# A Nonlinear Implicit Fractional Equation with Caputo Derivative

Authors: Ameth Ndiaye
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we study a nonlinear implicit differential equation with initial conditions. The considered problem involves the fractional Caputo derivatives under some conditions on the order. We prove an existence and uniqueness analytic result by application of Banach principle. Then, another result that deals with the existence of at least one solution is delivered and some sufficient conditions for this result are established by means of the fixed point theorem of Schaefer. At the end, we discuss two examples to illustrate the applicability of the main results.


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## Body

## 1. Introduction

The theory of differential equations of fractional order and fractional calculus is very important since they can be used in analyzing and modeling real word phenomena. Recently, several researchers are interested in the important progress of differential equations of fractional order. For more information on these works and their applications, one can consult the references [1–9]. In particular, research on the existence of unique solutions for fractional differential equations is of big importance since it helps physicians to better understand the behaviour of real phenomena. See, for more details, the references [10–14].The motivation for this work arises from both the development of the theory of fractional calculus itself and its wide applications to various fields of science, such as physics, chemistry, biology, electromagnetism of complex media, robotics, and economics.Much attention has been paid to the existence and uniqueness of solutions of fractional dynamical systems [15–18] due to the fact that existence is the fundamental problem and a necessary condition for considering some other properties for fractional dynamical systems, such as controllability and stability. Chai [19] provided sufficient conditions for the existence of solutions to a class of antiperiodic boundary value problems for fractional differential equations, while Sheng and Jiang [20] considered a class of initial value problems for fractional differential systems. There are several operators studied in the field of fractional calculus, for example, see [21–26], but the difference in this work is that the operator considered is in the sense of Caputo derivative.Motivated by the works of Benchohra et al. [27], we will establish in this paper existence and uniqueness results of the solutions of the fractional dynamical system with Caputo fractional derivative(1)Dαxt−ADβxt=ft,xt,Dβxt,Dαxt,t∈I=0,1,x0=x0,x′0=x0′,x″0=x0″,x″′0=x0″′,where Dα is in the sense of Caputo, f:I×ℝn×ℝn⟶ℝn is a given function, x0,x0′,x0″∈ℝn, A is an n×n matrix, and 1<β<2, 3<α<4, with β+2<α.Rest of the paper is organised as follows: in Section2, we recall some results and definitions which we use for the proof of our main results. In Section 3, we give and prove the main theorems of this paper, and we discuss some illustrative examples.


## 2. Preliminaries

In this section, we introduce some definitions, lemmas, and preliminaries facts which are used throughout this paper. See [7] for more information. Let . be a suitable norm in ℝn and . be the matrix norm. We denote by CI,ℝn the Banach space of continuous functions from I to ℝn with the norm x∞=supx,x∈I. We denote by L1I,ℝn the space of Lebesgue-integrable function x:I⟶ℝn with the norm(2)xL1=∫01xtdt.Let(3)X=x∈CJ,ℝn,x″∈CI;ℝn,with the norm(4)xX=x∞+x″∞.Definition 1.
The Riemann–Liouville integral of orderα>0 for a continuous function φ∈L10,1,ℝ is given by(5)Iαφt=1Γα∫0tt−τα−1φτdτ,∀t∈0,1,with Γα:=∫0∞e−uuα−1du.Definition 2.
Ifφ∈Cn0,1,ℝ and n−1<α≤n, then the Caputo fractional derivative is given by(6)Dαφt=In−αdndtnφt=1Γn−α∫0tt−sn−α−1φnsds.Lemma 1.
Letn∈ℕ∗ and n−1<α<n, then the general solution of Dαut=0 is given by(7)ut=∑i=0n−1citi,such that ci∈ℝ,i=0,1,2,…,n−1.Lemma 2.
Takingn∈ℕ∗ and n−1<α<n, we have(8)IαDαut=ut+∑k=0n−1uk0k!tk,with t>0, n−1<α<n.Lemma 3.
Let1<β<2 and 3<α<4. Then, it holds(9)IαDβut=Iα−βut−u0tα−βΓα−β+1−u′0tα−β+1Γα−β+2.Proof.
For this proof, we use the same method in [28]. We have(10)IαDβut=1ΓαΓ2−β∫0tt−τα−1∫0tτ−s1−βu″sdsdτ,=1ΓαΓ2−β∫0tu″sds∫stt−τα−1τ−s1−βdτ.
With the change of variableτ=s+t−sη, we have(11)∫stt−τα−1τ−s1−βdτ=ΓαΓ2−βΓα−β+2t−sα−β+1.
Now, we get(12)IαDβut=1Γα−β+2∫0tt−sα−β+1u″sds=Iα−βut−u0tα−βΓα−β+1−u′0tα−β+1Γα−β+2.Definition 3.
LetX be a Banach space. Then, a map T:X⟶Xis called a contraction mapping on X if there exists q∈0,1 such that(13)Tx−Ty≤qx−y,for all x,y∈X.Theorem 1.
(Banach’s fixed point theorem, see [29]). Let Ω be a nonempty closed subset of a Banach space X. Then, any contraction mapping T of Ω into itself has a unique fixed point.Theorem 2.
(Schaefer’s fixed point theorem, see [29]). Let X be a Banach space, and let N:X⟶X be a completely continuous operator. If the set E=y∈X:y=λNyfor someλ∈0,1 is bounded, then N has fixed points.


