# On Hilfer-Type Fractional Impulsive Differential Equations

Authors: Chanisara Metpattarahiran; Kulandhaivel Karthikeyan; Panjaiyan Karthikeyann; Thanin Sitthiwirattham
Journal: International Journal of Differential Equations
 (2022)

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## Abstract

Using the Schauder fixed point theorem, we prove the existence of impulsive fractional differential equations using Hilfer fractional derivative and nearly sectorial operators in this paper. We’ve gone over the two scenarios where the related semigroup is compact and noncompact for this purpose. We also go over an example to back up the main points.


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## Body

## 1. Introduction

We consider the following impulsive fractional differential equations involving Hilfer fractional derivative and almost sectorial operators(1)D§,ϰ£t+A£t=Yt,£t,t∈0,T=J,Δ£|t=tk=ℑk£tk−,k=1,2,3,…,m,I0+1−λ1−ϰ£0=£0,where D0+§,ϰ Hilfer fractional derivative of order §∈0,1 and type ϰ∈0,1 and A is an almost sectorial operator in Y having norm ⋅ and Δ£|t=tk denotes the jump of £t at t=tk, i.e., Δ£|t=tk=£tk+−£tk−, where £tk+ and £tk−. ℑk:U⟶Dℶ and £0∈Dℶ. Y:J×Y⟶Y is a function which is defined later.In fact, rapid changes in the dynamics of evolutionary processes might occur due to shocks, harvesting, or natural disasters, among other things. These short-term disturbances are frequently handled as impulses. Hilfer derivative fractional differential equations have recently gained a lot of attention in the literature. [1–12] appears to be the case.In [3], Anjali Jaiswal and Bahuguna studied the equations of Hilfer fractional deritaive with almost sectorial operator in the abstract sense.(2)D§,ϰt+A£t=Yt,£t,t∈0,T,I0+1−§1−ϰ£0=£0.We also refer to Hamdy M. Ahmed et.al studied in [13], which looked at the existence of nonlinear Hilfer fractional derivative differential equations with control. Sufficient circumstances have been found where the Hilfer derivative is the time fractional derivative. In [14], Yong Zhoy et.al studied the factional cauchy problems with almost sectorial operators of the form(3)D§£t=A£t+ℱt,£t,t∈0,T,I0+1−§£0=£0,where D§ is Riemann–Liouville derivative of order §, I1−§ is Riemann–Liouville integral of order 1−§, 0<§<1. A is an almost sectorial operator on a complex Banach space, and ℱ is a given function.The paper is structured as follows: we have presented some information in section2 about Hilfer derivative, almost sectorial operators, measure of non-compactness, mild solutions of equations 1.1−1.3 along with some basic definitions, results and lemmas. We discuss the main results for mild solutions for the equations 1.1−1.3 in section 3. In section 4 5, we have discussed the two cases if associated semigroup is compact and noncompact respectively. Finally, an example is discussed to illustrate the main result.The following sections describes the supporting results of the given problem and also generalizes the results in [14–16].


## 2. Preliminaries

Definition 1.
(see [17]). For §>0, the fractional integral of order § of a function ft is defined by(4)I0+§ft=1ϰ§∫0tfrt−r§−1dr.Definition 2.
(see [17]). For 0<§<1, the Riemann-Liouville (R-L) fractional derivative with order § of a function ft is defined by(5)D§ft=1ϰ1−§ddt∫0tfrt−r§dr.Definition 3.
(see [17]). For 0<§<1, the Caputo fractional derivative with order § of a function ft is defined by(6)D§cft=1ϰ1−§∫0tf′rt−r§dr.Definition 4.
(see [11]). Let 0<§<1 and 0≤v≤1. The Hilfer fractional derivative derivative of order § and type v is defined by(7)D0+§,ϰ£t=I0+v1−§ddtI0+1−ϰ1−§£t.


