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. 2018 Sep 25;2018(1):261. doi: 10.1186/s13660-018-1855-z

Noninstantaneous impulsive inequalities via conformable fractional calculus

Surang Sitho 1, Sotiris K Ntouyas 2,3, Praveen Agarwal 4, Jessada Tariboon 5,
PMCID: PMC6182428  PMID: 30363841

Abstract

We establish some new noninstantaneous impulsive inequalities using the conformable fractional calculus.

Keywords: Conformable fractional derivative, Conformable fractional integral, Impulsive inequalities

Introduction and preliminaries

The subject of fractional differential equations has evolved as an interesting and important field of research in view of numerous applications in physics, mechanics, chemistry, engineering (like traffic, transportation, logistic, etc.), and so forth [13]. The tools of fractional calculus play a key role in improving the mathematical modeling of many real-world processes based on classical calculus. For some recent development on the topic, see [412] and the references therein.

Various types of fractional derivatives were introduced: Riemann–Liouville, Caputo, Hadamard, Erdélyi–Kober, Grünwald–Letnikov, Marchaud, and Riesz, to just name a few. Commonly, all they are defined as integrals with different singular kernels, that is, they have a nonlocal structure. Due to this fact, there are many inconsistencies of the existing fractional derivatives with classical derivative. Thus they do not obey the familiar product rule, the quotient rule for two functions, and the chain rule. Also, the fractional derivatives do not have a corresponding Rolle’s theorem or a corresponding mean value theorem.

On the other hand, a recently introduced definition of the so-called conformable fractional derivative involves a limit instead of an integral; see [13, 14]. This local definition enables us to prove many properties analogous to those of integer-order derivatives. The authors in [14] showed that the conformable fractional derivative obeys the product and quotient rules and has results similar to the Rolle theorem and the mean value theorem in classical calculus.

For recent works on conformable derivatives, we refer to [1519] and the references therein.

Let us recall the definition of the conformable fractional derivative and integral.

Definition 1.1

Let 0<α1. The conformable fractional derivative starting from a point ϕ of a function f:[ϕ,)R is defined by

Dαϕ(t)=limϵ0f(t+ϵ(tϕ)1α)f(t)ϵ,t>ϕ, 1.1

and Dαϕf(ϕ)=limtϕ+ϕDαf(t).

Note that if f is differentiable, then

Dαϕf(t)=(tϕ)1αdf(t)dt. 1.2

Definition 1.2

Let 0<α1. The conformable fractional integral of a function f:[ϕ,)R from a point ϕ is defined by

Iαϕ(t)=ϕt(sϕ)α1f(s)ds. 1.3

The impulsive differential equations have been used to describe processes that have sudden changes in their states at certain moments. Many mathematical models in physical phenomena that have short-term perturbations at fixed impulse points tk, k=1,2,3, , caused by external interventions during their evolution appeared in population dynamics, biotechnology processes, chemistry, physics, engineering, and medicine; see [2022]. In [23, 24], the authors introduced a new class of noninstantaneous impulsive differential equations with initial conditions to describe some certain dynamic changes of evolution processes in the pharmacotherapy. This kind of impulsive differential equations can be distinguished from the usual one as the changing processes containing no ordinary or fractional derivatives of their states work over intervals (tk,sk], whereas the usual does at points tk, k=1,2,3, . There are some papers on existence and stability theory of this kind of impulsive ordinary or fractional differential equations [2536]. To the best of our knowledge, there is no literature on noninstantaneous impulsive inequalities. The main goal of the paper is to establish some new noninstantaneous impulsive inequalities using the conformable fractional calculus. The main results are presented in Sect. 2. In Sect. 3, the maximum principle and boundedness of solutions for noninstantaneous impulse problems are illustrated.