## 3. Main Results

We begin this section by some results that help us for solving the problem considered in (1).Lemma 4.
For anyx∈X and 1<β<2, we have(14)Dβx∞≤1Γ3−βx″∞≤1Γ3−βx″X.Proof.
By the definition of the operatorDβ, we have(15)Dβxt∞=1Γ2−β∫0tt−s1−βx″sds≤x″∞1Γ2−β∫011−s1−βds≤1Γ3−βx″∞.Lemma 5.
Let1<β<2,3<α<4, and G∈CI,ℝn. Then, we can state that the problem,(16)Dαxt−ADβxt=Gt,t∈I=0,1,x0=x0,x′0=x0′,x″0=x0″,x″′0=x0″′,has for solution the following function(17)xt=x0+x0′t+12x0″t2+16x0″′t3−Atα−βΓα−β+1x0−Atα−β+1Γα−β+2x0′+AΓα−β∫0tt−sα−β−1xsds+1Γα∫0tt−sα−1Gsds.Proof.
By applyingIα to both sides of equation (16), we have(18)IαDαxt−AIαDβxt=IαGt,and using the property established in Lemmas 2 and 3, we find that(19)xt=x0+x′0t+12x″0t2+16x″′0t3−Atα−βΓα−β+1x0−Atα−β+1Γα−β+2x′0+AΓα−β∫0tt−sα−β−1xsds+1Γα∫0tt−sα−1Gsds.
Some of the initial conditions allow us to have the result.
Conversely, assume thatxt satisfy the equation (16), then we see easily the initial conditions.
We use the factDαIαGt=Gt and DαC=0, where C is a constant; we get(20)Dαxt−ADβxt=Gt,t∈I=0,1.
Let us now transform the above problem to a fixed point one. Consider the nonlinear operatorT:X⟶X defined by(21)Txt=x0+x0′t+12x0″t2+16x0″′t3−Atα−βΓα−β+1x0−Atα−β+1Γα−β+2x0′+AΓα−β∫0tt−sα−β−1xsds+1Γα∫0tt−sα−1ft,xt,Dβxt,Dαxtds.
To prove the main results, we need to work with the following hypotheses:(H1)
The functionf defined on I×ℝ3n is continuous.(H2)
There exist nonnegative constantsc1,c2, and c3<1 such that, for any t∈I, x1,x2,x3,x1∗,x2∗,x3∗∈ℝn,(22)ft,x1,x2,x3−ft,x1∗,x2∗,x3∗≤c1x1−x1∗+c2x2−x2∗+c3x3−x3∗.
Also, we consider the quantities(23)D1=AΓα−β+1+c1Γ3−β+c3A+c21−c3Γα+1Γ3−β,D2=AΓα−β−1+c1Γ3−β+c3A+c21−c3Γα−1Γ3−β.
The first main result deals with the existence of a unique solution for (1). It is based on the application of Banach fixed point theorem for contraction mappings.Theorem 3.
If the conditions (H1) and (H2) are satisfied andD<1D:=D1+D2, then problem (1) has a unique solution on I.Proof.
It is sufficient for us to prove thatH is a contraction mapping.
Letx,y∈X2. Then, we can write(24)Txt−Tyt≤AΓα−β∫0tt−sα−β−1xs−ysds+1Γα∫0ts−tα−1gs−hsds,where g,h∈CI,ℝn defined by gt=ft,xt,Dβxt,gt+ADβxt and ht=ft,yt,Dβyt,ht+ADβyt.
From (H2) for eacht∈I, we have(25)gt−ht≤c1xt−yt+c2Dβxt−yt+c3gt−ht+c3Axt−yt,and using Lemma 4, we have(26)gt−ht≤c1Γ3−β+c3A+c21−c3Γ3−βx−yX.
Therefore, we have for eacht∈I,(27)Tx−Ty∞≤AΓα−β+1+c1Γ3−β+c3A+c21−c3Γα+1Γ3−βx−yX≤D1x−yX.
On the other hand, we have(28)Tx″t=x0″+x0″′t−Atα−β−2Γα−β−1x0−Atα−β−1Γα−βx0′+AΓα−β−2∫0tt−sα−β−3xsds+1Γα−2∫0tt−sα−3ft,xt,Dβxt,Dαxtds,which is clear in CI,ℝn.
Then, with the same arguments as before, we have(29)Tx″−Ty″∞≤AΓα−β−1+c1Γ3−β+c3A+c21−c3Γα−1Γ3−βx−yX≤D2x−yX.
Thus, we have(30)Tx−TyX≤D1+D2x−yX.
SinceD<1, then the operator T is contraction. Hence, by Banach’s contraction principle, T has a unique fixed point which is the unique solution of problem (1).
The following main result deals with the existence of at least one solution of the studied problem.Theorem 4.
Under the hypotheses (H1) and (H2), problem (1) has at least one solution ut,t∈I.Proof.
