Existence of a solution for a nonlinear integral equation by nonlinear contractions involving simulation function in partially ordered metric space

Nonlinear Analysis: Modelling and Control, Vol. 28, No. 3, 578–596 https://doi.org/10.15388/namc.2023.28.32119 Press Existence of a solution for a nonlinear integral equation by nonlinear contractions involving simulation function in partially ordered metric space* Parvaneh Lo0lo0a, Maryam Shamsb,1, Manuel De la Senc,d,2 aDepartment of Mathematics, Behbahan Khatam Alanbia University of Technology, Behbahan 6361647189, Iran [email protected] bDepartment of Pure Mathematics, University of Shahrekord, Shahrekord 88186-34141, Iran [email protected] cInstitute of Research and Development of Processes, University of Basque Country, Campus of Leioa, Bizkaia dDepartment of Electricity and Electronics, Faculty of Science and Technology, University of The Basque Country (UPV/EHU), 48080 Bilbao, Spain [email protected] Received: June 28, 2022 / Revised: February 24, 2023 / Published online: April 26, 2023 Abstract. In a recent paper, Khojasteh et al. presented a new collection of simulation functions, said Z-contraction. This form of contraction generalizes the Banach contraction and makes different types of nonlinear contractions. In this article, we discuss a pair of nonlinear operators that applies to a nonlinear contraction including a simulation function in a partially ordered metric space. For this pair of operators with and without continuity, we derive some results about the coincidence and unique common fixed point. In the following, many known and dependent consequences in fixed point theory in a partially ordered metric space are deduced. As well, we furnish two interesting examples to explain our main consequences, so that one of them does not apply to the principle of Banach contraction. Finally, we use our consequences to create a solution for a particular type of nonlinear integral equation. Keywords: simulation functions, coincidence point, compatible, partially ordered metric space, integral equation. *This research was supported by Basque Government, grant No. 1555-22. 1The author was supported by Shahrekord University and the Center of Excellence for Mathematics. 2Corresponding author. © 2023 The Author(s). Published by Vilnius University Press This is an Open Access article distributed under the terms of the Creative Commons Attribution Licence, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited. Existence of a solution for a nonlinear integral equation 579 1 Introduction Fixed point theory is a clear subject, which affords beneficial techniques and senses for dealing with diverse problems. Particularly, we mention the being of solutions of mathematical questions diminishable to equivalent fixed point problems. Thus, we remember that the Banach contraction principle [5] is based on this theory. Nevertheless, the fixed point theory has been able to attract many researchers. In 2018, Vetro [31] proved the existence and uniqueness of a fixed point in the setting of ordered metric spaces by introducing the notion of ordered S-G-contraction. Hoc and his colleagues [13] provided some new fixed point theorems in compact metric space. In 2022, Kim [18] studied the existence of a coupled fixed point in Hilbert space. Also, Gautam et al. [11] introduced the notion of interpolative Matkowski-type contraction, and they obtained the solution for the nonlinear matrix equations. Therefore, there are many achievements for enthusiasts, look, for example, [6,8–10,19–22, 28, 29, 33]. Recently, lots of conclusions became apparent linked to fixed point theorems in an ordered metric space. Run and Reurings [27] expressed the first conclusion