Sehgal Type Contractions on Dislocated Spaces

: In this paper, we investigate the contractive type inequalities for the iteration of the mapping at a given point in the setting of dislocated metric space. We consider an example to illustrate the validity of the given result. Further, as an application, we propose a solution for a boundary value problem of the second order differential equation.


Introduction and Preliminaries
In 1968, Bryant relaxed the assumption of Banach contraction mapping principle by using an iteration of the mapping. Theorem 1 ([1]). Let T be a self mapping on the complete metric space (M, d), and m a positive integer. Suppose that there exists q ∈ [0, 1) such that T satisfies the inequality d(T m v, T m w) ≤ qd(v, w), for all v, w ∈ M, where T m denotes the mth iterate of T. Then, there exists exactly one fixed point of T.
After then a number of authors deepen the research by considering an iteration of the mapping, see e.g., [2][3][4][5][6][7]. We recollect some significant results in this direction. One of the pioneer report in this way was given Seghal [4]. Theorem 2 ([4]). Let (M, d) be a complete metric space, T a continuous self-mapping of M which satisfies the condition that there exists a real number q, 0 < q < 1 such that, for each v ∈ M there exists a positive integer m(v) such that, for each w ∈ M, d(T m(v) v, T m(v) w) ≤ qd(v, w).
Then T has a unique fixed point in M.
Guseman [2] extended this result by removing the condition of continuity of T and later, other extensions for a single mapping were discussed in several papers, see e.g., Iseki [8], Matkowski [3], Singh [5] and the reference therein. One of the most interesting results for mappings which satisfy a general contractive conditions were announced by Singh. Theorem 3 ([5]). Let (M, d) be a complete metric space and T : M → M be a mapping such that for all v, w ∈ M we can find a positive integer m(v) such that d (T m(v) where q(v, w), r(v, w), s(v, w), t(v, w), p(v, w) are nonnegative functions such that sup {2t(v, w) + q(v, w) + r(v, w) + s(v, w) + p(v, w)} = λ < 1.
Then T has a unique fixed point v * .
In this paper, we consider more general contractive condition in the setting of dislocated metric space. For sake of completeness, we shall recollect some basic notions and fundamental results.
The space (M, D) is said to be a dislocated metric space (DMS). It is obvious that any metric space is a dislocated metric space, but conversely this is not true. (a) convergent to a point v ∈ M if the following limit exists and is finite exists and is finite.
Proposition 1 ([9]). Let (M, D) be a DMS. For any v, w ∈ M we have the following Definition 5. By a comparison function we mean a function ϕ : [0, ∞) → [0, ∞) with the following properties: We denote by Φ the class of the comparison function ϕ Next we list some basic properties of the comparison functions.
Proposition 2 ( [10,11]). If ϕ is a comparison function then: We denote by Φ c the family of c−comparison functions.
It can be shown that every c−comparison function is a comparison function. Throughout this paper we denote by Ψ the collection of all c−comparison functions ψ : [0, ∞) → [0, ∞) that satisfy the following condition In the following we recall the concept of α-admissible mappings. A function T : M → M is said to be α-admissible if Later, the notion of α-admissible mapping and triangular α-admissible mappings are refined by Popescu [12], as follows: Definition 7 ([12]). Let T : M → M and α : M × M → [0, ∞) . We say that T is an α−orbital admissible mapping if for all v ∈ M we have Every α−admissible mapping is an α-orbital admissible mapping, for more details on admissible mapping, see e.g., [13][14][15][16][17][18][19][20][21][22][23][24]. At the end of this section, we present two further concepts that will be essential in our next considerations.
A set M is regular with respect to mapping α : M × M → [0, ∞) if the following condition is satisfied:

Main Results
We are now prepared to establish the main result of this paper. Suppose that for all v ∈ M we can find a positive integer m(v) such that for any w ∈ M Suppose also that: (i) T is triangular α-orbital admissible; (ii) there exists v 0 in M such that α(v 0 , Tv 0 ) ≥ 1; (iii) either T is continuous, or (iv) the M space is regular and α satisfies the condition (U).
Then the function T has exactly one fixed point.
Proof. Consider the initial value v 0 ∈ M and define a sequence {v n } as follows: If we denote m k = m(v k ) for any k ∈ N, then we can write v k+1 = T m k v k . Now, T is α-orbital admissible and α(v 0 , Tv 0 ) ≥ 1. Thus, from condition (O), we have α(Tv 0 , T 2 v 0 ) ≥ 1 and so forth Taking into account (TO) and (8) we easily infer that Recursively, we can conclude that for all m ∈ {1, 2, ...}.
In the initial inequality (6) letting v = v k−1 , w = T m k v k−1 and using (9) we can find a positive integer, m k−1 such that Since ψ ∈ Ψ, the condition (c f 1 ) is satisfied and applying (D3) we obtain Let Then from (11) together with (c f iii ) we get that Using the same arguments, we can find a positive integer m k−2 such that where Very easily we can see from (12), (13) and taking into account (cc f 1), that Since ψ is monotone increasing, by continuing this process, we find that for p 1 , p 2 , ..., p k ∈ N. On one hand the inequality (15) shows us, taking into account (c f 2) from On the other hand using triangle inequality, for l ∈ N, we have We should focus our attention on the set D(v 0 , T i v 0 ), i ∈ N . More precisely, we will show that this set is bounded. In order to prove that, we mention first that by hypothesis there exists a positive integer It is clear then that x 0 ≤ l < a and we will show that x i < a for all i ∈ N. We suppose the contrary, that there exists k ∈ N such that x k < a ≤ x k+1 . Note that (according to (6), (9) and triangle inequality) But, Since ψ is increasing, from (19) we get which contradicts (18). This contradiction shows that our assumption was false. Thus, for all i ∈ N We have thus demonstrated that the set D(v 0 , T im 0 +s v 0 ) : i ∈ N is bounded, and also, varying With this observation, we return to (17) and we get The series ∑ ∞ j=0 ψ j (r(v 0 )) is convergent due to (cc f 2 ) and its sequence of partial sums, denoted by {Sn}, is convergent at S. Then as k → ∞, and, therefore {v k } is a 0-Cauchy sequence. By completeness of (M, D), there is some point v * ∈ M such that lim From the continuity of T it easily follows that and by the uniqueness of the limit, we get Tv * = v * .
We claim now that v * is a fixed point of T under the hypothesis (iv). The first step in our proof (6), there exists m k−1 such that for Using triangle inequality, we have Then, from (25) it follows that Repeating this process and keeping in mind the properties (c f 1 ), (c f 2 ) we find that Since v k → v * as k → ∞ and the space M is regular, by triangle inequality we have Letting k → ∞ in the above inequality, and taking (24) respectively (27) into account, we find that Let w * ∈ M another point such that T m(v * ) w * = w * and v * = w * . Since T satisfies (6) and the function α satisfies the condition (U) we get But the above inequality is possible only if D(v * , w * ) = 0, that is v * = w * . This is a contradiction. From the uniqueness of the fixed point we can conclude that v * is a fixed point for T. Indeed, shows that Tv * is also fixed point of T m(v * ) . But, T m(v * ) has a unique fixed point v * . Hence, Tv * = v * .
On the other hand, we can note that: For this reason, there exists m(a) = 3 such that for any w ∈ M the condition (6) is satisfied (since D(a, a) = 0).
The conclusion is that T satisfies all the assumptions of Theorem 4. Therefore T has exactly one fixed point, v = a.
Taking, in Theorem 4, α(v, w) = 1 we get the following result: Then the function T has exactly one fixed point.
Then the function T has exactly one fixed point.
If we take ψ(x) = qx, q ∈ [0, 1) in Corollary 1 respectively in Corollary 2 we find the following consequences: Then the function T has exactly one fixed point.
Then the function T has exactly one fixed point.
If we take m(v) = 1 in Theorem 4 we get: Suppose also that: (i) T is triangular α-orbital admissible; (ii) there exists v 0 in M such that α(v 0 , Tv 0 ) ≥ 1; (iii) either T is continuous, or (iv) the space M is regular and the condition (U) is satisfied.
Then the function T has exactly one fixed point.
Let us first notice that v n = T n v = v 3 n → 0 for any v ∈ A and T n 1 = 1 3 n−1 → 0. Since α(v n , 0) = 2 we get that assumption (iv) of Theorem 4 is satisfied. Also, since α(0, 0) = 2 ≥ 1 by simple calculation we can conclude that the assumptions (ii) and (iv) are satisfied. We remark that if v = 1 3 and w = 1 2 then T 1 3 = 1 9 , , which shows us that T does not satisfy the contraction condition of Banach, neither condition (36) of Corollary 6. We must discuss the next cases: 2. If v = w = 1 2 we can choose m(v) = 4. Then, T 4 1 2 = 1 27 and ).
The other cases are not interesting since α(v, w) = 0. Therefore v = 0 is the unique fixed point for T.
Inspired by Proposition 3 from [7] we will establish a new fixed point result for a T function on a DMS, not necessarily complete. and for any a 1 , a 2 , a 3 ∈ R 0 and a 1 + 4a 2 + 4a 3 < 1.
Replacing v = v * in (38), using triangle inequality and keeping in mind (39) we have or, since a 1 + 4a 2 + 4a 3 < 1 we get This is a contradiction. Hence D(v * , Tv * ) = 0 and v * a is fixed point of T. Proof. There exists exactly one point v * ∈ M such that Tv * = v * , which means that v * is a unique solution of fixed point Equation (41). Let w * ∈ M. There exists m(w * ) ∈ {1, 2, ...} such that (42) holds. Keeping in mind the properties of function ψ, the condition imposed on the alpha function and using the triangle inequality we obtain

Ulam-Stability
Taking into account the definition of the function β we have From the assumption, η is continuous and strictly increasing. Thus, η −1 is also continuous and increasing, with η −1 (0) = 0 Therefore, the Equation (41) is generalized Ulam-stable.

Application to Boundary Value Problem
Here we consider the following two point boundary value problems for the second order differential equation.
Clearly, (M, D) is a complete DMS.
Author Contributions: All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Funding: This research received no external funding.