# New Fixed Point Theorems and Application of Mixed Monotone Mappings in Partially Ordered Metric Spaces

Authors: Li Liu; Anmin Mao; Yingying Shi
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We consider the existence of a coupled fixed point for mixed monotone mappingF:X×X→X satisfying a new contractive inequality which involves an altering distance function in partially ordered metric spaces. We also establish some uniqueness results for coupled fixed points, as well as the existence of fixed points of mixed monotone operators. The presented results generalize and develop some existing results. In addition to an example as well as an application, we establish some uniqueness results for a system of integral equations.


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## Body

## 1. Introduction and Preliminaries

In this paper we aim to establish coupled fixed point theorems for a mixed monotone mapping in a metric space endowed with partial order. The concept of the mixed monotone operator was introduced by Guo and Lakshmikantham in [1]. Existence of fixed points in a metric space has been studied for a long time (see [2–10]). The Banach contraction principle, which plays a very important role in nonlinear analysis, is the most famous tool in the study of a fixed point theorem. In the last decade, there has been a trend to weaken the requirement on the contraction by considering metric spaces endowed with partial order. It is of interest to determine if it is still possible to establish the existence of a unique fixed point assuming that the operator considered is monotone in such a setting.Recently, there have been a lot of generalizations of the Banach contraction-mapping principle in the literature (see [11–28]). Investigation of the existence of fixed points for single-valued mappings in partially ordered metric spaces was initially considered by Ran and Reurings in [29] who proved the existence of a unique fixed point.Following basically the same approach as the one in [29], Bhaskar and Lakshmikantham in [30] proved a fixed point theorem for a mixed monotone mapping in a metric space endowed with partial order. Bhaskar and Lakshmikantham extended [29, Theorem 2.1] to mixed monotone operators so that they can enlarge, in a unified manner, the class of problems that can be investigated. The authors in [30] also established some uniqueness results for coupled fixed points, as well as the existence of fixed points of F.Based on the works in [30], Zhao in [31] obtained a more general coupled fixed point theorem for mixed monotone operators. Another generalization of the contraction principle was suggested by Alber and Guerre-Delabriere. In 1997, Alber and Guerre-Delabriere [32] introduced the concept of weak contractions in Hilbert spaces. This concept was extended to metric spaces by Rhoades in [33].Definition 1.
A mappingT:X→X, where (X,d) is a metric space, is said to be weakly contractive if (1)dTx,Ty≤dx,y-ψdx,y,where x,y∈X and ψ is an altering distance function.
The notion of altering distance function was introduced by Khan et al. [34]. An altering distance function is a control function that alters distance between two points in a metric space.
It was shown in [32] that, for Hilbert spaces, weakly contractive maps possess a unique fixed point, without any additional assumptions, and it was noted that the same is true at least for uniformly smooth and uniformly convex Banach spaces. In [33], Rhoades proved that the theorem remains true in arbitrary complete metric spaces, which was improved and extended by Dutta and Choudhury in [35]. Recently, a new version of the context of ordered metric spaces has been proved by Harjani and Sadarangani in [12]. We refer the readers to [36–41] for more related works.
Motivated by the papers mentioned above, we aim to establish coupled fixed point results for mixed monotone operatorF which isweakly contractive in partially ordered complete metric spaces. Our main results are Theorems 3, 6, and 8. To the best of our knowledge, there are no similar results in the literature on the existence of the coupled fixed point. Compared with the results obtained in the [32, 33, 35], ψ in our results is not necessary to be an altering distance function, which means that F satisfies a more general contractive condition. On the other hand, our result is still valid for F not necessarily continuous if we require that the underlying metric space X has an additional property. Our main results will generalize and develop the results given in [12].In Section2, we give the proof of our main results. In Section 3, as an application of our theorems, we consider the existence of a unique solution to a system of integral equations.


## 2. Coupled Fixed Point Theorems

Let(X,≤) be a partially ordered set and let d be a metric on X such that (X,d) is a complete metric space. Further, we endow the product space X×X with the following partial order: (2)forx,y,u,v∈X×X,u,v≤x,y⇔x≥u,y≤v.Definition 2 (see [34]).
