# Integral-Type Fractional Equations with a Proportional Riemann–Liouville Derivative

Authors: Nabil Mlaiki
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we present the necessary conditions where integral-type fractional equations with a proportional Riemann–Liouville derivative have a unique solution. Also, we give an example to illustrate our work.


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## Body

## 1. Introduction

Lately, many researchers have been focusing on the study of various types of fractional problems; we refer the reader to [1–17]. The fixed point and the monotone iterative techniques can be very useful tools to prove the existence and uniqueness of a solution to this type of problems; see [1]. In this manuscript, inspired by the work of Jankowski in [1], we investigate the existence and uniqueness of a solution to the following problem:(1)Dα,ρξt=gt,ξt,∫0tKt,τξτdτ≡ℱξt,t∈J0=0,a;a>0,ξ˜0=p,where Dα,ρξt denotes a proportional Riemann–Liouville fractional derivative for ρ∈0,1 and 0<α<1. Also, g∈CJ×ℝ×ℝ,ℝ,J=0,a, and ξ˜0=t1−αeρ−1/ρtξt|t⟶0+. Now, we remind the reader of the definition of the proportional Riemann–Liouville fractional integral and derivative.Definition 1 (see [18]).
Letα∈ℂ;Reα≥0, 0<ρ≤1, and t>0.(i)
The following integral is called the proportional Riemann–Liouville fractional integral:(2)Iα,ρft≔1ραΓα∫0teρ−1/ρt−τt−τα−1fτdτ.(ii)
The following derivative is called the proportional Riemann–Liouville fractional derivative:(3)Dα,ρft=Dn,ρIn−α,ρft=Dtn,ρρn−αΓn−α∫0teρ−1/ρt−τt−τn−α−1fτdτ,where n=Reα+1 and Dt1,ρ=1−ρft+ρf′t.Next, we present the following proposition.Proposition 1 (see [18]).
Ifα,γ∈ℂ, where Reα>0 and Reγ>0, then for any 0<ρ≤1, we have Iα,ρtγ−1eρ−1/ρtx=Γγ/ραΓα+γxγ+α−1eρ−1/ρx.In Section2, we prove the existence and uniqueness of a solution to problem (1) using the fixed point technique. In Section 3, we prove the existence and uniqueness of a solution to problem (1) using the monotone iterative method. In the conclusion, we present an open question.


