# Existence and Uniqueness of Solutions for Fractional Boundary Value Problems under Mild Lipschitz Condition

Authors: Imed Bachar; Habib Mâagli; Hassan Eltayeb
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This paper deals with the following boundary value problemDαut=ft,ut,t∈0,1,u0=u1=Dα−3u0=u′1=0, where 3<α≤4,Dα is the Riemann-Liouville fractional derivative, and the nonlinearity f, which could be singular at both t=0 and t=1, is required to be continuous on 0,1×ℝ satisfying a mild Lipschitz assumption. Based on the Banach fixed point theorem on an appropriate space, we prove that this problem possesses a unique continuous solution u satisfying ut≤cωt,fort∈0,1andc>0, where ωt≔tα−21−t2.


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## Body

## 1. Introduction

Higher order fractional differential equations subject to two-point boundary value problems occur naturally when modeling various phenomena in the applied sciences (see, for example, [1–4] and references therein). The study of existence, uniqueness, and qualitative properties of the solutions of such problems subject to various type of boundary conditions become an active area of research (see, for instance, [5–13] and references therein).In [10], the authors considered the following problem
(1)Dαut=ft,ut,t∈0,1,u0=u1=u′0=u1=0,where 3<α≤4,Dα is the standard Riemann–Liouville fractional derivative.By reducing problem (1) to an equivalent Fredholm integral equation and using some fixed-point theorems, they have proved the existence, multiplicity, and uniqueness of positive solutions. In their approach, properties of the corresponding Green’s function are used.In [6], the authors proved the existence and uniqueness of positive solutions of the problem
(2)Dαut=ptuσt,t∈0,1,u0=u1=Dα−3u0=u′1=0,where σ∈−1,1,3<α≤4,Dα is the standard Riemann–Liouville fractional derivative. Their approach relies on properties of Karamata regular variation functions and the Schauder fixed point theorem.Recently, in [13], Zou and He, by using the Banach fixed point theorem on an appropriate space, they have proved the existence and uniqueness of a solution to the following problem
(3)Dαut+ft,ut=0,t∈0,1,u0=u′0=u1=0,where 2<α≤3,Dα denotes the standard Riemann–Liouville fractional derivative and f satisfying:
(C1)
f∈C0,1×ℝ,ℝ and fs,0∈L10,1.(C2)
There existsp∈C0,1,0,∞ such that
(4)ft,u−ft,v≤ptu−v,∀t∈0,1,u,v∈ℝ0<∫01psds<∞.In this paper, following the approach used in [13], we will address the existence and uniqueness of a solution to the following fractional problem involving fractional boundary derivatives
(5)Dαut=ft,ut,t∈0,1,u0=u1=Dα−3u0=u′1=0,where 3<α≤4 and Dα denotes the standard Riemann–Liouville fractional derivative. The nonlinear term f is required to satisfy:
(H1)
f∈C0,1×ℝ,ℝ and ∫01s1−sα−3fs,0ds<∞.(H2)
There existsp∈C0,1,0,∞ such that
(6)ft,u−ft,v≤ptu−v,∀t∈0,1,u,v∈ℝ0<Lp,α≔1Γα∫01sα−11−sα−1psds<∞.Note that the nonlinearityf is allowed to be singular at t=0 or/and t=1.Remark 1.
It is worth mentioning that problem (5) is different form problem (3) and that conditions (H1) and (H2) are weaker then (C1) and (C2). Indeed, for example, the function ft,u≔1−t−α/3cosu with α∈3,4 satisfies (H1) and (H2) (see Example 8) but not (C1) and (C2).To simplify our statement, forα∈3,4, we fix the following notation:
(i)
Gαt,s denotes the Green’s function of the operator u → Dαu with boundary conditions u0=u1=Dα−3u0=u′1=0, which is given (see [6], Lemma 5) for t,s∈0,1 by(7)Gαt,s=1Γαtα−21−sα−2s−t+α−2s1−t+maxt−s,0α−1.(ii)
LetL be the minimum positive constant such that(8)∫01Gαt,sωtpsds≤Lωt,where
(9)ωt≔tα−21−t2,fort∈0,1.We will prove thatL satisfies
(10)αα−2Lp,α+1≤L≤γLp,α,where γ≔maxα−22,α−1.(iii)
Cω0,1≔u∈C0,1 there is σ>0 such that ut≤σωt,t∈0,1Note thatCω0,1 is a Banach space equipped with the .ω-norm
(11)uω≔infσ>0:ut≤σωt,t∈0,1=supt∈0,1utωt.Our main result is the following.Theorem 2.
Assume that (H1) and (H2) hold and L<1. Then, problem (5) has a unique solution u in Cω0,1.
In addition, for anyu0∈Cω0,1, the iterative sequence ukt≔∫01Gαt,sfs,uk−1sds converges to u with respect to the .ω-norm, and we have
(12)uk−uω≤Lk1−Lu1−u0ω.Remark 3.
Assume that (H1) and (H2) hold. If γLp,α<1, then from (10), it follows that L<1.
Note that the conditionγLp,α<1 is satisfied by a large class of functionsp including singular ones. As examples, we have
(i)
Ifp is a positive continuous function on 0,1 with supt∈0,1pt≤1, then(13)γLp,α≤γΓαBα,α<1.(ii)
Ifps≡1−s−α/3, then(14)γLp,α=γΓαBα,2α3<1.(iii)
Ifps≡s−α/31−s−α/3, then(15)γLp,α=γΓαB2α3,2α3<1,where B denotes the Beta function.The paper is organized as follows. In Section2, we recall basic properties of the Green’s function Gαt,s, and we prove that L satisfies the range estimates (10). In Section 3, we prove our main result. To illustrate our existence result, some examples and approximations are given.