A functionφ:[0,+∞)→[0,+∞) is called an altering distance function if the following conditions are satisfied:
(i)φ is continuous and nondecreasing
(ii)φ(t)=0⇔t=0
Next we introduce a set of functions(3)Ψ≔ψ∈C0,+∞,0,+∞∣ψ0=0,and  for  anyt>0,ψt>0.It follows from Definition2 that if φ is an altering distance function, φ∈Ψ.The first main result is the following coupled fixed point result.Theorem 3.
Assume(H1)
ψ∈Ψ (ψ is not necessary to be an altering distance function)(H2)
F:X×X→X being a mixed monotone mapping, there exist a constant k∈(0,1) such that(4)φdFu,v,Fx,y+dFv,u,Fy,x≤kφdu,x+dv,y-ψkdu,x+dv,yand, for eachu≥x,v≤y, φ is an altering distance function which satisfies(5)φt+s≤φt+φs,∀t,s∈0,+∞(H3)
there exist(u0,v0)∈X×X such that u0≤F(u0,v0) and v0≥F(v0,u0)(H4)
(a)
F is continuous or(b)
X has the following properties:
(i)
If a nondecreasing sequence{un}→u, then un≤u,∀n(ii)
If a nonincreasing sequence{vn}→v, then vn≥v,∀n
Then there existu,v∈X, such that(6)u=Fu,vandv=Fv,u.Proof.
Since(7)u0≤Fu0,v0andv0≥Fv0,u0,let F(u0,v0)=u1, F(v0,u0)=v1, F(u1,v1)=u2, and F(v1,u1)=v2; we have u0≤u1v0≥v1. Denote (8)F2u0,v0=FFu0,v0,Fv0,u0=Fu1,v1=u2,F2v0,u0=FFv0,u0,Fu0,v0=Fv1,u1=v2.Note that u0≤u1, v0≥v1; it follows from the mixed monotone property of F that(9)Fu1,v1≥Fu0,v1≥Fu0,v0,Fv1,u1≤Fv0,u1≤Fv0,u0,which implies (10)u2=F2u0,v0=Fu1,v1≥Fu0,v0=u1,v2=F2v0,u0=Fv1,u1≤Fv0,u0=v1.For n=1,2,⋯, let (11)un+1=Fn+1u0,v0=FFnu0,v0,Fnv0,u0,vn+1=Fn+1v0,u0=FFnv0,u0,Fnu0,v0.i.e.,(12)un+1=Fun,vn,vn+1=Fvn,un.It is easy to see that (13)u0≤Fu0,v0=u1≤F2u0,v0=u2≤···≤Fn+1u0,v0≤···,v0≥Fv0,u0=v1≥F2v0,u0=v2≥···≥Fn+1v0,u0≥···,i.e., (14)u0≤u1≤···≤un≤···,v0≥v1≥···≥vn≥···,n=1,2,⋯.By (4) (12), we have(15)φdun+1,un+dvn+1,vn=φdFun,vn,Fun-1,vn-1+dFvn,un,Fvn-1,un-1≤kφdun,un-1+dvn,vn-1-ψkdun,un-1+dvn,vn-1,and thus(16)φdun+1,un+dvn+1,vn≤kφdun,un-1+dvn,vn-1.From k∈[0,1), we have(17)φdun+1,un+dvn+1,vn≤φdun,vn-1+dvn,vn-1.Since φ is continuous and nondecreasing, (18)dun+1,un+dvn+1,vn≤dun,un-1+dvn,vn-1.Let ξn=d(un+1,un)+d(vn+1,vn). From 0≤ξn+1≤ξn, {ξn} is a Cauchy sequence, and thus there exist ξ≥0 such that(19)limn→∞⁡dun+1,un+dvn+1,vn=ξ.
Now we claimξ=0. In fact, by (14) and (H1), one has(20)φξ=limn→∞⁡φξn=limn→∞⁡φdun+1,un+dvn+1,vn≤klimn→∞⁡φdun,un-1+dvn,vn-1-limn→∞⁡ψkdun,un-1+dvn,vn-1=kφξ-limn→∞⁡ψkξn-1=kφξ-limt→kξ⁡ψt≤kφξwhich implies φ(ξ)=0, so is ξ.
Next we will show{un} and {vn} are Cauchy sequences.