### 2.1. Measure of Non-compactness

Letℶ⊂Y also bounded. The Hausdorff measure of non-compactness is considered as follows(8)Ξℶ=infζ>0that isℶ⊂∪j=1mBζxjwherexjÎY,m∈N.The Kurtawoski measure of noncompactnessΦ on a bounded set ℬ⊂Y is considered as follows(9)Φℶ=infϵ>0impliesℶ⊂∪j=1mMjalso diamMj≤ϵ,with the following properties1.
ℶ1⊂ℶ2 gives Ξℶ1≤Ξℶ2 where ℶ1,ℶ2 are bounded subsets of Y2.
Ξℶ=0 iff ℶ is relatively compact in Y3.
Ξz∪ℶ=Ξℶ for all z∈Yℶ⊆Y4.
Ξℶ1∪ℶ2≤maxΞℶ1,Ξℶ25.
Ξℶ1+ℶ2≤Ξℶ1+Ξℶ26.
Ξrℶ≤rΞℶ for r∈R.Letℸ⊂CI,Y and ℸr=υr∈Y|υ∈ℸ. We define(10)∫0tℸrdr=∫0tυrdr|υ∈ℸ,t∈J.


### 2.2. Almost Sectorial Operators

Let0<μ<π and −1<β<0. We define Sμ0=υ∈C,0,that is,argυ<μ and its closure by Sμ, that is Sμ=υ∈C,0:argυ<μ∪0.Definition 5.
(see [18]). For −1<β<0,0<ω<π/2, we define ⊙ωβ as a family of all closed and linear operators A:DA⊂Y⟶Y this implies1.
σA is contained in Sω.2.
For allμ∈ω,π, there exists Mμ implies(11)ℛz,ALX≤Mμzβ,where ℛz,A=zI−A−1 is the resolvent operator and A∈⊙ωβ is said to be an almost sectorial operator on Y.
We assume the following Wright-type function [17](12)M§θ=∑n∈ℕ−θn−1ϰ1−§nn−1!,θ∈ℂ.
For−1<σ<∞,r>0,(A1)
M§θ≥0,t>0;(A2)
∫0∞θσM§dθ=ϰ1+σ/ϰ1+§σ;(A3)
∫0∞§/θ§+1e−rθM§1/θ§dθ=e−r§.
We have defineS§t|t∈Sπ/2−w0, Q§t|t∈Sπ/2−w0 by(13)S§t=∫0∞M§ζϱt§ζdζ,Q§t=∫0∞§ζM§ζϱt§ζdζ.Theorem 1 (Theorem 4.6.1 in [17]).
For each fixedt∈Sπ/2−ω0,S§t and Q§t are bounded and linear operators on Y. Also(14)S§t≤ℭst−§1+β,Q§t≤ℭpt−§1+β,t>0,where ℭs and ℭp are constants.We introduce the following hypotheses to obtain our main results.(H1)
Fort∈J′,Yt,.:Y⟶Y is continuous function and for each £∈ℭJ′,Y,Y.,£,ℬ£:J′⟶Y is strongly measurable.(H2)
∃ a function k∈L1J′,ℝ+ satisfying(15)I0+−§βk∈ℭJ′,ℝ,limt1+§β1−ϰI0+−§βkt=0.(H3)
(16)sup0,Tt1+§β1−ϰS§,ϰtu0+t1+§β1−ϰ∫0tt−r−§β−1krdr≤r,for r>0 and u0∈DAθ,θ>1+β, where S§,ϰt=I0+ϰ1−§t−1Q§t.(H4)
∃ constants ωk such that ℑk£≤ωk,k=1,2,…,m for each £∈Y.Lemma 1.
The fractional Cauchy problem1.1−1.3 is equivalent to an integral equation given by(17)£t=£0ϰv1−§+§t1−§v−1+1ϰ§∫0tt−r§−1−A£r+Yr,£rdr+∑0<tk<tS§,ϰt−tkℑk£tk−,t∈J.Lemma 2.
Let £ is a solution of the integral equation given in (2.3), then £ satisfies(18)£t=S§,ϰt£0+∫0tK§t−rYr,£rdr+∑0<tk<tS§,ϰt−tkℑk£tk−,where S§,ϰt=I0+ϰ1−§K§t,K§=t§−1Q§t.Definition 6.
By a mild solution of the Cauchy problem1.1−1.3 we mean a function £∈ℂJ′,X that satisfies(19)£t=S§,ϰt£0+∫0tK§t−rYr,£rdr+∑0<tk<tS§,ϰt−tkℑk£tk−,t∈J′.
Now, we define an operatorP:BrJ′⟶BrJ′ as(20)B£t=S§,ϰt£0+∫0tt−r§−1Q§t−rYr,£rdr+∑0<tk<tS§,ϰt−tkℑk£tk−.Lemma 3 (see [3]).
K§t and S§,ϰt are bounded linear operators on Y, for every fixed t∈Sπ/2−ω0. Also for t>0.(21)K§tx≤ℭpt−1−§βx,S§,ϰtx≤ϰ−§βϰϰ1−§−§βℭptϰ1−§−§β−1x.Proposition 1 (see [3]).
K§t and S§,ϰt are strongly continuous, for t>0.


