Fixed Points for 𝜓-Graphic Contractions with Application to Integral Equations

Abstract

The aim of this paper is to define modified weak 𝛼-𝜓-contractive mappings and to establish fixed point results for such mappings defined on partial metric spaces using the notion of triangular 𝛼-admissibility. As an application, we prove new fixed point results for graphic weak 𝜓-contractive mappings. Moreover, some examples and an application to integral equation are given here to illustrate the usability of the obtained results.

1. Introduction and Preliminaries

The concept of partial metric space was introduced by Matthews [1] in 1994. Partial metric space is a generalized metric space in which each object does not necessarily have to have a zero distance from itself. A motivation behind introducing the concept of a partial metric was to obtain appropriate mathematical models in the theory of computation and, in particular, to give a modified version of the Banach contraction principle [2, 3]. Subsequently, several authors studied the problem of existence and uniqueness of a fixed point for mappings satisfying different contractive conditions on partial metric spaces (e.g., [4–10]). For a recent survey on the existence of fixed points in different spaces with generalized distance functions, the reader may check [11–18]. We start by recalling some definitions and properties of partial metric spaces.

Definition 1. A partial metric on a nonempty set 𝑋 is a function 𝑝∶𝑋×𝑋→ℝ⁺ such that, for all 𝑥,𝑦,𝑧∈𝑋,(p1)𝑥=𝑦⇔𝑝(𝑥,𝑥)=𝑝(𝑥,𝑦)=𝑝(𝑦,𝑦); (p2)𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑦); (p3)𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥); (p4)𝑝(𝑥,𝑧)≤𝑝(𝑥,𝑦)+𝑝(𝑦,𝑧)−𝑝(𝑦,𝑦).
A partial metric space is a pair (𝑋,𝑝) such that 𝑋 is nonempty set and 𝑝 is a partial metric on 𝑋.

From the above definition, if 𝑝(𝑥,𝑦)=0, then 𝑥=𝑦. But if 𝑥=𝑦, 𝑝(𝑥,𝑦) may not be 0 in general. A trivial example of a partial metric space is the pair (ℝ⁺,𝑝), where 𝑝∶ℝ⁺×ℝ⁺→ℝ⁺ is defined as 𝑝(𝑥,𝑦)=max{𝑥,𝑦}. For more examples of partial metric spaces, we refer to [6, 7].

Each partial metric 𝑝 on 𝑋 generates a 𝑇₀ topology 𝜏_𝑝 on 𝑋 which has as a base the family of open 𝑝-balls {𝐵_𝑝(𝑥,𝜖)∶𝑥∈𝑋,𝜖>0}, where 𝐵_𝑝(𝑥,𝜖)={𝑦∈𝑋∶𝑝(𝑥,𝑦)<𝑝(𝑥,𝑥)+𝜖} for all 𝑥∈𝑋 and 𝜖>0. A sequence {𝑥_𝑛} in 𝑋 converges to a point 𝑥∈𝑋, with respect to 𝜏_𝑝, if and only if 𝑝(𝑥,𝑥)=lim_𝑛→∞𝑝(𝑥,𝑥_𝑛). A sequence {𝑥_𝑛} in 𝑋 is called Cauchy sequence if lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚) exists and is finite. A partial metric space (𝑋,𝑝) is said to be complete if every Cauchy sequence {𝑥_𝑛} in 𝑋 converges, with respect to 𝜏_𝑝, to a point 𝑥∈𝑋 such that 𝑝(𝑥,𝑥)=lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).

A sequence {𝑥_𝑛} in 𝑋 is called 0-Cauchy sequence if lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚)=0. A partial metric space (𝑋,𝑝) is said to be 0-complete if every 0-Cauchy sequence {𝑥_𝑛} in 𝑋 converges, with respect to 𝜏_𝑝, to a point 𝑥∈𝑋 such that 𝑝(𝑥,𝑥)=0.

The partial metric space (ℚ∩[0;∞),𝑝), where ℚ denotes the set of rational numbers and the partial metric 𝑝 is given by 𝑝(𝑥,𝑦)=max{𝑥,𝑦}, provides an example of a 0-complete partial metric space which is not complete.

If 𝑝 is a partial metric on 𝑋, then the function 𝑝^𝑠∶𝑋×𝑋→ℝ⁺ given by 𝑝^𝑠𝑥,𝑦=2𝑝𝑥,𝑦−𝑝(𝑥,𝑥)−𝑝𝑦,𝑦 (1) is a metric on 𝑋. A partial metric space (𝑋,𝑝) is complete if and only if the metric space (𝑋,𝑝^𝑠) is complete. Furthermore, lim_𝑛→∞𝑝^𝑠(𝑥_𝑛,𝑧)=0 if and only if 𝑝(𝑧,𝑧)=lim_𝑛→∞𝑝(𝑥_𝑛,𝑧)=lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚). Some more properties of partial metric spaces can be found in [3, 7, 19].

Samet et al. [20] defined the notion of 𝛼-admissible mappings and proved the following result.

Definition 2. Let 𝑇 be a self-mapping on 𝑋 and 𝛼∶𝑋×𝑋→[0,+∞) a function. One says that 𝑇 is an 𝛼-admissible mapping if 𝑥,𝑦∈𝑋,𝛼𝑥,𝑦≥1⟹𝛼𝑇𝑥,𝑇𝑦≥1. (2)

Denote with Ψ the family of nondecreasing functions 𝜓∶[0,+∞)→[0,+∞) such that ∑^∞_𝑛=1𝜓^𝑛(𝑡)<+∞ for all 𝑡>0, where 𝜓^𝑛 is the 𝑛th iterate of 𝜓.

Theorem 3. Let (𝑋,𝑑) be a complete metric space and 𝑇 be 𝛼-admissible mapping. Assume that 𝛼𝑥,𝑦𝑑𝑇𝑥,𝑇𝑦≤𝜓𝑑𝑥,𝑦, (3) where 𝜓∈Ψ. Also, suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(ii)either 𝑇 is continuous or for any sequence {𝑥_𝑛} in 𝑋 with 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ∪{0} and 𝑥_𝑛→𝑥 as 𝑛→+∞, one has 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛∈ℕ∪{0}.
Then, 𝑇 has a fixed point.

Very recently, Salimi et al. [21] modified the notions of 𝛼-𝜓-contractive mappings and 𝛼-admissible mappings as follows.

Definition 4 (see [21]). Let 𝑇 be a self-mapping on 𝑋 and 𝛼,𝜂∶𝑋×𝑋→[0,+∞) two functions. One says that 𝑇 is an 𝛼-admissible mapping with respect to 𝜂 if .𝑥,𝑦∈𝑋,𝛼𝑥,𝑦≥𝜂𝑥,𝑦⟹𝛼𝑇𝑥,𝑇𝑦≥𝜂𝑇𝑥,𝑇𝑦 (4) Note that if we take 𝜂(𝑥,𝑦)=1, then this definition reduces to Definition 2. Also, if we take, 𝛼(𝑥,𝑦)=1, then we say that 𝑇 is an 𝜂-subadmissible mapping.

