# Some Weakly Singular Integral Inequalities and Their Applications to Tempered Fractional Differential Equations

Authors: Abdellatif Ben Makhlouf; Djalal Boucenna; A.M. Nagy; Lassaad Mchiri
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we establish several new weakly singular integral inequalities that generalize several previously known ones. Several applications for fractional differential equations in the Caputo context have been derived using tempered fractional differential equations.


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## Body

## 1. Introduction

Integral inequalities are important in the qualitative study of differential and integral equations, see [1, 2]. The Gronwall inequality defines explicit bounds on the solutions of a class of integral inequalities. This inequality has been refined and used in a variety of contexts [1, 3–5]. However, none of the preceding inequalities are directly applicable to the analysis of integral equations with weakly singular kernels. The author in [6] proposed a new approach for solving linear Integral Inequalities with weakly Singular Kernels (IISKs) in 1981. Medved developed a novel technique for solving Henry–Gronwall type integral inequalities and their Bihari version [7] as well as global solutions to semilinear evolution equations [8]. IISKs have recently received increased research attention due to the growth of fractional differential equations, see for example [3, 5–12]. Indeed, Medved and Ma as well as Pecaric have investigated the following integral inequalities in [7, 11]:(1)gξ≤aξ+bξ∫0ξξ−κγ−1vκgκdκ.Medved has studied the Henry-type integral inequalities in [7, 8]:(2)gξ≤aξ+∫0ξξ−κγ−1vκwgκdκ.Later, in [5], Zhu has described the integral inequalities (1) and (2) and, as an application, the existence of solutions to fractional differential equations with Caputo derivative has been studied.Tempered fractional calculus is a broadening of fractional calculus defined by integrals in which the kernel begins a fractional power function multiplied by an exponential factor [13, 14]. There are more applications for tempered differential equations and tempered fractional derivatives, such as finance [15], poroelasticity [16] and geophysical flows [17].The main contribution of this work, which was inspired by Zhu’s technique [5], is the description of certain weakly singular integral inequalities as well as the Ulam stability of the following issue as an application,(3)DC0+δ,λuξ=fξ,uξ,δ∈0,1,λ>0,t∈0,T,u0=u0,where DC0+δ,λ is the Caputo tempered fractional derivative and the function f:0,T×ℝ⟶ℝ, satisfying some conditions will be specified later.To the best of our knowledge, there are no other works in the literature dealing with the same subject. The following is the paper’s structure. Section2 introduces the preliminary. Section 3 presents several weakly singular integral inequalities. Section 4 deals with some applications of fractional differential equations. In the final section, an example is provided.


## 2. Preliminaries

Definition 1 (see [13, 18]) (Riemann–Liouville tempered fractional integral).
Letu∈L10,d,δ>0,λ≥0. The Riemann–Liouville tempered fractional integral of u of order δ is given by(4)I0+δ,λut=exp−λtI0+δexpλtut=exp−λtΓδ∫0texpλst−sδ−1usds=1Γδ∫0texp−λt−st−sδ−1usds,where I0+δ denotes the Riemann–Liouville fractional integral of order δ.Definition 2 (see [13, 18]) (Riemann–Liouville tempered fractional derivative).
Let0<δ<1 and λ≥0. The Riemann–Liouville tempered fractional derivative of u of order δ is given by(5)D0+δ,λut=exp−λtD0+δexpλtut=exp−λtΓ1−δddt∫0tt−s−δexpλsusds,where D0+δ denotes the Riemann–Liouville fractional derivative of order δ.Definition 3 (see [13, 18]) (Caputo tempered fractional derivative).
Let0<δ<1 and λ≥0. The Caputo tempered fractional derivative of u of order δ is given by(6)DC0+δ,λut=D0+δ,λut−exp−λtu0.


