# Controllability of Mild Solution of Nonlocal Conformable Fractional Differential Equations

Authors: Mohamed Hannabou; Mohamed Bouaouid; Khalid Hilal
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

In many research works Bouaouid et al. have proved the existence of mild solutions of an abstract class of nonlocal conformable fractional Cauchy problem of the form:dαxt/dtα=Axt+ft,xt,x0=x0+gx,t∈0,τ. The present paper is a continuation of these works in order to study the controllability of mild solution of the above Cauchy problem. Precisely, we shall be concerned with the controllability of mild solution of the following Cauchy problem dαxt/dtα=Axt+ft,xt+But,x0=x0+gx,t∈0,τ, where dα./dtα is the vectorial conformable fractional derivative of order α∈0,1 in a Banach space X and A is the infinitesimal generator of a semigroup Ttt≥0 on X. The element x0 is a fixed vector in X and f, g are given functions. The control function u is an element of L20,τ,U with U is a Banach space and B is a bounded linear operator from U into X.


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## Body

## 1. Introduction

Mathematical models based on factional derivatives with respect to time have been the focus of many studies due to their recent applications in various areas of science [1–5]. Many concrete applications prove that the fractional derivative is a very good approaches to deal better with modeling of dynamical systems with memories [6–17]. Regarding to the literature of fractional calculus, it is well known that there are many approaches to define fractional derivatives including the Riemann-Liouville and Caputo definitions. Unfortunately, these definitions have some shortcomings. For example, they do not satisfy derivative formulas for the product and quotient of two functions. In consequence, many researchers have paid attention to propose a best and simple definition of fractional derivative [18, 19]. For example in the work [18], the authors have proposed a new definition of fractional derivative named conformable fractional derivative. This novel fractional derivative is very simple and verifies all the properties of the classical derivative. Actually, the conformable fractional derivative becomes the subject of many research contributions [20–39].For example in [20–22], the authors have proved the existence of mild solution for the following nonlocal conformable fractional Cauchy problem:
(1)dαxtdtα=Axt+ft,xt,x0=x0+gx,t∈0,τ,where dα./dtα represents the conformable fractional derivative of order α∈0,1, and A is the infinitesimal generator of a semigroup Ttt≥0 on a Banach space X,. ([40]). The element x0 is a fixed vector in X and f:0,τ×X⟶X, g:C⟶X are given functions, with C is the Banach space of continuous functions x. defined from 0,τ into X equipped with the norm xc=supt∈0,τxt. The expression x0=x0+gx means the so-called nonlocal condition, which can be applied in physics with better effects than the classical initial condition [41–43].Recently, the study of control problems has attracted the attention of many mathematicians and physicists in various fields of science [44–50]. For example, in theory of differential equations, the controllability consists to control evolution systems from the initial position to the desired position. Motivated by the fact that the controllability is a most important qualitative behavior of a dynamical system, we will be concerned with the controllability of the Cauchy problem (1). Precisely, we will prove a controllability result for the following Cauchy problem
(2)dαxtdtα=Axt+ft,xt+But,x0=x0+gx,t∈0,τ,where the control function u. is an element of L20,τ,U with U is a Banach space and B is a bounded linear operator from U into X.The rest of this paper is organized as follows. In Section2, we briefly recall some tools related to the conformable fractional calculus. In Section 3, we present the main result. Section 4 is devoted to a concert application.


