# Fixed Point Results forG-α-Contractive Maps with Application to Boundary Value Problems

Authors: Nawab Hussain; Vahid Parvaneh; Jamal Rezaei Roshan
Journal: The Scientific World Journal
 (2014)

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## Abstract

We unify the concepts ofG-metric, metric-like, and b-metric to define new notion of generalized b-metric-like space and discuss its topological and structural properties. In addition, certain fixed point theorems for two classes of G-α-admissible contractive mappings in such spaces are obtained and some new fixed point results are derived in corresponding partially ordered space. Moreover, some examples and an application to the existence of a solution for the first-order periodic boundary value problem are provided here to illustrate the usability of the obtained results.


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## Body

## 1. Introduction and Mathematical Preliminaries

The concept of ab-metric space was introduced by Czerwik [1]. After that, several interesting results about the existence of fixed point for single-valued and multivalued operators in (ordered) b-metric spaces have been obtained (see, e.g., [2–11]).Definition 1 (see [1]).
LetX be a (nonempty) set and s≥1 a given real number. A function d:X×X→ℝ+ is a b-metric on X if, for all x,y,z∈X, the following conditions hold:(b1)  d(x,y)=0 if and only if x=y,(b2)  d(x,y)=d(y,x),(b3)  d(x,z)≤s[d(x,y)+d(y,z)].
In this case, the pair(X,d) is called a b-metric space.The concept of a generalized metric space, or aG-metric space, was introduced by Mustafa and Sims [12].Definition 2 (see [12]).
LetX be a nonempty set and G:X×X×X→ℝ+ a function satisfying the following properties:(G1)
G(x,y,z)=0 if and only if x=y=z;(G2)
0<G(x,x,y), for all x,y∈X with x≠y;(G3)
G(x,x,y)≤G(x,y,z), for all x,y,z∈X with y≠z;(G4)
G(x,y,z)=G(p{x,y,z}), where p is any permutation of x,y,z (symmetry in all three variables);(G5)
G(x,y,z)≤G(x,a,a)+G(a,y,z), for all x,y,z,a∈X (rectangle inequality).
Then, the functionG is called a G-metric on X and the pair (X,G) is called a G-metric space.Definition 3 (see [13]).
A metric-like on a nonempty setX is a mapping σ:X×X→ℝ+ such that, for all x,y,z∈X, the following hold:(σ1)  σ(x,y)=0 implies x=y;(σ2)  σ(x,y)=σ(y,x);(σ3)  σ(x,y)≤σ(x,z)+σ(z,y).
The pair(X,σ) is called a metric-like space.Below, we give some examples of metric-like spaces.Example 4 (see [14]).
LetX=[0,1]. Then, the mapping σ1:X×X→ℝ+ defined by σ1(x,y)=x+y-xy is a metric-like on X.Example 5 (see [14]).
LetX=ℝ; then the mappings σi:X×X→ℝ+(i∈{2,3,4}) defined by
(1)σ2(x,y)=|x|+|y|+a,σ3(x,y)=|x-b|+|y-b|,σ4(x,y)=x2+y2
are metric-likes on X, where a≥0 and b∈ℝ.Definition 6 (see [15]).
LetX be a nonempty set and s≥1 a given real number. A function σb:X×X→ℝ+ is a b-metric-like if, for all x,y,z∈X, the following conditions are satisfied:(σb1)  σb(x,y)=0 implies x=y;(σb2)  σb(x,y)=σb(y,x);(σb3)  σb(x,y)≤s[σb(x,z)+σb(z,y)].
Ab-metric-like space is a pair (X,σb) such that X is a nonempty set and σb is a b-metric-like on X. The number s is called the coefficient of (X,σb).In ab-metric-like space (X,σb) if x,y∈X and σb(x,y)=0, then x=y, but the converse may not be true and σb(x,x) may be positive for all x∈X. It is clear that every b-metric space is a b-metric-like space with the same coefficient s but not conversely in general.Example 7 (see [8]).
LetX=ℝ+, let p>1 be a constant, and let σb:X×X→ℝ+ be defined by
(2)σb(x,y)=(x+y)p∀x,y∈X.
Then,(X,σb) is a b-metric-like space with coefficient s=2p-1.The following propositions help us to construct some more examples ofb-metric-like spaces.Proposition 8 (see [8]).
Let(X,σ) be a metric-like space and σb(x,y)=[σ(x,y)]p, where p>1 is a real number. Then, σb is a b-metric-like with coefficient s=2p-1.From the above proposition and Examples4 and 5, we have the following examples of b-metric-like spaces.Example 9 (see [8]).
LetX=[0,1]. Then, the mapping σb1:X×X→ℝ+ defined by σb1(x,y)=(x+y-xy)p, where p>1 is a real number, is a b-metric-like on X with coefficient s=2p-1.Example 10 (see [8]).
LetX=ℝ. Then, the mappings σbi:X×X→ℝ+(i∈{2,3,4}) defined by
(3)σb2(x,y)=(|x|+|y|+a)p,σb3(x,y)=(|x-b|+|y-b|)p,σb4(x,y)=(x2+y2)p
are b-metric-like on X, where p>1, a≥0, and b∈ℝ.Eachb-metric-like σb on X generates a topology τσb on X whose base is the family of all open σb-balls {Bσb(x,ɛ):x∈X,ɛ>0}, where Bσb(x,ɛ)={y∈X:|σb(x,y)-σb(x,x)|<ɛ} for all x∈X and ɛ>0.Now, we introduce the concept of generalizedb-metric-like space, or Gσb-metric space, as a proper generalization of both of the concepts of b-metric-like spaces and G-metric spaces.Definition 11.
LetX be a nonempty set. Suppose that a mapping Gσb:X×X×X→ℝ+ satisfies the following:(Gσb1)
Gσb(x,y,z)=0 implies x=y=z;(Gσb2)
Gσb(x,y,z)=Gσb(p{x,y,z}), where p is any permutation of x,y,z (symmetry in all three variables);(Gσb3)
Gσb(x,y,z)≤s[Gσb(x,a,a)+Gσb(a,y,z)] for all x,y,z,a∈X (rectangle inequality).Then,Gσb is called a Gσb-metric and (X,Gσb) is called a generalized b-metric-like space.The following proposition will be useful in constructing examples of a generalizedb-metric-like space.Proposition 12.
Let(X,σb) be a b-metric-like space with coefficient s. Then,
(4)Gσbm(x,y,z)=max⁡{σb(x,y),σb(y,z),σb(z,x)},Gσbs(x,y,z)=σb(x,y)+σb(y,z)+σb(z,x)
are two generalized b-metric-like functions on X.Proof.
