Abstract

Firstly, the concept of a new triangular -orbital admissible condition is introduced, and two fixed point theorems for Sehgal-Guseman-type mappings are investigated in the framework of rectangular -metric spaces. Secondly, some examples are presented to illustrate the availability of our results. At the same time, we furnished the existence and uniqueness of solution of an integral equation.

1. Introduction

In nonlinear analysis, the most famous result is the Banach contraction principle, which is established by Banach [1] in 1922. After that, there are a large number of excellent results for fixed point in metric spaces. On recent development on fixed point theory in metric spaces, one can consult [2] the related references involved. Branciari [3] introduced a new concept, that is, the definition of rectangular metric spaces, and established an analogue of the Banach fixed point theorem in such a space. Then, a lot of fixed point theorems for a wide range of contractions on rectangular metric spaces had emerged in a blowout manner. In such type space, Lakzian and Samet [4] gave some results involving weakly contraction. Furthermore, several common fixed point results about -weakly contractions were obtained by Bari and Vetro [5]. In [6], George and Rajagopalan considered common fixed points of a new class of contractions. By use of -functions, Budhia et al. furnished several fixed point results in [7].

In [8], Czerwik put forward firstly the definition of -metric space, an extension of a metric space. Since then, this result has been extended in different angles. In a -metric space, in [9], Mitrovic provided a new method to prove Czerwik’s fixed point theorem. By using of increased range of the Lipschitzian constants, Hussain et al. [10] provided a proof of the Fisher contraction theorem. Mustafa et al. [11] gave several fixed point theorems for some new classes of -Chatterjea-contraction and -Kannan-contraction. Recently, also in this type spaces, Mitrovic et al. [12] presented some new versions of existing theorems. Savanović et al. [13] constructed some new results for multivalued quasicontraction. Furthermore, in [14], Aydi et al. obtained the existence of fixed point for --Geraghty contractions. In [15], several fixed point theorems of set valued interpolative Hardy-Rogers type contractions were studied. In [16], George et al. put forward the concept of rectangular -metric mapping. Meanwhile, they gave some fixed point theorems. Lately, Gulyaz-Ozyurt [17], Zheng et al. [18], and Guan et al. [19] also studied fixed point theory in such spaces and obtained some excellent results. In 2021, Hussain [20] presented some fractional symmetric --contractions and built up some new fixed point theorems for these types of contractions in -metric spaces. Recently, Arif et al. [21] introduced an ordered implicit relation and investigated the existence of the fixed points of contractive mapping dealing with implicit relation in a cone -metric space. Lately, in [22], some fixed point theorems of two new classes of multivalued almost contractions in a partial -metric spaces were established by Anwar et al.

On the other hand, in 1969, Sehgal [23] formulated an inequality that can be considered an extension of the renowned Banach fixed point theorem in a metric space. Matkowski [24] generalized some previous results of Khazanchi [25] and Iseki [26]. In 2012, the definition of -admissible mappings was given by Samet et al. [27]. Later, the notion of triangular -admissible mappings was introduced by Popescu [28]. Recently, Lang and Guan [29] studied the common fixed point theory of --Geraghty contraction and --Geraghty contractions in a -metric space.

In this paper, inspired by [30], we established two fixed point theorems for Sehgal-Guseman-type mappings in a rectangular -metric space. Also, we present two examples to illustrate the usability of established results.

2. Preliminaries

Definition 1 (see [8]). Suppose is a nonempty set and . We call a -metric if (i)(ii)(iii)where is constant.

It is usual that is called a -metric space with parameter .

Definition 2 (see [3]). Suppose is a nonempty set and . We call a triangular metric if (i)(ii)(iii)Usually, is called a rectangular metric space.

Definition 3 (see [16]). Suppose is a nonempty set and . We call a rectangular -metric if (i)(ii)(iii)where is constant.

In general, is called a rectangular -metric space with parameter .

Remark 4. A rectangular metric space is a rectangular -metric space, so is a -metric space. Moreover, the converse is not true.

Example 1. Suppose , where and . For , define with and

Thus, is a rectangular -metric space with . Furthermore, one can obtain the following: (1) is not a -metric with , since (2) is not a rectangular metric, since (3) is not a metric, since

Definition 5 (see [16]). Suppose is a rectangular -metric space with . Assume that in is a sequence and (i) is convergent to iff (ii) is Cauchy iff as (iii) is complete iff each Cauchy sequence is convergent

Remark 6. In a rectangular -metric space, a convergent sequence does not possess unique limit and a convergent sequence is not necessarily a Cauchy sequence. However, one can find that the limit of a Cauchy sequence is unique. In fact, suppose the sequence is Cauchy and converges to and with . It follows that for all . Let ; we get that Hence, , a contradiction.