Main results

Assume that the independent variable t is the time defined on the half-line R+=[0,). Let {ti}i=1 and {si}i=0 be two increasing sequences such that

0=s0<t1s1<t2s2<t3<tisi<ti+1

for i=1,2, and limktk=limksk=. In addition, we define subsets of R+ by Usk=k=0(sk,tk+1], Utk=k=1(tk,sk] and U=UskUtk. Note that U{0}=R+. Set PC(Usk,R) = {x:UskR;x(t) is continuous on Usk, and x(sk+) exists for k=0,1,2,}, PC(Utk,R) = {x:UtkR; x(t) is continuous for tUtk, and x(tk+) exists for k=1,2,3,}, PCskα(Usk,R) = {xPC(Usk,R):skDαx(t) is continuous everywhere for tUsk, and Dαskx(sk+) exists for k=0,1,2,}, PCtkα(Utk,R) = {xPC(Utk,R):tkIαx(t) is continuous everywhere for all tUtk, and Iαtkx(tk+) exists for k=1,2,3,}, and PCα(U,R)=PCskα(Usk,R)PCtkα(Utk,R).

Let the maximums of impulsive points less than or equal to t be defined by

sm=max{sk:skt,k=0,1,2,}andtm=max{tk:tkt,k=1,2,3,}. 2.1

In addition, we define

Hk=edk(sktk)αα;Qk=esk1tkp(ξ)(ξsk1)α1dξ;Gk=QkHk;Pk=QkHk1.

Note that

HmGm+1Gm+2Gm+3Gn1Qn=Pm+1Pm+2Pn1Pn 2.2

and

PmPm+1Pm+2Pn1PnHn=Hm1GmGm+1Gn1Gn, 2.3

where m<n are positive integers.

Throughout this paper, we assume that the unknown function uPCα(U,R) is left-continuous at sk and tk (k=1,2,3,). Now, we are in the position to establish noninstantaneous impulsive differential inequalities.

Theorem 2.1

Let bk, ck, dk be given constants such that bk,ck0 and dk>0, k=1,2,3, . Suppose that p,qPC(Usk,R) and

{Dαsku(t)p(t)u(t)+q(t),t(sk,tk+1],k=0,1,2,,u(t)cku(tk)+dktkt(ξtk)α1u(ξ)dξ+bk,t(tk,sk],k=1,2,. 2.4

Then

u(t)esmtp(ξ)(ξsm)α1dξ{u(s0)0<kmckGk+0<km(k<jmcjGj)Hkbk+0<km(k<jmcjGj)ckHksk1tkq(η)(ηsk1)α1eηtkp(ξ)(ξsk1)α1dξdη}+smtq(η)(ηsm)α1eηtp(ξ)(ξsm)α1dξdη,t(sk,tk+1],k=0,1,2,, 2.5

and

u(t)edm(ttm)αα{u(s0)cmQm0<k<mckGk+0<km(k<jmcjPj)bk+0<km(k<jmcjPj)cksk1tkq(η)(ηsk1)α1eηtkp(ξ)(ξsk1)α1dξdη},t(tk,sk],k=1,2,3,. 2.6

Proof

For t(s0,t1], the conformable fractional differential inequality can be written as

Dαs0[u(t)es0tp(ξ)(ξs0)α1dξ]q(t)es0tp(ξ)(ξs0)α1dξ.

By taking the conformable fractional integral of order α from s0 to t(s0,t1],

Iαs0s0Dα[u(t)es0tp(ξ)(ξs0)α1dξ]s0Iα[q(t)es0tp(ξ)(ξs0)α1dξ],

we obtain

u(t)u(s0)es0tp(ξ)(ξs0)α1dξ+s0tq(η)(ηs0)α1eηtp(ξ)(ξs0)α1dξdη,t(s0,t1], 2.7

which implies that (2.5) holds for k=0.

For t(t1,s1], we define the function

z(t)=t1t(ξt1)α1u(ξ)dξ. 2.8

Note that z(t1)=0 and

u(t)c1u(t1)+d1z(t)+b1,t(t1,s1].

Then, taking the derivative with respect to t, we have

z(t)=(tt1)α1u(t)(tt1)α1[c1u(t1)+b1]+d1(tt1)α1z(t).

Multiplying this inequality by the integrating factor ed1(tt1)αα, we get

ddt[z(t)ed1(tt1)αα][c1u(t1)+b1](tt1)α1ed1(tt1)αα,

which implies that

z(t)[c1u(t1)+b1]ed1(tt1)ααt1t(ηt1)α1ed1(ηt1)ααdη=1d1[c1u(t1)+b1][ed1(tt1)αα1].

