# A Fixed Point Technique for Solving an Integro-Differential Equation Using Mixed-Monotone Mappings

Authors: Hasanen A. Hammad; Rashwan A. Rashwan; Manuel De la Sen
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The objective of this manuscript is to present new tripled fixed point results for mixed-monotone mappings by a pivotal lemma in the setting of partially ordered complete metric spaces. Our outcomes sum up, enrich, and generalize several results in the current writing. Moreover, some examples have been discussed to strengthen and support our theoretical results. Finally, the theoretical results are applied to study the existence and uniqueness of the solution to an integro-differential equation.


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## Body

## 1. Brief Introduction

The fixed point (FP) technique is considered a fundamental pillar and a powerful tool in nonlinear analysis because of its many vital applications in many disciplines such as computer science, engineering, economics, biology, chemistry, and physics. In mathematics, this technique plays a prominent role in the study of statistical models, dynamical systems, game-theoretic models, differential equations, and many others. More clearly, for example, this method is mainly applied in finding the analytical solution to some differential and integral equations, fractional equations, integro-differential equations (IDEs), and functional analysis which facilitates the way to find numerical solutions to such problems. These problems were addressed by Fredholm [1], Rus [2], Hammad and De La Sen [3, 4], Ameer et al. [5], Hussain et al. [6, 7] and Younis et al. [8–10]In [11], the concepts of the coupled FP and a mixed-monotone mapping were initiated, and some exciting work in partially ordered metric spaces (POMSs) have been discussed by the same authors. This idea was investigated by many authors such as Berinde [12], Choudhury and Maity [13], and Aydi et al. [14]. Moreover, in abstract spaces, this concept has many applications in integral and functional equations; see the papers of Cirić et al. [15], Ding et al. [16], Hammad et al. [17, 18], Luong and Thuan [19], Choudhury and Kundu [20], Agarwal et al. [21], Radenović [22], and Hammad et al. [23].In 2011, coupled FP notions are generalized to tripled fixed points (TFPs) concepts by Berinde and Borcut [24] in the setting of POMSs. Via the mentioned spaces, Borcut and Berinde [25, 26], Karapnar et al. [27] presented pivotal results about TFP theorems and the applications in this direction introduced by Mustafa et al. [28] and Hammad and De la Sen [29, 30].Definition 1 [25].
We say that a trioω,κ,ν∈ℵ3 (where ℵ×ℵ×ℵ=ℵ3) is a TFP of a self-mapping ℧:ℵ3⟶ℵ if ω=℧ω,κ,ν, κ=℧κ,ν,ω, and ν=℧ν,ω,κ.Definition 2 [26].
A trioω,κ,ν∈ℵ3 on a nonempty set ℵ is called a tripled coincidence point of the two self-mappings ℧:ℵ3⟶ℵ and ℷ:ℵ⟶ℵ if ℷω=℧ω,κ,ν, ℷκ=℧κ,ν,ω, and ℷν=℧ν,ω,κ.Definition 3 [26].
Assume thatℵ≠∅ is a set. A trio ω,κ,ν∈ℵ3 is called a tripled common FP of ℧:ℵ3⟶ℵ and ℷ:ℵ⟶ℵ, if ω=ℷω=℧ω,κ,ν, κ=ℷκ=℧κ,ν,ω, and ν=ℷν=℧ν,ω,κ.Definition 4 [24].
Assume thatℵ,≼ is a partially ordered set on the product space ℵ3 defined as follows:
(1)ω,κ,ν,ω∗,κ∗,ν∗∈ℵ3,ω,κ,ν≼ω∗,κ∗,ν∗⇔ω≼ω∗,κ∗≼κ,ν≼ν∗.Under this partial order, we state the following definitions.Definition 5 [24].
A mapping℧:ℵ3⟶ℵ on a partially ordered set ℵ,≼ has a mixed-monotone property, if for any ω,κ,ν∈ℵ, we have
(2)ω1,ω2∈ℵ,ω1≼ω2⇒℧ω1,κ,ν≼℧ω2,κ,ν,κ1,κ2∈ℵ,κ1≼κ2⇒℧ω,κ1,ν≽℧ω,κ2,ν,ν1,ν2∈ℵ,ν1≼ν2⇒℧ω,κ,ν1≼℧ω,κ,ν2.Definition 6 [14].
A mapping℧:ℵ3⟶ℵ on a partially ordered set ℵ,≼ has a mixed ℷ-monotone property, where ℷ:ℵ⟶ℵ, if for any ω,κ,ν∈ℵ, we have
(3)ω1,ω2∈ℵ,ℷω1≼ℷω2⇒℧ω1,κ,ν≼℧ω2,κ,ν,κ1,κ2∈ℵ,ℷκ1≼ℷκ2⇒℧ω,κ1,ν≽℧ω,κ2,ν,ν1,ν2∈ℵ,ℷν1≼ℷν2⇒℧ω,κ,ν1≼℧ω,κ,ν2.Definition 7 [27].
Assume thatℵ is a nonempty set. We say that the mappings ℧:ℵ3⟶ℵ and ℷ:ℵ⟶ℵ are commutative if ℷ℧ω,κ,ν=℧ℷω,ℷκ,ℷν, for all ω,κ,ν∈ℵ.The first contribution of the TFP for a mixed-monotone mapping in a partially ordered set was presented as follows:Theorem 8 [24].
Letℵ,ζ,≼ be a complete partially ordered metric space (CPOMS). Assume that ℧:ℵ3⟶ℵ, so that
(i)
℧ has a mixed-monotone property(ii)
Either℧ is continuous or ℵ has the following properties:
(a)
ωn≼ω, if the nondecreasing sequence ωn⟶ω(b)
νn≽ν, if the nonincreasing sequence νn⟶ν, for all n(iii)
There isα,β,γ≥0 with α+β+γ<1 such that(4)ζ℧ω,κ,ν,℧ω∗,κ∗,ν∗≤αζω,ω∗+βζκ,κ∗+γζν,ν∗,for any ω,κ,ν,ω∗,κ∗,ν∗∈ℵ, for which ω≼ω∗, κ∗≼κ, and ν≼ν∗. If there are ω0,κ0,ν0∈ℵ so that ω0≼℧ω0,κ0,ν0, κ0≽℧κ0,ν0,ω0, and ν0≼℧ν0,ω0,κ0. Then ℧ has a TFP.In this manuscript, we utilize a pivotal lemma to obtain new TFP results for mixed-monotone mappings in CPOMSs. Our results unify, extend, and generalize the papers [19, 31, 32]. Also, some examples and a corollary are given. Later on, we apply the theoretical results to obtain the solution of a system of IDEs as an application.