Arguing indirectly we suppose{un} is not a Cauchy sequence. Thus, there exists a positive constant ϵ such that, for any K>0, there exist nk>mk>K such that (21)dunk,umk+dvnk,vmk≥ϵ.For mk, let nk be the smallest integer satisfying nk≥mk and (21). Thus, one has (22)dunk-1,umk+dvnk-1,vmk<ϵ,which implies(23)dunk,umk+dvnk,vmk≤dunk,unk-1+dunk-1,umk+dvnk,vnk-1+dvnk-1,vmk≤dunk,unk-1+dvnk,vnk-1+ϵ=ξnk-1+ϵ,and, by (21) and ξn→0 (n→∞), we have (24)sk≔dunk,umk+dvnk,vmk→ϵ,k→∞.
The triangular inequality gives us(25)sk≤dunk,unk+1+dunk+1,umk+1+dumk+1,umk+dvnk,vnk+1+dvnk+1,vmk+1+dvmk+1,vmk=ξnk+ξmk+dunk+1,umk+1+dvnk+1,vmk+1,and thus(26)φsk=φξnk+ξmk+dunk+1,umk+1+dvnk+1,vmk+1≤φξnk+ξmk+φdunk+1,umk+1+φdvnk+1,vmk+1.By (14), (27)φdunk+1,umk+1+φdvnk+1,vmk+1=φdFunk,vnk,Fumk,vmk+dFvnk,unk,Fvmk,umk≤kφdunk,umk+dvnk,vmk-ψkdunk,umk+dvnk,vmk=kφsk-ψksk.From (26) (27), one has(28)φsk≤φξnk+ξmk+kφsk-ψsk≤φξnk+ξmk+φsk-ψsk,and, letting k→∞ in the above inequality,(29)limk→∞⁡φsk≤limk→∞⁡φξnk+ξmk+φsk-limk→∞⁡ψsk.From (H1), note that ξn→0, sk→ϵ; we have (30)φϵ≤φ0+φϵ-limk→∞⁡ψsk=φϵ-limk→∞⁡ψsk<φϵ,which is a contradiction. This shows that {un} and {vn} are Cauchy sequences. Since X is a complete metric space, there exist u,v∈X such that(31)un→u,vn→v,n→∞.
Case (a). Assume F is continuous. Then(32)u=limn→∞⁡un=limn→∞⁡Fun-1,vn-1=Flimn→∞⁡un-1,limn→∞⁡vn-1=Fu,v,v=limn→∞⁡vn=limn→∞⁡Fvn-1,un-1=Flimn→∞⁡vn-1,limn→∞⁡un-1=Fv,u,and, thus, (u,v)∈X×X is a coupled fixed point of F.
Case (b). Assume X has the following properties:
(i) If a nondecreasing sequence{un}→u, then un≤u,∀n
(ii) If a nonincreasing sequence{vn}→v, then vn≥v,∀n
By (14) (31), we have (33)un≥u,vn≤v,n=1,2,⋯.Again, the triangular inequality gives us(34)du,Fu,v≤du,un+1+dun+1,Fu,v=du,un+1+dFun,vn,Fu,v.From (H1)(H2), observing that φ is nondecreasing and using un≤u, vn≥v, we get(35)φdu,Fu,v≤φdu,un+1+dun+1,Fu,v=φdu,un+1+dFun,vn,Fu,v≤φdu,un+1+φdFun,vn,Fu,v≤φdu,un+1+φdFun,vn,Fu,v+dFvn,un,Fv,u≤φdu,un+1+kφdun,u+dvn,v-ψkdun,u+dvn,v→0asn→∞and we get φ(d(u,F(u,v)))=0. Then u=F(u,v). Similarly, we get v=F(v,u). The proof is completed.Corollary 4.
Let(X,≤) be a partially ordered set and let d be a metric on X such that (X,d) is a complete metric space. Assume F:X×X→X is a fixed monotone operator and(f1)
there exist(u0,v0)∈X×X such that u0≤F(u0,v0) and v0≥F(v0,u0)(f2)
there existψ∈Ψ,k∈(0,1), for u,v,x,y∈X with u≥x, v≤y,(36)dFu,v,Fx,y+dFv,u,Fy,x≤kdu,x+dv,y-ψkdu,x+dv,y(f3)
(a)
F is continuous or(b)
X has the following properties:
(i)
If a nondecreasing sequence{un}→u, then un≤u,∀n(ii)
If a nonincreasing sequence{vn}→v, then vn≥v,∀n
Then there existu,v∈X such that(37)u=Fu,vandv=Fv,u.Proof.