By (2.7) with t=t1 we have

u(t)c1ed1(tt1)αα[u(s0)es0t1p(ξ)(ξs0)α1dξ+s0t1q(η)(ηs0)α1eηt1p(ξ)(ξs0)α1dξdη]+b1ed1(tt1)αα,t(t1,s1].

This shows that the bound in (2.6) is true for k=1.

Now, we assume that inequality (2.5) holds for t(sn,tn+1], n>0. By mathematical induction we will show that (2.6) is true for t(tn+1,sn+1]. Let

w(t)=tn+1t(ξtn+1)α1u(ξ)dξ,t(tn+1,sn+1].

Then w(tn+1)=0 and u(t)cn+1u(tn+1)+dn+1w(t)+bn+1. Using the above method, we have

w(t)(ttn+1)α1[cn+1u(tn+1)+bn+1]+dn+1(ttn+1)α1w(t),

which leads to

w(t)1dn+1[cn+1u(tn+1)+bn+1][edn+1(ttn+1)αα1].

Substituting the bound of w(t) and inequality (2.5) with t=tn+1, it follows that

u(t)cn+1u(tn+1)+dn+1w(t)+bn+1cn+1u(tn+1)edn+1(ttn+1)αα+bn+1edn+1(ttn+1)ααcn+1[esntn+1p(ξ)(ξsn)α1dξ{u(s0)0<knckGk+0<kn(k<jncjGj)Hkbk+0<kn(k<jncjGj)ckHksk1tkq(η)(ηsk1)α1eηtkp(ξ)(ξsk1)α1dξdη}+sntn+1q(η)(ηsn)α1eηtn+1p(ξ)(ξsn)α1dξdη]edn+1(ttn+1)αα+bn+1edn+1(ttn+1)αα=edn+1(ttn+1)αα{u(s0)cn+1Qn+10<k<n+1ckGk+0<kn+1(k<jn+1cjPj)bk+0<kn+1(k<jn+1cjPj)cksk1tkq(η)(ηsk1)α1eηtkp(ξ)(ξsk1)α1dξdη}

by using formula (2.2). Therefore (2.6) is satisfied for t(tn+1,sn+1].

Finally, we suppose that estimate (2.6) is fulfilled for t(tn,sn], where n>1. Next, we will prove that inequality (2.5) holds for (sn,tn+1]. By using the above method, we get the inequality

u(t)u(sn)esntp(ξ)(ξsn)α1dξ+sntq(η)(ηsn)α1eηtp(ξ)(ξsn)α1dξdη,t(sn,tn+1].

Using (2.6) with t=sn and applying (2.3), we obtain

u(t)esntp(ξ)(ξsn)α1dξ[edn(sntn)αα{u(s0)cnQn0<k<nckGk+0<kn(k<jncjPj)bk+0<kn(k<jncjPj)cksk1tkq(η)(ηsk1)α1eηtkp(ξ)(ξsk1)α1dξdη}]+sntq(η)(ηsn)α1eηtp(ξ)(ξsn)α1dξdη=esntp(ξ)(ξsn)α1dξ[u(s0)0<knckGk+0<kn(k<jncjGj)Hkbk+0<kn(k<jncjGj)ckHksk1tkq(η)(ηsk1)α1eηtkp(ξ)(ξsk1)α1dξdη]+sntq(η)(ηsn)α1eηtp(ξ)(ξsn)α1dξdη.

Therefore inequality (2.5) is valid on (sn,tn+1]. This completes the proof. □

The following corollary can be obtained by replacing the given functions p(t) and q(t) by constants M and N, respectively.

