Applications to Boundary Value Problems and Homotopy Theory Via Tripled Fixed Point Techniques in Partially Metric Spaces

mathematics

Article Applications to Boundary Value Problems and Homotopy Theory via Tripled Fixed Point Techniques in Partially Metric Spaces

Hasanen A. Hammad 1 , Praveen Agarwal 2,3,4,5,6,* and Juan L. G. Guirao 7,8

1 Department of Mathematics, Faculty of Science, Sohag University, Sohag 82524, Egypt; [email protected] 2 Department of Mathematics, Anand International College of Engineering, Jaipur 302012, India 3 Nonlinear Dynamics Research Center (NDRC), Ajman University, Ajman AE 346, United Arab Emirates 4 Institute of Mathematical Modeling, Almaty 050010, Kazakhstan 5 International Center for Basic and Applied Sciences, Jaipur 302029, India 6 Harish-Chandra Research Institute (HRI), Allahabad 211019, India 7 Departamento de Matematica Aplicada y Estadistica, Universidad Politecnica de Cartagena, Hospital de Marina, 30203 Murcia, Spain; [email protected] or [email protected] 8 Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia * Correspondence: [email protected] or [email protected]

Abstract: In this manuscript, some tripled fixed point results were derived under (ϕ, ρ, `)-contraction in the framework of ordered partially metric spaces. Moreover, we furnish an example which supports our theorem. Furthermore, some results about a homotopy results are obtained. Finally, theoretical results are involved in some applications, such as finding the unique solution to the Citation: Hammad, H.A.; Agarwal, boundary value problems and homotopy theory. P.; Guirao, J.L.G. Applications to Boundary Value Problems and Keywords: tripled fixed point; boundary value problem; homotopy theory; partially ordered Homotopy Theory via Tripled Fixed metric space Point Techniques in Partially Metric Spaces. Mathematics 2021, 9, 2012. MSC: 54H25; 47H10; 34B24 https://doi.org/10.3390/math9162012

Academic Editors: Christopher Goodrich and Hsien-Chung Wu 1. Introduction Fixed point theory is one of the important and indispensable branches of non-linear Received: 9 June 2021 analysis due to the proliferation of its applications in many disciplines such as engineer- Accepted: 14 August 2021 Published: 23 August 2021 ing, computer science, physics, economics, biology, chemistry, etc. In mathematics, this technique is credited with clarifying and studying the behavior of dynamical systems, sta-

Publisher’s Note: MDPI stays neutral tistical methods, game theory models, differential equations, and many others. Specifically, with regard to jurisdictional claims in this technique studies the existence and uniqueness of the solution to many integral and published maps and institutional affil- fractional equations, which facilitates the way to find numerical solutions to such problems, iations. see [1–11]. Homotopy theory is a fundamental branch of algebraic topology where topological objects are studied up to homotopy equivalence. In the last century, strong links have emerged between this theory and many branches of mathematics. For example, this trend plays a prominent role in strengthening ties between homotopy theory and Copyright: © 2021 by the authors. Licensee MDPI, Basel, Switzerland. category theory (higher-dimensional), which have received considerable attention in re- This article is an open access article cent years, see [12–15]. Furthermore, it is useful in quantum mechanics for dealing with distributed under the terms and Hamiltonian manifolds. conditions of the Creative Commons The idea of a partially metric space (PMS) was presented by Matthews [16] as a part Attribution (CC BY) license (https:// of the study of denotational semantics of data flow networks. In fact, it is widely shared creativecommons.org/licenses/by/ that PMSs play a prominent role in model building in computational and field theory in 4.0/). computer science, see [17–22].

Mathematics 2021, 9, 2012. https://doi.org/10.3390/math9162012 https://www.mdpi.com/journal/mathematics Mathematics 2021, 9, 2012 2 of 22

Fixed point results for a single mapping in partially ordered metric space (POMSs) were introduced by Matthews [16,23], Oltra and Valero [24] and Altun et al. [25]. For more papers in common fixed point consequences in abstract spaces, we mention [26–32]. In [33], the notion of a coupled fixed point was presented and some pivotal results from it in the setting of PMSs are studied. Later, coupled fixed and coupled common fixed point theorems are proved by [34–37]. In 2011, Berinde and Borcut [38] generalized the idea of a coupled fixed point to a tripled fixed point (TFP) in the setting of POMSs. Under this space, Borcut [39,40], Karapnar et al. [41], Radenović[42] and Aydi et al. [43] introduced some circular theorems concerning TFP theorems. Moreover, more applications in this line are presented by Mustafa et al. [44] and Hammad and De la Sen [45,46]. The purpose of this manuscript is to prove some TFP consequences via (ϕ, ρ, `)- contraction in the setting of ordered PMSs. In addition, to support our theoretical results, we give an example. Moreover, as applications, the existence and uniqueness of the solution to an initial value problem (IVP) and a homotopy theory are discussed.

2. Preliminaries This part is devoted to recall the standard deﬁnition of a homotopy and some elemen- tary properties for PMSs. A homotopy between two functions is deﬁned as follows: consider two continuous functions from a topological space to another; the two functions are considered homotopic if one can be continuously deformed into the other. This deformity is called a homotopy between the two functions. It can be formulated as follows:

Deﬁnition 1 ([12]). Assume that U, V : D → E are continuous mappings deﬁned on topological spaces D and E. Then the continuous function H : D × [0, 1] → E so that H(ϑ, 0) = Uϑ and H(ϑ, 1) = Vϑ is called a homotopy from U to V. Furthermore, U and V are called homotopic mappings.

+ Deﬁnition 2 ([16]). Let k 6= ∅. The function Λ : k × k → R is called a partial metric on k if for all ϑ1, ϑ2, ϑ3 ∈ k, the hypotheses below hold:

(Λ1) ϑ1 = ϑ2 iff Λ(ϑ1, ϑ1) = Λ(ϑ1, ϑ2) = Λ(ϑ2, ϑ2); (Λ2) Λ(ϑ1, ϑ1) ≤ Λ(ϑ1, ϑ2) and Λ(ϑ2, ϑ2) ≤ Λ(ϑ1, ϑ2); (Λ3) Λ(ϑ1, ϑ2) = Λ(ϑ2, ϑ1); (Λ4) Λ(ϑ1, ϑ2) ≤ Λ(ϑ1, ϑ3) + Λ(ϑ3, ϑ2) − Λ(ϑ3, ϑ3). We say that (k, Λ) is a PMS.

+ It should be noted that if Λ is partial metric on k, then the function ΩΛ : k × k → R , described by ΩΛ(ϑ1, ϑ2) ≤ 2Λ(ϑ1, ϑ2) − Λ(ϑ1, ϑ1) − Λ(ϑ2, ϑ2), is a metric on k.

Example 1 ([23]). Let k = [0, ∞). The pair (k, Λ) is a PMS under the distance Λ(ϑ1, ϑ2) = max{ϑ1, ϑ2}. It is obvious that Λ is not a usual metric and in this case ΩΛ(ϑ1, ϑ2) = |ϑ1 − ϑ2|.

Example 2. Assume k = {[ϑ1, ϑ2] : ϑ1, ϑ2 ∈ R, ϑ1 ≤ ϑ2} and deﬁne

Λ([ϑ1, ϑ3], [ϑ2, ϑ4]) = max{ϑ3, ϑ4} − min{ϑ1, ϑ2}.

Then (k, Λ) is a PMS.

Every partial metric Λ on k produces a Υ0 topology χΛ on k. The base of a topology χΛ is a family of open Λ-balls {OΛ(z, e), z ∈ k, e > 0}, where OΛ(z, e) = {ϑ ∈ k : Λ(z, ϑ) < Λ(z, z) + e}, ∀z ∈ k and e > 0. Mathematics 2021, 9, 2012 3 of 22

Now, we state some topological properties on PMSs.

Deﬁnition 3 ([16]). A sequence {ϑω} in a PMS (k, Λ) is called: ω (1) converges to the limit ϑ iff Λ(ϑ, ϑ) = limω→∞ Λ(ϑ, ϑ ). ν ω (2) a Cauchy sequence if limν,ω→∞ Λ(ϑ , ϑ ) exists and is ﬁnite.

Deﬁnition 4 ([16]). (i) A PMS (k, Λ) is called complete if every Cauchy sequence {ϑω} in k ν ω converges (with respect to χΛ) to a point ϑ ∈ k so that Λ(ϑ, ϑ) = limν,ω→∞ Λ(ϑ , ϑ ). (ii) A mapping ξ : k → k is said to be continuous at ϑ0 ∈ k, if ∀e > 0, ∃δ > 0 so that 0 0 ξ OΛ(ϑ , δ) ⊆ OΛ(ξϑ , e).

The following lemmas are very important in the following.

Lemma 1 ([16]). (i) We say that {ϑω} is a Cauchy sequence in the PMS (k, Λ) if it is a Cauchy sequence in the metric space (k, ΩΛ). (ii) If the metric space (k, ΩΛ) is complete, the PMS (k, Λ) is too. Furthermore,

ω ω υ ω lim ΩΛ(ϑ, ϑ ) = 0 ⇔ Λ(ϑ, ϑ) = lim ΩΛ(ϑ, ϑ ) = lim ΩΛ(ϑ , ϑ ). ω→∞ ω→∞ ω,υ→∞

ω ω Lemma 2 ([47]). Suppose that (k, Λ) is a PMS. If {ϑ } ∈ k so that limω→∞ ϑ = r and ω Λ(r, r) = 0. Then limω→∞ Λ(ϑ , s) = limω→∞ Λ(r, s) for every r, s ∈ k.

Lemma 3 ([47]). Assume that (k, Λ) is a PMS. Then

(a) If Λ(ϑ1, ϑ2) = 0, then ϑ1 = ϑ2; (b) If ϑ1 6= ϑ2, then Λ(ϑ1, ϑ2) > 0.

Remark 1 ([47]). If ϑ1 = ϑ2, then Λ(ϑ1, ϑ2) may not be 0.

Deﬁnition 5 ([46]). A mapping ξ : k3 → k (where k × k × k = k3) on a partially ordered set (POS) (k, ), has a mixed-monotone property, if for any ϑ, θ, ω ∈ k,

ϑ1, ϑ2 ∈ k, ϑ1 ϑ2 ⇒ ξ(ϑ1, θ, ω) ξ(ϑ2, θ, ω), θ1, θ2 ∈ k, θ1 θ2 ⇒ ξ(ϑ, θ1, ω) ξ(ϑ, θ2, ω), ω1, ω2 ∈ k, ω1 ω2 ⇒ ξ(ϑ, θ, ω1) ξ(ϑ, θ, ω2).

