t2 )u0 − Tα (t1 )u0 k + t1
X
0
k(Tα (t2 − tk ) − Tα (t1 − tk ))Ik (u− k )k
ED
k(t1 − s)α−1 Sα (t1 − s) − (t2 − s)α−1 Sα (t2 − s)k Z s · kf (s, u(s)) + q(s − τ )g(τ, u(τ ))dτ kds 0 Z t2 Z s α−1 + (m + 1) (t2 − s) kSα (t2 − s)k · kf (s, u(s)) + q(s − τ )g(τ, u(τ ))dτ kds + (m + 1)
PT
0
t1
CE
≤ kTα (t2 )u0 − Tα (t1 )u0 k +
X
0
0
k(Tα (t2 − tk ) − Tα (t1 − tk ))Ik (u− k )k + J1 + J2 ,
AC
where
J1 = (m+1)
Z
0
t1
α−1
k(t1 −s)
∗
α−1
Sα (t1 −s)−(t2 −s)
≤ (m + 1)(N1 + N2 q )
Z
0
t1
Z Sα (t2 −s)kkf (s, u(s))+
0
s
q(s−τ )g(τ, u(τ ))dτ kds
k(t1 − s)α−1 Sα (t1 − s) − (t2 − s)α−1 Sα (t2 − s)kds.
Since (t1 − s)α−1 Sα (t1 − s) − (t2 − s)α−1 Sα (t2 − s) → 0 as t1 → t2 because Sα (·) is strongly
10
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continuous this implies that limt1 →t2 J1 = 0 and Z t2 Z α−1 J2 = (m + 1) (t2 − s) kSα (t2 − s)k · kf (s, u(s)) + t1
0
q(s − τ )g(τ, u(τ ))dτ kds
M (m + 1)(N1 + N2 q ∗ )(t2 − t1 )α Γ(α + 1)
→0
CR IP T
≤
s
as t2 − t2 → 0.
As a result, k(Qu)(t2 ) − (Qu)(t1 )k tends to 0 independently of u ∈ Ω as t2 − t1 → 0, which
means that Q : Ω → Ω is equicontinuous.
Let B = coQ(Ω). Then it is easy to verify that Q maps B into itself and B ⊂ P C(I, E)
is equicontinuous. Now, we prove that Q : B → B is a condensing operator. For any
AN US
D ⊂ B, by Lemma 2.2, there exists a countable set D0 = {un } ⊂ D, such that ω(Q(D)) ≤ 2ω(Q(D0 )).
(3)
By the equicontinuity of B, we know that D0 ⊂ B is also equicontinuous. Therefore, by
Z
M
Lemma 2.4, the assumption (H4), we have n X Z tk ω(Q(D0 )(t)) = ω Tα (t)u0 + (tk − s)α−1 Sα (t − s) 0
i q(s − τ )g(τ, un (τ ))dτ ds f (s, un (s)) + 0 Z t Z s h i α−1 + (t − s) Sα (t − s) f (s, un (s)) + q(s − τ )g(τ, un (τ ))dτ ds tk 0 o X − + Tα (t − tk )Ik (u(tk )) s
PT
ED
h
0
Z s n h io Sα (t − s) f (s, un (s)) + q(s − τ )g(τ, un (τ ))dτ ds 0 0 Z 2αM (m + 1)(L1 + L2 q ∗ ) t ≤ (t − s)α−1 ω(D0 (s))ds Γ(α + 1) 0 2M (m + 1)(L1 + L2 q ∗ )τ α · ω(D). (4) ≤ Γ(α + 1)
AC
CE
≤ 2(m + 1)
Z
t
(t − s)α−1 ω
Since Q(D0 ) ⊂ B is bounded and equicontinuous, we know from Lemma 2.3 that ω(Q(D0 )) = max ω(Q(D0 )(t)). t∈I
Therefore, from (2),(3),(4) and (5), we know that ω(Q(D)) ≤
4M (m + 1)(L1 + L2 q ∗ )τ α · ω(D) ≤ ω(D). Γ(α + 1) 11
(5)
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Thus, Q : B → B is a condensing operator. It follows from Lemma 2.5 that Q has at least
one fixed point u(t0 ) ∈ B, which is just a mild solution of the problem (1) on the interval [0, τ1 ].
