JMATHJournal of Mathematics2314-47852314-4629Hindawi10.1155/2021/99904399990439Research ArticleIntegral-Type Fractional Equations with a Proportional Riemann–Liouville Derivativehttps://orcid.org/0000-0002-7986-886XMlaikiNabilCrescenzoAntonio DiDepartment of Mathematics and General SciencesPrince Sultan University RiyadhRiyadh 11586Saudi Arabiapsu.edu.sa202123620212021932021116202123620212021Copyright © 2021 Nabil Mlaiki.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

Prince Sultan UniversityRG-DES-2017-01-17
1. Introduction

Lately, many researchers have been focusing on the study of various types of fractional problems; we refer the reader to . 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 . In this manuscript, inspired by the work of Jankowski in , we investigate the existence and uniqueness of a solution to the following problem:(1)Dα,ρξt=gt,ξt,0tKt,τξτdτξt,tJ0=0,a;a>0,ξ˜0=p,where Dα,ρξt denotes a proportional Riemann–Liouville fractional derivative for ρ0,1 and 0<α<1. Also, gCJ××,,J=0,a, and ξ˜0=t1αeρ1/ρtξt|t0+. Now, we remind the reader of the definition of the proportional Riemann–Liouville fractional integral and derivative.

Definition 1 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

Let α;Reα0, 0<ρ1, and t>0.

The following integral is called the proportional Riemann–Liouville fractional integral:(2)Iα,ρft1ραΓα0teρ1/ρtτtτα1fτdτ.

The following derivative is called the proportional Riemann–Liouville fractional derivative:(3)Dα,ρft=Dn,ρInα,ρft=Dtn,ρρnαΓnα0teρ1/ρtτtτnα1fτdτ,

where n=Reα+1 and Dt1,ρ=1ρft+ρft.

Next, we present the following proposition.

Proposition 1 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

If α,γ, where Reα>0 and Reγ>0, then for any 0<ρ1, we have Iα,ρtγ1eρ1/ρtx=Γγ/ραΓα+γxγ+α1eρ1/ρx.

In Section 2, 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, let C1αJ,=fC0,a,|t1αfCJ,. Now, define the following two weighted norms:(4)f=max0,at1αft,f=max0,at1αeλtft for fixed λ>0.

Theorem 1.

Let 0<α<1, 0<ρ1, and gCJ××,,KCJ×J,×. Let β:=ρ1/ρ. Also, assume the following two hypotheses:

There exist nonnegative constants H,V, and W such that Kt,s<H and(5)gt,u1,u2gt,v1,v2Vv1u1+Wv2u2.

baα/Γ2αραV+HWa/2α<1, for α0,1/2.

Then, initial value problem (1) has a unique solution.

Proof.

First, let Sξt=t1αeρ1/ρtp+1/ραΓα0teρ1/ρtτtτα1ξτdτ. 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,YC1αJ,; we have two cases:

Case 1: α0,1/2.(6)Sξt=t1αeβtp+1ραΓα0teβtτtτα1ξτdτ,SξSY=1ραΓαmaxtJt1α0teβtτtτα1ξτℱYτdτ1ραΓαmaxtJt1α0teβtτtτα1VξτYτ+W0τKt,sξsYsdsdτ1ραΓαξYmaxtJt1α0teβtτtτα1Vτα1+HW0τsα1dsdτ=1ραΓαξYmaxtJt1α0teβtτtτα1Vτα1+HWτααdτ1ραΓαξYmaxtJt1αeβt0teβtτtτα1Veβττα1+HWeβττααdτ=1ραΓαξYmaxtJt1αeβtIα,ρVtα1eβt+Iα,ρHWαtαeβt=ΓαaαΓαΓ2αραV+HWa2αξY=bξY.

Hence, S is a contraction map. Therefore, S has a unique fixed point as desired.

Case 2: α1/2,1; in this case, we use with the positive constant λ>0 such that

(7)λβ>b1eβaVα+HWaΓ2α1a2α1αραΓαΓ22α1.

It is not difficult to see the following:

0te2λβτdτe2λβt/2λβ

t1α0ttτ2α1τ2α1dτ=Γ2α1a2α1/Γ22α1

Also, recall the Schwarz inequality for integrals:(8)0tfτgτdτ0tf2τdτ0tg2τdτ,SξSY=1ραΓαmaxtJt1αeλt0teβtτtτα1ξτℱYτdτ1ραΓαξYmaxtJt1αeλt0teβtτtτα1Veλττα1+HWeλτ0τsα1dsdτ1ραΓαξYmaxtJt1αeλt0teβtτtτα1Vτα1eλτ+HWτααeλτdτVα+HWaαραΓαξYmaxtJt1αeλt0teβtτeλτtτα1τα1dτ=Vα+HWaαραΓαξYmaxtJt1αeλt0teβtτ+λτtτα1τα1dτVα+HWaαραΓαξYmaxtJt1αeλt×0ttτ2α1τ2α1dτ0te2βtτ+2λτdτVα+HWaαραΓαξYmaxtJt1αeλt×0ttτ2α1τ2α1dτ0te2λβτdτb1λβξY.

Thus, S is a contraction map. Therefore, S has a unique fixed point as required.

As an application to Theorem 1, consider the following problem:(9)Dα,ρyt=Ltyt+ξt,ξ˜0=r.

