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Preprint submitted on 28 Apr 2020
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equations systems
José Vanterler da Costa Sousa, Donal O’Regan, E Capelas de Oliveira
To cite this version:
José Vanterler da Costa Sousa, Donal O’Regan, E Capelas de Oliveira. On attractivity for ψ-Hilfer fractional differential equations systems. 2020. �hal-02556357�
J. Vanterler da C. Sousa 1 [email protected]
Donal O’Regan2 [email protected]
E. Capelas de Oliveira3 [email protected]
1,3 Department of Applied Mathematics, Imecc-Unicamp, 13083-859, Campinas, SP, Brazil.
3 School of Mathematics, Statistics and Applied Mathematics, National University of Ireland, H91 TK33 Galway, Ireland.
Abstract
In this paper, we investigate the existence of a class of globally attractive solutions of the Cauchy fractional problem with theψ-Hilfer fractional derivative using the measure of noncompactness. An example is given to illustrate our theory.
Key words: Fractional differential equations, ψ-Hilfer fractional derivative, attractivity, measure noncompactness.
2010 Mathematics Subject Classification: 26A33, 34A12, 34A08, 34G20, 34GXX.
1 Introduction
Consider the following Cauchy fractional problem
( HD0ν,η;ψ+ θ(t) = u(t, θ(t)), t∈(0,∞) I01−γ;ψ+ θ(0) = θ0
(1.1) whereHD0ν,η;ψ+ θ(·) is theψ-Hilfer fractional derivative of order 0< ν <1 and type 0≤η≤1, I01−γ;ψ+ θ(·) is the fractional integral of order γ, with 0 ≤ γ < 1 with respect to another function, u : [0,∞)×Ω→ Ω is a continuous function satisfying some conditions and θ0 is a element of the Banach space.
The theory of fractional differential equations can be found in for example [1, 5, 12, 18, 19, 22, 24, 27, 31]. Existence, uniqueness and Ulam-Hyers stabilities of solutions of differ- ential and integrodifferential equations was studied using the ψ-Hilfer fractional derivative in [17, 23, 18, 19, 20, 22, 24, 26]. In 2013 Hernandez et al. [14], proposed a different approach to abstract fractional differential equations if one considers the existence of non- local mild solutions. In 2018, Sousa and Oliveira [18], investigated the Ulam-Hyers stability of an fractional integrodifferential equation using the Banach fixed point theorem and in 2019, Liu et al. [15] considered the ψ-Hilfer fractional derivative, and investigated the Ulam-Hyers stability of a fractional delay differential equation. We also refer the reader
1
to [4, 3, 11, 13, 27, 26, 25]. Attractivity of mild solutions of fractional differential and integrodifferential equations was considered in [1, 2, 10, 11, 14, 16, 28]. Chang et al. [8], investigated the asymptotic decay of some operators via fixed point theorems and they considered the existence and uniqueness for a class of mild solutions of Sobolev fractional differential equations. In 2008 Banas and O’Regan [6] investigated the existence and local attractivity of solutions of a quadratic Volterra integral equation of fractional order in Ba- nach spaces and in 2012 Chen et al. [9] considered the global attractivity of solutions of fractional differential equations in the Riemann-Liouville fractional derivative sense, using the Krasnoselskii fixed-point theorem and the Schauder fixed point theorem. Motivated by the above we will investigate the existence of globally attractivity solutions to the ψ-Hilfer Cauchy fractional problem (1.1) (paying attention to some particular cases of the ψ-Hilfer fractional derivative) .
This paper is organized as follows. In section 2 we present the definitions of the ψ- Riemann-Liouville fractional integral and the ψ-Hilfer fractional derivative and some im- portant results. In section 3 we investigate the globally attractivity existence of solutions of the Cauchy fractional problem. An example is given to illustrate our results.
2 Preliminaries
Let J = [a, b] (−∞< a < b <+∞) be a finite interval of Rand Ω a Banach space.
The space C(J,Ω) of continuous functionsθ on J has the norm given by [18, 21]
kθk:= sup
t∈J
|θ(t)|. We have n-times absolutely continuous functions given by
ACn(J,Ω) ={u:J 7−→Ω, u(n−1) ∈ AC(J,Ω)}.
In particular, AC1(J,Ω) =AC(J,Ω).