## 2. Preliminaries

Recalling some preliminary facts on the conformable fractional calculus.Definition 1 (see [18]).
Forα∈0,1, the conformable fractional derivative of order α of a function x.:0,+∞⟶ℝ is defined as
(3)dαxtdtα=limε⟶0xt+εt1−α−xtεfort>0anddαx0dtα=limt⟶0+dαxtdtα,provided that the limits exist.
The conformable fractional integralIα. of a function x. is defined by
(4)Iαxt=∫0tsα−1xsds,fort>0.Theorem 2 (see [21]).
Ifx. is a continuous function in the domain of Iα., then, we have
(5)dαIαxtdtα=xt.Theorem 3 (see [23]).
Ifx. is a differentiable function, then, we have
(6)Iαdαx.dtαt=xt−x0.Definition 4 (see [23]).
The conformable fractional Laplace transform of orderα∈0,1 of a function x. is defined as follows
(7)Lαxtλ≔∫0+∞tα−1e−λtα/αxtdt,λ>0.
The following proposition gives us the actions of the conformable fractional integral and the conformable fractional Laplace transform on the conformable fractional derivative, respectively.Proposition 5 (see [23]).
Ifx. is a differentiable function, then, we have the following results
(8)Iαdαx.dtαt=xt−x0,(9)Lαdαxtdtαλ=λLαxtλ−x0.According to [28], we have the following remark.Remark 6.
For two functionsx. and y., we have
(10)Lαxtααλ=L1xtλ,(11)Lα∫0tsα−1xtα−sααysdsλ=L1xtλLαytλ,provided that the both terms of each equality exist.