The following result properly contains Theorem 3 and Theorems 2.3 and 2.4 of [22].

Theorem 5 (see [21]). Let (𝑋,𝑑) be a complete metric space and 𝑇 be 𝛼-admissible mapping with respect to 𝜂. Assume that 𝑀,𝑥,𝑦∈𝑋,𝛼𝑥,𝑦≥𝜂𝑥,𝑦⟹𝑑𝑇𝑥,𝑇𝑦≤𝜓𝑥,𝑦 (5) where 𝜓∈Ψ and 𝑀𝑑,𝑑()𝑥,𝑦=max𝑥,𝑦𝑥,𝑇𝑥+𝑑𝑦,𝑇𝑦2,𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥2. (6) Also, suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥𝜂(𝑥₀,𝑇𝑥₀);(ii)either 𝑇 is continuous or for any sequence {𝑥_𝑛} in 𝑋 with 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥𝜂(𝑥_𝑛,𝑥_𝑛+1) for all 𝑛∈ℕ∪{0} and 𝑥_𝑛→𝑥 as 𝑛→+∞, one has 𝛼(𝑥_𝑛,𝑥)≥𝜂(𝑥_𝑛,𝑥) for all 𝑛∈ℕ∪{0}. Then, 𝑇 has a fixed point.

In fact, Banach contraction principle and Theorem 5 hold for the following example, but Theorem 3 does not hold.

Example 6 (see [21]). Let 𝑋=[0,∞) be endowed with the usual metric 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈𝑋, and let 𝑇∶𝑋→𝑋 be defined by 𝑇𝑥=(1/4)𝑥. Also, define, 𝛼∶𝑋²→[0,∞) by 𝛼(𝑥,𝑦)=3 and 𝜓∶[0,∞)→[0,∞) by 𝜓(𝑡)=(1/2)𝑡.

In this paper, we define modified weak 𝛼-𝜓-contractive mappings and establish fixed point results for such mappings defined on ordinary as well as ordered partial metric spaces using the notion of triangular 𝛼-admissibility. As an application, we prove new fixed point results for graphic weak 𝜓-contractive mappings. Moreover, some examples and an application to integral equation are given here to illustrate the usability of the obtained results.

2. Modified Weak 𝛼-𝜓-Contractions

Recently, Karapınar et al. [23] introduced the notion of triangular 𝛼-admissible mapping as follows.

Definition 7 (see [23]). Let 𝑇∶𝑋→𝑋 and 𝛼∶𝑋×𝑋→(−∞,+∞). One says that 𝑇 is a triangular 𝛼-admissible mapping if(T1)𝛼(𝑥,𝑦)≥1 implies 𝛼(𝑇𝑥,𝑇𝑦)≥1,𝑥,𝑦∈𝑋;(T2)𝛼(𝑥,𝑧)≥1,𝛼(𝑧,𝑦)≥1 imply 𝛼(𝑥,𝑦)≥1.

Lemma 8 (see [23]). Let 𝑇∶𝑋→𝑋 be a triangular 𝛼-admissible mapping. Assume that there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. Define sequence {𝑥_𝑛} by 𝑥_𝑛=𝑇^𝑛𝑥₀. Then, 𝛼𝑥_𝑚,𝑥_𝑛≥1∀𝑚,𝑛∈ℕ𝑤𝑖𝑡ℎ𝑚<𝑛. (7)

Motivated by Samet et al. [20] and Salimi et al. [21], we introduce the following mapping.

Definition 9. Let (𝑋,𝑝) be a partial metric space and 𝑇∶𝑋→𝑋, 𝛼∶𝑋×𝑋→[0,∞) two mappings. If there exists an upper semicontinuous from the right nondecreasing function 𝜓∶ℝ₊→ℝ₊ with 𝜓(𝑡)<𝑡 for all 𝑡>0 such that 𝑀𝑥,𝑦∈𝑋,𝛼𝑥,𝑦≥1⟹𝑝𝑇𝑥,𝑇𝑦≤𝜓_𝑝𝑥,𝑦, (8) where 𝑀_𝑝𝑝(),𝑝,𝑝𝑥,𝑦=max𝑥,𝑦,𝑝𝑥,𝑇𝑥𝑦,𝑇𝑦𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥2. (9) Then, we say that 𝑇 is a modified weak 𝛼-𝜓-contractive mapping.

Definition 10. Let (𝑋,𝑝) be a partial metric space. Let 𝛼∶𝑋×𝑋→(−∞,+∞) and 𝑇∶𝑋→𝑋. One says that 𝑇 is an 𝛼-continuous function on (𝑋,𝑝^𝑠) if for given 𝑥 and {𝑥_𝑛} in 𝑋, 𝑥_𝑛⟶𝑥as𝛼𝑥𝑛⟶∞,_𝑛,𝑥_𝑛+1≥1∀𝑛∈ℕimply𝑇𝑥_𝑛⟶𝑇𝑥. (10)

Example 11. Let 𝑋=[0,∞) and 𝑝(𝑥,𝑦)=max{𝑥,𝑦}. Assume that 𝑇∶𝑋→𝑋 and 𝛼∶𝑋²→(−∞,+∞) defined by 𝑇𝑥=3𝑥²,if𝑥∈[0,1]5𝑥+1,if(),𝛼=1,∞𝑥,𝑦1,if𝑥∈[0,1]−1,otherwise. (11) Clearly, 𝑇 is not continuous, but 𝑇 is an 𝛼-continuous on (𝑋,𝑝^𝑠). Indeed, if 𝑥_𝑛→𝑥 as 𝑛→∞ and 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1, then 𝑥_𝑛∈[0,1] and so lim_𝑛→∞𝑇𝑥_𝑛=lim_𝑛→∞3𝑥²_𝑛=3𝑥²=𝑇𝑥.

Theorem 12. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑇 a modified weak 𝛼-𝜓-contractive and triangular 𝛼-admissible mapping. Suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(ii)𝑇 is an 𝛼-continuous function on (𝑋,𝑝^𝑠). Then, 𝑇 has a fixed point.

