#### Abstract

This paper establishes and proves a fixed point theorem for Boyd and Wong type contraction in ordered partial metric spaces. In doing so, we have extended several existing results into ordered complete partial metric spaces. An illustrative example is given to demonstrate the validity of our results. Finally, the existence of the solution of nonlinear integral equation is discussed as an application of the main result.

#### 1. Introduction and Preliminaries

Banach’s fixed point theorem [1] has been extensively studied to solve the problems in nonlinear analysis since many years. This theorem provides the existence and uniqueness of the solution. It states that, if is a complete metric space and is a self-contractive mapping, then has a unique fixed point . Due to its usefulness and applications, this theorem has been massively investigated and generalized by different researchers. In 1969, Boyd and Wong [2] gave an important generalization of the Banach fixed point theorem by the application of control function in the Banach contraction condition. Boyd and Wong [2] took into account the condition as follows:whereby is a complete metric space, and a mapping is upper semicontinuous from the right on such that , . Consequently, has a unique fixed point and as , .

Imdad and Kumar [3] extended the existing results by relaxing “continuity” and lightening the “commutativity” requirement besides increasing the number of involved maps from “two” to “four.” Several other researchers extended these results in different directions. Some of them are [4–6] and the references therein. A beautiful survey of the fixed point theory was given by Kumar [7]. Naziku and Kumar [8, 9] proved results using the Boyd and Wong type contractive condition. Recently, Kumar [10] established and proved a fixed point theorem for Boyd and Wong type contraction for a pair of maps in complete metric spaces. This theorem gives conditions for a pair of mappings that possess a fixed point but not continuous at the fixed point and can be applied for both continuous and discontinuous mappings.

In the last few decades, fixed point results in a partially ordered set have been revealed as a very important area of interest to many researchers. In particular, the existence of the fixed point in partially ordered sets has been massively considered in [11–16] and others as they appear in the literature.

In literature [17], Matthews introduced the study of partial metric spaces as the important subject in the approach of formalizing the meaning of programming languages by formulating mathematical objects called “denotations.” Partial metric was introduced to ensure that partial order semantics should have a metric-based tools for program verification in which the notion of size of data object in a domain is used in quantifying how data object is well defined in the domain.

*Definition 1 (see [18]). *Let be a nonempty set. A function is called a partial metric on if it satisfies the followings: (PM0): (non-negativity and small self distance). (PM1): (indistancy implies equality). (PM2): (symmetric). (PM3): (triangularity), for all . is called a partial metric space.

Note that implies (by through ); the converse is always not true. Therefore, a metric space is a partial metric space with all self distances zero.

Several researchers generalized the results of metric fixed point theory using partial metric space setting in different directions. Some of them are [19–22] and the references therein.

Now, we will recall some definitions and lemmas which will be utilized in the proof of main results of this paper.

*Definition 2 (see [18]). *Let be a sequence in a partial metric space ; then,(i)A sequence if and only if .(ii)A sequence is called a Cauchy sequence if there exists such that for all , we have for some integers ; that is exists and it is finite.(iii)A partial metric space is complete if every Cauchy sequence converges to a point such that .

*Definition 3 (see [18]). *A contraction on a partial metric space is a function such that there exist a constant for all satisfies that

*Definition 4 (see [17]). *Let be a nonempty set. Partial ordering is a relation such that (PO 1) for all , (reflexive). (PO 2) for all , , and (antisymmetric). (PO 3) for all , , and (transitivity).

*Definition 5 (see [17]). *For each partial metric space , is a binary relation such that for all , .

Note that for each partial metric space , is a partial ordering.

*Definition 6 (see [23]). *Let be a metric space. Let a mapping to be injective (one to one) and continuous (ICS) mapping with the property that if is convergent, then the sequence is also convergent for all sequences .

*Definition 7 (see [23]). *Let be the set of functions satisfying,(i) for all .(ii) is an upper semicontinuous from right; that is, for any sequence such that as as , we have .Aydi and Karapinar [24] generalized results of Harjani et al. [11] and Luong and Thun [13] by using an ICS mapping and involved Boyd and Wong type contractive condition and provided the following theorem:

Theorem 1 (see [23]). *Let be a partially ordered set. Suppose there exists a metric such that is a complete metric space. Let be a mapping such that is an ICS mapping and is a nondecreasing mapping satisfying,** for all with where and*

Also assume that either(i) is continuous, or(ii)If the sequence is nondecreasing in such that , then .

If there exists a point such that , then has a unique fixed point.

In the next section, the letter will be used to refer to the set of all positive integer numbers.