∗ Deﬁnition 6 ([43]). A mapping ξ : k3 → k on a POS (k, ) has a mixed ξ −monotone property ∗ where ξ : k → k, if for any ϑ, θ, ω ∈ k,

∗ ∗ ϑ1, ϑ2 ∈ k, ξ ϑ1 ξ ϑ2 ⇒ ξ(ϑ1, θ, ω) ξ(ϑ2, θ, ω), ∗ ∗ θ1, θ2 ∈ k, ξ θ1 ξ θ2 ⇒ ξ(ϑ, θ1, ω) ξ(ϑ, θ2, ω), ∗ ∗ ω1, ω2 ∈ k, ξ ω1 ξ ω2 ⇒ ξ(ϑ, θ, ω1) ξ(ϑ, θ, ω2).

Deﬁnition 7 ([39]). We say that a trio (ϑ, θ, ω) ∈ k3 is a TFP of the self-mapping ξ : k3 → k if ϑ = ξ(ϑ, θ, ω), θ = ξ(θ, ω, ϑ) and ω = ξ(ω, ϑ, θ).

Deﬁnition 8 ([40]). A trio (ϑ, θ, ω) ∈ k3 is called a tripled coincidence point (TCP) of the ∗ ∗ ∗ two self-mappings ξ : k3 → k and ξ : k → k if ξ ϑ = ξ(ϑ, θ, ω), ξ θ = ξ(θ, ω, ϑ) and ξ∗ω = ξ(ω, ϑ, θ).

Deﬁnition 9 ([40]). Assume that k 6= ∅ is a set, a trio (ϑ, θ, ω) ∈ k3 is called a common ∗ ∗ ∗ TFP of ξ : k3 → k and ξ : k → k, if ϑ = ξ ϑ = ξ(ϑ, θ, ω), θ = ξ θ = ξ(θ, ω, ϑ) and ω = ξ∗ω = ξ(ω, ϑ, θ). Mathematics 2021, 9, 2012 4 of 22

∗ Deﬁnition 10 ([48]). The mappings ξ : k3 → k and ξ : k → k are called w−compatible if ξ∗(ξ(ϑ, θ, ω)) = ξ(ξ∗ϑ, ξ∗θ, ξ∗ω), ξ∗(ξ(θ, ω, ϑ)) = ξ(ξ∗θ, ξ∗ω, ξ∗ϑ) and ξ∗(ξ(ω, ϑ, θ)) = ξ(ξ∗ω, ξ∗ϑ, ξ∗θ), whenever ξ∗ϑ = ξ(ϑ, θ, ω), ξ∗θ = ξ(θ, ω, ϑ) and ξ∗ω = ξ(ω, ϑ, θ).

3. Theorems and Discussion We begin this part with the following deﬁnition:

∗ Deﬁnition 11. Assume that (k, Λ) is a PMS, ξ : k3 → k and ξ : k → k are mappings. We say that ξ veriﬁes a (ϕ, ρ, `)−contraction w.r.t ξ∗ if there are ϕ, ρ, ` : [0, ∞) → [0, ∞) verifying the assertions below:

(a1) ϕ is continuous and monotonically non-decreasing, ρ is continuous and ` is lower semi continuous (LSC); (a2) ϕ(ν) = 0 iff ν = 0, ρ(0) = `(0) = 0; (a3) ϕ(ν) − ρ(ν) + `(ν) > 0; ∗ ∗ ∗ ∗ ∗ ∗ (a4) for all θ1, θ2, θ3, ϑ1, ϑ2, ϑ3 ∈ k with ξ θ1 ξ ϑ1, ξ θ2 ξ ϑ2 and ξ θ3 ξ ϑ3, we have

ϕ(Λ(ξ(θ1, θ2, θ3), ξ(ϑ1, ϑ2, ϑ3))) ≤ ρ(i(θ1, θ2, θ3, ϑ1, ϑ2, ϑ3)) − `(i(θ1, θ2, θ3, ϑ1, ϑ2, ϑ3)),

where

i(θ1, θ2, θ3, ϑ1, ϑ2, ϑ3)  ∗ ∗ ∗ ∗ ∗ ∗  Λ(ξ θ1, ξ ϑ1), Λ(ξ θ2, ξ ϑ2), Λ(ξ θ3, ξ ϑ3),  ∗ ∗ ∗   Λ(ξ θ1, ξ(θ1, θ2, θ3)), Λ(ξ θ2, ξ(θ2, θ3, θ1)), Λ(ξ θ3, ξ(θ3, θ1, θ2)),   ∗ ∗ ∗   Λ(ξ ϑ1, ξ(ϑ1, ϑ2, ϑ3)), Λ(ξ ϑ2, ξ(ϑ2, ϑ3, ϑ1)), Λ(ξ ϑ3, ξ(ϑ3, ϑ1, ϑ2)),   ∗ ∗   Λ(ξ θ1,ξ(θ1,θ2,θ3))Λ(ξ θ2,ξ(θ2,θ3,θ1))  1+Λ(ξ∗θ ,ξ∗ϑ )+Λ(ξ∗θ ,ξ∗ϑ )+Λ(ξ(θ ,θ ,θ ),ξ(ϑ ,ϑ ,ϑ )) , = max 1 1∗ 2 2 ∗ 1 2 3 1 2 3 .  Λ(ξ θ1,ξ(θ1,θ2,θ3))Λ(ξ θ3,ξ(θ3,θ1,θ2))   1+Λ(ξ∗θ ,ξ∗ϑ )Λ(ξ∗θ ,ξ∗ϑ )+Λ(ξ(θ ,θ ,θ ),ξ(ϑ ,ϑ ,ϑ )) ,   1 ∗ 1 3 3 ∗ 1 2 3 1 2 3   Λ(ξ ϑ1,ξ(ϑ1,ϑ2,ϑ3))Λ(ξ ϑ2,ξ(ϑ2,ϑ3,ϑ1))   1+Λ(ξ∗θ ,ξ∗ϑ )+Λ(ξ∗θ ,ξ∗ϑ )+Λ(ξ(θ ,θ ,θ ),ξ(ϑ ,ϑ ,ϑ )) ,   1 ∗1 2 2 ∗ 1 2 3 1 2 3   Λ(ξ ϑ1,ξ(ϑ1,ϑ2,ϑ3))Λ(ξ ϑ3,ξ(ϑ3,ϑ1,ϑ2))   ∗ ∗ ∗ ∗  1+Λ(ξ θ1,ξ ϑ1)+Λ(ξ θ2,ξ ϑ2)+Λ(ξ(θ1,θ2,θ3),ξ(ϑ1,ϑ2,ϑ3))

Theorem 1. Let (k, ) be a POS and Λ be a partial metric so that (k, Λ) is a PMS. Assume that ∗ ξ : k3 → k and ξ : k → k are mappings which satisfy ∗ (Bi) ξ veriﬁes a (ϕ, ρ, `)−contraction with respect to ξ ; 3 ∗ ∗ (Bii) ξ k ⊆ ξ (k) and ξ (k) is a complete subspace of k; ∗ (Biii)ξ has a mixed ξ −monotone property; n n n n (Biv)(1) if non-decreasing sequences {θ1 } → θ1 and {θ3 } → θ3, then θ1 θ1 and θ3 θ3 for all n; n n (2) if a non-increasing sequence {θ2 } → θ2, then θ2 θ2 , for all n. 0 0 0 ∗ 0 0 0 0 ∗ 0 0 0 0 ∗ 0 If there are θ1, θ2, θ3 ∈ k so that ξ θ1 ξ θ1, θ2, θ3 , ξ θ2 ξ θ2, θ3, θ1 and ξ θ3 0 0 0 ∗ 3 ξ θ3, θ1, θ2 , then ξ and ξ have a TCP in k .

0 0 0 3 ∗ 0 0 0 0 ∗ 0 0 0 0 Proof. Assume that θ1, θ2, θ3 ∈ k such that ξ θ1 ξ θ1, θ2, θ3 , ξ θ2 ξ θ2, θ3, θ1 and ∗ 0 0 0 0 3 ∗ 1 1 1 ξ θ3 ξ θ3, θ1, θ2 . Because ξ k ⊆ ξ (k), we select θ1, θ2, θ3 ∈ k so that

∗ 0 0 0 0 ∗ 1 ∗ 0 0 0 0 ∗ 1 ∗ 0 0 0 0 ∗ 1 ξ θ1 ξ θ1, θ2, θ3 = ξ θ1, ξ θ2 ξ θ2, θ3, θ1 = ξ θ2 and ξ θ3 ξ θ3, θ1, θ2 = ξ θ3,

2 2 2 and select θ1, θ2, θ3 ∈ k so that

1 1 1 ∗ 2 1 1 1 ∗ 2 1 1 1 ∗ 2 ξ θ1, θ2, θ3 = ξ θ1, ξ θ2, θ3, θ1 = ξ θ2 and ξ θ3, θ1, θ2 = ξ θ3.

Because ξ has the mixed ξ∗-monotone property, we obtain

∗ 0 ∗ 1 ∗ 2 ∗ 0 ∗ 1 ∗ 2 ∗ 0 ∗ 1 ∗ 3 ξ θ1 ξ θ1 ξ θ1, ξ θ2 ξ θ2 ξ θ2 and ξ θ3 ξ θ3 ξ θ3. Mathematics 2021, 9, 2012 5 of 22

ω ω ω Continuing with the same scenario, we build the sequences {θ1 }, {θ2 } and {θ3 } in k so that

∗ ω+1 ω ω ω ∗ ω+1 ω ω ω ∗ ω+1 ω ω ω ξ θ1 = ξ(θ1 , θ2 , θ3 ), ξ θ2 = ξ(θ2 , θ3 , θ1 ) and ξ θ3 = ξ(θ3 , θ1 , θ2 ), ω = 0, 1, 2, . . . with

∗ 0 ∗ 1 ∗ 2 ξ θ1 ξ θ1 ξ θ1 · · · , ∗ 0 ∗ 1 ∗ 2 ξ θ2 ξ θ2 ξ θ2 · · · , ∗ 0 ∗ 1 ∗ 3 and ξ θ3 ξ θ3 ξ θ3 · · · .

∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 Now, if ξ θ1 = ξ θ1 , ξ θ2 = ξ θ2 and ξ θ3 = ξ θ3 , for some ω, then a ω ω ω 3 ∗ ω ∗ ω+1 trio (θ1 , θ2 , θ3 ) is a TCP in k and nothing proof. So, we assume that ξ θ1 = ξ θ1 or ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ξ θ2 = ξ θ2 or ξ θ3 = ξ θ3 , for all ω. Because ξ θ1 ξ θ1 , ξ θ2 ξ θ2 and ∗ ω ∗ ω+1 ξ θ3 ξ θ3 , from assertion (a4), we can write

∗ ω ∗ ω+1 ω−1 ω−1 ω−1 ω ω ω ϕ Λ ξ θ1 , ξ θ1 = ϕ Λ ξ θ1 , θ2 , θ3 , ξ(θ1 , θ2 , θ3 ) ω−1 ω−1 ω−1 ω ω ω ω−1 ω−1 ω−1 ω ω ω ≤ ρ i θ1 , θ2 , θ3 , θ1 , θ2 , θ3 − ` i θ1 , θ2 , θ3 , θ1 , θ2 , θ3 ,

where

ω−1 ω−1 ω−1 ω ω ω i θ1 , θ2 , θ3 , θ1 , θ2 , θ3  − − −   Λ ξ∗θω 1, ξ∗θω , Λ ξ∗θω 1, ξ∗θω , Λ ξ∗θω 1, ξ∗θω ,   1 1 2 2 3 3   − − − − − − − −   Λ ξ∗θω 1, ξ θω 1, θω 1, θω 1 , Λ ξ∗θω 1, ξ θω 1, θω 1, θω 1 ,   1 1 2 3 2 2 3 1   − − − −   Λ ξ∗θω 1, ξ θω 1, θω 1, θω 1 , Λ ξ∗θω, ξ θω, θω, θω ,   3 3 1 2 1 1 2 3   ∗ ω ω ω ω ∗ ω ω ω ω   Λ ξ θ2 , ξ θ2 , θ3 , θ1 , Λ ξ θ3 , ξ θ3 , θ1 , θ2 ,   − − − − − − − −   Λ(ξ∗θω 1,ξ(θω 1,θω 1,θω 1))Λ(ξ∗θω 1,ξ(θω 1,θω 1,θω 1))  = max 1 1 2 3 2 2 3 1 , 1+Λ(ξ∗θω−1,ξ∗θω )+Λ(ξ∗θω−1,ξ∗θω )+Λ(ξ(θω−1,θω−1,θω−1),ξ(θω,θω,θω ))  1 1 2 2 1 2 3 1 2 3   Λ(ξ∗θω−1,ξ(θω−1,θω−1,θω−1))Λ(ξ∗θω−1,ξ(θω−1,θω−1,θω−1))   1 1 2 3 3 3 1 2 ,   1+Λ ξ∗θω−1,ξ∗θω +Λ ξ∗θω−1,ξ∗θω +Λ ξ θω−1,θω−1,θω−1 ,ξ θω,θω,θω   ( 1 1 ) ( 3 3 ) ( ( 1 2 3 ) ( 1 2 3 ))   ∗ ω ω ω ω ∗ ω ω ω ω   Λ(ξ θ1 ,ξ(θ1 ,θ2 ,θ3 ))Λ(ξ θ2 ,ξ(θ2 ,θ3 ,θ1 ))   ω−1 ω−1 ω−1 ω−1 ω−1 ,   1+Λ(ξ∗θ ,ξ∗θω )+Λ(ξ∗θ ,ξ∗θω )+Λ(ξ(θ ,θ ,θ ),ξ(θω,θω,θω ))   1 1 2 2 1 2 3 1 2 3   Λ ξ∗θω,ξ θω,θω,θω Λ ξ∗θω,ξ θω,θω,θω   ( 1 ( 1 2 3 )) ( 3 ( 3 1 2 ))   ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ω−1 ω−1 ω−1 ω ω ω  1+Λ(ξ θ1 ,ξ θ1 )+Λ(ξ θ2 ,ξ θ2 )+Λ(ξ(θ1 ,θ2 ,θ3 ),ξ(θ1 ,θ2 ,θ3 ))   ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω  Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,     ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,     ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,   ∗ ω−1 ∗ ∗ ω−1 ∗   Λ(ξ θ ,ξ θω )Λ(ξ θ ,ξ θω )   1 1 2 2  ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω ∗ ω+1 , = max 1+Λ(ξ θ1 ,ξ θ1 )+Λ(ξ θ2 ,ξ θ2 )+Λ(ξ θ1 ,ξ θ1 ) .  Λ(ξ∗θω−1,ξ∗θω )Λ(ξ∗θω−1,ξ∗θω )   1 1 3 3 ,   1+Λ ξ∗θω−1,ξ∗θω +Λ ξ∗θω−1,ξ∗θω +Λ ξ∗θω,ξ∗θω+1   ( 1 1 ) ( 3 3 ) ( 1 1 )   Λ(ξ∗θω,ξ∗θω+1)Λ(ξ∗θω,ξ∗θω+1)   1 1 2 2 ,   1+Λ(ξ∗θω−1,ξ∗θω )+Λ(ξ∗θω−1,ξ∗θω )+Λ(ξ∗θω,ξ∗θω+1)   1 1 2 2 1 1   Λ ξ∗θω,ξ∗θω+1 Λ ξ∗θω,ξ∗θω+1   ( 1 1 ) ( 3 3 )   ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω ∗ ω+1  1+Λ(ξ θ1 ,ξ θ1 )+Λ(ξ θ3 ,ξ θ3 )+Λ(ξ θ1 ,ξ θ1 )

However, Mathematics 2021, 9, 2012 6 of 22

Λ ξ∗θω−1, ξ∗θω Λ ξ∗θω−1, ξ∗θω 1 1 2 2 ∗ ω ∗ ω+1 ∗ ω−1 ∗ ω ≤ max{Λ ξ θ1 , ξ θ , Λ ξ θ , ξ θ1 }, ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω ∗ ω+1 1 1 1 + Λ ξ θ1 , ξ θ1 + Λ ξ θ2 , ξ θ2 + Λ ξ θ1 , ξ θ1 Λ ξ∗θω−1, ξ∗θω Λ ξ∗θω−1, ξ∗θω 1 1 3 3 ∗ ω ∗ ω+1 ∗ ω−1 ∗ ω ≤ max{Λ ξ θ1 , ξ θ , Λ ξ θ , ξ θ3 }, ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω ∗ ω+1 1 3 1 + Λ ξ θ1 , ξ θ1 + Λ ξ θ3 , ξ θ3 + Λ ξ θ1 , ξ θ1 Λ ξ∗θω, ξ∗θω+1 Λ ξ∗θω, ξ∗θω+1 1 1 2 2 ∗ ω ∗ ω+1 ≤ Λ ξ θ2 , ξ θ , ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω ∗ ω+1 2 1 + Λ ξ θ1 , ξ θ1 + Λ ξ θ2 , ξ θ2 + Λ ξ θ1 , ξ θ1

and Λ ξ∗θω, ξ∗θω+1 Λ ξ∗θω, ξ∗θω+1 1 1 3 3 ∗ ω ∗ ω+1 ≤ Λ ξ θ3 , ξ θ . ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω ∗ ω+1 3 1 + Λ ξ θ1 , ξ θ1 + Λ ξ θ3 , ξ θ3 + Λ ξ θ1 , ξ θ1 Then   ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω  Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,  θω−1, θω−1, θω−1, θω, θω, θω = max . i 1 2 3 1 2 3 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1  Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3  Hence

   ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,  ϕ Λ ξ∗θω, ξ∗θω+1 ≤ ρ max 1 1  ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3     ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,  −` max .  ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 

Analogously, for the second and third components, we can write

   ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,  ϕ Λ ξ∗θω, ξ∗θω+1 ≤ ρ max 2 2  ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3     ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,  −` max ,  ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 

and

   ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,  ϕ Λ ξ∗θω, ξ∗θω+1 ≤ ρ max 3 3  ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3     ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,  −` max .  ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3  Mathematics 2021, 9, 2012 7 of 22

∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 Set ∇ω = max{Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 }. Let us consider for all ω ≥ 1, ∇ω 6= 0. Moreover, let, if possible for some ω, ∇ω−1 < ∇ω. Now

∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ϕ(∇ω) = ϕ max{Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 } ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 = max{ϕ Λ ξ θ1 , ξ θ1 , ϕ Λ ξ θ2 , ξ θ2 , ϕ Λ ξ θ3 , ξ θ3 }    ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,  ≤ ρ max  ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3     ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω ∗ ω−1 ∗ ω   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,  −` max  ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 

= ρ(max{∇ω−1, ∇ω}) − `(max{∇ω−1, ∇ω}) = ρ(∇ω) − `(∇ω).

It follows from (a2) and (a3) that ∇ω = 0, a contradiction. Hence ∇ω−1 ≤ ∇ω. This proves that the sequence {∇ω} is a non-increasing and must converge to a real number v(say)≥ 0. Furthermore,

ϕ(∇ω) ≤ ρ(∇ω−1) − `(∇ω−1).

Passing ω → ∞, we have

ϕ(v) ≤ ρ(v) − `(v).

Based on assertions (a2) and (a3), we have v = 0. Thus lim max{Λ ξ∗θω, ξ∗θω+1 , Λ ξ∗θω, ξ∗θω+1 , Λ ξ∗θω, ξ∗θω+1 } = 0, ω→∞ 1 1 2 2 3 3

this leads to lim Λ ξ∗θω, ξ∗θω+1 = 0, lim Λ ξ∗θω, ξ∗θω+1 = 0 and lim Λ ξ∗θω, ξ∗θω+1 = 0. (1) ω→∞ 1 1 ω→∞ 2 2 ω→∞ 3 3

By (Λ2), one can write

lim Λ(ξ∗θω, ξ∗θω) = 0, lim Λ(ξ∗θω, ξ∗θω) = 0 and lim Λ(ξ∗θω, ξ∗θω) = 0. (2) ω→∞ 1 1 ω→∞ 2 2 ω→∞ 3 3

It follows from (1), (2) and the deﬁnition of ΩΛ that ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 lim ΩΛ ξ θ , ξ θ = 0, lim ΩΛ ξ θ , ξ θ = 0 and lim ΩΛ ξ θ , ξ θ = 0. (3) ω→∞ 1 1 ω→∞ 2 2 ω→∞ 3 3 ∗ ω ∗ ω Now, using the contradiction method, we’re going to prove that {ξ θ1 }, {ξ θ2 } ∗ ω ∗ ω ∗ ω ∗ ω and {ξ θ3 } are Cauchy sequences. Assume that {ξ θ1 } or {ξ θ2 } or {ξ θ3 } is not ∗ υ ∗ ω ∗ υ ∗ ω ∗ υ ∗ ω Cauchy. This means ΩΛ ξ θ1 , ξ θ1 9 0 or ΩΛ(ξ θ2 , ξ θ2 ) 9 0 or ΩΛ ξ θ3 , ξ θ3 9 0 as ω, υ → ∞. Consequently,

∗ υ ∗ ω ∗ υ ∗ ω ∗ υ ∗ ω max{ΩΛ(ξ θ1 , ξ θ1 ), ΩΛ(ξ θ2 , ξ θ2 ), ΩΛ(ξ θ3 , ξ θ3 )} 9 0, as ω, υ → ∞.