If(10)aαραΓ2αmaxtJLt<1,for 0<α12,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,tJ0,ξ˜0=r.

Now, we introduce the following hypothesis.

Hypothesis 1.

(H1)

Lt=L,tJ or

The function L is nonconstant on J and

(12)aαραΓ2αmaxtJLt<1 only ifα0,12.

The following lemma is a consequence of Theorem 1.

Lemma 1.

If α0,1,LCJ,,zC1αJ,, and hypothesis H1 holds, then problem (11) has a unique solution.

We would like to bring to the reader’s attention that, in , in the hypothesis ρ should be as follows: ρTq/Γ2qK+WLT/2q which he used to prove the case where q0,1/2. 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)

Lt=L,tJ or

The function L is nonconstant, and if Lt is negative, then there exists L¯ which is nondecreasing, where LtL¯t on J and for every xJ, we have

(13)eβxραΓα0aaτα1eβaτL¯τdτ<1.

Now, for our purpose, we prove the following useful lemma.

Lemma 2.

Let α0,1 and LCJ,0, or LCJ,,0. Also, denote by β:=ρ1/ρ. Assume that qC1αJ, is a solution to the following problem:(14)Dα,ρqtLtqt,tJ0,q˜0<0.

If (H2) holds, then qt0 for all tJ.

Proof.

Assume that our lemma is false, that is, there exist x,y0,a such that qx=0,qy>0, and qt0 for t0,x; qt>0 for tx,y. Let x0 be the first maximal point of q on x,y.

Case 1: assume that Lt0 for all tJ. Thus, Dα,ρqt0 for tx,y. Hence,(15)xx0Dα,ρqt0.

Therefore, BIρ,1αqx0Iρ,1αqx0, but(16)B=1ρ1αΓ1α0x0eβx0τx0ταqτdτ0xeβxτxταqτdτ=1ρ1αΓ1α0xeβx0τx0ταeβxτxταqτdτ+xx0eβx0τx0ταqτdτ>1ρ1αΓ1αxx0eβx0τx0ταqτdτ>0,

Case 2: assume that Lt0 for all tJ, and consider L˜ to be nondecreasing on J. Now, if we apply Iα,ρ on problem (14), we obtain

(17)qtq˜0eβttα1ρα1ΓαIα,ρLtqt,for tx,x0.

Notice that q˜0eβttα1/ρα1Γα0 which is due to the fact that q˜00. Thus,(18)qx01ραΓα0x0x0τα1eβx0τLτqτdτ=1ραΓα0x0x0τα1eβx0τLτqτdτ+xx0x0τα1eβx0τLτqτdτ<qx0ραΓα0x0x0τα1eβx0τLτdτ, let σ=τx0=qx0eβx0x0αραΓα011σα1eβ1σLσx0dσqx0eβx0x0αραΓα011σα1eβ1σL˜σadσ=qx0eβx0+1/ax0αραΓαaα0aaτα1eβaτL˜τdτqx0eβx0+1/aραΓα0aaτα1eβaτL˜τdτ.

Hence, qx01eβx0+1/a/ραΓα0aaτα1eβaτL˜τdτ0. Using hypothesis H2 implies that qx00, which leads us to a contradiction, and this concludes our proof.

We say that y is a lower solution of problem (1) if(19)Dα,ρytyt,tJ0,y˜00,and we say that y is an upper solution of problem (1) if(20)Dα,ρytyt,tJ0,y˜00.

Next, the following hypothesis is defined.

Hypothesis 3.

(H3). There exists a function LCJ, where(21)gt,u1,u2gt,v1,v2Ltv1u1whenever x0u1v1y0 and  u2v2.

Theorem 2.

Assume that x0 is a lower solution of problem (1) and y0 is an upper solution of problem (1), where x0,y0C1αJ,. Moreover, assume that hypotheses H1,H2, and H3 hold; problem (1) has solutions in x0,y0=yC1αJ,|x0tyty0t,tJ0,x˜00y˜0y˜00.

Proof.

Using Lemmas 1 and 2, the proof is similar to the proof of Theorem 2 in .

Now, we present the following example.

Example 1.

Let 0<α<1, 0<ρ1, β=ρ1/ρ, and A,BC0,1,0, such that AtBt for t0,1. Now, consider the following problem:(22)Dα,ρξtξt,tJ0=0,1,ξ˜0=0,where(23)ξt=ραeβtt1αΓ2α+Atteβt1ξt3+βeβt1Bt0tsintτ4ξτdτ.

Now, let x0t=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)y0t=ραeβtt1αΓ2αAt+βeβt1Bt0tsintτ4τeβτdτραeβtt1αΓ2αAt+βeβt1Bt0teβτdτ=ραeβtt1αΓ2αAt+βeβt1Bteβtβ1β<ραeβtt1αΓ2α=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 α1/2,1, and for α0,1/2, we need to assume that 1/ραΓ2αmaxt0,1At<1.

4. Conclusion

In closing, note that the results of Jankowski  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?

Data Availability

No data were used to support this study.

Conflicts of Interest

The author declares that there are no conflicts of interest.

Acknowledgments

The author would like to thank Prince Sultan University for funding this work through the research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) (group no. RG-DES-2017-01-17).

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