#### 2. Main Results

We now present an extension of Definition 6 in partial metric spaces.

*Definition 8. *Let be a partial metric space. Let a mapping to be injective (one to one) and continuous (ICS) mapping with the property that if is convergent, then is also convergent for all sequences .

Corresponding to Theorem 1, we state and prove our main results and then provide an illustrative example to demonstrate our results.

Theorem 2. *Let be a partially ordered set (Poset). Let be a partial metric such that is a complete partial metric space. Also, let be an ICS mapping, and be a nondecreasing mapping satisfyingwith for all distinct , where and*

Furthermore, we assume that either(i)a mapping is continuous, or(ii)If a sequence is a nondecreasing sequence such that as , then .

Therefore, if there exists such that , then has a unique fixed point with .

Proof. Let a point such that . We define a sequence asfor all integers . Since is a nondecreasing mapping and is a Poset, then we can havefor all integers . By induction, we can have that

If we suppose that there exists such that , then has a fixed point , which ends the proof.

Now suppose that for all integers , then (8) becomes

From (4), we can have thatwhere

Suppose that for some integers , then (10) becomes,

From Definition 7 (i), we see that (12) becomeswhich is a contradiction. Hence, for all integers ; hence (10) becomes

Therefore, from the above equation, we can observe that the sequence is a decreasing sequence, and it is bounded below.

Let , for all integers . Therefore, there exists a real number such that .

We claim that . In contrary, we suppose that , then by the semicontinuity property of and considering (10) above, we can have thatwhich is a contradiction. Hence,

Now, we need to prove that the sequence is a Cauchy sequence. For the sake of contradiction, we suppose that and the sequence of integers for some such that

Furthermore, suppose that is chosen as the smallest integer such that (17) above holds so that we can have

Thus, by triangle inequality, we obtain

As in (19) above and considering (16), we obtain

Thus,

Similarly, we have

Also, as in (22) and again considering (26), we obtain

Thus,

Now from (1) and for all positive integers , we obtainwhere

As in (25) and in (26) and by considering (16), (21), and (24), then (25) becomeswhich is a contradiction. Hence, the sequence is a Cauchy sequence.

Since is a complete partial metric space; therefore, there exists a point such that a sequence converges to a point .

Given that is an ICS mapping, and a sequence converges, then there exists a point such that

Furthermore, since is also continuous, then

Now, we need to prove that is a fixed point for a mapping .(i)We suppose that the first assumption of Theorem 2 holds; that is, is a continuous mapping. Therefore, Hence, is a fixed point for a mapping .(ii)Now, we suppose that the second assumption of Theorem 2 holds. Given that a sequence is a nondecreasing sequence such that as and , then for all integers , we have that .

Since is a nondecreasing mapping, consequently we obtain that , And respectively we havefor all integers .

Since and as ; hence,

Now, we construct a new sequence such that as and which is defined as follows:.

Since the mapping is continuous, then

Consequently, since and . Similar to the above discussion, we can conclude that a sequence is a Cauchy sequence.

Since , then . Therefore, from (32), we obtainfor all integers . If we suppose , thenand hence, , which ends the proof that is .

Otherwise, suppose that , then since is an injective map. Therefore, .

From (1), we havewhere

Considering (16), (29), and (34) and by letting in (37) and (38), then (37) becomeswhich is a contradiction; thus , and therefore we have ; hence .

Therefore, is a unique fixed point of the mapping .

*Remark 1. *If we let for all and in Theorem 2, we get the following corollary:

Corollary 1. *Let be the same as in Theorem 2 such thatwith for all distinct , and is defined as in Theorem 2.*

Also assume either(i) is continuous, or(ii)If a sequence is a nondecreasing sequence such that as , then .

Therefore, if there exists a point such that , then has a unique fixed point.

*Remark 2. *If we take in Corollary 1 with such that , we get the following corollary:

Corollary 2. *Let be the same as in Theorem 2 such thatfor all with .*

Assume either(i) is continuous, or(ii)If a sequence is a nondecreasing sequence such that as , then .

Therefore, if there exists a point such that , then has a unique fixed point.

*Remark 3. *If we take the mapping in Corollary 2, we get an extension to the work of Harjani et al. [11] in partial metric spaces which runs as follows:

Corollary 3. *Let be the same as in Theorem 2 such thatfor all with .*

Assume either(i) is continuous, or(ii)If a sequence is a nondecreasing sequence such that as , then .

Therefore, if there exists a point such that , then has a unique fixed point.