Proof. Let 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. Define a sequence {𝑥_𝑛} in 𝑋 by 𝑥_𝑛=𝑇^𝑛𝑥₀=𝑇𝑥_𝑛−1 for all 𝑛∈ℕ. Then, by Lemma 8, we have 𝛼𝑥_𝑚,𝑥_𝑛≥1∀𝑚,𝑛∈ℕwith𝑚<𝑛. (12) If 𝑥_𝑛+1=𝑥_𝑛 for some 𝑛∈ℕ, then 𝑥=𝑥_𝑛 is a fixed point for 𝑇, and the result is proved. Hence, we suppose that 𝑥_𝑛+1≠𝑥_𝑛 for all 𝑛∈ℕ. Now, by (8) and (12) with 𝑥=𝑥_𝑛−1, 𝑦=𝑥_𝑛, we get 𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜓𝑀𝑥_𝑛−1,𝑥_𝑛 (13) On the other hand, max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1≤𝑀_𝑝𝑥_𝑛−1,𝑥_𝑛=max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛−1,𝑇𝑥_𝑛−1,𝑝𝑥_𝑛,𝑇𝑥_𝑛,𝑝𝑥_𝑛−1,𝑇𝑥_𝑛+𝑝𝑥_𝑛,𝑇𝑥_𝑛−12=max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑥_𝑛−1,𝑥_𝑛+1+𝑝𝑥_𝑛,𝑥_𝑛2≤max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑥_𝑛−1,𝑥_𝑛+𝑝𝑥_𝑛,𝑥_𝑛+12≤max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1 (14) and then, 𝑀_𝑝(𝑥_𝑛−1,𝑥_𝑛)=max{𝑝(𝑥_𝑛−1,𝑥_𝑛),𝑝(𝑥_𝑛,𝑥_𝑛+1)}. So, from (13), we have 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝜓max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1. (15) Now, if max{𝑝(𝑥_𝑛−1,𝑥_𝑛),𝑝(𝑥_𝑛,𝑥_𝑛+1)}=𝑝(𝑥_𝑛,𝑥_𝑛+1) for some 𝑛∈ℕ, then 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝜓max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1=𝜓𝑝𝑥_𝑛,𝑥_𝑛+1<𝑝𝑥_𝑛,𝑥_𝑛+1, (16) which is a contradiction. Hence, for all 𝑛∈ℕ, we have 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝜓𝑝𝑥_𝑛−1,𝑥_𝑛<𝑝𝑥_𝑛−1,𝑥_𝑛. (17)
This implies that the sequence {𝑝(𝑥_𝑛−1,𝑥_𝑛)} is decreasing, and so, by (17), there is 𝑠≥0 such that lim_𝑛→+∞𝑝𝑥_𝑛,𝑥_𝑛+1=lim_𝑛→+∞𝜓𝑝𝑥_𝑛−1,𝑥_𝑛=𝑠. (18) Now, we show that 𝑠 must be equal to 0.
In fact, if 𝑠>0, then we get 𝑠=limsup_𝑛→+∞𝜓𝑝𝑥_𝑛−1,𝑥_𝑛≤𝜓(𝑠)<𝑠, (19) which is a contradiction. Hence, lim_𝑛→+∞𝑝𝑥_𝑛,𝑥_𝑛+1=0. (20) We prove that {𝑥_𝑛} is a 0-Cauchy sequence. Suppose, to the contrary, that {𝑥_𝑛} is not a 0-Cauchy sequence. Then, there is 𝜀>0 and sequences {𝑚(𝑘)} and {𝑛(𝑘)} such that, for all positive integers 𝑘, 𝑛(𝑘)(𝑘)𝑝𝑥>𝑚>𝑘,_𝑛(𝑘),𝑥_𝑚(𝑘)𝑝𝑥≥𝜀,_𝑛(𝑘),𝑥_{𝑚(𝑘)−1}<𝜀. (21) Now, for all 𝑘∈ℕ, we have 𝜀≤𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)≤𝑝𝑥_𝑛(𝑘),𝑥_{𝑚(𝑘)−1}+𝑝𝑥_{𝑚(𝑘)−1},𝑥_𝑚(𝑘)<𝜀+𝑝𝑥_{𝑚(𝑘)−1},𝑥_𝑚(𝑘). (22) Taking the limit as 𝑘→+∞ in the above inequality and using (20), we get lim_𝑘→+∞𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)=𝜀. (23) Again, from 𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)≤𝑝𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1+𝑝𝑥_𝑚(𝑘)+1,𝑥_𝑛(𝑘)+1+𝑝𝑥_𝑛(𝑘)+1,𝑥_𝑛(𝑘)𝑝𝑥_𝑛(𝑘)+1,𝑥_𝑚(𝑘)+1≤𝑝𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1+𝑝𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+𝑝𝑥_𝑛(𝑘)+1,𝑥_𝑛(𝑘), (24) taking the limit as 𝑘→+∞ and by (20) and (23), we deduce lim_𝑘→+∞𝑝𝑥_𝑛(𝑘)+1,𝑥_𝑚(𝑘)+1=𝜀. (25) Also, since 𝜀≤𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)≤𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)+1+𝑝𝑥_𝑚(𝑘)+1,𝑥_𝑚(𝑘), (26) then by taking limit as 𝑛→∞ in the last inequality and applying (20) and (23), we deduce lim_𝑘→+∞𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)+1=𝜀. (27) Similarly, lim_𝑘→+∞𝑝𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+1=𝜀. (28) Then, by (8) and (12), we obtain 𝑝𝑥_𝑛(𝑘)+1,𝑥_𝑚(𝑘)+1=𝑝𝑇𝑥_𝑛(𝑘),𝑇𝑥_𝑚(𝑘)≤𝜓𝑀_𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘), (29) where 𝑀_𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)=max𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘),𝑝𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘),𝑝𝑥_𝑚(𝑘),𝑇𝑥_𝑚(𝑘),𝑝𝑥_𝑛(𝑘),𝑇𝑥_𝑚(𝑘)+𝑝𝑥_𝑚(𝑘),𝑇𝑥_𝑛(𝑘)2=max𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘),𝑝𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1,𝑝𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1,𝑝𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)+1+𝑝𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+12. (30) Taking limit supremum as 𝑛→∞ in the above inequality and applying (20), (23), (25), (27), and (28), we get 𝜀≤𝜓(𝜀)<𝜀, (31) which is a contradiction. Hence, {𝑥_𝑛} is a 0-Cauchy sequence.
Since 𝑇 is orbitally 𝛼-continuous on (𝑋,𝑝^𝑠), 𝑥_𝑛→𝑧 as 𝑛→∞ and 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1, then we have 𝑇𝑧=lim_𝑛→∞𝑇𝑥_𝑛=lim_𝑛→∞𝑥_𝑛+1=𝑧. (32) So, 𝑧 is a fixed point of 𝑇.

Theorem 13. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑇∶𝑋→𝑋 a modified weak 𝛼-𝜓-contractive and 𝛼-admissible mapping. Suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(ii)𝛼(𝑥,𝑥)≥1 for all 𝑥∈𝑋 and if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛∈ℕ. Then, 𝑇 has a fixed point.

