Existence of Solutions to Impulsive Fractional Sobolev-Type ......34 D. N. Chalishajar, D. Senthil Raja, P. Sundararajan and K. Karthikeyan De nition 2.2. ([2, 19]) The Riemann-Liouville

Nonlinear Analysis and Differential Equations, Vol. 9, 2021, no. 1, 31 - 56HIKARI Ltd, www.m-hikari.com

https://doi.org/10.12988/nade.2021.91135

Existence of Solutions to Impulsive Fractional

Sobolev-Type Integro-Differential Equations in

Banach Spaces with Operator Pairs and

State Dependent Delay

D. N. Chalishajar

Department of Applied Mathematics, Virginia Military Institute435 Mallory Hall, Lexington, VA-24450, USA

D. Senthil Raja

Department of Mathematics, K.S. Rangasamy College of TechnologyTiruchengode - 637 215, Tamil Nadu, India

P. Sundararajan

Department of Mathematics, Arignar Anna Government Arts CollegeNamakkal - 637 002, Tamil Nadu, India

K. Karthikeyan

Department of Mathematics, KPR Institute of Engineering and TechnologyCoimbatore - 641 407, Tamil Nadu, India

This article is distributed under the Creative Commons by-nc-nd Attribution License.

Copyright c© 2021 Hikari Ltd.

Abstract

We are concerned with impulsive fractional Sobolev-type integro-differentialequations in Banach spaces with operator pairs and state dependent delay, where theoperator pairs generates propagation families. With the help of the theory of prop-agation family and Laplace transforms as well as an estimate for a special sequenceimproved in this paper, we introduce a definition of mild solutions to the impulsiveproblem for these abstract fractional Sobolev-type integro-differential equations and

32 D. N. Chalishajar, D. Senthil Raja, P. Sundararajan and K. Karthikeyan

establish general existence theorems and a continuous dependence theorem, whichextend essentially some previous conclusions. In our results, the operator B couldbe unbounded, and the existence of an operator B−1 is not necessarily needed.Moreover, an example is given to illustrate our main results.

Mathematical Subject Classification: 34K37, 46B50, 47A50, 47D06, 47D99,47N20

Keywords: Sobolev-type, Fractional integro-differential equations, Operator pairs,Propagation family, Impulsive conditions, State dependent delay

1 Introduction

Consider the following fractional Sobolev-type integro-differential equations in a Banach spaceX with operator pairs (A,B) and impulsive conditions

cDq(Bu)(t) = Au(t) +Bf

(t, u(t),

∫ t

0ρ(t, s)h(t, s, uρ(s, u(s)))ds

), t ∈ J ′,

∆u|t=tk = Ik(u(tk)), k = 1, 2, . . . ,m,

u(0) = u0,

(1.1)

where cDq, q ∈ (0, 1), is the Caputo fractional derivative of order q with the lower limit zero, Aand B are closed linear operators with domains contained in X, and the pair (A,B) generates apropagation family W (t)t≥0, f : J ×X ×X → D(B) with J = [0, T ] and T > 0 is a constant,ρ : Υ→ R is continuous with Υ = (t, s) ∈ J × J : t ≥ s, h : Υ×X → X, J ′ = J \ 0,

0 = t0 < t1 < t2 < · · · < tm < tm+1 = T,

∆u|t=tk denotes the jump of u(t) at t = tk, i.e., ∆u|t=tk = u(t+k )−u(t−k ), where u(t+k ) and u(t−k )represent the right and left limits of u(t) at t = tk, respectively, and the impulsive functionsIk : X → D(B) are continuous and u0 ∈ D(B).For brevity, we set

% := max(t,s)∈Υ

|ρ(t, s)|.

In resent decades, fractional integro-differential equations as well as impulsive differentialequations of integer order or fractional order have been studied by many researchers and a lotof results have been obtained. For the theory and developments of the related topics, we referthe reader to, e.g., [2, 3, 6, 7, 8, 9, 10, 11, 12, 14, 19, 21, 26, 27, 28] and the references therein.Especially, for contributions about Sobolev-type equations, we refer the reader to, e.g., [1, 18, 20]and the references cited in. Stimulated by these works as well as the papers [20, 22, 24], weinvestigate here the definition, existence and continuous dependence of mild solutions of thefractional Sobolev-type integro-differential equations in a Banach space X with operator pairsand impulsive conditions (1.1), where the operator pairs generates propagation families. As youcan see, we study the system (1.1) without assuming B has bounded (or compact) inverse aswell as without any assumption on the relation between D(A) and D(B).

Impulsive fractional sobolev-type integro-differential equations 33

The rest of this paper is organized as follows. In section 2, we introduce some notations,recall some basic known results, and present a definition of mild solutions for the fractionalSobolev-type integro-differential equations in a Banach space X with operator pairs and impul-sive conditions (1.1), where the operator pairs generates propagation families. In section 3, wediscuss the existence of mild solutions for the system (1.1) in the Lipschitz case and further ina more general case. Moreover, we prove the continuous dependence of solutions to the initialvalues, which implies that the solution is unique in the Lipschitz case. All our results are evennew in the case of B = I. In section 4, an example is given to illustrate abstract results.

2 Preliminaries and definition of mild solutions

We begin this part with some notations. Throughout this paper, X is a Banach space withnorm ‖ · ‖. We denote by C(J,X) the space of all X-valued continuous functions on J with thenatural norm ‖x‖C(J,X) = sup

t∈J‖x(t)‖. Set

J1 = [0, t1], Jk = (tk−1, tk], k = 2, 3, . . . ,m, Jm+1 = (tm, T ].

Let PC(J,X) = u : J → X|u(t) is continuous at t 6= tk, and left continuous at t = tk, and u(t+k )exists, k = 1, 2, . . . ,m. Obviously, PC(J,X) is a Banach space with norm ‖u‖PC = sup

t∈J‖u(t)‖.

The beta function is defined by

B(p, q) =

∫ 1

0tp−1(1− t)q−1dt, p, q > 0.

