# Generalized Ulam–Hyers–Rassias Stability Results of Solution for Nonlinear Fractional Differential Problem with Boundary Conditions

Authors: A. Naimi; B. Tellab; Y. Altayeb; A. Moumen
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

The problem of existence and generalized Ulam–Hyers–Rassias stability results for fractional differential equation with boundary conditions on unbounded interval is considered. Based on Schauder’s fixed point theorem, the existence and generalized Ulam–Hyers–Rassias stability results are proved, and then some examples are given to illustrate our main results.


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## Body

## 1. Introduction and Position of Problem

There are various, not equivalent, definitions of fractional derivatives according to Grunwald Letnikov, Weil, Caputo, and Riemann–Liouville, etc. Ordinary and partial differential order equations (with fractional derivatives of Caputo and Riemann–Liouville) have awakened in recent years with considerable interest both in mathematics and in applications. Let us describe the abstract Cauchy problem:(1)Dtγut=Aut,m−1<γ≤m∈ℕ,dκdtκu0=ξκ,κ=0,…,m−1,where the corresponding solutions are represented through the Mittag–Leffler function. In mathematical papers on fractional differential equations, the Riemann–Liouville approach to the concept of a fractional order derivative γ≥0 is usually used as follows:(2)Dγut=ddtm1Γm−γ∫0tt−rm−γ−1urdr,m−1<γ≤m∈ℕ.The fractional Riemann-Liouville derivative is the left inverse to the corresponding fractional integral, which is a natural generalization of the Cauchy formula for the antiderivative functionut. The initial conditions, of the initial value problem for ordinary differential equations of fractional order γ with fractional derivatives in the Riemann–Liouville form, are given in terms of fractional integrals:(3)1Γm−κ−γ∫00+0+−rm−κ−γ−1uκrdr,κ=0,…,m−1.To satisfy the physical requirements, Caputo introduced an alternative definition of the fractional differential derivative. It was adopted by Caputo and Mainardi as(4)Dtγcut=1Γm−γ∫0tt−rm−γ−1umrdr,m−1<γ≤m.The advantage of this definition is a more natural solution for the problem of initial conditions for solving integro-differential equations of noninteger orders.The cases of Caputo derivative for0<γ<1 was called the regularized fractional derivative of order γ.This paper concerns the existence with Ulam stability for the following equation:(5)D0+βut+ft,ut+θtgut=0,t≥0.For a continuous functionut together with boundary conditions,(6)u0=0,u′0=0,D0+β−1u+∞=buξ+λ∫0σusds,where 2<β≤3,0≤λ,b<∞, we fix 0≤ξ<σ<∞, the functions f:ℝ+×ℝ⟶ℝ and g:ℝ⟶ℝ are continuous and θ is a continuous decreasing positive function such that 0<θt≤1, for all t∈0,+∞. D0+β is the standard Riemann–Liouville fractional derivative of order β.


## 2. Literature Overview

Fractional differential equations, which are often encountered in mathematical modeling of various processes in natural and technical sciences, play an important role in describing many phenomena in physics, bioengineering, and engineering applications. The properties of such equations were investigated in many reviews (among them, we refer [1–6]).Regarding the existence, we mention the work by Zhao and Ge [7], where the authors used the well-known Leray Schauder nonlinear alternative theorem to prove the existence of positive solutions to the problem(7)D0+αut+ft,ut=0,t∈0,+∞,1<α≤2,u0=0,D0+α−1u+∞=βuξ,where f∈C0,+∞×ℝ,0,+∞,0≤ξ,β<+∞. Next, Wang et al. [8] extended the above results and discussed the question of existence for solutions of (7) with condition:(8)D0+α−1u+∞=λ∫0τusds,where 0≤λ,τ<+∞. Shen et al. [9] considered the existence of solution for boundary value problem of nonlinear multipoint fractional differential equation:(9)D0+γut=ft,ut,D0+γut,t∈0,+∞,u0=0,u′0=0,D0+γ−1u+∞=∑i=1m−2βiuξi,where(10)2<γ≤3,f∈Cℝ+×ℝ×ℝ,ℝ,Γβ−∑i=1m−2βiξiγ−1.SM Ulam in 1940 was the first to raise the question of stability for functional equations. After his lecture, this question became popular for many specialists in mathematical analysis. It became an area of in-depth research (see for more details [10–12]). Next, many mathematicians turned in their studies to two types of stability—according to Ulam–Hyers and according to Ulam–Hyers–Rassias. This kind of study has become one of the central and most important in the fields of fractional differential equations. Details of recent advances in Ulam–Hyers sustainability and according to Ulam–Hyers for differential equations can be found in [13, 14] and in articles [15–17]. However, as far as we know, most authors discussed Ulam stability of some fractional differential problem on bounded/unbounded intervals, while the present paper discusses the existence of solutions and stability in the sense of Ulam–Hyers–Rassias for nonlinear fractional differential equations boundary conditions, for which research is just beginning, please see [18–23].