Letφ(t)=t,t∈[0,+∞). The proof is finished by Theorem 3.Note that, in (36), ψ is not necessary to be an altering distance function.Corollary 5.
Let(X,≤) be a partially ordered set and let d be a metric on X such that (X,d) is a complete metric space. Assume F:X×X→X is a fixed monotone operator and(f1)
there exist(u0,v0)∈X×X such that u0≤F(u0,v0) and v0≥F(v0,u0)(f2)
there existψ∈Ψ,k∈(0,1), for u,v,x,y∈X with u≥x, v≤y,(38)dFu,v,Fx,y+dFv,u,Fy,x≤k2du,x+dv,y(f3)
(a)
F is continuous or(b)
X has the following property:
(i)
If a nondecreasing sequence{un}→u, then un≤u,∀n(ii)
If a nonincreasing sequence{vn}→v, then vn≥v,∀n
Then there existu,v∈X such that(39)u=Fu,vandv=Fv,u.Proof.
Letψ(t)=t/2,t∈[0,+∞). The proof is finished by Corollary 4.Next, we discuss the uniqueness of the coupled fixed point. Since the contractivity assumption is made only on comparable elements inX×X, Theorem 3 cannot guarantee the uniqueness of the coupled fixed point. Before stating our uniqueness results, we require that the product space X×X endowed with the partial order mentioned earlier have the following property:( H 5 ) For every ( u , v ) , ( s , w ) ∈ X × X , there exists ( x , y ) ∈ X × X which is comparable to ( u , v ) and ( s , w ).Note that(H5) is equivalent to (see [42]) the following:( H 5 ′ ) Every pair of elements in X × X has either a lower bound or an upper bound.Next, we state our second main result.Theorem 6.
Assume(H1)-(H4) and (H5) (or (H5′)) hold. Then the coupled fixed point of F is unique.Proof.
By(H1)-(H4), applying Theorem 3, F has a coupled fixed point. Let (u,v),(w,s) be the coupled fixed points of F, i.e.,(40)u=Fu,v,v=Fv,u,w=Fw,s,s=Fs,w.To show the uniqueness, we need to prove u=w,v=s.
By(H5), there exists (x,y)∈X×X which is comparable to (u,v) and (s,w). Let (41)x0=x,y0=y,xn+1=Fxn,yn,yn+1=Fxn,yn.Since (x,y) is comparable to (u,v) with respect to the ordering in X×X, we suppose (42)x0,y0=x,y≤u,v.Then, for n=1,2,…, (43)xn,yn≤u,v.In fact, (42) implies that (44)x0=x≤u,y0=y≥v.For n=1, by the mixed monotone property, we have (45)x1=Fx0,y0≤Fu,y0≤Fu,v=u,y1=Fy0,x0≥Fv,x0≥Fv,u=v,i.e.,x1,y1≤u,v.For n=k, we suppose (xk,yk)≤(u,v). Therefore, for n=k+1, we have (46)xk+1=Fxk,yk≤Fu,yk≤Fu,v=u,yk+1=Fyk,xk≥Fv,xk≥Fv,u=v,i.e.,xk+1,yk+1≤u,v,which implies (43) holds.
From (41) (H2),(47)φdu,xn+1+dyn+1,v=φdFu,v,Fxn,yn+dFyn,xn,Fv,u≤kφdu,xn+dv,yn-ψkdu,xn+dv,yn≤φdu,xn+dv,ynwhich implies(48)du,xn+1+dyn+1,v≤du,xn+dv,yn.Set ξn≔d(u,xn)+d(yn,v),n∈N. By (48), we have 0≤ξn+1≤ξn. Then, there exists ξ∈R+=[0,+∞) such that (49)ξ=limn→∞⁡ξn=limn→∞⁡du,xn+dyn,v.If ξ>0, by (H2),(50)φξ=limn→∞⁡φξn≤limn→∞⁡kφξn-limn→∞⁡ψkξn=kφξ-limn→∞⁡ψkξn<kφξ<φξwhich is a contradiction. So ξ=0. We have(51)0=limn⁡ξn=limn→∞⁡du,xn+dyn,v=0,and then(52)limn→∞⁡du,xn=limn→∞⁡dyn,v=0.Similarly, we also have (53)limn→∞⁡dw,xn=limn→∞⁡dyn,s=0.It follows from (52) (53) that (54)u,v=w,s.Corollary 7.