Proof. Condition (ii) implies property (T2) in definition of triangular 𝛼-admissible map. Indeed, if 𝛼(𝑥,𝑦)≥1 and 𝛼(𝑦,𝑧)≥1, then applying (ii) to (𝑥_𝑛) defined by 𝑥₁∶=𝑥,𝑥₂∶=𝑦,𝑥_𝑛∶=𝑧for𝑛≥3, (33) we get 𝛼(𝑥_𝑛,𝑧)≥1 for 𝑛∈ℕ, and hence, 𝛼(𝑥,𝑧)≥1. Thus, as in Theorem 12, we obtain a 0-Cauchy sequence {𝑥_𝑛} such that 𝑥_𝑛→𝑧 as 𝑛→∞. Since, 𝛼𝑥_𝑛,𝑥_𝑛+1≥1 (34) for all 𝑛∈ℕ and 𝑥_𝑛→𝑧 as 𝑛→∞, then from (ii), we have 𝛼𝑥_𝑛,𝑧≥1 (35) for all 𝑛∈ℕ. Then, from (8), we get 𝑝𝑥_𝑛+1,𝑇𝑧≤𝜓𝑀_𝑝𝑥_𝑛,𝑧, (36) where 𝑀_𝑝𝑥_𝑛𝑝𝑥,𝑧=max_𝑛,𝑧,𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝(),𝑝𝑥𝑧,𝑇𝑧^𝑛,𝑇𝑧+𝑝𝑧,𝑥_𝑛+12. (37) By taking limit supremum as 𝑛→∞ in the above inequality, we have 𝑝(𝑧,𝑇𝑧)=limsup_𝑛→∞𝑝𝑥_𝑛+1,𝑇𝑧≤limsup_𝑛→∞𝜓𝑀_𝑝𝑥_𝑛,𝑧≤𝜓limsup_𝑛→∞𝑀_𝑝𝑥_𝑛𝑝()(),,𝑧=𝜓𝑧,𝑇𝑧<𝑝𝑧,𝑇𝑧 (38) which is a contradiction. Hence, 𝑝(𝑧,𝑇𝑧)=0. That is, 𝑧=𝑇𝑧.

Example 14. Let 𝑋=[0,∞) be endowed with the partial metric 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑋, and let 𝑇∶𝑋→𝑋 be defined by 1𝑇𝑥=2𝑥³if[]𝑥𝑥∈0,1₃+2ln𝑥if𝑥∈(1,∞). (39) Define 𝛼∶𝑋×𝑋→(−∞,+∞) and 𝜓∶[0,∞)→[0,∞) by 𝛼=2𝑥,𝑦if𝑥,𝑦∈[0,1]or𝑥=𝑦−2otherwise𝜓(𝑡)=12𝑡. (40)
Clearly, (𝑋,𝑝) is a 0-complete partial metric space. We show that 𝑇 is a triangular 𝛼-admissible mapping. Let 𝑥,𝑦∈𝑋, if 𝛼(𝑥,𝑦)≥1, then 𝑥,𝑦∈[0,1] or 𝑥=𝑦. On the other hand, for all 𝑥∈[0,1], we have 𝑇𝑥≤1. It follows that 𝛼(𝑇𝑥,𝑇𝑦)≥1. Also, if 𝛼(𝑥,𝑧)≥1, and 𝛼(𝑧,𝑦)≥1 then 𝑥,𝑦,𝑧∈[0,1]. That is, 𝛼(𝑥,𝑦)≥1. Hence, the assertion holds. In reason of the above arguments, 𝛼(0,𝑇0)≥1.
Now, if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ∪{0} and 𝑥_𝑛→𝑥 as 𝑛→+∞, then {𝑥_𝑛}⊂[0,1], and hence, 𝑥∈[0,1]. This implies that 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛∈ℕ.
Let 𝛼(𝑥,𝑦)≥1. Then, 𝑥,𝑦∈[0,1]. We get 𝑝=1𝑇𝑥,𝑇𝑦2max𝑥³,𝑦³≤12≤1max𝑥,𝑦2𝑀_𝑝𝑀𝑥,𝑦=𝜓_𝑝𝑥,𝑦. (41) That is, 𝛼𝑥,𝑦≥1⟹𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑀_𝑝𝑥,𝑦. (42) Hence, all conditions of Theorem 13 hold and 0 is a fixed point of 𝑇.

Corollary 15. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑇∶𝑋→𝑋 be such that 𝛼(𝑥₀,𝑇𝑥₀)≥1 for some 𝑥₀∈𝑋. Assume that 𝛼𝑥,𝑦𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑀_𝑝𝑥,𝑦 (43) hold for all 𝑥,𝑦∈𝑋. Also, suppose that one of the following assertions holds:(i)𝑇 is triangular 𝛼-admissible and 𝛼-continuous on (𝑋,𝑝^𝑠);(ii)𝑇 is 𝛼-admissible, 𝛼(𝑥,𝑥)≥1 for all 𝑥∈𝑋 and if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛∈ℕ. Then, 𝑇 has a fixed point.

Corollary 16. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑇∶𝑋→𝑋 be such that 𝛼(𝑥₀,𝑇𝑥₀)≥1 for some 𝑥₀∈𝑋. Assume that 𝛼𝑥,𝑦+ℓ^{𝑝(𝑇𝑥,𝑇𝑦)}≤(1+ℓ)^{𝜓(𝑀𝑝(𝑥,𝑦))} (44) hold for all 𝑥,𝑦∈𝑋 where ℓ>0. Also, suppose that one of the following assertions holds:(i)𝑇 is triangular 𝛼-admissible and 𝛼-continuous on (𝑋,𝑝^𝑠);(ii)𝑇 is 𝛼-admissible, 𝛼(𝑥,𝑥)≥1 for all 𝑥∈𝑋 and if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛∈ℕ. Then, 𝑇 has a fixed point.

Proof. Let 𝛼(𝑥,𝑦)≥1. Then, |𝛼(𝑥,𝑦)|≥1. Hence, by (44), we have (1+ℓ)^{𝑝(𝑇𝑥,𝑇𝑦)}≤𝛼𝑥,𝑦+ℓ^{𝑝(𝑇𝑥,𝑇𝑦)}≤(1+ℓ)^{𝜓(𝑀𝑝(𝑥,𝑦))}. (45) Thus, 𝑝(𝑇𝑥,𝑇𝑦)≤𝜓(𝑀_𝑝(𝑥,𝑦)). Hence, conditions of Corollary 15 hold, and 𝑇 has a fixed point.

Similarly, we have the following corollary.

Corollary 17. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑇∶𝑋→𝑋 be such that 𝛼(𝑥₀,𝑇𝑥₀)≥1 for some 𝑥₀∈𝑋. Assume that 𝑝𝑇𝑥,𝑇𝑦+ℓ^{|𝛼(𝑥,𝑦)|}≤𝜓𝑀_𝑝𝑥,𝑦+ℓ (46) hold for all 𝑥,𝑦∈𝑋 where ℓ>0. Also, suppose that one of the following assertions holds:(i)𝑇 is triangular 𝛼-admissible and 𝛼-continuous on (𝑋,𝑝^𝑠);(ii)𝑇 is 𝛼-admissible, 𝛼(𝑥,𝑥)≥1 for all 𝑥∈𝑋 and if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛∈ℕ.
Then, 𝑇 has a fixed point.