## 2. Main Results

We begin this part with the pivotal lemma below.Lemma 9.
Assume thatℵ,≼ is a partially ordered set and ℧:ℵ3⟶ℵ and ℷ:ℵ⟶ℵ are two mappings. Suppose that the following assumptions hold:
(s1) There is ⅁0∈ℵ, so that for ρ1n⅁0,ρ2n⅁0,ρ3n⅁0∈ℵ, ρ1,ρ2,ρ3∈0,1, and ρ1<ρ3<ρ2, n∈ℕ(5)℧⅁0,⅁0,⅁0≽ℷρ1n⅁0ρ1n⅁0,℧⅁0,⅁0,⅁0≼ℷρ2n⅁0ρ2n⅁0,℧⅁0,⅁0,⅁0≽ℷρ3n⅁0ρ3n⅁0.
(s2) There is Δ:0,∞⟶0,∞ with Δρ∈ρ1,ρ2, so that for ρ1n⅁0,ρ2n⅁0,ρ3n⅁0∈ℵ, n∈ℕ(6)℧ρ1n⅁0,ρ2n⅁0,ρ3n⅁0≽Δρ℧ρ1n−1⅁0,ρ2n−1⅁0,ρ3n−1⅁0,(7)℧ρ2n⅁0,ρ3n⅁0,ρ1n⅁0≼Δρ℧ρ2n−1⅁0,ρ3n−1⅁0,ρ1n−1⅁0,(8)℧ρ3n⅁0,ρ1n⅁0,ρ2n⅁0≽Δρ℧ρ3n−1⅁0,ρ1n−1⅁0,ρ2n−1⅁0.
Then, there isω0,κ0,ν0∈ℵ, so that ℧ω0,κ0,ν0≽ℷω0, ℧κ0,ν0,ω0≼ℷκ0, and ℧ν0,ω0,κ0≽ℷν0.Proof.
We split the proof into three steps:(Step 1)
SinceΔρ>ρ1, then there is a nonnegative integer ℓ=ln1/ρ1/lnΔρ/ρ1+1, such that for n≥ℓ, we get Δρ/ρ1n≥1/ρ1, i.e., Δρn≥ρ1n−1. Take ω0=ρ1n⅁0, κ0=ρ2n⅁0, and ν0=ρ3n⅁0. By Stipulation (s1) and (5), we have(9)℧ω0,κ0,ν0=℧ρ1n⅁0,ρ2n⅁0,ρ3n⅁0≽Δρ℧ρ1n−1⅁0,ρ2n−1⅁0,ρ3n−1⅁0≽Δρ2℧ρ1n−2⅁0,ρ2n−2⅁0,ρ3n−2⅁0≽⋯≽Δρn℧⅁0,⅁0,⅁0≽ρ1n−1℧⅁0,⅁0,⅁0≽ℷρ1n⅁0=ℷω0.(Step 2)
SinceΔρ≤ρ2, then for each n∈ℕ, we obtain ρ2≤ρ2/Δρn, i.e., Δρn≥ρ2n−1. Apply Stipulation (s1) and (6), we get(10)℧κ0,ν0,ω0=℧ρ2n⅁0,ρ3n⅁0,ρ1n⅁0≼Δρ℧ρ2n−1⅁0,ρ3n−1⅁0,ρ1n−1⅁0≼Δρ2℧ρ2n−2⅁0,ρ3n−2⅁0,ρ1n−2⅁0≼⋯≼Δρn℧⅁0,⅁0,⅁0≼ρ2n−1℧⅁0,⅁0,⅁0≼ℷρ2n⅁0=ℷκ0.(Step 3)
Similar to Step 1, sinceΔρ>ρ3, then we can write Δρn≥ρ3n−1. By Stipulation (s1) and (7), we have(11)℧ν0,ω0,κ0=℧ρ3n⅁0,ρ1n⅁0,ρ2n⅁0≽Δρ℧ρ3n−1⅁0,ρ1n−1⅁0,ρ2n−1⅁0≽Δρ℧ρ3n−1⅁0,ρ1n−1⅁0,ρ2n−1⅁0≽⋯≽Δρn℧⅁0,⅁0,⅁0≽ρ3n−1℧⅁0,⅁0,⅁0≽ℷρ3n⅁0=ℷν0.
This completes the proof.Remark 10.
The results (5)-(7) of Lemma 9 still hold if we reserve the symbols “≽” and “≼,” that is, ω0,κ0,ν0∈ℵ so that ℧ω0,κ0,ν0≼ℷω0, ℧κ0,ν0,ω0≽ℷκ0, and ℧ν0,ω0,κ0≼ℷν0.Theorem 11.
Letℵ,ζ,≼ be a CPOMS and ϑ be a zero element in ℵ. Assume that ℧:ℵ3⟶ℵ is mixed ℷ-monotone mapping, ℷ:ℵ⟶ℵ is self-mapping, and Δ:0,∞⟶0,∞, so that Δρ≤ρ for any ρ≥0. Suppose that the hypotheses below hold:
(i)
℧ℵ3⊂ℷℵ(ii)
ℷ and ℧ are continuous and commute(iii)
℧ω,κ,ν verifies stipulations (s1) and (s2) of Lemma 9(iv)
For anyω,κ,ν,ω∗,κ∗,ν∗∈ℵ with ℷω≽ℷω∗, ℷκ≼ℷκ∗, ℷν≽ℷν∗, and we have(12)ζ℧ω,κ,ν,℧ω∗,κ∗,ν∗≤Δmax12ζℷω,℧ω,κ,ν,ζℷω∗,℧ω∗,κ∗,ν∗.
Then, the following conclusions are fulfilled:
(C1) For a triplet ω0,κ0,ν0∈ℵ, construct three sequences ωn, κn, and νn in ℵ verifying
(13)ℷωn=℧ωn−1,κn−1,νn−1,ℷκn=℧κn−1,νn−1,ωn−1,ℷνn=℧νn−1,ωn−1,κn−1,for all n∈ℕ. Then, ℷωn⟶ω″, ℷκn⟶κ″, and ℷνn⟶ν″, as n⟶∞
(C2) ℧ and ℷ have a tripled coincidence point ω″,κ″,ν″. Moreover, assume that ℷω0, ℷκ0, and ℷν0 are comparable, and for each ω,κ,ν∈ℵ, ℷω0,ℷκ0,ℷν0 is comparable to ℧ω,κ,ν,℧κ,ν,ω,℧ν,ω,κ
(C3) ℧ and ℷ have a unique common fixed point ℷϑ, that is, ℷϑ=ℷℷϑ=℧ℷϑ,ℷϑ,ℷϑProof.