## 3. Main Result

Lemma 7.
Ifx∈C is a solution of Cauchy problem (2), then, the function x. satisfies the following integral equation
(12)xt=Ttααx0+gx+∫0tsα−1Ttα−sααfs,xs+Busds.The proof of this result is essentially based on the conformable fractional Laplace transform. For the complete proof, one can see the works [20–22].Definition 8 (see [20–22]).
A functionx∈C is called a mild solution of Cauchy problem (2) if
(13)xt=Ttααx0+gx+∫0tsα−1Ttα−sααfs,xs+Busds.
Now, we deal with the controllability of Cauchy problem (2).Definition 9.
The Cauchy problem (2) is said to be controllable on 0,τ, if for every x1∈X, there exists a control u∈L20,τ,U such that the mild solution x. of (2) satisfies xτ=x1.
In the sequel of this paper, we will need the following assumptions:
(H1) The function ft,.:X⟶X is continuous and there exist positive constants L, K such that ft,x≤Lx and ft,y−ft,x≤Ky−x for all x, y∈X.
(H2) The function f.,x:0,τ⟶X is continuous for all x∈X.
(H3) The function g:C⟶X is continuous.
(H4) There exist positive constants M and N such that
(14)gx≤Mxcandgy−gx≤Ny−xcforallx,y∈C.
(H5) The bounded linear operator W:L20,τ,U⟶X defined by
(15)Wu=∫0τsα−1Tτα−sααBusds,has an induced inverse operator W~−1, which takes values in L20,τ,U/KerW, and there exist positive constants R1, R2 such that B≤R1 and W~−1≤R2.Theorem 10.
Assume thatH1−H5 hold, then Cauchy problem (2) is controllable on 0,τ, provided that
(16)supt∈0,τTtαα1+R1R2τααsupt∈0,τTτααmaxM+τααL,N+τααK<1.Proof.
By using hypothesisH5 for an arbitrary function x., we can define a control ux. as follows
(17)ux.=W~−1x1−Tτααx0+gx−∫0τsα−1Tτα−sααfs,xsds..
For this control, we define the operatorΨ:C⟶C by
(18)Ψxt=Ttααx0+gx+∫0tsα−1Ttα−sααfs,xs+Buxsds.We also introduce for a radiusr>0 the ball Br≔x∈C,xc≤r, and we denote by . the norm in the space LX of bounded operators defined from X into itself.We will show that the operatorΨ has a fixed point, which is a mild solution of the control problem (2). To do so, we will give the proof in two steps.Step 1.
Prove that there exists a radiusδ>0 such that Γ:Bδ⟶Bδ.
Forx∈C and t∈0,τ, we have
(19)Ψxt=Ttααx0+gx+∫0tsα−1Ttα−sααfs,xs+Buxsds.
Then, one has(20)Ψxt≤supt∈0,τTtααx0+gx+∫0tsα−1fs,xs+Buxsds.
By using hypothesisH1, H4, and H5, we obtain
(21)Ψxt≤supt∈0,τTtααx0+Mxc+Lxc+R1ux2∫0τsα−1ds≤supt∈0,τTtααx0+Mxc+Lxc+R1ux2ταα.∗.
On the other hand, we have known that(22)ux=W~−1x1−Tτααx0+gx−∫0τsα−1Tτα−sααfs,xsds.
In view of assumptionsH1, H4, and H5, we obtain
(23)ux2≤R2x1−Tτααx0+gx−∫0τsα−1Tτα−sααfs,xsds≤R2x1+supt∈0,τTtααx0+gx+∫0τsα−1fs,xsds≤R2x1+supt∈0,τTtααx0+Mxc+Lxc∫0τsα−1ds≤R2x1+supt∈0,τTtααx0+Mxc+Lxcταα≤R2x1+supt∈0,τTtααx0+M+Lτααxc.
By replacing this estimate in∗, we get
(24)Ψxt≤supt∈0,τTtααx0+Mxc+Lxc+R1R2x1+supt∈0,τTtααx0+M+Lτααxcταα.
Separating the terms containing the expressionxc, one has
(25)Ψxt≤supt∈0,τTtααM+Lταα+R1R2τααsupt∈0,τTtααM+Lτααxc+supt∈0,τTtααx0+τααR1R2x1+τααR1R2supt∈0,τTtααx0.
By using a simple factorization, we obtain(26)Ψxt≤supt∈0,τTtααM+Lταα1+R1R2τααsupt∈0,τTtααxc+supt∈0,τTtαα1+τααR1R2supt∈0,τTtααx0+τααR1R2x1.
Hence, it suffices to considerδ as a solution in r of the following inequality
(27)supt∈0,τTtααM+Lταα1+R1R2τααsupt∈0,τTtααr+supt∈0,τTtαα1+τααR1R2supt∈0,τTtααx0+τααR1R2x1≤r.
Precisely, we can chooseδ such that
(28)δ≥supt∈0,τTtα/α1+τα/αR1R2supt∈0,τTtα/αx0+τα/αR1R2x11−supt∈0,τTtα/αM+Lτα/α1+R1R2τα/αsupt∈0,τTtα/α.Step 2.
We show thatΨ is a contraction operator on Bδ.
Fory, x∈C, we have
(29)Ψyt−Ψxt=Ttααgy−gx+∫0tsα−1Ttα−sααfs,ys−fs,xs+Buy−uxsds.
According toH1, H4, and H5, we obtain
(30)Ψyt−Ψxt≤supt∈0,τTtααgy−gx+∫0tsα−1fs,ys−fs,xs+Buy−uxsds≤supt∈0,τTtααNy−xc+Ky−xc+R1uy−ux2∫0tsα−1ds≤supt∈0,τTtααNy−xc+τααKy−xc+R1uy−ux2.∗∗.
In the other hand, we know that(31)uy−ux=W~−1−Tτααgy−gx−∫0τsα−1Tτα−sααfs,ys−fs,xsds.
Then, one has(32)uy−ux2≤R2supt∈0,τTtααgy−gx+∫0τsα−1fs,ys−fs,xsds≤R2supt∈0,τTtααN+Kτααy−xc.
By replacing this estimate in∗∗, we obtain
(33)Ψyt−Ψxt≤supt∈0,τTtααNy−xc+τααKy−xc+R1R2supt∈0,τTtααN+Kτααy−xc≤supt∈0,τTtααN+τααK+τααR1R2supt∈0,τTtααN+Kτααy−xc≤supt∈0,τTtααN+τααK1+τααR1R2supt∈0,τTtααy−xc.
Taking the supremum, we get(34)Ψyt−Ψxc≤supt∈0,τTtααN+τααK1+τααR1R2supt∈0,τTtααy−xc.
Sincesupt∈0,τTtα/αN+tα/αK1+tα/αR1R2supt∈0,τTtα/α<1, then, Ψ is a contraction operator on Bδ. Hence, there exists a unique element xδ.∈Bδ such that Ψxδt=xδt for all t∈0,τ.
It remains to show that the mild solutionxδ is controllable. To this end, we have
(35)xδτ=Ψxδτ≔Tτααx0+gxδ+∫0τsα−1Tτα−sααfs,xδs+Buxδsds=Tτααx0+gxδ+∫0τsα−1Tτα−sααfs,xδsds+∫0τsα−1Tτα−sααBuxδsds=−Wxδ+x1+∫0τsα−1Tτα−sααBuxδsds=−Wxδ+x1+Wxδ=x1.
Thus, Cauchy problem (2) is controllable on 0,τ.