The gamma function is defined by

Γ(p) =

∫ ∞0

tp−1e−tdt, p > 0.

It is well known that

B(p, q) =Γ(p)Γ(q)

Γ(p+ q), Γ(p+ 1) = pΓ(p).

A binomial coefficient is defined by

Cmn =n!

m!(n−m)!,

satisfying

Cmn + Cm−1n = Cmn+1,

where n! = n(n− 1)(n− 2) . . . 1.

Definition 2.1. ([2, 19]) The fractional integral of order q with the lower limit zero for afunction f ∈ AC[0,∞) is defined as

Iqf(t) =1

Γ(q)

∫ t

0(t− s)q−1f(s)ds, t > 0, 0 < q < 1.


Definition 2.2. ([2, 19]) The Riemann-Liouville derivative of order q with the lower limit zerofor a function f ∈ AC[0,∞) can be written as

RLDqf(t) =1

Γ(1− q)d

dt

∫ t

0

f(s)

(t− s)qds, t > 0, 0 < q < 1.

Definition 2.3. ([2, 19]) The Caputo derivative of order q with the lower limit zero for afunction f ∈ AC[0,∞) can be written as

cDqf(t) = RLDq(f(t)− f(0)), t > 0, 0 < q < 1.

Remark 2.4.(1) If f(t) ∈ C1[0,∞), then

cDqf(t) =1

Γ(1− q)

∫ t

0

f ′(s)

(t− s)qds = I1−qf ′(t), t > 0, 0 < q < 1.

(2) The Caputo derivative of a constant is equal to zero.

Next, we introduce the Kuratowski measure of noncompactness µ(·) defined on each boundedsubset B in the Banach space X byµ(B) = infd > 0|B can be covered by a finite number of sets of diameter < d.

Some basic properties of µ(·) are listed in the following lemma.

Lemma 2.5. ([5]) Let B,C ⊂ X be bounded sets, then we have that:(i) µ(B) = 0 if and only if B is relatively compact in X;(ii) µ(B) = µ(B) = µ(coB), where coB is the closed convex hull of B ;(iii) µ(B) ≤ µ(C) when B ⊆ C;(iv) µ(B + C) ≤ µ(B) + µ(C);(v) µ(B ∪ C) ≤ maxµ(B), µ(C);(vi) µ(B(0, r)) = 2r where B(0, r) = x ∈ X| ‖x‖ ≤ r, if dimX = +∞.

Lemma 2.6. ([23]) Let X be a Banach space, and let D ⊂ X be bounded. Then there exists acountable set D0 ⊂ D, such that α(D) ≤ 2α(D0).

Lemma 2.7. ([16]) Let X be a Banach space, and let Ω ⊂ PC(J,X) be equicontinuous andbounded. Then α(Ω(t)) ∈ PC(J,R+), and α(Ω) = max

t∈Jα(Ω(t)).

Lemma 2.8. ([23, 17]) Let un∞n=1 be a sequence of Bochner integrable functions from J intoX with ‖un(t)‖ ≤ m(t) for almost all t ∈ J and every n ≥ 1, where m ∈ L1(J,R+), then thefunction ψ(t) = µ(un∞n=1) belongs to L1(J,R+) and satisfies

µ

(∫ t

0un(s)ds : n ≥ 1

)≤ 2

∫ t

0ψ(s)ds.

The following Sadovskii fixed point theorem will be used later.


Theorem 2.9. ([15]) Let X be a Banach space. Assume that D ⊂ X is a bounded closed andconvex set on X and Q : D → D is a condensing mapping. Then Q has one fixed point on D.

For the following abstract degenerate Cauchy problem ([22])ddtBu(t) = Au(t), t ∈ J, t ≥ 0,

Bu(0) = Bu0,(2.1)

where A and B are closed linear operators in a sequentially complete locally convex space.Definition 2.10. ([22]) A strongly continuous operator family W (t)t≥0 of D(B) to a BanachspaceX, satisfying that W (t)t≥0 is exponentially bounded, which means that for any x ∈ D(B)there exist a > 0, M > 0 such that

‖W (t)x‖ ≤Meat‖x‖, t ≥ 0,

is called an exponentially bounded propagation family for (2.1) if for λ > a,

(λB −A)−1Bx =

∫ ∞0

e−λtW (t)xdt, x ∈ D(B). (2.2)

In this case, we also say that (2.1) has an exponentially bounded propagation family W (t)t≥0.Moreover, if (2.2) holds, we also say that the pair (A,B) generates an exponentially bounded

propagation family W (t)t≥0.In this paper, we assume that W (t)t≥0 is a norm continuous family for t > 0 and ‖W (t)‖ ≤

M . Definition 2.11. ([20]) By the mild solution of the following systemcDq(Bu)(t) = Au(t) +Bg(t), t ∈ J ′,u(0) = u0,

we mean that the function u ∈ C(J,X) which satisfies the following integral equation

u(t) = Q(t)u0 +

∫ t

0(t− s)q−1R(t− s)g(s)ds, t ∈ J,

where

Q(t) =

∫ ∞0

ξq(σ)W (tqσ)dσ, R(t) = q

∫ ∞0

σξq(σ)W (tqσ)dσ,

ξq(σ) =1

qσ−1− 1

q$q(σ− 1

q ) ≥ 0,

$q(σ) =1

π

∞∑n=1

(−1)n−1σ−qn−1 Γ(nq + 1)

n!sin(nπq), σ ∈ (0,∞),

ξq is a probability density function defined on (0,∞), that is

ξq(σ) ≥ 0, σ ∈ (0,∞) and

∫ ∞0

ξq(σ)dσ = 1.


By [20], we know that

‖Q(t)‖ ≤M, ‖R(t)‖ ≤ M

Γ(q), t ≥ 0. (2.3)

The proof of the following lemma is obvious, so we omit it.

Lemma 2.12. Q(t)t≥0 and R(t)t≥0 are also norm continuous for t > 0.