## 3. Preliminaries

Here, we present some notations, definitions, auxiliary lemmas concerning fractional calculus, fixed point theorems, and some preliminary concepts of fractional calculus.Definition 1.
(see [4, 24]). The Riemann–Liouville fractional integral of order β for a function f is defined as(11)I0+βft=1Γβ∫0tt−sβ−1fsds,provided the right side is pointwise defined on 0;+∞.Lemma 1 (see [25]).
LetU⊂X be a bounded set. U is said to be relatively compact in a space E if(i)
∀u∈U, the function ut/1+tβ−1 is equicontinuous on any compact subinterval of J.(ii)
∀ε>0, there exists a constant T=Tε>0 such that(12)ut21+t2β−1−ut11+t1β−1<ε,t1,t2≥Tandu∈U.Lemma 2 (see [4, 24]).
Assume thatx∈CJ∩L1J with a fractional derivative of order β>0. Then,(13)I0+βD0+βxt=xt−C1tβ−1−C2tβ−2−⋯−Cntβ−n,where C1,C2,…,CN∈ℝ with n=β+1.Lemma 3.
Let us define the following space:(14)E=u∈C0,+∞:supt≥0ut1+tβ−1<+∞,equipped with the norm(15)uE=supt≥0ut1+tβ−1.Then, clearlyE,⋅E is a Banach space.Lemma 4.
u is a solution of the problem (5)-(6) if and only if u satisfies the following integral equation:(16)ut=∫0+∞Ht,sfs,us+θsgusds,where(17)Ht,s=−t−sβ−1Γβ+Γβ+1−bβξ−sβ−1−λσ−sβtβ−1ΓβΓβ+1−λσβ−bβξβ−1,s≤t,s≤ξ,Γβ+1−bβξ−sβ−1−λσ−sβtβ−1ΓβΓβ+1−λσβ−bβξβ−1,t≤s≤ξ≤σ,−t−sβ−1Γβ+Γβ+1−λσ−sβtβ−1ΓβΓβ+1−λσβ−bβξβ−1,ξ≤s≤t,s≤σ,Γβ+1−λσ−sβtβ−1ΓβΓβ+1−λσβ−bβξβ−1,ξ≤s<σ,t≤s−t−sβ−1Γβ+Γβ+1tβ−1ΓβΓβ+1−λσβ−bβξβ−1,ξ≤σ≤s≤t,βtβ−1Γβ+1−λσβ−bβξβ−1,s≥t,s≥σ.Proof.
Using Lemma2, we have(18)ut=−I0+βft,ut+θtgut+c1tβ−1+c2tβ−2+c3tβ−3.
By the first and second conditions, we get(19)c3=0,c2=0.
Consequently,(20)ut=−I0+βft,ut+θtgut+c1tβ−1.
From the third boundary condition, it follows that(21)D0+β−1ut=−I0+β−β+1ft,ut+θtgut+c1Γβ=−∫0tfs,us+θsgusds+c1Γβ.
On the other hand, we have(22)buξ+λ∫0σusds=−bI0+βfξ,uξ+θξguξ+c1bξβ−1−λ∫0σI0+βfs,us+θsgusds+λ∫0σc1sβ−1ds=−bI0+βfξ,uξ+θξguξ+c1bξβ−1−λI0+β+1fσ,uσ+θσguσ+λσββc1.
Then, we deduce(23)c1=βΓβ+1−λσβ−bβξβ−1∫0+∞fs,us+θsgusds−bI0+βfξ,uξ+guξ−λI0+β+1fσ,uσ+θσguσ.
By substituting the values ofc1,c2, and c3 in (18), we get the following integral equation:(24)ut=−∫0tt−sβ−1Γβfs,us+θsgusds+βtβ−1Γβ+1−λσβ−bβξβ−1∫0+∞fs,us+θsgusds−bβtβ−1Γβ+1−λσβ−bβξβ−1∫0ξξ−sβ−1Γβfs,us+θsgusds−λβtβ−1Γβ+1−λσβ−bβξβ−1∫0σσ−sβΓβ+1fs,us+θsgusds.
Then, we get (16).
Conversely, suppose that (16) is satisfied. To get (5), we use the following appropriate relationships:(25)D0+βI0+βft,ut+θtgut=ft,ut+θtgut,D0+βtβ−1=0.
The present paper is organized as follows. In Section4, we prove the existence of the solution for problem (5)-(6) in the Banach space. The generalized Ulam–Hyers stable is stated and proved in Section 5. Finally, an illustrative example is given.