The weighted space Cγ,ψ(J,Ω) of functions θ on (a, b] is defined by [18, 21]
Cγ,ψ(J,Ω) ={θ: (a, b]→Ω,(ψ(t)−ψ(a))γθ(t)∈ C(J,Ω)}
with 0≤γ <1 and the norm is given by kθkC
γ,ψ(J,Ω) = k(ψ(t)−ψ(a))γθ(t)kC
γ,ψ(J,Ω)
= max
t∈J |(ψ(t)−ψ(a))γθ(t)| · The weighted space Cγ,ψn (J,Ω) of functions θ on (a, b] is defined by
Cγ,ψn (J,Ω) ={θ: (a, b]→Ω, θ(t)∈ Cn−1(J,Ω), θ(n) ∈ Cγ,ψ(J,Ω)}
with 0≤γ <1 and the norm is given by kθkCn
γ,ψ(J,Ω)=
n−1
X
k=0
θ(k)
C(J,Ω)+ θ(n)
C
γ,ψ(J,Ω)
·
For n= 0, we haveCγ,ψ0 (J,Ω) =Cγ,ψ(J,Ω) and Cγ,ψν,η(J,Ω) =n
θ∈ Cγ,ψ(J,Ω), HDaν,η;ψ+ θ∈ Cγ,ψ(J,Ω)o with γ=ν+η(1−ν).
Let (a, b) (−∞ ≤a < b≤+∞) and ν >0. Also letψ(t) be an increasing and positive monotone function on (a, b], having a continuous derivativeψ0(t) on (a, b). Theψ-Riemann- Liouville fractional integral (left-sided) of a function θ with respect to another function ψ on [a, b] is defined by [17, 23]
Iaν,ψ+ θ(t) = 1 Γ(ν)
Z t a
Θν−1ψ (s, t)θ(s) ds. (2.1) where Θν−1ψ (s, t) :=ψ0(s)(ψ(t)−ψ(s))ν−1. Similarly one can define theψ-Riemann-Liouville fractional integral (right-sided).
Let n−1 < ν < n, with n∈N, J = [a, b] is an interval such that−∞ ≤ a < b≤+∞
and θ, ψ ∈ Cn(J,R) are two functions such that ψ is increasing andψ(t) 6= 0, for all t∈J. Theψ-Hilfer fractional derivative (left-sided), denoted byHDaν,η;ψ+ (·) of a functionθof order ν and type 0≤η ≤1, is defined by [17, 23]
HDaν,η;ψ+ θ(t) =Iaη(n−ν);ψ+ 1
ψ0(t) d dt
n
I(1−η)(n−ν);ψ
a+ θ(t) (2.2)
where Iaν;ψ+ (·) is the fractional integral given in (2.1). Similarly one can define the ψ-Hilfer fractional derivative (right-sided).
Proposition 2.1 [17] Let ν >0 and δ >0. If θ(t) = (ψ(t)−ψ(0))δ−1 then Iν;ψ
0+ θ(t) = Γ(δ)
Γ(ν+δ)(ψ(t)−ψ(0))ν+δ−1· (2.3) Proposition 2.2 [17] Let ν >0, then
HD0ν,η;ψ+ (ψ(t)−ψ(0)) = 0 (2.4)
with HD0ν,η;ψ+ (·) is theψ-Hilfer fractional derivative.
Assumes that the operator u : [0,∞)×Ω → Ω is continuous. The Cauchy fractional problem (1.1) is equivalent to the integral Volterra equation,
θ(t) = (ψ(t)−ψ(0))γ−1
Γ(γ) θ0+ 1 Γ(ν)
Z t 0
ψ0(s)(ψ(t)−ψ(s))ν−1u(s, θ(s)) ds (2.5) with t >0.
Let Cγ,ψ0 ([t0,∞),Ω) ={θ∈ Cγ,ψ([t0,∞),Ω); limt→∞|θ(t)|= 0} · Note Cγ,ψ0 ([0,∞),Ω) is a Banach space. We need also the following generalized Arzel`a-Ascoli theorem [29].
Lemma 2.3 [29]The setH ⊂ C0([0,∞),Ω)is relatively compact if and only if the following conditions hold:
1. for any T >0, the function inH are equicontinuous on[0, T];
2. for any t∈[0,∞), H(t) ={θ(t) :θ∈ H}is relatively compact in Ω;
3. lim
t→∞|θ(t)|= 0 uniformly forθ∈ H.
Lemma 2.4 [12, 33] The noncompact measure µ(·) satisfies:
1. If for all bounded subsets B1, B2 of Ωimplies µ(B1)≤µ(B2);
2. If µ({x} ∪B) =µ(B) for everyx∈Ωand every nonempty subset B ⊆Ω;
3. µ(B) = 0 if and only if B is relatively compact inΩ;
4. µ(B1+B2)≤µ(B1) +µ(B2), where B1+B2 ={x+y;x∈B1, y ∈B2};
5. µ(B1∪B2)≤max{µ(B1), µ(B2)};
6. µ(λB)≤ |λ|µ(B) for anyλ∈R. For any W ⊂C(J,Ω), we define
Z t 0
W(s)ds= Z t
0
u(s)ds:u∈W
, for t∈J. (2.6)
Property 2.5 [12, 33]IfW ⊂ C(J,Ω)is bounded and equicontinuous, themcoW ⊂ C(J,Ω) is also bounded and equicontinuous.