For the rest of this section, we shall derive a definition of mild solutions to the system (1.1).We first consider the following impulsive fractional system:

cDq(Bu)(t) = Au(t) +Bg(t), t ∈ J ′,∆u|t=tk = yk, k = 1, 2, . . . ,m,

u(0) = u0,

(2.4)

where g(t) ∈ PC(J,D(B)), yk ∈ D(B), u0 ∈ D(B).

We can decompose u(·) which is a solution of the system (2.4) to v(·) +w(·), where v is themild solution to

cDq(Bv)(t) = Av(t) +Bg(t), t ∈ J ′,v(0) = u0,

(2.5)

on J , and w is the mild solution tocDq(Bw)(t) = Aw(t), t ∈ J ′,∆w|t=tk = yk, k = 1, 2, . . . ,m,

w(0) = 0,

(2.6)

By Definition 2.5, a mild solution of (2.5) is given by

v(t) = Q(t)u0 +

∫ t

0(t− s)q−1R(t− s)g(s)ds, t ∈ J.

Now we rewrite (2.6) in the following integral equation:

Bw(t) =m∑i=1

χi(t)Byi +1

Γ(q)

∫ t

0(t− s)q−1Aw(s)ds, t ∈ J, (2.7)

where

χi(t) =

0, t ∈ [0, ti),

1, t ∈ [ti, T ].

We apply the Laplace transform for (2.7) to get

Bw(λ) =

m∑i=1

e−tiλ

λByi +

1

λqAw(λ),


which implies

w(λ) =m∑i=1

e−tiλλq−1(λqB −A)−1Byi.

By [20], we know that the Laplace transform of Q(t)yi is λq−1(λqB − A)−1Byi. Hence wederive the mild solution of (2.6) as

w(t) =m∑i=1

χi(t)Q(t− ti)yi =∑

0<tk<t

Q(t− tk)yk.

Summarizing, the mild solution of (2.4) is given by

u(t) = Q(t)u0 +

∫ t

0(t− s)q−1R(t− s)g(s)ds+

∑0<tk<t

Q(t− tk)yk.

According to the analysis above, we introduce the following definition of the mild solutionto the system (1.1).

Definition 2.13. By a mild solution of the system (1.1), we mean a function u ∈ PC(J,X)satisfying the following integral equation with state dependent delay

u(t) =Q(t)u0 +

∫ t

0(t− s)q−1R(t− s)f

(s, u(s),

∫ s

0ρ(s, τ)h(s, τ, uρ(τ, u(τ)))dτ

)ds

+∑

0<tk<t

Q(t− tk)Ik(u(tk)).

3 Main Results

We first consider the Lipschitz case for the system (1.1).In order to obtain more general result, we first prove the following estimate for a special

sequence Sn, which generalizes a result in [24], will be used in the proof of our main results.

Theorem 3.1. Suppose that 0 < a, q < 1, b > 0 are constants. Let

Sn = an +C1na

n−1b

Γ(q + 1)+C2na

n−2b2

Γ(2q + 1)+ · · ·+ bn

Γ(nq + 1), n ∈ N.

Then

Sn = o

(1

nd+1

), n→∞,

for any real constants d > 0.Proof. It is not difficult to verify that

limm→∞

(am−1m

(m

m− 1

)m−1) 1m

= a < 1.


Therefore we can choose a positive integer M > 2, which is large enough such that(aM−1M

(M

M − 1

)M−1) 1M

≡ w < 1. (3.1)

For any positive integer n > 2M , set

n = Mj + p (0 ≤ p < M),

where M is given above. Clearly, j = [n/M ] < [n/2] Therefore, for any large enough positiveinteger n > 2M , it follows from the Stirling formula

n! =

(n

e

)n√2πn

(1 +O

(1

n

))and the equality (3.1) that

Sn ≡ C0na

n +C1na

n−1b1

Γ(q + 1)+C2na

n−2b2

Γ(2q + 1)+ · · ·+ Cjnan−jbj

Γ(jq + 1)

≤ Cjnan−j(

1 +b1

Γ(q + 1)+

b2

Γ(2q + 1)+ · · ·+ bj

Γ(jq + 1)

)= an−j

n!

j!(n− j)!O(1)

=O(1)an−jnn

√2πn

jj√

2πj√

2π(n− j)(n− j)n−j

= O

(M j

√j

)(aM

M − 1

)(M−1)j

= O

((aM−1M( M

M−1)M−1)j√j

)= O

(wMj

√j

)= O

(wn√n

).

On the other hand, without loss of generality, we can suppose that b > 1. By the Stirling formula

Γ(z + 1) =√

2πz

(z

e

)z(1 +O

(1

z

)), z → +∞

and

C[n2

]n = O(2n/

√n)

(see [25]), we obtain

˜Sn ≡Cj+1n an−j−1bj+1

Γ((j + 1)q + 1)+ · · ·+ Cnna

n−nbn

Γ(nq + 1)


≤ 1

Γ((j + 1)q + 1)C

[n2

]n (an−j−1bj+1 + · · ·+ an−nbn)

=O(

2n√n

)e(j+1)q(an−j−1bj+1 + · · ·+ an−nbn)√

2π(j + 1)q((j + 1)q)(j+1)q(

1 +O(

1(j+1)q

))≤O(

2n√n

)e(j+1)q(1 + b+ b2 + · · ·+ bj+1 + · · ·+ bn)√

2π(j + 1)q((j + 1)q)(j+1)q

≤ O(1)2ne(j+1)qbn+1

√n√

(j + 1)q((j + 1)q)(j+1)q

≤O(1)2n

(eq

)(j+1)qbn+1

j(j+1)q+1

= o

(1

jd+1

)= o

(1

nd+1

)(n→∞),

where d > 0 can be any real constant.

Therefore,

Sn = Sn + ˜Sn

= O

(wn√n

)+ o

(1

nd+1

)= o

(1

nd+1

)(n→∞),

where d > 0 can be any real constant. This ends the proof of this theorem.