## 2.1. Measure of Non-compactness

Letℶ⊂Y also bounded. The Hausdorff measure of non-compactness is considered as follows(8)Ξℶ=infζ>0that isℶ⊂∪j=1mBζxjwherexjÎY,m∈N.The Kurtawoski measure of noncompactnessΦ on a bounded set ℬ⊂Y is considered as follows(9)Φℶ=infϵ>0impliesℶ⊂∪j=1mMjalso diamMj≤ϵ,with the following properties1.
ℶ1⊂ℶ2 gives Ξℶ1≤Ξℶ2 where ℶ1,ℶ2 are bounded subsets of Y2.
Ξℶ=0 iff ℶ is relatively compact in Y3.
Ξz∪ℶ=Ξℶ for all z∈Yℶ⊆Y4.
Ξℶ1∪ℶ2≤maxΞℶ1,Ξℶ25.
Ξℶ1+ℶ2≤Ξℶ1+Ξℶ26.
Ξrℶ≤rΞℶ for r∈R.Letℸ⊂CI,Y and ℸr=υr∈Y|υ∈ℸ. We define(10)∫0tℸrdr=∫0tυrdr|υ∈ℸ,t∈J.


## 2.2. Almost Sectorial Operators

Let0<μ<π and −1<β<0. We define Sμ0=υ∈C,0,that is,argυ<μ and its closure by Sμ, that is Sμ=υ∈C,0:argυ<μ∪0.Definition 5.
(see [18]). For −1<β<0,0<ω<π/2, we define ⊙ωβ as a family of all closed and linear operators A:DA⊂Y⟶Y this implies1.
σA is contained in Sω.2.
For allμ∈ω,π, there exists Mμ implies(11)ℛz,ALX≤Mμzβ,where ℛz,A=zI−A−1 is the resolvent operator and A∈⊙ωβ is said to be an almost sectorial operator on Y.
We assume the following Wright-type function [17](12)M§θ=∑n∈ℕ−θn−1ϰ1−§nn−1!,θ∈ℂ.
For−1<σ<∞,r>0,(A1)
M§θ≥0,t>0;(A2)
∫0∞θσM§dθ=ϰ1+σ/ϰ1+§σ;(A3)
∫0∞§/θ§+1e−rθM§1/θ§dθ=e−r§.
We have defineS§t|t∈Sπ/2−w0, Q§t|t∈Sπ/2−w0 by(13)S§t=∫0∞M§ζϱt§ζdζ,Q§t=∫0∞§ζM§ζϱt§ζdζ.Theorem 1 (Theorem 4.6.1 in [17]).
For each fixedt∈Sπ/2−ω0,S§t and Q§t are bounded and linear operators on Y. Also(14)S§t≤ℭst−§1+β,Q§t≤ℭpt−§1+β,t>0,where ℭs and ℭp are constants.We introduce the following hypotheses to obtain our main results.(H1)
Fort∈J′,Yt,.:Y⟶Y is continuous function and for each £∈ℭJ′,Y,Y.,£,ℬ£:J′⟶Y is strongly measurable.(H2)
∃ a function k∈L1J′,ℝ+ satisfying(15)I0+−§βk∈ℭJ′,ℝ,limt1+§β1−ϰI0+−§βkt=0.(H3)
(16)sup0,Tt1+§β1−ϰS§,ϰtu0+t1+§β1−ϰ∫0tt−r−§β−1krdr≤r,for r>0 and u0∈DAθ,θ>1+β, where S§,ϰt=I0+ϰ1−§t−1Q§t.(H4)
∃ constants ωk such that ℑk£≤ωk,k=1,2,…,m for each £∈Y.Lemma 1.
The fractional Cauchy problem1.1−1.3 is equivalent to an integral equation given by(17)£t=£0ϰv1−§+§t1−§v−1+1ϰ§∫0tt−r§−1−A£r+Yr,£rdr+∑0<tk<tS§,ϰt−tkℑk£tk−,t∈J.Lemma 2.
Let £ is a solution of the integral equation given in (2.3), then £ satisfies(18)£t=S§,ϰt£0+∫0tK§t−rYr,£rdr+∑0<tk<tS§,ϰt−tkℑk£tk−,where S§,ϰt=I0+ϰ1−§K§t,K§=t§−1Q§t.Definition 6.
By a mild solution of the Cauchy problem1.1−1.3 we mean a function £∈ℂJ′,X that satisfies(19)£t=S§,ϰt£0+∫0tK§t−rYr,£rdr+∑0<tk<tS§,ϰt−tkℑk£tk−,t∈J′.
Now, we define an operatorP:BrJ′⟶BrJ′ as(20)B£t=S§,ϰt£0+∫0tt−r§−1Q§t−rYr,£rdr+∑0<tk<tS§,ϰt−tkℑk£tk−.Lemma 3 (see [3]).
K§t and S§,ϰt are bounded linear operators on Y, for every fixed t∈Sπ/2−ω0. Also for t>0.(21)K§tx≤ℭpt−1−§βx,S§,ϰtx≤ϰ−§βϰϰ1−§−§βℭptϰ1−§−§β−1x.Proposition 1 (see [3]).
K§t and S§,ϰt are strongly continuous, for t>0.