Property 2.6 [12, 33] If W ⊂ C(J,Ω) is bounded and equicontinuous, then t→ µ(W(t)) is continuous on J, and
µ(W) = max
t∈Jµ(W (t)), µ Z t
0
W(s)ds
≤ Z t
0
µ(W(s))ds, for t∈J.
Property 2.7 [12, 33]Let {un}∞n=1 be a sequence of Bochener integrable functions from J in to Ω with |un(t)| ≤ me (t) for almost all t ∈ J and every n ≥ 1, where me ∈ L(J,R+), then the function Φ (t) =e µ({un(t)}n=1∞ ) belongs toL(J,R+) and satisfies
µ Z t
0
un(s)ds:n≥1
≤2 Z t
0
Φ (s)e ds.
Property 2.8 [12, 33]IfW is bounded, then for eachε >0, there is a sequence{un}∞n=1 ⊂ W, such that
µ(W)≤µ({un(t)}∞n=1) +ε.
3 Attractivity with ψ-Hilfer fractional derivative
In this section, we will first discuss two important results, namely, Lemma 3.1 and Lemma 3.4. Then we investigate the existence of attractive solutions of the Cauchy fractional problem.
Now we introduce the following hypothesis:
(C1)|u(t, θ)| ≤L(ψ(t)−ψ(0))−ξ1|θ|ξ2 fort∈(0,∞) andθ∈Ω, L≥0,ν < ξ1 <1 and ξ2 ∈R;
(C2) There exists a constant k >0 such that for any bounded set E⊂Ω µ(u(t, E))≤kµ(E);
here µ(·) denotes the Hausdorff measure of non compactness.
For allθ∈ Cγ,ψ([0,∞),Ω) and for a given n∈N+, we define the operatorT by T (θ) (t) =T1(θ) (t) +T2(θ) (t)
where
T1(θ) (t) =
(ψ(t)−ψ(0)) + 1 n
γ−1
θ0
Γ (γ) (3.1)
and
T2(θ) (t) = 1 Γ (ν)
Z t
0
Θν−1ψ (s, t)u(s, θ(s))ds, (3.2) with t∈[0,∞).
As 0 < ν < ξ2 < 1, we can choose ξ > 0 small enough such that ν +ξ −1 < 0, 1−ξ1−ξξ2>0 andν+ξ−ξ1−ξξ2<0. Note that
|(Tθ)(t)|
=
(ψ(t)−ψ(0)) +n1γ−1
Γ(γ) θ0+ 1
Γ(ν) Z t
0
Θν−1ψ (s, t)u(s, θ(s))ds
(3.3)
≤
(ψ(t)−ψ(0)) +n1γ−1
Γ(γ) |θ0|+ 1 Γ(ν)
Z t 0
Θν−1ψ (s, t)|u(s, θ(s))|ds
≤
(ψ(t)−ψ(0)) + 1 n
γ−1
|θ0| Γ(γ) + 1
Γ(ν) Z t
0
Θν−1ψ (s, t)L(ψ(s)−ψ(0))−ξ1|θ(s)|ξ2ds
≤
(ψ(t)−ψ(0)) + 1 n
γ−1 |θ0| Γ(γ) + 1
Γ(ν) Z t
0
Θν−1ψ (s, t)L(ψ(s)−ψ(0))−ξ1−ξξ2ds
≤
(ψ(t)−ψ(0)) + 1 n
γ+ξ−1
|θ0| Γ(γ) + L
Γ(ν) Z t
0
Θν−1ψ (s, t)(ψ(s)−ψ(0))−ξ1−ξξ2ds .
Choosing δ=ξξ2−ξ1+ 1 and substituting in (2.3) (Proposition 2.1), we get I0ν,ψ+ θ(t) = Γ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2)(ψ(t)−ψ(0))ν−ξ1−ξξ2·
Then, choosing T >0 sufficiently large, from (3.3), we have
|T(θ)(t)|
≤
(ψ(t)−ψ(0)) + 1 n
γ+ξ−1
|θ0|
Γ(γ) +L Γ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2)(ψ(t)−ψ(0))ν−ξ1−ξξ2
≤
(ψ(t)−ψ(0)) + 1 n
γ+ξ−1
|θ0|
Γ(γ) +L Γ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2)(ψ(t)−ψ(0))ν−ξ1−ξξ2+ξ
≤ 1 (3.4)
for all t≥T.