Theorem 3.2. Assume that the following conditions hold:(I1) f : J ×X ×X → D(B) is continuous, and there exist positive constants l1, l2 such that

‖f(t, x1, x2)−f(t, y1, y2)‖ ≤ l1‖x1 − y1‖+ l2‖x2 − y2‖,t ∈ J, xi, yi ∈ X, i = 1, 2;

(I2) h : Υ×X → X is continuous, and there exists a positive constant l3 such that

‖h(t, s, x)− h(t, s, y)‖ ≤ l3‖x− y‖, (t, s) ∈ Υ, x, y ∈ X;

(I3) there exists a positive constants ck (k = 1, 2, . . . ,m) such that

‖Ik(x)− Ik(y)‖ ≤ ck‖x− y‖, k = 1, 2, . . . ,m, x, y ∈ X;

(I4)

Mm∑k=1

ck < 1.


Then the system (1.1) has a unique mild solution on J.Proof. Define an operator P by

(Pu)(t) =Q(t)u0 +

∫ t

0(t− s)q−1R(t− s)f

(s, u(s),

∫ s


)ds

+∑

0<tk<t

Q(t− tk)Ik(u(tk)). (3.2)

It is not difficult to verify that P maps PC(J,X) into PC(J,X).We want to prove the following (3.3) by induction:For any t ∈ J, u, v ∈ PC(J,X),

‖(Pnu)(t)− (Pnv)(t)‖ ≤Mnn∑i=0

Cin

( m∑k=1

ck

)n−i(l1 + %T l2l3)i(tq)i

Γ(iq + 1)‖u− v‖PC ,

n = 1, 2, . . . (3.3)

For n = 1, by (2.3), we get

‖(Pu)(t)− (Pv)(t)‖ ≤∑

0<tk<t

MCk‖u(tk)− v(tk)‖

+M

Γ(q)

∫ t

0(t− s)q−1

[l1‖u(s)− v(s)‖+ l2l3%

∫ s

0‖u(τ)− v(τ)‖dτ

]ds

≤M[ m∑k=1

ck +1

Γ(q + 1)(l1 + %T l2l3)tq

]‖u− v‖PC .

So, (3.3) holds for n = 1.Suppose that (3.3) holds for n = l, that is,

‖(P lu)(t)− (P lv)(t)‖ ≤M ll∑

i=0

Cil

( m∑k=1

ck

)l−i(l1 + %T l2l3)i(tq)i

Γ(iq + 1)‖u− v‖PC .

Then, we have

‖(P l+1u)(t)− (P l+1v)(t)‖ ≤∑

0<tk<t

MCk‖(P lu)(tk)− (P lv)(tk)‖

+M

Γ(q)

∫ t

0(t− s)q−1

[l1‖(P lu)(s)− (P lv)(s)‖

+ l2l3%

∫ s

0‖(P lu)(τ)− (P lv(τ)‖dτ

]ds

≤Mm∑k=1

ckMl

l∑i=0

Cil

( m∑k=1

ck

)l−i(l1 + %T l2l3)i(tq)i

Γ(iq + 1)‖u− v‖PC


+M(l1 + %T l2l3)

Γ(q)

∫ t

0(t− s)q−1.

M ll∑

i=0

Cil

( m∑k=1

ck

)l−i(l1 + %T l2l3)i(sq)i

Γ(iq + 1)‖u− v‖PCds.

In view of ∫ t

0(t− s)q−1smqds = t(m+1)qB(q,mq + 1)(m ∈ N),

and

B(p, q) =Γ(p)Γ(q)

Γ(p+ q)(p, q > 0),

we obtain

‖(P l+1u)(t)− (P l+1v)(t)‖ ≤M l+1l∑

i=0

Cil

( m∑k=1

ck

)l+1−i(l1 + %T l2l3)i(tq)i


+M l+1l∑

i=0

Cil

( m∑k=1

ck

)l−i(l1 + %T l2l3)i+1(tq)i+1

Γ((i+ 1)q + 1)‖u− v‖PC

= M l+1

[( m∑k=1

ck)l+1

+l∑

i=1

Cil

( m∑k=1

ck

)l−i+1

(l1 + %T l2l3)i(tq)i

Γ(iq + 1)

]‖u− v‖PC

+M l+1l∑

i=0

Cil

( m∑k=1

ck

)l−i(l1 + %T l2l3)i+1(tq)i+1

Γ((i+ 1)q + 1)‖u− v‖PC

= M l+1

[( m∑k=1

ck)l+1

+

l∑i=1

Cil

( m∑k=1

ck

)l−i+1


Γ(iq + 1)

]‖u− v‖PC

+M l+1l+1∑i=1

Ci−1l

( m∑k=1

ck

)l−i+1



= M l+1

[( m∑k=1

ck)l+1

+

l∑i=1

(Cil + Ci−1l )

( m∑k=1

ck

)l−i+1


Γ(iq + 1)

]‖u− v‖PC


+M l+1 (l1 + %T l2l3)l+1(tq)l+1

Γ((l + 1)q + 1)‖u− v‖PC .

Since

Cim + Ci−1m = Cim+1,

we have

‖(P l+1u)(t)− (P l+1v)(t)‖ ≤M l+1

[( m∑k=1

ck)l+1

+l∑

i=1

Cil+1

( m∑k=1

ck

)l−i+1


Γ(iq + 1)

]‖u− v‖PC

+M l+1 (l1 + %T l2l3)l+1(tq)l+1

Γ((l + 1)q + 1)‖u− v‖PC

= M l+1l+1∑i=0

Cil+1

( m∑k=1

ck

)l+1−i(l1 + %T l2l3)i(tq)i

Γ(iq + 1)

]‖u− v‖PC .

Hence, (3.3) holds for n = l + 1.Consequently, we see that

‖(Pnu)(t)− (Pnv)(t)‖ ≤Mnn∑i=0

Cin

( m∑k=1

ck


Γ(iq + 1)‖u− v‖PC , n = 1, 2, . . .

Therefore,

‖Pnu− Pnv‖PC ≤Mnn∑i=0

Cin

( m∑k=1

ck


Γ(iq + 1)‖u− v‖PC , n = 1, 2, . . .