## 4. Existence Result

In order to prove the existence of the solution for problem (5)-(6), we transform problem (5)-(6) into the fixed point problem Pu=u, where P is an operator defined on(26)Br=u∈E,uE≤r,by(27)Put=∫0+∞Ht,sfs,us+θsgusds.Theorem 1.
Letf:0,+∞×ℝ⟶ℝ and g:ℝ⟶ℝ are two functions such thatA1Γβ+1>λσβ+bβξβ−1.A2 There exist a nonnegative measurable function ψ1 defined on 0,+∞ and a real constant L>0 such that:(28)ft,ut−ft,vt≤ψ1tut−vt,u,v∈ℝ,gut−gvt≤Lut−vt,u,v∈ℝ,β∫0+∞1+tβ−1ψ1t+ψ2tdt<Γβ+1−λσβ−bβξβ−1with(29)ψ2t=θtL,for eacht∈0,+∞.A3 Let ϕ1t=ft,0 and ϕ2t=θtg0, t∈0,+∞ such that(30)∫0+∞ϕ1t+ϕ2tdt<+∞.Then, problem (5)-(6) has at least one solution in 0,+∞.Lemma 5.
IfA1 holds, then the Green function Ht,s satisfiesfor all ξ,σ,s,t∈0,+∞, we have(31)Ht,s1+tβ−1≤βΓβ+1−λσβ−bβξβ−1.Proof.
Ifs≤t, and s≤ξ, we get(32)Ht,s1+tβ−1=−t−sβ−11+tβ−1Γβ+Γβ+1−bβξ−sβ−1−λσ−sβtβ−1ΓβΓβ+1−λσβ−bβξβ−11+tβ−1≤Γβ+1tβ−1ΓβΓβ+1−λσβ−bβξβ−11+tβ−1≤βΓβ+1−λσβ−bβξβ−1.
All other cases ofHt,s are simple. This completes the proof of Lemma 5.Proof.
of Theorem1. We shall use Schauder’s fixed point theorem, which is divided into three steps.Step 1.
Letr>0 such that(33)r≥β∫0+∞ϕ1p+ϕ2pdpΓβ+1−λσβ−bβξβ−1−β∫0+∞1+pβ−1ψ1p+ψ2pdp.
Ifu is a continuous function on J, then Pu∈CJ. In order to show PBr⊂Br, let u∈Br,t∈ℝ+. Then,(34)Put1+tβ−1=∫0+∞Ht,s1+tβ−1fs,us+θsgusds≤βΓβ+1−λσβ−bβξβ−1∫0+∞fs,us+θsgusds≤βΓβ+1−λσβ−bβξβ−1×∫0+∞fs,us−fs,0+θsgus−g0+fs,0+θsg0ds≤βΓβ+1−λσβ−bβξβ−1∫0+∞ψ1sus+θsLus+ϕ1s+ϕ2sds≤βΓβ+1−λσβ−bβξβ−1∫0+∞ϕ1s+ϕ2sds+βuEΓβ+1−λσβ−bβξβ−1∫0+∞1+sβ−1ψ1s+ψ2sds≤βΓβ+1−λσβ−bβξβ−1∫0+∞ϕ1s+ϕ2sds+βrΓβ+1−λσβ−bβξβ−1∫0+∞1+sβ−1ψ1s+ψ2sds≤r.
Therefore,PE≤r; thus, PBr∈Br.Step 2.
P:Br⟶Br is continuous. Let un be a sequence which converges to u in Br. Then, for all t∈0,+∞,(35)Punt−Put1+tβ−1=∫0+∞Ht,s1+tβ−1fs,uns−fs,us+θsguns−gusds≤βΓβ+1−λσβ−bβξβ−1∫0+∞fs,uns−fs,us+θsguns−gusds≤βΓβ+1−λσβ−bβξβ−1∫0+∞fs,uns−fs,us+θguns−gusds≤βun−uEΓβ+1−λσβ−bβξβ−1∫0+∞ψ1s+ψ2s1+sβ−1ds<un−uE.
So, we conclude thatPun−PuE⟶0 as n⟶+∞. Hence, P is a continuous operator on E.Step 3.
We have two claims to verify thatPBr is a relatively compact set.
First claim: letI⊂J be a compact interval, t1,t2∈I with t1<t2. Then, for any u∈Br, we have(36)Put21+t2β−1−Put11+t1β−1≤∫0+∞Ht2,s1+t2β−1−Ht1,s1+t1β−1fs,us+θsgusds≤∫0+∞Ht2,s1+t2β−1−Ht1,s1+t1β−1ψ1s+ψ2s1+sβ−1uE+ϕ1s+ϕ2sds.
Since it is continuous onJ×J, we have that Ht,s/1+tβ−1 is a uniformly continuous function on the compact set I×I.
Fors≥t, the function depends only on t, then it is uniformly continuous on I×J/I. Therefore, we have ∀s∈J and t1,t2∈I; the next property holds.
∀ε>0, there is δε>0 such that, if t1−t2<δ, then(37)Ht2,s1+t2β−1−Ht1,s1+t1β−1≤ε.
This property, together with (36) and the fact that(38)∫0+∞1+sβ−1ψ1s+ψ2sr+ϕ1s+ϕ2sds<∞,means that Put/1+tβ−1 is equicontinuous on I.
Second claim: in order to achieve (ii) of Lemma1, we use(39)limt⟶+∞Ht,s1+tβ−1=1ΓβΓβ+1−λσβ−bβξβ−1×λσβ−σ−sβ+bβξβ−1−ξ−sβ−1,s≤t,s≤ξ,Γβ+1−bβξ−sβ−1−λσ−sβ,t≤s≤ξ≤σ,λσβ−σ−sβ+bβξβ−1,ξ≤s≤σ,Γβ+1−λσ−sβ,ξ≤t,s<σ,λσβ+bβξβ−1,ξ≤σ≤s≤t,βΓβ,s≥σ,s≥t.
From (39), it is not hard to see, ∀ε>0, there exists a dependent constant T=Tε>0 such that, for t1,t2≥T and s∈J, we have(40)Ht2,s1+t2β−1−Ht1,s1+t1β−1≤ε.
Now, from (36) and (38), the same property holds for Put/1+tβ−1, uniformly for u∈Br. Then, PBr is equiconvergent at ∞.
Thus, Lemma1 implies that PBr is relatively compact.
Therefore, the operatorP has a fixed point on Br. Then, from Schauder’s fixed point theorem, we conclude that problem (5)-(6) has at least one solution.