## 4. Application

LetX=U=L20,1 be equipped with the inner product and norm defined by <u,v>=∫01usvs¯ds and u=∫01us2ds. Define the operator A by
(36)Aφ=φ¨,(37)DA=φ∈X,φ,φ̇areabsolutelycontinousandφ¨∈X,φ0=φ1=0.As well known, the operatorA has a discrete spectrum, and the eigenvalues are −n2,n∈ℕ with the corresponding normalized eigenvectors xny=2sinny, n=1,2,⋯. The operator A generates a contraction semigroup Ttt≥0 given explicitly by
(38)Ttx=∑n=1+∞e−n2t<x,xn>xn,x∈X.Then, we have(39)Ax=−∑n=1+∞n2<x,xn>xn,x∈DA.Next, define the control operatorB as follows
(40)Bu=∑n=1+∞e−1/n2+1<u,xn>xn,u∈U.We have(41)Bu=∑n=1+∞e−2/n2+1<u,xn>2≤∑n=1+∞<u,xn>2≤u.Then,B≤1 and thus the operator B is bounded. Now return back to the operator W, we obtain
(42)Wu=∫01sα−1Tτα−sααBusds=∑n=1+∞1−e−n2/αe−1/n2+1n2<u,xn>xn.Hence, the right inverse of the operatorW may defined as follows
(43)W−1:DA⟶L20,1,L20,1,(44)u↦W−1u=∑n=1+∞n2e1/n2+11−e−n2/α<u,xn>xn.For the operatorW−1, we get
(45)W−1u≤e1/21−e−1/α∑n=1+∞n4<u,xn>xn=e1/21−e−1/αAu=e1/21−e−1/αuDA.Define the functionsf:0,1×X⟶X and g:X⟶X by
(46)ft,xt=e−txt20+et1+xt,(47)gx=∑i=1naixtiwhere∑i=1nai≤120and0<t1<t2<⋯<tn<1.For the functionf, we have
(48)ft,x−ft,y=e−t20+etx1+x−y1+y≤e−t20+etx−y≤120x−y.Here in this application example, we haveτ=1, L=1/20, K=1/20, M=1/20, N=1/20, R1=1, R2=e1/2/1−e−1/α and supt∈0,1Ttα/α≤1. Then, the contraction condition assumed in Theorem 10 becomes
(49)supt∈0,τTtαα1+R1R2τααsupt∈0,τTτααmaxM+τααL,N+τααK≤1+α20α2α+e1/21−e−1/α.Forα=1/2 in the last contraction condition, we get 1+α/20α2α+e1/2/1−e−1/α≃0.72<1. Thus, by using Theorem 10, we conclude that the following Cauchy problem
(50)∂1/2xt∂t1/2=Axt+e−txt20+et1+xt+Bu,t∈0,1,x0=∑i=1naixti.Has a unique controllable mild solution.


## 5. Conclusion and Comments

The existence of mild solutions of a Cauchy problem of nonlocal differential equations with conformable fractional derivative is largely studied in several works [20–22]. Our contribution in this present work is the study of the controllability of mild solutions for such Cauchy problems by means of the Banach fixed point theorem combined with theory of semigroups of linear operators. We notice that the constants of increases of the norms of the bounded operators W and W−1 in the previous application are given directly in a simple way in terms of the exponential function, however, for the Caputo fractional derivative in the application of the nice work [51] are given in terms of the so-called Mittag-Leffler function.



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