Advances on the fixed point results via simulation function involving rational terms

In this paper, we propose two new contractions via simulation function that involves rational expression in the setting of partial b-metric space. The obtained results not only extend, but also generalize and unify the existing results in two senses: in the sense of contraction terms and in the sense of the abstract setting. We present an example to indicate the validity of the main theorem.


Introduction and preliminaries
The origin of the fixed point theory goes back a century, to the pioneer work of Banach.
Since the first study of Banach, researchers have been extended, improved, and generalized this very simple stated but at the same time very powerful theorem. For this purpose, the terms of the contraction inequality and the abstract structure of Banach's theorem have been investigated. In this paper, we shall combine these two trends and introduce two new type contraction via simulation functions involving rational terms in the more general setting, partial-b-metric space.
For the sake of the completeness of the manuscript, we shall recall some basic results and concepts here.

2)
for all distinct v, ω ∈ A, then O possesses a unique fixed point in A.
Every partial metric ρ on A generates a T 0 topology on A, that has a base of the set of all open balls B ρ (v), where an open ball for a partial metric ρ on A is defined [23] as for each v ∈ A and e > 0. If (A, ρ) is a partial-metric space and {v m } a sequence in A, then: exists and is finite. Moreover, we say that the partial-metric space (A, ρ) is complete if every Cauchy sequence {v m } in A converges to a point u ∈ A, that is, Remark 1 The limit in a partial metric space may not be unique. For a sequence {v m } on (A, ρ), we denote by L({v m }) the set of limit points (if there exist any), We recall some results in the context of partial-metric spaces, necessary in our following considerations. (1.5) On a partial-metric space (A, ρ), a mapping O : A → A is continuous at v 0 if and only if for every e > 0, there exists δ > 0 such that

Definition 3 ([26]) Let
A be a non-empty set and s ≥ 1. A function ρ b : . In this case, we say that (A, ρ b , s) is a partial b-metric space.
Example 1 ( [26]) Let A be a non-empty set and v, ω ∈ A.
• if ρ is a partial metric on A, then the function ρ b defined as is a partial b-metric on A, with s = 2 λ-1 , for λ > 1.
• if b is a b-metric and ρ is a partial metric on A, then the function is a partial b-metric on A. (1.8) Remark 2 In [27] it is proved that a partial b-metric induces a b-metric, say δ b , with On the other hand, in [28], the notion of 0-ρ b -completeness was introduced and the relation between 0-ρ b -completeness and ρ b -completeness of a partial b-metric was established. (1.11) The next result is important in our future considerations.
for each m ∈ N, then the sequence {v m } is 0-ρ b -Cauchy.

Main results
We start with the definition of simulation function for partial b-metric spaces.
We shall denote by Z ψ b the family of all b-ψ-simulation functions.
Obviously, (η b1 ) holds. Now, considering two sequences {r n } and {t n } in (0, +∞) such that (2.1) holds, we have With the purpose to simplify the demonstrations, we prefer in the sequel, to discuss separately, the cases Then O admits exactly one fixed point.
Proof Let v 0 ∈ A be an arbitrary but fixed point and {v m } be the sequence in A defined as follows: (2.5) Moreover, by (2.3) we get Now, taking into account (η b1 ), the above inequality yields or, equivalently, Consequently, due to the monotony of the function ψ, we obtain , which is a contradiction (because s > 1). Therefore, for any m ∈ N we have Thus, by Lemma 7 we see that the sequence{v n } is a 0-ρ b -Cauchy sequence on the ρ b -complete partial bmetric space. Since by Lemma 5, the space is also 0-ρ b -complete, it follows that there Now, we claim that Assuming the contrary, we can find m 0 ∈ N such that which is a contradiction. Thus, there exists a subsequence {v m(l) } of {v m } such that . Therefore, letting l → +∞ and keeping (2.8) in mind we get On one hand, without loss of generality, we assume that v m = u, for infinitely many m ∈ N. Thus, which by (η b1 ) leads us to Taking into account the non-decreasing property of ψ On the other hand, Letting m → +∞ in the above inequality and keeping in mind (2.8) and (2.9) we get Therefore, s p lim m→+∞ ρ b (Ov m , Ou) = ρ b (u, Ou). Thus, letting r m = ρ b (Ov m , Ou) and t m = D A (v m , u), by (η b2 ) it follows lim sup m→+∞ η b (s p r m , t m ) < 0, which is a contradiction. Then As a last step, we establish uniqueness of the fixed point. Indeed, if we can find another point, z ∈ A, z = u such that z = Oz, which is a contradiction. Thus, u = z. It is easy to see that η b ∈ Z ψ b (by taking γ (t) = 15 16 in Example 2). We have and shall consider the following cases: 1.
and since the function ψ is non-decreasing, we get, for any m ∈ N, we get a contradiction, and then it follows that and by Lemma (7), we conclude that {v m } is a 0-ρ b -Cauchy on a ρ b -complete b-partialmetric space, and there exists u ∈ A such that lim m→+∞ v m = u. Taking into account the continuity of the mapping O, we have that is, u is a fixed point of the mapping O.
We claim that the fixed point of O is unique. Let u, z ∈ A be two different fixed point of O. Then which is a contradiction. Therefore, ρ b (u, z) = 0, that is (by Lemma 6), u = z.
Removing the condition 1 in Theorem 3, respectively, Theorem 4, we immediately obtain the next results.  (2.4). Then O has a unique fixed point. by (2.11). Then O has a unique fixed point.
which implies then O admits a unique fixed point.
in Theorem 3 and take into account Example 2.
in Theorem 4 and take into account Example 3.