in this orientation, where they expanded the Banach contraction principle in metric space equipped with a partial order. Subsequently, Nieto and Rodríguez-López [24] generalized the previous results and used them to find a unique solution for a specific type of ordinary differential equation. More progress in the above-argued results is detected in [2,3,12, 23, 25, 30]. Recently, the concept of simulation function was introduced and studied by Khojasteh et al. [17]. By using the simulation functions Vetro [32] investigated the existence of a common fixed point and coincidence point in both metric space and partial metric space. In this article, we presume a pair of nonlinear operators satisfying in nonlinear contractions including a simulation function in a metric space with a partial order. We generalize some results Khojaste et al. [17] to obtain coincidence and common fixed point results for this pair of operators with and without continuity. Also, we process two interesting examples to explain our main results, so that one of them does not apply to the principle of Banach contraction. Then we exploit our achievements to create a solution for a particular type of nonlinear integral equation. 2 Preliminaries The following definition was given by Argoubi et al. [4]. Definition 1. Let (X, d)be a metric space, and let ζ: [0,∞)×[0,∞)→Rsatisfies the following conditions: (ζ1)ζ(p, q)< q −pfor all p, q > 0; (ζ2) If {pn}and {qn}are sequences in (0,∞)such that limn→∞ pn= limn→∞ qn= l > 0, then lim sup n→∞ ζ(pn, qn)<0. Then ζis a simulation function. Nonlinear Anal. Model. Control, 28(3):578–596, 2023 580 P. Lo0lo0et al. Remark 1. Initially, Khojasteh et al. [17] defined the simulation function as a mapping ζ: [0,∞)×[0,∞)→Rsatisfying ζ(0,0) = 0 and conditions (ζ1) and (ζ2) of Definition 1. In the following, we will use the modified definition by Argoubi et al. [4]. Before starting the main results of this research, we render many examples that highlight their possible applicability to the field of fixed point theory. Example 1. (See [17].) Let ζi: [0,∞)×[0,∞)→R,i= 1,2,...,6, be defined by (i) ζ1(p, q) = ψ(q)−φ(p)for all p, q ∈[0,∞), where ψ, φ : [0,∞)→[0,∞)are two continuous functions such that ψ(t) = φ(t) = 0 if only if t= 0 and ψ(t)< t⩽φ(t)for all t > 0. (ii) ζ2(p, q) = αq −pfor all p, q ∈[0,∞)is a particular case of ζ1with φ(t) = t and ψ(t) = αt for all t⩾0and α∈[0,1). (iii) ζ3(p, q) = q−ϕ(q)−pfor all p, q ∈[0,∞), where ϕ: [0,∞)→[0,∞)is a lower semicontinuous function such that φ−1(0) = {0}. (iv) ζ4(p, q) = qϕ(q)−pfor all p, q ∈[0,∞), where ϕ: [0,∞)→[0,1) is a function such that lim supt→r+ϕ(t)<1for all r > 0. (v) ζ5(p, q) = q−(f(p, q)/g(p, q))pfor all p, q ∈[0,∞), where f, g : [0,∞)→ (0,∞)are two continuous functions with respect to each variable such that f(p, q)> g(p, q)for all p, q > 0. (vi) ζ6(p, q) = q−Rp 0φ(u) dufor all p, q ∈[0,∞), where φ: [0,∞)→[0,∞)is a function such that R 0φ(u) duexists, and R 0φ(u) du>for each  > 0. Definition 2. Let (X, d)be a metric space and S, T :X→X. If v=Su =Tu for some uin X, then uis called a coincidence point of Sand T. Definition 3. (See [15].) Let (X, d)be a metric space. The mappings S, T :X→X are compatible if and only if for any sequence {un}in Xsuch that limn→∞ Sun= limn→∞ Tun,limn→∞ d(STun, T Sun)=0. Definition 4. (See [16].) Let (X, d)be a metric space. The mappings S, T :X→Xare weakly compatible if and only if Su =Tu for some u∈Ximplies that STu =T Su or Sand Tcommute at their coincidence points. If Sand Tare compatible, then Sand Tare weakly compatible. Definition 5. (See [7].) Let (X, 4)is a partially ordered set and S, T :X→X.Sis said to be T-nondecreasing if for u, v ∈X, Tu 4Tv =⇒Su 4Sv. 3 Main result The following main