Global existence
CR IP T
4
2
In this section we consider the global existence of mild solution to the problem (1).
Theorem 4.1. Assume the hypotheses of Theorem 3.2. Let f, g : [0, ∞] × X → X are
continuous and maps bounded sets in [0, ∞] × X into bounded sets in X, for every u0 ∈ X,
tmax < ∞ then limt→tmax kuk = ∞.
AN US
then the problem (1) has a mild solution u on a maximal interval of existence [0, tmax ). If
Proof. From the local existence of mild solutions we have just proved it follows that there exists an interval [0, τ0 ] such that the problem (1) has a mild solution u ∈ P C([0, τ0 ], E),
and it can be extended to a large interval [0, τ0 + h] with h > 0 by defining u(t) = v(t) on
M
[τ0 , τ0 + h], where v(t) is the mild solution of the initial value problem c α Dt v(t) + Av(t) = f (t, v(t)), t ∈ [τ0 , τ0 + h], ∆v|t=tk = Ik (v(t− k = 1, . . . , m, k )), v(0) = u(h ),
ED
0
where h depends only on ku(h0 )k and R. Therefore, repeating the above procedure and
PT
using the methods of steps, we can prove that there exists a maximal interval [0, tmax ) such that u ∈ P C([0, tmax ), E) is a mild solution of the problem (1).
Next, we show that if tmax < ∞ then limt→t− ku(t)k = ∞. To do so we first prove that max
CE
tmax < ∞ implies lim supt→t− ku(t)k = ∞. Indeed, if tmax < ∞ and lim supt→t− ku(t)k < max max ∞, then there exists a constant 0 < R(tmax) < ∞ such that sup0≤t
AC
Denote N1 (tmax ) = sup{kf (t, u(t))k : ku(t)k ≤ R(tmax ), 0 ≤ t ≤ tmax + 1}. For 0 < t1 <
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t2 < tmax , if 0 < ρ1 < t1 < t2 < tmax , then we have
0
Z
s
q(s − τ )g(τ, u(τ ))dτ kds × kf (s, u(s)) + 0 Z X Z tk k[(tk − s)α−1 Sα (t2 − s)[f (s, u(s)) + + t1
0
tk−1
q(s − τ )g(τ, u(τ ))dτ ]kds
α−1
Sα (t2 − s) − (t1 − s)
Z
0
s
q(s − τ )g(τ, u(τ ))dτ kds
AN US
Sα (t1 − s)]k × kf (s, u(s)) + 0 Z t2 Z s + k[(t2 − s)α−1 Sα (t2 − s)[f (s, u(s)) + q(s − τ )g(τ, u(τ ))dτ ]kds t1 0
X
X
− + Tα (t2 − tk )Ik (u(t− )) − T (t − t )I (u(t )) α 1 k k k k
+
k[(t1 − s)
α−1
s
CR IP T
ku(t2 ) − u(t1 ) ≤ kTα (t2 )u0 − Tα (t1 )u0 k X Z tk k[(tk − s)α−1 Sα (t2 − s) − (t1 − s)α−1 Sα (t1 − s)]k +
0
0
M
Z αM (m + 1)(N1 + N2 q ∗ ) t2 ≤ kTα (t2 )u0 − Tα (t1 )u0 k + (t2 − s)α−1 ds Γ(α + 1) 0 Z Z αM (m + 1)(N1 + N2 q ∗ ) t1 αM (m + 1)(N1 + N2 q ∗ ) t2 + (t1 − s)α−1 ds + (t2 − s)α−1 ds Γ(α + 1) Γ(α + 1) 0 t1
X
X
− + (Tα (t2 − tk ) − Tα (t1 − tk ))Ik (u(tk )) + Tα (t2 − tk )Ik (u(t− )) k t
ED
0
t
[tα2 − tα1 ] + mρ1 kTα (t2 ) − Tα (t1 )kkuk
PT
M (m + 1)(N1 + N2 ≤ kTα (t2 )u0 − Tα (t1 )u0 k + Γ(α + 1)
X
+ Tα (t2 − tk )Ik (u(t− k )) kuk.