Write

a = M

m∑k=1

ck < 1, b = M(l1 + %T l2l3)T q.

By Theorem 3.1, we know that there exists a positive integer n0 such that

Mn0

n0∑i=0

Cin0

( m∑k=1

ck

)n0−i(l1 + %T l2l3)i(T q)i

Γ(iq + 1)< 1,

that is, Pn0 is a contraction mapping on PC(J,X). Then by a well-known extension of theBanach contraction mapping theorem, P has a unique fixed point u(t) on PC(J,X), which is


the unique mild solution of the system (1.1).

Next, we consider the continuous dependence of a mild solution to the system (1.1).

Theorem 3.3. Suppose that the conditions (I1) - (I4) hold. Let u(t), v(t) be the unique mildsolutions of the system (1.1) and the following system (3.4), respectively,


(t, u(t),

∫ t


), t ∈ J ′,

∆u|t=tk = Ik(u(tk)), k = 1, 2, . . . ,m,

u(0) = v0

(3.4)

where v0 ∈ D(B). Then there exists a constant M > 0 such that

‖u− v‖PC ≤ M‖u0 − v0‖.

Proof. Let u(t) and v(t) satisfy the following two equations respectively:

u(t) =Q(t)u0 +

∫ t

0(t− s)q−1R(t− s)f

(s, u(s),

∫ s


)ds

+∑

0<tk<t

Q(t− tk)Ik(u(tk)),

v(t) =Q(t)v0 +

∫ t

0(t− s)q−1R(t− s)f

(s, v(s),

∫ s

0ρ(s, τ)h(s, τ, vρ(τ, v(τ)))dτ

)ds

+∑

0<tk<t

Q(t− tk)Ik(v(tk)).

We want to prove the following (3.5) by induction:

‖u(t)− v(t)‖ ≤M[1 +

n∑j=1

M jj∑i=0

Cij

( m∑k=1

ck

)j−i(l1 + %T l2l3)i(tq)i

Γ(iq + 1)

]‖u0 − v0‖

+Mn+1n+1∑i=0

Cin+1

( m∑k=1

ck

)n+1−i(l1 + %T l2l3)i(tq)i

Γ(iq + 1)‖u− v‖PC , n = 1, 2, . . .

(3.5)

First, we have

‖u(t)− v(t)‖ ≤M‖u0 − v0‖+M

Γ(q)

∫ t

0(t− s)q−1(l1 + %T l2l3)‖u− v‖PCds+M

m∑k=1

ck‖u− v‖PC

= M‖u0 − v0‖+M

[ m∑k=1

ck +(l1 + %T l2l3)tq

Γ(q + 1)

]‖u− v‖PC .


For n = 1,

‖u(t)− v(t)‖ ≤M‖u0 − v0‖+M

Γ(q)

∫ t

0(t− s)q−1(l1 + %T l2l3)

M‖u0 − v0‖

+M

[ m∑k=1

ck +(l1 + %T l2l3)sq

Γ(q + 1)

]‖u− v‖PC

ds

+Mm∑k=1

ck

M‖u0 − v0‖+M

[ m∑k=1


Γ(q + 1)

]‖u− v‖PC

≤M[1 +M

( m∑k=1


Γ(q + 1)

)]‖u0 − v0‖

+M2

[( m∑k=1

ck

)2

+

2

( m∑k=1

ck

)(l1 + %T l2l3)tq

Γ(q + 1)+

(l1 + %T l2l3)2(tq)2

Γ(2q + 1)

]‖u− v‖PC .

So, (3.5) holds for n = 1.Suppose that (3.5) holds for n = l, that is,

‖u(t)− v(t)‖ ≤M[1 +

l∑j=1

M jj∑i=0

Cij

( m∑k=1

ck


Γ(iq + 1)

]‖u0 − v0‖

+M l+1l+1∑i=0

Cil+1

( m∑k=1

ck

)l+1−i(l1 + %T l2l3)i(tq)i

Γ(iq + 1)‖u− v‖PC .

Then we obtain

‖u(t)− v(t)‖ ≤M‖u0 − v0‖+M

Γ(q)

∫ t

0(t− s)q−1(l1 + %T l2l3)

M

[1

+l∑

j=1

M jj∑i=0

Cij

( m∑k=1

ck

)j−i(l1 + %T l2l3)i(sq)i

Γ(iq + 1)

]‖u0 − v0‖

+M l+1l+1∑i=0

Cil+1

( m∑k=1

ck

)l+1−i(l1 + %T l2l3)i(sq)i


ds

+Mm∑k=1

ck

M

[1 +

l∑j=1

M jj∑i=0

Cij

( m∑k=1

ck


Γ(iq + 1)

]‖u0 − v0‖


+M l+1l+1∑i=0

Cil+1

( m∑k=1

ck

)l+1−i(l1 + %T l2l3)i(tq)i


= M

[1 +

l+1∑j=1

M jj∑i=0

Cij

( m∑k=1

ck


Γ(iq + 1)

]‖u0 − v0‖

+M l+2l+2∑i=0

Cil+2

( m∑k=1

ck

)l+2−i(l1 + %T l2l3)i(tq)i

Γ(iq + 1)‖u− v‖PC .

Hence, (3.5) holds for n = l + 1.

We have thus proved (3.5) by induction.Therefore, we see that

‖u(t)− v(t)‖PC ≤M[1 +

n∑j=1

M jj∑i=0

Cij

( m∑k=1

ck

)j−i(l1 + %T l2l3)i(T q)i

Γ(iq + 1)

]‖u0 − v0‖

+Mn+1n+1∑i=0

Cin+1

( m∑k=1

ck

)n+1−i(l1 + %T l2l3)i(T q)i

Γ(iq + 1)‖u− v‖PC , n = 1, 2, . . .

By Theorem 3.1, if we set

M1 =∞∑n=1

Mnn∑i=0

Cin

( m∑k=1

ck

)n−i(l1 + %T l2l3)i(T q)i

Γ(iq + 1)

then we get

limn→∞

Mn+1n+1∑i=0

Cin+1

( m∑k=1

ck

)n+1−i(l1 + %T l2l3)i(T q)i

Γ(iq + 1)= 0.

