# A Nonlinear Fractional Problem with Mixed Volterra-Fredholm Integro-Differential Equation: Existence, Uniqueness, H-U-R Stability, and Regularity of Solutions

Authors: Somia Khaldi; Rachid Mecheraoui; Aiman Mukheimer
Journal: Journal of Function Spaces
 (2020)

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## Abstract

This paper considers nonlinear fractional mixed Volterra-Fredholm integro-differential equation with a nonlocal initial condition. We propose a fixed-point approach to investigate the existence, uniqueness, and Hyers-Ulam-Rassias stability of solutions. Results of this paper are based on nonstandard assumptions and hypothesis and provide a supplementary result concerning the regularity of solutions. We show and illustrate the wide validity field of our findings by an example of problem with nonlocal neutral pantograph equation, involving functional derivative andψ-Caputo fractional derivative.


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## Body

## 1. Introduction

Over the last few decades, there has been significant development in the area of ordinary and partial fractional differential equations with boundary integral conditions. This is reflected by various state-of-the-art papers, e.g., [1–15]. Problems with initial integral conditions have many important applications. An example is when a direct measurement quantity is impossible but their mean values are known. The fractional derivatives and integrals appear to be a very efficient tool to model various physical phenomena: kinetic theories, statistical mechanics, dynamics in complex media, control theory, signal processing, bioengineering and biomedical applications, and many others. We refer the reader to [16–18].In an attempt to formulate different problems, distinct definitions of the fractional derivative and fractional integral are available in the literature. We mention some of them.Letα>0, f an integrable function defined on a,b, and ψ∈C1a,b an increasing function with ψ′t≠0 for all t∈a,b.
(i)
The Riemann–Liouville fractional derivative:(1)Dαfx=1Γm−αddxm∫axx−tm−α−1ftdtifm−1<α<mddxm−1fxifα=m−1(ii)
The Caputo fractional derivative:(2)CDαfx=1Γm−α∫axx−tm−α−1fmtdtifm−1<α<mddxm−1fxifα=m−1(iii)
Theψ-Riemann–Liouville fractional integral:(3)Iα,ψfx=1Γα∫axψ′tψx−ψtα−1ftdt(iv)
Theψ-Riemann–Liouville fractional derivative:(4)Dα,ψfx=1Γn−α1ψ′tddxm∫axψ′tψx−ψtm−α−1ftdt,wherem=α+1Very recently, the author introduced in [10] the so-called ψ-Caputo-type fractional integral and derivative with respect to another function as follows:Definition 1 (ψ-Caputo fractional derivative).
Letn−1<α<n, f∈Cna,b, and ψ∈C1a,b be an increasing function with ψ′t≠0 for all t∈a,b. The ψ-Caputo fractional derivative operator of order α of a function f is defined by
(5)CDα,ψfx=aIxn−αfnx=1Γn−α∫axψ′tψx−ψtn−α−11ψ′tddxnftdt,where n=α+1. In particular, when α∈0,1, we have Dα,ψc(6)CDα,ψfx=1Γ1−α∫axψx−ψt−αf′tdt.
For some special cases ofψ, we obtain the Caputo fractional derivative, the Caputo–Hadamard fractional derivative, and the Caputo–Erdélyi–Kober fractional derivative.
The relationship between theψ-Caputo and the ψ-Riemann–Liouville integrals can be written as follows:Lemma 2 [10].
Letf∈Cna,b and α>0. Then, we have
(7)Iα,ψDα,ψCft=ft−∑k=1n−1fka+k!ψt−ψak.
In particular, givenα∈0,1, we have
(8)Iα,ψDα,ψCft=ft−fa.
In this paper, we state and prove a weak form of Banach contraction principal. Together with the Schauder’s fixed-point theorem, we discuss the existence, uniqueness, Hyers-Ulam-Rassias stability, and regularity of solutions for the following nonlinear fractional mixed Volterra-Fredholm integro-differential equation with nonlocal initial condition:(9)cDα,ψut=Ft,u,∫atkt,s,usds,∫aβht,s,usds,(10)ua=∫aλgsusds,where cDα,ψ is the ψ-Caputo derivative of order α, 0<α<1, t∈I=a,b, β≤λ∈a,b, g∈L1I,ℝ+, ∫aλgtdt<1, u∈Y=CI,X is a continuous function on I with values in Banach space X, uY=maxt∈IutX, F:I×Y×X2→X, and k,h:I2×X→X are continuous X valued functions. For the sake of simplicity, we denote
(11)Kut≔∫atkt,s,usds,Hut≔∫aβht,s,usds.The main contributions in this paper reside in the following:(i)
This work extends, improves, and generalizes several recent state-of-the-art results, including [19–24](ii)
Assumptions (A2)-(A6) are not standard; they have been considerably weakened. In fact, they are not supposed to hold on the overall space, but only on a subspace. Example in last section shows the importance of this new form of assumptions(iii)
Results in Section2 are obtained on the basis of a weak form of the Banach contraction principal, which is stated and proved in this paper(iv)
We discuss the Hyers-Ulam-Rassias stability of problems (9) and (10) with respect to a function φ. Generally, in the existing literature, function φ is supposed to be bounded, continuous, and verifies infφt>0. In our result, function φ is supposed to be neither continuous nor bounded nor positive. It is just an integrable locally bounded function(v)
In addition to the existence, uniqueness, and stability of solutions, the main theorem in Section3 provides a supplementary result concerning the regularity of solutions(vi)
We give an Hyers-Ulam-Rassias stability result on unbounded intervals(vii)
Last, in this paper, we prove the existence, uniqueness, Ulam-Hyers-Rassias stability, and regularity of solutions for an example of problems with nonlocal neutral pantograph equation to show the large validity field of our resultsThe remainder of this paper is structured as follows. In Section2, we briefly recall some basic definitions and some preliminary concepts about fractional calculus and auxiliary results used in the following sections. In Section 3, we discuss the existence and uniqueness of mild solutions for problems (9) and (10), using a weak form of Banach contraction principal and Schauder’s fixed-point theorem. Section 4 is devoted to present an Ulam-Hyers-Rassias stability result for problems (9) and (10), and we give a regularity result of solutions for this problem. Finally, we supply in Section 5 a well-suited example to illustrate the application and the validity of our results. In this example, we discuss the existence, uniqueness, Ulam-Hyers-Rassias stability, and regularity of solutions for a problem with nonlocal neutral pantograph equation involving a functional derivative and ψ-Caputo fractional derivative.