Corollary 18 (Matthews [1]). Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑇∶𝑋→𝑋 a given mapping satisfying 𝑝𝑇𝑥,𝑇𝑦≤𝑘𝑝𝑥,𝑦 (47) for all 𝑥,𝑦∈𝑋, where 𝑘∈[0,1). Then, 𝑇 has a unique fixed point.

3. Fixed Point Results in Partially Ordered Partial Metric Spaces

Fixed point theorems for monotone operators in partially ordered metric spaces are widely investigated and have found various applications in differential and integral equations (see [24, 25] and references therein).

Theorem 19 (see [24]). Let (𝑋,𝑑,⪯) be a partially ordered complete metric space and 𝑇∶𝑋→𝑋 a continuous increasing self-mapping such that 𝑥₀⪯𝑇𝑥₀ for some 𝑥₀∈𝑋. Assume that 𝑑𝑇𝑥,𝑇𝑦≤𝑟𝑑𝑥,𝑦 (48) hold for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦, where 0≤𝑟<1. Then, 𝑇 has a fixed point.

Theorem 20. Let (𝑋,𝑝,⪯) be a partially ordered 0-complete partial metric space and 𝑇∶𝑋→𝑋 an increasing self-mapping such that 𝑥₀⪯𝑇𝑥₀ for some 𝑥₀∈𝑋. Assume that 𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑀_𝑝𝑥,𝑦 (49) hold for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦. Now, if 𝑇 is a continuous mapping on (𝑋,𝑝^𝑠), then 𝑇 has a fixed point.

Proof. Define, 𝛼∶𝑋²→(−∞,+∞) by 𝛼𝑥,𝑦=1,if𝑥⪯𝑦0,otherwise. (50) At first we prove that 𝑇 is a triangular 𝛼-admissible mapping. Let 𝛼(𝑥,𝑦)≥1, then 𝑥⪯𝑦. As 𝑇 is an increasing mapping, we have 𝑇𝑥⪯𝑇𝑦. That is, 𝛼(𝑇𝑥,𝑇𝑦)≥1. Also, let 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑧,𝑦)≥1, then 𝑥⪯𝑧 and 𝑧⪯𝑦. So, from transitivity, we have 𝑥⪯𝑦. That is, 𝛼(𝑥,𝑦)≥1. Thus, 𝑇 is a triangular 𝛼-admissible mapping. Also, there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀ which implies 𝛼(𝑥₀,𝑇𝑥₀)≥1. Let 𝛼(𝑥,𝑦)≥1, then 𝑥⪯𝑦. Now, from (49), we have 𝑝(𝑇𝑥,𝑇𝑦)≤𝜓(𝑀_𝑝(𝑥,𝑦)). That is, 𝛼𝑥,𝑦≥1⟹𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑀_𝑝𝑥,𝑦. (51) Hence, all conditions of Theorem 12 are satisfied, and 𝑇 has a fixed point.

Theorem 21 (see [25]). Let (𝑋,𝑑,⪯) be a partially ordered complete metric space and 𝑇∶𝑋→𝑋 an increasing mapping such that 𝑑𝑇𝑥,𝑇𝑦≤𝑟𝑑𝑥,𝑦 (52) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦, where 0≤𝑟<1. Suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀;(ii)if {𝑥_𝑛} is a sequence in 𝑋 such that 𝑥_𝑛⪯𝑥_𝑛+1 for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞, then 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ. Then, 𝑇 has a fixed point.

Theorem 22. Let (𝑋,𝑝,⪯) be a partially ordered 0-complete partial metric space and 𝑇∶𝑋→𝑋 an increasing mapping such that 𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑀_𝑝𝑥,𝑦 (53) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦. Suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀;(ii)if {𝑥_𝑛} is a sequence in 𝑋 such that 𝑥_𝑛⪯𝑥_𝑛+1 for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞, then 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ. Then, 𝑇 has a fixed point.

Proof. Define 𝛼∶𝐴×𝐴→(−∞,+∞) as in proof of Theorem 20. Assume 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ such that 𝑥_𝑛→𝑥 as 𝑛→∞. Then, 𝑥_𝑛⪯𝑥_𝑛+1 for all 𝑛∈ℕ. Hence, by (ii), we get 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ and so 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛∈ℕ. Proceeding as in proof of Theorem 20, we can deduce that 𝑇 is a modified weak 𝛼-𝜓-contractive and 𝛼-admissible mapping and there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. Hence, all conditions of Theorem 13 hold, and 𝑇 has a fixed point.

Remark 23. Similarly, we may obtain more fixed point results on ordered partial metric spaces as immediate consequences of Corollaries 15–17.

4. Fixed Point Results for Graphic Contractions

Consistent with Jachymski [26], let (𝑋,𝑝) be a partial metric space, and let Δ denote the diagonal of the Cartesian product 𝑋×𝑋. Consider a directed graph 𝐺 such that the set 𝑉(𝐺) of its vertices coincides with 𝑋 and the set 𝐸(𝐺) of its edges contains all loops, that is, 𝐸(𝐺)⊇Δ. We assume that 𝐺 has no parallel edges, so we can identify 𝐺 with the pair (𝑉(𝐺),𝐸(𝐺)). Moreover, we may treat 𝐺 as a weighted graph (see [26]) by assigning to each edge the distance between its vertices. If 𝑥 and 𝑦 are vertices in a graph 𝐺, then a path in 𝐺 from 𝑥 to 𝑦 of length 𝑁(𝑁∈ℕ) is a sequence {𝑥_𝑖}^𝑁_𝑖=0 of 𝑁+1 vertices such that 𝑥₀=𝑥, 𝑥_𝑁=𝑦 and (𝑥_𝑛−1,𝑥_𝑛)∈𝐸(𝐺) for 𝑖=1,…,𝑁. A graph 𝐺 is connected if there is a path between any two vertices. 𝐺 is weakly connected if 𝐺 is connected (see for details [11, 26–28]).

Recently, some results have appeared providing sufficient conditions for a mapping to be a Picard operator if (𝑋,𝑑) is endowed with a graph. The first result in this direction was given by Jachymski [26].

