# Existence of Solution for a Conformable Fractional Cauchy Problem with Nonlocal Condition

Authors: Khalid Hilal; Ahmed Kajouni; Najat Chefnaj
Journal: International Journal of Differential Equations
 (2022)

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## Abstract

In this work, we prove the existence and uniqueness of mild solution of the fractional conformable Cauchy problem with nonlocal condition. We obtained these results by applying the fixed point theorems precisely to the fixed point theorem of Krasnoselskii and Banach’s fixed point theorem. At the end, we provide application.


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## Body

## 1. Introduction

Many dynamic processes in physics, biology, economics, and other fields of application can be governed by differential evolution equations of neutral type of the following form:(1)ddtxt−Ft,xh1t=−Axt−Ft,xh1t+Gt,xh2t.We replace the partial derivative by a fractional derivative [1, 2] because fractional derivatives have been proven to be a very good way to model many phenomena with memory in various fields of science and engineering [1–9]. Consequently, several researchers are working to form the best definition of the fractional derivative. A new definition called conformal fractional derivative is introduced in [7]. This new fractional derivative becomes the subject of many contributions in several areas of science [9]. Motivated by the better effect of the fractional derivative and simple properties of the conformable fractional derivative, we consider model (1) in this work; we are going to study Cauchy problem with fractional derivative. Precisely, we consider fractional Cauchy problem of the following form:(2)dαdtαxt−Ft,xh1t=−Axt−Ft,xh1t+Gt,xh2t.1.1,x0+gx=x0∈X=DAα,where dα/dtα is the conformable fractional derivative of α∈0,1. A is a sectorial operator which generates a strongly analytic semigroup (Tt) on a Banach space X. For more details about semigroup theory, we refer to [7]. We denote by C(0,a,DAα the banach space of continuous function from 0,a into D(Aα with the norm uα=sups∈0,ausα. For our study, the functions F:0,a×DAα⟶DAα and G:0,a×DAα⟶DAα and g:C0,a,DAα⟶DAα and also h1,h2∈C0,a,DAα.x0+gx=x0 is nonlocal condition; this notion has been a hot topic in recent years. Their association to classical problems has brought a lot of improvement at the level of modeling, thus making it more realistic. The nonlocal condition joined the main equation instead of the classical initial condition which is necessary to model well and write mathematically, physical phenomena like in electronics, in mechanics of materials, or in biomathematics in the way closest to the reality of many phenomena in multiple disciplines. The nonlocal condition means that the initial condition depends on some future times.In this paper, we prove the existence of mild solution of conformable fractional differential equations with nonlocal condition. The main results are based on semigroup theory combined with the Krasnoselskii fixed point theorem.The content of this paper is organized as follows. In Section2, we recall some preliminary facts on the conformable fractional calculus and Section 3 is devoted to prove the main result.In this section, we recall some concepts on conformable fractional calculus.


## 2. Preliminaries

Definition 1.
The conformable fractional derivative ofx of order α at t>0 is defined as(3)dαxtdtα=limh⟶0xt+heα−1t−xth.
When the limit exists, we say thatx is α-differentiable at t.
Ifx is α-differentiable and limt⟶0+dαxt/dtα exists, then define(4)dαx0dtα=limt⟶0+dαxtdtα,
Theα-fractional integral of a function x is defined by(5)Iαft=∫0tsα−1fsds.Theorem 1.
Ifx is a continuous function in the domain of Iα, then(6)dα(Iα(xt))dtα=xt.Defintion 2.
The fractional Laplace transform of orderα of x is defined by(7)ℒαxtλ=∫0+∞tα−1e−λtααxtdt.
The fractional Laplace transform of conformable fractional derivative is given by the following proposition.Proposition 1.
Ifxt is differentiable, then(8)ℒαdαxtdtαλ=λℒαxtλ−x0.
Fractional powers of an operator.Definition 3.
LetA be a sectorial operator defined on a Banach space X, such that ℛeσA>0; for α>0, we note by A−α the operator defined by(9)A−α=1Γα∫0+∞tα−1Ttdt.Definition 4.
LetA be a sectorial operator defined on a Banach space X, such that ℛeσA>0. We define the family of operators Aαα≥0 as follows: A0=IX, and for α>0,(10)Aα=A−α−1,DAα=ImA−α.Theorem 2.
If−A is the infinitesimal generator of an analytic semigroup (Tt)t≥0 and if 0∈ρA, then(1)
DAα is a Banach space with the norm xα=Aαx for every x∈DAα.(2)
Tt:X⟶DAα for all t>0 and α≥0.(3)
For everyx∈DAα, we have TtAαx=AαTtx.We assume that M is a closed bounded convex subset of a Banach space E:(i)
Ax+By∈M for each x,y∈M.(ii)
A is continuous and compact.(iii)
B is contraction.Then, there existsy∈M such that y=Ay+By.We end these preliminaries with the notion of sectorial operator.Definition 3.
A:DA⊂X⟶X is said to be sectorial operator of type M,ω,Θ if there exists M>0, ω∈ℝ, and 0<Θ<π/2 as follows:(1)
A is closed and linear operator.(2)
∀λ∉w+Sθ, the resolvent λI−A−1 of A exists.(3)
∀λ∉w+Sθ, λI−A−1≤M/λ−w,where w+Sθ:=w+λ/λ∈ℂavecArg−λ<θ.Theorem 2.
A densely sectorial operator generates a strongly analytic semigroup (Tt)t≥0. Moreover,(11)Tt=12πi∫γeλtλI−A−1dλ,with gamma being a suitable path λ∉w+Sθ.
Now, we give the main contribution results.