It is clear thatGσbm and Gσbs satisfy conditions Gσb1 and Gσb2 of Definition 11. So, we only show that Gσb3 is satisfied by Gσbm and Gσbs. Let x,y,z,a∈X. Then, using the triangular inequality in b-metric-like spaces, we have
(5)Gσbm(x,y,z)=max⁡{σb(x,y),σb(y,z),σb(z,x)}≤max⁡{s(σb(x,a)+σb(a,y)),σb(y,z),s(σb(z,a)+σb(a,x))}≤max⁡{s(σb(x,a)+σb(a,y)),s(σb(y,z)+σb(a,a)),s(σb(z,a)+σb(a,x))}=smax⁡{σb(x,a),σb(a,a),σb(a,x)}+smax⁡{σb(a,y),σb(y,z),σb(z,a)}=s(Gσbm(x,a,a)+Gσbm(a,y,z)).
Also,
(6)Gσbs(x,y,z)=σb(x,y)+σb(y,z)+σb(z,x)≤s(σb(x,a)+σb(a,y))+σb(y,z)+s(σb(z,a)+σb(a,x))≤s(σb(x,a)+σb(a,y))+s(σb(y,z)+σb(a,a))+s(σb(z,a)+σb(a,x))=s(σb(x,a)+σb(a,a)+σb(a,x))+s(σb(a,y),σb(y,z),σb(z,a))=s(Gσbs(x,a,a)+Gσbs(a,y,z)).According to the above proposition, we provide some examples of generalizedb-metric-like spaces.Example 13.
LetX=ℝ+, let p>1 be a constant, and let Gσbm,Gσbs:X×X×X→ℝ+ be defined by
(7)Gσbm(x,y,z)=max⁡{(x+y)p,(y+z)p,(z+x)p},Gσbs(x,y,z)=(x+y)p+(y+z)p+(z+x)p,
for all x,y,z∈X. Then, (X,Gσbm) and (X,Gσbs) are generalized b-metric-like spaces with coefficient s=2p-1. Note that, for x=y=z>0, Gσbm(x,x,x)=(2x)p>0 and Gσbs(x,x,x)=3(2x)p>0.Example 14.
LetX=[0,1]. Then, the mappings Gσb1m,Gσb1s:X×X×X→ℝ+ defined by
(8)Gσb1m(x,y,z)=max⁡{(x+y-xy)p,(y+z-yz)p,(z+x-zx)p},Gσb1s(x,y,z)=(x+y-xy)p+(y+z-yz)p+(z+x-zx)p,
where p>1 is a real number, are generalized b-metric-like spaces with coefficient s=2p-1.By some straight forward calculations, we can establish the following.Proposition 15.
LetX be a Gσb-metric space. Then, for each x,y,z,a∈X, it follows that:(1)
Gσb(x,y,y)>0forx≠y;(2)
Gσb(x,y,z)≤s(Gσb(x,x,y)+Gσb(x,x,z));(3)
Gσb(x,y,y)≤2sGσb(y,x,x);(4)
Gσb(x,y,z)≤sGσb(x,a,a)+s2Gσb(y,a,a)+s2Gσb(z,a,a).Definition 16.
Let(X,Gσb) be a Gσb-metric space. Then, for any x∈X and r>0, the Gσb-ball with center x and radius r is
(9)BGσb(x,r)={y∈X∣|Gσb(x,x,y)-Gσb(x,x,x)|<+r}.The family of allGσb-balls
(10)Ϝ={BGσb(x,r)∣x∈X,r>0}
is a base of a topology τ(Gσb) on X, which we call it Gσb-metric topology.Definition 17.
Let(X,Gσb) be a Gσb-metric space. Let {xn} be a sequence in X. Consider the following.(1)
A pointx∈X is said to be a limit of the sequence {xn}, denoted by xn→x, if lim⁡n,m→∞Gσb(x,xn,xm)=Gσb(x,x,x).(2)
{xn} is said to be a Gσb-Cauchy sequence, if lim⁡n,m→∞Gσb(xn,xm,xm) exists (and is finite).(3)
(X,Gσb) is said to be Gσb-complete if every Gσb-Cauchy sequence in X is Gσb-convergent to an x∈X such that
(11)lim⁡n,m→∞Gσb(xn,xm,xm)=lim⁡n→∞Gσb(xn,x,x)=Gσb(x,x,x).Using the above definitions, one can easily prove the following proposition.Proposition 18.
Let(X,Gσb) be a Gσb-metric space. Then, for any sequence {xn} in X and a point x∈X, the following are equivalent:(1)
{xn}isGσb-convergent tox;(2)
Gσb(xn,xn,x)→Gσb(x,x,x), asn→∞;(3)
Gσb(xn,x,x)→Gσb(x,x,x), asn→∞;Definition 19.
Let(X,Gσb) and (X′,Gσb′) be two generalized b-metric like spaces and let f:(X,Gσb)→(X′,Gσb′) be a mapping. Then, f is said to be Gσb-continuous at a point a∈X if, for a given ɛ>0, there exists δ>0 such that x∈X and |Gσb(a,a,x)-Gσb(a,a,a)|<δ imply that |Gσb′(f(a),f(a),f(x))-Gσb′(f(a),f(a),f(a))|<ɛ. The mapping f is Gσb-continuous on X if it is Gσb-continuous at all a∈X. For simplicity, we say that f is continuous.Proposition 20.
Let(X,Gσb) and (X′,Gσb′) be two generalized b-metric like spaces. Then, a mapping f:X→X′ is Gσb-continuous at a point x∈X if and only if it is Gσb-sequentially continuous at x; that is, whenever {xn} is Gσb-convergent to x, {f(xn)} is Gσb-convergent to f(x).We need the following simple lemma about theGσb-convergent sequences in the proof of our main results.Lemma 21.
Let(X,Gσb) be a Gσb-metric space and suppose that {xn}, {yn}, and {zn} are Gσb-convergent to x, y, and z, respectively. Then, we have
(12)1s3Gσb(x,y,z)-1s2Gσb(x,x,x)-1sGσb(y,y,y)-Gσb(z,z,z)≤liminf⁡n→∞Gσb(xn,yn,zn)≤limsup⁡n→∞Gσb(xn,yn,zn)≤s3Gσb(x,y,z)+sGσb(x,x,x)+s2Gσb(y,y,y)+s3Gσb(z,z,z).
In particular, if {yn}={zn}=a are constant, then
(13)1sGσb(x,a,a)-Gσb(x,x,x)≤liminf⁡n→∞Gσb(xn,a,a)≤limsup⁡n→∞Gσb(xn,a,a)≤sGσb(x,a,a)+sGσb(x,x,x).Proof.