## 3. Main Results

Theorem 1.
Let0<T≤∞,λ>0,δ∈0,1,aξ,uξandwξare nonnegative, continuous functions on0,Twith(7)uξ≤aξ+1Γδ∫0ξexp−λξ−κξ−κδ−1wκuκdκ.Then,(8)uξ≤ℛξ+exp−λαξ∫0ξexpλακGκℛκexp∫κξGςdςdκα,ξ∈0,T.Ifaξ is nondecreasing on 0,T, then the inequality (8) becomes(9)uξ≤ℛξexp∫0ξGκdκα.Ifaξ≡0 on 0,T, we find(10)uξ≡0,where ℛξ=21/α−1a1/αξ,Gξ=21/α−1/Γ1/αδΓδ−α/1−αΓ1−δ/1−α1−α/αξδ−α/αw1/αξ, and 0<α<δ<1.Proof.
Based on inequality (7) and Holder inequality, we get(11)uξ≤aξ+exp−λξΓδ∫0ξξ−κδ−1κα−δexpλκκδ−αwκuκdκ≤aξ+exp−λξΓδ∫0ξξ−κδ−1κα−δ1/1−αdκ1−α×∫0ξexpλκκδ−αwκuκ1/αdκα≤aξ+exp−λξΓδ∫0ξξ−κδ−1/1−ακα−δ/1−αdκ1−α×∫0ξexpλακκδ−α/αw1/ακu1/ακdκα≤aξ+exp−λξΓδΓδ−α1−αΓ1−δ1−α1−α×∫0ξexpλακκδ−α/αw1/ακu1/ακdκα.
Taking into consideration thatx1+x2σ≤2σ−1x1σ+x2σ for all x1,x2∈ℝ+2 and σ≥1, we get(12)u1/αξ≤21/α−1a1/αξ+exp−λ/αξΓ1/αδΓδ−α1−αΓ1−δ1−α1−α/α×∫0ξexpλακκδ−α/αw1/ακu1/ακdκ.
Therefore,(13)expλαξu1/αξ≤21/α−1expλαξa1/αξ+1Γ1/αδΓδ−α1−αΓ1−δ1−α1−α/α×∫0ξexpλακκδ−α/αw1/ακu1/ακdκ.
By takingθξ=expλ/αξu1/αξ, the above inequality becomes(14)θξ≤expλαξℛξ+∫0ξGκθκdκ.
Using Lemma 2.2 in [5], the inequality (8) is obtained. The rest of the proof is obvious.Theorem 2.
Let0<T≤∞,λ,δ>0,aξ,bξ,uξandwξare nonnegative, continuous functions on0,Tsuch that(15)uξ≤aξ+bξΓδ∫0ξexp−λξ−κξ−κδ−1wκuκdκ.Then,(16)uξ≤aξ+ℱξ∫0ξGκℋκapκdκ1/p1−1−ℋξ1/p,ξ∈0,T,where Gξ=expλξwpξ, ℱξ=bξexp−λξ/ΓδΓγξγE1,γ+1λξ1/q, ℋξ=exp−∫0ξGκℱpκdκ, E1,γ+1 is the general Mittag–Leffler function and p,q∈0,∞ with 1/q+δ>1, 1/q+1/p=1 and γ=δ−1q+1.Proof.
Using Holder inequality and the fact that(17)1Γβ∫0ξexpλκξ−κβ−1dκ=ξβE1,β+1λξ,forβ>0,we get(18)uξ≤aξ+bξΓδ∫0ξexp−λξ−κξ−κδ−1wκuκdκ≤aξ+bξexp−λξΓδ∫0ξexpλκξ−κδ−1qdκ1/q×∫0ξexpλκwκuκpdκ1/p≤aξ+bξexp−λξΓδΓγξγE1,γ+1λξ1/q×∫0ξexpλκwpκupκdκ1/p.
So,(19)uξ≤aξ+ℱξ∫0ξGκupκdκ1/p.
By using Lemma 2.3 in [5] the inequality (16) is obtained.Theorem 3.
