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).