Example 2 (see [16]). Let , where and . Define with and Here, is a positive number. Thus, is a rectangular -metric with . However, we have that is convergent to 2 and 3. Moreover, ; therefore, is not a Cauchy sequence.

Definition 7 (see [28]). Suppose is a nonempty set and and are two mappings. We call -orbital admissible mapping if

Definition 8 (see [28]). Assume that and . We call a triangular -orbital admissible mapping if (i) and imply (ii) is -orbital admissible

Lemma 9 (see [24]). Assume is an increasing mapping. Then, .

3. Main Results

In this part, two fixed point results of injective mappings will be presented on rectangular -metric spaces.

Definition 10. Suppose is a nonempty set, and are two constants, and , . We call orbital admissible mapping if

Definition 11. Suppose is a nonempty set, and are two constants, and , . We call triangular orbital admissible mapping if (i) and imply (ii) is orbital admissible

Lemma 12. Suppose is a nonempty set and , are mappings satisfying which is triangular orbital admissible, . Suppose there has a with . Define in by . Then, ,

Proof. Since and is triangular orbital admissible, we have Similarly, since , we get Applying the above argument repeatedly, one can deduce that for all . Since implies and implies , we can obtain . Based on this conclusion, we deduce that . Repeatedly using the above discussion, we have for all .

Define . That is, if with , we have

Furthermore, we set for .

Theorem 13. Suppose is a complete rectangular -metric space with . Suppose is a continuous injectivity, and . Assume that for any , there is a positive number satisfying where and (1)(2)Suppose that (i)there has a in such that (ii) is triangular orbital admissible(iii)if in satisfies and , then one can choose a subsequence of with (iv) with , we have for any Then, possesses a unique fixed point . Further, for each , the iteration converges to

Proof. By condition (i), one can choose an such that . If is a fixed point of and is the other one, then and . From condition (iv), we have . It follows from (13) that From Lemma 9, we have . Thus, which is contradiction. From this, we get that is the unique fixed point. After that, in the subsequent discussion, we assume that . Now we define in by .
First, we shall show that the orbit is bounded. For this purpose, we fix an integer . Let Since , there has such that . It is obvious that . Assume that there has a positive number with . Evidently, one may suppose that . Let be different from each other. Otherwise, we consider six cases.
Case 1. One can get that It follows that is a constant which implies that is bounded.
Case 2. We deduce that Hence, It follows that is bounded.
Case 3. . Obviously, As the argument of Case 1, we get that is bounded.
Case 4. . In this case, we obtain that , a contradiction.
Case 5. . It follows that It is a contradiction.
Case 6. . It is obvious that a contradiction.
It is easy to get from Lemma 12. By using triangle inequality and (16), we have That is, , which is impossible. Therefore, for . It follows that is bounded.
If there exists some satisfying , then is a fixed point of . Assume there is such that and , by condition (iv), we have and which is contradiction. From this, possesses the unique fixed point . Since , we have because of the uniqueness of . Subsequently, we assume that , .
Next, we show that is Cauchy. Suppose and are two positive numbers. It is obvious that . Then, For each , we have According to (27) and (28), we deduce That is, is Cauchy. In light of the completeness of , one can find an with . We might as well let and . Otherwise, we have according to the continuity of . In view of triangle inequality, one deduce On the other hand, From the continuity of , . Thereupon, by the use of (30) and (31), one can obtain as . Assume there exists satisfying and we have according to the condition (iv). Then, impossible. After that, has the unique fixed point . Since , we deduce . That is, has a fixed point.
Now we show that if condition (iv) is met. So possesses a unique fixed point. Assume is another one; from condition (iv), one can obtain . In view of (13), we have Lemma 9 ensures that . Thus, which is impossible. It follows that is the unique fixed point of .
Finally, we prove the last part. To show this statement, we fix an integer , , and let for . If there exists satisfying , we have If , one can obtain that , which is a contradiction. Hence, . It follows that .
Now we suppose that , . Therefore, we obtain If for some , , we deduce , which is a contradiction. Hence, we get . That is, for , the sequence converges to for any . Consequently, one can obtain that the sequences are convergent to the point . It follows that we get converges to the point for .

Example 3. Let be the same as it is in Example 1. Define as Define mapping by Define for all , and it follows that Let for all . For such that , we get that . It follows that we consider the following two cases: (i) and That is, .(ii). Let and with One can obtain that The above inequalities imply that Thus, all conditions of Theorem 13 are fulfilled with . As a result, possesses a unique fixed point 0. Meanwhile, for each , converges to the point .