Define a set Qξ;ψ as follows Qξ;ψ =
n
θ(t)|θ∈ Cγ,ψ([0,∞),Ω);
(ψ(t)−ψ(0))ξθ(t)
≤1, t≥T o
. (3.5)
Note that by choosingψ(t) =t in (3.5), we have Qξ =Qξ,t =n
θ(t)|θ∈ Cγ,t([0,∞),Ω);
tξθ(t)
≤1, t≥To
and these sets are particular cases of fractional derivatives (namely Riemann-Liouville and Caputo).
Choosing ψ(t) =tρ, ρ >0 in (3.5), we get Qξ=Qξ,tρ =n
θ(t)|θ∈ Cγ,tρ([0,∞),Ω);
tξρθ(t)
≤1, t≥To
and these sets are particular cases of fractional derivatives (namely Katugampola and Caputo-type).
Note Qξ;ψ 6=∅ andQξ;ψ is a closed convex subset of Cγ,ψ0 ([0,∞),Ω).
Lemma 3.1 Assume(C1)holds. Then,{Tθ;θ∈ Qξ,ψ}is equicontinuous andlimt→∞|Tθ(t)|= 0 uniformly for θ∈ Qξ,ψ.
Proof: Asν−ξ1−ξξ2 <0, then there exists eε >0 and T1 >0 large enough such that
(ψ(t)−ψ(0)) + 1 n
γ−1 |θ0| Γ(γ) < eε
4 and
L Γ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2)(ψ(t)−ψ(0))ν1−ξ1−ξξ2 < eε 4 fort≥T1.
For eachθ∈ Qξ,ψ and t1, t2 ≥T1, we have
|(Tθ)(t2)−(Tθ)(t1)|
=
(ψ(t2)−ψ(0)) + 1 n
γ−1 |θ0| Γ(γ) + 1
Γ(ν) Z t2
0
Θν−1ψ (s, t2)u(s, θ(s))ds
−
(ψ(t1)−ψ(0)) + 1 n
γ−1
|θ0| Γ(γ)− 1
Γ(ν) Z t1
0
Θν−1ψ (s, t1)u(s, θ(s))ds
≤
(ψ(t2)−ψ(0)) + 1 n
γ−1
|θ0| Γ(γ) +
(ψ(t1)−ψ(0)) + 1 n
γ−1
|θ0| Γ(γ)+ + 1
Γ(ν) Z t2
0
Θνψ−1(s, t2)|u(s, θ(s))|ds+ 1 Γ(ν)
Z t1
0
Θν−1ψ (s, t1)|u(s, θ(s))|ds
≤
(ψ(t2)−ψ(0)) + 1 n
γ−1 |θ0| Γ(γ) +
(ψ(t1)−ψ(0)) + 1 n
γ−1 |θ0| Γ(γ)+ + 1
Γ(ν) Z t2
0
Θνψ−1(s, t2) (ψ(s)−ψ(0))−ξ1−ξξ2ds + 1
Γ(ν) Z t1
0
Θνψ−1(s, t1) (ψ(s)−ψ(0))−ξ1−ξξ2ds
≤
(ψ(t2)−ψ(0)) + 1 n
γ−1
|θ0| Γ(γ) +
(ψ(t1)−ψ(0)) + 1 n
γ−1
|θ0| Γ(γ)+ +L Γ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2)(ψ(t2)−ψ(0))ν−ξ1−ξξ2 + +L Γ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2)(ψ(t1)−ψ(0))ν−ξ1−ξξ2
≤
(ψ(t2)−ψ(0)) + 1 n
γ−1 |θ0| Γ(γ) +
(ψ(t1)−ψ(0)) + 1 n
γ−1 |θ0| Γ(γ)+ +L Γ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2) h
(ψ(t2)−ψ(0))ν−ξ1−ξξ2+ (ψ(t1)−ψ(0))ν−ξ1−ξξ2i
< εe 4+ eε
4+ eε 4 +εe
4 =ε,e (3.6)
and then, we have
|(Tθ)(t2)−(Tθ) (t1)|<eε.
For 0≤t1< t2 ≤T1 we have
|(Tθ)(t2)−(Tθ)(t1)|
≤
(ψ(t2)−ψ(0)) + 1 n
γ−1
−
(ψ(t1)−ψ(0)) + 1 n
γ−1!