Using the rectangle inequality, we obtain(14)Gσb(x,y,z)≤sGσb(x,xn,xn)+s2Gσb(y,yn,yn)+s3Gσb(z,zn,zn)+s3Gσb(xn,yn,zn),Gσb(xn,yn,zn)≤sGσb(xn,x,x)+s2Gσb(yn,y,y)+s3Gσb(zn,z,z)+s3Gσb(x,y,z).
Taking the lower limit as n→∞ in the first inequality and the upper limit as n→∞ in the second inequality, we obtain the desired result.
If{yn}={zn}=a, then
(15)Gσb(x,a,a)≤sGσb(x,xn,xn)+sGσb(xn,a,a),Gσb(xn,a,a)≤sGσb(xn,x,x)+sGσb(x,a,a).
Again taking the lower limit as n→∞ in the first inequality and the upper limit as n→∞ in the second inequality, we obtain the desired result.


## 2. Main Results

Samet et al. [16] defined the notion of α-admissible mappings and proved the following result.Definition 22.
LetT be a self-mapping on X and α:X×X→[0,∞) a function. We say that T is an α-admissible mapping if
(16)x,y∈X,α(x,y)≥1⟹α(Tx,Ty)≥1.Denote withΨ′ the family of all nondecreasing functions ψ:[0,∞)→[0,∞) such that ∑n=1∞‍ψn(t)<∞ for all t>0, where ψn is the nth iterate of ψ.Theorem 23.
Let(X,d) be a complete metric space and T an α-admissible mapping. Assume that
(17)α(x,y)d(Tx,Ty)≤ψ(d(x,y)),
where ψ∈Ψ′. Also, suppose that the following assertions hold:(i)
there existsx0∈X such that α(x0,Tx0)≥1;(ii)
eitherT is continuous or, for any sequence {xn} in X with α(xn,xn+1)≥1 for all n∈ℕ∪{0} such that xn→x as n→∞, we have α(xn,x)≥1 for all n∈ℕ∪{0}.
Then,T has a fixed point.For more details onα-admissible mappings, we refer the reader to [17–20].Definition 24 (see [21]).
Let(X,G)   be a G-metric space, let f be a self-mapping on X, and let α:X3→[0,∞) be a function. We say that f is aG-α-admissible mapping if
(18)x,y,z∈X,α(x,y,z)≥1⟹α(fx,fy,fz)≥1.Motivated by [22], let ℱ denote the class of all functions β:[0,∞)→[0,1/s) satisfying the following condition:
(19)limsup⁡n→∞β(tn)=1s⟹limsup⁡n→∞tn=0.Definition 25.
Letf:X→X and α:X×X×X→ℝ. We say that f is a rectangular G-α-admissible mapping if(T1)
α(x,y,z)≥1 implies α(fx,fy,fz)≥1,x,y,z∈X;(T2)
{α(x,y,y)≥1α(y,z,z)≥1 implies α(x,z,z)≥1,x,y,z∈X.From now on, letα:X3→[0,∞) be a function and
(20)Ms(x,y,z)=max⁡{Gσb(x,fx,fx)Gσb(y,fy,fy)Gσb(z,fz,fz)1+s2Gσb2(fx,fy,fz)Gσb(x,y,z),Gσb(x,fx,fx)Gσb(y,fy,fy)Gσb(z,fz,fz)1+s2Gσb2(fx,fy,fz)}.Theorem 26.
Let(X,Gσb) be a Gσb-complete generalized b-metric-like space and let f:X→X be a rectangular G-α-admissible mapping. Suppose that
(21)sα(x,y,z)Gσb(fx,fy,fz)≤β(Gσb(x,y,z))Ms(x,y,z),
for all x,y,z∈X.
Also, suppose that the following assertions hold:(i)
there existsx0∈X such that α(x0,fx0,fx0)≥1;(ii)
f is continuous and, for any sequence {xn} in X with α(xn,xn+1,xn+1)≥1 for all n∈ℕ∪{0} such that xn→x as n→∞, we have α(x,x,x)≥1 for all n∈ℕ∪{0}.
Then,f has a fixed point.Proof.
Letx0∈X  be such that α(x0,fx0,fx0)≥1. Define a sequence {xn} by xn=fnx0 for all n∈ℕ. Since f is a G-α-admissible mapping and α(x0,x1,x1)=α(x0,fx0,fx0)≥1, we deduce that α(x1,x2,x2)=α(fx0,fx1,fx1)≥1. Continuing this process, we get α(xn,xn+1,xn+1)≥1 for all n∈ℕ∪{0}.
Step
I. We will show that lim⁡n→∞Gσb(xn,xn+1,xn+1)=0. If xn=xn+1 for some n∈ℕ, then xn=fxn. Thus, xn is a fixed point of f. Therefore, we assume that xn≠xn+1 for all n∈ℕ.
Sinceα(xn,xn+1,xn+1)≥1 for each n∈ℕ, then we can apply (21) which yields
(22)sGσb(xn,xn+1,xn+1)=sGσb(fxn-1,fxn,fxn)≤β(Gσb(xn-1,xn,xn))Ms(xn-1,xn,xn)<1sGσb(xn-1,xn,xn)≤Gσb(xn-1,xn,xn),(23)Ms(xn-1,xn,xn)=max⁡{(1+s2Gσb2(fxn-1,fxn,fxn))-1Gσb(xn-1,xn,xn),(Gσb(xn-1,fxn-1,fxn-1)×Gσb(xn,fxn,fxn)Gσb(xn,fxn,fxn))×(1+s2Gσb2(fxn-1,fxn,fxn))-1}=max⁡{(1+s2Gσb2(xn,xn+1,xn+1))-1Gσb(xn-1,xn,xn),(Gσb(xn-1,xn,xn)×Gσb(xn,xn+1,xn+1)Gσb(xn,xn+1,xn+1))×(1+s2Gσb2(xn,xn+1,xn+1))-1}=Gσb(xn-1,xn,xn).
Therefore, {Gσb(xn,xn+1,xn+1)} is a decreasing and bounded sequence of nonnegative real numbers. Then, there exists r≥0 such that lim⁡n→∞Gσb(xn,xn+1,xn+1)=r. Letting n→∞ in (22), we have
(24)sr≤r.
Since s>1, we deduce that r=0, that is
(25)lim⁡n→∞Gσb(xn,xn+1,xn+1)=0.
By Proposition15(2), we conclude that
(26)lim⁡n→∞Gσb(xn,xn,xn+1)=0.
Step
II. Now, we prove that the sequence {xn} is a Gσb-Cauchy sequence. For this purpose, we will show that
(27)limsup⁡m,n→∞Gσb(xn,xm,xm)=0.