Remark 14. (1)Since rectangular metric spaces can be seen as rectangular -metric spaces with parameter , one can get the corresponding conclusions of Sehgal-Guseman-type mappings in rectangular metric spaces(2)Since -metric spaces with parameter can be seen as rectangular -metric spaces with parameter , one can obtain the corresponding conclusions of Sehgal-Guseman-type mappings in -metric spaces(3)If , one can get the generalized -Sehgal-Guseman-type contractive mappings in rectangular -metric spaces

Theorem 15. Suppose is a complete rectangular -metric space with . Suppose is a continuous injectivity and satisfying that for any ; there is a positive number satisfying where Then, possesses a unique fixed point . Furthermore, for each , the iteration is convergent to .

Proof. Let . Consider a sequence in by . If for an , then becomes to a fixed point of . Assume there exists with and ; then, where From this, we get which is impossible. Therefore, is the unique fixed point of . Since , we have because of the uniqueness of . Subsequently, we assume that , .
For , set . We first prove that for all . Assume is a positive number satisfying and . We suppose that are four distinct elements. Otherwise, the conclusion is true. Thus, where By (46), (47), and (48), we deduce Hence, one can conclude that by induction. Indeed, when , we have . If , we get . If , we get . We assume ; then, . Hence, .
Next, we prove that . By contractive condition (42), we have where It is obvious that , so For each , we have where It means . So we deduce That is, .
For the sequence , we consider by the following cases. For the sake of convenience, set .
If is odd, assume , If is even, assume , In view of (56) and (57), one can get that is Cauchy. By the completeness of , one can choose a point with . We might as well let and . Otherwise, we have according to the continuity of . And from that, one can deduce where It follows that Since is a continuous mapping, . Therefore, This means that . Now, where Hence, we get , i.e., . Assume there has a satisfying and ; then, and which is impossible. So possesses the unique fixed point .
At the end, we prove the last part. To do this, we fix an integer , , and ; we put . Then, ; we have where If , then . According to (65), we have It follows that as If , one can get that Similarly, where If , then that is, Since is a fixed point of , one get . Consequently, as If , then We continue to calculate according to this method; if there exists satisfying , then as Otherwise, one can conclude that Therefore, for each , the iteration is convergent to .

Example 4. Let and Obviously, is a complete rectangular -metric space with . Define with Define mappings and , . One has That is, .

Thus, all hypotheses of Theorem 15 are fulfilled. So possesses the unique common fixed point 0. Furthermore, for each , the iteration is convergent to 0.

4. Application

In this part, we will prove the solvability of this initial value problem:

where and are constants and is a continuous mapping.

Obviously, problem (78) is related to the integral equation:

where is defined as where is a constant.

Next, by using Theorem 13 and Theorem 15, we shall present the solvability of the integral equation:

Let . For define

Hence, is a complete rectangular -metric space with .

In the following, define by

Suppose is a given function that satisfies the following condition:

Theorem 16. Assume that (i) is continuous(ii)there has an satisfying for all (iii) and , imply (iv)if satisfies , , and , then we can choose a subsequence of such that , (v)for each with , we have for any (vi)there is a continuous mapping satisfyingThen, (81) possesses a unique solution .

Proof. Set by One can check that is triangular orbital admissible. In view of (i)-(vi), for , we obtain which implies that where , , and . After that, all hypotheses of Theorem 13 are fulfilled. Hence, has a unique fixed point . That is, is the unique solution of integral equation (81).

Remark 17. If , ; then, (78) has a unique solution by Theorem 16.

Theorem 18. Suppose that (i) is continuous(ii)there is a continuous mapping satisfyingThen, (81) possesses a unique solution .

Proof. For , according to the conditions (i)-(ii), one can get where is the same as in Theorem 15. Thus, all the hypotheses of Theorem 15 are fulfilled with and . It follows that possesses a unique fixed point , and so is a solution of (81).

5. Conclusions

In rectangular -metric spaces, we introduced a new triangular -orbital admissible condition and established two fixed point results for mappings with a contractive iterate at a point. Further, we provided two examples that elaborated the usability of presented results. At the same time, we proved the existence and uniqueness of solution of an integral equation.

Data Availability

No data were used to support this study.

Conflicts of Interest

No potential conflicts of interest are declared by the authors.

Authors’ Contributions

All authors contributed equally in writing this article. All authors approved the final manuscript.

Acknowledgments

This work was financially supported by the Science and Research Project Foundation of Liaoning Province Education Department (No. LJC202003).