theorem is a generalized coincidence point theorem for maps that are not necessarily continuous. https://www.journals.vu.lt/nonlinear-analysis Existence of a solution for a nonlinear integral equation 581 Theorem 1. Let (X, 4)be a partially ordered set and suppose that there exists a metric don Xsuch that (X, d)is complete metric space. Suppose that there exist a simulation function ζand S, T :X→Xsuch that ζd(Sx, Sy), d(Tx, Ty)⩾0∀x, y ∈X:Tx 4T y, (1) and suppose the following hypotheses: (i) SX ⊆TX and TX is closed; (ii) Sis T-nondecreasing; (iii) If {Txn} ⊂ Xis a nondecreasing sequence, which converges to Tu in TX, then Txn4Tu for all n⩾0. If there exists x0∈Xsuch that Tx04Sx0, then Sand Thave a coincidence point, that is, there exists v∈Xsuch that Sv =Tv. Proof. Using the theorem condition, we have x0∈Xsuch that T x04Sx0. Since SX ⊆TX, then there exists x1∈Xsuch that Tx1=Sx0and Tx04Sx0=Tx1. Since Sis T-nondecreasing, we have Sx04Sx1. Continuing this process, we construct the sequence {xn}with the following conditions: Sxn=Txn+1 ∀n⩾0,(2) and (3) Tx04Sx0=Tx14Sx1=T x24Sx24· · · 4Sxn−1=Txn4Sxn=Txn+1 4· · · . If two consecutive members of the sequences {Sxn}or {Txn}are equal, then the conclusion of the theorem follows. So we have d(Sxn, Sxn+1)6= 0, d(Txn, Txn+1)6= 0 ∀n⩾0.(4) If for some n∈N, we assume that d(Txn−1, Txn)< d(T xn, Txn+1), then by property (ζ1) of simulation function and (2)–(4) we have 0⩽ζd(Sxn−1, Sxn), d(Txn−1, Txn) =ζd(Txn, Txn+1), d(Txn−1, Txn) < d(Txn−1, Txn)−d(Txn, T xn+1)<0. This contradiction shows that d(Txn, Txn+1)⩽d(Txn−1, Txn). This implies that the sequence {d(Txn−1, Txn)}is a monotone decreasing sequence of nonnegative real numbers, and consequently, there exists r⩾0such that the sequence {d(Txn−1, Txn)}converges to r. Nonlinear Anal. Model. Control, 28(3):578–596, 2023 582 P. Lo0lo0et al. Suppose r > 0. By (3) we know that the elements Txnand T xn+1 are comparable, so using property (ζ2) of a simulation function with pn=d(Sxn, Sxn+1)and qn= d(Sxn−1, Sxn), we have 0⩽lim sup n→∞ ζd(Sxn−1, Sxn), d(Txn−1, Txn) = lim sup n→∞ ζd(Txn, Txn+1), d(Txn−1, Txn)<0, which is a contradiction, and hence, lim n→∞ d(Txn−1, Txn) = 0. The next step is to show that the sequence {T xn}is Cauchy. By contradiction and by Lemma 2.1 of [14] there exist an  > 0and {Txm(k)},{Txn(k)}⊂{Txn}with n(k)> m(k)⩾kfor all k∈Nsuch that lim k→∞ d(Txm(k), Txn(k)) = lim k→∞ d(Txm(k)+1, Txn(k)+1) = , (5) d(Txm(k), Txn(k))⩾. (6) Then we can assume that d(Txm(k)+1, Txn(k)+1)>0∀k∈N.(7) Again, by (3) we know that the elements Txm(k)and Txn(k)are comparable, so using (5)–(7) and property (ζ2) of a simulation function with pn=d(Txm(k)+1, Txn(k)+1)and qn=d(Txm(k), Txn(k)), we have 0⩽lim sup k→∞ ζd(Sxm(k), Sxn(k)), d(Txm(k), Txn(k)) = lim sup k→∞ ζd(Txm(k)+1, Txn(k)+1), d(Txm(k), Txn(k))<0, which is a contradiction. We conclude that the sequence {Txn}is a Cauchy sequence, and hence, {Txn}is convergent in the complete metric space (X, d).TX is closed, therefore, by (2) there exists u∈Xsuch that lim n→∞ Sxn= lim n→∞ Txn=Tu. (8) From (3) and (8) we know that {T xn}is a nondecreasing sequence in TX such that Txn→Tu, then by condition (iii) and (4) we have Txn≺Tu. (9) Again, by (3), (4) and since Sis T-nondecreasing, we have Sxn≺Su. (10) https://www.journals.vu.lt/nonlinear-analysis Existence of a solution for a nonlinear integral equation 583 Using property (ζ1) of a simulation function, (9) and (10), we have 0⩽ζd(Sxn, Su), d(Txn, Tu)< d(T xn, Tu)−d(Sxn, Su)∀n∈N. Taking n→ ∞ in the above inequality, we have limn→∞ Sxn=Su. Then Su = lim n→∞ Sxn= lim n→∞ Txn=Tu. (11) This completes the proof. Now, we will prove the existence and uniqueness theorem of a common fixed point. Theorem 2. If in Theorem 1, it is