We shall prove Conclusion (C1). By Condition (iii), there are ω0=ρ1n⅁0, κ0=ρ2n⅁0, and ν0=ρ3n⅁0, so that ℧ω0,κ0,ν0≽ℷω0, ℧κ0,ν0,ω0≼ℷκ0, and ℧ν0,ω0,κ0≽ℷν0, where ρ1,ρ2,ρ3∈0,1. Since ℧ℵ3⊂ℷℵ, by Condition (i), this yields that there exists ω1,κ1,ν1∈ℵ, so that ℷω1=℧ω0,κ0,ν0, ℷκ1=℧κ0,ν0,ω0, and ℷν1=℧ν0,ω0,κ0. Generally, we can build the three sequences ωn, κn, and νn in ℵ, so that
(14)ℷωn=℧ωn−1,κn−1,νn−1,ℷκn=℧κn−1,νn−1,ωn−1,ℷνn=℧νn−1,ωn−1,κn−1,forn∈ℕ.
Since℧ is a mixed ℷ-monotone, then we get
(15)ℷω2=℧ω1,κ1,ν1≽℧ω0,κ0,ν0=ℷω1,ℷκ2=℧κ1,ν1,ω1≼℧κ0,ν0,ω0=ℷκ1,ℷν2=℧ν1,ω1,κ1≽℧ν0,ω0,κ0=ℷν1.
By induction forn∈ℕ, one can write
(16)ℷωn≽ℷωn−1,ℷκn≼ℷκn−1,ℷνn≽ℷνn−1.
It follows from (12) and (13) that for n∈ℕ, we have
(17)ζℷωn+1,ℷωn=ζ℧ωn,κn,νn,℧ωn−1,κn−1,νn−1≤Δmax12ζℷωn,℧ωn,κn,νn,ζℷωn−1,℧ωn−1,κn−1,νn−1≤max12ζℷωn,℧ωn,κn,νn,ζℷωn−1,℧ωn−1,κn−1,νn−1=max12ζℷωn,ℷωn+1,ζℷωn−1,ℷωn.
Ifζℷωn−1,ℷωn≤ζℷωn,ℷωn+1, then by (16), we find directly that
(18)ζℷωn+1,ℷωn≤12ζℷωn+1,ℷωn,this is a contradiction. So, we should take ζℷωn+1,ℷωn=0. It is clear that for each ε>0 and σ∈ℕ, we have
(19)ζℷωn,ℷωn+σ<ε.
Sinceζℷωn−1,ℷωn>1/2ζℷωn,ℷωn+1, then by (16), we obtain that
(20)ζℷωn+1,ℷωn≤ζℷωn,ℷωn−1,forn∈ℕ.
Set∇n=ζℷωn+1,ℷωn. Using (12)–(17), one can obtain
(21)∇n≤∇n−1≤∇n−2≤⋯≤∇0,forn∈ℕ∪0,thus, we have
(22)ζℷωn+1,ℷωn≤ζℷω1,ℷω0.
According to the proof of Lemma9 and Condition (ii) (the continuity), there is ⅁0∈ℵ, ρ1,ρ2,ρ3∈0,1, so that
(23)limn⟶∞ℷω0=limn⟶∞ℷρ1n⅁0≼limn⟶∞℧ρ1n⅁0,ρ2n⅁0,ρ3n⅁0=limn⟶∞℧ω0,κ0,ν0,limn⟶∞ℷκ0=limn⟶∞ℷρ2n⅁0≽limn⟶∞℧ρ2n⅁0,ρ3n⅁0,ρ1n⅁0=limn⟶∞℧κ0,ν0,ω0,limn⟶∞ℷν0=limn⟶∞ℷρ3n⅁0≼limn⟶∞℧ρ3n⅁0,ρ1n⅁0,ρ2n⅁0=limn⟶∞℧ν0,ω0,κ0,that is,
(24)ℷϑ≼℧ϑ,ϑ,ϑ,ℷϑ≼℧ϑ,ϑ,ϑ.
Therefore,ℷϑ=℧ϑ,ϑ,ϑ. By the triangle inequality and (21), one can write for any positive integers m,n with m>n, so we have
(25)ζℷωm,ℷωn≤ζℷωm,ℷωm−1+ζℷωm−1,ℷωm−2+⋯+ζℷωn+1,ℷωn≤ζℷω1,ℷω0+ζℷω1,ℷω0+⋯+ζℷω1,ℷω0=ζ℧ω0,κ0,ν0,ℷω0+ζ℧ω0,κ0,ν0,ℷω0+⋯+ζ℧ω0,κ0,ν0,ℷω0=ζ℧ρ1n⅁0,ρ2n⅁0,ρ3n⅁0,ℷρ1n⅁0+⋯+ζ℧ρ1n⅁0,ρ2n⅁0,ρ3n⅁0,ℷρ1n⅁0.
It follows thatlimm,n⟶∞ζℷωm,ℷωn=0. Similarly, we can show that limm,n⟶∞ζℷκm,ℷκn=0 and limm,n⟶∞ζℷνm,ℷνn=0. This illustrates that ωn, κn, and νn are Cauchy sequences. The completeness of ℵ leads to the conclusion that there are ω″,κ″,ν″∈ℵ so that
(26)limn⟶∞ℷωn=ω″,limn⟶∞ℷκn=κ″,limn⟶∞ℷνn=ν″.
Next, we shall show ConclusionC2. Since ℷ is continuous, then by (25), we get
(27)limn⟶∞ℷℷωn=ℷω″,limn⟶∞ℷℷκn=ℷκ″,limn⟶∞ℷℷνn=ℷν″.
Also, by the commutativity ofℷ and ℧, we have
(28)ℷℷωn=ℷ℧ωn−1,κn−1,νn−1=℧ℷωn−1,ℷκn−1,ℷνn−1,ℷℷκn=ℷ℧κn−1,νn−1,ωn−1=℧ℷκn−1,ℷνn−1,ℷωn−1,ℷℷνn=ℷ℧νn−1,ωn−1,κn−1=℧ℷνn−1,ℷωn−1,ℷκn−1.
By (26) and (27), we deduce that
(29)ℷω″=limn⟶∞ℷℷωn=limn⟶∞℧ℷωn−1,ℷκn−1,ℷνn−1=℧ω″,κ″,ν″,ℷκ″=limn⟶∞ℷℷκn=limn⟶∞℧ℷκn−1,ℷνn−1,ℷωn−1=℧κ″,ν″,ω″,ℷν″=limn⟶∞ℷℷνn=limn⟶∞℧ℷνn−1,ℷωn−1,ℷκn−1=℧ν″,ω″,κ″.