Using the rectangular inequality with (21) (as α(xn,xm,xm)≥1, since f is a rectangular G-α-admissible mapping), we have
(28)Gσb(xn,xm,xm)≤sGσb(xn,xn+1,xn+1)+s2Gσb(xn+1,xm+1,xm+1)+s2Gσb(xm+1,xm,xm)≤sGσb(xn,xn+1,xn+1)+sβ(Gσb(xn,xm,xm))Ms(xn,xm,xm)+s2Gσb(xm+1,xm,xm).
Taking limit asm,n→∞ in the above inequality and applying (25) and (26), we have
(29)limsup⁡m,n→∞Gσb(xn,xm,xm)≤slimsup⁡n,m→∞β(Gσb(xn,xm,xm))limsup⁡n,m→∞Ms(xn,xm,xm).
Here,
(30)Gσb(xn,xm,xm)≤Ms(xn,xm,xm)=max⁡{Gσb(xn,fxn,fxn)[Gσb(xm,fxm,fxm)]21+s2Gσb2(fxn,fxm,fxm)Gσb(xn,xm,xm),Gσb(xn,fxn,fxn)[Gσb(xm,fxm,fxm)]21+s2Gσb2(fxn,fxm,fxm)}=max⁡{Gσb(xn,xn+1,xn+1)[Gσb(xm,xm+1,xm+1)]21+s2Gσb2(xn+1,xm+1,xm+1)Gσb(xn,xm,xm),Gσb(xn,xn+1,xn+1)[Gσb(xm,xm+1,xm+1)]21+s2Gσb2(xn+1,xm+1,xm+1)}.
Letting m,n→∞ in the above inequality, we get
(31)limsup⁡m,n→∞Ms(xn,xm,xm)=limsup⁡m,n→∞Gσb(xn,xm,xm).
Hence, from (29) and (31), we obtain
(32)limsup⁡m,n→∞Gσb(xn,xm,xm)≤slimsup⁡m,n→∞β(Gσb(xn,xm,xm))limsup⁡m,n→∞Gσb(xn,xm,xm).
If limsup⁡m,n→∞Gσb(xn,xm,xm)≠0, then we get
(33)1s≤limsup⁡m,n→∞β(Gσb(xn,xm,xm)).
Since β∈ℱ, we deduce that
(34)limsup⁡m,n→∞Gσb(xn,xm,xm)=0,
which is a contradiction. Consequently, {xn} is a Gσb-Cauchy sequence in X. Since (X,Gσb) is Gσb-complete, there exists u∈X such that xn→u, as n→∞. Now, from (34) and Gσb-completeness of X,
(35)lim⁡n→∞Gσb(u,xn,xn)=lim⁡m,n→∞Gσb(xn,xm,xm)=Gσb(u,u,u)=0.
Step
III. Now, we show that u is a fixed point of f.
Using the rectangle inequality, we get(36)Gσb(u,fu,fu)≤sGσb(u,fxn,fxn)+sGσb(fxn,fu,fu).
Letting n→∞ and using the continuity of f and (35), we obtain
(37)Gσb(u,fu,fu)≤slim⁡n→∞Gσb(u,fxn,fxn)+slim⁡n→∞Gσb(fxn,fu,fu)=sGσb(fu,fu,fu).
Note that, from (21), as α(u,u,u)≥1, we have
(38)sGσb(fu,fu,fu)≤β(Gσb(u,u,u))Ms(u,u,u),
where, by (37),
(39)Ms(u,u,u)=max⁡{Gσb(u,fu,fu)Gσb(u,fu,fu)Gσb(u,fu,fu)1+s2Gσb2(fu,fu,fu)Gσb(u,u,u),Gσb(u,fu,fu)Gσb(u,fu,fu)Gσb(u,fu,fu)1+s2Gσb2(fu,fu,fu)}<Gσb(u,fu,fu).
Hence, as β(t)≤1 for all t∈[0,∞), we have sGσb(fu,fu,fu)≤Gσb(u,fu,fu). Thus, by (37), we obtain that sGσb(fu,fu,fu)=Gσb(u,fu,fu). But then, using (38), we get that
(40)Gσb(u,fu,fu)=sGσb(fu,fu,fu)≤β(Gσb(u,u,u))Ms(u,u,u)<1sGσb(u,fu,fu),
which is a contradiction. Hence, we have fu=u. Thus, u is a fixed point of f.We replace condition (ii) in Theorem26 by regularity of the space X.Theorem 27.
Under the same hypotheses of Theorem26, instead of condition (ii), assume that whenever {xn} in X is a sequence such that α(xn,xn+1,xn+1)≥1 for all n∈ℕ∪{0} and xn→x as n→∞, one has α(xn,x,x)≥1 for all n∈ℕ∪{0}. Then, f has a fixed point.Proof.
Repeating the proof of Theorem26, we can construct a sequence {xn} in X such that α(xn,xn+1,xn+1)≥1 for all n∈ℕ∪{0} and xn→u∈X for some u∈X. Using the assumption on X, we have α(xn,u,u)≥1 for all n∈ℕ∪{0}. Now, we show that u=fu. By Lemma 21 and (35),
(41)s[1sGσb(u,fu,fu)-Gσb(u,u,u)]≤slimsup⁡n→∞Gσb(xn+1,fu,fu)≤limsup⁡n→∞(β(Gσb(xn,u,u))Ms(xn,u,u))≤1slimsup⁡n→∞Ms(xn,u,u),
where
(42)lim⁡n→∞Ms(xn,u,u)=lim⁡n→∞max⁡{[Gσb(xn,fxn,fxn),Gσb(u,fu,fu)]21+s2Gσb2(fxn,fu,fu)Gσb(xn,u,u),[Gσb(xn,fxn,fxn)][Gσb(u,fu,fu)]21+s2Gσb2(fxn,fu,fu)}=lim⁡n→∞max⁡{[Gσb(xn,xn+1,xn+1),Gσb(u,fu,fu)]21+s2Gσb2(xn+1,fu,fu)Gσb(xn,u,u),[Gσb(xn,xn+1,xn+1)][Gσb(u,fu,fu)]21+s2Gσb2(xn+1,fu,fu)}=max⁡{Gσb(u,u,u),0}=0(see(25)and(35)).
Therefore, we deduce that Gσb(u,fu,fu)≤0. Hence, we have u=fu.A mappingφ:[0,∞)→[0,∞) is called acomparison function if it is increasing and φn(t)→0, as n→∞ for any t∈[0,∞) (see, e.g., [23, 24] for more details and examples).Definition 28.