Then there is an e > 0 and monotone increasing sequences {υj} and {ωj} so that ωj > υj > j,

n ∗ υj ∗ ωj ∗ υj ∗ ωj ∗ υj ∗ ωj o max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3 ≥ e, (4) Mathematics 2021, 9, 2012 8 of 22

and

n ∗ υj ∗ ωj−1 ∗ υj ∗ ωj−1 ∗ υj ∗ ωj−1o max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3 < e. (5)

From (4) and (5), we obtain that

n ∗ υj ∗ ωj ∗ υj ∗ ωj ∗ υj ∗ ωj o e ≤ max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ υj ∗ ωj−1 ∗ υj ∗ ωj−1 ∗ υj ∗ ωj−1o ≤ max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ ωj−1 ∗ ωj ∗ ωj−1 ∗ ωj ∗ ωj−1 ∗ ωj o + max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ ωj−1 ∗ ωj ∗ ωj−1 ∗ ωj ∗ ωj−1 ∗ ωj o < e + max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3 .

As j → ∞ and applying (3), one sees that

n ∗ υj ∗ ωj ∗ υj ∗ ωj ∗ υj ∗ ωj o lim max ΩΛ ξ θ , ξ θ , ΩΛ ξ θ , ξ θ , ΩΛ ξ θ , ξ θ = e. (6) j→∞ 1 1 2 2 3 3

According to the deﬁnition of ΩΛ and (2), one can obtain

n υ ω υ ω υ ω o e lim max Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j = . (7) j→∞ 1 1 2 2 3 3 2

From (4), we ﬁnd that

n ∗ υj ∗ ωj ∗ υj ∗ ωj ∗ υj ∗ ωj o e ≤ max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ υj ∗ υj−1 ∗ υj ∗ υj−1 ∗ υj ∗ υj−1o ≤ max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj o + max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ υj ∗ υj−1 ∗ υj ∗ υj−1 ∗ υj ∗ υj−1o ≤ 2 max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj o + max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3 . (8)

Taking j → ∞ and applying (3), (6) and (8), we have

n ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj o lim max ΩΛ ξ θ , ξ θ , ΩΛ ξ θ , ξ θ , ΩΛ ξ θ , ξ θ = e. (9) j→∞ 1 1 2 2 3 3

Hence, we obtain

n υ −1 ω υ −1 ω υ −1 ω o e lim max Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j = . (10) j→∞ 1 1 2 2 3 3 2

From (5), we obtain that Mathematics 2021, 9, 2012 9 of 22

n ∗ υj ∗ ωj ∗ υj ∗ ωj ∗ υj ∗ ωj o e ≤ max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ υj ∗ υj−1 ∗ υj ∗ υj−1 ∗ υj ∗ υj−1o ≤ max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ υj−1 ∗ ωj+1 ∗ υj−1 ∗ ωj+1 ∗ υj−1 ∗ ωj+1o + max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ ωj+1 ∗ ωj ∗ ωj+1 ∗ ωj ∗ ωj+1 ∗ ωj o + max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ υj ∗ υj−1 ∗ υj ∗ υj−1 ∗ υj ∗ υj−1o ≤ 2 max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3 n ∗ υj ∗ ωj ∗ υj ∗ ωj ∗ υj ∗ ωj o + max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ ωj+1 ∗ ωj ∗ ωj+1 ∗ ωj ∗ ωj+1 ∗ ωj o +2 max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3 . (11)

Letting j → ∞ in (11), using (3) and (9), we have

n ∗ υj−1 ∗ ωj+1 ∗ υj−1 ∗ ωj+1 ∗ υj−1 ∗ ωj+1o lim max ΩΛ ξ θ , ξ θ , ΩΛ ξ θ , ξ θ , ΩΛ ξ θ , ξ θ = e. j→∞ 1 1 2 2 3 3

Hence, we obtain

n υ −1 ω +1 υ −1 ω +1 υ −1 ω +1o e lim max Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j = . j→∞ 1 1 2 2 3 3 2

Again, from (4), one can obtain

n ∗ υj ∗ ωj ∗ υj ∗ ωj ∗ υj ∗ ωj o e ≤ max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ υj ∗ ωj+1 ∗ υj ∗ ωj+1 ∗ υj ∗ ωj+1o ≤ max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

n ∗ ωj+1 ∗ ωj ∗ ωj+1 ∗ ωj ∗ ωj+1 ∗ ωj o + max ΩΛ ξ θ1 , ξ θ1 , ΩΛ ξ θ2 , ξ θ2 , ΩΛ ξ θ3 , ξ θ3

Setting j → ∞, from(2) and (3), one sees that

n ∗ υj ∗ ωj+1 ∗ υj ∗ ωj+1 ∗ υj ∗ ωj+1o e ≤ lim max ΩΛ ξ θ , ξ θ , ΩΛ ξ θ , ξ θ , ΩΛ ξ θ , ξ θ + 0 j→∞ 1 1 2 2 3 3

 υj ωj+1 υj υj ωj+1 ωj+1   2Λ ξ∗θ , ξ∗θ − Λ ξ∗θ , ξ∗θ − Λ ξ∗θ , ξ∗θ ,   1 1 1 1 1 1   υ ω +1 υ υ ω +1 ω +1  ≤ lim max 2Λ ξ∗θ j , ξ∗θ j − Λ ξ∗θ j , ξ∗θ j − Λ ξ∗θ j , ξ∗θ j , j→∞ 2 2 2 2 2 2  + + +   ∗ υj ∗ ωj 1 ∗ υj ∗ υj ∗ ωj 1 ∗ ωj 1   2Λ ξ θ3 , ξ θ3 − Λ ξ θ3 , ξ θ3 − Λ ξ θ3 , ξ θ3  n υ ω +1 υ ω +1 υ ω +1o = 2 lim max Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j . j→∞ 1 1 2 2 3 3

Thus e n υ ω +1 υ ω +1 υ ω +1o ≤ lim max Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j . 2 j→∞ 1 1 2 2 3 3

By the deﬁnition of ϕ,

e n υ ω +1 υ ω +1 υ ω +1o ϕ ≤ lim ϕ max Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j , Λ ξ∗θ j , ξ∗θ j (12) 2 j→∞ 1 1 2 2 3 3 n υ ω +1 υ ω +1 υ ω +1o = lim max ϕ Λ ξ∗θ j , ξ∗θ j , ϕ Λ ξ∗θ j , ξ∗θ j , ϕ Λ ξ∗θ j , ξ∗θ j . j→∞ 1 1 2 2 3 3

Consider Mathematics 2021, 9, 2012 10 of 22

∗ υj ∗ ωj+1 υj−1 υj−1 υj−1 ωj ωj ωj ϕ Λ ξ θ1 , ξ θ1 = ϕ Λ ξ θ1 , θ2 , θ3 , ξ θ1 , θ2 , θ3

υj−1 υj−1 υj−1 ωj ωj ωj υj−1 υj−1 υj−1 ωj ωj ωj ≤ ρ i θ1 , θ2 , θ3 , θ1 , θ2 , θ3 − ` i θ1 , θ2 , θ3 , θ1 , θ2 , θ3   ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,    − − −    ∗ υj 1 ∗ υj ∗ υj 1 ∗ υj ∗ υj 1 ∗ υj    Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,    + + +    ∗ ωj ∗ ωj 1 ∗ ωj ∗ ωj 1 ∗ ωj ∗ ωj 1    Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,      υj−1 υj υj−1 υj    Λ ξ∗θ ,ξ∗θ Λ ξ∗θ ,ξ∗θ    1 1 2 2    ,   υj−1 ωj υj−1 ωj υj ωj+1    1+Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ    1 1 2 2 1 1    υ − υ υ − υ    ∗ j 1 ∗ j ∗ j 1 ∗ j  = ρmax Λ ξ θ1 ,ξ θ1 Λ ξ θ3 ,ξ θ3   ,    υj−1 ωj υj−1 ωj υj ωj+1   1+Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ    1 1 3 3 1 1       ωj ωj+1 ωj ωj+1    Λ ξ∗θ ,ξ∗θ Λ ξ∗θ ,ξ∗θ    1 1 2 2   ,    υj−1 ωj υj−1 ωj υj ωj+1    1+Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ    1 1 2 2 1 1   ω ω + ω ω +    ∗ j ∗ j 1 ∗ j ∗ j 1    Λ ξ θ1 ,ξ θ1 Λ ξ θ3 ,ξ θ3       υ −1 ω υ −1 ω υ ω +1   ∗ j ∗ j ∗ j ∗ j ∗ j ∗ j   1+Λ ξ θ1 ,ξ θ1 +Λ ξ θ3 ,ξ θ3 +Λ ξ θ1 ,ξ θ1 

  ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj ∗ υj−1 ∗ ωj   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,    − − −    ∗ υj 1 ∗ υj ∗ υj 1 ∗ υj ∗ υj 1 ∗ υj    Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,    + + +    ∗ ωj ∗ ωj 1 ∗ ωj ∗ ωj 1 ∗ ωj ∗ ωj 1    Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,      υj−1 υj υj−1 υj    Λ ξ∗θ ,ξ∗θ Λ ξ∗θ ,ξ∗θ    1 1 2 2    ,   υj−1 ωj υj−1 ωj υj ωj+1    1+Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ    1 1 2 2 1 1    υ − υ υ − υ    ∗ j 1 ∗ j ∗ j 1 ∗ j  −`max Λ ξ θ1 ,ξ θ1 Λ ξ θ3 ,ξ θ3 .  ,    υj−1 ωj υj−1 ωj υj ωj+1   1+Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ    1 1 3 3 1 1       ωj ωj+1 ωj ωj+1    Λ ξ∗θ ,ξ∗θ Λ ξ∗θ ,ξ∗θ    1 1 2 2   ,    υj−1 ωj υj−1 ωj υj ωj+1    1+Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ +Λ ξ∗θ ,ξ∗θ    1 1 2 2 1 1   ω ω + ω ω +    ∗ j ∗ j 1 ∗ j ∗ j 1    Λ ξ θ1 ,ξ θ1 Λ ξ θ3 ,ξ θ3       υ −1 ω υ −1 ω υ ω +1   ∗ j ∗ j ∗ j ∗ j ∗ j ∗ j   1+Λ ξ θ1 ,ξ θ1 +Λ ξ θ3 ,ξ θ3 +Λ ξ θ1 ,ξ θ1 

Passing j → ∞, we can write

υ ω +1 e e lim ϕ Λ ξ∗θ j , ξ∗θ j ≤ ρ − ` . j→∞ 1 1 2 2

Analogously, we obtain

υ ω +1 e e lim ϕ Λ ξ∗θ j , ξ∗θ j ≤ ρ − ` , j→∞ 2 2 2 2

and υ ω +1 e e lim ϕ Λ ξ∗θ j , ξ∗θ j ≤ ρ − ` . j→∞ 3 3 2 2 So, from (12), one can write e e e ρ ≤ ρ − ` . 2 2 2 Mathematics 2021, 9, 2012 11 of 22

e ∗ ω It follows from hypotheses (a2) and (a3) that 2 = 0, a contradiction. Hence, {ξ θ1 }, ∗ ω ∗ ω {ξ θ2 } and {ξ θ3 } are Cauchy sequences in the metric space (k, ΩΛ). Therefore, ∗ υ ∗ ω ∗ υ ∗ ω ∗ υ ∗ ω ΩΛ ξ θ1 , ξ θ1 → 0, ΩΛ(ξ θ2 , ξ θ2 ) → 0 and ΩΛ ξ θ3 , ξ θ3 → 0 as υ, ω → ∞. Thus, by (2) and the deﬁnition of ΩΛ, we have

lim Λ(ξ∗θυ, ξ∗θω) = 0, lim Λ(ξ∗θυ, ξ∗θω) = 0 and lim Λ(ξ∗θυ, ξ∗θω) = 0. (13) υ,ω→∞ 1 1 υ,ω→∞ 2 2 υ,ω→∞ 3 3