Corollary 2.1

Let bk,ck0 and dk>0, k=1,2,3, , be constants. If M>0, NR, and

{Dαsku(t)Mu(t)+N,t(sk,tk+1],k=0,1,2,,u(t)cku(tk)+dktkt(ξtk)α1u(ξ)dξ+bk,t(tk,sk],k=1,2,, 2.9

then

u(t)eM(tsm)αα{u(s0)0<kmckGk+0<km(k<jmcjGj)Hkbk+NM0<km(k<jmcjGj)ckHk(eM(tksk1)αα1)}+NM(eM(tsm)αα1),t(sk,tk+1],k=0,1,2,, 2.10

and

u(t)edm(ttm)αα{u(s0)cmQm0<k<mckGk+0<km(k<jmcjPj)bk+NM0<km(k<jmcjPj)ck(eM(tksk1)αα1)},t(tk,sk],k=1,2,3,, 2.11

where Qk=eM(tksk1)αα, Gk=QkHk, and Pk=QkHk1.

Let H(t) be the Heaviside function. We define two functions

φ(t)=i=0H(tsi)H(tti+1+)={0,t(tk,sk],k=1,2,3,,1,t(sk,tk+1],k=0,1,2,,

and

ψ(t)=i=1H(tti)H(tsi+)={0,t(sk,tk+1],k=0,1,2,,1,t(tk,sk],k=1,2,3,.

Next, we establish some new noninstantaneous impulsive integral inequalities.

Theorem 2.2

Let pPC(Usk,R+), constants ck,bk0, dk>0, k=1,2,3, , and AR. If

u(t)(A+smt(ξsm)α1p(ξ)u(ξ)dξ)φ(t)+(cmu(tm)+dmtmt(ξtm)α1u(ξ)dξ+bm)ψ(t),tR+, 2.12

where sm and tm are defined by (2.1), then we have

u(t)esmtp(ξ)(ξsm)α1dξ(A0<kmckGk+0<km(k<jmcjGj)Hkbk) 2.13

for t(sk,tk+1], k=0,1,2, , and

u(t)edm(ttm)αα(AcmQm0<k<mckGk+0<km(k<jmcjPj)bk) 2.14

for t(tk,sk], k=1,2,3, .

Proof

To prove inequalities (2.13) and (2.14), for tR+, we define the function

v(t)=(A+smt(ξsm)α1p(ξ)u(ξ)dξ)φ(t)+(cmu(tm)+dmtmt(ξtm)α1u(ξ)dξ+bm)ψ(t),

which yields u(t)v(t) for all tR+ and v(s0)=A. For any t(sk,tk+1], k=0,1,2, , we get

v(t)=A+smt(ξsm)α1p(ξ)u(ξ)dξ.

Also, taking the conformable fractional derivative of order α, we have

Dαsmv(t)=p(t)u(t)p(t)v(t). 2.15

For t(tk,sk], k=1,2,3, , we obtain

v(t)=cmu(tm)+dmtmt(ξtm)α1u(ξ)dξ+bmcmv(tm)+dmtmt(ξtm)α1v(ξ)dξ+bm. 2.16

An application of Theorem 2.1 to (2.15) and (2.16) yields

v(t)esmtp(ξ)(ξsm)α1dξ(A0<kmckGk+0<km(k<jmcjGj)Hkbk)

for t(sk,tk+1], k=0,1,2, , and

v(t)edm(ttm)αα(AcmQm0<k<mckGk+0<km(k<jmcjPj)bk)

for t(tk,sk], k=1,2,3, . From u(t)v(t), tR+, we get the desired results in (2.13) and (2.14). The proof is completed. □

Theorem 2.3

Let pPC(Usk,R+), let h be a positive fractional integrable function of order α, and let ck,bk0 and dk>0, k=1,2,3, , be constants. If

u(t)h(t)+(smt(ξsm)α1p(ξ)u(ξ)dξ)φ(t)+(cmu(tm)+dmtmt(ξtm)α1u(ξ)dξ+bm)ψ(t),tR+, 2.17

where sm and tm are defined by (2.1), then we have

u(t)h(t)+esmtp(ξ)(ξsm)α1dξ{0<km(k<jmcjGj)Hk(bk+ckh(tk)+dkKk)+0<km(k<jmcjGj)ckHksk1tkp(η)h(η)(ηsk1)α1eηtkp(ξ)(ξsk1)α1dξdη}+smtp(η)h(η)(ηsm)α1eηtp(ξ)(ξsm)α1dξdη,t(sk,tk+1],k=0,1,2,, 2.18

and

u(t)h(t)+edm(ttm)αα{0<km(k<jmcjPj)(bk+ckh(tk)+dkKk)+0<km(k<jmcjPj)cksk1tkp(η)h(η)(ηsk1)α1eηtkp(ξ)(ξsk1)α1dξdη},t(tk,sk],k=1,2,3,, 2.19

where the constants Kk, k=1,2,3, , are defined by Kk=tksk(ξtk)α1h(ξ)dξ.