In addition to the hypothesis of Corollary4, suppose (H5)(or(H5′)) holds. Then F has a unique coupled fixed point.Next, we state our third main result.Theorem 8.
In addition to the hypothesis of Theorem6, suppose one of the followings holds:
(i) Elements of the coupled fixed point(u,v) are comparable
(ii)u0 and v0 are comparable
(iii) Every pair of elements of X has an upper bound or a lower bound inX
Then, we haveu=v; that is, F has a unique fixed point:(55)Fu,u=u.Proof.
From Theorem6, F has a unique coupled fixed point (u,v).
Case (i). Since u and v are comparable, we have u≥v or u≤v. Suppose u≥v; then u=F(u,v) and v=F(v,u) are comparable. By (4) in (H2), one obtains(56)φ2du,v=φdu,v+dv,u=φdFu,v,Fv,u+dFv,u,Fu,v≤kφdu,v+dv,u-ψkdu,v+dv,u≤kφ2du,v.Noting k∈(0,1), it follows from (56) that d(u,v)=0, i.e., u=v.
Case (ii). Since u0 and v0 are comparable, we have u0≥v0 or u0≤v0. Suppose we are in the first case, u0≥v0. Let(57)un+1=Fun,vn,vn+1=Fvn,un,n=0,1,2⋯.From Theorem 3, the coupled fixed point (u,v) satisfies (58)u=limn→∞⁡un=limn→∞⁡Fun,vn,v=limn→∞⁡vn=limn→∞⁡Fvn,un.By u0≥v0 and the mixed monotone property of F, one has (59)u1=Fu0,v0≥Fv0,u0=v1,and, hence, by induction one obtains (60)un≥vn,n=0,1,2···.On the one hand, it follows from the continuity of the distance d that(61)du,v=dlimn→∞⁡Fun,vn,limn→∞⁡Fvn,un=limn→∞⁡dFun,vn,Fvn,un=limn→∞⁡dun+1,vn+1.On the other hand, by (4) in (H2), one gets(62)φ2dun+1,vn+1=φdun+1,vn+1+dvn+1,un+1=φdFun,vn,Fvn,un+dFvn,un,Fun,vn≤kφdun,vn+dvn,un-ψkdun,vn+dvn,un≤kφ2dun,vn≤k·kφ2dun-1,vn-1≤knφ2du1,v1,which implies (63)0=limn→∞⁡φ2dun+1,vn+1=φ2du,vand therefore d(u,v)=0, which finishes the proof.Case (iii). If u,v are comparable, see Case (i). If u,v are not comparable, then there exists w∈X comparable to u and v. Without loss of generality, we suppose u≤w,v≤w (the other case is similar). In view of the order of X×X (or X2 for short), one has (64)u,v≥u,w,u,w≤w,u,w,u≥v,u;that is, (u,v),(u,w),(u,w),(w,u),(w,u),(v,u) are comparable in X2.Inspired by [31], we consider the functional d2:X2×X2→R+=[0,+∞) defined by (65)d2Y,V=12dx,u+dy,v,forY=x,y,V=u,v∈X2.d2 is a metric on X2 and, moreover, if (X,d) is complete, then (X2,d2) is a complete metric space, too. Define another operator T:X2→X2 as follows:(66)TV=Fu,v,Fv,uforV=u,v∈X2.For Y=(x,y),V=(u,v)∈X2, let Y¯=(y,x),V¯=(v,u); one has(67)d2Y,V=dx,u+dy,v2,d2TY,TV=dFx,y,Fu,v+dFy,x,Fv,u2=dFY,FV+dFY¯,FV¯2.By contractive condition (4) in (H2), we get a Banach type contraction condition: (68)φ2d2TY,TV=φdFx,y,Fu,v+dFy,x,Fv,u≤kφ2d2Y,V,and thus(69)φ2d2TTY,TTV≤kφ2d2TY,TV≤k2φ2d2Y,V,and, hence, by induction we have (70)φ2d2TnY,TnV≤knφ2d2Y,V,n=1,2,3,⋯.Now, applying (70) to the comparable pairs Y=(u,v),V=(u,w),Y=(u,w),V=(w,u),Y=(w,u),V=(v,u), one obtains(71)φ2d2Tnu,v,Tnu,w≤knφ2d2u,v,u,w=knφdv,w,(72)φ2d2Tnu,w,Tnw,u≤knφ2d2u,w,w,u=knφ2du,w,(73)φ2d2Tnw,u,Tnv,u≤knφ2d2w,u,v,u=knφdw,v.Now, forY=(u,v),V=(v,u), note that u=F(u,v)=F(Y),v=F(v,u)=F(V); we get (74)TY=Fu,v,Fv,u=u,v=Y,du,v=dFY,FV=dFu,v,Fv,u=dFV¯,FY¯.So, using the triangle inequality and (4) (71)-(74), we have(75)φ2du,v=φ2×dFY,FV+dFV¯,FY¯2=φ2d2Tnu,v,Tnv,u≤φ2d2Tnu,v,Tnu,w+φ2d2Tnu,w,Tnw,u+φ2d2Tnw,u,Tnv,u≤knφdv,w+φ2du,w+φdw,v→0asn→∞,which implies d(u,v)=0, so u=v.Corollary 9.