|θ0| Γ(γ)
+
1 Γ(ν)
Z t2
0
Θν−1ψ (s, t2)u(s, θ(s))ds− Z t1
0
Θν−1ψ (s, t1)u(s, θ(s))ds
≤
(ψ(t2)−ψ(0)) + 1 n
γ−1
−
(ψ(t1)−ψ(0)) + 1 n
γ−1
! |θ0| Γ(γ) +
1 Γ(ν)
Z t2
0
ψ0(s)[(ψ(t2)−ψ(s))ν−1−(ψ(t1)−ψ(s))ν−1]|u(s, θ(s))|ds + 1
Γ(ν) Z t2
t1
Θν−1ψ (s, t2)|u(s, θ(s))|ds
≤
(ψ(t2)−ψ(0)) + 1 n
γ−1
−
(ψ(t1)−ψ(0)) + 1 n
γ−1
! |θ0| Γ(γ)
+ M
Γ(ν) Z t2
0
ψ0(s)[(ψ(t2)−ψ(s))ν−1−(ψ(t1)−ψ(s))ν−1]ds
+ M
Γ(ν) Z t2
t1
Θνψ−1(s, t2)ds
≤
(ψ(t2)−ψ(0)) + 1 n
γ−1
−
(ψ(t1)−ψ(0)) + 1 n
γ−1
! |θ0| Γ(γ)
+ M
Γ(ν)
(ψ(t2)−ψ(t1))ν −(ψ(t2)−ψ(0))ν
ν +(ψ(t1)−ψ(0))ν
ν
+ M
Γ(ν)
(ψ(t2)−ψ(t1))ν ν
which goes to zero as t2→t1; here
M = sup
t∈[0,t2] x∈Sξ,ψ
|u(t, θ(t))|.
Similarly, for t1 < T1 < T2 we have
|(Tθ)(t2)−(Tθ)(t1)| = |Tθ(t2)− Tθ(t1) +Tθ(T1)− Tθ(t1)|
≤ |Tθ(t2)− Tθ(T1)|+|Tθ(T1)− Tθ(t1)| →0 ast2 →t1.
Thus {Tθ:θ∈ Qξ,ψ} equicontinuous.
Now, we show limt→∞|(Tθ)(t)|= 0 uniformly for θ∈ Qξ,ψ. We have,
|Tθ(t)|
=
(ψ(t)−ψ(0)) + 1 n
γ−1
|θ0| Γ(γ) + 1
Γ(ν) Z t
0
Θν−1ψ (s, t)u(s, θ(s))ds
≤
(ψ(t)−ψ(0)) + 1 n
γ−1
|θ0| Γ(γ) + 1
Γ(ν) Z t
0
Θν−1ψ (s, t) (ψ(s)−ψ(0))−ξ1−ξξ2ds
≤
(ψ(t)−ψ(0)) + 1 n
γ−1 |θ0|
Γ(γ) + LΓ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2)(ψ(t)−ψ(0))ν−ξ1−ξξ2 which goes to zero for t→ ∞ (note 0< γ <1 andν−ξ1−ξξ2+ξ <0).
Thus lim
t→∞|Tθ(t)|= 0 uniformly forθ∈ Qξ,ψ, which concludes the proof. 2 Lemma 3.2 Assume (C1) holds with ψ(t) = t. Then, {Tθ;θ ∈ Qξ,t} is equicontinuous and limt→∞|Tθ(t)|= 0 uniformly for θ∈ Qξ,t.
Proof: This follows immediately from Lemma 3.1. 2
Lemma 3.3 Assume (C1) holds with ψ(t) = tρ. Then {Tθ;θ ∈ Qξ,tρ} is equicontinuous and limt→∞|Tθ(t)|= 0 uniformly for θ∈ Qξ,tρ.
Proof: This follows immediately from Lemma 3.1. 2
Lemma 3.4 Suppose (C1)holds. Then,T takesQξ,ψ intoQξ,ψ and is continuous on Qξ,ψ. Proof: First we prove T takes Qξ,ψ into Qξ,ψ. For θ∈ Qξ,ψ and from Lemma 2.3, we see that Tθ∈ Cγ,ψ([0,∞),Ω).
Using the inequality (3.4) we have
(ψ(t)−ψ(0))ξ(Tθ)(t) =
(ψ(t)−ψ(0))ξ
"
(ψ(t)−ψ(0) + 1 n
γ−1
|θ0| Γ(γ) + 1
Γ(ν) Z t
0
Θν−1ψ (s, t)u(s, θ(s))ds
≤ (ψ(t)−ψ(0))ξ
"
(ψ(t)−ψ(0) + 1 n
γ−1
|θ0| Γ(γ) + 1
Γ(ν) Z t
0
Θν−1ψ (s, t1)|u(s, θ(s))|ds
≤ (ψ(t)−ψ(0))ξ
"
(ψ(t)−ψ(0) + 1 n
γ−1
|θ0| Γ(γ) 1
Γ(ν) Z t
0
Θν−1ψ (s, t1) (ψ(s)−ψ(0))−ξξ2−ξ1ds
≤
(ψ(t)−ψ(0) + 1 n
ξ+γ−1
|θ0| Γ(γ) + LΓ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2)(ψ(t)−ψ(0))ν+ξ−ξ1−ξξ2
≤ 1 fort≥T.
Thus, we have
(ψ(t)−ψ(0))ξTθ(t)
≤1, soT takes Qξ,ψ intoQξ,ψ (T Qξ,ψ⊂ Qξ,ψ).