Therefore, the trioω″,κ″,ν″ is a tripled coincidence point of ℧ and ℷ.
Finally, to prove ConclusionC2, assume that ℷω0≼ℷκ0, ℷκ0≼ℷν0, ℷν0≼ℷω0, and ℷω0,ℷκ0,ℷν0≼℧ω,κ,ν,℧κ,ν,ω,℧ν,ω,κ, for ω,κ,ν∈ℵ3. Since ω″,κ″,ν″,κ″,ν″,ω″,ν″,ω″,κ″∈ℵ3 and by Definition 4, we have
(30)ℷω0,ℷκ0,ℷν0≼℧ω″,κ″,ν″,℧κ″,ν″,ω″,℧ν″,ω″,κ″,which yields
(31)ℷω0≼℧ω″,κ″,ν″=ℷω″,ℷκ0≽℧κ″,ν″,ω″=ℷκ″,ℷν0≼℧ν″,ω″,κ″=ℷν″.
It follows from (13) and (30) and mixed ℷ-monotonicity of ℧ that
(32)℧ω0,κ0,ν0≼℧ω″,κ″,ν″,℧κ0,ν0,ω0≽℧κ″,ν″,ω″,℧ω0,κ0,ν0≼℧ν″,ω″,κ″.
This leads to(33)ℷω1≼ℷω″,ℷκ1≽ℷκ″,ℷν1≼ℷν″.
By induction, forn∈ℕ, we have
(34)ℷωn≼ℷω″,ℷκn≽ℷκ″,ℷνn≼ℷν″.
Applying (12), we observe that
(35)ζℷω″,ℷωn+1=ζ℧ω″,κ″,ν″,℧ωn,κn,νn≤Δmax12ζℷω″,℧ω″,κ″,ν″,ζℷωn,℧ωn,κn,νn=max12ζℷω″,ℷω″,ζℷωn,ℷωn+1≤ζℷωn,ℷωn+1.
In the same manner, we have(36)ζℷκ″,ℷκn+1≤ζℷκn,ℷκn+1,ζℷν″,ℷνn+1≤ζℷνn,ℷνn+1.
From another direction, sinceℷω0≼ℷκ0, and the mixed ℷ-monotonicity of ℧, then ℧ω0,κ0,ν0≼℧κ0,ν0,ω0, i.e., ℷω1≼ℷκ1; similarly, one can obtain ℷκ1≼ℷν1 and ℷν1≼ℷω1. By the proof of Conclusion C1, we have
(37)ℷω0≼⋯≼ℷωn≼ℷκn≼⋯≼ℷκ0,ℷκ0≼⋯≼ℷκn≼ℷνn≼⋯≼ℷν0,ℷν0≼⋯≼ℷνn≼ℷωn≼⋯≼ℷω0.
Taking the limit asn⟶∞, we get
(38)limn⟶∞ℷωn=limn⟶∞ℷκn=limn⟶∞ℷνn=ℷϑ.
This implies together with (25) that
(39)ω″=κ″=ν″=ℷϑ.
Hence, by (34), (35), (37), and (38), we have
(40)limn⟶∞ζℷω″,ℷωn+1=limn⟶∞ζℷκ″,ℷκn+1=limn⟶∞ζℷν″,ℷνn+1=0.
With the help of (39) and the triangle inequality, we can write
(41)ζℷω″,ℷϑ≤ζℷω″,ℷωn+1+ζℷωn+1,ℷϑ⟶0asn⟶∞,ζℷκ″,ℷϑ≤ζℷκ″,ℷκn+1+ζℷκn+1,ℷϑ⟶0asn⟶∞,ζℷν″,ℷϑ≤ζℷν″,ℷνn+1+ζℷνn+1,ℷϑ⟶0asn⟶∞.
Thus,ℷω″=ℷκ″=ℷν″=ℷϑ. According to (39), we obtain ℷϑ=ℷℷϑ=℧ℷϑ,ℷϑ,ℷϑ, which proves that ℷϑ is a common FP of ℧ and ℷ.
To discuss the uniqueness, suppose that℘ is another common FP of ℧ and ℷ, thus ℘=ℷ℘=℧℘,℘,℘. By the above results and Definition 4, we get
(42)ℷω0,ℷκ0,ℷν0≼℧℘,℘,℘,℧℘,℘,℘,℧℘,℘,℘.
That is, forω0=ρ1n⅁0, κ0=ρ2n⅁0, and ν0=ρ3n⅁0, where ρ1,ρ2,ρ3∈0,1, we have
(43)ℷω0≼℧℘,℘,℘≼ℷκ0,ℷν0≼℧℘,℘,℘≼ℷκ0.
This leads to(44)limn⟶∞ℷω0=limn⟶∞ℷκ0=limn⟶∞ℷν0=ℷϑ.
Hence,℧℘,℘,℘=ℷϑ, which leads to ℘=ℷϑ. Therefore, ℷϑ is a unique common FP of ℧ and ℷ. This finishes the proof.Examples below verify the assumptions of Theorem11.Example 1.