Let0<T≤∞,λ,δ>0,aξ,bξ,uξandwξare nonnegative, continuous functions on0,Tsuch that(20)uξ≤aξ+bξΓδ∫0ξexp−λξ−κξ−κδ−1wκuκdκ.Then,(21)uξ≤ℛξ+ℱξ∫0ξGκℛκexp∫κξGςℱςdςdκ1/p,x∈0,T,where ℛξ=2p−1apξ, ℱξ=2p−1bξexp−λξ/ΓδΓγξγE1,γ+1λξ1/qp, Gξ=expλξwpξ and p,q∈0,∞ with 1/q+δ>1, 1/q+1/p=1 and γ=δ−1q+1.Proof.
From the above theorem, we have(22)uξ≤aξ+bξΓδ∫0ξexp−λξ−κξ−κδ−1wκuκdκ≤aξ+bξexp−λξΓδΓγξγE1,γ+1λξ1/q×∫0ξexpλκwpκupκdκ1/p.
Then,(23)upξ≤2p−1apξ+bξexp−λξΓδΓγξγE1,γ+1λξ1/qp×∫0ξexpλκwpκupκdκ.
By takingvξ=upξ, the last above inequality becomes(24)vξ≤ℛξ+ℱξ∫0ξGκvκdκ.
From Martyniuk et al. [4], we obtain inequality (21).Theorem 4.
Letδ∈0,1,λ>0,aξbe a nondecreasingC1function on0,Twithat>0for allt∈0,T,wξ,uξare nonnegative, continuous functions on0,Tandh:0,∞⟶0,∞be an increasing, continuous function with(25)uξ≤aξ+1Γδ∫0ξexp−λξ−κξ−κδ−1wκhexpλκuκdκ.Then,(26)uξ≤exp−λξΩ−1Ωℛξ+∫0ξGκdκα,ξ∈0,T1,where 0<α<δ<1, ℛξ=21/α−1expλ/αξa1/αξ, Gξ=21/α−1/Γ1/αδΓδ−α/1−αΓ1−δ/1−α1−α/αexpλ/αξξδ−α/αw1/αξ, Ωy=∫t0y1/Hκdκ,Hξ=h1/αξα, t0>0, Ω−1 is the inverse function of Ω, and T1∈0,T is such that Ωℛξ+∫0ξGκdκ∈DomΩ−1 for all ξ∈0,T1.Proof.
From (25) and Holder inequality, we get(27)uξ≤aξ+exp−λξΓδ∫0ξξ−κδ−1κα−δexpλκκδ−α×wκhexpλκuκdκ≤aξ+exp−λξΓδ∫0ξξ−κδ−1/1−ακα−δ/1−αdκ1−α×∫0ξexpλακκδ−α/αw1/ακh1/αexpλκuκdκα≤aξ+exp−λξΓδΓδ−α1−αΓ1−δ1−α1−α×∫0ξexpλακκδ−α/αw1/ακh1/αexpλκuκdκα.
Then,(28)u1/αξ≤21/α−1a1/αξ+exp−λ/αξΓ1/αδΓδ−α1−αΓ1−δ1−α1−α/α×∫0ξexpλακκδ−α/αw1/ακh1/αexpλκuκdκ.
By takingQξ=expλ/αξu1/αξ, the last above inequality becomes(29)Qξ≤ℛξ+∫0ξGκHQκdκ.
LetVξ be the right-hand side of the last above inequality. Then,(30)V′ξHVξ=ℛ′ξ+GξHQξHVξ≤ℛ′ξHℛξ+Gξ,
or(31)ddξΩVξ≤ddξΩℛξ+Gξ.
By integrating both sides of the last inequality from 0 toξ, we obtain(32)ΩVξ≤Ωℛξ+∫0ξGκdκ,and since Ω is an increasing function, so we get(33)Qξ≤Vξ≤Ω−1Ωℛξ+∫0ξGκdκ.
This achieves the proof.