Now we prove T is continuous in Qξ,ψ.
Now, for θm, θ ∈ Qξ,ψ, m = 1,2,3, . . . with lim
m→∞θm = θ, we prove Tθm → Tθ, as m→ ∞. For ∀eε >0, there existsT2 >0 large enough such that
(ψ(T2)−ψ(0))γ <
r εe 2 with γ=ν+η(1−ν), and
LΓ(1−ξ1−ξξ2)
Γ(1 +ν−ξ1−ξξ2)(ψ(T2)−ψ(0))ν−ξ1−ξξ2 <
r eε 2·
Then, for t > T2, we have
|(ψ(t)−ψ(0))γ(Tθm(t)− Tθ(t))|
=
(ψ(t)−ψ(0))γ 1
Γ(ν) Z t
0
Θν−1ψ (s, t)u(s, θm(s))ds− 1 Γ(ν)
Z t 0
Θν−1ψ (s, t)u(s, θ(s))ds
≤ (ψ(t)−ψ(0))γ 1 Γ(ν)
Z t
0
Θν−1ψ (s, t1) (|u(s, θm(s))|+|u(s, θ(s))|)ds
≤ 2L(ψ(t)−ψ(0))γ Γ(ν)
Z t 0
Θν−1ψ (s, t1) (ψ(s)−ψ(0))ξξ2−ξ1ds
≤ 2L(ψ(T2)−ψ(0))γ(ψ(T2)−ψ(0))ν−ξ1−ξξ2 Γ(1−ξ1−ξξ2) Γ(1 +ν−ξ1+ξξ2)
< 2 r
εe 2
r eε 2 =ε .e For 0< t≤T2 we have
|(ψ(t)−ψ(0))γ(Txm(t)− Tθ(t))| (3.7)
=
(ψ(t)−ψ(0))γ 1
Γ(ν) Z t
0
Θν−1ψ (s, t)u(s, xm(s))ds−
1 Γ(ν)
Z t 0
Θν−1ψ (s, t)u(s, θ(s))ds
≤ (ψ(t)−ψ(0))γ 1
Γ(ν) Z t
0
Θν−1ψ (s, t)|u(s, θm(s))−u(s, θ(s))|ds
.
Taking the limit as m → ∞ on both sides of (3.7) and using the Lebesgue dominated convergence theorem, we get
|(ψ(t)−ψ(0))γ[Tθm(t)− Tθ(t)]| →0·
Thus kTθm− TθkC
γ,ψ → 0 as m → ∞ so T is continuous, which concludes the proof.
2
Lemma 3.5 Assume (C1) holds with ψ(t) =t. Then, T takes Qξ,t into Qξ,t and is con- tinuous on Qξ,t.
Proof: This follows immediately from Lemma 3.4. 2
Lemma 3.6 Assume (C1) holds with ψ(t) = tρ. Then, T takes Qξ,tρ into Qξ,tρ and is continuous on Qξ,tρ.
Proof: This follow immediately from Lemma 3.4. 2
Theorem 3.7 Assume (C1) and (C2) hold. Then, the Cauchy fractional problem (1.1) admits at least one attractive solution.
Proof: Note T :Qξ,ψ→ Qξ,ψ is bounded, continuous (see Lemma 3.4). Also {Tθ:θ∈ Qξ,ψ} is equicontinuous and lim
t→∞|Tθ(t)|= 0 uniformly for x ∈ Qξ,ψ (see Lemma 3.1), in particular {T2θ:θ∈ Qξ,ψ}.
Let’s check that for any t∈ [0,∞), {(Tθ)(t) :θ∈ Qξ,ψ} is relatively compact in Ω by using (C2). For each bounded subset Q0⊂ Qξ,ψ, set
T1(Q0) =T2(Q0), Tn(Q0) =T2 co Tn−1(Q0)
, n= 2,3, ...
where cois closure convex hull [7].
Using the condition (C2), Property 2.8 and Property 2.7, for any eε > 0, there is a sequence n
θ(1)n
o∞
n=1 such that µ T1(Q0(t))
= µ(T2(Q0))
≤ 2µ 1
Γ (ν) Z t
0
Θν−1ψ (s, t)u
s, n
θn(1)(s) o∞
n=1
ds
+εe
≤ 4
Γ (ν) Z t
0
Θν−1ψ (s, t)µ u
s,n
θ(1)n (s)o∞ n=1
ds+eε
≤ 4k
Γ (ν)µ(Q0) Z t
0
ψ0(s) (ψ(t)−ψ(s))ν−1ds+eε
= 4k
Γ (ν+ 1)µ(Q0) (ψ(t)−ψ(0))ν+ε.e Since ε >e 0 is arbitrary, we get
µ T1(Q0(t))
≤ 4k
Γ (ν+ 1)µ(Q0) (ψ(t)−ψ(0))ν.