*Remark 4. *If we set and set to be the identity mapping in Theorem 2, then (4), which isbecomeswhich leads to the extension of the work of Ran and Reurings [15] and Nieto and Rodriguez-Lopez [14] in partial metric spaces which runs as follows:

Corollary 4. *Let be the same as in Theorem 2 such thatfor all with .*

Assume either(i) is continuous, or(ii)If a sequence is a nondecreasing sequence such that as , then .

Therefore, if there exists a point such that , then has a unique fixed point.

The following theorem results are from the additional assumption made on the assumptions to the hypotheses of Theorem 2.

Theorem 3. *Let be a partially ordered set, and be a partial metric on such that is a complete partial metric space. Also, let be an ICS mapping, and be a nondecreasing mapping satisfyingwith for all distinct , where , and*

Furthermore, assume that either(i)A mapping is continuous, or(ii)If a sequence is a nondecreasing sequence such that as , then (iii)For every , there exists , that is comparable to and

Therefore, if there exists such that , then has a unique fixed point .

Proof. In contrary, suppose that a mapping has two distinct fixed points, say and .

From assumption (iii) of Theorem 3, there exists which is comparable to and . Without the lose of generality, we choose .

Now, we construct a sequence as follows:

for all such that .

Since is a nondecreasing mapping, then implies that

. Inductively, we obtain .

If we suppose that there exists such that , then , for all .

This implies that . Similarly, , which completes the proof.

Now, if we suppose that , for all integers , then since the mapping is one to one.

From (46), we havewhere

Analogously to Theorem 2, we have

Thus, (48) becomes

We observe that the sequence is the decreasing sequence which is bounded below. Therefore, there exists a constant such that . We claim that . In contrary, suppose that .

From (51), we can have thatwhich leads towhich is a contradiction since is an upper semicontinuous mapping. Hence, . Therefore, we conclude that

Similarly,

Since and are distinct fixed points of and is one to one mapping, thenwhere

Therefore, from (56), we havewhich is a contradiction. Hence, , which ends the proof.

*Example 1. *Let be a set equipped with a partial metric for all . Let the order be defined by , for all .

It is easy to check that is a partially ordered set (Poset) by proving through from Definition 4.

, for all , is defined. Also,is defined. It is easy to see that a mapping is an ICS mapping as it is injective and continuous. Also, we observe that a function is continuous and nondecreasing .

Now, we show that inequality (4) holds.

From left hand side of (4), for all , we haveFrom the right hand side of (4), we havewhereSimilarly,and .

*Case 1. *Suppose, .

Then, . Hence, .

*Case 2. *Suppose, .

Then, . Hence, .

From left hand side of (4), for all , we haveFrom the right hand side of (4), we havewhereSimilarly,and .

In the similar way, we find that .

Now, if we define , for all , we obtain that . Therefore, (4) holds, and is the fixed point of a mapping .

#### 3. The Existence Solution of Nonlinear Integral Equations

In this section, we studied the existence of solutions for nonlinear integral equations, as an application to the fixed point theorems proved in the previous section.

The following integral equations were inspired by [25, 26]. An unknown function is considered. Now, an application of Theorem 3 is presented as a study of the existence and uniqueness of solution to nonlinear integral equations.where and are given continuous functions.

Let be the set of real continuous function on and let be given by

It is easy to see that is a partial metric and that is a complete partial metric space.

Next, we prove a theorem to establish the existence of a common fixed point for a pair of self mappings:

Theorem 4. *Let us consider the integral equation (68) as above. Also, suppose that it satisfies the following conditions:*(i)*for and , there exists a nondecreasing function such that the following inequality holds:* *where*(ii)

Then, the integral equation (68) has a unique common solution .

Proof. Let be a mapping defined byThis implies that and possess a fixed point . To prove the existence of the fixed point of , we prove that is a contraction. On contrary, we assume that , for all . Using conditions (i) and (ii) of Theorem 4, we haveFor *b* = 1, *a* = 0, we havewhich is a contradiction. Hence, is a common fixed of and , also a solution to integral equation (74). Thus, Theorem 3 is satisfied.

#### 4. Conclusion

The main contribution of this paper to fixed point theory is Definition 8, Theorems 2, and 3. Here, the results have proved for Boyd and Wong type contraction in ordered partial metric spaces. Several existing results in the literature are generalized and extended into ordered complete partial metric spaces. Suitable examples are given to demonstrate the validity of the results. Finally, the existence of the solution of nonlinear integral equation is discussed as an application of the main result.

#### Data Availability

No data were used to support this study.

#### Ethical Approval

Not applicable.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Authors’ Contributions

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