## 4. Applications to Fractional Differential Equations

In this section, we give the existence, uniqueness and Ulam stability results of (3).For0<T<∞, we introduce the following assumptions.(A)f:0,T×ℝ⟶ℝ is a continuous function and there exists g:0,∞⟶0,∞ a continuous, increasing function and g0=0 such that(34)fξ,u−fξ,v≤gexpλξu−v,for all λ>0,u,v∈ℝ and ξ∈0,T.Note that if (A) is satisfied then there exists a positive continuous functionkξ such that(35)fξ,u≤gexpλξu+kξ,for all u∈ℝ and ξ∈0,T.


### 4.1. The Existence and Uniqueness

In the initial value problem (IVP) (3), the solution is given by [13](36)ut=u0exp−λξ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ.We will verify the existence and uniqueness using Theorem4.Theorem 5.
Assume that (A) is satisfied. If(37)limy⟶0+∫1y1Hκdκ=−∞,and(38)limy⟶+∞∫1y1Hκdκ=+∞,whereHs=g1/αsα,0<α<δ. Then the(IVP) (3) has a unique solution.Proof.
Consider the operatorG:C0,T,ℝ⟶C0,T,ℝ defined by(39)Guξ=u0exp−λξ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ.
We shall use Schaefer’s fixed point theorem to prove thatG has a fixed point. The proof will be given in several steps.Step 1.
G is continuous.
Letun be a sequence such that un⟶u in C0,T,ℝ. Then, for each ξ∈0,T, we have(40)Gunξ−Guξ≤1Γδ∫0ξξ−κδ−1fκ,unκ−fκ,uκdκ≤TδΓδ+1gexpλTun−u∞,and this implies that(41)Gun−Gu∞≤TδΓδ+1gexpλTun−u∞.
Thus,G is continuous.Step 2.
G maps bounded sets into bounded sets in C0,T,ℝ.
It is enough to show that for anyr1>0 there exists a positive constant r2>0 such that for each u∈Br1=u∈C0,T,ℝ,u∞≤r1 we have Gu∈Br2. For each ξ∈0,T we have(42)Guξ=u0exp−λξ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤u0+1Γδ∫0ξξ−κδ−1fκ,uκdκ≤u0+1Γδ∫0ξξ−κδ−1gexpλκuκdκ+1Γδ∫0ξξ−κδ−1kκdκ≤u0+TδΓδ+1gr1+k∗,where k∗=supt∈0,Tkt and gr1=gr1expλT; thus,(43)Gu∞≤u0+TδΓδ+1gr1+k∗≕r2.Step 3.
G maps bounded sets into equicontinuous sets of C0,T,ℝ.
Letr1>0 and u∈Br1. For 0≤ξ1<ξ2<T, we have(44)Guξ2−Guξ1≤u0exp−λξ2−exp−λξ1+1Γδ|∫0ξ2exp−λξ2−κξ2−κδ−1fκ,uκdκ−∫0ξ1exp−λξ1−κξ1−κδ−1fκ,uκdκ≤u0exp−λξ2−exp−λξ1+1Γδ∫0ξ1|exp−λξ2−κξ2−κδ−1−exp−λξ1−κξ1−κδ−1fκ,uκdκ+1Γδ∫ξ1ξ2exp−λξ2−κξ2−κδ−1fκ,uκdκ≤u0exp−λξ2−exp−λξ1+1Γδ∫0ξ1exp−λξ2−κξ1−κδ−1−ξ2−κδ−1fκ,uκdκ+1Γδ∫0ξ1exp−λξ2−κ−exp−λξ1−κξ1−κδ−1fκ,uκdκ+1Γδ∫ξ1ξ2exp−λξ2−κξ2−κδ−1fκ,uκdκ≤λu0ξ2−ξ1+gr1+k∗Γδ+1ξ1δ−ξ2δ+ξ2−ξ1δ+λgr1+k∗TδΓδ+1ξ2−ξ1+gr1+k∗Γδ+1ξ2−ξ1δ≤λu0+gr1+k∗TδΓδ+1ξ2−ξ1+gr1+k∗Γδ+1ξ1δ−ξ2δ+2ξ2−ξ1δ
Asξ2⟶ξ1, the right-hand side of the above inequality tends to zero. As a consequence of Steps 1 to 3 together with the Arzelá–Ascoli theorem, we conclude that the operator G is completely continuous.Step 4.
A priori bounds.
It remains to show that the set(45)A=u∈C0,T,ℝ,u=λ˜Gu,forλ˜∈0,1,is bounded. Taking 0<α<δ<1, and η=δ−α/1−α. Let u∈A, then(46)uξ=λ˜u0exp−λξ+λ˜Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤exp−λξu0+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤exp−λξu0+1Γδ∫0ξexp−λξ−κξ−κδ−1kκdκ+1Γδ∫0ξexp−λξ−κξ−κδ−1gexpλκuκdκ≤exp−λξu0+exp−λξΓδΓηξηE1,η+1λξ1−α×∫0ξexpλκk1/ακdκα+1Γδ∫0ξexp−λξ−κξ−κδ−1gexpλκuκdκ.
Thus, for eachϵ>0, we obtain(47)uξ≤ɛ+exp−λξu0+exp−λξΓδΓηξηE1,η+1λξ1−α×∫0ξexpλκk1/ακdκα+1Γδ∫0ξexp−λξ−κξ−κδ−1gexpλκuκdκ.
According to (38) and Theorem 4, we get(48)uξ≤exp−λξΩ−1Ωℛξ+∫0ξGκdκα,ξ∈0,T,where the formula of Ωξ,ℛξ and Gξ are defined as in Theorem 4, with aξ=ε+exp−λξu0+exp−λξ/ΓδΓηξηE1,η+1λξ1−α∫0ξexpλκk1/ακdκα, hξ=gξ and wξ=1. As a consequence of Schaefer’s fixed point theorem, we deduce that the operator G has at least one fixed point in C0,T,ℝ which is a solution of the (IVP) (3).
For the uniqueness, we suppose thatu1ξ, u2ξ are two solutions of (IVP) (3). Then,(49)u1ξ−u2ξ=1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,u1κ−fκ,u2κdκ≤1Γδ∫0ξexp−λξ−κξ−κδ−1gexpλξu1κ−u2κdκ.
Thus, for eachε>0, we obtain(50)u1ξ−u2ξ≤ϵ+1Γδ∫0ξexp−λξ−κξ−κδ−1×gexpλξu1κ−u2κdκ.
According to (37), (38) and Theorem 4, we get(51)u1ξ−u2ξ≤exp−λξΩ−1Ωℛξ+∫0ξGκdκα,ξ∈0,T,where the formula of Ωξ,ℛξ and Gξ are defined as in Theorem 4, with aξ=ε,hξ=gξ and wξ=1. Thus, u1ξ=u2ξ for all ξ∈0,T.