∗ ∗ ω ∗ ω ∗ ω Because ξ (k) is a complete subspace of k and {ξ θ1 }, {ξ θ2 } and {ξ θ3 } are Cauchy ∗ ∗ ω ∗ ω ∗ ω sequences in a complete metric space (ξ (k), ΩΛ), then {ξ θ1 }, {ξ θ2 } and {ξ θ3 } con- ∗ verges to some $1, $2 and $3 in ξ (k), respectively. Thus,

∗ ω ∗ ω ∗ ω lim ΩΛ(ξ θ , $1) = 0, lim ΩΛ(ξ θ , $2) = 0 and lim ΩΛ(ξ θ , $3) = 0. ω→∞ 1 ω→∞ 2 ω→∞ 3

∗ ∗ ∗ for some $1, $2, $3 ∈ ξ (k). Since $1, $2, $3 ∈ ξ (k), there are θ1, θ2, θ3 ∈ k so that $1 = ξ θ1, ∗ ∗ ∗ ω ∗ ω ∗ ω $2 = ξ θ2 and $3 = ξ θ3. Because {ξ θ1 }, {ξ θ2 } and {ξ θ3 } are Cauchy sequences, ∗ ω ∗ ω ∗ ω ∗ ω+1 ∗ ω+1 then {ξ θ1 } → $1, {ξ θ2 } → $2, {ξ θ3 } → $3, {ξ θ1 } → $1, {ξ θ2 } → $2 and ∗ ω+1 {ξ θ3 } → $3. Applying Lemma1 (ii) and (13), we obtain

∗ ω Λ($1, $1) = lim Λ(ξ θ , $1) = Λ($2, $2) ω→∞ 1 ∗ ω ∗ ω = lim Λ(ξ θ , $2) = Λ($3, $3) = lim Λ(ξ θ , $3) = 0. (14) ω→∞ 2 ω→∞ 3

Next, we want to show that

∗ ω lim Λ(ξ(θ1, θ2, θ3), ξ θ ) = Λ(ξ(θ1, θ2, θ3), $1). ω→∞ 1

Based on deﬁnition of ΩΛ, we obtain

∗ ω ∗ ω ΩΛ(ξ(θ1, θ2, θ3), ξ θ1 ) = 2Λ(ξ(θ1, θ2, θ3), ξ θ1 ) ∗ ω ∗ ω −Λ(ξ(θ1, θ2, θ3), ξ(θ1, θ2, θ3)) − Λ(ξ θ1 , ξ θ1 ),

Passing ω → ∞ and using (2),

∗ ω ΩΛ(ξ(θ1, θ2, θ3), $1) = 2 lim Λ(ξ(θ1, θ2, θ3), ξ θ ) − Λ(ξ(θ1, θ2, θ3), ξ(θ1, θ2, θ3)) − 0 ω→∞ 1

According to deﬁnition of ΩΛ and (13), one can write

∗ ω lim Λ(ξ(θ1, θ2, θ3), ξ θ ) = Λ(ξ(θ1, θ2, θ3), $1), ω→∞ 1

Similarly ∗ ω lim Λ(ξ(θ2, θ3, θ1), ξ θ ) = Λ(ξ(θ2, θ3, θ1), $2), ω→∞ 2 and ∗ ω lim Λ(ξ(θ3, θ1, θ2), ξ θ ) = Λ(ξ(θ3, θ1, θ2), $3). ω→∞ 3

From (Λ4), we obtain

∗ ω+1 ∗ ω+1 Λ($1, ξ(θ1, θ2, θ3)) ≤ Λ $1, ξ θ1 + Λ ξ θ1 , ξ(θ1, θ2, θ3) .

Setting ω → ∞, we have

ω ω ω Λ($1, ξ(θ1, θ2, θ3)) ≤ 0 + lim Λ(ξ(θ , θ , θ ), ξ(θ1, θ2, θ3)). (15) ω→∞ 1 2 3 Mathematics 2021, 9, 2012 12 of 22

From (a1), (a2) and (15), one can obtain

ω ω ω ϕ(Λ($1, ξ(θ1, θ2, θ3))) ≤ lim ϕ(Λ(ξ(θ , θ , θ ), ξ(θ1, θ2, θ3))) ω→∞ 1 2 3 ω ω ω ω ω ω ≤ lim [ρ(i(θ , θ , θ , θ1, θ2, θ3)) − `(i(θ , θ , θ , θ1, θ2, θ3))], ω→∞ 1 2 3 1 2 3

where

ω ω ω i(θ1 , θ2 , θ3 , θ1, θ2, θ3)  ∗ ω ∗ ω ∗ ω  Λ ξ θ , $ , Λ(ξ θ , $2), Λ ξ θ , $3 ,  1 1 2 3   ∗ ω ∗ ω+1 ∗ ω ∗ ω+1 ∗ ω ∗ ω+1   Λ ξ θ1 , ξ θ1 , Λ ξ θ2 , ξ θ2 , Λ ξ θ3 , ξ θ3 ,     Λ($1, ξ(θ1, θ2, θ3)), Λ($2, ξ(θ2, θ3, θ1)), Λ($3, ξ(θ3, θ1, θ2)),   + +   Λ(ξ∗θω,ξ∗θω 1)Λ(ξ∗θω,ξ∗θω 1)   1 1 2 2 ,   1+Λ ξ∗θω,$ +Λ ξ∗θω,$ +Λ ξ∗θω+1,ξ(θ ,θ ,θ )  = max ( 1 1) ( 2 2) ( 1 1 2 3 ) Λ(ξ∗θω,ξ∗θω+1)Λ(ξ∗θω,ξ∗θω+1)  1 1 3 3 ,   1+Λ(ξ∗θω,$ )+Λ(ξ∗θω,$ )+Λ(ξ∗θω+1,ξ(θ ,θ ,θ ))   1 1 3 3 1 1 2 3   Λ($1,ξ(θ1,θ2,θ3))Λ($2,ξ(θ2,θ3,θ1))   + ,   1+Λ(ξ∗θω,$ )+Λ(ξ∗θω,$ )+Λ(ξ∗θω 1,ξ(θ ,θ ,θ ))   1 1 2 2 1 1 2 3   Λ($2,ξ(θ2,θ3,θ1))Λ($3,ξ(θ3,θ1,θ2))   ∗ ω ∗ ω ∗ ω+1  1+Λ(ξ θ1 ,$1)+Λ(ξ θ3 ,$3)+Λ(ξ θ1 ,ξ(θ1,θ2,θ3)) → max{Λ($1, ξ(θ1, θ2, θ3)), Λ($2, ξ(θ2, θ3, θ1)), Λ($3, ξ(θ3, θ1, θ2))} as ω → ∞.

Hence     Λ($1, ξ(θ1, θ2, θ3)),  ϕ(Λ($1, ξ(θ1, θ2, θ3))) ≤ ρmax Λ($2, ξ(θ2, θ3, θ1)),    Λ($3, ξ(θ3, θ1, θ2))     Λ($1, ξ(θ1, θ2, θ3)),  −`max Λ($2, ξ(θ2, θ3, θ1)), ,   Λ($3, ξ(θ3, θ1, θ2))

Analogously,     Λ($1, ξ(θ1, θ2, θ3)),  ϕ(Λ($2, ξ(θ2, θ3, θ1))) ≤ ρmax Λ($2, ξ(θ2, θ3, θ1)),    Λ($3, ξ(θ3, θ1, θ2))     Λ($1, ξ(θ1, θ2, θ3)),  −`max Λ($2, ξ(θ2, θ3, θ1)), ,   Λ($3, ξ(θ3, θ1, θ2))

and     Λ($1, ξ(θ1, θ2, θ3)),  ϕ(Λ($3, ξ(θ3, θ2, θ1))) ≤ ρmax Λ($2, ξ(θ2, θ3, θ1)),    Λ($3, ξ(θ3, θ1, θ2))     Λ($1, ξ(θ1, θ2, θ3)),  −`max Λ($2, ξ(θ2, θ3, θ1)), .   Λ($3, ξ(θ3, θ1, θ2)) Mathematics 2021, 9, 2012 13 of 22

Therefore       Λ($1, ξ(θ1, θ2, θ3)),   ϕ(Λ($1, ξ(θ1, θ2, θ3))),  ϕmax Λ($2, ξ(θ2, θ3, θ1)),  = max ϕ(Λ($2, ξ(θ2, θ3, θ1))),     Λ($3, ξ(θ3, θ2, θ1)) ϕ(Λ($3, ξ(θ3, θ2, θ1)))     Λ($1, ξ(θ1, θ2, θ3)),  ≤ ρmax Λ($2, ξ(θ2, θ3, θ1)),    Λ($3, ξ(θ3, θ1, θ2))     Λ($1, ξ(θ1, θ2, θ3)),  −`max Λ($2, ξ(θ2, θ3, θ1)), .   Λ($3, ξ(θ3, θ1, θ2))

This implies that

max{Λ($1, ξ(θ1, θ2, θ3)), Λ($2, ξ(θ2, θ3, θ1)), Λ($3, ξ(θ3, θ1, θ2))} = 0.

So $1 = ξ(θ1, θ2, θ3), $2 = ξ(θ2, θ3, θ1) and $3 = ξ(θ3, θ1, θ2). This leads to ξ(θ1, θ2, θ3) = ∗ ∗ ∗ ∗ ξ θ1 = $1, ξ(θ2, θ3, θ1) = ξ θ2 = $2 and ξ(θ3, θ1, θ2) = ξ θ3 = $3. Therefore, ξ and ξ have a coincidence point in k3. The following theorem gives the uniqueness of a TFP:

Theorem 2. Adding to hypotheses of Theorem1 the following hypothesis: 3 3 Let for each (θ1, θ2, θ3), (ϑ1, ϑ2, ϑ3) ∈ k there is a trio ($1, $2, $3) ∈ k so that a trio (ξ($1, $2, $3), ξ($2, $3, $1), ξ($3, $1, $2)) is comparable to (ξ(θ1, θ2, θ3), ξ(θ2, θ3, θ1), ξ(θ3, θ1, θ2)) and (ξ(ϑ1, ϑ2, ϑ3), ξ(ϑ2, ϑ3, ϑ1), ξ(ϑ3, ϑ1, ϑ2)). If (θ1, θ2, θ3) and (ϑ1, ϑ2, ϑ3) are TCPs of ξ and ξ∗ then

∗ ∗ ξ(θ1, θ2, θ3) = ξ θ1 = ξ ϑ1 = ξ(ϑ1, ϑ2, ϑ3), ∗ ∗ ξ(θ2, θ3, θ1) = ξ θ2 = ξ ϑ2 = ξ(ϑ2, ϑ3, ϑ1), ∗ ∗ and ξ(θ3, θ1, θ2) = ξ θ3 = ξ ϑ3 = ξ(ϑ3, ϑ1, ϑ2).