By means of the Property 2.7 and Property 2.8, for any eε > 0, there is a sequence n
θ(2)n
o∞ n=1
⊂co T1(Q0)
such that µ T2(Q0(t))
= µ T2 co T1(Q0(t))
≤ 2µ 1
Γ (ν) Z t
0
Θν−1ψ (s, t)u
s, n
θ(2)n (s) o∞
n=1
ds
+εe
≤ 4
Γ (ν) Z t
0
Θν−1ψ (s, t)µ u
s,n
θ(2)n (s)o∞ n=1
ds+εe
≤ 4k2
µ(Q0) Γ (ν) Γ (ν+ 1)
Z t 0
ψ0(s) (ψ(t)−ψ(s))ν−1(ψ(s)−ψ(0))νds+εe (letu=ψ(s)−ψ(0))
= 4k2
µ(Q0) Γ (ν) Γ (ν+ 1)
Z ψ(t)−ψ(0) 0
(ψ(t)−ψ(0)−u)ν−1uνdu+eε
= 4k2
µ(Q0)
Γ (ν) Γ (ν+ 1)(ψ(t)−ψ(0))ν−1
Z ψ(t)−ψ(0) 0
1− u
ψ(t)−ψ(0) ν−1
uνdu+εe
letp= u
ψ(t)−ψ(0)
= 4k2
µ(Q0)
Γ (ν) Γ (ν+ 1)(ψ(t)−ψ(0))2ν−1 Z 1
0
(1−p)ν−1pνdp+εe
= 4k2
µ(Q0)
Γ (2ν+ 1) (ψ(t)−ψ(0))2ν−1+eε
≤ 4k2
µ(Q0)
Γ (2ν+ 1) (ψ(t)−ψ(0))2ν+eε.
By mathematical induction, for every ne∈N, we have µ
Ten(Q0(t))
≤ 4ken
(ψ(t)−ψ(0))νne
Γ (νne+ 1) µ(Q0). Since
lim
n→∞e
4k(ψ(a)−ψ(0))νne
Γ (νen+ 1) = 0, there existsm∈Z+ such that
4km
(ψ(t)−ψ(0))νm
Γ (νm+ 1) ≤
4k(ψ(a)−ψ(0))νm
Γ (νm+ 1) =eq <1.
Then
µ(Tm(Q0(t)))≤eqµ(Q0).
We know from Property 2.5, Tm(Q0(t)) is bounded and equicontinuous. Then, by Property 2.6, we get
µ(Tm(Q0)) = max
t∈[0,a]µ(Tm(Q0(t))). Hence
µ(Tm(Q0))≤qµe (Q0). We will prove that ∃De ⊂ Qξ,ψ, such that µ
T2 De
= 0, i.e., T2 De
is relatively compact.
Let D0 =Qξ,ψ, D1 =co(Tm(D)), ..., Dn=co(Tm(Dn−1)), n= 2,3, ....
So, we can get
1. D0⊃D1 ⊃D2⊃ · · · ⊃Dn−1 ⊃Dn⊃ · · ·;
2. lim
n→∞µ(Dn) = 0.
Then De =
∞
T
n=0
Dn is a nonempty, compact and convex subset in Qξ,ψ. We will proveT2
De
⊂D. Firstly, we prove,e
T2(Dn)⊂Dn, n= 0,1,2, .... (3.8) FromT1(D0) =T2(D0)⊂D0, we know co T1(D0)
⊂D0. ThereforeT2(D0) =T2 co T1(D0)
⊂ T2(D0) =T1(D0),T3(D0) =T2 co T2(D0)
⊂ T2 co T1(D0)
=T2(D0),T4(D0) =T2 co T3(D0)
⊂ T2 co T2(D0)
=T3(D0).
Performing this procedure, m-times, we have Tm(D0) =T2 co Tm−1(D0)
⊂ T2 co Tm−2(D0)
=Tm−1(D0). Hence, D1 = co(Tm(D0)) ⊂ co Tm−1(D0)
, so T (D1) ⊂ T co Tm−1(D0)
= Tm(D0)⊂co(Tm(D0)) =D1.
Employing the same method, we can prove T2(Dn) ⊂Dn(n= 0,1,2, ...). By (3.8), we get
T2 De
⊂
∞
\
n=0
T2(Dn)⊂
∞
\
n=0
Dn=D.e
Then T2 De
is compact. Hence,µ T2
De
= 0,i.e., T2 De
is relatively compact.