## 2. Preliminaries

Throughout this paper, we will use the following notations:For the sake of simplicity, we used the same symbol,·, for all norms. The closure of a set C will be denoted by C¯. F,k,h,g are given functions (problems (9) and (10)), and L2,N2,C2,μ denote the constants:
(12)L2=maxt∈IFt,0,0,0,N2=maxt,s∈I×Ikt,s,0,C2=maxt,s∈I×Iht,s,0,μ=∫aλgsds.Define operatorsT:Y→Y and G:Y×Y→Y by
(13)Tut=11−μΓα∫aλψ′τFτ,u,Kuτ,Huτ∫τλgsψs−ψτα−1dsdτ+1Γα∫atψ′τψt−ψτα−1Fτ,u,Kuτ,Huτdτ,(14)Gv,ut=11−μΓα∫aλψ′τFτ,v,Kvτ,Hvτ∫τλgsψs−ψτα−1dsdτ+1Γα∫atψ′τψt−ψτα−1Fτ,u,Kuτ,Huτdτ.For anyu∈Y, we set:
(15)Wu=Tnu;n∈ℕandVu=Gv1,v2;v1,v2∈Y×Wu¯∪Wu¯×Y,where Tn denotes T composed with itself n times.Now, we present some important theorems and lemmas in obtaining the main results.Theorem 3 (Weak form of the Banach contraction principal).
LetX,d and Y,d be two complete metric spaces, C a nonempty subset of X∩Y, and T:X→Y such that TC⊂C. Let dC (dC¯, resp.) denotes the induced metric on C (C¯, resp.). If T:C¯,dC¯→Y,d is continuous and T:C,dC→C,dC is a contraction mapping, i.e., there exists 0<λ<1 such that
(16)dTu,Tv≤λdu,v,for all u,v∈C. Then, T has a unique fixed point u∗ in the closure of C. Further, the sequence Tnun converges tou∗ for all u∈C.Proof.
For allu,v∈C and n∈ℕ, using (16) and the triangle inequality, we get on one hand that
(17)dTnu,Tnv≤λndu,v,and on the other that
(18)du,v≤11−λdu,Tu+dv,Tv.
Now, for allu∈C and n,m∈ℕ, by virtue of inequalities (17) and (18), we obtain
(19)dTnu,Tmu≤11−λdTnu,Tn+1u+dTmu,Tm+1u≤λn+λm1−λdu,Tu⟶n,m→∞0.
This means thatTnun is a Cauchy sequence in the complete metric space X. Then, it converges to a point u∗ of C¯. Reusing (19), we obtain
(20)dTTnu,Tmu≤λn+1+λm1−λdu,Tu.
Lettingn,m tend to infinity, taking into account continuity of mapping T, we deduce that u∗ is a fixed point of T. If v∗∈C¯ is a fixed point of T, there exits a sequence vnn in C that converges to v∗. By (16), we get for all u∈C and n∈ℕ,
(21)dTvn,Tn+1u≤λdvn,Tnu.
Lettingn tends to infinity, by the continuity of T in C¯,dC¯, we deduce that dTv∗,Tu∗≤λdv∗,u∗. But u∗,v∗ are two fixed points of T so v∗=u∗. This achieves the proof.Theorem 4 (Schauder’s fixed-point theorem).
LetF be a closed convex set in a Banach space X and assume that T:F⟶F is a continuous mapping such that TF is a relatively compact subset of F. Then, T has a fixed point.Theorem 5 (Arzela-Ascoli theorem).
Assume thatK is a compact set in ℝn, n≥1. Then, a set S⊂CK is relatively compact in CK if the functions in S are uniformly bounded and equicontinuous on K.Lemma 6.
Letf:a,b→ℝ+ be a nondecreasing function and ψ∈C1a,b an increasing function with ψ′t≠0 for all t∈a,b, then
(i)
Functionf is Riemann integrable and bounded(ii)
There exist positive constantsP1 and P2 such that(22)∫atfξdξ≤P1ft,∫atψ′ξψt−ψξα−1fξdξ≤P2ft,for all t∈a,b.Proof.
f is a nondecreasing function on a,b; then, it is Riemann integrable, and it attains its maximum and minimum at points b and a, respectively. Let t∈a,b, and clearly,
(23)∫atfξdξ≤b−aft,∫atψ′ξψt−ψξα−1fξdξ≤1αψb−ψaαft.
This achieves the proof.Lemma 7.
Letf:a,b→ℝ+ a continuous function, φ:a,b→ℝ+ an integrable function, and P1,P2 two positive constants. Suppose that
(i)
There existst0∈a,b;φt0=0(ii)
Inequality∫atφξdξ≤P1φt holds for all t∈a,b(iii)
Inequalityft≤P2φt holds for all t∈a,b