q∗)
By using a similar method which we used in proving the equicontinuous of the operator Q,
CE
we can prove that ku(t2 ) − u(t1 )k → 0 as t1 , t2 → t− max . Therefore, by Cauchy criteria we know that limt→t− u(t) exists. Let limt→t− u(t) = x1 . By the local existence of mild max max
AC
solutions we proved above we know that the fractional evolution equation c α Dt u(t) + Au(t) = f (t, u(t)) t ≥ tmax , ∆u|t=tk = Ik (u(t− k )), u(t max ) = x1
k = 1, . . . , m,
has a mild solution on [tmax , tmax + h], where h > 0 is constant. This means that the mild solution u of the problem (1) can be extended beyond tmax , which contradicts with u ∈ P C([0, tmax ), E) being a mild solution of the problem (1). Therefore, the assumption
ku(t)k = ∞. To conclude the proof we show that tmax < ∞ implies that lim supt→t− max 13
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tmax < ∞ implies limt→t− ku(t)k = ∞. If this is not true, then there exists a constant max
R > 0 and a sequence tn → tmax such thatku(tn )k ≤ R for all n. Let C = sup{kf (t, u(t))k :
ku(t)k ≤ M (R + 1), 0 ≤ t ≤ tm ax}. By the fact that t → ku(t)k is continuous and lim supt→t− ku(t)k = ∞ one can find a sequence {hn }(0 < hn < tmax − tn ) such that max calculations, we have
M (R + 1) = ku(tn + hn )k ≤ kTα (hn )u(tn )k +
× kf (s, u(s)) +
q(s − τ )g(τ, u(τ ))dτ kds +
0
Z
0
tn
s
s
Z
tk
q(s − τ )g(τ, u(τ ))dτ kds + q∗)
tn +hn
tk
tk−1
k(tk − s)α−1 Sα (tn + hn − s)k
k(tn + hn − s)α−1 Sα (tn + hn − s)k
X
tn
M (m + 1)(N1 + N2 hαn + M mρ1 kuk Γ(α + 1) M (m + 1)(N1 + N2 q ∗ ) α ≤ M R(1 + mρ1 ) + hn Γ(α + 1) m M (m + 1)(N1 + N2 q ∗ ) α ≤ MR 1 + + hn 2mR Γ(α + 1) M M (m + 1)(N1 + N2 q ∗ ) α = MR + + hn 2 Γ(α + 1) M → MR + as n → ∞. 2
)) Tα (tn + hn − tk )Ik (u(t− k
ED
M
≤ M kuk +
Z
AN US
× kf (s, u(s)) +
Z
X
CR IP T
ku(t)k ≤ M (R + 1) for tn ≤ t ≤ tn + hn and ku(tn + hn )k = M (R + 1). By direct
PT
From this we conclude that M ≤ 0, which contradicts with the definition of M. Therefore,
we have proved that if tmax < ∞ then limt→t− ku(t)k = ∞. This completes the proof of max Theorem 4.1. 2
CE
Next, we discuss the existence of a global mild solution for the problem (1). To this end, we need to replace the assumption (H1) by the following assumption
AC
(H1’) There exist two functions C1 , C2 ∈ C([0, ∞), [0, ∞)) such that kf (t, u)k ≤ C1 (t) + C2 (t)kuk for 0 ≤ t < ∞, u ∈ X.
