Fixed point for F⊥-weak contraction

In this paper, we establish some fixed point results for F⊥weak contraction in orthogonal metric space and we give an application for the solution of second order differential equation.


Introduction
In 2017, Gordeji et al. [3] introduced the idea of orthogonal metric spaces and gave the following definitions. Definition 1 ( [3]). Let X be a non-empty set and ⊥⊆ X × X be a binary relation. If ⊥ satisfies the condition there exists x 0 ; ((for all y; y ⊥ x 0 ) or (for all y; x 0 ⊥ y)), then (X, ⊥) is called an orthogonal set (briefly O-set).
Example 1 ( [3]). Let X be the collection of people in world. Define orthogonality as x ⊥ y if x can donate blood to y. As we know blood group AB + is universal acceptor and O − is universal donor so X is an O-set where orthogonal element is not unique.
Type You can give blood to You can recive blood to for all x ⊥ y and 0 < λ < 1.
It is clear that for x, y ∈ X such that T x = T y, the inequalities d(T x, T y) ≤ e −τ d(x, y) also holds, i.e. T is Banach contraction.
Using the concept of F -contraction, Wardowski [8] generalized the Banach Contraction Principle and proved that every F -contraction in a complete metric space has a unique fixed point. Later in 2014, Wardowski [9] introduced the concept of F -weak contraction and established some fixed point results.
Recently, Sawangsup et al. [6] generalized F -contraction in orthogonal metric space and proved some fixed point results for F ⊥ -contraction mappings.

Main Results
Now, we generalize the condition of Wardowski [9] and prove some results in orthogonal metric spaces.
Theorem 1. Let (X, ⊥, d) be an O−complete metric space and T be a ⊥preserving, F ⊥ -weak contraction, ⊥-continuous on X. Then, T has unique fixed point in X.
Proof. From the definition of orthogonality, it follows that If max{d(x n−1 , x n ), d(x n , x n+1 )} = d(x n , x n+1 ), for some n, using equation (3), we have which is a contradiction. Therefore, for all n ∈ N, we have Hence, we get Taking limit as n → ∞, we get Hence, by the (F 2) property By property (F 3), there exist j ∈ (0, 1) such that Taking limit and using equations (5) and (6), we get lim n→∞ n(d(x n , x n+1 )) j = 0.
Then, there exists n 1 ∈ N s.t. n(d(x n , x n+1 )) j ≤ 1, for all n ≥ n 1 , we get Using tiangular inequality and equation (7), for all m, n ∈ N s.t. m > n ≥ n 1 , we get Neeraj Garakoti, Mahesh C. Joshi, Rohit Kumar Hence, z is a fixed point of T . For the uniqueness of fixed point, if w is another fixed point of T , then from equation (3), we get a contradiction that τ ≤ 0. So, d(w, z) = 0, i.e., w = z. Define the binary relation ⊥ on X by x ⊥ y if xy = 0 or x. It is easy to see that 0 ⊥ y for all y ∈ X. Hence, (X, ⊥) is an O-set. As we know the set of Q with Euclidian metric is not a complete metric space. Hence, X is not a complete metric space but it is O-complete. Let T be a map defined as It is easy to see that T is ⊥-continuous and ⊥-preserving. But, T is not F ⊥ -contraction because for x = 1 and y = 3/4, where 1 is orthogonal for each y ∈ [0, 1] ∩ Q, Hence, by equation (2), we get τ < 0 which is contradiction. Now, we prove that T satisfies the condition of F ⊥ -weak contraction. Since, 0 is orthgonal for each y ∈ [0, 1] ∩ Q. Therefore, for x = 0 and y ∈ (0, 1) ∩ Q, Also, 1 is orthogonal for each y ∈ [0, 1] ∩ Q. For x = 1 and y ∈ (0, 1) ∩ Q Hence, T is F ⊥ -weak contraction for F (x) = ln x and τ = ln 2. By Theorem (1), T has a unique fixed point, i.e., x = 0.
Proof. For all x, y ∈ X, with x ⊥ y, we have Following the proof of theorem (1) we get the required result.
The following result is generalization of the result given in [5] where the existence of the fixed point for the mappings satisfying a contraction condition for the points of a closed ball was considered.
Theorem 2. Let (X, ⊥, d) be an O-complete metric space and T : X → X. If T is F ⊥ -weak contraction for all x, y ∈ B(x 0 , r), ⊥-preserving and ⊥continuous on X. Moreover, for any r > 0, Then there exists a point z ∈ B(x 0 , r) s.t. T z = z.
Proof. We construct a sequence x n such that x n = T x n−1 and show that it is O-Cauchy sequence. First we claim that x n ∈ B(x 0 , r) for all n. We have, Suppose, x 2 , x 3 , . . . , x n ∈ B(x 0 , r). Following the proof of Theorem (1), we have Hence, x n ∈ B(x 0 , r), for all n ∈ N. Again, following the proof of theorem (1), sequence {x n } is O-Cauchy sequence and {x n } → z ∈ B(x 0 , r). Since T is ⊥-continuous, we have T z = z.

Remark 2.
Taking (X, d) a complete metric space in Theorem 2, we get the following result.
T satisfies all the criteria for F (x) = ln(x) of Theorem 1. Hence, we get a unique solution of equation (9).