Then,ft=0 for all t∈a,t0.Proof.
Lett1∈a,t0, and by items (5) and (9), we have ∫at1φξdξ≤∫at0φξdξ≤P1φt0=0. Then, φt=0 a.a. t∈a,t1. So, there exits a sequence tnn⊂a,b converges to t1 and verifies φtn=0, for all n∈ℕ. But, by item (10), we deduce that ftn≤P2φtn. Keeping in mind that f is continuous, it yields ft1=0, and this concludes the proof.Lemma 8.
Letu0∈Y. The following statements hold true:
(i)
OperatorsT,Gu0,·:Y→Y are continuous(ii)
Find a solution of (9) and (10) is equivalent to find a fixed point of T, that is, an element u∈Y such that(24)ut=11−μΓα∫aλψ′τFτ,u,Kuτ,Huτ∫τλgsψs−ψτα−1dsdτ+1Γα∫atψ′sψt−ψsα−1Fs,u,Kus,Husds(i)
Ifu0 is a solution of problems (9) and (10), then for all function f∈Y, we have(25)∫aλgsTfsds=Tfa∫aλgsGu0,fsds=Gu0,ffaProof.
To prove thatT is continuous, let un be a sequence such that un→u in Y. For each t∈I, we have
(26)Tunt−Tut≤1Γα∫atψ′sψt−ψsα−1×sups∈IFs,un,Kuns,Huns−Fs,u,Kus,Husds+11−μΓα∫aλψ′τsupτ∈IFτ,un,Kunτ,Hunτ−Fτ,u,Kuτ,Huτ×∫τλgsψs−ψτα−1dsdτ.
ButF,u,un,k,h are the continuous functions on the compact I; hence, there exists t0∈I verifying the following:
(27)Ft0,un,Kunt0,Hunt0−Ft0,u,Kut0,Hut0=supξ∈IFξ,un,Kunξ,Hunξ−Fξ,u,Kuξ,Huξ.
So, by (26),
(28)Tunt−Tut≤ψb−ψaα1−μΓα+1Ft0,un,Kunt0,Hunt0−Ft0,u,Kut0,Hut0.
Therefore, sinceF:I×Y×X2→X is continuous, operator T is continuous. Similarly, we prove that T0 is continuous.
For point2, let u be a solution of ψ-fractional integro-differential problems (9) and (10). Applying the fractional integral operator Iα,ψ on both sides of (9), we get
(29)ut=ua+1Γα∫atψ′sψt−ψsα−1Fs,u,Kus,Husds.
On the other hand, keeping in mind initial condition (10), we have
(30)ua=∫aλgsusds=∫aλgsua+1Γα∫asψ′τψs−ψτα−1Fτ,u,Kuτ,Huτdτds=∫aλgsuads+1Γα∫aλ∫asgsψ′τψs−ψτα−1Fτ,u,Kuτ,Huτdτds,by Dirichlet’s formula, and it follows that
(31)ua1−∫aλgsds=1Γα∫aλψ′τFτ,u,Kuτ,Huτ∫τλgsψs−ψτα−1dsdτ.
Substituting this latter in (29) and putting ∫aλgsds=μ, we obtain
(32)ut=11−μΓα∫aλψ′τFτ,u,Kuτ,Huτ∫τλgsψs−ψτα−1dsdτ+1Γα∫atψ′sψt−ψsα−1Fs,u,Kus,Husds.
Conversely, letu in X verifying Tu=u. We have cDα,ψut=1/1−μΓαCDα,ψ∫aλψ′τFτ,u,Kuτ,Huτ∫τλgsψs−ψτα−1dsdτ+cDα,ψ1/Γα∫atψ′τψt−ψτα−1Fτ,u,Kuτ,Huτdτ=CDα,ψIα,ψFt,u,Kut,Hut=Ft,u,Kut,Hut.
Now, reusing the Dirichlet’s formula, (24) gives ua=∫aλgsusds, and this achieves the proof of point 2. For point 3, let f∈Y, and we have
(33)1−μTfa=1Γα∫aλgs∫asψ′τψs−ψτα−1Fτ,f,Kfτ,Hfτdτds=∫aλgsTfs−Tfads.
Therefore,(34)Tfa=∫aλgsTfsds.
Similarly, we get(35)T0fa=∫aλgsT0fsds.
QED.
Letu∈Y, and the following assumptions are also used:
A1 There exists a positive constant M such that
(36)supt,u∈I×YFt,ut,Kut,Hut≔M<∞.
A2u There exists a positive constant L1 such that, for all u1,u2∈Vu, x1,y1,x2,y2∈X, and t∈I:
(37)Ft,u1,x1,y1−Ft,u2,x2,y2≤L1u1t−u2t+x1−x2+y1−y2.
A3u There exists a positive constant N1 such that, for all u1,u2∈Vu and t,s∈I:
(38)kt,s,u1s−kt,s,u2s≤N1u1t−u2t.
A4u There exists a positive constant C1 such that, for all u1,u2∈Vu and t,s∈I:
(39)ht,s,u1s−ht,s,u2s≤C1u1t−u2t.
A5u With M1≔L11+b−aN1+C1, we have
(40)0≤M1ψb−ψaαΓα+11−μ<1.
A6u For all u1,u2∈Vu, if Fa,u1,Ku1a,Hu1a=Fa,u2,Ku2a,Hu2a=0, then u1a=u2a.Lemma 9.
Letu∈Y, and if A2u and A3u are satisfied, then the following statements hold true.
(41)Ku1t≤t−aN1u1+N2,Ku1t−Ku2t≤N1t−au1−u2,Hu1t≤β−aC1u1+C2,Hu1t−Hu2t≤C1β−au1−u2,for any t∈I, and u1,u2∈Vu.Proof. Immediate.