### 4.2. Generalised Ulam Stability

We consider the following inequality(52)D0Cδ,λvξ−fξ,vξ≤ε,forλ,ε>0,δ∈0,1,ξ∈0,T.Definition 4.
The equation (3) is Ulam-Hyers stable if there exists a function ψ:0,∞⟶0,∞ such that for each ε>0, and for each solution v∈C0,T,ℝ of the inequality (51) there exists a solution u∈C0,T,ℝ of (3) with(53)vξ−uξ≤ψε,where limε⟶0ψε=0.
Taking into consideration thatE1,δ+1λξ≤1/Γ3/2expλξ, we obtain the following remark:Remark 1.
Letv be a solution of the inequality (51) then v is a solution of the following integral inequality(54)vξ−v0exp−λξ−1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,vκdκ≤εTδΓ3/2.Theorem 6.
Assume that (A), (37) and (38) are satisfied. Then (3) is Ulam-Hyers stable.Proof.
Letv be a solution of (52) and u the unique solution of the following Cauchy problem(55)DC0+δ,λuξ=fξ,uξ,δ∈0,1,λ>0,ξ∈0,T,u0=v0.
Therefore,(56)uξ=v0exp−λξ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ.
From Remark1, we find(57)vξ−uξ≤vξ−v0exp−λξ−1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤vξ−v0exp−λξ−1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,vκdκ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,vκdκ−1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤εTδΓ3/2+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,vκ−fκ,uκdκ≤εTδΓ3/2+1Γδ∫0ξexp−λξ−κξ−κδ−1×gexpλκvκ−uκdκ.
According to (38) and Theorem 4, we get(58)vξ−uξ≤exp−λξΩ−1Ωℛξ+∫0ξGκdκα,ξ∈0,T,where the formula of Ωξ,ℛξ and Gξ are defined as in Theorem 4 with aξ=εTδ/Γ3/2,hξ=gξ and wξ=1. Thus,(59)vξ−uξ≤ψε,where(60)ψε=Ω−1ΩℛT+∫0TGκdκα.
It follows from (37) we find that ψε⟶0 as ε⟶0. The proof is completed.Example 1.
Let us consider the following problem(61)DC0+1/2,1/2uξ=fξ,uξ,ξ∈0,π4,u0=1,where fξ,u=expξarctanu+ξ. For all u,v∈ℝ and ξ∈0,π/4, we have(62)fξ,u−fξ,v≤expξu−v.
Obviously, the assumption (A) and (37) and (38) are satisfied. As a consequence of Theorems 5 and 6 we deduce that the (IVP) (61) has a unique solution on 0,π/4 and this solution is Ulam–Hyers stable.