Furthermore, if ξ and ξ∗ are w−compatible, then there is a unique common TFP of ξ and ξ∗ in k3.

Proof. The proof follows immediately from Theorem1 and the concept of comparability.

The result below follows from Theorem1 and it is important in the next section.

Corollary 1. Let (k, ) be a POS and Λ be a partial metric so that (k, Λ) is a PMS. Assume that ξ : k3 → k is a mapping so that

ϕ(Λ(ξ(θ1, θ2, θ3), ξ(ϑ1, ϑ2, ϑ3))) ≤ ρ(max{Λ(θ1, ϑ1), Λ(θ2, ϑ2), Λ(θ3, ϑ3)})

−`(max{Λ(θ1, ϑ1), Λ(θ2, ϑ2), Λ(θ3, ϑ3)}), (16)

for all θ1, θ2, θ3, ϑ1, ϑ2, ϑ3 ∈ k with θ1 ϑ1, θ2 ϑ2 and θ3 ϑ3, where ϕ, ρ and ` are described in Deﬁnition 11 and n n n n (i) If non-decreasing sequences {θ1 } → θ1 and {θ3 } → θ3, then θ1 θ1 and θ3 θ3 for all n; n n (ii) If a non-increasing sequence {θ2 } → θ2, then θ2 θ2 , for all n. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 If there are θ1, θ2, θ3 ∈ k so that θ1 ξ θ1, θ2, θ3 , θ2 ξ θ2, θ3, θ1 and θ3 ξ θ3, θ1, θ2 , then ξ has a TCP in k3.

Example 3. Suppose that k = [0, 1]. Describe a partially ordered on k as

θ1 θ2 ⇐⇒ θ1 ≤ θ2. Mathematics 2021, 9, 2012 14 of 22

2 2 2 3 θ1 +θ2 +θ3 Deﬁne the mapping ξ : → by ξ(θ1, θ2, θ3) = and Λ : × → [0, ∞) k k 8(θ1+θ2+θ3+1) k k by Λ(θ1, θ2) = max{θ1, θ2}. It is clear that (k, Λ) is a PMS. Deﬁne ϕ, ρ, ` : [0, ∞) → [0, ∞) by 5ν ν ϕ(ν) = ν, ρ(ν) = 8 and `(ν) = 4 . Consider

Λ(ξ(θ1, θ2, θ3), ξ(ϑ1, ϑ2, ϑ3)) ( ) θ2 + θ2 + θ2 ϑ2 + ϑ2 + ϑ2 = max 1 2 3 , 1 2 3 8(θ1 + θ2 + θ3 + 1) 8(ϑ1 + ϑ2 + ϑ3 + 1) ( ) 1 θ2 ϑ2 = max 1 , 1 8 θ1 + θ2 + θ3 + 1 ϑ1 + ϑ2 + ϑ3 + 1 ( ) θ2 ϑ2 + max 2 , 2 θ1 + θ2 + θ3 + 1 ϑ1 + ϑ2 + ϑ3 + 1 ( )! θ2 ϑ2 + max 3 , 3 θ1 + θ2 + θ3 + 1 ϑ1 + ϑ2 + ϑ3 + 1 ( ) ( ) ( )! 1 θ2 ϑ2 θ2 ϑ2 θ2 ϑ2 ≤ max 1 , 1 + max 2 , 2 + max 3 , 3 8 θ1 + 1 ϑ1 + 1 θ2 + 1 ϑ2 + 1 θ3 + 1 ϑ3 + 1 1 θ ϑ θ ϑ θ ϑ ≤ max 1 , 1 + max 2 , 2 + max 3 , 3 8 θ1 + 1 ϑ1 + 1 θ2 + 1 ϑ2 + 1 θ3 + 1 ϑ3 + 1 1 ≤ (max{θ , ϑ } + max{θ , ϑ } + max{θ , ϑ }) 8 1 1 2 2 3 3 1 = (Λ(θ , ϑ ) + Λ(θ , ϑ ) + Λ(θ , ϑ )) 8 1 1 2 2 3 3 3 ≤ max{Λ(θ , ϑ ), Λ(θ , ϑ ), Λ(θ , ϑ )} 8 1 1 2 2 3 3 5 1 = max{Λ(θ , ϑ ), Λ(θ , ϑ ), Λ(θ , ϑ )} − max{Λ(θ , ϑ ), Λ(θ , ϑ ), Λ(θ , ϑ )} 8 1 1 2 2 3 3 4 1 1 2 2 3 3 = ρ(max{Λ(θ1, ϑ1), Λ(θ2, ϑ2), Λ(θ3, ϑ3)}) − `(max{Λ(θ1, ϑ1), Λ(θ2, ϑ2), Λ(θ3, ϑ3)}).

Therefore, all assertions of Corollary1 are fulﬁlled and (0, 0, 0) is a unique TFP of ξ on k3.

4. Application to IVPs In the setting of PMSs, this part is devoted to discussing the existence of a uniqueness solution to the IVP below: 0 θ1(ν) = Ξ(ν, θ1(ν), θ1(ν), θ1(ν)), ν ∈ χ = [0, 1], 0 (17) θ1(0) = θ1,

0 0 0 0 θ1 θ1 θ1 θ1 0 where Ξ : χ × [ 5 , ∞) × [ 5 , ∞) × [ 5 , ∞) → [ 5 , ∞) is a continuous function for θ1 ∈ R. Now, we state and prove our main theorem in this part.

0 0 0 θ1 θ1 θ1 Theorem 3. Assume that the IVP (17) with Ξ ∈ C χ × [ 5 , ∞) × [ 5 , ∞) × [ 5 , ∞) and

 ν  R 16θ0  1 Ξ(γ, θ (γ), θ (γ), θ (γ))dγ − 1 ,   5 1 1 1 25  ν  0  Z  ν 0  1 R 16θ1 Ξ(γ, θ1(γ), θ2(γ), θ3(γ))dγ ≤ max 5 Ξ(γ, θ2(γ), θ2(γ), θ2(γ))dγ − 25 , .  0  0  ν   16θ0   1 R Ξ(γ, θ (γ), θ (γ), θ (γ))dγ − 1   5 3 3 3 25  0 Mathematics 2021, 9, 2012 15 of 22

0 θ1 Then the IVP (17) has a unique solution in C χ, [ 4 , ∞) .

Proof. The IVP (17) is equivalent to the following integral equation:

ν Z 0 θ1(ν) = θ1 + Ξ(γ, θ1(γ), θ2(γ), θ3(γ))dγ. (18) 0

0 0 0 0 θ1 θ1 θ1 θ1 Assume that k = C χ, [ 5 , ∞) and Λ(θ1, θ2) = max θ1 − 5 , θ2 − 5 , θ3 − 5 for 4ν 3ν θ1, θ2 ∈ k. Deﬁne ϕ, ρ, ` : [0, ∞) → [0, ∞) by ϕ(ν) = ν, ρ(ν) = 5 and `(ν) = 5 . Describe the mapping ξ : k3 → k as

ν Z 0 ξ(θ1, θ2, θ3)(ν) = θ1 + Ξ(γ, θ1(γ), θ2(γ), θ3(γ))dγ. 0

Now

Λ(ξ(θ1, θ2, θ3), ξ(ϑ1, ϑ2, ϑ3)) ( ) θ0 θ0 = max ξ(θ , θ , θ ) − 1 , ξ(ϑ , ϑ , ϑ ) − 1 1 2 3 5 1 2 3 5  ν ν  0 Z 0 Z  4θ1 4θ1  = max + Ξ(γ, θ1(γ), θ2(γ), θ3(γ))dγ, + Ξ(γ, ϑ1(γ), ϑ2(γ), ϑ3(γ))dγ  5 5  0 0   ν   16θ0   1 R ( ( ) ( ) ( )) − 1     5 Ξ γ, θ1 γ , θ1 γ , θ1 γ dγ 25 ,     0     ν    4θ0  R 16θ0    1 + 1 Ξ(γ, θ (γ), θ (γ), θ (γ))dγ − 1 ,   5 max 5 2 2 2 25 ,    0     ν     16θ0     1 R Ξ(γ, θ (γ), θ (γ), θ (γ))dγ − 1     5 3 3 3 25   0 ≤ max  ν  16θ0   1 R ( ( ) ( ) ( )) − 1     5 Ξ γ, ϑ1 γ , ϑ1 γ , ϑ1 γ dγ 25 ,     0     ν    4θ0  16θ0    1 1 R ( ( ) ( ) ( )) − 1   5 + max 5 Ξ γ, ϑ2 γ , ϑ2 γ , ϑ2 γ dγ 25 ,    0     ν     16θ0     1 R (γ ϑ (γ) ϑ (γ) ϑ (γ))dγ − 1     5 Ξ , 3 , 3 , 3 25   0  n θ (ν) θ0 ϑ (ν) θ0 o   max 1 − 1 , 1 − 1 ,   5 25 5 25   n ( ) θ0 ( ) θ0 o  = θ2 ν 1 ϑ2 ν 1 max max 5 − 25 , 5 − 25 ,  n 0 0 o   θ3(ν) θ1 ϑ3(ν) θ1   max 5 − 25 , 5 − 25   n θ0 θ0 o   max θ (ν) − 1 , ϑ (ν) − 1 ,   1 5 1 5  1  n θ0 θ0 o  = max max θ (ν) − 1 , ϑ (ν) − 1 , 5 2 5 2 5  n 0 0 o   θ1 θ1   max θ3(ν) − 5 , ϑ3(ν) − 5  1 = max{Λ(θ , ϑ ), Λ(θ , ϑ ), Λ(θ , ϑ )} 5 1 1 2 2 3 3 4 3 = max{Λ(θ , ϑ ), Λ(θ , ϑ ), Λ(θ , ϑ )} − max{Λ(θ , ϑ ), Λ(θ , ϑ ), Λ(θ , ϑ )} 5 1 1 2 2 3 3 5 1 1 2 2 3 3 = ρ(max{Λ(θ1, ϑ1), Λ(θ2, ϑ2), Λ(θ3, ϑ3)}) − `(max{Λ(θ1, ϑ1), Λ(θ2, ϑ2), Λ(θ3, ϑ3)}).