(H2’) There exist two functions C3 , C4 ∈ C([0, ∞), [0, ∞)) such that kg(t, u)k ≤ C3 (t) + C4 (t)kuk for 0 ≤ t < ∞, u ∈ X. Theorem 4.2. Let E be a real Banach space, A : D(A) ⊂ E → E be a closed linear oper-
ator and −A generates an equicontinuous C0 -semigroup T (t)(t ≥ 0) of uniformly bounded 14
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operators in X. Assume that the nonlinear term f, g : [0, ∞) × X → X is continuous and satisfies the assumptions (H1’),(H2’),(H3) and (H4), then for every u0 ∈ X the problem (1) has a global mild solution u ∈ P C([0, ∞), X).
Proof. It is obvious that (H1’) ⇒ (H1), (H2’) ⇒ (H2). Therefore, by Theorem 4.1 we know
CR IP T
that the problem (1) has a mild solution u on a maximal interval of existence [0, tmax ). If tmax = ∞, then the result has been proved. If tmax < ∞, by the proof process of Theorem 4.1 one can see that the problem (1) has a global mild solution if u(t) is bounded for every
t in the interval of existence of u. Therefore, in order to prove Theorem 4.2, we only need to prove that u(t) is bounded for every t ∈ [0, tmax ), (tmax < ∞). Let sup
C1 (t),
C2 =
t∈[0,tmax ]
sup t∈[0,tmax ]
C2 (t), C3 =
sup
C3 (t),
t∈[0,tmax ]
AN US
C1 =
C4 =
sup
C4 (t).
t∈[0,tmax ]
By Definition 3.1, Lemma 2.1(i) and the assumptions (H1)’ and (H2)’ we know that for any t ∈ [0, tmax ),
M (m + 1)(C1 + C3 q ∗ )tαmax Γ(α + 1) Z αM (m + 1)(C2 + C4 q ∗ ) t + (t − s)α−1 ku(s)kds + M mρ1 Γ(α + 1) 0 Z t := K1 + K2 (t − s)α−1 ku(s)kds,
M
k(u)(t)k ≤ M ku0 k +
K1 = M ku0 k +
ED
0
where
M (m + 1)(C1 + C3 q ∗ )tαmax + M mρ1 , Γ(α + 1)
K2 =
αM (m + 1)(C2 + C4 q ∗ ) . Γ(α + 1)
PT
By Lemma 2.6, we have
ku(t)k ≤ K1 Eα (K2 Γ(α)tα ),
CE
which implies that u(t) is bounded for every t ∈ [0, tmax ). This completes the proof of
Theorem 4.2.
Applications
AC
5
2
In this section, we present two examples, which illustrate the applicability of our main
results.
Throughout this section, we let X = L2 ([0, π]) and consider the operator A : D(A) ⊆
X → X defined by
n o ∂2 ∂u ∂ 2 u , 2 ∈ X, u(0) = u(π) = 0 , Au = − 2 u. D(A) = u ∈ X : ∂x ∂x ∂x It is well known that −A generates a uniformly bounded C0 semigroup (T (t))t > 0 in X. 15
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CR IP T
Example 5.1. We consider the following impulsive fractional differential equation α R t t−s 2e−t ku(x,t)k ∂2 es ∂ ds, t ∈ [0, 1], x ∈ (0, π), t 6= 21 , ∂tα u(x, t) − ∂x2 u(x, t) = 1+ku(x,t)k + 0 e 5+ku(x,s)k u(0, t) = u(π, t), t ≥ 0, 1 1− ∆u| ku(x, )k , 1 = sin t= 7 2 2 u(x, 0) = u (x), 0
(6)
where
∂α ∂tα is
L2 [0, π].
by
the Caputo fractional order partial derivative of order α, 0 < α ≤ 1, u0 ∈
Let u, v ∈ C([0, 1], X), define the operators f : I ×X → X, g : I ×X → X and q : I → X 2e−t ku(x, t)k , 1 + ku(x, t)k
AN US
et , 5 + ku(x, t)k 1 q(t − s) = et−s , Ik (u) = sin ku(x, t)k , 7 then the impulsive fractional differential equation (6) can be transformed into the abstract f (t, u) =
form of problem (1).