## 3. Main Results

Before presenting our main results, we introduce the following assumptions:(H1)
F:0,a×DAα⟶DAα is continuous and there exists a constant L1>0 such that Ft,x1−Ft,x2α≤L1x1−x2α for all x1,x2∈DAα and for 0≤t≤a.(H2)
G:0,a×DAα⟶DAα is continuous and for all r>0, there exists a function μr∈ℒ∞0,a,ℝ+ such that sup‖x‖≤rGt,xα≤μrt.(H3)
g:C0,a,DAα⟶DAα; there exists a constant L3>0 such that gu1−gu2α≤L3u1−u2α, for all u1,u2∈C(0,a,DAα with uα=sups∈0,ausα for u∈C(0,a,DAα.(H4)
h1,h2∈C(0,a,DAα.H(5)
There exists a constantL2>0 such that(12)Gt,x1−Gt,x2α≤L2x1−x2α,for all x1,x2∈DAα and for 0≤t≤a.Existence of mild solution:Applying the Laplace transform to equation (2), we obtain(13)ℒαdαdtαxt−Ft,xh1tλ=−ℒαAxt−Ft,xh1t+ℒαGt,xh2tλλℒαxt−Ft,xh1tλ−x0−F0,xh10=−Aℒαxt−Ft,xh1tλ+ℒαGt,xh2tλλ+Aℒαxt−Ft,xh1tλ=x0−F0,xh10+ℒαGt,xh2tλ.Then,(14)ℒαxt−Ft,xh1tλ=λ+A−1x0−F0,xh10+λ+A−1ℒαGt,xh2tλ.Hence,(15)ℒαxt−Ft,xh1tλ=λ+A−1x0−gx−F0,xh10+λ+A−1ℒαGt,xh2tλ.Then,(16)λ+A−1x0−F0,xh10=ℒα(Ttααλx0−gx−F0,xh10.Hence,(17)λ+A−1ℒαGt,xh2tλ=ℒα∫0tα/αTtαα−sGαs1/α,xh2αs1/α)dsλ=ℒα∫0tsα−1Ttαα−sααGs,xh2sdsλ.Therefore,(18)ℒαxt−Ft,xh1tλ=ℒα(Ttαα(x0−gx−F0,xh10λ+ℒα∫0tsα−1Ttαα−sααGs,xh2sdsλ.According to inverse fractional Laplace transform, we find the formula(19)xt−F(t,xh1t=Ttααx0−gx−F(0,xh10+ℒα∫0tsα−1Ttαα−sααGs,xh2sdsλ.Then, we obtain(20)xt=F(t,xh1t+Ttααx0−gx−F(0,xh10+∫0tsα−1Ttαα−sααGs,xh2sds.Defintion 1.
We say thatx∈C0,a,DAα is a mild solution of equation (2) if the following assertion is true:(21)xt=F(t,xh1t+Ttααx0−gx−F(0,xh10+∫0tsα−1Ttα−sααGh1s,xvds.Theorem 3.
If(Tt)t>0 is compact and H1−H4 are satisfied, then problem (2) has at least one mild solution, provided that(22)L1+supt∈0,aTtααL3+L1<1.Proof.
Choosing(23)r≥F(t,x(h1t))α+supt∈0,aTtα/αaα/αμrℒ∞+x0α+gxα+Ft,xh1tα1−L1+supt∈0,aTtα/αL3+L1,let Br=x∈Xxα≤r, for x∈Br, and t∈0,a define the operators Γ1 and Γ2 by(24)Γ1xt=F(t,xh1t+Ttααx0−gx−F(0,xh10,Γ2xt=∫0tsα−1Ttαα−sααGs,xh2sds.Claim 1.
We prove thatΓ1 is contraction on Br. We have(25)Γ1xt−Γ1ytα≤AαF(t,xh1t−F(t,yh1t+TtααAαgy−gx+TtααAαF0,yh10−F0,yh10≤L1sup0≤s≤axs−ysα+supt∈0,aTtααL3+L1sup0≤s≤axs−ysα≤L1+supt∈0,aTtααL3+L1sup0≤s≤axs−ysα.
Then,(26)Γ1x−Γ1yα≤L1+supt∈0,aTtααL3+L1x−yα.
SinceL1+supt∈0,aTtα/αL3+L1<1, so Γ1 is a contraction on Br.Claim 2.
We prove thatΓ1x+Γ1y∈Br for every x,y∈Br.(27)Γ1(x)t+Γ2ytα≤Ft,xh1tα+aααsupt∈0,aTtααμrℒ∞+supt∈0,aTtααx0α+gxα+F0,xh10α≤Ft,xh1tα+supt∈0,aTtααx0α+gxα+F0,xh10α+aααμrℒ∞≤r1−L1+supt∈0,a]TtααL3+L1.Claim 3.