Proof

For tR+, setting

y(t)=(smt(ξsm)α1p(ξ)u(ξ)dξ)φ(t)+(cmu(tm)+dmtmt(ξtm)α1u(ξ)dξ+bm)ψ(t),

we have

Dαsmy(t)=p(t)u(t),y(s0)=0,

for t(sk,tk+1], k=0,1,2, , and

y(t)=cmu(tm)+dmtmt(ξtm)α1u(ξ)dξ+bm

for t(tk,sk], k=1,2,3, . Since u(t)h(t)+y(t), tR+, this reduces to

Dαsmy(t)p(t)y(t)+p(t)h(t),y(s0)=0, 2.20

and

y(t)cmy(tm)+dmtmt(ξtm)α1y(ξ)dξ+(bm+cmh(tm)+dmKm). 2.21

Now Theorem 2.1, together with the inequality u(t)h(t)+y(t), yields estimates (2.18) and (2.19), completing the proof. □

Next, we obtain the following corollary by putting constant values h(t)B>0 and p(t)M>0.

Corollary 2.2

Let constants ck,bk0 and dk>0, k=1,2,3, . If

u(t)B+(Msmt(ξsm)α1u(ξ)dξ)φ(t)+(cmu(tm)+dmtmt(ξtm)α1u(ξ)dξ+bm)ψ(t),tR+, 2.22

where sm and tm are defined by (2.1), then we have

u(t)BeM(tsm)αα+eM(tsm)αα{0<km(k<jmcjGj)HkZk+B0<km(k<jmcjGj)ckHk(eM(tksk1)αα1)}t(sk,tk+1],k=0,1,2,, 2.23

and

u(t)B+edm(ttm)αα{0<km(k<jmcjPj)Zk+B0<km(k<jmcjPj)ck(eM(tksk1)αα1)},t(tk,sk],k=1,2,3,, 2.24

where Gj and Pj are defined as in Corollary 2.1, and Zk=bk+Bck+Bdk(sktk)α/α.

Applications

In this section, we establish two applications of noninstantaneous impulsive differential and integral inequalities. Let J=[0,T] with tn+1=T and J=[0,T] with sn+1=T for some n1. The first purpose is accomplished by considering two problems that have the end points at tn+1 and sn+1, respectively. Now, we consider

{Dαsku(t)Mu(t)+a(t)0,t(sk,tk+1],k=0,1,2,,n,u(t)cku(tk)+dktkt(ξtk)α1u(ξ)dξ,t(tk,sk],k=1,2,3,,n,u(0)=u(T)+λ, 3.1

and

{Dαskv(t)Mv(t)+a(t)0,t(sk,tk+1],k=0,1,2,,n,v(t)ckv(tk)+dktkt(ξtk)α1v(ξ)dξ,t(tk,sk],k=1,2,3,,n+1,v(0)=v(T)+λ, 3.2

where M>0, a(t)C[R+,R+], ck0, and dk>0. Let us state the following conditions:

(H1)

eM(Tsn)ααk=1nckG<1,

(H2)

λeM(Tsn)ααsnTa(η)(ηsn)α1eM(ηsn)ααdη,

(H3)

k=1n+1ckGk<1,

(H4)
λedn+1(Ttn+1)ααk=1n+1(k<jn+1cjPj)ckDk, where Dk is defined by
Dk=eM(tksk1)ααsk1tka(η)(ηsk1)α1eM(ηsk1)ααdη,k=1,2,,n+1.