## 5. Stability Result

Before stating and proving our main stability results, let us consider the following integration formula:(41)vt=∫0+∞Ht,sfs,vs+θsgvsds,for continuous function H,θ. Here, we suppose that v∈C0,+∞,E has a fractional derivative of order β, where 2<β≤3, f:0,+∞×E⟶ℝ and g:E⟶ℝ are two continuous functions, and let us define the following nonlinear continuous operator:(42)ℱ:0,+∞×E⟶ℝ,ℱvt=D0+βvt+ft,vt+θtgvt,Definition 2 (see [26, 27]).
For anyϵ>0 and for each solution v of (5)–(6), such that(43)ℱvE≤ϵ,problem (5)-(6) is said to be Ulam–Hyers stable, if we can find a positive real number Cf>0 and a solution u∈C0,+∞,E of (5)-(6) satisfying the inequality(44)u−vE≤ϵCf.Definition 3 (see [26, 27]).
LetCf∈Cℝ+,ℝ+ with ξ0=0, for each solution v of (5)–(6), we can find a solution u∈C0,+∞,E of (5)-(6) such that(45)u−vE≤Cfϵ.
Then, problem (5)-(6), is called generalized Ulam–Hyers stable.Definition 4 (see [26, 27]).
For anyϵ>0 for each solution v of (5)-(6), problem (5)-(6) is called Ulam–Hyers–Rassias stable with respect to Φ∈C0,+∞,ℝ+ if(46)ℱvE≤ϵΦtt∈0,+∞,and there exists a real number Cf,Φ>0 and a solution u∈C0,+∞,E of (5)-(6) such that(47)u−vE≤ϵCf,ΦΦt,t∈0,+∞.Definition 5 (see [26, 27]).
For anyϵ>0 and for each solution v of (5)-(6), problem (5)-(6) is called generalized Ulam–Hyers–Rassias stable with respect to Φ∈C0,+∞,ℝ+ if(48)ℱvE≤Φtt∈0,+∞,and there exists Cf,Φ>0 and a solution u∈C0,+∞,E of (5)–(6) such that(49)u−vE≤Cf,ΦΦt,t∈0,+∞.Theorem 2.
If the assumptionsA1 and A2 hold, then problems (5)-(6) are generalized Ulam–Hyers stable.Proof.
By the equivalence between the operatorsId−P and ℱ and the assumptions A1,A2, we find(50)vt−ut≤vt−Pvt+Pvt−ut=Id−Pvt+Pvt−Put=ℱvt+|∫0+∞Ht,sfs,vs−fs,usds,−∫0+∞Ht,sθsgvs−gusds|≤ℱvt+∫0+∞Ht,sfs,vs−fs,usds+∫0+∞Ht,sθsgvs−gusds.
Then,(51)v−uE≤ℱvE+βv−uEΓβ+1−λσβ−bβξβ−1×∫0+∞ψ1s+ψ2s1+s1−βds≤ϵ+βv−uEΓβ+1−λσβ−bβξβ−1×∫0+∞ψ1s+ψ2s1+s1−βds.
Consequently,(52)v−uE≤Γβ+1−λσβ−bβξβ−1Γβ+1−λσβ−bβξβ−1−β∫0+∞ψ1s+ψ2s1+sβ−1dsε.