## 2. Preliminaries

For the convenient of the reader, we recall the following definition.Definition 4 ([2, 14]).
The Riemann-Liouville fractional derivative of orderα>0 for a measurable function g:0,∞→ℝ is defined as
(16)Dαgt=1Γn−αddtn∫0tt−sn−α−1gsds,provided that the right-hand side is pointwise defined on 0,∞, where n=α+1 and α denotes the integer part of α.The next key lemma is useful. For the proof, we refer the reader to [6].Lemma 5.
LetGαt,s be the Green’s function given by (7). Then
(i)
Gαt,s is a nonnegative continuous function on 0,1×0,1(ii)
For allt,s∈0,1, we have(17)α−2Hαt,s≤Gαt,s≤γHαt,s,where Hαt,s≔1/Γαtα−31−ts1−sα−3mint,s1−maxt,s.(iii)
Letg be a function such that the map t→t21−tα−2gt is continuous and integrable on 0,1. Then, Vgt≔∫01Gαt,sgsds is the unique solution in C0,1 of the boundary value problem(18)Dαut=gt,t∈0,1,u0=u1=Dα−3u0=u′1=0.Lemma 6.
Letp∈C0,1,0,∞ such that 0<Lp,α<∞. Then
(19)αα−2Lp,α+1≤L≤γLp,α,where γ≔maxα−22,α−1, and L is the constant defined in (8).Proof.
Consider the set(20)E=a>0:∫01Gαt,sωspsds≤aωt,t∈0,1,where ωt≔tα−21−t2,t∈0,1.From the upper inequality in (17), we obtain
(21)∫01Gαt,sωspsds≤γ∫01Hαt,sωspsds≤γLp,αωt.It follows thatE≠∅ and L≤γLp,α.On the other hand, from the lower inequality in (17), we deduce for any a∈E that
(22)aωt≥α−2Γαtα−31−t∫01sα−11−sα−1mint,s1−maxt,spsds≥α−2Γαtα−31−t∫01sα−11−sα−1ts1−t1−spsds=αα−2ωtLp,α+1.Hence, for eacha∈E,(23)a≥αα−2Lp,α+1.Therefore,L≥αα−2Lp,α+1. That is L∈αα−2Lp,α+1,γLp,α.