A functionφ:[0,∞)→[0,∞) is said to be a (c)-comparison function if(c1)
φ is increasing,(c2)
there existsk0∈ℕ, a∈(0,1), and a convergent series of nonnegative terms ∑k=1∞vk such that φk+1(t)≤aφk(t)+vk for k≥k0 and any t∈[0,∞).Later, Berinde [5] introduced the notion of (b)-comparison function as a generalization of (c)-comparison function.Definition 29 (see [5]).
Lets≥1 be a real number. A mapping φ:[0,∞)→[0,∞) is called a (b)-comparison function if the following conditions are fulfilled:(1)
φ is increasing;(2)
there existk0∈ℕ, a∈(0,1), and a convergent series of nonnegative terms ∑k=1∞vk such that sk+1φk+1(t)≤askφk(t)+vk for k≥k0 and any t∈[0,∞).LetΨb be the class of all (b)-comparison functions φ:[0,∞)→[0,∞). It is clear that the notion of (b)-comparison function coincides with (c)-comparison function for s=1.Lemma 30 (see [25]).
Ifφ:[0,∞)→[0,∞) is a (b)-comparison function, then we have the following:(1)
the series∑k=0∞skφk(t) converges for any t∈ℝ+;(2)
the functionbs:[0,∞)→[0,∞) defined by bs(t)=∑k=0∞skφk(t), t∈[0,∞), is increasing and continuous at 0.Remark 31.
It is easy to see that ifφ∈Ψb, then we have φ(0)=0 and φ(t)<t for each t>0 and φ is continuous at 0.In the next example, we present a class of(b)-comparison functions.Example 32.
Any function of the formψ(t)=ln⁡((a/s)t+1) for all t∈[0,∞) where 0<a<1 is a (b)-comparison function.Proof.
From the part(1) of Lemma 30, the necessary condition is that the series ∑k=0∞skφk(t) converges for any t∈ℝ+. But, for each t>0 and k≥1, we have
(43)skφk(t)=skφ(φk-1(t))=skln⁡(asφk-1(t)+1)≤skasφk-1(t)≤⋯≤sk(as)kt=akt.
So, according to the comparison test of the series, we should have a<1. On the other hand, we have
(44)sk+1φk+1(t)=sk+1φ(φk(t))=sk+1ln⁡(asφk(t)+1)≤sk+1asφk(t)=skaφk(t).
Therefore, for any convergent series of nonnegative terms ∑k=1∞‍vk and each k≥k0=1, we have
(45)sk+1φk+1(t)≤askφk(t)≤askφk(t)+vk.For example, fors=2 and a=1/2, the function ψ(t)=ln⁡(t/4+1) is a (b)-comparison function.Theorem 33.
Let(X,Gσb) be a Gσb-complete generalized b-metric-like space and let f:X→X be a G-α-admissible mapping. Suppose that
(46)sα(x,y,z)Gσb(fx,fy,fz)≤ψ(Ms(x,y,z)),
for all x,y,z∈X where ψ∈Ψb and
(47)Ms(x,y,z)=max⁡{Gσb(x,x,fx)Gσb(y,y,fy)Gσb(z,z,fz)1+2s2Gσb2(fx,fy,fz)Gσb(x,y,z),Gσb(x,x,fx)Gσb(y,y,fy)Gσb(z,z,fz)1+2s2Gσb2(fx,fy,fz)}.Also, suppose that the following assertions hold:(i)
there existsx0∈X such that α(x0,fx0,f2x0)≥1;(ii)
(a)
f is continuous and, for any sequence {xn} in X with α(xn,xn+1,xn+2)≥1 for all n∈ℕ∪{0} such that xn→x as n→∞, one has α(x,x,x)≥1 for all n∈ℕ∪{0};(b)
assume that whenever{xn} in X is a sequence such that α(xn,xn+1,xn+2)≥1 for all n∈ℕ∪{0} and xn→x as n→∞, one has α(xn,x,x)≥1 for all n∈ℕ∪{0}.Then,f has a fixed point.Proof.
Letx0∈X  be such that  α(x0,fx0,f2x0)≥1. Define a sequence {xn} by xn=fnx0 for all n∈ℕ. Since f is a G-α-admissible mapping and α(x0,x1,x2)=α(x0,fx0,f2x0)≥1, we deduce that α(x1,x2,x3)=α(fx0,fx1,fx2)≥1. Continuing this process, we get α(xn,xn+1,xn+2)≥1 for all n∈ℕ∪{0}.
If there existsn0∈ℕ such that xn0=xn0+1, then xn0=fxn0 and so we have nothing to prove. Hence, for all n∈ℕ, we assume that xn≠xn+1.
Step
I (Cauchyness of{xn}). As α(xn,xn+1,xn+2)≥1 for all n≥0, using condition (46), we obtain
(48)sGσb(xn,xn+1,xn+2)≤sα(xn-1,xn,xn+1)Gσb(xn,xn+1,xn+2)=sα(xn-1,xn,xn+1)Gσb(fxn-1,fxn,fxn+1)≤ψ(Ms(xn-1,xn,xn+1)).
Using Proposition15(2) as xn≠xn+1, we get
(49)Ms(xn-1,xn,xn+1)=max⁡{(1+2s2Gσb2(fxn-1,fxn,fxn+1))-1Gσb(xn-1,xn,xn+1),(Gσb(xn-1,xn-1,fxn-1)×Gσb(xn,xn,fxn)×Gσb(xn+1,xn+1,fxn+1))×(1+2s2Gσb2(fxn-1,fxn,fxn+1))-1}=max⁡{(1+2s2Gσb2(xn,xn+1,xn+2))-1Gσb(xn-1,xn,xn+1),(Gσb(xn-1,xn-1,xn)×Gσb(xn,xn,xn+1)×Gσb(xn+1,xn+1,xn+2))×(1+2s2Gσb2(xn,xn+1,xn+2))-1}≤max⁡{2sGσb(xn-1,xn,xn+1)Gσb2(xn,xn+1,xn+2)1+2s2Gσb2(xn,xn+1,xn+2)Gσb(xn-1,xn,xn+1),2sGσb(xn-1,xn,xn+1)Gσb2(xn,xn+1,xn+2)1+2s2Gσb2(xn,xn+1,xn+2)}=Gσb(xn-1,xn,xn+1).
Hence,(50)Gσb(xn,xn+1,xn+2)≤sGσb(xn,xn+1,xn+2)≤ψ(Gσb(xn-1,xn,xn+1)).
By induction, sincexn≠xn+1, we get that
(51)Gσb(xn,xn+1,xn+2)≤ψ(Gσb(xn-1,xn,xn+1))≤ψ2(Gσb(xn-2,xn-1,xn))≤⋯≤ψn(Gσb(x0,x1,x2)).