In addition to the hypothesis of Corollary7, suppose one of the followings holds:
(i) Elements of the coupled fixed point(u,v) are comparable
(ii)u0 and v0 are comparable
(iii) Every pair of elements of X has an upper bound or a lower bound inX
Then, we haveu=v; that is, F has a unique fixed point: (76)Fu,u=u.Remark 10.
Corollary9 includes theorems in [31].Example 11.
LetX=R,d(x,y)=|x-y| and F:X×X→X be defined by (77)Fx,y=16x-2y,x,y∈X2=X×X.Then,
(i)F has a coupled fixed point
(ii)F has a unique coupled fixed point
(iii)F has a unique fixed pointProof.
Letφ(t)=4t, ψ(t)=t, t∈[0,+∞). Since (78)Fx,y=16x-2y,x,y∈X2,one gets, for (u,v),(m,n)∈X2,(79)dFu,v,Fm,n=16u-2v-m+2n=16u-m+2n-v≤16u-m+13n-v,(80)dFv,u,Fn,m=16v-2u-n+2m=16v-n+2m-u≤16n-v+13u-m,and (81)φdFu,v,Fm,n+dFv,u,Fn,m≤2u-m+2n-v.Note that(82)du,m+dv,n=u-m+n-v,and, for k∈[2/3,1), (83)kφdu,m+dv,n-ψkdu,m+dv,n=3ku-m+3kn-v,and, hence, (81) (83) imply that (84)φdFu,v,Fm,n+dFv,u,Fn,m≤kφdu,m+dv,n-ψkdu,m+dv,n.
On the other hand,F is continuous and u0=-1, v0=1 satisfy(85)Fu0,v0=16u0-2v0=-12>-1=u0,Fv0,u0=16v0-2u0=12<1=v0.
(i) As mentioned above, by Theorem3, F has a coupled fixed point
(ii) SinceX2 has property (H5) or (H5′), by Theorem 6, the uniqueness of the coupled fixed point is obtained
(iii) Noting thatu0 and v0 are comparable, by Theorem 8 (ii), F has a unique fixed point


## 3. Applications to Integral Equations

As an application to the fixed point theorem proved in Section2, we shall study a class of integral equation (86)ux=∫mnq1x,y+q2x,yfy,uy+gy,uydy+px,where x∈I=[m,n]. We will establish existence and uniqueness results. It is well known that some boundary value problems are equivalent to an integral equation or a system of integral equations.LetX=C(I,R) be a partially ordered set such that, for u,v∈X, (87)u≤v⇔ux≤vx,x∈I.X is endowed with the sup metric:(88)du,v=supx∈I⁡ux-vx,u,v∈X,so (X,d) is a complete metric space. The corresponding metric d1 on X2 is defined by (89)d1u1,v1,u2,v2=12supx∈I⁡u1x-u2x+supx∈I⁡v1x-v2x,and then consider on X2 the partial order relation: (90)u1,v1≤u2,v2⇔u1x≤u2xandv1x≥v2x,x∈I.A pair (α,β)∈X2 is called a coupledlower-upper solution of (86) if (91)αx≤∫mnq1x,yfy,αy+gy,βydy+∫mnq2x,yfy,βy+gy,αydy+px,βx≥∫mnq1x,yfy,βy+gy,αydy+∫mnq2x,yfy,αy+gy,βydy+px.Ifα=β, α will be a solution of (86).Letθ:[0,+∞)→[0,+∞) be a nondecreasing function such that (92)θu=u2-ψu2,u∈0,+∞,where ψ∈Ψ satisfies ψ(t1+t2)≤ψ(t1)+ψ(t2), t1,t2∈[0,+∞).We make the following assumptions:(i)q1∈C(I×I,[0,+∞)), q2∈C(I×I,(-∞,0])(ii)p∈C(I,R)(iii)f,g∈C(I×R,R)(iv) there exist constantsλ,μ>0 such that, for u,v∈R,u≥(93)0≤fx,u-fx,v≤λθu-v,-μθu-v≤gx,u-gx,v≤0,max⁡λ,μsupx∈I⁡∫mnq1x,y-q2x,ydy≤14,(v)(α,β)∈X2 is a coupledlower-upper solution of (86)Theorem 12.