Definition 24 (see [26]). One says that a mapping 𝑇∶𝑋→𝑋 is a Banach 𝐺-contraction or simply 𝐺-contraction if 𝑇 preserves edges of 𝐺, that is, (𝐺)⟹𝑇(𝑥)𝑦(𝐺),∀𝑥,𝑦∈𝑋𝑥,𝑦∈𝐸,𝑇∈𝐸 (54) and 𝑇 decreases weights of edges of 𝐺 in the following way: ()(𝐺)𝑇(𝑥)𝑦.∃𝛼∈0,1,∀𝑥,𝑦∈𝑋𝑥,𝑦∈𝐸⟹𝑑,𝑇≤𝛼𝑑𝑥,𝑦 (55)

Definition 25 (see [26]). A mapping 𝑇∶𝑋→𝑋 is called 𝐺-continuous if given 𝑥∈𝑋 and sequence {𝑥_𝑛}𝑥_𝑛⟶𝑥as𝑥𝑛⟶∞,_𝑛,𝑥_𝑛+1∈𝐸(𝐺)∀𝑛∈ℕimply𝑇𝑥_𝑛⟶𝑇𝑥. (56)

Definition 26. Let (𝑋,𝑝) be a partial metric space endowed with a graph 𝐺 and 𝑇∶𝑋→𝑋 a self-mapping. If there exists an upper semicontinuous from the right function 𝜓∶ℝ₊→ℝ₊ with 𝜓(𝑡)<𝑡 for all 𝑡>0 such that (𝐺)⟹𝑇(𝑥)𝑦(𝐺)(𝐺),𝑀∀𝑥,𝑦∈𝑋,𝑥,𝑦∈𝐸,𝑇∈𝐸∀𝑥,𝑦∈𝑋,𝑥,𝑦∈𝐸⟹𝑝𝑇𝑥,𝑇𝑦≤𝜓_𝑝𝑥,𝑦, (57) where 𝑀_𝑝𝑝(),𝑝,𝑝𝑥,𝑦=max𝑥,𝑦,𝑝𝑥,𝑇𝑥𝑦,𝑇𝑦𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥2. (58) Then, 𝑇 is called a weak 𝜓-graphic contractive mapping.

Theorem 27. Let (𝑋,𝑝) be a 0-complete partial metric space endowed with a graph 𝐺 and 𝑇∶𝑋→𝑋 a weak 𝜓-graphic contractive mapping. Suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that (𝑥₀,𝑇𝑥₀)∈𝐸(𝐺);(ii)𝑇 is 𝐺-continuous on (𝑋,𝑝^𝑠);(iii)(𝑥,𝑧)∈𝐸(𝐺) and (𝑧,𝑦)∈𝐸(𝐺) imply (𝑥,𝑦)∈𝐸(𝐺) for all 𝑥,𝑦,𝑧∈𝑋, that is, 𝐸(𝐺) is a quasi-order [26]. Then, 𝑇 has a fixed point.

Proof. Define 𝛼∶𝑋²→(−∞,+∞) by 𝛼𝑥,𝑦=1,if𝑥,𝑦∈𝐸(𝐺)0,otherwise. (59) At first we prove that 𝑇 is a triangular 𝛼-admissible mapping. Let 𝛼(𝑥,𝑦)≥1, then (𝑥,𝑦)∈𝐸(𝐺). As 𝑇 is a weak 𝜓-graphic contractive mapping, we have (𝑇𝑥,𝑇𝑦)∈𝐸(𝐺). That is, 𝛼(𝑇𝑥,𝑇𝑦)≥1. Also, let 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑧,𝑦)≥1, then (𝑥,𝑧)∈𝐸(𝐺) and (𝑧,𝑦)∈𝐸(𝐺). So, from (iii), we have (𝑥,𝑦)∈𝐸(𝐺). That is, 𝛼(𝑥,𝑦)≥1. Thus, 𝑇 is a triangular 𝛼-admissible mapping. Let 𝑇 be 𝐺-continuous on (𝑋,𝑝^𝑠). Then, 𝑥_𝑛⟶𝑥as𝑥𝑛⟶∞,_𝑛,𝑥_𝑛+1∈𝐸(𝐺)∀𝑛∈ℕimply𝑇𝑥_𝑛⟶𝑇𝑥. (60) That is, 𝑥_𝑛⟶𝑥as𝛼𝑥𝑛⟶∞,_𝑛,𝑥_𝑛+1≥1∀𝑛∈ℕimply𝑇𝑥_𝑛⟶𝑇𝑥 (61) which implies 𝑇 is 𝛼-continuous on (𝑋,𝑝^𝑠). From (i), there exists 𝑥₀∈𝑋 such that (𝑥₀,𝑇𝑥₀)∈𝐸(𝐺). That is, 𝛼(𝑥₀,𝑇𝑥₀)≥1.
If 𝛼(𝑥,𝑦)≥1, then (𝑥,𝑦)∈𝐸(𝐺). Now, since 𝑇 is a weak 𝜓-graphic contractive mapping, so 𝑝(𝑇𝑥,𝑇𝑦)≤𝜓(𝑀_𝑝(𝑥,𝑦)). That is, 𝛼𝑥,𝑦≥1⟹𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑀_𝑝𝑥,𝑦. (62) Hence, all conditions of Theorem 12 are satisfied, and 𝑇 has a fixed point.

If 𝐺 is a connected graph, then condition (iii) of Theorem 27 is automatically satisfied. Thus, we have the following result.

Corollary 28. Let (𝑋,𝑝) be a 0-complete partial metric space endowed with a graph 𝐺 and 𝑇 a weak 𝜓-graphic contractive mapping. Suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that (𝑥₀,𝑇𝑥₀)∈𝐸(𝐺);(ii)𝑇 is 𝐺-continuous on (𝑋,𝑝^𝑠); (iii)𝐺 is a connected graph. Then, 𝑇 has a fixed point.

Theorem 29. Let (𝑋,𝑝) be a 0-complete partial metric space endowed with a graph 𝐺 and 𝑇 a weak 𝜓-graphic contractive mapping. Suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that (𝑥₀,𝑇𝑥₀)∈𝐸(𝐺);(ii)if {𝑥_𝑛} is a sequence in 𝑋 such that (𝑥_𝑛,𝑥_𝑛+1)∈𝐸(𝐺) for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞, then (𝑥_𝑛,𝑥)∈𝐸(𝐺) for all 𝑛∈ℕ. Then, 𝑇 has a fixed point.

Proof. Define, 𝛼∶𝑋²→(−∞,+∞) as in proof of Theorem 27. Condition (ii) implies that 𝐸(𝐺) is a quasi-order, that is, (𝑥,𝑧)∈𝐸(𝐺) and (𝑧,𝑦)∈𝐸(𝐺) imply (𝑥,𝑦)∈𝐸(𝐺) for all 𝑥,𝑦,𝑧∈𝑋 (see Remark 3.1 [26]). Let, 𝑥_𝑛→𝑥 as 𝑛→∞ and 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ. Then, {𝑥_𝑛} is a sequence in 𝑋 such that (𝑥_𝑛,𝑥_𝑛+1)∈𝐸(𝐺) for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞. So, by (ii), we have (𝑥_𝑛,𝑥)∈𝐸(𝐺) for all 𝑛∈ℕ. That is, 𝛼(𝑥_𝑛,𝑥)≥1. All other conditions of Theorem 13, follow similarly as in proof of Theorem 27 and consequently 𝑇 has a fixed point.