Let ɛ>0 be arbitrary. Then, there exists a natural number N such that
(52)∑n=N∞‍snψn(Gσb(x0,x1,x2))<ɛ2s.
Letm>n≥N. Then, by the rectangular inequality and Proposition 15(2) as xn≠xn+1, we get
(53)Gσb(xn,xm,xm)≤sGσb(xn,xn+1,xn+1)+s2Gσb(xn+1,xn+2,xn+2)+⋯+sm-n-1Gσb(xm-1,xm,xm)≤2s(sGσb(xn,xn,xn+1)+s2Gσb(xn+1,xn+1,xn+2)+⋯+sm-n-1Gσb(xm-1,xm-1,xm))≤2s(sGσb(xn,xn+1,xn+2)+s2Gσb(xn+1,xn+2,xn+3)+⋯+sm-n-1Gσb(xm-1,xm,xm+1))≤2s∑k=nm-2‍sk-n+1ψk(Gσb(x0,x1,x2))≤2s∑k=n∞‍skψk(Gσb(x0,x1,x2))<ɛ.
Consequently, {xn} is a Gσb-Cauchy sequence in X. Since (X,Gσb) is Gσb-complete, so there exists u∈X such that
(54)lim⁡n→∞Gσb(u,xn,xn)=lim⁡m,n→∞Gσb(xn,xm,xm)=Gσb(u,u,u)=0.
Step
II. Now, we show that u is a fixed point of f. Suppose to the contrary, that is, fu≠u, then, we have Gσb(u,u,fu)>0.
Let the part (a) of (ii) holds.
Using the rectangle inequality, we get(55)Gσb(u,u,fu)≤sGσb(fu,fxn,fxn)+sGσb(fxn,u,u).
Letting n→∞ and using the continuity of f, we get
(56)Gσb(u,u,fu)≤sGσb(fu,fu,fu).
From (46) and part (a) of condition (ii), we have
(57)sGσb(fu,fu,fu)≤ψ(Ms(u,u,u)),
where, by using (56), we have
(58)Ms(u,u,u)=max⁡{Gσb(u,u,fu)Gσb(u,u,fu)Gσb(u,u,fu)1+2s2Gσb2(fu,fu,fu)Gσb(u,u,u),Gσb(u,u,fu)Gσb(u,u,fu)Gσb(u,u,fu)1+2s2Gσb2(fu,fu,fu)}≤Gσb(u,u,fu).
Hence, from properties of ψ, sGσb(fu,fu,fu)≤Gσb(u,u,fu). Thus, by (56), we obtain that
(59)Gσb(u,u,fu)=sGσb(fu,fu,fu).
Moreover, (57) yields that Gσb(u,u,fu)≤ψ(Ms(u,u,u))≤ψ(Gσb(u,u,fu)). This is impossible, according to our assumptions on ψ. Hence, we have fu=u. Thus, u is a fixed point of f.
Now, let part (b) of (ii) holds.
As{xn} is a sequence such that α(xn,xn+1,xn+2)≥1 for all n∈ℕ∪{0} and xn→x as n→∞, we have α(xn,x,x)≥1 for all n∈ℕ∪{0}.
Now, we show thatu=fu. By (46), we have
(60)sGσb(fu,fu,xn)≤sα(u,u,xn-1)Gσb(fu,fu,fxn-1)≤ψ(Ms(u,u,xn-1)),
where
(61)Ms(u,u,xn-1)=max⁡{[Gσb(u,u,fu)]2Gσb(xn-1,xn-1,fxn-1)1+[sGσb(fu,fu,fxn-1)]2Gσb(u,u,xn-1),[Gσb(u,u,fu)]2Gσb(xn-1,xn-1,fxn-1)1+[sGσb(fu,fu,fxn-1)]2}.
Letting n→∞ in the above inequality and using (54) and Lemma 21, we get
(62)lim⁡n→∞Ms(u,u,xn-1)=0.
Again, taking the upper limit as n→∞ in (60) and using (62) and Lemma 21, we obtain
(63)s[1sGσb(u,fu,fu)-Gσb(u,u,u)]≤slimsup⁡n→∞Gσb(xn,fu,fu)≤limsup⁡n→∞ψ(Ms(u,u,xn-1))=ψ(limsup⁡n→∞Ms(u,u,xn-1))=ψ(0)=0.
So, we get Gσb(u,fu,fu)=0. That is, fu=u.Let(X,Gσb,⪯) be a partially ordered Gσb-metric-like space. We say that T is an increasing mapping on X if ∀xy∈X,x⪯y⇒T(x)⪯T(y) [26]. Fixed point theorems for monotone operators in ordered metric spaces are widely investigated and have found various applications in differential and integral equations (see [27–30] and references therein). From the results proved above, we derive the following new results in partially ordered Gσb-metric-like space.Theorem 34.
Let(X,Gσb,⪯) be a partially ordered Gσb-complete generalized b-metric-like space and let f:X→X be an increasing mapping. Suppose that
(64)sGσb(fx,fy,fz)≤β(Gσb(x,y,z))Ms(x,y,z),
for all x,y,z∈X with x⪯y⪯z, where
(65)Ms(x,y,z)=max⁡{Gσb(x,fx,fx)Gσb(y,fy,fy)Gσb(z,fz,fz)1+s2Gσb2(fx,fy,fz)Gσb(x,y,z),Gσb(x,fx,fx)Gσb(y,fy,fy)Gσb(z,fz,fz)1+s2Gσb2(fx,fy,fz)}.
Also, suppose that the following assertions hold:(i)
there existsx0∈X such that x0⪯fx0;(ii)
f is continuous or assume that whenever {xn} in X is an increasing sequence such that xn→x as n→∞, one has xn⪯x for all n∈ℕ∪{0}.
Then,f has a fixed point.Proof.
Defineα:X×X×X→[0,+∞) by
(66)α(x,y,z)={1,ifx⪯y⪯z0,otherwise.
First, we prove that f is a triangular α-admissible mapping. Hence, we assume that α(x,y,z)≥1. Therefore, we have x⪯y⪯z. Since f is increasing, we get fx⪯fy⪯fz; that is, α(fx,fy,fz)≥1. Also, let α(x,y,y)≥1 and α(y,z,z)≥1; then x⪯y and y⪯z. Consequently, we deduce that x⪯z; that is, α(x,z,z)≥1. Thus, f is a triangular α-admissible mapping. Since f satisfies (64) so, by the definition of α, we have
(67)sα(x,y,z)Gσb(fx,fy,fz)≤β(Gσb(x,y,z))Ms(x,y,z)
for all x,y,z∈X. Therefore, f satisfies the contractive condition (21). From (i), there exists x0∈X such that x0⪯fx0; that is, α(x0,fx0,fx0)≥1. According to (ii), we conclude that all the conditions of Theorems 26 and 27 are satisfied and so f has a fixed point.Similarly, using Theorem33, we can prove following result.Theorem 35.