We will prove thatΓ2 is continuous on Br
Letxn∈Br such that limn⟶∞xn=x.
We will prove thatlimn⟶∞Γ2xn=Γ2x.
We have(28)Γ2xn−Γ2xα≤∫0tsα−1Ttα−sααGs,xnh2s−Gs,xh2sαds≤supt∈0,aTtαα∫0tsα−1Gs,xnh2s−G(s,x(h2s))αds.
ByH2, we have sα−1Gs,xnh2v−G(s,x(h2s))α≤2μrsα−1 and limn⟶∞Gs,xnh2s=G9s,xh2s. According to Lebesgue dominated convergence theorem, we obtain(29)limn⟶∞Γ2xn−Γ2xα=0.Claim 4.
We prove thatΓ2 is compact.Step 1.
We prove thatΓ2xt|x∈Br is relativement compact in X.
For some fixedt∈]0,a[, let ε∈]0,t[ and define the operator Γ2ε by(30)Γ2εxt=∫0tα−εα1/αsα−1Ttα−sααGs,xh2sds=Tεαα∫0tα−εα1/αsα−1Ttα−sα−εααGs,xh2sds.
According to compactness of(Tt)t>0, the set Γ2εxt|x∈Br is relatively compact in X.
We have(31)Γ2εxt−Γ2xtα≤supt∈0,aTtααμrL∞0,a,ℝ+εαα.
Therefore,Γ2xt|x∈Br is relatively compact in X. It is clear that Γ2x0|x∈Br is compact.
Finally,Γ2xt|x∈Br is relativement compact in X for all t∈0,a.Step 2.
We prove thatΓ2Br is equicontinuous.
Lett1,t2∈]0,a] such that t1<t2. We have(32)Γ2xt2−Γ2xt1=∫0t2sα−1Tt2α−sααGs,xh2sds−∫0t1sα−1Tt1α−sααGs,xh2sds=∫0t1sα−1Tt2α−sαα−Tt1α−sααGs,xh2sds+∫t1t2sα−1Tt2α−sααGs,xh2sds.
Then,(33)Γ2xt2−Γ2xt1=Tt2α−t1αα−I∫0t1sα−1Tt1α−sααGs,xh2sds+∫t1t2sα−1Tt2α−sααGs,xh2sds.
Then,(34)Γ2xt2−Γ2xt1α≤Tt2α−t1αα−I∫0τsα−1Tt1α−sααG(s,x(h2s))αds+∫t1t2sα−1Tt2α−sααG(s,x(h2s))αds≤supt∈0,aTtααμL∞0,a,ℝ+Tt2α−t1αα−Iaαα+t2α−t1αα≤supt∈0,aTtα/αμL∞0,a,ℝ+αTt2α−t1αα−Iaα+t2α−t1α.
We conclude thatΓ2x, x∈Br are equicontinuous at t∈0,a. By using Arzela Ascoli theorem, we obtain that Γ2 is compact. Finally, the Krasnoselskii theorem helps us to complete the proof.
Uniqueness of mild solution.Theorem 4.
Assume thatH1−H5 hold, then the Cauchy problem (2) has a unique solution, provided that(35)L1+L1+L3+L2aααsupt∈0,aTtαα<1.Proof.
Define operatorP:C(0,a,DAα⟶C(0,a,DAα(36)Pxt=F(t,xh1t+Ttααx0−gx−F(0,xh10+∫0tsα−1Ttα−sααGs,xh2sds.
Next, letx,y∈C(0,a,DAα, we have(37)Pxt−Pyt=F(t,xh1t−F(t,yh1t+Ttααgy−gx+F(0,yh10−F(0,xh10+∫0tsα−1Ttα−sααGs,xh2s−(Gs,yh2sds.
Then,(38)Pxt−Pytα≤AαF(t,xh1t−F(t,yh1t+TtααAαgy−gx+TtααAαF(0,yh10−F(0,xh10+∫0tsα−1Ttα−sααAαGs,xh2s−(Gs,yh2sds.
Hence,(39)Pxt−Pytα≤1+supt∈0,aTtααL1sup0≤s≤axs−ysα+L3supt∈0,aTtααsup0≤s≤axs−ysα+supt∈0,aTtααaααL2sup0≤s≤axs−ysα.
Therefore,(40)Px−Pyα≤L1+L1+L3+L2aααsupt∈0,aTtααx−yα.
SinceL1+L1+L3+L2aα/αsupt∈0,aTtα/α<1, as a consequence, P has a unique fixed point in C(0,a,DAα.