## 3. Existence and Uniqueness Results

Our first result is based on the weak form of the Banach contraction principle.Theorem 10.
Letu∈Y. If assumptions A2u−A5u are satisfied, then the fractional integro-differential problems (9)-(10) have a unique solution u∗ continuous on I. Furthermore, the sequence Tnun converges to u∗ in Y.Proof.
Letu∈Y where assumptions A2u−A5u hold. Clearly, TWu⊂Wu. We shall prove that T is a contraction on Vu. For this, let u1,u2∈Vu. By (13), we get
(42)Tu1t−Tu2t≤11−μΓα∫aλψ′τfτ,u1,Ku1τ,Hu1τ−fτ,u2,Ku2τ,Hu2τ×∫τλgsψs−ψτα−1dsdτ+1Γα∫atψ′sψt−ψsα−1×fs,u1,Ku1s,Hu1s−fs,u2,Ku2s,Hu2sds.
By means of assumptionA2u, we obtain
(43)Tu1t−Tu2t≤11−μΓα∫aλψ′τL1u1τ−u2τ+Ku1τ−Ku2τ+Hu1τ−Hu2τ×∫τλgsψs−ψτα−1dsdτ+1Γα∫atψ′sψt−ψsα−1L1u1s−u2sKu1s−Ku2sHu1s−Hu2sds,and by virtue of Lemma 9, it follows that
(44)Tu1t−Tu2t≤L1b−aC1+N1+11−μΓαu1−u2∞∫aλψ′τ∫τλgsψs−ψτα−1dsdτ+L1b−aC1+N1+1Γαu1−u2∞∫atψ′sψt−ψsα−1ds.
On the other hand, a simple calculation shows that(45)∫aλψ′τ∫τλgsψs−ψτα−1dsdτ≤μαψλ−ψaα,(46)∫atψ′sψt−ψsα−1ds≤1αψb−ψaα.
Substituting estimates (45) and (46) in relation (44), we deduce that
(47)Tu1t−Tu2t≤M1ψb−ψaαΓα+11−μu1−u2∞,or, keeping in mind condition A5u, operator T is a contraction on Vu and consequently continuous on Wu¯,dWu¯. Therefore, by the weak form of the Banach contraction principal, T has a unique fixed point u∗=Tu∗ in the closure of Wu, which is a solution of (24) and hence, by item 2 in Lemma 8, a solution of problems (9) and (10). Note that if v∗ is a solution of (9) and (10), by (9), we get fa,v∗,Kv∗a,Hv∗a=fa,u∗,Ku∗a,Hu∗a=0. Assumption (A6) implies that v∗a=u∗a; then, Gu∗,v∗=Gv∗,v∗=v∗, so v∗∈Vu. By (47), we have
(48)u∗t−v∗t=Tu∗t−Tv∗t≤M1ψb−ψaαΓα+11−μu∗−v∗∞,and this leads to the uniqueness of solutions. As a recap,
(i)
Problems (9) and (10) have a unique solution u∗(ii)
The sequenceTnun converges to u∗
This achieves the proof.In the next theorem, we shall use Schauder’s fixed-point theorem to establish the existence of solutions for problems (9) and (10), with less conditions.Theorem 11.
Under assumption (A1), problems (9) and (10) have at least one solution on I.Proof.
LetBr denotes the closed ball in Y of radius r: Br=u∈Y:u≤r⊆Y, with
(49)r≥ψb−ψaαΓα+11−μsupt,u∈I×YFt,ut,Kut,Hut,and defines the operator T on the Banach space Y by
(50)Tut=1Γα∫atψ′sψt−ψsα−1Fs,u,Kus,Husds+11−μΓα∫aλψ′τFτ,u,Kuτ,Huτ∫τλgsψs−ψτα−1dsdτ.
Clearly,Tu∈Br whenever u∈Br, i.e., T:Br→Br, and item 2 in Lemma 8 assures that T is continuous. Now, we shall prove that TBr is an equicontinuous set of Y. Let t1,t2∈I and u∈Br, we have
(51)Tut1−Tut2=1Γα∫at1ψ′sψt1−ψsα−1Fs,u,Kus,Husds−∫at2ψ′sψt2−ψsα−1Fs,u,Kus,Husds≤1Γα∫t2t1ψ′sψt1−ψsα−1Fs,u,Kus,Husds+1Γα∫at2ψ′sψt2−ψsα−1−ψt1−ψsα−1Fs,u,Kus,Husds.
Taking into account assumption (A1), we obtain(52)Tut1−Tut2≤MΓα∫t2t1ψ′sψt1−ψsα−1ds+∫at2ψ′sψt2−ψsα−1−ψt1−ψsα−1ds.
But,(53)∫t2t1ψ′sψt1−ψsα−1ds=1αψt1−ψt2α∫at2ψ′sψt2−ψsα−1−ψt1−ψsα−1ds=1α−ψt2−ψaα−ψt1−ψt2α+ψt1−ψaα.
Therefore,(54)Tut1−Tut2≤MΓα+1ψt1−ψaα−ψt2−ψaα,and consequently, supu∈BrTut1−Tut2→0 as t1→t2, which means that TBr is an equicontinuous set of Y.
ButT:Br⊂CI,X→Br, so TBr is uniformly bounded, and the Arzela-Ascoli theorem implies that TBr is relatively compact. Therefore, T:Br→Br is continuous on the closed convex set Br, and TBr is relatively compact. According to the Schauder’s fixed-point theorem, mapping T has at least a fixed point. Item 2 in Lemma 8 achieves the proof.