Letℵ=0,∞ be endowed with
(45)ζω,κ=κ−ω,κ,ω∈ℵ.Define the order relation≤ by
(46)κ,ω∈ℵ,ω≼κ⇔ω≤κ.It is obvious thatℵ,ζ,≼ is a CPOMS. Define the mappings ℧:ℵ3⟶ℵ and ℷ:ℵ⟶ℵ by
(47)℧ω,κ,ν=ω+κ−ν4,ifω≥κ≥ν,0,otherwise,ℷω=2ω,respectively. It is clear that ℧ℵ3⊂ℷℵ, ℷ and ℧ are continuous, and ℧ have a mixed ℷ-monotone property.Now, let us verify Condition (12) of Theorem 11 for all ω,κ,ν,ω∗,κ∗,ν∗∈ℵ with ω≥ω∗, κ≤κ∗, and ν≥ν∗. Consider that the function Δ:0,+∞⟶0,+∞ is given by
(48)Δmaxω,κ=ω+κ2,ω,κ∈0,∞.Now, we consider the cases below:(⋆1) If ω≥κ≥ν and ω∗≥κ∗≥ν∗, then ω≥ω∗≥κ∗≥κ≥ν, and we have
(49)ζ℧ω,κ,ν,℧ω∗,κ∗,ν∗=ω+κ−ν4−ω∗+κ∗−ν∗4≤ω+κ−ν4+ω∗+κ∗−ν∗4≤ω+ω∗+κ+κ∗4≤2ω+2κ∗4≤ω+ω∗2≤127ω8+ν−κ8+7ω∗4+ν∗−κ∗4=1212ζℷω,℧ω,κ,ν+ζℷω∗,℧ω∗,κ∗,ν∗=Δmax12ζℷω,℧ω,κ,ν,ζℷω∗,℧ω∗,κ∗,ν∗.(⋆2) If ω≥κ≥ν and ω∗<κ∗<ν∗, then ω≥ν≥ν∗≥ω∗≥κ, and we have
(50)ζ℧ω,κ,ν,℧ω∗,κ∗,ν∗=ω+κ−ν4−0≤ω+κ4≤ω+ω∗4≤127ω8+ν−κ8+2ω∗=1212ζℷω,℧ω,κ,ν+ζℷω∗,℧ω∗,κ∗,ν∗=Δmax12ζℷω,℧ω,κ,ν,ζℷω∗,℧ω∗,κ∗,ν∗.(⋆3) If ω<κ<ν and ω∗≥κ∗≥ν∗, then ν≥ω≥ω∗≥κ∗, and we have
(51)ζ℧ω,κ,ν,℧ω∗,κ∗,ν∗=0−ω∗+κ∗−ν∗4=ω∗+κ∗−ν∗4≤ω∗+κ∗4≤ω+ω∗4≤122ω+7ω∗4+ν∗−κ∗4=1212ζℷω,℧ω,κ,ν+ζℷω∗,℧ω∗,κ∗,ν∗=Δmax12ζℷω,℧ω,κ,ν,ζℷω∗,℧ω∗,κ∗,ν∗.(⋆4) If ω<κ<ν and ω∗<κ∗<ν∗, then we have
(52)ζ℧ω,κ,ν,℧ω∗,κ∗,ν∗=0≤12ω+2ω∗=1212ζℷω,℧ω,κ,ν+ζℷω∗,℧ω∗,κ∗,ν∗=Δmax12ζℷω,℧ω,κ,ν,ζℷω∗,℧ω∗,κ∗,ν∗.The four cases indicate that the requirements of Theorem11 are fulfilled and 0,0,0 is a unique TFP.Example 2.
Assume that the first requirements of Example1 hold with the usual order “≤.” Define the mappings ℧:ℵ3⟶ℵ and ℷ:ℵ⟶ℵ by
(53)℧ω,κ,ν=ω+κ+ν3,ℷω=ω,respectively. It is obvious that ℵ,ζ,≤ is a CPOMS, ℧ℵ3⊂ℷℵ, ℷ and ℧ are continuous, and ℧ have a mixed ℷ-monotone property.Now, let us verify Condition (12) of Theorem 11 for all ω,κ,ν,ω∗,κ∗,ν∗∈ℵ with ω≤ω∗, κ≥κ∗, and ν≤ν∗. Let Δ:0,+∞⟶0,+∞ be a function defined by
(54)Δmaxω,κ=ω+κ2,forallω,κ∈0,∞.Now, consider(55)ζ℧ω,κ,ν,℧ω∗,κ∗,ν∗=ω+κ+ν3−ω∗+κ∗+ν∗3≤ω−ω∗3+κ−κ∗3+ν−ν∗3≤κ−κ∗3≤κ+κ∗2≤122κ6+ω+ν6+2κ∗3+ω∗+ν∗3=1212ζℷκ,℧κ,ν,ω+ζℷκ∗,℧κ∗,ν∗,ω∗=Δmax12ζℷκ,℧κ,ν,ω,ζℷκ∗,℧κ∗,ν∗,ω∗.Hence, all conditions of Theorem11 are satisfied and ℧ and ℷ have a unique common TFP in ℵ for all ω=κ=ν.If we setℷ=Iℵ (the identity mapping on ℵ) in Theorem 11, we deduce the result below:Corollary 12.
Letℵ,ζ,≼ be a CPOMS and ϑ be a zero element in ℵ. Assume that ℧:ℵ3⟶ℵ is a mixed-monotone mapping and Δ:0,∞⟶0,∞ is so that Δρ≤ρ for any ρ≥0. Suppose that the assumptions below are satisfied:
(i)
℧ is continuous(ii)
℧ω,κ,ν verifies stipulations (s1) and (s2) of Lemma 9(iii)
For anyω,κ,ν,ω∗,κ∗,ν∗∈ℵ with ω≽ω∗, κ≼κ∗, and ν≽ν∗, and we have(56)ζ℧ω,κ,ν,℧ω∗,κ∗,ν∗≤Δmax12ζω,℧ω,κ,ν,ζω∗,℧ω∗,κ∗,ν∗.