## 3. Main results

Theorem 2.
LetA∈⊙ωβ for −1<β<0 and 0<ω<π/2. Assuming H1−H4 are satisfied, the operators Fy:y∈BrJ is equicontinuous provided £0∈DAθ with θ>1+β.Proof.
Fory∈BrJ and t1=0<t2≤T, we gave(22)Fyt2−Fy0=t21+§μ1−ϰS§,ϰt2u0+∫0t2t2−r§−1Q§t2−r×Yr,£rdr+∑0<tk<t2S§,ϰt2−tkℑk£tk−≤t21+§μ1−ϰS§,ϰt2u0+t21+§μ1−ϰ∫0t2t2−r§−1Q§t2−rYr,£rdr+t21+§μ1−ϰ∑0<tk<t2S§,ϰt2−tkℑk£tk−.
⟶0, as t2⟶0.
Now, let0<t1<t2≤T,(23)Fyt2−Fyt1≤t21+§μ1−ϰS§,ϰt2u0−t11+§μ1−ϰS§,ϰt1u0+t21+§μ1−ϰ∫0t2t2−r§−1Q§t2−rYr,£rdr−t11+§μ1−ϰ∫0t1t1−r§−1Q§t1−rYr,£rdr+t21+§μ1−ϰ∑0<tk<t2S§,ϰt2−tkℑk£tk−−t11+§μ1−ϰ∑0<tk<t1S§,ϰt1−tkℑk£tk−,here, using the triangle inequality, we have(24)Fyt2−Fyt1≤t21+§μ1−ϰS§,ϰt2u0−t11+§μ1−ϰS§,ϰt1u0+t21+§μ1−ϰ∫t1t2t2−r§−1Q§t2−rYr,£rdr+t21+§μ1−ϰ∫0t1t2−r§−1Q§t2−rYr,£rdr−t11+§μ1−ϰ∫0t1t1−r§−1Q§t2−rYr,£rdr+t11+§μ1−ϰ∫0t1t1−r§−1Q§t2−rYr,£rdr−t11+§μ1−ϰ∫0t1t1−r§−1Q§t1−rYr,£rdr,+t21+§μ1−ϰ∑0<tk<t2S§,ϰt2−tkℑk£tk−−t11+§μ1−ϰ∑0<tk<t1S§,ϰt1−tkℑk£tk−=ℑ1+ℑ2+ℑ3+ℑ4+ℑ5.
Since the strong continuous ofS§,ϰt, we get ℑ1⟶0 as t2⟶t1. Also(25)ℑ2≤ℭpt21+§β1−ϰ∫t1t2t2−r−§β−1κrdr≤ℭpt21+§β1−ϰ∫0t2t2−r−§β−1κrdr−t21+§β1−ϰ×∫0t1t1−r−§β−1κrdr≤ℭp∫0t1t11+§β1−ϰt1−r−§β−1−t21+§β1−ϰt2−r−§β−1κrdr.
Thenℑ2⟶0 as t2⟶t1, by using H2 and the dominated convergence theorem. Since(26)ℑ3≤ℭp∫0t1t2−r−§−§β×t21+§β1−ϰt2−r§−1−t11+§β1−ϰt1−r§−1κrdrt2−r−§−§βt21+§β1−ϰt2−r§−1−t11+§β1−ϰt1−r§−1κr≤t21+§β1−ϰt2−r§−1κr+t11+§β1−ϰt1−r§−1κr≤2t11+§β1−ϰt1−r§−1κr,and ∫0t12t11+§β1−ϰt1−r§−1κr exists, i.e, I3⟶0 as t2⟶t1.