additionally assumed that Sand Tare weakly compatible and Tu 4TTu, where uis a coincidence point of Sand T, then Sand Thave a common fixed point in X. Moreover, if a set of fixed points of Tis totally ordered, then Sand Thave a unique common fixed point. Proof. We prove v=Su =Tu. Since Sand Tare weakly compatible, by (11) we have STu =TSu. Then Tv =TTu=TSu =STu =SSu =Sv. (12) If Tv =vor Sv =v, then vis a common fixed point. Otherwise, i.e., if T v 6=vand Sv 6=v, by property (ζ1) of a simulation function with Tu 4TTu 0⩽ζd(v, Sv), d(v, Tv)=ζd(Su, SSu), d(Tu, TTu) < d(Tu, TT u)−d(Su, SSu). Using (11) and (12) in the above inequality, we have d(Su, SSu)< d(Tu, T Tu) = d(Su, SSu), which is a contradiction. Therefore, Tv =vor Sv =v, and we conclude that v=Sv = Tv. Now, suppose that the set of fixed points of Tis totally ordered. Assume on the contrary that v=Sv =Tv and v0=Sv0=Tv0but v6=v0. Since vand v0contain a set of fixed points of T, without loss of generality, we assume that T v 4Tv0. If Sv =Sv0 or Tv =Tv0, then v=v0, which is a contradiction. Otherwise, i.e., if Sv 6=Sv0and Tv 6=Tv0, by property (ζ1) of a simulation function we have 0⩽ζd(Sv, Sv0), d(T v, Tv0)=ζd(v, v0), d(v, v0) < d(v, v0)−d(v, v0) = 0, which is a contradiction. Therefore, Sand Thave a unique common fixed point. In the next theorem, we will omit condition (iii) of Theorem 1, and we will assume that S, T :X→Xare continuous and compatible. Nonlinear Anal. Model. Control, 28(3):578–596, 2023 584 P. Lo0lo0et al. Theorem 3. Let (X, 4)be a partially ordered set, and let there exists a metric don X such that (X, d)is complete metric space. Suppose that there exist a simulation function ζand S, T :X→Xsuch that ζd(Sx, Sy), d(Tx, Ty)⩾0∀x, y ∈X:Tx 4T y. We suppose the following hypotheses: (i) SX ⊆TX; (ii) Sis T-nondecreasing; (iii) Sand Tare continuous; (iv) The pair {S, T }is compatible. If there exists x0∈Xsuch that Tx04Sx0, then Sand Thave a coincidence point, that is, there exists u∈Xsuch that Su =Tu. Further, if Tu 4TTu and the set of fixed points of Tis totally ordered, then Sand Thave a unique common fixed point. Proof. Following the proof of Theorem 1, we have that {T xn}is a Cauchy sequence in the complete metric space (X, d). Then there exists u∈Xsuch that lim n→∞ Sxn= lim n→∞ Txn=u. (13) Since Sand Tare compatible, this implies that lim n→∞ S(Txn), T(Sxn)= 0.(14) From (13) and the continuity of Sand Twe have lim n→∞ T(Txn) = Tu, lim n→∞ S(Txn) = Su. (15) By the triangular inequality we have d(Su, Tu)⩽Su, S(T xn)+dS(T xn), T(Sxn)+dT(Txn+1), Tu. By (14) and (15) and letting n→ ∞, we obtain: d(Su, Tu)⩽0, therefore, Su =Tu, that is, uis the coincidence point of Sand T. Finally, because Sand Tare compatible (therefore, they are weakly compatible) and, on the other hand, Tu 4TTu and set of fixed points of Tis totally ordered, then by Theorem 2, Sand Thave a unique common fixed point. If T:X→Xis the identity mapping, we can deduce easily the following fixed point results. It is an immediate consequence of Theorem 1. https://www.journals.vu.lt/nonlinear-analysis Existence of a solution for a nonlinear integral equation 585 Theorem 4. Let (X, 4)be a partially ordered set and suppose that there exists a metric don Xsuch that (X, d)is complete metric space. Suppose that there exist a simulation function ζand S:X→Xsuch that ζd(Sx, Sy), d(x, y)⩾0∀x, y ∈X:x4y. We suppose the following hypotheses: (i) Sis a nondecreasing function; (ii) If {un}is a nondecreasing sequence, which converges to uin X, then un4u for all n⩾0. If there exists x0∈Xsuch that x04Sx0, then Shas a fixed point. The following result is an immediate consequence of Theorem 3. Theorem 5. Let (X, 4)be a partially ordered set and suppose that there exists a metric don Xsuch that (X, d)is complete metric space. Suppose that there exist a simulation function ζand S:X→Xsuch that ζd(Sx, Sy), d(x, y)⩾0∀x, y ∈X:x4y. We suppose the following hypotheses: (i) Sis a nondecreasing function; (ii) Sis continuous. If there exists x0∈Xsuch that x04Sx0, then Shas a fixed point. 