## 2. Fixed Point Approach

First of all, letC1−αJ,ℝ=f∈C0,a,ℝ|t1−αf∈CJ,ℝ. Now, define the following two weighted norms:(4)f∗=max0,at1−αft,f∗=max0,at1−αe−λtftfor fixedλ>0.Theorem 1.
Let0<α<1,0<ρ≤1, and g∈CJ×ℝ×ℝ,ℝ,K∈CJ×J,×ℝ. Let β:=ρ−1/ρ. Also, assume the following two hypotheses:(1)
There exist nonnegative constantsH,V,andW such that Kt,s<H and(5)gt,u1,u2−gt,v1,v2≤Vv1−u1+Wv2−u2.(2)
b≡aα/Γ2αραV+HWa/2α<1,forα∈0,1/2.
Then, initial value problem (1) has a unique solution.Proof.
First, letSξt=t1−αeρ−1/ρtp+1/ραΓα∫0teρ−1/ρt−τt−τα−1ℱξτdτ. Note that if S has a unique fixed point and that is Sξt=ξt, then initial value problem (1) has a unique solution, i. e., it will be enough to show that S is a contraction map. So, let ξ,Y∈C1−αJ,ℝ; we have two cases:Case 1:α∈0,1/2.(6)Sξt=t1−αeβtp+1ραΓα∫0teβt−τt−τα−1ℱξτdτ,Sξ−SY∗=1ραΓαmaxt∈Jt1−α∫0teβt−τt−τα−1ℱξτ−ℱYτdτ≤1ραΓαmaxt∈Jt1−α∫0teβt−τt−τα−1Vξτ−Yτ+W∫0τKt,sξs−Ysdsdτ≤1ραΓαξ−Y∗maxt∈Jt1−α∫0teβt−τt−τα−1Vτα−1+HW∫0τsα−1dsdτ=1ραΓαξ−Y∗maxt∈Jt1−α∫0teβt−τt−τα−1Vτα−1+HWτααdτ≤1ραΓαξ−Y∗maxt∈Jt1−αe−βt∫0teβt−τt−τα−1Veβττα−1+HWeβττααdτ=1ραΓαξ−Y∗maxt∈Jt1−αe−βtIα,ρVtα−1eβt+Iα,ρHWαtαeβt=ΓαaαΓαΓ2αραV+HWa2αξ−Y∗=bξ−Y∗.Hence,S is a contraction map. Therefore, S has a unique fixed point as desired.Case 2:α∈1/2,1; in this case, we use ⋅∗ with the positive constant λ>0 such that(7)λ−β>b1≡e−βaVα+HWaΓ2α−1a2α−1αραΓαΓ22α−1.
It is not difficult to see the following:(1)
∫0te2λ−βτdτ≤e2λ−βt/2λ−β(2)
t1−α∫0tt−τ2α−1τ2α−1dτ=Γ2α−1a2α−1/Γ22α−1
Also, recall the Schwarz inequality for integrals:(8)∫0tfτgτdτ≤∫0tf2τdτ∫0tg2τdτ,Sξ−SY∗=1ραΓαmaxt∈Jt1−αe−λt∫0teβt−τt−τα−1ℱξτ−ℱYτdτ≤1ραΓαξ−Y∗maxt∈Jt1−αe−λt∫0teβt−τt−τα−1Veλττα−1+HWeλτ∫0τsα−1dsdτ≤1ραΓαξ−Y∗maxt∈Jt1−αe−λt∫0teβt−τt−τα−1Vτα−1eλτ+HWτααeλτdτ≤Vα+HWaαραΓαξ−Y∗maxt∈Jt1−αe−λt∫0teβt−τeλτt−τα−1τα−1dτ=Vα+HWaαραΓαξ−Y∗maxt∈Jt1−αe−λt∫0teβt−τ+λτt−τα−1τα−1dτ≤Vα+HWaαραΓαξ−Y∗maxt∈Jt1−αe−λt×∫0tt−τ2α−1τ2α−1dτ∫0te2βt−τ+2λτdτ≤Vα+HWaαραΓαξ−Y∗maxt∈Jt1−αe−λt×∫0tt−τ2α−1τ2α−1dτ∫0te2λ−βτdτ≤b1λ−βξ−Y∗.
Thus,S is a contraction map. Therefore, S has a unique fixed point as required.As an application to Theorem1, consider the following problem:(9)Dα,ρyt=−Ltyt+ξt,ξ˜0=r.If(10)aαραΓ2αmaxt∈JLt<1,for0<α≤12,then it is not difficult to see that, by using Theorem 1, problem (9) has a unique solution. In closing of this section, the following linear problem is considered:(11)Dα,ρξt=−Ltξt+zt,t∈J0,ξ˜0=r.Now, we introduce the following hypothesis.Hypothesis 1.
(H1)(1)
Lt=L,t∈J or(2)
The functionL is nonconstant on J and(12)aαραΓ2αmaxt∈JLt<1only ifα∈0,12.The following lemma is a consequence of Theorem1.Lemma 1.
Ifα∈0,1,L∈CJ,ℝ,z∈C1−αJ,ℝ, and hypothesis H1 holds, then problem (11) has a unique solution.We would like to bring to the reader’s attention that, in [1], in the hypothesis ρ should be as follows: ρ≡Tq/Γ2qK+WLT/2q which he used to prove the case where q∈0,1/2. This way, his result will be stronger or he can just change the last equality to the inequality.


