#### 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.

#### 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:where denotes a proportional Riemann–Liouville fractional derivative for and . Also, , and . Now, we remind the reader of the definition of the proportional Riemann–Liouville fractional integral and derivative.

*Definition 1 (see [18]). *Let , , and .(i)The following integral is called the proportional Riemann–Liouville fractional integral:(ii)The following derivative is called the proportional Riemann–Liouville fractional derivative:where and .

Next, we present the following proposition.

Proposition 1 (see [18]). *If , where and , then for any , we have .*

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 . Now, define the following two weighted norms:

Theorem 1. *Let , and . Let . Also, assume the following two hypotheses:*(1)*There exist nonnegative constants such that and*(2)*.**Then, initial value problem (1) has a unique solution.*

*Proof. *First, let . Note that if has a unique fixed point and that is , then initial value problem (1) has a unique solution, i. e., it will be enough to show that is a contraction map. So, let ; we have two cases: Case 1: . Hence, is a contraction map. Therefore, has a unique fixed point as desired. Case 2: ; in this case, we use with the positive constant such thatIt is not difficult to see the following:(1)(2)Also, recall the Schwarz inequality for integrals:Thus, is a contraction map. Therefore, has a unique fixed point as required.

As an application to Theorem 1, consider the following problem:

Ifthen 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:

Now, we introduce the following hypothesis.

*Hypothesis 1. *()(1) or(2)The function is nonconstant on and

The following lemma is a consequence of Theorem 1.

Lemma 1. *If , and hypothesis holds, then problem (11) has a unique solution.*

We would like to bring to the reader’s attention that, in [1], in the hypothesis should be as follows: which he used to prove the case where . 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. *()(1) or(2)The function is nonconstant, and if is negative, then there exists which is nondecreasing, where on and for every , we have

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

Lemma 2. *Let and or . Also, denote by . Assume that is a solution to the following problem:**If () holds, then for all .*

*Proof. *Assume that our lemma is false, that is, there exist such that , and for ; for . Let be the first maximal point of on . Case 1: assume that for all . Thus, for . Hence, Therefore, , but which leads us to a contradiction given the fact that . Case 2: assume that for all , and consider to be nondecreasing on . Now, if we apply on problem (14), we obtainNotice that which is due to the fact that . Thus,Hence, . Using hypothesis implies that , which leads us to a contradiction, and this concludes our proof.

We say that is a lower solution of problem (1) ifand we say that is an upper solution of problem (1) if

Next, the following hypothesis is defined.

*Hypothesis 3. *(). There exists a function where

Theorem 2. *Assume that is a lower solution of problem (1) and is an upper solution of problem (1), where . Moreover, assume that hypotheses , and hold; problem (1) has solutions in .*

*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 , , , and such that for . Now, consider the following problem:whereNow, let and ; first, note that is a lower solution of problem (22). Next, we show that is an upper solution of problem (22):Thus, 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 if , and for , we need to assume that .

#### 4. Conclusion

In closing, note that the results of Jankowski [1] are a special case of our work which is by taking . 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 is not constant , but it is a function of $t$ say , so that the problem involves

#### 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).