## 4. Ulam-Hyers-Rassias Stability

In this section, we discuss the Ulam-Hyers-Rassias stability of problems (9) and (10). First, we introduce a basic definition and some notations and hypotheses for this section.Definition 12 (Ulam-Hyers-Rassias stability).
If for each functionu satisfying
(55)Dα,ψut−Ft,u,Kut,Hutc≤ψt,where ψ is a nonnegative function, there exists a solution u∗ of the fractional differential problems (9) and (10) and a constant C>0 independent of u and u∗ such that
(56)ut−u∗t≤Cψt,for all t; then, we say that problems (9) and (10) are the Hyers-Ulam-Rassias stability.Let (H) denotes the following hypothesis:We denote byΦ the set of all functions φ∈L1I,ℝ+ verifying φ is locally bounded, inft∈Iφt=0, and there exist two positive constants P1,P2 such that, for all t∈I(57)∫atφτdτ≤P1φt,∫atψ′τψt−ψτα−1φτdτ≤P2φt.Letφ∈Φ, P3∈ℝ+, and C:I→ℝ+ a function.Letu:I→X be a continuously differential function. Assume that assumptions A2u−A6u hold together with one of the following:
(i)
For allt∈I we have(58)1αψt−ψaα≤P3φt(ii)
For allu1,u2∈Vu and t∈I we have(59)Hu1t−Hu2t≤Ct∫aλgsu1s−u2sdsFinally, we denote byγ the constant
(60)γ=0;Ifinequality18holds1;Otherwise,and we assume that
(61)Ω≔L1P21+N1P1+P3γC1∫aβφτdτΓα∈0,1.Now, we state and prove the main result of this section.Theorem 13.
If the functionu satisfies
(62)Dα,ψut−Ft,u,Kut,Hutc≤φt,(63)ua=∫aλgsusds,for all t∈I. Then, there exists a unique function u∗:I→ℝ solution of problems (9) and (10), with
(64)ut−u∗t≤P21−ΩΓαφt,forallt∈I.Remark 14.
Here, we give some important remarks:(i)
If functiont↦Ft,0,∫atkt,s,0ds,∫aβht,s,0ds is member of Φ we get(65)u∗t≤P21−ΩΓαFt,0,∫atkt,s,0ds,∫aβht,s,0ds,for all t∈I(ii)
Using Lemma6, we can substitute space Φ by the space of nondecreasing functions(iii)
The caseinfξ∈Iφξ>0 is obvious, so we have omitted this case, supposing thatinfξ∈Iφξ=0Proof.
For allf1,f2∈Y, we denote by Ωf1,f2 the set
(66)Ωf1,f2≔C∈0,∞∣f1t−f2t≤Cφt,forallt∈I.
Easily, we can prove thatd:Y×Y⟶ℝ defined by
(67)df1,f2=infΩf1,f2ifΩf1,f2≠Φ+∞ifΩf1,f2=Φis a generalized metric on Y verifying the following assumptions:
(i)
If there exists a sequencetn→t0 in I such that limφtn=0, then df1,f2=+∞ for all f1,f2∈Y satisfying f1t0≠f2t0(ii)
For allf1,f2∈Y, we have(68)f1t−f2t≤df1,f2φt,for any t∈I
We claim thatY,d is complete. In fact, let vn be a Cauchy sequence in Y,d, and having in mind definition (67), we have
(69)∀ε>0,∃Nε>0,∀n,m≥Nε∀t∈I:vmt−vnt≤εφt.
SinceX is complete, (69) implies that vnt converges for each t∈I. Let us show that function v:I⟶X defined by vt=limn→∞vnt is continuous or belongs to Y. Passing to the limit with respect to m in the latter inequality, it follows that
(70)∀ε>0,∃Nε>0,∀n≥Nε∀t∈I:vt−vnt≤εφt.
But,φ must be bounded because it is locally bounded on the compact I. So,
(71)∀ε>0,∃Nε>0,∀n≥Nεsupt∈Ivt−vnt≤εsupt∈Iφt.
Hence,vn converges uniformly to v, and consequently, v is continuous. Further, if we consider (67) and (69), we may conclude that
(72)∀ε>0,∃Nε>0,∀n≥Nεdv,vn≤ε.
This means, the Cauchy sequencevn converges to v in Y,d; thus, Y,d is complete. On the other hand, Theorem 10 assures the existence of a unique function u0∈Y solution of problems (9) and (10), verifying limnTnu=u0. We define an operator T0:Y→Y by T0f=Gu0,f, i.e.,