Therefore, ξ veriﬁes the stipulation (16) of Corollary1. Thus, ξ has a unique TFP (θ1, θ2, θ3) with θ1 = θ2 = θ3, which is a unique solution of the IVP (17). Mathematics 2021, 9, 2012 16 of 22

5. Application to a Homotopy Here, we discuss a unique solution to homotopy theory.

Theorem 4. Suppose that (k, Λ) is a complete PMS, A is an open subset of k and A is a closed subset of k so that A ⊆ A. Assume that H : A × A × A × [0, 1] → k is an operator satisﬁes the hypotheses below:

(ℵ1) θ1 6= H(θ1, θ2, θ3, σ), θ2 6= H(θ2, θ3, θ1, σ) and θ3 6= H(θ3, θ1, θ2, σ), for each θ1, θ2, θ3 ∈ ∂A (here ∂A refer to the boundary of A in k) and σ ∈ [0, 1]; (ℵ2)

ϕ(Λ(H(θ1, θ2, θ3, σ), H(ϑ1, ϑ2, ϑ3, σ))) ≤ ρ(max{Λ(θ1, ϑ1), Λ(θ2, ϑ2), Λ(θ3, ϑ3)})

−`(max{Λ(θ1, ϑ1), Λ(θ2, ϑ2), Λ(θ3, ϑ3)}),

for all θ1, θ2, θ3, ϑ1, ϑ2, ϑ3 ∈ A and σ ∈ [0, 1], where ϕ, ρ : [0, ∞) → [0, ∞) are continuous and non-decreasing and ` : [0, ∞) → [0, ∞) is an LSC with ϕ(ν) − ρ(ν) + `(ν) > 0, for ν > 0; (ℵ3) There exists C ≥ 0 so that

∗ ∗ Λ(H(θ1, θ2, θ3, σ), H(θ1, θ2, θ3, σ )) ≤ C|σ − σ |,

∗ for each θ1, θ2, θ3 ∈ A and σ, σ ∈ [0, 1]. Then H(., 0) has a TFP, whenever H(., 1) has a TFP.

Proof. Deﬁne the set

f = {σ ∈ [0, 1] : (θ1, θ2, θ3) = H(θ1, θ2, θ3, σ) for some θ1, θ2, θ3 ∈ A}.

Because H(., 0) has a TFP in ∇, we obtain 0 ∈ f, this proves that f 6= ∅. We claim that f is open and closed in [0, 1] so by the connectedness, we obtain f = [0, 1]. Consequently, H(., 1) has a TFP in ∇. Initially, we shall show that f is open and ω ∞ ω closed in [0, 1]. To do this, assume {σ }ω=1 ⊆ f with σ → σ ∈ [0, 1] as ω → ∞. It must be shown that σ ∈ f. ω ω ω ω ω ω ω ω Because σ ∈ f, for ω ≥ 1, there are θ1 , θ2 , θ3 ∈ A with $ = θ1 , θ2 , θ3 = ω ω ω ω H θ1 , θ2 , θ3 , σ . Consider

ω ω+1 ω ω ω ω ω+1 ω+1 ω+1 ω+1 Λ θ1 , θ1 = Λ H(θ1 , θ2 , θ3 , σ ), H θ1 , θ2 , θ3 , σ ω ω ω ω ω+1 ω+1 ω+1 ω ≤ Λ H(θ1 , θ2 , θ3 , σ ), H θ1 , θ2 , θ3 , σ ω+1 ω+1 ω+1 ω ω+1 ω+1 ω+1 ω+1 +Λ H θ1 , θ2 , θ3 , σ , H θ1 , θ2 , θ3 , σ ω+1 ω+1 ω+1 ω ω+1 ω+1 ω+1 ω −Λ H θ1 , θ2 , θ3 , σ , H θ1 , θ2 , θ3 , σ ≤ ( ω ω ω ω) ω+1 ω+1 ω+1 ω + ω − ω+1 Λ H θ1 , θ2 , θ3 , σ , H θ1 , θ2 , θ3 , σ C σ σ .

As ω → ∞ in the above inequality, we have lim Λ θω, θω+1 ≤ lim Λ H(θω, θω, θω, σω), H θω+1, θω+1, θω+1, σω + 0, ω→∞ 1 1 ω→∞ 1 2 3 1 2 3

Since ϕ is non-decreasing and continuous, we obtain

h i lim ϕ Λ θω, θω+1 ≤ lim ϕ Λ H(θω, θω, θω, σω), H θω+1, θω+1, θω+1, σω ω→∞ 1 1 ω→∞ 1 2 3 1 2 3  n ω ω+1 ω ω+1 ω ω+1o  ρ max Λ θ1 , θ1 , Λ θ2 , θ2 , Λ θ3 , θ3 ≤ lim  n o . ω→∞ ω ω+1 ω ω+1 ω ω+1 −` max Λ θ1 , θ1 , Λ θ2 , θ2 , Λ θ3 , θ3 Mathematics 2021, 9, 2012 17 of 22

Analogously,  n ω ω+1 ω ω+1 ω ω+1o  + ρ max Λ θ1 , θ1 , Λ θ2 , θ2 , Λ θ3 , θ3 lim ϕ Λ θω, θω 1 ≤ lim  n o , ω→∞ 2 2 ω→∞ ω ω+1 ω ω+1 ω ω+1 −` max Λ θ1 , θ1 , Λ θ2 , θ2 , Λ θ3 , θ3 and  n ω ω+1 ω ω+1 ω ω+1o  + ρ max Λ θ1 , θ1 , Λ θ2 , θ2 , Λ θ3 , θ3 lim ϕ Λ θω, θω 1 ≤ lim  n o . ω→∞ 3 3 ω→∞ ω ω+1 ω ω+1 ω ω+1 −` max Λ θ1 , θ1 , Λ θ2 , θ2 , Λ θ3 , θ3 This implies that lim Λ θω, θω+1 = 0, lim Λ θω, θω+1 = 0 and lim Λ θω, θω+1 = 0. (19) ω→∞ 1 1 ω→∞ 2 2 ω→∞ 3 3

It follows from (Λ2) that

lim Λ(θω, θω) = 0, lim Λ(θω, θω) = 0 and lim Λ(θω, θω) = 0. (20) ω→∞ 1 1 ω→∞ 2 2 ω→∞ 3 3

From the deﬁnition of ΩΛ, we can write

ω ω+1 ω ω+1 ω ω+1 lim ΩΛ θ , θ = 0, lim ΩΛ θ , θ = 0 and lim ΩΛ θ , θ = 0. (21) ω→∞ 1 1 ω→∞ 2 2 ω→∞ 3 3

ω ω ω ω In order to prove that {θ1 }, {θ2 } and {θ3 } are Cauchy sequences, assume that {θ1 } ω ω or {θ2 } or {θ3 } is not a Cauchy. Then there is an e > 0 and monotone increasing sequences {υj} and {ωj} so that ωj > υj > j,

n υj ωj υj ωj υj ωj o max ΩΛ θ1 , θ1 , ΩΛ θ2 , θ2 , ΩΛ θ3 , θ3 ≥ e, (22)

and n υj ωj−1 υj ωj−1 υj ωj−1o max ΩΛ θ1 , θ1 , ΩΛ θ2 , θ2 , ΩΛ θ3 , θ3 < e. (23) Using (22) and (23), we obtain

n υj ωj υj ωj υj ωj o e ≤ max ΩΛ θ1 , θ1 , ΩΛ θ2 , θ2 , ΩΛ θ3 , θ3

n υj ωj−1 υj ωj−1 υj ωj−1o ≤ max ΩΛ θ1 , θ1 , ΩΛ θ2 , θ2 , ΩΛ θ3 , θ3

n ωj−1 ωj ωj−1 ωj ωj−1 ωj o + max ΩΛ θ1 , θ1 , ΩΛ θ2 , θ2 , ΩΛ θ3 , θ3

n ωj−1 ωj ωj−1 ωj ωj−1 ωj o < e + max ΩΛ θ1 , θ1 , ΩΛ θ2 , θ2 , ΩΛ θ3 , θ3 .

Letting j → ∞ and using (21), we have

n υj ωj υj ωj υj ωj o lim max ΩΛ θ , θ , ΩΛ θ , θ , ΩΛ θ , θ = e. (24) j→∞ 1 1 2 2 3 3

Based on the deﬁnition of ΩΛ and by (2), one can obtain

n υ ω υ ω υ ω o e lim max Λ θ j , θ j , Λ θ j , θ j , Λ θ j , θ j = . (25) j→∞ 1 1 2 2 3 3 2

Taking j → ∞ and applying (24) and (21) in

υ ω +1 υ ω ω ω +1 j j − j j ≤ j j ΩΛ θ1 , θ1 ΩΛ θ1 , θ1 ΩΛ θ1 , θ1 ,

we have υj ωj+1 lim ΩΛ θ , θ = e, j→∞ 1 1 Mathematics 2021, 9, 2012 18 of 22

hence, we obtain υ ω +1 e lim Λ θ j , θ j = . j→∞ 1 1 2 By the same manner, one can obtain

υ ω +1 e lim Λ θ j , θ j = , j→∞ 2 2 2

and υ ω +1 e lim Λ θ j , θ j = . j→∞ 3 3 2 Let

υ ω +1 υ υ υ ω +1 ω +1 ω +1 j j j j j υj j j j ωj+1 ΩΛ θ1 , θ1 = Λ H θ1 , θ2 , θ13, σ , H θ1 , θ2 , θ3 , σ υ υ υ υ υ υ j j j υj j j j ωj+1 ≤ Λ H θ1 , θ2 , θ3 , σ , H θ1 , θ2 , θ3 , σ υ υ υ ω +1 ω +1 ω +1 j j j ωj+1 j j j ωj+1 +Λ H θ1 , θ2 , θ3 , σ , H θ1 , θ2 , θ3 , σ υ υ υ υ υ υ j j j ωj+1 j j j ωj+1 −Λ H θ1 , θ2 , θ3 , σ , H θ1 , θ2 , θ3 , σ υ υ υ ω +1 ω +1 ω +1 ≤ υj − ωj+1 + j j j ωj+1 j j j ωj+1 C σ σ Λ H θ1 , θ2 , θ3 , σ , H θ1 , θ2 , θ3 , σ .