Let us prove the result by considering the following steps:Continuous ofT: if the proof is trivial, then it is omitted (we just apply the fact that f is continuous.Uniform boundness ofT: let us take r>0 and consider the (bounded) ball Br:=x∈X;xX≤r. For y∈Br, we can write(31)Tyt≤x0+12x0″+16x0″+Ax0Γα−β+1+Ax0′Γα−β+2+AΓα−β−1y∞+1Γα+1g∞.With a simple calculus, we get(32)g∞≤c1Γ3−β+c3A+c21−c3Γ3−βyX+m∗,wherem∗=supt∈Ift,0,0,0.Then, we have(33)Ty∞≤x0+12x0″+16x0″+Ax0Γα−β+1+Ax0′Γα−β+2+m∗Γα+1+D1r<+∞,and also we have(34)Ty″∞≤x0″+|x0″′+Ax0Γα−β−1+Ax0′Γα−β+m∗Γα−1+D2r<+∞.The above two inequalities show thatTyX<+∞.Consequently,T is uniformly bounded.Equicontinuity ofT: we prove that, for any bounded set Br for instance, we obtain that TBr is an equicontinuous set of X.Taket1,t2∈0,1,t1<t2 and consider the above (bounded) ball Br of X. So, by considering y∈Br, we can state that(35)Tyt2−Tyt1≤x0′t2−t1+12x0″t22−t12+16x0″′t23−t13+Ax0′Γα−β+1t2α−β−t1α−β+Ax0″Γα−β+2t2α−β+1−t1α−β+1+ArΓα−β+1t2α−β−t1α−β+MΓα+1t2α−t1α,whereM=c1Γ3−β+c3A+c2/1−c3Γ3−βr+m∗.Ast2⟶t1, the right-hand side of the above inequality tends to zero, and we have also(36)Ty″t2−Ty″t1≤x0″′t2−t1+Ax0′Γα−β−1t2α−β−2−t1α−β−2+Ax0″Γα−βt2α−β−1−t1α−β−1+ArΓα−β−1t2α−β−2−t1α−β−2+MΓα−1t2α−2−t1α−2.Ast2⟶t1, the right-hand side of the above inequality tends to zero. From a consequence of the Ascoli-Arzela's theorem, we conclude that T is completely continuous.Boundness ofAγ: the set Aγ:=x∈X:x=γTx,γ∈0,1 is bounded.Lety∈Aγ. Then, we have y=γTy for some 0<γ<1. Hence, we can write(37)y∞≤γx0+12x0″+16x0″+Ax0Γα−β+1+Ax0′Γα−β+2+m∗Γα+1+D1r,(38)y″∞≤γx0″+x0″′+Ax0Γα−β−1+Ax0′Γα−β+m∗Γα−1+D2r.
From (37) and (38), we state that yX<∞. The set is thus bounded.
Consequently, thanks to Schaefer fixed point theorem, we deduce thatT has at least one fixed point. Thus, problem (1) has a solution.Example 1.
Let us consider the following example:(39)Dαxt−ADβxt=ft,xt,Dβxt,Dαxt,t∈I=0,1,x0=12,0,x′0=0,12,x″0=12,0,x″′0=0,12,where(40)f:0,1×ℝ2×ℝ2×ℝ2⟶ℝ2,t,u,v,w↦110et+51+u+v+w,110et+71+u+v+w,with u=maxx1,x2, u=x1,x2. We take A=1/20000,α=15/4, and β=3/2.
We can see clearly that the functionf is continuous.
For anyu,v,w,u¯,v¯,w¯∈ℝ2 and t∈0,1,(41)110et+51+u¯+v¯+w¯−110et+51+u¯+v¯+w¯≤110e5u−u¯+v−v¯+w−w¯,110et+71+u¯+v¯+w¯−110et+71+u¯+v¯+w¯≤110e7u−u¯+v−v¯+w−w¯,which give(42)ft,u,v,w−ft,u¯,v¯,w¯≤c1u−u¯+c2v−v¯+c3w−w¯,where c1=c2=c3=1/10e5.
Hence, the hypotheses (H1) and (H2) are satisfied.
With a simple computation, we getD1=0,0807 and D2=0,24, which imply D<1.
Thus, all the assumptions from (H1)–(H3) are satisfied. From Theorem 3, we conclude that equation (1) has a unique solution.


## 4. Conclusion

In this work, we consider a nonlinear implicit fractional differential equation and we use the Caputo derivative operator. We prove two theorems and an example to illustrate our results. In the first theorem, we prove the existence and uniqueness of the solution and the second theorem deals with the existence of at least one solution. The methods used are the Banach’s fixed point theorem and Schaefer’s fixed point theorem. Here, two Caputo derivative operators of different fractional orders were used in the considered equation and it would be relevant to generalize this idea by considering several Caputo operators of different fractional orders.



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