Theorem  3.2(2°) in [26] and Theorem 2.3(2) in [29] are extended to weak 𝜓-graphic contractive maps defined on a 0-complete partial metric space as follows.

Corollary 30. Let (𝑋,𝑝) be a 0-complete partial metric space endowed with a graph 𝐺 and 𝑇 a weak 𝜓-graphic contractive mapping. Suppose that the following assertions hold:(i) there exists 𝑥₀∈𝑋 such that (𝑥₀,𝑇𝑥₀)∈𝐸(𝐺);(iis)𝐸(𝐺) is a quasi-order and if {𝑥_𝑛} is a sequence in 𝑋 such that (𝑥_𝑛,𝑥_𝑛+1)∈𝐸(𝐺) for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞, then there is a subsequence {𝑥_𝑘𝑛} with (𝑥_𝑘𝑛,𝑥)∈𝐸(𝐺) for all 𝑛∈ℕ. Then, 𝑇 has a fixed point.

Proof. Condition (iis) implies that of (ii) in Theorem 29 (see Remark 3.1 [26]). Now, the conclusion follows from Theorem 29.

Corollary 31. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝜖-chainable for some 𝜖>0, that is, given 𝑥,𝑦∈𝑋, there is 𝑁∈ℕ and a sequence {𝑥_𝑖}^𝑁_𝑖=0 such that 𝑥₀=𝑥, 𝑥_𝑁=𝑦 and 𝑝(𝑥_𝑖−1,𝑥_𝑖)<𝜖 for 𝑖=1,…,𝑁. Suppose that 𝑇∶𝑋→𝑋 is a mapping satisfying 𝑝.∀𝑥,𝑦∈𝑋,𝑝𝑥,𝑦<𝜖,⟹𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑥,𝑦 (63) Then, 𝑇 has a fixed point.

Proof. Consider the graph 𝐺 with 𝑉(𝐺)=𝑋 and 𝐸(𝐺)={(𝑥,𝑦)∈𝑋×𝑋∶𝑝(𝑥,𝑦)<𝜖}. Then, 𝜖-chainability of (𝑋,𝑝) means 𝐺 is connected, and hence, 𝐸(𝐺) is quasi-order. If (𝑥,𝑦)∈𝐸(𝐺), then 𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑝𝑥,𝑦<𝑝𝑥,𝑦<𝜖, (64) so (𝑇𝑥,𝑇𝑦)∈𝐸(𝐺), hence 𝑇 is a (𝐺,𝜓)-contraction. Let {𝑥_𝑛} be in 𝑋 with 𝑥_𝑛→𝑥, then 𝑝(𝑥_𝑛,𝑥)<𝜖 for sufficiently large 𝑛, so there is {𝑥_𝑘𝑛} such that (𝑥_𝑘𝑛,𝑥)∈𝐸(𝐺). Thus, by Corollary 30, 𝑇 has a fixed point.

Definition 32. Let (𝑋,𝑝,⪯) be a partially ordered partial metric space endowed with a graph 𝐺 and 𝑇∶𝑋→𝑋 a self-mapping. If there exists an upper semicontinuous from the right function 𝜓∶ℝ₊→ℝ₊ with 𝜓(𝑡)<𝑡 for all 𝑡>0 such that, for all 𝑥,𝑦∈𝑋, (𝑥,𝑦)∈𝐸(𝐺) with 𝑥⪯𝑦⇒(𝑇(𝑥),𝑇(𝑦))∈𝐸(𝐺), where 𝑇𝑥⪯𝑇𝑦𝑥,𝑦∈𝐸(𝐺)with𝑀𝑥⪯𝑦,⟹𝑝𝑇𝑥,𝑇𝑦≤𝜓_𝑝𝑥,𝑦, (65) where 𝑀_𝑝𝑝(),𝑝,𝑝𝑥,𝑦=max𝑥,𝑦,𝑝𝑥,𝑇𝑥𝑦,𝑇𝑦𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥2. (66) Then, we say 𝑇 is an ordered weak 𝜓-graphic contractive mapping.

Theorem 33. Let (𝑋,𝑝,⪯) be a partially ordered 0-complete partial metric space endowed with a graph 𝐺 and 𝑇 an ordered weak 𝜓-graphic contractive mapping. Suppose that the following assertions hold:(i)there exists 𝑥₀∈𝑋 such that (𝑥₀,𝑇𝑥₀)∈𝐸(𝐺) with 𝑥₀⪯𝑇𝑥₀,(ii)either 𝑇 is 𝐺-continuous in (𝑋,𝑝^𝑠) and (𝑥,𝑧)∈𝐸(𝐺) and (𝑧,𝑦)∈𝐸(𝐺) imply (𝑥,𝑦)∈𝐸(𝐺) or; (iii)if {𝑥_𝑛} is a sequence in 𝑋 such that (𝑥_𝑛,𝑥_𝑛+1)∈𝐸(𝐺) with 𝑥_𝑛⪯𝑥_𝑛+1 for all 𝑛∈ℕ∪{0} and 𝑥_𝑛→𝑥 as 𝑛→+∞, then (𝑥_𝑛,𝑥)∈𝐸(𝐺) with 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ∪0. Then, 𝑇 has a fixed point.