Corollary 5 Let
then O admits a unique fixed point.
We will prove below results similar to those stated in Theorems 3, 4 that can be formulated for the case s = 1. (A, ρ) be a ρ b -complete partial-metric space and O : A → A be a mapping. If there exists a function η ∈ Z ψ such that

Theorem 5 Let
then O admits exactly one fixed point.
First of all, we claim that lim n→+∞ ρ(v m , v m+1 ) = 0. From (2.13), we have Consequently, we get which, since ψ is non-decreasing, implies Therefore, the sequence {ρ(v m , v m+1 )} is decreasing, so, we can find θ ≥ 0 such that lim m→+∞ ρ(v m , v m+1 ) = θ . On the other hand, it is easy to see that lim m→+∞ D 1 A (v m-1 , v m ) = θ , as well. Assuming that θ > 0, from (η 2 ) and (2.13) it follows that which is a contradiction. So, we found that We claim that {v m } is a Cauchy sequence. If we suppose that lim m,q→+∞ ρ(v m , v q ) = 0, there exist two subsequences {v m l }, {v q l } of the sequence {v m } and a number e > 0 such that ρ(v m l , v q l ) > e.
Moreover, by Lemma 1, we have Looking on the definition of the function D 1 A , we have and keeping in mind (2.15) and (2.16) we get On the other hand, by (2.15), we have (2.20) Thus, by the triangle inequality and taking into account (2.20), we get and then e 2 < ρ(v m l , v q l -1 ). Therefore, We shall prove that u = Ou. By (ρ b2 ), we get Thus, by the non-decreasing property of ψ, we obtain and using (2.21) we get ρ(u, Ou) = 0. Thus, u = Ou and u is a fixed point of O.
In order to show the uniqueness of the fixed point, let u, z ∈ A such that u = Ou and z = Oz. We have which is a contradiction. Thus, we conclude that u is the unique fixed point of O. Proof Let v 0 ∈ A and consider the sequence {v m }, with v m = Ov m-1 . We assume that ρ(v m , v m=1 ) > 0 for each m ∈ N because we remark that, on the contrary, if there exits l 0 such that v l 0 = v l 0 +1 = Ov l 0 , that is v l 0 is a fixed point for the mapping O, then by (2.23), for any terms v = v m and ω = v m+1 we have On the other hand, by (2.22), But ψ ∈ and then If for some m, max{ρ(v m+1 , v m+2 ), ρ(v m , v m+1 )} = ρ(v m+1 , v m+2 ) then (2.24) becomes ρ(v m+1 , v m+2 ) < ρ(v m+1 , v m+2 ), which is a contradiction. Then, for each m ≥ 0, max{ρ(v m+1 , v m+2 ), ρ(v m , v m+1 )} = ρ(v m , v m+1 ), the inequality (2.24) yields Thus, the sequence {ρ(v m , v m+1 )} is decreasing, so it is convergent (being bounded from below). In this case, we can find a real number u ≥ 0 such that lim m→+∞ ρ(v m , v m+1 ) = u. Assume that u > 0, let r m = ρ(v m+1 , v m+2 ) and t m = D 1 B (v m , v m+1 . Since Therefore, for l > max{n 1 , n 2 } we have and we can conclude e 2 ≤ ρ(v m l -1 , v q l ). Thus,