(73)T0ft=11−μΓα∫aλψ′τFτ,u0,Ku0τ,Hu0τ∫τλgsψs−ψτα−1dsdτ+1Γα∫atψ′sψt−ψsα−1Fs,f,Kfs,Hfsds,
for allf∈Y, and we discuss its contractivity. Note that T0Wu⊂Vu for all u∈Y. Let u:I→X be a continuously differential function satisfies (62) and (63). By virtue of Lemma 7 and relation (62), we obtain
(74)Dα,ψua−Fa,u,Kua,Huac=0,and by (5), we deduce that
(75)cDα,ψua=cDα,ψu0a=Fa,u0,Ku0a,Hu0a=0.
Thus, keeping assumptionA6u, we deduce that ua=u0a, and consequently, by item 3 in Lemma 8,
(76)∫aλgsfsds=u0aforallf∈Γu,where Γu≔T0nu;n∈ℕ. By definition, Γu⊂Vu. So, taking into account assumption A2u, for all f1,f2∈Γu, we obtain
(77)T0f1t−T0f2t≤1Γα∫atψ′sψt−ψsα−1Fs,f1,Kf1s,Hf1s−Fs,f2,Kf2s,Hf2sds≤L1Γα∫atψ′τψt−ψτα−1f1τ−f2τ+Kf1τ−Kf2τHf1τ−Hf2τdτ.
By virtue of assumptionA3u and relations (57) and (68), we immediately get
(78)Kf1s−Kf2s≤N1P1φsdf1,f2,foralls∈I,and by relation (76), assumption A4u, and hypotheses (H), we obtain
(79)Hf1s−Hf2s≤maxγC1∫aβf1τ−f2τdτ,∫aλgτf1τ−f2τdτ=γC1df1,f2∫aβφτdτ.
Substituting estimates (78) and (79) in (77), it yields
(80)T0f1t−T0f2t≤L11+NP1Γαdf1,f2∫atψ′sψt−ψsα−1φsds+L1γC1∫aβφτdτΓαdf1,f2∫atψ′sψt−ψsα−1ds≤L1P21+N1P1Γαdf1,f2φt+L1γP3C1∫aβφτdτΓαdf1,f2φt.
Or, withΩ=L1P21+N1P1+P3γC1∫aβφτdτ/Γα and using the continuity of T0,
(81)dT0f1,T0f2≤Ωdf1,f2,for all f1,f2∈Γu¯. Obviously, T0Γu⊂Γu, and by induction, we get
(82)dT0mf1,T0mf2≤Ωmdf1,f2.
Keeping in mind assumptionA5u, from (81) we get
(83)df1,f2≤11−Ωdf1,T0f1+df2,T0f2.
By the two last inequalities, these yield(84)dT0nu,T0mu≤Ωn+Ωm1−Ωdu,T0u,for all n,m∈ℕ. Applying the fractional integration operator Iα,ψ to both sides of (62), taking into account the fact that ua=T0ua, we obtain
(85)ut−T0ut≤P2Γαφt,for all t∈I. This implies that du,T0u<∞, and consequently, dT0nu,T0mu→0 as n and m tend to infinity. Space Y is complete; then, this Cauchy sequence converges to a point u∗ of Y. By Lemma 8, operator T0 is continuous; so, using inequality (84), we deduce that u∗ is a fixed point of T0. But u0 is a fixed point for T0, and inequality (81) assures that u∗ is none other than u0. Letting m tends to infinity in (84), with n=0, we get
(86)du,u∗≤11−Ωdu,T0u,and consequently,
(87)ut−u∗t≤P21−ΩΓαφt,for all t∈I. This achieves the proof.Corollary 15.
Assume thatI=a,+∞ is an unbounded interval. If the function u satisfies
(88)Dα,ψut−Ft,u,Kut,Hutc≤φt,(89)ua=∫aλgsusds,for all t∈I. Then, there exists a unique function u∗:I→ℝ solution of problems (9) and (10), with
(90)ut−u∗t≤P21−ΩΓαφt,for all t∈I.Proof.
Letu:I→ℝ be a continuously differential function satisfies (88) and (89) and n0>λ a positive integer. For all n≥n0, by Theorem 13., there exists a unique continuous function un:In→ℝ satisfying
(91)unt=11−μΓα∫anλψ′τFτ,un,Kunτ,Hunτ∫τλgsψs−ψτα−1dsdτ+1Γα∫antψ′sψt−ψsα−1Fs,un,Kuns,Hunsds,(92)ut−unt≤P21−ΩΓαφt,for all t∈In. By virtue of the uniqueness of the solution un, we deduce that unt=un+jt for all j∈ℕ and all t∈In. We define nt∈ℕ as
(93)nt=minn∈ℕ∣t∈Inand the function v0:I→ℝ by
(94)v0t=untt.
Easily, by (91)-(94), we can prove that
(95)v0t=11−μΓα∫aλψ′τFτ,v0,Kv0τ,Hv0τ∫τλgsψs−ψτα−1dsdτ+1Γα∫atψ′sψt−ψsα−1Fs,v0,Kv0s,Hv0sds,v0t−ut≤P21−ΩΓαφt,forallt∈I.
This achieves the proof.