Suppose(i)-(v) hold. Then (86) has a unique solution.Proof.
In order to obtain the (unique) solution of (86), define for x∈[m,n] the operator F:X×X→X by(94)Fu,vx=∫mnq1x,yfy,uy+gy,vydy+∫mnq2x,yfy,vy+gy,uydy+px,∀x∈m,n.If (u,v)∈X2 is a coupled fixed point of F, then we have (95)ux=Fu,vx,vx=Fv,ux.It is obvious that the fixed point of F is the solution of (86). In what follows, we will show that F has a unique fixed point.
(1) We will show that the operatorF has a unique coupled fixed point in X2
Firstly, by (v), we have(96)α≤Fα,β,Fβ,α≤β.
Secondly, we checkF is mixed monotone for u1,u2∈X such that u1≤u2. From (i) and (iv), we have(97)Fu1,vx-Fu2,vx=∫mnq1x,yfy,u1y-fy,u2ydy+∫mnq2x,ygy,u1y-gy,u2ydy≤0.Similarly, for v1,v2∈X such that v1≤v2, we have(98)Fu,v1x-Fu,v2x=∫mnq1x,ygy,v1y-gy,v2ydy+∫mnq2x,yfy,v1y-fy,v2ydy≥0,which yields that F is mixed monotone.
Thirdly, we show thatF verifies the contraction condition. Let us consider u,v,a,b∈X with u≥a, v≤b; we have(99)
d
F
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.Similarly, one obtains (100)dFv,u,Fb,a≤max⁡λ,μsupx∈I⁡∫mnq1x,y-q2x,yθuy-ay+θby-vydy.Since(101)θuy-ay≤θdu,a,θby-vy≤θdb,v,which together with (iv) imply that(102)dFu,v,Fa,b+dFv,u,Fb,a=2max⁡λ,μsupx∈I⁡∫mnq1x,y-q2x,yθuy-ay+θby-vydy≤θdu,a+θdb,v=12du,a+db,v-ψ12du,a+ψ12db,v≤du,a+db,v2-ψdu,a+db,v2,which proves that F verifies contraction condition (36) in Corollary 4.
Next, we consider a monotone nondecreasing sequence{un}⊂X converging to u∈X. Then, for every x∈I, the sequence of real numbers (103)u1x≤u2x≤⋯≤unx≤⋯,converges to u(x). So, for all x∈I,n∈N,un(x)≤u(x) Hence, un≤u, for all n. Similarly, we can verify that the limit v(x) of monotone nonincreasing sequence {vn} in X is a lower bound for all the elements in the sequence. That is, v≤vn for all n.
Therefore, it follows from Corollary4 that F has a coupled fixed point (u0,v0)∈X2, i.e.,(104)u0=Fu0,v0andv0=Fv0,u0.
On the other hand,X has property (H5′) since, for any g,h∈X, M(x)=max(g(x),h(x)), m(x)=min(g(x),h(x)), for each x∈I, are in X and are the upper and lower bounds of g,h, respectively. Then, by Corollary 7, F has a unique coupled fixed point.
(2) Noting that(α,β) is a coupled lower-upper solution of (86), one obtains α(x)≤β(x) for all x∈I. So, α and β are comparable. By Corollary 9, F has a unique fixed point



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