Then, the following conclusions are fulfilled:
(C1) For a triplet ω0,κ0,ν0∈ℵ, construct three sequences ωn, κn, and νn in ℵ verifying
(57)ωn=℧ωn−1,κn−1,νn−1,κn=℧κn−1,νn−1,ωn−1,νn=℧νn−1,ωn−1,κn−1,for n∈ℕ. Then, ωn⟶ω″, κn⟶κ″, and νn⟶ν″, as n⟶∞
(C2) ℧ has a TFP ω″,κ″,ν″. Moreover, assume that ω0, κ0, and ν0 are comparable and for each ω,κ,ν∈ℵ, ω0,κ0,ν0 is comparable to ℧ω,κ,ν,℧κ,ν,ω,℧ν,ω,κ
(C3) ℧ has a unique FP ϑ, that is ϑ=℧ϑ,ϑ,ϑ


## 3. An Application to Integro-Differential Equation

In fact, this part is a fundamental pillar of our paper, where the theoretical results presented in the above section are involved in order to obtain the existence of the solution to an IDE of the following form:(58)ωρ=12∫ℓ1ρΘρ,σℵ1σ,ωσ,ω′σ+R1σ,ωσ+ℵ2σ,κσ,κ′σ+R2σ,κσ+ℵ3σ,νσ,ν′σ+R3σ,νσdσ,for ρ∈ℓ1,ℓ2, where Θ:ℓ1,ℓ2×ℓ1,ℓ2⟶1,∞, ℵi:ℓ1,ℓ2×0,∞×0,∞⟶0,∞, and Ri:ℓ1,ℓ2×0,∞⟶0,∞ (for i=1,2,3) are given continuous functions.Assume thatωρ, κρ, νρ, ω′ρ, κ′ρ, and ν′ρ are nonnegative real continuous functions which are differentiable on ℓ1,ℓ2, where ω′ρ, κ′ρ, and ν′ρ are the first derivative of ωρ, κρ, and νρ with respect to ρ, respectively.Also, suppose thatΘ is continuously differentiable with respect to its first variable, where ∂Θ/∂ρ>0.In order to find the existence solution of Problem (57), we shall derive the hypotheses below:(‡i) ℵi:ℓ1,ℓ2×0,∞×0,∞⟶0,∞ and Ri:ℓ1,ℓ2×0,∞⟶0,∞ (for i=1,2,3) are continuous functions so that
(59)R1ρ,ωρ≤R1ρ,ω∗ρ,R2ρ,κρ≥R2ρ,κ∗ρ,R3ρ,νρ≤R3ρ,ν∗ρ,ℵ1ρ,ωρ,κ′ρ≤ℵ1ρ,ω∗ρ,κ∗′ρ,ℵ2ρ,κρ,ν′ρ≥ℵ2ρ,κ∗ρ,ν∗′ρ,ℵ3ρ,νρ,ω′ρ≤ℵ3ρ,ν∗ρ,ω∗′ρ,for any fixed ρ∈ℓ1,ℓ2(‡ii) Define the mappings ℧ and ℷ by
(60)℧ω,κ,νρ=∫ℓ1ρΘρ,σℵ1σ,ωσ,κ′σ+ℵ2σ,κσ,ν′σ+ℵ3σ,νσ,ω′σdσ,(61)ℷωρ=∫ℓ1ρΘρ,σR1σ,ωσ+R2σ,κσ+R3σ,νσdσ,such that
(62)R1ρ,℧ω,κ,νρ=ℵ1ρ,ℷωρ,ℷκ′ρ,R2ρ,℧ω,κ,νρ=ℵ2ρ,ℷκρ,ℷν′ρ,R3ρ,℧ω,κ,νρ=ℵ3ρ,ℷνρ,ℷω′ρ,for any fixed ρ∈ℓ1,ℓ2(‡iii) For any fixed ρ∈ℓ1,ℓ2, there is β∈0,1/2 so that
(63)ℵ1ρ,ω∗ρ,κ∗′ρ≥R1ρ,ωρ,ℵ2ρ,κ∗ρ,ν∗′ρ≥R2ρ,κρ,ℵ3ρ,ν∗ρ,ω∗′ρ≥R3ρ,νρ,(64)ℵ1ρ,ωρ,κ′ρ−ℵ1ρ,ω∗ρ,κ∗′ρ≤β2ℵ1ρ,ωρ,κ′ρ−R1ρ,ωρ,ℵ2ρ,κρ,ν′ρ−ℵ2ρ,κ∗ρ,ν∗′ρ≤β2ℵ2ρ,κρ,ν′ρ−R2ρ,κρ,ℵ3ρ,νρ,ω′ρ−ℵ3ρ,ν∗ρ,ω∗′ρ≤β2ℵ3ρ,νρ,ω′ρ−R3ρ,νρ,(‡iv) there is ⅁0ρ∈0,∞, so that for any fixed ρ∈ℓ1,ℓ2, ρ1,ρ2,ρ3∈0,1,
(65)Rρ,ρ1j⅁0ρ≤ℵρ,ρ1j⅁0ρ,ρ2j⅁0′ρ≤Rρ,ρ2j⅁0ρ,Rρ,ρ3j⅁0ρ≤ℵρ,ρ3j⅁0ρ,ρ2j⅁0′ρ≤Rρ,ρ2j⅁0ρ.Now, our main theorem of this section is stated as follows:Theorem 13.
Under hypotheses (‡i)-(‡iv), System (57) has a unique solution ℷϑ.Proof.
The proof is splitting into the following steps:
(St1) Construct a CPOMS. Assume that ℵ=Cℓ1,ℓ2,ℝ+ is the set of all nonnegative real continuous functions on ℓ1,ℓ2, ϑ∈ℵ. Define a metric ζ:ℵ×ℵ⟶0,∞ on ℵ by
(66)ζω,κ=ωρ−κρ12,∀ω,κ∈ℵ,ρ∈ℓ1,ℓ2,where ωρ12=ωρ+ω′ρ. Define the partial ordered ≼ by
(67)ω≼κ⇔ωρ≤κρ,∀ρ∈ℓ1,ℓ2.Then, a trio ℵ,ζ,≼ is a CPOMS if ω≼ω∗, κ∗≼κ, and ν≼ν∗, whenever ωρ≤ω∗ρ, κ∗ρ≤κρ, and νρ≤ν∗ρ, for all ω,κ,ν,ω∗,κ∗,ν∗∈ℵ
(St2) Construct the mappings ℧′ and ℷ′. For this purpose, we involve a derivative with respect to ρ on both sides of (59) and (60), respectively, for ω,κ,ν∈ℵ, ρ∈ℓ1,ℓ2, we get
(68)℧′ω,κ,νρ=Θρ,ρℵ1ρ,ωρ,κ′ρ+ℵ2ρ,κρ,ν′ρ+ℵ3ρ,νρ,ω′ρ+∫ℓ1ρ∂∂ρΘρ,σℵ1ρ,ωρ,κ′ρ+ℵ2ρ,κρ,ν′ρ+ℵ3ρ,νρ,ω′ρdσ,(69)ℷω′ρ=Θρ,ρR1ρ,ωρ+R2ρ,κρ+R3ρ,νρ+∫ℓ1ρ∂∂ρΘρ,σR1σ,ωσ+R2σ,κσ+R3σ,νσdσ.
(St3) Show that ℧ is a mixed ℷ-monotone and ℧ and ℷ are commuting. If ℷκ′ρ≥ℷκ∗′ρ, then by (68), we have
(70)Θρ,ρR2ρ,κρ−R2ρ,κ∗ρ+R3ρ,νρ−R3ρ,ν∗ρ+R1ρ,ωρ−R1ρ,ω∗ρ+∫ℓ1ρ∂∂ρΘρ,σR2ρ,κρ−R2ρ,κ∗ρ+R3ρ,νρ−R3ρ,ν∗ρ+R1ρ,ωρ−R1ρ,ω∗ρdσ≥0,since Θρ,ρ≥1, and ∂/∂ρΘρ,σ>0, then, we get R2ρ,κρ≥R1ρ,κ∗ρ. Moreover by hypothesis (‡i), for any fixed ρ∈ℓ1,ℓ2, ω,κ,ν,ω∗,κ∗,ν∗∈ℵ. If
(71)ℷωρ≤ℷω∗ρ,ℷκ′ρ≥ℷκ∗′ρ,ℷνρ≤ℷν∗ρ,then, we have
(72)℧ω,κ′,νρ≼℧ω∗,κ′∗,ν∗ρ.