Hence,

‖u− v‖PC ≤M(1 +M1)‖u0 − v0‖

This implies the conclusion of Theorem 3.3.

For the general case, we will require the following assumptions.


(H1) (i) f : J ×X ×X → D(B) ⊂ X satisfies f(·, ν, ω) : J → D(B) ⊂ X is measurable forall (ν, ω) ∈ X ×X and f(t, ·, ·) : X ×X → D(B) ⊂ X is continuous for a.e. t ∈ J , and thereexists a function µ1(·) ∈ Lp(J,R+)(p > 1

q > 1) and a continulus function µ1(·) such that

‖f(t, ν, ω)‖ ≤ µ1(t)‖ν‖+ µ2(t)‖ω‖

for almost all t ∈ J .(ii) There exists a function η(·) ∈ Lp(J,R+) such that for any bounded sets D1, D2 ⊂ X,

α(f(t,D1, D2)) ≤ η(t)(α(D1) + α(D2)), a.e. t ∈ J.

(H2) (i) The function h(t, s, ·) : X → X is continulus for a.e. (t, s) ∈ ∆, and for each u ∈ X,the function h(·, ·, u) : ∆→ X is measurable.

Moreover, there exists a function m : ∆→ R+ with

supt∈J

∫ t

0m(t, s)ds := m∗ <∞

such that

‖h(t, s, u)‖ ≤ m(t, s)‖u‖, u ∈ X.

(ii) For any bounded set D ⊂ X and 0 ≤ s ≤ t ≤ T , there exists a function ζ : ∆ → R+

such that

α(h(t, s,D)) ≤ ζ(t, s)α(D),

where

supt∈J

∫ t

0ζ(t, s)ds := ζ∗ <∞.

(H3) (i) There exists positive constants Lk, Nk (k = 1, 2, . . . ,m) such that

‖Ik(x)‖ ≤ Lk‖x‖+Nk, x ∈ X.

(ii) There exists positive constant Mk (k = 1, 2, . . . ,m) such that for any bounded set D ⊂ X,

α(Ik(D)) ≤Mkα(D).

(H4)

M

Γ(q)

(lp,qT

q− 1p ‖µ1‖Lp +

T q%m∗µ∗2q

)+M

m∑k=1

Lk < 1, (3.6)

and

2M

Γ(q)

(lp,qT

q− 1p ‖ η‖Lp +

2%ζ∗T q

q

)+M

m∑k=1

Mk <1

2, (3.7)


where

lp,q :=

(p− 1

pq − 1

) p−1p

, µ∗2 = supt∈J

µ2(t).

Theorem 3.4. Assume that (H1)− (H4) are satisfied. Then the system (1.1) has at least onemild solution on J.Proof. Consider the operator P defined in (3.2). P maps PC(J,X) into PC(J,X). Set

Br = u ∈ PC(J,X) : ‖u‖PC ≤ r.

Step 1. We show that there exists some r > 0 such that P (Br) ⊂ Br.Suppose this is not true. Then for each r > 0, there exists ur(·) ∈ Br and some t ∈ J such that

‖(Pur)(t)‖ > r.

It follows from (H1)(i), (H2)(i), (H3)(i) that

r < ‖Q(t)u0‖+

∫ t

0(t− s)q−1

∥∥∥∥R(t− s)f(s, ur(s),

∫ s

0ρ(s, τ)h(s, τ, uρ(τ, ur(τ)))dτ

)ds

∥∥∥∥+∑

0<tk<t

‖Q(t− tk)Ik(ur(tk))‖

≤M‖u0‖+M

Γ(q)

∫ t

0(t− s)q−1

[µ1(s)r + µ2(s)%r

∫ s

0m(s, τ)dτ

]ds+M

m∑k=1

(rLk +Nk)

≤M‖u0‖+Mr

Γ(q)

(∫ t

0(t− s)q−1µ1(s)ds+

T q%m∗µ∗2q

)+Mr

m∑k=1

Lk +Mm∑k=1

Nk

Moreover, by Holder inequality, we have∫ t

0(t− s)q−1µ1(s)ds ≤ t

pq−1p lp,q‖µ1‖Lp ≤ lp,qT q−

1p ‖µ1‖Lp .

Thus

r ≤M‖u0‖+Mr

Γ(q)

(lp,qT

q− 1p ‖µ1‖Lp +

T q%m∗µ∗2q

)+Mr

m∑k=1

Lk +Mm∑k=1

Nk. (3.8)

Dividing both sides of (3.8) by r, and taking r →∞, we get

M

Γ(q)

(lp,qT

q− 1p ‖µ1‖Lp +

T q%m∗µ∗2q

)+M

m∑k=1

Lk ≥ 1.

This contradicts (3.6). Hence for some positive number r, PBr ⊂ Br.Step 2. We show that P is continuous from Br into Br.Let un ∈ Br, n = 1, 2, . . . , be a sequence such that un → u ∈ Br in PC(J,X) as n→∞. By

(H1)(i) and (H2)(i), we see that for almost every t ∈ J ,

f

(t, un(t),

∫ t

0ρ(t, s)h(t, s, uρ(s, un(s)))ds

)→ f

(t, u(t),

∫ t


),


as n → ∞. Noting that un → u in PC(J,X), we infer that there exists ε > 0 such that‖un − u‖PC ≤ ε for k sufficiently large. Therefore, we have∥∥∥∥f(t, un(t),

∫ t


)− f

(t, u(t),

∫ t


)∥∥∥∥≤ µ1(t)(‖un(t)‖+ ‖u(t)‖) + µ2(t)

(∫ t

0‖ρ(t, s)h(t, s, uρ(s, un(s)))‖ds+

∫ t

0‖ρ(t, s)h(t, s, uρ(s, u(s)))‖ds

)≤ µ1(t)(‖un(t)− u(t)‖+ 2‖u(t)‖) + µ2(t)%

∫ t

0m(t, s)(‖un(s)− u(s)‖+ 2‖u(s)‖)ds

≤ (µ1(t) + µ∗2%m∗)(ε+ 2 sup

t∈J‖u(t)‖).