4 Consequences In this section, as applications, we obtain some results of Theorem 1 in fixed point theory in partially ordered metric space via specific choices of simulation functions. Let (X, 4)be a partially ordered set and suppose that there exists a metric don X such that (X, d)is a complete metric space. Corollary 1. Let S, T :X→Xbe mappings such that there exist two continuous functions φ, ψ : [0,∞)→[0,∞)verifying ψ(t) = φ(t)=0if and only if t= 0, ψ(t)< t ⩽φ(t)for all t > 0, and φd(Sx, Sy)⩽ψd(Tx, Ty)∀x, y ∈X:Tx 4T y. We suppose the following hypotheses: (i) SX ⊆TX and TX is closed; (ii) Sis T-nondecreasing; (iii) If {Txn}⊂Xis a nondecreasing sequence converges to Tu in TX, then Txn4 Tu for all n⩾0. Nonlinear Anal. Model. Control, 28(3):578–596, 2023 586 P. Lo0lo0et al. If there exists x0∈Xsuch that Tx04Sx0, then Sand Thave a coincidence point, that is, there exists u∈Xsuch that Su =T u. Further, if Sand Tare weakly compatible, Tu 4TTu, and the set of fixed points of Tis totally ordered, then Sand Thave a unique common fixed point. Proof. The result follows from Theorems 1 and 2 by taking as simulation function ζ1(p, q) = ψ(q)−φ(p)∀p, q ⩾0, which was introduced in Example 2. Corollary 2 [Banach type]. Let S, T :X→Xbe mappings such that there exists α∈[0,1) verifying d(Sx, Sy)⩽αd(Tx, Ty)∀x, y ∈X:T x 4Ty. We suppose the following hypotheses: (i) SX ⊆TX and TX is closed; (ii) Sis T-nondecreasing; (iii) If {Txn}⊂Xis a nondecreasing sequence converges to Tu in TX, then Txn4 Tu for all n⩾0. If there exists x0∈Xsuch that Tx04Sx0, then Sand Thave a coincidence point, that is, there exists u∈Xsuch that Su =T u. Further, if Sand Tare weakly compatible, Tu 4TTu, and the set of fixed points of Tis totally ordered, then Sand Thave a unique common fixed point. Proof. The result follows from Theorems 1 and 2 by taking as simulation function ζ2(p, q) = αq −p∀p, q ⩾0, which was introduced in Example 2. Corollary 3. Let S, T :X→Xbe mappings such that there exists a lower semicontinuous function ϕ: [0,∞)→[0,∞)verifying ϕ−1({0}) = {0}and d(Sx, Sy)⩽d(Tx, Ty)−ϕd(Tx, T y)∀x, y ∈X:Tx 4Ty. We suppose the following hypotheses: (i) SX ⊆TX and TX is closed; (ii) Sis T-nondecreasing; (iii) If {Txn} ⊂ Xis a nondecreasing sequence, which converges to Tu in T X, then Txn4Tu for all n⩾0. If there exists x0∈Xsuch that Tx04Sx0, then Sand Thave a coincidence point, that is, there exists u∈Xsuch that Su =T u. Further, if Sand Tare weakly compatible, Tu 4TTu, and the set of fixed points of Tis totally ordered, then Sand Thave a unique common fixed point. https://www.journals.vu.lt/nonlinear-analysis Existence of a solution for a nonlinear integral equation 593 which implies kx−yk⩽1 1− |δ|P2N2 kTx −Tyk.(18) From (17) and (18) we get kSx −Syk⩽||P1N1 1− |δ|P2N2 kTx −Tyk, and, since α=||P1N1/(1 − |δ|P2N2)<1, if we define ζ(p, q) = αq −pfor all p, q ∈[0,∞), then we have ζd(Sx, Sy), d(Tx, Ty)⩾0 for all x, y ∈C(I)with Tx 4Ty. Thus, condition (1) is trivially satisfied. Next, we can show that S(C(I)) ⊆T(C(I)). Indeed, by (iii) for x(t)∈C(I)we have TSx(t) + f1(t) =Sx(t) + f1(t)−f1(t)−δ T Z 0 n2(t, s)k2s, Sx(s) + f1(s)ds =Sx(t)−δ T Z 0 n2(t, s)k2 s,  s Z 0 n1(s, v)k1v, x(v)dv+f1(s)−f2(s)!ds =Sx(t). Clearly, hypothesis (iv) means that Sis T-nondecreasing. Next, by (v) we get x0−f1(t)−δ T Z 0 n2(t, s)k2s, x0(s)ds⩽−f2(t) +  t Z 0 n1(t, s)k1s, x0(s)ds, that is, Tx04Sx0. 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