Solutions of boundary value problems on extended-Branciari b-distance

In this paper, we consider a new distance structure, extended Branciari b-distance, to combine and unify several distance notions and obtain fixed point results that cover several existing ones in the corresponding literature. As an application of our obtained result, we present a solution for a fourth-order differential equation boundary value problem.


Introduction
The distance notion is as old as the history of finding the writing. On the other hand, the abstract formulation of the notion of distance is relatively new. It was formulated by Fréchet in 1905 for the distance of sets. In the setting of points, it was discovered by Pompeiu and Hausdorff in 1914 under the name of metric. After that it has been improved, generalized, and extended in several ways. Among them, we mention the distance that was proposed by Branciari.
On the other hand, we recall the notion of extended b-metric space (simply, d e -metric space) introduced by Kamran et al. [32], which is the most general form of the concept of metric. We recollect the definition as well.

Definition 2 ([32]
) For a non-empty set S and a mapping ω : for all x, y, z ∈ S. The symbol (X, ρ e ) denotes a ρ e -metric space.
We shall combine these two notions, extended b-metric and Branciari distance, under the name of an extended Branciari b-distance space by the following definition.
for all x, y ∈ S and all distinct u, v ∈ S \ {x, y}. The couple of symbols (S, d e ) denotes an extended Branciari b-distance space (shortly, d e -metric space).
The function ω : S ×S → [1, ∞) is defined by ω(x, y) = 2 1 p for all x, y ∈ S. Then d e satisfies all the conditions of an extended Branciari b-distance space. Indeed if p = 1, the quadrilateral inequality trivially holds. So, let p > 1 and x = (x n ) n≥1 ; y = (y n ) n≥1 ; z = (z n ) n≥1 ; w = (w n ) n≥1 be sequences in S with y, w ∈ S \ {x, z}. Consider |x nz n | p ≤ p|x nz n | p = p|x nz n ||x nz n | p-1 = p |x ny n + y nw n + w nz n | |x nz n | p-1 ≤ p |x ny n | + |y nw n | + |w nz n | |x nz n | p-1 = p |x ny n ||x nz n | p-1 + |y nw n ||x nz n | p-1 Thus d e (x, z) ≤ ω(x, z)[d e (x, y) + d e (y, w) + d e (w, z)] and d e is an extended Branciari bdistance space.
We will prove the extended quadrilateral inequality only as the other conditions are clear.
Remark 1 If θ (x, y) = s for s ≥ 1, then we obtain the definition of Branciari b-distance and s = 1 yields the standard Branciari distance. As is known well, b-metric does not need to be continuous. Consequently, an extended Branciari b-distance is not necessarily continuous either. In this paper, we presume that each extended Branciari b-distance is continuous.
Motivated and inspired by the above concerns, we have organized the article as follows: • In Sect. 1, the concept of an extended Branciari b-distance space is introduced and needed definitions are presented. • In Sect. 2, various topics called Θ-Branciari contraction, Ćirić-Reich-Rus type Θ-Branciari contraction, and interpolative-Θ-Branciari contraction are introduced. By using these new contractions, we formulate and prove some fixed point theorems in the setting of extended Branciari b-distance spaces. • A supporting example is presented by using various sequences. • In Sect. 3, as an application, we present a solution for a fourth-order differential equation boundary value problem.

Main results
Now, we start this section by introducing the concept of Θ-Branciari contraction.
Definition 5 Let (S, d e ) be an extended Branciari b-distance space and T : X → X be a mapping. Then T is said to be a Θ-Branciari contraction if there exists a function θ ∈ Θ such that