It follows from the continuity of Ik(k = 1, 2, . . . ,m) and the Lebesgue’s Dominated ConvergenceTheorem that

‖(Pun)(t)− (Pu)(t)‖ ≤∫ t

0(t− s)q−1

∥∥∥∥R(t− s)[f

(t, un(t),

∫ t


)− f

(t, u(t),

∫ t


)]∥∥∥∥ds+∑

0<tk<t

‖Q(t− tk)[Ik(un(tk))− Ik(u(tk))]‖

≤ M

Γ(q)

∫ t

0(t− s)q−1

∥∥∥∥f(t, un(t),

∫ t


)− f

(t, u(t),

∫ t


)∥∥∥∥ds+M

∑0<tk<t

‖Ik(un(tk))− Ik(u(tk))]‖

→ 0, n→∞.

Therefore, we obtain

limn→∞

‖Pun − Pu‖PC = 0.

this means that P is continuous.

Step 3. We prove that Pu : u ∈ Br is a family of equicontinuous functions.

Since W (t) is strongly continuous on D(B) for t ≥ 0, we know that Q(·)u0 : · ∈ J isequicontinuous.

Without loss of generality, we suppose that tj ≤ r2 < r1 < tj+1, u ∈ Br. Then we obtain∥∥∥∥∫ r1

0(r1 − s)q−1R(r1 − s)f

(s, u(s),

∫ s


)ds

−∫ r2

0(r2 − s)q−1R(r2 − s)f

(s, u(s),

∫ s


)ds

∥∥∥∥


≤∫ r2

0

∥∥∥∥[(r1 − s)q−1R(r1 − s)− (r2 − s)q−1R(r2 − s)]

× f(s, u(s),

∫ s


)∥∥∥∥ds+

∫ r1

r2

(r1 − s)q−1‖R(r1 − s)‖∥∥∥∥f(s, u(s),

∫ s


)∥∥∥∥ds=: I1 + I2.

I1 ≤∫ r2

0|(r1 − s)q−1 − (r2 − s)q−1|‖R(r1 − s)‖ ×

∥∥∥∥f(s, u(s),

∫ s


)∥∥∥∥ds+

∫ r2

0(r2 − s)q−1‖R(r1 − s)−R(r2 − s)‖ ×

∥∥∥∥f(s, u(s),

∫ s


)∥∥∥∥ds≤ M

Γ(q)

∫ r2

0|(r1 − s)q−1 − (r2 − s)q−1|

(µ1(s)‖u(s)‖+ µ∗2%m

∗ supτ∈[0,s]

‖u(τ)‖)ds

+

∫ r2

0(r2 − s)q−1‖R(r1 − s)−R(r2 − s)‖

(µ1(s)‖u(s)‖+ µ∗2%m

∗ supτ∈[0,s]

‖u(τ)‖)ds

≤ Mr

Γ(q)

∫ r2

0|(r1 − s)q−1 − (r2 − s)q−1|µ1(s)ds

+ r

∫ r2

0µ1(s)(r2 − s)q−1‖R(r1 − s)−R(r2 − s)‖ds

+M%m∗µ∗2r

Γ(q)

∫ r2

0|(r1 − s)q−1 − (r2 − s)q−1|ds

+ %m∗µ∗2r

∫ r2

0(r2 − s)q−1‖R(r1 − s)−R(r2 − s)‖ds

=: I3 + I4 + I5 + I6.

Clearly, I3 tends to 0 as r2 → r1. I4 tends to 0 as r2 → r1 as a consequence of the continuityof R(t) in the uniform operator topology for t > 0. Similarly, I5 and I6 tends to 0 as r2 → r1,respectively.

For I2, we have

I2 ≤Mr

Γ(q)

[ ∫ r1

r2

(r1 − s)q−1µ1(s)ds+ %m∗µ∗2

∫ r1

r2

(r1 − s)q−1ds

]→ 0, as r2 → r1.

In conclusion,

‖(Pu)(r2)− (Pu)(r1)‖ → 0, as r2 → r1,

which means that the operator P : Br → Br is equicontinuous.Let H = coQ(Br). Then it is easy to see that P maps H into itself and H ⊂ Br is

equicontinuous.Step 4. Now we prove that P : H → H is a condensing operator.For any D ⊂ H, by Lemma 2.6, there exists a countable set D0 = un ⊂ D, such that

α(P (D)) ≤ 2α(P (D0)).


By the equicontinuity of H, we know that D0 ⊂ D is also equicontinuous.For t ∈ J , by Lemma 2.5 and Lemma 2.8, we have

α(P (D0)(t)) = α

[(Q(t)u0 +

∫ t

0(t− s)q−1R(t− s)f

(s, un(s),

∫ s

0ρ(s, τ)h(s, τ, uρ(τ, un(τ)))dτ

)ds

+∑

0<tk<t

Q(t− tk)Ik(un(tk))

)]

≤ 2M

Γ(q)

∫ t

0(t− s)q−1α

[(f

(s, un(s),

∫ s

0ρ(s, τ)h(s, τ, uρ(τ, un(τ)))dτ

))]ds

+M∑

0<tk<t

α

[(Ik(un(tk))

)]

≤ 2M

Γ(q)

∫ t

0(t− s)q−1

[η(s)

(α(D0) + 2%

∫ s

0ζ(s, τ)dτα(D0)

)]ds+M

m∑k=1

Mkα(D0)

≤[

2M

Γ(q)

∫ t

0(t− s)q−1

(η(s) + 2%ζ∗

)ds+M

m∑k=1

Mk

]α(D0)

≤[

2M

Γ(q)

(lp,qT

q− 1p ‖η‖Lp +

2%ζ∗T q

q

)+M

m∑k=1

Mk

]α(D).

Since P (D0) ⊂ H is bounded and equicontinuous, we know from Lemma 2.7 that

α(P (D0)) = maxt∈J

α(P (D0)(t)).