Proof. Define 𝛼∶𝑋²→(−∞,+∞) by 𝛼𝑥,𝑦=1,if𝑥,𝑦∈𝐸(𝐺)with𝑥⪯𝑦0,otherwise. (67) At first, we prove that 𝑇 is a triangular 𝛼-admissible mapping. Let 𝛼(𝑥,𝑦)≥1, then (𝑥,𝑦)∈𝐸(𝐺) with 𝑥⪯𝑦. As 𝑇 is an ordered weak 𝜓-graphic contractive mapping, we have (𝑇𝑥,𝑇𝑦)∈𝐸(𝐺) where 𝑇𝑥⪯𝑇𝑦. That is, 𝛼(𝑇𝑥,𝑇𝑦)≥1. Also, let 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑧,𝑦)≥1, then (𝑥,𝑧)∈𝐸(𝐺) with 𝑥⪯𝑧 and (𝑧,𝑦)∈𝐸(𝐺) with 𝑧⪯𝑦. So from (ii), we have (𝑥,𝑦)∈𝐸(𝐺). Also, 𝑥⪯𝑧 and 𝑧⪯𝑦 implies 𝑥⪯𝑦. Hence, 𝛼(𝑥,𝑦)≥1. Thus, 𝑇 is a triangular 𝛼-admissible mapping. Let 𝑇 be 𝐺-continuous on (𝑋,𝑝^𝑠). Then, 𝑥_𝑛⟶𝑥as𝑥𝑛⟶∞,_𝑛,𝑥_𝑛+1∈𝐸(𝐺)∀𝑛∈ℕimply𝑇𝑥_𝑛⟶𝑇𝑥. (68) That is, 𝑥_𝑛⟶𝑥as𝛼𝑥𝑛⟶∞,_𝑛,𝑥_𝑛+1≥1∀𝑛∈ℕimply𝑇𝑥_𝑛⟶𝑇𝑥 (69) which implies 𝑇 is 𝛼-continuous on (𝑋,𝑝^𝑠). From (i), there exists 𝑥₀∈𝑋 such that (𝑥₀,𝑇𝑥₀)∈𝐸(𝐺). That is, 𝛼(𝑥₀,𝑇𝑥₀)≥1.
Let 𝛼(𝑥,𝑦)≥1, then (𝑥,𝑦)∈𝐸(𝐺) with 𝑥⪯𝑦. Now, since 𝑇 is an ordered weak 𝜓-graphic contractive mapping, then 𝑝(𝑇𝑥,𝑇𝑦)≤𝜓(𝑀_𝑝(𝑥,𝑦)). That is, 𝛼𝑥,𝑦≥1⟹𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑀_𝑝𝑥,𝑦. (70) Hence, all conditions of Theorem 12 (or 13) are satisfied, and 𝑇 has a fixed point.

Remark 34. All our results established above are new even in the setting of complete metric spaces.

5. Application to Existence of Solutions of Integral Equations

Fixed point theorems for monotone operators in ordered metric spaces are widely investigated and have found various applications in differential and integral equations (see [30, 31] and references therein). In this section, we apply our result to the existence of a solution of an integral equation. Let 𝑋=𝐶([0,𝑇],ℝ) be the set of real continuous functions defined on [0,𝑇], and let 𝑑∶𝑋×𝑋→ℝ₊ be defined by 𝑑𝑥,𝑦=𝑥−𝑦_∞ (71) for all 𝑥,𝑦∈𝑋. Then, (𝑋,𝑑) is a complete metric space. Also, assume this metric space endowed with a graph 𝐺.

Consider the integral equation 𝑥(𝑡)=𝑝(𝑡)+^𝑇₀𝑆(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠 (72) and let 𝐹∶𝑋→𝑋 be defined by 𝐹(𝑥)(𝑡)=𝑝(𝑡)+^𝑇₀𝑆(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (73) We assume that (A)𝑓∶[0,𝑇]×ℝ→ℝ is continuous; (B)𝑝∶[0,𝑇]→ℝ is continuous; (C)𝑆∶[0,𝑇]×ℝ→[0,+∞) is continuous; (D)there exists an upper semicontinuous from the right nondecreasing function 𝜓∶ℝ₊→ℝ₊ with 𝜓(𝑡)<𝑡 for all 𝑡>0 such that for all 𝑠∈[0,𝑇], (𝐺)⟹𝐹(𝑥)𝑦(𝐺)(𝐺)((𝑠(𝑠)𝑥(𝑠)(𝑠,|𝑥(𝑠)(𝑥(𝑠,𝑦(𝑠)𝑦(𝑠),1∀𝑥,𝑦∈𝑋𝑥,𝑦∈𝐸,𝐹∈𝐸∀𝑥,𝑦∈𝑋𝑥,𝑦∈𝐸⟹0≤𝑓𝑠,𝑥))−𝑓𝑠,𝑦≤𝜓max−𝑦)−𝐹))|−𝐹2𝑥(𝑠)𝑦(𝑠+𝑦(𝑠)(𝑥(𝑠;−𝐹)−𝐹)) (74)(F) there exist 𝑥₀∈𝑋 such that (𝑥₀,𝐹(𝑥₀))∈𝐸(𝐺);(G) if {𝑥_𝑛} is a sequence in 𝑋 such that (𝑥_𝑛,𝑥_𝑛+1)∈𝐸(𝐺) for all 𝑛∈ℕ and 𝑥_𝑛→𝑥 as 𝑛→+∞, then (𝑥_𝑛,𝑥)∈𝐸(𝐺) for all 𝑛∈ℕ;(H) assume sup_{[]𝑡∈0,𝑇}^𝑇₀𝑆(𝑡,𝑠)𝑑𝑠≤1. (75)

Theorem 35. Under assumptions (𝐴)--(𝐻), the integral equation (72) has a solution in 𝑋=𝐶([0,𝑇],ℝ).

Proof. Consider the mapping 𝐹∶𝑋→𝑋 defined by (73). Let (𝑥,𝑦)∈𝐸(𝐺). Then, from (D), we deduce 𝐹(𝑥𝑡)𝑦(𝑡=)(−𝐹)_𝑇0𝑆(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))−𝑓𝑠,𝑦(𝑠)𝑑𝑠≤^𝑇₀𝑆(𝑓((𝑠(𝑠)𝑥(𝑠)(𝑠,|𝑥(𝑠)(𝑥(𝑠,𝑦(𝑠)𝑦(𝑠),1𝑡,𝑠)𝑠,𝑥))−𝑓𝑠,𝑦𝑑𝑠≤𝜓max−𝑦)−𝐹))|−𝐹2𝑥(𝑠)𝑦(𝑠+𝑦(𝑠)(𝑥(𝑠𝑥(𝑠)(𝑠,‖𝑥(𝑠)(𝑥(𝑠,𝑦(𝑠)𝑦(𝑠,1−𝐹)−𝐹))≤𝜓max−𝑦)−𝐹))‖−𝐹)2𝑥(𝑠)𝑦(𝑠)+𝑦(𝑠)(𝑥(𝑠.−𝐹−𝐹)) (76) Then, 𝐹𝑥−𝐹𝑦_∞𝑥(𝑠)(𝑠,‖𝑥(𝑠)(𝑥(𝑠,𝑦(𝑠)𝑦(𝑠),1≤𝜓max−𝑦)−𝐹))‖−𝐹2𝑥(𝑠)𝑦(𝑠+𝑦(𝑠)(𝑥(𝑠.−𝐹)−𝐹)) (77) That is, (𝑥,𝑦)∈𝐸(𝐺) implies 𝐹𝑥−𝐹𝑦_∞≤𝜓max𝑥−𝑦_∞,‖𝑥−𝐹(𝑥)‖_∞,𝑦−𝐹𝑦_∞,12𝑥−𝐹𝑦_∞+𝑦−𝐹(𝑥)_∞. (78) Thus, all the hypotheses of Theorem 29 are satisfied, and hence, the mapping 𝐹 has a fixed point, that is, a solution in 𝑋=𝐶([0,𝑇],ℝ) of the integral equation (72).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This paper was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The first and second authors acknowledge with thanks DSR, KAU, for financial support.