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-2017-01-171. 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 [<xref ref-type="bibr" rid="B18">18</xref>]).

Let α∈ℂ;Reα≥0, 0<ρ≤1, and t>0.

The following integral is called the proportional Riemann–Liouville fractional integral:(2)Iα,ρft≔1ραΓα∫0teρ−1/ρt−τt−τα−1fτdτ.

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

If α,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 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 C1−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−λtft for fixed λ>0.

Theorem 1.

Let 0<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:

There exist nonnegative constants H,V, and W such that Kt,s<H and(5)gt,u1,u2−gt,v1,v2≤Vv1−u1+Wv2−u2.

b≡aα/Γ2αραV+HWa/2α<1, for α∈0,1/2.

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

Proof.

First, let Sξt=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:

∫0te2λ−βτdτ≤e2λ−βt/2λ−β

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

If(10)aαραΓ2αmaxt∈JLt<1,for 0<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)

Lt=L,t∈J or

The function L is nonconstant on J and

(12)aαραΓ2αmaxt∈JLt<1 only ifα∈0,12.

The following lemma is a consequence of Theorem 1.

Lemma 1.

If α∈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)

Lt=L,t∈J or

The function L 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.

Let α∈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 exist x,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 that Lt≥0 for all t∈J. Thus, Dα,ρqt≤0 for t∈x,y. Hence,(15)∫xx0Dα,ρqt≤0.

which leads us to a contradiction given the fact that B≤0.

Case 2: assume that Lt≤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,for t∈x,x0.

Notice that q˜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, let σ=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 that y 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−u1whenever x0≤u1≤v1≤y0 and u2≤v2.

Theorem 2.

Assume that x0 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 Lemmas 1 and 2, the proof is similar to the proof of Theorem 2 in [1].

Now, we present the following example.

Example 1.

Let 0<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, 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)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?

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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