Theorem 1 Let (S, d e ) be a complete extended Branciari b-distance and T : X → X be a Θ-Branciari contraction. Then T has a unique fixed point in S.
Proof For an arbitrary point x 0 ∈ S, we construct an iterative sequence {x n } 0 as follows: Suppose that if T n * x = T n * +1 x for some n * ∈ N, then T n * x is clearly a fixed point of T.
Hence, without loss of generality, we may assume that d e (T n x, T n+1 x) > 0 for all n ∈ N. From Definition 5, we have Recursively, we find that Accordingly, we obtain that Similarly, we can easily deduce that Suppose that l < ∞. In this case, let B = l 2 > 0. Using limit definition, we pick n 0 ∈ N such that for all n ≥ n 0 .
Then, we derive that Suppose that l = ∞. Let B > 0 be an arbitrary positive number.
Using the limit definition, we find n 0 ∈ N such that Thus, in all cases, there exist 1 B > 0 and n 0 ∈ N such that If we let n → ∞ in the above inequality, then we obtain Let N = max{n 0 , n 1 }. Due to the modified triangle inequality, we have two cases. For all n ≥ 1, we have two cases as follows. Case 1: Let x n = x m for some integers n = m. So, for m > n, we have T m-n (x n ) = x n . Choose y = x n and p = mn. Then T p y = y, that is, y is a periodic point of T. Thus, d e (y, Ty) = d e (T p y, T p+1 y). Thus, by the above argument, we can easily deduce that d e (y, Ty) = 0, so y = Ty, that is, y is a fixed point of T.
Case 2: Suppose that T n x = T m x for all integers n = m. Let n < m be two natural numbers. To show that {x n } is a Cauchy sequence, we need to consider two subcases as follows.
Subcase 1: We claim that if nm is odd, then d e (x n , x m ) converges to 0 as n, m → ∞ To prove this, we may assume that m = n + 2p + 1. Thus, This can be written as follows: which is convergent to 0 as n, m → ∞ and 1 q > 1. Subcase 2: We claim that if nm is even, then d e (x n , x m ) converges to 0 as n, m → ∞.
To prove this, we may assume that m = n + 2p. Thus, Hence, by the fact that sup m≥1 lim n→∞ ω(x n , x m ) < 1 r , subcase 1, (2.4), and (2.5), we can easily deduce that d e (x n , x m ) converges to 0 as n, m → ∞. Thus, the sequence {x n } in S is a Cauchy sequence.
Since (S, d e ) is a complete extended Branciari b-distance, there exists a point η in S such that {x n } converges to η.
This contradicts the assumption that T does not have a periodic point. Hence, assume that η is a periodic point of T with period q.
Suppose that the set of fixed points of T is empty.
Then we have d e (z, Tz) > 0 for all z ∈ S and d e (z, T q z) = 0 for q > 1.
Using Definition 5, we get Tz) , which leads to a contradiction. So, there exists a point η ∈ S such that Tη = η.
Suppose that f has another fixed point ζ such that η = ζ . Then clearly d e (η, ζ ) = d e (f η, f ζ ) = 0. Now, using condition (2.1), we get Therefore, η = ζ . This claims that T has a unique fixed point in S.
If we take ω(x, y) = b > 1 in Theorem 1, then we get the following corollary. (X, d). Suppose that there exist ϑ ∈ Θ and r ∈ (0, 1) such that

Then T has a unique fixed point in S.
If we take ω(x, y) = 1 in the above theorem, then we get the following corollary. Proof As in Theorem 1, we construct an iterative sequence {x n } 0 ∞ by starting an arbitrary point x 0 ∈ S as follows:

Corollary 2 Let T be a self-map on a complete
x n = f n x 0 for all n ∈ N. Now, using condition (2.11), we get a contradiction. Accordingly, we have η = ζ . Thus f has a unique fixed point in S.

Definition 7
Let (S, d e ) be an extended Branciari b-distance and f : S → S be a mapping.
Then f is said to be an interpolative-Θ-Branciari contraction if there exist a function θ ∈ Θ and non-negative real numbers r 1 , r 2 , r 3 with r 1 + r 2 + r 3 < 1 such that for all x, y ∈ S, where lim sup n,m→∞ ω(x n , x m ) < 1 r , here x n = f n x 0 for x 0 ∈ S and r ∈ (0, 1).

Theorem 3 Let (S, d e ) be a complete extended Branciari b-distance where d e is a continuous functional. If f : S → S is an interpolative-Θ-Branciari contraction, then f possesses a unique fixed point in S.
We skip the proof since θ d e (x, y) r 1 θ d e (x, fx) r 2 θ d e (y, fy) r 3 ≤ M θ,f (x, y) r 1 +r 2 +r 3 .
Thus, it is sufficient to choose r := r 1 + r 2 + r 3 < 1 in Theorem 2 to conclude the theorem above.
In Theorem 3, if we take r 2 = 0, r 3 = 0, then the above theorem reduces to the following. for all x, y ∈ S, (2.12) where r ∈ [0, 1) and lim sup m,n→∞ ω(x n , x m ) < 1 r , then f has a unique fixed point in S.
In Theorem 3, if we take r 1 = 0, r 2 = 0, r 3 = 0, then the above theorem reduces to the following one.

Existence of a solution of fourth-order differential equation
We consider the problem where g : [0, 1] × R 3 × R → R is a continuous function. This problem known as a boundary value problem (shortly, BVP) is employed to model such phenomena as deformations of an elastic beam in its equilibrium state, where one end-point is free while the other is fixed. In the discipline of mechanics, boundary value problem is said to be a Cantilever beam equation. Due to its significance in mathematics, the existence of solutions for such a problem plays a vital role. With this inspiration, we shall employ the fixed point technique to find the existence of solution of BVP.
Note that the space S = (C[0, 1], d e ) is a complete extended Branciari b-distance space. Now, we consider the above fourth-order ordinary differential equation boundary value problem. Then the problem BVP can be written in the following integral form:
Then the problem BVP has a solution in S.