## 3. Monotone Iterative Method

First of all, we start by introducing the following hypothesis.Hypothesis 2.
(H2)(1)
Lt=L,t∈J or(2)
The functionL is nonconstant, and if Lt is negative, then there exists L¯ which is nondecreasing, where −Lt≤L¯t on J and for every x∈J, we have(13)eβxραΓα∫0aa−τα−1eβa−τL¯τdτ<1.Now, for our purpose, we prove the following useful lemma.Lemma 2.
Letα∈0,1 and L∈CJ,0,∞ or L∈CJ,−∞,0. Also, denote by β:=ρ−1/ρ. Assume that q∈C1−αJ,ℝ is a solution to the following problem:(14)Dα,ρqt≤−Ltqt,t∈J0,q˜0<0.
If (H2) holds, then qt≤0 for all t∈J.Proof.
Assume that our lemma is false, that is, there existx,y∈0,a such that qx=0,qy>0, and qt≤0 for t∈0,x; qt>0 for t∈x,y. Let x0 be the first maximal point of q on x,y.Case 1: assume thatLt≥0 for all t∈J. Thus, Dα,ρqt≤0 for t∈x,y. Hence,(15)∫xx0Dα,ρqt≤0.Therefore,B≡Iρ,1−αqx0−Iρ,1−αqx≤0, but(16)B=1ρ1−αΓ1−α∫0x0eβx0−τx0−τ−αqτdτ−∫0xeβx−τx−τ−αqτdτ=1ρ1−αΓ1−α∫0xeβx0−τx0−τ−α−eβx−τx−τ−αqτdτ+∫xx0eβx0−τx0−τ−αqτdτ>1ρ1−αΓ1−α∫xx0eβx0−τx0−τ−αqτdτ>0,which leads us to a contradiction given the fact thatB≤0.Case 2: assume thatLt≤0 for all t∈J, and consider L˜ to be nondecreasing on J. Now, if we apply Iα,ρ on problem (14), we obtain(17)qt−q˜0eβttα−1ρα−1Γα≤−Iα,ρLtqt,fort∈x,x0.
Notice thatq˜0eβttα−1/ρα−1Γα≤0 which is due to the fact that q˜0≤0. Thus,(18)qx0≤−1ραΓα∫0x0x0−τα−1eβx0−τLτqτdτ=−1ραΓα∫0x0x0−τα−1eβx0−τLτqτdτ+∫xx0x0−τα−1eβx0−τLτqτdτ<−qx0ραΓα∫0x0x0−τα−1eβx0−τLτdτ,letσ=τx0=−qx0eβx0x0αραΓα∫011−σα−1eβ1−σLσx0dσ≤qx0eβx0x0αραΓα∫011−σα−1eβ1−σL˜σadσ=qx0eβx0+1/ax0αραΓαaα∫0aa−τα−1eβa−τL˜τdτ≤qx0eβx0+1/aραΓα∫0aa−τα−1eβa−τL˜τdτ.
Hence,qx01−eβx0+1/a/ραΓα∫0aa−τα−1eβa−τL˜τdτ≤0. Using hypothesis H2 implies that qx0≤0, which leads us to a contradiction, and this concludes our proof.We say thaty is a lower solution of problem (1) if(19)Dα,ρyt≤ℱyt,t∈J0,y˜0≤0,and we say that y is an upper solution of problem (1) if(20)Dα,ρyt≥ℱyt,t∈J0,y˜0≤0.Next, the following hypothesis is defined.Hypothesis 3.
(H3). There exists a function L∈CJ,ℝ where(21)gt,u1,u2−gt,v1,v2≤Ltv1−u1wheneverx0≤u1≤v1≤y0andu2≤v2.Theorem 2.
Assume thatx0 is a lower solution of problem (1) and y0 is an upper solution of problem (1), where x0,y0∈C1−αJ,ℝ. Moreover, assume that hypotheses H1,H2, and H3 hold; problem (1) has solutions in x0,y0=y∈C1−αJ,ℝ|x0t≤yt≤y0t,t∈J0,x˜00≤y˜0≤y˜00.Proof.
Using Lemmas1 and 2, the proof is similar to the proof of Theorem 2 in [1].Now, we present the following example.Example 1.
Let0<α<1, 0<ρ≤1, β=ρ−1/ρ, and A,B∈C0,1,0,∞ such that At≤Bt for t∈0,1. Now, consider the following problem:(22)Dα,ρξt≡ℱξt,t∈J0=0,1,ξ˜0=0,where(23)ℱξt=ραeβtt1−αΓ2−α+Atteβt−1−ξt3+βeβt−1Bt∫0tsintτ4ξτdτ.
Now, letx0t=0 and y0t=teβt; first, note that x0t is a lower solution of problem (22). Next, we show that y0t is an upper solution of problem (22):(24)ℱy0t=ραeβtt1−αΓ2−α−At+βeβt−1Bt∫0tsintτ4τeβτdτ≤ραeβtt1−αΓ2−α−At+βeβt−1Bt∫0teβτdτ=ραeβtt1−αΓ2−α−At+βeβt−1Bteβtβ−1β<ραeβtt1−αΓ2−α=Dα,ρy0t.
Thus,y0t is an upper solution of problem (22). Now, it is not difficult to see that all the hypotheses of Theorem 2 are satisfied. Therefore, problem (22) has solutions in x0,y0 if α∈1/2,1, and for α∈0,1/2, we need to assume that 1/ραΓ2αmaxt∈0,1At<1.


## 4. Conclusion

In closing, note that the results of Jankowski [1] are a special case of our work which is by taking ρ=1. Also, we would like to bring to the reader attention the following open question.What are the necessary and sufficient conditions for problem (1) to have a unique solution if ρ is not constant , but it is a function of $t$ say gt, so that the problem involves Dα,gt?



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