Similarly, if(73)ℷω′ρ≤ℷω∗′ρ,ℷκρ≥ℷκ∗ρ,ℷνρ≤ℷν∗ρ⇒℧ω′,κ,νρ≤℧ω′∗,κ∗,ν∗ρ,
and if(74)ℷωρ≤ℷω∗ρ,ℷκρ≥ℷκ∗ρ,ℷν′ρ≤ℷν∗′ρ⇒℧ω,κ,ν′ρ≤℧ω∗,κ∗,ν′∗ρ,then this implies that ℧ is a mixed ℷ-monotone. By the definition of ℧ and ℷ, we can write
(75)℧ℷω,ℷκ,ℷνρ=∫ℓ1ρΘρ,σℵ1σ,ℷωσ,ℷκ′σ+ℵ2σ,ℷκσ,ℷν′σ+ℵ3σ,ℷνσ,ℷω′σdσ=∫ℓ1ρΘρ,σR1ρ,℧ω,κ,νρ+R2ρ,℧ω,κ,νρ+R3ρ,℧ω,κ,νρdσ=ℷ℧ω,κ,νρ.
(St4) Fulfill Condition (12) of Theorem 11. It follows from (60), (63), and (68) that
(76)℧ω,κ,νρ−℧ω∗,κ∗,ν∗ρ=∫ℓ1ρΘρ,σℵ1σ,ωσ,κ′σ−ℵ1σ,ω∗σ,κ∗′σ+ℵ2σ,κσ,ν′σ−ℵ2σ,κ∗σ,ν∗′σ+ℵ3σ,νσ,ω′σ−ℵ3σ,ν∗σ,ω∗′σdσ≤∫ℓ1ρΘρ,σβ2R1σ,ωσ−ℵ1σ,ωσ,κ′σ+β2R2σ,κσ−ℵ2σ,κσ,ν′σ+β2R3σ,νσ−ℵ3σ,νσ,ω′σdσ=β2∫ℓ1ρΘρ,σR1σ,ωσ−ℵ1σ,ωσ,κ′σ+R2σ,κσ−ℵ2σ,κσ,ν′σ+R3σ,νσ−ℵ3σ,νσ,ω′σdσ,(77)℧′ω,κ,νρ−℧′ω∗,κ∗,ν∗ρ=Θρ,ρℵ1ρ,ωρ,κ′ρ−ℵ1ρ,ω∗ρ,κ∗′ρ+ℵ2ρ,κρ,ν′ρ−ℵ2ρ,κ∗ρ,ν∗′ρ+ℵ3ρ,νρ,ω′ρ−ℵ3ρ,ν∗ρ,ω∗′ρ+∫ℓ1ρ∂∂ρΘρ,σℵ1ρ,ωρ,κ′ρ−ℵ1ρ,ω∗ρ,κ∗′ρ+ℵ2ρ,κρ,ν′ρ−ℵ2ρ,κ∗ρ,ν∗′ρ+ℵ3ρ,νρ,ω′ρ−ℵ3ρ,ν∗ρ,ω∗′ρdσ≤Θρ,ρβ2R1ρ,ωρ−ℵ1ρ,ωρ,κ′ρ+β2R2ρ,κρ−ℵ2ρ,κρ,ν′ρ+β2R3ρ,νρ−ℵ3ρ,νρ,ω′ρ+∫ℓ1ρ∂∂ρΘρ,σβ2R1σ,ωσ−ℵ1σ,ωσ,κ′σ+β2R2σ,κσ−ℵ2σ,κσ,ν′σ+β2R3σ,νσ−ℵ3σ,νσ,ω′σdσ=β2Θρ,ρR1ρ,ωρ−ℵ1ρ,ωρ,κ′ρ+R2ρ,κρ−ℵ2ρ,κρ,ν′ρ+R3ρ,νρ−ℵ3ρ,νρ,ω′ρ+∫ℓ1ρ∂∂ρΘρ,σR1σ,ωσ−ℵ1σ,ωσ,κ′σ+R2σ,κσ−ℵ2σ,κσ,ν′σ+R3σ,νσ−ℵ3σ,νσ,ω′σdσ.
From (59), (60), (67)–(76) and by definition of .12, we conclude that
(78)℧ω,κ,νρ−℧ω∗,κ∗,ν∗ρ=∫ℓ1ρΘρ,σℵ1σ,ωσ,κ′σ−ℵ1σ,ω∗σ,κ∗′σ+ℵ2σ,κσ,ν′σ−ℵ2σ,κ∗σ,ν∗′σ+ℵ3σ,νσ,ω′σ−ℵ3σ,ν∗σ,ω∗′σdσ≤∫ℓ1ρΘρ,σβ2R1σ,ωσ−ℵ1σ,ωσ,κ′σ+β2R2σ,κσ−ℵ2σ,κσ,ν′σ+β2R3σ,νσ−ℵ3σ,νσ,ω′σdσ=β2∫ℓ1ρΘρ,σR1σ,ωσ−ℵ1σ,ωσ,κ′σ+R2σ,κσ−ℵ2σ,κσ,ν′σ+R3σ,νσ−ℵ3σ,νσ,ω′σdσ,that is
(79)ζ℧ω,κ,νρ,℧ω∗,κ∗,ν∗ρ≤β×12ζℷωρ,℧ω,κ,νρ≤βmax12ζℷωρ,℧ω,κ,νρ,ζℷω∗ρ,℧ω∗,κ∗,ν∗ρ.
SetΔω=βω for ω∈0,∞,β∈0,1/2. It is clear that Δω≤ω for ω≥0. Through Inequality (78), Hypothesis (iv) of Theorem 11 is verified. Suppose that ω0ρ=ρ1n⅁0, κ0ρ=ρ2n⅁0, and ν0ρ=ρ3n⅁0, then by ((65)), we have
(80)∫ℓ1ρΘρ,σRρ,ω0ρdσ≤∫ℓ1ρΘρ,σℵρ,ω0ρ,κ0′ρdσ≤∫ℓ1ρΘρ,σRρ,κ0ρdσ,∫ℓ1ρΘρ,σRρ,ν0ρdσ≤∫ℓ1ρΘρ,σℵρ,ν0ρ,κ0′ρdσ≤∫ℓ1ρΘρ,σRρ,κ0ρdσ,that is
(81)ℷω0ρ≤℧ω0,κ0,ν0ρ≤ℷκ0ρ,ℷν0ρ≤℧ω0,κ0,ν0ρ≤ℷκ0ρ.