M
Next, we calculate
g(t, u) =
u v ku − vk
− ≤ 2e−t ku − vk ≤ 2ku − vk, kf (t, u) − f (t, v)k = 2e−t
= 2e−t 1+u 1+v (1 + u)(1 + v)
ED
hence the condition H1 holds with N1 = 2.
1 1 et ku − vk e
kg(t, u) − g(t, v)k = et − ≤ ku − vk,
= 5+u 5+v (5 + u)(5 + v) 25
and
CE
PT
e hence the condition H2 holds with N2 = 25 . Then we have
1 1
1 kIk (u) − Ik (v)k = sin u − sin v ≤ ku − vk 7 7 7
AC
1
1 kIk (u)k = sin u ≤ kuk. 7 7 Hence the condition (H3) holds with µ = ρ1 = 71 . Let t ∈ I, then we have Z t Z t q ∗ = sup |q(t − s)|ds = e(t−s) ds = et − 1 ≤ e − 1. t∈I
0
0
Then by Theorem 3.2 the problem (1) has a unique solution on [0, 1]. From the definition of nonlinear terms f, g , we can easily verify that the assumptions (H1’),(H2’),(H3) and (H4) are satisfied with C1 (t) = C2 (t) = 2 and C3 (t) = C4 (t) =
e 5.
Therefore, by Theorem 4.2, we deduce that equation (6) has a global solution u ∈ P C([0, 1]×
[0, ∞)).
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Example 5.2. Consider the following impulsive fractional differential equation R t −|t−s| cos(tx) ∂2 ∂α cos(u(s, x))ds, t ∈ [0, 1], x ∈ [0, π], t 6= 21 , ∂t α u(x, t) − ∂x2 u(x, t) = 1+ku(x,t)k + 0 e u(0, t) = u(π, t), t ≥ 0, 1−
where
∂α ∂tα is
L2 [0, π].
(7)
the Caputo fractional order partial derivative of order α, 0 < α ≤ 1, u0 ∈
Let u, v ∈ C([0, 1], X), define the operators f : I ×X → X, g : I ×X → X and q : I → X f (t, u) =
AN US
by
CR IP T
|u(x, 2 )| ∆u|t= 1 = − , 20+|u(x, 12 )| 2 u(x, 0) = u (x), 0
cos(tx) , 1 + ku(x, t)k
q(t − s) = e
−|t−s|
,
g(t, u) = cos(u(s, x)),
Ik (u) =
−
|u(x, 21 )|
−
20 + |u(x, 21 )|
,
then the impulsive fractional differential equation (7) can be transformed into the abstract
M
form of problem (1).
From the definition of nonlinear terms f, g , we can easily verify that the assumptions
ED
(H1’) and (H2’) are satisfied with C1 (t) = C2 (t) = C3 (t) = C4 (t) = π. Moreover, kIk (u) − Ik (v)k = |
PT
and
v 20|u − v| 1 u − |= ≤ |u − v| 20 + u 20 − v (20 + u)(20 − v) 20 kIk (u)k = |
u 1 |≤ u + 20 20
AC
CE
Hence the condition (H3) holds with µ = ρ1 = q1 =
Z
0
1
1 20 .
Let t ∈ I, then we have
1 e−|t| dt = 1 − . e
Therefore, by Theorem 4.2, we deduce that equation (7) has a global solution u ∈ P C([0, 1]×
[0, ∞)).
Acknowledgment The authors are grateful to the editorial board members and the referee for some useful suggestions.
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M
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ED
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PT
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CE
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AC
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delay in Banach spaces, J. Franklin Inst. 349 (2012) 1-24. [20] J. Wang, Y. Zhou, M. Feckan, Abstract Cauchy problem for fractional differential
PT
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