Forε>0, we have(27)ℑ4=∫0t1t11+§β1−ϰQ§t2−r−Q§t1−rt1−r§−1Yr,£rdr≤∫0t1−ϵt11+§β1−ϰQ§t2−r−Q§t1−rLXt1−r§−1κrdr+∫t1t1t11+§β1−ϰQ§t2−r−Q§t1−rLXt1−r§−1κrdr≤t11+§β1−ϰ∫0t1t1−r§−1κrdr.sups∈0,t1−ϵQ§t2−r−Q§t1−rLX+ℭp∫t1t1t11+§β1−ϰt2−r−§−§β+t1−r−§−§βt1−r§−1κrdr≤t11+§β1−ϰ+§1+β∫0t1t1−r−§β−1κrdr.sups∈0,t1−ϵQ§t2−r−Q§t1−rLX+2ℭp∫t1−ϵt1t11+§β1−ϰt1−r−§β−1κrdr.
SinceQ§t is uniformly continuous and limt2⟶t1ℑ2=0, then ℑ4⟶0 as t2⟶t1 i.e. independent of y∈BrJ.
Clearly, since the strongly continuous of ofS§,ϰt, we get(28)ℑ5=t21+§μ1−ϰ∑0<tk<t2S§,ϰt2−tkℑk£tk−−t11+§μ1−ϰ∑0<tk<t1S§,ϰt1−tkℑk£tk−⟶0ast2⟶t1.
HenceFyt2−Fyt1⟶0 independently of y∈BrJ as t2⟶t1 therefore Fy:y∈BrJ is equicontinuous.Theorem 3.
Let−1<β<0 and 0<ω<π/2 and A∈⊙ωβ. Then under consideration H1−H3 the operator Fy:y∈BrJ is continuous and bounded provided £0∈DAθ with θ>1+β.Proof.
We verify thatF maps BrJ. Taking y∈BrJ and define £t=t−1+§β1−ϰyt, we have £∈BrYJ′. Let t∈0,T,(29)F≤t1+§β1−ϰS§,ϰtu0+t1+§β1−ϰ×∫0tt−r§−1Q§t−rYr,£rdr.
FromH2−H3, we get(30)Fyt≤t1+§β1−ϰS§,ϰtu0+t1+§β1−ϰ×∫0tt−r−§β−1κrdr≤sup0,Tt1+§β1−ϰ∫0tt−r−§β−1κrdr≤r.
HenceFy≤r, for any y∈ℬrI.
Now, to verifyF is continuous in ℬrI, let yn,y∈ℬI,n=1,2,…, with limn⟶∞yn=y. That is limn⟶∞ynt=yt and limn⟶∞t−1+§β1−ϰynt=t−1+§β1−ϰyt and limn⟶∞t−1+§β1−ϰynt=t−1+§β1−ϰyt, on J′.
H1 implies(31)Yt,£nt=Yt,t−1+§β1−ϰynt,t−1+§β1−ϰ⟶Yt,t−1+§β1−ϰyt,t−1+§β1−ϰ,as n⟶∞.
FromH2 to obtain the inequality t−r−§β−1|Yr,£nr|≤2t−r−§β1−ϰκr,
i.e.,(32)∫0tt−r−§β−1Yr,£nr−Yr,£rdr⟶0,asn⟶∞.
Lett∈0,T. Now,(33)Fynt−Fyt≤t1+§β1−ϰ×∫0tt−r§−1Q§t−rYr,£nr−Yr,£rdr.
Applying Theorem1, we have(34)Fynt−Fyt≤ℭpt1+§β1−ϰ×∫0tt−r−§β−1Yr,£nr−Yr,£rdr.
⟶0 as n⟶∞, i.e, Fyn⟶Fy pointwise on J. Also Theorem 2 implies that Fyn⟶Fy uniformly on J as n⟶∞. That is F is continuous.