Thus, we get the Ulam–Hyers stability of (5)–(6). Then, if we take Cfε equal to the right hand side of (52), we obtain the generalized Ulam–Hyers stability of (5)-(6).Theorem 3.
Assume that the hypothesesA1 and A2 hold. In addition, the following hypotheses hold:A4 There exist two positive constants p and q such that(53)ψ1t≤pv−uEΦt,ψ2t≤qv−uEΦt.A5 There exists a positive real number CΦ such that, for each t∈0,+∞, we have(54)Φt≤∫0∞1+sβ−1Φsds≤CΦΦt.Then, problems (5) and (6) are generalized Ulam–Hyers–Rassias stable.Proof.
By exploiting the assumptionsA2,A3,A4, and A5, then we get(55)vt−ut≤vt−Pvt+Pvt−ut≤ℱvt+Pvt−Put≤ℱvt+∫0+∞Ht,sfs,vs−fs,usds+∫0+∞Ht,sθsgvs−gusds.
Then,(56)u−vE≤ℱvE+∫0∞Ht,s1+t1−βfs,vs−fs,usds+∫0∞Ht,s1+t1−βθsgvs−gusds≤Φt+βΓβ+1−λσβ−bβξβ−1∫0+∞fs,vs−fs,usds+βΓβ+1−λσβ−bβξβ−1∫0+∞θsgvs−gusds≤∫0∞1+sβ−1Φsds+βΓβ+1−λσβ−bβξβ−1∫0+∞fs,vs−fs,usds+βΓβ+1−λσβ−bβξβ−1∫0+∞θsgvs−gusds≤CΦΦt+βp+qΓβ+1−λσβ−bβξβ−1∫0+∞1+s1−βΦsds≤1+βp+qΓβ+1−λσβ−bβξβ−1CΦΦt=Cf,ΦΦt.
Hence, problems (5) and (6) are generalized Ulam–Hyers–Rassias stable.Example 1.
(57)D0+5/2ut=e−t+sinut1001+t21+t3/2+1+sinut1001+t21+t3/2t∈0,+∞,u0=0,u′0=0,D0+3/2u+∞=bu1+λ∫02usds.
In this example, we have(58)ft,ut=e−t+sinut1001+t21+t,gut=1+sinut,ft,ut−ft,vt≤11001+t21+t3/2ut−vt,gut−gvt≤ut−vt,L=1,θt=11001+t21+t3/2,ψ1t=11001+t21+t3/2,ψ2t=11001+t21+t3/2.
Sinceξ and σ are fixed, then λ and b are chosen so that hypothesis A1 is satisfied. So, we have(59)b<Γβ+1−λσββξβ−1,λ<Γβ+1σβ.
In our example, we haveβ=5/2, ξ=1, σ=2. Then,(60)b<2Γ7/2−425≈0.19,λ<Γ7/242≈0.58.
So, we can chooseλ=1/2 and b=1/10.
By simple computation, we get(61)β∫0+∞1+tβ−1ψ1t+ψ2tdt=5200∫0+∞dt1+t2+5200∫0+∞dt1+t2=5200arctant−11+t0+∞=5200π2+1≈0.06,Γβ+1−λσβ−bβξβ−1≈0.25.
Thus,A2 is satisfied.
Now, it remains to verifyA3. We have(62)∫0+∞ϕ1t+ϕ2tdt≤∫0+∞11+t2+11+t2dt=π2+1<+∞.
All hypotheses of Theorem1 are satisfied. Therefore, boundary value problems (5)-(6) has at least one solution in ℝ.



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