Let(X,Gσb,⪯) be an ordered Gσb-complete generalized b-metric-like space and let f:X→X be an increasing mapping. Suppose that
(68)sGσb(fx,fy,fz)≤ψ(Ms(x,y,z)),
for all x,y,z∈X with x⪯y⪯z where ψ∈Ψb and
(69)Ms(x,y,z)=max⁡{Gσb(x,x,fx)Gσb(y,y,fy)Gσb(z,z,fz)1+2s2Gσb2(fx,fy,fz)Gσb(x,y,z),Gσb(x,x,fx)Gσb(y,y,fy)Gσb(z,z,fz)1+2s2Gσb2(fx,fy,fz)}.
Also, suppose that the following assertions hold:(i)
there existsx0∈X such that x0⪯fx0;(ii)
(a)f is continuous;(b) assume that whenever {xn} in X is an increasing sequence such that xn→x as n→∞, one has xn⪯x for all n∈ℕ∪{0}.
Then,f has a fixed point.We conclude this section by presenting some examples that illustrate our results.Example 36.
LetX=[0,(-1+5)/2] be endowed with the usual ordering on ℝ. Define the generalized b-metric-like function Gσb given by
(70)Gσb(x,y,z)=max⁡{(x+y)2,(y+z)2,(z+x)2}
with s=2. Consider the mapping f:X→X defined by f(x)=(1/4)xe-(x+((-1+5)/2))2 and the function β∈ℱ given by β(t)=(1/2)e-t,  t>0, and β(0)∈[0,1/2). It is easy to see that f is an increasing function on X. We show that f is Gσb-continuous on X. By Proposition 20, it is sufficient to show that f is Gσb-sequentially continuous on X. Let {xn} be a sequence in X such that lim⁡n→∞Gσb(xn,x,x)=Gσb(x,x,x), so we have max⁡{(lim⁡n→∞xn+x)2,4x2}=4x2  and, equivalently, lim⁡n→∞xn=α≤x. On the other hand, we have
(71)lim⁡n→∞Gσb(fxn,fx,fx)=lim⁡n→∞max⁡{(fxn+fx)2,4f2x}=lim⁡n→∞max⁡{116(xne-(xn+(-1+5)/2)2+xe-(x+(-1+5)/2)2)2,14x2e-(x+(-1+5)/2)2e-(xn+(-1+5)/2)2}=max⁡{116(lim⁡n→∞xne-(lim⁡n→∞xn+(-1+5)/2)2+xe-(x+(-1+5)/2)2)2,14x2e-(x+(-1+5)/2)2116(lim⁡n→∞xne-(lim⁡n→∞xn+(-1+5)/2)2}=max⁡{116(αe-(α+(-1+5)/2)2+xe-(x+(-1+5)/2)2)2,14x2e-(x+(-1+5)/2)2αe-(α+(-1+5)/2)2}=14x2e-(x+(-1+5)/2)2=Gσb(fx,fx,fx).
So, f is Gσb-sequentially continuous on X.
For all elementsx,y,z∈X, and the fact that g(x)=x2e-x2 is an increasing function on X, we have
(72)sGσb(fx,fy,fz)=2max⁡{(fx+fy)2,(fy+fz)2,(fz+fx)2}=2max⁡{+14xe-(x+(-1+5)/2)2)2(14xe-(x+(-1+5)/2)2+14ye-(y+(-1+5)/2)2)2,(14ye-(y+(-1+5)/2)2+14ze-(z+(-1+5)/2)2)2,(14ze-(z+(-1+5)/2)2+14xe-(x+(-1+5)/2)2)2}≤2max⁡{(14xe-(x+y)2+14ye-(y+x)2)2,(14ye-(y+z)2+14ze-(z+y)2)2,(14ze-(z+x)2+14xe-(x+z)2)2}=18max⁡{(x+y)2e-(x+y)2,(y+z)2e-(y+z)2,(z+x)2e-(z+x)2}=18e-max⁡{(x+y)2,(y+z)2,(z+x)2}×max⁡{(x+y)2,(y+z)2,(z+x)2}≤12e-max⁡{(x+y)2,(y+z)2,(z+x)2}×max⁡{(x+y)2,(y+z)2,(z+x)2}=β(Gσb(x,y,z))Gσb(x,y,z)≤β(Gσb(x,y,z))Ms(x,y,z).
Hence, f satisfies all the assumptions of Theorem 34 and thus it has a fixed point (which is u=0).Example 37.
LetX=[0,1] with the usual ordering on ℝ. Define the generalized b-metric-like function Gσb given by Gσb(x,y,z)=max⁡{(x+y)2,(y+z)2,(z+x)2} with s=2. Consider the mapping f:X→X defined by f(x)=ln⁡(1+xe-x/4) and the function ψ∈Ψb given by ψ(t)=(1/8)t,t≥0. It is easy to see that f is increasing function. Now, we show that f is a Gσb-continuous function on X.
Let{xn} be a sequence in X such that lim⁡n→∞Gσb(xn,x,x)=Gσb(x,x,x), so we have max⁡{(lim⁡n→∞xn+x)2,4x2}=4x2  and, equally, lim⁡n→∞xn=α≤x. On the other hand, we have
(73)lim⁡n→∞Gσb(fxn,fx,fx)=lim⁡n→∞max⁡{(fxn+fx)2,4f2x}=max⁡{(ln⁡(1+lim⁡n→∞xne-lim⁡n→∞xn4)+ln⁡(1+xe-x4)(1+lim⁡n→∞xne-lim⁡n→∞xn4))2,4(ln⁡(1+xe-x4))2}=max⁡{(ln⁡(1+αe-α4)+ln⁡(1+xe-x4))2,4(ln⁡(1+xe-x4))2}=4(ln⁡(1+xe-x4))2=Gσb(fx,fx,fx).
So, f is Gσb-sequentially continuous on X.
For all elementsx,y,z∈X, we have
(74)sGσb(fx,fy,fz)=2max⁡{(ln⁡(1+xe-x4)+ln⁡(1+ye-y4))2,(ln⁡(1+ye-y4)+ln⁡(1+ze-z4))2,(ln⁡⁡(1+ze-z4)+ln⁡⁡(1+xe-x4))2}≤2max⁡{(xe-x4+ye-y4)2,(ye-y4+ze-z4)2,(ze-z4+xe-x4)2}≤18max⁡{(x+y)2,(y+z)2,(z+x)2}=ψ(Gσb(x,y,z))≤ψ(Ms(x,y,z)).