On the other hand, for any θ1, θ2∈De and t∈J, we have
|(ψ(t)−ψ(0))γ[(T1θ1) (t)−(T1θ2) (t)]|
=
(ψ(t)−ψ(0))γ (
(ψ(t)−ψ(0)) + 1 n
γ−1
θ0 Γ (γ) −
(ψ(t)−ψ(0)) +1 n
γ−1
θ0 Γ (γ)
)
= 0
which implies that kT1θ1− T1θ2kC
γ,ψ = 0.Thus, we obtain that µ
T1 De
= 0.
So, we have
µ T
De
≤µ T2
De +µ
T1 De
= 0 implies µ
T De
= 0, thereforeT De
is relatively compact. By means of Arzel`a-Ascoli theorem (see Lemma 2.3), T is relatively compact. Therefore, by Schauder’s fixed point theorem guarantees that T has a fixed point θn ∈ Qξ,ψ with θn(t) → 0 as t→ ∞. Using the idea as in Lemma 3.1, we know that {θn(t)} is uniformly bounded and equicontinuous on [0,∞) and for all t∈[0,∞),{θn(t)} is relatively compact.
As {(Tθ)(t) : θ ∈ Qξ,ψ} is relatively compact for any t ∈ [0,∞), then, every se- quence{θn}inQξ,ψadmit a uniformly convergent subsequence{θnk}inCγ,ψ0 (J,Ω) (Qξ,ψ ⊂ Cγ,ψ0 (J,Ω)) by Arzel`a-Ascoli theorem.
Furthermore, {θnk} satisfies θnk(t) =
(ψ(t)−ψ(0) + 1 nk
γ−1
θ0
Γ(γ) + 1 Γ(ν)
Z t 0
Θν−1ψ (s, t)u(s, θnk(s)) ds, (3.9)
with t∈[0,∞).
Let θ∗(t) = lim
k→∞θnk(t) (t 6= 0). The Lebesgue dominated convergence theorem with (3.9) yields
θ∗(t) =
(ψ(t)−ψ(0) + 1 n
γ−1
θ0
Γ(γ)+ 1 Γ(ν)
Z t 0
Θν−1ψ (s, t)u(s, θ∗(s))ds,
with t∈[0,∞), soθ∗(t) is an attractive solution for the Cauchy fractional problem. 2
Corollary 3.8 Assume (C1) holds. Then, the Cauchy fractional problem (1.1) admits at least one attractive solution.
We consider the following problem, a fractional differential equation and an initial con- dition, in R
( HD0ν,η;ψ+ θ(t) = (ψ(t)−ψ(0))−ξ1, t∈(0,∞)
I01−γ;ψ+ θ(0) = 0 (3.10)
with 0< ν < ξ1<1, 0≤η≤1 and γ =ν+η(1−ν).
From Corollary 3.8, the problem given in (3.10) has an attractive solution since (C1) is valid. Indeed, this can be proved directly, since the solution of (3.10) has an exact solution, given by
θ(t) = Γ(1−ξ1)
Γ(ν+ 1−ξ1)(ψ(t)−ψ(0))ν−ξ1· (3.11) Using (2.5), i.e.,
θ(t) = (ψ(t)−ψ(0))γ−1
Γ(γ) I01−γ,ψ+ θ(0) + 1 Γ(ν)
Z t 0
Θν−1ψ (s, t)u(s, θ(s))ds, and taking u(t, θ(t)) = (ψ(t)−ψ(0))−ξ1 we have
θ(t) =Iν,ψ
0+ [(ψ(t)−ψ(0))−ξ1] = Γ(1−ξ1)
Γ(ν+ 1−ξ1)(ψ(t)−ψ(0))ν−ξ1 which is attractively global.
Remark 3.9 As a particular case of (3.10), we takeψ(t) =t and η→0. Then θ(t) = Γ(1−ξ1)
Γ(ν+ 1−ξ1)tν−ξ1 (3.12)
which is the solution of (3.10), in the Riemann-Liouville fractional derivative sense.
Also, taking ψ(t) =tρ (ρ >0) and η→1 in (3.10), we get θ(t) = Γ(1−ξ1)
Γ(ν+ 1−ξ1)tρ(ν−ξ1) (3.13)
which is the solution of (3.10), in the Caputo-type fractional derivative sense.
The following graph is for (3.13). We choose t∈[0,1],ρ= 0.5 andξ1 = 0.06.
Taking ψ(t) =t andν →1 on both sides of (3.10), we obtain the problem
θ0(t) = t−ξ, t∈(0,∞) θ(0) = = 0.
Consequently, the solution is given by θ(t) = Γ(1−ξ1)
Γ(2−ξ1)t1−ξ1 = t1−ξ1 1−ξ1
with θ(t)→ ∞ as t→ ∞.
Acknowledgment
JVCS acknowledges the financial support of a PNPD-CAPES (no88882.305834/2018-01) scholarship of the Postgraduate Program in Applied Mathematics of IMECC-Unicamp.
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