## 4.ℑt is Compact

Theorem 4.
Let−1<β<0,0<ω<π/2 and A∈Ξωβ. If Ttt>0 is compact and H1−H4 hold then ∃ a mild solution of 1.1−1.3 in ℬrYI′ for every £0∈DAθ with θ>1+β.Proof.
Since we have assumedℑt is compact, then the equicontinuity of ℑtt>0. Moreover, by Theorem 2 and 3, F:BrYJ′⟶BrYJ′ is continuous and bounded and ε:BrJ⟶BrJ is bounded, continuous and εy:y∈BrJ equicontinuous. We can write ϵ:BrJ⟶BrJ by(35)ϵyt=ϵ1yt+ϵ2yt,where(36)ϵ1yt=t1+§β1−ϰS§,ϰt£0=t1+§β1−ϰℐ0+ϰ1−§t§−1Q§t£0=t1+§β1−ϰϰϰ1−§∫0tt−rϰ1−§−1r§−1×∫0∞§θM§θℑr§θ£0dθdr=§t1+§β1−ϰϰϰ1−§∫0t∫0∞t−rϰ1−§−1r§−1θM§θℑr§θ×£0dθdr,ε2yt=t1+§β1−ϰ∫0tt−r§−1Q§t−rYr,£rdr+∑0<tk<tS§,ϰt−tkℑk£tk−.
Forσ>0 and ζ∈0,t, we define an operator εζ,σ1 on BrJ by(37)εζ,σ1yt=t1+§β1−ϰϰϰ1−§∫ζt∫σ∞t−r1−§ϰ−1r§−1θM§θℑr§θ×£0dθdr=§t1+§β1−ϰϰϰ1−§Tζ§σ∫ζt∫σ∞t−r1−§ϰ−1r§−1θM§θ×ℑr§θ−ζ§σ£0dθdr.
SinceTϵ§δ is compact Vζ,σ1t=εζ,σ1yt,y∈BrJ is precompact in Y∀ζ∈0,t and δ>0. Moreover, for any y∈BrJ(38)ε1yt−εζ,σ1yt≤K§,ϰt1+§β1−ϰ∫0t∫0σt−rϰ1−§−1r§−1θM§θℑr§θ£0dθdr+K§,ϰt1+§β1−ϰ∫0ζ∫σ∞t−rϰ1−§−1r§−1θM§θℑr§θ£0dθdr≤K§,ϰt1+§β1−ϰ∫0t∫0σt−rϰ1−§−1r§−1θM§θr−§ϰ−§£0θ−β−1dθdr+K§,ϰt1+§β1−ϰ∫0ζ∫σ∞t−rϰ1−§−1r§−1θM§θr−§β−§θ−β−1£0dθdr=K§,ϰt1+§β1−ϰ∫0tt−rϰ1−§−1r−§β−1£0dr∫0σθ−βM§θdθ+K§,ϰt1+§β1−ϰ∫0ζt−rϰ1−§−1r−§β−1£0dr∫η∞θ−βM§θdθ≤Kt−§ϰ1+β£0∫0ηθ−βM§θdθ+Kt−§ϰ1+β£0∫0ζ1−sϰ1−§−1r−§β−1dr∫η∞θ−βM§θdθ,⟶0,asζ⟶0,σ⟶0,where, K§,ϰ=§/ϰϰ1−§.
Therefore,Vζ,σ1t=εζ,σ1yt,y∈BrJ are arbitrarily close to V1t=ε1yt,y∈BrJ , for t>0. Hence V1t, for t>0, is precompact in Y.
Forζ∈0,t, σ>0, we can present an operator εζ,σ2 on BrJ by(39)εζ,σ2yt=§t1+§β1−ϰ∫0t−ζ∫σ∞θM§θt−r§−1ℑt−r§θYr,£rdθdr+∑0<tk<tS§,ϰt−tkℑk£tk−=§t1+§β1−ϰTζ§σ∫0t−ζ∫σ∞θM§θt−r§−1×ℑt−r§θ−ζ§σYr,£rdθdr+∑0<tk<tS§,ϰt−tkℑk£tk−.
Hence Vζ,σ2t=εζ,σ2yt,y∈BrJ is precompact in Y∀ζ∈0,t and σ>0 due to the compactness of ℑζ§σ. For every y∈BrJ, we get(40)ε2yt−εζ,σ2yt≤§t1+§β1−ϰ∫0t∫0σθM§θt−r§−1ℑt−r§θYr,£rdθdr+∑0<tk<tS§,ϰt+σ−tkℑk£tk−+§t1+§β1−ϰ∫t−ζt∫σ∞t−r§−1θM§θℑt−r§θYr,£rdθdr+∑0<tk<tS§,ϰt+ζ−tkℑk£tk−≤§ℭ0t1+§β1−ϰ∫0tt−r−§β−1krdr∫0σθ−βM§θdθ+∑0<tk<σωk+§ℭ0t1+§β1−ϰ∫t−ζtt−r−§β−1krdr∫0∞θ−βM§θdθ+∑0<tk<ζωk×§ℭ0t1+§β1−ϰ∫0tt−r−§β−1krdr×∫0σθ−βM§θdθ+∑0<tk<ηωk+§ℭ0ϰ1−βϰ1−§βt1+§β1−ϰ∫t−ζtt−r−§β−1krdr+∑0<tk<ζωk,⟶0asσ⟶0,σ⟶0.
Therefore,Vζ,σ2t=εζ,σ2yt,y∈BrJ are arbitrarily close to V2t=ε2yt,y∈BrJ,t>0. That is, εy,y∈BrJ is relatively compact by using the Arzela-Ascoli Theorem. Then £∗ is a mild solution of (1.1)-(1.3).