Therefore, by (3.7) we have

α(P (D)) ≤ 2

[2M

Γ(q)

(lp,qT

q− 1p ‖η‖Lp +

2%ζ∗T q

q

)+M

m∑k=1

Mk

]α(D) < α(D).

Thus P : H → H is a condensing mapping. It follows from Theorem 2.9 that P has at least onefixed point u ∈ H, which is just a mild solution of the system (1.1).

When B = I, then D(B) = X. We assume that A generates a norm continuous semigroupW (t)t≥0 of uniformly bounded linear operators on X. Then we have the following theorems.

Theorem 3.5. Assume that (I1)− (I4) are satisfied. Then the systemcDqu(t) = Au(t) + f

(t, u(t),

∫ t


), t ∈ J ′,

∆u|t=tk = Ik(u(tk)), k = 1, 2, . . . ,m,

u(0) = u0, u0 ∈ X,

(3.9)

has a unique mild solution on J.


Theorem 3.6. Assume that (I1)− (I4) are satisfied. Let u(t),v(t) be the unique mild solutionsof the system (3.9) and the following system (3.10), respectively,

cDqu(t) = Au(t) + f

(t, u(t),

∫ t

0ρ(t, s)h(t, s, uρ(s, us)))ds

), t ∈ J ′,

∆u|t=tk = Ik(u(tk)), k = 1, 2, . . . ,m,

u(0) = v0, v0 ∈ X.

(3.10)

Then there exists a constant M > 0 such that

‖u− v‖PC ≤ M‖u0 − v0‖.

Theorem 3.7. Assume that (I1) − (I4) are satisfied. Then the system (3.9) has at least onemild solution on J.

4 An application

In this section, we give an example to illustrate the main results obtained above.

Example 4.1. Consider the following impulsive Cauchy problem for fractional partial-integraldifferential equations

cDq(u−∆u)(t) = ∆u+ f

(t, u(t),

∫ t

0ρ(t, s)h(t, s, uρ(s, us)))ds

)−∆f, t ∈ J ′,

∆u|t=tk = Ik(u(tk)), k = 1, 2, . . . ,m,

u(0) = u0,

(4.1)

where

q =1

2, ∆ =

∂2

∂x21

+∂2

∂x22

+∂2

∂x23

, ρ(t, s) = 1,

h(t, s, u(s, x)) =1

5s2 · sin u(s, x)

t, f(t, u, v) =

1

5k · k√tu+

t2

5kv,

m = 3, t1 =1

4, t2 =

1

2, t3 =

3

4,

I1(u) =1

50u, I2(u) =

1

50sinu, I3(u) ≡ u1,

where

1k√t∈ Lp([0, 1], R+)(p > 2), u0, u1 ∈ H2(R3).

Take X = L2(R3) and define

A = ∆, B = I −∆,


with

D(A) = D(B) = H2(R3).

Then the problem (4.1) is transformed into the following fractional Sobolev-type integro-differentialequations in the Banach space X with operator pairs (A,B) and impulsive conditions


(t, u(t),

∫ t


), t ∈ J ′,

∆u|t=tk = Ik(u(tk)), k = 1, 2, . . . ,m,

u(0) = u0,

(4.2)

By [22], we know that the pair (A,B) generates a propagation family W (t) of uniformly bounded.Moreover, by the arguments similar to those in the proof of (2.15), (2.16) and (2.17) in [22], wededuce that W (t)t≥0 is norm continuous for t > 0 and ‖W (t)‖ ≤ 1.

We have

‖f(t, u, v)‖ ≤ 1

5k · k√t‖u‖+

t2

5k‖v‖

:= µ1(t)‖u‖+ µ2(t)‖v‖,

and for any bounded sets D1, D2 ⊂ X,

f(t,D1, D2) ≤ 1

5k · k√t(α(D1) + α(D2)) := η(t)(α(D1 + α(D2)), t ∈ (0, 1].

Moreover,

‖h(t, s, u)‖ ≤ s2

5t‖u‖ := m(t, s)‖u‖

and

supt∈[0,1]

∫ t

0m(t, s)ds = sup

t∈[0,1]

∫ t

0

s2

5tds =

1

15:= m∗.

For any u1, u2 ∈ X,

‖h(t, s, u1)− h(t, s, u2)‖ ≤ s2

5t‖u1 − u2‖.

So, for any bounded set D ⊂ X,

α(h(t, s,D)) ≤ s2

5tα(D) := ζ(t, s)α(D)

and

supt∈[0,1]

∫ t

0ζ(t, s)ds = sup

t∈[0,1]

∫ t

0

s2

5tds =

1

15:= ζ∗.


If we take p = 4, k = 5, then

‖µ1‖Lp([0,1],R+) = ‖η‖Lp([0,1],R+) =

(1

5

) 74

, µ∗2 =1

25.

Noting that

Γ

(1

2

)=√π, lp,q :=

(p− 1

pq − 1

) p−1q

= 334

L1 = M1 =1

50, L2 = M2 =

1

50, L3 = M3 = 0,

we have

M

Γ(q)

(lp,qT

q− 1p ‖µ1‖Lp +

T q%m∗µ∗2q

)+M

m∑k=1

Lk =1√π

(1

5

(3

5

) 34 +

2

375

)+

1

25

≈ 0.43

< 1,

2M

Γ(q)

(lp,qT

q− 1p ‖η‖Lp +

2%ζ∗T q

q

)+M

m∑k=1

Mk =2√π

(1

5

(3

5

) 34 +

4

15

)+

1

25

≈ 0.49

< 0.5.

Hence (H1)− (H4) are satisfied. This means that the problem (4.2), i.e., the problem (4.1) has

a mild solution by Theorem 3.4.

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Received: April 21, 2021; Published: June 5, 2021

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Existence of Solutions to Impulsive Fractional Sobolev-Type ......34 D. N. Chalishajar, D. Senthil Raja, P. Sundararajan and K. Karthikeyan De nition 2.2. ([2, 19]) The Riemann-Liouville