Letting j → ∞ in the above and since {σωj } is Cauchy, we obtain

e υ υ υ ω +1 ω +1 ω +1 ≤ lim Λ H θ j , θ j , θ j , σωj+1 , H θ j , θ j , θ j , σωj+1 2 j→∞ 1 2 3 1 2 3

Because ϕ is non-decreasing and continuous, we have

e h υ υ υ ω +1 ω +1 ω +1 i ϕ ≤ lim ϕ Λ H θ j , θ j , θ j , σωj+1 , H θ j , θ j , θ j , σωj+1 2 j→∞ 1 2 3 1 2 3

 n υj ωj+1 υj ωj+1 υj ωj+1o  ρ max Λ θ1 , θ1 , Λ θ2 , θ2 , Λ θ3 , θ3 ≤ lim  + + +  ω→∞ υj ωj 1 υj ωj 1 υj ωj 1 −` Λ θ1 , θ1 , Λ θ2 , θ2 , Λ θ3 , θ3 e e ≤ ρ − ` , 2 2 ω this implies that e ≤ 0, which is a contradiction. Hence, {θ1 } is Cauchy sequence. Sim- ω ω υ ω υ ω ilarly, {θ2 } and {θ3 } are too in (k, ΩΛ) and ΩΛ θ1 , θ1 → 0, ΩΛ(θ2 , θ2 ) → 0 and υ ω ΩΛ θ3 , θ3 → 0 as υ, ω → ∞. Thus, by (20) and the deﬁnition of ΩΛ, we have

lim Λ(θυ, θω) = 0, lim Λ(θυ, θω) = 0 and lim Λ(θυ, θω) = 0. υ,ω→∞ 1 1 υ,ω→∞ 2 2 υ,ω→∞ 3 3

ω ω ω It follows from Lemma1 (i) that {θ1 }, {θ2 } and {θ3 } are Cauchy sequences in (k, Λ). Because (k, Λ) is a complete, from Lemma1 (ii), there exist $1, $2, $3 ∈ k with

ω ω+1 ω υ Λ($1, $1) = lim Λ(θ , $1) = lim Λ θ , $1 = lim Λ(θ , θ ), ω→∞ 1 ω→∞ 1 ω,υ→∞ 1 1 ω ω+1 ω υ Λ($2, $2) = lim Λ(θ , $2) = lim Λ θ , $2 = lim Λ(θ , θ ), ω→∞ 2 ω→∞ 2 ω,υ→∞ 2 2 ω ω+1 ω υ Λ($3, $3) = lim Λ(θ , $3) = lim Λ θ , $3 = lim Λ(θ , θ ), ω→∞ 3 ω→∞ 3 ω,υ→∞ 3 3

Using Lemma2, we have

ω lim Λ(θ , H($1, $2, $3, σ)) = Λ($1, H($1, $2, $3, σ)). ω→∞ 1 Mathematics 2021, 9, 2012 19 of 22

Now

ω ω ω ω ω Λ(θ1 , H($1, $2, $3, σ)) = Λ(H(θ1 , θ2 , θ3 , σ ), H($1, $2, $3, σ)) ω ω ω ω ω ω ω ≤ Λ(H(θ1 , θ2 , θ3 , σ ), H(θ1 , θ2 , θ3 , σ)) ω ω ω +Λ(H(θ1 , θ2 , θ3 , σ), H($1, $2, $3, σ)) ω ω ω ω ω ω −Λ(H(θ1 , θ2 , θ3 , σ), H(θ1 , θ2 , θ3 , σ)) ω ω ω ω = C|σ − σ| + Λ(H(θ1 , θ2 , θ3 , σ), H($1, $2, $3, σ)).

Passing ω → ∞, we obtain

ω ω ω Λ($1, H($1, $2, $3, σ)) ≤ lim Λ(H(θ , θ , θ , σ), H($1, $2, $3, σ)). ω→∞ 1 2 3

Because ϕ is non-decreasing and continuous, we obtain

ω ω ω ϕ(Λ($1, H($1, $2, $3, σ))) ≤ lim ϕ[Λ(H(θ , θ , θ , σ), H($1, $2, $3, σ))] ω→∞ 1 2 3 ρΛθω, $ , Λ(θω, $ ), Λθω, $ ≤ lim 1 1 2 2 3 3 → ω ω ω ω ∞ −` Λ θ1 , $1 , Λ(θ2 , $2), Λ θ3 , $3 = 0.

This implies that Λ($1, H($1, $2, $3, σ)) = 0. Thus, $1 = H($1, $2, $3, σ). Analogously, $2 = H($2, $3, $1, σ) and $3 = H($3, $1, $2, σ). Hence, σ ∈ f, this proves that f is closed in [0, 1]. 0 0 0 0 0 0 0 0 0 Assume that σ ∈ f. Then there are θ1, θ2, θ3 ∈ A with θ1 = H(θ1, θ2, θ3, σ ). Because 0 0 0 f is open, then there exists z > 0 so that OΛ(θ1, z) ⊆ f. Select σ ∈ (σ − e, σ + e) so that 0 1 σ − σ ≤ Cω < e. 0 0 0 0 Then θ1 ∈ OΛ(θ1, z) = {θ1 ∈ k/Λ(θ1, θ1) ≤ z + Λ(θ1, θ1). We obtain

0 0 0 0 0 Λ(H(θ1, θ2, θ3, σ), θ1) = Λ(H(θ1, θ2, θ3, σ), H(θ1, θ2, θ3, σ )) 0 ≤ Λ(H(θ1, θ2, θ3, σ), H(θ1, θ2, θ3, σ )) 0 0 0 0 0 +Λ(H θ1, θ2, θ3, σ , H(θ1, θ2, θ3, σ )) 0 0 −Λ(H θ1, θ2, θ3, σ , H(θ1, θ2, θ3, σ )) 0 0 0 0 0 0 ≤ C σ − σ + Λ(H θ1, θ2, θ3, σ , H(θ1, θ2, θ3, σ )) 1 ≤ + Λ(H θ , θ , θ , σ0 , H(θ0, θ0, θ0, σ0)). Cω−1 1 2 3 1 2 3 Letting ω → ∞, we have

0 0 0 0 0 0 Λ(H(θ1, θ2, θ3, σ), θ1) ≤ Λ(H θ1, θ2, θ3, σ , H(θ1, θ2, θ3, σ )).

Because ϕ is non-decreasing and continuous, we obtain

0 h 0 0 0 0 0 i ϕ Λ(H(θ1, θ2, θ3, σ), θ1) ≤ ϕ Λ H θ1, θ2, θ3, σ , H(θ1, θ2, θ3, σ ) 0 0 0 0 0 0 ≤ ρ Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 − ` Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 .

By the same scenario, one can write 0 0 0 0 0 0 0 ϕ Λ(H(θ2, θ3θ1, σ), θ2) ≤ ρ Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 − ` Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 , and 0 0 0 0 0 0 0 ϕ Λ(H(θ3, θ1, θ2, σ), θ3) ≤ ρ Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 − ` Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 . Mathematics 2021, 9, 2012 20 of 22

Hence

n 0 0 0 o ϕ max Λ(H(θ1, θ2, θ3, σ), θ1), Λ(H(θ2, θ3θ1, σ), θ2), Λ(H(θ3, θ1, θ2, σ), θ3) n 0 0 0o n 0 0 0o ≤ ρ max Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 − ` max Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 n 0 0 0o ≤ ϕ max Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 .

Since ϕ is non-decreasing, we obtain

n 0 0 0 o max Λ(H(θ1, θ2, θ3, σ), θ1), Λ(H(θ2, θ3θ1, σ), θ2), Λ(H(θ3, θ1, θ2, σ), θ3) n 0 0 0o ≤ max Λ θ1, θ1 , Λ θ2, θ2 , Λ θ3, θ3 n 0 0 0 0 0 0o ≤ max z + Λ θ1, θ1 , z + Λ θ2, θ2 , z + Λ θ3, θ3 .

0 0 0 0 Therefore, for each σ ∈ (σ − e, σ + e), we have H : OΛ(θ1, z) → OΛ(θ1, z). Moreover, because assertion (ℵ2) holds and ϕ, ρ : [0, ∞) → [0, ∞) are continuous and non-decreasing and ` : [0, ∞) → [0, ∞) is LSC with ϕ(ν) − ρ(ν) + `(ν) > 0, for ν > 0. Then, all hypotheses of Corollary1 are fulfilled. Hence, we conclude that H(., σ) has a TFP in A. Since this TFP must be contained in A since (ℵ1) is satisfied. Thus, σ ∈ f for any σ ∈ (σ0 − e, σ0 + e). Hence, (σ0 − e, σ0 + e) ⊆ f and therefore f is open in [0, 1]. For the reverse implication, we use the same strategy. This finishes the proof.

Corollary 2. Suppose that (k, Λ) is a complete PMS, A is an open subset of k and H : A × A × A × [0, 1] → k with hypotheses below:

(i) θ1 6= H(θ1, θ2, θ3, σ), θ2 6= H(θ2, θ3, θ1, σ) and θ3 6= H(θ3, θ1, θ2, σ), for each θ1, θ2, θ3 ∈ ∂A (here ∂A refer to the boundary of A in k) and σ ∈ [0, 1]; (ii) There are θ1, θ2, θ3, ϑ1, ϑ2, ϑ3 ∈ A and σ ∈ [0, 1], K ∈ [0, 1) so that

Λ(H(θ1, θ2, θ3, σ), H(ϑ1, ϑ2, ϑ3, σ)) ≤ K max{Λ(θ1, ϑ1), Λ(θ2, ϑ2), Λ(θ3, ϑ3)},

(iii) There exist C ≥ 0 so that

∗ ∗ Λ(H(θ1, θ2, θ3, σ), H(θ1, θ2, θ3, σ )) ≤ C|σ − σ |,

∗ for each θ1, θ2, θ3 ∈ A and σ, σ ∈ [0, 1]. Then H(., 0) has a TFP, whenever H(., 1) has a TFP.

Proof. The proof follows immediately from Theorem4 by putting ϕ(ν) = ν, ρ(ν) = Kν − ν with K ∈ [0, 1) and `(ν) = ν, for ν > 0.

Author Contributions: H.A.H.: Writing–original draft; J.L.G.G.: Methodology; P.A. Writing–review and editing. All authors contributed equally to this article. All authors have read and agreed to the published version of the manuscript. Funding: This paper has been partially supported by Ministerio de Ciencia, Innovacion y Univer- sidades grant number PGC2018-0971-B-100 and Fundacion Seneca de la Region de Murcia grant number 20783/PI/18. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: The data used to support the ﬁndings of this study are included within the article. Acknowledgments: Juan L.G. Guirao is thankful to the Ministerio de Ciencia, Innovacion y Univer- sidades grant number PGC2018-0971-B-100 and Fundacion Seneca de la Region de Murcia grant Mathematics 2021, 9, 2012 21 of 22

number 20783/PI/18 for parrtially support this research. Praveen Agarwal was very thankful to the SERB (project TAR/2018/000001), DST (project DST/INT/DAAD/P-21/2019, INT/RUS/RFBR/308) and NBHM (project 02011/12/ 2020NBHM(R.P)/R&D II/7867) for their necessary support. Conﬂicts of Interest: The authors declare that they have no competing interests.

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