## 5.ℑt is Noncompact

We consider as follows,H5∃ a constant k>0 satisfies the following(41)ΞYt,E1,E2≤kΞE1,E2fora.et∈0,T,for every bounded subset E1,E2⊂Y.Theorem 5.
Let−1<β<0,0<ω<π/2, and A∈Ξωβ. Suppose H1−H5 hold. Then, 1.1−1.3 has a mild solution in BrYJ′ for every u0∈DYθ with θ>1+β.Proof.
By Theorem2 and 3, we get ε:BrJ⟶BrJ is continuous, bounded, and εy:y∈BrJ is equicontinuous. Also, we prove that there is a subset of BrJ such that ε is campact in it.
For any bounded setℙ0⊂BrJ, set(42)ε1ℙ0=εℙ0,εnℙ0=εco¯εn−1ℙ0,n=2,3,….
For anyϵ>0, we can obtain from Propositions 2.2−2.4, a subsequence yn1n=1∞⊂ℙ0 satisfying(43)Ξε1ℙ0t≤2Ξt1+§β1−ϰ∫0tt−r§−1Q§t−rYr,r−1+§β1−ϰyn1rn=1∞dr+∑0<tk<tωk≤4ℭpt1+§β1−ϰ∫0tt−r−§β−1ΞYr,r−1+§β1−ϰyn1rn=1∞dr+∑0<tk<tωk≤4ℭpkt1+§β1−ϰΞP0∫0tt−r−§β−1r−1+§β1−ϰdr+∑0<tk<tωk=4ℭpkt−§βΞP0ϰ−§βϰ−§β+ϰ1+§βϰ−2§β+ϰ1+§β+∑0<tk<tωk.
Fromϵ is arbitrary,(44)Ξε1ℙ0t≤4ℭpkt−§βΞℙ0ϰ−§βϰ−§β+ϰ1+§βϰ−2§β+ϰ1+§β+∑0<tk<tωk.
Again for anyϵ>0, we can get from Propositions 2.2−2.4 a subsequence yn2,ℬyn2n=1∞⊂co¯ε1ℙ0 implies that(45)Ξε2ℙ0t=Ξεco¯ε1ℙ0t≤2Ξt1+§β1−ϰ∫0tt−r§−1Q§t−r×Yr,r−1+§β1−ϰyn2rn=1∞dr+∑0<tk<tωk≤4ℭpt1+§β1−ϰ∫0tt−r−§β−1Ξ×Yr,r−1+§β1−ϰyn2rn=1∞dr+∑0<tk<tωk≤4ℭpkt1+§β1−ϰ∫0tt−r−§β−1×Ξr−1+§β1−ϰyn2rn=1∞dr+∑0<tk<tωk≤4ℭpkt1+§β1−ϰ∫0tt−r−§β−1r−1+§β1−ϰ×Ξyn2rn=1∞dr+∑0<tk<tωk≤4ℭpk2t1+§β1−ϰϰ−§βϰ−§β+ϰ1+§βϰ−2§β+ϰ1+§β×Ξℙ0∫0tt−r−§β−1r−1+§β1−ϰ−§βdr+∑0<tk<tωk=4ℭpk2t−2§βϰ2−§βϰ−§β+ϰ1+§βϰ−3§β+ϰ1+§β+∑0<tk<tωkΞℙ0.
Now,(46)Ξεnℙ0t≤4ℭpknt−n§βϰn−§βϰ−§β+ϰ1+§βϰ−n+1§β+ϰ1+§βΞℙ0,n∈ℕ.
LetM=4ℭpkT−§βϰ−§β. We can find m,k∈ℕ big enough such that 1/k<§β<1/k−1 and n+1/k>2 for n>m.ϰ−n+1§β+ϰ1+§β>ϰn+1/k. That is(47)4ℭpknT−n§βϰn−§βϰ−§β+ϰ1+§βϰ−n+1§β+ϰ1+§β<4ℭpknT−n§βϰn−§βϰ−§β+ϰ1+§βϰn+1/k.
Replacen+1 by j+1k. Then R.H.S of inequality given above becomes(48)Mj+1k−1ϰ−§β+ϰ1+§βϰj+1=MkjMk−1ϰ−§β+ϰ1+§βj!⟶0asj⟶∞.
Therefore,∃ a constant n0∈ℕ that is(49)4ℭpknt−n§βϰn−§βϰ−§β+ϰ1+§βϰ−n+1§β+ϰ1+§β≤4ℭpkn0T−n0§βϰn0−§βϰ−§β+ϰ1+§βϰ−n0+1§β+ϰ1+§β=p<1.
Now(50)Ξεn0ℙ0t≤pΞℙ0.
Fromεn0ℙ0t is bounded and equicontinuous, applying Proposition (2.2), we get(51)Ξεn0ℙ0=maxt∈0,TΞεn0ℙ0t.
Hence,(52)Ξεn0ℙ0≤pΞℙ0,where p<1. By applying the Schauder fixed point theorem, we obtain a fixed point y∗ in BrJ of ε. Let £∗t=Ψt1+§φϰ−1y∗t. Then, £∗t is a mild solution of 1.1−1.3s.