## 4.1. The Existence and Uniqueness

In the initial value problem (IVP) (3), the solution is given by [13](36)ut=u0exp−λξ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ.We will verify the existence and uniqueness using Theorem4.Theorem 5.
Assume that (A) is satisfied. If(37)limy⟶0+∫1y1Hκdκ=−∞,and(38)limy⟶+∞∫1y1Hκdκ=+∞,whereHs=g1/αsα,0<α<δ. Then the(IVP) (3) has a unique solution.Proof.
Consider the operatorG:C0,T,ℝ⟶C0,T,ℝ defined by(39)Guξ=u0exp−λξ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ.
We shall use Schaefer’s fixed point theorem to prove thatG has a fixed point. The proof will be given in several steps.Step 1.
G is continuous.
Letun be a sequence such that un⟶u in C0,T,ℝ. Then, for each ξ∈0,T, we have(40)Gunξ−Guξ≤1Γδ∫0ξξ−κδ−1fκ,unκ−fκ,uκdκ≤TδΓδ+1gexpλTun−u∞,and this implies that(41)Gun−Gu∞≤TδΓδ+1gexpλTun−u∞.
Thus,G is continuous.Step 2.
G maps bounded sets into bounded sets in C0,T,ℝ.
It is enough to show that for anyr1>0 there exists a positive constant r2>0 such that for each u∈Br1=u∈C0,T,ℝ,u∞≤r1 we have Gu∈Br2. For each ξ∈0,T we have(42)Guξ=u0exp−λξ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤u0+1Γδ∫0ξξ−κδ−1fκ,uκdκ≤u0+1Γδ∫0ξξ−κδ−1gexpλκuκdκ+1Γδ∫0ξξ−κδ−1kκdκ≤u0+TδΓδ+1gr1+k∗,where k∗=supt∈0,Tkt and gr1=gr1expλT; thus,(43)Gu∞≤u0+TδΓδ+1gr1+k∗≕r2.Step 3.
G maps bounded sets into equicontinuous sets of C0,T,ℝ.
Letr1>0 and u∈Br1. For 0≤ξ1<ξ2<T, we have(44)Guξ2−Guξ1≤u0exp−λξ2−exp−λξ1+1Γδ|∫0ξ2exp−λξ2−κξ2−κδ−1fκ,uκdκ−∫0ξ1exp−λξ1−κξ1−κδ−1fκ,uκdκ≤u0exp−λξ2−exp−λξ1+1Γδ∫0ξ1|exp−λξ2−κξ2−κδ−1−exp−λξ1−κξ1−κδ−1fκ,uκdκ+1Γδ∫ξ1ξ2exp−λξ2−κξ2−κδ−1fκ,uκdκ≤u0exp−λξ2−exp−λξ1+1Γδ∫0ξ1exp−λξ2−κξ1−κδ−1−ξ2−κδ−1fκ,uκdκ+1Γδ∫0ξ1exp−λξ2−κ−exp−λξ1−κξ1−κδ−1fκ,uκdκ+1Γδ∫ξ1ξ2exp−λξ2−κξ2−κδ−1fκ,uκdκ≤λu0ξ2−ξ1+gr1+k∗Γδ+1ξ1δ−ξ2δ+ξ2−ξ1δ+λgr1+k∗TδΓδ+1ξ2−ξ1+gr1+k∗Γδ+1ξ2−ξ1δ≤λu0+gr1+k∗TδΓδ+1ξ2−ξ1+gr1+k∗Γδ+1ξ1δ−ξ2δ+2ξ2−ξ1δ
Asξ2⟶ξ1, the right-hand side of the above inequality tends to zero. As a consequence of Steps 1 to 3 together with the Arzelá–Ascoli theorem, we conclude that the operator G is completely continuous.Step 4.
A priori bounds.
It remains to show that the set(45)A=u∈C0,T,ℝ,u=λ˜Gu,forλ˜∈0,1,is bounded. Taking 0<α<δ<1, and η=δ−α/1−α. Let u∈A, then(46)uξ=λ˜u0exp−λξ+λ˜Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤exp−λξu0+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤exp−λξu0+1Γδ∫0ξexp−λξ−κξ−κδ−1kκdκ+1Γδ∫0ξexp−λξ−κξ−κδ−1gexpλκuκdκ≤exp−λξu0+exp−λξΓδΓηξηE1,η+1λξ1−α×∫0ξexpλκk1/ακdκα+1Γδ∫0ξexp−λξ−κξ−κδ−1gexpλκuκdκ.
Thus, for eachϵ>0, we obtain(47)uξ≤ɛ+exp−λξu0+exp−λξΓδΓηξηE1,η+1λξ1−α×∫0ξexpλκk1/ακdκα+1Γδ∫0ξexp−λξ−κξ−κδ−1gexpλκuκdκ.
According to (38) and Theorem 4, we get(48)uξ≤exp−λξΩ−1Ωℛξ+∫0ξGκdκα,ξ∈0,T,where the formula of Ωξ,ℛξ and Gξ are defined as in Theorem 4, with aξ=ε+exp−λξu0+exp−λξ/ΓδΓηξηE1,η+1λξ1−α∫0ξexpλκk1/ακdκα, hξ=gξ and wξ=1. As a consequence of Schaefer’s fixed point theorem, we deduce that the operator G has at least one fixed point in C0,T,ℝ which is a solution of the (IVP) (3).
For the uniqueness, we suppose thatu1ξ, u2ξ are two solutions of (IVP) (3). Then,(49)u1ξ−u2ξ=1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,u1κ−fκ,u2κdκ≤1Γδ∫0ξexp−λξ−κξ−κδ−1gexpλξu1κ−u2κdκ.
Thus, for eachε>0, we obtain(50)u1ξ−u2ξ≤ϵ+1Γδ∫0ξexp−λξ−κξ−κδ−1×gexpλξu1κ−u2κdκ.
According to (37), (38) and Theorem 4, we get(51)u1ξ−u2ξ≤exp−λξΩ−1Ωℛξ+∫0ξGκdκα,ξ∈0,T,where the formula of Ωξ,ℛξ and Gξ are defined as in Theorem 4, with aξ=ε,hξ=gξ and wξ=1. Thus, u1ξ=u2ξ for all ξ∈0,T.