This means℧ω0,κ0,ν0≽ℷω0, ℧κ0,ν0,ω0≼ℷκ0, and ℧ν0,ω0,κ0≽ℷν0. Thus, all requirements of Theorem 11 are fulfilled. So, there is a unique common FP for the mappings ℧ and ℷ in the form of ℷϑ∈ℵ, so that ℷϑ=1/2℧ℷϑ,ℷϑ,ℷϑ+ℷϑ, which is the unique solution to Problem (57).The example below supports our application.Example 3.
Consider the following problem:(82)ωρ=132∫0ρρeρσe2ω+eω+e2κ+eκ+e2ν+eνdσ.Problem (80) is a special form of Problem (57), with the following constraints:
(83)ℵ1ρ,ωρ,ω′ρ=ρ14eω14eω=116ρe2ω,R1ρ,ωρ=1414ρeω=116ρeω,ℵ2ρ,κρ,κ′ρ=116ρe2κ,R2ρ,κρ=116ρeκ,ℵ3ρ,νρ,ν′ρ=116ρe2ν,R3ρ,νρ=116ρeν.Θρ,σ=eρσ, for all ρ∈0,1 and ℓ1=0, ℓ2=1.Here, we consideredωρ=1/4eω, κρ=1/4eκ, and νρ=1/4eν. Thus, ω′ρ=1/4eω, κ′ρ=1/4eκ, and ν′ρ=1/4eν, respectively. It is clear that the function Θ:0,1×0,1⟶1,∞ is continuously differentiable with respect to its first variable, where ∂Θ/∂ρ=σeρσ>0. Moreover, 1/4eω, 1/4eκ, and 1/4eν are nonnegative real continuous functions on 0,1. Also, the two mappings take the following form:
(84)℧ω,κ,νρ=116∫01ρeρσe2ω+e2κ+e2νdσ,(85)ℷωρ=116∫01ρeρσeω+eκ+eνdσ.Now, we are going to satisfy the hypotheses of Theorem13.
(1)
Forω≤ω∗, κ≥κ∗, and ν≤ν∗, the functions ℵi:0,1×0,∞×0,∞⟶0,∞ and Ri:0,1×0,∞⟶0,∞ (for i=1,2,3) are continuous functions, so that(86)R1ρ,ωρ=116ρeω≤116ρeω∗=R1ρ,ω∗ρ,R2ρ,κρ=116ρeκ≥116ρeκ∗=R2ρ,κ∗ρ,R3ρ,νρ=116ρeν≤116ρeν∗=R3ρ,ν∗ρ.This obviously leads to
(87)ℵ1ρ,ωρ,κ′ρ=116ρeω+κ≤116ρeω∗+κ∗=ℵ1ρ,ω∗ρ,κ∗′ρ.Similarly, one can write
(88)ℵ2ρ,κρ,ν′ρ≥ℵ2ρ,κ∗ρ,ν∗′ρ,ℵ3ρ,νρ,ω′ρ≤ℵ3ρ,ν∗ρ,ω∗′ρ.(2)
From (82) and (83), and assuming that ℷκ′=ℷκ, we have(89)ℵ1ρ,ℷωρ,ℷκ′ρ=116ρeℷωeℷκ=116ρeℷω+ℷκ=116ρe1/16∫01ρeρσe2ω+e2κ+e2νdσ=116ρe℧ω,κ,ν=R1ρ,℧ω,κ,νρ.Similarly, one can write
(90)R2ρ,℧ω,κ,νρ=ℵ2ρ,ℷκρ,ℷν′ρ,R3ρ,℧ω,κ,νρ=ℵ3ρ,ℷνρ,ℷω′ρ.(3)
For any fixedρ∈0,1, there is β=1/4, so that(91)ℵ1ρ,ω∗ρ,κ∗′ρ=116ρeω∗+κ∗≥1414ρeω=R1ρ,ωρ.In the same manner, we have
(92)ℵ2ρ,κ∗ρ,ν∗′ρ≥R2ρ,κρ,ℵ3ρ,ν∗ρ,ω∗′ρ≥βR3ρ,νρ.Also, we have(93)ℵ1ρ,ωρ,κ′ρ−ℵ1ρ,ω∗ρ,κ∗′ρ=116ρeω+κ−eω∗+κ∗≤18ρeω+κ−eωsinceeω∗+κ∗≥eω=β2ℵ1ρ,ωρ,κ′ρ−R1ρ,ωρ.Similarly, withβ=1/4, one can write
(94)ℵ2ρ,κρ,ν′ρ−ℵ2ρ,κ∗ρ,ν∗′ρ≤β2ℵ2ρ,κρ,ν′ρ−R2ρ,κρ,ℵ3ρ,νρ,ω′ρ−ℵ3ρ,ν∗ρ,ω∗′ρ≤β2ℵ3ρ,νρ,ω′ρ−R3ρ,νρ.(4)
There is⅁0ρ=eρ∈0,∞, so that for any fixed ρ=1∈0,1, ρ1=1/16, ρ2=1/2, and ρ3=1/20, we have(95)164e≤132e2≤18e⇒14116eρ≤116×12eρ≤1412eρ⇒Rρ,116eρ≤ℵρ,116eρ,12eρ≤Rρ,12eρ⇒Rρ,ρ1n⅁0ρ≤ℵρ,ρ1n⅁0ρ,ρ2n⅁0′ρ≤Rρ,ρ2n⅁0ρ,180e≤140e2≤18e⇒14120eρ≤120×12eρ≤1412eρ⇒Rρ,120eρ≤ℵρ,120eρ,12eρ≤Rρ,12eρ⇒Rρ,ρ3n⅁0ρ≤ℵρ,ρ3n⅁0ρ,ρ2n⅁0′ρ≤Rρ,ρ2n⅁0ρ.Hence, we find that all hypotheses of Theorem13 are fulfilled. So, the Problem (80) has a unique solution.


## 4. Conclusion

Due to the multiple applications of fixed point theory, it has become widespread in many scientific disciplines, especially in nonlinear analysis. It contributes significantly to the study of the existence and uniqueness of the solution to many differential and integral equations, as well as integro-differential equations. So, the main objectives of this paper have been to present some new tripled fixed point results for mixed monotone mappings in the framework of partially ordered metric spaces, and these new results have extended to a lot of papers in the literature. Furthermore, to support the proposed results, some illustrative examples have been given and the existence and uniqueness of the solution to the integro-differential equation have been obtained.



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