Hence, f satisfies all the assumptions of Theorem 35 and thus it has a fixed point (which is u=0).


## 3. Application

In this section, we present an application of our results to establish the existence of a solution to a periodic boundary value problem (see [30, 31]).LetX=C([0,T]) be the set of all real continuous functions on [0,T]. We first endow X with the b-metric-like
(75)σb(u,v)=sup⁡t∈[0,T](|u(t)|+|v(t)|)p
for all u,v∈X where p>1 and then we endow it with the generalized b-metric-like Gσb defined by
(76)Gσb(u,v,w)=max⁡{sup⁡t∈[0,T](|u(t)|+|v(t)|)p,sup⁡t∈[0,T](|v(t)|+|w(t)|)p,sup⁡t∈[0,T](|w(t)|+|u(t)|)p}.
Clearly, (X,Gσb) is a complete generalized b-metric-like space with parameter K=2p-1. We equip C([0,T]) with a partial order given by
(77)x⪯yiffx(t)≤y(t)∀t∈[0,T].
Moreover, as in [30], it is proved that (C([0,T]),⪯) is regular; that is, whenever {xn} in X is an increasing sequence such that xn→x as n→∞, we have xn⪯x for all n∈ℕ∪{0}.Consider the first-order periodic boundary value problem(78)x′(t)=f(t,x(t)),x(0)=x(T),
where t∈I=[0,T], with T>0, and f:I×ℝ→ℝ is a continuous function.A lower solution for (78) is a function α∈C1[0,T] such that
(79)α′(t)≤f(t,α(t)),α(0)≤α(T),
where t∈I=[0,T].Assume that there existsλ>0 such that, for all x,y∈C[0,T], we have
(80)|f(t,x(t))+λx(t)|+|f(t,y(t))+λy(t)|≤λ2p-1ln⁡(a2p-1(|x(t)|+|y(t)|)p+1)p,
where 0≤a<1. Then, the existence of a lower solution for (78) provides the existence of a solution of (78).Problem (78) can be rewritten as
(81)x′(t)+λx(t)=f(t,x(t))+λx(t),x(0)=x(T).
Consider
(82)x′(t)+λx(t)=δx(t)=F(t,x(t)),x(0)=x(T),
where t∈I.Using variation of parameters formula, we get(83)x(t)=x(0)e-λt+∫0t‍e-λ(t-s)δx(s)ds
which yields
(84)x(T)=x(0)e-λT+∫0T‍e-λ(T-s)δx(s)ds.
Since x(0)=x(T), we get
(85)x(0)[1-e-λT]=e-λT∫0T‍eλ(s)δx(s)ds
or
(86)x(0)=1eλT-1∫0T‍eλsδx(s)ds.
Substituting the value of x(0) in (83), we arrive at
(87)x(t)=∫0T‍G(t,s)δx(s)ds
where
(88)G(t,s)={eλ(T+s-t)eλT-1,0≤s≤t≤T,eλ(s-t)eλT-1,0≤t≤s≤T.
Now, define the operator S:C[0,T]→C[0,T] as
(89)Sx(t)=∫0T‍G(t,s)F(s,x(s))ds.
The mapping S is increasing [31]. Note that if u∈C[0,T] is a fixed point of S, then u∈C1[0,T] is a solution of (78).Letx,y,z∈C[0,T]. Then, we have
(90)2p-1[|Sx(t)|+|Sy(t)|]=2p-1[∑11|∫0T‍G(t,s)F(s,x(s))ds|+|∫0T‍G(t,s)F(s,y(s))ds|∑11]≤2p-1∫0T‍|G(t,s)|×[|F(s,x(s))|+|F(s,y(s))|]ds≤2p-1∫0T‍|G(t,s)|λ2p-1×ln⁡(a2p-1(|x(t)|+|y(t)|)p+1)pds≤λln⁡(a2p-1σb(x,y)+1)p×[∫0t‍eλ(T+s-t)eλT-1ds+∫tT‍eλ(s-t)eλT-1ds]=λln⁡(a2p-1σb(x,y)+1)p×[1λ(eλT-1)(eλ(T+s-t)|0t+eλ(s-t)|tT)]=λln⁡(a2p-1σb(x,y)+1)p×[1λ(eλT-1)(eλT-eλ(T-t)+eλ(T-t)-1)]=ln⁡(a2p-1σb(x,y)+1)p≤ln⁡(a2p-1Gσb(x,y,z)+1)p≤ln⁡(a2p-1M(x,y,z)+1)p.
Similarly,
(91)2p-1|Sy(t)|+|Sz(t)|≤ln⁡(a2p-1M(x,y,z)+1)p,2p-1|Sx(t)|+|Sz(t)|≤ln⁡⁡(a2p-1M(x,y,z)+1)p,
where
(92)M(x,y,z)=max⁡{Gσb(x,Sx,Sx)Gσb(y,Sy,Sy)Gσb(z,Sz,Sz)1+22p-1Gσb2(Sx,Sy,Sz)Gσb(x,y,z),Gσb(x,Sx,Sx)Gσb(y,Sy,Sy)Gσb(z,Sz,Sz)1+22p-1Gσb2(Sx,Sy,Sz)}.
Equivalently,
(93)(2p-1|Sx(t)|+|Sy(t)|)p≤ln⁡(a2p-1M(x,y,z)+1),(2p-1|Sy(t)|+|Sz(t)|)p≤ln⁡(a2p-1M(x,y,z)+1),(2p-1|Sx(t)|+|Sz(t)|)p≤ln⁡(a2p-1M(x,y,z)+1),
which yields that
(94)2p-1σb(Sx,Sy)≤ln⁡(a2p-1M(x,y,z)+1),2p-1σb(Sy,Sz)≤ln⁡(a2p-1M(x,y,z)+1),2p-1σb(Sx,Sz)≤ln⁡(a2p-1M(x,y,z)+1).
Finally, it is easy to obtain that
(95)2p-1Gσb(Sx,Sy,Sz)≤ln⁡(a2p-1M(x,y,z)+1).
Finally, since α is a lower solution for (78), so it is easy to show that α⪯S(α) [31].Hence, the hypotheses of Theorem35 are satisfied, with ψ(t)=ln⁡((a/2p-1)t+1) where 0≤a<1. Hence, there exists a fixed point x^∈C[0,T] such that Sx^=x^ which is a solution to periodic boundary value problem (78).



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*Source: 585964-2014-05-07.xml*