## 4. Application

We consider the conformable fractional Cauchy problem of the following form:(41)∂1/2∂t1/2ut,ξ−∫0πbt,ξ,σusint,σdσ=∂2.∂x2ut,ξ−∫0πbt,ξ,σusint,σdσ+ψt,∂ut,ξ∂ξut,0=ut,π=0u0,ξ−∫0πkξ,σu0,σdσ=u0ξ,0≤ξ≤1..WithX=L20,π, we define operator A by Af=−f” and DA=H020,π=f∈X:f′,f″∈Xandf0=fπ=0 with the norm 2.(i)
b:0,1×0,π×0,π⟶ℝ is class C1 and bt,.,0=bt,.,π.(ii)
ψ:0,1×ℝ⟶ℝ is continuous and there exists a constant a0>0 such that(42)ψt,ξ1−ψt,ξ2≤a0ξ1−ξ2,ξ1,ξ2∈ℝ,0≤t≤1.(iii)
k:0,π×0,π⟶ℝ is class C1 and k.,0=k.,π.Forα=1/2,t∈0,1,v∈DA1/2 with DA1/2=f∈X/∑0∞nf,enen and w∈E=C0,1,DA1/2, we define Ft,v.=∫0πbt,.,σvσdσ, Gt,v.=Gt,v′, gw.=∫0πk.,σwσdσ, and h1t=h2t=sintA=∂2./∂x2.The problem takes the following form:(43)dαdtαxt−Ft,xh1t=−Axt−Ft,xh1t+Gt,xh2tx0+gx=x0.F:0,1×DA1/2⟶DA1/2 and G:0,a×DAα⟶X, on the other hand,(44)Ft,v1/22=∫0π∫0π∂∂ξbt,ξ,σvσdσ2dξ≤∫0π∫0π∂∂ξbt,ξ,σ2dσdξ∫0πv2σdσ.From the equality of Poincare, we have(45)Ft,v1/22≤sup0≤t≤1∫0π∫0π∂∂ξbt,ξ,σ2dσdξv1/22,which implies that hypothesis H1 is verifiedLetv1,v2∈DA1/2 and ξ∈0,π. We know that(46)Gt,v1−Gt,v22≤t,v1ξ−Gt,v2ξ|2dξ≤a02∫0πv1′ξ−v2′ξ2dξ≤a02v1′−v2′1/22,and hypothesis H5 is verified.Letv∈∈DA1/2; we have gv1/22=A1/2gv2=∂/∂ξgvξ2,(47)gv1/22=`∫0π∫0π∂∂ξkξ,σvσdσ2dξ≤∫0π∫0π∂∂ξkξ,σ2dσdξ∫0πvσ2dσ.From the equality of Poincare, we have(48)gv1/22≤∫0π∫0π∂∂ξkξ,σ2dσdξv1/22.Then,H3 is verified.Moreover,sup0≤t≤1∫0π∫0π∂/∂ξbt,ξ,σ2dσdξ+∫0π∫0π∂/∂ξkξ,σ2dσdξ+2a0)M≤1. As a consequence, (41) has a unique fixed point in C(0,a,DAα.



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