## 4.2. Generalised Ulam Stability

We consider the following inequality(52)D0Cδ,λvξ−fξ,vξ≤ε,forλ,ε>0,δ∈0,1,ξ∈0,T.Definition 4.
The equation (3) is Ulam-Hyers stable if there exists a function ψ:0,∞⟶0,∞ such that for each ε>0, and for each solution v∈C0,T,ℝ of the inequality (51) there exists a solution u∈C0,T,ℝ of (3) with(53)vξ−uξ≤ψε,where limε⟶0ψε=0.
Taking into consideration thatE1,δ+1λξ≤1/Γ3/2expλξ, we obtain the following remark:Remark 1.
Letv be a solution of the inequality (51) then v is a solution of the following integral inequality(54)vξ−v0exp−λξ−1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,vκdκ≤εTδΓ3/2.Theorem 6.
Assume that (A), (37) and (38) are satisfied. Then (3) is Ulam-Hyers stable.Proof.
Letv be a solution of (52) and u the unique solution of the following Cauchy problem(55)DC0+δ,λuξ=fξ,uξ,δ∈0,1,λ>0,ξ∈0,T,u0=v0.
Therefore,(56)uξ=v0exp−λξ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ.
From Remark1, we find(57)vξ−uξ≤vξ−v0exp−λξ−1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤vξ−v0exp−λξ−1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,vκdκ+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,vκdκ−1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,uκdκ≤εTδΓ3/2+1Γδ∫0ξexp−λξ−κξ−κδ−1fκ,vκ−fκ,uκdκ≤εTδΓ3/2+1Γδ∫0ξexp−λξ−κξ−κδ−1×gexpλκvκ−uκdκ.
According to (38) and Theorem 4, we get(58)vξ−uξ≤exp−λξΩ−1Ωℛξ+∫0ξGκdκα,ξ∈0,T,where the formula of Ωξ,ℛξ and Gξ are defined as in Theorem 4 with aξ=εTδ/Γ3/2,hξ=gξ and wξ=1. Thus,(59)vξ−uξ≤ψε,where(60)ψε=Ω−1ΩℛT+∫0TGκdκα.
It follows from (37) we find that ψε⟶0 as ε⟶0. The proof is completed.Example 1.
Let us consider the following problem(61)DC0+1/2,1/2uξ=fξ,uξ,ξ∈0,π4,u0=1,where fξ,u=expξarctanu+ξ. For all u,v∈ℝ and ξ∈0,π/4, we have(62)fξ,u−fξ,v≤expξu−v.
Obviously, the assumption (A) and (37) and (38) are satisfied. As a consequence of Theorems 5 and 6 we deduce that the (IVP) (61) has a unique solution on 0,π/4 and this solution is Ulam–Hyers stable.


## 5. Conclusion

In this paper, we created new weakly singular integral inequalities that generalize some previously known inequalities. Using the tempered fractional derivative, we discovered several applications for fractional differential equations in the Caputo context.



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