## 6. Illustrate an application for main results

We consider the following impulsive Hilfer derivative fractional system:(53)D0+§,ϰ£t,x−∂x2£t,x=Yt,£t,x,t∈0,T,x∈0,a,£t,0=£t,a=0ont∈0,T,ℐ0+1−§1−ϰk£0,x=£0x,x∈0,a,Δ£|t=1/2=ℑ1£12−,in Banach space Y=C§0,a0<§<1 of all Holder continuous functions, where §=1/4,ϰ=1/2,Yt,£,ℬ£=t−1/5cos2£. Here, we can convert the above problem 1.1−1.3 in abstract form as(54)D§,ϰ£t+A£t=Yt,£t,t∈0,T=J,Δ£|t=tk=ℑk£tk−,k=1,2,3,…,m,I0+1−§1−ϰk£0=£0.HereA=−∂x2 with DA=£∈C2+§0,athat is£t,0=0=£t,a. It follows from the work in [18] ∃ constants δ,ϵ>0, implies A+δ∈⊙π/2−ϵπ/2−1Y. To verify the compactness of semigroup ℑt, it is enough to prove that ℛ§,−A+δ is compact for every §>0. By DA⊂C2+§0,a and C2+§0,a is compactly embedded in C§0,a, the compactness of the resolvent operator for every §>0. We take lt=t−1/5.(55)r=sup0,Tt1+§β1−ϰS§,ϰtu0+T17/20ϰ−β/4ϰ4/5ϰ4/5−β/4.Then, hypothesesH1−H5 are satisfied. According to Theorem 4, the problem (6.1) has mild solution in BrY0,T.


## 7. Conclusion

The main objective of in this paper, we discussed the mild solutions for Hilfer fractional derivative differential equations involving jump conditions and almost sectorial operator when associated semigroup is compact and noncompact using Schauder fixed point theorem. Our theorems guarantee the effectiveness of existence results It is the results of the equations concerned. we discussed an example to verify the existence results. We will find to to investigate stability of similar problem in our future research work.



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