## 3. Main Results

Assume that (H1) and (H2) hold and L<1. We prove that problem (5) has a unique solution u in Cω0,1. In addition, for any u0∈Cω0,1, the iterative sequence ukt≔∫01Gαt,sfs,uk−1sds converges to u with respect to the .ω-norm, and we have
(24)uk−uω≤Lk1−Lu1−u0ω.Consider the operatorT defined by
(25)Tut≔∫01Gαt,sfs,usds,t∈0,1,u∈Cω0,1.We claim thatT is a contraction operator from Cω0,1,.ω into itself.Indeed, letu∈Cω0,1 and σ>0 such that ut≤σωt, for all t∈0,1.Since by Lemma5 (ii), 0≤Gαt,s≤γ/Γαs1−sα−3, it follows from (H2) that
(26)Gαt,sfs,us≤γΓαs1−sα−3fs,us−fs,0+fs,0≤γΓαs1−sα−3psus+fs,0≤γΓασsα−11−sα−1ps+s1−sα−3fs,0.Using the fact thatGαt,s is continuous on 0,1×0,1, we deduce by (H1),(H2), and the dominated convergence theorem that Tu∈C0,1.On the other hand, from Lemma6 (ii), we have
(27)0≤Gαt,s≤γΓαs1−sα−3ωt.By using (27) and similar arguments as above, we obtain
(28)Tut≤γσLp,α+1Γα∫01s1−sα−3fs,0dsωt.So,TCω0,1⊂Cω0,1.Letu,v∈Cω0,1. By using (H2) and (8), we obtain for t∈0,1,(29)Tut−Tvt≤∫01Gαt,sfs,us−fs,vsds≤∫01Gαt,spsus−vsds≤u−vω∫01Gαt,spsωsds≤Lu−vωωt.Hence(30)Tu−Tvω≤Lu−vω.SinceL<1, we deduce that T is a contraction operator in Cω0,1. Hence, there exists a unique u∈Cω0,1 satisfying
(31)ut=∫01Gαt,sfs,usds,t∈0,1.We need to prove thatu is a solution of problem (5). Indeed, it is clear that s→s21−sα−2fs,us is continuous on 0,1.On the other hand, by writing(32)fs,us≤fs,us−fs,0+fs,0≤psus+fs,0,we deduce from (H1), (H2), and u∈Cω0,1 that the function s→s21−sα−2fs,us is integrable on 0,1.Hence, from (31) and Lemma 5 (iii), we conclude that u is the unique solution of problem (5).Finally, it is well known that for anyu0∈Cω0,1, the iterative sequence uk≔Tuk−1 converges to u, and we have
(33)uk−uω≤Lk1−Lu1−u0ω.Corollary 7.
Let3<α≤4 and f be a function satisfying (H1) and (H3) for all t∈0,1 and u,v∈ℝ(34)ft,u−ft,v≤L0u−v,where 0<L0<Γ2α/maxα−22,α−1Γα.
Then, problem (5) has a unique solution u in Cω0,1.Proof.
The conclusion follows from Lemma6 and Theorem 2.Example 8.
Let3<α≤4,γ=maxα−22,α−1 and consider the following singular fractional problem:
(35)Dαut=1−t−α/3cosu,t∈0,1,u0=u1=Dα−3u0=u′1=0.
To verify that hypotheses (H1) and (H2) are satisfied, set ft,u≔1−t−α/3cosu, for all t∈0,1 and u∈ℝ.
We havef∈C0,1×ℝ,ℝ and ∫01s1−sα−3fs,0ds=∫01s1−s2α/3−3ds<∞. So condition (H1) is satisfied.
On the other hand, we have for allt∈0,1 and u,v∈ℝ(36)ft,u−ft,v≤ptu−v,where pt=1−t−α/3 and
(37)0<Lp,α=1Γα∫01sα−11−s2α/3−1ds<∞.
Furthermore, since by Lemma6 and Remark 3 (ii), L≤γLp,α<1, we deduce from Theorem 2 that problem (35) has a unique solution u∈Cω0,1.
In particular, forα=7/2, this solution is approximated (see Figure 1) by the iterative sequence ukt≔∫01G7/2t,s1−s−7/6cosuk−1sds with u0t=t3/21−t2,t∈0,1.
The sequences of functionsu0t,u1t,u2t,u3t, and u4t are illustrated in Figure 1.Example 9.
Consider the following singular fractional problem:(38)D10/3ut=t−10/91−t−10/91+u,t∈0,1,u0=u1=D1/3u0=u′1=0.
By using similar arguments as in the previous example, we verify that conditions (H1) and (H2) are fulfilled.
Hence, by applying Theorem2, problem (38) has a unique solution u∈Cω0,1.
Furthermore, the iterative sequence defined byukt≔∫01G10/3t,ss−10/91−s−10/91+uk−1sds and u0t≔t4/31−t2 converges to u. Some iterations are depicted in Figure 2.Figure 1
The approximation of the solution of problem (35).Figure 2
The approximation of the solution of problem (38).



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