Corollary 3.1

Let u and v be unknown functions satisfying (3.1) and (3.2), respectively. If (H1)(H2) hold, then u(t)0 for tJ. If (H3)(H4) hold, then v(t)0 for tJ.

Proof

Applying Theorem 2.1 to the first two inequalities in problem (3.1), we have

u(t)u(0)eM(tsm)αα0<kmckGkeM(tsm)αα0<km(k<jmcjGj)ckHkDkeM(tsm)ααsmta(η)(ηsm)α1eM(ηsm)ααdη,t(sk,tk+1],k=0,1,2,,n.

Since a(t)0 for all tR+ and all constants are positive, it is sufficient to show that u(0)0. At the end point t=T, we obtain

u(T)u(0)eM(Tsn)ααk=1nckGkeM(Tsn)ααk=1n(k<jncjGj)ckHkDkeM(Tsn)ααsnTa(η)(ηsn)α1eM(ηsn)ααdη.

By conditions (H1)–(H2) we have

u(0)(1eM(Tsn)ααk=1nckGk)λeM(Tsn)ααk=1n(k<jncjGj)ckHkDkeM(Tsn)ααsnTa(η)(ηsn)α1eM(ηsn)ααdη0,

which yields u(0)0. Therefore u(t)0 for t[0,T].

Next, we will show that v(t)0 for tJ. The application of Theorem 2.1 for the first two inequalities in problem (3.2) leads to

v(t)edm(ttm)αα{v(0)cmQm0<k<mckGk0<km(k<jmcjPj)ckDk},t(tk,sk],k=1,2,3,,n+1.

Substituting the end point at t=T, we have

v(T)v(0)k=1n+1ckGkedn+1(Ttn+1)ααk=1n+1(k<jn+1cjPj)ckDk,

which implies

v(0)(1k=1n+1ckGk)λedn+1(Ttn+1)ααk=1n+1(k<jn+1cjPj)ckDk0,

by conditions (H3)–(H4). This means that v(0)0. In the same way, we can conclude that v(t)0 for tJ. The proof is completed. □

Finally, we apply the noninstantaneous impulsive inequality to the initial value problem of the form

{Dαsku(t)=f(t,u(t)),t(sk,tk+1],k=0,1,2,,u(t)=cku(tk)+dktkt(ξtk)α1u(ξ)dξ,t(tk,sk],k=1,2,,u(0)=u0, 3.3

where 0<α1, ck0, dk>0, u0R, and the given function fPC(Usk×R,R) satisfies

(H5)

|f(t,u)|M|u|, M>0, for all tUsk.

Corollary 3.2

If (H5) holds, then the solution u(t) of problem (3.3) is estimated as

|u(t)|eM(tsm)αα|u0|0<kmckGk,t(sk,tk+1],k=0,1,2,, 3.4

and

|u(t)|edm(ttm)αα|u0|cmQm0<k<mckGk,t(tk,sk],k=1,2,3,. 3.5

Proof

Taking the conformable fractional integral of order α to the first equation of problem (3.3), we obtain

u(t)=u(sk)+skt(ξsk)α1f(ξ,u(ξ))dξ,t(sk,tk+1],k=0,1,2,.

From condition (H5) it follows that

|u(t)||u(sk)|+skt(ξsk)α1|f(ξ,u(ξ))|dξ,|u(sk)|+Mskt(ξsk)α1|u(ξ)|dξ.

Since u(s0)=u0, by Theorem 2.2 inequalities (3.4)–(3.5) hold, and the proof is completed. □

Availability of data and materials

Data sharing not applicable to this article as no data sets were generated or analyzed during the current study.

Authors’ contributions

All authors contributed equally to this work. All authors read and approved the final manuscript.

Funding

This research was funded by College of Industrial Technology, King Mongkut’s University of Technology North Bangkok, Thailand.

Competing interests

The authors declare that they have no competing interests.

Footnotes

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Contributor Information

Surang Sitho, Email: srg@kmutnb.ac.th.

Sotiris K. Ntouyas, Email: sntouyas@uoi.gr

Praveen Agarwal, Email: praveen.agarwal@anandice.ac.in.

Jessada Tariboon, Email: jessada.t@sci.kmutnb.ac.th.

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