## 5. Example

We consider the following nonlocal problem with neutral pantograph equation involving functional derivative andψ-Caputo fractional derivative:
(96)Dα,ψut=3lnt+1+ut+∫01u2/3sds3π+∫0tln1+ssu2′sdsexpt+ut+∫01u2/3sds,Cu0=12∫01uss+sds,where α=2/3 and ψ:I=0,1→ℝ+ such that ψt=ln2t+1.We claim that problem (96) has a unique solution u∗ verifying the following:
(97)u∗t≤10π10π−3ln2ln1+t,for all t∈0,1.We shall prove that if a continuously differential functionu:I→ℝ satisfies
(98)CDα,ψut−3lnt+1+∣ut∣+∫01u2/3sds3π+∫0tln1+ssu2′sdsexpt+∣ut∣+∫01u2/3sds≤ln1+t,(99)u0=14∫01uss+sds,for all t∈I. Then, there exists a unique continuous function u∗:I→ℝ solution of problem (96) with
(100)ut−u∗t≤10π10π−3ln2φt,for all t∈I.Letu0∈Y≔C10,1,ℝ verifying (98) and (99), and we put φt≔ln1+t. It is seen that problem (96) is equivalent to the following:
(101)CDα,ψut=ft,u,∫0tkt,s,usds,∫02/3ht,s,usdsu0=∫01gsusds,where
(102)I=0,1,kt,s,x=0,ht,s,x=32x,gt=12t+t,ft,u,x,y=3lnt+1+∣ut∣+∣y∣3π+∫0tln1+ssu2′sdsexpt+∣ut∣+∣y∣,for all t,s∈I, u∈C1I,ℝ, and x,y∈ℝ. Further, we have the following:(B1) f is uniformly bounded, in fact
(103)supt∈I,u∈Y,x,y∈ℝ∣ft,u,x,y∣<13π.B2 For all u1,u2∈Y, if f0,u1,Ku10,Hu10=f0,u2,Ku20,Hu20=0, then u10=u20=0.B3f:I×Vu0×ℝ2satisfies
(104)∣ft,u1,x1,y1−ft,u2,x2,y2∣≤325∣u1t−u2t∣+∣x1−x2∣+∣y1−y2∣.In fact, for allt∈I, x1,x2,y1,y2∈ℝ and u1,u2∈Vu0, we have
(105)∣ft,u1,x1,y1−ft,u2,x2,y2∣≤13π+∫0tln1+ssu12′sds3lnt+1+∣u1t∣+∣y1∣expt+∣u1t∣+∣y1∣−3lnt+1+∣u2t∣+∣y2∣expt+∣u2t∣+∣y2∣+13π+∫0tln1+ssu12′sds−13π+∫0tln1+ssu22′sds×3lnt+1+∣u2t∣+∣y2∣expt+∣u2t∣+∣y2∣.On the other hand, we have(106)3lnt+1+∣u1t∣+∣y1∣expt+∣u1t∣+∣y1∣−3lnt+1+∣u2t∣+∣y2∣expt+∣u2t∣+∣y2∣≤supξ∈I∣3lnξ+1−expξ∣1expt+∣u1t∣+∣y1∣−1expt+∣u2t∣+∣y2∣≤∣u1t−u2t∣+∣y1−y2∣,(107)13π+∫0tln1+ssu12′sds−13π+∫0tln1+ssu22′sds≤19π2∫0tln1+ssu12−su22′sds≤ln29π2tu12t−u22t.But, usingB1, for all u,v∈Vu0, gets
(108)∣Gu,vt∣≤2ln23ππ1−ln2.Therefore, remembering thatu1,u2∈Vu0, we obtain
(109)13π+∫0tln1+su1′sds−1π+∫0tln1+su2′sds≤2ln227π2∣u1t−u2t∣Substituting (106) and (109) in (105), we get, immediately, the desired inequality.B4k:I×I×ℝ→ℝ is continuous and satisfies the Lipschitz condition:
(110)kt,s,x−kt,s,y≤0x−y,for all t,s∈Iandf,g∈ℝ.B5h:I×I×ℝ→ℝ is continuous and satisfies the Lipschitz condition:
(111)Hu1s−Hu2s=∫02/3ht,s,u1s−ht,s,u2sds=32∫02/3u1s−u2sds.for all t,s∈I and x,y,x1,y1,z1,x2,y2,z2∈ℝ.B6φ∈L1I,ℝ+ verifying φ is bounded, inft∈Iφt=0, and for all t∈I, we have ∫0tln1+sds≤P1ln1+t, ∫0tψ′sψt−ψs−1/2ln1+sds≤P2ln1+t, and ψt−ψ01/2≤P3ln1+t, with P1=1, P2=2ln2 and P3=1.B7 With M1≔L11+b−aN1+C1, L1=3/25, N1=0, C1=3/2, α=1/2, a=0, b=λ=1, P1=1, P2=2ln2, and P3=1, we have
(112)0≤M1ψb−ψaαΓα+11−μ<1,0<Ω≔L1P21+N1P1+P3C1∫aλφτdτΓα<1.By assumptionsB1−B7 and Theorems 10 and 13., we deduce that problem (96) has, exactly, one solution u∗ and if a continuously differential function u:I→ℝ satisfies
(113)CDα,ψut−3lnt+1+∣ut∣+∫01u2/3sds3π+∫0tln1+su′sdsexpt+∣ut∣+∫01u2/3sds≤ln1+t,u0=12∫01uss+sds,for all t∈I. Then,
(114)∣ut−u∗t∣≤10π10π−3ln2ln1+tforallt∈I.With functionu≡0, we get
(115)CDα,ψut−3lnt+1+∣ut∣+∫01u2/3sds3π+∫0tln1+su′sdsexpt+∣ut∣+∫01u2/3sds=ln1+tπexpt≤ln1+t,and
(116)u0=12∫01uss+sds=0.Therefore, by (109),
(117)u∗t≤10π10π−3ln2ln1+t,for all t∈0,1.


## 6. Conclusions

In this paper, using a fixed-point approach and by means of theψ-Caputo fractional derivative, we have studied the stability of solutions of a nonlinear fractional mixed Volterra-Fredholm integro-differential equation with an integral initial condition.In the first part of this paper, a weak form of the Banach contraction principal and some useful lemmas are stated and proved. In the second part, we established conditions that assure the existence and uniqueness of solutions for the considered problem, basing on the weak form of the Banach contraction principal and the Schauder’s fixed-point theorem. The study of the Ulam-Hyers-Rassias stability is with respect to a functionφ. When we impose more restrictions onφ, it will make the result less interesting in the reality world and vice versa. In the third part of this work, we have discussed of Ulam-Hyers-Rassias with respect to a function with very less restrictions. Two tough restrictions were omitted, and the continuity and that infφt must be positive. Results in this paper provide a supplementary result concerning the regularity of solutions. Finally, the whole analysis has been demonstrated by a suitable example.



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